problem_id,problem_statement,final_answer,difficulty_tier,reasoning_trace 175c2b0c2ff3,"# Task No. 1.2 ## Condition: Five friends - Katya, Polina, Alyona, Lena, and Svetlana - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation. Polina: I stood next to Alyona. Alyona: I was the first in line! Lena: No one was after me. Katya: There were five people in front of me. Svetlana: Only one person was after me. The girls are friends, so they don't lie to each other. How many people were between Katya and Polina?",See reasoning trace,easy,"# Answer: 3 Exact match of the answer - 1 point Solution by analogy with task №1.1 #" b6beed7241f7,"1. M. V. Lomonosov spent one coin on bread and kvass. When prices increased by $20 \%$, he could buy half a loaf of bread and kvass with that same coin. Would that same coin be enough to buy at least kvass if prices rise by another $20 \%$?","$ $=1.44 y$. Since $1.5 y>1.44 y$, the money will be enough for the kvass.",medium,"1. Let the price of bread be $x$, and the price of kvass be $y$ (before the price increase). Then, according to the condition: money $=x+y$. After the price increase, bread costs $1.2 x$, and kvass costs $1.2 y$. Therefore, $$ \text { money }=0.5 \text { bread }+ \text { kvass }=0.6 x+1.2 y . $$ From these equations, we get: $$ x+y=0.6 x+1.2 y $$ Simplifying, we obtain $2 x=y$. Express the money in terms of $y$: $$ \text { money }=x+y=1.5 y \text {. } $$ After the second price increase, kvass costs $1.2 \cdot 1.2 y=$ $=1.44 y$. Since $1.5 y>1.44 y$, the money will be enough for the kvass." da09ce469236,11.4. Consider all acute-angled triangles with a given side $a$ and angle $\alpha$. What is the maximum value of the sum of the squares of the lengths of sides $b$ and $c$?,"a^{2}+2 b c \cos \alpha$. Since $2 b c \leqslant b^{2}+c^{2}$ and $\cos \alpha>0$, then $b^{2}+c^{2}",easy,"11.4. By the Law of Cosines, $b^{2}+c^{2}=a^{2}+2 b c \cos \alpha$. Since $2 b c \leqslant b^{2}+c^{2}$ and $\cos \alpha>0$, then $b^{2}+c^{2} \leqslant a^{2}+\left(b^{2}+c^{2}\right) \cos \alpha$, i.e., $b^{2}+c^{2} \leqslant a^{2} /(1-\cos \alpha)$. Equality is achieved if $b=c$." 36a597627f1c,"Example 6.35. Find the confidence interval for estimating the mathematical expectation $a$ of a normal distribution with a reliability of 0.95, knowing the sample mean $\bar{X}_{\mathrm{B}}=10.43$, the sample size $n=100$, and the standard deviation $\sigma=5$.",See reasoning trace,medium,"Solution. We will use the formula $$ \bar{x}_{\mathrm{B}}-t \cdot \frac{\sigma}{\sqrt{n}}p$. (D) $m5 k+5 l, & \text { if } kl$ are possible. Hence, the answer is $(E)$." 7b4348b80a99,"2.2. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}-4 a x+1$ takes values, the modulus of which does not exceed 3, at all points of the interval $[0 ; 4]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong.",1,easy,Answer: 1.5 (the set of values for $a$: $\left[-\frac{1}{2} ; 0\right) \cup(0 ; 1]$ ). bdcf61914236,3.126. $1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)$.,$\frac{2 \sqrt{2} \cos ^{2} \frac{3 \alpha}{2} \sin \left(\frac{\pi}{4}-3 \alpha\right)}{\cos 3 \alpha}$,medium,"## Solution. $$ 1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)= $$ $$ \begin{aligned} & =1-\sin 3 \alpha+\cos 3 \alpha-\operatorname{tg} 3 \alpha=1-\sin 3 \alpha+\cos 3 \alpha-\frac{\sin 3 \alpha}{\cos 3 \alpha}= \\ & =\frac{(\cos 3 \alpha-\sin 3 \alpha) \cos 3 \alpha+\cos 3 \alpha-\sin 3 \alpha}{\cos 3 \alpha}=\frac{(\cos 3 \alpha-\sin 3 \alpha)(\cos 3 \alpha+1)}{\cos 3 \alpha}= \\ & =\frac{\left(\cos 3 \alpha-\cos \left(\frac{\pi}{2}-3 \alpha\right)\right)(\cos 3 \alpha+\cos 2 \pi)}{\cos 3 \alpha}= \\ & =\frac{-2 \sin \frac{\pi}{4} \sin \left(3 \alpha-\frac{\pi}{4}\right) \cdot 2 \cos \left(\pi+\frac{3 \alpha}{2}\right) \cos \left(\pi-\frac{3 \alpha}{2}\right)}{\cos 3 \alpha}= \\ & =\frac{-2 \cdot \frac{\sqrt{2}}{2} \cdot \sin \left(3 \alpha-\frac{\pi}{4}\right) \cdot 2 \cdot\left(-\cos \frac{3 \alpha}{2}\right)\left(-\cos \frac{3 \alpha}{2}\right)}{\cos 3 \alpha}= \\ & =\frac{2 \sqrt{2} \cos ^{2} \frac{3 \alpha}{2} \sin \left(\frac{\pi}{4}-3 \alpha\right)}{\cos 3 \alpha} . \end{aligned} $$ Answer: $\frac{2 \sqrt{2} \cos ^{2} \frac{3 \alpha}{2} \sin \left(\frac{\pi}{4}-3 \alpha\right)}{\cos 3 \alpha}$." 47ce8566c007,"Given a square $A B C D$ with side $4 \sqrt{2}$. Point $O$ is chosen in the plane of the square such that $O B=10, O D=6$. Find the angle between the vector $\overrightarrow{O B}$ and the vector directed from point $O$ to the farthest vertex of the square from it.",$\arccos \frac{23}{5 \sqrt{29}}$,easy,"From the condition, it is clear that $B D=8$. Therefore, triangle $B D O$ is a right triangle. Let points $O$ and $C$ lie on opposite sides of line $B D$. Then $O C^{2}=(6+8 / 2)^{2}+4^{2}=116>O B^{2}$ (that is, vertex $C$ is the farthest from $O$). By the cosine rule, $$ \cos \angle B O C=\frac{O B^{2}+O C^{2}-B C^{2}}{2 O B \cdot O C}=\frac{100+116-32}{40 \sqrt{29}}=\frac{23}{5 \sqrt{29}} $$ ## Answer $\arccos \frac{23}{5 \sqrt{29}}$" 0f5076c0ce8b,,See reasoning trace,medium," Solution: Denote the required number by $f_{n}$. It is easy to see that $f_{1}=1, f_{2}=2, f_{3}=4$. Divide the subsets of $S_{n}$ having no two consecutive numbers into two groups-those that do not contain the element $n$ and those that do. Obviously the number of subsets in the first group is $f_{n-1}$. Let $T$ be a set of the second group. Therefore either $T=\{n\}$ or $T=\left\{a_{1}, \ldots, a_{k-1}, n\right\}$, where $k>1$. It is clear that $a_{k-1} \neq n-1$, so $\left\{a_{1}, \ldots a_{k-1}\right\} \subset S_{n-2}$, whence the number of sets in the second group is $f_{n-2}+1$. Therefore $$ f_{n}=f_{n-1}+f_{n-2}+1 $$ After substituting $u_{n}=f_{n}+1$ we get $$ u_{1}=2, u_{2}=3, \quad u_{n}=u_{n-1}+u_{n-2} $$ Therefore the sequence $\left\{u_{n}\right\}_{n=1}^{\infty}$ coincides with the Fibonacci sequence from its third number onwards. Thus we obtain $$ f_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right)-1 $$ " 2a04f9a7fc63,"In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids? [asy] size((70)); draw((0,0)--(7.5,13)--(15,0)--(0,0)); draw((1.88,3.25)--(9.45,3.25)); draw((11.2,0)--(7.5,6.5)); draw((9.4,9.7)--(5.6,3.25)); [/asy] $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$",\textbf{(C),easy,The area outside the small triangle but inside the large triangle is $16-1=15$. This is equally distributed between the three trapezoids. Each trapezoid has an area of $15/3 = \boxed{\textbf{(C)}\ 5}$. 65e6e546f07e,"In a spatial rectangular coordinate system, a sphere centered at the origin has a radius of 3 units. How many lattice points lie on the surface of the sphere?",See reasoning trace,medium,"Let's take a point $P$ on the surface of the sphere, with coordinates $x, y, z$. We know that the radius of the sphere is 3 units, so using the Pythagorean theorem, we can write the following equation: $$ x^{2}+y^{2}+z^{2}=9 $$ We need to find the integer solutions to this equation. The equation holds if one of the coordinates is +3 or -3, and the other two are 0, that is, $$ x= \pm 3, y=0, z=0, y= \pm 3, z=0, x=0, z= \pm 3, x=0, y=0 $$ This gives us a total of 6 possibilities. If none of the coordinates is 3, then the absolute value of two of them must be 2, and the third coordinate must be +1 or -1. The two coordinates with an absolute value of 2 can be chosen in 3 different ways. For example, if $x=y= \pm 2$ (this gives 4 possibilities) and $z= \pm 1$, this results in 8 solutions. Similarly, it is possible that $y=z= \pm 2$ and $x= \pm 1$ or $x=z= \pm 2$ and $y= \pm 1$, which gives a total of 24 cases. Thus, a total of 30 lattice points fall on the surface of the sphere. Mult Attila (Békéscsaba, Rózsa F. Gimn., 9. o.t.)" b92cd324382c,"A5. Given a stack of boxes, it is stated that a quarter of the boxes are empty. We open a quarter of the boxes and see that one fifth of them are not empty. What fraction of the unopened boxes is empty? A) $\frac{4}{15}$ B) $\frac{1}{4}$ C) $\frac{1}{15}$ D) $\frac{1}{16}$ E) $\frac{1}{20}$",See reasoning trace,easy,"A5. C) $\frac{1}{15}$ For convenience, we can assume there are 20 boxes; this does not affect the ratios. Of the 20 boxes, 5 are empty. Of the 5 boxes that are opened, one fifth are not empty, exactly 1 box. Since 4 of the opened boxes are empty, there is exactly 1 empty box left among the 15 unopened boxes." b22272b246cc,"4. A. As shown in Figure $1, \odot O$'s diameter $A B$ intersects with chord $C D$ at point $P$, forming an angle of $45^{\circ}$. If $P C^{2}+P D^{2}=8$, then the radius of $\odot O$ is ( ). (A) $\sqrt{2}$ (B) 2 (C) $2 \sqrt{2}$ (D) 4",See reasoning trace,easy,"4. A. B. Draw $O H \perp C D$, with the foot of the perpendicular at $H$. Without loss of generality, assume point $H$ lies on segment $C P$. Then $\angle O P H=45^{\circ}$. Thus, $O H=H P=\frac{1}{2}(P C-P D)$. By $C H=\frac{1}{2}(P C+P D)$, in the right triangle $\triangle O H C$, applying the Pythagorean theorem, we get $$ \begin{array}{l} O C^{2}=O H^{2}+C H^{2}=\frac{1}{2}\left(P C^{2}+P D^{2}\right)=4 \\ \Rightarrow O C=2 \end{array} $$" da668a841921,"9-3. Arina wrote down all the numbers from 71 to 81 in a row without spaces, forming a large number 717273...81. Sofia started appending the next numbers to it (i.e., she first appended 82, then 83, ...). She stopped when the large number became divisible by 12. The last number she appended was $N$. What is $N$?",88,medium,"Answer: 88. Solution. A number is divisible by 12 if and only if it is divisible by 3 and by 4. For a number to be divisible by 4, the number formed by its last two digits must also be divisible by 4. Therefore, the last number that Sofia writes must be divisible by 4. The nearest number that is divisible by 4 is 84, but the number 71727374...84 has a sum of digits equal to 158, which means it is not divisible by 3. The next number that is divisible by 4 is 88. The sum of the digits of the number 71727374...88 is 216, which means the entire number is divisible by 3." 3c8afebe8b1b,"2. Given an integer $n \geqslant 3$, let $a_{1}, a_{2}, \cdots, a_{2 n} ; b_{1}, b_{2}, \cdots, b_{2 n}$ be $4 n$ non-negative real numbers, satisfying $a_{1}+a_{2}+\cdots+a_{2 n}=b_{1}+b_{2}+\cdots+b_{2 n}>0$, and for any $i=1,2, \cdots, 2 n$, $a_{i} a_{i+2} \geqslant b_{i}+b_{i+1}$ (here $a_{2 n+1}=a_{1}, a_{2 n+2}=a_{2}, b_{2 n+1}=b_{1}$ ). Find the minimum value of $a_{1}+a_{2}+\cdots+a_{2 n}$.","1}^{n} a_{2 k-1} a_{2 k+1} \geqslant \sum_{k=1}^{2 n} b_{k}=S \Rightarrow S \geqslant 16$, equality ",medium,"Let $S=\sum_{k=1}^{2 n} a_{k}=\sum_{k=1}^{2 n} b_{k}, T=\sum_{k=1}^{n} a_{2 k-1}$, without loss of generality, assume $T \leqslant \frac{S}{2}$. When $n=3$, since $\left(a_{1}+a_{3}+a_{5}\right)^{2}-3\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right)$ $=a_{1}^{2}+a_{3}^{2}+a_{5}^{2}-\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right)=\frac{1}{2}\left[\left(a_{1}-a_{3}\right)^{2}+\left(a_{3}-a_{5}\right)^{2}+\left(a_{5}-a_{1}\right)^{2}\right] \geqslant 0$, thus $\frac{S^{2}}{4} \geqslant T^{2} \geqslant 3\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right) \geqslant 3\left(b_{1}+b_{2}+b_{3}+b_{4}+b_{5}+b_{6}\right)=3 S$ $\Rightarrow S \geqslant 12$, equality holds when $a_{k}=b_{k}=2, k=1,2, \cdots, 6$. When $n \geqslant 4$, if $n$ is even, then $$ \begin{array}{l} \sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-3}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-1}\right) \\ \leqslant\left(\frac{a_{1}+a_{3}+a_{5}+a_{7}+\cdots+a_{2 n-3}+a_{2 n-1}}{2}\right)^{2}=\frac{T^{2}}{4}, \end{array} $$ if $n$ is odd, without loss of generality, assume $a_{1} \leqslant a_{3}$, then $\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \leqslant \sum_{k=1}^{n-1} a_{2 k-1} a_{2 k+1}+a_{2 n-1} a_{3}$ $$ \leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-1}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-3}\right) \leqslant \frac{T^{2}}{4} \text {. } $$ Thus $\frac{S^{2}}{16} \geqslant \frac{T^{2}}{4} \geqslant \sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \geqslant \sum_{k=1}^{2 n} b_{k}=S \Rightarrow S \geqslant 16$, equality holds when, in summary, when $n=3$, the minimum value of $S$ is 12; when $n \geqslant 4$, the minimum value of $S$ is 16." 77061566bd1b,"3. If point $P\left(x_{0}, y_{0}\right)$ is such that the chord of contact of the ellipse $E: \frac{x}{4}+y^{2}=1$ and the hyperbola $H: x^{2}-\frac{y^{2}}{4}=1$ are perpendicular to each other, then $\frac{y_{0}}{x_{0}}=$ $\qquad$",See reasoning trace,easy,"3. $\pm 1$. Point $P\left(x_{0}, y_{0}\right)$ has the tangent chord line equations for ellipse $E$ and hyperbola $H$ respectively as $$ \frac{x x_{0}}{4}+y y_{0}=1, x x_{0}-\frac{y y_{0}}{4}=1 . $$ From the fact that these two lines are perpendicular, we have $$ \begin{array}{l} \left(\frac{x_{0}}{4}, y_{0}\right) \cdot\left(x_{0},-\frac{y_{0}}{4}\right)=0 \\ \Rightarrow x_{0}^{2}-y_{0}^{2}=0 \Rightarrow \frac{y_{0}}{x_{0}}= \pm 1 \end{array} $$" 2476cdb3f4e5,"4. Two classes are planting trees. Class one plants 3 trees per person, and class two plants 5 trees per person, together planting a total of 115 trees. The maximum sum of the number of people in the two classes is $\qquad$ .",】 37 people,easy,"【Key Points】Combinations, Maximum Value Problems 【Difficulty】ふ 【Answer】 37 people. 【Analysis】Let class one have $a$ people, and class two have $b$ people, then we have $3 a+5 b=115$. To find the maximum number of people in both classes, the equation can be transformed into: $3(a+b)+2 b=115$, maximizing $a+b$ while minimizing $b$. When $b=2$, $a+b=37$. The maximum total number of people in both classes is 37." 13f1eff8ad2a,"Example 3. Find the derivative of the function $$ y=\operatorname{arctg} \frac{x}{a}+\frac{1}{2} \ln \left(x^{2}+a^{2}\right) $$",See reasoning trace,medium,"Solution. Differentiate term by term and use formula $\left(2.18^{\prime}\right):$ $$ \begin{gathered} {\left[\operatorname{arctg} \frac{x}{a}+\frac{1}{2} \ln \left(x^{2}+a^{2}\right)\right]^{\prime}=\left(\operatorname{arctg} \frac{x}{a}\right)^{\prime}+\frac{1}{2}\left[\ln \left(x^{2}+a^{2}\right)\right]^{\prime}=} \\ =\frac{1}{1+\left(\frac{x}{a}\right)^{2}}\left(\frac{x}{a}\right)^{\prime}+\frac{1}{2} \cdot \frac{1}{x^{2}+a^{2}}\left(x^{2}+a^{2}\right)^{\prime}= \\ =\frac{a}{x^{2}+a^{2}}+\frac{x}{x^{2}+a^{2}}=\frac{x+a}{x^{2}+a^{2}} \end{gathered} $$" 3219ea656a26,"Three, (10 points) Given $$ \frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z} \text {. } $$ Find the value of $\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}$.",See reasoning trace,medium,"$$ \begin{array}{l} \frac{x+y+z+u}{y+z+u}=\frac{x+y+z+u}{z+u+x} \\ =\frac{x+y+z+u}{u+x+y}=\frac{x+y+z+u}{x+y+z} . \end{array} $$ (1) If the numerator $x+y+z+u \neq 0$, then from the denominators we get $x=y=z=u$. At this time, $$ \begin{array}{l} \frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z} \\ =1+1+1+1=4 . \end{array} $$ (2) If the numerator $x+y+z+u=0$, then $$ x+y=-(z+u), y+z=-(u+x) \text {. } $$ At this time, $\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}$ $$ =(-1)+(-1)+(-1)+(-1)=-4 . $$" 4780ab4cc4ff,"## Task 4 - 140524 Students of a 5th grade class held a 14-day chess tournament. Exactly 6 games were played each of the 14 days. The number of participating boys was greater than that of the participating girls. Each girl played against every other girl and each boy against every other boy exactly twice. None of the girls played against a boy. Determine the number of girls and the number of boys who participated in this tournament!",See reasoning trace,medium,"According to the problem, because $14 \cdot 6 = 84$, a total of 84 games were played in this tournament. This number is the sum of the number of games played between the girls and the number of games involving the boys. For $n$ players, where each player plays exactly two games against each of the other $(n - 1)$ players, the total number of games is $n(n-1)$. Thus, the following table can be constructed: | Number of Players | Number of Games | Complement to 84 | | :---: | :---: | :---: | | 2 | 2 | 82 | | 3 | 6 | 78 | | 4 | 12 | 72 | | 5 | 20 | 64 | | 6 | 30 | 54 | | 7 | 42 | 42 | | 8 | 56 | 28 | | 9 | 72 | 12 | | $\geq 10$ | $\geq 90$ | - | Since a total of 84 games were played according to the problem, the number of participating boys or girls could not have been greater than 9. The summands can only be the numbers determined in the table, and they must be exactly two numbers of the form $n(n-1)$ whose sum is 84. This is only possible for $42 + 42 = 84$ and $12 + 72 = 84$, as a comparison of the numbers in the second and third columns of the table shows. Since the number of participating boys was greater than the number of girls, the solution must be: 4 girls and 9 boys participated in the tournament mentioned in the problem. Solutions of the II. Round 1974 taken from [5] ### 2.16 XV. Olympiad 1975 ### 2.16.1 I. Round 1975, Class 5" 5f6b3708ad56,"3. Alice and Bob are independently trying to figure out a secret password to Cathy's bitcoin wallet. Both of them have already figured out that: - it is a 4-digit number whose first digit is 5; - it is a multiple of 9 ; - The larger number is more likely to be a password than a smaller number. Moreover, Alice figured out the second and the third digits of the password and Bob figured out the third and the fourth digits. They told this information to each other but not actual digits. After that the conversation followed: Alice: ""I have no idea what the number is."" Bob: ""I have no idea too."" After that both of them knew which number they should try first. Identify this number.","0$ or $Z=9$. Similarly, $X=0$ or $X=9$. So the largest possibility is 5949 .",easy,"Solution. Let the password be $5 X Y Z$. As this number is a multiple of 9 , we know that $5+X+Y+Z$ is a multiple of 9 . Alice knows $5+X+Y$ and since she cannot identify the number it must be that $Z=0$ or $Z=9$. Similarly, $X=0$ or $X=9$. So the largest possibility is 5949 ." 8a067a096d09,Compute \[\left\lfloor \dfrac{2007!+2004!}{2006!+2005!}\right\rfloor.\] (Note that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.),2006,medium,"1. Start with the given expression: \[ \left\lfloor \dfrac{2007! + 2004!}{2006! + 2005!} \right\rfloor \] 2. Factor out \(2004!\) from both the numerator and the denominator: \[ \dfrac{2007! + 2004!}{2006! + 2005!} = \dfrac{2004! \cdot (2005 \cdot 2006 \cdot 2007 + 1)}{2004! \cdot (2005 \cdot 2006 + 2005)} \] 3. Simplify the expression by canceling \(2004!\): \[ \dfrac{2005 \cdot 2006 \cdot 2007 + 1}{2005 \cdot 2006 + 2005} \] 4. Factor out \(2005\) from the denominator: \[ 2005 \cdot 2006 + 2005 = 2005 \cdot (2006 + 1) = 2005 \cdot 2007 \] 5. Substitute this back into the expression: \[ \dfrac{2005 \cdot 2006 \cdot 2007 + 1}{2005 \cdot 2007} \] 6. Separate the fraction: \[ \dfrac{2005 \cdot 2006 \cdot 2007}{2005 \cdot 2007} + \dfrac{1}{2005 \cdot 2007} \] 7. Simplify the first term: \[ \dfrac{2005 \cdot 2006 \cdot 2007}{2005 \cdot 2007} = 2006 \] 8. Combine the terms: \[ 2006 + \dfrac{1}{2005 \cdot 2007} \] 9. Since \(\dfrac{1}{2005 \cdot 2007}\) is a very small positive number, the floor function of the sum will be: \[ \left\lfloor 2006 + \dfrac{1}{2005 \cdot 2007} \right\rfloor = 2006 \] Conclusion: \[ \boxed{2006} \]" 30c8b7cf0395,"8. If the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3,3 z_{1}-2 z_{2}=\frac{3}{2}-\mathrm{i}$, then $z_{1} \cdot z_{2}=$",6[\cos (\alpha+\beta)+\mathrm{i} \sin (\alpha+\beta)]=-\frac{30}{13}+\frac{72}{13} \mathrm{i}$.,medium,"Let $z_{1}=2(\cos \alpha+\mathrm{i} \sin \alpha), z_{2}=3(\cos \beta+\mathrm{i} \sin \beta)$, then we have $$ \begin{array}{l} \left\{\begin{array} { l } { 6 ( \cos \alpha - \cos \beta ) = \frac { 3 } { 2 } , } \\ { 6 ( \sin \alpha - \sin \beta ) = - 1 } \end{array} \Rightarrow \left\{\begin{array}{l} \cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=\frac{1}{4}, \\ \sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=-\frac{1}{6} \end{array}\right.\right. \\ \Rightarrow \tan \frac{\alpha+\beta}{2}=\frac{3}{2} \Rightarrow\left\{\begin{array}{l} \cos (\alpha+\beta)=\frac{1-\frac{9}{4}}{1+\frac{9}{4}}=-\frac{5}{13}, \\ \sin (\alpha+\beta)=\frac{3}{1+\frac{9}{4}}=\frac{12}{13} . \end{array}\right. \end{array} $$ Therefore, $z_{1} \cdot z_{2}=6[\cos (\alpha+\beta)+\mathrm{i} \sin (\alpha+\beta)]=-\frac{30}{13}+\frac{72}{13} \mathrm{i}$." fdf59eda05b7,"1. Each of the six houses on one side of the street is connected by cable lines to each of the eight houses on the opposite side. How many pairwise intersections do the shadows of these cables form on the surface of the street, if no three of them intersect at the same point? Assume that the light causing these shadows falls vertically downward.",$C_{6}^{2} \cdot C_{8}^{2}=420$,medium,"# Solution Let's take an arbitrary pair of houses on one side of the street and an arbitrary pair on the other. They are the vertices of a convex quadrilateral (since two sides of the quadrilateral, coming from each chosen pair, lie on one side of the line, i.e., the angles do not exceed $180^{\circ}$), so its diagonals intersect. Each pairwise intersection of shadows (cables) is a point of intersection of the diagonals of such a quadrilateral. Thus, it remains to find their number, which is equal to the product of the ways to choose an ordered pair of houses on each side of the street. Answer: $C_{6}^{2} \cdot C_{8}^{2}=420$." 4965402fb749,"2. $1^{2}, 2^{2}, 3^{2}, \cdots, 123456789^{2}$ The unit digit of the sum is $\qquad$ .","45$, so the required digit is 5.",easy,"2. 5 Solution: Since $123456789=10 \times 12345678+9$, the required digit is equal to $$ (1+4+9+6+5+6+9+4+1+0) \times 12345678 $$ $+(1+4+9+6+5+6+9+4+1)$, which is the unit digit of $5 \times 8+5=45$, so the required digit is 5." 6824919673c0,"3. Given the equation about $x$: $x^{2}+a|x|+a^{2}-3=0(a \in \mathbf{R})$ has a unique real solution, then $a=$","x^{2}+a|x|+a^{2}-3$ is an even function, the only real solution must be 0, so $a^{2}-3=0$ and $a>0$.",easy,"3. $\sqrt{3}$ Detailed Explanation: Since $f(x)=x^{2}+a|x|+a^{2}-3$ is an even function, the only real solution must be 0, so $a^{2}-3=0$ and $a>0$. Therefore, $a=\sqrt{3}$." b18dff8cd75a,"Solve and discuss the following parametric equation: $$ \frac{a x-3}{b x+1}=2 $$","-3 b$, then after the transformations, the root obtained is false, as substituting this into the den",easy,"In solving the task, many only managed to reach the point that the condition $a \neq 2 b$ must be satisfied. Fewer noticed that this condition is not sufficient, since if $a=-3 b$, then after the transformations, the root obtained is false, as substituting this into the denominator on the left side of the fraction results in a value of 0. Therefore, the equation has a unique solution only if the conditions $a \neq 2 b$ and $a \neq-3 b$ are satisfied; otherwise, there is no solution." 9f15e38506cd,"2. If $\alpha, \beta \in \mathbf{R}$, then $\alpha+\beta=90^{\circ}$ is a ( ) condition for $\sin \alpha+\sin \beta>1$. (A) Sufficient but not necessary (B) Necessary but not sufficient (C) Sufficient and necessary (D) Neither sufficient nor necessary","\sqrt{3}>1$, but $\alpha+\beta \neq 90^{\circ}$.",easy,"2. D. When $\alpha=0, \beta=90^{\circ}$, $\sin \alpha+\sin \beta=1$. When $\alpha=\beta=60^{\circ}$, $\sin \alpha+\sin \beta=\sqrt{3}>1$, but $\alpha+\beta \neq 90^{\circ}$." 05500519329e,"6. Given $a_{1}, a_{2}, \cdots, a_{2011}$ is a sequence of distinct positive integers. If the order of these 2011 numbers is arbitrarily changed and denoted as $b_{1}, b_{2}, \cdots, b_{2011}$. Then the value of the number $$ M=\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \cdots\left(a_{2011}-b_{2011}\right) $$ must be $(\quad)$. (A) 0 (B) 1 (C) odd number (D) even number",See reasoning trace,easy,"6. D. Assume $M$ is odd. Then $$ a_{i}-b_{i}(i=1,2, \cdots, 2011) $$ must all be odd, and their sum is also odd; $$ \begin{array}{l} \text { but }\left(a_{1}-b_{1}\right)+\left(a_{2}-b_{2}\right)+\cdots+\left(a_{2011}-b_{2011}\right) \\ =\left(a_{1}+a_{2}+\cdots+a_{2011}\right)- \\ \left(b_{1}+b_{2}+\cdots+b_{2011}\right) \\ = \end{array} $$ is even, a contradiction. Therefore, $M$ is even." 0bcb5a89e95c,"2. Let $a, b>0, \theta \in\left(0, \frac{\pi}{2}\right)$. Find $$ f(\theta)=\frac{a}{\sin \theta}+\frac{b}{\cos \theta} $$ the minimum value.",See reasoning trace,medium,"$$ \begin{array}{l} \left(\frac{a}{\sin \theta}+\frac{b}{\cos \theta}\right)^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ \geqslant\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right)^{3} \\ \Rightarrow \frac{a}{\sin \theta}+\frac{b}{\cos \theta} \geqslant\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} . \end{array} $$ By Hölder's inequality, we have $$ \begin{array}{l} \left(\frac{a}{\sin \theta}+\frac{b}{\cos \theta}\right)^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ \geqslant\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right)^{3} \\ \Rightarrow \frac{a}{\sin \theta}+\frac{b}{\cos \theta} \geqslant\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} . \end{array} $$" baeca84360a3,"6. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=3$, and $a_{n+1}=a_{n}^{2}-$ $(3 n-1) a_{n}+3$. Then the sum of the first 11 terms of the sequence $\left\{a_{n}\right\}$, $S_{11}=(\quad)$. (A) 198 (B) 55 (C) 204 (D) 188","3$ yields $a_{2}=6$, substituting $a_{n}=3 n$ yields $a_{n+1}=3(n+1)$.",easy,"6. A. Substituting $a_{1}=3$ yields $a_{2}=6$, substituting $a_{n}=3 n$ yields $a_{n+1}=3(n+1)$." 553fef1e02cb,"B3. In triangle ABC, the angle $\alpha=58^{\circ}$ and the angle $\beta=84^{\circ}$. What is the measure of angle $x$ between the angle bisector of $\gamma$ and the altitude to side $c$?",52&width=1639&top_left_y=2327&top_left_x=274),medium,"B3. The angle $x$ can be calculated in two ways, namely $x=\frac{\gamma}{2}-$ $\left(90^{\circ}-\beta\right)=19^{\circ}-6^{\circ}=13^{\circ}$ or $x=90^{\circ}-\alpha-\frac{\gamma}{2}=32^{\circ}-19^{\circ}=$ $13^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=448&width=511&top_left_y=1718&top_left_x=1406) The appropriate sketch ........................................................................................................................................ ![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=48&width=1637&top_left_y=2232&top_left_x=278) ![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=48&width=1642&top_left_y=2283&top_left_x=273) ![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=52&width=1639&top_left_y=2327&top_left_x=274)" a85abb6e5b89,"From a barrel, we pour out 4 liters of wine and replace it with 4 liters of water. From the resulting mixture, we again pour out 4 liters and replace it with 4 liters of water. We repeat this process a total of three times, and in the final result, there is 2.5 liters more water than wine. How many liters of wine were originally in the barrel?",See reasoning trace,medium,"Let there be $x$ liters of wine in the barrel initially. After the first tapping, $x-4$ liters remain. After replacing the 4 liters with water, the wine content in 1 liter of the mixture is: $\frac{x-4}{x}$ liters. Thus, the second tapping lets out $4 \cdot \frac{x-4}{x}$ liters of wine from the barrel, and in the barrel remains $$ x-4-4 \cdot \frac{x-4}{x}=\frac{(x-4)^{2}}{x} $$ liters of wine. After replenishing with more water, 1 liter of the mixture contains $\frac{(x-4)^{2}}{x^{2}}$ liters of pure wine. Thus, after the third tapping, in the barrel remains $$ \frac{(x-4)^{2}}{x}-4 \cdot \frac{(x-4)^{2}}{x^{2}}=\frac{(x-4)^{3}}{x^{2}} $$ liters of wine. The amount of water that has entered is $x-\frac{(x-4)^{3}}{x^{2}}$ liters. According to the problem, this is 2.5 liters more than the amount of wine: $$ \frac{(x-4)^{3}}{x^{2}}=x-\frac{(x-4)^{3}}{x^{2}}-2.5 $$ Multiplying through by $x^{2}$ and rearranging: $$ 2(x-4)^{3}-x^{3}+2.5 x^{2}=0 $$ from which $$ x^{3}-21.5 x^{2}+96 x-128=0 $$ We need to find the roots of the obtained equation. The equation can be transformed as follows: $$ \begin{aligned} & x^{3}-16 x^{2}-5.5 x^{2}+88 x+8 x-128= \\ = & x^{2}(x-16)-5.5 x(x-16)+8(x-16)= \\ = & (x-16)\left(x^{2}-5.5 x+8\right)=0 . \end{aligned} $$ One root of the equation is $x=16$, and the other two roots are provided by the quadratic factor. Since the discriminant of the resulting quadratic equation is negative, the equation has no more real solutions. Therefore, there were originally 16 liters of wine in the barrel. We can verify by calculation that the obtained solution is indeed correct." 03bdb43c02cc,"Higher Secondary P4 If the fraction $\dfrac{a}{b}$ is greater than $\dfrac{31}{17}$ in the least amount while $b<17$, find $\dfrac{a}{b}$.",\frac{11,medium,"1. We start with the given inequality: \[ \frac{a}{b} > \frac{31}{17} \] We need to find the smallest fraction \(\frac{a}{b}\) such that \(b < 17\). 2. Rewrite \(\frac{a}{b}\) in the form: \[ \frac{a}{b} = 2 - \frac{c}{b} \] where \(c\) is a positive integer. The condition \(\frac{31}{17} < \frac{a}{b}\) translates into: \[ \frac{31}{17} < 2 - \frac{c}{b} \] Simplifying this inequality, we get: \[ \frac{31}{17} < 2 - \frac{c}{b} \implies \frac{31}{17} - 2 < -\frac{c}{b} \implies -\frac{3}{17} < -\frac{c}{b} \implies \frac{3}{17} > \frac{c}{b} \] 3. To minimize \(\frac{a}{b}\), we need to maximize \(\frac{c}{b}\). We analyze different values of \(c\) to find the maximum \(\frac{c}{b}\) under the constraint \(b < 17\). 4. Consider \(c = 3\): \[ \frac{c}{b} = \frac{3}{b} \] For \(\frac{3}{b} > \frac{3}{17}\), we need \(b < 17\). However, since \(b < 17\) is already given, this does not provide a contradiction. But we need to check if this value of \(c\) provides the smallest \(\frac{a}{b}\). 5. Consider \(c = 2\): \[ \frac{c}{b} = \frac{2}{b} \] For \(\frac{2}{b} > \frac{3}{17}\), we need: \[ b < \frac{2 \cdot 17}{3} = \frac{34}{3} \approx 11.33 \] Hence, \(b \geq 12\). The smallest possible value for \(\frac{c}{b}\) in this case is: \[ \frac{2}{12} = \frac{1}{6} \] 6. Consider \(c = 1\): \[ \frac{c}{b} = \frac{1}{b} \] For \(\frac{1}{b} > \frac{3}{17}\), we need: \[ b < \frac{17}{3} \approx 5.67 \] Hence, \(b \geq 6\). The smallest possible value for \(\frac{c}{b}\) in this case is: \[ \frac{1}{6} \] 7. Summarizing, the largest possible value for \(\frac{c}{b}\) is \(\frac{1}{6}\), and hence the smallest possible value for \(\frac{a}{b}\) is: \[ \frac{a}{b} = 2 - \frac{1}{6} = \frac{12}{6} - \frac{1}{6} = \frac{11}{6} \] The final answer is \(\boxed{\frac{11}{6}}\)." ca3b89890459,"3. (HUN) Let $x$ be an angle and let the real numbers $a, b, c, \cos x$ satisfy the following equation: $$ a \cos ^{2} x+b \cos x+c=0 . $$ Write the analogous quadratic equation for $a, b, c, \cos 2 x$. Compare the given and the obtained equality for $a=4, b=2, c=-1$. Second Day","4, b=2$, and $c=-1$ we get $4 \cos ^{2} x+2 \cos x-1=0$ and $16 \cos ^{2} 2 x+8 \cos 2 x-4=0 \Righta",medium,"3. Multiplying the equality by $4\left(a \cos ^{2} x-b \cos x+c\right)$, we obtain $4 a^{2} \cos ^{4} x+$ $2\left(4 a c-2 b^{2}\right) \cos ^{2} x+4 c^{2}=0$. Plugging in $2 \cos ^{2} x=1+\cos 2 x$ we obtain (after quite a bit of manipulation): $$ a^{2} \cos ^{2} 2 x+\left(2 a^{2}+4 a c-2 b^{2}\right) \cos 2 x+\left(a^{2}+4 a c-2 b^{2}+4 c^{2}\right)=0 $$ For $a=4, b=2$, and $c=-1$ we get $4 \cos ^{2} x+2 \cos x-1=0$ and $16 \cos ^{2} 2 x+8 \cos 2 x-4=0 \Rightarrow 4 \cos ^{2} 2 x+2 \cos 2 x-1=0$." 9928adfbbf19,"6. Write a sequence of consecutive positive integers starting with 1 on the blackboard, and erase one of the numbers. The arithmetic mean of the remaining numbers is $35 \frac{7}{17}$. The number erased is A. 7 B. 8 C. 9 D. Uncertain","35 \frac{7}{17}$, solving this gives $x=7$ as the solution.",medium,"6. Let $n$ be the largest integer written on the blackboard. If 1 is the number that is erased, then the largest arithmetic mean is $\frac{2+3+\cdots+n}{n-1}=\frac{(n+2)(n-1)}{2(n-1)}=\frac{n+2}{2}$; if $n$ is the number that is erased, then the smallest arithmetic mean is $\frac{1+2+\cdots+(n-1)}{n-1}=\frac{n(n-1)}{2(n-1)}=\frac{n}{2}$. Therefore, we have $\frac{n}{2} \leqslant 35 \frac{7}{17} \leqslant \frac{n+2}{2}$, solving this gives $68 \frac{14}{17} \leqslant n \leqslant 70 \frac{14}{17}$. Since $n$ is an integer, $n=69$ or 70. Since $35 \frac{7}{17}$ is the arithmetic mean of $n-1$ integers, $35 \frac{7}{17}(n-1)$ must be an integer. When $n=70$, this is not satisfied, so $n=69$. Let $x$ be the number that is erased, then $\frac{70 \cdot 69-2 x}{2 \cdot 68}=35 \frac{7}{17}$, solving this gives $x=7$ as the solution." 9dabf0f5d9e4,What could be the largest prime divisor of a number in the decimal system of the form $\overline{x y x y x y}$?,See reasoning trace,medium,"The prime factors of the number $\overline{x y x y x y}$ will be deduced from the following product representation: $$ \overline{x y x y x y}=10101 \cdot(10 x+y)=3 \cdot 7 \cdot 13 \cdot 37 \cdot(10 x+y) $$ From this, it follows that the largest prime factor of $\overline{x y x y x y}$ is 37, or it comes from the prime factors of $10 x+y$. The largest prime factor of numbers in the form of $10 x+y$ could be the largest two-digit prime number, which is 97. Therefore, the sought largest prime factor is 97, and it belongs to the number 979797." 7d6105ad6535,"Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which the arable land can decrease on average each year.",See reasoning trace,medium,"Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then $$ \begin{array}{l} \frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\ \geqslant \frac{M \times 10^{4}}{p} \times(1+10 \%) . \end{array} $$ Simplifying, we get $$ \begin{aligned} x \leqslant & 10^{3} \times\left[1-\frac{1.1 \times(1+0.01)^{10}}{1.22}\right] \\ = & 10^{3} \times\left[1-\frac{1.1}{1.22}\left(1+\mathrm{C}_{10}^{1} \times 0.01\right.\right. \\ & \left.\left.+\mathrm{C}_{10}^{2} \times 0.01^{2}+\cdots\right)\right] \\ \approx & 10^{3} \times\left[1-\frac{1.1}{1.22} \times 1.1045\right] \approx 4.1 . \end{aligned} $$ Therefore, the maximum average annual reduction of arable land in hectares is 4." 94e27098a189,"Example 1. The random variable $\xi$ is distributed according to the normal law with parameters $a$ and $\sigma^{2}$. From the sample $x_{1}, \ldots, x_{n}$ of values of $\xi$, the empirical moments $M_{1}^{*}=\bar{x}=2.3$ and $M_{2}^{*}=\overline{x^{2}}=8.7$ have been determined. Using these moments, find the parameters $a$ and $\sigma^{2}$.",. $\quad a=2,medium,"## Solution. 1. We calculate the theoretical moments $M \xi^{k}$ for $k=1,2$, expressing them in terms of the unknown parameters $a$ and $\sigma$. Since $\xi$ is normally distributed with parameters $a$ and $\sigma^{2}$, we have $$ \begin{aligned} & M \xi=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{-\infty}^{+\infty} x e^{-(x-a)^{2} /\left(2 \sigma^{2}\right)} d x=a \\ & M \xi^{2}=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{-\infty}^{+\infty} x^{2} e^{-(x-a)^{2} /\left(2 \sigma^{2}\right)} d x=a^{2}+\sigma^{2} \end{aligned} $$ (In this case, $a_{1}=a, a_{2}=\sigma^{2}, m_{1}\left(a_{1}, a_{2}\right)=a_{1}, m_{2}\left(a_{1}, a_{2}\right)=$ $\left.=a_{1}^{2}+a_{2}.\right)$ 2. By equating the found theoretical moments $M \xi, M \xi^{2}$ to the given empirical moments $M_{1}^{*}=2.3$ and $M_{2}^{*}=8.7$, we obtain a system of 2 equations $$ \left\{\begin{array}{l} a=2.3 \\ a^{2}+\sigma^{2}=8.7 \end{array}\right. $$ 3. We solve the system of equations (2) and obtain $a=2.3, \sigma^{2}=3.41$. Answer. $\quad a=2.3, \quad \sigma^{2}=3.41$." 1363a7293b39,"11. Let $A$ and $B$ be two distinct points on the parabola $C: y^{2}=4 x$, and $F$ be the focus of the parabola $C$. If $\overrightarrow{F A}=-4 \overrightarrow{F B}$, then the slope of the line $A B$ is (). (A) $\pm \frac{2}{3}$ (B) $\pm \frac{3}{2}$ (C) $\pm \frac{3}{4}$ (D) $\pm \frac{4}{3}$",\pm \frac{4}{3}$.,easy,"11. D. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. From $\overrightarrow{F A}=-4 \overrightarrow{F B} \Rightarrow y_{1}=-4 y_{2}$. Let $l_{A B}: y=k(x-1)$, and combine with the parabola to get $$ \begin{array}{l} k y^{2}-4 y-4 k=0 \\ \Rightarrow y_{1}+y_{2}=\frac{4}{k}, y_{1} y_{2}=-4 . \end{array} $$ Combining equations (1) and (2) yields $k= \pm \frac{4}{3}$." 9dde7e83c5b7,"Example 2 (2006 National High School Mathematics Competition Question) Express 2006 as the sum of 5 positive integers $x_{1}, x_{2}, x_{3}, x_{4}$, $x_{5}$. Let $S=\sum_{1 \leqslant i2 x+a-1 \text {, } $$ which is equivalent to $(x-1) a+x^{2}-2 x+1>0$. Let $f(a)=(x-1) a+x^{2}-2 x+1$. $\because f(a)$ is always greater than 0 for $a \in[-1,1]$, $$ \therefore\left\{\begin{array} { l } { f ( - 1 ) > 0 , } \\ { f ( 1 ) > 0 , } \end{array} \text { i.e., } \left\{\begin{array}{l} x^{2}-3 x+2>0, \\ x^{2}-x>0 . \end{array}\right.\right. $$ Solving these, we get $x>2$ or $x<1$ and $x>1$ or $x<0$. Combining the results, we get $x>2$ or $x<0$." f6b3c5f2f1de,"Example 5. Find two sets of smaller positive integer solutions for the system of equations $$ \left\{\begin{array}{l} 2 x+3 y-4 z=8, \\ 5 x-7 y+6 z=7 \end{array}\right. $$","0,1$, we get two sets of smaller positive integer solutions for the system of equations: $(3,2,1)$ a",medium,"Eliminate $z$: (1) $\times 3+$ (2) $\times 2$, we get $$ 16 x-5 y=38 \text{. } $$ Solve equation (3) for $y$, we get $$ y=\frac{9 x-38}{5}=3 x-7+\frac{-3}{5}. $$ Since $x, y$ in (4) are integers, we can let $\frac{x-3}{5}=k$ (an integer), i.e., $\square$ $$ x=5 k+3 \text{. } $$ Substitute (5) into (4), we get $$ y=16 k+2 \text{. } $$ Substitute (4) and (6) into (1), we get $$ z=\frac{1}{4}(58 k+4)=14 k+1+\frac{k}{2}. $$ Since $z$ is an integer, $k$ must be even, so we can let $k=2 n$ (where $n$ is any integer). Substitute (8) into (5), (6), and (7), we get $$ x=10 n+3, y=32 n+2, z=29 n+1 \text{. } $$ When $n=0,1$, we get two sets of smaller positive integer solutions for the system of equations: $(3,2,1)$ and $(13,34,30)$." 98c7063ce798,"2. (9-11) Kolya and Zhenya agreed to meet at the metro in the first hour of the day. Kolya arrives at the meeting place between noon and one o'clock, waits for 10 minutes, and leaves. Zhenya does the same. a) (1b) What is the probability that they will meet? b) (1b) How will the probability of meeting change if Zhenya decides to arrive before half past twelve? And Kolya, as before, between noon and one o'clock? c) (2b) How will the probability of meeting change if Zhenya decides to arrive at any time from 12:00 to 12:50, and Kolya, as before, between 12:00 and 13:00?",\(\frac{11}{36}\),medium,"# Solution. a) Let's graphically represent the arrival times. We will plot Zhenya's arrival time on the X-axis and Kolya's arrival time on the Y-axis. ![](https://cdn.mathpix.com/cropped/2024_05_06_0cd9b72133447262cb4bg-01.jpg?height=619&width=971&top_left_y=1932&top_left_x=591) The figure outlined in black is the set of arrival times for both that result in a meeting. The probability will be calculated as the ratio of the times of arrival that satisfy the meeting condition to the set of possible arrivals. Thus, we need to divide the area of the highlighted part by the area of the square \(6 \times 6\), if we count the area in cells. Answer: \(\frac{11}{36}\). b) Let's graphically represent the arrival times, as in part a). However, in this case, the possible arrival time for Zhenya will be in the interval from 0 to 30 minutes. The area of the highlighted figure relative to the area of the entire figure \(6 \times 3\) is \(\frac{5.5}{18} = \frac{11}{36}\). Thus, the probability remains unchanged. ![](https://cdn.mathpix.com/cropped/2024_05_06_0cd9b72133447262cb4bg-02.jpg?height=631&width=985&top_left_y=707&top_left_x=587) Answer: \(\frac{11}{36}\). c) Kolya, min ![](https://cdn.mathpix.com/cropped/2024_05_06_0cd9b72133447262cb4bg-02.jpg?height=600&width=822&top_left_y=1556&top_left_x=751) \[ \frac{9.5}{30} = \frac{19}{60} > \frac{11}{36} \] Answer: \(\frac{19}{60}\)." 0ff5261f9b59,"5.15. Given the vector $\bar{a}(1; -2; 5)$. Find the coordinates of the vector $\bar{b}$, lying in the $x Q y$ plane and perpendicular to the vector $\bar{a}$, if $|\vec{b}|=2 \sqrt{5}$",$(4 ; 2 ; 0)$ or $(-4 ;-2 ; 0)$,easy,"5.15. Since the vector $\bar{b} \epsilon=x O y$, it has coordinates $(x ; y ; 0)$. Using the conditions $|\vec{b}|=2 \sqrt{5}, \bar{b} \perp \bar{a}$, we can form the system of equations $$ \left\{\begin{array}{l} x^{2}+y^{2}=20 \\ x-2 y=0 \end{array}\right. $$ Solving it, we get two vectors $(4 ; 2 ; 0)$ and $(-4 ;-2 ; 0)$. Answer: $(4 ; 2 ; 0)$ or $(-4 ;-2 ; 0)$." 66d1f2045eb3,"## Given the numbers $1,2,3, \ldots, 1000$. Find the largest number $m$ with the property that by removing any $m$ numbers from these 1000 numbers, among the $1000-m$ remaining numbers, there exist two such that one divides the other. Selected Note: a) The actual working time is 3 hours. b) All c) Each ## NATIONAL MATHEMATICS OLYMPIAD Local stage - 15.02.2014 ## Grade IX ## Grading Rubric","499$ has the required property. 1 point We show that among any 501 numbers from 1 to 1000, there is ",medium,"## Problem 4 If $m \geq 500$ then, for example, by removing the numbers from 1 to 500, among the remaining numbers, from 501 to 1000, there is no pair of numbers that can be divisible by each other, because their ratio is $<2$. 2 points We will show that $m=499$ has the required property. 1 point We show that among any 501 numbers from 1 to 1000, there is one that divides another. Each of the 501 numbers is of the form $2^{k}(2 t+1), 0 \leq k \leq 9, t$ is a natural number; to each such number, we associate $2 t+1$. There are 500 odd numbers less than 1000, so two of the 501 numbers will correspond to the same odd number. Among these two numbers, one will be obtained from the other by multiplying by a power of 2. 4 points" b1d34df6fd63,"2・ 10 (1) For what natural numbers $n>2$, is there a set of $n$ consecutive natural numbers such that the largest number in the set is a divisor of the least common multiple of the other $n-1$ numbers? (2) For what $n>2$, is there exactly one set with the above property?","3$, which proves that $\{3,4,5,6\}$ is the only set that satisfies the requirement.",medium,"[Solution] (1) First, we point out that $n \neq 3$. Otherwise, let $\{r, r+1, r+2\}$ have the property required by the problem: $(r+2) \mid r(r+1)$. Since $(r+1, r+2)=1$, it follows that $(r+2) \mid r$, which is a contradiction. Below, we discuss the cases for $n \geqslant 4$ by considering whether $n$ is odd or even. (i) If $n=2k$, since $n-1$ is odd, the set of $n$ consecutive natural numbers $\{n-1, n, n+1, \cdots, 2(n-1)\}$ satisfies the requirement. (ii) If $n=2k+1$, then $n-2$ is odd, so the set of $n$ consecutive natural numbers $\{n-3, n-2, n-1, \cdots, 2(n-2)\}$ satisfies the requirement. From (i) and (ii), we can see that when $n \geqslant 4$, there is always a set of $n$ consecutive natural numbers that satisfies the requirement. (2) We first prove that when $n \geqslant 5$, there are at least two sets of $n$ consecutive natural numbers that satisfy the requirement. $\square$ (i) When $n$ is an even number not less than 6, in addition to the set given in (i), the set $\{n-5, n-4, \cdots, 2(n-3)\}$ also satisfies the requirement. (ii) When $n \geqslant 9$ is an odd number, in addition to the set given in (1), the set $\{n-7, n-6, \cdots, 2(n-4)\}$ also satisfies the requirement. (iii) When $n=5$, both $\{2,3,4,5,6\}$ and $\{8,9,10,11,12\}$ satisfy the requirement; when $n=7$, both $\{4,5,6,7,8,9,10\}$ and $\{6,7,8,9,10,11,12\}$ satisfy the requirement. Finally, we prove that when $n=4$, the only set $\{3,4,5,6\}$ satisfies the requirement. Suppose $\{k, k+1, k+2, k+3\}$ satisfies the requirement, i.e., $k+3$ divides the least common multiple of $k, k+1, k+2$. Since $(k+2, k+3)=1$, it follows that $k+3$ divides the least common multiple of $k, k+1$. Since $(k, k+1)=1$, it follows that $(k+3) \mid k(k+1)$. If $k+3$ is odd, then $k+1$ and $k+3$ are two consecutive odd numbers, so $(k+1, k+3)=1$. Therefore, $(k+3) \mid k(k+1)$ implies $(k+3) \mid k$, which is a contradiction. If $k+3$ is even, let $k+3=2m$, then we have $2m \mid 2(m-1)(2m-3)$. This implies $m \mid 2m-3$, hence $m \mid 3$. Since $m \neq 1$, it must be that $m=3$, which proves that $\{3,4,5,6\}$ is the only set that satisfies the requirement." 6b1b50ea6672,"7. The fractional part, integer part, and the number itself of a positive number form a geometric sequence, then this positive number is","x(n+x)$. From $0 < x < 1$ and $n > 0$ we know $x=\frac{\sqrt{5}-1}{2}$, so the required positive num",easy,"7. $\frac{1}{2}(\sqrt{5}+1)$ Let the decimal part of this positive number be $x$, and the integer part be $n(0 \leqslant x < 1, n > 0)$; (2) $n^{2}=x(n+x)$. From $0 < x < 1$ and $n > 0$ we know $x=\frac{\sqrt{5}-1}{2}$, so the required positive number is $n+x=\frac{\sqrt{5}+1}{2}$." 9d0e6c7b7faa,"1. Given $a, b>0, a \neq 1$, and $a^{b}=\log _{a} b$, then the value of $a^{a^{b}}-\log _{a} \log _{a} b^{a}$ is",See reasoning trace,easy,"1. -1 Explanation: From $a^{b}=\log _{a} b$ we know: $b=a^{a^{b}}, b=\log _{a} \log _{a} b$, $$ a^{a^{b}}-\log _{a} \log _{a} b^{a}=b-\log _{a}\left(a \log _{a} b\right)=b-\log _{a} a-\log _{a} \log _{a} b=b-1-b=-1 $$" 77e6987a55a0,Let $f(n) = \sum^n_{d=1} \left\lfloor \frac{n}{d} \right\rfloor$ and $g(n) = f(n) -f(n - 1)$. For how many $n$ from $1$ to $100$ inclusive is $g(n)$ even?,90,medium,"1. **Define the function \( f(n) \):** \[ f(n) = \sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor \] This function sums the floor of the division of \( n \) by each integer from 1 to \( n \). 2. **Express \( f(n) \) in terms of the number of divisors:** We claim that: \[ f(n) = \sum_{k=1}^n \tau(k) \] where \( \tau(k) \) is the number of divisors of \( k \). To understand why this is true, consider the following steps: \[ \sum_{k=1}^n \tau(k) = \sum_{k=1}^n \sum_{j \mid k} 1 \] Here, \( j \mid k \) means \( j \) is a divisor of \( k \). We can rewrite this as: \[ \sum_{k=1}^n \sum_{j \mid k} 1 = \sum_{j=1}^n \sum_{k=1}^n [j \mid k] \] where \( [j \mid k] \) is 1 if \( j \) divides \( k \) and 0 otherwise. This can be further simplified to: \[ \sum_{j=1}^n \sum_{k=1}^n [j \mid k] = \sum_{j=1}^n \left\lfloor \frac{n}{j} \right\rfloor \] Therefore, we have: \[ f(n) = \sum_{j=1}^n \left\lfloor \frac{n}{j} \right\rfloor \] 3. **Define the function \( g(n) \):** \[ g(n) = f(n) - f(n-1) \] Using the expression for \( f(n) \), we get: \[ g(n) = \sum_{k=1}^n \tau(k) - \sum_{k=1}^{n-1} \tau(k) = \tau(n) \] Thus, \( g(n) = \tau(n) \), which is the number of divisors of \( n \). 4. **Determine when \( g(n) \) is even:** \( g(n) \) is even if \( \tau(n) \) is even. The number of divisors \( \tau(n) \) is even if and only if \( n \) is not a perfect square. This is because a perfect square has an odd number of divisors (since one of the divisors is repeated). 5. **Count the number of non-perfect squares from 1 to 100:** The perfect squares between 1 and 100 are \( 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 \). There are 10 perfect squares in this range. Therefore, the number of non-perfect squares from 1 to 100 is: \[ 100 - 10 = 90 \] Conclusion: \[ \boxed{90} \]" b924305c260a,"9. (16 points) Let the real number $t \in [0, \pi]$. If the equation $\cos (x+t)=1-\cos x$ has a solution for $x$, find the range of values for $t$.",See reasoning trace,medium,"II. 9. The original equation is equivalent to $$ \cos \left(x+\frac{t}{2}\right) \cdot \cos \frac{t}{2}=\frac{1}{2} \text {. } $$ When $t=\pi$, the left side of equation (1) is 0, which clearly has no solution. When $t \in[0, \pi)$, equation (1) further simplifies to $\cos \left(x+\frac{t}{2}\right)=\frac{1}{2 \cos \frac{t}{2}}$. Note that $\cos \left(x+\frac{t}{2}\right) \in[-1,1]$, and $\cos \frac{t}{2}>0$. Therefore, equation (1) has a solution if and only if $0<\frac{1}{2 \cos \frac{t}{2}} \leqslant 1$, which implies $\frac{1}{2} \leqslant \cos \frac{t}{2} \leqslant 1 \Rightarrow 0 \leqslant t \leqslant \frac{\pi}{3}$. In summary, $t \in\left[0, \frac{2 \pi}{3}\right]$." d80cd99e8a37,"Example 6. Independent random variables $X$ and $Y$ have the following densities: $$ \begin{aligned} & p_{1}(x)= \begin{cases}0 & \text { if } x1, \\ 0.5 & \text { if }-12 \\ 0.5 & \text { if } 0 1$ $$ F_{1}(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{-1} 0 d t+\int_{-1}^{1} 0.5 d t+\int_{1}^{x} 0 d t=\int_{-1}^{1} 0.5 d t=0.5 t \bigg|_{-1}^{1}=1 $$ Therefore, $$ F_{1}(x)= \begin{cases}0 & \text { if } x \leq-1 \\ 0.5(x+1) & \text { if } -1 < x \leq 1 \\ 1 & \text { if } x > 1\end{cases} $$ $$ F_{2}(y)= \begin{cases}0 & \text { if } y \leq 0 \\ 0.5 y & \text { if } 0 < y \leq 2 \\ 1 & \text { if } y > 2\end{cases} $$ Since the random variables $X$ and $Y$ are independent, equation (2.8.19) must hold, and by finding the product $F_{1}(x) \cdot F_{2}(y)$, we obtain the distribution function $F(x, y)$ of the random variable $(X, Y):$ $$ F(x, y)= \begin{cases}0 & \text { if } x \leq-1 \text { and } y \leq 0 \\ 0.25(x+1) y & \text { if } -1 < x \leq 1, 0 < y \leq 2 \\ 0.5 y & \text { if } x > 1, 0 < y \leq 2 \\ 1 & \text { if } x > 1, y > 2\end{cases} $$ Since the random variables $X$ and $Y$ are independent, equation (2.8.20) must also hold, and by finding the product $p_{1}(x) \cdot p_{2}(y)$, we obtain the density function of the random variable $(X, Y)$: $$ p(x, y)= \begin{cases}0 & \text { if } x \leq -1 \text { or } y \leq 0 \text { or } x > 1, y > 2 \\ 0.25 & \text { if } -1 < x \leq 1, 0 < y \leq 2\end{cases} $$" 91f690490f86,Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$.,See reasoning trace,medium,"Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of $$ \underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2} $$ The unit digits follow a periodic pattern: $$ 1,4,9,6,5,6,9,4,1 \text {, } $$ The unit digit of the sum of each group is 5, so the unit digit of the sum of 199 groups is also 5. Considering the unit digit sum of the remaining four numbers is 0, the unit digit of the required sum is 5." 4bc518feec36,"Let $R$ be a square region and $n\ge4$ an integer. A point $X$ in the interior of $R$ is called $n\text{-}ray$ partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional? $\textbf{(A)}\,1500 \qquad\textbf{(B)}\,1560 \qquad\textbf{(C)}\,2320 \qquad\textbf{(D)}\,2480 \qquad\textbf{(E)}\,2500$",2320,medium,"1. **Assume the area of the square is 1.** - This simplifies calculations since the total area is normalized to 1. 2. **Identify the requirement for $n$-ray partitional points.** - A point $X$ in the interior of the square $R$ is $n$-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. 3. **Determine the conditions for 100-ray partitional points.** - Since the square must be divided into triangles, rays must emanate from $X$ to all four corners of the square. - The four rays from $X$ to the corners of the square divide the square into four triangles. - Each of these four triangles must be further divided into smaller triangles with area $\frac{1}{100}$. - Let $h$ be the length of the altitude from $X$ to the bottom side of the square. The area of the bottom triangle is $\frac{h}{2}$. - Similarly, the area of the top triangle is $\frac{1 - h}{2}$. - Both areas must be multiples of $\frac{1}{100}$. 4. **Set up the equations for the areas.** \[ \frac{h}{2} = \frac{m}{100} \quad \text{and} \quad \frac{1 - h}{2} = \frac{n}{100} \] \[ \implies h = \frac{2m}{100} = \frac{m}{50} \quad \text{and} \quad 1 - h = \frac{2n}{100} = \frac{n}{50} \] \[ \implies m = 50h \quad \text{and} \quad n = 50 - 50h \] 5. **Determine the possible values of $h$.** - $h$ must be a multiple of $\frac{1}{50}$ and $h \ne 0, 1$. - Therefore, $h = \frac{i}{50}$ where $i \in \{1, 2, ..., 49\}$. - There are 49 possible values of $h$. 6. **Repeat the process for the altitude from $X$ to the right side of the square.** - Similarly, there are 49 possible values of $h'$. - Since $X$ is uniquely determined by $h$ and $h'$, there are $49^2$ points in the square that are 100-ray partitional. 7. **Determine the conditions for 60-ray partitional points.** - Using the same logic, $X$ is 60-ray partitional if $h = \frac{i}{30}$ where $i \in \{1, 2, ..., 29\}$. - For a point to be both 100-ray partitional and 60-ray partitional, there must exist positive integers $k \in \{1, 2, ..., 29\}$ and $l \in \{1, 2, ..., 49\}$ such that: \[ \frac{k}{30} = \frac{l}{50} \implies \frac{k}{3} = \frac{l}{5} \] - From the domains of $k$ and $l$, there are 9 such pairs $(k, l)$. 8. **Count the number of points that are both 100-ray partitional and 60-ray partitional.** - There are 9 possible values of $h$ and 9 possible values of $h'$. - Therefore, there are $9^2$ points that are both 100-ray partitional and 60-ray partitional. 9. **Calculate the final number of points that are 100-ray partitional but not 60-ray partitional.** \[ 49^2 - 9^2 = (49 + 9)(49 - 9) = 58 \cdot 40 = 2320 \] The final answer is $\boxed{2320}$" b9bd0b90579f,* Let $[\tan x]$ denote the greatest integer not exceeding the real number $\tan x$. Then the solution to the equation $[\tan x]=2 \cos ^{2} x$ is $\qquad$,"0$, $\cos x=0$, at this point $\tan x$ is undefined. When $[\tan x]=2$, $\cos ^{2} x=1$, at this poi",easy,"Given $0 \leqslant 2 \cos ^{2} x \leqslant 2$, hence $[\tan x]$ can only take the values 0, 1, 2. When $[\tan x]=0$, $\cos x=0$, at this point $\tan x$ is undefined. When $[\tan x]=2$, $\cos ^{2} x=1$, at this point $\tan x=0$, which is impossible. When $[\tan x]=1$, $\cos x= \pm \frac{\sqrt{2}}{2}$, i.e., $x= \pm \frac{\pi}{4}+k \pi, k \in \mathbf{Z}$, note that $[\tan x]=$ 1, so it can only be $x=\frac{\pi}{4}+k \pi, k \in \mathbf{Z}$." 38fe3150ec8e,"In a bag, there are 8 balls, labeled from 1 to 8. After stirring these 8 balls sufficiently, two balls are drawn from the bag at a time. Let denote $ S$ the sum of the two balls. (1) Find the probability such that $ S$ doesn't exceed in eight. (2) Let $ f(n)$ be the probability such that $ S$ takes value $ n$. Find the value of $ n$ for which $ f(n)$ is maximal.",\frac{3,medium,"### Part 1: Probability that \( S \) doesn't exceed 8 1. **Calculate the total number of ways to draw 2 balls from the bag:** - There are 8 balls labeled from 1 to 8. - The number of ways to choose 2 balls out of 8 is given by the combination formula: \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \] 2. **Identify the pairs of balls whose sum \( S \) does not exceed 8:** - List all possible pairs and their sums: \[ \begin{aligned} & (1,2) = 3, \quad (1,3) = 4, \quad (1,4) = 5, \quad (1,5) = 6, \quad (1,6) = 7, \quad (1,7) = 8, \\ & (2,3) = 5, \quad (2,4) = 6, \quad (2,5) = 7, \quad (2,6) = 8, \\ & (3,4) = 7, \quad (3,5) = 8 \end{aligned} \] - The pairs that sum to 8 or less are: \[ (1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5) \] - There are 12 such pairs. 3. **Calculate the probability:** - The probability that the sum \( S \) does not exceed 8 is: \[ P(S \leq 8) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{28} = \frac{3}{7} \] ### Part 2: Value of \( n \) for which \( f(n) \) is maximal 1. **Calculate the probability \( f(n) \) for each possible sum \( n \):** - The possible sums \( S \) range from 3 to 15. - List the pairs and count the frequency of each sum: \[ \begin{aligned} & S = 3: (1,2) \quad \text{(1 pair)} \\ & S = 4: (1,3) \quad \text{(1 pair)} \\ & S = 5: (1,4), (2,3) \quad \text{(2 pairs)} \\ & S = 6: (1,5), (2,4) \quad \text{(2 pairs)} \\ & S = 7: (1,6), (2,5), (3,4) \quad \text{(3 pairs)} \\ & S = 8: (1,7), (2,6), (3,5) \quad \text{(3 pairs)} \\ & S = 9: (1,8), (2,7), (3,6), (4,5) \quad \text{(4 pairs)} \\ & S = 10: (2,8), (3,7), (4,6) \quad \text{(3 pairs)} \\ & S = 11: (3,8), (4,7), (5,6) \quad \text{(3 pairs)} \\ & S = 12: (4,8), (5,7) \quad \text{(2 pairs)} \\ & S = 13: (5,8), (6,7) \quad \text{(2 pairs)} \\ & S = 14: (6,8) \quad \text{(1 pair)} \\ & S = 15: (7,8) \quad \text{(1 pair)} \end{aligned} \] 2. **Determine the sum \( n \) with the highest frequency:** - The sum \( S = 9 \) has the highest frequency with 4 pairs. 3. **Calculate the probability \( f(9) \):** - The probability \( f(9) \) is: \[ f(9) = \frac{\text{Number of pairs summing to 9}}{\text{Total number of pairs}} = \frac{4}{28} = \frac{1}{7} \] Conclusion: \(\boxed{\frac{3}{7}}\)" ec808032e6e9,"368. Which of the three functions \[ \begin{aligned} & y=100 x \\ & y=1000+100 x \\ & y=10000+99 x \end{aligned} \] increases faster than the others?",See reasoning trace,medium,"368. $100(x+h)-100 x=100 h$ $$ [1000+100(x+h)]-(1000+100 x)=100 h \cdot .(2) $$ $$ [10000+99(x+h)]-(10000+99 x)=99 h . .(3) $$ The first two increase at the same rate and faster than the third. 369. $\quad[2(x+h)+3]-(2 x+3)=2 h$ $$ \begin{array}{r} {\left[\frac{1}{2}(x+h)+18\right]-\left[\frac{1}{2} x+18\right]=\frac{1}{2} h} \\ 4(x+h)-4 x=4 h \\ {[4(x+h)-4]-(4 x-4)=4 h} \end{array} $$ Thus: the functions (3) and (4) increase the fastest and at the same rate, while (2) increases the slowest." c01bcc836762,"6. Given that $\overrightarrow{O A} \perp \overrightarrow{O B}$, and $|\overrightarrow{O A}|=|\overrightarrow{O B}|=24$. If $t \in[0,1]$, then $$ |t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right| $$ the minimum value is ( ). (A) $2 \sqrt{193}$ (B) 26 (C) $24 \sqrt{2}$ (D) 24",See reasoning trace,medium,"6. B. Construct a square $O A C B$, connect the diagonal $A B$, and let $D$ and $E$ be points on the diagonal $A B$ and the side $O B$ respectively, such that $$ \begin{array}{l} t \overrightarrow{A B}-\overrightarrow{A O}=\overrightarrow{O D}, \\ \frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}=\overrightarrow{E D}, \\ E B=10, O D=D C . \\ \text { Therefore }|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right| \\ =|\overrightarrow{E D}|+|\overrightarrow{D C}| \leqslant|\overrightarrow{E C}|=26 . \end{array} $$" 5bceefa40ebc,"Find the smallest positive integer $k$ such that \[(16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) < k(a^2 + b^2 + c^2)^2,\] for some ordered triple of positive integers $(a,b,c)$.",k = 1297,medium,"1. **Apply the Cauchy-Schwarz Inequality:** The Cauchy-Schwarz Inequality states that for any sequences of real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\), \[ \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n y_i^2 \right) \geq \left( \sum_{i=1}^n x_i y_i \right)^2. \] In our case, we can apply this inequality to the sequences \((4a, 6b, 9c)\) and \((9a, 6b, 4c)\): \[ (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) \geq (36a^2 + 36b^2 + 36c^2)^2. \] 2. **Simplify the Right-Hand Side:** \[ (36a^2 + 36b^2 + 36c^2)^2 = 36^2(a^2 + b^2 + c^2)^2 = 1296(a^2 + b^2 + c^2)^2. \] Therefore, we have: \[ (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) \geq 1296(a^2 + b^2 + c^2)^2. \] This implies that \(k > 1296\). 3. **Check if \(k = 1297\) works:** We need to show that: \[ (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) < 1297(a^2 + b^2 + c^2)^2 \] for some positive integers \(a, b, c\). Let's test with \(a = 1\), \(c = 1\), and vary \(b\). 4. **Substitute \(a = 1\) and \(c = 1\):** \[ (16 \cdot 1^2 + 36b^2 + 81 \cdot 1^2)(81 \cdot 1^2 + 36b^2 + 16 \cdot 1^2) < 1297(1^2 + b^2 + 1^2)^2. \] Simplify the left-hand side: \[ (16 + 36b^2 + 81)(81 + 36b^2 + 16) = (97 + 36b^2)(97 + 36b^2). \] Simplify the right-hand side: \[ 1297(1 + b^2 + 1)^2 = 1297(2 + b^2)^2. \] 5. **Compare the expressions:** \[ (97 + 36b^2)^2 < 1297(2 + b^2)^2. \] Let \(x = b^2\). Then we need to show: \[ (97 + 36x)^2 < 1297(2 + x)^2. \] Expand both sides: \[ 9409 + 2 \cdot 97 \cdot 36x + 36^2x^2 < 1297(4 + 4x + x^2). \] Simplify: \[ 9409 + 6984x + 1296x^2 < 5188 + 5188x + 1297x^2. \] Combine like terms: \[ 4221 + 1796x < x^2. \] This inequality holds for sufficiently large \(x\) (i.e., \(b\)). Therefore, \(k = 1297\) works. \(\blacksquare\) The final answer is \( \boxed{ k = 1297 } \)." 31c2cff4b649,"3. Let the set $X=\{1,2, \cdots, 20\}, A$ be a subset of $X$, the number of elements in $A$ is at least 2, and all elements of $A$ can be arranged as consecutive positive integers. Then the number of such sets $A$ is $\qquad$.",See reasoning trace,easy,"If $|A|=2$, then there are 19 sets $A$; if $|A|=3$, then there are 18 sets $A$; $\cdots$; if $|A|=20$, then there is 1 set $A$. Therefore, the number of sets $A$ that satisfy the condition is $$ 1+2+\cdots+19=\frac{19 \times(1+19)}{2}=190 \text {. } $$" a1eb291ae8e8,"11.7 Find all values of the parameter \(a\), for each of which the system of equations \[ \left\{ \begin{array}{l} x^{2}+y^{2}=2 a, \\ x+\log _{2}\left(y^{2}+1\right)=a \end{array} \right. \] has a unique solution.",$a=0$,medium,"Answer: $a=0$. ![](https://cdn.mathpix.com/cropped/2024_05_06_7a0ab87e7d282e4dcdc0g-1.jpg?height=636&width=671&top_left_y=1241&top_left_x=378) Solution. Note that if $(x, y)$ is a solution to the system, then $(x, -y)$ is also a solution to this system. Therefore, if for a parameter $a$ the system has a unique solution, the value of $y$ must be zero. Then from the second equation, we have $x=a$, and substituting $x=a$, $y=0$ into the first equation, we get $a^2 = 2a$. Thus, $a=0$ or $a=2$. Now we need to check these values of $a$ (note that they are derived as a consequence of the uniqueness of the solution to the system, but do not yet guarantee uniqueness). For $a=0$, the first equation is equivalent to the equalities $x=y=0$, and for these zero values of the unknowns, the second equation is satisfied, i.e., the value $a=0$ works. For $a=2$, the system has the solution $x=2, y=0$, but for the system, this solution is not unique. Indeed, the curve $x=2-\log_2(y^2+1)$ intersects the circle $x^2+y^2=4$ not only at the point $(2; 0)$ but also at two symmetric points in the second and third quadrants. (see figure). This follows from the fact that the points $(1; 1)$ and $(1; -1)$ lie on the given curve and are inside the circle $x^2+y^2=4$ (or one can verify that the points $(0; \pm \sqrt{3})$ on the y-axis will also be such points), and for $|y|>2$, the corresponding points on the curve are outside the circle." 16c85acc8b42,6.50. $\lim _{x \rightarrow 1} \frac{x^{4}-1}{\ln x}$.,\lim _{x \rightarrow 1} \frac{\left(x^{4}-1\right)^{\prime}}{(\ln x)^{\prime}}=4$.,medium,"Solution. Let's check the conditions for applying L'Hôpital's rule: 1) there is an indeterminate form $$ \left.\frac{x^{4}-1}{\ln x}\right|_{x=1}=\frac{0}{0} $$ 2) the functions $x^{4}-1$ and $\ln x$ are differentiable in a neighborhood of the point $\left.x=1 ; 3)(\ln x)_{x=1}^{\prime} \neq 0 ; 4\right)$ the limit exists $$ \lim _{x \rightarrow 1} \frac{\left(x^{4}-1\right)^{\prime}}{(\ln x)^{\prime}}=\lim _{x \rightarrow 1} \frac{4 x^{3}}{1 / x}=\lim _{x \rightarrow 1} 4 x^{4}=4 $$ Therefore, $\lim _{x \rightarrow 1} \frac{x^{4}-1}{\ln x}=\lim _{x \rightarrow 1} \frac{\left(x^{4}-1\right)^{\prime}}{(\ln x)^{\prime}}=4$." beafae36b26b,"5. A, B, C, and D four people put up the same amount of money, to partner in buying ducks. A, B, and C each got 3, 7, and 14 more items of goods than D. At the final settlement, B paid (A) $2 x ;$ (b) 56 (C) 70 (D) 12","$ 6 (items), that is, Yi only has 1 more item of goods than the average, which means each item costs",easy,"5. C (Hint: Since $(3+7+14) \div 4=$ 6 (items), that is, Yi only has 1 more item of goods than the average, which means each item costs 14 yuan, and Bing has $14-6=8$ (items). But Jia is short of $6-3=3$ (items), and Yi has already paid 1 item's worth of money to Ding, so, Bing should pay Jia 3 items' worth of money, and pay Ding 5 items' worth of money, which is $5 \times 14=70$ (yuan))" ac4b5c077c3d,"1. One of the following numbers is prime. Which is it? A $2017-2$ B $2017-1$ C 2017 D $2017+1$ E $2017+2$",See reasoning trace,easy,"SolUtion C We see that: $2017-2=2015$ and 2015 is a multiple of 5. So $2017-2$ is not prime. $2017-1=2016$ and 2016 is a multiple of 2. So $2017-1$ is not prime. $2017+1=2018$ and 2018 is a multiple of 2. So $2017+1$ is not prime. $2017+2=2019$ and 2019 is a multiple of 3. So $2017+2$ is not prime. We are told that one of the given options is prime. We may therefore deduce that the remaining option, 2017 , is prime." 82ee4ba44fa7,"12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that $$ a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. } $$ Then $1992 a c-1811 b c-181 a b=$",See reasoning trace,easy,"12. 0 . Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have $$ \begin{array}{l} a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\ a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\ a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d . \end{array} $$ Therefore, $1992 a c-1811 b c-181 a b$ $$ =\frac{a-c}{d}+\frac{c-b}{d}+\frac{b-a}{d}=0 . $$" bbee4b36180d,"A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.",621,easy,"From the given information, $121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. Since we need to minimize the value of $n$, we want to minimize $a$, so we have $a = 5$. Then we know $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$. ~ JHawk0224" 4b48bc0fa255,"2. The product of two two-digit numbers is 4032. The second number is written with the same digits as the first, but in reverse order. What are these numbers?","4032=2^{6} \cdot 3^{2} \cdot 7$. From this, it follows that 3 is a divisor of both numbers $\overlin",medium,"Solution. Let's denote the given two-digit numbers as $\overline{\mathrm{ab}}$ and $\overline{\mathrm{ba}}$. Then, from the condition in the problem, we have $\overline{\mathrm{ab}} \cdot \overline{\mathrm{ba}}=4032=2^{6} \cdot 3^{2} \cdot 7$. From this, it follows that 3 is a divisor of both numbers $\overline{a b}$ and $\overline{\mathrm{ba}}$ (since if one of them is divisible by 9, then the other will also be divisible by 9, meaning the product of the two numbers is divisible by 81, which is incorrect). Furthermore, 7 is a divisor of only one of the numbers, for example, $\overline{\mathrm{ab}}$. This means 21 is a divisor of $\overline{\mathrm{ab}}$, i.e., $\overline{\mathrm{ab}} \in\{21,42,84\}$. By checking, it is found that the required numbers are 84 and 48." 9c22d01518e8,"Example 3 Find the minimum value of the function $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}$ with real numbers $x, y$ as variables. (1991 ""Hope Cup"" Invitational Competition Question)",(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$.,medium,"Solve $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2$. From the above equation, we can consider points $P_{1}\left(x, \frac{q}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate axes as asymptotes; when $y \in \mathbf{R},|y| \leqslant \sqrt{2}$, the trajectory of $P_{2}$ is the lower half of a circle centered at the origin with a radius of $\sqrt{2}$. From the graph (omitted), it is evident that the minimum value of $\left|P_{1} P_{2}\right|$ is $3 \sqrt{2}-\sqrt{2}$. Therefore, $u(x, y)_{\text {min }}=(3 \sqrt{2}-\sqrt{2})^{2}-2=6$ is the desired result. Note: A slight variation of this example is the 1983 Putnam Competition problem: Find the minimum value of $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$." c1207ea83edd,"1. $\sqrt{2}-\sqrt{8}+\sqrt{18}$ equals ( ). (A) 0 (B) $\sqrt{2}$ (C) $2 \sqrt{2}$ (D) $3 \sqrt{2}$",See reasoning trace,easy,"1. C Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 811b69cf1a06,"Question 52, in $\triangle ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $\sin A \sin 2A=$ $(1-\cos A)(1-\cos 2A)$, if the area of $\triangle ABC$ is $S=\frac{\sqrt{3}}{12}\left(8 b^{2}-9 a^{2}\right)$, find the value of $\cos B$.",See reasoning trace,medium,"The 52nd, Solution: According to the conditions, we have: $$ \sin \mathrm{A} \sin 2 \mathrm{~A}=2 \sin ^{2} \mathrm{~A} \cos \mathrm{A}=(1-\cos \mathrm{A})(1-\cos 2 \mathrm{~A})=(1-\cos \mathrm{A}) \cdot 2 \sin ^{2} \mathrm{~A} $$ $\Rightarrow \cos A=\frac{1}{2} \Rightarrow A=60^{\circ}$ Thus, we know: $$ \begin{array}{l} S=\frac{\sqrt{3}}{12}\left(8 \mathrm{~b}^{2}-9 \mathrm{a}^{2}\right)=\frac{b c}{2} \cdot \sin A=\frac{\sqrt{3 b c}}{4} \\ \Rightarrow 8 \mathrm{~b}^{2}-9 \mathrm{a}^{2}=3 \mathrm{bc} \end{array} $$ According to the cosine rule, we have: $$ \begin{array}{l} \mathrm{a}^{2}=\mathrm{b}^{2}+\mathrm{c}^{2}-2 \mathrm{bc} \cos \mathrm{A}=\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{bc} \\ \text { From (1) and (2), we know: } \end{array} $$ From (1) and (2), we know: $$ \begin{array}{l} 8 b^{2}-9\left(b^{2}+c^{2}-b c\right)=3 b c \\ \Rightarrow(b-3 c)^{2}=0 \\ \Rightarrow b=3 c \end{array} $$ From (2) and (3), we know: $$ \begin{array}{l} a^{2}=b^{2}+c^{2}-b c=7 c^{2} \\ \Rightarrow a=\sqrt{7} c \end{array} $$ From (3), (4), and the cosine rule, we know: $$ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 b c}=\frac{(\sqrt{7} c)^{2}+c^{2}-(3 c)^{2}}{2 \cdot \sqrt{7} \cdot c}=-\frac{\sqrt{7}}{14} $$ In conclusion, the value of $\cos B$ is $-\frac{\sqrt{7}}{14}$." 73268740f15d,"An integer is between $0$ and $999999$ (inclusive) is chosen, and the digits of its decimal representation are summed. What is the probability that the sum will be $19$?",\frac{7623,medium,"1. **Determine the total number of integers in the range:** The range of integers is from \(0\) to \(999999\) inclusive. Therefore, the total number of integers is: \[ 999999 - 0 + 1 = 1000000 \] 2. **Construct the generating function:** The generating function for the sum of the digits of the decimal representation of integers in the range \([0, 999999]\) is: \[ (1 + x + x^2 + \ldots + x^9)^6 \] This can be rewritten using the formula for the sum of a geometric series: \[ (1 + x + x^2 + \ldots + x^9)^6 = \left(\frac{1 - x^{10}}{1 - x}\right)^6 \] Simplifying further: \[ \left(\frac{1 - x^{10}}{1 - x}\right)^6 = \frac{(1 - x^{10})^6}{(1 - x)^6} = (1 - x^{10})^6 (1 - x)^{-6} \] 3. **Find the coefficient of \(x^{19}\):** We need to find the coefficient of \(x^{19}\) in the expansion of \((1 - x^{10})^6 (1 - x)^{-6}\). This can be done using the binomial theorem and the negative binomial series expansion: \[ (1 - x)^{-6} = \sum_{k=0}^{\infty} \binom{k+5}{5} x^k \] Therefore: \[ (1 - x^{10})^6 (1 - x)^{-6} = \left( \sum_{k=0}^{\infty} \binom{k+5}{5} x^k \right) - 6 \left( \sum_{k=0}^{\infty} \binom{k+5}{5} x^{k+10} \right) + \ldots \] 4. **Extract the coefficient of \(x^{19}\):** The coefficient of \(x^{19}\) in \((1 - x)^{-6}\) is \(\binom{24}{5}\): \[ [x^{19}] (1 - x)^{-6} = \binom{24}{5} \] The coefficient of \(x^9\) in \((1 - x)^{-6}\) is \(\binom{14}{5}\): \[ [x^9] (1 - x)^{-6} = \binom{14}{5} \] Therefore: \[ [x^{19}] \left((1 - x^{10})^6 (1 - x)^{-6}\right) = \binom{24}{5} - 6 \binom{14}{5} \] 5. **Calculate the binomial coefficients:** \[ \binom{24}{5} = \frac{24!}{5!(24-5)!} = 42504 \] \[ \binom{14}{5} = \frac{14!}{5!(14-5)!} = 2002 \] Therefore: \[ [x^{19}] \left((1 - x^{10})^6 (1 - x)^{-6}\right) = 42504 - 6 \cdot 2002 = 42504 - 12012 = 30492 \] 6. **Calculate the probability:** The probability that the sum of the digits of a randomly chosen integer between \(0\) and \(999999\) is \(19\) is: \[ \frac{30492}{1000000} = \frac{7623}{250000} \] The final answer is \(\boxed{\frac{7623}{250000}}\)." acb4cf0a4382,"Example 12 Given $a, b, c, d \in \mathbf{R}^{+}$, try to find $f(a, b, c, d)=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{a+b+d}+\frac{d}{a+b+c}$ the minimum value.","b=c=d$, hence the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$.",medium,"Solution 1 Since directly eliminating the denominator is complicated, consider changing the problem. Let $b+c+d=A, c+d+a=B, d+a+b=C, a+b+c=D$. Adding these four equations, we get $a+b+c+d=\frac{1}{3}(A+B+C+D)$. Thus, we can find that $a=\frac{1}{3}(B+C+D-2 A), b=\frac{1}{3}(C+D+A-2 B)$, $$ c=\frac{1}{3}(D+A+B-2 C), d=\frac{1}{3}(A+B+C-2 D). $$ Therefore, the original expression $=\frac{B+C+D-2 A}{3 A}+\frac{C+D+A-2 B}{3 B}+\frac{D+A+B-2 C}{3 C}+$ $$ \frac{A+B+C-2 D}{3 D}, $$ which is $3 \cdot f(a, b, c, d)+12$ $$ \begin{array}{l} =\frac{A+B+C+D}{A}+\frac{A+B+C+D}{B}+\frac{A+B+C+D}{C}+\frac{A+B+C+D}{D} \\ =(A+B+C+D) \cdot\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \geqslant 16, \end{array} $$ i.e., $3 \cdot f(a, b, c, d)+12 \geqslant 16$, hence $f(a, b, c, d) \geqslant \frac{4}{3}$, where equality holds if and only if $A=$ $B=C=D$, i.e., $a=b=c=d$. Therefore, the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$. Solution 2 Noting the weighted sum inequality, we have $$ \begin{aligned} f(a, b, c, d) & =\frac{a^{2}}{a(b+c+d)}+\frac{b^{2}}{b(a+c+d)}+\frac{c^{2}}{c(a+b+d)}+\frac{d^{2}}{d(a+b+c)} \\ & \geqslant \frac{(a+b+c+d)^{2}}{2(a b+a c+a d+b c+b d+c d)} \\ & =\frac{(a+b+c+d)^{2}}{(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)} \end{aligned} $$ and $a^{2}+b^{2}+c^{2}+d^{2}=\frac{a^{2}}{1}+\frac{b^{2}}{1}+\frac{c^{2}}{1}+\frac{d^{2}}{1} \geqslant \frac{(a+b+c+d)^{2}}{4}$, thus $$ f(a, b, c, d) \geqslant \frac{(a+b+c+d)^{2}}{(a+b+c+d)^{2}-\frac{1}{4}(a+b+c+d)^{2}}=\frac{4}{3} . $$ where equality holds if and only if $a=b=c=d$, hence the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$. Solution 3 Noting the Cauchy inequality or the weighted sum inequality, we have $$ \begin{aligned} f(a, b, c, d) & =\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+ \\ & \frac{a+b+c+d}{a+b+c}-4 \\ & \geqslant(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\right. \\ & \left.\frac{1}{a+b+c}\right)-4 \\ & \geqslant(a+b+c+d) \cdot \frac{4^{2}}{3(a+b+c+d)}-4=\frac{4}{3} . \end{aligned} $$ where equality holds if and only if $a=b=c=d$, hence the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$." 75b630b17b96,"Given an integer $n\geq 2$, determine the maximum value the sum $x_1+\cdots+x_n$ may achieve, as the $x_i$ run through the positive integers, subject to $x_1\leq x_2\leq \cdots \leq x_n$ and $x_1+\cdots+x_n=x_1 x_2\cdots x_n$.",S = 2n,medium,"To determine the maximum value of the sum \( x_1 + x_2 + \cdots + x_n \) subject to the conditions \( x_1 \leq x_2 \leq \cdots \leq x_n \) and \( x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n \), we can proceed as follows: 1. **Define the sum and product:** Let \( S_i = \sum_{j=1}^i x_j \) and \( P_i = \prod_{j=1}^i x_j \). We need to find the maximum \( S_n \) such that \( S_n = P_n \). 2. **Express \( x_n \) in terms of \( S_{n-1} \) and \( P_{n-1} \):** Since \( S_n = P_n \), we can write: \[ x_n = \frac{S_{n-1}}{P_{n-1} - 1} \] because \( S_n = S_{n-1} + x_n \) and \( P_n = P_{n-1} \cdot x_n \). 3. **Ensure \( x_n \geq 2 \):** Given \( x_n \geq 2 \), we have: \[ S_{n-1} \geq 2(P_{n-1} - 1) \geq P_{n-1} \] This implies \( S_{n-1} \geq P_{n-1} \). 4. **Inductive step:** For \( x_i \geq 2 \), we need \( S_{i-1} \geq P_{i-1} \). If \( x_1 = 2 \), then for \( n = 2 \), \( x_2 = 2 \). For other solutions, we have \( x_1 = x_2 = \cdots = x_k = 1 \) and \( 2 \leq x_{k+1} \leq x_{k+2} \leq \cdots \leq x_n \). 5. **Factorization approach:** For a given \( S = S_n = P_n \), we can factorize \( S \) as \( S = (y_1 + 1)(y_2 + 1) \cdots (y_l + 1) \) where \( 1 \leq y_1 \leq y_2 \leq \cdots \leq y_l \). We find \( k = S - (y_1 + 1) - (y_2 + 1) - \cdots - (y_l + 1) \) and \( n = k + l = S - y_1 - y_2 - \cdots - y_l \). 6. **Prime consideration:** \( S \) is not prime. If \( l = 2 \), then \( S = (a + 1)(b + 1) \) and \( n = (a + 1)(b + 1) - a - b = ab + 1 \). Thus, \( S \leq 2n \). 7. **General case:** If \( l \geq 3 \), then \( S = (y_1 + 1)(y_2 + 1) \cdots (y_l + 1) \) and \( n = S - y_1 - y_2 - \cdots - y_l \). We have: \[ S = n \frac{1}{1 - x}, \quad x = \frac{\sum_i y_i}{\prod_i (1 + y_i)} \leq \frac{1}{2} \implies S \leq 2n \] \( S = 2n \) only if \( x_n = n, x_{n-1} = 2, x_i = 1 \) for \( i < n - 1 \). The final answer is \( \boxed{ S = 2n } \)." a6148b2b21c6,"5. The solution set of the inequality $\left|1+x+\frac{x^{2}}{2}\right|<1$ is A. $\{x \mid-1V_{\text {prism }}$. Assuming the side length of the equilateral triangle paper is 2, then the bases of both the regular tetrahedron and the regular triangular prism are equilateral triangles with side length 1, and their area is $\frac{\sqrt{3}}{4}$. Moreover, their heights are: Thus, $V_{\text {tetrahedron }}>V_{\text {prism }}$." afddd4c0bb26,"4. A water pool has 5 inlet and outlet water pipes numbered (1), (2), ..., (5). It is known that the time it takes to fill the pool when certain pipes are opened is as shown in the table below. If all 5 pipes are opened simultaneously, ( ) hours will be needed to fill the pool. (A.) $\frac{1}{5}$ (B) $\frac{1}{2}$ (C.) 1 (D) 2 \begin{tabular}{c|c|c|c|c|c} Pipe Number & (1)(2) & (2)(3) & (3)(4) & (4)(5) & (5)(1) \\ \hline Time (hours) & 2 & 4 & 7 & 14 & 28 \\ \hline \end{tabular}",See reasoning trace,easy,"4. (D). Consider the sum of the fractions of the pool that can be filled in one hour by doubling the rate of all the pipes: $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=1 . $$ This shows that the pool is exactly filled because the combined rate of the 5 pipes doubled for one hour" 297e51baa73d,"Example 13 Find the sum of all negative roots of the equation $x^{3}-\frac{3}{2} \sqrt[3]{6} x^{2}+3=0$. (2nd Hope Cup Mathematics Competition)",See reasoning trace,medium,"Solution: The original equation can be transformed into $$ 2 x^{3}-3 \sqrt[3]{6} x^{2}+6=0 \text {. } $$ Let $\sqrt[3]{6}=y$, then the original equation can be transformed into $$ 2 x^{3}-3 x^{2} y+y^{3}=0 \text {, } $$ which is $(x-y)^{2}(2 x+y)=0$. Thus, $x=y$ or $x=-\frac{1}{2} y$, i.e., $x=\sqrt[3]{6}$ or $x=-\frac{1}{2} \sqrt[3]{6}$. $\therefore$ The sum of the negative roots of the original equation is $-\frac{1}{2} \sqrt[3]{6}$. (4) Using the average value substitution. When the sum of several algebraic expressions in the equation differs by a constant, set their average value as a new unknown, which can simplify the equation." 36f315cdbb6d,"8.394. $\left\{\begin{array}{l}\sin x+\cos y=0, \\ \sin ^{2} x+\cos ^{2} y=\frac{1}{2} .\end{array}\right.$",See reasoning trace,medium,"## Solution. Let's write the given system of equations in the form $$ \begin{aligned} & \left\{\begin{array} { l } { \sin x + \cos y = 0 , } \\ { (\sin x + \cos y)^2 - 2 \sin x \cos y = \frac { 1 } { 2 } } \end{array} \Leftrightarrow \left\{\begin{array}{l} \sin x + \cos y = 0, \\ \sin x \cos y = -\frac{1}{4} \end{array} \Rightarrow\right.\right. \\ & \Rightarrow \cos y = -\sin x, \sin x \cdot (-\sin x) = -\frac{1}{4}, \sin^2 x = \frac{1}{4}, \sin x = \pm \frac{1}{2} \end{aligned} $$ Thus, the given system is equivalent to the following system of two systems of equations: a) $\left\{\begin{array}{l}\sin x + \cos y = 0 \\ \sin x = -\frac{1}{2}\end{array}\right.$ b) $\left\{\begin{array}{l}\sin x + \cos y = 0 \\ \sin x = \frac{1}{2}\end{array}\right.$ Solving these systems, we find $$ \left\{\begin{array} { l } { x _ { 1 } = ( - 1 ) ^ { k + 1 } \frac { \pi } { 6 } + \pi k , } \\ { y _ { 1 } = \pm \frac { \pi } { 3 } + 2 \pi n , } \end{array} \left\{\begin{array}{l} x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k \\ y_{2}= \pm \frac{2}{3} \pi + 2 \pi n, \end{array} k, n \in \mathbb{Z}\right.\right. $$ Answer: $x_{1}=(-1)^{k+1} \frac{\pi}{6}+\pi k, y_{1}= \pm \frac{\pi}{3}+2 \pi n$, $$ x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k, \quad y_{2}= \pm \frac{2}{3} \pi + 2 \pi n, \text { where } k \text { and } n \in \mathbb{Z} $$" 2389cac9e30e,"4.1. Let $p$ be an odd number with exactly $n$ distinct prime divisors. How many solutions does the equation $p^{2}+b^{2}=c^{2}$ have with coprime $b$ and $c$ (i.e., primitive Pythagorean triples $(p, b, c)$)?",to the problem,easy,"4.1. Let's rewrite the equation as $p^{2}=(c-b)(c+b)$. If we denote $c-b=u, c+b=v$, then $c=(u+v) / 2, b=(v-u) / 2$, so $u$ and $v$ must be coprime, meaning each of the $n$ prime divisors of $p$ divides exactly one of the numbers $u, v$. The number of ways to partition $n$ prime divisors into two subsets is $2^{n-1}$, which is the answer to the problem." 5d94d7af9c29,"5. Given real numbers $x, y$ satisfy $$ \left(2015+x^{2}\right)\left(2015+y^{2}\right)=2^{22} \text {. } $$ Then the maximum value of $x+y$ is",See reasoning trace,easy,"5. $2 \sqrt{33}$. Notice that, $2^{22}=2048^{2}=(2015+33)^{2}$. By the Cauchy-Schwarz inequality, we have $$ \begin{array}{l} {\left[(2015-33)+33+x^{2}\right]\left[(2015-33)+y^{2}+33\right]} \\ \geqslant[(2015-33)+\sqrt{33} y+x \sqrt{33}]^{2} \\ \Rightarrow(2015+33)^{2} \\ \quad \geqslant[(2015-33)+\sqrt{33}(x+y)]^{2} \\ \Rightarrow x+y \leqslant 2 \sqrt{33} . \end{array} $$ When $x=y=\sqrt{33}$, the equality holds. Therefore, the maximum value of $x+y$ is $2 \sqrt{33}$." 9e31cb947243,"A [i]T-tetromino[/i] is formed by adjoining three unit squares to form a $1 \times 3$ rectangle, and adjoining on top of the middle square a fourth unit square. Determine the least number of unit squares that must be removed from a $202 \times 202$ grid so that it can be tiled using T-tetrominoes.",4,medium,"1. **Coloring the Grid:** - Color the $202 \times 202$ grid in a chessboard pattern, where each cell is either black or white, and adjacent cells have different colors. - A T-tetromino covers 3 cells of one color and 1 cell of the other color. Therefore, to tile the entire grid, we need an equal number of T-tetrominoes that cover 3 black and 1 white cells and those that cover 3 white and 1 black cells. 2. **Counting the Cells:** - The total number of cells in the $202 \times 202$ grid is $202 \times 202 = 40804$. - Since each T-tetromino covers 4 cells, the number of T-tetrominoes needed to cover the grid is $\frac{40804}{4} = 10201$. - However, 10201 is an odd number, which means we cannot have an equal number of T-tetrominoes of each type (3 black, 1 white and 3 white, 1 black). Therefore, it is impossible to tile the entire $202 \times 202$ grid with T-tetrominoes without removing some cells. 3. **Removing Cells:** - To make the grid tileable by T-tetrominoes, we need to remove cells such that the remaining number of cells is divisible by 4. - Since $40804$ is already divisible by 4, we need to remove at least 4 cells to make the grid tileable. 4. **Choosing Cells to Remove:** - Number the columns and rows from $0$ to $201$. - Remove the cells at positions $(200,0)$, $(0,200)$, $(201,200)$, and $(201,201)$. - After removing these 4 cells, the grid is divided into a $200 \times 200$ grid and two strips of width 2 along the top and right edges, with some cells removed. 5. **Tiling the $200 \times 200$ Grid:** - A $200 \times 200$ grid can be tiled completely by T-tetrominoes because $200 \times 200 = 40000$, which is divisible by 4. 6. **Tiling the Strips:** - Define $L_k$ as a structure obtained by deleting the upper rightmost cell and bottom leftmost cell from a $4k+1 \times 2$ board. - The rightmost part of the grid can be covered by $L_{50}$, and the uppermost part can also be covered by $L_{50}$. - It is easy to see that $L_k$ structures can be covered by T-tetrominoes. Conclusion: By removing 4 specific cells, the $202 \times 202$ grid can be tiled using T-tetrominoes. The final answer is $\boxed{4}$." edd33c8db29f,"The remainder can be defined for all real numbers $x$ and $y$ with $y \neq 0$ by \[\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor\]where $\left \lfloor \tfrac{x}{y} \right \rfloor$ denotes the greatest integer less than or equal to $\tfrac{x}{y}$. What is the value of $\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )$? $\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}$",\textbf{(B),medium,"The value, by definition, is \begin{align*} \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right) &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ &= \frac{3}{8}-\frac{2}{5}\\ &= \boxed{\textbf{(B) } -\frac{1}{40}}. \end{align*}" c891f9002b93,"One. (20 points) Given the function $$ f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. } $$ (1) Find the intervals where the function $f(x)$ is monotonically increasing; (2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ all lie on the graph of the function $y=f(x)$, and satisfy $$ x_{1}=\frac{\pi}{6}, x_{n+1}-x_{n}=\frac{\pi}{2}\left(n \in \mathbf{Z}_{+}\right) . $$ Find the value of $y_{1}+y_{2}+\cdots+y_{2018}$.",0$.,medium,"(1) From the problem, we have $$ f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. } $$ Therefore, the monotonic increasing interval of $f(x)$ is $$ \left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. } $$ (2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\frac{\pi}{6}=\frac{\pi}{2}$. Then $t_{n+1}-t_{n}=\pi \Rightarrow t_{n}=\left(n-\frac{1}{2}\right) \pi$. Thus, $y_{n}=2 \sin \frac{2 n-1}{2} \pi$ $$ =\left\{\begin{array}{ll} 2, & n=2 k-1 ; \\ -2, & n=2 k \end{array}(k \in \mathbf{Z}) .\right. $$ Therefore, $y_{1}+y_{2}+\cdots+y_{2018}=0$." 4f6bafe79349,"【Question 10】 A total of 100 students from grades three, four, and five participated in a tree planting activity, planting a total of 566 trees. It is known that the number of students in grades three and four is equal, with third-grade students planting an average of 4 trees per person, fourth-grade students planting an average of 5 trees per person, and fifth-grade students planting an average of 6.5 trees per person. The third-grade students planted $\qquad$ trees in total.",84$ trees.,medium,"【Analysis and Solution】 (Method One) If all 100 people were fifth-grade students, then they would plant $6.5 \times 100=650$ trees; This is $650-566=84$ trees more than the actual number; Replacing 2 fifth-grade students with 1 third-grade student and 1 fourth-grade student would result in $6.5 \times 2-4-5=4$ fewer trees; The number of third-grade students is $84 \div 4=21$; The third-grade students planted a total of $4 \times 21=84$ trees. (Method Two) Let the number of third and fourth-grade students be $x$ each, and the number of fifth-grade students be $y$; $\left\{\begin{array}{c}2 x+y=100 \\ 4 x+5 x+6.5 y=566\end{array} ;\right.$ Solving this gives $\left\{\begin{array}{l}x=21 \\ y=58\end{array}\right.$; The third-grade students planted a total of $4 \times 21=84$ trees." 2e26ed8e88a7,"1.91 Find all natural numbers greater than 3, such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}$ divides $2^{2000}$. (13th China High School Mathematics Winter Camp, 1998)",7$ and $n=23$.,medium,"[Solution] Since 2 is a prime number, this problem is equivalent to finding natural numbers $n>3$ such that $$ \begin{aligned} 1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} & =2^{k}(k \in \mathbb{N}, k \leqslant 2000). \\ 1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} & =1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\ & =\frac{(n+1)\left(n^{2}-n+6\right)}{6} \end{aligned} $$ Thus, we have $(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1}$. By substituting $m=n+1$, we get $$ m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1}. $$ We now classify and discuss $m$. (1) If $m=2^{c}(m>4, s \geqslant 3)$, then $$ m^{2}-3 m+8=3 \times 2^{t}(t \in \mathbb{N}). $$ If $s \geqslant 4$, then $m^{2}-3 m+8=3 \times 2^{t} \equiv 8(\bmod 16)$. This implies that $t=3$, thus $$ m^{2}-3 m+8=24, \text{ i.e., } m(m-3)=16. $$ This is impossible. Therefore, only $s=3, m=8$, i.e., $n=7$. (2) If $m=3 \times 2^{u}(m>4, u \geqslant 1)$, then $$ m^{2}-3 m+8=2^{v}(v \in \mathbb{N}). $$ If $u \geqslant 4$, then $$ m^{2}-3 m+8=2^{v} \equiv 8(\bmod 16), $$ This implies that $v=3$. Hence $m(m-3)=0$. This is also impossible. For $u=1$ and $u=2$, i.e., $m=3 \times 2$ and $m=3 \times 2^{2}$, neither can satisfy $m^{2}-3 m+8=2^{v}$. When $m=3 \times 2^{3}=24$, we can find $n=23$. Since $\quad 1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=64=2^{6} \leqslant 2^{2000}$, and $1+C_{23}^{1}+C_{23}^{2}+C_{23}^{3}=2^{11} \leqslant 2^{2000}$. Therefore, the solutions to this problem are $n=7$ and $n=23$." c6eb23ff9a66,"2. Vasya has coins worth 49 tugriks, and Petya has coins worth 99 tugriks (each has a sufficient number of coins). Vasya owes Petya one tugrik. Will they be able to settle the debt?",they can,medium,"Solution. We need to find a solution in non-negative integers for the equation $49 n-99 m=1$. Notice that $99-49 \cdot 2=1$ or $49 \cdot(-2)-99 \cdot(-1)=1(*)$. This solution is not suitable for us due to the negativity of the coefficients. Consider the homogeneous equation $49 x-99 y=0$. It has a series of solutions $x=99 k, y=49 k$, where $k$ runs through the set of integers. By adding the solutions of the homogeneous equation to equation (*), we obtain a series of solutions for the non-homogeneous equation, among which we can choose those with positive coefficients, for example, $97 \cdot 49-48 \cdot 99=1$. Answer: they can. Comment. A correct example is given -7 points." ab3695d6791b,"12.143 $a$ is a real number greater than zero. It is known that there exists a unique real number $k$, such that the quadratic equation in $x$, $x^{2}+\left(k^{2}+a k\right) x+1999+k^{2}+a k=0$, has two roots that are both prime numbers. Find the value of $a$. (China Junior High School Mathematics League, 1999)","a^{2}-4 \times 502=0$, yielding $a=2 \sqrt{502}$.",medium,"［Solution］Let the two prime roots of the equation be $p, q$. By the relationship between the roots and coefficients of a quadratic equation, we have $$ \begin{array}{c} p+q=-k^{2}-a k, \\ p q=1999+k^{2}+a k . \end{array} $$ Adding (1) and (2) gives $$ p+q+p q=1999 \text {. } $$ That is, $$ (p+1)(q+1)=2000=2^{4} \times 5^{3} . $$ From (3), we know that $p$ and $q$ cannot both be 2, so $p$ and $q$ are odd primes. If $\frac{p+1}{2}$ is odd, it must be that $\frac{p+1}{2}=5^{r} (r=1,2,3)$. Then $p=2 \cdot 5^{r}-1$, which means $p=9$ or 49, 249, all of which are composite numbers, so $\frac{p+1}{2}$ must be even. Similarly, $\frac{q+1}{2}$ is also even. Therefore, $\frac{p+1}{4}, \frac{q+1}{4}$ are both integers, and $\frac{p+1}{4} \cdot \frac{q+1}{4}=5^{3}$. Assume without loss of generality that $p \leqslant q$, then $\frac{p+1}{4}=1$ or 5, which gives $p=3, q=499$, or $p=19, q=99$. But $q=99$ is not a prime number. Thus, only $p=3, q=499$. Substituting into (1) gives $3+499=-k^{2}-a k$, which simplifies to $k^{2}+a k+502=0$. According to the problem, this equation has a unique real solution. Therefore, $\Delta=a^{2}-4 \times 502=0$, yielding $a=2 \sqrt{502}$." 4f62aea5db11,"2.4. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}+4 a x-1$ takes values, the modulus of which does not exceed 4, at all points of the interval $[-4 ; 0]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong.",. 2 (the set of desired values of $a: \left[-\frac{5}{4} ; 0\right) \cup\left(0 ; \frac{3}{4}\right]$),easy,Answer. 2 (the set of desired values of $a: \left[-\frac{5}{4} ; 0\right) \cup\left(0 ; \frac{3}{4}\right]$). 7c7fb16d3834,"Let $k \geqslant 0$ be given. Determine the minimum of the expression $A=\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}}+\frac{z}{1+t^{2}}+$ $\frac{t}{1+z^{2}}$ under the condition $x+y+z+t=k$ and $x, y, z, t$ positive.","1+y^{2}$ and $Y=1+x^{2}$, which gives $\left(\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}}\right)((x+y)(1+x y)",medium,"We apply $\left(\frac{x}{X}+\frac{y}{Y}\right)(x X+y Y) \geqslant(x+y)^{2}$ to $X=1+y^{2}$ and $Y=1+x^{2}$, which gives $\left(\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}}\right)((x+y)(1+x y)) \geqslant(x+y)^{2}$. If $(x, y) \neq(0,0)$, this simplifies to $\frac{x}{1+y^{2}}+$ $\frac{y}{1+x^{2}} \geqslant \frac{x+y}{1+x y^{2}}$, an inequality that is always true for $x=y=0$. Since $x y \leqslant(x+y)^{2} / 4 \leqslant k^{2} / 4$, we deduce that $\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}} \geqslant \frac{4(x+y)}{4+k^{2}}$, hence $A \geqslant \frac{4 k}{4+k^{2}}$, a bound that is achieved for $x=y=k / 2$ and $z=t=0$ for example." 33999a4709b1,"19. Next year's 'leap day', 29th February 2024 , may be written in 'ddmmyy' format as the number 290224. What is the sum of all the two-digit factors of 290224 ?",397$.,easy,"Solution 397 The prime factorisation of 290224 is $290224=2 \times 2 \times 2 \times 2 \times 11 \times 17 \times 97$. Since the product of any two of $11,16=2^{4}, 17$ and 97 will not be a two-digit number, we need only consider 16 ; as well as $11,17,97$ and any multiples of each of these by the powers of two. The only such two-digit multiples are $22,44,88 ; 34,68$. The sum of these numbers is $11+16+17+97+22+44+88+34+68=397$." 127e17e0eba0,"Determine all integers $n$ for which $$ \sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}} $$ is an integer.",\left(\frac{p^{2}-25}{2}\right)^{2}$ and obviously $p$ is an odd number not less than 5. If $p \geq ,easy,"Let $$ p=\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}=\sqrt{25+2 \sqrt{n}} . $$ Then $n=\left(\frac{p^{2}-25}{2}\right)^{2}$ and obviously $p$ is an odd number not less than 5. If $p \geq 9$ then $n>\frac{625}{4}$ and the initial expression would be undefined. The two remaining values $p=5$ and $p=7$ give $n=0$ and $n=144$ respectively." 14f6ccbae298,"4. $[x]$ represents the greatest integer not exceeding the real number $x$. If $$ \left[\log _{3} 6\right]+\left[\log _{3} 7\right]+\cdots+\left[\log _{3} n\right]=2009 \text {, } $$ determine the value of the positive integer $n$.","2}^{4} 2 \times 3^{k} k+5\left(n-3^{5}+1\right)=2009$, solving for $n$ gives $n=474$.",medium,"4. For $3^{k} \leqslant i \leqslant 3^{k+1}-1$, we have $\left[\log _{3} i\right]=k$. Therefore, $\sum_{i=3^{k}}^{3 k+1}\left[\log _{3} i\right]=2 \times 3^{k} k$. Also, $3+\sum_{k=2}^{4} 2 \times 3^{k} k<2009<3+\sum_{k=2}^{5} 2 \times 3^{k} k$, hence $3^{5}0$. Since $A D=7 k$ and $F$ divides $A D$ in the ratio $4: 3$, then $A F=4 k$ and $F D=3 k$. Drop a perpendicular from $D$ to $P$ on $B C$ extended, and let the height of trapezoid $A B C D$ be $5 h$, for some real number $h>0$. Draw a line through $E$ parallel to $A D$ and $B C$ meeting $D P$ at $Q$. Note that $E Q$ is perpendicular to $D P$. Now $\triangle D Q E$ is similar to $\triangle D P C$ since each is right-angled, and the two triangles share a common angle at $D$. Since $C E: E D=2: 3$, then $D E: D C=3: 5$. By similarity, $D Q: D P=3: 5$; since $D P=5 h$, then $D Q=3 h$. This means that $Q P=D P-D Q=2 h$. ![](https://cdn.mathpix.com/cropped/2024_04_30_264fdb419065898dfdf2g-08.jpg?height=350&width=572&top_left_y=2270&top_left_x=836) Trapezoid $A D B C$ has parallel sides $A D=7 k$ and $B C=5 k$, and height $5 h$; thus, its area is $\frac{1}{2}(7 k+5 k)(5 h)=30 h k$. $\triangle F D E$ has base $F D=3 k$ and height equal to the distance from $F D$ to $E Q$, which is $3 h$; thus, its area is $\frac{1}{2}(3 k)(3 h)=\frac{9}{2} h k$. $\triangle B C E$ has base $B C=5 k$ and height equal to the distance from $B C$ to $E Q$, which is $2 h$; thus, its area is $\frac{1}{2}(5 k)(2 h)=5 h k$. We are given that the area of quadrilateral $A B E F$ is 123 . Trapezoid $A B C D$ is made up of quadrilateral $A B E F, \triangle F D E$, and $\triangle B C E$. Thus, the area of trapezoid $A B C D$ equals the sum of the areas of quadrilateral $A B E F, \triangle F D E$, and $\triangle B C E$. This gives $$ \begin{aligned} 30 h k & =123+\frac{9}{2} h k+5 h k \\ \frac{41}{2} h k & =123 \\ h k & =6 \end{aligned} $$ Since the area of trapezoid $A B C D$ equals $30 h k$, then this area is $30(6)$ or 180 . ANSWER: 180 ## Individual Problems" 106504799e37,"7. Given $\operatorname{tg} 2 \alpha=\frac{3}{4}, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, when the function $f(x)=\sin (x+2)+$ $\sin (\alpha-x)-2 \sin \alpha$ has a minimum value of 0, find the values of $\cos 2 \alpha$ and $\operatorname{tg} \frac{\alpha}{2}$.","-\frac{4}{5}, \sin \alpha=-\frac{3 \sqrt{10}}{10}, \operatorname{tg} \frac{\alpha}{2}=\frac{1-\sqrt{",medium,"7. From $\operatorname{tg} 2 \alpha=\frac{3}{4}, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we get $\operatorname{tg} \alpha=-3$ or $\operatorname{tg} \alpha=\frac{1}{3}$. Given the function $f(x)=2 \sin \alpha(\cos x-1)$, when $\cos x=1$, $f(x)=0$ is the minimum value. Therefore, $\sin \alpha<0$, which means $-\frac{\pi}{2}<\alpha<0$, hence $\operatorname{tg} \alpha=-3$. Thus, $\cos 2 \alpha=-\frac{4}{5}, \sin \alpha=-\frac{3 \sqrt{10}}{10}, \operatorname{tg} \frac{\alpha}{2}=\frac{1-\sqrt{10}}{3}$." 9a1a3165cec0,"1. Solve the inequality $\frac{\sqrt{\frac{x}{\gamma}+(\alpha+2)}-\frac{x}{\gamma}-\alpha}{x^{2}+a x+b} \geqslant 0$. In your answer, specify the number equal to the number of integer roots of this inequality. If there are no integer roots, or if there are infinitely many roots, enter the digit 0 in the answer sheet. $$ \alpha=3, \gamma=1, a=-15, b=54 . $$",7,medium,"# Answer: 7. Solution. Given the numerical values, we solve the inequality: $$ \frac{\sqrt{x+5}-x-3}{x^{2}-15 x+54} \geqslant 0 \Leftrightarrow \frac{\sqrt{x+5}-x-3}{(x-6)(x-9)} \geqslant 0 $$ Since $\sqrt{x+5} \geqslant x+3 \Leftrightarrow\left[\begin{array}{c}-5 \leqslant x \leqslant-3, \\ x+5 \geqslant(x+3)^{2}\end{array}\right.$, the expression in the numerator is non-negative for $x \in[-5 ;-1]$, and negative for $x>-1$. It is also true that $x^{2}-15 x+54<0$ for $x \in(6 ; 9)$, otherwise this expression is non-negative. Therefore, $x \in[-5 ;-1] \bigcup(6 ; 9)$, and the number of integer solutions is $5+2=7$." 546fb186f199,"7.2. Find the GCD of all numbers obtained by all possible permutations of the digits of the number 202120222023 Solution. By the divisibility rule, all these numbers are divisible by 9 (the sum of the digits is 18). A sufficient condition to prove that there are no other numbers is that the difference between any two such numbers is also divisible by the GCD. For example, 222222100032 - 222222100023 = 9 is divisible by 9 and cannot be greater than 9.",9,easy,"Answer: 9. ## Criteria 7 points - complete solution; 3 points - GCD found, but not proven that there can be no other divisors." f0acc0fa689f,"Raskina I.V. Every day the sheep learns the same number of languages. By the evening of his birthday, he knew 1000 languages. By the evening of the first day of the same month, he knew 820 languages, and by the last day of the month - 1100 languages. When is the sheep's birthday?",February 19,medium,"In the month when the sheep's birthday occurs, not counting the first day, the sheep learned $1100-820=280$ languages. A month can have 28, 29, 30, or 31 days. Therefore, the month minus one day has 27, 28, 29, or 30 days. Since the sheep learns the same number of languages each day, 280 must be divisible by the number of days without a remainder. The only number that satisfies this condition is 28, so the sheep learns $280: 28=10$ languages per day. From the second day of the month until the birthday, the sheep learned $1000-820=180$ languages. Therefore, 18 days have passed, which means the sheep's birthday is on February 19. ## Answer February 19. Send a comment" 227fa5113d22,"G8.4 If $m n p, n m p, m m p$ and $n n p$ are numbers in base 10 composed of the digits $m, n$ and $p$, such that: $m n p-n m p=180$ and $m m p-n n p=d$. Find $d$.",180 \\ 100(m-n)-10(m-n)=180 \\ m-n=2 \\ d=m m p-n n p \\ \quad=100 m+10 m+p-(100 n+10 n+p) \\ =110(m,easy,$\begin{array}{l}100 m+10 n+p-(100 n+10 m+p)=180 \\ 100(m-n)-10(m-n)=180 \\ m-n=2 \\ d=m m p-n n p \\ \quad=100 m+10 m+p-(100 n+10 n+p) \\ =110(m-n) \\ =220\end{array}$ 32fd6a825887,"22. A rectangular prism with its top and bottom faces as squares is cut into two equal-volume cubes, and the surface area increases by 98 square centimeters. Then the original volume of the rectangular prism is $\qquad$ cubic centimeters.",686,easy,Reference answer: 686 ef415cd5891e,"2. For a sequence $P=\left(p_{1}, p_{2}\right.$, $\cdots, p_{n}$ ), the ""Cesàro sum"" of $P$ (Cesàro is a mathematician) is defined as $\frac{S_{1}+S_{2}+\cdots+S_{n}}{n}$, where, $$ S_{k}=p_{1}+p_{2}+\cdots+p_{k}(1 \leqslant k \leqslant n) . $$ If the Cesàro sum of the sequence $\left(p_{1}, p_{2}, \cdots, p_{2006}\right)$ is 2007, then the Cesàro sum of the sequence $\left(1, p_{1}, p_{2}, \cdots, p_{2006}\right)$ is ( ). (A) 2007 (B) 2008 (C) 2006 (D) 1004",See reasoning trace,medium,"2.A. Let $S_{k}^{\prime}=1+p_{1}+p_{2}+\cdots+p_{k-1}=1+S_{k-1}$, then $$ \begin{array}{l} \frac{\sum_{k=1}^{2007} S_{k}^{\prime}}{2007}=\frac{2007+\sum_{k=1}^{2006} S_{k}}{2007} \\ =\frac{2007+2007 \times 2006}{2007}=2007 . \end{array} $$" 9e0315e1ba41,"30.3. In a trapezoid with bases $x$ and 5, express the distance between the midpoints of the diagonals as a function of $x$. $$ \text { (8-9 grades) } $$",See reasoning trace,medium,"30.3. Let in trapezoid $ABCD$ points $K$ and $M$ be the midpoints of the lateral sides $AB$ and $CD$, respectively. Denote by $P$ and $Q$ the points of intersection of the midline $KM$ with the diagonals $AC$ and $BD$, respectively. Let $AD = x$, $BC = 5$, and consider two cases: 1) Suppose $x > 5$ (Fig. 26a). Then, taking into account that $KP = QM = 0.5 \cdot BC = 2.5$, we get: $$ PQ = KM - KP - QM = \frac{x + 5}{2} - 5 = \frac{x - 5}{2} $$ 2) Suppose $x < 5$ (Fig. 26b). Since $KP = QM = 0.5 \cdot AD = 0.5x$, we have: $$ PQ = KM - KP - QM = \frac{x + 5}{2} - x = \frac{5 - x}{2} $$ ![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-41.jpg?height=314&width=830&top_left_y=241&top_left_x=610) Fig. 26 Combining both cases and denoting $PQ$ by $y$, we get: $$ y = \frac{|x - 5|}{2} $$" ecd11e7580cc,"Find the maximum value of $M =\frac{x}{2x + y} +\frac{y}{2y + z}+\frac{z}{2z + x}$ , $x,y, z > 0$",1,medium,"To find the maximum value of \( M = \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \) for \( x, y, z > 0 \), we will use the method of inequalities. 1. **Initial Setup:** We need to show that: \[ \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1 \] 2. **Using the AM-GM Inequality:** Consider the following application of the AM-GM (Arithmetic Mean-Geometric Mean) inequality: \[ \frac{x}{2x + y} \leq \frac{x}{2x} = \frac{1}{2} \] Similarly, \[ \frac{y}{2y + z} \leq \frac{y}{2y} = \frac{1}{2} \] and \[ \frac{z}{2z + x} \leq \frac{z}{2z} = \frac{1}{2} \] 3. **Summing the Inequalities:** Adding these inequalities, we get: \[ \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \] However, this approach does not directly help us to prove the desired inequality. We need a different approach. 4. **Using the Titu's Lemma (Cauchy-Schwarz in Engel Form):** Titu's Lemma states that for non-negative real numbers \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\), \[ \frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{b_1 + b_2 + \cdots + b_n} \] Applying Titu's Lemma to our problem, we set \(a_1 = \sqrt{x}\), \(a_2 = \sqrt{y}\), \(a_3 = \sqrt{z}\), \(b_1 = 2x + y\), \(b_2 = 2y + z\), \(b_3 = 2z + x\): \[ \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \geq \frac{(x + y + z)^2}{2x + y + 2y + z + 2z + x} = \frac{(x + y + z)^2}{3(x + y + z)} = \frac{x + y + z}{3} \] 5. **Conclusion:** Since \( \frac{x + y + z}{3} \leq 1 \) for all positive \(x, y, z\), we have: \[ \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1 \] Therefore, the maximum value of \( M \) is \(1\). The final answer is \(\boxed{1}\)." 9d7f49e3d859,"Find the largest four-digit number, all digits of which are different and which is divisible by 2, 5, 9, and 11. #",8910,medium,"The numbers $2, 5, 9$ and 11 have no common divisors, so if a number is divisible by each of them, it is also divisible by their product. That is, the number we are looking for is divisible by $2 \cdot 5 \cdot 9 \cdot 11=990$. Let's list all four-digit numbers that are divisible by 990: 1980, 2970, 3960, 4950, 5940, 6930, 7920, 8910, 9900. The largest of these is 9900, but it has repeating digits. The largest number with all distinct digits is 8910. ## Answer 8910. Problem 108965 Topics: $\left.\begin{array}{l}\text { Motion problems } \\ {[\text { Arithmetic. Mental calculation, etc. ]] }\end{array}\right]$ Difficulty: $2+$ Grades: 7,8 A motorboat set off upstream at 9 o'clock, and at the moment of its departure, a ball was thrown from the boat into the river. At 9:15, the boat turned and started moving downstream. At what time will the boat catch up with the ball, given that its own speed remained constant? ## Solution The speed of the river current can be disregarded, as it affects the movement of both the boat and the ball equally. Therefore, the distance between the boat and the ball changes only at the speed of the boat itself. Since the boat was moving away from the ball for 15 minutes, it will catch up with the ball 15 minutes after it turned back. ## Answer At 9:30." 29035d4b2298,"1. a) For which values of $m$ and $n$ does the expression $$ \frac{1}{4 m^{2}+12 m n+9 n^{2}+1} $$ have the greatest value? b) Sketch the graph of the function $$ y=(4 m+6 n-2)|x|+2 m+3 n $$ for all values of $m$ and $n$ for which the expression $\frac{1}{4 m^{2}+12 m n+9 n^{2}+1}$ has the greatest value.","0$, we get that the function is $y=-2|x|$ and its graph is shown in the above diagram.",medium,"Solution. a) The expression $$ \frac{1}{4 m^{2}+12 m n+9 n^{2}+1} $$ has the largest value if $$ 4 m^{2}+12 m n+9 n^{2}+1 $$ has the smallest value. Since $$ 4 m^{2}+12 m n+9 n^{2}+1=(2 m+3 n)^{2}+1 \geq 1 $$ ![](https://cdn.mathpix.com/cropped/2024_06_05_5cbf21d72fe8685988dag-04.jpg?height=353&width=589&top_left_y=436&top_left_x=871) the smallest value of $4 m^{2}+12 m n+9 n^{2}+1$ is obtained if $2 m+3 n=0$. b) Since $2 m+3 n=0$, we get that the function is $y=-2|x|$ and its graph is shown in the above diagram." 7339255f3014,"B3. The expression is $Z=5 a^{-x}\left(1-a^{-x}\right)^{-1}-3 a^{-x}\left(1+a^{-x}\right)^{-1}-2 a^{x}\left(a^{2 x}-1\right)^{-1}$, where $a^{x} \neq 0,1,-1$. a) Simplify the expression $Z$. b) Calculate the value of the expression $Z$ for $a=9^{b+c} \cdot 3^{2 b+c}: 27^{\frac{4}{3} b+c+\frac{1}{3}}$ and $x=1$. 19th Knowledge Competition in mathematics for students of secondary technical and vocational schools National Competition, April 13, 2019 ## Time for solving: 120 minutes. In section A, we will award three points for each correct answer, and deduct one point for each incorrect answer. Write your answers for section A in the left table, leave the right table blank. ![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-10.jpg?height=165&width=319&top_left_y=657&top_left_x=640) | B1 | B2 | B3 | | :--- | :--- | :--- | | | | |",137&width=220&top_left_y=357&top_left_x=929),medium,"B3. We simplify each term separately. The first term is transformed into $$ 5 a^{-x}\left(1-a^{-x}\right)^{-1}=\frac{5}{a^{x}}\left(\frac{a^{x}-1}{a^{x}}\right)^{-1}=\frac{5}{a^{x}-1} $$ Similarly, the second term is transformed into $$ 3 a^{-x}\left(1+a^{-x}\right)^{-1}=\frac{3}{a^{x}+1} $$ We factor the denominator of the third term into a product of a sum and a difference $$ 2 a^{x}\left(a^{2 x}-1\right)^{-1}=\frac{2 a^{x}}{a^{2 x}-1}=\frac{2 a^{x}}{\left(a^{x}-1\right)\left(a^{x}+1\right)} $$ We observe that this is also the common denominator for the first two terms. We extend the first two terms, simplify the expression, and obtain the result $$ \frac{5}{a^{x}-1}-\frac{3}{a^{x}+1}-\frac{2 a^{x}}{\left(a^{x}-1\right)\left(a^{x}+1\right)}=\frac{8}{\left(a^{x}-1\right)\left(a^{x}+1\right)}=\frac{8}{a^{2 x}-1} $$ We calculate the value of the expression $$ a=9^{b+c} \cdot 3^{2 b+c}: 27^{\frac{4}{3} b+c+\frac{1}{3}}=3^{2 b+2 c} \cdot 3^{2 b+c}: 3^{3\left(\frac{4}{3} b+c+\frac{1}{3}\right)}=\frac{1}{3} . $$ We calculate the value of the expression \( Z \) for the obtained values of \( a \) and \( x: \frac{8}{\left(\frac{1}{3}\right)^{2}-1}=-9 \). Simplified first term \(\frac{5}{a^{x}-1}\) ..... 1 point Simplified second term \(\frac{3}{a^{x}+1}\) ..... 1 point Simplified third term \(\frac{2 a^{x}}{a^{2 x-1}}\) ..... 1 point Extending all terms to a common denominator ..... 1 point Transforming the expression into \(\frac{8}{a^{2 x}-1}\) or \(\frac{8}{\left(a^{x}-1\right)\left(a^{x}+1\right)}\) ..... 1 point Calculated value of the expression \( a=\frac{1}{3} \) ..... \(1^*+1\) points Calculated value -9 of the expression \( Z \) for \( a=\frac{1}{3} \) and \( x=1 \) ..... 1 point ## 19th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools National Competition, April 13, 2019 ## Solutions and Scoring Guide (April 5, 2019, 10:30) A competitor who arrives at the solution by any correct method (even if the scoring guide does not provide for it) receives all possible points. A correct method is considered any procedure that: - sensibly takes into account the wording of the problem, - leads to the solution of the problem, - is mathematically correct and complete. If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided. If, however, the solution is guessed (it is not possible to arrive at it without calculation), even if it is correct, it is scored with 0 points. A competitor who has only partially solved the problem, from a generally correct solving procedure but without a clear path to the final solution of the problem, cannot receive more than half of the possible points. The mark '*' next to the points means that the point or points can be awarded to the competitor for a correct procedure, even if the calculation is incorrect. ## Third Year ![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=137&width=220&top_left_y=357&top_left_x=929)" 4141fee4e07c,"Find the maximal value of the following expression, if $a,b,c$ are nonnegative and $a+b+c=1$. \[ \frac{1}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{1}{c^2 -4c+9} \]",\frac{7,medium,"1. **Claim the Maximum Value**: We claim that the maximum value of the given expression is \(\boxed{\frac{7}{18}}\). This is achieved when \((a, b, c) = (1, 0, 0)\) and its permutations. 2. **Expression to Prove**: We need to prove that: \[ \frac{1}{a^2 - 4a + 9} \leq \frac{a + 2}{18} \] for \(a \in [0, 1]\). 3. **Multiply Both Sides**: Multiply both sides by \(18(a^2 - 4a + 9)\): \[ 18 \leq (a + 2)(a^2 - 4a + 9) \] 4. **Expand and Simplify**: Expand the right-hand side: \[ 18 \leq a^3 - 4a^2 + 9a + 2a^2 - 8a + 18 \] Simplify the expression: \[ 18 \leq a^3 - 2a^2 + a + 18 \] 5. **Subtract 18 from Both Sides**: \[ 0 \leq a^3 - 2a^2 + a \] 6. **Factor the Polynomial**: Factor the right-hand side: \[ 0 \leq a(a^2 - 2a + 1) \] \[ 0 \leq a(a - 1)^2 \] 7. **Analyze the Inequality**: Since \(a \geq 0\) and \((a - 1)^2 \geq 0\), the product \(a(a - 1)^2 \geq 0\) is always true for \(a \in [0, 1]\). 8. **Summing Up**: Since \(\frac{1}{a^2 - 4a + 9} \leq \frac{a + 2}{18}\) holds for \(a \in [0, 1]\), we can sum the inequalities for \(a, b, c\): \[ \frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9} \leq \frac{a + 2}{18} + \frac{b + 2}{18} + \frac{c + 2}{18} \] Since \(a + b + c = 1\), we have: \[ \frac{a + 2}{18} + \frac{b + 2}{18} + \frac{c + 2}{18} = \frac{(a + b + c) + 6}{18} = \frac{1 + 6}{18} = \frac{7}{18} \] Thus, the maximum value of the given expression is \(\boxed{\frac{7}{18}}\)." 705c2612f1ef,"## Task 3 - 300613 Lina buys 7 pencils and 8 notebooks and notices that 7 pencils are more expensive than 8 notebooks. Which is more expensive: 10 pencils and 2 notebooks or 11 notebooks and 2 pencils? Justify your answer!",10 pencils and 2 notebooks are more expensive than 11 notebooks and 2 pencils,easy,"Answer: 10 pencils and 2 notebooks are more expensive than 11 notebooks and 2 pencils. Reasoning: If $b$ is the price of a pencil and $h$ is the price of a notebook, then according to Lina's observation, $7 b > 8 h$. (1) From this, it follows that $7 b > 7 h$, (2) thus $b > h$. (3) From (1) and (3), it follows that $8 b > 9 h$. (4) If we now increase both $8 b$ and $9 h$ by $2 b + 2 h$, it follows that $10 b + 2 h > 2 b + 11 h$, as stated in the answer." 0cc9b465f4a4,2. (2 points) Point $M$ lies on the side of a regular hexagon with side length 10. Find the sum of the distances from point $M$ to the lines containing the other sides of the hexagon.,$30 \sqrt{3}$,medium,"Answer: $30 \sqrt{3}$. Solution: The distance from a point on one side to the opposite side is equal to the distance between these sides and, consequently, the distance between two vertices of the hexagon located one apart. This distance is obviously equal to $2 \cdot \frac{\sqrt{3}}{2} \cdot 12=10 \sqrt{3}$. Also, if we drop perpendiculars from our point to the lines containing two opposite sides of the hexagon, they form one segment, and its length will also be equal to the distance between these lines, that is, $10 \sqrt{3}$. Thus, we get the answer $30 \sqrt{3}$." 451f6ba8b986,"One of the receipts for a math tournament showed that $72$ identical trophies were purchased for $\$$-$99.9$-, where the first and last digits were illegible. How much did each trophy cost?",11.11,medium,"1. **Understanding the Problem:** We need to determine the cost of each trophy given that 72 identical trophies were purchased for a total price of $\$ -99.9-$, where the first and last digits are illegible. 2. **Convert the Total Price to Cents:** Let the total price be $x99.9y$ dollars, where $x$ and $y$ are the unknown digits. Converting this to cents, we get: \[ 100 \times (x99.9y) = 1000x + 9990 + y \] Therefore, the total price in cents is: \[ 1000x + 9990 + y \] 3. **Divisibility by 72:** The total price in cents must be divisible by 72. Since 72 is the product of 8 and 9, the total price must be divisible by both 8 and 9. 4. **Divisibility by 8:** For a number to be divisible by 8, its last three digits must be divisible by 8. The last three digits of our number are $99y$. We need to find $y$ such that $99y$ is divisible by 8. \[ 990 + y \equiv 0 \pmod{8} \] Testing values for $y$: \[ 990 \equiv 6 \pmod{8} \implies 6 + y \equiv 0 \pmod{8} \implies y \equiv 2 \pmod{8} \] Therefore, $y = 2$. 5. **Divisibility by 9:** For a number to be divisible by 9, the sum of its digits must be divisible by 9. The sum of the digits of $x99.92$ is: \[ x + 9 + 9 + 9 + 2 = x + 29 \] We need $x + 29$ to be divisible by 9. Testing values for $x$: \[ x + 29 \equiv 0 \pmod{9} \] \[ x \equiv -29 \pmod{9} \implies x \equiv 7 \pmod{9} \] Therefore, $x = 7$. 6. **Total Price:** The total price is $799.92$ dollars. 7. **Cost per Trophy:** Dividing the total price by the number of trophies: \[ \frac{799.92}{72} = 11.11 \] Therefore, the cost of each trophy is $\$11.11$. The final answer is $\boxed{11.11}$." 572489f57690,Consider a regular hexagon with an incircle. What is the ratio of the area inside the incircle to the area of the hexagon?,\frac{\pi \sqrt{3,medium,"1. **Assume the side length of the regular hexagon is 1.** - Let the side length of the hexagon be \( s = 1 \). 2. **Calculate the area of the regular hexagon.** - The formula for the area of a regular hexagon with side length \( s \) is: \[ A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} s^2 \] - Substituting \( s = 1 \): \[ A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} \cdot 1^2 = \frac{3\sqrt{3}}{2} \] 3. **Determine the radius of the incircle.** - The radius \( r \) of the incircle of a regular hexagon with side length \( s \) is given by: \[ r = \frac{s \sqrt{3}}{2} \] - Substituting \( s = 1 \): \[ r = \frac{1 \cdot \sqrt{3}}{2} = \frac{\sqrt{3}}{2} \] 4. **Calculate the area of the incircle.** - The area of a circle with radius \( r \) is: \[ A_{\text{incircle}} = \pi r^2 \] - Substituting \( r = \frac{\sqrt{3}}{2} \): \[ A_{\text{incircle}} = \pi \left( \frac{\sqrt{3}}{2} \right)^2 = \pi \cdot \frac{3}{4} = \frac{3\pi}{4} \] 5. **Find the ratio of the area of the incircle to the area of the hexagon.** - The ratio is given by: \[ \text{Ratio} = \frac{A_{\text{incircle}}}{A_{\text{hexagon}}} = \frac{\frac{3\pi}{4}}{\frac{3\sqrt{3}}{2}} = \frac{3\pi}{4} \cdot \frac{2}{3\sqrt{3}} = \frac{3\pi \cdot 2}{4 \cdot 3\sqrt{3}} = \frac{6\pi}{12\sqrt{3}} = \frac{\pi}{2\sqrt{3}} = \frac{\pi \sqrt{3}}{6} \] The final answer is \(\boxed{\frac{\pi \sqrt{3}}{6}}\)." 2fb07c7016a5,"(2) Let the internal angles $A, B, C$ of $\triangle ABC$ have opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is $\qquad$.","\frac{3}{5} c$, solving these equations simultaneously gives $a \cos B=\frac{4}{5} c, b \cos A=\frac",medium,"(2) 4 Hint: Method one From the given and the cosine rule, we get $$ a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c, $$ which means $a^{2}-b^{2}=\frac{3}{5} c^{2}$. Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}}$ $$ =\frac{c^{2}+a^{2}-b^{2}}{b^{2}+c^{2}-a^{2}}=\frac{\frac{8}{5} c^{2}}{\frac{2}{5} c^{2}}=4 \text {. } $$ Method two As shown in the figure, draw $C D \perp A B$, with the foot of the perpendicular at $D$, then $$ a \cos B=D B, b \cos A=A D \text {. } $$ From the given, we have $D B-A D=\frac{3}{5} c$. Also, $D B+D A=c$, solving these equations simultaneously gives $A D=$ $\frac{1}{5} c, D B=\frac{4}{5} c$. $$ \text { Therefore, } \frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{C D}{D B}}=\frac{D B}{A D}=4 \text {. } $$ Method three From the projection theorem, we get $a \cos B+b \cos A=c$. Also, $a \cos B-b \cos A=\frac{3}{5} c$, solving these equations simultaneously gives $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$. Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cos B}{b \cos A}=\frac{\frac{4}{5} c}{\frac{1}{5} c}=4$." 6d05277368d0,,8,medium,"Answer: 8. Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that $V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and thus are equal, and their volumes are also equal. Therefore, the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$. We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=B_{1} N=\frac{3}{5} B B_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{N N_{1}}{B B_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{1}{3} \frac{3}{5} 40=8$. ![](https://cdn.mathpix.com/cropped/2024_05_06_08868d1266709e60d590g-18.jpg?height=634&width=1232&top_left_y=684&top_left_x=413) ## Variant IV" fde06a9a858e,"Task 3. Given the digits $0,1,3,4,5$. Using these digits, write down all the five-digit numbers with different digits that are divisible by 4 but not divisible by 5.",See reasoning trace,medium,"Solution. Let the form of the numbers be $\overline{a b c d e}$, where $a, b, c, d, e \in\{0,1,3,4,5\}$ and these digits are mutually distinct. Since the number $\overline{a b c d e}$ must be a five-digit number, it follows that $a \neq 0$. From $4 \mid \overline{a b c d e}$, it follows that $4 \mid \overline{d e}$, i.e., $\overline{d e} \in\{04,40\}$. From $5 \nmid \overline{a b c d e}$, it follows that $e \notin\{0,5\}$. Therefore, $\overline{d e}=04$, i.e., $d=0$ and $e=4$. The remaining digits are $1,3,5$ from which the three-digit beginning must be formed. All possible ways to do this are $\overline{a b c} \in\{135,153,315,351,513,531\}$. Thus, the required five-digit numbers are 13504, 15304, 31504, 35104, $51304,53104$." 8b4c5cfb323c,"Let $a_1 = 2,$ and for $n\ge 1,$ let $a_{n+1} = 2a_n + 1.$ Find the smallest value of an $a_n$ that is not a prime number.",95,medium,"To find the smallest value of \(a_n\) that is not a prime number, we start by calculating the first few terms of the sequence defined by \(a_1 = 2\) and \(a_{n+1} = 2a_n + 1\). 1. Calculate \(a_1\): \[ a_1 = 2 \] Since 2 is a prime number, we move to the next term. 2. Calculate \(a_2\): \[ a_2 = 2a_1 + 1 = 2 \cdot 2 + 1 = 5 \] Since 5 is a prime number, we move to the next term. 3. Calculate \(a_3\): \[ a_3 = 2a_2 + 1 = 2 \cdot 5 + 1 = 11 \] Since 11 is a prime number, we move to the next term. 4. Calculate \(a_4\): \[ a_4 = 2a_3 + 1 = 2 \cdot 11 + 1 = 23 \] Since 23 is a prime number, we move to the next term. 5. Calculate \(a_5\): \[ a_5 = 2a_4 + 1 = 2 \cdot 23 + 1 = 47 \] Since 47 is a prime number, we move to the next term. 6. Calculate \(a_6\): \[ a_6 = 2a_5 + 1 = 2 \cdot 47 + 1 = 95 \] We need to check if 95 is a prime number. To do this, we check for divisibility by prime numbers less than \(\sqrt{95}\). - 95 is odd, so it is not divisible by 2. - The sum of the digits of 95 is \(9 + 5 = 14\), which is not divisible by 3, so 95 is not divisible by 3. - 95 ends in 5, so it is divisible by 5. Since 95 is divisible by 5, it is not a prime number. Therefore, the smallest value of \(a_n\) that is not a prime number is \(a_6 = 95\). The final answer is \(\boxed{95}\)." 327909a55ced,"Let $a_1,a_2,\ldots,a_n$ be a permutation of the numbers $1,2,\ldots,n$, with $n\geq 2$. Determine the largest possible value of the sum \[ S(n)=|a_2-a_1|+ |a_3-a_2| + \cdots + |a_n-a_{n-1}| . \] [i]Romania[/i]",\left\lfloor \frac{n^2,medium,"To determine the largest possible value of the sum \[ S(n) = |a_2 - a_1| + |a_3 - a_2| + \cdots + |a_n - a_{n-1}|, \] we need to consider the arrangement of the permutation \(a_1, a_2, \ldots, a_n\) that maximizes the sum of the absolute differences between consecutive terms. 1. **Understanding the Problem:** - We are given a permutation of the numbers \(1, 2, \ldots, n\). - We need to maximize the sum of the absolute differences between consecutive terms in this permutation. 2. **Strategy for Maximization:** - To maximize the sum of absolute differences, we should place the numbers in such a way that the difference between consecutive terms is as large as possible. - This can be achieved by alternating between the largest and smallest remaining numbers. 3. **Constructing the Permutation:** - For an odd \(n\), we can arrange the numbers as follows: \[ \left(\frac{n+1}{2}, n, 1, n-1, 2, n-2, \ldots, \frac{n+1}{2} + 1\right) \] - For an even \(n\), we can arrange the numbers as follows: \[ \left(\frac{n}{2}, n, 1, n-1, 2, n-2, \ldots, \frac{n}{2} + 1\right) \] 4. **Calculating the Sum \(S(n)\):** - For both odd and even \(n\), the sum \(S(n)\) can be calculated by summing the absolute differences between consecutive terms in the constructed permutation. 5. **Example Calculation:** - For \(n = 5\) (odd): \[ \left(3, 5, 1, 4, 2\right) \] \[ S(5) = |5 - 3| + |1 - 5| + |4 - 1| + |2 - 4| = 2 + 4 + 3 + 2 = 11 \] - For \(n = 4\) (even): \[ \left(2, 4, 1, 3\right) \] \[ S(4) = |4 - 2| + |1 - 4| + |3 - 1| = 2 + 3 + 2 = 7 \] 6. **General Formula:** - The general formula for the maximum sum \(S(n)\) is: \[ S(n) = \left\lfloor \frac{n^2}{2} \right\rfloor - 1 \] The final answer is \( \boxed{ \left\lfloor \frac{n^2}{2} \right\rfloor - 1 } \)." 0adbf21c9102,"Task B-2.6. Solve the following system of equations in the set $\mathbb{C}$ $$ \begin{aligned} & (x+2 y)^{2}-x-2 y-12=0 \\ & x \cdot y=2 \end{aligned} $$","(2,1)$.",medium,"Solution. Let $x+2 y=t$. Then $t^{2}-t-12=0, t_{1}=-3, t_{2}=4$. a) $x+2 y=-3, x \cdot y=2$. We reduce the system to the quadratic equation $2 y^{2}+3 y+2=0$, (1 point) whose solutions are $y_{1,2}=\frac{-3 \pm \sqrt{7} i}{4}$. (1 point) Then the solutions of the system are the ordered pairs $(x, y)$: $$ \left(\frac{-3+i \sqrt{7}}{2}, \frac{-3-i \sqrt{7}}{4}\right),\left(\frac{-3-i \sqrt{7}}{2}, \frac{-3+i \sqrt{7}}{4}\right) $$ b) $x+2 y=4, x \cdot y=2$. We reduce the system to the quadratic equation $y^{2}-2 y+1=0$, which has the only solution $y=1$. The solution of this system is the ordered pair $(x, y)=(2,1)$." 0c5469c77aa2,"Find the invertibles modulo 8, then their inverses. Do the same modulo 9.","5$, and $4^{-1}=7$.",easy,"Modulo 8, the invertibles are $1,3,5$ and 7, and they are their own inverses. Modulo 9, the invertibles are $1,2,4,5,7$ and 8, 1 and 8 are their own inverses, $2^{-1}=5$, and $4^{-1}=7$." 5a095534d153,"One degree on the Celsius scale is equal to 1.8 degrees on the Fahrenheit scale, while $0^{\circ}$ Celsius corresponds to $32^{\circ}$ Fahrenheit. Can a temperature be expressed by the same number of degrees both in Celsius and Fahrenheit?",Can,easy,"From the condition, it follows that the temperature in Fahrenheit is expressed through the temperature in Celsius as follows: $T_{F}=1.8 T_{C}+32^{\circ}$. If $T_{F}=T_{C}$, then $0.8 T_{C}+32=0$, that is, $T_{C}=-40$. ## Answer Can. " f286c08fd989,,See reasoning trace,medium,"Solution. We will prove that the maximum number of colored points is 10. The points that divide a given segment in the ratio $3: 4$ (or $4: 3$) are two; hence, on one segment, there are at most two colored points. On the other hand, each colored point lies on at least two segments. Therefore, the number of colored points is at most $\frac{10 \cdot 2}{2} = 10$. The diagram below shows one of the possible ![](https://cdn.mathpix.com/cropped/2024_06_03_a96bf614c78521f03e19g-1.jpg?height=394&width=405&top_left_y=1473&top_left_x=1245) examples with 10 colored points." 11f8f1a0a224,33 (1275). Find all prime numbers $p$ and $q$ for which $p^{2} - 2 q^{2}=1$.,"$p=3, q=2$",medium,"Solution. From the equality $p^{2}-2 q^{2}=1$ we have $p^{2}-1=2 q^{2}$, from which $\frac{(p-1)(p+1)}{2}=q \cdot q$. Since in the right-hand side of the equality we have the product of equal prime numbers ( $q$ has no other divisors except $q$ and 1), then in the left-hand side of the equality we also have the product of equal numbers, i.e., $p-1=\frac{p+1}{2}$, from which $p=3$. Therefore, $9-2 q^{2}=1, q^{2}=4, q=2$. Answer: $p=3, q=2$. Remark. Another way to solve this problem is given in $\$ 4$, p. 28." 123499be52f4,"10.074. One of the two parallel lines is tangent to a circle of radius $R$ at point $A$, while the other intersects this circle at points $B$ and $C$. Express the area of triangle $A B C$ as a function of the distance $x$ between the lines.",$x \sqrt{2Rx - x^2}$,easy,"## Solution. Let the distance between the parallel lines be $x$ (Fig. 10.72); then the area $S$ of triangle $ABC$ is $0.5 BC \cdot x$. Since $AM \perp BC$, then $BM = MC$ and $BM \cdot MC = AM \cdot MD = x(2R - x)$. Therefore, $0.25 BC^2 = x(2R - x)$, from which $S = 0.5 x \cdot 2 \sqrt{2Rx - x^2} = x \sqrt{2Rx - x^2}$. Answer: $x \sqrt{2Rx - x^2}$." 50f64cf7d529,"5. In $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{A B}$. Then the maximum value of $\frac{|\overrightarrow{A C}|+|\overrightarrow{A B}|}{|\overrightarrow{B C}|}$ is (). (A) $\frac{\sqrt{5}}{2}$ (B) $\sqrt{3}$ (C) 2 (D) $\sqrt{5}$",See reasoning trace,medium,"5. B. From the given conditions, $$ \begin{array}{l} a^{2}+c^{2}-b^{2}=3\left(b^{2}+c^{2}-a^{2}\right) \\ \Rightarrow 2 a^{2}=2 b^{2}+c^{2} . \\ \text { Then } (b+c)^{2}=\left(\frac{\sqrt{2}}{2} \cdot \sqrt{2} b+1 \cdot c\right)^{2} \\ \leqslant\left(\left(\frac{\sqrt{2}}{2}\right)^{2}+1^{2}\right)\left(2 b^{2}+c^{2}\right)=3 a^{2} . \\ \text { Therefore, } \frac{b+c}{a} \leqslant \sqrt{3} . \end{array} $$ Equality holds if and only if $\frac{\sqrt{2} b}{\frac{\sqrt{2}}{2}}+\frac{c}{1}, 2 a^{2}=2 b^{2}+c^{2}$, i.e., $a: b: c=\sqrt{3}: 1: 2$. Thus, the maximum value sought is $\sqrt{3}$." 06d842047945,"Let $\text{o}$ be an odd whole number and let $\text{n}$ be any whole number. Which of the following statements about the whole number $(\text{o}^2+\text{no})$ is always true? $\text{(A)}\ \text{it is always odd} \qquad \text{(B)}\ \text{it is always even}$ $\text{(C)}\ \text{it is even only if n is even} \qquad \text{(D)}\ \text{it is odd only if n is odd}$ $\text{(E)}\ \text{it is odd only if n is even}$",\text{E,medium,"Solution 1 We can solve this problem using logic. Let's say that $\text{n}$ is odd. If $\text{n}$ is odd, then obviously $\text{no}$ will be odd as well, since $\text{o}$ is odd, and the product of two odd numbers is odd. Since $\text{o}$ is odd, $\text{o}^2$ will also be odd. And adding two odd numbers makes an even number, so if $\text{n}$ is odd, the entire expression is even. Let's say that $\text{n}$ is even. If $\text{n}$ is even, then $\text{no}$ will be even as well, because the product of an odd and an even is even. $\text{o}^2$ will still be odd. That means that the entire expression will be odd, since the sum of an odd and an even is odd. Looking at the multiple choices, we see that our second case fits choice E exactly. $\boxed{\text{E}}$ Solution 2 We are given that $\text{o}\equiv 1\pmod{2}$, so in mod $2$ we have \[1^2+1(n) = n+1\] which is odd only if $\text{n}$ is even $\rightarrow \boxed{\text{E}}$" d02db5c12c0d,"3. Let $\alpha, \beta \in(0, \pi), \cos \alpha, \cos \beta$ be the roots of the equation $5 x^{2}-3 x-1=0$, then the value of $\sin \alpha \sin \beta$ is $\qquad$",See reasoning trace,medium,"$$ \begin{array}{l} \left\{\begin{array}{l} \cos \alpha + \cos \beta = \frac{3}{5}, \\ \cos \alpha \cos \beta = -\frac{1}{5} \end{array} \Rightarrow \cos^2 \alpha + \cos^2 \beta = (\cos \alpha + \cos \beta)^2 - 2 \cos \alpha \cos \beta = \frac{19}{25}\right. \\ \Rightarrow \sin^2 \alpha \sin^2 \beta = \left(1 - \cos^2 \alpha\right)\left(1 - \cos^2 \beta\right) = 1 - \left(\cos^2 \alpha + \cos^2 \beta\right) + \cos^2 \alpha \cos^2 \beta \\ = 1 - \frac{19}{25} + \frac{1}{25} = \frac{7}{25} \text{, and } \alpha, \beta \in (0, \pi) \text{, so } \sin \alpha \sin \beta = \frac{\sqrt{7}}{5} \text{. } \\ \end{array} $$" 3eac7a1fb471,"12. Find the largest natural number $n$ which satisfies the inequality $$ n^{6033}<2011^{2011} \text {. } $$",See reasoning trace,easy,"12. Answer: 12 Since $6033=3 \times 2011$, we have $$ \begin{aligned} n^{6003}2011$. Thus, the largest possible natural number $n$ satisfying the given inequality is 12 ." 4a985c13384d,"22. Let $a, b, c, d$ be positive integers, $(a, b)=24$, $(b, c)=36$, $(c, d)=54$, $70<(d, a)<100$. Then the factor of $a$ is ( ). (A) 5 (B) 7 (C) 11 (D) 13 (E) 17",13$ fits the option.,easy,"22. D. Notice that, $(a, b)=24=2^{3} \times 3$, $(b, c)=36=2^{2} \times 3^{2},(c, d)=54=2 \times 3^{3}$. Thus, $\left(2^{2} \times 3^{2}\right) 1 b$. Let $a=2^{3} \times 3 a_{1}, c=2 \times 3^{2} a_{2}$, where $a_{i} \in \mathbf{Z}_{+}(i=1,2)$. Therefore, $(a, d)=2 \times 3 k\left(k \in \mathbf{Z}_{+}\right)$. Also, $70<(d, a)<100$, hence, $$ 70<6 k<100\left(k \in \mathbf{Z}_{+}\right) \Rightarrow 12 \leqslant k \leqslant 16 \text {. } $$ Therefore, only $k=13$ fits the option." 8cb1895d6728,SG. 1 The acute angle between the 2 hands of a clock at 3:30 p.m. is $p^{\circ}$. Find $p$.,See reasoning trace,easy,"At 3:00 p.m., the angle between the arms of the clock $=90^{\circ}$ From 3:00 p.m. to 3:30 p.m., the hour-hand had moved $360^{\circ} \times \frac{1}{12} \times \frac{1}{2}=15^{\circ}$. The minute hand had moved $180^{\circ}$. $$ p=180-90-15=75 $$" 837f4a97f0c6,"16. 2.3 * In $\{1000,1001, \cdots, 2000\}$, how many pairs of consecutive integers can be added without carrying over?",156$ pairs.,medium,"Let the decimal representation of $n$ be $\overline{1 a b c}$. If one of $a, b, c$ is $5, 6, 7$ or 8, then adding $n$ and $n+1$ requires a carry; if $b=9, c \neq 9 ; b \neq 9$ or $c \neq 9, a=9$, then adding $n$ and $n+1$ also requires a carry. Therefore, $n$ must be one of the following forms: $\overline{1 a b c}, \overline{1 a b 9}, \overline{1 a 99}, \overline{1999}$, where $a, b, c \in\{0,1,2,3,4\}$. For such $n$, adding $n$ and $n+1$ does not require a carry. Therefore, there are $5^{3}+5^{2}+5+1=156$ pairs." 7e2162f32bd5,"5. In a $4 \times 4$ grid, fill each cell with 0 or 1, such that the sum of the four numbers in each $2 \times 2$ subgrid is odd. There are $\qquad$ different ways to do this.",See reasoning trace,easy,$128$ 66c8477ed49a,114. Find the geometric locus of points $M$ on the coordinate plane such that two tangents drawn from $M$ to the parabola $y=x^{2}$ are perpendicular to each other.,the line \(y = -\frac{1}{4}\),medium,"114. Find the geometric locus of points \(M\) on the coordinate plane such that two tangents drawn from \(M\) to the parabola \(y = x^2\) are perpendicular to each other. Answer: the line \(y = -\frac{1}{4}\). Hint: Let \((a; b)\) be the coordinates of point \(M\), and \(M_{1}(x_{1}; x_{1}^{2}), M_{2}(x_{2}; x_{2}^{2})\) be the points of tangency with the parabola of the two tangents from point \(M\). The equation of the tangent at point \(M_{1}\) is \(y - x_{1}^{2} = 2 x_{1}(x - x_{1})\) (here we consider that the slope of the tangent is \((x^2)' = 2x\)). Substituting the coordinates of point \(M\), we get \(x_{1}^{2} - 2 x_{1} a + b = 0\). We will get the same equation with \(x_{1}\) replaced by \(x_{2}\) for point \(M_{2}\), i.e., \(x_{1}, x_{2}\) are the roots of the equation \(x^{2} - 2 a x + b = 0\), and by Vieta's theorem \(b = x_{1} x_{2}\). From the condition of perpendicularity of the two tangents, we have \((2 x_{2}) = -\frac{1}{2 x_{1}} \Leftrightarrow x_{1} x_{2} = -\frac{1}{4}\). Therefore, \(b = x_{1} x_{2} = -\frac{1}{4}\)." dd73db5c51e8,"5. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and $M, N$ are the midpoints of edges $A_{1} D_{1}$ and $C C_{1}$, respectively, then the volume of the tetrahedron $O-M N B_{1}$ is $\qquad$ .",\frac{1}{3} S_{\triangle NQB_{1}} \cdot A_{1}B_{1}=\frac{7}{48}$.,medium,"(5) $\frac{7}{48}$ Hint: Draw a line through $O$ parallel to $AB$, intersecting edges $AD$ and $BC$ at points $E$ and $F$ respectively, and connect $BE$, taking the midpoint of $BF$ as $Q$. It is easy to see that $OQ \parallel BE \parallel B_{1}M$. Therefore, $OQ \parallel$ plane $MNB_{1}$. Hence $$ V_{O-MNB_{1}}=V_{Q-MNB_{1}}=V_{M-NQB_{1}}. $$ Also, $$ \begin{aligned} S_{\triangle NQB_{1}}= & S_{BCC_{1}B_{1}}-S_{\triangle NC_{1}B_{1}} \\ & -S_{\triangle QCN}-S_{\triangle QBB_{1}} \\ = & 1-\frac{1}{4}-\frac{3}{16}-\frac{1}{8}=\frac{7}{16}. \end{aligned} $$ Therefore, $V_{Q-MNB_{1}}=\frac{1}{3} S_{\triangle NQB_{1}} \cdot A_{1}B_{1}=\frac{7}{48}$." 74a447131210,"6. Pete was given a new electric jigsaw on his birthday, with a feature to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of 50 cm and cut it into squares with sides of 10 cm and 20 cm. How many squares in total were obtained, if the electric jigsaw shows a total cut length of $2 \mathbf{m} 80 \mathrm{~cm}$?",16,medium,"Answer: 16. Solution. The perimeter of the figures must be taken into account, in addition to the perimeter of the original square, from which we get that the total perimeter of the resulting squares is $280 \cdot 2 + 200 = 760$. Now, we can denote the number of squares through $x$ and $y$ respectively and solve the system $\left\{\begin{array}{c}10^{2} \cdot x + 20^{2} \cdot y = 50^{2}, \\ 4 \cdot 10 \cdot x + 4 \cdot 20 \cdot y = 760,\end{array}\right.$ or proceed with the following reasoning. If all the squares had a side of 10, there would be $6 \mathbf{~} 25$ of them, then the total perimeter would be $25 \cdot 40 = 1000$. If we replace four small squares $10 \times 10$ with one square $20 \times 20$, the perimeter would decrease by $160 - 80 = 80$ cm. To get 760 from 1000, this needs to be done 3 times. Therefore, there will be 3 squares $20 \times 20$ and 13 squares $10 \times 10$—a total of 16 squares. ## 2015/2016 Academic Year CRITERIA FOR DETERMINING WINNERS AND PRIZE WINNERS ${ }^{2}$ of the Lomonosov Mathematics Olympiad for school students 5-9 grades ELIMINATION STAGE WINNER: From 91 points inclusive and above. PRIZE WINNER: From 70 points to 90 points inclusive. ## FINAL STAGE WINNER (Diploma I degree): Not to be awarded. PRIZE WINNER (Diploma II degree): 100 points inclusive. PRIZE WINNER (Diploma III degree): From 80 points to 99 points inclusive. [^0]: ${ }^{2}$ Approved at the meeting of the jury of the Lomonosov Mathematics Olympiad for school students." 0d4f190a8c6a,"1. Arrange the consecutive natural numbers from 1 to 99 in sequence to form a large number: 1234567891011…979899, By extracting four consecutive digits, you can get a four-digit number, such as 5678, 1011, etc. Among the four-digit numbers obtained by this method, the largest is . $\qquad$",See reasoning trace,easy,$9909$ cdb26f490071,,"3$, then the product has decreased by $600-525=75$. This means that the number is $75: 3=25$. The se",easy,"Solution. From the condition, we have that when the first number is reduced by $4-1=3$, then the product has decreased by $600-525=75$. This means that the number is $75: 3=25$. The second number will be obtained as $600: 25=24$. Therefore, the student multiplied the numbers 25 and 24." d3b3c26729b7,"Example 6 In the quadrilateral pyramid $S-ABCD$, the side length of the square base $ABCD$ is $a$, and the side edges $SA=SB=SC=SD=2a$, $M$ and $N$ are the midpoints of edges $SA$ and $SC$ respectively. Find the distance $d$ between the skew lines $DM$ and $BN$ and the cosine value of the angle $\alpha$ they form.",See reasoning trace,medium,"Solution: Taking the center $O$ of the square base as the origin, establish a rectangular coordinate system as shown in Figure 5, then $$ \begin{array}{l} B\left(\frac{a}{2},-\frac{a}{2}, 0\right), \\ D\left(-\frac{a}{2}, \frac{a}{2}, 0\right), \\ M\left(-\frac{a}{4},-\frac{a}{4}, \frac{\sqrt{14} a}{4}\right), \\ N\left(\frac{a}{4}, \frac{a}{4}, \frac{\sqrt{14} a}{4}\right) . \end{array} $$ Thus, $B N=\left(-\frac{a}{4},-\frac{3 a}{4}, \frac{\sqrt{14} a}{4}\right)$, $$ D M=\left(\frac{a}{4},-\frac{3 a}{4}, \frac{\sqrt{14} a}{4}\right) . $$ Let $\boldsymbol{n}=(x, y, z)$ be a direction vector of the common perpendicular to $D M$ and $B N$. From $\boldsymbol{B N} \cdot \boldsymbol{n}=0, \boldsymbol{D M} \cdot \boldsymbol{n}=0$, we get $\boldsymbol{n}=(3,1,0)$. Since $M N=\left(\frac{a}{2}, \frac{a}{2}, 0\right)$, the distance $d$ between the skew lines $D M$ and $B N$ is $d=\frac{|M \boldsymbol{M} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{\sqrt{10} a}{5}$. Also, $|B N|=|D M|=\frac{\sqrt{6} a}{2}$, therefore, $$ \cos \alpha=\frac{|B N \cdot D M|}{|B N||D M|}=\frac{1}{6} . $$" 1700ea4d228c,"Example 3 In the Cartesian coordinate system $x O y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $x$-axis. When $\angle M P N$ takes the maximum value, find the x-coordinate of point $P$.","1$, $\angle M P N=\frac{\pi}{4} ;$ when $x_{0}=-1$, $\angle M P N=\arctan \frac{1}{2}\angle M P^{\pr",medium,"Solution 1: Let $P\left(x_{0}, 0\right)$, and let $\theta$ be the angle from line $N P$ to line $M P$, then when $x_{0} \neq \pm 1$, $k_{P \mathrm{M}}=\frac{2}{-1-x_{0}}, k_{P \mathrm{~V}}=\frac{4}{1-x_{0}}$, Thus, $$ \tan \theta=\frac{\frac{2}{-1-x_{0}}-\frac{4}{1-x_{0}}}{1+\frac{8}{x_{0}^{2}-1}}=\frac{2 x_{0}+6}{x_{0}^{2}+7} . $$ By the discriminant method, it is easy to find that $-\frac{1}{7} \leqslant \tan \theta \leqslant 1$. And the equality $\tan \theta=1$ is achieved when $x_{0}=1$, i.e., when $x_{0} \neq \pm 1$, the maximum value of $\angle M P N$ is $\frac{\pi}{4}$. Furthermore, when $x_{0}=1$, $\angle M P N=\frac{\pi}{4} ;$ when $x_{0}=-1$, $\angle M P N=\arctan \frac{1}{2}\angle M P^{\prime} N$, hence the point $P(1,0)$ is the solution, so the x-coordinate of point $P$ is 1." 32d929c7356f,"Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$.",See reasoning trace,medium,"Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then $$ S_{1}=a x+b y=5 \text {, } $$ $$ \begin{array}{l} S_{0}=a x^{2}+b y^{2}=9 . \\ S=a x^{3}+b y^{3}=17 . \\ S_{1}=a x^{4}+b y^{\prime}=33 . \end{array} $$ Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore, $$ \begin{array}{l} x^{2}-A x+B=0, \\ y^{2}-A y+B=0 . \end{array} $$ a. $x^{n-2} \cdot$ (3) $+b y^{n-2} \cdot$ (1), we get $$ S_{n}=A S_{n-1}-B S_{n-2} \text {. } $$ From the problem, we have $\left\{\begin{array}{l}S_{3}=A S_{2}-B S_{1}, \\ S_{4}=A S_{3}-B S_{2},\end{array}\right.$ which gives $\left\{\begin{array}{l}17=9 A-5 B, \\ 33=17 A-9 B .\end{array}\right.$ Solving these, we get $\left\{\begin{array}{l}A=3, \\ B=2 .\end{array}\right.$. Thus, $\left\{\begin{array}{l}x+y=3 \\ x y=2\end{array}\right.$ Substituting $x=1, y=2$ into (1) and (2), we get $a=1, b=2$. Similarly, when $x=2, y=1$, we get $a=2, b=1$. (1) When $x=1, y=2, a=1, b=2$, $$ x^{4}+y^{b}=1^{1}+2^{2}=5 \text {. } $$ (2) When $x=2, y=1, a=2, b=1$, $$ x^{2}+y^{3}=1^{1}+2^{2}=5 \text {. } $$ （From Yu Yuanzheng, ""Zhonghua Yinghao School, Conghua, Guangzhou, 510960）" 329c897f2dfe,"Example 2. If the difference between the two roots of the equation $\mathrm{x}^{2}-5 \mathrm{x} \cdot \sin \theta+1=0$ is $2 \sqrt{3}$, and $180^{\circ}<\theta<270^{\circ}$, find the value of $\operatorname{ctg} \theta$.","-\frac{4}{5}$, at this time $\cos \theta=-\sqrt{1-\left(-\frac{4}{5}\right)^{2}}=-\frac{3}{5}$. Ther",easy,"$$ =\sqrt{2} \sin ^{2} \theta-4=2 \sqrt{3} \text {. } $$ Given: $\sin ^{2} \theta=\frac{13}{25}$. Also, $180^{\circ}<\theta<270^{\circ}$, so take $\sin \theta=-\frac{4}{5}$, at this time $\cos \theta=-\sqrt{1-\left(-\frac{4}{5}\right)^{2}}=-\frac{3}{5}$. Therefore, $\operatorname{ctg}=\frac{\cos \theta}{\sin \theta}=\frac{3}{4}$." 4c8030e9d145,"14. Let $A B C D$ be a cyclic quadrilateral, and suppose that $B C=C D=2$. Let $I$ be the incenter of triangle $A B D$. If $A I=2$ as well, find the minimum value of the length of diagonal $B D$.","A D / D X=2$, which follows from the exterior angle bisector theorem; if $I_{A}$ is the $A$-excenter",medium,"Answer: $2 \sqrt{3}$ Let $T$ be the point where the incircle intersects $A D$, and let $r$ be the inradius and $R$ be the circumradius of $\triangle A B D$. Since $B C=C D=2, C$ is on the midpoint of $\operatorname{arc} B D$ on the opposite side of $B D$ as $A$, and hence on the angle bisector of $A$. Thus $A, I$, and $C$ are collinear. We have the following formulas: $$ \begin{aligned} A I & =\frac{I M}{\sin \angle I A M}=\frac{r}{\sin \frac{A}{2}} \\ B C & =2 R \sin \frac{A}{2} \\ B D & =2 R \sin A \end{aligned} $$ The last two equations follow from the extended law of sines on $\triangle A B C$ and $\triangle A B D$, respectively. Using $A I=2=B C$ gives $\sin ^{2} \frac{A}{2}=\frac{r}{2 R}$. However, it is well-known that $R \geq 2 r$ with equality for an equilateral triangle (one way to see this is the identity $1+\frac{r}{R}=\cos A+\cos B+\cos D$ ). Hence $\sin ^{2} \frac{A}{2} \leq \frac{1}{4}$ and $\frac{A}{2} \leq 30^{\circ}$. Then $$ B D=2 R\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)=B C \cdot 2 \cos \frac{A}{2} \geq 2\left(2 \cdot \frac{\sqrt{3}}{2}\right)=2 \sqrt{3} $$ with equality when $\triangle A B D$ is equilateral. Remark: Similar but perhaps simpler computations can be made by noting that if $A C$ intersects $B D$ at $X$, then $A B / B X=A D / D X=2$, which follows from the exterior angle bisector theorem; if $I_{A}$ is the $A$-excenter of triangle $A B C$, then $A I_{A} / X I_{A}=2$ since it is well-known that $C$ is the circumcenter of cyclic quadrilateral $B I D I_{A}$." f33b70024fa1,"7. Let $r(n)$ denote the sum of the remainders when $n$ is divided by $1, 2, \cdots, n$. Find all positive integers $m (1 < m \leqslant 2014)$ such that $r(m) = r(m-1)$.","2^{s}\left(s \in \mathbf{Z}_{+}, s \leqslant 10\right)$.",medium,"7. Let the remainder of $n$ divided by $k(1 \leqslant k \leqslant n)$ be $r_{k}(n)$. Thus, $r_{k}(n)=n-k\left[\frac{n}{k}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. $$ \begin{array}{l} \text { Then } r(m)=\sum_{k=1}^{m} r_{k}(m) \\ =\sum_{k=1}^{m}\left(m-k\left[\frac{m}{k}\right]\right)=m^{2}-\sum_{k=1}^{m} k\left[\frac{m}{k}\right] . \\ \text { Hence } r(m)=r(m-1) \\ \Leftrightarrow \sum_{k=1}^{m} k\left[\frac{m}{k}\right]-\sum_{k=1}^{m-1} k\left[\frac{m-1}{k}\right]=2 m-1 \\ \Leftrightarrow m+\sum_{k=1}^{m-1} k\left(\left[\frac{m}{k}\right]-\left[\frac{m-1}{k}\right]\right)=2 m-1 \\ \Leftrightarrow m+\sum_{\substack{k \mid m \\ 1 \leqslant k \leqslant m-1}} k=2 m-1 \\ \Leftrightarrow \sum_{\substack{k \mid m \\ 1 \leqslant k \leqslant m}} k=2 m-1 \text {. } \\ \end{array} $$ The left side of equation (1) is the sum of all positive divisors of $m$, denoted as $\sigma(m)$. Let $m=\prod_{i=1}^{t} p_{i}^{\alpha_{i}}$ (where $p_{1}1$ is a prime number), and by $130 \\ \Rightarrow \mathrm{k} \geq 31 $$ So $\mathrm{k} \geq 31$, and the equality can only hold when $r=3$. Therefore, $n=4 k+r \geq 4 \times 31+3=127$. It is also easy to verify that the first three decimal places of $\frac{32}{127}$ are $\overline{251}$, so the smallest value of $\mathrm{n}$ that satisfies the condition is 127." 793d3fa398c9,"3. Given positive numbers $a, b, c$ satisfying $a+b+c=1$, then $\sqrt{\frac{(a+b c)(b+c a)}{c+a b}}+\sqrt{\frac{(b+c a)(c+a b)}{a+b c}}+\sqrt{\frac{(c+a b)(a+b c)}{b+c a}}=$ $\qquad$",See reasoning trace,easy,$2$ 04284e873820,"5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows: $$ \begin{array}{l} a_{0}=0, a_{1}=1, \\ a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) . \end{array} $$ Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive integers $p, q$, such that $a_{l}+k a_{0}=a_{m}+k a_{q^{*}} \quad$ (Xiong Bin)",See reasoning trace,medium,"When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient. For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$. $$ \begin{array}{l} \text { then } a_{2 n} \equiv a_{0}=0(\bmod k), \\ a_{2 n+1} \equiv a_{1}=1(\bmod k)(n=0,1, \cdots) . \end{array} $$ If there exist $l, m \in \mathbf{N}, p, q \in \mathbf{N}_{+}, l \neq m$, satisfying $a_{l}+k a_{p}=a_{m}+k a_{q}$, without loss of generality, assume $l0$, then $$ \begin{array}{l} a_{l}+k a_{p} \leqslant k a_{p}+a_{p-1}=a_{p+1} \\ \leqslant a_{m}\frac{a_{l}+k a_{p}-a_{m}}{k}=a_{q} \text {. } \end{array} $$ From $k a_{q}+a_{m}=k a_{p}+a_{l} \geqslant k a_{p}$, we know $a_{q} \geqslant a_{p}-\frac{a_{m}}{k} \geqslant a_{p}-\frac{a_{p}}{k}=\frac{k-1}{k} a_{p}$. Noting that $a_{p} \geqslant k a_{p-1}$. Thus, $a_{p}>a_{q} \geqslant \frac{k-1}{k} a_{p} \geqslant(k-1) a_{p-1} \geqslant a_{p-1}$. From equation (4), we know $a_{q}=a_{p-1}$. Therefore, the equalities in equations (1), (2), and (3) must all hold. From equations (2) and (3), we get $m=p, p=2$. Thus, $a_{q}=a_{p-1}=a_{1}=1$. From equation (1), we know $l=0$. Therefore, from $a_{l}+k a_{p}=a_{m}+k a_{q}$, we get $k^{2}=k+k$, i.e., $k=2$, which is a contradiction. Thus, $k \geqslant 3$ does not satisfy the problem's conditions. Hence, $k$ can only be 2." 05b5846bc8e1,"$17 \cdot 135$ In an isosceles triangle, the altitude to the base is 8, and the perimeter is 32. Then the area of the triangle is (A) 56. (B) 48. (C) 40. (D) 32. (E) 24. (9th American High School Mathematics Examination, 1958)",See reasoning trace,easy,"[Solution] Let the base of the isosceles triangle be $b$, then the legs are $\frac{1}{2}(32-b)$. By the Pythagorean theorem, we have $$ \left[\frac{1}{2}(32-b)\right]^{2}=\left(\frac{b}{2}\right)^{2}+8^{2} \text {, } $$ Therefore, the correct choice is (B). $$ b=12, \quad S_{\triangle}=\frac{1}{2} \cdot 12 \cdot 8=48 \text {. } $$" d49b19dd8a74,"Example 5 Arrange all powers of 3 and the sums of distinct powers of 3 in an increasing sequence: $$ 1,3,4,9,10,12,13, \cdots \text {. } $$ Find the 100th term of this sequence.",See reasoning trace,medium,"Analysis: Write the known sequence in the form of the sum of powers of 3: $$ \begin{array}{l} a_{1}=3^{0}, a_{2}=3^{1}, a_{3}=3^{1}+3^{0}, \\ a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\ a_{7}=3^{2}+3^{1}+3^{0}, \cdots \end{array} $$ It is easy to find that the terms correspond exactly to the binary representation of the natural number sequence: $$ \begin{array}{l} 1=2^{0}, 2=2^{1}, 3=2^{1}+2^{0}, \\ 4=2^{2}, 5=2^{2}+2^{0}, 6=2^{2}+2^{1}, \\ 7=2^{2}+2^{1}+2^{0}, \cdots \end{array} $$ Since $100=(1100100)_{2}=2^{6}+2^{5}+2^{2}$, the 100th term of the original sequence is $$ 3^{6}+3^{5}+3^{2}=981 \text {. } $$ Explanation: This example demonstrates the superiority of base $p$ $(p \neq 0)$. Generally, each base system has its strengths and weaknesses. Depending on the specific problem, flexibly choosing the base system often yields surprising results. Please see another example." 68a5b81b9c95,"Let's determine all the triples of numbers $(x, y, m)$ for which $$ -2 x + 3 y = 2 m, \quad x - 5 y = -11 $$ and $x$ is a negative integer, $y$ is a positive integer, and $m$ is a real number.",See reasoning trace,medium,"From the second equation, $x=5 y-11$. Since $x$ is negative according to our condition, the equality can only hold if $y<\frac{11}{5}$. From the constraint on $y$, we get $y=2$ or $y=1$, and therefore $x=-1$ or $x=-6$. Now we can determine the values of $m$ from the first equation: \[ \begin{aligned} \text { if } y=2 \text { and } x=-1, \quad \text { then } m=4 \\ \text { and if } y=1 \text { and } x=-6, \quad \text { then } m=7.5 \end{aligned} \] Remark. During the solution, we only assumed that $x$ is negative, no constraint is needed for $m$, and for any real number pair $(x ; y)$, $-2 x+3 y$ is always real." acdaa060a415,"## Subject IV. (20 points) Consider the sequences $\left(a_{n}\right)_{n>1}$ and $\left(b_{n}\right)_{n>1}$, where: $a_{n}=\sum_{k=1}^{n} \frac{k+2}{k!+(k+1)!+(k+2)!}$ and $b_{n}=\sum_{k=1}^{n} k!(k+2)$ Required: a) $\lim _{n \rightarrow \infty} a_{n}$; b) $\lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}$. Prof. Alb Nicolae, Lic. T. “O. Goga” Huedin All questions are mandatory. 10 points are awarded by default. Time allowed for work - 3 hours. ## Grading Scale for 11th Grade (OLM 2013 - local stage) ## Default: 10 points",See reasoning trace,medium,"## Subject IV a) $a_{n}=\sum_{k=1}^{n} \frac{k+2}{k![1+k+1+(k+1)(k+2)]}=\sum_{k=1}^{n} \frac{1}{k!(k+2)}=\sum_{k=1}^{n} \frac{1}{k!(k+2)}=\sum_{k=1}^{n}\left[\frac{1}{(k+1)!}-\frac{1}{(k+2)!}\right]=\frac{1}{2}-\frac{1}{(n+2)!} \lim _{n \rightarrow \infty} a_{n}=\frac{1}{2}$. (10 points) b) Clearly $\lim _{n \rightarrow \infty} b_{n}=+\infty$ and thus the limit we are asked to find is in the case $1^{\infty}$. $$ \begin{aligned} & \lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}=\lim _{n \rightarrow \infty}\left[1-\frac{2}{(n+2)!}\right]^{b_{n}}=e^{\lim _{n \rightarrow \infty}\left[\frac{2}{(n+2)!} \cdot b_{n}\right]} . \\ & \lim _{n \rightarrow \infty} \frac{b_{n}}{(n+2)!}=\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n} k!(k+2)}{(n+2)!} \stackrel{\text { Stolz } C}{=} \lim _{n \rightarrow \infty} \frac{(n+1)!(n+3)}{(n+3)!-(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n+1)!(n+3)}{(n+2)!(n+2)}=\lim _{n \rightarrow \infty} \frac{(n+3)}{(n+2)^{2}}=0 . \end{aligned} $$ Thus: $\lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}=e^{0}=1$ (10 points)" 155af38b084f,"[ Diameter, main properties ] [ Signs and properties of the tangent ] A circle passes through the midpoints of the hypotenuse $A B$ and the leg $B C$ of a right triangle $A B C$ and is tangent to the leg $A C$. In what ratio does the point of tangency divide the leg $A C$.","$1: 3$, measured from vertex $C$",medium,"Use the theorem about the diameter of a circle perpendicular to a chord and the theorem about the midline of a triangle. ## Solution Let $M$ be the midpoint of the hypotenuse $AB$, $N$ be the midpoint of the leg $BC$, $K$ be the point of tangency of the given circle with the line $AC$, and $P$ be the midpoint of the midline $MN$ of triangle $ABC$. The perpendicular to $AC$ through point $K$ passes through the center of the circle and bisects the chord $MN$ perpendicular to it, i.e., it also passes through point $P$. Therefore, $$ CK = NP = \frac{1}{2} MN = \frac{1}{2} \cdot \frac{1}{2} AC = \frac{1}{4} AC. $$ Thus, $CK : AK = 1 : 3$. ## Answer $1: 3$, measured from vertex $C$." 2f505895349f,"The decimal representation of $\frac{1}{11}$ is $0.09090909 \ldots$ Another way to write this decimal representation is $0 . \overline{09}$. Similarly, $0 . \overline{125}$ represents the number $0.125125125 \ldots$ The decimal representation of $\frac{1}{7}$ is $0 . \overline{142857}$. In the decimal representation of $\frac{1}{7}$, the 100th digit to the right of the decimal is (A) 1 (B) 4 (C) 2 (D) 8 (E) 5",See reasoning trace,easy,"The digits to the right of the decimal place in the decimal represenation of $\frac{1}{7}$ occur in blocks of 6 , repeating the block of digits 142857 . Since $16 \times 6=96$, then the 96th digit to the right of the decimal place is the last in one of these blocks; that is, the 96th digit is 7 . This means that the 97 th digit is 1 , the 98 th digit is 4 , the 99 th digit is 2 , and the 100th digit is 8 ." 8f1883c7c414,"Example 5 Find the maximum constant $k$, such that $\frac{k a b c}{a+b+c} \leqslant(a+b)^{2}+(a+b+4 c)^{2}$ holds for all positive real numbers $a, b, c$.",100$.,medium,"Let $a=b=2c$, then $k \leqslant 100$. Also, $$ \begin{aligned} & \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+b+4 c)^{2}\right] \\ = & \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+2 c+b+2 c)^{2}\right] \\ \geqslant & \frac{a+b+c}{a b c} \cdot\left[4 a b+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}\right] \\ = & \frac{a+b+c}{a b c} \cdot(4 a b+8 a c+8 b c+16 \sqrt{a b} \cdot c) \\ = & (a+b+c) \cdot\left(\frac{4}{c}+\frac{8}{a}+\frac{8}{b}+\frac{16}{\sqrt{a b}}\right) \\ = & \left(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\right) \cdot\left(\frac{4}{c}+\frac{8}{a}+\frac{8}{b}+\frac{8}{\sqrt{a b}}+\frac{8}{\sqrt{a b}}\right) \\ \geqslant & \left(5 \sqrt[5]{\frac{a^{2} b^{2} c}{2^{4}}}\right) \cdot\left(5 \sqrt[5]{\frac{2^{14}}{a^{2} b^{2} c}}\right)=100 . \end{aligned} $$ Therefore, $k_{\max }=100$." 08d64e78b5a1,"3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $$ \begin{array}{l} a=\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{6}}},}{2016 \text { nested radicals }}, \\ b=\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{6}}},}{2 \text { 2017 nested radicals }}, \end{array} $$ then $[a+b]=$ . $\qquad$",4$.,easy,"3. 4 . Notice that, $2.4<\sqrt{6}0$), we get $$ 2\left(a^{2}+k a b+b^{2}\right) \geqq(k+2)(a+b) \sqrt{a b} $$ and by dividing both sides by the positive number $b^{2}$, we have $$ 2\left(\frac{a^{2}}{b^{2}}+k \frac{a}{b}+1\right) \geqq(k+2)\left(\frac{a}{b}+1\right) \sqrt{\frac{a}{b}} . $$ Let $\sqrt{a / b}=x$. By a suitable choice of positive numbers $a, b$, $x$ can take any positive value. Therefore, it suffices to consider the inequality $$ 2\left(x^{4}+k x^{2}+1\right) \geqq(k+2)\left(x^{2}+1\right) x $$ and find the largest $k$ such that the above inequality holds for every positive $x$. After simple equivalent manipulations aimed at isolating $k$, we get $$ \begin{aligned} k\left(\left(x^{2}+1\right) x-2 x^{2}\right) & \leqq 2\left(x^{4}+1-\left(x^{2}+1\right) x\right), \\ k\left(x^{3}-2 x^{2}+x\right) & \leqq 2\left(x^{4}-x^{3}-x+1\right), \\ k x\left(x^{2}-2 x+1\right) & \leqq 2\left(x^{3}(x-1)-(x-1)\right), \\ k x(x-1)^{2} & \leqq 2(x-1)^{2}\left(x^{2}+x+1\right) . \end{aligned} $$ For $x=1$, the last inequality is always satisfied. For $x \neq 1$, we divide the inequality by the positive expression $x(x-1)^{2}$ and obtain a direct estimate for $k$: $$ k \leqq \frac{2\left(x^{2}+x+1\right)}{x}=2+2\left(x+\frac{1}{x}\right) $$ For positive $x$, $x+1 / x \geqq 2$ with equality only for $x=1$. For $x \neq 1$, the expression $x+1 / x$ takes all values in the interval $(2, \infty)$, and the right-hand side (1) thus takes all values in the interval $(6, \infty)$. From this, it is clear that the largest $k$ such that (1) holds for all positive $x \neq 1$ is $k=6$." e89f2ffe6870,"Find all the three digit numbers $\overline{a b c}$ such that $$ \overline{a b c}=a b c(a+b+c) $$",135 and 144,medium,"We will show that the only solutions are 135 and 144. We have \(a > 0, b > 0, c > 0\) and \[ 9(11a + b) = (a + b + c)(abc - 1) \] - If \(a + b + c \equiv 0 \pmod{3}\) and \(abc - 1 \equiv 0 \pmod{3}\), then \(a \equiv b \equiv c \equiv 1 \pmod{3}\) and \(11a + b \equiv 0 \pmod{3}\). It follows now that \[ a + b + c \equiv 0 \pmod{9} \text{ or } abc - 1 \equiv 0 \pmod{9} \] - If \(abc - 1 \equiv 0 \pmod{9}\) we have \(11a + b = (a + b + c)k\), where \(k\) is an integer and it is easy to see that we must have \(19\). Now we will deal with the case when \(a + b + c \equiv 0 \pmod{9}\) or \(a + b + c = 9l\), where \(l\) is an integer. - If \(l \geq 2\) we have \(a + b + c \geq 18, \max\{a, b, c\} \geq 6\) and it is easy to see that \(abc \geq 72\) and \(abc(a + b + c) > 1000\), so the case \(l \geq 2\) is impossible. - If \(l = 1\) we have \[ 11a + b = abc - 1 \text{ or } 11a + b + 1 = abc \leq \left(\frac{a + b + c}{3}\right)^3 = 27 \] So we have only two cases: \(a = 1\) or \(a = 2\). - If \(a = 1\), we have \(b + c = 8\) and \(11 + b = bc - 1\) or \(b + (c - 1) = 7\) and \(b(c - 1) = 12\) and the solutions are \((a, b, c) = (1, 3, 5)\) and \((a, b, c) = (1, 4, 4)\), and the answer is 135 and 144. - If \(a = 2\) we have \(b(2c - 1) = 23\) and there is no solution for the problem." 27c3500bed3b,"## Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$. $M_{1}(-2 ; 0 ;-4)$ $M_{2}(-1 ; 7 ; 1)$ $M_{3}(4 ;-8 ;-4)$ $M_{0}(-6 ; 5 ; 5)$",See reasoning trace,medium,"## Solution Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$: $$ \left|\begin{array}{ccc} x-(-2) & y-0 & z-(-4) \\ -1-(-2) & 7-0 & 1-(-4) \\ 4-(-2) & -8-0 & -4-(-4) \end{array}\right|=0 $$ Perform the transformations: $$ \begin{aligned} & \left|\begin{array}{ccc} x+2 & y & z+4 \\ 1 & 7 & 5 \\ 6 & -8 & 0 \end{array}\right|=0 \\ & (x+2) \cdot\left|\begin{array}{cc} 7 & 5 \\ -8 & 0 \end{array}\right|-y \cdot\left|\begin{array}{cc} 1 & 5 \\ 6 & 0 \end{array}\right|+(z+4) \cdot\left|\begin{array}{cc} 1 & 7 \\ 6 & -8 \end{array}\right|=0 \\ & (x+2) \cdot 40-y \cdot(-30)+(z+4) \cdot(-50)=0 \end{aligned} $$ $40 x+80+30 y-50 z-200=0$ $40 x+30 y-50 z-120=0$ $4 x+3 y-5 z-12=0$ The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$: $d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$ Find: $$ d=\frac{\mid 4 \cdot(-6)+3 \cdot 5-5 \cdot 5-12) \mid}{\sqrt{4^{2}+3^{2}+(-5)^{2}}}=\frac{|-24+15-25-12|}{\sqrt{16+9+25}}=\frac{46}{\sqrt{50}}=\frac{46}{5 \sqrt{2}}=\frac{23 \sqrt{2}}{5} $$ ## Problem Kuznetsov Analytic Geometry $8-24$" 8a0144c5d6b5,"3. As shown in Figure 1, in a regular hexagon $A B C D E F$ with side length 10, $H$ is the midpoint of side $D E$, and $G$ is a point on side $B C$ such that $\angle A G B = \angle C G H$. Then the area of pentagon $A F E H G$ is $\qquad$",See reasoning trace,medium,"3. $\frac{205 \sqrt{3}}{2}$. As shown in Figure 3, draw a line through point $H$ parallel to $B E$, intersecting $G C$ at point $K$. It is easy to see that $K H=15$, $B K=K C=5$. Let $B G=x$. Then $$ G K=5-x \text {. } $$ Since $\triangle K G H \backsim \triangle B G A$ $$ \begin{array}{l} \Rightarrow \frac{K H}{A B}=\frac{G K}{B G} \Rightarrow \frac{15}{10}=\frac{5-x}{x} \\ \Rightarrow x=2, G K=3 . \end{array} $$ Then $S_{\text {hexagon }}=(10+20) \times \frac{\sqrt{3}}{2} \times 10=150 \sqrt{3}$, $S_{\text {trapezoid } C K H D}=\frac{1}{2}(10+15) \times \frac{5 \sqrt{3}}{2}=\frac{125 \sqrt{3}}{4}$, $S_{\triangle K G H}=\frac{1}{2} \times 3 \times 15 \times \frac{\sqrt{3}}{2}=\frac{45 \sqrt{3}}{4}$, $S_{\triangle B G A}=\frac{1}{2} \times 2 \times 10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3}$. Thus, $S_{\text {pentagon }}$. $A E H G$ $$ \begin{array}{l} =150 \sqrt{3}-\frac{125 \sqrt{3}}{4}-\frac{45 \sqrt{3}}{4}-5 \sqrt{3} \\ =\frac{205 \sqrt{3}}{2} . \end{array} $$" cea8292d3d2d,"12. Gnomes and the Thunderstorm. In a terrible thunderstorm, n gnomes are climbing up a rope ladder in a chain. If there is a thunderclap, each gnome, independently of the others, may fall with probability $p(0n>n-m$. From this and the factorization $98=2 \cdot 7 \cdot 7$ into prime factors, we obtain the system of equations $n-m=2, n^{2}+n m+m^{2}=49$. The only solution to this system in natural numbers is the pair $n=5, m=3$. Remark. The correct answer with verification is worth up to 3 points." c2d4c9b0ad42,"2. Solve the equation $\sqrt{8 x+5}+2\{x\}=2 x+2$. Here $\{x\}$ is the fractional part of the number $\mathrm{x}$, i.e., $\{x\}=x-[x]$. In the answer, write the sum of all solutions.",0,easy,"Solution. The equation is equivalent to the following $\sqrt{8 x+5}-2[x]-2=0, \quad[x]=n \in \mathbb{N}$, $n \leq x0$. Thus we conclude that for all $x \neq 0, f(x)>0$ must be satisfied. Now, we show the following fact $$ k>0, f(k)=1 \Leftrightarrow k=1 . $$ Let $k>0$ for which $f(k)=1$. We have $f\left(k^{2}\right)=f(f(k))=f(1)$, so by $(1), f(1 / k)=f(k)=$ 1 , so we may assume $k \geq 1$. By applying (2) with $x=\sqrt{k^{2}-1}$ and $y=k$, and using $f(x) \geq 0$, we get $$ f\left(k^{2}-1+f(k)\right)=f\left(k^{2}-1\right)+f\left(k^{2}\right)+2 f\left(k \sqrt{k^{2}-1}\right) \geq f\left(k^{2}-1\right)+f\left(k^{2}\right) . $$ This simplifies to $0 \geq f\left(k^{2}-1\right) \geq 0$, so $k^{2}-1=0$ and thus $k=1$. Next we focus on showing $f(1)=1$. If $f(1)=m \leq 1$, then we may proceed as above by setting $x=\sqrt{1-m}$ and $y=1$ to get $m=1$. If $f(1)=m \geq 1$, now we note that $f(m)=f(f(1))=f\left(1^{2}\right)=f(1)=m \leq m^{2}$. We may then proceed as above with $x=\sqrt{m^{2}-m}$ and $y=1$ to show $m^{2}=m$ and thus $m=1$. We are now ready to finish. Let $x>0$ and $m=f(x)$. Since $f(f(x))=f\left(x^{2}\right)$, then $f\left(x^{2}\right)=$ $f(m)$. But by (1), $f\left(m / x^{2}\right)=1$. Therefore $m=x^{2}$. For $x0$ for $x \neq 0$ as in the previous solution, we may also proceed as follows. We claim that $f$ is injective on the positive real numbers. Suppose that $a>b>0$ satisfy $f(a)=f(b)$. Then by setting $x=1 / b$ in (1) we have $f(a / b)=f(1)$. Now, by induction on $n$ and iteratively setting $x=a / b$ in (1) we get $f\left((a / b)^{n}\right)=1$ for any positive integer $n$. Now, let $m=f(1)$ and $n$ be a positive integer such that $(a / b)^{n}>m$. By setting $x=$ $\sqrt{(a / b)^{n}-m}$ and $y=1$ in (2) we obtain that $$ f\left((a / b)^{n}-m+f(1)\right)=f\left((a / b)^{n}-m\right)+f\left(1^{2}\right)+2 f\left(\sqrt{\left.(a / b)^{n}-m\right)}\right) \geq f\left((a / b)^{n}-m\right)+f(1) \text {. } $$ Since $f\left((a / b)^{n}\right)=f(1)$, this last equation simplifies to $f\left((a / b)^{n}-m\right) \leq 0$ and thus $m=$ $(a / b)^{n}$. But this is impossible since $m$ is constant and $a / b>1$. Thus, $f$ is injective on the positive real numbers. Since $f(f(x))=f\left(x^{2}\right)$, we obtain that $f(x)=x^{2}$ for any real value $x$." f936fd77eda9,"4. Let the base of the quadrilateral pyramid $P-ABCD$ not be a parallelogram. If a plane $\alpha$ cuts this quadrilateral pyramid such that the cross-section is a parallelogram, then the number of such planes $\alpha$ ( ). (A) does not exist (B) there is only 1 (C) there are exactly 4 (D) there are infinitely many",See reasoning trace,easy,"4. D. Let the intersection lines of two sets of non-adjacent lateral faces of a quadrilateral pyramid be $m$ and $n$. The lines $m$ and $n$ determine the plane $\beta$. Construct a plane $\alpha$ parallel to plane $\beta$ that intersects the lateral faces of the quadrilateral pyramid. The resulting quadrilateral is a parallelogram. There are infinitely many such planes $\alpha$." db3da4293ef2,"A cuboid has an integer volume. Three of the faces have different areas, namely $7, 27$, and $L$. What is the smallest possible integer value for $L$?",21,medium,"1. Let the sides of the cuboid be \(a\), \(b\), and \(c\). The volume of the cuboid is given by \(abc\), and the areas of the three different faces are \(ab\), \(ac\), and \(bc\). 2. We are given that the areas of three faces are \(7\), \(27\), and \(L\). Therefore, we have: \[ ab = 7, \quad ac = 27, \quad bc = L \] 3. To find the volume \(abc\), we multiply the areas of the three faces and take the square root: \[ \sqrt{(ab)(ac)(bc)} = \sqrt{(abc)^2} = abc \] 4. Substituting the given values, we get: \[ \sqrt{7 \cdot 27 \cdot L} = abc \] 5. Simplifying inside the square root: \[ \sqrt{189L} = abc \] 6. For \(abc\) to be an integer, \(\sqrt{189L}\) must be an integer. This implies that \(189L\) must be a perfect square. 7. The prime factorization of \(189\) is: \[ 189 = 3^3 \cdot 7 \] 8. For \(189L\) to be a perfect square, \(L\) must include the factors necessary to make all exponents in the prime factorization even. Therefore, \(L\) must include at least one factor of \(3\) and one factor of \(7\) to balance the exponents: \[ 189L = 3^3 \cdot 7 \cdot L \] 9. The smallest \(L\) that makes \(189L\) a perfect square is when \(L\) includes one additional factor of \(3\) and one additional factor of \(7\): \[ L = 3 \cdot 7 = 21 \] 10. Therefore, the smallest possible integer value for \(L\) is \(21\). The final answer is \(\boxed{21}\)." b264504cc1d9,"Given a rhombus $ABCD$. A point $M$ is chosen on its side $BC$. The lines, which pass through $M$ and are perpendicular to $BD$ and $AC$, meet line $AD$ in points $P$ and $Q$ respectively. Suppose that the lines $PB,QC,AM$ have a common point. Find all possible values of a ratio $\frac{BM}{MC}$. [i]S. Berlov, F. Petrov, A. Akopyan[/i]",\frac{BM,medium,"1. **Identify the given elements and relationships:** - We have a rhombus \(ABCD\). - A point \(M\) is chosen on side \(BC\). - Lines through \(M\) perpendicular to \(BD\) and \(AC\) meet line \(AD\) at points \(P\) and \(Q\) respectively. - Lines \(PB\), \(QC\), and \(AM\) are concurrent at a point \(R\). 2. **Use the concurrency condition:** - Suppose \(PB\), \(QC\), and \(AM\) concur at \(R\). - By the properties of the rhombus and the given conditions, we need to find the ratio \(\frac{BM}{MC}\). 3. **Relate the segments using the properties of the rhombus:** - Since \(P\) and \(Q\) are on \(AD\) and the lines through \(M\) are perpendicular to \(BD\) and \(AC\), we can use the properties of perpendiculars and the symmetry of the rhombus. - Let \(\frac{BM}{MC} = x\). 4. **Use the properties of the parallelogram:** - In the rhombus, diagonals \(AC\) and \(BD\) bisect each other at right angles. - Since \(APMC\) is a parallelogram, \(PA = MC\). - Similarly, in parallelogram \(BMQD\), \(BM = DQ\). 5. **Set up the ratio using the concurrency condition:** - Given that \(PB\), \(QC\), and \(AM\) are concurrent, we can use the fact that the ratio of segments created by the concurrency point is equal. - Therefore, \(\frac{BM}{MC} = \frac{PA}{AQ}\). 6. **Express the ratio in terms of \(x\):** - Since \(PA = MC\) and \(BM = DQ\), we have: \[ \frac{BM}{MC} = \frac{MC}{2BM + MC} \] - Let \(BM = x\) and \(MC = 1\) (assuming \(MC\) as the unit length for simplicity), then: \[ x = \frac{1}{2x + 1} \] 7. **Solve the quadratic equation:** - Multiply both sides by \(2x + 1\): \[ x(2x + 1) = 1 \] - Simplify to get the quadratic equation: \[ 2x^2 + x - 1 = 0 \] - Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \] - This gives two solutions: \[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \] 8. **Interpret the solutions:** - Since \(x = \frac{BM}{MC}\) represents a ratio of lengths, it must be positive. - Therefore, the valid solution is \(x = \frac{1}{2}\). Conclusion: \[ \boxed{\frac{BM}{MC} = \frac{1}{2}} \]" 49f16d376b54,"5. Let $f(x)=a x^{5}+b x^{3}+c x+10$, and $f(3)$ $=3$. Then $f(-3)=$","20$. Therefore, $f(-3)=17$.",easy,"5.17. Notice that for any $x$, $f(x)+f(-x)=20$. Therefore, $f(-3)=17$." 462bb5cbab2c,"2.1. Vasya loves to collect mushrooms. He calculated that over the autumn he collected a number of mushrooms that is expressed by a three-digit number, the sum of the digits of which is 14. Then Vasya calculated that 8% of the collected mushrooms were white, and 14% were boletus. How many mushrooms did Vasya collect?",950,easy,"Answer: 950. Solution. For $14 \%$ of the number of mushrooms to be an integer, it is necessary for the total number of mushrooms to be a multiple of 50. Then the last two digits are 00 or 50. But if a three-digit number ends with two zeros, the sum of the digits cannot be greater than 9, so the last two digits must be 50, i.e., the first digit is $14-5-0=9$." 6b30722fc858,"Let $\Delta$ denote the set of all triangles in a plane. Consider the function $f: \Delta\to(0,\infty)$ defined by $f(ABC) = \min \left( \dfrac ba, \dfrac cb \right)$, for any triangle $ABC$ with $BC=a\leq CA=b\leq AB = c$. Find the set of values of $f$.","\left[1, \frac{\sqrt{5",medium,"1. **Define the function and variables:** Let \( \Delta \) denote the set of all triangles in a plane. Consider the function \( f: \Delta \to (0, \infty) \) defined by \( f(ABC) = \min \left( \frac{b}{a}, \frac{c}{b} \right) \), for any triangle \( ABC \) with \( BC = a \leq CA = b \leq AB = c \). 2. **Express the ratios:** Define \( \frac{b}{a} = m \) and \( \frac{c}{b} = n \). Given \( a \leq b \leq c \), it follows that \( m, n \geq 1 \). 3. **Triangle inequality constraints:** The triangle inequality \( a + b > c \) translates to: \[ a + b > c \implies a + b > \frac{c}{b} \cdot b = n \cdot b \implies 1 + \frac{b}{a} > n \implies 1 + m > mn \] 4. **Case analysis:** - **Case i: \( m \leq n \)** \[ m \leq n < 1 + \frac{1}{m} \] Therefore, \( \min\{m, n\} = m \). - **Case ii: \( n \leq m \)** \[ n \leq m < \frac{1}{n-1} \] Therefore, \( \min\{m, n\} = n \). 5. **Combining the cases:** In both cases, we have: \[ \min\{m, n\} < \frac{\sqrt{5} + 1}{2} \] This is because the maximum value of \( \min\{m, n\} \) occurs when \( m = n \), and solving \( 1 + m = m^2 \) gives \( m = \frac{\sqrt{5} + 1}{2} \). 6. **Range of \( f \):** We can show that for any \( 0 < \epsilon < \frac{\sqrt{5} - 1}{2} \), there exist positive reals \( m \) and \( n \) such that \( \min\{m, n\} = \frac{\sqrt{5} + 1}{2} - \epsilon \). 7. **Conclusion:** Therefore, the set of values of \( f \) is: \[ \left[1, \frac{\sqrt{5} + 1}{2}\right) \] The final answer is \( \boxed{ \left[1, \frac{\sqrt{5} + 1}{2}\right) } \)" 0ac3891bae30,"1. Find all values of $p$, for each of which the numbers $p-2$, $3 \cdot \sqrt{p}$, and $-8-p$ are respectively the first, second, and third terms of some geometric progression.",$p=1$,easy,"Answer: $p=1$. Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(p-2)(-8-p)$, from which $$ \left\{\begin{array}{l} p>0 \\ p^{2}+15 p-16=0 \end{array} \Leftrightarrow p=1\right. $$" 5fcaf5c4e21e,"Three numbers form a geometric progression. If we add 8 to the second number, an arithmetic progression is formed. If we add 64 to the third term of this arithmetic progression, we again get a geometric progression. What are the original numbers?",See reasoning trace,medium,"Let the three numbers be: $a, a q, a q^{2}$. According to the problem: $$ 2(a q+8)=a+a q^{2} $$ and $$ (a q+8)^{2}=a\left(a q^{2}+64\right) $$ (1) can also be written as $$ a\left(1-2 q+q^{2}\right)=16 $$ Simplifying and rearranging (2): $$ a(4-q)=4 $$ The ratio of (3) and (4) is $$ \frac{1-2 q+q^{2}}{4-q}=4 $$ from which $$ q^{2}+2 q-15=0 $$ hence $$ q_{1}=3, q_{2}=-5 $$ and thus $$ a_{1}=4 \quad \text { and } \quad a_{2}=\frac{4}{9} $$ Therefore, the three numbers are 4, 12, 36 or $\frac{4}{9},-\frac{20}{9}, \frac{100}{9}$. Balatoni Ferenc (Bp. III., Árpád g. III. o. t.)" fca854a71066,"2. There is a $19 \times 19$ square chessboard. If two horizontal lines and two vertical lines are randomly selected, the probability that the enclosed shape is a square is $\qquad$ .",See reasoning trace,easy,"2. $\frac{13}{190}$. Squares with side length 1 have $19^{2}$, squares with side length 2 have $18^{2}$, ... squares with side length 19 have $1^{2}$. Therefore, the total number of squares is $$ 1^{2}+2^{2}+\cdots+19^{2}=\frac{19 \times 20 \times 39}{6} \text { (squares) } . $$ The required probability is $$ P=\frac{19 \times 20 \times 39}{6} \div \mathrm{C}_{20}^{2} \mathrm{C}_{20}^{2}=\frac{13}{190} . $$" 297b54c48752,"## Calculate the definite integral: $$ \int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}} $$",See reasoning trace,medium,"## Solution $$ \int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}}= $$ Substitution: $$ \begin{aligned} & x=3 \sin t \Rightarrow d x=3 \cos t d t \\ & x=0 \Rightarrow t=\arcsin \frac{0}{3}=0 \\ & x=\frac{3}{2} \Rightarrow t=\arcsin \frac{\left(\frac{3}{2}\right)}{3}=\arcsin \frac{1}{2}=\frac{\pi}{6} \end{aligned} $$ We get: $$ \begin{aligned} & =\int_{0}^{\pi / 6} \frac{9 \sin ^{2} t \cdot 3 \cos t d t}{\sqrt{9-9 \sin ^{2} t}}=\int_{0}^{\pi / 6} \frac{9 \sin ^{2} t \cdot 3 \cos t}{3 \sqrt{1-\sin ^{2} t}} d t=9 \cdot \int_{0}^{\pi / 6} \frac{\sin ^{2} t \cdot \cos t}{\cos t} d t= \\ & =9 \cdot \int_{0}^{\pi / 6} \sin ^{2} t d t=9 \cdot \int_{0}^{\pi / 6} \frac{1-\cos 2 t}{2} d t=\frac{9}{2} \cdot \int_{0}^{\pi / 6}(1-\cos 2 t) d t= \\ & =\left.\frac{9}{2}\left(t-\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 6}=\frac{9}{2}\left(\frac{\pi}{6}-\frac{1}{2} \sin \frac{\pi}{3}\right)-\frac{9}{2}\left(0-\frac{1}{2} \sin 0\right)= \\ & =\frac{9}{2}\left(\frac{\pi}{6}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right)-0=\frac{3 \pi}{4}-\frac{9 \sqrt{3}}{8} \end{aligned} $$ Source — ""http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_12-31"" Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals - Content is available under CC-BY-SA 3.0." 7920f747745c,"The two numbers $0$ and $1$ are initially written in a row on a chalkboard. Every minute thereafter, Denys writes the number $a+b$ between all pairs of consecutive numbers $a$, $b$ on the board. How many odd numbers will be on the board after $10$ such operations? [i]Proposed by Michael Kural[/i]",683,medium,"1. **Reduction to modulo 2**: We start by reducing all numbers to $\pmod{2}$. This is valid because the parity (odd or even nature) of the numbers will determine the number of odd numbers on the board. 2. **Define variables**: Let $a_n$ be the number of $00$, $b_n$ be the number of $01$, $c_n$ be the number of $10$, and $d_n$ be the number of $11$ in adjacent pairs after $n$ operations. 3. **Lemma 1**: \[ a_n + b_n + c_n + d_n = 2^n \] **Proof**: The number of numbers on the board after $n$ operations satisfies the recurrence $A_n = 2A_{n-1} - 1$. This is because each number generates a new number between each pair, doubling the count and then subtracting 1 for the overlap. Solving this recurrence with $A_1 = 3$, we get $A_n = 2^n + 1$. Therefore, the number of adjacent pairs is $A_n - 1 = 2^n$. 4. **Lemma 2**: \[ a_n = 0, \quad b_{n+1} = b_n + d_n, \quad c_{n+1} = c_n + d_n, \quad d_{n+1} = b_n + c_n \] **Proof**: The operations map as follows: \[ 00 \mapsto 000, \quad 01 \mapsto 011, \quad 10 \mapsto 110, \quad 11 \mapsto 101 \] From these mappings, we see that $a_{n+1} = 2a_n$. Since $a_1 = 0$, we have $a_n = 0$ for all $n$. The other recurrences follow from the mappings without overcounts. 5. **Lemma 3**: Let the number of $1$s after $n$ operations be $X_n$. Then, \[ X_n = X_{n-1} + b_{n-1} + c_{n-1} \] **Proof**: The mappings $01 \mapsto 011$ and $10 \mapsto 110$ each add one more $1$. Thus, the number of $1$s increases by the number of $01$ and $10$ pairs. 6. **Lemma 4**: Let $k_n = b_n + c_n$. Then, \[ k_n = \frac{2}{3} \cdot 2^n + \frac{1}{3} \cdot (-1)^n \] **Proof**: We have the recurrence: \[ k_n = b_n + c_n = b_{n-1} + c_{n-1} + 2d_{n-1} = k_{n-1} + 2k_{n-2} \] The characteristic equation for this recurrence is: \[ x^2 - x - 2 = 0 \] Solving this, we get the roots $x = 2$ and $x = -1$. Using initial conditions $k_1 = 1$ and $k_2 = 3$, we find: \[ k_n = \frac{2}{3} \cdot 2^n + \frac{1}{3} \cdot (-1)^n \] 7. **Calculate $X_{10}$**: We have $X_1 = 2$ and: \[ X_n = X_{n-1} + \frac{2}{3} \cdot 2^{n-1} + \frac{1}{3} \cdot (-1)^{n-1} \] Solving for $X_{10}$: \[ X_{10} = 2 + \frac{2}{3} \cdot (2 + 2^2 + \cdots + 2^9) + \frac{1}{3} \cdot (-1 + 1 - \cdots - 1) \] The sum of the geometric series is: \[ 2 + 2^2 + \cdots + 2^9 = 2(1 + 2 + 2^2 + \cdots + 2^8) = 2(2^9 - 1) = 2 \cdot 511 = 1022 \] Thus: \[ X_{10} = 2 + \frac{2}{3} \cdot 1022 - \frac{1}{3} = 2 + \frac{2044}{3} - \frac{1}{3} = 2 + \frac{2043}{3} = 2 + 681 = 683 \] The final answer is $\boxed{683}$." cb6fb7769b8d,Let $m$ be a positive integer for which there exists a positive integer $n$ such that the multiplication $mn$ is a perfect square and $m- n$ is prime. Find all $m$ for $1000\leq m \leq 2021.$,"\{1156, 1296, 1369, 1600, 1764\",medium,"1. Let \( m \) be a positive integer such that there exists a positive integer \( n \) for which \( mn \) is a perfect square and \( m - n \) is prime. We need to find all \( m \) in the range \( 1000 \leq m \leq 2021 \). 2. Since \( mn \) is a perfect square, we can write \( m \) and \( n \) in the form \( m = s q^2 \) and \( n = s r^2 \), where \( s \) is a common factor and \( q \) and \( r \) are integers. This ensures that \( mn = s^2 q^2 r^2 \) is a perfect square. 3. Given \( m - n \) is prime, we have: \[ p = m - n = s q^2 - s r^2 = s (q^2 - r^2) = s (q - r)(q + r) \] Since \( p \) is prime, the factors \( s \), \( q - r \), and \( q + r \) must be such that one of them is 1. The only feasible solution is \( s = 1 \), \( q - r = 1 \), and \( q + r = p \). This simplifies to: \[ q = \frac{p + 1}{2} \quad \text{and} \quad r = \frac{p - 1}{2} \] 4. Therefore, we have: \[ m = q^2 = \left( \frac{p + 1}{2} \right)^2 \quad \text{and} \quad n = r^2 = \left( \frac{p - 1}{2} \right)^2 \] 5. We need to find \( m \) such that \( 1000 \leq m \leq 2021 \). This translates to: \[ 1000 \leq \left( \frac{p + 1}{2} \right)^2 \leq 2021 \] 6. Taking the square root of the inequality: \[ \sqrt{1000} \leq \frac{p + 1}{2} \leq \sqrt{2021} \] \[ 31.62 \leq \frac{p + 1}{2} \leq 44.94 \] 7. Multiplying through by 2: \[ 63.24 \leq p + 1 \leq 89.88 \] 8. Subtracting 1 from all parts: \[ 62.24 \leq p \leq 88.88 \] 9. Since \( p \) must be a prime number, we consider the prime numbers in the range \( 63 \leq p \leq 87 \): \[ p \in \{67, 71, 73, 79, 83\} \] 10. For each \( p \), we calculate \( m \): - For \( p = 67 \): \[ m = \left( \frac{67 + 1}{2} \right)^2 = 34^2 = 1156 \] - For \( p = 71 \): \[ m = \left( \frac{71 + 1}{2} \right)^2 = 36^2 = 1296 \] - For \( p = 73 \): \[ m = \left( \frac{73 + 1}{2} \right)^2 = 37^2 = 1369 \] - For \( p = 79 \): \[ m = \left( \frac{79 + 1}{2} \right)^2 = 40^2 = 1600 \] - For \( p = 83 \): \[ m = \left( \frac{83 + 1}{2} \right)^2 = 42^2 = 1764 \] 11. All these values of \( m \) lie within the range \( 1000 \leq m \leq 2021 \). The final answer is \( \boxed{\{1156, 1296, 1369, 1600, 1764\}} \)" ab9e57b28fc7,"3.40 A red pencil costs 27 kopecks, a blue one - 23 kopecks. No more than 9 rubles 40 kopecks can be spent on purchasing pencils. It is necessary to purchase the maximum possible total number of red and blue pencils. At the same time, the number of red pencils should be as few as possible, but the number of blue pencils should not differ from the number of red pencils by more than 10. How many red and how many blue pencils should be purchased under the given conditions",14 red and 24 blue pencils,medium,"3.40 Let $x$ red and $y$ blue pencils be bought. According to the conditions, $27 x+23 y \leqslant 940$ and $y-x \leqslant 10$. Let's construct the lines ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-108.jpg?height=597&width=625&top_left_y=113&top_left_x=287) \[ \begin{aligned} & 27 x+23 y=940 \\ & y-x=10 \end{aligned} \] From Fig. 3.8, it is clear that these lines intersect at point $A$, the coordinates of which satisfy equations (1) and (2), and at this point, the maximum possible sum $x+y$ is achieved. Solving the system (1), (2) and considering that the numbers $x$ and $y$ are natural, we get $x=14, y=24$. Answer: 14 red and 24 blue pencils." 63ae4a5c32d2,"Let $n$ be a nonzero natural number, and $x_1, x_2,..., x_n$ positive real numbers that $ \frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}= n$. Find the minimum value of the expression $x_1 +\frac{x_2^2}{2}++\frac{x_3^3}{3}+...++\frac{x_n^n}{n}$.",1 + \frac{1,medium,"1. **Given Condition**: We are given that \( n \) is a nonzero natural number and \( x_1, x_2, \ldots, x_n \) are positive real numbers such that \[ \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = n. \] We need to find the minimum value of the expression \[ x_1 + \frac{x_2^2}{2} + \frac{x_3^3}{3} + \cdots + \frac{x_n^n}{n}. \] 2. **Initial Calculation**: Let's consider the case where \( x_1 = x_2 = \cdots = x_n = 1 \). Then, \[ \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = \frac{1}{1} + \frac{1}{1} + \cdots + \frac{1}{1} = n, \] which satisfies the given condition. The expression becomes: \[ x_1 + \frac{x_2^2}{2} + \frac{x_3^3}{3} + \cdots + \frac{x_n^n}{n} = 1 + \frac{1^2}{2} + \frac{1^3}{3} + \cdots + \frac{1^n}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}. \] 3. **Using Cauchy-Schwarz Inequality**: By the Cauchy-Schwarz inequality, we have: \[ (x_1 + x_2 + \cdots + x_n) \left( \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} \right) \geq n^2. \] Given that \( \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = n \), it follows that: \[ (x_1 + x_2 + \cdots + x_n) \cdot n \geq n^2 \implies x_1 + x_2 + \cdots + x_n \geq n. \] 4. **Using AM-GM Inequality**: By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have: \[ \frac{x_k^k}{k} + \frac{1}{k} + \frac{1}{k} + \cdots + \frac{1}{k} \geq x_k, \] where there are \( k \) terms of \( \frac{1}{k} \). Simplifying, we get: \[ \frac{x_k^k}{k} + \frac{k-1}{k} \geq x_k \implies \frac{x_k^k}{k} \geq x_k - \frac{k-1}{k}. \] 5. **Summing the Inequalities**: Summing these inequalities for \( k = 1, 2, \ldots, n \), we get: \[ \sum_{k=1}^{n} \frac{x_k^k}{k} \geq \sum_{k=1}^{n} \left( x_k - \frac{k-1}{k} \right). \] Simplifying the right-hand side, we have: \[ \sum_{k=1}^{n} \left( x_k - \frac{k-1}{k} \right) = \sum_{k=1}^{n} x_k - \sum_{k=1}^{n} \frac{k-1}{k}. \] Since \( \sum_{k=1}^{n} x_k \geq n \) and \( \sum_{k=1}^{n} \frac{k-1}{k} = \sum_{k=1}^{n} \left( 1 - \frac{1}{k} \right) = n - \sum_{k=1}^{n} \frac{1}{k} \), we get: \[ \sum_{k=1}^{n} \frac{x_k^k}{k} \geq n - \left( n - \sum_{k=1}^{n} \frac{1}{k} \right) = \sum_{k=1}^{n} \frac{1}{k}. \] 6. **Conclusion**: Therefore, the minimum value of the expression \( x_1 + \frac{x_2^2}{2} + \frac{x_3^3}{3} + \cdots + \frac{x_n^n}{n} \) is achieved when \( x_1 = x_2 = \cdots = x_n = 1 \), and the minimum value is: \[ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}. \] The final answer is \(\boxed{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\)." cd30d4fc55aa,"Example 1-21 Laying a $1 \times 7$ module with $1 \times 1, 1 \times 2, 1 \times 3$ blocks, how many patterns are there?",44$ types of patterns.,medium,"(1) Using 7 pieces of $1 \times 1$ finished products, the pattern is 1 type; (2) Using 5 pieces of $1 \times 1$ and 1 piece of $1 \times 2$ finished products, the pattern is $6!/ 5!=6$ types; (3) Using 4 pieces of $1 \times 1$ and 1 piece of $1 \times 3$ finished products, the pattern is $5!/ 4!=5$ types; (4) Using 3 pieces of $1 \times 1$ and 2 pieces of $2 \times 2$ finished products, the pattern is $\frac{5!}{3!2!}=10$ types; (5) Using 2 pieces of $1 \times 1$, 1 piece of $1 \times 2$, and 1 piece of $1 \times 3$ finished products, the pattern is $4!/ 2!=12$ types; (6) Using 1 piece of $1 \times 1$ and 3 pieces of $1 \times 2$ finished products, the pattern is $4!/ 3!=4$ types; (7) Using 1 piece of $1 \times 1$ and 2 pieces of $1 \times 3$ finished products, the pattern is $3!/ 2!=3$ types; (8) Using 2 pieces of $1 \times 2$ and 1 piece of $1 \times 3$ finished products, the pattern is $3!/ 2!=3$ types. According to the addition rule, there are a total of $1+6+5+10+12+4+3+3=44$ types of patterns." 01ace483dcf6,"6) In a rectangle with an area of $150 \mathrm{~m}^{2}$, the length of the base is equal to $\frac{3}{2}$ of the height. What is the perimeter of the rectangle? (A) $50 \mathrm{~m}$, (B) $54 \mathrm{~m}$, (C) $60 \mathrm{~m}$, (D) $64 \mathrm{~m}$, (E) $70 \mathrm{~m}$.",$(\mathbf{A})$,medium,"6. The answer is $(\mathbf{A})$. The rectangle in question can be divided into 6 squares with side length $A H$, where $\overline{A H}=\frac{1}{3} \overline{A B}=\frac{1}{2} \overline{B C}$. The area of the rectangle is therefore $\overline{A B} \cdot \overline{B C}=6 \overline{A H}^{2}=150 \mathrm{~m}^{2}$, which means $\overline{A H}=5 \mathrm{~m}$. The perimeter of the rectangle is thus $2 \overline{A B}+2 \overline{B C}=6 \overline{A H}+4 \overline{A H}=10 \overline{A H}=50 \mathrm{~m}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_3d973c0a184ce9fc7311g-2.jpg?height=320&width=423&top_left_y=545&top_left_x=1499) Alternatively, if we denote the length and height of the rectangle by $x$ and $y$ respectively, we have $x y=150 \mathrm{~m}^{2}, x=\frac{3}{2} y$ from which we obtain $y=10 \mathrm{~m}$ and $x=15 \mathrm{~m}$. The perimeter is finally given by $2 x+2 y=50 \mathrm{~m}$." d13eeaee0a0e,"A3. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x)= \begin{cases}x ; & x \geq 2 \\ f(4-x) ; & 0 \leq x<2 \\ f(x+2) ; & x<0\end{cases} $$ What is $f(-5)$? (A) -3 (B) -1 (C) 1 (D) 3 (E) 5",See reasoning trace,easy,"A3. Taking into account the properties of the function $f$, we calculate $$ f(-5)=f(-5+2)=f(-3)=f(-3+2)=f(-1)=f(-1+2)=f(1)=f(4-1)=f(3)=3 $$" 55b8d6e4dfa6,"The rectangle $A B C D$ with dimensions $A B=240 \mathrm{~cm}$ and $B C=288 \mathrm{~cm}$ represents a piece of paper that will be folded along the segment $E F$, where $E$ belongs to $A D$ and $F$ belongs to $B C$, such that point $C$ will be on the midpoint of $A B$. ![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-06.jpg?height=597&width=973&top_left_y=542&top_left_x=423) a) What is the length of $C C^{\prime}$? b) What is the length of $E F$?",See reasoning trace,medium,"Solution The operation of folding the paper is equivalent to producing a reflection along the fold. To see this, notice that corresponding segments have the same length, for example, $C^{\prime} F=C F$ and $C^{\prime} E=C E$. Thus, $E F$ bisects the segment $C C^{\prime}$ and is perpendicular to it. ![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-06.jpg?height=581&width=507&top_left_y=1753&top_left_x=656) a) Since $\triangle C B C^{\prime}$ is a right triangle at $B$, by the Pythagorean Theorem, we have $C C^{\prime}=\sqrt{120^{2}+288^{2}}=$ $312 \mathrm{~m}$. b) Since $E F$ is perpendicular to $C C^{\prime}$, it follows that $$ \angle C C^{\prime} B=90^{\circ}-\angle C^{\prime} C B=\angle E F C $$ Taking $G \in B C$ such that $E G \perp B C$, we can conclude that $\triangle E F G$ is similar to $\triangle C C^{\prime} B$. Thus $$ \begin{aligned} \frac{E F}{E G} & =\frac{C C^{\prime}}{B C} \\ \frac{E F}{240} & =\frac{312}{288} \\ E F & =260 \mathrm{~m} \end{aligned} $$" 329900e1b999,"$\begin{aligned} & {\left[\begin{array}{l}\text { Sum of the angles of a triangle. Theorem about the exterior angle. } \\ \text { [ Criteria and properties of an isosceles triangle. }\end{array}\right]}\end{aligned}$ In triangle $ABC$, angle $B$ is $20^{\circ}$, angle $C$ is $40^{\circ}$. The bisector $AD$ is 2. Find the difference between sides $BC$ and $AB$.",2,easy,"On side $B C$, lay off segment $B M$ equal to $A B$. ## Solution On side $B C$, we lay off segment $B M$ equal to $A B$. In the isosceles triangle $A B M$, the angles at the base $A M$ are each $80^{\circ}$, so $\angle C A M = \angle A M D - \angle A C B = 40^{\circ} = \angle A C M$. Moreover, $\angle A D M = \angle A B C + \angle B A D = 80^{\circ} = \angle A M D$. Thus, triangles $A M C$ and $A M D$ are isosceles. Therefore, $B C - A B = C M = A M = A D = 2$. ## Answer 2." 9235328474c3,"Tomongho A.K. Find all increasing arithmetic progressions with a finite number of terms, the sum of which is 1, and each term has the form $1 / k$, where $k$ is a natural number.",## Polynomials (miscellaneous),medium,"Let $1 / m$ be the largest term of the progression. Then the progression has more than $m$ terms (otherwise the sum is less than 1). On the other hand, the common difference $d \geq 1 / m - 1 / (m+1) = 1 / m(m+1)$, so there are no more than $m+1$ terms. Thus, the number of terms is $m+1$, and the arithmetic mean of all terms is $1 / (m+1)$. Since only one term of the progression is greater than the arithmetic mean, the progression has no more than three terms. Therefore, $m+1=3$, and the progression itself is $1 / 6, 1 / 3, 1 / 2$. ## Answer ## Polynomials (miscellaneous). $\quad]$ Problem 64668 Topics: [ Examples and counterexamples. Constructions ] [Identical transformations (trigonometry).] Does there exist a polynomial $f(x)$ of degree 6 such that for any $x$ the equality $f(\sin x) + f(\cos x) = 1$ holds? ## Solution Example 1. $f(x) = (4 x^{3} - 3 x)^{2}$. Indeed, $f(\sin x) + f(\cos x) = \sin^{2} 3x + \cos^{2} 3x = 1$. Example 2. $f(x) = -2 x^{6} + 3 x^{4}$. Indeed, $f(\sin x) + f(\cos x) = -2 \sin^{6} x + 3 \sin^{4} x - 2 \cos^{6} x + 3 \cos^{4} x = $ $= -2 (\sin^{6} x + \cos^{6} x) + 3 (\sin^{4} x + \cos^{4} x) = -2 (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x) + 3 ((\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x) = $ $= -2 (1 - 3 \sin^{2} x \cos^{2} x) + 3 (1 - 2 \sin^{2} x \cos^{2} x) = 1$. ## Answer There exists." a22f2e5ed603,"(a) Does there exist a polynomial $ P(x)$ with coefficients in integers, such that $ P(d) \equal{} \frac{2008}{d}$ holds for all positive divisors of $ 2008$? (b) For which positive integers $ n$ does a polynomial $ P(x)$ with coefficients in integers exists, such that $ P(d) \equal{} \frac{n}{d}$ holds for all positive divisors of $ n$?",n = 1,medium,"Let's break down the problem into two parts and provide a detailed solution for each. ### Part (a) We need to determine if there exists a polynomial \( P(x) \) with integer coefficients such that \( P(d) = \frac{2008}{d} \) for all positive divisors \( d \) of 2008. 1. **Identify the divisors of 2008:** \[ 2008 = 2^3 \times 251 \] The positive divisors of 2008 are \( 1, 2, 4, 8, 251, 502, 1004, 2008 \). 2. **Set up the condition for the polynomial:** For each divisor \( d \) of 2008, we need: \[ P(d) = \frac{2008}{d} \] 3. **Check the consistency of the polynomial:** Consider two distinct divisors \( a \) and \( b \) of 2008. We must have: \[ \frac{P(a) - P(b)}{a - b} \in \mathbb{Z} \] Substituting \( P(a) = \frac{2008}{a} \) and \( P(b) = \frac{2008}{b} \), we get: \[ \frac{\frac{2008}{a} - \frac{2008}{b}}{a - b} = \frac{2008(b - a)}{ab(a - b)} = \frac{2008}{ab} \in \mathbb{Z} \] This implies that for any distinct divisors \( a \) and \( b \) of 2008, \( ab \) must also be a divisor of 2008. 4. **Check if \( ab \) is a divisor of 2008 for all pairs of divisors:** - For \( a = 1 \) and \( b = 2 \), \( ab = 2 \), which is a divisor. - For \( a = 1 \) and \( b = 4 \), \( ab = 4 \), which is a divisor. - For \( a = 2 \) and \( b = 4 \), \( ab = 8 \), which is a divisor. - For \( a = 2 \) and \( b = 251 \), \( ab = 502 \), which is a divisor. - For \( a = 4 \) and \( b = 251 \), \( ab = 1004 \), which is a divisor. - For \( a = 8 \) and \( b = 251 \), \( ab = 2008 \), which is a divisor. Since \( ab \) is not always a divisor of 2008 for all pairs of divisors, there is no polynomial \( P(x) \) with integer coefficients that satisfies \( P(d) = \frac{2008}{d} \) for all positive divisors \( d \) of 2008. ### Part (b) We need to determine for which positive integers \( n \) there exists a polynomial \( P(x) \) with integer coefficients such that \( P(d) = \frac{n}{d} \) for all positive divisors \( d \) of \( n \). 1. **Condition for the polynomial:** For any distinct divisors \( a \) and \( b \) of \( n \), we must have: \[ \frac{P(a) - P(b)}{a - b} \in \mathbb{Z} \] Substituting \( P(a) = \frac{n}{a} \) and \( P(b) = \frac{n}{b} \), we get: \[ \frac{\frac{n}{a} - \frac{n}{b}}{a - b} = \frac{n(b - a)}{ab(a - b)} = \frac{n}{ab} \in \mathbb{Z} \] This implies that for any distinct divisors \( a \) and \( b \) of \( n \), \( ab \) must also be a divisor of \( n \). 2. **Determine the values of \( n \):** - If \( n \) is a prime number, its only divisors are 1 and \( n \). For these values, \( ab = n \), which is a divisor of \( n \). - If \( n = 1 \), its only divisor is 1, and the condition is trivially satisfied. Therefore, \( n \) must be either 1 or a prime number. 3. **Construct the polynomial for \( n = 1 \) or a prime number:** - For \( n = 1 \), the polynomial \( P(x) = 1 \) satisfies \( P(1) = 1 \). - For a prime number \( p \), the polynomial \( P(x) = -x + p + 1 \) satisfies: \[ P(1) = -1 + p + 1 = p \quad \text{and} \quad P(p) = -p + p + 1 = 1 \] Thus, \( P(d) = \frac{p}{d} \) for \( d = 1 \) and \( d = p \). Conclusion: \(\blacksquare\) The final answer is \( \boxed{ n = 1 } \) or \( n \) is a prime number." c16fe0d34bb6,"2. As shown in Figure 1, in the square $A_{1} A_{2} A_{3} A_{4}$ with side length 1, points $A_{5}$ and $A_{6}$ are taken on the sides $A_{2} A_{3}$ and $A_{3} A_{4}$ respectively, such that $A_{3} A_{5}=A_{3} A_{6}=p$, and the lines $A_{1} A_{2}$ and $A_{4} A_{5}$, $A_{1} A_{4}$ and $A_{2} A_{6}$, $A_{1} A_{5}$ and $A_{4} A_{3}$, $A_{1} A_{6}$ and $A_{2} A_{3}$ intersect at points $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$ which are collinear. Then the value of $p$ is $(\quad)$. (A) $\frac{\sqrt{5}+2}{3}$ (B) $\frac{\sqrt{5}-2}{2}$ (C) $\frac{3+\sqrt{5}}{3}$ (D) $\frac{3-\sqrt{5}}{2}$","0$. Since $0\frac{66}{5}$, i.e., $u \geqslant 14$. Also, $2 \times u$ and $3 \times u$, so $u \geqslant 17$. By $v \geqslant u$ and (4), we get $\frac{66 u-1}{5 u-66} \geqslant u$, i.e., $5 u^{2}-132 u+1 \leqslant 0$. Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$, therefore $17 \leqslant u \leqslant 26$. Furthermore, by $2 \chi_{u}$ and $3 X_{u}$, $u$ can only take $17,19,23,25$. When $u=17$, from (4), we get $v=\frac{1121}{19}=59$. When $u=19$, from (4), we get $v=\frac{1253}{29}$ which is not a positive integer, so it should be discarded. When $u=23$, from (4), we get $v=\frac{1517}{49}$ which is not a positive integer, so it should be discarded. When $u=25$, from (4), we get $v=\frac{1649}{59}$ which is not a positive integer, so it should be discarded. Therefore, only when $p=u+v=76$, the two roots of equation (*) are both positive integers, and the three roots of the original equation are all positive integers. Solution 2 Using the notation in Solution 1, from (1)(2), we get $5 u v=66(u+v)-1$, i.e., $5^{2} u v=5 \times 66(u+v)-5$. The above equation can be factored into $(5 u-66)(5 v-66)=19 \times 229$. The above indeterminate equation can be factored into $\left\{\begin{array}{l}5 v-66=229, \\ 5 u-66=19 ;\end{array}\right.$ or $\left\{\begin{array}{l}5 v-66=19 \times 229, \\ 5 u-66=1 .\end{array}\right.$ The latter has no positive integer solutions, the former's solution is $v=59, u=17$. Therefore, when $p=u+v=17+59=76$, the two roots of equation (*) are both positive integers, i.e., when $p=76$, the three roots of the original cubic equation are all positive integers. Comment When $p=76$, the two roots of the quadratic equation (*) are 59 and 17; the three positive integer roots of the original cubic equation are 1, 59, and 17." 0f237d00ee4c,"## Let $G$ be the set of 3x3 square matrices whose elements are equal to -1 or 1. a) How many elements does the set $G$ have? b) Determine the set $\{\operatorname{det}(A) \mid A \in G\}$.",See reasoning trace,medium,"## Problem 3. Let G be the set of 3x3 square matrices whose elements are equal to -1 or 1. a) How many elements does the set G have? b) Determine the set $\{\operatorname{det} A \mid A \in G\}$. ## Solution a) A matrix in G has 9 elements, each of which can take 2 values (-1 or 1). The set G has $2^{9}=512$ elements. b) Let $A \in G$. From the definition of a determinant of order 3, we obtain that det(A) is a sum of 6 terms of the form $( \pm 1) \cdot( \pm 1) \cdot( \pm 1)$, so $-6 \leq \operatorname{det}(A) \leq 6$. By adding the first row to the second and third rows, we get $\operatorname{det}(A)=\operatorname{det}(B)$ where $B$ is a matrix that has -1 or 1 on the first row, and 0, -2, or 2 on the second and third rows. We factor out the common factor 2 from the second and third rows, and then $\operatorname{det}(A)=4 \operatorname{det}(C)$, where $C$ is a matrix that has -1 or 1 on the first row, and 0, -1, or 1 on the second and third rows. Since $\operatorname{det}(C) \in \mathbb{Z}$, it follows that 4 divides $\operatorname{det}(A)$, hence $\operatorname{det}(A) \in\{-4,0,4\}$. 3p $$ \begin{aligned} & \text { Let } A_{1}=\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right) \text { and } A_{2}=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & 1 & -1 \end{array}\right) . \text { Clearly } A_{1}, A_{2},-A_{2} \in G \quad \text { and } \operatorname{det}\left(A_{1}\right)=0, \operatorname{det}\left(A_{2}\right)=4 \text {, } \\ & \operatorname{det}\left(-A_{2}\right)=-4 \text {. Therefore, }\{\operatorname{det}(\mathbf{A}) \mid \mathbf{A} \in G\}=\{-4, \mathbf{0}, \mathbf{4}\} . \end{aligned} $$" 8ff32b15782b,"$4 . a 、 b 、 c 、 d$ are pairwise distinct positive integers, and $a+b=c d, a b=c+d$. Then the number of quadruples $a 、 b 、 c 、 d$ that satisfy the above requirements is $(\quad)$. (A) 4 (B) 6 (C) 8 (D) 10",See reasoning trace,medium,"4. (C). (1) If $a, b, c, d$ are all not equal to 1, and $a \neq b, c \neq d$, then $(a-1)(b-1) \geqslant 2$, i.e., $a b > a + b$, $(c-1)(d-1) \geqslant 2$, i.e., $c d > c + d$. But $a + b = c d > c + d, c + d = a b > a + b$, which is a contradiction. Therefore, there is no solution in this case. (2) If one of $a, b, c, d$ is 1, without loss of generality, let $a=1$, then $$ \begin{array}{l} \left\{\begin{array}{l} 1 + b = c d, \\ b = c + d \end{array} \Rightarrow c d = c + d + 1\right. \\ \Rightarrow (c-1)(d-1) = 2 \\ \Rightarrow\left\{\begin{array}{l} c - 1 = 1, \\ d - 1 = 2 \end{array} \text { or } \left\{\begin{array}{l} c - 1 = 2, \\ d - 1 = 1, \end{array}\right.\right. \end{array} $$ i.e., $\left\{\begin{array}{l}c=2, \\ d=3\end{array}\right.$ or $\left\{\begin{array}{l}c=3, \\ d=2 \text {. }\end{array}\right.$ In this case, $b=5$. Similarly, when $b=1, c=1, d=1$, we can get 6 more solutions. Therefore, there are a total of 8 solutions: $$ \begin{array}{l} (1,5,2,3),(1,5,3,2),(5,1,2,3),(5,1,3,2), \\ (2,3,1,5),(2,3,5,1),(3,2,1,5),(3,2,5,1) . \end{array} $$" 887edf4aedc0,"A truck started from city $A$ to city $B$ at midnight, and a car started from city $B$ to city $A$ (on the same route) at $t_{1}$. They met at $t_{2}$. The car arrived at its destination $r$ hours later than the truck. - After finishing their business, they started back, and met again at $t_{3}$, and finally arrived home at the same time. When did they arrive? - Numerical data: $t_{1}=2^{\mathrm{h}} 40^{\mathrm{m}}, t_{2}=4^{\mathrm{h}}, r=0^{\mathrm{h}} 40^{\mathrm{m}}, t_{3}=14^{\mathrm{h}}$.","8 / 3$ hours, and $x=16.4$ hours.",medium,"In the solutions, we will use the following notations: Assuming the speeds of the vehicles are constant, as is customary in such problems, let the speed of the truck be $v$, the speed of the car be $V$, the distance between $A$ and $B$ be $s$, and the common time of arrival at home be $x$. I. Solution. Up to the first meeting, the truck travels $v t_{2}$, and the car travels $V\left(t_{2}-t_{1}\right)$, and together this is the entire distance $$ s=v t_{2}+V\left(t_{2}-t_{1}\right) $$ The travel time of the truck is $t_{1}-r$ hours more than that of the car, because the latter left $A$ $t_{1}$ hours later and arrived at $B$ $r$ hours later than the truck, i.e., $$ \frac{s}{v}=\frac{s}{V}+t_{1}-r $$ Eliminating the distance $s$ from these two equations, $$ t_{2}+\frac{V}{v}\left(t_{2}-t_{1}\right)=\frac{v}{V} t_{2}+t_{2}-r $$ and rearranging this for the ratio $\frac{v}{V}$, we get the equation $$ t_{2}\left(\frac{v}{V}\right)^{2}-r\left(\frac{v}{V}\right)-\left(t_{2}-t_{1}\right)=0 $$ The positive root of this equation is $$ \frac{v}{V}=\frac{1}{2 t_{2}}\left(r+\sqrt{r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)}\right) $$ For $t_{2}>t_{1}(\geq 0)$, the root is always real. The negative root is not meaningful. After the second meeting, both vehicles were on the road for $x-t_{3}$ more time, and during this time, they collectively covered the distance $s$ again, i.e., $$ v\left(x-t_{3}\right)+V\left(x-t_{3}\right)=s=v t_{2}+V\left(t_{2}-t_{1}\right) $$ Dividing by $V+v$ and rearranging, $$ x=t_{3}+t_{2}-\frac{1}{\frac{v}{V}+1} \cdot t_{1} $$ With the numerical data $$ \frac{v}{V}=\frac{2}{3} \quad \text { and } \quad x=16 \frac{2}{5} \text { hours, } $$ i.e., both vehicles arrived home at 16 hours and 24 minutes. Remarks. 1. The following consideration leads directly to a slightly rearranged form of equation (1). Up to the first meeting, the truck traveled $v t_{2}$, and the car traveled $V\left(t_{2}-t_{1}\right)$, and at that point, each had as much distance left as the other had already traveled. The truck took $\frac{V\left(t_{2}-t_{1}\right)}{v}$ time to cover the remaining distance, and the car took $\frac{v t_{2}}{V}$ time, but the latter arrived at the destination $r$ hours later, so $$ \frac{V\left(t_{2}-t_{1}\right)}{v}+r=\frac{v t_{2}}{V} $$ 2. Similarly, $x$ can also be determined. From the start of the return trip to the second meeting, the truck traveled the distance $V\left(x-t_{3}\right)$ in front of the car. This took $\frac{V\left(x-t_{3}\right)}{v}$ time. Similarly, the car started the return trip $\frac{v\left(x-t_{3}\right)}{V}$ time before the meeting. The difference between these two times is also the difference in the times required to travel the entire distance. According to the data for the outbound trip, this time difference is $t_{1}-r$, i.e., $$ \frac{V\left(x-t_{3}\right)}{v}-\frac{v\left(x-t_{3}\right)}{V}=t_{1}-r $$ from which $x$ can be calculated knowing the ratio $\frac{v}{V}$. II. Solution. Let the first meeting point be $C$. The ratio of the travel times of the car and the truck on the $A B$ route is the same as on the $A C$ route. Let the travel times on the $A B$ route be $T$ and $t$, respectively, so $$ T: t=\left(t_{2}-t_{1}\right):\left(t-t_{2}\right) $$ Furthermore, we have seen that $$ t-T=t_{1}-r $$ Eliminating $T$, $$ t^{2}-\left(2 t_{2}-r\right) t+t_{2}\left(t_{1}-r\right)=0 $$ from which the larger root is $$ t=t_{2}+\frac{1}{2}\left(-r+\sqrt{r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)}\right) $$ (since it is obvious that $t>t_{2}$, the other root cannot give the solution to the problem). Now we can calculate the time $x-t_{3}$ from the second meeting to the arrival home. In this time, the truck covers the fraction $\left(x-t_{3}\right) / t$ of the entire distance, and the car covers the fraction $\left(x-t_{3}\right) / T$ of the entire distance, and together they cover the entire distance (1 times the distance), i.e., $$ \begin{gathered} \frac{x-t_{3}}{t}+\frac{x-t_{3}}{T}=1 \\ x=t_{3}+\frac{1}{\frac{1}{t}+\frac{1}{T}}=t_{3}+\frac{t}{1+\frac{t-t_{2}}{t_{2}-t_{1}}}=t_{3}+\frac{t\left(t_{2}-t_{1}\right)}{t-t_{1}} \end{gathered} $$ With the numerical data, $t=6, x=16.4$ is obtained. III. Solution. We can also solve the problem using a calculation based on a sketch of the time-distance graph of the movements. ![](https://cdn.mathpix.com/cropped/2024_05_02_0d0c9579a5365d70b052g-2.jpg?height=413&width=1023&top_left_y=1062&top_left_x=537) Each car's round trip (separately) is represented by segments with equal slopes, so the endpoints of these segments, as well as the points representing the meetings, form similar triangles, and the projections of the meeting points cut off proportional segments. With the notations of the first figure, $$ \frac{B_{1} B_{2}}{B_{1} B_{3}}=\frac{A_{2} A_{1}}{A_{2} A}=\frac{B_{7} B_{6}}{B_{7} B_{5}}=\frac{B_{7} B_{6}}{B B_{3}} $$ Let $A_{1} A_{2}=u$ be the time it takes for the car to reach $A$ after the first m" 634ca3abaeb3,9. Let $A=\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$. Determine the value of $100 A$.,\frac{3}{2}$ and $100 A=75$.,easy,"9. Answer: 75 Solution. Let $$ B=\sin ^{2} 10^{\circ}+\sin ^{2} 50^{\circ}-\cos 40^{\circ} \cos 80^{\circ} . $$ Then $$ A+B=2-\cos 40^{\circ} $$ and $$ \begin{aligned} A-B=\cos 20^{\circ}+\cos 100^{\circ} & +\cos 120^{\circ}=2 \cos 60^{\circ} \cos 40^{\circ}+\cos 120^{\circ} \\ & =\cos 40^{\circ}-\frac{1}{2} \end{aligned} $$ Thus $2 A=\frac{3}{2}$ and $100 A=75$." 89335487a6fb,"## Calculate the limit of the function: $\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}$",\frac{\lim _{x \rightarrow 0} \frac{5 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightar,medium,"## Solution $\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)}{\sin 2 x-\sin x}=$ $=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$ $=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 2 x-\sin x)}=$ $=\left(\lim _{x \rightarrow 0} \frac{e^{5 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$ Using the substitution of equivalent infinitesimals: $e^{5 x}-1 \sim 5 x$, as $x \rightarrow 0(5 x \rightarrow 0)$ $e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$ $\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$ $\sin x \sim x$, as $x \rightarrow 0$ We get: $=\frac{\lim _{x \rightarrow 0} \frac{5 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 5-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{5-3}{2-1}=2$" 7e3646653fef,"1. Eight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one squares left or right, or (2) two squares left or right followed by one square up or down. Find the probability that every square, occupied or not, is attacked by some knight.",See reasoning trace,easy,"Solution: 0 . Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares as well." a520b9f36032,"3. Given that $f(x)$ is an even function, for any $x \in \mathbf{R}^{+}$, we have $f(2+x)=-2 f(2-x)$, and $f(-1)=4, f(-3)$ equals $\qquad$ .","1$, then $f(3)=-2 f(1)$. Since $f(x)$ is an even function, then $f(-3)=f(3)=-2 f(1)=$ $-2 f(-1)=-8$",easy,"3. -8 . Let $x=1$, then $f(3)=-2 f(1)$. Since $f(x)$ is an even function, then $f(-3)=f(3)=-2 f(1)=$ $-2 f(-1)=-8$" 82fb213ddd52,What is the maximum number of a) rooks; b) queens that can be placed on an $8 \times 8$ chessboard so that each of these pieces is attacked by no more than one of the others?,"a) 10 rooks, b) 10 queens",medium,"a) Let $k$ rooks be placed while satisfying the condition. On each field where a rook stands, we write the number 0. In each of the $n$ columns, we perform the following operation: if there are two numbers in the column, we add 1 to both; if there is one number, we add 2 to it (we do nothing in an empty column). Then we perform the same operation for each row. Clearly, on the position of each of the $k$ rooks, the result will be either 3 or 4, so the sum $S$ of all written numbers is at least $3k$; on the other hand, since in each of the eight columns and then in each of the eight rows we added no more than 2, we have $S \leq 32$. Therefore, $3k \leq S \leq 4n$, from which $k \leq 10$. An example of placing 10 rooks that satisfy the condition is shown in the left figure. ![](https://cdn.mathpix.com/cropped/2024_05_06_de7bb834a1e75b584fa6g-23.jpg?height=732&width=1762&top_left_y=594&top_left_x=136) b) Clearly, the number of queens that can be placed is no more than the number of rooks. The right figure shows how to place 10 queens. ## Answer a) 10 rooks, b) 10 queens." 0a6a218a7f7f,1. $2010+2.6 \times 26-\frac{7}{5} \times 14=$ $\qquad$,: 2058,easy,"【Solution】Solve: $2010+2.6 \times 26-\frac{7}{5} \times 14$ $$ \begin{array}{l} =2010+67.6-19.6 \\ =2077.6-19.6 \\ =2058 \end{array} $$ Therefore, the answer is: 2058." 73c89bc7f81a,"$9 \cdot 43$ Find a permutation $a_{1}, a_{2}, a_{3}, \cdots, a_{1962}$ of $1,2,3, \cdots, 1962$ such that the following sum is maximized: $$ \left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1961}-a_{1962}\right|+\left|a_{1962}-a_{1}\right| . $$",See reasoning trace,medium,"[Solution] Let $S=|a_{1}-a_{2}|+|a_{2}-a_{3}|+\cdots+|a_{1962}-a_{1}|$. If we remove the absolute value signs, $S$ can be written as the sum of $2 \times 1962$ numbers, half of which are positive and half are negative. The absolute values of these numbers cover $1,2,3, \cdots, 1962$ and each number is taken twice. Therefore, we have $$ S \leqslant 2(982+983+\cdots+1962)-2(1+2+\cdots+981) . $$ Thus, the permutation of $1,2,3, \cdots, 1962$ in the following form maximizes $S$: $$ 1962,1,1961,2,1960,3, \cdots, 982,981 \text {. } $$" ec808f486358,"Berolov s.l. The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $X$ and $Y$ respectively. Point $K$ is the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$ (not containing point $C$). It turns out that line $XY$ bisects segment $AK$. What can the angle $BAC$ be?",$120^{\circ}$,medium,"Let segments $X Y$ and $A I$ intersect at point $S$. As known, $K I = K A$ (see problem $\underline{53119}$), which means the height $K T$ of triangle $A K I$ is also its median. Since $X Y \perp A I$, then $X Y \| K T$, and because $X Y$ bisects side $A K$, $X Y$ is the midline of triangle $A K T$. Therefore, $A S: S T = 1: 3$; and since $X S$ is the height of the right triangle $A X I$, we have $A S / S T = \left( \frac{A X}{X I} \right)^2$, $\frac{X I}{A X} = \sqrt{3}$. Thus, $\angle X A I = 60^{\circ}$ and $\angle B A C = 2 \angle X A I = 120^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-27.jpg?height=563&width=1512&top_left_y=1805&top_left_x=273) ## Answer $120^{\circ}$." 81053cfe55f6,"4. In a certain country, there are 47 cities. Each city has a bus station from which buses run to other cities in the country and possibly abroad. A traveler studied the schedule and determined the number of internal bus routes departing from each city. It turned out that if the city of Lake is not considered, then for each of the remaining 46 cities, the number of internal routes departing from it differs from the number of routes departing from other cities. Find out how many cities in the country have direct bus connections with the city of Lake. The number of internal bus routes for a given city is the number of cities in its own country that can be reached from the given city by a direct bus, without transfers. The routes are symmetric: if you can travel from city $A$ to city $B$, then you can also travel from city $B$ to city $A$.",. Lake City is connected to 23 cities,medium,"Solution. Note that external lines are not considered in this problem. There are a total of 47 variants of the number of internal lines - from 0 to 46. Note that the existence of a city with 46 lines excludes the existence of a city with 0 lines and vice versa. Suppose there is a city with 46 lines. Then the smallest number of lines leading out of one city is one, and all numbers from 1 to 46 are encountered without gaps (cities excluding Lake City - 46). The city with one bus line cannot be connected to Lake City, as its only line must lead to the city with 46 lines (otherwise, the 46th pair would not be found). We have organized the pair 1-46. Now consider the city with 45 lines. We will match it with the city with two lines. Note that the city with two lines cannot be connected to Lake City. Continuing this reasoning, we will organize pairs 3-44, 4-43, and so on, up to the pair 23-24. The city with the smaller number of lines in each pair is not connected to Lake City. Note that Lake City does not enter these pairs, as a connection with Lake City is necessary for a larger number of lines in each pair. Thus, all cities are divided into pairs, and each number of lines from 1 to 46 occurs exactly once. Since in each pair exactly one city is connected to Lake City, Lake City is connected to 23 cities, the number of pairs. Suppose the maximum number of lines is 45. Then there is a city with 0 lines (buses from it go only abroad), and all numbers from 0 to 45 are encountered without gaps (cities excluding Lake City - 46). Again, we will form pairs 0-45, 1-44, and so on to 22-23. Similarly to the previous case, in the pair 0-45, the first city is not connected to Lake City, while the second is connected (otherwise, it would not be possible to get 45 lines from it). In the pair 1-44, the city with one line is not connected to Lake City (its connection is occupied by the line to the city with 44 lines), while the city with 44 lines is connected to Lake City to reach the required number of lines. Further, in the pair 2-43, the city with two lines is not connected to Lake City (its connection is occupied by lines to the cities with 44 and 43 lines), while the city with 43 lines is connected to Lake City to reach the required number of lines. We have 23 pairs. Lake City is connected to exactly one city in each pair, so it is connected to 23 cities. Answer. Lake City is connected to 23 cities." c761af78c5d5,"The graph of $y=x^2+2x-15$ intersects the $x$-axis at points $A$ and $C$ and the $y$-axis at point $B$. What is $\tan(\angle ABC)$? $\textbf{(A)}\ \frac{1}{7} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{3}{7} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{4}{7} \qquad$",\textbf{(E),medium,"First, find $A=(-5,0)$, $B=(0,-15)$, and $C=(3,0)$. Create vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}.$ These can be reduced to $\langle -1, 3 \rangle$ and $\langle 1, 5 \rangle$, respectively. Then, we can use the dot product to calculate the cosine of the angle (where $\theta=\angle ABC$) between them: \begin{align*} \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. \end{align*} Thus, \[\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.\] ~Indiiiigo" 3e3b814210e4,,"187 \mathrm{~kg}$ of apples. For this, they received $4340+8750=13090$ denars. Therefore, the price ",easy,"Solution. The two stores together sold: $365-(102+76)=187 \mathrm{~kg}$ of apples. For this, they received $4340+8750=13090$ denars. Therefore, the price of $1 \mathrm{~kg}$ of apples was: $13090: 187=70$ denars. At the first store, initially there were $4340: 70+102=164 \mathrm{~kg}$ of apples, and at the second store there were $365-164=201 \mathrm{~kg}$ of apples." 78fc3a6055e5,"9. Given a quadratic equation with real coefficients $a x^{2}+2 b x$ $+c=0$ has two real roots $x_{1} 、 x_{2}$. If $a>b>c$, and $a$ $+b+c=0$, then $d=\left|x_{1}-x_{2}\right|$ the range of values is $\qquad$",See reasoning trace,easy,9. $\sqrt{3}b>0$, then $a^{2}>b^{2}$. The number of correct propositions among the above is (A) 1. (B) 2. (C) 3. (D) 4. (""Zu Chongzhi Cup"" Junior High School Mathematics Invitational Competition, 1988)",$(A)$,easy,"[Solution] If $a, b$ are two irrational numbers that are opposites of each other, then (1) does not hold. If $k<0$, then (2) does not hold. If $a \neq 1, a \neq 0 ; \quad b$ is a rational number not equal to 0, and greater than -1 and less than 1, such as $a=2, b=\frac{1}{2}$, then (3) does not hold. If $x<0, y<0$, then (4) does not hold. According to the uniqueness of the choice, (5) is correct. Therefore, the answer is $(A)$." 9b54bdb62d7e,"## Task B-4.2. Determine the equation of the circle that is tangent to the line $x-3y+4=0$, and its center lies on the lines $2x-3y-9=0$ and $y+1=0$.",10$.,easy,"## Solution. The center of the circle is at the intersection of the lines $2 x-3 y-9=0$ and $y+1=0$, which is the solution to the given system, and that point is $(3,-1)$. (1 point) The radius is equal to the distance from the center $S(3,-1)$ to the line $x-3 y+4=0$ $$ r=\frac{|3+3+4|}{\sqrt{1+9}}=\sqrt{10} $$ Solution: $(x-3)^{2}+(y+1)^{2}=10$." 09ccfc1df5fd,"11.1. [7-8.6 (20 points)] On the bisector of angle $B A C$ of triangle $A B C$, a point $M$ is marked, and on the extension of side $A B$ beyond point $A$, a point $N$ is marked such that $A C=A M=1$ and $\angle A N M=\angle C N M$. Find the length of segment $A N$.",1,medium,"Answer: 1. Solution. Let's place point $K$ on the extension of segment $NC$ beyond point $C$. Point $M$ lies on the bisectors of angles $BAC$ and $ANC$, which means it is equidistant from lines $AC$ and $NC$, and therefore lies on the bisector of angle $ACK$. Then, considering the equality of angles $M$ and $C$ in the isosceles triangle $AMC$, from the equality of alternate interior angles $CMA$ and $MCK$, we get that lines $AM$ and $NC$ are parallel. From this, it follows that alternate interior angles $MAC$ and $ACN$ are equal to $\angle BAC / 2$, and thus $\angle ANC = \angle BAC - \angle ACN = \angle BAC / 2$. We have obtained that triangle $NAC$ is isosceles and $AN = AC = AM = 1$." 98fd0a052946,"A cube-shaped cake is completely covered with chocolate, including the bottom. We want to share the cake among $K$ people so that everyone gets the same amount of dough and the same amount of chocolate coating. We want to achieve this by dividing the cake into $N \times N \times N$ identical small cubes, and giving each person the same number of pieces, ensuring that the total area of the chocolate-coated sides of the pieces received by each person is the same. Is this possible for any $K$? What is the minimum number of pieces the cake must be divided into if $K=1997$?",421&width=469&top_left_y=993&top_left_x=817),medium,"The cake must be divided among $K$ people, with each person receiving $\frac{N^{3}}{K}$ small cubes and $\frac{6 N^{2}}{K}$ $1 \times 1$ chocolate strips. These fractions must be integers, so if $K$ is prime, then $K$ must be a divisor of $N$. First, we show that the desired distribution is possible for every natural number $K \geq 2$. Assume for now that $K \geq 4$, and let $K=N$. Cut the cake with vertical planes into $L$ ""vertical slices."" The vertical slice on the left side of the diagram is drawn with a thick line. The two outer slices contain $4 K + K^{2}$ strips, while the others (inner slices) contain $4 K$ strips. Each ""vertical slice"" can be cut into $K$ vertical columns. These columns can be divided into three groups: corner columns, marked with $S$ on their top, edge-center columns, marked with $E$; and inner columns, marked with $B$. An $E$-marked column has $K$ more strips than a $B$-marked one, and an $S$-marked column has $K$ more strips than an $E$-marked one. Then, the cake is divided among the $K$ people such that each person gets one ""vertical slice,"" and the two who get the outer slices exchange their $K-2$ $E$-marked columns for $B$-marked ones with those who received inner slices, ensuring that two columns are exchanged in each inner slice. The exchange is possible because with $K \geq 4$, there are $K-2$ inner slices, and each inner slice has at least two $B$-marked columns. After the exchanges, each person will have $6 K$ strips. Indeed, for the two outer slices, the number of strips decreases by $K(K-2)$, while for the inner slices, it increases by $2 K$. If $K=2$, cutting the cake into 8 small cubes ensures that each person gets 4 identical pieces. The case $K=3$ is also easily solvable. Since 1997 is prime, and as we established, $K$ must be a divisor of $N$, in the case $K=1997$, the cake must be cut into at least $1997^{3}$ pieces. Tamás Birkner (Budapest, Deutsche Schule, 5th grade) ![](https://cdn.mathpix.com/cropped/2024_05_02_9f30a844912a27437061g-1.jpg?height=421&width=469&top_left_y=993&top_left_x=817)" ba4fbf47e576,"Example 6 Let $x y=1$, and $x>y>0$. Find the minimum value of $\frac{x^{2}+y^{2}}{x-y}$. untranslated text is retained in its original format and directly output the translation result.",See reasoning trace,easy,"Explanation: Since $x>y>0$, we can set $x=y+\Delta y$ $(\Delta y>0)$, then $$ \frac{x^{2}+y^{2}}{x-y}=\frac{(x-y)^{2}+2 x y}{x-y}=\frac{(\Delta y)^{2}+2}{\Delta y} \geqslant 2 \sqrt{2} \text{. } $$ Equality holds if and only if $\Delta y=\sqrt{2}$, i.e., $$ x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=\frac{\sqrt{6}-\sqrt{2}}{2} $$ In this case, the minimum value of $\frac{x^{2}+y^{2}}{x-y}$ is $2 \sqrt{2}$." 80c61c4162bd,"A rational number written in base eight is $\underline{ab} . \underline{cd}$, where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$. Find the base-ten number $\underline{abc}$.",321,medium,"First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12. $11_{12}=15_8$ $22_{12}=32_8$ $33_{12}=47_8$ $44_{12}=64_8$ $55_{12}=101_8$ We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number. Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$. Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$. When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$. Since $d\neq0$, the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$" 8134d997f34a,"$33 \cdot 65$ If the number $121_{b}$ (i.e., the number in base $b$) is the square of an integer, then $b$ is (A) only $b=10$. (B) only $b=10, b=5$. (C) $2 \leqslant b \leqslant 10$. (D) $b>2$. (E) no such $b$ exists. (13th American High School Mathematics Examination, 1962)",$(D)$,easy,"［Solution］From the given and the definition of base $b$, we have $121_{b}=1 \cdot b^{2}+2 \cdot b+1=(b+1)^{2}$, which is a perfect square. However, in base $b$, the largest digit can only be $b-1$, so $b$ cannot be 1 or 2, i.e., $b>2$. Therefore, the answer is $(D)$." c8adf5f8de40,80. Find: a) $C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+C_{n}^{6}+\ldots ;$ b) $C_{n}^{1}+C_{n}^{3}+C_{n}^{5}+\ldots$,C_{n}^{1}+C_{n}^{3}+C_{n}^{5}+\ldots=2^{n-1}$.,easy,"80. If in the identity $$ \begin{aligned} & (1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+C_{n}^{2} x^{2}+\ldots+ \\ & \quad+C_{n}^{n-1} x^{n-1}+C_{n}^{n} x^{n} \end{aligned} $$ we set $x=1$, we get $$ 2^{n}=C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n-1}+C_{n}^{n} $$ For $x=-1$ we get $$ 0=C_{n}^{0}-C_{n}^{1}+C_{n}^{2}+\ldots+(-1)^{n} C_{n}^{n} $$ By adding and subtracting these equations term by term, we obtain $C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+\ldots=C_{n}^{1}+C_{n}^{3}+C_{n}^{5}+\ldots=2^{n-1}$." 6fdf994b670b,"Example 6 Let $a>0$ be a constant, solve the inequality $\sqrt{x^{2}+1}-a x \leqslant 1$.",See reasoning trace,medium,"Solve the inequality by transforming it into $\sqrt{x^{2}+1} \leqslant a x+1$. Let $y=\sqrt{x^{2}+1}(y \geqslant 0)$, which represents the upper branch of the hyperbola $y^{2}-x^{2}=1$. $y=a x+1$ represents a line passing through the fixed point $P(0,1)$ with a slope of $a$. See Figure 5-5. When $02 \times 1200=2400$, and the smallest $n$ of this form is $2^{12}=4096$. - If $n$ is of the form $3^{b}$, then we have $n>3 \times 1200=3600$, and the smallest $n$ of this form is $3^{8}=6561$. - If $n$ is of the form $2^{a} 3^{b}$, then we have $n>3 \times 1200=3600$, and after some trial (for example, by first fixing $a$ ) we find that the smallest $n$ of this form is $2^{4} \times 3^{5}=3888$. - If $n$ contains a prime factor greater than or equal to 5 , then we have $n>5 \times 1200=6000$. Combining the above cases, the answer is 3888 ." 944497650dd0,"We create a random sequence using the digits $0, 1, 2$. For sequences of what length will the probability be at least $61 \%$ that all three digits appear in the sequence?",See reasoning trace,medium,"Solution. The number of sequences of length $n$ using a single digit is 3, and the number of sequences using exactly two different digits is $\binom{3}{2}\left(2^{n}-2\right)$; thus, $$ p_{n}=\frac{3+3\left(2^{n}-2\right)}{3^{n}}=\frac{2^{n}-1}{3^{n-1}} $$ is the probability that in a sequence of length $n$, at least one digit does not appear. The $$ p_{n}-p_{n+1}=\frac{2^{n}-1}{3^{n-1}}-\frac{2^{n+1}-1}{3^{n}}=\frac{3 \cdot 2^{n}-3-2^{n+1}+1}{3^{n}}=\frac{2^{n}-2}{3^{n}} \geq 0 $$ relation shows that the sequence of probabilities $p_{n}$ is monotonically decreasing. Since $$ p_{4}=\frac{5}{9}>\frac{39}{100}>\frac{31}{81}=p_{5} $$ the requirement of the problem is satisfied for sequences of length at least 5." 12753a2bf502,"## Calculate the area of the figure bounded by the lines given by the equations. $$ \begin{aligned} & \left\{\begin{array}{l} x=2 \sqrt{2} \cdot \cos ^{3} t \\ y=\sqrt{2} \cdot \sin ^{3} t \end{array}\right. \\ & x=1(x \geq 1) \end{aligned} $$",See reasoning trace,medium,"## Solution Let's find the points of intersection: ![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-39.jpg?height=911&width=1491&top_left_y=1115&top_left_x=431) $$ \begin{aligned} & x=2 \sqrt{2} \cos ^{3} t=1 ; \Rightarrow \\ & \cos ^{3} t=\frac{1}{2 \sqrt{2}} \\ & \cos t=\frac{1}{\sqrt{2}} \\ & t= \pm \frac{\pi}{4}+2 \pi k, k \in \mathbb{Z} \end{aligned} $$ Since the functions \( x=2 \sqrt{2} \cdot \cos ^{3} t \) and \( y=\sqrt{2} \cdot \sin ^{3} t \) are periodic (with a period of \( 2 \pi \)), we can take any interval of length \( 2 \pi \). Let's take \([- \pi ; \pi]\). Then: $$ t=-\frac{\pi}{4} \text { or } t=\frac{\pi}{4} $$ $$ x \geq 1 \text { on the interval }\left[-\frac{\pi}{4} ; \frac{\pi}{4}\right] $$ From the graph, it is clear that the region is symmetric with respect to the \( x \)-axis, and its area can be calculated using the formula: $$ \begin{aligned} & S=2 \cdot S_{0} ; x \in\left[\frac{\pi}{4} ; 0\right] \\ & S=2 \cdot \int_{\pi / 4}^{0} y(t) \cdot x^{\prime}(t) d t= \\ & =2 \cdot \int_{\pi / 4}^{0} \sqrt{2} \sin ^{3} t \cdot\left(2 \sqrt{2} \cos ^{3} t\right)^{\prime} d t= \\ & =8 \cdot \int_{\pi / 4}^{0} \sin ^{3} t \cdot 3 \cos ^{2} t \cdot(-\sin t) d t= \end{aligned} $$ $$ \begin{aligned} & =-24 \cdot \int_{\pi / 4}^{0} \sin ^{4} t \cdot \cos ^{2} t d t=\left|\begin{array}{c} \sin t \cos t=\frac{1}{2} \sin 2 t, \\ \sin ^{2} t=\frac{1}{2}(1-\cos 2 t), \\ \int_{a}^{a} f(x) d x=-\int_{b} f(x) d x \end{array}\right|= \\ & =24 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 2 t) \cdot\left(\frac{1}{2} \sin 2 t\right)^{2} d t=3 \cdot \int_{0}^{\pi / 4}(1-\cos 2 t) \sin ^{2} 2 t d t= \\ & =3 \cdot \int_{0}^{\pi / 4} \sin ^{2} 2 t d t-3 \cdot \int_{0}^{\pi / 4} \cos 2 t \cdot \sin ^{2} 2 t d t= \\ & =3 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 4 t) d t-3 \cdot \int_{0}^{\pi / 4} \frac{1}{2} \cdot \sin ^{2} 2 t d(\sin 2 t)= \\ & =\left.\frac{3}{2} \cdot\left(t-\frac{1}{4} \sin 4 t\right)\right|_{0} ^{\pi / 4}-\left.\frac{3}{2} \cdot\left(\frac{1}{3} \sin ^{3} 2 t\right)\right|_{0} ^{\pi / 4}= \\ & =\frac{3}{2} \cdot\left(\frac{\pi}{4}-\frac{1}{4} \sin \frac{4 \pi}{4}\right)-\frac{3}{2} \cdot\left(0-\frac{1}{4} \sin ^{2}\right)-\frac{1}{2} \cdot\left(\sin ^{3} \frac{2 \pi}{4}-\sin ^{3} 0\right)= \\ & =\frac{3}{2} \cdot \frac{\pi}{4}-\frac{3}{2} \cdot \frac{1}{4} \cdot \sin \pi-0-\frac{1}{2} \cdot \sin ^{3} \frac{\pi}{2}+0=\frac{3}{8} \cdot \pi-\frac{3}{8} \cdot 0-\frac{1}{2} \cdot 1=\frac{3}{8} \cdot \pi-\frac{1}{2} \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-19$ » Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals - Last edited: 05:46, June 21, 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 15-20 ## Material from PlusPi" d7f21cdaca82,"10. (20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b>0), c^{2}+b^{2}=a^{2}$, a line passing through the focus intersects the ellipse at points $M, N$, and the length of segment $\mathrm{MN}$ is in the range $[3,4]$. (1) Find the equation of the ellipse. (2) Let $\mathrm{D}, \mathrm{E}$ be two points on the ellipse that are symmetric with respect to the $\mathrm{x}$-axis. For any two points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$ on the $\mathrm{x}$-axis, when $D A$ is not tangent to the ellipse, it always makes the intersection point $C$ of lines $D A, B E$ lie on the ellipse. Find the relationship between $x_{1}, x_{2}$.",\frac{2 b^{2} n^{2} x_{1}^{2}-a^{2} b^{2} n^{2}}{n\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\ri,medium,"(1) From the knowledge of ellipses, it is easy to know that the shortest length is when the line is perpendicular to the $x$-axis, which is $\frac{2 b^{2}}{a}$. When the line is the $x$-axis, it is the longest, which is $2a$. Solving for $a=2, b=\sqrt{3}, c=1$, the equation is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. (2) $x_{1} x_{2}=a^{2}$. Let $D(m, n), E(m,-n)$, then the equation of line $DA$ is $y=\frac{n}{m-x_{1}}\left(x-x_{1}\right)$, and combining it with the ellipse equation and eliminating $y$ gives $\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right] x^{2}-2 a^{2} n^{2} x_{1} x+a^{2} n^{2} x_{1}^{2}-a^{2} b^{2}\left(m-x_{1}\right)^{2}=0$. Let the other intersection point of line $DA$ with the ellipse be $C_{1}\left(x_{3}, y_{3}\right)$, then $x_{3}=\frac{2 a^{2} n^{2} x_{1}}{a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}}-m=\frac{2 a^{2} n^{2}}{b^{2}\left(x_{1}-2 m+x_{2}\right)}-m$. Similarly, let the other intersection point of line $BE$ with the ellipse be $C_{2}\left(x_{4}, y_{4}\right)$, $x_{4}=\frac{2 a^{2} n^{2}}{b^{2}\left(x_{1}-2 m+x_{2}\right)}-m$, so $x_{3}=x_{4}$. Combining the equation of line $DA$ with the ellipse equation and eliminating $x$ gives $$ \left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right] y^{2}+2 b^{2} n x_{1}\left(m-x_{1}\right) y+b^{2} n^{2} x_{1}^{2}-a^{2} b^{2} n^{2}=0 $$ $y_{3}=\frac{2 b^{2} n^{2} x_{1}^{2}-a^{2} b^{2} n^{2}}{n\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right]}=\frac{n\left(x_{1}-x_{2}\right)}{x_{1}-2 m+x_{2}}$, similarly, $y_{4}=y_{3}$. Therefore, $C_{1}$ and $C_{2}$ coincide. Thus, the intersection point $C$ of lines $AD$ and $BE$ is still on the ellipse." 2ac462d9b0f6,"20) Knowing that between 200 and 300 (inclusive) there are exactly 13 multiples of the integer $n$, what is the value of $n$? (A) $\leq$ (B) (C) 8 (D) 9 (E) $\geq 10$",(C),medium,"20) The answer is (C). Let $\alpha n \geq 200$ be the first multiple and $(\alpha+12) n \leq 300$ be the last. To ensure there are no more than 13, we must have $\alpha n - n \leq 300$. The conditions that $n$ must satisfy are then: - $\alpha n \geq 200$ - $(\alpha-1) n \leq 300$ Subtracting the first inequality from the third gives $12 n \leq 100$, which implies $n \leq 8$. Subtracting the second inequality from the fourth gives $14 n > 100$, which means $n > 7$. It follows that $n$ must be equal to 8. The 13 numbers 200, 208, ..., 296 are all and only the multiples of 8 between 200 and 300 (inclusive)." eeb7363a3419,"13.238. The shooter at the shooting range was offered the following conditions: each hit on the target is rewarded with five tokens, but for each miss, three tokens are taken away. The shooter was not very accurate. After the last ($n$-th) shot, he had no tokens left. How many shots were in the series and how many were successful, if $10 0, x = 4 - 2 y = 2 > 0 \] solution of the system \((2 ; 1)\); 2) \(\left\{\begin{array}{l}x0 - \text{no solutions;}\end{array}\right.\) ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-357.jpg?height=250&width=729&top_left_y=1590&top_left_x=162) 4) \(\left\{\begin{array}{l}x0, \\ 2 x - 3 y = 1, \\ -x + 2 y = 4|2| ;\end{array} \quad\left\{\begin{array}{l}x0, \\ 2 x - 3 y = 1, \\ -2 x + 4 y = 8 ;\end{array}\right.\right.\) \(y = 9 ; x = 2 y - 4 = 14 > 0\) - no solutions. Answer: \((2 ; 1)\). ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-358.jpg?height=370&width=344&top_left_y=536&top_left_x=255) a) ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-358.jpg?height=358&width=421&top_left_y=527&top_left_x=659) b) Fig. 11.9" 1198a90151df,5th USAMO 1976,"2a 1 , b = 2b 1 , c = 2c 1 . Then a 1 2 + b 1 2 + c 1 2 = square. Repeating, we find that a, b, c mu",easy,": 0, 0, 0. Squares must be 0 or 1 mod 4. Since the rhs is a square, each of the squares on the lhs must be 0 mod 4. So a, b, c are even. Put a = 2a 1 , b = 2b 1 , c = 2c 1 . Then a 1 2 + b 1 2 + c 1 2 = square. Repeating, we find that a, b, c must each be divisible by an arbitrarily large power of 2. So they must all be zero. 5th USAMO 1976 © John Scholes jscholes@kalva.demon.co.uk 11 Jul 2002" 4fa65321b68f,"The measure of angle $ABC$ is $50^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$. The measure of angle $ADC$ is $\text{(A)}\ 90^\circ \qquad \text{(B)}\ 100^\circ \qquad \text{(C)}\ 115^\circ \qquad \text{(D)}\ 122.5^\circ \qquad \text{(E)}\ 125^\circ$",C,easy,"Let $\angle CAD = \angle BAD = x$, and let $\angle ACD = \angle BCD = y$ From $\triangle ABC$, we know that $50 + 2x + 2y = 180$, leading to $x + y = 65$. From $\triangle ADC$, we know that $x + y + \angle D = 180$. Plugging in $x + y = 65$, we get $\angle D = 180 - 65 = 115$, which is answer $\boxed{C}$." b5a35ba8c355,"376. Given a triangle $A B C$, the angles of which are $\alpha, \beta$ and $\gamma$. Triangle $D E F$ is circumscribed around triangle $A B C$ such that vertices $A, B$ and $C$ are located on sides $E F$, $F D$ and $D E$ respectively, and $\angle E C A = \angle D B C = \angle F A B = \varphi$. Determine the value of the angle $\varphi$ for which the area of triangle $E F D$ reaches its maximum value.",\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma$. Note. The angle $\ome,medium,"376. Let's describe circles around triangles $A B F, B C D$ and $C A E$. They have a common point $M$. Since the angles of triangle $DEF$ are constant, $\angle D=\gamma, \angle E=\alpha, \angle F=\beta$, the constructed circles and point $M$ do not depend on $\varphi$. The side $D F$ (and therefore $E F$ and $E D$) will be the largest when $D F$ is perpendicular to $B M$. Let $\varphi_{0}$ be the angle corresponding to this position. Then $\angle M B C=\angle M C A=\angle M A B=90^{\circ}-\varphi_{0}$. Extend $C M$ to intersect the circumcircle of triangle $A M B$ at point $F_{1}$. It can be found that $\angle F_{1} B A=\alpha$, $\angle F_{1} A B=\beta ; F_{1} B$ turns out to be parallel to $A C$. Drop perpendiculars $F_{1} N$ and $B L$ from $F_{1}$ and $B$ to $A C$. Since $\left|F_{1} N\right|=|B L|$, then $$ \operatorname{tg} \varphi_{0}=\operatorname{ctg}\left(90^{\circ}-\varphi_{0}\right)=\frac{|C N|}{\left|F_{1} N\right|}=\frac{|A N|}{\left|F_{1} N\right|}+\frac{|A L|}{|B L|}+\frac{|C L|}{|B L|}=\operatorname{ctg} \beta+ $$ $+\operatorname{ctg} \alpha+\operatorname{ctg} \gamma$. Thus, $\operatorname{tg} \varphi_{0}=\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma$. Note. The angle $\omega=90^{\circ}-\varphi_{0}$ is called the Brocard angle, and point $M$ is the Brocard point. For each triangle, there are two Brocard points. The position of the second point $M_{1}$ is determined by the condition $\angle M_{1} B A=\angle M_{1} A C=\angle M_{1} C B$." 31277ba81003,"Blinkov A.D: On the side $AB$ of rectangle $ABCD$, a point $M$ is chosen. Through this point, a perpendicular to the line $CM$ is drawn, which intersects the side $AD$ at point $E$. Point $P$ is the foot of the perpendicular dropped from point $M$ to the line $CE$. Find the angle $APB$.",$90^{\circ}$,easy,"Since $$ \angle M A E=\angle M P E=\angle C P M=\angle M B C=90^{\circ} $$ (see figure), quadrilaterals $A E P M$ and $B M P C$ are cyclic. Therefore, $\angle B P A=\angle B P M+\angle M P A=\angle B C M+\angle M E A=$ $=180^{\circ}-\angle E M A-\angle C M B=\angle E M C=90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_b0b755a355dbd05a96cdg-11.jpg?height=760&width=600&top_left_y=1052&top_left_x=729) ## Answer $90^{\circ}$." 32df926c43c8,,"384$ cm, so the three extra horizontal cuts have a total length of 54 cm, each of which is 18. This ",easy,"# Answer: 9 Solution. Note that the sum of eight vertical and eight horizontal cuts is $8 \square 48=384$ cm, so the three extra horizontal cuts have a total length of 54 cm, each of which is 18. This is the length of the horizontal side of the original rectangle. The vertical side of it is $48-18=30 \text{~cm}$. After the cuts, the resulting rectangle has a horizontal side of $18 / 9=2$ cm, and a vertical side of $30 / 12=2.5$ cm, so its perimeter is 9 cm." 53effdf853a4,4. Calculate: $\frac{1}{11 \times 13 \times 15}+\frac{1}{13 \times 15 \times 17}+\cdots+$ $\frac{1}{29 \times 31 \times 33}=$ $\qquad$,$\frac{20}{13299}$,easy,"Hint: Original expression $\times 4$ $$ \begin{aligned} = & \left(\frac{1}{11 \times 13}-\frac{1}{13 \times 15}\right)+\left(\frac{1}{13 \times 15}-\frac{1}{15 \times 17}\right)+ \\ & \cdots+\left(\frac{1}{29 \times 31}-\frac{1}{31 \times 33}\right) \\ = & \frac{1}{11 \times 13}-\frac{1}{31 \times 33}=\frac{80}{13299} . \end{aligned} $$ Answer: $\frac{20}{13299}$." e95318a453ca,"A lattice point is a point $(x, y)$ where both $x$ and $y$ are integers For how many different integer values of $k$ will the two lines $k x-5 y+7=0$ and $k^{2} x-5 y+1=0$ intersect at a lattice point? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5",(B),medium,"Subtracting the first equation from the second, $$ \begin{aligned} k^{2} x-k x-6 & =0 \\ \left(k^{2}-k\right) x & =6 \\ {[k(k-1)] x } & =6 \end{aligned} $$ Since we want both $k$ and $x$ to be integers, then $k(k-1)$ is a factor of 6 , ie. is equal to one of $\pm 1, \pm 2, \pm 3, \pm 6$. Now $k(k-1)$ is the product of two consecutive integers, so on this basis we can eliminate six of these eight possibilities to obtain $$ k(k-1)=2 \quad \text { or } \quad k(k-1)=6 $$ which yields $$ k^{2}-k-2=0 \quad \text { or } \quad k^{2}-k-6=0 $$ and so $k=2,-1,3,-2$. We now make a table of values of $k, x$ and $y$ to check when $y$ is also an integer. (We note that from the first equation, $y=\frac{1}{5}(k x+7)$.) | $k$ | $x$ | $y$ | | :---: | :---: | :---: | | 2 | 3 | $\frac{13}{5}$ | | -1 | 3 | $\frac{4}{5}$ | | 3 | 1 | 2 | | -2 | 1 | 1 | So there are two values of $k$ for which the lines intersect at a lattice point. ANSWER: (B)" f4d3e3dde2b1,"3. Given an integer $n \geqslant 2$. Find the smallest positive real number $c$, such that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have $$ \left|\sum_{i=1}^{n} z_{i}\right|+c \sum_{1 \leqslant i0$, we can divide through by $2 a b$. Thus, $$ \sin \gamma+1=\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1 $$ Notice that the fraction on the right is nothing but $\cos \gamma$ (from the cosine rule), so the equation simplifies to $$ \sin \gamma=\cos \gamma, \quad \text { where } \quad 0^{\circ}<\gamma<180^{\circ} $$ The equality holds only when $\gamma=45^{\circ}$, so the angle of the triangle we are looking for is $45^{\circ}$. Since we performed equivalent transformations throughout, this is indeed a solution to the original equation (which we can also verify by substitution)." a214cd0fcb7f,,See reasoning trace,easy,"Solution. First method. We will start solving in the reverse direction. To the number 5, which is obtained as the result, we add 5, then multiply by 5, and add 5 again to get Stojan's age. Specifically, Stojan is $(5+5) \cdot 5+5=55$ years old. Second method. Let $x$ be the number of years Stojan has. Then from the condition of the problem we have $$ \begin{aligned} & (x-5): 5-5=5 \\ & (x-5): 5=10 \\ & x-5=50 \end{aligned} $$ So Stojan is 55 years old." d0398fb1075d,"2. If $f(x)=\frac{1+x}{1-3 x}, f_{1}(x)=f[f(x)], f_{2}(x)=f\left[f_{1}(x)\right], \cdots, f_{n+1}(x)=f\left[f_{n}(x)\right]$, then $f_{2017}(-2)=$ A. $-\frac{1}{7}$ B. $\frac{1}{7}$ C. $-\frac{3}{5}$ D. $\frac{3}{5}$",$D$,easy,"Solve $f_{1}(x)=\frac{1+\frac{1+x}{1-3 x}}{1-3 \cdot \frac{1+x}{1-3 x}}=\frac{x-1}{3 x+1}, f_{2}(x)=\frac{1+\frac{x-1}{3 x+1}}{1-3 \cdot \frac{x-1}{3 x+1}}=x$, thus $f_{2017}(-2)=f_{1}(-2)=\frac{3}{5}$, so the answer is $D$." 040c1d04016e,,121,medium,"Answer: 121. Solution. The sum of numbers from 4 to 16 is 130. If at least one number is erased, the sum of the remaining numbers does not exceed 126. Let's sequentially consider the options: - if the sum is 126, then Masha could have erased only the number 4; then the remaining numbers can be divided into two groups with a sum of 63: $$ 16+15+14+13+5=12+11+10+9+8+7+6 $$ - if the sum is 125, then Masha could have erased only the number 5; then the remaining numbers can be divided into five groups with a sum of 25: $$ 16+9=15+10=14+11=13+12=8+7+6+4 $$ - if the sum is 124, then Masha could have erased only the number 6; then the remaining numbers can be divided into two groups with a sum of 62: $$ 16+15+14+13+4=12+11+10+9+8+7+5 $$ - if the sum is 123, then Masha could have erased only the number 7; then the remaining numbers can be divided into three groups with a sum of 41: $$ 16+15+10=14+13+9+5=12+11+8+6+4 $$ - if the sum is 122, then Masha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 61: $$ 16+15+14+12+4=13+11+10+9+7+6+5 $$ - if Masha erased the number 9, then the numbers left on the board have a sum of 121: they could be divided either into 11 groups with a sum of 11, or into 121 groups with a sum of 1; but some group will include the number 16, so the sum in this group will be at least 16; therefore, in this case, it is impossible to divide the numbers into groups with the same sum." db21841f49ef,"If the equation $$ x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0 $$ has exactly one real solution, then the value of $p$ is $\qquad$",$-\frac{9}{4}$,easy,"Hint: Use factorization to transform the original equation into $x^{2}+3 x-p=0$ or $x^{2}+3 x-2 p=0$. From the conditions, we have $$ \left\{\begin{array} { l } { \Delta _ { 1 } = 9 + 4 p = 0 , } \\ { \Delta _ { 2 } = 9 + 8 p < 0 } \end{array} \text { or } \left\{\begin{array}{l} \Delta_{1}=9+4 p<0, \\ \Delta_{2}=9+8 p=0 . \end{array}\right.\right. $$ Answer: $-\frac{9}{4}$." 5bfd8bf0197d,669. For which numbers $x$ does the equation $x+|x|=0$ hold?,"-x$, which occurs when $x \leqslant 0$. (Note that the boundary point 0 is also included.) $\triangl",easy,"$\triangleright$ This equality means that $|x|=-x$, which occurs when $x \leqslant 0$. (Note that the boundary point 0 is also included.) $\triangleleft$" 1bd797fba709,It is known that the equation$ |x - 1| + |x - 2| +... + |x - 2001| = a$ has exactly one solution. Find $a$.,1001000,medium,"1. **Identify the nature of the function:** The given function is \( f(x) = |x - 1| + |x - 2| + \cdots + |x - 2001| \). This function is a sum of absolute values, which is piecewise linear and changes slope at each of the points \( x = 1, 2, \ldots, 2001 \). 2. **Determine the median:** The function \( f(x) \) achieves its minimum value when \( x \) is the median of the set \(\{1, 2, \ldots, 2001\}\). Since there are 2001 points, the median is the middle value, which is \( x = 1001 \). 3. **Calculate the value of the function at the median:** To find \( a \), we need to evaluate \( f(1001) \): \[ f(1001) = |1001 - 1| + |1001 - 2| + \cdots + |1001 - 2001| \] This can be split into two sums: \[ f(1001) = \sum_{i=1}^{1000} (1001 - i) + \sum_{i=1002}^{2001} (i - 1001) \] 4. **Evaluate the first sum:** \[ \sum_{i=1}^{1000} (1001 - i) = \sum_{i=1}^{1000} 1001 - \sum_{i=1}^{1000} i = 1000 \cdot 1001 - \frac{1000 \cdot 1001}{2} = 1000 \cdot 1001 - 500 \cdot 1001 = 500 \cdot 1001 \] 5. **Evaluate the second sum:** \[ \sum_{i=1002}^{2001} (i - 1001) = \sum_{i=1}^{1000} i = \frac{1000 \cdot 1001}{2} = 500 \cdot 1001 \] 6. **Combine the results:** \[ f(1001) = 500 \cdot 1001 + 500 \cdot 1001 = 1000 \cdot 1001 = 1001000 \] Conclusion: \[ a = 1001000 \] The final answer is \(\boxed{1001000}\)." f73e219ec03e,"10. $[8] A B C D$ is a convex quadrilateral such that $A B=2, B C=3, C D=7$, and $A D=6$. It also has an incircle. Given that $\angle A B C$ is right, determine the radius of this incircle.",$\frac{1+\sqrt{13}}{3}$,easy,"Answer: $\frac{1+\sqrt{13}}{3}$. Note that $A C^{2}=A B^{2}+B C^{2}=13=C D^{2}-D A^{2}$. It follows that $\angle D A C$ is right, and so $$ [A B C D]=[A B C]+[D A C]=2 \cdot 3 / 2+6 \cdot \sqrt{13} / 2=3+3 \sqrt{13} $$ On the other hand, if $I$ denotes the incenter and $r$ denotes the inradius, $$ [A B C D]=[A I B]+[B I C]+[C I D]+[D I A]=A B \cdot r / 2+B C \cdot r / 2+C D \cdot r / 2+D A \cdot r / 2=9 r $$ Therefore, $r=(3+3 \sqrt{13}) / 9=\frac{1+\sqrt{13}}{3}$." f5e778b13a52,"Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$. Which of the following statements is not true? $\mathrm{(A) \ } x \heartsuit y = y \heartsuit x$ for all $x$ and $y$ $\mathrm{(B) \ } 2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$ $\mathrm{(C) \ } x \heartsuit 0 = x$ for all $x$ $\mathrm{(D) \ } x \heartsuit x = 0$ for all $x$ $\mathrm{(E) \ } x \heartsuit y > 0$ if $x \neq y$",\mathrm{(C),easy,"We start by looking at the answers. Examining statement C, we notice: $x \heartsuit 0 = |x-0| = |x|$ $|x| \neq x$ when $x<0$, but statement C says that it does for all $x$. Therefore the statement that is not true is $\boxed{\mathrm{(C)}\ x\heartsuit 0=x\ \text{for all}\ x}$" b3147e504b45,"4. (7 points) On the board, 49 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 49 minutes?",1176,medium,"Answer: 1176. Solution: Let's represent 49 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups "" $x$ "" and "" $y$ "" will be connected by $x y$ line segments. This is the same number of candies that Karlson eats. After 49 minutes, all points will be connected. In total, $\frac{49 \cdot(49-1)}{2}=1176$ line segments will be drawn. Therefore, Karlson will eat 1176 candies." cd29c648dc51,"1. (6 points) Today is January 31, 2015, welcome to the 2015 ""Spring Cup"" Final. The calculation result of the expression $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$ is",: 6500,easy,"【Solution】Solve: $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$ $$ \begin{array}{l} =\frac{\left(\frac{2015}{1}+\frac{2015}{0.31}\right) \times 0.31}{(1+0.31) \times 0.31} \\ =\frac{(1+0.31) \times 2015}{(1+0.31) \times 0.31} \\ =6500 ; \end{array} $$ Therefore, the answer is: 6500." a4dbc4045975,6.005. $\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12}$.,"$x_{1,2} \in \varnothing, x_{3}=-3, x_{4}=1$",medium,"Solution. Domain of definition: $x \neq 0, x \neq-1, x \neq-2$. $\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12} \Leftrightarrow \frac{1}{x^{2}+2 x}-\frac{1}{x^{2}+2 x+1}=\frac{1}{12}$. Let $x^{2}+2 x=z$, then $\frac{z^{2}+z+12}{z(z+1)}=0 \Rightarrow z^{2}+z+12=0 \Rightarrow z_{1}=-4, z_{2}=3$. To find $x$, solve the two equations: $x^{2}+2 x=-4$ or $x^{2}+2 x=3$. Solving them, we find: $x_{1,2} \in \varnothing (D<0), x_{3}=-3, x_{4}=1$. Answer: $x_{1,2} \in \varnothing, x_{3}=-3, x_{4}=1$." 473975c15948,"Three. (50 points) Given that $n$ is a positive integer greater than 10, and set $A$ contains $2n$ elements. If the family of sets $$ \left\{A_{i} \subseteq A \mid i=1,2, \cdots, m\right\} $$ satisfies the following two conditions, it is called ""suitable"": (1) For any $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\right)=n$; (2) For any $1 \leqslant i < j \leqslant m, \operatorname{Card}\left(A_{i} \cap A_{j}\right) \leqslant 10$, Given $n>10$, find the maximum positive integer $m$, such that there exists a suitable family of sets containing $m$ sets.",4$.,medium,"The maximum positive integer $m=4$. Take two non-complementary $n$-element subsets $A_{1}, A_{2}$ of $A$, and let $A_{3}, A_{4}$ be the complements of $A_{1}, A_{2}$, respectively. From these four sets, any three sets will always have two sets that are complementary, hence the intersection of these three sets is the empty set. Assume the family of sets $\left\{A_{i} \subseteq A\right\}$ contains 5 subsets of $A$. The 5 subsets $A_{i}(i=1,2, \cdots, 5)$ contain a total of $5 n$ elements. (1) If there is 1 element that appears 5 times, then the other elements appear at most 2 times. Thus, the 5 subsets contain at most $5 + 2(2 n-1) = 4 n + 3$ elements, which is a contradiction. (2) If there are 2 elements each appearing 4 times, then these 2 elements must appear together in at least 3 subsets, which is a contradiction. (3) If exactly 1 element appears 4 times, and let the number of elements appearing 3 times be $s$, then $$ \begin{array}{l} 4 + 3 s + 2(2 n - s - 1) \geq 5 n \\ \Rightarrow s \geq n - 2 . \end{array} $$ (i) When $s > n - 2$, at least $n$ elements each appear at least 3 times; (ii) When $s = n - 2$, there is 1 element appearing 4 times, and $n - 2$ elements each appearing 3 times. (4) If no element appears 4 times, and let the number of elements appearing 3 times be $s$. Then $$ 3 s + 2(2 n - s) \geq 5 n \Rightarrow s \geq n . $$ In summary, there are only two cases: (1) At least $n$ elements each appear at least 3 times; (2) One element appears 4 times, and $n - 2$ elements each appear 3 times. For (1), since the intersection of any 3 subsets out of the 5 subsets contains at most 1 element, the number of elements appearing at least 3 times, $n \leq \mathrm{C}_{5}^{3} = 10$, which is a contradiction. For (2), similarly, by the number of elements appearing at least 3 times, $n - 1 \leq \mathrm{C}_{5}^{3} = 10$, we know $n = 11$ and the intersection of any 3 subsets out of the 5 subsets is exactly a singleton set. These 10 singleton sets are all different, but for the element appearing 4 times, it appears in 4 singleton sets, which is a contradiction. Therefore, there does not exist a suitable family of sets containing 5 subsets. In conclusion, the maximum positive integer $m=4$." 0518991969dc,"Let $ABCD$ be an inscribed trapezoid such that the sides $[AB]$ and $[CD]$ are parallel. If $m(\widehat{AOD})=60^\circ$ and the altitude of the trapezoid is $10$, what is the area of the trapezoid?",100\sqrt{3,medium,"1. **Identify the given information and properties:** - $ABCD$ is an inscribed trapezoid with $AB \parallel CD$. - $m(\widehat{AOD}) = 60^\circ$. - The altitude (height) of the trapezoid is $10$. 2. **Use the property of inscribed trapezoids:** - Since $ABCD$ is an inscribed trapezoid, the non-parallel sides $AD$ and $BC$ are equal, i.e., $AD = BC$. 3. **Analyze the given angle:** - The angle $m(\widehat{AOD}) = 60^\circ$ is given. Since $O$ is the center of the circle, $\widehat{AOD}$ is a central angle. - This implies that the arc $AD$ subtended by $\widehat{AOD}$ is $60^\circ$. 4. **Determine the lengths of the diagonals:** - Since $ABCD$ is an inscribed trapezoid, the diagonals $AC$ and $BD$ are equal. Let $AC = BD = d$. 5. **Use the altitude to find the area:** - The altitude (height) of the trapezoid is given as $10$. - The area of a trapezoid is given by the formula: \[ \text{Area} = \frac{1}{2} \times (AB + CD) \times \text{height} \] 6. **Calculate the lengths of $AB$ and $CD$:** - Since $m(\widehat{AOD}) = 60^\circ$, the triangle $AOD$ is an equilateral triangle (as $AD = AO = OD$). - Let $r$ be the radius of the circle. Then $AD = r$. - The height of the equilateral triangle $AOD$ is $\frac{\sqrt{3}}{2}r$. - Given that the height of the trapezoid is $10$, we have: \[ \frac{\sqrt{3}}{2}r = 10 \implies r = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \] 7. **Calculate the area of the trapezoid:** - The length of $AD$ (or $BC$) is $r = \frac{20\sqrt{3}}{3}$. - The area of the trapezoid is: \[ \text{Area} = \frac{1}{2} \times (AB + CD) \times 10 \] - Since $AB$ and $CD$ are parallel and $AD = BC$, we need to find $AB$ and $CD$ in terms of $r$. 8. **Final calculation:** - Since $AB$ and $CD$ are parallel and the height is $10$, we can use the properties of the circle and the trapezoid to find the exact lengths of $AB$ and $CD$. - However, the problem does not provide enough information to determine the exact lengths of $AB$ and $CD$ without additional assumptions or information. The final answer is $\boxed{100\sqrt{3}}$." 6f47eeb83fed,44. Solve the inequality $4 \log _{16} \cos 2 x+2 \log _{4} \sin x+\log _{2} \cos x+$ $+3<0$.,See reasoning trace,easy,"44. We transition to logarithms with base 2. The given inequality reduces to the inequality $\log _{2} 2 \sin 4 x<0$, i.e., $0<\sin 4 x<\frac{1}{2}$. But $0 \sqrt{961} = 31 $$ Thus, $m+n > 64$. From this, we conclude that $m+n \geqslant 65$. It remains to show that it is possible for there to be $m+n=65$ competitors at the tournament. We will construct an example for the case $m=30$ and $n=35$. Divide the girls into six groups (call them $X_{1}, \ldots, X_{6}$), and the boys into seven groups (call them $Y_{1}, \ldots, Y_{7}$), each with 5 people. For each pair of groups $X_{i}$ and $Y_{j}$, let the girl in position $k$ in group $X_{i}$ win the boys in positions $k, k+1$, and $k+2$ in group $Y_{j}$, considering positions cyclically (the sixth position is the first, the seventh is the second, the eighth is the third). ![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-24.jpg?height=314&width=151&top_left_y=2256&top_left_x=881) Then each girl has won $7 \cdot 3 = 21$ boys, and each boy has won $6 \cdot 2 = 12$ girls. ![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-25.jpg?height=329&width=1119&top_left_y=338&top_left_x=400) Note: From the inequality $(m-12)(n-21) \geqslant 252$, we cannot directly conclude that $m+n$ will be the smallest when equality holds. For example, $a b = 7 + 11$. Therefore, we need to use some form of the inequality between the arithmetic and geometric means or reasoning as in the second solution. Note: The example from the solution can also be described in a different way. Divide the girls into five groups of 6 each (call them $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$), and the boys into five groups of 7 each (call them $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$). We have 5 groups of girls and 5 groups of boys (1 point). Let all boys in group $B_{i}$ win the girls in groups $A_{i}$ and $A_{i+1}$, for $1 \leqslant i \leqslant 4$, and $B_{5}$ win the girls in $A_{5}$ and $A_{1}$. They lose to all other girls. (2 points) This means that each boy has won girls from exactly two groups of girls, i.e., in exactly $2 \cdot 6 = 12$ matches. Each girl has won boys from exactly three of their groups, i.e., in exactly $3 \cdot 7 = 21$ matches. (1 point)" 732214211a93,1. Solve the equation: $\frac{8 x+13}{3}=\frac{19-12 x}{2}-\left(16 x-\frac{7-4 x}{6}\right)$.,See reasoning trace,easy,"1. $\frac{8 x+13}{3}=\frac{19-12 x}{2}-\left(16 x-\frac{7-4 x}{6}\right)$ $\frac{8 x+13}{3}=\frac{19-12 x}{2}-16 x+\frac{7-4 x}{6} / \cdot 6$ 1 BOD $2 \cdot(8 x+13)=3 \cdot(19-12 x)-96 x+7-4 x \quad 1$ BOD $16 x+26=57-36 x-96 x+7-4 x \quad 1$ BOD $16 x+36 x+96 x+4 x=57+7-26$ $152 x=38 \quad /: 152$ 1 BOD $x=\frac{38}{152}$ 1 BOD $x=\frac{1}{4}$ 1 BOD TOTAL 6 POINTS Note: If the student has not fully simplified the fraction, the solution should be evaluated with 5, not 6 points." efa89925b3f0,,See reasoning trace,medium," Solution. The discriminant of $f(x)$ is $D=4\left(4 p^{2}-p+1\right)$ and $D>0$ for all real $p$. Consequently $f(x)$ has two real roots $x_{1}$ and $x_{2}$, i.e. $f(x)$ intersects the $x$-axis in two different points $-A$ and $B$. The vertex $C$ of the parabola has coordinates $2 p$ and $h=f(2 p)=4 p^{2}-p+1>0$. We have $$ A B=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}-x_{2}\right)^{2}}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=2 \sqrt{4 p^{2}-p+1} $$ Now we find $S=S_{A B C}=\frac{A B \cdot h}{2}=\left(4 p^{2}-p+1\right)^{\frac{3}{2}}$. Denote $q=4 p^{2}-p+1$. Since $q$ is rational and $q^{3}=S^{2}$ is an integer, then $q$ is an integer too. Then $\frac{S}{q}$ is rational and $\left(\frac{S}{q}\right)^{2}=q$ is integer, thus $\frac{S}{q}$ is integer too. Therefore $q=n^{2}$, where $n$ is a positive integer, i.e. $4 p^{2}-p+1-n^{2}=0$. The quadratic equation (with respect to $p$ ) has a rational root exactly when its discriminant $16 n^{2}-15$ is a square of a rational number. Consequently $16 n^{2}-15=m^{2}$, and we can consider $m$ to be a positive integer. From the equality $(4 n-m)(4 n+m)=15$ we get $4 n-m=1$, $4 n+m=15$ or $4 n-m=3,4 n+m=5$. From here $n=2, m=7$ or $n=1, m=1$. The rational numbers we are looking for are $0,1, \frac{1}{4},-\frac{3}{4}$. " 218ced041a41,"12.209. Find the lateral surface area and volume of a right parallelepiped if its height is $h$, the diagonals form angles $\alpha$ and $\beta$ with the base, and the base is a rhombus.",$2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta} ; \frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$,medium,"Solution. Let the rhombus $ABCD$ be the base of a right prism ![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0956.jpg?height=458&width=677&top_left_y=1370&top_left_x=582) Fig. 12.76 $ABCD A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.76), $AA_{1}=BB_{1}=h, O$ - the point of intersection of the diagonals $AC$ and $BD$ of the rhombus, $\angle A_{1}CA=\alpha, \angle B_{1}DB=\beta$. In $\triangle B_{1}BD\left(\angle B_{1}BD=90^{\circ}\right): BD=BB_{1} \operatorname{ctg} \angle B_{1}DB=h \operatorname{ctg} \beta$. From $\triangle A_{1}AC\left(\angle A_{1}AC=90^{\circ}\right): AC=AA_{1} \operatorname{ctg} \angle A_{1}CA=h \operatorname{ctg} \alpha$. The area of the base of the prism $S=\frac{1}{2} AC \cdot BD=\frac{1}{2} h^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta \Rightarrow$ the volume of the prism $V=S h=\frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$. $$ OA=\frac{1}{2} AC=\frac{1}{2} h \operatorname{ctg} \alpha, OD=\frac{1}{2} BD=\frac{1}{2} h \operatorname{ctg} \beta $$ In $\triangle AOD\left(\angle AOD=90^{\circ}\right): AD=\sqrt{AO^{2}+OD^{2}}=\frac{1}{2} h \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$. The lateral surface area of the prism: $S_{\sigma}=4 AD \cdot h=2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$. Answer: $2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta} ; \frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$. ![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0957.jpg?height=499&width=668&top_left_y=905&top_left_x=92) Fig. 12.77" 3e480280b0c0,"1. Given arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ are both permutations of $1,2, \cdots, n$. Then $$ a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} $$ the maximum value is",See reasoning trace,easy,"$$ -1 . \frac{n(n+1)(2 n+1)}{6} \text {. } $$ Notice that, $1<2<\cdots b$ can then be found by swapping $a$ and $b$. We examine the possibilities for $a$. Case $a \geqslant 3$: Since $3 \leqslant a \leqslant b$, both $a!$ and $b!$ are divisible by 3, so $a! + b!$ is also divisible by 3. Since $2^n$ is not divisible by 3 for any value of $n$, there are no solutions. Case $a = 1$: For $b$, it must hold that $b! = 2^n - 1$. Since $2^n$ is even (because $n \geqslant 1$), $b!$ must be odd. Since $b!$ is divisible by 2 for all $b \geqslant 2$, only $b = 1$ remains. We find $1! = 2^n - 1$, so $n = 1$. The only solution is thus $(a, b, n) = (1, 1, 1)$. Case $a = 2$: For $b \geqslant 4$, there are no solutions. Indeed, since $b!$ is then divisible by 4, $2^n = b! + 2$ cannot be a multiple of 4, so $2^n = 2$ is the only possibility. But this contradicts the fact that $2^n = b! + 2 \geqslant 24 + 2$. For $b = 2$, we find $2^n = 2 + 2 = 4$. Thus, $n = 2$ and $(a, b, n) = (2, 2, 2)$ is the only solution. For $b = 3$, we find $2^n = 2 + 6 = 8$. Thus, $n = 3$ and $(a, b, n) = (2, 3, 3)$ is the only solution. By swapping $a$ and $b$, we also find the solution $(a, b, n) = (3, 2, 3)$. In total, there are four solutions: $(1, 1, 1)$, $(2, 2, 2)$, $(2, 3, 3)$, and $(3, 2, 3)$." a0dc8376006f,A3. What is the value of $\left(\frac{4}{5}\right)^{3}$ as a decimal?,See reasoning trace,easy,"$\begin{array}{ll}\text { Solution } & \mathbf{0 . 5 1 2}\end{array}$ $$ \left(\frac{4}{5}\right)^{3}=\frac{4^{3}}{5^{3}}=\frac{64}{125}=\frac{64 \times 8}{125 \times 8}=\frac{512}{1000}=0.512 \text {. } $$" b6b42fae07c0,9. Parallelogram $A E C F$ is inscribed in square $A B C D$. It is reflected across diagonal $A C$ to form another parallelogram $A E^{\prime} C F^{\prime}$. The region common to both parallelograms has area $m$ and perimeter $n$. Compute the value of $\frac{m}{n^{2}}$ if $A F: A D=1: 4$.,"7: 1$. Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \",easy,"Solution: By symmetry, the region is a rhombus, AXCY, centered at the center of the square, $O$. Consider isoceles right triangle $A C D$. By the technique of mass points, we find that $D O: Y O=7: 1$. Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \sqrt{2}$. The perimeter of the rhombus is $20 \sqrt{2} N$, and the area is $14 N^{2}$. The required ratio is thus $\frac{7}{400}$." 1bd095c2a0ca,"91. A discrete random variable $X$ has three possible values: $x_{1}=1, x_{2}$, and $x_{3}$, with $x_{1}7^{2013}$, we obtain that $b+11^{2014}>b+7^{2013}$, thus $f_{2}>1$. From $f_{1}<1$ and $f_{2}>1$ we conclude that $f_{1}q!$ and hence $p>q$. Therefore $$ \frac{p!}{q!}=\frac{1 \times 2 \times \cdots \times q \times(q+1) \times \cdots \times p}{1 \times 2 \times \cdots \times q}=(q+1) \times(q+2) \times \cdots \times(p-1) \times p \text {. } $$ Thus $840=\frac{p!}{q!}$ is the product of the consecutive integers $q+1, q+2, \ldots, p-1, p$, where $p \leq 9$. Since 840 is not a multiple of $9, p \neq 9$. Since 840 is a multiple of $7, p \geq 7$. Now $5 \times 6 \times 7 \times 8=1680>840$, while $6 \times 7 \times 8=336<840$. Hence 840 is not the product of consecutive integers of which the largest is 8 . We deduce that $p=7$. It is now straightforward to check that $$ 840=4 \times 5 \times 6 \times 7=\frac{7!}{3!} . $$ Therefore $p=7$ and $q=3$. Hence $p+q=7+3=10$." c376f20614ec,"2. In rectangle $A B C D$, points $E$ and $F$ lie on sides $A B$ and $C D$ respectively such that both $A F$ and $C E$ are perpendicular to diagonal $B D$. Given that $B F$ and $D E$ separate $A B C D$ into three polygons with equal area, and that $E F=1$, find the length of $B D$.","D F=\frac{1}{3} A B$. Let $C E \cap B D=X$, then $\frac{E X}{C X}=\frac{B E}{C D}=\frac{1}{3}$, so t",easy,"Answer: $\sqrt{3}$ Observe that $A E C F$ is a parallelogram. The equal area condition gives that $B E=D F=\frac{1}{3} A B$. Let $C E \cap B D=X$, then $\frac{E X}{C X}=\frac{B E}{C D}=\frac{1}{3}$, so that $B X^{2}=E X \cdot C X=3 E X^{2} \Rightarrow B X=\sqrt{3} E X \Rightarrow$ $\angle E B X=30^{\circ}$. Now, $C E=2 B E=C F$, so $C E F$ is an equilateral triangle and $C D=\frac{3}{2} C F=\frac{3}{2}$. Hence, $B D=\frac{2}{\sqrt{3}} \cdot \frac{3}{2}=\sqrt{3}$." 7e8e6b03d695,2nd Australian 1981,"(√5-1)/2 = 0.618... . Then for x k, then x 2 > 1-x, so it is worse for A if B shoots first. Moreove",medium,"If the first player to shoot hits the target then he wins. If he misses, then the other player waits until he is up against the target and cannot miss, so if the first player to shoot misses, then he loses. So if A shoots first at x, then he has a prob x 2 of winning, whereas if B shoots first at x, then A has a prob 1-x of winning. Put k = (√5-1)/2 = 0.618... . Then for x k, then x 2 > 1-x, so it is worse for A if B shoots first. Moreover, the longer he waits the worse his position if B manages to shoot first. Thus A's strategy is as follows. If B shoots at x < k, then he waits until he reaches the target before shooting. If B has not shot at x = k, then A shoots. s are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust, can be ordered by email. 2nd Aus 1981 © John Scholes jscholes@kalva.demon.co.uk 28 Oct 2003 Last updated/corrected 28 Oct 03" 753d92a240df,"Someone says that seven times their birth year, when divided by 13, leaves a remainder of 11, and thirteen times their birth year, when divided by 11, leaves a remainder of 7. In which year of their life will this person be in 1954?",See reasoning trace,medium,"Let the birth year be denoted by $x$, then according to the problem, $$ \begin{aligned} 7 x & =13 y+11 \\ 13 x & =11 z+7 \end{aligned} $$ Subtracting 7 times (2) from 13 times (1), we get $$ 169 y-77 z+94=0 $$ Thus, $$ \begin{aligned} & z=\frac{169 y+94}{77}=2 y+1+\frac{15 y+17}{77}=2 y+1+u \\ & y=\frac{77 u-17}{15}=5 u-1+\frac{2 u-2}{15}=5 u-1+2 v \\ & u=15 v+1 \end{aligned} $$ By substituting back, $$ y=75 v+5-1+2 v=77 v+4 $$ and thus $$ x=\frac{13 y+11}{7}=\frac{13 \cdot 77 v+52+11}{7}=143 v+9 $$ Given the nature of the problem, $$ 1750a_{n}$ For any $n \geqslant 2$, try to find the range of values for $A$ such that the inequality $\left|\sum_{i=1}^{n} a_{i} x_{i}\right| \leqslant A\left(a_{1}-a_{n}\right)$ always holds.",See reasoning trace,medium,"When $n=2$, $x_{1}=\frac{1}{2}, x_{2}=-\frac{1}{2}$ or $x_{1}=-\frac{1}{2}, x_{2}=\frac{1}{2}$. Thus, the original inequality becomes $$ \frac{1}{2}\left|a_{1}-a_{2}\right|=\frac{1}{2}\left(a_{1}-a_{2}\right) \leqslant A\left(a_{1}-a_{2}\right) \text {. } $$ Therefore, $\frac{1}{2} \leqslant A$. Next, we prove that for any $n \geqslant 2$, when $A \geqslant \frac{1}{2}$, the original inequality holds. Let $S_{k}=\sum_{i=1}^{k} x_{i}(k=1,2, \cdots, n)$. By the problem statement, $S_{n}=0$, and $$ \left|S_{k}\right|=\left|\sum_{j=1}^{k} x_{j}\right|=\left|-\sum_{j=k+1}^{n} x_{j}\right|=\left|\sum_{j=k+1}^{n} x_{j}\right| \text {. } $$ Then $2\left|S_{k}\right|=\left|\sum_{j=1}^{k} x_{j}\right|+\left|\sum_{j=k+1}^{n} x_{j}\right| \leqslant \sum_{j=1}^{n}\left|x_{j}\right|=1$. So, $\left|S_{k}\right| \leqslant \frac{1}{2}$. Applying Theorem 2, we get $$ \begin{array}{l} \left|\sum_{i=1}^{n} a_{i} x_{i}\right|=\left|a_{n} S_{n}+\sum_{i=1}^{n-1} S_{i}\left(a_{i}-a_{i+1}\right)\right| \\ =\left|\sum_{i=1}^{n-1} S_{i}\left(a_{i}-a_{i+1}\right)\right| \\ \leqslant \frac{1}{2}\left|\sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)\right| \\ =\frac{1}{2}\left|a_{1}-a_{n}\right|=\frac{1}{2}\left(a_{1}-a_{n}\right) . \end{array} $$" 7334869e3763,4.58 Find the integers $m$ and $n$ that satisfy the equation $(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n}$.,n=0$.,easy,"[Solution] Let $(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n}=A+B \sqrt{2}$, where $A, B$ are rational numbers. If $B=0$, then it must be that $m=n=0$. If $B \neq 0$, then $$ (5-3 \sqrt{2})^{m}=(3-5 \sqrt{2})^{n}=A-B \sqrt{2} $$ On the other hand, since $0<5-3\sqrt{2}<1$ and $0<3-5\sqrt{2}<1$, it follows that $(5-3\sqrt{2})^m < 1$ and $(3-5\sqrt{2})^n < 1$. Therefore, (1) cannot hold. This is a contradiction. Hence, the equation holds only when $m=n=0$." 32288093e86d,"8. A deck of playing cards has a total of 54 cards, with 13 cards each of Spades, Hearts, Clubs, and Diamonds, plus 2 jokers. At least how many cards must be drawn to ensure that there are 2 cards of each of the 4 suits?","】If there are 12 cards of the same color left, it is impossible to ensure that there are 2 cards of each of the 4 suits, so at least 43 cards must be drawn",easy,"43 【Answer】If there are 12 cards of the same color left, it is impossible to ensure that there are 2 cards of each of the 4 suits, so at least 43 cards must be drawn." 2c444949ea17,"Find all triplets of strictly positive integers $(a, b, c)$ such that $6^{a}=1+2^{b}+3^{c}$",See reasoning trace,medium,"We notice that 3 divides $2^{b}+1$, so $b$ is odd. If $b=1,1+3^{c-1}=2 \cdot 6^{a-1}$, so $a=1$. Indeed, if $a>1,3$ divides $1+3^{c-1}$, which is absurd. Therefore, if $b=1$, we have the solution $(1,1,1)$. Otherwise, $b \geqslant 3$, so $a \geqslant 2$ (since $6<1+2^{3}$). Modulo 4, we get $1+3^{c} \equiv 0[4]$, so $c$ is odd. Modulo 8, we then have $1+2^{b}+3^{c} \equiv 4[8]$, so $6^{a} \equiv 4[8]$, so $a=2$. Thus $2^{b}+3^{c}=35$, and an exhaustive study gives the solutions $(2,3,3)$ and $(2,5,1)$. Sometimes when we cannot find a modulo or factorization that allows us to advance on an arithmetic problem, we can resort to inequalities. This method partly relies on the following proposition: ## Proposition 1. If $a$ and $b$ are integers such that $a$ divides $b$, then $|a| \leqslant|b|$ This property, though obvious, is nonetheless essential in solving many difficult exercises. Indeed, it can often be useful to combine several inequalities of this type to obtain an enclosure on a variable or an absurdity. An interesting point to note in this regard is that the property can always be refined: for example, if we know that $a \neq \pm b$, then $|a| \leqslant 2|b|$. A second observation that can be made and which seems obvious but is equally useful is the following: there is no other square between two consecutive squares. It can of course be adapted with cubes, etc. Without further ado, here are some application exercises. In this section, exercises 3, 4, and 5 have been treated." 6272ae9bca85,"2.44 Write down the two-digit numbers from 19 to 80 in succession. Ask: Is the resulting number $$ 19202122 \cdots 787980 $$ divisible by 1980?",See reasoning trace,medium,"［Solution］Let $A=19202122 \cdots 7980$. Obviously, $20 \mid A$. Since $100^{k}=(99+1)^{k}=99 M+1$. Where $M$ is a positive integer, $k$ is a positive integer. Therefore, $$ 100^{k} \equiv 1 $$ $(\bmod 99)$. Thus, we have $$ \begin{aligned} A & =19 \cdot 100^{61}+20 \cdot 100^{60}+\cdots+79 \cdot 100+80 \\ & \equiv(19+20+21+\cdots+79+80) \\ & \equiv 31 \cdot 99 \\ & \equiv 0 \quad(\bmod 99) . \end{aligned} $$ Therefore, $99 \mid A$. Since $(20,99)=1$, we have $$ 20 \cdot 99=1980 \mid A \text {. } $$" 8432d6bc58c4,"Find all pairs of integers $(x, y)$ such that $y^{2}=x^{3}+16$ (Factorization)","x^{3}$. Let $y-4=a$. If $2 \mid a$, then $2 \mid x$, so $8 \mid x^{3}=a(a+8)$, hence $a \equiv 0,4[8",medium,"We have $(y-4)(y+4)=x^{3}$. Let $y-4=a$. If $2 \mid a$, then $2 \mid x$, so $8 \mid x^{3}=a(a+8)$, hence $a \equiv 0,4[8]$. If $a \equiv 4[8], 4 \mid a$, thus $4^{3} \mid x^{3}=a(a+8)$, so $a(a+8) \equiv 0[64]$, but $a=4+8 a^{\prime}$, thus $a(a+8)=64 a^{\prime 2}+64 a+16+32+64 a^{\prime} \equiv 48[64]$. This is absurd. Therefore, $8 \mid a$. Otherwise, $2 \nmid a$. Thus, $a \wedge(a+8)=a \wedge 8=8$ or 1. We have, if $a=8 b$: then there exists $y$ such that $x=4 y$, hence $b(b+1)=y^{3}$ and $b \wedge(b+1)=1$, so $b$ and $b+1$ are both cubes: necessarily $b=0$, thus $(x, y)=(0,4)$. If $a \wedge(a+8)=1$: then $a$ and $a+8$ are both cubes. Necessarily $a=0$, which is absurd. The only solution is therefore $(x, y)=(0,4)$." d44b71aa361e,"Find all integers $x, y, z \geq 0$ such that $5^{x} 7^{y}+4=3^{z}$.","7^{y}$ and $3^{n}+2=5^{x}$. Then $4=5^{x}-7^{y}$. Suppose, for the sake of contradiction, that $x, y",medium,"It is clear that $(x, y) \neq(0,0)$ and that $z \geq 1$. First, $(x, y, z)=(1,0,2)$ is indeed a solution. If $y=0$ (and $x \geq 1$), by looking at the equation modulo 5, we find $z \equiv 2 \bmod 4$. If $y \geq 1$, by looking modulo 7, we find that $z \equiv 4 \bmod 6$. In all cases, $z$ is even. We can therefore write $z=2 n$. Then $5^{x} 7^{y}=\left(3^{n}-2\right)\left(3^{n}+2\right)$. Since $3^{n}-2$ and $3^{n}+2$ are coprime, two cases arise: Case 1: $3^{n}-2=5^{x}$ and $3^{n}+2=7^{y}$. By looking at the second equation modulo 3, we see that there are no solutions. Case 2: $3^{n}-2=7^{y}$ and $3^{n}+2=5^{x}$. Then $4=5^{x}-7^{y}$. Suppose, for the sake of contradiction, that $x, y \geq 1$. By looking modulo 7, we see $x \equiv 2 \bmod 6$ and thus $x$ is even. By looking modulo 5, we see that $y \equiv 0 \bmod 4$ and thus $y$ is even. By writing $x=2 x^{\prime}$ and $y=2 y^{\prime}$, we have $4=\left(5^{x^{\prime}}-7^{y^{\prime}}\right)\left(5^{x^{\prime}}+7^{y^{\prime}}\right)$, which does not yield solutions. Therefore, we have $x=0$ or $y=0$, and we find the only solution $(x, y, z)=(1,0,2)$." 3bee392577d9,"Five. (12 points) Try to find the largest and smallest seven-digit numbers that can be divided by 165, formed by the 7 digits $0,1,2,3,4,5,6$ without repetition (require the reasoning process to be written out). Find the largest and smallest seven-digit numbers that can be formed using the digits $0,1,2,3,4,5,6$ without repetition, which are divisible by 165 (the reasoning process must be included).",See reasoning trace,medium,"Five, since $165=3 \times 5 \times 11$, the number we are looking for must be a multiple of 3, 5, or 11. Also, $0+1+2+3+4+5+6=21$. Therefore, the sum of these seven digits is always a multiple of 3. Let the sum of the four digits in the odd positions be $A$, and the sum of the three digits in the even positions be $B$. Then $|A-B|=11k$ (where $k$ is an integer greater than or equal to 0). Also, $A+B=21$ and $A, B$ are both positive. Thus, $|A-B|<21$. Since $A+B$ is odd, $A-B \neq 0$. Therefore, $k \neq 0$, so $k=1$. Thus, one of $A$ and $B$ is 16, and the other is 5. Since $0+1+2+3=6$, $A$ cannot be 5, so $A$ is 16, and $B$ is 5. Among these seven numbers, the only pairs that sum to 5 are $0,1,4$ or $0,2,3$. (1) $B=0+1+4, A=2+3+5+6$, (2) $B=0+2+3, A=1+4+5+6$. Since 0 is always in one of the three groups (i.e., even positions), 0 cannot be the leading digit. Also, since the number we are looking for is a multiple of 5, the last digit must be 5. Therefore, the smallest number is 1042635, and the largest number is 6431205. (Provided by Liu Hanwen, Hubei)" 42a5434e38a4,"9. In the diagram, $P Q=P R=Q S$ and $\angle Q P R=20^{\circ}$. What is $\angle R Q S$ ? A $50^{\circ}$ B $60^{\circ}$ C $65^{\circ}$ D $70^{\circ}$ E $75^{\circ}$","Q S$, triangle $P S Q$ is isosceles and hence $\angle P S Q=20^{\circ}$. Since the angles in a trian",medium,"9. B Since $P Q=Q S$, triangle $P S Q$ is isosceles and hence $\angle P S Q=20^{\circ}$. Since the angles in a triangle add to $180^{\circ}$, we have $20^{\circ}+20^{\circ}+\angle S Q P=180^{\circ}$ and hence $\angle S Q P=140^{\circ}$. Since $P Q=P R$, triangle $P R Q$ is isosceles and hence $\angle P R Q=\angle R Q P$. Also $\angle P R Q+\angle R Q P+20^{\circ}=180^{\circ}$ and hence $\angle R Q P=80^{\circ}$. Since $\angle R Q S=\angle S Q P-\angle R Q P$, the size of $\angle R Q S$ is $140^{\circ}-80^{\circ}=60^{\circ}$." 9819539c0b95,"If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to $\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}$",\text{D,easy,"Draw Diagram later We can let $AB=s$ and $AC=r$. We can also say that the area of a triangle is $\frac{1}{2}rs\sin A$, which we know, is $64$. We also know that the geometric mean of $r$ and $s$ is 12, so $\sqrt{rs}=12$. Squaring both sides gives us that $rs=144$. We can substitute this value into the equation we had earlier. This gives us \[\frac{1}{2}\times144\times\sin A=64\] \[72\sin A=64\] \[\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}\] -edited for readability" 2554a3cf5156,"Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$, the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$, and the resulting line is reflected in $l_2^{}$. Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$. Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$, find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$.",945,medium,"Let $l$ be a line that makes an angle of $\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$. The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$. Thus, $l'$ makes an angle of $\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta$ with the positive $x$-axis. Similarly, since the angle between $l'$ and $l_2$ is $\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}$, the angle between $l''$ and the positive $x$-axis is $\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta$. Thus, $R(l)$ makes an $\frac{8\pi}{945} + \theta$ angle with the positive $x$-axis. So $R^{(n)}(l)$ makes an $\frac{8n\pi}{945} + \theta$ angle with the positive $x$-axis. Therefore, $R^{(m)}(l)=l$ iff $\frac{8m\pi}{945}$ is an integral multiple of $\pi$. Thus, $8m \equiv 0\pmod{945}$. Since $\gcd(8,945)=1$, $m \equiv 0 \pmod{945}$, so the smallest positive integer $m$ is $\boxed{945}$." 91b4bd770812,"Example 18 (2003 National High School Competition Question) Given that $x$ and $y$ are both in the interval $(-2,2)$, and $xy=-1$, then the minimum value of the function $u=\frac{4}{4-x^{2}}+$ $\frac{9}{9-y^{2}}$ is ( ). A. $\frac{8}{5}$ B. $\frac{24}{11}$ C. $\frac{12}{7}$ D. $\frac{12}{5}$","\frac{4}{x^{2}}$, i.e., $x^{2}=\frac{2}{3}$, the value of $9 x^{2}+\frac{4}{x^{2}}$ is the smallest,",easy,"Solution: Choose D. Reason: From the given, we have $y=-\frac{1}{x}$, hence $$ u=\frac{4}{4-x^{2}}+\frac{9 x^{2}}{9 x^{2}-1}=1+\frac{35}{37-\left(9 x^{2}+\frac{4}{x^{2}}\right)} \text{. } $$ And $x \in\left(-2,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right)$, when $9 x^{2}=\frac{4}{x^{2}}$, i.e., $x^{2}=\frac{2}{3}$, the value of $9 x^{2}+\frac{4}{x^{2}}$ is the smallest, at which point $u$ has the minimum value $\frac{12}{5}$." 387a55fcc3a9,"11. (15 points) Let $f(x)$ be a function defined on the domain $D$. If for any real number $\alpha \in(0,1)$ and any two numbers $x_{1}, x_{2}$ in $D$, it always holds that $$ f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \leqslant \alpha f\left(x_{1}\right)+(1-\alpha) f\left(x_{2}\right), $$ then $f(x)$ is called a $C$ function defined on $D$. (1) Given that $f(x)$ is a $C$ function on $\mathbf{R}$, and $m$ is a given positive integer. Let $a_{n}=f(n)(n=0,1, \cdots, m)$, and $a_{0}=0, a_{m}=2 m$, and denote $S_{f}=a_{1}+a_{2}+\cdots+a_{m}$. For any function $f(x)$ that satisfies the conditions, try to find the maximum value of $S_{f}$. (2) If $f(x)$ is a function defined on $\mathbf{R}$ with the smallest positive period $T$, prove that $f(x)$ is not a $C$ function on $\mathbf{R}$.",See reasoning trace,medium,"11. (1) For any $n(0 \leqslant n \leqslant m)$, take $$ x_{1}=m, x_{2}=0, \alpha=\frac{n}{m} \in[0,1] \text {. } $$ Since $f(x)$ is a $C$ function on $\mathbf{R}$, $a_{n}=f(n)$, and $a_{0}=0, a_{m}=2 m$, we have $$ \begin{array}{l} a_{n}=f(n)=f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \\ \leqslant \alpha f\left(x_{1}\right)+(1-\alpha) f\left(x_{2}\right)=\frac{n}{m} \times 2 m=2 n . \end{array} $$ Thus, $S_{f}=a_{1}+a_{2}+\cdots+a_{m}$ $$ \leqslant 2(1+2+\cdots+m)=m^{2}+m \text {. } $$ We can prove that $f(x)=2 x$ is a $C$ function, and it makes $a_{n}=$ $2 n(n=0,1, \cdots, m)$ all hold. In this case, $$ S_{f}=m^{2}+m \text {. } $$ In summary, the maximum value of $S_{f}$ is $m^{2}+m$. (2) Assume $f(x)$ is a $C$ function on $\mathbf{R}$. If there exists $mf(n)$, let $x_{1}=n, x_{2}=n-T$, $\alpha=1-\frac{n-m}{T}$, a contradiction can also be derived. Therefore, $f(x)$ is a constant function on $[0, T)$. Since $f(x)$ is a periodic function with period $T$, $f(x)$ is a constant function on $\mathbf{R}$, which contradicts the fact that the smallest positive period of $f(x)$ is $T$. Hence, $f(x)$ is not a $C$ function on $\mathbf{R}$." 30c17f7a51b9,"Three. (Total 25 points) Find all integer solutions to the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{3}+y^{3}+z^{3}=3\end{array}\right.$.",See reasoning trace,medium,"$$ \begin{array}{l} \text { 3. } x+y=3-z, \\ x^{3}+y^{3}=3-z^{3} . \end{array} $$ (1) - (2) gives $xy=\frac{8-9z+3z^{2}}{3-z}$. Knowing that $x, y$ are the roots of $t^{2}-(3-z)t+\frac{8-9z+3z^{2}}{3-z}=0$. Solving gives $t_{1,2}=\frac{(3-z) \pm \sqrt{(3-z)^{2}-4 \times \frac{8-9z+3z^{2}}{3-z}}}{2}$. $$ \begin{aligned} \text { Also, } & (3-z)^{2}-4 \times \frac{8-9z+3z^{2}}{3-z}=(z-1)^{2} \cdot \frac{z+5}{z-3} \\ & =(z-1)^{2} \cdot\left(1+\frac{8}{z-3}\right) . \end{aligned} $$ When $z-1=0$ i.e., $z=1$, $x=y=1$; When $z-3=1$ i.e., $z=4$, $x=4$ or $-5$, $y=-5$ or $4$; When $z-3=-8$ i.e., $z=-5$, $x=y=4$. Thus, there are four solutions: $(1,1,1),(4,-5,4),(-5,4,4)$, $(4,4,-5)$." 5ebb72fb5a6f,"50. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$. Try to find the integer part of $E=x_{1}+$ $\frac{x_{2}}{\sqrt{1-x_{1}^{2}}}+\frac{x_{3}}{\sqrt{1-\left(x_{1}+x_{2}\right)^{2}}}+\cdots+\frac{x_{n}}{\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{n-1}\right)^{2}}}$.",$ $\frac{2 \sin \frac{\alpha_{i+1}+\alpha_{i}}{2} \sin \frac{\alpha_{i}-\alpha_{i+1}}{2}}{\sin \alph,medium,"50. Since $\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{i}\right)^{2}} \leqslant 1(1 \leqslant i \leqslant n)$, we have $E \geqslant x_{1}+ x_{2}+\cdots+x_{n}=1$ Let $\alpha_{1}=\frac{\pi}{2}, \alpha_{i}=\arccos \left(x_{1}+x_{2}+\cdots+x_{i-1}\right) \quad(2 \leqslant i \leqslant n)$, which means $x_{i}=\cos \alpha_{i+1}-\cos \alpha_{i}$, so $$E=\frac{\cos \alpha_{2}-\cos \alpha_{1}}{\sin \alpha_{1}}+\frac{\cos \alpha_{3}-\cos \alpha_{2}}{\sin \alpha_{2}}+\cdots+\frac{\cos \alpha_{n+1}-\cos \alpha_{n}}{\sin \alpha_{n}}$$ Since $\alpha_{i+1}<\alpha_{i}, \sin x\frac{10000}{2000}=5$ and, since $N<10$, it follows that $N \in\{6,7,8,9\}$. Let $N=9$. Then from (4) we get $\frac{10000}{9}<\overline{defg}<\frac{10020}{9}$, which implies $1111<\overline{defg} \leq 1113$, i.e., $\overline{defg} \in\{1112,1113\}$. Considering $\overline{defg}=1112$, from relation (3) we get $1112=\overline{abc} \cdot 8$, i.e., $\overline{abc}=139$. Therefore, $\overline{abcdefg}=1391112$ and $N_{\max }=9$." 8b92bdd6298f,"Let's determine all functions $f$ defined on the set of positive real numbers that satisfy the equation $f(x)+2 f\left(\frac{1}{x}\right)=3 x+6$ for all $x$ in their domain. --- Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. --- Determine all functions $f$ defined on the set of positive real numbers that satisfy the equation $f(x)+2 f\left(\frac{1}{x}\right)=3 x+6$ for all $x$ in their domain.",See reasoning trace,medium,"If in the given equation, $\frac{1}{x}$ is substituted for $x$ everywhere in the set of positive numbers, we get the following relationship: $$ f\left(\frac{1}{x}\right) + 2 f(x) = 3 \cdot \frac{1}{x} + 6 $$ Together with the original condition, we have a system of two equations with the unknowns $f(x)$ and $f\left(\frac{1}{x}\right)$: $$ \begin{aligned} & \quad f(x) + 2 f\left(\frac{1}{x}\right) = 3 x + 6 \\ & (2) \quad f\left(\frac{1}{x}\right) + 2 f(x) = \frac{3}{x} + 6 \end{aligned} $$ Subtract twice the second equation from the first equation: $$ -3 f(x) = 3 x - \frac{6}{x} - 6 $$ Thus, $$ f(x) = \frac{2}{x} - x + 2 $$ Then, $f\left(\frac{1}{x}\right) = 2 x - \frac{1}{x} + 2$, and substituting this back, it can be seen that the obtained function satisfies the condition. Therefore, there is exactly one function that satisfies the equation given in the problem. Orbán Zsolt (Nagykanizsa, Batthyány Lajos Gymnasium, 8th grade)" f182bbec29c0,2.204. $\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$.,1,medium,"Solution. $$ \begin{aligned} & \sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{(2+\sqrt{2+\sqrt{2+\sqrt{3}}})(2-\sqrt{2+\sqrt{2+\sqrt{3}}})}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2^{2}-(\sqrt{2+\sqrt{2+\sqrt{3}}})^{2}}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{4-2-\sqrt{2+\sqrt{3}}}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2-\sqrt{2+\sqrt{3}}}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{(2+\sqrt{2+\sqrt{3}})(2-\sqrt{2+\sqrt{3}})}=\sqrt{2+\sqrt{3}} \cdot \sqrt{4-2-\sqrt{3}}= \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2-\sqrt{3}}=\sqrt{(2+\sqrt{3})(2-\sqrt{3})}=\sqrt{2^{2}-(\sqrt{3})^{2}}=\sqrt{4-3}= \\ & =\sqrt{1}=1 . \end{aligned} $$ Answer: 1." 398b618b8b7c,"3. Given that the perimeter of a triangle is less than 15, the lengths of its three sides are all prime numbers, and one of the sides has a length of 3. How many such triangles are there? (A) 4 (B) 5 (C) 6 (D) 7",See reasoning trace,easy,$3 . B$ 08b73452e31e,"II. (40 points) Given the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ satisfies $f(f(n))=2016 n-215 f(n)$. Find $f(n)$.","9 a_{0}$, i.e., $f(n)=9 n$.",medium,"Let $f^{(n)}(x)=a_{n}$. Then $a_{n}>0$, and $a_{n+2}=2016 a_{n}-215 a_{n+1}$. Hence $a_{n}-9 a_{n-1}=-224\left(a_{n-1}-9 a_{n-2}\right)$. Let $b_{n}=a_{n}-9 a_{n-1}$. Then $b_{n}=-224 b_{n-1}=(-224)^{n-1} b_{1}$. Let $m=-224$, we get $a_{n}-9 a_{n-1}=m^{n-1} b_{1}$ $\Rightarrow m \cdot \frac{a_{n}}{m^{n}}-9 \cdot \frac{a_{n-1}}{m^{n-1}}=b_{1}$. Let $c_{n}=\frac{a_{n}}{m^{n}}$, we get $m c_{n}-9 c_{n-1}=b_{1}$. Solving for the general term, we get $$ \begin{array}{l} c_{n}=\left(\frac{9}{m}\right)^{n-1} c_{1}+\frac{b_{1}}{m} \cdot \frac{\left(\frac{9}{m}\right)^{n-1}-1}{\frac{9}{m}-1} \\ \Rightarrow a_{n}=\frac{9^{n}-m^{n}}{9-m} a_{1}-\frac{9^{n} m-9 m^{n}}{9-m} a_{0}>0 . \end{array} $$ When $n$ is even, $$ a_{1}\frac{9^{n} m-9 m^{n}}{9^{n}-m^{n}} a_{0}, $$ as $n \rightarrow+\infty$, $a_{1} \geqslant 9 a_{0}$. In summary, $a_{1}=9 a_{0}$, i.e., $f(n)=9 n$." 0db9f48d4048,"11. A tennis tournament was played on a knock-out basis. The following list is of all but one of the last seven matches (the quarter-finals, the semi-finals and the final), although not correctly ordered: Bella beat Ann; Celine beat Donna; Gina beat Holly; Gina beat Celine; Celine beat Bella; and Emma beat Farah. Which result is missing? A Gina beat Bella B Celine beat Ann C Emma beat Celine D Bella beat Holly E Gina beat Emma",See reasoning trace,medium,"$\mathbf{E}$ The most matches that any player could play is three. Any player who wins twice will play in the final. Hence Celine and Gina must be the two finalists, and Gina beat Celine. This means Gina must have won three matches altogether, but only two are recorded. Hence the missing result must be that Gina beat someone. All the other players must have lost once, but Emma has no loss recorded, so the missing result must be that Gina beat Emma. Indeed, from the given information we can deduce that the pairings must have been as shown (using players' initials instead of their full names):" 9f1b4e8da171,"Given that the answer to this [i]Proposed by Ankit Bisain[/i]",203010,medium,"1. We start with the given expression $a \cdot b \cdot c$, where $a$, $b$, and $c$ are pairwise relatively prime positive integers, and $b = 10$. 2. We need to compute $1000a + 100b + 10c$. 3. Given the equation $10ac = 1000a + 10c + 1000$, we can simplify it as follows: \[ 10ac = 1000a + 10c + 1000 \] Dividing the entire equation by 10: \[ ac = 100a + c + 100 \] 4. Rearrange the equation to isolate $c$: \[ ac - c = 100a + 100 \] Factor out $c$ on the left-hand side: \[ c(a - 1) = 100a + 100 \] Solve for $c$: \[ c = \frac{100a + 100}{a - 1} \] 5. Since $a$, $b$, and $c$ are pairwise relatively prime and $b = 10$, $a$ must be relatively prime to 10. This means $a$ cannot be divisible by 2 or 5. 6. We test values of $a$ that are relatively prime to 10 and check if $c$ is an integer: - For $a = 3$: \[ c = \frac{100 \cdot 3 + 100}{3 - 1} = \frac{300 + 100}{2} = \frac{400}{2} = 200 \] Here, $c = 200$, but $a = 3$ and $c = 200$ are not pairwise relatively prime. - For $a = 9$: \[ c = \frac{100 \cdot 9 + 100}{9 - 1} = \frac{900 + 100}{8} = \frac{1000}{8} = 125 \] Here, $c = 125$, but $a = 9$ and $c = 125$ are not pairwise relatively prime. - For $a = 11$: \[ c = \frac{100 \cdot 11 + 100}{11 - 1} = \frac{1100 + 100}{10} = \frac{1200}{10} = 120 \] Here, $c = 120$, but $a = 11$ and $c = 120$ are not pairwise relatively prime. - For $a = 21$: \[ c = \frac{100 \cdot 21 + 100}{21 - 1} = \frac{2100 + 100}{20} = \frac{2200}{20} = 110 \] Here, $c = 110$, but $a = 21$ and $c = 110$ are not pairwise relatively prime. - For $a = 26$: \[ c = \frac{100 \cdot 26 + 100}{26 - 1} = \frac{2600 + 100}{25} = \frac{2700}{25} = 108 \] Here, $c = 108$, but $a = 26$ and $c = 108$ are not pairwise relatively prime. - For $a = 41$: \[ c = \frac{100 \cdot 41 + 100}{41 - 1} = \frac{4100 + 100}{40} = \frac{4200}{40} = 105 \] Here, $c = 105$, but $a = 41$ and $c = 105$ are not pairwise relatively prime. - For $a = 51$: \[ c = \frac{100 \cdot 51 + 100}{51 - 1} = \frac{5100 + 100}{50} = \frac{5200}{50} = 104 \] Here, $c = 104$, but $a = 51$ and $c = 104$ are not pairwise relatively prime. - For $a = 101$: \[ c = \frac{100 \cdot 101 + 100}{101 - 1} = \frac{10100 + 100}{100} = \frac{10200}{100} = 102 \] Here, $c = 102$, but $a = 101$ and $c = 102$ are not pairwise relatively prime. - For $a = 201$: \[ c = \frac{100 \cdot 201 + 100}{201 - 1} = \frac{20100 + 100}{200} = \frac{20200}{200} = 101 \] Here, $c = 101$, and $a = 201$, $b = 10$, and $c = 101$ are pairwise relatively prime. 7. Therefore, the values $a = 201$, $b = 10$, and $c = 101$ satisfy the conditions. 8. Compute $1000a + 100b + 10c$: \[ 1000 \cdot 201 + 100 \cdot 10 + 10 \cdot 101 = 201000 + 1000 + 1010 = 203010 \] The final answer is $\boxed{203010}$." e746356303a2,"10. Given $P$ is any point on the circumcircle of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$. (1) Find the maximum and minimum values of $\sum_{k=1}^{n}\left|P A_{k}\right|$; (2) When $P$ is not restricted to the circle, what is the minimum value of $\sum_{k=1}^{n}\left|P A_{k}\right|$?","0$, $\sum_{k=1}^{n} \mid P A_{k}$ has a minimum value of $n$.",medium,"10. Solution (1) Let $A_{k}=\varepsilon^{k}(k=1,2, \cdots, n), \varepsilon=e^{\frac{2 x}{n}}$. As shown in the figure, $P=e^{i \theta}$, $\left(0 \leqslant \theta \leqslant \frac{2 \pi}{n}\right)$. It is easy to see that $\angle A_{1} P A_{k}=\frac{\pi}{n}(k-1)$, so $\arg \left[\overrightarrow{P A_{k}} \cdot e^{-\frac{x}{n}(k-1) i}\right]=\arg \left(\overrightarrow{P A_{1}}\right), k=1$, $2, \cdots, n$. $$ \text { Hence } \begin{aligned} \sum_{k=1}^{n}\left|P A_{k}\right| & =\sum_{k=1}^{n}\left|e^{-\frac{\pi}{n}(k-1) i} \cdot\left(A_{k}-P\right)\right| \\ & =\left|\sum_{k=1}^{n} P \cdot \exp \left(-\frac{(k-1) \pi}{n} i\right)-\exp \left(\frac{(k+1) \pi}{n} i\right)\right| \\ & =\left|\frac{2 P}{1-e^{-\frac{\pi}{n} i}}-\frac{2 e^{\frac{2 x_{i}}{n}}}{1-e^{\frac{x_{i}}{n}}}\right|=2 \cdot\left|\frac{P+e^{\frac{x_{i}}{n}}}{1-e^{-\frac{\pi}{n} i}}\right|=\frac{2 \cos \left(\frac{\theta}{2}-\frac{\pi}{2 n}\right)}{\sin \frac{\pi}{2 n}} . \end{aligned} $$ Since $\left|\frac{\theta}{2}-\frac{\pi}{2 n}\right| \leqslant \frac{\pi}{2 n}$, when $P$ is the midpoint of $\overparen{A_{1} A_{n}}$, $\sum_{k=1}^{n}\left|P A_{k}\right|$ is maximized, with the maximum value being $2 \csc \frac{\pi}{2 n}$; when $P$ is at $A_{1}$ or $A_{n}$, $\sum_{k=1}^{n}\left|P A_{k}\right|$ is minimized, with the minimum value being $2 \cot \frac{\pi}{2 n}$. (2) If $P$ is not restricted to the circle, then $$ \sum_{k=1}^{n}\left|P A_{k}\right|=\sum_{k=1}^{n}\left|A_{k}-P\right|=\sum_{k=1}^{n}\left|A_{k}^{-1}\left(A_{k}-P\right)\right| \geqslant\left|\sum_{k=1}^{n} A_{k}^{-1}\left(A_{k}-P\right)\right|=\left|n-P \cdot \sum_{k=1}^{n} A_{k}^{-1}\right|=n . $$ Thus, when $P=0$, $\sum_{k=1}^{n} \mid P A_{k}$ has a minimum value of $n$." 9e23cbe3e054,"Exercise 10. Find all triplets ( $\mathbf{p , q , r} \mathbf{)}$ ) of prime numbers such that the 3 differences $$ |p-q|,|q-r|,|r-p| $$ are also prime numbers.","3$. Similarly, some said that $|p-q|=|q-r|=|r-p|=2$ is impossible without justifying it, which was q",medium,"Solution to Exercise 10: Note that the three numbers must be pairwise distinct since 0 is not a prime number. We can therefore assume, without loss of generality, that $\mathrm{p}>\mathrm{q}>\mathrm{r}$. A prime number is either odd or equal to 2. Suppose that $p, q$, and $r$ are all odd. Then $p-q, q-r$, and $r-p$ are even. Since their absolute values are prime, these numbers must all be 2. Thus, the integers $p, p+2$, and $p+4$ are all prime. If $p$ is divisible by 3, then $p=3$ and $\mathrm{q}=5$ and $\mathrm{r}=7$. However, the triplet $(3,5,7)$ is not a solution to the problem: $7-3=4$ is not prime. If $p$ is not divisible by 3, then $p$ is of the form $3k+1$ or $3k+2$. The first case implies that $p+2$ is divisible by 3, so $p+2=3$ but $p=1$ is not a prime number. The second case implies that $p+4$ is divisible by 3, but $p+4=3$ does not yield a positive solution. Suppose that $r=2$. Then $p$ and $q$ are odd and $p-q$ is even and prime, so it must be 2. It follows that $\mathrm{q}+2$, $\mathrm{q}$, and $\mathrm{q}-2$ are all prime. From the previous case, this implies that $q-2=3$ so $p=7$. Conversely, the triplet $(p, q, r)=(2,5,7)$ and its permutations are indeed solutions to the problem. The only triplets that are solutions are therefore $(2,5,7)$ and its permutations. Graders' Comments Many students had the main ideas but lost points for poor writing. It is always important to verify that the solutions obtained satisfy the statement of the problem. Some students claimed without any justification that if $p, p+2$, and $p+4$ are prime, then $p=3$. Similarly, some said that $|p-q|=|q-r|=|r-p|=2$ is impossible without justifying it, which was quite clear (we can assume $p>q>r$ for example). It is better to detail the important points a bit more to avoid losing points." 36bfe178e874,"Example 44 Let $n$ be an integer greater than 2, and $a_{n}$ be the largest $n$-digit number that is neither the sum of two perfect squares nor the difference of two perfect squares. (1) Find $a_{n}$ (expressed as a function of $n$). (2) Find the smallest value of $n$ such that the sum of the squares of the digits of $a_{n}$ is a perfect square.",66$.,medium,"(1) $a_{n}=10^{n}-2$. First, prove the maximality. Among $n$-digit decimal integers, only $10^{n}-1>10^{n}-2$. But $10^{n}-1=9 \times \frac{10^{n}-1}{9}$ $$ \begin{array}{l} =\left(\frac{9+\frac{10^{n}-1}{9}}{2}+\frac{\frac{10^{n}-1}{9}-9}{2}\right)\left(\frac{9+\frac{10^{n}-1}{9}}{2}-\frac{\frac{10^{n}-1}{9}-9}{2}\right) \\ =\left(\frac{9+\frac{10^{n}-1}{9}}{2}\right)^{2}-\left(\frac{\frac{10^{n}-1}{9}-9}{2}\right)^{2} . \end{array} $$ Since $\frac{10^{n}-1}{9}$ is odd, $10^{n}-1$ can be expressed as the difference of two perfect squares. This contradicts the problem's condition. Next, prove that $10^{n}-2$ satisfies the condition. If $10^{n}-2$ can be expressed as the difference of two perfect squares, then it must be congruent to 0, 1, or 3 modulo 4. But $10^{n}-2 \equiv 2 \pmod{4}$, so $10^{n}-2$ cannot be expressed as the difference of two perfect squares. If $10^{n}-2$ can be expressed as the sum of two perfect squares, then it must be divisible by 4 or congruent to 2 modulo 8. But $10^{n}-2$ is not divisible by 4 and is congruent to 6 modulo 8 (since $n>2$). Therefore, $10^{n}-2$ cannot be expressed as the sum of two perfect squares. (2) From $9^{2}(n-1)+64=k^{2}$, we get $9^{2}(n-1)=(k-8)(k+8)$. Since $n \geqslant 3$ and $-8 \neq 8 \pmod{9}$, we have $81 \mid (k-8)$ or $81 \mid (k+8)$. If $81 \mid (k-8)$, then $k_{\text{min}}=89, n=98$. If $81 \mid (k+8)$, then $k_{\text{min}}=73, n=66$. Therefore, $n_{\min}=66$." 92ef123ae889,"The product of three positive integers is 42 . The sum of two of these integers is 9 . The third integer is (A) 1 (B) 7 (C) 6 (D) 3 (E) 2",(D),medium,"Solution 1 If the sum of two positive integers is 9 , the possible pairs are 1 and 8,2 and 7,3 and 6 , and 4 and 5 . Of these pairs, the only one in which each number is a divisor of 42 is 2 and 7 . Since the three positive integers have a product of 42 , then two of them must be 2 and 7 , so the third is $42 \div(2 \times 7)=42 \div 14=3$. ## Solution 2 The possible sets of three positive integers which multiply to give 42 are $\{1,1,42\},\{1,2,21\}$, $\{1,3,14\},\{1,6,7\}$, and $\{2,3,7\}$. The only one of these sets that contains two integers which add to 9 is $\{2,3,7\}$. Therefore, the third number must be 3 . ANSWER: (D)" fe778db8a702,"Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y \in \mathbb{R}$, we have: $$ f(x+y)=f(x) f(y) $$","\exp (c x)$. We verify this solution, which also works.",medium,"Let $x=y$, then we have: $f(2 x)=f(x)^{2} \geq 0$. As $x$ runs through $\mathbb{R}$, $2 x$ also runs through $\mathbb{R}$. Therefore, $\forall x \in \mathbb{R}, f(x) \geq 0$. If there exists $x_{0}$ such that $f\left(x_{0}\right)=0$, let $x=x_{0}$. Then, $f\left(x_{0}+y\right)=f\left(x_{0}\right) f(y)=0$ for all $y \in \mathbb{R}$. As $y$ runs through $\mathbb{R}$, $x_{0}+y$ also does. Therefore, $\forall y \in \mathbb{R}, f(y)=0$. We verify that the zero function is indeed a solution. If for all $x \in \mathbb{R}, f(x) \neq 0$, then $\forall x \in \mathbb{R}, f(x)>0$. We can therefore compose both sides of the equality with the natural logarithm function. We then have: $$ \ln (f(x+y))=\ln (f(x) f(y))=\ln (f(x))+\ln (f(y)) $$ Let $g=\ln \circ f$, then $g$ is a solution to Cauchy's equation. Since $g$ is the composition of two continuous functions, it is also continuous. Therefore, according to Cauchy's equation, $g(x)=c x$. To find $f$, we compose both sides with the exponential function, yielding: $f(x)=\exp (c x)$. We verify this solution, which also works." b5e836567fe8,"2. The side length of an equilateral triangle is $\mathrm{a}, \mathrm{PA} \perp$ plane $A B C$. If $P A=h$, find the distance from $P$ to $B C$.",See reasoning trace,easy,$\frac{1}{2} \sqrt{3 a^{2}+4 h^{2}}$ 02c14f43cf65,"## Task 6 Calculate the numbers. - A is four times the smallest three-digit number, - B is half of the smallest four-digit number, - C is the sum of D and E, - D is $\mathrm{B}$ minus $\mathrm{A}$, - E is the sum of B and D.",See reasoning trace,easy,"smallest three-digit number 100, i.e., $\mathrm{A}=400$, analogously $\mathrm{B}=500$. Therefore, $\mathrm{D}=100, \mathrm{E}=600, \mathrm{C}=700$ ### 12.18 Further Problems, Grade 3 The following problems cannot yet be assigned to a specific Olympiad year." 627fec992636,"When a positive integer $N$ is divided by 60 , the remainder is 49 . When $N$ is divided by 15 , the remainder is (A) 0 (B) 3 (C) 4 (D) 5 (E) 8",(C),medium,"When a positive integer $N$ is divided by 60 , the remainder is 49 . When $N$ is divided by 15 , the remainder is (A) 0 (B) 3 (C) 4 (D) 5 (E) 8 ## Solution This problem can be done in a number of ways. The easiest way is to consider that if $N$ is divided by 60 to achieve a remainder of 49 , it must be a number of the form, $60 k+49, k=0,1,2, \ldots$. This implies that the smallest number to meet the requirements is 49 itself. If we divide 49 by 15 we get a remainder of 4 . Or, if $k=1$ in our formula then the next number to satisfy the requirements is 109 which when divided by 15 gives 4 as the remainder. ANSWER: (C)" d79f70359d39,"6. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, and common ratio $q=-\frac{1}{2}$, let $\pi_{n}$ denote the product of its first $n$ terms. The maximum value of $\pi_{n}$ is $\qquad$ .","1536 \times\left(-\frac{1}{2}\right)^{n-1}, \pi_{n}=(1536)^{n} \times\left(-\frac{1}{2}\right)^{\fra",medium,"6. $\pi_{12}$. From the problem, we have $a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1}, \pi_{n}=(1536)^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}},\left|\pi_{n}\right|=(1536)^n \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}}$, and $\frac{\left|\pi_{n+1}\right|}{\left|\pi_{n}\right|}=\left|a_{n+1}\right|=1536 \cdot\left(\frac{1}{2}\right)^{n}$. When $n \leqslant 10,\left|\pi_{n+1}\right|>\left|\pi_{n}\right| ; n \geqslant 11$, $\left|\pi_{n+1}\right|<\left|\pi_{n}\right| ; \left|\pi_{12}\right|>\left|\pi_{13}\right| \cdots$. Also, $\pi_{11}>0, \pi_{10}>0, \pi_{13}>0, \pi_{12}>\pi_{13}$, so it can only be that $\pi_{12}$ and $\pi_{9}$ are the largest. And $\frac{\pi_{12}}{\pi_{9}}=\frac{27}{8}>1, \pi_{12}>\pi_{9}$, hence $\pi_{12}$ is the largest." 213b077e3cc8,"2. Given $\sqrt{24-a}-\sqrt{8-a}=2$. Then $\sqrt{24-a}+\sqrt{8-a}=(\quad)$. (A) 7 (B) 8 (C) 9 (D) 10",\frac{(\sqrt{24-a})^{2}-(\sqrt{8-a})^{2}}{\sqrt{24-a}-\sqrt{8-a}}=8 .\end{array}$,easy,"$\begin{array}{l}\text { 2. B. } \\ \sqrt{24-a}+\sqrt{8-a} \\ =\frac{(\sqrt{24-a})^{2}-(\sqrt{8-a})^{2}}{\sqrt{24-a}-\sqrt{8-a}}=8 .\end{array}$ The translation is as follows: $\begin{array}{l}\text { 2. B. } \\ \sqrt{24-a}+\sqrt{8-a} \\ =\frac{(\sqrt{24-a})^{2}-(\sqrt{8-a})^{2}}{\sqrt{24-a}-\sqrt{8-a}}=8 .\end{array}$" 3697d93e1874,"2. Find all pairs of real numbers $a$ and $b$ such that for all real numbers $x$ and $y$ the equality $$ \lfloor a x+b y\rfloor+\lfloor b x+a y\rfloor=(a+b)\lfloor x+y\rfloor \text {. } $$ holds. (For a real number $t$, $\lfloor t\rfloor$ denotes the greatest integer not greater than $t$.)",See reasoning trace,medium,"2. For $y=-x$, the given condition reduces to $\lfloor(a-$ $b) x\rfloor+\lfloor(b-a) x\rfloor=0$. If the numbers $a$ and $b$ were different, this would not be satisfied, for example, for $x=\frac{1}{2(a-b)}$ (then it would hold $\lfloor(a-b) x\rfloor+\lfloor(b-a) x\rfloor=\left\lfloor\frac{1}{2}\right\rfloor+\left\lfloor-\frac{1}{2}\right\rfloor=$ ![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-23.jpg?height=416&width=365&top_left_y=407&top_left_x=1151) Op 20191 A 1 $0+(-1)=-1 \neq 0)$. Therefore, it must hold that $a=b$, and the condition of the problem becomes $\lfloor a(x+y)\rfloor=a\lfloor x+y\rfloor$. Choosing $0 \leqslant x+y<1$. Since for $0 \leqslant x+y<1$ it must hold that $\lfloor a(x+y)\rfloor=a\lfloor x+y\rfloor=0$, i.e., $0 \leqslant a(x+y)<1$, it follows that $x+y<\frac{1}{a}$; conversely, if we choose $0 \leqslant x+y<\frac{1}{a}$, the given condition reduces to $a\lfloor x+y\rfloor=\lfloor a(x+y)\rfloor=0$, from which it follows that $x+y<1$. In other words, the conditions $0 \leqslant x+y<1$ and $0 \leqslant x+y<\frac{1}{a}$ are equivalent, from which it follows that $a=1$. Therefore, the only solutions are the pairs $(a, b)=(0,0)$ and $(a, b)=(1,1)$. ![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-23.jpg?height=346&width=351&top_left_y=1286&top_left_x=192) Op 2019 1A 3" 01667c479f5e,"A football team's coach believes that his players kick the penalty with 95% certainty. What is the probability that out of five players, exactly three will miss?",See reasoning trace,medium,"Solution. Since the performance of the five players is independent of each other, we can apply the formula for the probability of the simultaneous occurrence of independent events $P\left(A_{1} ; A_{2} ; \ldots ; A_{n}\right)=P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot \ldots \cdot P\left(A_{n}\right)$. Let the five players be denoted by $A, B, C, D, E$. According to the above formula, the probability that exactly $C, D$, and $E$ miss their penalty kicks is $0.95 \cdot 0.95 \cdot 0.05 \cdot 0.05 \cdot 0.05$ (since if the probability that someone scores is 0.95, then the probability of the negation of this event, or that they miss the penalty, is $0.05$. Since we can choose 3 people from 5 in more than one way, we need to multiply the obtained probability by $\binom{5}{3}$: $\binom{5}{3} \cdot 0.95^{2} \cdot 0.05^{3} \approx 0.00113$ Therefore, the probability that three out of five will miss their penalty kicks is approximately $0.00113$." 0d0d73399a0d,8. Solve the system $\left\{\begin{array}{l}3^{y} \cdot 81=9^{x^{2}} ; \\ \lg y=\lg x-\lg 0.5 .\end{array}\right.$,$x=2$,easy,"Solution: $\left\{\begin{array}{l}3^{y} \cdot 81=9^{x^{2}}, \\ \lg y=\lg 2 x,\end{array} 3^{2 x} \cdot 9^{2}=9^{x^{2}}, x+2=x^{2}, x^{2}-x-2=0\right.$, $\left\{\begin{array}{l}x=-1-\text { n.s., } \\ x=2,\end{array} \Rightarrow x=2 \Rightarrow y=4\right.$. Verification: $\left\{\begin{array}{l}3^{4} \cdot 81=9^{4}, \\ \lg 4=\lg (2 \cdot 2) .\end{array}\right.$ Answer: $x=2$." bb3add4d2a39,"10. In the complex plane, take any three different roots of the equation $z^{100}-1=0$ as vertices to form a triangle, then the number of different acute triangles is $\qquad$ .",39200$.,medium,"As shown in the figure, select $z_{1}$ and draw the diameter $z_{1} z_{51}$. Then select $z_{j}$ and draw the diameter. At this point, since the center is inside the acute triangle, the third point can only be chosen on the minor arc $\overparen{z_{1} z_{j}}$ opposite to it, and there are $j-2$ ways to do so. Thus, for $z_{1}$, the number of ways to form an acute triangle is $0+1+2+\cdots+48=49 \times 24$. The same applies to the other points. Note that each acute triangle is counted 3 times, so the number of different acute triangles is $\frac{49 \times 24 \times 100}{3}=39200$." 2eac526e3ed4,"There are 15 girls in a class of 27 students. The ratio of boys to girls in this class is (A) $4: 5$ (B) $5: 3$ (C) $3: 4$ (D) $4: 9$ (E) $9: 5$",(A),easy,"If 15 of 27 students in the class are girls, then the remaining $27-15=12$ students are boys. The ratio of boys to girls in the class is $12: 15=4: 5$. ANSWER: (A)" 9a22c9d99bd9,"Three. (25 points) Find all positive integer triples $(x, y, z)$ such that $1+2^{x} \times 3^{y}=5^{z}$ holds. (Zhang Lei)","(3,1,2)$.",medium,"Three, taking the original equation modulo 3 yields that $z$ is even. Let $z=2 r\left(r \in \mathbf{N}_{+}\right)$. Then $\left(5^{r}-1\right)\left(5^{r}+1\right)=2^{x} \times 3^{y}$. Since $\left(5^{\prime}-1,5^{\prime}+1\right)=\left(5^{\prime}-1,2\right)=2$, we have $\left(\frac{5^{r}-1}{2}, \frac{5^{\prime}+1}{2}\right)=1$. Thus, $\frac{5^{\prime}-1}{2} \cdot \frac{5^{\prime}+1}{2}=2^{x-2} \times 3^{y}(x \geqslant 2)$. Also, $\frac{5^{r}-1}{2}$ is even, and $\frac{5^{r}+1}{2} \geqslant 3$, so $\frac{5^{r}-1}{2}=2^{x-2}, \frac{5^{r}+1}{2}=3^{y}$. Hence, $3^{y}-2^{x-2}=1$. When $x=2$, $y$ is not an integer. When $x=3$, $y=1, z=2$. When $x \geqslant 4$, taking equation (1) modulo 4 shows that $y$ is even. Let $y=2 t\left(t \in \mathbf{N}_{+}\right)$. Then $\left(3^{t}-1\right)\left(3^{t}+1\right)=2^{x-2}$. Since $\left(3^{t}-1,3^{t}+1\right)=\left(3^{t}-1,2\right)=2$, we have $\left(\frac{3^{t}-1}{2}, \frac{3^{t}+1}{2}\right)=1$. Therefore, $\frac{3^{t}-1}{2} \cdot \frac{3^{t}+1}{2}=2^{x-4}(x \geqslant 4)$. Since $\frac{3^{t}-1}{2}<\frac{3^{t}+1}{2}$, we have $\frac{3^{t}-1}{2}=1$. Solving gives $t=1, y=2$. From $\frac{5^{\prime}+1}{2}=3^{y}$, we know that $r$ is not an integer, so the original equation has no positive integer solutions in this case. In summary, the positive integer solution to the original equation is $(x, y, z)=(3,1,2)$." a80feccc1fb4,"21. Given a square with side length $x$ inscribed in a right-angled triangle with side lengths $3, 4, 5$, and one vertex of the square coincides with the right-angle vertex of the triangle; a square with side length $y$ is also inscribed in the right-angled triangle with side lengths $3, 4, 5$, and one side of the square lies on the hypotenuse of the right-angled triangle. Then the value of $\frac{x}{y}$ is $(\quad)$. (A) $\frac{12}{13}$ (B) $\frac{35}{37}$ (C) 1 (D) $\frac{37}{35}$ (E) $\frac{13}{12}$",See reasoning trace,medium,"21. D. As shown in Figure 4, let $A D=x$, and the square $A F E D$ is inscribed in the right triangle $\triangle A B C$. Then $$ \begin{array}{l} \triangle A B C \backsim \triangle D E C \Rightarrow \frac{D E}{A B}=\frac{C D}{A C} \\ \Rightarrow \frac{x}{4}=\frac{3-x}{3} \Rightarrow x=\frac{12}{7} . \end{array} $$ As shown in Figure 5, let $Q R=y$, and the square $S T Q R$ is inscribed in the right triangle $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then $$ \begin{array}{l} \triangle A^{\prime} B^{\prime} C^{\prime} \subset \triangle R B^{\prime} Q \backsim \triangle S T C^{\prime} \\ \Rightarrow \frac{4}{3} y+y+\frac{3}{4} y=5 \Rightarrow y=\frac{60}{37} . \\ \text { Hence } \frac{x}{y}=\frac{37}{35} . \end{array} $$" ebcd23114217,"Example 1. If the equation $m x^{2}-2(m+2) x+m$ $+5=0$ has no real roots with respect to $x$, then the number of real roots of the equation $(m-5) x^{2}$ $-2(m+2) x+m=0$ with respect to $x$ is ( ). (A) 2 (B) 1 (C) 0 (D) Uncertain (1989, National Junior High School Mathematics Competition)",(D),easy,"Solve: For $m x^{2}-2(m+2) x+m+5=0$ to have no real roots, then $$ \Delta_{1}=4(m+2)^{2}-4 m(m+5)4$. Also, the discriminant of $(m-5) x^{2}-2(m+2) x+m=0$ is $$ \Delta_{2}=4(m+2)^{2}-4 m(m-5)=36 m+16 \text {. } $$ When $m>4$ and $m \neq 5$, $\Delta_{2}>0$, the equation has two real roots; when $m=5$, the equation is a linear equation, with only one real root. Therefore, the number of real roots of the equation is uncertain. Hence, the answer is (D)." 0e32127fead6,"2. Given that $x$ is a positive integer. Then, the number of triangles that can be formed with sides $3$, $x$, and $10$ is ( ). (A) 2 (B) 3 (C) 5 (D) 7",See reasoning trace,easy,"2. C Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 1951f2328576,"130 The complex number $z$ satisfies $|z+5-12i|=3$, then the maximum value of $|z|$ is $(\quad)$ A. 3 B. 10 C. 20 D. 16",3$ is a circle with center at $-5+12 i$ and radius 3. The maximum distance from the origin to a poin,easy,"130 D. $|z+5-12 i|=3$ is a circle with center at $-5+12 i$ and radius 3. The maximum distance from the origin to a point on the circle, which is the maximum value of $|z|$, is $\sqrt{5^{2}+12^{2}}+3=16$." 2a0bc5bb836d,"## [ Auxiliary area. The area helps to solve the Pythagorean Theorem (direct and inverse)In triangle $A B C$, heights $A E$ and $C D$ are drawn. Find the side $A B$, if $B D=18, B C=30, A E=20$.",See reasoning trace,medium,"$A B \cdot C D=B C \cdot A E$ ## Solution By the Pythagorean theorem from the right triangle $B D C$, we find that $$ C D^{2}=B C^{2}-B D^{2}=30^{2}-18^{2}=(30-18)(30+18)=12 \cdot 48=12^{2} \cdot 2^{2} $$ thus, $C D=24$. Let $S$ be the area of the triangle. Then $$ S=\frac{1}{2} A B \cdot C D \text { and } S=\frac{1}{2} B C \cdot A E $$ therefore, $A B \cdot C D=B C \cdot A E$, from which $$ A B=\frac{B C \cdot A E}{C D}=\frac{30 \cdot 20}{24}=25 $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-46.jpg?height=111&width=1875&top_left_y=242&top_left_x=0) In the isosceles triangle $A B C$, the lateral sides $B C$ and $A C$ are twice the length of the base $A B$. The angle bisectors at the base intersect at point $M$. What fraction of the area of triangle $A B C$ is the area of triangle $A M B$? ## Hint The angle bisector divides the base of the triangle into segments proportional to the lateral sides. ## Solution Let $A A_{1}$ and $B B_{1}$ be the angle bisectors at the base $A B$ of triangle $A B C$. Since $$ \frac{B A_{1}}{A_{1} C}=\frac{A B}{A C}=\frac{1}{2}, B A_{1}=\frac{1}{3} B C $$ $$ S_{\triangle \mathrm{ABA}_{1}}=\frac{1}{3} S_{\Delta \mathrm{ABC}} $$ Since $B M$ is the angle bisector of triangle $B A A_{1}$, then $$ \frac{A_{1} M}{A M}=\frac{B A_{1}}{A B}=\frac{2}{3}, A M=\frac{3}{5} A A_{1} $$ Therefore, $$ S_{\Delta \mathrm{AMB}}=\frac{3}{5} S_{\Delta \mathrm{ABA}_{1}}=\frac{1}{5} S_{\Delta \mathrm{ABC}} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-46.jpg?height=412&width=558&top_left_y=1592&top_left_x=750) ## Answer" 8fc98d4fbdbf,"1. The capacities of cubic vessels are in the ratio $1: 8: 27$, and the volumes of the liquid poured into them are in the ratio $1: 2: 3$. After this, a certain amount of liquid was transferred from the first to the second vessel, and then from the second to the third, so that the level of liquid in all three vessels became the same. After this, $128 \frac{4}{7}$ liters were transferred from the first vessel to the second, and then from the second back to the first so that the height of the liquid column in the first vessel became twice that in the second. It turned out that there was 100 liters less in the first vessel than initially. How much liquid was there initially in each vessel?",". 350,700 and 1,050 liters",easy,"67.1. Answer. 350,700 and 1,050 liters." 503c2de3a218,"1. Petya and Vasya competed in a 60 m race. When Petya finished, Vasya was 9 m behind him. During the second race, Petya started exactly 9 m behind Vasya. Who finished first in the second race and by how many meters did he outpace his opponent? (Assume that each boy ran at a constant speed both times).",Petya outpaced Vasya by 1,easy,"Answer: Petya outpaced Vasya by 1.35 m. Solution. In the second race, Petya will eliminate the lag from Vasya by running 60 m (Vasya will run 51 m during this time). 9 m remain to the finish line - which is $20 / 3$ times less than at the start of the 1st race. Therefore, the lead will also be $20 / 3$ times less, i.e., $9 \cdot \frac{3}{20}=1.35$ m. Evaluation. Full solution is worth 11 points." 239e8c9a53c1,Initial 26. Find all natural numbers of the form $\overline{30 x 070 y 03}$ that are divisible by 37.,See reasoning trace,medium,"Solve $\overline{30 x 070 y 03}=300070003+10^{6} x+10^{2} y$ $$ \begin{aligned} = & 37(8110000+27027 x+2 y) \\ & +(3+x-11 y) . \end{aligned} $$ Therefore, from the given conditions, $3+x-11 y$ is a multiple of 37. And $0 \leqslant x, y \leqslant 9, -96 \leqslant 3+x-11 y \leqslant 12$. Thus, the value of $3+x-11 y$ can only be $0, -37$ or -74. (1) If $3+x-11 y=0$, then by $0 \leqslant x, y \leqslant 9$ we get $x=8, y=1$. (2) If $3+x-11 y=-37$, the integer solution for this equation within 0 to 9 is $x=4, y=4$. (3) If $3+x-11 y=-74$, then the corresponding values of $x, y$ are $x=0, y=7$. In summary, the required natural numbers are 3080701 C, 304070403 or 300070703. (Liu Jinyi, Hebei Yongqing Normal School, 3026:0)" 8a03fabdc92d,"Given a positive integer $n$. Find the largest real number $c$, such that: if the sum of the reciprocals of a set of integers greater than 1 (which can be the same) is less than $c$, then it is always possible to divide this set of numbers into no more than $n$ groups, so that the sum of the reciprocals of the numbers in each group is less than 1.",See reasoning trace,medium,"II. $c=\frac{n+1}{2}$. Take this set of numbers as $a_{1}, a_{2}, \cdots, a_{4}$, if $a_{1}=a_{2}=\cdots=a_{s}=2$, then $c \leqslant \frac{n+1}{2}$. We will prove the following by mathematical induction on $n$: If $\sum_{i=1}^{s} \frac{1}{a_{i}}<\frac{n+1}{2}$, then $a_{1}, a_{2}, \cdots, a_{s}$ can be divided into no more than $n$ groups such that the sum of the reciprocals in each group is less than 1. (1) When $n=1$, the conclusion holds. (2) Assume the conclusion holds for $n-1$. Now consider the case for $n$. Note that $a_{1} \geqslant 2$, i.e., $\frac{1}{a_{1}} \leqslant \frac{1}{2}$. Let $t$ be the largest positive integer such that $\sum_{i=1}^{t-1} \frac{1}{a_{i}}<\frac{1}{2}$. If $t=s+1$, then the conclusion holds. If $t \leqslant s$, then $\sum_{i=1}^{t-1} \frac{1}{a_{i}}<\frac{1}{2} \leqslant \sum_{i=1}^{t} \frac{1}{a_{i}}$. Also, $\sum_{i=1}^{t} \frac{1}{a_{i}}<\frac{1}{2}+\frac{1}{a_{i}} \leqslant 1$, so when $t=s$, the conclusion holds. Now assume $tx+1$, i.e., $z=x+1+k$ for some natural number $k$. Then $$ 2^{x}+2^{y}+2^{z}=2^{x+1}\left(1+2^{k}\right) $$ is not equal to 2336 for the same reasons as in 2). Similarly, it can be shown that $x \neq z$ and $y \neq z$. Let $xc), \angle B A C=60^{\circ}$, its orthocenter is $H$, circumcenter is $O, O H$ intersects $A B$ and $A C$ at points $D$ and $E$ respectively. Then the length of $D E$ is $($. (A) $\frac{b+c}{2}$ (B) $\frac{b+c}{3}$ (C) $2(b-c)$ (D) $3(b-c)$",\frac{1}{3}(A C+A B)=\frac{b+c}{3}$.,medium,"4. B. As shown in Figure 4, it is easy to see that $$ \begin{array}{l} \angle B O C \\ =2 \angle B A C \\ =120^{\circ}, \\ \angle B H C=\angle G H F \\ =180^{\circ}-\angle A \\ =120^{\circ} . \end{array} $$ Therefore, points $B, H, O, C$ are concyclic. Hence, $\angle D H C=\angle O B C=30^{\circ}$. Also, $\angle D C H=90^{\circ}-\angle B A C=30^{\circ}$. Thus, $D H=D C$. Similarly, $E H=E B$. Therefore, $D C+E B=D E$. Given that $\angle A D E=\angle D C H+\angle D H C=60^{\circ}$, we know $A D =A E=D E$. Thus, $D E=\frac{1}{3}(A C+A B)=\frac{b+c}{3}$." 32e8aba9e901,"The value of $0.9+0.09$ is (A) 1.08 (B) 0.909 (C) 1.8 (D) 0.99 (E) 0.18",(D),easy,"Adding, we get $0.9+0.09=0.99$. ANSWER: (D)" 17d4f482f2ec,"4. If $|\mathrm{a}| \geqslant 1$, then the range of $\operatorname{arctg} \frac{1}{a}$ is",See reasoning trace,easy,"4. $\left[-\frac{\pi}{4}, 0\right) \cup\left(0, \frac{\pi}{4}\right]$," 9e43aa6beae5,"Given that real numbers $a$, $b$, and $c$ satisfy $ab=3$, $ac=4$, and $b+c=5$, the value of $bc$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$. [i]2021 CCA Math Bonanza Team Round #2[/i]",349,medium,"1. We start with the given equations: \[ ab = 3, \quad ac = 4, \quad b + c = 5 \] 2. We add and factor the equations involving \(a\): \[ a(b + c) = ab + ac = 3 + 4 = 7 \] Since \(b + c = 5\), we substitute this into the equation: \[ a \cdot 5 = 7 \implies a = \frac{7}{5} \] 3. Next, we use the value of \(a\) to find \(b\) and \(c\). From \(ab = 3\): \[ \left(\frac{7}{5}\right)b = 3 \implies b = \frac{3 \cdot 5}{7} = \frac{15}{7} \] 4. Using \(b + c = 5\) to find \(c\): \[ \frac{15}{7} + c = 5 \implies c = 5 - \frac{15}{7} = \frac{35}{7} - \frac{15}{7} = \frac{20}{7} \] 5. Now, we calculate \(bc\): \[ bc = \left(\frac{15}{7}\right) \left(\frac{20}{7}\right) = \frac{15 \cdot 20}{7 \cdot 7} = \frac{300}{49} \] 6. The fraction \(\frac{300}{49}\) is already in its simplest form since 300 and 49 have no common factors other than 1. Therefore, \(m = 300\) and \(n = 49\). 7. Finally, we compute \(m + n\): \[ m + n = 300 + 49 = 349 \] The final answer is \(\boxed{349}\)." c2576f08a589,"G2.3 Let $a \oplus b=a b+10$. If $C=(1 \oplus 2) \oplus 3$, find the value of $C$.",2+10=12 ; C=12 \oplus 3=36+10=46$,easy,$1 \oplus 2=2+10=12 ; C=12 \oplus 3=36+10=46$ 0460c642a3e1,"VIII-3. Given a right triangle ABC ( $\angle \mathrm{C}=90^{\circ}$ ), whose median to the hypotenuse is equal to $20 \mathrm{~cm}$. From the midpoint $D$ of the hypotenuse, a perpendicular to the hypotenuse is drawn, intersecting one of the legs at point $E$, and $\overline{D E}=15 \mathrm{~cm}$. Calculate the legs of the triangle.",283&width=339&top_left_y=1678&top_left_x=1109),medium,"Solution: If $\overline{C D}=20 \mathrm{~cm}$, then $\overline{A B}=40 \mathrm{~cm}$. From $\triangle A D E: \overline{\mathrm{AE}}^{2}=\overline{\mathrm{DE}}^{2}+\overline{\mathrm{AD}}^{2}$, $\overline{A E}=\sqrt{15^{2}+20^{2}}=25 \mathrm{~cm}, \triangle A B C \sim \triangle A D E, \alpha$ is a common angle, so $\overline{\mathrm{AB}}: \overline{\mathrm{AE}}=\overline{\mathrm{AC}}: \overline{\mathrm{AD}} ; 40: 25=\overline{\mathrm{AC}}: 20$; $\overline{A C}=32 \mathrm{~cm}$, and $\overline{B C}=24 \mathrm{~cm}$. ![](https://cdn.mathpix.com/cropped/2024_06_05_4e46949c190613f78f8fg-3.jpg?height=283&width=339&top_left_y=1678&top_left_x=1109)" ad940d6fd2fb,"8.1. On the coordinate plane, a triangle $O A B$ is drawn, where the point of intersection of the medians is at $\left(\frac{19}{3}, \frac{11}{3}\right)$, and points $A$ and $B$ have natural coordinates. Find the number of such triangles. (Here, $O$ denotes the origin - the point $(0,0)$; two triangles with the same set of vertices are considered the same, i.e., $O A B$ and $O B A$ are considered the same triangle.)",90,medium,"Answer: 90. Let $M$ be the midpoint of $AB$. Then, by the property of the median, $OM = \frac{3}{2} OG$, where $G$ is the centroid. Therefore, $M$ has coordinates $\left(\frac{19}{2}, \frac{11}{2}\right)$. ![](https://cdn.mathpix.com/cropped/2024_05_06_dd1d5c49961b7f585dbeg-10.jpg?height=660&width=942&top_left_y=561&top_left_x=503) Let $\left(x_{a}, y_{a}\right)$ and $\left(x_{b}, y_{b}\right)$ be the coordinates of points $A$ and $B$. Since $M$ is the midpoint of segment $AB$, we have $x_{a} + x_{b} = 19$ and $y_{a} + y_{b} = 11$. Then, for $x_{a}$, there are 18 possible natural values: $1, 2, \ldots, 18$, each of which uniquely corresponds to a natural $x_{b} = 19 - x_{a}$. Similarly, for $y_{a}$, there are 10 possibilities. In total, there are $18 \cdot 10 = 180$ possibilities for choosing point $A$, and for each of them, a suitable point $B$ is uniquely determined. According to our agreement (triangles $OAB$ and $OBA$ are considered the same), we have $180 / 2 = 90$ variants. Note that in all the specified variants, point $A$ (and $B$) does not lie on the line $OM$, i.e., triangle $OAB$ does not degenerate (otherwise, $y_{a} / x_{a} = 11 / 19$ would hold). Thus, all the found variants are indeed valid." eaa2a8a1db4a,"3. Misha calculated the products $1 \times 2, 2 \times 3$, $3 \times 4, \ldots, 2017 \times 2018$. For how many of them is the last digit zero?",is provided: 1 point,medium,"Solution. The last digit of the product depends on the last digits of the factors. In the sequence of natural numbers, the last digits repeat every ten. In each ten, in the sequence of products, four products end in zero: ... $4 \times \ldots 5, \ldots 5 \times \ldots 6, \ldots 9 \times \ldots 0, \ldots 0 \times \ldots 1$. ... There are 201 complete tens. And at the end, there are seven more products, two of which end in zero. Thus, a total of $4 \times 201 + 2 = 806$ products end in zero. Criteria. Any correct solution: 7 points. In the solution, two zeros from the last incomplete ten are lost: 4 points. Only the correct answer is provided: 1 point." 510746d2bcac,"9. (6 points) A hollow cone and a hollow cylinder, with base inner radii and heights of $12 \mathrm{~cm}, 20 \mathrm{~cm}$ respectively, are combined to form a vertically placed container as shown in the figure. Some fine sand is poured into this container, filling the cone and part of the cylinder. It is measured that the sand in the cylinder is $5 \mathrm{~cm}$ high. If the container is inverted, the height of the sand will be $\qquad$ $\mathrm{cm}$.",The height of the sand is $11 \frac{2}{3}$ cm,medium,"9. (6 points) A hollow cone and a hollow cylinder, with base inner radii and heights of $12 \mathrm{~cm}, 20 \mathrm{~cm}$ respectively, are combined to form a vertically placed container as shown in the figure. Some fine sand is poured into this container, filling the cone and part of the cylinder. It is measured that the sand in the cylinder is $5 \mathrm{~cm}$ high. If the container is inverted, the height of the sand is $-11 \frac{2}{3} \_\mathrm{cm}$. 【Solution】Solution: According to the analysis, the height of the sand is: $5+20 \div 3=11 \frac{2}{3}$ (cm); Answer: The height of the sand is $11 \frac{2}{3}$ cm. Therefore, the answer is: $11 \frac{2}{3}$. Note: The negative sign in the problem statement is likely a typographical error and should be disregarded." 7ea1dbe5d27d,1. Let $a_{m+n}=A$ and $a_{m-n}=B$ be terms of an arithmetic progression. Express the terms $a_{n}$ and $a_{m}$ in terms of $A$ and $B$.,See reasoning trace,easy,"Solution. We have $$ a_{m+n}=a_{1}+(m+n-1) d=A $$ and $$ a_{m-n}=a_{1}+(m-n-1) d=B $$ which gives $$ a_{m}+n d=A \text { and } a_{m}-n d=B, $$ hence $d=\frac{A-B}{2 n}$. Now from $a_{m}-n d=B$ and $a_{n}+m d=A$ it follows that $$ a_{m}=B+n d=B+n \frac{A-B}{2 n}=\frac{A+B}{2} \text { and } a_{n}=A-m d=A-m \frac{A-B}{2 n}=\frac{(2 n-m) A+m B}{2 n} $$" 6f2db29ba8a3,"(2) Among the four functions $y=\sin |x|, y=\cos |x|, y=|\tan x|, y=$ $-\ln |\sin x|$, the one that has a period of $\pi$, is monotonically decreasing in $\left(0, \frac{\pi}{2}\right)$, and is an even function is ( ). (A) $y=\sin |x|$ (B) $y=\cos |x|$ (C) $y=|\tan x|$ (D) $y=-\ln |\sin x|$","\sin |x|$ is not a periodic function, A is incorrect; since $y=$ $\cos |x|$ is a periodic function w",easy,"(2) D Hint: Since $y=\sin |x|$ is not a periodic function, A is incorrect; since $y=$ $\cos |x|$ is a periodic function with the smallest positive period of $2 \pi$, B is incorrect; since $y=|\tan x|$ is monotonically increasing in $\left(0, \frac{\pi}{2}\right)$, C is incorrect; the function $y=-\ln |\sin x|$ meets the requirements of the problem." f32b10923367,"5. (10 points) Figure $ABCD$ is a parallelogram, $M$ is the midpoint of $DC$, $E$ and $F$ are on $AB$ and $AD$ respectively, and $EF$ is parallel to $BD$. If the area of triangle $MDF$ is 5 square centimeters, then the area of triangle $CEB$ is ( ) square centimeters. A. 5 B. 10 C. 15 D. 20",The area of triangle $E B C$ is 10 square centimeters,medium,"【Analysis】Connect $F C, D E, F B$, in trapezoid $F B C D$, $S_{\triangle F D B}$ and $S_{\triangle F D C}$ have the same base and height, so their areas are equal; in trapezoid $E B C D$, $S_{\triangle E D B}$ and $S_{\triangle E B C}$ have the same base and height, so their areas are equal; in trapezoid $F E B D$, $S_{\triangle F D B}$ and $S_{\triangle E D B}$ have the same base and height, so their areas are equal; therefore, $S_{\triangle F D C}=S_{\triangle E B C}$. Since $M$ is the midpoint of $D C$, according to the property that the area of a triangle is proportional to its base when the height is constant, $S_{\triangle E B C}=2 \times 5=10 \mathrm{~cm}^{2}$. 【Solution】Solution: As shown in the figure, connect $F C, D E, F B$, In trapezoid $F B C D$, $S_{\triangle F D B}=S_{\triangle F D C}$, In trapezoid $E B C D$, $S_{\triangle E D B}=S_{\triangle E B C}$, In trapezoid $F E B D$, $S_{\triangle F D B}=S_{\triangle E D B}$, Therefore, $S_{\triangle F D C}=S_{\triangle E B C}$, Since $M$ is the midpoint of $D C$, So $S_{\triangle E B C}=2 \times 5=10$ (square centimeters). Thus, $S_{\triangle E B C}=10$ square centimeters, Answer: The area of triangle $E B C$ is 10 square centimeters. Therefore, the correct choice is: B." b20012b714e6,"Which of the numbers lies between $2 / 5$ and $3 / 4$? (a) $1 / 6$ (b) $4 / 3$ (c) $5 / 2$ (d) $4 / 7$ (e) $1 / 4$",See reasoning trace,medium,"The correct option is (d). Remember that the order between fractions made up of positive integers is determined by the cross product of the integers, that is, $\frac{a}{b}<\frac{c}{d}$ is equivalent to the statement $a \times d4 x+p-3$ always holds. Try to find the range of values for $x$.","(x-1) p+\left(x^{2}-4 x+3\right)$, then $f(p)$ is a linear function of $p$. It is easy to see that t",medium,"Analysis We take $p$ as the main variable, transforming it into a linear function problem, thereby simplifying the calculation. Solution The original inequality is equivalent to $p(x-1)+\left(x^{2}-4 x+3\right)>0$. Let $f(p)=(x-1) p+\left(x^{2}-4 x+3\right)$, then $f(p)$ is a linear function of $p$. It is easy to see that the original inequality is equivalent to $\left\{\begin{array}{l}f(0)>0 \\ f(4)>0\end{array}\right.$, which is $\left\{\begin{array}{l}x^{2}-4 x+3>0 \\ x^{2}-1>0\end{array}\right.$. Solving this, we get $x>3$ or $x<-1$." 3fd1d94ba286,"5. Determine the unknown number $x$ and explain. | 9 | 22 | 13 | 16 | | :---: | :---: | :---: | :---: | | 144 | 176 | 143 | 192 | | 16 | 8 | 11 | $x$ | Tasks worth 10 points",See reasoning trace,easy,"5. Since $144: 9=16$, $176: 22=8$ and $143: 13=11$, then it must also be so in the last column. 2 points Therefore, $x=192$ : 16, which means $x=12$. 2 points TOTAL 4 POINTS" 7450ef000525,"1. Given $a$ is a real number. Then the number of subsets of the set $M=\left\{x \mid x^{2}-3 x-a^{2}+2=0, x \in \mathbf{R}\right\}$ is A. 1 B. 2 C. 4 D. Uncertain",$C$,easy,"Solve: $\Delta=9+4\left(a^{2}-2\right)=4 a^{2}+1>0$, then the equation $x^{2}-3 x-a^{2}+2=0$ has 2 distinct real roots, and the number of subsets of set $A$ is 4. Therefore, the answer is $C$." 78344ad8c7b3,"1. Let $x^{2}+y^{2}-2 x-2 y+1=0(x, y \in$ $\mathbf{R})$. Then the minimum value of $F(x, y)=\frac{x+1}{y}$ is $\qquad$ .",See reasoning trace,medium,"II. $1 . \frac{3}{4}$. The original expression is transformed into $(x-1)^{2}+(y-1)^{2}=1$. Let $x-1=\cos \theta, y-1=\sin \theta$, then $x=1+\cos \theta, y=1+\sin \theta$. Therefore, $\frac{x+1}{y}=\frac{2+\cos \theta}{1+\sin \theta}$, let this be $\mu$. Then $\mu+\mu \sin \theta=2+\cos \theta$, which means $\mu \sin \theta-\cos \theta=2-\mu$. So, $$ \sqrt{\mu^{2}+1} \sin (\theta-\varphi)=2-\mu . $$ (Where, $\left.\cos \varphi=\frac{\mu}{\sqrt{\mu^{2}+1}}, \sin \varphi=\frac{1}{\sqrt{\mu^{2}+1}}\right)$ $$ \sin (\theta-\varphi)=\frac{2-\mu}{\sqrt{\mu^{2}}}+\left|\frac{2-\mu}{\sqrt{\mu^{2}}+1}\right| \leqslant 1 . $$ Solving this, we get $\mu \geqslant \frac{3}{4}$, so the minimum value of $F(x, y)$ is $\frac{3}{4}$." dc0049cd3413,"6. Let $f(x)$ be a quadratic function, $g(x)=2^{x} f(x), g(x+1)-g(x)=2^{x+1} \cdot x^{2}$, then $f(x)=$","a x^{2}+b x+c$, substitute into the conditions and proceed.",easy,"6. $2 x^{2}-8 x+12$ Hint: Use the method of undetermined coefficients, we can assume $f(x)=a x^{2}+b x+c$, substitute into the conditions and proceed." 6a8044051e49,"## Calculate the definite integral: $$ \int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{2+\cos x} $$",See reasoning trace,medium,"## Solution We will use the universal substitution: $$ t=\operatorname{tg} \frac{x}{2} $$ From which: $$ \begin{aligned} & \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\ & x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\ & x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1 \end{aligned} $$ Substitute: $$ \begin{aligned} & \int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{2+\cos x}=\int_{0}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{2+\frac{1-t^{2}}{1+t^{2}}}=\int_{0}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}\right)^{2} \cdot \frac{2+2 t^{2}+1-t^{2}}{1+t^{2}}}= \\ & =2 \cdot \int_{0}^{1} \frac{1-t^{2}}{\left(1+t^{2}\right) \cdot\left(3+t^{2}\right)} d t= \end{aligned} $$ Decompose the proper rational fraction into elementary fractions using the method of undetermined coefficients: $$ \begin{aligned} & \frac{1-t^{2}}{\left(1+t^{2}\right) \cdot\left(3+t^{2}\right)}=\frac{A t+B}{t^{2}+1}+\frac{C t+D}{t^{2}+3}= \\ & =\frac{(A t+B)\left(t^{2}+3\right)+(C t+D)\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(t^{2}+3\right)}= \\ & =\frac{A t^{3}+3 A t+B t^{2}+3 B+C t^{3}+C t+D t^{2}+D}{\left(t^{2}+1\right)\left(t^{2}+3\right)}= \end{aligned} $$ $$ \begin{aligned} & =\frac{(A+C) t^{3}+(B+D) t^{2}+(3 A+C) t+(3 B+D)}{\left(t^{2}+1\right)\left(t^{2}+3\right)} \\ & \left\{\begin{array}{l} A+C=0 \\ B+D=-1 \\ 3 A+C=0 \\ 3 B+D=1 \end{array}\right. \end{aligned} $$ Subtract the first equation from the third: $$ \begin{aligned} & \left\{\begin{array}{l} A+C=0 \\ B+D=-1 \\ 2 A=0 \end{array}\right. \\ & \left\{\begin{array}{l} C=0 \\ 3 B+D=1 \\ A=0 \\ 3 B+D=-1 \end{array}\right. \\ & \left\{\begin{array}{l} C=1 \end{array}\right. \end{aligned} $$ Subtract the second equation from the fourth: $$ \begin{aligned} & \left\{\begin{array}{l} C=0 \\ B+D=-1 \\ A=0 \\ 2 B=2 \end{array}\right. \\ & \left\{\begin{array}{l} C=0 \\ D=-2 \\ A=0 \\ B=1 \end{array}\right. \\ & \frac{1-t^{2}}{\left(1+t^{2}\right) \cdot\left(3+t^{2}\right)}=\frac{1}{t^{2}+1}-\frac{2}{t^{2}+3} \end{aligned} $$ Then: $$ \begin{aligned} & =2 \cdot \int_{0}^{1}\left(\frac{1}{t^{2}+1}-\frac{2}{t^{2}+3}\right) d t=\left.2 \cdot\left(\operatorname{arctg} t-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{t}{\sqrt{3}}\right)\right|_{0} ^{1}= \\ & =2 \cdot\left(\operatorname{arctg} 1-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{1}{\sqrt{3}}\right)-2 \cdot\left(\operatorname{arctg} 0-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{0}{\sqrt{3}}\right)= \\ & =2 \cdot\left(\frac{\pi}{4}-\frac{2}{\sqrt{3}} \cdot \frac{\pi}{6}\right)-2 \cdot\left(0-\frac{2}{\sqrt{3}} \cdot 0\right)=2 \cdot\left(\frac{9 \pi}{36}-\frac{4 \sqrt{3} \pi}{36}\right)=\frac{(9-4 \sqrt{3}) \pi}{18} \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-2$ » Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals Ukrainian Banner Network - Last edited: 11:00, April 23, 2009. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals $8-3$ ## Material from PlusPi" ced43476cdf4,"Exercise 5 Calculate using a simple method. 1. $682+325$ $573+198$ 2. $897+234$ $788+143$ 3. $694+367$ $595+698$",See reasoning trace,easy,"Practice 5 $$ \begin{array}{llllll} 1.1007 & 771 & 2.1131 & 931 & 3.1061 & 1293 \end{array} $$" eda7d1d87bbc,"A2. The sum $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}$ is equal to (A) $\frac{3}{1110}$ (B) $\frac{3}{1000}$ (C) $\frac{111}{1000}$ (D) $\frac{111}{1110}$ (E) $\frac{3}{111}$",See reasoning trace,easy,"A2. The sum is equal to $$ \frac{1}{10}+\frac{1}{100}+\frac{1}{1000}=\frac{100}{1000}+\frac{10}{1000}+\frac{1}{1000}=\frac{111}{1000} $$" c0adb4c2b75d,"10,11 The base of a right prism is an isosceles trapezoid $ABCD$, where $AB=CD=13, BC=11, AD=$ 21. The area of the diagonal section of the prism is 180. Find the area of the complete surface of the prism.",906,medium,"Let $A B C D A 1 B 1 C 1 D 1$ be the given prism, $S$ - the area of its total surface. The diagonal section of the prism is a rectangle $A C C 1 A 1$. Dropping a perpendicular $C K$ from vertex $C$ of the smaller base of the trapezoid to the larger base $A D$. Then $$ D K=\frac{1}{2}(A D-B C)=\frac{1}{2}(21-11)=5 $$ $$ A K=\frac{1}{2}(A D+B C)=\frac{1}{2}(21+11)=16 $$ By the Pythagorean theorem from right triangles $C K D$ and $А K C$ we find that $$ C K=\sqrt{C D^{2}-D K^{2}}=\sqrt{169-25}=12 $$ $$ A C=\sqrt{A K^{2}+C K^{2}}=\sqrt{256+144}=20 $$ According to the problem, $S_{A C C} 1 A 1=A C \cdot A A 1=180$. Therefore, $$ A A 1=\frac{180}{A C}=\frac{180}{20}=9 $$ Thus, $$ \begin{gathered} S=2 S_{A B C D}+S_{A B B} 1 A 1+S_{B C C} 1 B 1+S_{C D D} 1 C 1+S_{A D D} 1 A 1= \\ =2 \cdot \frac{1}{2}(A D+B C) C K+(A B+B C+C D+A D) A A 1= \\ =2 \cdot 16 \cdot 12+58 \cdot 9=906 \end{gathered} $$ ## Answer 906.00" 86b4e35f7028,"4. The even function $f(x)$ defined on $\mathbf{R}$ satisfies $f(x+1)=-f(x)$, and is increasing in the interval $[-1,0]$, then (). (A) $f(3)1, b>1, a \neq b$. Assume the common root of equations (1) and (2) is $x_{0}$. Then $$ \begin{array}{l} (a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\ (b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 . \end{array} $$ (3) $\times(b-1)$ - (4) $\times(a-1)$ and simplifying, we get $$ (a-b)(a b-a-b-2)\left(x_{0}-1\right)=0 \text {. } $$ Since $a \neq b$, we have $x_{0}=1$ or $a b-a-b-2=0$. If $x_{0}=1$, substituting into equation (3) yields $a=1$, which is a contradiction. Therefore, $x_{0} \neq 1$. Hence, $a b-a-b-2=0$ $\Rightarrow(a-1)(b-1)=3$ $\Rightarrow a=2, b=4$ or $a=4, b=2$. Substituting into the required algebraic expression, we get $$ \frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=\frac{2^{4}+4^{2}}{2^{-4}+4^{-2}}=256 . $$" 76d2e4d37f44,"5. For any positive integer $n$, let $a_{n}$ denote the number of triangles with integer side lengths and the longest side length being $2 n$. (1) Find the expression for $a_{n}$ in terms of $n$; (2) Let the sequence $\left\{b_{n}\right\}$ satisfy $$ \sum_{k=1}^{n}(-1)^{n-k} \mathrm{C}_{n}^{k} b_{k}=a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {, } $$ Find the number of positive integers $n$ such that $b_{n} \leqslant 2019 a_{n}$.",See reasoning trace,medium,"5. (1) Let the lengths of the three sides be integers, the longest side has a length of $2n$, and the shortest side has a length of $k(k \leqslant 2n)$, the number of such triangles is $a_{n}(k)$. Thus, $a_{n}=\sum_{k=1}^{2n} a_{n}(k)$. When $k \leqslant n$, the length of the third side can be $$ 2n+1-k, 2n+2-k, \cdots, 2n \text{, } $$ there are $k$ possibilities; When $nt . \end{array}\right. \\ \text{By } \sum_{k=1}^{n}(-1)^{n-k} \mathrm{C}_{n}^{k} b_{k}=a_{n} \\ \Rightarrow \sum_{k=1}^{n} \mathrm{C}_{n}^{k} a_{k}=\sum_{k=1}^{n} \mathrm{C}_{n}^{k}\left(\sum_{t=1}^{k}(-1)^{k-t} \mathrm{C}_{k}^{t} b_{t}\right) \\ =\sum_{t=1}^{n}\left(\sum_{k=t}^{n}(-1)^{k-t} \mathrm{C}_{n}^{k} \mathrm{C}_{k}^{t}\right)=b_{n} \\ \Rightarrow b_{n}=\sum_{k=1}^{n} \mathrm{C}_{n}^{k} a_{k}=\sum_{k=1}^{n} \mathrm{C}_{n}^{k} k(k+1) \\ =n(n+1) \times 2^{n-2}+n \times 2^{n-1} \\ =n(n+3) \times 2^{n-2} \text{. } \\ \text{By } \frac{b_{n}}{a_{n}} \leqslant 2019 \\ \Rightarrow n(n+3) \times 2^{n-2} \leqslant 2019 n(n+1) \text{. } \\ \end{array} $$ When $n \geqslant 2$, by $$ \begin{array}{l} \frac{n(n+3)}{n(n+1)} 2^{n-2}=\frac{n+3}{n+1} 2^{n-2} \\ \leqslant 2^{n-1} \leqslant 2019 \\ \Rightarrow n \leqslant 12 . \end{array} $$ Thus, the number of positive integers $n$ such that $\frac{b_{n}}{a_{n}} \leqslant 2019$ is 12." 33d92a171bf2,"3. Find all integer sequences $a_{1}, a_{2}, \cdots, a_{n}$, such that: (1) For any $1 \leqslant i \leqslant n$, we have $$ 1 \leqslant a_{i} \leqslant n ; $$ (2) For any $1 \leqslant i, j \leqslant n$, we have $$ \left|a_{i}-a_{j}\right|=|i-j| . $$","n$, $a_{i}=n+1-i(2 \leqslant i \leqslant n)$.",medium,"3. The sequences that satisfy the conditions are only $$ a_{i}=i(1 \leqslant i \leqslant n) $$ and $a_{i}=n+1-i(1 \leqslant i \leqslant n)$. According to condition (2), for $2 \leqslant i \leqslant n$, $$ \left|a_{i}-a_{1}\right|=i-1 . $$ Then $\left|a_{i}-a_{1}\right|$ has $n-1$ different values. For any $1 \leqslant j \leqslant n$, the integer $a_{j}$ satisfies $1 \leqslant a_{i} \leqslant n$, so only when $a_{1}=1$ or $n$, $\left|a_{i}-a_{1}\right|$ can take $n-1$ different values. Thus, when $a_{1}=1$, $a_{i}=i(2 \leqslant i \leqslant n)$; when $a_{1}=n$, $a_{i}=n+1-i(2 \leqslant i \leqslant n)$." e2a51b643b11,"9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer.",-27,easy,Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$. 6d40c13f850c,"How many integers $n$ with $0\leq n < 840$ are there such that $840$ divides $n^8-n^4+n-1$? $ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 $",2,medium,"1. First, we note that \( 840 = 3 \cdot 5 \cdot 7 \cdot 8 \). We need to find the integers \( n \) such that \( 0 \leq n < 840 \) and \( 840 \) divides \( n^8 - n^4 + n - 1 \). This means we need \( n^8 - n^4 + n - 1 \equiv 0 \pmod{840} \). 2. We will consider the congruences modulo \( 3 \), \( 5 \), \( 7 \), and \( 8 \) separately. 3. **Modulo 3:** - Since \( n^2 \equiv 1 \pmod{3} \) for \( n \) such that \( \gcd(n, 3) = 1 \), we have: \[ n^8 \equiv 1 \pmod{3}, \quad n^4 \equiv 1 \pmod{3} \] \[ n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{3} \] Therefore, \( n \equiv 1 \pmod{3} \). 4. **Modulo 5:** - Since \( n^4 \equiv 1 \pmod{5} \) for \( n \) such that \( \gcd(n, 5) = 1 \), we have: \[ n^8 \equiv 1 \pmod{5}, \quad n^4 \equiv 1 \pmod{5} \] \[ n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{5} \] Therefore, \( n \equiv 1 \pmod{5} \). 5. **Modulo 8:** - Since \( \varphi(8) = 4 \), we have \( n^4 \equiv 1 \pmod{8} \) for \( n \) such that \( \gcd(n, 8) = 1 \), we have: \[ n^8 \equiv 1 \pmod{8}, \quad n^4 \equiv 1 \pmod{8} \] \[ n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{8} \] Therefore, \( n \equiv 1 \pmod{8} \). 6. **Modulo 7:** - Since \( n^6 \equiv 1 \pmod{7} \), we have: \[ n^8 \equiv n^2 \pmod{7}, \quad n^4 \equiv n^2 \pmod{7} \] \[ n^8 - n^4 + n - 1 \equiv n^2 - n^2 + n - 1 \equiv n - 1 \pmod{7} \] Therefore, \( n \equiv 1 \pmod{7} \) or \( n \equiv 3 \pmod{7} \). 7. **Combining the congruences using the Chinese Remainder Theorem:** - For \( n \equiv 1 \pmod{3} \), \( n \equiv 1 \pmod{5} \), \( n \equiv 1 \pmod{8} \), and \( n \equiv 1 \pmod{7} \): \[ n \equiv 1 \pmod{840} \] This gives us \( n = 1 \). - For \( n \equiv 1 \pmod{3} \), \( n \equiv 1 \pmod{5} \), \( n \equiv 1 \pmod{8} \), and \( n \equiv 3 \pmod{7} \): Using the Chinese Remainder Theorem, we find another solution \( n \) in the interval \( 0 \leq n < 840 \). 8. Therefore, there are two solutions in total. The final answer is \(\boxed{2}\)." b88ce77fe400,"4. Let $n>0, n=2^{k} m$ (where $m$ is odd). Then the greatest common divisor of $\mathrm{C}_{2 n}^{1}, \mathrm{C}_{2 n}^{3}$, $\cdots, \mathrm{C}_{2 n}^{2 n-1}$ is $\qquad$",See reasoning trace,medium,"4. $2^{k+1}$. First, $2^{k+1} \| \mathrm{C}_{2 n}^{1}$. It is easy to prove that $2^{k+1} \| \mathrm{C}_{2 n}^{i} (i=3,5, \cdots, 2 n-1)$. This is because $i \mathrm{C}_{2 n}^{i}=2 n \mathrm{C}_{2 n-1}^{i-1}, (i, 2 n)=1$. Therefore, the greatest common divisor of $\mathrm{C}_{2 n}^{1}, \mathrm{C}_{2 n}^{3}, \cdots, \mathrm{C}_{2 n}^{2 n-1}$ is $2^{k+1}$." 7aad4fa7acb5,"## Task 2 - 300822 A rectangle whose side lengths are in the ratio $1: 2$ is to be divided into eight congruent isosceles right triangles. a) Draw and describe such a division! Explain why the eight triangles formed according to your description are isosceles right triangles and congruent to each other! b) Determine the length of a leg of these triangles in terms of the smaller of the two side lengths of the rectangle!",See reasoning trace,medium,"a) The figure shows a possible decomposition of the required type for a rectangle $A B C D$. ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0802.jpg?height=237&width=332&top_left_y=1600&top_left_x=862) Description: If $E, F$ are the midpoints of $A B$ and $C D$ respectively, the decomposition is created by drawing the segments $E F, A F, D E, B F$, and $C E$. Justification: Since $A B C D$ is a rectangle and $E, F$ bisect the segments $A B$ and $C D$ respectively, it follows from the condition $A D: A B = 1: 2$. The segments $A E$ and $D F$ are of equal length and parallel to each other, and the segments $A E$ and $A D$ are of equal length and perpendicular to each other; thus, $A E F D$ is a square. Similarly, it follows that $E B C F$ is a square. Each of these squares is divided into four triangles by the drawn diagonals. Since in each square the diagonals are of equal length and perpendicular to each other and bisect each other, all eight resulting triangles are isosceles right triangles with equal leg lengths, and thus also congruent to each other. b) In dependence on $a = A D$, the square $A E F D$ has an area of $a^{2}$. Each of its four mutually (congruent, thus) area-equal sub-triangles therefore has an area of $\frac{1}{4} a^{2}$. If $x$ is the sought leg length of these triangles, then the area is also equal to $\frac{1}{2} x^{2}$. Therefore, we have $$ \frac{1}{2} x^{2} = \frac{1}{4} a^{2} \quad ; \quad x = \sqrt{\frac{1}{2} a^{2}} = \frac{1}{2} a \sqrt{2} $$" 6043af188e9c,"Example 6 Let $x y=1$, and $x>y>0$, find the minimum value of $\frac{x^{2}+y^{2}}{x-y}$. untranslated text is retained in its original form.","\sqrt{2}$, i.e., $x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=\frac{\sqrt{6}-\sqrt{2}}{2}$.",easy,"Given $x>y>0$, we can set $x=y+\Delta y(\Delta y>0)$, then $\frac{x^{2}+y^{2}}{x-y}=\frac{(x-y)^{2}+2 x y}{x-y}=\frac{(\Delta y)^{2}+2}{\Delta y} \geqslant 2 \sqrt{2}$, the equality holds if and only if $\Delta y=\sqrt{2}$, i.e., $x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=\frac{\sqrt{6}-\sqrt{2}}{2}$." bb47973c26c3,"Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$",191,medium,"For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions. First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions. Together, the requested probability is \[\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$ ~MRENTHUSIASM" 0a161a9f3cbf,"\section*{Exercise 1 - 081221} Give all prime numbers \(p\) for which both \(p+10\) and \(p+14\) are also prime numbers!","13, p+14=17\) and since these are prime numbers, there is exactly one solution, namely \(p=3\).",easy,"} Let \(a \in \{0,1,2\}\) be the remainder of \(p\) when divided by 3. If \(a=1\), then \(p+14\) is divisible by 3 and since \(p+14>3\), it is not a prime number. If \(a=2\), then similarly \(p+10\) is divisible by 3 and is not a prime number. Therefore, \(p\) must be divisible by 3, so \(p=3\). Then \(p+10=13, p+14=17\) and since these are prime numbers, there is exactly one solution, namely \(p=3\)." 6cc5f2e61c41,"13. Vector $\vec{a}=(1, \sin \theta), \vec{b}=(\cos \theta, \sqrt{3}), \theta \in R$, then the range of $|\vec{a}-\vec{b}|$ is",See reasoning trace,easy,"$[1,3]$ $\qquad$ Answer: $$ \begin{aligned} |\vec{a}-\vec{b}| & =\sqrt{(1-\cos \theta)^{2}+(\sin \theta-\sqrt{3})^{2}}=\sqrt{5-2(\cos \theta-\sqrt{3} \sin \theta)} \\ & =\sqrt{5-4 \sin \left(\frac{\pi}{6}-\theta\right)} \text {, its maximum value is } 3 \text {, minimum value is } 1 \text {, the range of values is } [1,3]。 \end{aligned} $$" 71d1f7ca6b7d,"On a examination of $n$ questions a student answers correctly $15$ of the first $20$. Of the remaining questions he answers one third correctly. All the questions have the same credit. If the student's mark is $50\%$, how many different values of $n$ can there be? $ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the",1,medium,"1. Let \( n \) be the total number of questions on the examination. 2. The student answers 15 out of the first 20 questions correctly. 3. For the remaining \( n - 20 \) questions, the student answers one third correctly. Therefore, the number of correctly answered questions in the remaining part is \( \frac{1}{3}(n - 20) \). 4. The total number of correctly answered questions is: \[ 15 + \frac{1}{3}(n - 20) \] 5. The student's mark is 50%, which means the student answered half of the total questions correctly. Therefore, we set up the equation: \[ \frac{15 + \frac{1}{3}(n - 20)}{n} = \frac{1}{2} \] 6. To solve for \( n \), we first clear the fraction by multiplying both sides by \( n \): \[ 15 + \frac{1}{3}(n - 20) = \frac{n}{2} \] 7. Simplify the equation: \[ 15 + \frac{n - 20}{3} = \frac{n}{2} \] 8. Multiply through by 6 to clear the denominators: \[ 6 \cdot 15 + 6 \cdot \frac{n - 20}{3} = 6 \cdot \frac{n}{2} \] \[ 90 + 2(n - 20) = 3n \] 9. Distribute and simplify: \[ 90 + 2n - 40 = 3n \] \[ 50 + 2n = 3n \] 10. Subtract \( 2n \) from both sides: \[ 50 = n \] 11. Therefore, the total number of questions \( n \) is 50. The final answer is \( \boxed{1} \)" 7d00d5fd4ebc,"## Calculate the limit of the function: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}+2 x-3}{x^{2}+4 x-5}\right)^{\frac{1}{2-x}} $$",See reasoning trace,easy,"## Solution $\lim _{x \rightarrow 1}\left(\frac{x^{2}+2 x-3}{x^{2}+4 x-5}\right)^{\frac{1}{2-x}}=\lim _{x \rightarrow 1}\left(\frac{(x-1)(x+3)}{(x-1)(x+5)}\right)^{\frac{1}{2-x}}=$ $=\lim _{x \rightarrow 1}\left(\frac{x+3}{x+5}\right)^{\frac{1}{2-x}}=\left(\frac{1+3}{1+5}\right)^{\frac{1}{2-1}}=\left(\frac{4}{6}\right)^{1}=\frac{2}{3}$ ## Problem Kuznetsov Limits 20-29" a963403cb785,"7. Let $a_{1}, a_{2}, \cdots, a_{21}$ be a permutation of $1,2, \cdots, 21$, satisfying $$ \left|a_{20}-a_{21}\right| \geqslant\left|a_{19}-a_{21}\right| \geqslant\left|a_{18}-a_{21}\right| \geqslant \cdots \geqslant\left|a_{1}-a_{21}\right| \text {. } $$ The number of such permutations is $\qquad$ .",See reasoning trace,medium,"When $a_{21}=1$, there is only one case: $a_{1}=2, a_{2}=3, \cdots, a_{20}=21$, and similarly for $a_{21}=21$; when $a_{21}=2$, $a_{1}, a_{2}$ can be 1,3, and the rest must be in order, and similarly for $a_{21}=20$; when $a_{21}=3$, $a_{1}, a_{2}$ can be 2,4, and $a_{3}, a_{4}$ can be 1,5, and the rest must be in order, and similarly for $a_{21}=19$; $\cdots$; When $a_{21}=11$, $a_{1}, a_{2}$ can be 10,12, $a_{3}, a_{4}$ can be 9,13, $\cdots$, $a_{19}, a_{20}$ can be 1,21. Therefore, the number of permutations that satisfy the condition is $$ 2\left(1+2+2^{2}+\cdots+2^{9}\right)+2^{10}=2\left(2^{10}-1\right)+2^{10}=3 \cdot 2^{10}-2=3070 $$" c1151fae3a3f,"# Task 4. (12 points) Solve the equation $x^{2}+y^{2}+1=x y+x+y$. #",$x=y=1$,easy,"# Solution $$ \begin{gathered} x^{2}+y^{2}+1=x y+x+y \Leftrightarrow(x-y)^{2}+(x-1)^{2}+(y-1)^{2}=0 \Leftrightarrow \\ x=y=1 . \end{gathered} $$ Answer: $x=y=1$." 1c9dd11fa897,"A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds? [asy] unitsize(2mm); defaultpen(linewidth(.8pt)); fill((0,0)--(0,5)--(5,5)--cycle,gray); fill((25,0)--(25,5)--(20,5)--cycle,gray); draw((0,0)--(0,5)--(25,5)--(25,0)--cycle); draw((0,0)--(5,5)); draw((20,5)--(25,0)); [/asy] $\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$",\frac15,easy,"Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{\frac15}$. The answer is $\mathrm{(C)}$." 1d91da3d90e5,"17. Let $f:(0,1) \rightarrow(0,1)$ be a differentiable function with a continuous derivative such that for every positive integer $n$ and odd positive integer $a<2^{n}$, there exists an odd positive integer $b<2^{n}$ such that $f\left(\frac{a}{2^{n}}\right)=\frac{b}{2^{n}}$. Determine the set of possible values of $f^{\prime}\left(\frac{1}{2}\right)$.","$\{-1,1\}$ The key step is to notice that for such a function $f, f^{\prime}(x) \neq 0$ for any $x$",medium,"Answer: $\{-1,1\}$ The key step is to notice that for such a function $f, f^{\prime}(x) \neq 0$ for any $x$. Assume, for sake of contradiction that there exists $0\frac{1}{2}$. This contradicts the assumption that $\left|f^{\prime}(x)\right| \leq \frac{1}{2}$ for all $x \in(c, d)$. Since $f^{\prime}(x) \neq 0$, and $f^{\prime}$ is a continuous function, $f^{\prime}$ is either always positive or always negative. So $f$ is either increasing or decreasing. $f\left(\frac{1}{2}\right)=\frac{1}{2}$ always. If $f$ is increasing, it follows that $f\left(\frac{1}{4}\right)=\frac{1}{4}, f\left(\frac{3}{4}\right)=\frac{3}{4}$, and we can show by induction that indeed $f\left(\frac{a}{2^{n}}\right)=\frac{a}{2^{n}}$ for all integers $a, n$. Since numbers of this form are dense in the interval $(0,1)$, and $f$ is a continuous function, $f(x)=x$ for all $x$. It can be similarly shown that if $f$ is decreasing $f(x)=1-x$ for all $x$. So the only possible values of $f^{\prime}\left(\frac{1}{2}\right)$ are $-1,1$ Query: if the condition that the derivative is continuous were omitted, would the same result still hold?" f53762b32485,"Example 1 (Question from the 13th ""Hope Cup"" Invitational Competition) The real roots of the equations $x^{5}+x+1=0$ and $x+\sqrt[5]{x}+1=0$ are $\alpha, \beta$ respectively, then $\alpha+\beta$ equals ( ). A. -1 B. $-\frac{1}{2}$ C. $\frac{1}{2}$ D. 1","f(-1-\beta)$, which means $\alpha+\beta=-1$.",medium,"Solution: Choose A. Reason: Consider the function $f(x)=x^{5}+x+1$, which is an increasing function on $\mathbf{R}$, when $x^{5}+x+1=0$, we get $x=\sqrt[5]{-x-1}$, substituting $-x-1$ for $x$ yields $-x-1=\sqrt[5]{x}$. From (2), we have $(-x-1)^{5}=x$, which is also $(-x-1)^{5}+(-x-1)+1=0$. Let $\alpha$ be a root of equation (1), i.e., $f(\alpha)=0$. Let $\beta$ be a root of equation (2), i.e., $-1-\beta=\sqrt[5]{\beta}$, thus from (3) we know that $f(-1-\beta)=0$. Therefore, we have $f(\alpha)=f(-1-\beta)$, which means $\alpha+\beta=-1$." 7fca1683b32c,"7. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $$ \left\{\frac{2015!}{2017}\right\}= $$ $\qquad$",\frac{1}{2017} .\end{array}$,easy,$\begin{array}{l}\text { 7. } \frac{1}{2017} \cdot \\ \text { From } 2016! \equiv -1 \equiv 2016(\bmod 2017) \\ \Rightarrow 2017!2016(2015!-1) \\ \Rightarrow 2017 \text { ! } 2015!-1) \\ \Rightarrow\left\{\frac{2015!}{2017}\right\}=\frac{1}{2017} .\end{array}$ 5aaa8ee8be2f,"2. If $3 \leqslant x+y \leqslant 5, -1 \leqslant x-y \leqslant 1$, then the maximum value of $3 x+y$ is",See reasoning trace,easy,$11$ 804cdbdac3ad,"7. Let the line $l$ passing through the fixed point $M(a, 0)$ intersect the parabola $y^{2}=4 x$ at points $P$ and $Q$. If $\frac{1}{|P M|^{2}}+\frac{1}{|Q M|^{2}}$ is a constant, then the value of $a$ is $\qquad$ .",2$.,medium,"7. 2 . Let the parametric equation of line $l$ be $$ \left\{\begin{array}{l} x=a+t \cos \alpha, \\ y=t \sin \alpha, \end{array}\right. $$ where $t$ is the parameter, $\alpha$ is the inclination angle, and $\alpha \in(0, \pi)$. Substituting into the parabola equation yields $$ t^{2} \sin ^{2} \alpha-4 t \cos \alpha-4 a=0 \text {. } $$ Let the two roots of this equation be $t_{1}$ and $t_{2}$. Then $$ \begin{array}{l} t_{1}+t_{2}=\frac{4 \cos \alpha}{\sin ^{2} \alpha}, t_{1} t_{2}=-\frac{4 a}{\sin ^{2} \alpha} . \\ \text { Hence } \frac{1}{|P M|^{2}}+\frac{1}{|Q M|^{2}}=\frac{1}{t_{1}^{2}}+\frac{1}{t_{2}^{2}} \\ =\frac{t_{1}^{2}+t_{2}^{2}}{\left(t_{1} t_{2}\right)^{2}}=\left(\frac{t_{1}+t_{2}}{t_{1} t_{2}}\right)^{2}-\frac{2}{t_{1} t_{2}} \\ =\frac{2 \cos ^{2} \alpha+a \sin ^{2} \alpha}{2 a^{2}} \text { (constant). } \end{array} $$ Therefore, $a=2$." 2402665ad68d,"9. (14 points) Given $f(x)=\frac{x^{4}+k x^{2}+1}{x^{4}+x^{2}+1}, k, x \in \mathbf{R}$, (1) Find the maximum and minimum values of $f(x)$; (2) Find all real numbers such that for any three real numbers $a$, $b$, $c$, there exists a triangle with side lengths $f(a)$, $f(b)$, $f(c)$.","\frac{k+2}{3}, f(x)_{\min }=1$; when $kf(x)_{\max } \Rightarrow-\frac{1}{2}0$ and both $p$ and $q$ are prime numbers.","(m, p, q) = (1, 11, 3)",medium,"We are given the equation: \[2^m p^2 + 1 = q^5\] where \(m > 0\) and both \(p\) and \(q\) are prime numbers. We need to find all integer triples \((m, p, q)\) that satisfy this equation. 1. **Rearrange and Factorize:** \[2^m p^2 = q^5 - 1\] We can factorize the right-hand side using the difference of powers: \[q^5 - 1 = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\] Therefore, we have: \[2^m p^2 = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\] 2. **GCD Analysis:** Note that \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) \mid 5\). This is because \(\frac{q^5 - 1}{q - 1}\) is a polynomial in \(q\) with integer coefficients, and the only prime factors of \(q^5 - 1\) that are not factors of \(q - 1\) are those dividing 5. 3. **Case 1: \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) = 1\)** - **Subcase 1.1: \(p = 2\)** \[2^{m+2} = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\] Since \(q^5 - 1 > (q - 1)^2\), we must have \(q - 1 = 1 \Rightarrow q = 2\). This gives: \[\frac{2^5 - 1}{2 - 1} = 31\] However, \(2^{m+2} = 31\) has no integer solutions for \(m\). - **Subcase 1.2: \(p \neq 2\)** We have two possibilities: - \(q - 1 = 2^m\) and \(\frac{q^5 - 1}{q - 1} = p^2\) - \(q - 1 = p^2\) and \(\frac{q^5 - 1}{q - 1} = 2^m\) For the first possibility: \[q - 1 = 2^m \Rightarrow q = 2^m + 1\] Substituting back, we get: \[2^m p^2 = (2^m + 1)^5 - 1\] Simplifying, we get: \[p^2 = \frac{(2^m + 1)^5 - 1}{2^m}\] For the second possibility: \[q - 1 = p^2 \Rightarrow q = p^2 + 1\] Substituting back, we get: \[2^m = \frac{(p^2 + 1)^5 - 1}{p^2}\] - If \(m\) is odd: \[q = 2^m + 1\] For \(m = 1\): \[q = 2 + 1 = 3\] \[2^1 p^2 + 1 = 3^5 \Rightarrow 2p^2 + 1 = 243 \Rightarrow 2p^2 = 242 \Rightarrow p^2 = 121 \Rightarrow p = 11\] This gives the solution \((m, p, q) = (1, 11, 3)\). - If \(m\) is even: \[q = 4^k + 1\] For \(k \geq 2\): \[p^2 = \frac{(4^k + 1)^5 - 1}{4^k}\] This yields a contradiction since \(p^2 \equiv 5 \pmod{8}\) is not possible. 4. **Case 2: \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) = 5\)** Clearly, we must have \(p = 5\): \[q - 1 = 5 \Rightarrow q = 6\] This is a contradiction since \(q\) must be a prime number. We conclude from our search that \(\boxed{(m, p, q) = (1, 11, 3)}\) is the only solution." 3f705acbd1a9,"9. (24th American Mathematics Competition) There are 1990 piles of stones, with the number of stones in each pile being $1, 2, \cdots, 1990$. Perform the following operation: each time, you can choose any number of piles and remove the same number of stones from each chosen pile. How many operations are required at minimum to remove all the stones?","1024$ stones from each pile that has enough; the second time, take away $2^{9}=512$ stones from each",medium,"9. Since $1990=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5}+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0}$, and writing $1,2, \cdots$, 1989 in binary form, the operation is as follows: The first time, take away $2^{10}=1024$ stones from each pile that has enough; the second time, take away $2^{9}=512$ stones from each pile that has enough, $\cdots$, finally, take away the stones from the piles that have only $2^{0}=1$ stone left. In this way, a total of 11 times are used, because the pile with one stone must have an operation to take one stone. If the remaining operations take away more than 2 stones each time, then the pile with exactly two stones cannot be taken away. Therefore, 2 operations can take at most $1+2=3$ stones, $\cdots$, 10 operations can take at most $1+2+2^{2}+\cdots+2^{9}=1023$ stones, so 1990 piles of stones must undergo at least 11 operations." d0a0cbee1cf7,"Find whether among all quadrilaterals, whose interiors lie inside a semi-circle of radius $r$, there exist one (or more) with maximum area. If so, determine their shape and area.",\frac{3\sqrt{3,medium,"1. **Existence of Maximum Area Quadrilateral:** - The existence of a maximum area quadrilateral is guaranteed by the compactness argument. Since the semi-circle is a closed and bounded set, any continuous function (such as the area function) defined on it will attain its maximum value. 2. **Convexity and Vertices on Boundary:** - A quadrilateral achieving maximum area must be convex and have all its vertices on the boundary of the semi-circle. This is because a non-convex quadrilateral or one with vertices inside the semi-circle would have a smaller area compared to a convex quadrilateral with vertices on the boundary. 3. **Vertices on the Diameter:** - Assume two vertices lie on the diameter of the semi-circle. If not, consider the side with the least distance from the center of the circle of radius \( r \) and extend it to a full chord. This ensures the quadrilateral is inscribed in the semi-circle with diameter \( \leq 2r \). 4. **Symmetrization:** - Symmetrize the quadrilateral with respect to the diameter. The result is a polygon with 6 vertices, of double the area, contained in the circle of radius \( r \). 5. **Steiner's Result:** - By Steiner's result, the largest area of an \( n \)-gon contained in a circle is realized by the regular \( n \)-gon. Therefore, in our case, the maximum area is given by a half regular hexagon. 6. **Area Calculation:** - A regular hexagon inscribed in a circle of radius \( r \) has an area given by: \[ A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} r^2 \] - Since we are considering a half regular hexagon, the area of the quadrilateral is: \[ A_{\text{quadrilateral}} = \frac{1}{2} \times \frac{3\sqrt{3}}{2} r^2 = \frac{3\sqrt{3}}{4} r^2 \] The final answer is \( \boxed{ \frac{3\sqrt{3}}{4} r^2 } \)" 37bd069ee59f,"Task B-2.1. Draw in the complex plane the set of points $(x, y)$, associated with complex numbers $z=x+yi$, for which $$ \operatorname{Re} z \cdot \operatorname{Im} z<0 \quad \mathrm{and} \quad|z|^{2} \leq \operatorname{Im}\left(z^{2}\right)+1 $$ Calculate the area of this set of points.",See reasoning trace,medium,"## Solution. From the first inequality, it follows that $x \cdot y < 0$, which means that the desired points lie in the second or fourth quadrant. From the second inequality, it follows that $$ x^{2} + y^{2} \leq 2 x y + 1 $$ or $(x-y)^{2} \leq 1$ or $|x-y| \leq 1$. From this, we have $-1 \leq x-y \leq 1$, or $y \leq x+1$ and $y \geq x-1$. The set of desired points is a part of the plane between the lines $y=x+1$ and $y=x-1$, and within the second and fourth quadrants, as shown in the figure. ![](https://cdn.mathpix.com/cropped/2024_05_30_4f4efb97ca3723b253dag-06.jpg?height=919&width=919&top_left_y=1454&top_left_x=576) From the figure, it is clear that $$ P=2 \cdot \frac{1 \cdot 1}{2}=1 $$" b10fd0b39dd0,"3.176. Given: $\operatorname{ctg} \alpha=4, \operatorname{ctg} \beta=\frac{5}{3}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$.",$\alpha+\beta=\frac{\pi}{4}$,medium,"Solution. $$ \operatorname{ctg} \alpha=4, \frac{\cos \alpha}{\sin \alpha}=4, \frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}=16, \frac{1-\sin ^{2} \alpha}{\sin ^{2} \alpha}=16 $$ from which $\sin ^{2} \alpha=\frac{1}{17}$, hence for $0<\alpha<\frac{\pi}{2}$ we have $$ \sin \alpha=\frac{1}{\sqrt{17}} ; \operatorname{ctg} \beta=\frac{\cos \beta}{\sin \beta}=\frac{5}{3}, \frac{\cos ^{2} \beta}{\sin ^{2} \beta}=\frac{25}{9}, \frac{1-\sin ^{2} \beta}{\sin ^{2} \beta}=\frac{25}{9} $$ from which $\sin ^{2} \beta=\frac{9}{34}$, therefore, for $0<\beta<\frac{\pi}{2}$ we have $\sin \beta=\frac{3}{\sqrt{34}}$. Then $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}=\frac{\sin \beta \cos \alpha+\sin \alpha \cos \beta}{\sin \alpha \sin \beta}=4+\frac{5}{3}=\frac{17}{3}$. $\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}=\frac{17}{3}$. Using the found values $\sin \alpha=\frac{1}{\sqrt{17}}$ and $\sin \beta=\frac{3}{\sqrt{34}}$, we have $$ \frac{\sin (\alpha+\beta)}{\frac{1}{\sqrt{17}} \cdot \frac{3}{\sqrt{34}}}=\frac{17}{3}, \quad \frac{17 \sqrt{2} \sin (\alpha+\beta)}{3}=\frac{17}{3} $$ from which $\sin (\alpha+\beta)=\frac{\sqrt{2}}{2}$. Hence $\alpha+\beta=\frac{\pi}{4}$ for $0<\alpha<\frac{\pi}{2}$ and $0<\beta<\frac{\pi}{2}$. Answer: $\alpha+\beta=\frac{\pi}{4}$." a078169ba954,"(Projective Algebra) Solve in $\mathbb{C}^{3}$ the system $$ \left\{\begin{array}{l} a^{2}+a b+c=0 \\ b^{2}+b c+a=0 \\ c^{2}+c a+b=0 \end{array}\right. $$ ## - Correction of exercises -",See reasoning trace,medium,"We must keep in mind Exercise 3: quadratic equations can be put in the form of a homography, which is also a projective transformation. We transform the first equation into $a=\frac{-c}{a+b-c}$. By setting $S=a+b+c$, we have $a=\frac{c}{c-S}=f(c)$ where $f$ is a projective transformation. Similarly, we find $b=f(a)$ and $c=f(b)$. Let's study $f \circ f \circ f$: $c^{\prime}$ is a projective transformation that has fixed points $A, B$, and $C$. However, a projective transformation has only two distinct fixed points, so $a=b$ without loss of generality. Thus, we can solve our equation and find the triplets $(0,0,0)$ and $\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$. ## 5 Pot-pourri ## 1 IMO 2019 ## - Day 1 -" 3294ceb5e31f,,See reasoning trace,easy,"Solution. In the school, there are $65\%$ boys, which means there are $30\%$ more boys than girls. If $30\%$ is 252 children, then $10\%$ is 84 children. Therefore, we conclude that there are 840 students in the school, of which 294 are female, and 546 are male." 6e2dd02fd5d4,"10. (10 points) As shown in the figure, in triangle $ABC$, $AD=2BD$, $AD=EC$, $BC=18$, the area of triangle $AFC$ is equal to the area of quadrilateral $DBEF$. What is the length of $AB$?",The length of $A B$ is 9,medium,"【Analysis】This problem can be approached by finding the lengths of $A D$ and $B D$ to determine the length of $A B$. Since $A D = E C$, we only need to find the length of $E C$ to know the length of $A D$. Given that the area of triangle $A F C$ is equal to the area of quadrilateral $D B E F$, we can find that $\frac{\mathrm{BE}}{\mathrm{EC}} = \frac{2}{1}$, which allows us to determine $E C = \frac{1}{3} B C = \frac{1}{3} \times 18 = 6$; then, using $A D = 2 B D$, we find that $2 B D = 6$, so $B D = 3$. Therefore, we can find the length of $A B$ and solve the problem. 【Solution】Solution: Let $S_{\triangle A F C} = S_{\text{quadrilateral } D B E F} = 1, S \triangle C E F = x, S \triangle A D F = y$, Since $A D = 2 B D$, we have $\frac{S \triangle_{\mathrm{ADC}}}{S_{\mathrm{SCBD}}} = \frac{\mathrm{AD}}{\mathrm{BD}} = \frac{2}{1}$, which means $\frac{S_{\triangle \mathrm{ADF}} + S_{\triangle \mathrm{AFC}}}{S \text{ quadrilateralDBEF} + S_{\triangle C E F}} = \frac{y + 1}{x + 1} = \frac{2}{1}$, Solving this, we get $y = 2x + 1$. Then, $\frac{S_{\triangle A B E}}{S_{\triangle A E C}} = \frac{S \triangle A D F + S \text{ quadrilateral } \mathrm{DAEF}}{S \triangle A F C + S \triangle E F C} = \frac{(2x + 1) + 1}{x + 1} = \frac{2}{1}$, So, $\frac{\mathrm{BE}}{\mathrm{EC}} = \frac{2}{1}$. Therefore, $E C = \frac{1}{1+2} \times B C = \frac{1}{3} \times 18 = 6$, and $A D = E C = 6$. Since $A D = 2 B D$, we have $B D = A D \div 2 = 6 \div 2 = 3$. In summary, $A B = A D + B D = 6 + 3 = 9$. Answer: The length of $A B$ is 9." c64051f6b17a,"(4) Given a regular tetrahedron $P-ABC$ with the side length of the equilateral triangle base being 1, and the center $O$ of its circumscribed sphere satisfies $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}$, then the volume of this regular tetrahedron is $\qquad$.",See reasoning trace,medium,"(4) Since $\overrightarrow{O C}=-(\overrightarrow{O A}+\overrightarrow{O B})$, point $C$ lies on the plane determined by points $O, A, B$, i.e., points $O, A, B, C$ are coplanar, thus point $O$ is on the base plane $A B C$. Also, since $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}$, point $O$ is the centroid of $\triangle A B C$. Since $\triangle A B C$ is an equilateral triangle, point $O$ is its center. As shown in the figure, connect $C O, P O$, then $$ P O=O C=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 1=\frac{\sqrt{3}}{3}, $$ Therefore, the volume of the regular tetrahedron is $$ \frac{1}{3} \times S_{\triangle A B C} \times P O=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 1^{2} \times \frac{\sqrt{3}}{3}=\frac{1}{12} . $$" 08e09b5f77f7,"2. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ be any two points on the graph of the function $f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}$, and $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$. Given that the x-coordinate of point $M$ is $\frac{1}{2}$, then the y-coordinate of point $M$ is $\qquad$.","\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$, $M$ is the midpoint of $A B$. Let $M(x, y)$",medium,"2. $\frac{1}{2}$ Analysis: Since $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$, $M$ is the midpoint of $A B$. Let $M(x, y)$. From $\frac{1}{2}\left(x_{1}+x_{2}\right)=x=\frac{1}{2}$, we get $x_{1}+x_{2}=1$. And $y_{1}+y_{2}=\frac{1}{2}+\log _{2} \frac{x_{1}}{1-x_{1}}+\frac{1}{2}+\log _{2} \frac{x_{2}}{1-x_{2}}=1+\log _{2}\left(\frac{x_{1}}{1-x_{1}} \cdot \frac{x_{2}}{1-x_{2}}\right)=1+$ $\log _{2} \frac{x_{1} \cdot x_{2}}{x_{1} \cdot x_{2}}=1$, so $y=\frac{y_{1}+y_{2}}{2}=\frac{1}{2}$." 8948472c48f9,"\section*{ Determine all natural numbers \(t\) for which \(\sqrt{t+24 \sqrt{t}}\) is rational!",See reasoning trace,medium,"} Let \(r^{2}=t+24 \sqrt{t}\) with a rational number \(r\). Then \(\sqrt{t}=\frac{r^{2}-t}{24}\) is also a rational number, since \(t\) is a natural number. The square root of a natural number \(t\) is rational if and only if \(t\) is a perfect square, i.e., there exists a natural number \(n\) such that \(t=n^{2}\). Then \(r^{2}=n^{2}+24 n\) is a natural number, and thus \(r\) is also a natural number. Therefore, \(r^{2}+144=n^{2}+2 \cdot n \cdot 12 + 12^{2}=(n+12)^{2}\), so with \(m:=n+12>0\), we have \(144=m^{2}-r^{2}=(m+r)(m-r)\). Since \(m>0\) and \(r>0\), \(m+r>0\) and thus \(m-r>0\), \(m+r\) and \(m-r\) are positive divisors and their respective counterparts of 144. Since \(r>0\), \(m+r\) is the larger and \(m-r\) is the smaller divisor, and since \((m+r)-(m-r)=2 r\), the two divisors differ by a multiple of 2. Since 144 is divisible by 2, both factors must be divisible by 2. Since 144 has exactly the numbers 2, 4, 6, and 8 as even divisors that are less than \(\sqrt{144}=12\), the following solutions arise: - For \(m-r=2\), \(m+r=72\), so \(r=35, m=37, n=m-12=25\), and \(t=n^{2}=625\). Indeed, \(\sqrt{625+24 \cdot \sqrt{625}}=\sqrt{625+24 \cdot 25}=\sqrt{1225}=35\) is rational. - For \(m-r=4\), \(m+r=36\), so \(r=16, m=20, n=m-12=8\), and \(t=n^{2}=64\). Indeed, \(\sqrt{64+24 \cdot \sqrt{64}}=\sqrt{64+24 \cdot 8}=\sqrt{256}=16\) is rational. - For \(m-r=6\), \(m+r=24\), so \(r=9, m=15, n=m-12=3\), and \(t=n^{2}=9\). Indeed, \(\sqrt{9+24 \sqrt{9}}=\sqrt{9+24 \cdot 3}=\sqrt{81}=9\) is rational. - And for \(m-r=8\), \(m+r=18\), so \(r=5, m=13\), and \(n=t=1\). Indeed, \(\sqrt{1+24 \cdot \sqrt{1}}=\sqrt{25}=5\) is rational. Thus, there are exactly four solutions \(t \in\{1,9,64,625\}\)." a1d9f22e298f,"4. (3 points) On the side $B C$ of triangle $A B C$ as a diameter, a circle with a radius of 20 cm is constructed. This circle intersects sides $A B$ and $A C$ at points $X$ and $Y$ respectively. Find $B X \cdot A B + C Y \cdot A C$.",\frac{2 A C^{2} - 2 A B^{2} + 3200}{4 A C}$. Then $B X \cdot A B + C Y \cdot A C = \frac{6400}{4} = ,medium,"Answer: 1600 Solution: Let point $M$ be the midpoint of side $B C$, which is also the center of the circle. Then $A B \cdot A X = A C \cdot A Y = (A M + 20)(A M - 20)$. Therefore, $A X = \frac{A M^{2} - 400}{A B} = \frac{2 A B^{2} + 2 A C^{2} - 1600 - 1600}{4 A B}$. Accordingly, $B X = A B - \frac{2 A B^{2} + 2 A C^{2} - 3200}{4 A B} = \frac{2 A B^{2} - 2 A C^{2} + 3200}{4 A B}$. Similarly, $C Y = \frac{2 A C^{2} - 2 A B^{2} + 3200}{4 A C}$. Then $B X \cdot A B + C Y \cdot A C = \frac{6400}{4} = 1600$." e99ddfbfea35,"If $ x,y$ are positive real numbers with sum $ 2a$, prove that : $ x^3y^3(x^2\plus{}y^2)^2 \leq 4a^{10}$ When does equality hold ? Babis",4a^{10,medium,"1. Let \( x + y = 2a \) and \( xy = b \). By the AM-GM inequality, we have: \[ x + y \geq 2\sqrt{xy} \implies 2a \geq 2\sqrt{b} \implies a^2 \geq b \] 2. We need to prove that: \[ x^3 y^3 (x^2 + y^2)^2 \leq 4a^{10} \] 3. Express \( x^3 y^3 \) and \( x^2 + y^2 \) in terms of \( a \) and \( b \): \[ x^3 y^3 = b^3 \] \[ x^2 + y^2 = (x + y)^2 - 2xy = 4a^2 - 2b \] 4. Substitute these into the inequality: \[ b^3 (4a^2 - 2b)^2 \leq 4a^{10} \] 5. Expand and simplify the left-hand side: \[ b^3 (4a^2 - 2b)^2 = b^3 (16a^4 - 16a^2 b + 4b^2) \] \[ = 16a^4 b^3 - 16a^2 b^4 + 4b^5 \] 6. We need to show: \[ 16a^4 b^3 - 16a^2 b^4 + 4b^5 \leq 4a^{10} \] Divide both sides by 4: \[ 4a^4 b^3 - 4a^2 b^4 + b^5 \leq a^{10} \] 7. Rearrange the inequality: \[ a^{10} \geq 4a^4 b^3 - 4a^2 b^4 + b^5 \] 8. By the AM-GM inequality, we have: \[ a^{10} + 3a^2 b^4 \geq 4a^4 b^3 \quad \text{(1)} \] \[ a^2 b^4 \geq b^5 \quad \text{(2)} \] 9. Combining (1) and (2): \[ a^{10} + 3a^2 b^4 \geq 4a^4 b^3 \] \[ a^2 b^4 \geq b^5 \] Adding these inequalities: \[ a^{10} + 4a^2 b^4 \geq 4a^4 b^3 + b^5 \] This completes the proof. Equality holds when \( x = y \). In this case, \( x = y = a \), and we have: \[ x^3 y^3 (x^2 + y^2)^2 = a^3 a^3 (a^2 + a^2)^2 = a^6 (2a^2)^2 = a^6 \cdot 4a^4 = 4a^{10} \] The final answer is \( \boxed{ 4a^{10} } \)." a1a60c7ffb5c,"How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$. $\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$",\textbf{(C),medium,"The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$, we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since: $acbd$ is not allowed because of the $cb$, and $acdb$ is not allowed because of the $cd$. We get the same problem if we start with a $d$, since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$. If we start with a $b$, the next letter would have to be a $d$, and since we can put an $a$ next to it and then a $c$ after that, this configuration works. The same approach applies if we start with a $c$. So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$, giving us: \[1 + 1 = \boxed{\textbf{(C)}\ 2}.\]" eb3eb20ab958,"A college math class has $N$ teaching assistants. It takes the teaching assistants $5$ hours to grade homework assignments. One day, another teaching assistant joins them in grading and all homework assignments take only $4$ hours to grade. Assuming everyone did the same amount of work, compute the number of hours it would take for $1$ teaching assistant to grade all the homework assignments.",20,medium,"1. Let \( W \) be the total amount of work required to grade all the homework assignments. 2. If there are \( N \) teaching assistants, and it takes them 5 hours to grade all the homework, then the work done by each teaching assistant per hour is: \[ \frac{W}{N \cdot 5} \] 3. When another teaching assistant joins, making it \( N+1 \) teaching assistants, it takes them 4 hours to grade all the homework. Thus, the work done by each teaching assistant per hour is: \[ \frac{W}{(N+1) \cdot 4} \] 4. Since the total work \( W \) remains the same, we can set up the equation: \[ \frac{W}{N \cdot 5} = \frac{W}{(N+1) \cdot 4} \] 5. Simplifying the equation by canceling \( W \) from both sides, we get: \[ \frac{1}{N \cdot 5} = \frac{1}{(N+1) \cdot 4} \] 6. Cross-multiplying to solve for \( N \): \[ 4N = 5(N + 1) \] 7. Expanding and simplifying: \[ 4N = 5N + 5 \implies 4N - 5N = 5 \implies -N = 5 \implies N = -5 \] 8. Since \( N \) must be a positive integer, we re-evaluate the equation: \[ 4N = 5(N + 1) \implies 4N = 5N + 5 \implies N = 5 \] 9. Therefore, there are \( N = 4 \) teaching assistants in the class. 10. If 4 teaching assistants take 5 hours to grade the homework, the total work \( W \) can be calculated as: \[ W = 4 \cdot 5 = 20 \text{ hours} \] 11. Thus, it would take one teaching assistant 20 hours to grade all the homework assignments. The final answer is \( \boxed{20} \) hours." 0ea097a91c7d,"Example 18. Solve the equation $$ \log _{x / 2} x^{2}-14 \log _{16 x} x^{3}+40 \log _{4 x} \sqrt{x}=0 $$",See reasoning trace,medium,"Solution. The first method. Let's find the domain of the equation (16). It is defined by the system of inequalities $$ x>0 ; \quad x \neq 1 / 16 ; \quad x \neq 1 / 4 ; \quad x \neq 2 $$ The equation (16) on this domain is equivalent to the equation $$ 2 \log _{x / 2} x-42 \log _{16 x} x+20 \log _{4 x} x=0 $$ It is easy to see that $x_{1}=1$ is a root of equation (17), and therefore, of equation (16). Now let $x$ belong to the domain and $x \neq 1$. For such values of $x$, equation (17) is equivalent to the equation $$ \frac{1}{\log _{x} \frac{x}{2}}-\frac{21}{\log _{x} 16 x}+\frac{10}{\log _{x} 4 x}=0 \Leftrightarrow $$ $$ \Leftrightarrow \frac{1}{1-\log _{x} 2}-\frac{21}{1+4 \log _{x} 2}+\frac{10}{1+2 \log _{x} 2}=0 $$ By setting $t=\log _{x} 2$, we arrive at the equation $$ \frac{1}{1-t}-\frac{21}{1+4 t}+\frac{10}{1+2 t}=0 $$ This equation is equivalent to the system $$ \left\{\begin{array}{l} (1+4 t)(1+2 t)-21(1-t)(1+2 t)+10(1-t)(1+4 t)=0 \\ (1-t)(1+4 t)(1+2 t) \neq 0 \end{array}\right. $$ i.e., the system $$ \left\{\begin{array}{l} (t+2)(t-1 / 2)=0 \\ (1-t)(1+4 t)(1+2 t) \neq 0 \end{array}\right. $$ This system has two solutions: $t_{1}=-2$ and $t_{2}=1 / 2$. Therefore, equation (16) on its domain and for $x \neq 1$ is equivalent to the system of two equations: $$ \log _{x} 2=-2, \quad \log _{x} 2=1 / 2 $$ From this system, we find the numbers $x_{2}=\sqrt{2} / 2$ and $x_{3}=4$, which are solutions to equation (16). All transitions were equivalent; therefore, the set of all solutions to equation (16) consists of three numbers: $$ x_{1}=1, \quad x_{2}=\sqrt{2} / 2, \quad x_{3}=4 $$ If the case $x=1$ had not been considered separately when transitioning from equation (16) to equation (18), the root $x=1$ would have been lost. Indeed, for $x=1$, $\log _{x / 2} x$ exists and equals 0, but the expression $\frac{1}{\log _{x}(x / 2)}$ is meaningless. Whenever it is necessary to apply the formula for changing the base, it is best to switch to a base equal to some number. Specific examples suggest which base to use. For example, in equation (16), it would be best to switch to base 2. The second method. The domain of the equation is defined by the system of inequalities: $$ x>0, \quad x \neq 1 / 16, \quad x \neq 1 / 4, \quad x \neq 2 $$ In equation (16), we switch to logarithms with base 2. On its domain, it is equivalent to the equation $$ \frac{2 \log _{2} x}{\log _{2} x-1}-\frac{42 \log _{2} x}{\log _{2} x+4}+\frac{20 \log _{2} x}{\log _{2} x+2}=0 $$ By setting $y=\log _{2} x$, we arrive at the equation $$ \frac{2 y}{y-1}-\frac{42 y}{y+4}+\frac{20 y}{y+2}=0 $$ which is equivalent to the system $$ \left[\begin{array}{l} y=0 \\ \frac{1}{y-1}-\frac{21}{y+4}+\frac{10}{y+2}=0 \end{array}\right. $$ The second equation of this system is equivalent to the equation $$ \begin{aligned} \frac{y+4-21(y-1)}{(y-1)(y+4)}+\frac{10}{y+2} & =0 \Leftrightarrow \frac{-2 y^{2}+3 y+2}{(y-1)(y+4)(y+2)}=0 \Leftrightarrow \\ & \Leftrightarrow \frac{-2(y-2)\left(y+\frac{1}{2}\right)}{(y-1)(y+2)(y+4)}=0 \Leftrightarrow\left[\begin{array}{l} y=-1 / 2, \\ y=2 . \end{array}\right. \end{aligned} $$ Thus, equation (16) on its domain is equivalent to the system of equations ![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-115.jpg?height=185&width=700&top_left_y=1701&top_left_x=675) The found numbers $x_{1}=1, x_{2}=\sqrt{2} / 2, x_{3}=4$ belong to the domain of equation (16); therefore, they constitute the set of all its solutions. The application of the main logarithmic identity $$ a^{\log a b}=b, \quad b>0, \quad a>0, \quad a \neq 1 $$ can lead to the appearance of extraneous roots if the conditions of its applicability are not monitored." a559bb61758c,"1. (7 points) The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the $2019-$th term of the sequence.",. 130,easy,"Solution. Let's find the first few terms of the sequence: $a_{1}=934 ; a_{2}=16 \cdot 13=208 ; a_{3}=10 \cdot 13=130 ; a_{4}=4 \cdot 13=52 ;$ $a_{5}=7 \cdot 13=91 ; a_{6}=10 \cdot 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a period of 3 from this point on. The number 2019 is a multiple of 3, so $a_{2019}=a_{3}=130$. Answer. 130." 890c74ada5fa,"14.18 Indicate all points $(x, y)$ in the $(x, y)$ plane that satisfy $[x]^{2}+[y]^{2}=4$. (7th Canadian Mathematics Competition, 1975)",See reasoning trace,easy,"[Solution] Since $[x]$ and $[y]$ are both integers, there are the following four possibilities: (1) $[x]=2, [y]=0$, i.e., $$ 2 \leqslant x<3, \quad 0 \leqslant y<1 ; $$ (2) $[x]=0, \quad [y]=2$, i.e., $$ 0 \leqslant x<1, \quad 2 \leqslant y<3 ; $$ (3) $[x]=-2, \quad [y]=0$, i.e., $$ -2 \leqslant x<-1, \quad 0 \leqslant y<1 ; $$ (4) $[x]=0, \quad [y]=-2$, i.e., $$ 0 \leqslant x<1, \quad -2 \leqslant y<-1 . $$ From this, we can obtain the shaded regions as shown in the figure on the right." 927d09027e0f,"Find all prime number $p$ such that the number of positive integer pair $(x,y)$ satisfy the following is not $29$. [list] [*]$1\le x,y\le 29$ [*]$29\mid y^2-x^p-26$ [/list]",2 \text{ and,medium,"1. **Understanding the problem**: We need to find all prime numbers \( p \) such that the number of positive integer pairs \((x, y)\) satisfying the conditions: - \(1 \leq x, y \leq 29\) - \(29 \mid y^2 - x^p - 26\) is not equal to 29. 2. **Analyzing the conditions**: The condition \(29 \mid y^2 - x^p - 26\) can be rewritten as: \[ y^2 \equiv x^p + 26 \pmod{29} \] This means that for each \( y \), \( y^2 - 26 \) must be a \( p \)-th power residue modulo 29. 3. **Using properties of prime numbers**: Since 29 is a prime number, we can use properties of residues and non-residues. Specifically, if \( p \nmid 28 \) (since \( 28 = 29 - 1 \)), then for each \( a \in \mathbb{Z}_{29} \), the equation \( x^p \equiv a \pmod{29} \) has a unique solution in \( \mathbb{Z}_{29} \). This implies that for each \( 1 \leq y \leq 29 \), there is exactly one \( x \in \{1, 2, \ldots, 29\} \) such that \( 29 \mid y^2 - x^p - 26 \). Hence, \( p \mid 28 \), i.e., \( p \in \{2, 7\} \). 4. **Checking \( p = 2 \)**: For \( p = 2 \), we need to check if \(-26\) is a quadratic residue modulo 29. Using the Legendre symbol: \[ \left( \frac{-26}{29} \right) = \left( \frac{3}{29} \right) = \left( \frac{29}{3} \right) = -1 \] Since \(\left( \frac{3}{29} \right) = -1\), \(-26\) is not a quadratic residue modulo 29. This means there are no solutions for \( y^2 \equiv -26 \pmod{29} \), and thus the number of pairs \((x, y)\) is not 29. 5. **Checking \( p = 7 \)**: For \( p = 7 \), we need to check the set of all elements of \( \mathbb{Z}_{29} \) expressible as \( x^7 \pmod{29} \). The set is \(\{0, 1, 12, 17, 28\}\), which can be verified by checking that \( 2^7 \equiv 12 \pmod{29} \) and the set, without the 0, contains \(\frac{29-1}{7} = 4\) elements. Since 26 is not in this set, for each \( z \in \{0, 1, 12, 17, 28\} \), the number of solutions to \( y^2 \equiv z + 26 \pmod{29} \) in terms of \( y \) (in \( \mathbb{Z}_{29} \)) is even. This implies that the number of pairs \((x, y)\) of solutions in this case is even, and thus not 29. 6. **Conclusion**: The only prime numbers \( p \) that satisfy the given conditions are \( p = 2 \) and \( p = 7 \). The final answer is \( \boxed{2 \text{ and } 7} \)" 1315ec1c739c,"The numbers $p$ and $q$ are prime and satisfy $$ \frac{p}{p+1}+\frac{q+1}{q}=\frac{2 n}{n+2} $$ for some positive integer $n$. Find all possible values of $q-p$. Origin. Luxembourg (Pierre Haas).",See reasoning trace,medium,"Rearranging the equation, $2 q n(p+1)=(n+2)(2 p q+p+q+1)$. The left hand side is even, so either $n+2$ or $p+q+1$ is even, so either $p=2$ or $q=2$ since $p$ and $q$ are prime, or $n$ is even. If $p=2,6 q n=(n+2)(5 q+3)$, so $(q-3)(n-10)=36$. Considering the divisors of 36 for which $q$ is prime, we find the possible solutions $(p, q, n)$ in this case are $(2,5,28)$ and $(2,7,19)$ (both of which satisfy the equation). If $q=2,4 n(p+1)=(n+2)(5 p+3)$, so $n=p n+10 p+6$, a contradiction since $n1$ if and only if $q \mid p+1$. Since $(\ell, \ell+1)$, we see that, if $q \nmid p+1$, then $\ell=p q+p+1$ and $\ell+1=q(p+1)$, so $q=p+2$ (and $\left(p, p+2,2\left(2 p^{2}+6 p+3\right)\right)$ satisfies the original equation). In the contrary case, suppose $p+1=r q$, so $\ell(p+1)=(\ell+1)(p+r)$, a contradiction since $\ell<\ell+1$ and $p+1 \leq p+r$. Thus the possible values of $q-p$ are 2,3 and 5." 224591e64836,"Task B-1.2. If $\left[\left(\frac{a}{b}-\frac{b}{a}\right):\left((a+b)\left(\frac{1}{b}-\frac{1}{a}\right)\right)-\frac{a}{b}\right]:\left(\frac{a}{b}+1\right)=\frac{1}{2014}$, what is $\frac{a}{b}$?",See reasoning trace,medium,"## Solution. First, we calculate the product $$ (a+b)\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{a}{b}-\frac{b}{a} $$ Then the given equation becomes $$ \left[\left(\frac{a}{b}-\frac{b}{a}\right):\left(\frac{a}{b}-\frac{b}{a}\right)-\frac{a}{b}\right]:\left(\frac{a}{b}+1\right)=\frac{1}{2014} $$ or equivalently $$ \frac{1-\frac{a}{b}}{\frac{a}{b}+1}=\frac{1}{2014} \quad\left(\text { or the form } \frac{b-a}{a+b}=\frac{1}{2014}\right) $$ It follows that $$ \begin{aligned} \frac{a}{b}+1 & =2014-2014 \cdot \frac{a}{b} \\ 2015 \cdot \frac{a}{b} & =2013 \\ \frac{a}{b} & =\frac{2013}{2015} \end{aligned} $$" 5b6a7d763930,"## Task B-4.1. In the set of complex numbers, solve the equation $$ (x+5)^{5}=x^{5}+5^{5} $$",See reasoning trace,medium,"## Solution. By applying the binomial theorem, the given equation can be written in the form $$ x^{5}+5 \cdot x^{4} \cdot 5+10 \cdot x^{3} \cdot 5^{2}+10 \cdot x^{2} \cdot 5^{3}+5 \cdot x \cdot 5^{4}+5^{5}=x^{5}+5^{5} $$ from which, after simplification, we obtain the equation $25 x^{4}+250 x^{3}+1250 x^{2}+3125 x=0$. By factoring the expression on the left side of the equation, we obtain the following sequence of equalities: $$ \begin{aligned} 25 x\left(x^{3}+10 x^{2}+50 x+125\right) & =0 \\ 25 x\left((x+5)\left(x^{2}-5 x+25\right)+10 x(x+5)\right) & =0 \\ 25 x(x+5)\left(x^{2}+5 x+25\right) & =0 \end{aligned} $$ From the last equation, we find that the solutions to the given equation are: $$ x_{1}=0, x_{2}=-5, x_{3}=\frac{-5+5 \sqrt{3} i}{2}, x_{4}=\frac{-5-5 \sqrt{3} i}{2} $$" 6ff30811b2d7,"39. If the solution of the equation $\frac{2 x-a}{x-1}+\frac{3}{1-x}=4$ with respect to $\mathrm{x}$ is non-negative, then the range of values for $a$ is",See reasoning trace,easy,"Answer: $a \leq 1$ and $a \neq -1$ Solution: $\frac{2 x-a}{x-1}+\frac{3}{1-x}=4$ can be transformed into $$ \frac{2 x-a-3}{x-1}=4, $$ Eliminating the denominator, we get $$ 2 x-a-3=4 x-4 \text {, } $$ Solving for $x$, we get $$ x=\frac{1-a}{2} \text {, } $$ Since the solution of the equation is non-negative, we have $$ \frac{1-a}{2} \geq 0 \text { and } \frac{1-a}{2} \neq 1, $$ Solving for $a$, we get $a \leq 1$ and $a \neq -1$." 0d391f289f1a,"4. The graph of the inverse function of $y=\frac{a-x}{x-a-1}$ is symmetric about the point $(-1,4)$. Then the real number $a$ equals ( ). A. 2 B. 3 C. -2 D. -3","\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, the center of symmetry is $((a+1),-1)$. The center of symme",easy,"4. B. Given $y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, the center of symmetry is $((a+1),-1)$. The center of symmetry of its inverse function is $(-1,(a+1)) \Rightarrow a+1=4 \Rightarrow a=3$." 3fdd972abe9c,"6. Given that the three sides of triangle $A B C$ are all positive integers, if $\angle A=2 \angle B$, and $C A=9$, then the minimum possible value of $B C$ is $\qquad$ .",\frac{9}{\sin B}=\frac{c}{\sin 3 B} \Rightarrow \frac{a}{2 \cos B}=\frac{c}{3-4 \sin ^{2} B}=9$ $\Ri,easy,"By the Law of Sines, $\frac{a}{\sin 2 B}=\frac{9}{\sin B}=\frac{c}{\sin 3 B} \Rightarrow \frac{a}{2 \cos B}=\frac{c}{3-4 \sin ^{2} B}=9$ $\Rightarrow\left\{\begin{array}{l}a=18 \cos B, \\ c=36 \cos ^{2} B-9\end{array} \Rightarrow \cos B>\frac{1}{2}\right.$ and the denominator of $\cos B$ is one of $2,3,6$, so $\cos B=\frac{2}{3}$ or $\frac{5}{6}$. Therefore, the minimum possible value of $B C$ is 12." bd955951e4f7,"6. Let's find the range of the function $f(x)=g\left(g^{2}(x)\right)$, where $g(x)=\frac{3}{x^{2}-4 x+5}$. The function $g(x)=\frac{3}{x^{2}-4 x+5}=\frac{3}{(x-2)^{2}+1}$ is defined for all real numbers and takes all values from the interval $(0 ; 3]$. The function $g(x)$ reaches its maximum value at the point $x=2$, $g_{\max }=g(2)=3$, on the interval $(-\infty ; 2)$ the function $g(x)$ is increasing, and on the interval $(2 ;+\infty)$ it is decreasing. The function $g^{2}(x)$ takes all values from the interval $(0 ; 9]$, since $t=g(x) \in(0 ; 3]$, and the function $t^{2}$ is increasing in this interval. To find the range of the function $f(x)=g\left(g^{2}(x)\right)$, it is sufficient to find the range of the function $g(x)$ on the interval $(0 ; 9]$. On this interval, $g(x)$ takes all values from the set $[g(9) ; g(2)]=[3 / 50 ; 3]$.",$E_{f}=\left[\frac{3}{50} ; 3\right]$,easy,Answer: $E_{f}=\left[\frac{3}{50} ; 3\right]$ b9164a14634f,"## Calculate the limit of the function: $\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}$",\sqrt{0+4 \cos 0}=\sqrt{4 \cdot 1}=2$,easy,"## Solution Since $2+\sin \left(\frac{1}{x}\right)_{\text { is bounded, then }}$ $x\left(2+\sin \left(\frac{1}{x}\right)\right) \rightarrow 0 \quad$, as $x \rightarrow 0$ Then: $\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}=\sqrt{0+4 \cos 0}=\sqrt{4 \cdot 1}=2$" c96a6ad3bb51,"Question 102: Let the complex number $z$ satisfy $z+\frac{1}{z} \in[1,2]$, then the minimum value of the real part of $z$ is $\qquad$ Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.","According to the condition, $\operatorname{Im}(z)=-\operatorname{Im}\left(\frac{1}{z}\right)$",medium,"Question 102, Answer: According to the condition, $\operatorname{Im}(z)=-\operatorname{Im}\left(\frac{1}{z}\right)$. And $\frac{1}{z}=\frac{z}{|z|^{2}}$, so either $|z|=1$, or $z \in$ R. If $z \in R$, since $1 \leq z+\frac{1}{z} \leq 2$, it can only be $z=1$, that is, $\operatorname{Re}(z)=z=1$; if $|z|=1$, then $z+\frac{1}{z}=2 \operatorname{Re}(z) \in[1,2]$, so $\operatorname{Re}(z) \geq \frac{1}{2}$. When $z=\frac{-1+\sqrt{3}}{2}$, it satisfies the condition, and $\operatorname{Re}(z)=\frac{1}{2}$." 43ecb882efa2,"Solve the following equation: $$ \sqrt[3]{x}+\sqrt[3]{1-x}=\frac{3}{2} $$","\frac{1}{2}$, but this is clearly not a root of (1'), which has no roots. (F. T.)",medium,"By cubing both sides of the equation and using the identity $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$, we get $$ \sqrt[3]{x(1-x)}(\sqrt[3]{x}+\sqrt[3]{1-x})=\frac{19}{24} $$ Here, according to the original equation, $\sqrt[3]{x}+\sqrt[3]{1-x}=\frac{3}{2}$. Substituting this into (2), we get $$ \frac{3}{2} \sqrt[3]{x(1-x)}=\frac{19}{24} $$ From this, after another cubing and rearrangement, we obtain the equation $$ x^{2}-x+\left(\frac{19}{36}\right)^{3}=0 $$ The roots of this equation are: $$ \begin{aligned} & x_{1}=\frac{1}{2}+\frac{31 \sqrt{5}}{216} \approx 0.8209 \\ & x_{2}=\frac{1}{2}-\frac{31 \sqrt{5}}{216} \approx 0.1791 \end{aligned} $$ It is clear that the roots of (1) are also roots of (4), so if (1) has roots, they can only be $x_{1}$ or $x_{2}$—or possibly both. However, based on the steps so far, we cannot yet conclude that $x_{1}$ and $x_{2}$ are roots of (1), since we substituted (1) into (2). Thus, the fact that, for example, $x_{1}$ is a root of (4), and thus of the equivalent (3), would imply that $x_{1}$ is a root of (2), and thus of (1), only if we already knew that $\sqrt[3]{x_{1}}+\sqrt[3]{1-x_{1}}=\frac{3}{2}$. The reverse implication would only follow from the fact that $x_{1}$ is a root of (1) if we already knew that $x_{1}$ is a root of (1). This is a circular argument. Recognizing that $x_{1}=\left(\frac{9+\sqrt{5}}{12}\right)^{3}$ and $x_{2}=1-x_{1}=\left(\frac{9-\sqrt{5}}{12}\right)^{3}$, the verification of the roots can be done by substitution: $$ \sqrt[3]{x_{1}}+\sqrt[3]{1-x_{1}}=\sqrt[3]{x_{2}}+\sqrt[3]{1-x_{2}}=\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}}=\frac{9+\sqrt{5}}{12}+\frac{9-\sqrt{5}}{12}=\frac{3}{2} $$ Thus, the roots of the equation are: $\frac{1}{2}+\frac{31 \sqrt{5}}{216}$ and $\frac{1}{2}-\frac{31 \sqrt{5}}{216}$. Remarks. 1. Many incomplete solutions were submitted for the problem. Almost every solver considered the substitution of $\sqrt[3]{x}+\sqrt[3]{1-x}$ as an equivalent transformation. Another common mistake occurred during the verification of the roots. Many used approximate values, but even if their result approximated $\frac{3}{2}$ to several significant digits, nothing can be concluded from this. 2. The following example shows that the ""critical"" transformation is not necessarily equivalent. It is not just a theoretical error that could lead to a mistake! If we start from the equation $$ \sqrt[3]{x}+\sqrt[3]{1-x}=\sqrt[3]{-\frac{1}{2}} $$ then the method followed in the solution now leads to the quadratic equation $$ 4 x^{2}-4 x+1=0 $$ The root of this equation is $x=\frac{1}{2}$, but this is clearly not a root of (1'), which has no roots. (F. T.)" 7a2afbef27ba,"Example 7 Given $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, and satisfies the conditions: (1) For any $x, y \in \mathbf{R}_{+}$, $f(x f(y))=y f(x)$; (2) As $x \rightarrow+\infty$, $f(x) \rightarrow 0$. Try to find the function $f(x)$.",See reasoning trace,medium,"Solution: Let $x=y$, then we have $f(x f(x))=x f(x)$. Obviously, $x f(x)$ is a fixed point of $f(x)$, and the set of fixed points is $\left\{x f(x) \mid x \in \mathbf{R}_{+}\right\}$. Now let $x=y=1$, substituting into condition (1) we get $f(f(1))=f(1)$. Next, let $x=1, y=f(1)$, substituting into the original equation we get $f(f(f(1)))=f^{2}(1)$. Therefore, $f^{2}(1)=f(1)$. Hence $f(1)=1$ (we discard $f(1)=0$), which means $x=1$ is a fixed point of $f(x)$. Thus, $x f(x)=1$. So, $f(x)=\frac{1}{x}\left(x \in \mathbf{R}_{+}\right)$. The following is a proof by contradiction that $x=1$ is the only fixed point of $f(x)$. Assume there exists $\alpha \neq 1$ and $\alpha=f(\alpha)$. (1) If $\alpha>1$, by $f(x f(x))=x f(x)$, let $x=\alpha$, we get $f(\alpha f(\alpha))=\alpha f(\alpha)$. Hence $f\left(\alpha^{2}\right)=\alpha^{2}$. Therefore, $f\left(\alpha^{4}\right)=\alpha^{4}, \cdots, f\left(\alpha^{2 n}\right)=\alpha^{2 n}$. This contradicts $x \rightarrow+\infty, f(x) \rightarrow 0$. (2) If $0<\alpha<1$, Similarly, we derive a contradiction. Thus, $f(x)$ has only one fixed point." 7fc2ea2a9ce1,"[ equations in integers ] [ Decimal number system ] a) In a three-digit number, the first digit from the left was erased, then the resulting two-digit number was multiplied by 7 and the original three-digit number was obtained. Find such a number. b) In a three-digit number, the middle digit was erased and the resulting number was 6 times smaller than the original. Find such a three-digit number. #","6(10 x + z)$, from which $8 x + 2 y = z$. The last equality is true only when $x = 1, y = 0, z = 8$.",easy,"Let $x, y, z$ be the digits of the desired number. a) $100 x + 10 y + z = 7(10 y + z)$, from which $50 x = 3(10 y + z)$. Therefore, $x$ is divisible by 3. Since $3(10 y + z) < 300$, then $x = 3, 10 y + z = 50$. b) $100 x + 10 y + z = 6(10 x + z)$, from which $8 x + 2 y = z$. The last equality is true only when $x = 1, y = 0, z = 8$." 9824fc7366d7,"$28 \cdot 5$ A cube is formed by welding 12 pieces of wire, each 3 inches long. A fly first lands on one vertex, then crawls along the wires. If it does not turn back and does not walk on the same path, the maximum distance it can crawl to return to the original landing point is (A) 24 inches. (B) 12 inches. (C) 30 inches. (D) 18 inches. (E) 36 inches. (8th American High School Mathematics Examination, 1957)",$(A)$,medium,"［Solution］As shown in the figure, the fly can travel along a maximum of eight edges, so the maximum distance it can crawl is $3 \cdot 8=24$ (inches). Otherwise, if the fly could travel along more than eight edges, then since each edge has two endpoints, the number of endpoints would be more than 16. However, a cube has 8 vertices, and each vertex is the intersection of three edges, so there must be a vertex from which the fly has crawled along all three edges, which is impossible. Therefore, the answer is $(A)$." a9c4134a4aee,"Find the largest positive real $ k$, such that for any positive reals $ a,b,c,d$, there is always: \[ (a\plus{}b\plus{}c) \left[ 3^4(a\plus{}b\plus{}c\plus{}d)^5 \plus{} 2^4(a\plus{}b\plus{}c\plus{}2d)^5 \right] \geq kabcd^3\]",k = 174960,medium,"1. We start by considering the given inequality: \[ (a+b+c) \left[ 3^4(a+b+c+d)^5 + 2^4(a+b+c+2d)^5 \right] \geq kabcd^3 \] We need to find the largest positive real \( k \) such that this inequality holds for any positive reals \( a, b, c, d \). 2. Let's test the inequality with specific values. Set \( a = b = c = \frac{1}{3} \) and \( d = 1 \). Then: \[ a + b + c = 1 \] Substituting these values into the inequality, we get: \[ 1 \left[ 3^4(1+1)^5 + 2^4(1+2)^5 \right] \geq k \left( \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 1^3 \right) \] Simplifying the right-hand side: \[ k \left( \frac{1}{27} \right) = \frac{k}{27} \] 3. Now, we simplify the left-hand side: \[ 3^4 \cdot 2^5 + 2^4 \cdot 3^5 = 81 \cdot 32 + 16 \cdot 243 \] Calculating each term: \[ 81 \cdot 32 = 2592 \] \[ 16 \cdot 243 = 3888 \] Adding these together: \[ 2592 + 3888 = 6480 \] 4. Therefore, the inequality becomes: \[ 6480 \geq \frac{k}{27} \] Solving for \( k \): \[ k \leq 6480 \cdot 27 = 174960 \] 5. To prove that this value of \( k \) is indeed the maximum, we use the AM-GM inequality. Assume \( a + b + c = 1 \). By the AM-GM inequality: \[ 3^4(1+d)^5 + 2^4(1+2d)^5 \geq 3^4(2\sqrt{d})^5 + 2^4(3\sqrt[3]{d^2})^5 \] Simplifying the right-hand side: \[ 3^4 \cdot 32d^{5/2} + 2^4 \cdot 243d^{10/3} \] Using the weighted AM-GM inequality: \[ 6^4 \cdot 5 \left( \frac{2}{5}d^{5/2} + \frac{3}{5}d^{10/3} \right) \geq 6^4 \cdot 5 \cdot d^3 \] Since \( a + b + c = 1 \), we have: \[ 6^4 \cdot 5 \cdot d^3 \geq 6^4 \cdot 5 \cdot \frac{abc}{27} d^3 \] Therefore: \[ k = \frac{6^4 \cdot 5}{27} = 174960 \] The final answer is \( \boxed{ k = 174960 } \)" 7a18e1c7d738,"Task 2. Determine all digits $a, b$ for which the number $\overline{a 13 b}$ is divisible by 45.","0$, then $\overline{a 130}$ is divisible by 9 if $a+1+3+0=a+4$ is divisible by 9 and $a=5$, so the d",easy,"Solution. The number $\overline{a 13 b}$ is divisible by 45 if it is divisible by 9 and 5. This means the digit $b$ can be 0 or 5. If $b=0$, then $\overline{a 130}$ is divisible by 9 if $a+1+3+0=a+4$ is divisible by 9 and $a=5$, so the desired number is 5130. If, however, $b=5$, then $\overline{a 135}$ is divisible by 9 if $a+1+3+5=a+9$. But $a \neq 0$, so $a=9$, i.e., the desired number is 9135." 0ca5ff1f9d43,"| How many different seven-digit telephone numbers exist (assuming that a number cannot start with zero)? #",$9 \cdot 10^{6}$ numbers,easy,"The first digit can be chosen in 9 ways, and each of the six remaining digits can be chosen in 10 ways. In total, $9 \cdot 10^{6}$ ways. ## Answer $9 \cdot 10^{6}$ numbers." 3a11767374bc,"7. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$. If for any $x \in[a, a+2]$, the inequality $f(x+a) \geqslant 2 f(x)$ always holds, then the range of the real number $a$ is $\qquad$",See reasoning trace,medium,"Solution: $[\sqrt{2},+\infty)$. From the problem, we know that $f(x)=\left\{\begin{array}{l}x^{2}(x \geqslant 0), \\ -x^{2}(x<0),\end{array}\right.$, then $2 f(x)=f(\sqrt{2} x)$. Therefore, the original inequality is equivalent to $f(x+a) \geqslant f(\sqrt{2} x)$. Since $f(x)$ is an increasing function on $\mathbf{R}$, we have $x+a \geqslant \sqrt{2} x$, which means $a \geqslant(\sqrt{2}-1) x$. Also, $x \in[a, a+2]$. Therefore, when $x=a+2$, $(\sqrt{2}-1) x$ reaches its maximum value of $(\sqrt{2}-1)(a+2)$. Thus, $a \geqslant(\sqrt{2}-1)(a+2)$, solving this gives $a \geqslant \sqrt{2}$. Hence, the range of values for $a$ is $[\sqrt{2},+\infty)$." 4072051a8e03,"2.1.1 * Let $x \in\left(-\frac{1}{2}, 0\right)$, then the size relationship of $a_{1}=\cos (\sin x \pi), a_{2}=\sin (\cos x \pi), a_{3}=$ $\cos (x+1) \pi$ is (). (A) $a_{3}0)$, draw two tangent lines $R A$ and $R B$ from the point $R(1,-1)$ to the parabola $C$, with the points of tangency being $A$ and $B$. Find the minimum value of the area of $\triangle R A B$ as $p$ varies.","0$. By combining the line $A B$ with the parabola equation, we get: $x^{2}-2 x-2 p=0$. By Vieta's fo",easy,"9. Analysis: It is easy to know that the equation of the chord of tangents $A B$ corresponding to $R$ is: $p y-x-p=0$. By combining the line $A B$ with the parabola equation, we get: $x^{2}-2 x-2 p=0$. By Vieta's formulas: $x_{A}+x_{B}=2, x_{A} \cdot x_{B}=-2 p$, hence we know $S_{\mathrm{VR} A B}=\frac{|1+2 p| \cdot\left|x_{A}-x_{B}\right|}{2}=\frac{|1+2 p| \cdot \sqrt{1+2 p}}{p} \geq 3 \sqrt{3}$" 3e68c2baf829,"Hagrid has 100 animals. Among these animals, - each is either striped or spotted but not both, - each has either wings or horns but not both, - there are 28 striped animals with wings, - there are 62 spotted animals, and - there are 36 animals with horns. How many of Hagrid's spotted animals have horns? (A) 8 (B) 10 (C) 2 (D) 38 (E) 26",(E),easy,"Each of the animals is either striped or spotted, but not both. Since there are 100 animals and 62 are spotted, then there are $100-62=38$ striped animals. Each striped animal must have wings or a horn, but not both. Since there are 28 striped animals with wings, then there are $38-28=10$ striped animals with horns. Each animal with a horn must be either striped or spotted. Since there are 36 animals with horns, then there are $36-10=26$ spotted animals with horns. ANSWER: (E)" d6ad0722da47,"1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{4}+3 x+\frac{253}{4}$.",11,easy,"Answer: 11. Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+3 x+\frac{253}{4}>0 \Leftrightarrow-\frac{1}{4}(x+11)(x-23)>0$, from which $-11 0)$, then $-(4-h) = -\frac{1}{25} x^{2}$. Therefore, $d = 2x = 10 \sqrt{4-h}$. (2) When the width of the water surface under the bridge is 18 meters, $18 = 10 \sqrt{4-h}$, $h = 0.76$. That is, when the water depth under the bridge exceeds 2.76 meters, it will affect the navigation of passing vessels under the bridge. !Comment This problem involves knowledge of parabolas in analytic geometry. The following problem is also related to analytic geometry and involves the knowledge of ellipses." 0674a09cfa00,"Question 144, As shown in the figure, plane \(ABDE \perp\) plane \(\mathrm{ABC}\), \(\triangle \mathrm{ABC}\) is an isosceles right triangle, \(A C = B C = 4\), quadrilateral \(\mathrm{ABDE}\) is a right trapezoid, \(\mathrm{BD} / / \mathrm{AE}, \mathrm{BD} \perp \mathrm{AB}, \mathrm{BD} = 2, \mathrm{AE} = 4\), points \(O\) and \(\mathrm{M}\) are the midpoints of \(\mathrm{CE}\) and \(\mathrm{AB}\) respectively. Find the sine value of the angle formed by line \(\mathrm{CD}\) and plane \(\mathrm{ODM}\).","2 \sqrt{5}$, the sine value of the angle formed by the line $C D$ and the plane $O D M$ is $\frac{\s",medium,"Question 144, Solution: As shown in the figure, according to the conditions, we can supplement the figure to form a cube EFGH$A N B C$ with a side length of 4. Taking the midpoint of $A C$ as $K$, then $O K$ is parallel and equal to $B D$, so quadrilateral $O D B K$ is a rectangle. Taking the midpoint of $A K$ as $\mathrm{B}_{2}$, then $\mathrm{ML} / / \mathrm{BK} / / 0 \mathrm{D}$. Since $S_{\triangle O L C}=\frac{L C \cdot O K}{2}=\frac{3 \times 2}{2}=3$, and the distance from $D$ to the plane $O L C$ is 4, thus $\mathrm{CO}-\mathrm{OLD}=\frac{3 \times 4}{3}=4$. Also, $\mathrm{OL}=\sqrt{5}, \mathrm{OD}=2 \sqrt{5}, \mathrm{LD}^{2}=\sqrt{29}$, so according to Heron's formula, $\mathrm{S}_{\triangle \mathrm{OLD}}=2 \sqrt{6}$. Therefore, we have $$ 4=V_{C-O L D}=\frac{2 \sqrt{6} \cdot d(C, O L D)}{3} \Rightarrow d(C, O L D)=\sqrt{6} $$ Noting that $C D=2 \sqrt{5}$, the sine value of the angle formed by the line $C D$ and the plane $O D M$ is $\frac{\sqrt{6}}{2 \sqrt{5}}=\frac{\sqrt{30}}{10}$." 0226dc3ddb5f,"Example 1 A city has $n$ high schools, the $i$-th high school sends $c_{i}$ students $\left(1 \leqslant c_{i} \leqslant 39\right)$ to watch a ball game at the gymnasium, where $\sum_{i=1}^{n} c_{i}=1990$. Each row in the stand has 199 seats, and it is required that students from the same school sit in the same row. Question: What is the minimum number of rows that need to be arranged so that all students can definitely be seated?",See reasoning trace,hard,"Analysis and Solution First, consider what the worst-case scenario is. The worst-case scenario refers to the situation where the number of empty seats in each row is maximized. Clearly, if the number of participants from each school varies, it is easier to arrange, as the remaining seats can be filled by students from schools with fewer participants. Therefore, a worse scenario is when each school sends a relatively ""uniform"" and larger number of participants. Thus, we can first assume that all schools send the same number of participants. Suppose each school sends $r$ people, we examine what value of $r$ would maximize the number of empty seats in each row. The estimation is listed as follows: \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline Number of participants $r$ from each school & 39 & 38 & 37 & 36 & 35 & 34 & 33 & 32 & 31 & 30 & 29 & $r \leqslant 28$ \\ \hline Number of schools that can be arranged in each row & 5 & 5 & 5 & 5 & 5 & 5 & 6 & 6 & 6 & 6 & 6 & $\ldots$ \\ \hline Number of empty seats $t$ in each row & 4 & 9 & 14 & 19 & 24 & 29 & 1 & 7 & 13 & 19 & 25 & $t \leqslant 28$ \\ \hline \end{tabular} From the table, we can see that when each school sends 34 participants, the number of empty seats in each row is maximized at 29. Note that $1990=34 \times 58+18$. Therefore, we first examine the scenario where 58 schools each send 34 participants, and one school sends 18 participants, to see how many rows are needed. In this case, each row can accommodate a maximum of 6 schools, and the row that accommodates the school sending 18 participants can only be one row. Thus, the minimum number of rows required is $1+\left[\frac{58-6}{5}\right]+1=12$. Finally, we prove that 12 rows are sufficient for any scenario. The most natural arrangement method is as follows: First, arrange the first row so that the number of empty seats $x_{1}$ in the first row is minimized. Then, arrange the second row: again, minimize the number of empty seats $x_{2}$. Clearly, $x_{1} \leqslant x_{2}$. Otherwise, if $x_{1}>x_{2}$, moving the students from the second row to the first row would reduce the number of empty seats in the first row, contradicting the minimality of $x_{1}$. Continue this process until the 11th row, making the number of empty seats $x_{11}$ in the 11th row the smallest. We prove that with such an arrangement, all remaining students can definitely be arranged in the 12th row. $\square$ In fact, suppose there are still $x$ people left unseated after arranging 12 rows, then $x>x_{12}$. Otherwise, if $x \leq x_{12}$, these $x$ people could be seated in the 12th row. Therefore, $x=1990-\sum_{i=1}^{12}\left(199-x_{i}\right) \geqslant x_{12}$, which implies $x_{1}+x_{2}+\cdots+x_{11}>398$, so $39810\) for which the number \(2^p\) also ends in the digits ...024! Explain why the program you have formulated solves this Hint: Note that for the numbers involved in the calculations, there are limitations on the number of digits when using typical computer usage.",See reasoning trace,medium,"} A BASIC program of the required type is, for example: \(10 \mathrm{P}=10\) \(20 \mathrm{Z}=24\) \(30 \mathrm{P}=\mathrm{P}+1\) \(40 \mathrm{Z}=\mathrm{Z} * 2\) 50 IF \(\mathrm{Z}>999\) THEN \(\mathrm{Z}=\mathrm{Z}-1000\) 60 IF \(\mathrm{Z}\) <> 24 THEN GOTO 30 70 PRINT P For values of the exponent \(p\), the last three digits of the power \(2^{p}\) are formed as an integer \(z\) with \(0 \leq z \leq 999\). Starting from the initial values \(p=10, z=024\) (lines 10, 20), the next values are found step by step: In each step, \(p\) is incremented by 1 (line 30) and \(z\) is doubled (line 40), and if a non-three-digit value initially results, only its last three digits are retained. This can be achieved by subtracting 1000 (line 50); because if for the previous value of \(z\) \(0 \leq z<1000\) held, then the initially resulting value \(2 \cdot z<2000\), and if for it \(1000 \leq 2 \cdot z\), the value resulting from subtracting 1000 again satisfies \(0 \leq 2 \cdot z - 1000 < 1000\). By this step-by-step reduction, the large numbers \(2^{p}\) are avoided, as required by the ""hint"". These steps are repeated until the sequence \(z=024\) is not reached again (line 60). Otherwise, the process ends with the output of the desired exponent \(p\) (line 70). The end must be reached (no ""infinite loop"" occurs). This can be established as a result of a ""risky test run"" (and thus find the desired exponent \(p=110\)); it can also be proven that for every \(p \geq 3\), the sequence of the last three digits of \(2^{p}\) must recur at a larger \(p\). Solutions of the 1st Round 1993 adopted from [5] \subsection*{6.35.2 2nd Round 1993, Class 9}" 5dd011cdcc4d,"Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real numbers $x$ and $y$, $$ f(x+y)=\max (f(x), y)+\min (f(y), x) . $$","is the identity function $f(x) \equiv x$, which clearly works",easy,"We claim the only answer is the identity function $f(x) \equiv x$, which clearly works. Let $(x, y)=(a, 0)$ and $(x, y)=(0, a)$ so that $$ \begin{aligned} & f(a)=\max (f(a), 0)+\min (f(0), a) \\ & f(a)=\max (f(0), a)+\min (f(a), 0) . \end{aligned} $$ Sum: $$ 2 f(a)=(f(a)+0)+(f(0)+a) $$ so $f(a)=a+f(0)$. It is easy to check from here $f$ is identity." 94fd2c62c832,2. Solve the equation $2 \log _{3} \operatorname{ctg} x=\log _{2} \cos x$.,"$x=\frac{\pi}{3}+2 \pi k, k \in \mathbb{Z}$",medium,"Solution 1. Rewrite the equation as $$ \log _{3}(\operatorname{ctg} x)=\log _{4}(\cos x)=t $$ Then $\cos x=4^{t}, \operatorname{ctg} x=3^{t}$ and $\sin x=\left(\frac{4}{3}\right)^{t}$, so $4^{2 t}+\left(\frac{4}{3}\right)^{2 t}=1$. The function on the right side of the obtained equation is increasing, so it has no more than one solution. It is clear that the solution is $t=-\frac{1}{2}$, from which $x=\frac{\pi}{3}+2 \pi k, k \in \mathbb{Z},($ since $\sin x>0)$. Answer: $x=\frac{\pi}{3}+2 \pi k, k \in \mathbb{Z}$." 96de4439c437,"How many numbers from the set $\{-5,-4,-3,-2,-1,0,1,2,3\}$ satisfy the inequality $-3 x^{2}<-14$ ? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5",(D),medium,"Solution 1 If $-3 x^{2}14$ or $x^{2}>\frac{14}{3}=4 \frac{2}{3}$. Since we are only looking at $x$ being a whole number, then $x^{2}$ is also a whole number. Since $x^{2}$ is a whole number and $x^{2}$ is greater than $4 \frac{2}{3}$, then $x^{2}$ must be at least 5 . Of the numbers in the set, the ones which satisfy this condition are $-5,-4,-3$, and 3 . ## Solution 2 For each number in the set, we substitute the number for $x$ and calculate $-3 x^{2}$ : | $x$ | $-3 x^{2}$ | | :---: | :---: | | -5 | -75 | | -4 | -48 | | -3 | -27 | | -2 | -12 | | -1 | -3 | | 0 | -3 | | 0 | -12 | As before, the four numbers $-5,-4,-3$, and 3 satisfy the inequality. ANSWER: (D)" ac9aad9aad3f,"9. Find the 12-combination number of the multiset $S=\{4 \cdot a, 3 \cdot b, 4 \cdot c, 5 \cdot d\}$.",See reasoning trace,easy,"9. 34. Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 2e9b15542949,Suppose the side lengths of triangle $ABC$ are the roots of polynomial $x^3 - 27x^2 + 222x - 540$. What is the product of its inradius and circumradius?,10,medium,"1. **Identify the roots of the polynomial:** The polynomial given is \(x^3 - 27x^2 + 222x - 540\). The roots of this polynomial are the side lengths \(a\), \(b\), and \(c\) of the triangle \(ABC\). 2. **Sum and product of the roots:** By Vieta's formulas, for the polynomial \(x^3 - 27x^2 + 222x - 540\): - The sum of the roots \(a + b + c = 27\). - The product of the roots \(abc = 540\). 3. **Inradius and circumradius formulas:** - The inradius \(r\) of a triangle is given by: \[ r = \frac{\text{Area}}{s} \] where \(s\) is the semi-perimeter of the triangle, \(s = \frac{a + b + c}{2}\). - The circumradius \(R\) of a triangle is given by: \[ R = \frac{abc}{4 \cdot \text{Area}} \] 4. **Product of inradius and circumradius:** - The product of the inradius and circumradius is: \[ r \cdot R = \left(\frac{\text{Area}}{s}\right) \cdot \left(\frac{abc}{4 \cdot \text{Area}}\right) \] Simplifying this expression: \[ r \cdot R = \frac{\text{Area} \cdot abc}{s \cdot 4 \cdot \text{Area}} = \frac{abc}{4s} \] 5. **Substitute the known values:** - The semi-perimeter \(s\) is: \[ s = \frac{a + b + c}{2} = \frac{27}{2} \] - The product of the roots \(abc = 540\). 6. **Calculate the product \(r \cdot R\):** \[ r \cdot R = \frac{abc}{4s} = \frac{540}{4 \cdot \frac{27}{2}} = \frac{540}{54} = 10 \] The final answer is \(\boxed{10}\)." d3cc1112551c,"3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$",126$.,medium,"3. 126. First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls. Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa. Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$." de6341b48251,,all the questions on the ticket,medium,"Solution. Consider the events: $A_{1}$ - the student knows the first question; $A^{2}$ - the student knows the second question; $A_{3}$ - the student will solve the problem; $A=A_{1} A_{2}$ - the student knows both questions; Then $A A_{3}$ - the student will answer all the questions on the ticket. Events $A_{1}$ and $A_{2}$ are dependent, $A$ and $A_{3}$ are independent. We have $p\left(A_{1}\right)=\frac{50}{56} ; p\left(A_{2} / A_{1}\right)=\frac{49}{55} ; p\left(A_{3}\right)=\frac{22}{28}=\frac{11}{14}$. By formulas (10) and (9) we get: $$ \begin{aligned} p\left(A A_{3}\right)=p(A) p\left(A_{3}\right)=p\left(A_{1}\right) p\left(A_{2} / A_{1}\right) p\left(A_{3}\right) & = \\ & =\frac{50}{56} \cdot \frac{49}{55} \cdot \frac{22}{28}=\frac{5}{8}=0.625 \end{aligned} $$ Answer. $0.625$. ## Exercises 1. A salvo is fired from two guns at a target. The probability of hitting with the first gun is 0.85; with the second 0.91. Find the probability of hitting the target. 2. In a lottery, 24 monetary and 10 material prizes are distributed among 1000 tickets. What is the probability of winning with two tickets: a) at least one ticket; b) one monetary prize and one material prize? 3. A company produces $95\%$ of standard products, of which $86\%$ are of the first grade. Find the probability that a randomly selected product, produced by this company, will be of the first grade. 4. In an urn, there is 1 black and 9 white balls. Three balls are drawn at once. What is the probability that all the drawn balls are white? 5. Three shots are fired at a target. The probability of hitting with each shot is 0.5. Find the probability that only one hit will occur as a result of these shots. 6. In the first urn, there is 1 white, 2 red, and 3 blue balls. In the second urn, there are 2 white, 6 red, and 4 blue balls. One ball is drawn from each urn. What is the probability that none of them are blue? 7. Among 20 electric bulbs, 3 are non-standard. Find the probability that two bulbs taken sequentially will be non-standard. 8. A student intends to take a needed book from one of two libraries. The probability that the book is in the first library is 0.75; in the second 0.85. Find the probability of the events: a) the book will be found only in one library; b) at least in one library. 9. Two shooters fire at a target. The probability of hitting with one shot for the first shooter is 0.75; for the second 0.85. Find the probability that in one salvo at the target: a) only one of the shooters will hit; b) only the first shooter will hit; c) both shooters will hit; d) neither will hit. 10. In an urn, there are 9 white and 3 red balls. Three balls are drawn sequentially without replacement. Find the probability that all the balls are white. 11. Four telephone sets reliably operate during the day with probabilities of 0.3, 0.4, 0.4, and 0.7, respectively. Find the probability that during the day: a) all 4 sets will operate reliably; b) only two sets will operate reliably; c) only one set will operate reliably; d) none will operate reliably. ## Answers 1. 0.9865. 2. a) 0.067; b) 0.00024. 3. 0.817. 4. 0.7. 5. 0.375. 6. $\frac{1}{3}$. 7. $\frac{3}{190}$. 8. a) 0.325; b) 0.9625. 9. a) 0.325; b) 0.1125; c) 0.6375; d) 0.0375. 10. $\frac{21}{35}=0.3818$. 11. a) 0.0336; b) 0.3876; c) 0.3096; d) 0.0756. ## § 5. Theorem of Addition of Probabilities of Joint Events Theorem 5. The probability of the occurrence of at least one of two joint events is equal to the sum of the probabilities of these events minus the probability of their joint occurrence: $$ p(A+B)=p(A)+p(B)-p(A B) $$ Corollary 1. If event $B$ depends on event $A$, then $p(A B)=p(A) p(B / A)$ and from (13) it follows: $$ p(A+B)=p(A)+p(B)-p(A) p(B / A) $$ Corollary 2. If events $A$ and $B$ are independent, then $p(A B)=$ $p(A) p(B)$ and from (13) it follows: $$ p(A+B)=p(A)+p(B)-p(A) p(B) $$ Corollary 3. If events $A$ and $B$ are mutually exclusive, then $p(A B)=0$, from which it follows: $$ p(A+B)=p(A)+p(B) $$ ## Problems with Solutions" 614ae8de298c,"2. Given is a square $A B C D$ with side length 4. Determine the largest natural number $k$ such that, for any arrangement of $k$ points strictly inside the square $A B C D$, there always exists a square with side length 1, contained within the square $A B C D$ (whose sides do not have to be parallel to the sides of the square $A B C D$), in whose strict interior none of the observed $k$ points lie. (Bojan Bašić)",$k=15$,easy,"2. The answer is $k=15$. It is clear that $k=15$ satisfies the conditions. Indeed, divide the square $A B C D$ into 16 unit squares; no matter how we arrange 15 points inside the square $A B C D$, at least one of these unit squares will not contain any of these points. Let's show that $ka$, then it contains at least one marked point." 15f4d5501b4d,1. What is the largest factor of 130000 that does not contain the digit 0 or 5 ? Proposed by: Farrell Eldrian Wu,26,easy,"Answer: 26 If the number is a multiple of 5 , then its units digit will be either 0 or 5 . Hence, the largest such number must have no factors of 5 . We have $130000=2^{4} \cdot 5^{4} \cdot 13$. Removing every factor of 5 , we get that our number must be a factor of $2^{4} \cdot 13=208$. If our number contains a factor of 13 , we cancel the factor of 2 from 208,104 , and 52 until we get 26 . Otherwise, the largest number we can have is $2^{4}=16$. We conclude that the answer is 26 ." bab40b124732,"6.2. Masha and the Bear ate a basket of raspberries and 40 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate both raspberries and pies 3 times faster than Masha. How many pies did Masha eat, if they ate the raspberries equally?",4 pies,easy,"Answer: 4 pies. Solution: The bear ate his half of the raspberries three times faster than Masha. This means Masha ate pies for three times less time than the bear. Since she eats three times slower, she ate 9 times fewer pies than the bear. Dividing the pies in the ratio of $9: 1$, we see that Masha got the 10th part, that is, 4 pies. Criterion: Correct answer - no less than 2 points, with full justification - 7 points." 5a7c5425129e,"10. If for all positive numbers $x, y$, the inequality $\sqrt{x}+\sqrt{y}$ $\leqslant a \sqrt{x+y}$ holds, then the minimum value of $a$ is $\qquad$","y$, the equality holds, hence the minimum value of $a$ is $\sqrt{2}$.",easy,"10. $\sqrt{2}$. From $\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1$, we have $\frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2}$. When $x=y$, the equality holds, hence the minimum value of $a$ is $\sqrt{2}$." 11f3ce668405,"1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is",See reasoning trace,easy,"From $2 a+b+c=5, b-c=1$, we can obtain $$ b=3-a, c=2-a \text {. } $$ Therefore, $a b+b c+c a$ $$ \begin{array}{l} =a(3-a)+(3-a)(2-a)+a(2-a) \\ =-a^{2}+6 \leqslant 6 . \end{array} $$" 06a9a678bfc9,"7. If the complex numbers $\left|z_{i}\right|=1,(i=1,2,3)$, then $\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=$",\left|\frac{\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}}{z_{1}+z_{2}+z_{3}}\right|=1$.,easy,7. $1 \quad\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=\left|\frac{\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}}{z_{1}+z_{2}+z_{3}}\right|=1$. a0714ee3e1ec,9.026. $\log _{0.3}(3 x-8)>\log _{0.3}\left(x^{2}+4\right)$.,$\quad x \in\left(\frac{8}{3} ; \infty\right)$,easy,"Solution. This inequality is equivalent to the system of inequalities $$ \left\{\begin{array} { l } { 3 x - 8 > 0 } \\ { 3 x - 8 \frac{8}{3} \\ x^{2}-3 x+12>0 \end{array}\right.\right. $$ Since $x^{2}-3 x+12>0$ for $x \in R$, the last system of inequalities is equivalent to the inequality $x>\frac{8}{3}$. Answer: $\quad x \in\left(\frac{8}{3} ; \infty\right)$" c8a989825870,"2.119 The number of values of $x$ that satisfy the equation $\frac{2 x^{2}-10 x}{x^{2}-5 x}=x-3$ is (A) 0. (B) 1. (C) 2. (D) 3. (E) An integer greater than 3. (17th American High School Mathematics Examination, 1966)",See reasoning trace,easy,"[Solution] From $x^{2}-5 x=x(x-5)$, when $x \neq 0,5$, the equation can be simplified to $2=x-3$, which means $x=5$, contradicting the previous assumption. Therefore, this equation has no solution. Hence, the correct choice is $(A)$." 11968877f621,"12. Fill in the $\square$ in the following equation with the numbers 1 to 9 respectively, so that the equation holds. $$ \square \div \square=\square \div \square=\square \square \square \div 7 \square $$",$4 \div 2=6 \div 3=158 \div 79$,easy,Reference answer: $4 \div 2=6 \div 3=158 \div 79$ 81817adaac14,"3. In astronomy, ""parsec"" is commonly used as a unit of distance. If in a right triangle $\triangle ABC$, $$ \angle ACB=90^{\circ}, CB=1.496 \times 10^{8} \text{ km}, $$ i.e., the length of side $CB$ is equal to the average distance between the Moon $(C)$ and the Earth $(B)$, then, when the size of $\angle BAC$ is 1 second (1 degree equals 60 minutes, 1 minute equals 60 seconds), the length of the hypotenuse $AB$ is 1 parsec. Therefore, 1 parsec = $\qquad$ km (expressed in scientific notation, retaining four significant figures).",See reasoning trace,easy,$\begin{array}{l}\text { 3. } 3.086 \times \\ 10^{13}\end{array}$ 602920d7b405,"Example 6.6. Two machines work independently of each other. The probability of the first machine operating without failure over a certain period of time $t$ is $p_{1}=0.9$, and for the second machine $-p_{2}=0.8$. What is the probability of both machines operating without failure over the specified period of time",See reasoning trace,medium,"Solution. Consider the following events: $A_{1}$ and $A_{2}$ - trouble-free operation of the first and second machines, respectively, over time $t$; $A$ - trouble-free operation of both machines over the specified time. Then the event $A$ is the conjunction of events $A_{1}$ and $A_{2}$, i.e., $A=A_{1} A_{2}$. Since events $A_{1}$ and $A_{2}$ are independent (the machines operate independently of each other), we obtain by formula (5): $$ P(A)=P\left(A_{1}\right) \cdot P\left(A_{2}\right)=0.9 \cdot 0.8=0.72 $$" 8dbc2017eefa,"## Task 7 - V00907 The sum of the digits of a two-digit number is 12. If you subtract from this number the number that contains the same digits in reverse order, you get 54. What is the number?","9$ and $b=3$. Since $93-39=54$ the check is correct, so 93 is the sought number.",easy,"Let the sought number $z$ be $z=10a+b$. Then for the digits $a$ and $b$ \[ \begin{aligned} a+5 & =12 \\ (10a+5)-(10b+a) & =54 \end{aligned} \] The system of equations has the solution $a=9$ and $b=3$. Since $93-39=54$ the check is correct, so 93 is the sought number." c74c4a2c0517,,$a=27$,easy,"Answer: $a=27$. Solution. We have $$ \sqrt{b}=\sqrt{52-30 \sqrt{3}}=\sqrt{27-2 \cdot 5 \cdot 3 \sqrt{3}+25}=3 \sqrt{3}-5 $$ Therefore, $\sqrt{a}-\sqrt{a-2}=3 \sqrt{3}-5, \frac{2}{\sqrt{a}+\sqrt{a-2}}=\frac{2}{3 \sqrt{3}+5}, \sqrt{a}+\sqrt{a-2}=\sqrt{27}+\sqrt{25}$. Since the function $f(a)=\sqrt{a}+\sqrt{a-2}$ is increasing and $f(27)=0$, the only solution to this equation is $a=27$. Answer to variant 2: $c=16$." c16718f945d5,"15. Huáshì Supermarket offers discounts to customers, with the following rules: (1) If a single shopping trip is less than 200 yuan, no discount is given; (2) If a single shopping trip is between 200 yuan and 500 yuan, a 10% discount is given; (3) If a single shopping trip exceeds 500 yuan, the first 500 yuan is given a 10% discount, and the amount exceeding 500 yuan is given a 20% discount. Xiaoming went to the supermarket twice and paid 198 yuan and 554 yuan respectively. Now Xiaoliang decides to buy the same amount of items in one trip that Xiaoming bought in two trips. How much does he need to pay?",See reasoning trace,medium,"Three, 15. The first payment of 198 yuan might be the actual price of the purchased items without any discount; it could also be the amount paid after a 10% discount. Therefore, we should discuss two scenarios. (1) When 198 yuan is the amount paid for shopping without a discount, the original price of the purchased items is 198 yuan. Also, $554=450+104$, where 450 yuan is the amount paid for 500 yuan worth of items at a 10% discount, and 104 yuan is the amount paid for items at a 20% discount, $104 \div 0.8 = 130$ (yuan). Therefore, the original price of the items purchased for 554 yuan is $130+500=630$ (yuan). Thus, for purchasing items originally priced at $198+630=828$ (yuan) in one go, Xiao Liang should pay $$ 500 \times 0.9 + (828 - 500) \times 0.8 = 712.4 \text{ (yuan). } $$ (2) When 198 yuan is the amount paid for shopping at a 10% discount, the original price of the purchased items is $198 \div 0.9 = 220$ (yuan). Following the discussion in (1), for purchasing items originally priced at $220 + 630 = 850$ (yuan) in one go, the payment should be $$ 500 \times 0.9 + (850 - 500) \times 0.8 = 730 \text{ (yuan). } $$ In summary, if Xiao Liang goes to the supermarket to purchase the same amount of items that Xiao Ming has already bought, he should pay 712.4 yuan or 730 yuan." 20b716f9d171,"1. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{2}=10, S_{5}=55$, then a direction vector of the line passing through points $P\left(n, a_{n}\right)$ and $Q(n+2$, $a_{n+2})$ can be ( ). (A) $\left(2, \frac{1}{2}\right)$ (B) $\left(-\frac{1}{2},-2\right)$ (C) $\left(-\frac{1}{2},-1\right)$ (D) $(-1,-1)$",\frac{a_{n+2}-a_{n}}{(n+2)-n}=d=4$.,easy,"-1.B. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with a common difference of $d$. Then $\left\{\begin{array}{l}S_{2}=2 a_{1}+d=10, \\ S_{5}=5 a_{1}+10 d=55\end{array} \Rightarrow\left\{\begin{array}{l}a_{1}=3, \\ d=4 .\end{array}\right.\right.$ Therefore, $k_{P Q}=\frac{a_{n+2}-a_{n}}{(n+2)-n}=d=4$." ec68c6bc97fb,"## Task B-4.1. Solve the inequality $5 \cdot\binom{n+2}{n-1}>\binom{n+4}{4}$ in the set of natural numbers.",See reasoning trace,easy,"## Solution. The inequality can be written as $5 \cdot\binom{n+2}{3}>\binom{n+4}{4}$, or $5 \cdot \frac{n(n+1)(n+2)}{6}>\frac{(n+1)(n+2)(n+3)(n+4)}{24}$. 1 point If we multiply by the denominator and divide by $(n+1)(n+2)$, we get $$ \begin{gathered} 20 n>(n+3)(n+4) \\ n^{2}-13 n+12<0 \end{gathered} $$ The solutions to this inequality are natural numbers $n \in\{2,3, \ldots, 11\}$." 263055c6e02c,"5. [5] The function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f\left(x^{2}\right) f^{\prime \prime}(x)=f^{\prime}(x) f^{\prime}\left(x^{2}\right)$ for all real $x$. Given that $f(1)=1$ and $f^{\prime \prime \prime}(1)=8$, determine $f^{\prime}(1)+f^{\prime \prime}(1)$.",6,medium,"Answer: 6. Let $f^{\prime}(1)=a$ and $f^{\prime \prime}(1)=b$. Then setting $x=1$ in the given equation, $b=a^{2}$. Differentiating the given yields $$ 2 x f^{\prime}\left(x^{2}\right) f^{\prime \prime}(x)+f\left(x^{2}\right) f^{\prime \prime \prime}(x)=f^{\prime \prime}(x) f^{\prime}\left(x^{2}\right)+2 x f^{\prime}(x) f^{\prime \prime}\left(x^{2}\right) . $$ Plugging $x=1$ into this equation gives $2 a b+8=a b+2 a b$, or $a b=8$. Then because $a$ and $b$ are real, we obtain the solution $(a, b)=(2,4)$." 40b88aacb626,"10. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2$, $AB=2$, and points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$.","\sqrt{3}$, let $O D=d$ $\Rightarrow(\sqrt{3}-d)^{2}=d^{2}+1 \Rightarrow d=\frac{\sqrt{3}}{3}$. There",easy,"$S A=S B=S C \Rightarrow S$ the projection of $S$ on the base $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $D$ of the hypotenuse $A B$, thus the center of the sphere $O$ lies on the line $S D$. $\triangle S A B$ is an equilateral triangle $\Rightarrow S D=\sqrt{3}$, let $O D=d$ $\Rightarrow(\sqrt{3}-d)^{2}=d^{2}+1 \Rightarrow d=\frac{\sqrt{3}}{3}$. Therefore, the distance from point $O$ to the plane $A B C$ is $\frac{\sqrt{3}}{3}$." a17452477847,"2. If the real number $x$ satisfies $\log _{2} \log _{2} x=\log _{4} \log _{4} x$, then $x=$ $\qquad$ .",See reasoning trace,easy,"2. $\sqrt{2}$. Let $y=\log _{a} \log _{a} x$. Then $a^{a^{y}}=x$. Let $a=2$ and $a=4$. According to the problem, $$ \begin{array}{l} 2^{2^{y}}=4^{4^{y}}=\left(2^{2}\right)^{\left(2^{2}\right) y}=2^{2^{2 y+1}} \\ \Rightarrow y=2 y+1 \Rightarrow y=-1 \\ \Rightarrow x=\sqrt{2} . \end{array} $$" ee328d5f4aa4,"183. Solve the equation in integers: $$ 7 x+5 y=62 $$","31-5 t$ and $y=7 t-31$, where $t$ is any integer common to both formulas.",easy,"183. $7 x+5 y=62, y=\frac{2(31-x)}{5}-x ; \quad$ therefore, $\frac{31-x}{5}$ is an integer; let $\frac{31-x}{5}=t$, then $x=31-$ $-5 t, y=2 t-(31-5 t)=7 t-31$. Thus, all integer solutions of the equation are represented by the formulas $x=31-5 t$ and $y=7 t-31$, where $t$ is any integer common to both formulas." 651411cff85c,"3. Determine all positive real numbers $a, b, c, d$ for which 1) $16abcd = \left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a b c+b c d+c d a+d a b)$ and 2) $2 a b+2 c d+a c+a d+b c+b d=8$.","4$, from which it follows that $a=b=c=d=1$.",medium,"Solution. From condition 1) we have $$ \begin{aligned} 1 & =\frac{16 a b c d}{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a b c+b c d+c d a+d a b)}=\frac{1}{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \cdot \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} \\ & \leq \frac{1}{\left(\frac{a+b+c+d}{4}\right)^{2}} \cdot \frac{a+b+c+d}{4}=\frac{4}{a+b+c+d} \end{aligned} $$ so $a+b+c+d \leq 4$. The equality holds if and only if $a=b=c=d=1$. From condition 2) we have $$ \begin{aligned} 16 & =4 a b+4 c d+2 a c+2 a d+2 b c+2 b d \\ & =2 a b+2 c d+2 a b+2 c d+2 a c+2 a d+2 b c+2 b d \\ & \leq a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 c d+2 a c+2 a d+2 b c+2 b d \\ & =(a+b+c+d)^{2} \end{aligned} $$ so $4 \leq a+b+c+d$. Therefore, $a+b+c+d=4$, from which it follows that $a=b=c=d=1$." 35a674060105,"9. [6] Newton and Leibniz are playing a game with a coin that comes up heads with probability $p$. They take turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips a heads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half of the time, what is the probability $p$ ?",$\frac{3-\sqrt{5}}{2}$ The probability that Newton will win on the first flip is $p$,medium,"Answer: $\frac{3-\sqrt{5}}{2}$ The probability that Newton will win on the first flip is $p$. The probability that Newton will win on the third flip is $(1-p) p^{2}$, since the first flip must be tails, the second must be heads, and the third flip must be heads. By the same logic, the probability Newton will win on the $(2 n+1)^{\text {st }}$ flip is $(1-p)^{n}(p)^{n+1}$. Thus, we have an infinite geometric sequence $p+(1-p) p^{2}+(1-p)^{2} p^{3}+\ldots$ which equals $\frac{p}{1-p(1-p)}$. We are given that this sum must equal $\frac{1}{2}$, so $1-p+p^{2}=2 p$, so $p=\frac{3-\sqrt{5}}{2}$ (the other solution is greater than 1 )." 754a867a057b,"1. For any real number $x$, the quadratic trinomial $x^{2}+3 m x+m^{2}-$ $m+\frac{1}{4}$ is a perfect square, then $m=$ $\qquad$.",9 m^{2}-4\left(m^{2}-m+\frac{1}{4}\right)=0$. Solving it yields $m=\frac{1}{5}$ or -1.,easy,二、1. $\Delta=9 m^{2}-4\left(m^{2}-m+\frac{1}{4}\right)=0$. Solving it yields $m=\frac{1}{5}$ or -1. fcae919faaf3,"Let $S=\{1,2, \ldots, n\}, A$ be an arithmetic sequence with at least two terms, a positive common difference, all of whose terms are in $S$, and such that adding any other element of $S$ to $A$ does not form an arithmetic sequence with the same common difference as $A$. Find the number of such $A$s. (Here, a sequence with only two terms is also considered an arithmetic sequence).",See reasoning trace,medium,"Solution 1 Let the common difference of $A$ be $d$, then $1 \leqslant d \leqslant n-1$. We discuss in two cases: (a) Suppose $n$ is even, then when $1 \leqslant d \leqslant \frac{n}{2}$, there are $d$ $A$s with a common difference of $d$; when $\frac{n}{2}+1 \leqslant d \leqslant n-1$, there are $n-d$ $A$s with a common difference of $d$. Therefore, when $n$ is even, the total number of such $A$s is $$ \left(1+2+\cdots+\frac{n}{2}\right)+\left\{1+2+\cdots+\left[n-\left(\frac{n}{2}+1\right)\right]\right\}=\frac{n^{2}}{4} \text { (cases). } $$ (b) Suppose $n$ is odd, then when $1 \leqslant d \leqslant \frac{n-1}{2}$, there are $d$ $A$s with a common difference of $d$; when $\frac{n+1}{2} \leqslant d \leqslant n-1$, there are $n-d$ $A$s with a common difference of $d$. Therefore, when $n$ is odd, the total number of such $A$s is $$ \left(1+2+\cdots+\frac{n-1}{2}\right)+\left(1+2+\cdots+\frac{n-1}{2}\right)=\frac{n^{2}-1}{4} \text { (cases). } $$ Both cases can be unified as: the number of such $A$s is $\left[\frac{n^{2}}{4}\right]$. Solution 2 For $n=2 k$, the sequence $A$ must have two consecutive terms, one in $\{1,2, \cdots, k\}$ and the other in $\{k+1, k+2, \cdots, n\}$. Conversely, by choosing any number from $\{1,2, \cdots, k\}$ and any number from $\{k+1, k+2, \cdots, n\}$, a sequence $A$ can be constructed with their difference as the common difference. This correspondence is one-to-one. Therefore, the number of such $A$s is $$ k^{2}=\frac{n^{2}}{4} . $$ For $n=2 k+1$, the situation is similar. Note that the set $\{k+1, k+2, \cdots, n\}$ contains $k+1$ numbers, so the number of such $A$s is $$ k(k+1)=\left(n^{2}-1\right) / 4 . $$ Both formulas can be unified as $\left[\frac{n^{2}}{4}\right]$." 0a2a293886ce,"24. Arrange the numbers $1,2,3,4,5$ in a circle. If there exists an integer $n(1 \leqslant n \leqslant 15)$, such that the sum of any consecutive numbers in the arrangement is not equal to $n$, then the arrangement is called a ""bad arrangement"". For circular arrangements obtained by rotation or reflection, they are considered the same arrangement. Then there are ( ) different bad arrangements. (A) 1 (B) 2 (C) 3 (D) 4 (E) 5",See reasoning trace,medium,"24. B. The number of circular permutations of the numbers $1, 2, 3, 4, 5$ arranged on a circle is $4!=24$. Next, verify which of these are bad permutations. If $n \in \{1,2,3,4,5\}$, then in any circular permutation, there is a number that equals $n$; If $n=15$, then the sum of all numbers in any circular permutation equals $n$; If $n \in \{10,11,12,13,14\}$, then in any circular permutation, removing the number $15-n$ results in a sum of the remaining numbers equal to $n$; If $n=6$, then a permutation is not bad if it contains the sequence of three consecutive numbers $1,2,3$, or if it contains the pairs $2, 4$ or $1, 5$. Since circular permutations that are rotations or reflections of each other are considered the same, there is only one bad permutation (as shown in Figure 10); If $n=9$, when a part of the consecutive numbers in a circular permutation sums to 6, the remaining part sums to 9, so the bad permutation is also as shown in Figure 10; Similarly, if $n=7$ or 8, there is also only one bad permutation (as shown in Figure 11). In summary, there are two ""bad permutations""." ec310f1d6c72,"2.132 The number of real roots of the equation $$ \sqrt{x^{2}+2 x-63}+\sqrt{x+9}-\sqrt{7-x}+x+13=0 $$ is (A) 0. (B) 1. (C) 2. (D) More than 2. (China Beijing High School Mathematics Competition, 1993)",$(B)$,easy,"［Solution］If the square root is meaningful, then we have $\left\{\begin{array}{l}x^{2}+2 x-63 \geqslant 0, \\ x+9 \geqslant 0, \\ 7-x \geqslant 0\end{array}\right.$ $\left\{\begin{array}{l}x \leqslant-9 \text { or } x \geqslant 7, \\ x \geqslant-9, \\ x \leqslant 7 .\end{array}\right.$ $\therefore x=7$ or -9. Substituting into the original equation, only $x=-9$ is suitable. Therefore, the answer is $(B)$." e62e0e43cb2e,"37. Find the smallest $a \in \mathbf{N}^{*}$, such that the following equation has real roots: $$ \cos ^{2} \pi(a-x)-2 \cos \pi(a-x)+\cos \frac{3 \pi x}{2 a} \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+2=0 . $$",See reasoning trace,medium,"$$ \begin{array}{l} \text { 37. Original equation } \Leftrightarrow[\cos \pi(a-x)-1]^{2}+\left[\cos \frac{3 \pi}{2 a} x \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+1\right]=0 \\ \Leftrightarrow[\cos \pi(a-x)-1]^{2}+\frac{1}{2}\left[\cos \left(\frac{2 \pi}{a} x+\frac{\pi}{3}\right)+\cos \left(\frac{\pi x}{a}-\frac{\pi}{3}\right)+2\right]=0 \\ \Leftrightarrow\left\{\begin{array} { l } { \cos \pi ( a - x ) = 1 } \\ { \cos ( \frac { 2 \pi } { a } x + \frac { \pi } { 3 } ) = - 1 } \\ { \cos ( \frac { \pi x } { a } - \frac { \pi } { 3 } ) = - 1 } \end{array} \Leftrightarrow \left\{\begin{array}{l} a-x=2 k_{1} \cdots \cdots \text { (1) } \\ \frac{2 x}{a}+\frac{1}{3}=2 k_{2}-1, k_{i} \in \mathbf{Z} \cdots \text { (2) } \\ \frac{x}{a}-\frac{1}{3}=2 k_{3}-1 \cdots \cdots \text { (3) } \end{array}\right.\right. \\ \Leftrightarrow\left\{\begin{array}{l} a-x=2 k_{1} \cdots \cdots \text { (1) } \\ \left(3 k_{2}-2\right) \cdot a=3 x \cdots \cdots \text { (2) } \\ 3 x-a=3 a\left(2 k_{3}-1\right) \cdots \cdots \text { (3) } \end{array}\right. \end{array} $$ From (1), we know: since $a \in \mathbf{N}^{*}$, so $x \in \mathbf{Z}$ and $x, a$ have the same parity. From (2), we know that $a$ is a multiple of 3. If $x, a$ are both odd, substituting into (3) leads to a contradiction, so $x, a$ must both be even. Therefore, $a$ is a multiple of 6, and since $a \in \mathbf{N}^{*}$, we have $a \geqslant 6$. When $a=6$, $x=8$ is a real solution. Therefore, $a_{\text {min }}=6$. Note 1: If the equation is changed to: $$ \cos ^{2} \pi(a-x)+2 \cos \pi(a-x)+\cos \frac{3 \pi x}{2 a} \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+2=0 $$ Using the same method, we get $a_{\text {min }}=3$, at which point $x=4$ is a real solution. Note 2: When simplifying $\cos \frac{3 \pi}{2 a} x \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+1=0$, we can let $\theta=\frac{\pi x}{2 a}+\frac{\pi}{3}$, then $3 \theta=\frac{3 \pi x}{2 a}+\pi$. Then the equation $\Leftrightarrow \cos (3 \theta-\pi) \cdot \cos \theta+1=0 \Leftrightarrow \cos 3 \theta \cos \theta-1=0 \Leftrightarrow\left(4 \cos ^{3} \theta-3 \cos \theta\right) \cos \theta-1=0 \Leftrightarrow\left(\cos ^{2} \theta-\right.$ 1) $\left(4 \cos ^{2} \theta+1\right)=0 \Leftrightarrow \cos ^{2} \theta=1$. $$" 35ffe6c806af,"2.111 If $a, b, c, d$ are the solutions of the equation $x^{4}-b x-3=0$, then the four solutions $$ \frac{a+b+c}{d^{2}}, \quad \frac{a+b+d}{c^{2}}, \quad \frac{a+c+d}{b^{2}}, \quad \frac{b+c+d}{a^{2}} $$ are the solutions of the equation (A) $3 x^{4}+b x+1=0$. (B) $3 x^{4}-b x+1=0$. (C) $3 x^{4}+b x^{3}-1=0$. (D) $3 x^{4}-b x^{3}-1=0$. (E) None of the above equations. (32nd American High School Mathematics Examination, 1981)",$(D)$,medium,"[Solution] Since the coefficient of the cubic term $x^{3}$ in the polynomial function $f(x)=x^{4}-b x-3$ is zero, by Vieta's formulas we have $$ a+b+c+d=0, $$ and $\quad \frac{a+b+c}{d^{2}}=\frac{a+b+c+d-d}{d^{2}}=\frac{-1}{d}$, $$ \frac{a+c+d}{b^{2}}=\frac{-1}{b}, \quad \frac{a+b+d}{c^{2}}=\frac{-1}{c}, \quad \frac{b+c+d}{a^{2}}=\frac{-1}{a} \text {. } $$ Since $a, b, c, d$ are the roots of $f(x)$, then $\frac{-1}{d}, \frac{-1}{b}, \frac{-1}{c}, \frac{-1}{a}$ are the roots of $f\left(-\frac{1}{x}\right)=0$, i.e., $\frac{a+b+c}{d^{2}}, \quad \frac{a+b+d}{c^{2}}, \quad \frac{a+c+d}{b^{2}}, \quad \frac{b+c+d}{a^{2}}$ are also the roots of $f\left(-\frac{1}{x}\right)=0$. This equation is $$ \frac{1}{x^{4}}+\frac{b}{x}-3=0 \text { or } 3 x^{4}-b x^{3}-1=0 \text {. } $$ Therefore, the answer is $(D)$." a00a2f00fc44,"## Write the decomposition of vector $x$ in terms of vectors $p, q, r$: $x=\{-5 ; 9 ;-13\}$ $p=\{0 ; 1 ;-2\}$ $q=\{3 ;-1 ; 1\}$ $r=\{4 ; 1 ; 0\}$",See reasoning trace,medium,"## Solution The desired decomposition of vector $x$ is: $x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$ Or in the form of a system: $$ \left\{\begin{array}{l} \alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\ \alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\ \alpha \cdot p_{3}+\beta \cdot q_{3}+\gamma \cdot r_{3}=x_{3} \end{array}\right. $$ We obtain: $$ \left\{\begin{array}{l} 3 \beta+4 \gamma=-5 \\ \alpha-\beta+\gamma=9 \\ -2 \alpha+\beta=-13 \end{array}\right. $$ Add the second row multiplied by -4 to the first row: $$ \left\{\begin{array}{l} -4 \alpha+7 \beta=-41 \\ \alpha-\beta+\gamma=9 \\ -2 \alpha+\beta=-13 \end{array}\right. $$ Add the third row multiplied by -2 to the first row: $$ \left\{\begin{array}{l} 5 \beta=-15 \\ \alpha-\beta+\gamma=9 \\ -2 \alpha+\beta=-13 \end{array}\right. $$ $$ \begin{aligned} & \left\{\begin{array}{l} \beta=-3 \\ \alpha-\beta+\gamma=9 \\ -2 \alpha+\beta=-13 \end{array}\right. \\ & \left\{\begin{array}{l} \beta=-3 \\ \alpha-(-3)+\gamma=9 \\ -2 \alpha+(-3)=-13 \end{array}\right. \\ & \left\{\begin{array}{l} \beta=-3 \\ \alpha+\gamma=6 \\ -2 \alpha=-10 \end{array}\right. \\ & \left\{\begin{array}{l} \beta=-3 \\ \alpha+\gamma=6 \\ \alpha=5 \end{array}\right. \\ & \left\{\begin{array}{l} \beta=-3 \\ 5+\gamma=6 \\ \alpha=5 \end{array}\right. \\ & \left\{\begin{array}{l} \beta=-3 \\ \gamma=1 \\ \alpha=5 \end{array}\right. \end{aligned} $$ The desired decomposition: $x=5 p-3 q+r$ ## Problem Kuznetsov Analytic Geometry $2-20$" b7e4c8d432e4,"For which integers $x$ is the quotient $\frac{x+11}{x+7}$ an integer? Find all solutions. (L. Hozová) Hint. Can you somehow simplify the given expression?",-7$ is related to the fact that the function is not defined at this point and its values are not con,medium,"The given expression can be rewritten as follows (""whole part plus remainder""): $$ \frac{x+11}{x+7}=\frac{(x+7)+4}{x+7}=\frac{x+7}{x+7}+\frac{4}{x+7}=1+\frac{4}{x+7} . $$ Thus, $\frac{x+11}{x+7}$ is an integer if and only if $\frac{4}{x+7}$ is an integer, i.e., if and only if $x+7$ is a divisor of the number 4. The number 4 has six divisors: $-4,-2,-1,1,2$ and 4. The corresponding values of $x$ are 7 less. The number $\frac{x+11}{x+7}$ is an integer for $x$ equal to $-11,-9,-8,-6,-5$ or -3. Another solution. To ensure the denominator is not zero, $x$ cannot be -7. By substituting integers near this value, we get: | $x$ | -6 | -8 | -5 | -9 | -4 | -10 | -3 | -11 | $\ldots$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\frac{x+11}{x+7}$ | 5 | -3 | 3 | -1 | $\frac{7}{3}$ | $-\frac{1}{3}$ | 2 | 0 | $\ldots$ | As the distance of $x$ from -7 increases, we get non-integer values on the second row between 2 and 0, which approach 1 but never reach it: if $\frac{x+11}{x+7}=1$, then $x+11=x+7$, which is impossible. All valid options are thus contained in the given table. The number $\frac{x+11}{x+7}$ is an integer for $x$ equal to $-11,-9,-8,-6,-5$ or -3. Note. Consider the previous argument as satisfactory, even though it is not perfect: to show correctly that for $x>-3$ or $x<-11$ the values of $\frac{x+11}{x+7}$ are between 2 and 0, probably exceeds the capabilities of solvers in this category. This discussion relates to the monotonic behavior of the function $x \mapsto \frac{x+11}{x+7}$. The apparent confusion in the values of the function near $x=-7$ is related to the fact that the function is not defined at this point and its values are not constrained. Curious solvers may consider the behavior of this function, sketch its graph, and compare it with the previous conclusions." b8c315b354e2,"12. A piggy bank contains 24 coins, all nickels, dimes, and quarters. If the total value of the coins is two dollars, what combinations of coins are possible?","(20,0,4),(17,4,3),(14,8,2),(11,12,1)$, $(8,16,0)$",easy,"12. $($ nickels, dimes, quarters $)=(20,0,4),(17,4,3),(14,8,2),(11,12,1)$, $(8,16,0)$" 946c6f931f20,"The radii of circles $S_{1}$ and $S_{2}$, touching at point $A$, are $R$ and $r(R>r)$. Find the length of the tangent line drawn to circle $S_{2}$ from point $B$, lying on circle $S_{1}$, if it is known that $A B=a$. (Consider the cases of internal and external tangency). #",$a \sqrt{1 \pm \frac{r}{R}}$,medium,"Apply the theorem of the tangent and secant. ## Solution Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency). Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore, $$ A X=\frac{r}{R} \cdot A B=\frac{a r}{R} $$ By the theorem of the tangent and secant, $$ B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right) $$ Thus, $B M=a \sqrt{1+\frac{r}{R}}$. In the case of internal tangency, similarly, we obtain that $$ B M=a \sqrt{1-\frac{r}{R}} $$ Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency). Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore, $$ A X=\frac{r}{R} \cdot A B=\frac{a r}{R} $$ By the theorem of the tangent and secant, $$ B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right) $$ Thus, $B M=a \sqrt{1+\frac{r}{R}}$. In the case of internal tangency, similarly, we obtain that $$ B M=a \sqrt{1-\frac{r}{R}} $$ Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency). Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore, $$ A X=\frac{r}{R} \cdot A B=\frac{a r}{R} $$ By the theorem of the tangent and secant, $$ B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right) $$ Thus, $B M=a \sqrt{1+\frac{r}{R}}$. In the case of internal tangency, similarly, we obtain that $$ B M=a \sqrt{1-\frac{r}{R}} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_7b30f89ee421e63350f8g-09.jpg?height=440&width=672&top_left_y=169&top_left_x=699) ## Answer $a \sqrt{1 \pm \frac{r}{R}}$." 372be3268b77,"1. The desired number is equal to $(9+1)+(99+1)+\ldots+(\underbrace{99 \ldots 9}_{2016 \text { times }}+1)+1=$ $=10+10^{2}+\cdots+10^{2016}+1=\underbrace{11 \ldots 1}_{2017 \text { times }}$.",$\underbrace{11 \ldots 1}_{2017 \text { times }}$,easy,Answer: $\underbrace{11 \ldots 1}_{2017 \text { times }}$. 1de7bf55abf8,"For all positive integers $n$ the function $f$ satisfies $f(1) = 1, f(2n + 1) = 2f(n),$ and $f(2n) = 3f(n) + 2$. For how many positive integers $x \leq 100$ is the value of $f(x)$ odd? $\mathrm{(A) \ } 4 \qquad \mathrm{(B) \ } 5 \qquad \mathrm {(C) \ } 6 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 10$",7,medium,"1. We start by analyzing the given function properties: - \( f(1) = 1 \) - \( f(2n + 1) = 2f(n) \) - \( f(2n) = 3f(n) + 2 \) 2. We need to determine for how many positive integers \( x \leq 100 \) the value of \( f(x) \) is odd. 3. First, let's consider the case when \( x \) is odd: - If \( x \) is odd, then \( x = 2n + 1 \) for some integer \( n \). - Using the function property \( f(2n + 1) = 2f(n) \), we see that \( f(2n + 1) \) is always even because it is twice some integer \( f(n) \). 4. Next, consider the case when \( x \) is even: - If \( x \) is even, then \( x = 2n \) for some integer \( n \). - Using the function property \( f(2n) = 3f(n) + 2 \), we need to determine when \( f(2n) \) is odd. - For \( f(2n) \) to be odd, \( 3f(n) + 2 \) must be odd. This implies that \( 3f(n) \) must be odd (since adding 2 to an odd number results in an odd number). - For \( 3f(n) \) to be odd, \( f(n) \) must be odd (since 3 times an odd number is odd). 5. We now need to find the values of \( n \) for which \( f(n) \) is odd: - We know \( f(1) = 1 \) is odd. - If \( n \) is a power of 2, say \( n = 2^k \), we can use the function property recursively: - \( f(2^k) = 3f(2^{k-1}) + 2 \) - Since \( f(1) = 1 \) is odd, we can check the next few values: - \( f(2) = 3f(1) + 2 = 3 \cdot 1 + 2 = 5 \) (odd) - \( f(4) = 3f(2) + 2 = 3 \cdot 5 + 2 = 17 \) (odd) - \( f(8) = 3f(4) + 2 = 3 \cdot 17 + 2 = 53 \) (odd) - \( f(16) = 3f(8) + 2 = 3 \cdot 53 + 2 = 161 \) (odd) - \( f(32) = 3f(16) + 2 = 3 \cdot 161 + 2 = 485 \) (odd) - \( f(64) = 3f(32) + 2 = 3 \cdot 485 + 2 = 1457 \) (odd) 6. We observe that for \( n = 2^k \), \( f(2^k) \) is always odd. We need to count the powers of 2 up to 100: - \( 2^0 = 1 \) - \( 2^1 = 2 \) - \( 2^2 = 4 \) - \( 2^3 = 8 \) - \( 2^4 = 16 \) - \( 2^5 = 32 \) - \( 2^6 = 64 \) 7. There are 7 powers of 2 less than or equal to 100. Conclusion: The number of positive integers \( x \leq 100 \) for which \( f(x) \) is odd is 7. The final answer is \(\boxed{7}\)." ab868682f02f,What is the value of $\frac{1^{3}+2^{3}+3^{3}}{(1+2+3)^{2}}$ ?,See reasoning trace,easy,"Evaluating, $$ \frac{1^{3}+2^{3}+3^{3}}{(1+2+3)^{2}}=\frac{1+8+27}{6^{2}}=\frac{36}{36}=1 $$ ANSWER: 1 #" 41e8cdd4f324,"Example 4. A discrete random variable $X$ has the following distribution: | $X$ | 3 | 4 | 5 | 6 | 7 | | :--- | :--- | :--- | :--- | :--- | :--- | | $P$ | $p_{1}$ | 0.15 | $p_{3}$ | 0.25 | 0.35 | Find the probabilities $p_{1}=P(X=3)$ and $p_{3}=P(X=5)$, given that $p_{3}$ is 4 times $p_{1}$.",0.05 ; p_{3}=0.20$.,easy,"Solution. Since $$ p_{2}+p_{4}+p_{5}=0.15+0.25+0.35=0.75 $$ then based on equality (2.1.2), we conclude that $$ p_{1}+p_{3}=1-0.75=0.25 $$ Since by the condition $p_{3}=4 p_{1}$, then $$ p_{1}+p_{3}=p_{1}+4 p_{1}=5 p_{1} . $$ Thus, $5 p_{1}=0.25$, from which $p_{1}=0.05$; therefore, $$ p_{3}=4 p_{1}=4 \cdot 0.05=0.20 $$ Therefore, $p_{1}=0.05 ; p_{3}=0.20$." 7afde00fdb33,Example - (2004 American Invitational Mathematics Examination) A circle of radius 1 is randomly placed inside a $15 \times 36$ rectangle $ABCD$. Find the probability that this circle does not intersect with the diagonal $AC$.,\frac{375}{442}$.,easy,"Analysis We use the area relationship of the figures. Solution If a circle with radius 1 is placed inside $ABCD$, its center must lie within a $13 \times$ 34 rectangle, with an area of 442. If the circle does not intersect with the diagonal $AC$, then the center of the circle lies within two right-angled triangular regions (the shaded parts in Figure 14-1), where the hypotenuse is 1 unit away from $AC$. The area of the shaded parts can be calculated as 375. Therefore, $P=\frac{375}{442}$." 6b6c6f06b247,"4. Let the side length of the equilateral $\triangle ABC$ be $2, M$ is the midpoint of side $AB$, $P$ is any point on side $BC$, and $PA + PM$ are denoted as $s$ and $t$ for their maximum and minimum values, respectively. Then $s^{2}-t^{2}=$ $\qquad$.",(2+\sqrt{3})^{2}-7=4 \sqrt{3}$.,medium,"$4.4 \sqrt{3}$. First, find $s$. $$ \begin{array}{l} \because P A \leqslant A C, P M \leqslant C M, \\ \therefore P A+P M \leqslant C A+C M=2+\sqrt{3} . \end{array} $$ When point $P$ is at vertex $C$, the equality holds. Thus, $s=2+\sqrt{3}$. Next, find $t$. As shown in Figure 4, construct the equilateral $\triangle A^{\prime} E C$. Let $M^{\prime}$ be the midpoint of $A^{\prime} B$, then $\triangle P B M \cong \triangle P B M^{\prime}$. $$ \begin{array}{c} \therefore P M=P M^{\prime} . \\ P A+P M=P A+P M^{\prime} \geqslant A M^{\prime} . \end{array} $$ Connect $C M^{\prime}$, then $\angle A C M^{\prime}=90^{\circ}$. $$ \begin{array}{l} \therefore A M^{\prime}=\sqrt{A C^{2}+\left(C M^{\prime}\right)^{2}}=\sqrt{4+3}=\sqrt{7} . \\ \therefore t=\sqrt{7} . \end{array} $$ Therefore, $s^{2}-t^{2}=(2+\sqrt{3})^{2}-7=4 \sqrt{3}$." 4d437102653c,"Third question Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn.",See reasoning trace,hard,"``` Solution 1 The maximum number of line segments that can be connected is 15, an example is shown in Figure 5. It is only necessary to prove that 16 line segments are not possible. If 16 line segments are connected, it will produce 32 endpoints. By the pigeonhole principle, there is a point with a degree of at least 4. First, we prove a lemma by contradiction. Lemma: If there are 16 line segments that meet the conditions and the maximum degree of each point is 4, then the number of points with a degree of 4 does not exceed two. Proof: Assume there are three points with a degree of 4, let these three points with a degree of 4 be }A, B, C\mathrm{. In this case, there are two scenarios. ``` (1) If points $A$ and $B$ are not connected, this can be further divided into two scenarios: the four points connected to $A$ and $B$ are not repeated; the four points connected to $A$ and $B$ have one repetition (if the points connected to $A$ and $B$ have two repetitions, it will form a quadrilateral). By the pigeonhole principle, among the four points connected to $C$, there must be two points in one of the above two groups, forming a triangle or a quadrilateral, which is a contradiction. (2) If points $A$ and $B$ are connected, assume point $A$ is connected to $A_{1}$, $A_{2}$, $A_{3}$, and point $B$ is connected to $B_{1}$, $B_{2}$, $B_{3}$, as shown in Figure 6. Similarly, by the pigeonhole principle, a contradiction is derived. The lemma is proved. For the original problem, we discuss two scenarios: (1) The maximum degree of all points is 4. By the lemma, there are two points with a degree of 4, and the rest of the points have a degree of 3. Assume these two points are $A$ and $B$. At this time, the degree of the points other than $A$ and $B$ is 3. (i) If points $A$ and $B$ are not connected, this can be further divided into two scenarios: the four points connected to $A$ and $B$ are not repeated; the four points connected to $A$ and $B$ have one repetition (if the points connected to $A$ and $B$ have two repetitions, it will form a quadrilateral). Additionally, there is at most one point $D$. But each point in the above groups needs to draw two more edges (not to the points in the same group, but to the other group, at most one edge). Therefore, the edges drawn to point $D$ are more than 4, which is a contradiction. (ii) If points $A$ and $B$ are connected, assume point $A$ is connected to $A_{1}$, $A_{2}$, $A_{3}$, and point $B$ is connected to $B_{1}$, $B_{2}$, $B_{3}$, as shown in Figure 6. The other two points are $C$ and $D$. Similarly, $A_{i}$ and $B_{i}$ ($i=1,2,3$) each draw one edge to point $C$ or $D$. Therefore, one of the points $C$ or $D$ draws three edges, forming a quadrilateral, which is a contradiction. (2) The maximum degree of a point is not less than 5, assume the maximum degree is $i$ ($i \geqslant 5$). Assume the point with a degree of $i$ is $A$, and the $i$ points connected to it are $A_{1}$, $A_{2}$, ..., $A_{i}$; the other $9-i$ points are $B_{1}$, $B_{2}$, ..., $B_{9-i}$. Then, point $B_{j}$ ($j=1,2, \cdots, 9-i$) can only be connected to $A_{1}$, $A_{2}$, ..., $A_{i}$. Among them, $B_{j}$ cannot be connected to two points at the same time (otherwise, it will form a quadrilateral). Therefore, $B_{1}$, $B_{2}$, ..., $B_{9-i}$ can connect at most $9-i$ line segments to $A_{1}$, $A_{2}$, ..., $A_{i}$. Now consider how many line segments can be connected between $B_{1}$, $B_{2}$, ..., $B_{9-i}$. According to the assumption, we need to connect $16-(9-i)-i=7$ more line segments. The maximum number of line segments that can be connected between $9-i$ points is $$ C_{9-i}^{2}=\frac{(9-i)(8-i)}{2} \text {. } $$ By $i \geqslant 5$, we know $C_{9-i}^{2} \leqslant 6<7$, which is a contradiction, so this scenario is excluded. In summary, the maximum number of line segments that can be connected is not 16. Therefore, the original proposition holds, and the maximum value is 15. (Du Yulin, Experimental Class 4, Tianjin Yaohua Middle School, 300040) Solution 2 Let the ten points be $A_{1}$, $A_{2}$, ..., $A_{10}$, and the edges between them form a graph $G(V, E)$. Then, the graph $G$ has exactly $\sum_{i=1}^{10} C_{d(1)}^{2}$ angles. According to the conditions, in any $\angle A_{i} A_{k} A_{j}$ in graph $G$, $A_{i}$ and $A_{j}$ cannot be connected; and in any two different $\angle A_{i_{1}} A_{k_{1}} A_{j_{1}}$ and $\angle A_{i_{2}} A_{k_{2}} A_{j_{2}}$ in graph $G$, we have $$ \left\{A_{i_{1}}, A_{j_{1}}\right\}=\left\{A_{i_{2}}, A_{j_{2}}\right\} . $$ Thus, the complement graph of $G(V, E)$ has at least $\sum_{i=1}^{10} C_{d(A, i)}^{2}$ edges. $$ \begin{array}{l} \text { Hence } C_{10}^{2} \geqslant \sum_{i=1}^{10} C_{d(A, i)}^{2}+|E| \\ =\frac{\sum_{i=1}^{10} d^{2}(A_{i})-\sum_{i=1}^{10} d(A_{i})}{2}+|E| \\ \geqslant \frac{\frac{1}{10}\left(\sum_{i=1}^{10} d(A_{i})\right)^{2}-\sum_{i=1}^{10} d(A_{i})}{2}+|E| \\ =\frac{|E|^{2}}" 6b826a521eed,"[b]p1.[/b] Compute the greatest integer less than or equal to $$\frac{10 + 12 + 14 + 16 + 18 + 20}{21}$$ [b]p2.[/b] Let$ A = 1$.$B = 2$, $C = 3$, $...$, $Z = 26$. Find $A + B +M + C$. [b]p3.[/b] In Mr. M's farm, there are $10$ cows, $8$ chickens, and $4$ spiders. How many legs are there (including Mr. M's legs)? [b]p4.[/b] The area of an equilateral triangle with perimeter $18$ inches can be expressed in the form $a\sqrt{b}{c}$ , where $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Find $a + b + c$. [b]p5.[/b] Let $f$ be a linear function so $f(x) = ax + b$ for some $a$ and $b$. If $f(1) = 2017$ and $f(2) = 2018$, what is $f(2019)$? [b]p6.[/b] How many integers $m$ satisfy $4 < m^2 \le 216$? [b]p7.[/b] Allen and Michael Phelps compete at the Olympics for swimming. Allen swims $\frac98$ the distance Phelps swims, but Allen swims in $\frac59$ of Phelps's time. If Phelps swims at a rate of $3$ kilometers per hour, what is Allen's rate of swimming? The answer can be expressed as $m/n$ for relatively prime positive integers $m, n$. Find $m + n$. [b]p8.[/b] Let $X$ be the number of distinct arrangements of the letters in ""POONAM,"" $Y$ be the number of distinct arrangements of the letters in ""ALLEN"" and $Z$ be the number of distinct arrangements of the letters in ""NITHIN."" Evaluate $\frac{X+Z}{Y}$ : [b]p9.[/b] Two overlapping circles, both of radius $9$ cm, have centers that are $9$ cm apart. The combined area of the two circles can be expressed as $\frac{a\pi+b\sqrt{c}+d}{e}$ where $c$ is not divisible by the square of any prime and the fraction is simplified. Find $a + b + c + d + e$. [b]p10.[/b] In the Boxborough-Acton Regional High School (BARHS), $99$ people take Korean, $55$ people take Maori, and $27$ people take Pig Latin. $4$ people take both Korean and Maori, $6$ people take both Korean and Pig Latin, and $5$ people take both Maori and Pig Latin. $1$ especially ambitious person takes all three languages, and and $100$ people do not take a language. If BARHS does not o er any other languages, how many students attend BARHS? [b]p11.[/b] Let $H$ be a regular hexagon of side length $2$. Let $M$ be the circumcircle of $H$ and $N$ be the inscribed circle of $H$. Let $m, n$ be the area of $M$ and $N$ respectively. The quantity $m - n$ is in the form $\pi a$, where $a$ is an integer. Find $a$. [b]p12.[/b] How many ordered quadruples of positive integers $(p, q, r, s)$ are there such that $p + q + r + s \le 12$? [b]p13.[/b] Let $K = 2^{\left(1+ \frac{1}{3^2} \right)\left(1+ \frac{1}{3^4} \right)\left(1+ \frac{1}{3^8}\right)\left(1+ \frac{1}{3^{16}} \right)...}$. What is $K^8$? [b]p14.[/b] Neetin, Neeton, Neethan, Neethine, and Neekhil are playing basketball. Neetin starts out with the ball. How many ways can they pass 5 times so that Neethan ends up with the ball? [b]p15.[/b] In an octahedron with side lengths $3$, inscribe a sphere. Then inscribe a second sphere tangent to the first sphere and to $4$ faces of the octahedron. The radius of the second sphere can be expressed in the form $\frac{\sqrt{a}-\sqrt{b}}{c}$ , where the square of any prime factor of $c$ does not evenly divide into $b$. Compute $a + b + c$. PS. You should use hide for answers. Collected [url=https://artof",4035,medium,"To solve the problem, we need to find the linear function \( f(x) = ax + b \) given the conditions \( f(1) = 2017 \) and \( f(2) = 2018 \). 1. **Set up the equations based on the given conditions:** \[ f(1) = a \cdot 1 + b = 2017 \implies a + b = 2017 \] \[ f(2) = a \cdot 2 + b = 2018 \implies 2a + b = 2018 \] 2. **Solve the system of linear equations:** Subtract the first equation from the second equation to eliminate \( b \): \[ (2a + b) - (a + b) = 2018 - 2017 \] \[ 2a + b - a - b = 1 \] \[ a = 1 \] 3. **Substitute \( a = 1 \) back into the first equation to find \( b \):** \[ a + b = 2017 \] \[ 1 + b = 2017 \] \[ b = 2016 \] 4. **Form the linear function \( f(x) \):** \[ f(x) = ax + b = 1x + 2016 = x + 2016 \] 5. **Calculate \( f(2019) \):** \[ f(2019) = 2019 + 2016 = 4035 \] The final answer is \( \boxed{4035} \)." a05466d313a1,"27. (BLR) Ten points such that no three of them lie on a line are marked in the plane. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors in such a way that for any $k$ of the ten points, there are $k$ segments each joining two of them with no two being painted the same color. Determine all integers $k, 1 \leq k \leq 10$, for which this is possible.",9$ (the addition being modulo 9 ). We can ignore the edges painted with the extra colors. Then the e,medium,"27. Since this is essentially a graph problem, we call the points and segments vertices and edges of the graph. We first prove that the task is impossible if $k \leq 4$. Cases $k \leq 2$ are trivial. If $k=3$, then among the edges from a vertex $A$ there are two of the same color, say $A B$ and $A C$, so we don't have all the three colors among the edges joining $A, B, C$. Now let $k=4$, and assume that there is a desired coloring. Consider the edges incident with a vertex $A$. At least three of them have the same color. say blue. Suppose that four of them, $A B, A C, A D, A E$, are blue. There is a blue edge, say $B C$, among the ones joining $B, C, D, E$. Then four of the edges joining $A, B, C, D$ are blue, and we cannot complete the coloring. So, exactly three edges from $A$ are blue: $A B, A C, A D$. Also, of the edges connecting any three of the 6 vertices other than $A, B, C, D$, one is blue (because the edges joining them with $A$ are not so). By a classical result, there is a blue triangle $E F G$ with vertices among these six. Now one of $E B, E C, E D$ must be blue as well, because none of $B C, B D, C D$ is. Let it be $E B$. Then four of the edges joining $B, E, F, G$ are blue, which is impossible. For $k=5$ the task is possible. Label the vertices $0,1, \ldots, 9$. For each color. we divide the vertices into four groups and paint in this color every edge joining two from the same group, as shown below. Then among any 5 vertices, 2 must belong to the same group, and the edge connecting them has the considered color. \begin{tabular}{lllll} yellow: & 011220 & 366993 & 57 & 48 \\ red: & 233442 & 588115 & 79 & 60 \\ blue: & 455664 & 700337 & 91 & 82 \\ green: & 677886 & 922559 & 13 & 04 \\ orange: & 899008 & 144771 & 35 & 26. \end{tabular} A desired coloring can be made for $k \geq 6$ as well. Paint the edge ij in the $(i+j)$ th color for $i2$, inscribe in the arena a regular $k$-gon, where $k=n(n-1) / 2$. Then we can establish a one-to-one correspondence between the segments cut off by the sides of the $k$-gon and the pairs of spotlights. Let each spotlight illuminate the entire $k$-gon and the segments corresponding to the pairs in which it is involved. It is easy to verify that this illumination has the required properties. ## Answer For all $n \geq 2$." 3c87b5843b75,"A3. The length of the aquarium is $50 \mathrm{~cm}$, the width is $20 \mathrm{~cm}$, and the height is $25 \mathrm{~cm}$. How many $\mathrm{cm}$ from the top edge of the aquarium will the water level be if we pour 19 liters of water into it? (A) $19 \mathrm{~cm}$ (B) $1,9 \mathrm{~cm}$ (C) $10,9 \mathrm{~cm}$ (D) $6 \mathrm{~cm}$ (E) $0,6 \mathrm{~cm}$",is D,easy,A3. We find that the volume of water is $19 \mathrm{dm}^{3}$. We write the data into the formula for the volume of water ( $x$ is the height of the water $\mathrm{v}$ in the aquarium) and get the equation $V=5 \cdot 2 \cdot x=19$. We calculate $x=1.9 \mathrm{dm}$ $=19 \mathrm{~cm}$. The water level will thus be $6 \mathrm{~cm}$ lower than the top edge of the aquarium. The correct answer is D. 2f6fd5a65f5c,"11. Choose three different digits from $0,1, \cdots, 9$ to form a four-digit number (one of the digits can appear twice), such as 5 224. Then the total number of such four-digit numbers is $\qquad$.",3888$ numbers.,medium,"11. 3888 . Discuss in three scenarios. Four-digit numbers that do not contain 0: $$ \mathrm{C}_{9}^{3} \mathrm{C}_{3}^{1} \mathrm{C}_{4}^{2} \mathrm{~A}_{2}^{2}=3024 \text { (numbers); } $$ Four-digit numbers that contain 0 and 0 appears only once: $$ \mathrm{C}_{9}^{2} \mathrm{C}_{2}^{1} \times 3 \times 3=648 \text { (numbers); } $$ Four-digit numbers that contain 0 and 0 appears twice: $$ \mathrm{C}_{9}^{2} \times 3 \mathrm{~A}_{2}^{2}=216 \text { (numbers). } $$ Thus, there are a total of $3024+648+216=3888$ numbers." e4b31f7e72b0,"7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$",\frac{D E}{D E}=1$.,easy,"$=, 7.1$. Connect $O A, D E$. $$ \begin{array}{l} \text { Since } \angle B O C=2 \angle B A C=120^{\circ} \\ \Rightarrow \angle C O E=60^{\circ}=\angle D A E \\ \Rightarrow A, D, O, E \text { are concyclic } \\ \Rightarrow \angle D E B=\angle D A O=\angle D B E \\ \Rightarrow D B=D E . \end{array} $$ Similarly, $C E=D E$. Thus, $\frac{B D}{C E}=\frac{D E}{D E}=1$." 660fb06eb2b3,"$$ (r+s-r s)(r+s+r s)=r s . $$ Find the minimum values of $r+s-r s$ and $r+s+r s$.",See reasoning trace,medium,"Solution. The given equation can be rewritten into $$ (r+s)^{2}=r s(r s+1) $$ Since $(r+s)^{2} \geq 4 r s$ for any $r, s \in \mathbb{R}$, it follows that $r s \geq 3$ for any $r, s>0$. Using this inequality, equation (1), and the assumption that $r$ and $s$ are positive, we have $$ \begin{aligned} r+s-r s=\sqrt{r s(r s+1)}-r s & =\frac{1}{\sqrt{1+\frac{1}{r s}}+1} \\ & \geq \frac{1}{\sqrt{1+\frac{1}{3}}+1}=-3+2 \sqrt{3} . \end{aligned} $$ Similarly, we also have $$ r+s+r s \geq 3+2 \sqrt{3} \text {. } $$ We show that these lower bounds can actually be attained. Observe that if $r=s=\sqrt{3}$, then $$ r+s-r s=-3+2 \sqrt{3} \text { and } r+s+r s=3+2 \sqrt{3} . $$ Therefore, the required minimum values of $r+s-r s$ and $r+s+r s$ are $-3+2 \sqrt{3}$ and $3+2 \sqrt{3}$, respectively." efdbe94b0ac1,149. Two equally matched opponents are playing chess. Find the most probable number of wins for any chess player if $2 N$ decisive (without draws) games will be played.,"2 N \cdot 1 / 2=N$ is an integer, the sought most probable number $k_{0}$ of games won is $N$.",medium,"Solution. It is known that if the product of the number of trials $\boldsymbol{n}$ and the probability $p$ of the event occurring in one trial is an integer, then the most probable number is $$ k_{0}=n p . $$ In the problem at hand, the number of trials $n$ is equal to the number of games played $2 N$; the probability of the event occurring is equal to the probability of winning in one game, i.e., $p=1 / 2$ (by the condition that the opponents are of equal strength). Since the product $n p=2 N \cdot 1 / 2=N$ is an integer, the sought most probable number $k_{0}$ of games won is $N$." 6145b7b3d27b,"5・142 Find all non-zero polynomials $P(x)$ that satisfy $$P\left(x^{2}-2 x\right) \equiv(P(x-2))^{2}, x \in R$$","(x+1)^{n}, n \in Z^{+}$.",easy,"[Solution] Let $y=x-1, Q(y)=P(y-1)$, then we have $$\begin{array}{l} \left(P(x-2)^{2}\right)=(P(y-1))^{2}=(Q(y))^{2} \\ P\left(x^{2}-2 x\right)=P\left(y^{2}-1\right)=Q\left(y^{2}\right) \end{array}$$ Thus, the original identity is transformed into $$Q\left(y^{2}\right) \equiv(Q(y))^{2}, y \in R$$ From the previous problem, we know that $Q(y)=y^{n}$. Therefore, $P(y)=(y+1)^{n}, n \in Z^{+}$. Hence, $P(x)=(x+1)^{n}, n \in Z^{+}$." 93d9e6efbda7,"## Task 2 - 340732 Imagine the numbers 1, 2, 3, 4, ... up to 100 written consecutively so that a number $z$ of the form $z=12345678910111213 \ldots .9899100$ is formed. a) How many digits does $z$ have? b) 100 digits of the number $z$ should be deleted so that the number $z^{\prime}$ represented by the remaining digits is as large as possible. The order of the digits remaining in $z^{\prime}$ from $z$ should not be changed. Determine which digits to delete, and give the first 10 digits of the new number $z^{\prime}$!",See reasoning trace,medium,"a) The number $z$ has its digits from the 9 single-digit numbers $1, \ldots, 9$, the 90 two-digit numbers 10, ..., 99, and the three-digit number 100. Thus, it has $9 + 90 \cdot 2 + 3 = 192$ digits. b) Of the 192 digits of the number $z$, exactly 92 digits should remain in $z'$. Of two 92-digit numbers that begin with a different number of nines, the one that begins with the larger number of nines is the larger number. Before the first nine that appears in $z$, there are 8 digits different from nine, before the second nine there are 19 more, and before the third, fourth, and fifth nines, there are again 19 more digits different from nine each. If these are removed, a total of $8 + 4 \cdot 19 = 84$ digits have already been removed; from the remaining number $$ 999995051525354555657585960 \ldots 9899100 $$ exactly 16 digits still need to be removed. These cannot be the 19 digits up to the first nine following 99999, nor the 17 digits up to the first eight following 99999. Of any two 92-digit numbers that begin with 99999 and have different digits at the sixth position, the one with the larger digit at the sixth position is always the larger number. The largest possibility here is to remove the first 15 digits following 99999 to reach the beginning 999997. After that, exactly one more digit needs to be removed. Of the two possibilities, to remove the five following 999997 or to leave it (and remove a later digit), removing the five yields the larger number. Thus, it has been determined which 100 digits from $z$ should be removed to obtain the largest possible number $z'$. The first ten digits of this number are 9999978596." aca9462b3b8d,"28. $x, y$ are both positive integers, the equation $[3.11 x]+[2.03 y]=30$ about $x, y$ has $\qquad$ solutions $(x, y)$. (Note: $[x]$ represents the greatest integer not exceeding $x$, for example $[2.1]=2,[3]=3$)",See reasoning trace,easy,$4$ f0e72fb1e59d,"[Permutations and Substitutions (other) $]$ [ Combinatorial Orbits $]$ Seventeen girls are forming a circle. In how many different ways can they stand in a circle? #",16! ways,medium,"The first method. We will fix one of the places in the circle. It is always possible to rotate the circle so that the first girl ends up in this place. The remaining 16 girls can arrange themselves in the remaining 16 places in 16! ways. The second method. 17 girls in 17 places can be arranged in 17! ways. We will divide all these arrangements into groups, combining into one group the arrangements that can be obtained from each other by rotations. Obviously, there are 17 arrangements in each group. Therefore, the number of groups (i.e., the number of ways to stand in a circle) is \( 17! / 17 = 16! \). ## Answer 16! ways." c9e4ef91e412,"14. Concept. In Anchuria, there are $K$ laws and $N$ ministers. The probability that a randomly chosen minister knows a randomly chosen law is $p$. One day, the ministers gathered for a council to write the Concept. If at least one minister knows a law, then this law will be considered in the Concept; otherwise, this law will not be considered in the Concept. Find: a) The probability that exactly $M$ laws will be considered in the Concept. b) The expected number of laws considered in the Concept.",a) $C_{K}^{M}\left(1-(1-p)^{N}\right)^{M}(1-p)^{N \cdot(K-M)} ; \boldsymbol{\text { b) }} K\left(1-(1-p)^{N}\right)$,medium,"Solution. a) Let's find the probability that no minister knows the first law: $(1-p)^{N}$. Therefore, the probability that the first law will be taken into account is $q=1-(1-p)^{N}$. The probability that exactly $M$ of the $K$ laws will be taken into account in the Concept can be easily found using the binomial probability formula: $$ \mathrm{P}_{M}=C_{K}^{M} q^{M}(1-q)^{K-M}=C_{K}^{M}\left(1-(1-p)^{N}\right)^{M}(1-p)^{N \cdot(K-M)} $$ b) Let's introduce an indicator - a random variable $I_{k}$, which equals 1 if the law numbered $k$ is taken into account and 0 otherwise. Such a variable is called an indicator. The expected value is $$ \mathrm{E} I_{k}=1 \cdot\left(1-(1-p)^{N}\right)+0 \cdot(1-p)^{N}=1-(1-p)^{N} $$ The number of laws taken into account $X$ is equal to the sum of these variables: $X=I_{1}+I_{2}+\ldots+I_{K}$. Therefore, $\mathrm{E} X=\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{K}=K \cdot\left(1-(1-p)^{N}\right)$. Answer: a) $C_{K}^{M}\left(1-(1-p)^{N}\right)^{M}(1-p)^{N \cdot(K-M)} ; \boldsymbol{\text { b) }} K\left(1-(1-p)^{N}\right)$." 98bba6702045,"A2. The coefficients of the polynomial $p$ of degree 3 are single-digit natural numbers, and it is also given that $p(\sqrt{10})=$ $12+34 \sqrt{10}$. What is $p(10)$? (A) 46 (B) 352 (C) 2022 (D) 3142 (E) 3494","1, d=2, a=3$ and $c=4$. Therefore, $p(x)=3 x^{3}+x^{2}+4 x+2$ and thus $p(10)=$ $3000+100+40+2=3142$",easy,"A2. Let's write $p(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c$ and $d$ are one-digit natural numbers. Then, $$ p(\sqrt{10})=a \cdot 10 \sqrt{10}+b \cdot 10+c \cdot \sqrt{10}+d=(10 b+d)+(10 a+c) \sqrt{10} $$ Since $a, b, c$ and $d$ are one-digit natural numbers, i.e., digits, we have $$ p(\sqrt{10})=\overline{b d}+\overline{a c} \sqrt{10}=12+34 \sqrt{10} $$ It follows that $b=1, d=2, a=3$ and $c=4$. Therefore, $p(x)=3 x^{3}+x^{2}+4 x+2$ and thus $p(10)=$ $3000+100+40+2=3142$." 8686df054be2,"7. How many different products that are divisible by 10 (order does not matter) can be formed from the numbers $2,3,5,7,9$? Numbers in the product do not repeat!",eight products,medium,"7. For a number to be divisible by 10, it must be divisible by 2 and 5. Thus, the product must necessarily include 2 and 5. We get the first product $-2 \times 5$. Now, let's add one more factor to this product: $$ 2 \times 5 \times 3, 2 \times 5 \times 7, 2 \times 5 \times 9 $$ Add one more factor to the obtained products: $$ 2 \times 5 \times 3 \times 7, 2 \times 5 \times 3 \times 9, 2 \times 5 \times 7 \times 9 $$ Add one more factor: $$ 2 \times 5 \times 3 \times 7 \times 9 $$ In total, we have eight products $1+3+3+1=8$. Answer: eight products. ## TWENTY-THIRD" d5e49e6a77e9,"10. There is a cup filled with a $16 \%$ saltwater solution. There are large, medium, and small iron balls, with their volume ratios being $10: 4: 3$. First, the small ball is submerged in the saltwater cup, causing $10 \%$ of the saltwater to overflow, and then the small ball is removed; next, the medium ball is submerged in the saltwater cup and then removed; then the large ball is submerged in the saltwater cup and removed; finally, pure water is added to the cup until it is full. What is the concentration of the saltwater in the cup at this point? (Round to one decimal place)","\frac{1}{3}$ of the saltwater, so the final amount of saltwater overflowed is $\frac{1}{3}$ of the c",easy,"Then, the ""10 parts"" volume of the large ball corresponds to $10 \% \div 3 \times 10=\frac{1}{3}$ of the saltwater, so the final amount of saltwater overflowed is $\frac{1}{3}$ of the cup's capacity. At this point, the concentration of saltwater in the cup is $16 \% \times\left(1-\frac{1}{3}\right) \div 1 \approx 10.7 \%$." 5c86bf4400ca,"## Task 1 - 280621 On the railway line from Pfiffigstadt to Knobelshausen, there are three more railway stations between these two towns: Adorf, Bedorf, and Cedorf. In each of these five stations, you can buy tickets to each of the other stations. André owns exactly one ticket for each of these possible connections. He does not yet have any other tickets in his collection. How many tickets does André have in total? (Hint: Outbound and return journeys are considered different connections, but there are no combined ""round-trip tickets"".)",20$ tickets due to this.,easy,"From each of the five places, there are exactly four train connections. Since all these train connections are different and André has exactly one ticket for each of them, he has a total of $5 \cdot 4=20$ tickets due to this." 4fbb50c08cf8,"131. Let $x=1, y, z$ be any three numbers; $x_{1}$, $y_{1}, z_{1}$ be the absolute values $|x-y|,|y-z|,|z-x|$ of the pairwise differences of these numbers; $x_{2}, y_{2}, z_{2}$ be the absolute values of the pairwise differences of the numbers $x_{1}, y_{1}, z_{1}$ (i.e., the numbers $\left.\left|x_{1}-y_{1}\right|,\left|y_{1}-z_{1}\right|,\left|z_{1}-x_{1}\right|\right) ; x_{3}, y_{3}, z_{3}$ be the absolute values of the differences of the numbers $x_{2}, y_{2}, z_{2}$, and so on. It is known that for some $n$ the triplet of numbers $x_{n}, y_{n}, z_{n}$ does not differ from the triplet of numbers $x, y, z$. What are the values of the numbers $y$ and $z$?","1,2,3, \ldots\). Since, obviously, \(x_i = x_{i-1} - z_{i-1}\) for all \(i \geqslant 1\) (where \(x_",medium,"131. First of all, let's determine for which numbers \(x, y, z\) the triples \((x_n, y_n, z_n)\) and \((x, y, z)\) can coincide. Since all triples of numbers starting from \(x_1, y_1, z_1\) are non-negative, the numbers \(x\), \(y\), and \(z\) must also be non-negative. Let's agree to consider that \(x \geqslant y \geqslant z\) and \(x_i \geqslant y_i \geqslant z_i\) for all \(i=1,2,3, \ldots\). Since, obviously, \(x_i = x_{i-1} - z_{i-1}\) for all \(i \geqslant 1\) (where \(x_0\) and \(z_0\) are understood to be the numbers \(x\) and \(z\)), it follows that \(x \geqslant x_1 \geqslant x_2 \geqslant x_3 \geqslant \ldots\), and if at least one number \(z_i (i=0,1,2, \ldots)\) is greater than zero, then \(x_{i+1}\) is also less than \(x\); therefore, if \(x_n = x\), then \(z_i = 0\) for \(i=0,1, \ldots, n-1\). Thus, it must be that \(z=0\) and \(z_1=0\), from which it follows that either \(y=x\) (and \(z_1 = x - y = 0\)), or \(y = z = 0\) (and \(z_1 = y - z = 0\)). Therefore, we see that for the triple of numbers \((x, y, z)\) to coincide with any triple \((x_n, y_n, z_n)\), it is necessary that the initial triple, when \(x=1\), either has the form \((1,1,0)\) or the form \((1,0,0)\). The second case can be immediately discarded, because from the triple \((1,0,0)\) we transition to the (different from the initial) triple \((1,1,0)\); therefore, if \(x=1\) and the triple \((x, y, z)\) coincides with \((x_n, y_n, z_n)\), then \((x, y, z)\) is the triple of numbers \((1,1,0)\) (and the triples \((x_n, y_n, z_n)\) for all \(n\) have the same structure):" c0ff8e92afe6,"57. If $a, b, c$ are all multiples of 5, $a\frac{1}{2} . $$",\sqrt{3-x}-\sqrt{x+1}$ is well-defined only for $-1 \leq x \leq 3$ and is decreasing (and obviously ,easy,2. We note that $f(x)=\sqrt{3-x}-\sqrt{x+1}$ is well-defined only for $-1 \leq x \leq 3$ and is decreasing (and obviously continuous) on this interval. We also note that $f(-1)=2>1 / 2$ and $f(1-\sqrt{31} / 8)=\sqrt{(1 / 4+\sqrt{31} / 4)^{2}}-$ $\sqrt{(1 / 4-\sqrt{31} / 4)^{2}}=1 / 2$. Hence the inequality is satisfied for $-1 \leq x<$ $1-\sqrt{31} / 8$ 0234ae2ff525,"$6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$. Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. $6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$.",$x=0$,medium,"Solution. $\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4} \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ x+x+9+2 \sqrt{x(x+9)}=x+1+x+4+2 \sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ 2+\sqrt{x(x+9)}=\sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 0 \\ 4+x^{2}+9 x+4 \sqrt{x(x+9)}=x^{2}+5 x+4\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ x+\sqrt{x(x+9)}=0\end{array} \Leftrightarrow x=0\right.$. Answer: $x=0$. ## $6.281 \sqrt[3]{x+5}+\sqrt[3]{x+6}=\sqrt[3]{2 x+11}$ Solution. Raise both sides to the third power: $2 x+11+3 \sqrt[3]{(x+5)(x+6)} \cdot(\sqrt[3]{x+5}+\sqrt[3]{x+6})=2 x+11 \Leftrightarrow$ $\sqrt[3]{(x+5)(x+6)} \cdot(\sqrt[3]{x+5}+\sqrt[3]{x+6})=0 \Leftrightarrow \sqrt[3]{(x+5)(x+6)(2 x+11)}=0 \Rightarrow\left[\begin{array}{l}x=-5, \\ x=-6, \\ x=-\frac{11}{2} .\end{array}\right.$ By checking, we confirm that all these values are roots of the original equation. Answer: $x_{1}=-6 ; x_{2}=-5.5 ; x_{3}=-5$." 944ca0a6ed33,314. Find the derivative of the function $y=\sin \left(x^{3}-3 x^{2}\right)$.,"x^{3}-3 x^{2}, y=\sin u$. We find $u_{x}^{\prime}=\left(x^{3}-\right.$ $\left.-3 x^{2}\right)^{\prim",easy,"Solution. Here $u=x^{3}-3 x^{2}, y=\sin u$. We find $u_{x}^{\prime}=\left(x^{3}-\right.$ $\left.-3 x^{2}\right)^{\prime}=3 x^{2}-6 x$. Then $y_{x}^{\prime}=y_{u}^{\prime} u_{x}^{\prime}=\left(3 x^{2}-6 x\right) \cos \left(x^{3}-3 x^{2}\right)$." 05b17f1ed503,"10.193. Inside a right angle, there is a point $M$, the distances from which to the sides of the angle are 4 and $8 \mathrm{~cm}$. A line passing through point $M$ cuts off a triangle from the right angle with an area of $100 \mathrm{~cm}^{2}$. Find the legs of the triangle.",40 and 5 cm or 10 and 20 cm,medium,"Solution. Given $\angle C=90^{\circ}, M P=4 \mathrm{~cm}, M Q=8 \mathrm{~cm}, S_{\triangle A B C}=100 \mathrm{~cm}^{2}$ (Fig. 10.3); we need to find $B C$ and $A C$. Let $B C=x, A C=y$; then $0.5 x y=100$, i.e., $x y=200$. Since $\triangle B P M \sim \triangle M Q A$, we have $\frac{M P}{A Q}=\frac{B P}{M Q}$ or $\frac{4}{y-4}=\frac{x-8}{8}$. We have the system of equations $$ \left\{\begin{array}{l} \frac{4}{y-4}=\frac{x-8}{8}, \\ x y=200 \end{array} \Leftrightarrow \left\{\begin{array}{l} x+2 y=50, \\ x y=200 \end{array} \Rightarrow y^{2}-25 y+100=0, y_{1}=5 \right.\right. \text { cm } $$ $y_{2}=20 \mathrm{~cm}$. We obtain two solutions: $x_{1}=40 \mathrm{~cm}, y_{1}=5 \mathrm{~cm} ; x_{2}=10 \mathrm{~cm}$, $y_{2}=20 \mathrm{~cm}$. Answer: 40 and 5 cm or 10 and 20 cm." 790d1c4ca67a,"10. Given the parabola $y^{2}=4 p(x+p)$ (where $p>0$), draw any two mutually perpendicular chords $A B$ and $C D$ through the origin $O$. Find the minimum value of $|A B|+|C D|$.","\frac{2 p}{1-\cos \theta}$. Given that $A B \perp C D$, we can set the polar coordinates of points $",medium,"10. Since the coefficient of $x$ in the parabola equation $y^{2}=4 p(x+p)$ is $4 p$, the distance from the focus to the vertex is $p$. Therefore, the origin $O$ is precisely the focus of the parabola. Taking a polar coordinate system with the origin $O$ as the pole and the positive half-axis of the $x$-axis as the polar axis, the polar equation of this parabola is $\rho=\frac{2 p}{1-\cos \theta}$. Given that $A B \perp C D$, we can set the polar coordinates of points $A, B, C, D$ to be $\left(\rho_{1}, \theta\right), \left(\rho_{2}, \theta+\pi\right), \left(\rho_{3}, \theta+\frac{\pi}{2}\right), \left(\rho_{4}, \theta+\frac{3 \pi}{2}\right)$. Thus, $\rho_{1}=\frac{2 p}{1-\cos \theta}, \rho_{2}=\frac{2 p}{1-\cos (\theta+\pi)}=\frac{2 p}{1+\cos \theta}, \rho_{3}=\frac{2 p}{1-\cos \left(\theta+\frac{\pi}{2}\right)}=\frac{2 p}{1+\sin \theta}, \rho_{4}=\frac{2 p}{1-\cos \left(\theta+\frac{3 \pi}{2}\right)}=\frac{2 p}{1-\sin \theta}$. Therefore, $|A B|=|O A|+|O B|=\rho_{1}+\rho_{2}=\frac{2 p}{1-\cos \theta}+\frac{2 p}{1+\cos \theta}=\frac{4 p}{\sin ^{2} \theta}$. Similarly, we get $|C D|=\rho_{3}+\rho_{4}=\frac{4 p}{\cos ^{2} \theta}$. Hence, $|A B|+|C D|=\frac{4 p}{\sin ^{2} \theta}+\frac{4 p}{\cos ^{2} \theta}=\frac{4 p}{\sin ^{2} \theta \cos ^{2} \theta}=\frac{16 p}{\left|\sin ^{2} 2 \theta\right|} \geqslant 16 p$. When $\theta=\frac{\pi}{4}$ or $\theta=\frac{3 \pi}{4}$, $|A B|+|C D|$ achieves its minimum value of $16 p$." c0c46f43f8dd,"5. $[a]$ represents the integer part of $a$, for example: [1.5]=1, [2]=2. Calculate: $\left[\frac{100}{2}\right]+\left[\frac{100}{2^{2}}\right]+\left[\frac{100}{2^{3}}\right]+\cdots+\left[\frac{100}{2^{100}}\right]=$",See reasoning trace,easy,$97$ 19829c09b774,"1. (17 points) Three cyclists, Dima, Misha, and Petya, started a 1 km race simultaneously. At the moment Misha finished, Dima still had to cover one tenth of the distance, and at the moment Dima finished, Petya still had to cover one tenth of the distance. How far apart (in meters) were Petya and Misha when Misha finished? (The cyclists are riding at constant speeds).",190,medium,"Answer: 190. Solution. Let $v_{1}$ be the speed of Misha, $v_{2}$ be the speed of Dima, and $v_{3}$ be the speed of Petya. Then $v_{2}=0.9 v_{1}$, since in the same time Misha traveled $1000 \mathrm{m}$, while Dima traveled $900 \mathrm{m}$, and $v_{3}=0.9 v_{2}$, since in the same time Dima traveled 1000 m, while Petya traveled 900 m. Since $v_{3}=$ $0.9 \cdot 0.9 v_{1}=0.81 v_{1}$, at the moment Misha finished, Petya had traveled 810 meters. The distance between Petya and Misha: $1000-810=190$" 49150adb6d85,"A [polynomial] \[P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0\] has [real] [coefficients] with $c_{2004}\not = 0$ and $2004$ distinct complex [zeroes] $z_k = a_k + b_ki$, $1\leq k\leq 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0$, and \[\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.\] Which of the following quantities can be a non zero number? $\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}$",\mathrm{E,medium,"We have to evaluate the answer choices and use process of elimination: $\mathrm{(A)}$: We are given that $a_1 = b_1 = 0$, so $z_1 = 0$. If one of the roots is zero, then $P(0) = c_0 = 0$. $\mathrm{(B)}$: By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), we know that $-\frac{c_{2003}}{c_{2004}}$ is the sum of all of the roots of $P(x)$. Since that is real, $\sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}$, and $\frac{c_{2003}}{c_{2004}}=0$, so $c_{2003}=0$. $\mathrm{(C)}$: All of the coefficients are real. For sake of contradiction suppose none of $b_{2\ldots 2004}$ are zero. Then for each complex root $z_k$, its [complex conjugate](https://artofproblemsolving.com/wiki/index.php/Complex_conjugate) $\overline{z_k} = a_k - b_k i$ is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that $b_1 = 0$.) This gives us the contradiction, and therefore the product is equal to zero. $\mathrm{(D)}$: We are given that $\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}$. Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero. There is, however, no reason to believe that $\boxed{\mathrm{E}}$ should be zero (in fact, that quantity is $P(1)$, and there is no evidence that $1$ is a root of $P(x)$)." 6f8a79ac5e4f,Example 4. Find the length of the common chord of the circles $x^{2}+y^{2}-10 x-10 y=0$ and $x^{2}+y^{2}+6 x+2 y-40=0$.,"2^{3} \cdot 1306 \mathrm{~m}=0\right)$, $10112=2^{7} \cdot 79(\mathrm{~m}=1)$, it is clear that they",medium,"Subtracting the two equations yields $4 x+3 y-10=0$. (Equation of the common chord) Completing the square for $x^{2}+y^{2}-10 x-10 y=0$, we get $(x-5)^{2}+(y-5)^{2}=50$. Thus, the center of the circle is $\mathbf{C}(5,5)$. The distance from the center of the circle to the common chord is $$ \mathrm{d}=\frac{|20+15-10|}{5}=5 \text {. } $$ By the Pythagorean theorem, the length of the chord is $$ \mathrm{l}=2 \sqrt{50-25}=10 \text {. } $$ $2^{n}$ cannot be divisible by 32. However, by assumption, $2^{n}(n \geqslant 6)$ can be divided by $2^{6}=64$, leading to a contradiction. Assume $2^{n}(n \geqslant 6)$ meets the problem's requirements and ends with the digit 2. In this case, the last few digits of $2^{\mathbf{n}}$ could be 02, 22, 012, or it could be of the form $$ 2^{\mathbf{n}}=10 \underbrace{0 \cdots 0112}_{\mathrm{m} \text { digits }} .(\mathrm{m} \geqslant 0) $$ Since 02 and 22 are not divisible by 4, and 012 is not divisible by 8, in these cases, $2^{n}$ cannot be divisible by 64. Therefore, the first two scenarios are impossible. For the last scenario, it is also impossible. Because $\left.1112=2^{3} \cdot 1306 \mathrm{~m}=0\right)$, $10112=2^{7} \cdot 79(\mathrm{~m}=1)$, it is clear that they are not powers of 2. When $\mathrm{m} \geqslant 2$, $10^{\mathrm{m}+3}$ is divisible by 32, but 112 is not divisible by 32. This also leads to a contradiction. In summary, only $n=5$ meets the requirements." fdfaad2b2ddf,"4. Given a unit cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, there is a point $M$ on the side $A A^{\prime} B^{\prime} B$ such that the distances from $M$ to the lines $A B$ and $B^{\prime} C^{\prime}$ are equal. Then the minimum distance from $M$ to point $C^{\prime}$ is $\qquad$",See reasoning trace,medium,"4. $\frac{\sqrt{5}}{2}$. Taking the midpoint $O$ of $B B^{\prime}$ as the origin, the direction of $B A$ as the $x$-axis, and the direction of $B B^{\prime}$ as the $y$-axis, establish a Cartesian coordinate system. Then, according to the problem, the trajectory of point $M$ is $$ y=\frac{1}{2} x^{2}(0 \leqslant x \leqslant 1), $$ The minimum distance from a point on this trajectory to point $B^{\prime}$ is $\frac{1}{2}$. Therefore, the minimum distance from the moving point $M$ to point $C^{\prime}$ is $\frac{\sqrt{5}}{2}$." e75d719c77cf,"15. Let the domain and range of $f(x)$ both be the set of non-negative real numbers, and satisfy the following conditions: (1) $f(x \cdot f(y)) \cdot f(y)=f(x+y), x, y \geqslant 0$; (2) $f(2)=0$; (3) $f(x) \neq 0$ for all $0 \leqslant x<2$. Find all functions $f(x)$ that satisfy the above conditions.","2$, then $f(x)=f((x-2)+2)=f((x-2) \cdot f(2)) \cdot f(2)=0$; when $0 \leqslant x<2$, then $f((2-x) \",medium,"15. When $x>2$, take $y=2$, then $f(x)=f((x-2)+2)=f((x-2) \cdot f(2)) \cdot f(2)=0$; when $0 \leqslant x<2$, then $f((2-x) \cdot f(x)) f(x)=f(2)=0$. Since $f(x) \neq 0$, then $f((2-x) \cdot f(x))=0, (2-x) f(x) \geqslant 2$, which gives $f(x) \geqslant \frac{2}{2-x}(00$, solving yields $a=\frac{1+\sqrt{5}}{2}$." 644882eac9af,"(4) Let $x, y, z \in \mathbf{R}^{+}, x+y+z=1$, then the maximum value of the function $f(x, y, z)=x y^{2} z^{3}$ is $\qquad$","\frac{y}{2}=\frac{z}{3}=\frac{1}{6}$, i.e., $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$.",medium,"(4) $\frac{1}{432}$ Hint: From $$ \begin{aligned} 1 & =x+y+z \\ & =x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3} \\ & \geqslant 6 \sqrt[6]{x \cdot \frac{y}{2} \cdot \frac{y}{2} \cdot \frac{z}{3} \cdot \frac{z}{3} \cdot \frac{z}{3},} \end{aligned} $$ Therefore, $$ \frac{1}{4 \cdot 27} x y^{2} z^{3} \leqslant\left(\frac{1}{6}\right)^{6}, $$ which means $$ x y^{2} z^{3} \leqslant \frac{1}{2^{4} \cdot 3^{3}}=\frac{1}{432}, $$ Equality holds when $x=\frac{y}{2}=\frac{z}{3}=\frac{1}{6}$, i.e., $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$." 261d28485097,"Example 1. Find the characteristic numbers and eigenfunctions of the integral equation $$ \varphi(x)-\lambda \int_{0}^{\pi}\left(\cos ^{2} x \cos 2 t+\cos 3 x \cos ^{3} t\right) \varphi(t) d t=0 $$",See reasoning trace,medium,"Solution. Imesm $$ \varphi(x)=\lambda \cos ^{2} x \int_{0}^{\pi} \varphi(t) \cos 2 t d t+\lambda \cos 3 x \int_{0}^{\pi} \varphi(t) \cos ^{3} t d t $$ Introducing the notations $$ C_{1}=\int_{0}^{\pi} \varphi(t) \cos 2 t d t, \quad C_{2}=\int_{0}^{\pi} \varphi(t) \cos ^{3} t d t $$ we will have $$ \varphi(x)=C_{1} \lambda \cos ^{2} x+C_{2} \lambda \cos 3 x $$ Substituting (6) into (5), we obtain a system of linear homogeneous equations $$ \left\{\begin{aligned} C_{1}\left(1-\lambda \int_{0}^{\pi} \cos ^{2} t \cos 2 t d t\right)-C_{2} \lambda \int_{0}^{\pi} \cos 3 t \cos 2 t d t & =0 \\ -C_{1} \lambda \int_{0}^{\pi} \cos ^{5} t d t+C_{2}\left(1-\lambda \int_{0}^{\pi} \cos ^{3} t \cos 3 t d t\right) & =0 \end{aligned}\right. $$ Since $$ \begin{aligned} & \int_{0}^{\pi} \cos ^{2} t \cos 2 t d t=\frac{\pi}{4}, \quad \int_{0}^{\pi} \cos 3 t \cos 2 t d t=0 \\ & \int_{0}^{\pi} \cos ^{5} t d t=0, \quad \int_{0}^{\pi} \cos ^{3} t \cos 3 t d t=\frac{\pi}{8} \end{aligned} $$ the system (7) will take the form $$ \left\{\begin{array}{l} \left(1-\frac{\lambda \pi}{4}\right) C_{1}=0 \\ \left(1-\frac{\lambda \pi}{8}\right) C_{2}=0 \end{array}\right. $$ The equation for finding the characteristic numbers: $$ \left|\begin{array}{cc} 1-\frac{\lambda \pi}{4} & 0 \\ 0 & 1-\frac{\lambda \pi}{8} \end{array}\right|=0 $$ Characteristic numbers: $\lambda_{1}=\frac{4}{\pi}, \lambda_{2}=\frac{8}{\pi}$. For $\lambda=\frac{4}{\pi}$ the system (8) takes the form $$ \left\{\begin{array}{l} 0 \cdot C_{1}=0 \\ \frac{1}{2} \cdot C_{2}=0 \end{array}\right. $$ from which $C_{2}=0, C_{1}$ is arbitrary. The eigenfunction will be $\varphi_{1}(x)=C_{1} \lambda \cos ^{2} x$, and, setting $C_{1} \lambda=1$, we get $\varphi_{1}(x)=\cos ^{2} x$. For $\lambda=\frac{8}{\pi}$ the system (8) will take the form $$ \left\{\begin{array}{r} (-1) \cdot C_{1}=0 \\ 0 \cdot C_{2}=0 \end{array}\right. $$ from which $C_{1}=0, C_{2}$ is arbitrary, and, therefore, the eigenfunction will be $\varphi_{2}(x)=$ $C_{2} \lambda \cos 3 x$, or, setting $C_{2} \lambda=1$, we get $\varphi_{2}(x)=\cos 3 x$. Thus, the characteristic numbers are: $$ \lambda_{1}=\frac{4}{\pi}, \quad \lambda_{2}=\frac{8}{\pi} $$ the corresponding eigenfunctions are: $$ \varphi_{1}(x)=\cos ^{2} x, \quad \varphi_{2}(x)=\cos 3 x $$ A homogeneous Fredholm integral equation may generally have no characteristic numbers and eigenfunctions, or it may have no real characteristic numbers and eigenfunctions." 64f86f4c340a,Example 5 Arrange $\frac{n(n+1)}{2}$ different numbers randomly into a triangular array as shown in Figure $12-1$. Let $M_{k}$ be the maximum number in the $k$-th row (from top to bottom). Try to find the probability that $M_{1}1\right\}, \\ N=\left\{(x, y) ; 00} \] This sequence indeed fits the recurrence relation, whatever \(k \in \mathbb{Z}_{\ge 0}\). - **Case \(c = 3\):** \[ a_1 = \frac{3(b - 1)}{2} \] This implies \(b\) is odd. We have: \[ a_2 = b, \quad a_3 = \frac{3(b + 1)}{2} \] \[ a_4 = b + 1 + \frac{b - 1}{3b + 1} \] This implies \(b = 1\), which is impossible since \(a_1 = 0\), not suitable. Conclusion: The only sequence that satisfies the given recurrence relation for all \(n \in \mathbb{Z}_{\ge 1}\) is: \[ \boxed{a_n = k + n \quad \forall n \in \mathbb{Z}_{>0}} \]" f01489ec1d7d,"Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$.","x_{1}+x_{2}$, the value of $y$ is 7.",easy,"Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have $$ \begin{array}{l} a x_{1}^{2}+b x_{1}+7=1994, \\ a x_{2}^{2}+b x_{2}+7=1994 . \end{array} $$ Subtracting the two equations and simplifying, we get $$ \left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 . $$ Since $P_{1}, P_{2}$ are two distinct points, i.e., $x_{1} \neq x_{2}$, we have $$ x_{1}+x_{2}=-\frac{b}{a}, $$ Thus, when $x=x_{1}+x_{2}$, the value of $y$ is 7." 61e60652feb5,"6. In the process of measuring a certain physical quantity, due to the errors of the instrument and observation, $n$ measurements yield $a_{1}, a_{2}, \cdots, a_{n}$, a total of $n$ data points. The ""best approximate value"" $a$ of the measured physical quantity is defined as a quantity such that, compared with other approximate values, the sum of the squares of the differences between $a$ and each data point is the smallest. According to this definition, the $a$ derived from $a_{1}, a_{2}, \cdots, a_{n}$ is equal to ( 1). A. $\sqrt[n]{a_{1} a_{2} \cdots a_{n}}$ B. $\frac{a_{1}+a_{n}}{2}$ C. $\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$ D. $\frac{1}{2} a_{n}$",See reasoning trace,easy,"6. C. $$ \begin{array}{l} \quad y=\left(a-a_{1}\right)^{2}+\left(a-a_{1}\right)^{2}+\cdots+\left(a-a_{n}\right)^{2}=n a^{2}-2\left(a_{1}+a_{n}+\cdots+a_{n}\right) a+\left(a_{1}^{2}+ \\ \left.a_{2}^{2}+\cdots+a_{n}^{2}\right) . \end{array} $$" bab7e4954469,"2. Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a continuous function and $M=\left\{\left.A(x)=\left(\begin{array}{cc}x & f(x) \\ 0 & x\end{array}\right) \right\rvert\, x \in(0, \infty)\right\}$. Determine the functions $f$ for which $M$ is a group with respect to matrix multiplication.",See reasoning trace,medium,"2. It is observed that $A(x) \cdot A(y)=\left(\begin{array}{cc}x y & x f(y)+y f(x) \\ 0 & x y\end{array}\right)$. Since $A(x) \cdot A(y)=A(x y)$, it follows that $f(x y)=x \cdot f(y)+y \cdot f(x), \forall x \in(0, \infty)$. With the notation $g(x)=\frac{f(x)}{x}, x>0$, we obtain that the continuous function $g$ satisfies the functional equation $g(x y)=g(x)+g(y), \forall x \in(0, \infty)$. Then $g(x)=k \log _{a} x$, hence $f(x)=k x \log _{a} x$, $\forall x \in(0, \infty)$, where $k \in \mathbb{R}, a \in(0,1) \cup(1, \infty)$. For functions $f$ of this form, the group axioms are immediately verified." 314322226115,"In the diagram, point $P$ is on the number line at 3 and $V$ is at 33 . The number line between 3 and 33 is divided into six equal parts by the points $Q, R, S, T, U$. ![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-026.jpg?height=117&width=683&top_left_y=2099&top_left_x=686) What is the sum of the lengths of $P S$ and $T V$ ? (A) 25 (B) 23 (C) 24 (D) 21 (E) 27",(A),easy,"The segment of the number line between 3 and 33 has length $33-3=30$. Since this segment is divided into six equal parts, then each part has length $30 \div 6=5$. The segment $P S$ is made up of 3 of these equal parts, and so has length $3 \times 5=15$. The segment $T V$ is made up of 2 of these equal parts, and so has length $2 \times 5=10$. Thus, the sum of the lengths of $P S$ and $T V$ is $15+10$ or 25 . ANSWER: (A)" 95e3c400abf4,"10.319. The diagonals of an isosceles trapezoid are perpendicular to each other, and its area is $a^{2}$. Determine the height of the trapezoid.",$a$,easy,"Solution. Let in trapezoid $A B C D$ (Fig. 10.108) $A B=C D, A C \perp B D, O$ - the intersection point of $A C$ and $B D, C K$ - the height of the trapezoid. Since the trapezoid is isosceles, then $A K=\frac{A D+B C}{2}, A O=D O$, and since $\angle A O D=90^{\circ}$, then $\angle O A D=45^{\circ}$. Therefore, in the right $\triangle A K C \quad A K=C K$. The area of the trapezoid $S=\frac{A D+B C}{2} \cdot C K=A K \cdot C K=C K^{2}=a^{2}$. Thus, $C K=a$. Answer: $a$." 3d395c31b06b,"II. (16 points) A plot of land can be covered by $n$ identical square tiles. If smaller identical square tiles are used, then $n+76$ such tiles are needed to cover the plot. It is known that $n$ and the side lengths of the tiles are integers. Find $n$.","\frac{19}{81}, n=324$.",medium,"Let the side length of the larger tile be $x$, and the side length of the smaller tile be $r$. Then $x, y$ are both natural numbers, and $$ n x^{2}=(n+76) y^{2} \text {. } $$ If $(x, y)=d$, let $x_{1}=\frac{x}{d}, y_{1}=\frac{y}{d}$, then $x_{1}, y_{1}$ are both natural numbers, and $\left(x_{1}, y_{1}\right)=1$. Thus, the previous equation can be transformed into $$ n x_{1}^{2}=(n+76) y_{1}^{2} \text {. } $$ We have $\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{n+76}{n}, \frac{x_{1}^{2}-y_{1}^{2}}{y_{1}^{2}}=\frac{76}{n}$. Since $\left(x_{1}^{2}-y_{1}^{2}, y_{1}^{2}\right)=1$, $\left(x_{1}-y_{1}\right)\left(x_{1}+y_{1}\right)$ is a divisor of 76 $=$ $2^{2} \times 19$. Noting that $x_{1}+y_{1}, x_{1}-y_{1}$ have the same parity, and $x_{1}+y_{1}>x_{1}-y_{1}$, we have $$ \left\{\begin{array} { l } { x _ { 1 } + y _ { 1 } = 1 9 , } \\ { x _ { 1 } - y _ { 2 } = 1 } \end{array} \text { or } \left\{\begin{array}{l} x_{1}+y_{1}=2 \times 19, \\ x_{1}-y_{1}=2 . \end{array}\right.\right. $$ Since $\left(x_{1}, y_{1}\right)=1$, we can only have $\left\{\begin{array}{l}x_{1}=10 \\ y_{1}=9 .\end{array}\right.$ Thus, $\frac{76}{n}=\frac{19}{81}, n=324$." 177fbc0830e0,"4. (2006 National High School Mathematics League Shaanxi Province Preliminary Test) If $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta, 0 \leqslant \theta<2 \pi$, then the range of angle 0 is A. $\left[0, \frac{\pi}{4}\right]$ B. $\left[\frac{\pi}{4}, \pi\right]$ C. $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$ D. $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right)$",See reasoning trace,easy,"4. C From $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta$, we get $(\sin \theta-\cos \theta)\left(2+\frac{1}{2} \sin ^{2} \theta\right) \geqslant 0$. Since $2+\frac{1}{2} \sin ^{2} \theta>0$, it follows that $\sin \theta-\cos \theta \geqslant 0$. Solving this, we get $\frac{\pi}{4} \leqslant \theta \leqslant \frac{5 \pi}{4}$." eaafa199412d,Find all integers $n \in \mathbb{Z}$ such that $n+1 \mid n^{2}+1$.,"2$. The divisors of 2 are $-2,-1,1,2$ so the solutions are $-3,-2,0,1$.",easy,"The integer $n$ satisfies the required property if and only if $n+1 \mid n^{2}+1-(n+1)(n-1)=2$. The divisors of 2 are $-2,-1,1,2$ so the solutions are $-3,-2,0,1$." 52a8a4245273,"4. Let $0a^{a}=x_{2}$, we get $x_{1}=a^{1}c$, then $\overline{\mathrm{abc}}=$ $\qquad$ .",: 452,medium,"【Solution】Solution: According to the analysis, the sum of the digits of $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are both: $a+b+c$, the remainder when the three-digit numbers $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are divided by the sum of their digits is 1, According to the principle of remainders, the difference between $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ can be divided by $a+b+c$. Let $\overline{\mathrm{abc}}-\overline{\mathrm{cba}}=\mathrm{n}(\mathrm{a}+\mathrm{b}+\mathrm{c})$. According to the principle of place value, we have: $$ \begin{array}{l} \overline{\mathrm{abc}}=100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c} ; \overline{\mathrm{cba}}=100 \mathrm{c}+10 \mathrm{~b}+\mathrm{a} \quad \text {, then: } \\ \overline{\mathrm{abc}}-\overline{\mathrm{cba}}=(100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c})-(100 \mathrm{c}+10 \mathrm{~b}+\mathrm{a}=n(a+b+c) \Rightarrow \\ 99 \times(a-c)=n(a+b+c) \Rightarrow 9 \times 11 \times(a-c)=n(a+b+c) ; \end{array} $$ $\therefore n$ and $(a+b+c)$ must have at least one that is a multiple of 11 and 9, and $\because a 、 b 、 c$ are all integers, $$ \text { and } 9 \geqslant a>c \geqslant 1,0 \leqslant b \leqslant 9, \quad \therefore 3=2+1+0 \leqslant a+b+c \leqslant 9+8+7=24 \text {, } $$ The possible values of $a+b+c$ are: $9,11,18,22$, (1) When $a+b+c=9$, according to the characteristic of being divisible by 9, the remainder when $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are divided by 9 is 0, which contradicts the problem statement; (2) When $a+b+c=11$, the three-digit number $\overline{a b c}$ is: $281, 362, 371, 452, 461, 524, 542, 614, 623, 632, 641, 731, 713, 812, 821, 704, 902, 605$, after checking, only 452 meets the condition; (3) When $a+b+c=22$, the three-digit number $\mathrm{abc}$ is: $985, 976, 967, 958, 796, 895$, the number that meets the condition is 0; (4) When $a+b+c=18$, the three-digit number $\overline{\mathrm{abc}}$ is: $954, 972, 945, 963, 981, 936, 927, 918, 891, 873, 864, 846, 837, 792, 783, 765, 695, 685, 693, 594$, after checking, the number that meets the condition is 0. In summary, $\overline{\mathrm{abc}}=452$, The answer is: 452." 3f4e06c96aee,"3. Given $O$ is the circumcenter of $\triangle A B C$, $|A B|=2,|A C|=1, \angle B A C=\frac{2}{3} \pi$, let $\overrightarrow{A B}=\boldsymbol{a}, \overrightarrow{A C}=\boldsymbol{b}$, if $\overrightarrow{A O}=\lambda_{1} a+\lambda_{2} \boldsymbol{b}$, then $\lambda_{1}+\lambda_{2}=$ $\qquad$","0$, which is $-\frac{5}{4} \times\left(-\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{4}-y\right) \times ",medium,"3. $\frac{13}{6}$, Detailed Explanation: Establish a Cartesian coordinate system, $A(0,0), B(2,0), C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Clearly, the midpoint of $A C$ is $M\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$, and we can set $O(1, y)$. Since $O M \perp A C$, we have $\overrightarrow{O M} \cdot \overrightarrow{A C}=0$, which is $-\frac{5}{4} \times\left(-\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{4}-y\right) \times \frac{\sqrt{3}}{2}=0$. Solving this, we get $y=\frac{2 \sqrt{3}}{3}$, so $\overrightarrow{A O}=\left(1, \frac{2 \sqrt{3}}{3}\right)$. From $\overrightarrow{A O}=\lambda_{1} a+\lambda_{2} b$, we have $1=2 \lambda_{1}+\left(-\frac{1}{2}\right) \lambda_{2}$, and $\frac{2 \sqrt{3}}{3}=\frac{\sqrt{3}}{2} \lambda_{2}$. Solving these, we get $\lambda_{2}=\frac{4}{3}, \lambda_{1}=\frac{5}{6}$, hence $\lambda_{1}+\lambda_{2}=\frac{13}{6}$." 5052f428855a,"4. Given the set $=\left\{x \mid x=a_{0}+a_{1} \times 8+a_{2} \times 8^{2}+a_{3} \times 8^{3}\right\}$, where $a_{i} \in\{0,1,2,3,4,5,6,7\}, i=0,1,2,3$ and $a_{3} \neq 0$, if positive integers $m, n \in A$, satisfy $m+n=2018$, and $m>n$, then the number of positive integers $m$ that meet the condition is $\qquad$",497$,easy,"Given: $\because m>n, m+n=2018, \therefore m \geq 1010$ $$ \because n \geq 1 \times 8^{3}=512, \quad m \leq 2018-512=1506, \quad \therefore 1010 \leq m \leq 1506 $$ The number of $m$ that satisfy the condition is $1506-1010+1=497$" 757d74da1f98,"10. In the following equation, the sum of 5 consecutive even numbers is exactly equal to the sum of 4 consecutive odd numbers. $$ 4+6+8+10+12=7+9+11+13 $$ Apart from the example given, in other equations that satisfy the condition, what is the smallest sum of 5 consecutive even numbers?",80,easy,Reference answer: 80 ad726ce3bc52,"$$ AK^4 - DH^4 = 15 $$","30^{\circ}$, and therefore, $A M=2$.",easy,"Instruction. Equation (1) has a unique solution. Let $\angle A B M=x$. Then $D H=2 \sin \left(60^{\circ}-x\right)$, $A K=4 \sin x$. Equation (1) takes the form: $$ 256 \sin ^{4} x-16 \sin ^{4}\left(60^{\circ}-x\right)=15 $$ Solving equation (2) on a microcalculator, we get $x=30^{\circ}$, and therefore, $A M=2$." 5b9cbd68e04c,"The lateral side of an isosceles trapezoid is equal to $a$, the midline is equal to $b$, and one angle at the larger base is $30^{\circ}$. Find the radius of the circle circumscribed around the trapezoid.",\(\sqrt{b^2 + \frac{a^2}{4}}\),medium,"The projection of the diagonal of an isosceles trapezoid onto the larger base is equal to the midline of the trapezoid. ## Solution Let $F$ be the projection of vertex $C$ of the smaller base $BC$ of the isosceles trapezoid $ABCD$ onto the larger base $AD$. Then the segment $AF$ is equal to the midline of the trapezoid, and since in the right triangle $CFD$ the angle $D$ is $30^{\circ}$, we have $CF = \frac{1}{2} CD$. From the right triangle $ACF$, we find that \[ AC = \sqrt{AF^2 + CF^2} = \sqrt{b^2 + \frac{a^2}{4}} \] If $R$ is the radius of the circle circumscribed around triangle $ACD$, then \[ R = \frac{AC}{2 \sin \angle D} = \sqrt{b^2 + \frac{a^2}{4}} \] It remains to note that the circle circumscribed around triangle $ACD$ coincides with the circle circumscribed around trapezoid $ABCD$. ## Answer \(\sqrt{b^2 + \frac{a^2}{4}}\)." aa5964a54095,"Find all functions $ f: \mathbb{Q}^{\plus{}} \rightarrow \mathbb{Q}^{\plus{}}$ which satisfy the conditions: $ (i)$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ for all $ x \in \mathbb{Q}^{\plus{}}$ $ (ii)$ $ f(x^2)\equal{}f(x)^2$ for all $ x \in \mathbb{Q}^{\plus{}}$.",f(x) = x,medium,"1. **Given Conditions:** - \( f(x + 1) = f(x) + 1 \) for all \( x \in \mathbb{Q}^+ \) - \( f(x^2) = f(x)^2 \) for all \( x \in \mathbb{Q}^+ \) 2. **Induction on \( f(x + n) \):** - Base case: For \( n = 1 \), \( f(x + 1) = f(x) + 1 \). - Inductive step: Assume \( f(x + k) = f(x) + k \) for some \( k \in \mathbb{N} \). Then, \[ f(x + (k + 1)) = f((x + k) + 1) = f(x + k) + 1 = (f(x) + k) + 1 = f(x) + (k + 1). \] - By induction, \( f(x + n) = f(x) + n \) for all \( n \in \mathbb{N} \). 3. **Analyzing \( f \left( \frac{a}{b} + b \right)^2 \):** - For positive integers \( a \) and \( b \), consider \( f \left( \left( \frac{a}{b} + b \right)^2 \right) \): \[ f \left( \left( \frac{a}{b} + b \right)^2 \right) = f \left( \frac{a^2}{b^2} + 2a + b^2 \right). \] - Using the induction result, we have: \[ f \left( \frac{a^2}{b^2} + 2a + b^2 \right) = f \left( \frac{a^2}{b^2} \right) + 2a + b^2. \] - Also, from the second condition: \[ f \left( \left( \frac{a}{b} + b \right)^2 \right) = f \left( \frac{a}{b} + b \right)^2. \] - Therefore: \[ f \left( \frac{a}{b} + b \right)^2 = f \left( \frac{a^2}{b^2} \right) + 2a + b^2. \] 4. **Simplifying the Equation:** - Let \( y = \frac{a}{b} \). Then: \[ f(y + b)^2 = f(y^2) + 2a + b^2. \] - Using the first condition: \[ f(y + b) = f(y) + b. \] - Substituting this into the equation: \[ (f(y) + b)^2 = f(y^2) + 2a + b^2. \] - Expanding and simplifying: \[ f(y)^2 + 2bf(y) + b^2 = f(y^2) + 2a + b^2. \] - Canceling \( b^2 \) from both sides: \[ f(y)^2 + 2bf(y) = f(y^2) + 2a. \] 5. **Solving for \( f(y) \):** - Since \( y = \frac{a}{b} \), we can rewrite \( 2a \) as \( 2by \): \[ f(y)^2 + 2bf(y) = f(y^2) + 2by. \] - For \( y = \frac{a}{b} \), we have: \[ f \left( \frac{a}{b} \right)^2 + 2b f \left( \frac{a}{b} \right) = f \left( \left( \frac{a}{b} \right)^2 \right) + 2a. \] - This implies: \[ f \left( \frac{a}{b} \right) = \frac{a}{b}. \] 6. **Conclusion:** - Since \( f \left( \frac{a}{b} \right) = \frac{a}{b} \) for all positive rational numbers \( \frac{a}{b} \), we conclude that: \[ f(x) = x \quad \forall x \in \mathbb{Q}^+. \] The final answer is \( \boxed{ f(x) = x } \) for all \( x \in \mathbb{Q}^+ \)." 1b6878a8aed0,"3. Let $X$ denote the set of all triples $(a, b, c)$ of integers. Define a function $f: X \rightarrow X$ by $$ f(a, b, c)=(a+b+c, a b+b c+c a, a b c) $$ Find all triples $(a, b, c)$ in $X$ such that $f(f(a, b, c))=(a, b, c)$.",See reasoning trace,medium," Solution: We show that the solutionset consists of $\{(t, 0,0) ; t \in \mathbb{Z}\} \cup\{(-1,-1,1)\}$. Let us put $a+b+c=d, a b+b c+c a=e$ and $a b c=f$. The given condition $f(f(a, b, c))=(a, b, c)$ implies that $$ d+e+f=a, \quad d e+e f+f d=b, d e f=c $$ Thus $a b c d e f=f c$ and hence either $c f=0$ or $a b d e=1$. Case I: Suppose $c f=0$. Then either $c=0$ or $f=0$. However $c=0$ implies $f=0$ and vice-versa. Thus we obtain $a+b=d, d+e=a, a b=e$ and $d e=b$. The first two relations give $b=-e$. Thus $e=a b=-a e$ and $d e=b=-e$. We get either $e=0$ or $a=d=-1$. If $e=0$, then $b=0$ and $a=d=t$, say. We get the triple $(a, b, c)=(t, 0,0)$, where $t \in \mathbb{Z}$. If $e \neq 0$, then $a=d=-1$. But then $d+e+f=a$ implies that $-1+e+0=-1$ forcing $e=0$. Thus we get the solution family $(a, b, c)=(t, 0,0)$, where $t \in \mathbb{Z}$. Case II: Suppose $c f \neq 0$. In this case $a b d e=1$. Hence either all are equal to 1 ; or two equal to 1 and the other two equal to -1 ; or all equal to -1 . Suppose $a=b=d=e=1$. Then $a+b+c=d$ shows that $c=-1$. Similarly $f=-1$. Hence $e=a b+b c+c a=1-1-1=-1$ contradicting $e=1$. Suppose $a=b=1$ and $d=e=-1$. Then $a+b+c=d$ gives $c=-3$ and $d+e+f=a$ gives $f=3$. But then $f=a b c=1 \cdot 1 \cdot(-3)=-3$, a contradiction. Similarly $a=b=-1$ and $d=e=1$ is not possible. If $a=1, b=-1, d=1$, $e=-1$, then $a+b+c=d$ gives $c=1$. Similarly $f=1$. But then $f=a b c=1 \cdot 1 \cdot(-1)=-1$ a contradiction. If $a=1, b=-1, d=-1, e=1$, then $c=-1$ and $e=a b+b c+c a=-1+1-1=-1$ and a contradiction to $e=1$. The symmetry between $(a, b, c)$ and $(d, e, f)$ shows that $a=-1, b=1, d=1, e=-1$ is not possible. Finally if $a=-1, b=1, d=-1$ and $e=1$, then $c=-1$ and $f=-1$. But then $f=a b c$ is not satisfied. The only case left is that of $a, b, d, e$ being all equal to -1 . Then $c=1$ and $f=1$. It is easy to check that $(-1,-1,1)$ is indeed a solution. ## Alternatively $c f \neq 0$ implies that $|c| \geq 1$ and $|f| \geq 1$. Observe that $$ d^{2}-2 e=a^{2}+b^{2}+c^{2}, \quad a^{2}-2 b=d^{2}+e^{2}+f^{2} $$ Adding these two, we get $-2(b+e)=b^{2}+c^{2}+e^{2}+f^{2}$. This may be written in the form $$ (b+1)^{2}+(e+1)^{2}+c^{2}+f^{2}-2=0 $$ We conclude that $c^{2}+f^{2} \leq 2$. Using $|c| \geq 1$ and $|f| \geq 1$, we obtain $|c|=1$ and $|f|=1$, $b+1=0$ and $e+1=0$. Thus $b=e=-1$. Now $a+d=d+e+f+a+b+c$ and this gives $b+c+e+f=0$. It follows that $c=f=1$ and finally $a=d=-1$. " 50e951ada44f,"Find all pairs of real numbers $(x, y)$, that satisfy the system of equations: $$\left\{\begin{matrix} 6(1-x)^2=\frac{1}{y},\\6(1-y)^2=\frac{1}{x}.\end{matrix}\right.$$",\left(\frac{3,medium,"To find all pairs of real numbers \((x, y)\) that satisfy the system of equations: \[ \left\{ \begin{matrix} 6(1-x)^2 = \frac{1}{y}, \\ 6(1-y)^2 = \frac{1}{x} \end{matrix} \right. \] 1. Start by rewriting the given equations: \[ 6(1-x)^2 = \frac{1}{y} \quad \text{and} \quad 6(1-y)^2 = \frac{1}{x} \] 2. Multiply both sides of each equation by \(y\) and \(x\) respectively: \[ 6y(1-x)^2 = 1 \quad \text{and} \quad 6x(1-y)^2 = 1 \] 3. Equate the two expressions: \[ 6y(1-x)^2 = 6x(1-y)^2 \] 4. Simplify the equation: \[ y(1-x)^2 = x(1-y)^2 \] 5. Expand both sides: \[ y(1 - 2x + x^2) = x(1 - 2y + y^2) \] 6. Rearrange the terms: \[ y - 2xy + yx^2 = x - 2xy + xy^2 \] 7. Combine like terms: \[ y + yx^2 = x + xy^2 \] 8. Rearrange to isolate terms involving \(x\) and \(y\): \[ yx^2 - xy^2 = x - y \] 9. Factor out common terms: \[ xy(x - y) = x - y \] 10. Since \(x - y \neq 0\) (otherwise, the terms would cancel out), divide both sides by \(x - y\): \[ xy = 1 \] 11. Substitute \(y = \frac{1}{x}\) into the first equation: \[ 6(1-x)^2 = x \] 12. Solve the quadratic equation: \[ 6(1 - 2x + x^2) = x \implies 6x^2 - 12x + 6 = x \implies 6x^2 - 13x + 6 = 0 \] 13. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -13\), and \(c = 6\): \[ x = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{13 \pm 5}{12} \] 14. Solve for \(x\): \[ x = \frac{18}{12} = \frac{3}{2} \quad \text{or} \quad x = \frac{8}{12} = \frac{2}{3} \] 15. Corresponding \(y\) values are: \[ y = \frac{1}{x} \implies y = \frac{2}{3} \quad \text{or} \quad y = \frac{3}{2} \] 16. Therefore, the pairs \((x, y)\) are: \[ \left(\frac{3}{2}, \frac{2}{3}\right) \quad \text{and} \quad \left(\frac{2}{3}, \frac{3}{2}\right) \] 17. Consider the case \(x = y\): \[ 6(1-x)^2 = \frac{1}{x} \] 18. Substitute \(x = y\) into the equation: \[ 6(1-x)^2 = \frac{1}{x} \implies 6x^3 - 12x^2 + 6x - 1 = 0 \] 19. Solve the cubic equation using the cubic formula or numerical methods: \[ x = \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}}) \] 20. Therefore, the pair \((x, y)\) is: \[ \left(\frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}}), \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}})\right) \] The final answer is \( \boxed{ \left(\frac{3}{2}, \frac{2}{3}\right) } \), \(\left(\frac{2}{3}, \frac{3}{2}\right)\), and \(\left(\frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}}), \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}})\right)\)." 6b2ea46cb9e3,"4. (ROM) Solve the equation $$ \cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1 . $$",1+\cos ^{2} x$ and $\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\fr,easy,"4. Since $\cos 2 x=1+\cos ^{2} x$ and $\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$, we have $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1 \Leftrightarrow \cos 2 x+\cos 4 x+2 \cos ^{2} 3 x=$ $2 \cos 3 x(\cos x+\cos 3 x)=0 \Leftrightarrow 4 \cos 3 x \cos 2 x \cos x=0$. Hence the solutions are $x \in\{\pi / 2+m \pi, \pi / 4+m \pi / 2, \pi / 6+m \pi / 3 \mid m \in \mathbb{Z}\}$." 7342fa55d9e3,"On Monday, $10 \%$ of the students at Dunkley S.S. were absent and $90 \%$ were present. On Tuesday, $10 \%$ of those who were absent on Monday were present and the rest of those absent on Monday were still absent. Also, $10 \%$ of those who were present on Monday were absent and the rest of those present on Monday were still present. What percentage of the students at Dunkley S.S. were present on Tuesday? (A) $81 \%$ (B) $82 \%$ (C) $90 \%$ (D) $91 \%$ (E) $99 \%$","820$ students present on Tuesday, or $\frac{820}{1000} \times 100 \%=82 \%$ of the whole student pop",medium,"Suppose that there are 1000 students at Dunkley S.S. On Monday, there were thus 100 students absent and 900 students present. On Tuesday, $10 \%$ of the 900 students who were present on Monday, or $0.1(900)=90$ students, were absent. The remaining $900-90=810$ students who were present on Monday were still present on Tuesday. Similarly, $10 \%$ of the 100 students who were absent on Monday, or $0.1(100)=10$ students, were present on Tuesday. The remaining $100-10=90$ students who were absent on Monday were still absent on Tuesday. Thus, there were $810+10=820$ students present on Tuesday, or $\frac{820}{1000} \times 100 \%=82 \%$ of the whole student population." 84a9aa4c15c8,"4. In how many ways can a bamboo trunk (a non-uniform natural material) 4 m long be sawn into three parts, the lengths of which are multiples of 1 dm, and from which a triangle can be formed? (12 points)",171,medium,"Solution. Let $A_{n}$ be the point on the trunk at a distance of $n$ dm from the base. We will saw at points $A_{p}$ and $A_{q}, p25$, so he cannot guarantee himself any win. c) Let Papa Carlo bet a total of $S$ gold coins, of which $x$ on the first horse, $y$ on the second, and $z$ on the third $(x+y+z=S)$. Suppose the first horse wins. Then Papa Carlo will receive $x+4x=5x$, and he needs this amount to be at least 5 more than $S$. We have the inequality $5x \geqslant S+5$, which means $x \geqslant \frac{S}{5}+1$. Now suppose the second horse wins; similar reasoning leads to the inequality $y \geqslant \frac{S}{4}+1.25$. Finally, assuming the third horse wins, we get the inequality $z \geqslant \frac{S}{2}+2.5$. Adding all three inequalities, we get $x+y+z \geqslant \frac{19S}{20}+4.75$. Since $x+y+z=S$, the last inequality is equivalent to the condition $S \geqslant 95$. An example where $S=95$ is obtained if we replace all inequalities in the previous reasoning with equalities. Then $x=20, y=25$, and $z=50$. It is easy to check that in any outcome of the races, Papa Carlo will win his 5 gold coins. d) The reasoning is similar to the reasoning in the previous section. Suppose Karabas-Barabas bet $S$ gold coins, of which $x$ on the first horse, $y$ on the second, and $z$ on the third. To guarantee a 6% win, it is necessary to bet on each horse so that the money received for its win (including the return of the bet) is at least $1.06S$. This is equivalent to the system of inequalities: $5x \geqslant 1.06S, 4y \geqslant 1.06S$, and $2z \geqslant 1.06S$. Multiplying the first inequality by 4, the second by 5, and the third by 10, and adding the left and right sides of all three inequalities, we get $20(x+y+z) \geqslant 1.06S(4+5+10)$, from which (considering that $x+y+z=S$), we get the impossible inequality $20 \geqslant 20.14$. Therefore, a 6% win cannot be guaranteed." 6faa39e3e7e6,"Prove that there exists a $k_0\in\mathbb{N}$ such that for every $k\in\mathbb{N},k>k_0$, there exists a finite number of lines in the plane not all parallel to one of them, that divide the plane exactly in $k$ regions. Find $k_0$.",k_0 = 5,medium,"1. **Understanding the problem**: We need to prove that there exists a natural number \( k_0 \) such that for every \( k \in \mathbb{N} \) with \( k > k_0 \), there exists a finite number of lines in the plane, not all parallel to one another, that divide the plane into exactly \( k \) regions. We also need to find the smallest such \( k_0 \). 2. **Analyzing the number of regions formed by lines**: - For \( n \) lines in general position (no two lines are parallel and no three lines are concurrent), the maximum number of regions \( R(n) \) formed is given by: \[ R(n) = \frac{n(n+1)}{2} + 1 \] - For example: - 1 line divides the plane into 2 regions. - 2 lines (if not parallel) divide the plane into 4 regions. - 3 lines (if no two are parallel and no three are concurrent) divide the plane into 7 regions. - 4 lines (in general position) divide the plane into 11 regions. - 5 lines (in general position) divide the plane into 16 regions. 3. **Special cases with fewer regions**: - If some lines are parallel or concurrent, the number of regions decreases. - For 5 lines: - If one pair is parallel, we get 15 regions. - If two pairs are parallel, we get 14 regions. - If one triple is parallel, we get 13 regions. - If two pairs are parallel and the fifth line intersects at two points, we get 12 regions. 4. **Generalizing for \( N \geq 12 \)**: - For \( N \geq 12 \), we can write \( N = 5k + m \) where \( m \in \{12, 13, 14, 15, 16\} \). - We can start with an arrangement of 5 lines that gives \( m \) regions. - By adding \( k \) lines parallel to one of the existing lines (which is not parallel to any other line), we can increase the number of regions by \( k \), thus achieving \( N \) regions. 5. **Finding the smallest \( k_0 \)**: - We need to check the smallest number of regions that can be formed with fewer lines: - With 4 lines, we can get up to 11 regions. - With 3 lines, we can get up to 7 regions. - With 2 lines, we can get up to 4 regions. - With 1 line, we get 2 regions. - We observe that we cannot get exactly 5 regions without having all lines parallel. - Therefore, the smallest \( k_0 \) such that for every \( k > k_0 \), there exists a finite number of lines not all parallel that divide the plane into exactly \( k \) regions is \( k_0 = 5 \). The final answer is \( \boxed{ k_0 = 5 } \)." e9b654bc2f10,"Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$. For how many integers with $1\le{n}\le{22}$ is $k_n1998$, Also, $44^{2}=1936<1998$; Thus, $n=45$, the odd number that was subtracted is $2025-1998=27$. Therefore, the answer is: 27." a3d7177814b8,"2. Given $$ \left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. } $$ Then the minimum value of $x+y$ is $\qquad$.","y=1$, the equality holds.",medium,"2. 2 . Notice, $$ \begin{array}{l} \left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\ \Rightarrow 2 x+\sqrt{4 x^{2}+1} \\ \geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} . \end{array} $$ When $x=\frac{1}{y}$, $$ \left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right)=1 . $$ And $f(x)=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing, so $x \geqslant \frac{1}{y}$. Thus, $x+y \geqslant y+\frac{1}{y} \geqslant 2$. When $x=y=1$, the equality holds." 709068148f21,"# 9. Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Find the largest possible value of the sum of the four numbers that Petya came up with. Points for the #",806,easy,"# 9. Problem 9.1* Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Find the largest possible value of the sum of the four numbers that Petya came up with. Points for the problem: 13. Answer: 806" a79ff6587241,"Example 4.29 Place $n(n \geqslant 1)$ distinct balls into 4 distinct boxes $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$, find the number of different ways to place the balls, $g_{n}$, such that $A_{1}$ contains an odd number of balls, and $A_{2}$ contains an even number of balls.",See reasoning trace,medium,"Let $n$ balls be denoted as $a_{1}, a_{2}, \cdots, a_{n}$, then any method of placing the balls that satisfies the problem corresponds to an $n$-repeated permutation of the 4-element set $A=\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$. The correspondence is as follows: if ball $a_{i}$ is placed in box $A_{k}$, then $A_{k}$ is placed in the $i$-th position. Thus, $g_{n}$ equals the number of $n$-repeated permutations of the 4-element set $A=\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ that satisfy the condition: “$A_{1}$ appears an odd number of times, $A_{2}$ appears an even number of times,” and thus $g_{n}$ is $$ \begin{aligned} E(t)= & \left(t+t^{3} / 3!+t^{5} / 5!+\cdots\right) \cdot\left(1+t^{2} / 2!+t^{4} / 4!+\cdots\right) \cdot \\ & \left(1+t+t^{2} / 2!+t^{3} / 3!+\cdots\right)^{2} \end{aligned} $$ The coefficient of $t^{n} / n!$ in the expansion, and $$ \begin{aligned} E(t) & =\frac{\mathrm{e}^{t}-\mathrm{e}^{-t}}{2} \cdot \frac{\mathrm{e}^{t}+\mathrm{e}^{-t}}{2} \cdot \mathrm{e}^{2 t} \\ & =\frac{1}{4}\left(\mathrm{e}^{4 t}-1\right)=\sum_{n=1}^{\infty} \frac{4^{n}}{4} t^{n} / n! \\ & =\sum_{n=1}^{\infty} 4^{n-1} t^{n} / n!, \end{aligned} $$ Therefore, $$ g_{n}=4^{n-1} . $$" 5e804289ef14,"11.1. For what real values of the parameter a does the graph of the function $f: \square \rightarrow \square$, $f(x)=x^{4}-8 x^{3}+14 x^{2}+$ ax have an axis of symmetry parallel to the line $x=0$?",$a=8$,medium,"## Solution. Since the graph of the function $f(x)$ has an axis of symmetry parallel to the line $x=0$, then the axis of symmetry has the equation $x=c$, where $c$ is a constant real value. Substituting $t=x-c$, we obtain that the graph of the function $f(t)$, as a function of $t$, will be symmetric with respect to the line $t=0$, i.e., the $O y$ axis. We have: $$ \begin{gathered} P(t)=x^{4}-8 x^{3}+14 x^{2}+a x=(t+c)^{4}-8(t+c)^{3}+14(t+c)^{2}+a(t+c)= \\ =t^{4}+4 t^{3} c+6 t^{2} c^{2}+4 t c^{3}+c^{4}-8\left(t^{3}+3 t^{2} c+3 t c^{2}+c^{3}\right)+14\left(t^{2}+2 t c+c^{2}\right)+a t+a c= \\ =t^{4}+t^{3}(4 c-8)+t^{2}\left(6 c^{2}-24 c+14\right)+t\left(4 c^{3}-24 c^{2}+28 c+a\right)+\left(c^{4}-8 c^{3}+14 c^{2}+a c\right) \end{gathered} $$ The graph of the function $f(t)$ is symmetric with respect to the $O y$ axis if and only if the function $f(t)$ is even, i.e., if and only if the coefficients of the function $f(t)$ corresponding to the odd powers of $t$ are equal to 0. We obtain $$ \left\{\begin{array} { c } { 4 c - 8 = 0 } \\ { 4 c ^ { 3 } - 2 4 c ^ { 2 } + 2 8 c + a = 0 } \end{array} \Leftrightarrow \left\{\begin{array} { c } { c = 2 } \\ { a = - 4 c ^ { 3 } + 2 4 c ^ { 2 } - 2 8 c } \end{array} \Leftrightarrow \left\{\begin{array}{c} c=2 \\ a=-4 \cdot 8+24 \cdot 4-28 \cdot 2=8 \end{array}\right.\right.\right. $$ Final answer: $a=8$." 97e5101380d4,"6、Let $x$ be an acute angle, then the maximum value of the function $y=\sin x \sin 2 x$ is $\qquad$ Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. 6、Let $x$ be an acute angle, then the maximum value of the function $y=\sin x \sin 2 x$ is $\qquad$",$\frac{4 \sqrt{3}}{9}$,easy,"Answer: $\frac{4 \sqrt{3}}{9}$. Solution: Given $y=2 \sin ^{2} x \cos x$, we get $y^{2}=4 \sin ^{4} x \cos ^{2} x=2\left(1-\cos ^{2} x\right)\left(1-\cos ^{2} x\right) \cdot 2 \cos ^{2} x$ $$ \leq 2\left(\frac{\left(1-\cos ^{2} x\right)+\left(1-\cos ^{2} x\right)+2 \cos ^{2} x}{3}\right)^{3}=2 \cdot\left(\frac{2}{3}\right)^{3}=\frac{16}{27} \text {, } $$ Therefore, $y \leq \frac{4 \sqrt{3}}{9}$, with equality when $\cos ^{2} x=\frac{1}{3}$." b5fdf5f5ded8,"3. The fields of a board with 25 rows and 125 columns are alternately colored black and white (like the fields on a chessboard), and knight figures are placed on it, which attack according to special rules. A knight on a white field attacks all white fields in the same row and all black fields in the same column, while a knight on a black field attacks all black fields in the same row and all white fields in the same column (even if there is another figure in between). Determine the maximum number of knights that can be placed on the board so that they do not attack each other.",1071&width=1725&top_left_y=1663&top_left_x=180),medium,"## Solution: First, note that in each row, there can be at most two knights, one on a white square and one on a black square - if there were three knights in one row, two of them would have to be on squares of the same color, thus attacking each other. The board has 25 rows, so it can have at most $2 \cdot 25 = 50$ knights. The problem will be solved if we show that it is indeed possible to place 50 knights on the board in the required manner. In a given column, all knights must be on squares of the same color. The image below shows one possible arrangement of 50 knights that do not attack each other: ![](https://cdn.mathpix.com/cropped/2024_05_30_4ba13bd558581b5f9252g-4.jpg?height=1071&width=1725&top_left_y=1663&top_left_x=180)" e713168b6975,"\section*{ How many ordered pairs of integers \((x, y)\) are there in total for which \(x \cdot y=1987\) holds?",See reasoning trace,medium,"} The number 1987 is not divisible by any of the numbers \[ 2,3,5,6,7,11,17,19,23,29,31,37,41,43 \] (This can be verified, for example, by calculating the 14 divisions \(1987: 2, 1987: 3, \ldots\) using a calculator). Since \(\sqrt{1987} < 44\), this means that 1987 is not divisible by any prime number \(p\) for which \(p \leq \sqrt{1987}\). Thus, it is proven: 1987 is a prime number. It follows that there are exactly the following four ordered pairs of integers \((x ; y)\) such that \(x \cdot y = 1987\): \[ (1,1987) \quad, \quad(-1,-1987) \quad, \quad(1987,1) \quad, \quad(-1987,-1) \]" bccee4439801,"76. Teacher Wang bought 132 pencils, 75 exercise books, and 37 pencil sharpeners, and divided them into several identical prize packages, with the same number of pencils, exercise books, and pencil sharpeners remaining. How many prize packages did Teacher Wang make at most?",19,easy,Reference answer: 19 8c4a95547dca,"## Calculate the limit of the function: $$ \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{-7 x}}{2 x-\tan x} $$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{-7 x}}{2 x-\operatorname{tg} x}=\lim _{x \rightarrow 0} \frac{\left(243^{x}-1\right)-\left(128^{-x}-1\right)}{2 x-\operatorname{tg} x}= \\ & =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 243}\right)^{x}-1\right)-\left(\left(e^{\ln 128}\right)^{-x}-1\right)}{2 x-\operatorname{tg} x}= \\ & =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)}{2 x-\operatorname{tg} x}= \\ & =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)\right)}{\frac{1}{x}(2 x-\operatorname{tg} x)}= \\ & =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 x-\operatorname{tg} x)}= \end{aligned} $$ $$ =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 243}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 128}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x}\right)= $$ Using the substitution of equivalent infinitesimals: $e^{x \ln 243}-1 \sim x \ln 243$, as $x \rightarrow 0(x \ln 243 \rightarrow 0)$ $e^{-x \ln 128}-1 \sim -x \ln 128$, as $x \rightarrow 0(-x \ln 128 \rightarrow 0)$ $\operatorname{tg} x \sim x$, as $x \rightarrow 0$ We get: $$ \begin{aligned} & =\frac{\lim _{x \rightarrow 0} \frac{x \ln 243}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 128}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 243-\lim _{x \rightarrow 0}-\ln 128}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}= \\ & =\frac{\ln 243+\ln 128}{2-1}=\ln (243 \cdot 128)=\ln \left(3^{5} \cdot 2^{7}\right) \end{aligned} $$ ## Problem Kuznetsov Limits 15-27" 5e7cb89cb42c,"3. Each square on an $8 \times 8$ checkers board contains either one or zero checkers. The number of checkers in each row is a multiple of 3 , the number of checkers in each column is a multiple of 5 . Assuming the top left corner of the board is shown below, how many checkers are used in total?",30 checkers,medium,"Solution. Answer: 30 checkers. The total number of checkers $T$, must be a multiple of 3 (adding all rows). It must also be a multiple of 5 (adding all columns). 3 and 5 are both prime, hence the only numbers which are multiples of both must be multiples of 15 , hence $T=15 k$ for some $k$. The depicted corner of the board has 2 checkers, hence the total number of checkers is $>0$. Every column has a multiple of 5 , but also has a maximum of 8 , hence the maximum possible number is 5 per column, hence $T \leq 5 \times 8=40$. Hence, $T=15$ or $T=30$. Currently we have two checkers in different columns. Each of these columns must have 5 checkers. Assuming we have 15 checkers in total, this indicates that we have only one remaining non-zero column but both the first and second row need to have two additional checkers in them to reach a multiple of three, leading to (at least) four non-zero columns, ie $T \geq 4 \times 5=20$. The only multiple of 15 in the appropriate range is 30 , therefore 30 checkers have been used." a520f12f175f,Putnam 1992,(y+1)(y+2) ... (y+1992)/1992! . Hence the integrand is simply the derivative of (y+1)(y+2) ... (y+19,medium,"Trivial. The usual Taylor series gives C(-y-1) = (y+1)(y+2) ... (y+1992)/1992! . Hence the integrand is simply the derivative of (y+1)(y+2) ... (y+1992)/1992! , so the integral evaluates to 1993 - 1 = 1992. Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 24 Dec 1998" 391b4a30a446,"## Task Condition Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line. $M(2, -1, 1)$ $$ \frac{x-4.5}{1}=\frac{y+3}{-0.5}=\frac{z-2}{1} $$",See reasoning trace,medium,"## Solution We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector: $\vec{n}=\vec{s}=\{1 ;-0.5 ; 1\}$ Then the equation of the desired plane is: $1 \cdot(x-2)-0.5 \cdot(y-(-1))+1 \cdot(z-1)=0$ $x-2-0.5 \cdot y-0.5+z-1=0$ $x-0.5 \cdot y+z-3.5=0$ We find the point $M_{0}$ of intersection of the line and the plane. We write the parametric equations of the line. $$ \begin{aligned} & \frac{x-4.5}{1}=\frac{y+3}{-0.5}=\frac{z-2}{1}=t \Rightarrow \\ & \left\{\begin{array}{l} x=4.5+t \\ y=-3-0.5 t \\ z=2+t \end{array}\right. \end{aligned} $$ Substitute into the equation of the plane: $(4.5+t)-0.5 \cdot(-3-0.5 t)+(2+t)-3.5=0$ $4.5+t+1.5+0.25 t+2+t-3.5=0$ $2.25 t+4.5=0$ $t=-2$ We find the coordinates of the point of intersection of the line and the plane: $$ \left\{\begin{array}{l} x=4.5+(-2)=2.5 \\ y=-3-0.5 \cdot(-2)=-2 \\ z=2+(-2)=0 \end{array}\right. $$ We get: $$ M_{0}(2.5 ;-2 ; 0) $$ Since $M_{0}$ is the midpoint of the segment $M M^{\prime}$, then $$ \begin{aligned} & x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2.5-2=3 \\ & y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-2)-(-1)=-3 \\ & z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0-1=-1 \end{aligned} $$ We get: $M^{\prime}(3 ;-3 ;-1)$" 5ad48ccc5a5f,"$2 \cdot 63$ If the roots of the equation $x^{2}+(m-1) x+4-m=0$ are both greater than 2, then the range of values for $m$ is (A) $m \leqslant-6$. (B) $-62$ and $x_{2}>2$. Let $y=x-2$, i.e., $x=y+2$ substituted into the original equation gives i.e., $$ \begin{array}{c} (y+2)^{2}+(m-1)(y+2)+4-m=0, \\ y^{2}+(m+3) y+m+6=0 . \end{array} $$ $\because$ The two roots of the original equation are both greater than 2, then the two roots of $\circledast$ are both greater than 0. Thus $\left\{\begin{array}{l}\Delta=(m+3)^{2}-4(m+6) \geqslant 0, \\ m+3>0 .\end{array}\right.$ Solving gives $\quad\left\{\begin{array}{l}m \geqslant 3 \text { or } m \leqslant-5, \\ m>-3 .\end{array}\right.$ $\therefore \quad-60)$. If $\triangle P O B$ is an isosceles triangle, then the coordinates of point $P$ are $\qquad$",See reasoning trace,medium,"4. $\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right)$ or $(2 \sqrt{3}, 2)$. From the problem, we know $O P=O A=4, O B=4 \sqrt{3}$. Since $\triangle P O B$ is an isosceles triangle, we have $B P=B O$ or $O P=P B$. When $B P=B O=4 \sqrt{3}$, As shown in Figure 3, draw $P F \perp O B$. Let $P(x, y)$. Then $$ \begin{array}{l} \left\{\begin{array}{l} 4^{2}=x^{2}+y^{2}, \\ (4 \sqrt{3})^{2}=(4 \sqrt{3}-x)^{2}+y^{2} \end{array}\right. \\ \Rightarrow P\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right) . \end{array} $$ Figure 3 When $O P=P B=4$, it is easy to see that, $$ \begin{array}{l} O F=F B=\frac{1}{2} O B=2 \sqrt{3} \\ \Rightarrow P F=\sqrt{O P^{2}-O F^{2}}=2 \\ \Rightarrow P(2 \sqrt{3}, 2) . \end{array} $$ Therefore, the required points are $P\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right)$ or $P(2 \sqrt{3}, 2)$." 9fe0bb7e68cc,"## Calculate the limit of the function: $\lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}$",See reasoning trace,medium,"## Solution Substitution: $$ \begin{aligned} & x=y+\pi \Rightarrow y=x-\pi \\ & x \rightarrow \pi \Rightarrow y \rightarrow 0 \end{aligned} $$ We get: $$ \begin{aligned} & \lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y+\pi}{2}\right)}{\pi-(y+\pi)}= \\ & =\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y}{2}+\frac{\pi}{2}\right)}{-y}=\lim _{y \rightarrow 0} \frac{1-\cos \left(\frac{y}{2}\right)}{-y}= \end{aligned} $$ Using the substitution of equivalent infinitesimals: $1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$ We get: $=\lim _{y \rightarrow 0} \frac{\frac{y^{2}}{2}}{-y}=\lim _{y \rightarrow 0}-\frac{y}{2}=-\frac{0}{2}=0$ ## Problem Kuznetsov Limits 13-24" c5c9b45482d9,"$11 \cdot 42$ For an integer $x \geqslant 1$, let $p(x)$ be the smallest prime that does not divide $x$, and $q(x)$ be the product of all primes less than $p(x)$. In particular, $p(1)=2$, and if there is some $x$ such that $p(x)=2$, then $q(x)=1$. The sequence $x_{0}, x_{1}, x_{2}, \cdots$ is defined by: $$ \begin{array}{c} x_{0}=1, \\ x_{n+1}=\frac{x_{n} p\left(x_{n}\right)}{q\left(x_{n}\right)} \end{array} $$ where $n \geqslant 0$, find all integers $n$ such that $x_{n}=1995$.",142$ is the integer $n$ such that $x_{n}=1995$.,medium,"[Proof] Clearly, from the definitions of $p(x)$ and $q(x)$, for any $x$, $q(x)$ divides $x$, thus $$ x_{n+1}=\frac{x_{n}}{q\left(x_{n}\right)} \cdot p\left(x_{n}\right) $$ which means that for any $n$, $x_{n+1}$ is a positive integer. Additionally, it is easy to prove by mathematical induction that for all $n$, $x_{n}$ has no square factors. Therefore, a unique code can be assigned to $x$ based on whether it is divisible by a certain prime number. The specific method is as follows: Let $p_{0}=2, p_{1}=3, p_{2}=5, \cdots$ be the sequence of all prime numbers in ascending order. Let $x>1$ be any number without square factors. Suppose $p_{m}$ is the largest prime that divides $x$. Divide $x$ by $p_{i}$. If $p_{i}$ divides $x$, set the code $S i=1$; otherwise, set $S i=0$. Then the code for $x$ is $\left(1, S_{m-1}, S_{m-2}, \cdots, S_{1}, S_{0}\right)$. Define $f(x)=\frac{x p(x)}{q(x)}$. If the last bit of $x$'s code is 0, then $x$ is odd, $p(x)=2, q(x)=1, f(x)=2 x$, and the code for $f(x)$ is the same as $x$'s code except that the last 0 is replaced by 1. If the last several bits of $x$'s code are $011 \cdots 1$, then the last several bits of $f(x)$'s code are $100 \cdots 0$. If the code traverses all binary numbers, then the code for $f(x)$ can be obtained by adding 1 to $x$'s code. Given $x_{1}=2$ and for $n \geqslant 2$, $x_{n+1}=f\left(x_{n}\right)$, the code for $x_{n}$ can directly equal the binary representation of $n$. Therefore, when $x_{n}=1995=3 \cdot 5 \cdot 7 \cdot 19$, there exists a unique $n$ such that the code for $x_{n}$ is 10001110 (the numbering from right to left is based on $2 \nmid 1995, 3 \mid 1995, 5 \mid 1995, 7 \mid 1995, 11 \nmid$ By $(10001110)_{2}=(142)_{10}$. Thus, $n=142$ is the integer $n$ such that $x_{n}=1995$." 87590b53fd76,"Example 5 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=2005$, find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$.","2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its minimum value. If there exist $x_{i},",medium,"Solve: The number of ways to write 2005 as the sum of 10 positive integers is finite, so there must be a way that maximizes or minimizes the sum of their squares. If the positive integers $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its maximum. Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$. If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right)$. $\left[\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2}\right]-\left(x_{1}^{2}+x_{2}^{2}\right)=-2 x_{1}+2 x_{2}+2=2\left(x_{2}-x_{1}\right)+2 \geqslant 2$. Therefore, replacing $x_{1}, x_{2}$ with $x_{1}-1, x_{2}+1$ keeps their sum unchanged but increases the sum of their squares. This contradicts the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is maximized, so $x_{1}=1$. Similarly, $x_{2}=x_{3}=\cdots=x_{9}=1, x_{10}=1996$. Thus, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is: $9+1996^{2}=3984025$. Now assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$, satisfying $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its minimum value. If there exist $x_{i}, x_{j}$ and $x_{j}-x_{i} \geqslant 2(j>i)$, then $\left[\left(x_{j}-1\right)^{2}+\left(x_{i}+1\right)^{2}\right]-\left(x_{j}^{2}+x_{i}^{2}\right)=-2 x_{j}+2 x_{i}+2=2\left(x_{i}-x_{j}\right)+2 \leqslant-2$. Therefore, replacing $x_{j}, x_{i}$ with $x_{j}-1, x_{i}+1$ keeps their sum unchanged but decreases the sum of their squares, contradicting the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is minimized. Thus, the absolute difference between any two of $x_{1}, x_{2}, \cdots, x_{10}$ is at most 1. Therefore, these 10 numbers can only be 5 instances of 200 and 5 instances of 201. Thus, the minimum value is $5 \times 200^{2}+5 \times 201^{2}=402005$." 5605cf86dc97,How many perfect squares are there between 2 and 150 ?,"1,2^{2}=4,12^{2}=144$, and $13^{2}=169$, then the perfect squares between 2 and 150 are $2^{2}$ thro",easy,"Since $1^{2}=1,2^{2}=4,12^{2}=144$, and $13^{2}=169$, then the perfect squares between 2 and 150 are $2^{2}$ through $12^{2}$, of which there are 11 ." 2f68ae6efb97,"2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is .",See reasoning trace,medium,"2. 7 . Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $S_{10}=55$ must be composite numbers, meaning that in $S_{1}, S_{2}, \cdots, S_{10}$, the primes are no greater than 7. The example in Table 1 shows that there can be 7 primes. Table 1 \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline$a_{i}$ & 7 & 6 & 4 & 2 & 9 & 1 & 3 & 5 & 10 & 8 \\ \hline$S_{i}$ & 7 & 13 & 17 & 19 & 28 & 29 & 32 & 37 & 47 & 55 \\ \hline \end{tabular} Therefore, in $S_{1}, S_{2}, \cdots, S_{10}$, there can be at most 7 primes." e6c6b9d864a8,"G9.4 Find the sum $d$ where $$ d=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{60}\right)+\left(\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+\cdots+\frac{2}{60}\right)+\left(\frac{3}{4}+\frac{3}{5}+\cdots+\frac{3}{60}\right)+\cdots+\left(\frac{58}{59}+\frac{58}{60}\right)+\frac{59}{60} $$",\frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\l,medium,$\begin{aligned} d & =\frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+\cdots+\left(\frac{1}{60}+\frac{2}{60}+\cdots+\frac{59}{60}\right) \\ & =\frac{1}{2}+\frac{\frac{3 \times 2}{2}}{3}+\frac{\frac{4 \times 3}{2}}{4}+\frac{\frac{5 \times 4}{2}}{5}+\cdots+\frac{\frac{60 \times 59}{2}}{60} \\ & =\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\cdots+\frac{59}{2} \\ & =\frac{1}{2}(1+2+3+4+\cdots+59) \\ & =\frac{1}{2} \times \frac{1}{2} \cdot 60 \cdot 59=885\end{aligned}$ ebecbe261fd2,"(Find the triplets of prime numbers $(p, q, r)$ such that $3 p^{4}-5 q^{4}-4 r^{2}=26$. )","5$. Then we look modulo 3, and similarly we get $q=3$. The calculation then yields $r=19$. The solut",easy,"We look modulo 5, knowing that the only squares are $1, -1$, and 0. If $p \neq 5$, it's impossible, so $p=5$. Then we look modulo 3, and similarly we get $q=3$. The calculation then yields $r=19$. The solution is $(5,3,19)$." cb2414b5470c,"11.5. A square plot of 14 by 14 cells needs to be paved with rectangular tiles of size $1 \times 4$. The tiles can only be laid along the grid (not diagonally), and the tiles cannot be broken. What is the maximum number of tiles required? Will there be any uncovered area left?",- 0 points,medium,"# Solution: Evaluation. The total number of cells on the plot is $14 \times 14=196$. Dividing by the number of cells in one tile, 196:4 = 49. Therefore, the number of tiles that can be cut from the $14 \times 14$ plot is no more than 49. We will color the cells of the plot in 4 colors, as shown in the diagram. Clearly, each rectangle of size $1 \times 4$ will contain one cell of each of the four colors. There are 49 cells of color 1, 50 cells of color 2, 49 cells of color 3, and only 48 cells of color 4. Therefore, no more than 48 tiles can be laid, and 4 cells will remain uncovered. | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | 3 | 4 | 1 | 2 | ![](https://cdn.mathpix.com/cropped/2024_05_06_5febab1abb19bb05583bg-2.jpg?height=745&width=714&top_left_y=1004&top_left_x=1022) Criteria: Only the answer - 0 points. Only the evaluation - 5 points. Only the example - 2 points. 7 points should be given if both the evaluation and the example are present." ebf24df96f78,"1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$. a) Find the angle $A B C$. b) Suppose additionally that $M P=\frac{1}{2}, N T=2, B D=\sqrt{3}$. Find the area of triangle $A B C$.",(a) $90^{\circ};$ (b) $\frac{5 \sqrt{13}}{3 \sqrt{2}}$,medium,"Answer: (a) $90^{\circ};$ (b) $\frac{5 \sqrt{13}}{3 \sqrt{2}}$. Solution. (a) Points $P$ and $T$ lie on the circle with diameter $BD$, so $\angle BPD = \angle BTD = 90^{\circ}$. Therefore, triangles $ADP$ and $DCT$ are right triangles; $PM$ and $TN$ are their medians. Since the median of a right triangle to the hypotenuse is half the hypotenuse, $TN = CN = DN$, $PM = AM = DM$. Let $\angle TCD = \gamma$. Since triangle $CTN$ is isosceles and $\angle CTN = \gamma$, $\angle TND = 2\gamma$ (as the external angle of $\triangle CTN$). Angles $\angle PMA$ and $\angle TND$ are equal due to the parallelism of lines $PM$ and $TN$. Since triangle $AMP$ is also isosceles, $\angle PAM = 90^{\circ} - \frac{1}{2} \angle PMA = 90^{\circ} - \frac{1}{2} \angle TND = 90^{\circ} - \gamma$. Therefore, the sum of angles $A$ and $C$ in triangle $ABC$ is $90^{\circ}$, and its third angle $\angle ABC$ is also $90^{\circ}$. (b) As stated above, $CD = 2NT = 4$, $AD = 2MP = 1$. Let $\angle ADB = \psi$. Then $\angle BDC = 180^{\circ} - \psi$. By the cosine theorem for triangles $ABD$ and $ACD$, we get $AB^2 = 1 + 3 - 2\sqrt{3} \cos \psi$, $BC^2 = 16 + 3 - 8\sqrt{3} \cos (180^{\circ} - \psi)$. By the Pythagorean theorem, $AB^2 + BC^2 = AC^2 = 25$, from which it follows that $4 - 2\sqrt{3} \cos \psi + 19 + 8\sqrt{3} \cos \psi = 25$, $\cos \psi = \frac{1}{3\sqrt{3}}$. Next, we find: $\sin \psi = \sqrt{1 - \cos^2 \psi} = \frac{\sqrt{26}}{3\sqrt{3}}$, $S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle BCD} = \frac{1}{2} DA \cdot DB \sin \psi + \frac{1}{2} DC \cdot DB \sin (180^{\circ} - \psi) = \frac{1}{2} AC \cdot BD \sin \psi = \frac{1}{2} \cdot 5 \cdot \sqrt{3} \cdot \frac{\sqrt{26}}{3\sqrt{3}} = \frac{5\sqrt{13}}{3\sqrt{2}}$." a70f0e1fa60c,"1. In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tournament.",5,medium,"(12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n-1) n$ games. The number of matches between teams from city $A$ and teams from city $B$ is $2 n^{2}$. Let $m$ be the number of wins in these matches by teams from city $A$, then teams from city $B$ won $2 n^{2}-m$ games in these matches. In total, teams from city $A$ won $\frac{(n-1) n}{2}+m$ games, and teams from city $B$ won $(2 n-1) n+2 n^{2}-m$ games. According to the condition, we have $\frac{\frac{(n-1) n}{2}+m}{(2 n-1) n+2 n^{2}-m}=\frac{4}{3}$, $3 n^{2}-3 n+6 m=32 n^{2}-8 n-8 m, 14 m=29 n^{2}-5 n, m=\frac{29 n^{2}-5 n}{14}, \frac{29 n^{2}-5 n}{14} \leq 2 n^{2}, n^{2}-5 n \leq 0$. The number $n$ can be $1,2,3,4,5$. By substituting into the equation $m=\frac{29 n^{2}-5 n}{14}$, we get that $m$ is an integer only when $n=5$. Answer: 5." f6961cf375c5,"3. Given $x-y=3, z=\sqrt{y^{2}-3 x^{2}}$. Then the maximum value of $z$ is $\qquad$ .",See reasoning trace,easy,3. $\frac{3}{2} \sqrt{6}$ 8d88351943f1,"1. In a right-angled triangle, the square of the hypotenuse is exactly equal to twice the product of the two legs. Then, the ratio of the three sides of this triangle is ( ). (A) $3: 4: 5$ (B) $1: 1: 1$ (C) $2: 3: 4$ (D) $1: 1: \sqrt{2}$","2 a b \Rightarrow a=b$. Therefore, the ratio of the three sides of this right triangle is $1: 1: \sq",easy,"$-1 . \mathrm{D}$ Let the two legs be $a$ and $b$. Then $a^{2}+b^{2}=2 a b \Rightarrow a=b$. Therefore, the ratio of the three sides of this right triangle is $1: 1: \sqrt{2}$." 92ad17c31877,"Example 2 The two lateral edges $SB$ and $SC$ of the tetrahedron $S-ABC$ are equal, and the projection of vertex $S$ on the base is the orthocenter $H$ of $\triangle ABC$. The base edge $BC=a$, and the dihedral angle between the lateral face $SBC$ and the base is $\alpha$. Find the volume of this tetrahedron.",See reasoning trace,medium,"Solution: Connect $S H, B H, A H, C H$, and extend $A H$ to intersect $B C$ at $M$, connect $S M$, as shown in Figure 16-2. From $S B=S C$, we get $H B=H C$, and since $H$ is the orthocenter, then $A M \perp B C$. Thus, $M B=M C=\frac{1}{2} B C=\frac{a}{2}$, which implies $A B=A C$. Since $H$ is the orthocenter, we have $M B^{2}=A M \cdot H M$, so $\left(\frac{a}{2}\right)^{2}=A M \cdot H M$. From $S B=S C, B M=C M$, we get $S M \perp B C$, and since $A M \perp B C$, then $\angle S M A=\alpha$. Therefore, from $\angle S H M=90^{\circ}$, we get $S H=\operatorname{tg} \alpha \cdot H M$. Thus, $S_{\triangle A B C}=\frac{1}{2} A M \cdot B C=\frac{a}{2} A M$. Hence, $$ \begin{aligned} V & =\frac{1}{3} S H \cdot S_{\triangle A B C}=\frac{1}{3} \operatorname{tg} \alpha \cdot H M \cdot \frac{1}{2} A M \cdot a \\ & =\frac{a}{6} \operatorname{tg} \alpha \cdot H M \cdot A M=\frac{a}{6} \operatorname{tg} \alpha \cdot \frac{a^{2}}{4}=\frac{a^{3}}{24} \operatorname{tg} \alpha . \end{aligned} $$" 57e9d00a0d5c,"82. In a circle of radius $R$, two intersecting perpendicular chords are drawn. a) Find the sum of the squares of the four segments of these chords, into which they are divided by the point of intersection. b) Find the sum of the squares of the lengths of the chords, if the distance from the center of the circle to their point of intersection is $d$.",(|AM| + |MB|)^{2} + (|CM| + |MD|)^{2} = |AM|^{2} + |MB|^{2} + |CM|^{2} + |MD|^{2} + 2|AM| \cdot |MB|,easy,"82. Let $AB$ and $CD$ be given chords, and $M$ be their point of intersection. a) The arcs $\overline{AC}$ and $\overline{BD}$ together form a semicircle; therefore, $|AC|^{2} + |BD|^{2} = 4R^{2}$; thus, $|AM|^{2} + |MC|^{2} + |MB|^{2} + |MD|^{2} = |AC|^{2} + |BD|^{2} = 4R^{2}$, b) $|AB|^{2} + |CD|^{2} = (|AM| + |MB|)^{2} + (|CM| + |MD|)^{2} = |AM|^{2} + |MB|^{2} + |CM|^{2} + |MD|^{2} + 2|AM| \cdot |MB| + 2|CM| \cdot |MD| = 4R^{2} + 2(R^{2} - d^{2}) + 2(R^{2} - d^{2}) = 4(2R^{2} - d^{2})$." 769c53d85b7a,"Find all polynomials $P$ with real coefficients such that $$ \frac{P(x)}{y z}+\frac{P(y)}{z x}+\frac{P(z)}{x y}=P(x-y)+P(y-z)+P(z-x) $$ for all nonzero real numbers $x, y, z$ obeying $2 x y z=x+y+z$.",\( P(x) = c (x^2 + 3) \) which all work,medium,"The given can be rewritten as saying that $$ \begin{aligned} Q(x, y, z) & :=x P(x)+y P(y)+z P(z) \\ & -x y z(P(x-y)+P(y-z)+P(z-x)) \end{aligned} $$ is a polynomial vanishing whenever \( x y z \neq 0 \) and \( 2 x y z = x + y + z \), for real numbers \( x, y, z \). **Claim**: This means \( Q(x, y, z) \) vanishes also for any complex numbers \( x, y, z \) obeying \( 2 x y z = x + y + z \). Define $$ R(x, y) := Q\left(x, y, \frac{x+y}{2 x y-1}\right) $$ which vanishes for any real numbers \( x \) and \( y \) such that \( x y \neq \frac{1}{2} \), \( x \neq 0 \), \( y \neq 0 \), and \( x + y \neq 0 \). This can only occur if \( R \) is identically zero as a rational function with real coefficients. If we then regard \( R \) as having complex coefficients, the conclusion then follows. **Remark (Algebraic geometry digression on real dimension)**: Note here we use in an essential way that \( z \) can be solved for in terms of \( x \) and \( y \). If \( s(x, y, z) = 2 x y z - (x + y + z) \) is replaced with some general condition, the result may become false; e.g., we would certainly not expect the result to hold when \( s(x, y, z) = x^2 + y^2 + z^2 - (x y + y z + z x) \) since for real numbers \( s = 0 \) only when \( x = y = z \)! The general condition we need here is that \( s(x, y, z) = 0 \) should have ""real dimension two"". Here is a proof using this language, in our situation. Let \( M \subset \mathbb{R}^3 \) be the surface \( s = 0 \). We first contend \( M \) is a two-dimensional manifold. Indeed, the gradient \( \nabla s = \langle 2 y z - 1, 2 z x - 1, 2 x y - 1 \rangle \) vanishes only at the points \( \left( \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}} \right) \) where the \( \pm \) signs are all taken to be the same. These points do not lie on \( M \), so the result follows by the regular value theorem. In particular, the topological closure of points on \( M \) with \( x y z \neq 0 \) is all of \( M \) itself; so \( Q \) vanishes on all of \( M \). If we now identify \( M \) with the semi-algebraic set consisting of maximal ideals \( (x - a, y - b, z - c) \) in \( \operatorname{Spec} \mathbb{R}[x, y, z] \) satisfying \( 2 a b c = a + b + c \), then we have real dimension two, and thus the Zariski closure of \( M \) is a two-dimensional closed subset of \( \operatorname{Spec} \mathbb{R}[x, y, z] \). Thus it must be \( Z = \mathcal{V}(2 x y z - (x + y + z)) \), since this \( Z \) is an irreducible two-dimensional closed subset (say, by Krull's principal ideal theorem) containing \( M \). Now \( Q \) is a global section vanishing on all of \( Z \), therefore \( Q \) is contained in the (radical, principal) ideal \( (2 x y z - (x + y + z)) \) as needed. So it is actually divisible by \( 2 x y z - (x + y + z) \) as desired. Now we regard \( P \) and \( Q \) as complex polynomials instead. First, note that substituting \( (x, y, z) = (t, -t, 0) \) implies \( P \) is even. We then substitute $$ (x, y, z) = \left(x, \frac{i}{\sqrt{2}}, \frac{-i}{\sqrt{2}}\right) $$ to get $$ \begin{aligned} & x P(x) + \frac{i}{\sqrt{2}} \left( P\left( \frac{i}{\sqrt{2}} \right) - P\left( \frac{-i}{\sqrt{2}} \right) \right) \\ = & \frac{1}{2} x \left( P\left( x - \frac{i}{\sqrt{2}} \right) + P\left( x + \frac{i}{\sqrt{2}} \right) + P(\sqrt{2} i) \right) \end{aligned} $$ which in particular implies that $$ P\left( x + \frac{i}{\sqrt{2}} \right) + P\left( x - \frac{i}{\sqrt{2}} \right) - 2 P(x) \equiv P(\sqrt{2} i) $$ identically in \( x \). The left-hand side is a second-order finite difference in \( x \) (up to scaling the argument), and the right-hand side is constant, so this implies \( \operatorname{deg} P \leq 2 \). Since \( P \) is even and \( \operatorname{deg} P \leq 2 \), we must have \( P(x) = c x^2 + d \) for some real numbers \( c \) and \( d \). A quick check now gives the answer \( P(x) = c (x^2 + 3) \) which all work." d5546ad128c8,"Let $a,b$ be positive integers. Prove that $$\min(\gcd(a,b+1),\gcd(a+1,b))\leq\frac{\sqrt{4a+4b+5}-1}{2}$$ When does the equality hold?","\min(\gcd(a, b+1), \gcd(a+1, b)) \leq \frac{\sqrt{4a + 4b + 5",medium,"1. Define \( d_1 = \gcd(a, b+1) \) and \( d_2 = \gcd(a+1, b) \). Note that both \( d_1 \) and \( d_2 \) divide \( a+b+1 \). This is because: \[ d_1 = \gcd(a, b+1) \implies d_1 \mid a \text{ and } d_1 \mid (b+1) \implies d_1 \mid (a + b + 1) \] Similarly, \[ d_2 = \gcd(a+1, b) \implies d_2 \mid (a+1) \text{ and } d_2 \mid b \implies d_2 \mid (a + b + 1) \] 2. Since \( d_1 \) and \( d_2 \) are both divisors of \( a+b+1 \), and \( \gcd(d_1, d_2) = 1 \) (as \( d_1 \) and \( d_2 \) are coprime), it follows that \( d_1 d_2 \) divides \( a+b+1 \). Therefore: \[ d_1 d_2 \leq a + b + 1 \] 3. Without loss of generality, assume \( d_2 \geq d_1 \). Then, we can write: \[ d_1 (d_1 + 1) \leq d_1 d_2 \leq a + b + 1 \] This implies: \[ d_1^2 + d_1 \leq a + b + 1 \] 4. Solving the quadratic inequality \( d_1^2 + d_1 - (a + b + 1) \leq 0 \), we use the quadratic formula: \[ d_1 \leq \frac{-1 + \sqrt{1 + 4(a + b + 1)}}{2} = \frac{-1 + \sqrt{4a + 4b + 5}}{2} \] 5. Therefore, we have: \[ \min(\gcd(a, b+1), \gcd(a+1, b)) \leq \frac{\sqrt{4a + 4b + 5} - 1}{2} \] 6. To determine when equality holds, assume \( \gcd(a, b+1) = d \) and \( \gcd(a+1, b) = d+1 \). Then: \[ a + b + 1 = d(d+1) \] 7. We have the congruences: \[ a \equiv 0 \pmod{d} \quad \text{and} \quad a \equiv -1 \pmod{d+1} \] Solving these, we get: \[ a \equiv d \pmod{d(d+1)} \] 8. Similarly, for \( b \): \[ b \equiv -1 \pmod{d} \quad \text{and} \quad b \equiv 0 \pmod{d+1} \] Solving these, we get: \[ b \equiv -d-1 \pmod{d(d+1)} \] 9. Therefore, \( a \geq d \) and \( b \geq d^2 - 1 \). Since \( a + b + 1 = d(d+1) \), it follows that: \[ (a, b) = (d, d^2 - 1) \] 10. This gives equality, and the answer is: \[ \{a, b\} = \{d, d^2 - 1\} \quad \text{for any } d \geq 2 \] The final answer is \( \boxed{ \min(\gcd(a, b+1), \gcd(a+1, b)) \leq \frac{\sqrt{4a + 4b + 5} - 1}{2} } \)" 6e40b04e2b92,What is the order of 2 modulo 101?,See reasoning trace,easy,"According to Euler's theorem, the order of 2 modulo 101 divides $100=2^{2} \cdot 5^{2}$. Let's calculate $2^{20}$ and $2^{50}$ modulo 101. To this end, note that $2^{10}=1024 \equiv 14 \bmod 101$. Thus, $$ 2^{20} \equiv 14^{2}=196 \equiv-6 \quad \bmod 101 $$ and $$ 2^{50} \equiv(-6)^{2} \cdot 14=504 \equiv-1 \quad \bmod 101 $$ Therefore, the order of 2 modulo 101 is 100." 6b1216fa97f7,"35. The fish soup prepared by the tourists on the seashore after a successful fishing trip was good, but due to lack of culinary experience, they put too little salt in it, so they had to add salt at the table. The next time they made the same amount of fish soup, they put in twice as much salt as the first time, but even then they had to add salt to the soup at the table, though they used only half as much salt as they did the first time. What fraction of the necessary amount of salt did the cook put in the fish soup the first time?","(a-2x) \cdot 2$, from which $x=\frac{a}{3}$. Therefore, the first time, $\frac{1}{3}$ of the necessa",easy,"35. Let the amount of salt that should have been added the first time be $a$ g, and the amount actually added be $x$ g, resulting in a shortage of $(a-x)$ g. Another time, $2x$ g of salt was added, leading to a shortage of $a-2x$. Since the amount of salt added the second time was half of what was added the first time, we have $a-x=(a-2x) \cdot 2$, from which $x=\frac{a}{3}$. Therefore, the first time, $\frac{1}{3}$ of the necessary amount of salt was added to the soup." 7670f7e32f10,"7.14 There is a $5 \times 5$ square grid chessboard, consisting of 25 $1 \times 1$ unit square cells. At the center of each unit square cell, a red dot is painted. Please find several straight lines on the chessboard that do not pass through any red dots, dividing the chessboard into several pieces (which may be of different shapes and sizes), such that each piece contains at most one red dot. How many straight lines do you need to draw at a minimum? Provide an example of such a drawing and prove your conclusion.",See reasoning trace,medium,"[Solution] As shown in the figure, after dividing the chessboard into several sections with 8 straight lines, there is at least one red point in each small section. Below, we use proof by contradiction to show that it is impossible to have fewer lines that satisfy the requirements of the problem. Assume that the number of lines drawn does not exceed 7, and they meet the requirements of the problem. At this point, we connect the 16 red points on the edge sequentially with unit-length line segments to form a square with a side length of 4. Since the 7 lines drawn intersect at most 14 of these small line segments, at least two unit-length small line segments do not intersect with any of these 7 lines, and they must entirely fall into one of the small regions divided by these lines, including their two endpoints (red points), which contradicts the problem's requirements. Therefore, at least 8 lines are needed." 7270da9ca2f4,14th Chinese 1999,576 possibilities. 14th Chinese 1999 (C) John Scholes jscholes@kalva.demon.co.uk 19 June 2002,medium,": 576. The key is that there is a bijection between admissible arrangements of the red cubes and 4 x 4 Latin squares. On the top face of the cube write 1, 2, 3 or 4 in each square according to how far down the red cube is. Note that there must be at least one red cube in the column under each square. There are 16 squares, and only 16 red cubes, so there must be exactly one red cube in each column. Now there must be a 1 in each row of the square, otherwise there would not be a red cube in the corresponding 1 x 1 x 4 cuboid. Similarly, there must be a 2, 3, and 4. So each row must be a permutation of 1, 2, 3, 4. Similarly each column. So the square must be a Latin square. Conversely, a Latin square corresponds to an admissible arrangement. So we have to count Latin squares. It is easy to check that there are just four Latin squares with the pattern shown at the left: 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2 x x x 2 3 4 1 2 1 4 3 2 1 4 3 2 4 1 3 3 x x x 3 4 1 2 3 4 2 1 3 4 1 2 3 1 4 2 4 x x x 4 1 2 3 4 3 1 2 4 3 2 1 4 3 2 1 In each case we can permute the columns arbitrarily (to give any permutation of 1, 2, 3, 4 as the first row) and then permute rows 2, 3, 4 arbitrarily. Conversely any Latin square can be put in the form shown at the left by such permutations. Hence there are just 4 x 24 x 6 = 576 possibilities. 14th Chinese 1999 (C) John Scholes jscholes@kalva.demon.co.uk 19 June 2002" f2dab1cb16f4,"# Task 1. Mom told Dima that he should eat 13 spoons of porridge. Dima told his friend that he had eaten 26 spoons of porridge. Each child, telling about Dima's feat, increased the amount of porridge Dima had eaten by 2 or 3 times. As a result, one of the children told Dima's mom about 33696 spoons of porridge. How many times in total did the children, including Dima, tell about the feat?",See reasoning trace,medium,"Answer: 9 Solution. $33696=2^{5} \cdot 3^{4} \cdot 13$. From this, it follows that the story of the feat was retold 5 times with the amount of porridge doubling and 4 times with it tripling, totaling 9 times. ## B-2 Dad persuaded Natasha to eat 11 spoons of porridge. Natasha told her friend that she had eaten 22 spoons of porridge. Then each child, retelling the story of Natasha's feat, increased the amount of porridge she had eaten by 2 or 3 times. As a result, one of the children told Natasha's dad about 21384 spoons of porridge. How many times in total did the children, including Natasha, retell the story? ## Answer: 8 ## B-3 Mom told Grisha that he had to eat 3 spoons of porridge. Grisha told his friend that he had eaten 9 spoons of porridge. Then each child, retelling the story of Grisha's feat, increased the amount of porridge he had eaten by 3 or 5 times. As a result, one of the children told Grisha's mom about 18225 spoons of porridge. How many times in total did the children, including Grisha, retell the story? Answer: 7 ## B-4 Dad persuaded Tanya to eat 7 spoons of porridge. Tanya told her friend that she had eaten 21 spoons of porridge. Then each child, retelling the story of Tanya's feat, increased the amount of porridge she had eaten by 2 or 3 times. As a result, one of the children told Tanya's dad about 27216 spoons of porridge. How many times in total did the children, including Tanya, retell the story? Answer: 9 ## Lomonosov Olympiad for Schoolchildren in Mathematics Preliminary Stage 2020/21 academic year for 5-6 grades #" 4d2c7c46dc29,"Suppose $\alpha,\beta,\gamma\in\{-2,3\}$ are chosen such that \[M=\max_{x\in\mathbb{R}}\min_{y\in\mathbb{R}_{\ge0}}\alpha x+\beta y+\gamma xy\] is finite and positive (note: $\mathbb{R}_{\ge0}$ is the set of nonnegative real numbers). What is the sum of the possible values of $M$?",\frac{13,medium,"1. Let \((a, b, c) := (\alpha, \beta, \gamma)\) for convenience. We need to find the maximum value of the expression \(ax + by + cxy\) over \(x \in \mathbb{R}\) and \(y \in \mathbb{R}_{\ge0}\). 2. First, we minimize \(ax + by + cxy\) over \(y \in \mathbb{R}_{\ge0}\). If \(b + cx < 0\), we can let \(y \rightarrow \infty\) so that \((b + cx)y \rightarrow -\infty\), which is absurd. Therefore, we must have \(b + cx \ge 0\), and the minimum value of \(y\) is \(0\). 3. Next, we maximize \(ax\) over \(x \in \mathbb{R}\) given that \(b + cx \ge 0\). 4. Consider the case when \(a > 0\): - If \(c > 0\), then \(x \rightarrow \infty\) in \(b + cx \ge 0\) is possible, which contradicts the requirement that \(M\) is finite. Hence, \(c < 0\). - Given \(c < 0\), we have \(x \le -\frac{b}{c}\). The expression \(ax\) is maximized when \(x = -\frac{b}{c}\), provided that \(-\frac{b}{c}\) is positive. If \(-\frac{b}{c}\) is not positive, then \(M\) is not positive. - For \((a, b, c) = (3, 3, -2)\), we have \(M = -\frac{ab}{c} = \frac{9}{2}\). 5. Consider the case when \(a < 0\): - If \(c < 0\), then \(x \rightarrow -\infty\) in \(b + cx \ge 0\) is possible, which contradicts the requirement that \(M\) is finite. Hence, \(c > 0\). - Given \(c > 0\), we have \(x \ge -\frac{b}{c}\). The expression \(ax\) is maximized when \(x = -\frac{b}{c}\), provided that \(-\frac{b}{c}\) is negative. If \(-\frac{b}{c}\) is not negative, then \(M\) is not positive. - For \((a, b, c) = (-2, 3, 3)\), we have \(M = -\frac{ab}{c} = 2\). 6. Summing the possible values of \(M\), we get \(\frac{9}{2} + 2 = \frac{13}{2}\). The final answer is \(\boxed{\frac{13}{2}}\)." 0517ec852a41,"If $2.4 \times 10^{8}$ is doubled, then the result is equal to (A) $2.4 \times 20^{8}$ (B) $4.8 \times 20^{8}$ (C) $4.8 \times 10^{8}$ (D) $2.4 \times 10^{16}$ (E) $4.8 \times 10^{16}$",(C),easy,"When $2.4 \times 10^{8}$ is doubled, the result is $2 \times 2.4 \times 10^{8}=4.8 \times 10^{8}$. ANswer: (C)" bf0ba2d37aac,"## Task Condition Write the canonical equations of the line. $$ \begin{aligned} & x+5 y-z+11=0 \\ & x-y+2 z-1=0 \end{aligned} $$",See reasoning trace,medium,"## Solution Canonical equations of a line: $\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$ where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector. Since the line belongs to both planes simultaneously, its direction vector $\vec{s}$ is orthogonal to the normal vectors of both planes. The normal vectors of the planes are: $\overrightarrow{n_{1}}=\{1 ; 5 ;-1\}$ $\overrightarrow{n_{2}}=\{1 ;-1 ; 2\}$ Find the direction vector $\vec{s}:$ $$ \begin{aligned} & \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc} i & j & k \\ 1 & 5 & -1 \\ 1 & -1 & 2 \end{array}\right|= \\ & =i \cdot\left|\begin{array}{cc} 5 & -1 \\ -1 & 2 \end{array}\right|-j \cdot\left|\begin{array}{cc} 1 & -1 \\ 1 & 2 \end{array}\right|+k \cdot\left|\begin{array}{cc} 1 & 5 \\ 1 & -1 \end{array}\right|= \\ & =9 i-3 j-6 k=\{9 ;-3 ;-6\} \end{aligned} $$ Find some point on the line $\left(x_{0} ; y_{0} ; z_{0}\right)$. Let $z_{0}=0$, then $$ \begin{aligned} & \left\{\begin{array}{l} x_{0}+5 y_{0}-0+11=0 \\ x_{0}-y_{0}+2 \cdot 0-1=0 \end{array}\right. \\ & \left\{\begin{array}{l} x_{0}+5 y_{0}+11=0 \\ x_{0}-y_{0}-1=0 \end{array}\right. \\ & \left\{\begin{array}{l} 6 y_{0}+12=0 \\ x_{0}-y_{0}-1=0 \end{array}\right. \\ & \left\{\begin{array}{l} y_{0}=-2 \\ x_{0}-y_{0}-1=0 \end{array}\right. \\ & \left\{\begin{array}{l} y_{0}=-2 \\ x_{0}-(-2)-1=0 \end{array}\right. \\ & \left\{\begin{array}{l} y_{0}=-2 \\ x_{0}=-1 \end{array}\right. \end{aligned} $$ Thus, the point $(-1 ;-2 ; 0)$ belongs to the line. We obtain the canonical equations of the line: $$ \frac{x+1}{9}=\frac{y+2}{-3}=\frac{z}{-6} $$ ## Problem Kuznetsov Analytic Geometry 13-22" 5b54139cb3a8,"6. A city's waterworks, the reservoir originally has 500 tons of water. In one day, while water is being added to the reservoir, water is also being supplied to users from the reservoir. $x(0 \leqslant x \leqslant 24)$ hours of water supplied to users is $80 \sqrt{24 x}$ tons. When 96 tons of water are added to the reservoir per hour, then the minimum amount of water in the reservoir in one day is ( ) tons. (A) 120 (B) 100 (C) 80 (D) 60","10$, the minimum value of $y$ is 100.",easy,"6. (B). From the given, we have $$ y=500+96 x-80 \sqrt{24 x}(0 \leqslant x \leqslant 24) \text {. } $$ Let $\sqrt{24 x}=t(0 \leqslant t \leqslant 24)$. Then $$ y=4 t^{2}-80 t+500=4(t-10)^{2}+100 \text {. } $$ Therefore, when $t=10$, the minimum value of $y$ is 100." 98f91c740773,"3. Given the sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$, for any $n \in \mathbf{Z}_{+}$, it holds that $a_{n}>b_{n}$. As $n \rightarrow+\infty$, the limits of the sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$ are $A 、 B$ respectively. Then ( ). (A) $A>B$ (B) $A \geqslant B$ (C) $A \neq B$ (D) The relationship between $A$ and $B$ is uncertain","\lim _{n \rightarrow+\infty} b_{n}=0$, so $A>B$ is not necessarily true, but it can ensure that $A \",easy,"3. B. If we take $a_{n}=\frac{1}{n}, b_{n}=\frac{1}{n+1}$, then $a_{n}>b_{n}$. And $\lim _{n \rightarrow+\infty} a_{n}=\lim _{n \rightarrow+\infty} b_{n}=0$, so $A>B$ is not necessarily true, but it can ensure that $A \geqslant B$." 8b5fa48a848a,"Find all positive integers $k$ such that the equation $$ x^{2}+y^{2}+z^{2}=k x y z $$ has positive integer solutions $(x, y, z)$.",1$ or $3$. $x=y=z=1$ is a solution to the equation $x^{2}+y^{2}+z^{2}=3 x y z$. $x=y=z=3$ is a solut,medium,"Let $(a, b, c)$ be a solution, and among all solutions, $(a, b, c)$ minimizes $x+y+z$. Assume $a \geqslant b \geqslant c$. By Vieta's formulas, we have $d=\frac{b^{2}+c^{2}}{a}=k b c-a$, which means $x^{2}+b^{2}+c^{2}=$ $k b c x$ (2) has positive integer solutions. Moreover, $(b, c, d)$ is a solution to (1). Thus, by the minimality of $(a, b, c)$, $$ \begin{array}{l} a \leqslant d, a^{2} \leqslant a d=b^{2}+c^{2}<(b+c)^{2}, a0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$, then the maximum value of the function $f(x)$ on $[-3,-2]$ is","2 x$, so the maximum value of $f(x)$ on $[-3,-2]$ is -4.",easy,"It is known that $f(x)=2 x$, so the maximum value of $f(x)$ on $[-3,-2]$ is -4." 2d84428e6a71,2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432.,2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this nu,easy,"Solution. The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0 ; 2]$, $\delta \in[0 ; 1], \varphi \in[0 ; 1]$. The total number of such divisors is $(8+1) \cdot 4 \cdot(2+1)(1+1)(1+1)=432$." 2df060716d87,"Az $x^{2}-(3 a+1) x+\left(2 a^{2}-3 a-2\right)=0$ egyenletben határozzuk meg $a$ értékét úgy, hogy a gyökök valósak legyenek és négyzetösszegük minimális legyen.",See reasoning trace,medium,"A szóban forgó gyökök akkor valósak, ha $$ (3 a+1)^{2}-4\left(2 a^{2}-3 a-2\right) \geq 0 $$ vagyis $$ a^{2}+18 a+9 \geq 0 $$ Ez akkor és csak akkor teljesül, ha $$ a \leq-9-6 \sqrt{2}=-17,489 $$ vagy $$ a \geq-9+6 \sqrt{2}=-0,515 $$ A gyökök (esetleg két egybeeső gyök) négyzetösszegét a gyökök és együtthatók közötti összefüggések alapján határozzuk meg. A két gyököt jelölje $x_{1}$ és $x_{2}$. Ekkor $$ \begin{gathered} x_{1}+x_{2}=3 a+1 \\ x_{1} x_{2}=2 a^{2}-3 a-2 \end{gathered} $$ Ezekból $$ x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5 a^{2}+12 a+5=5(a+1,2)^{2}-2,2 $$ Látható, hogy a kapott kifejezés értéke $a=-1,2$-nél a legkisebb. Az is megállapítható, hogy minél jobban eltér ettől $a$, annál nagyobb lesz a gyökök négyzetösszege. $a=-1,2$-nél a gyökök azonban nem valósak, az ehhez legközelebb levő megengedett érték $-9+6 \sqrt{2}$. Ez tehát a keresett szám. Megjegyzés. A dolgozatok elbírálása a következő szempontok alapján történt: Helyes a kifogástalan dolgozat. Hiányos az a dolgozat, amelyben a megoldás lényegét nem érintő kisebb számolási hiba van. Hibás az a dolgozat, amelyben 1. a megoldás jellegét is megváltoztató számolási hiba van; 2 . nem veszi figyelembe azt a kikötést, hogy a gyökök valósak legyenek és az $a=-1,2$-et adja meg eredményül; 3. azt mondja, hogy $a=-1,2$-nél a gyökök nem valósak, tehát a feladatnak nincs megoldása. " 280d80bc3e8d,"3. As shown in Figure 1, in the regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $D$ and $E$ are points on the side edges $B B_{1}$ and $C C_{1}$, respectively, with $E C=B C=2 B D$. Then the size of the dihedral angle formed by the section $A D E$ and the base $A B C$ is $\qquad$",See reasoning trace,easy,"3. $45^{\circ}$. Let $B C=2$. Then the area of $\triangle A B C$ is $\sqrt{3}$. Since $E C=B C=2 B D$, we have, $$ \begin{array}{l} E C=2, B D=1 \\ \Rightarrow A E=2 \sqrt{2}, A D=D E=\sqrt{5} \\ \Rightarrow S_{\triangle A D E}=\sqrt{6} \\ \Rightarrow \cos \theta=\frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{2}}{2} \Rightarrow \theta=45^{\circ} . \end{array} $$" e2a9b495a188,"## Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$. $A(-1 ; 2 ;-3), B(0 ; 1 ;-2), C(-3 ; 4 ;-5)$",See reasoning trace,medium,"## Solution Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$: $\overrightarrow{A B}=(0-(-1) ; 1-2 ;-2-(-3))=(1 ;-1 ; 1)$ $\overrightarrow{A C}=(-3-(-1) ; 4-2 ;-5-(-3))=(-2 ; 2 ;-2)$ We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$: $$ \begin{aligned} & \cos (\overrightarrow{A B, \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\ & =\frac{1 \cdot(-2)+(-1) \cdot 2+1 \cdot(-2)}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{(-2)^{2}+2^{2}+(-2)^{2}}}= \\ & =\frac{-2-2-2}{\sqrt{1+1+1} \cdot \sqrt{4+4+4}}=\frac{-6}{\sqrt{3} \cdot \sqrt{12}}=-1 \end{aligned} $$ Thus, the cosine of the angle: $\cos (\overrightarrow{A B, A C})=-1$ and consequently the angle $\widehat{A B,} \overrightarrow{A C}=\pi$ ## Problem Kuznetsov Analytic Geometry 4-24" d27ada55ac61,"Example X Given that $a, b, c$ are the sides of $\triangle ABC$, $a \leqslant b \leqslant c$, and $R$ and $r$ are the radii of the circumcircle and incircle of $\triangle ABC$, respectively. Let $f=a+b-2 R-2 r$, try to determine the sign of $f$ using the size of angle $C$.","\frac{\pi}{2}$, $f=0$; when $C \in\left(\frac{\pi}{2}, \pi\right)$, $f<0$.",medium,"Analysis We will unify the expression in terms of angles. Solution Since $S_{\triangle A B C}=\frac{1}{2} r(a+b+c)$ Therefore, $r=2 \frac{S_{\triangle A B C}}{a+b+c}=\frac{4 R^{2} \sin A \sin B \sin C}{2 R(\sin A+\sin B+\sin C)}$ $$ =2 R \frac{\sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} $$ Thus, $f=2 R\left(\sin A+\sin B-1-4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$ $$ \begin{array}{l} =2 R(\sin A+\sin B-\cos A-\cos B-\cos C) \\ =2 R\left[\left(2 \sin \frac{A+B}{2}-2 \cos \frac{A+B}{2}\right) \cos \frac{A-B}{2}-\left(\cos ^{2} \frac{C}{2}-\sin ^{2} \frac{C}{2}\right)\right] \\ =2 R\left(\cos \frac{C}{2}-\sin \frac{C}{2}\right)\left(2 \cos \frac{A-B}{2}-\cos \frac{C}{2}-\cos \frac{A+B}{2}\right) \end{array} $$ By $a \leqslant b \leqslant c$ and the monotonicity of the cosine function, we know: $$ \cos \frac{A-B}{2}>\cos \frac{C}{2}, \cos \frac{A-B}{2}>\cos \frac{A+B}{2} $$ Therefore, $2 \cos \frac{A-B}{2}-\cos \frac{C}{2}-\cos \frac{A+B}{2}>0$ If $\cos \frac{C}{2}1, C \in\left(\frac{\pi}{2}, \pi\right)$ then $f\sin \frac{C}{2}$, i.e., $\tan \frac{C}{2}0$. Also, since $c \geqslant b \geqslant a$, we know: $C \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$ if $\cos \frac{C}{2}=\sin \frac{C}{2}$, i.e., $\tan \frac{C}{2}=1, C=\frac{\pi}{2}$, then $f=0$ In summary: When $C \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$, $f>0$; when $C=\frac{\pi}{2}$, $f=0$; when $C \in\left(\frac{\pi}{2}, \pi\right)$, $f<0$." 91b079a38be9,"18. Let $a=1+3^{1}+3^{2}+3^{3}+3^{4}+\ldots+3^{999}+3^{1000}$, then the remainder when $a$ is divided by 4 is ( ). (A) 3. (B) 2. (C) 1. (D) 0.",$\mathbf{C}$,easy,"Answer: $\mathbf{C}$. Solution: Since $a=1+3(1+3)+3^{3}(1+3)+3^{5}(1+3)+\cdots+3^{999}(1+3)$, the remainder when $a$ is divided by 4 is 1. Therefore, the correct choice is C." 47e43b3d2ec7,,5,easy,"Answer: 5. Solution. Since there are 3 times more apples than non-apples, apples make up $\frac{3}{4}$ of the total number of fruits, i.e., there are $\frac{3}{4} \cdot 60=45$ apples. Since there are 5 times fewer pears than non-pears, pears make up $\frac{1}{6}$ of the total number of fruits, i.e., there are $\frac{1}{6} \cdot 60=10$ pears. Therefore, the total number of oranges is $60-45-10=5$." 1c03034ede18,"10. (2004 High School League Preliminary Contest, Liaoning Province) Given real numbers $x, y$ satisfy $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, find the maximum value of the function $U=x+y$. untranslated part: 将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 Note: The last part is a note and not part of the text to be translated. It is provided for context.","3 \cos \alpha \\ y=2 \sin \alpha\end{array}\right.$, then $u=3 \cos \alpha+2 \sin \alpha=\sqrt{13} \",easy,"10. Let $\left\{\begin{array}{l}x=3 \cos \alpha \\ y=2 \sin \alpha\end{array}\right.$, then $u=3 \cos \alpha+2 \sin \alpha=\sqrt{13} \sin (\alpha+\varphi)$, where $\varphi=\arctan \frac{3}{2}$, so $u=x+y \leqslant \sqrt{13}$" aefdea99a7e7,"6. Given point $P$ is in the plane of Rt $\triangle A B C$, $\angle B A C=90^{\circ}, \angle C A P$ is an acute angle, and $|\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1$. When $| \overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P} |$ is minimized, $\tan \angle C A P=$ $\qquad$ .","\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}$, i.e., $\tan \alpha=\frac{\sqrt{2}}{2}$, $|\overrightarro",medium,"6. $\frac{\sqrt{2}}{2}$. Let $\angle C A P=\alpha$. From the problem, we know $\angle B A P=\frac{\pi}{2}-\alpha$. $$ \begin{aligned} \text { Given } & |\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1 \\ \Rightarrow & |\overrightarrow{A C}|=\frac{1}{\cos \alpha},|\overrightarrow{A B}|=\frac{1}{2 \sin \alpha} \\ \Rightarrow & |\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|^{2} \\ & =|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+|\overrightarrow{A P}|^{2}+2 \overrightarrow{A B} \cdot \overrightarrow{A C}+ \\ & 2 \overrightarrow{A B} \cdot \overrightarrow{A P}+2 \overrightarrow{A C} \cdot \overrightarrow{A P} \\ = & \frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\cos ^{2} \alpha}+10 \\ = & \frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}+\frac{45}{4} \\ \geqslant & \geqslant \sqrt{\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha} \cdot \frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}}+\frac{45}{4}=\frac{49}{4}, \end{aligned} $$ When and only when $\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}$, i.e., $\tan \alpha=\frac{\sqrt{2}}{2}$, $|\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|$ achieves its minimum value $\frac{7}{2}$." de54bad4c454,"10.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all possible answers. (N. Agakhanov)",. The digit 6,medium,"Answer. The digit 6. Solution. By the condition, $M=3 N$, so the number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6. We will show that $N$ cannot end in 1. If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second-to-last digit. This means that the second-to-last digit of the number $A=2 N$ would be even, but it must be odd. This is a contradiction. Remark. Pairs of numbers $N$ and $M$ as described in the condition do exist, for example, $N=16, M=48$. Moreover, there are infinitely many such pairs. All suitable numbers $N$ can be described as follows: the first digit is 1 or 2, followed by several (possibly zero) digits, each of which is 5 or 6, and the last digit is 6. Comment. A correct answer and an example of a number $N$ ending in the digit 6 - 1 point. It is established that the last digit of the number $M$ is 2 more than the last digit of the number $N$ - 1 point. It is shown that the last digit of the number $N$ can only be 1 or 6 - 2 points. Points for various advancements are cumulative. Note that the problem does not require providing an example of such a number. It is sufficient to prove that no digit other than 6 can be the last digit." e2474cd1527f,235. The surface area of a spherical segment is $S$ (the spherical part of the segment is considered). Find the maximum value of the volume of this segment.,\sqrt{\frac{S}{2 \pi}}$; it will be equal to $\frac{S}{3} \sqrt{\frac{S}{2 \pi}}$,easy,"235. If $h$ is the height of the segment, then its volume is $\frac{1}{2} S h - \frac{1}{3} \pi h^{3}$. The maximum volume will be at $h = \sqrt{\frac{S}{2 \pi}}$; it will be equal to $\frac{S}{3} \sqrt{\frac{S}{2 \pi}}$" 806c593ab007,"![](https://cdn.mathpix.com/cropped/2024_05_06_1f1bf0225c3b69484645g-03.jpg?height=331&width=329&top_left_y=196&top_left_x=562) Then the teacher decided to complicate the task and asked to count the number of dots, but for the letter ""P"" obtained in the same way from a square with a side of 10 cm. How many dots will Zhenya have this time?",31,easy,"Answer: 31. Solution. Along each of the three sides of the letter ""П"", there will be 11 points. At the same time, the ""corner"" points are located on two sides, so if 11 is multiplied by 3, the ""corner"" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$." c69cde963957,"The yearly changes in the population census of a town for four consecutive years are, respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. The net change over the four years, to the nearest percent, is: $\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$",\textbf{(A),easy,"A 25% increase means the new population is $\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\frac{3}{4}$ of the original population. Thus, after four years, the population is $1 \cdot \frac{5}{4} \cdot \frac{5}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{225}{256}$ times the original population. Thus, the net change is -12%, so the answer is $\boxed{\textbf{(A)}}$." 7312c920f44d,"SUBIECTUL IV Consider the sequence: $\left(x_{n}\right)_{n \geq 1}$ of real numbers defined by $x_{1}=\sqrt{\frac{1}{2}}, \quad x_{n+1}=\sqrt{\frac{1+x_{n}}{2}}, \quad n \geq 1$. Calculate $\lim _{n \rightarrow \infty} x_{1} \cdot x_{2} \cdot \ldots . . x_{n}$ ## BAREM CLASA XI",See reasoning trace,easy,"## SUBJECT IV $$ \begin{aligned} & x_{1}=\cos \frac{\pi}{4}(1 p) ; x_{2}=\cos \frac{\pi}{8}(1 p) ; x_{n}=\cos \frac{\pi}{2^{n+1}}(1 p) ; x_{1} \cdot x_{2} \cdot \ldots \ldots \cdot x_{n}=\frac{1}{2^{n} \sin \frac{\pi}{2^{n+1}}}(2 p) ; \\ & \lim _{n \rightarrow \infty} x_{1} \cdot x_{2} \cdot \ldots . . \cdot x_{n}=\frac{2}{\pi}(2 p) \end{aligned} $$" d76e49814045,"4. Let $a$ be a real number such that the graph of the function $y=f(x)=a \sin 2x + \cos 2x + \sin x + \cos x$ is symmetric about the line $x=-\pi$. Let the set of all such $a$ be denoted by $S$. Then $S$ is ( ). (A) empty set (B) singleton set (contains only one element) (C) finite set with more than one element (D) infinite set (contains infinitely many elements)","0$ or $a \neq 0$, leading to a contradiction. Therefore, $S$ must be an empty set.",medium,"4. A. By symmetry, we have $$ f(-x)=f(x-2 \pi) \text {. } $$ Since $f(x)$ is a periodic function with a period of $2 \pi$, we have $$ f(x)=f(x-2 \pi) \text {. } $$ Therefore, $f(-x)=f(x)$. Thus, $-a \sin 2 x+\cos 2 x-\sin x+\cos x$ $$ =a \sin 2 x+\cos 2 x+\sin x+\cos x \text {, } $$ which simplifies to $\sin x(2 a \cos x+1)=0$ (for any $x \in \mathbf{R}$). Hence, $2 a \cos x=-1$ (for any $x \neq k \pi, k \in \mathbf{Z}$). However, the above equation cannot hold for $a=0$ or $a \neq 0$, leading to a contradiction. Therefore, $S$ must be an empty set." cf780d11b369,"1. [4] A hundred friends, including Petya and Vasya, live in several cities. Petya learned the distance from his city to the city of each of the remaining 99 friends and added these 99 numbers. Vasya did the same. Petya got 1000 km. What is the largest number Vasya could have obtained? (Consider the cities as points on a plane; if two people live in the same city, the distance between their cities is considered to be zero.) Boris Frenkin",99000 km,medium,"Answer: 99000 km. Solution. Estimate. Let Vasya live in city $V$, and Petya - in city $P$. Consider an arbitrary friend of Vasya (this could be Petya), let's say he lives in city $X$. By the triangle inequality, $V X \leq P V + P X$, and this sum is no more than Petya's total, i.e., 1000 km. Therefore, the sum of the distances from $V$ to the cities of all 99 of Vasya's friends is no more than $99 \cdot 1000$ km. Example. Everyone except Vasya lives in one city, and Vasya lives 1000 km away from them." e26af7424266,"In an arithmetic sequence with a common difference of $d$, $a_{1}=1$ and $a_{n}=81$. In a geometric sequence with a common ratio of $q$, $b_{1}=1$ and $b_{n}=81$. We also know that $\frac{q}{d}=0.15$. Find all such sequences.","1$, if $s>1$ this would not be an integer. Therefore, $s=1$, so $q$ is a positive integer. This impl",medium,"I. solution. Based on the formula for the $n$-th term of an arithmetic and geometric sequence: $$ 81=1+(n-1) d $$ and $$ 81=q^{n-1} $$ From (1), $\mathrm{d}=\frac{80}{n-1}$, and from $\frac{q}{d}=0.15$ we get $$ q=0.15 d=0.15 \cdot \frac{80}{n-1}=\frac{12}{n-1} $$ Substituting this into (2) we get $81=\left(\frac{12}{n-1}\right)^{n-1}$. Here, if $n-1 \geq 13$ then $\frac{12}{n-1} \leq 1$, so $\left(\frac{12}{n-1}\right)^{n-1} \leq 1$ would be true, and for $n=13$ we would have $\left(\frac{12}{n-1}\right)^{n-1}=1$. Therefore, $2 \leq n \leq 12$. Let's check these 11 cases: $$ \begin{array}{llll} \left(\frac{12}{1}\right)^{1}=12 ; & \left(\frac{12}{2}\right)^{2}=36 ; & \left(\frac{12}{3}\right)^{3}=64 ; & \left(\frac{12}{4}\right)^{4}=81 \\ \left(\frac{12}{5}\right)^{5} \approx 79.63 ; & \left(\frac{12}{6}\right)^{6}=64 ; & \left(\frac{12}{7}\right)^{7} \approx 43.51 ; & \left(\frac{12}{8}\right)^{8} \approx 25.6 \\ \left(\frac{12}{9}\right)^{9} \approx 13.32 ; & \left(\frac{12}{10}\right)^{10} \approx 6.19 ; & \left(\frac{12}{11}\right)^{11} \approx 2.60 . \end{array} $$ It can be seen that only for $n=5$ we get a solution, in which case $q=\frac{12}{4}=3$ and $d=\frac{80}{4}=20$. II. solution. Based on the definition of an arithmetic sequence, $$ (n-1) d=a_{n}-a_{1}=81-1=80 $$ so $d=\frac{80}{n-1}$. Since $\frac{q}{d}=0.15=\frac{3}{20}$, then $d=\frac{20}{3} q$, so $\frac{20}{3} q=\frac{80}{n-1}$. From this, $q=\frac{12}{n-1}$ follows. Since $n \geq 2$ and is an integer, $q$ is a positive rational number. Let $q=\frac{p}{s}$, where $(p, s)=1$. Then in the geometric sequence, $$ 81=q^{n-1}=\frac{p^{n-1}}{s^{n-1}} $$ Since $(p, s)=1$, if $s>1$ this would not be an integer. Therefore, $s=1$, so $q$ is a positive integer. This implies that $n-1 \mid 12$ and is positive, so the possible values of $n-1$ are $1,2,3,4,6,12$. Then the values of $q$, and $q^{n-1}$ are respectively $12,6,4,3,2,1$ and $12,36,64,81,64,1$. It can be seen that only $n-1=4$ gives a good solution, in which case $d=\frac{80}{4}=20, a_{n}=1+4 \cdot 20=81$ is also satisfied. Therefore, the only solution is: $n=5, q=3, d=20$." ebf43fad3c59,"4. If $2 x+y \geqslant 1$, then the minimum value of the function $u=y^{2}-2 y+x^{2}$ $+4 x$ is $\qquad$",-\frac{9}{5}$.,medium,"4. $u_{\min }=-\frac{9}{5}$. By completing the square from $u=y^{2}-2 y+x^{2}+4 x$, we get $$ (x+2)^{2}+(y-1)^{2}=u+5 \text {. } $$ The left side $(x+2)^{2}+(y-1)^{2}$ can be interpreted as the square of the distance between point $P(x, y)$ and the fixed point $A(-2,1)$. The constraint $2 x+y \geqslant 1$ indicates that point $P$ is in the region $G$ which is the right upper area of the line $l: 2 x+y=1$. Thus, the problem is transformed into finding the minimum distance between point $A(-2,1)$ and the points in region $G$, which is the distance $d$ from point $A$ to the line $l$: $$ d=\frac{|2 \times(-2)+1-1|}{\sqrt{2^{2}+1^{2}}}=\frac{4}{\sqrt{5}} . $$ Therefore, $d^{2}=\frac{16}{5}$. From $u+5=\frac{16}{5}$, we have $u=\frac{16}{5}-5=-\frac{9}{5}$, which means $u_{\text {min }}=-\frac{9}{5}$." 4d00e71cac67,"6. The integer 36 is divisible by its units digit. The integer 38 is not. How many integers between 20 and 30 are divisible by their units digit? A 2 B 3 C 4 D 5 E 6",See reasoning trace,easy,"SOLUTION $\mathbf{C}$ Note that 21 is divisible by 1,22 is divisible by 2,24 is divisible by 4 and 25 is divisible by 5 . However, $23,26,27,28$ and 29 are not divisible by $3,6,7,8$ and 9 respectively. Therefore there are four integers between 20 and 30 that are divisible by their units digit." 561be59ef122,"10.3. What is the maximum number of digits that a natural number can have, where all digits are different, and it is divisible by each of its digits?",7 digits,medium,"Answer: 7 digits. Evaluation. There are 10 digits in total. The number cannot contain the digit 0, so there are no more than 9. If all 9, then the digit 5 must be at the end of the number (divisibility rule for 5), but in this case, the number must also be divisible by 2. Therefore, the digit 5 is not in this number. If the number lacks only 0 and 5, then it contains 9, and the sum of the digits of this number will be 40, which contradicts divisibility by 9. Thus, there are no more than seven digits. An example of such a number is 9176328. Verification. The three-digit number formed by the last three digits is 328, which is divisible by 8, meaning the given number is divisible by its digits 2 and 8. The sum of the digits of this number is 36, so it is divisible by 3 and 9, and also by 6. It is easy to see that it is divisible by 7: $9176328: 7=1310904$. Remarks. An example of a seven-digit number with verification is provided - 3 points. It is proven that there are no more than seven digits, but an example of such a number is not provided - 3 points. If both are present, 7 points." aa5132df6d75,,45$.,easy,"Solution. The unit digit can be 1, 3, 5, 7 or 9, which means we have 5 possibilities. The tens digit is 4, which means we have 1 possibility. The hundreds digit can be: $1,2,3,4,5,6,7,8$ or 9, which means we have 9 possibilities. Finally, the number of desired numbers is $5 \cdot 1 \cdot 9=45$." 39fa1642d2b1,"The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is: $\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$",2\implies y=2-x$. Then $2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8$. We now notice that $x=0\implies,easy,"$x+y=2\implies y=2-x$. Then $2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8$. We now notice that $x=0\implies (2)(4)=8$, so $\fbox{C}$" d57f41eed66b,"Four (This sub-question is worth 50 points) $n$ chess players participate in a chess tournament, with each pair of players competing in one match. The winner gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. If it is found after the tournament that among any $m$ players, there is one player who has defeated the other $m-1$ players, and there is also one player who has lost to the other $m-1$ players, this situation is said to have property $P(m)$. For a given $m (m \geq 4)$, find the minimum value of $n$, denoted as $f(m)$, such that for any situation with property $P(m)$, all $n$ players have different scores. --- Please note that the translation preserves the original text's formatting and line breaks.",2 m-3$.,medium,"Proof: We first prove two lemmas. Lemma 1 When $n \geq m$, if the match results of $n$ players have property $P(m)$, then there must be a player who wins all their matches. When $n=m$, the proposition is obviously true. Assume the proposition holds for $n$, then for $n+1$ players, take any $n$ players. By the induction hypothesis, there must be a player who wins all their matches among these $n$ players, let's say $A_{1}, A_{2}, \ldots, A_{n}$, where $A_{1}$ wins all. For the other player $A_{n+1}$: If $A_{1}$ wins $A_{n+1}$, then among the $n+1$ players, $A_{1}$ wins all; If $A_{1}$ draws with $A_{n+1}$, consider the players $A_{1}, A_{2}, \ldots, A_{n-1}, A_{n+1}$, there is no one who wins all, which is impossible; if $A_{n+1}$ wins $A_{1}$, consider the players $A_{1}, A_{3}, A_{4}, \ldots, A_{n}, A_{n+1}$, the one who wins all can only be $A_{n+1}$, especially, $A_{n+1}$ wins $A_{3}, A_{4}, \ldots, A_{n}$. Similarly, $A_{n+1}$ also wins $A_{2}$, so among these $n+1$ players, $A_{n+1}$ wins all. By the principle of induction, the proposition holds for any $n \geq m$. Similarly, we can prove: Lemma 2 When $n \geq m$, if the match results of $n$ players have property $P(m)$, then there must be a player who loses all their matches. Back to the original problem. We will prove: When $n \geq 2 m-3$, the scores of all players must be different. By Lemma 1, there is a player $A_{1}$ who wins the other $n-1$ players, there is a player $A_{2}$ who wins the other $n-2$ players except $A_{1}$, ..., there is a player $A_{n-m+1}$ who wins the other $m-1$ players except $A_{1}, A_{2}, \ldots, A_{n-m}$. By Lemma 2, there is a player $A_{n}$ who loses to the other $n-1$ players, there is a player $A_{n-1}$ who loses to the other $n-2$ players except $A_{n}$, ..., there is a player $A_{n-m+3}$ who loses to the other $n-m+2$ players except $A_{n}, A_{n-1}, \ldots, A_{n-m+4}$, and there is another player $A_{n-m+2}$. Thus, these $n$ players $A_{1}, A_{2}, \ldots, A_{n}$, the players with smaller numbers defeat the players with larger numbers, so their scores are $n-1, n-2, \ldots, 1, 0$, all different. For $n=2 m-4$, assume the levels of the $n$ players are: $$ 1,2, \ldots, m-3, m-2, m-2, m-1, \ldots, 2 m-5, $$ where players with smaller level numbers win players with larger level numbers, and players with the same level number draw. Then for any $m$ players, there must be a minimum level number $i(1 \leq i \leq m-3)$, and another maximum level number $j(m-1 \leq j \leq 2 m-5)$, thus among these $m$ players, the player with level $i$ wins all, and the player with level $j$ loses all, so these $n$ players have property $P(m)$, but there are two players with the same score. In summary, $f(m)=2 m-3$." d03ca11ff07f,"3-ча 1. Solve the system: $$ \left\{\begin{aligned} x+y+z & =a \\ x^{2}+y^{2}+z^{2} & =a^{2} \\ x^{3}+y^{3}+z^{3} & =a^{3} \end{aligned}\right. $$","$(0,0, a),(0, a, 0)$ and $(a, 0,0)$",medium,"Solve 1. Answer: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$. The identity $(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+x z)$ shows that $$ x y+y z+x z=0 $$ The identity $(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)$ shows that $(x+y)(y+z)(z+x)=0$. Considering equation (1), we get $3 x y z=0$. If $x=0$, then equation (1) shows that $y z=0$. Therefore, either $y=0$ and $z=a$, or $z=0$ and $y=a$. The other cases are similarly analyzed. In the end, we get the following solutions: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$." 2b0faa49893d,"14.50. The two players are of equal strength, i.e., they have equal chances of winning each game. They agreed that the prize would go to the one who first wins 6 games. They had to stop the game after the first player won 5 games, and the second player won 3. In what ratio should the prize be fairly divided?",$7: 1$,easy,"14.50. Answer: $7: 1$. Let the players play three more games, i.e., the game is fictitiously continued even after the first player wins a prize. The second player will receive a prize if and only if he wins all three games. Since all $2^{3}=8$ outcomes of these three games are equally likely, the second player will receive a prize with a probability of $1 / 8$. Accordingly, the first player will receive a prize with a probability of $7 / 8$." 3cdcb3fa8101,"4. In $\triangle A B C$, $A B=15, A C=13$, and the altitude $A D=12$. Let the radius of the circle that can completely cover $\triangle A B C$ be $R$. Then the minimum value of $R$ is $\qquad$","4$, and $\triangle A B C$ is an obtuse triangle. In this case, the smallest circle that can complete",medium,"4. $\frac{65}{8}$ or $\frac{15}{2}$. As shown in figure (a), if $D$ is on $BC$, by the Pythagorean theorem, we get $$ \begin{array}{l} B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{15^{2}-12^{2}}=9, \\ D C=\sqrt{A C^{2}-A D^{2}}=\sqrt{13^{2}-12^{2}}=5 . \\ \therefore B C=14 . \\ \because B C^{2}+A C^{2}=14^{2}+13^{2}>15^{2}=A B^{2}, \end{array} $$ $\therefore \triangle A B C$ is an acute triangle. In this case, the smallest circle that can completely cover $\triangle A B C$ is the circumcircle of $\triangle A B C$. Thus, $R=\frac{A B \cdot A C}{2 A D}=\frac{65}{8}$. As shown in figure (b), if $D$ is on the extension of $BC$, then $B C=4$, and $\triangle A B C$ is an obtuse triangle. In this case, the smallest circle that can completely cover $\triangle A B C$ is the circle with the largest side $A B$ as its diameter, $R=$ $\frac{1}{2} A B=\frac{15}{2}$." c929572c98ee,"1. Given (1) $a>0$; (2) When $-1 \leqslant x \leqslant 1$, it satisfies $$ \left|a x^{2}+b x+c\right| \leqslant 1 \text {; } $$ (3) When $-1 \leqslant x \leqslant 1$, $a x+\dot{b}$ has a maximum value of 2. Find the constants $a, b, c$.","2$. Hence, $f(x)=2 x^{2}-1$.",easy,"1. From (1), we know that $y=a x^{2}+b x+c$ is a parabola opening upwards. From (1) and (3), we have $$ a+b=2 \text{. } $$ From (2), we have $$ \begin{array}{l} |a+b+c| \leqslant 1, \\ |c| \leqslant 1 . \end{array} $$ From (1) and (2), we have $$ |2+c| \leqslant 1 \text{. } $$ From (3) and (4), we get $c=-1$. Therefore, when $x=0$, $y=a x^{2}+b x+c$ reaches its minimum value. Thus, $-\frac{b}{2 a}=0, b=0$. From (1), we get $a=2$. Hence, $f(x)=2 x^{2}-1$." 2c970727b4a7,1st Centromerican 1999,See reasoning trace,medium,"8, eg BA, CA, DA, EA, AB, AC, AD, AE. 2n-2 Solution Consider the case of n people. Let N be the smallest number of calls such that after they have been made at least one person knows all the news. Then N ≥ n-1, because each of the other n-1 people must make at least one call, otherwise no one but them knows their news. After N calls only one person can know all the news, because otherwise at least one person would have known all the news before the Nth call and N would not be minimal. So at least a further n-1 calls are needed, one to each of the other n-1 people. So at least 2n-2 calls are needed in all. But 2n-2 is easily achieved. First everyone else calls X, then X calls everyone else. 1st OMCC 1999 © John Scholes jscholes@kalva.demon.co.uk 26 November 2003 Last corrected/updated 26 Nov 03" 2bd964f85fbc,"18. There are 30 students standing in a row, and from left to right, they are wearing hats in the order of ""red, yellow, blue, red, yellow, blue......"". The PE teacher asks the students to count off from left to right as ""1, 2, 1, 2, 1, 2....."", and those who count off 2 step forward. Among the remaining students, those wearing red hats step back. At this point, the students are divided into 3 rows, and in the middle row, there are $\qquad$ people wearing blue hats.",See reasoning trace,easy,$5$ 293a715818cf,"6. The sequence $a_{n}$ is defined as follows: $a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$.",499500,easy,"Answer: 499500. ## School Olympiad ""Conquer Sparrow Hills"" ## Tasks for 7-8 Grades ## Variant 1b (Zheleznovodsk)" 1bd717c6b6ee,"Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$ $\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$",\textbf{(C),easy,"Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\triangle AUV$ has $\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac 34\cdot [AMC].$ Thus, \[\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]\] \[72=\frac 34\cdot [AMC]\] \[[AMC]=96\rightarrow \boxed{\textbf{(C)}}.\]" c2be171d1af3,"Example 4 Given $A(-2,2)$, and $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, $F$ is the left focus. When $\mid A B \mid +\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of point $B$. (1999, National High School Mathematics Competition)",See reasoning trace,medium,"Analysis: Since the eccentricity of the ellipse $e=\frac{3}{5}$, we have $|A B|+\frac{5}{3}|B F|=|A B|+\frac{1}{e}|B F|$. And $\frac{|B F|}{e}$ is the distance from the moving point $B$ to the left directrix. Therefore, this problem is transformed into: Finding a point $B$ on the ellipse such that the sum of its distances to point $A$ and the left directrix is minimized. Let the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse be $a$, $b$, and $c$ respectively, then $a=5$, $b=4$, $c=3$, $e=\frac{3}{5}$, and the left directrix $l: x=-\frac{25}{3}$. Draw $B N \perp l$ at point $N$, and $A M \perp l$ at point $M$. By the definition of the ellipse, we have $|B N|=\frac{|B F|}{e}=\frac{5}{3}|B F|$, thus, $$ \begin{array}{l} |A B|+\frac{5}{3}|B F|=|A B|+|B N| \\ \geqslant|A N| \geqslant|A M| \end{array} $$ is a constant, with equality holding if and only if $B$ is the intersection of $A M$ and the ellipse, i.e., $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$. Therefore, when $|A B|+\frac{5}{3}|B F|$ is minimized, the coordinates of point $B$ are $\left(-\frac{5}{2} \sqrt{3}, 2\right)$." c45548bff4c6,"4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$. 保持源文本的换行和格式如下： 4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$.","7$, we get $x=7$.",easy,"[Solution] Let $x$ be the root of the given equation, and $[x]=n$. From the original equation, we have $$x^{2}+7=8 n$$ This indicates that $n>0$. Using $n \leqslant x < n+1$, we get $$\left\{\begin{array}{l} n^{2}+7<8(n+1) \\ n^{2}-8 n+7 \leqslant 0 \end{array}\right.$$ Solving this, we get $\quad 1 \leqslant n<2$ or $42 \cdot x^{2}$. We rearrange and factor the inequality to get $x(1-2 x)>0$. We read off the zeros $x_{1}=0$ and $x_{2}=\frac{1}{2}$. On the number line, we mark the signs and determine that the solution is the interval $\left(0, \frac{1}{2}\right)$." 313327c7ba3f,"8.15 If $2^{a}+2^{b}=3^{c}+3^{d}$, then the number of negative integers among $a, b, c, d$ is (A) 4. (B) 3. (C) 2. (D) 1. (E) 0. (14th American High School Mathematics Examination, 1963)",$(E)$,medium,"［Solution］We discuss the following cases: (1) If only one of $a, b, c, d$ is negative, then one side of the equation is a fraction, and the other side is an integer, which is not appropriate; (2) If exactly two of the four numbers are negative: (i) If $a\beta$ the equation can be transformed into: $$ \left(2^{\alpha-\beta}+1\right) 3^{\gamma+\delta}=\left(3^{\gamma}+3^{\delta}\right) 2^{\alpha}, $$ the left side of the above equation is odd, while the right side is even, which is not appropriate; (ii) If $\alpha=\beta$, then the previous equation becomes: $$ 2^{\alpha+1} 3^{\gamma+\delta}=\left(3^{\gamma}+3^{\delta}\right) 2^{2 \alpha} \text {. } $$ When $\alpha=1$, it becomes $3^{\gamma} \cdot 3^{\delta}=3^{\gamma}+3^{\delta}$, which is not true for all $\gamma, \delta$; When $\alpha>1$, then $\alpha-1>0$, the previous equation becomes: $$ 3^{\gamma+\delta}=2^{\alpha-1}\left(3^{\gamma}+3^{\delta}\right) \text {. } $$ The left side of the above equation is odd, while the right side is even, which is also not appropriate. In summary, the number of negative numbers among $a, b, c, d$ can only be 0. Therefore, the answer is $(E)$." cf4719f0b488,"(2) (2007 - Ningxia Hainan) Let the function $f(x)=\frac{(x+1)(x+a)}{x}$ be an odd function, then $a=$",See reasoning trace,easy,"From the given, we have $f(-x)=-f(x)$, which means $\frac{(-x+1)(-x+a)}{-x}=-\frac{(x+1)(x+a)}{x}$, or equivalently, $-(a+1) x=(a+1) x$ for any real number $x \neq 0$. Therefore, $a=-1$. " b99e42455c7d,"A point $D$ is chosen on side $B C$ of triangle $A B C$. The line tangent to the incircles of triangles $A B D$ and $A D C$ and different from $B C$ and $A D$ intersects segment $A D$ at $T$. If $A B=40 \text{ cm}, A C=50 \text{ cm}$, and $B C=60 \text{ cm}$, determine the value of the length of $A T$. ![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-07.jpg?height=539&width=1108&top_left_y=1079&top_left_x=546)",30 \text{ cm}$.,easy,"Solution We know that the lengths of the tangent segments drawn from an external point to a circle are congruent and that $X Y=H I$. Thus, $$ \begin{aligned} 2 A T & =(A F-T F)+(A J-T J) \\ & =A G+A K-(T X+T Y) \\ & =(A B-B G)+(A C-C K)-X Y \\ & =A B+A C-(B H+H I+I C) \\ & =A B+A C-B C \\ & =30 \end{aligned} $$ Therefore, $A T=30 \text{ cm}$." 3504f555065b,"1. Given sets $M=\left\{a_{1}, a_{2}, \cdots, a_{2 n+1}\right\}, N$ $=\left\{-2^{2 n},-2^{2 n-1}, \cdots,-2,0,2, \cdots, 2^{2 n}\right\}$. If the injection $f: M \rightarrow N$ satisfies $$ f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{2 n+1}\right)=0, $$ then the number of such injections $f$ is $(\quad)$. (A) $(2 n+1)!\mathrm{C}_{2 n}^{n}$ (B) $(2 n+1)!\mathrm{C}_{2 n+1}^{n}$ (C) $(2 n+1)!\mathrm{C}_{4 n+1}^{2 n+1}$ (D) $(2 n+1)!\mathrm{C}_{4 n}^{2 n}$","1,2, \cdots, 2 n+1\right\}$, the number of such injections is $(2 n+1)!\mathrm{C}_{n, n}^{n}$.",medium,"-.1.A. For any positive integer $m$, we have $$ 2^{m}>2^{m-1}+\cdots+2=2^{m}-2 \text {. } $$ Thus, the highest power of 2 in $\left|f\left(a_{j}\right)\right| j=1,2, \cdots, 2 n+1$ is the same as the highest power of -2. Otherwise, the sum would not be 0. By canceling these two terms, similarly, $2^{i}$ and $-2^{i}$ either both appear or do not appear in $\left|f\left(a_{j}\right)\right| j=1,2, \cdots, 2 n+1 \mid$ $(i=1,2, \cdots, 2 n)$. Since there are an odd number of terms, there exists an $i$ such that $f\left(a_{i}\right)=0$. Given that $\left\{2,2^{2}, \cdots, 2^{2 n}\left\{, \mid-2,-2^{2}, \cdots,-2^{2 n}\right\}\right.$ each appear $n$ times in $\left\{f\left(a_{j}\right) \mid j=1,2, \cdots, 2 n+1\right\}$, the number of such injections is $(2 n+1)!\mathrm{C}_{n, n}^{n}$." 421fc0507ae9,7.213. $\left|\log _{\sqrt{3}} x-2\right|-\left|\log _{3} x-2\right|=2$.,$\frac{1}{9} ; 9$,medium,"## Solution. Domain of definition: $x>0$. Switch to base 3. Then $\left|2 \log _{3} x-2\right|-\left|\log _{3} x-2\right|=2$. By opening the absolute values, we get three cases: 1) $\left\{\begin{array}{l}\log _{3} x<1, \\ -2 \log _{3} x+2+\log _{3} x-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3} x<1, \\ \log _{3} x=-2\end{array} \Rightarrow x_{1}=3^{-2}=\frac{1}{9}\right.\right.$; 2) $\left\{\begin{array}{l}1 \leq \log _{3} x<2, \\ 2 \log _{3} x-2+\log _{3} x-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}1 \leq \log _{3} x<2, \\ \log _{3} x=2 .\end{array}\right.\right.$ $\log _{3} x=2$ does not fit because $\log _{3} x<2$. 3) $\left\{\begin{array}{l}\log _{3} x \geq 2, \\ 2 \log _{3} x-2-\log _{3} x+2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3} x \geq 2, \\ \log _{3} x=2\end{array} \Rightarrow x_{2}=3^{2}=9\right.\right.$. Answer: $\frac{1}{9} ; 9$." c65322d1c622,"[ Geometry (miscellaneous).] A sphere of radius $\sqrt{5}$ with center at point $O$ touches all sides of triangle $ABC$. The point of tangency $N$ bisects side $AB$. The point of tangency $M$ divides side $AC$ such that $AM=\frac{1}{2} MC$. Find the volume of the pyramid $OABC$, given that $AN=NB=1$.",2,medium,"The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semi-perimeter. ## Solution Let the given sphere touch the side $BC$ of triangle $ABC$ at point $K$. Then $$ BK = BN = 1, AM = AN = 1, CM = 2 \cdot AM = 2, CK = CM = 2 $$ The section of the sphere by the plane of triangle $ABC$ is the circle inscribed in triangle $ABC$, and the center $O_1$ of this circle is the orthogonal projection of the center $O$ of the sphere onto the plane of triangle $ABC$. Therefore, $OO_1$ is the height of the pyramid $OABC$. Let $r$ be the radius of the circle inscribed in triangle $ABC$, $p$ be the semi-perimeter of the triangle, and $s$ be the area. Since triangle $ABC$ is isosceles, the segment $CN$ is its height. Then $$ \begin{gathered} CN = \sqrt{AC^2 - AN^2} = \sqrt{9 - 1} = 2\sqrt{2}, \\ s = \frac{1}{2} AB \cdot CN = 2\sqrt{2}, \quad r = s / p = 2\sqrt{2} / 4 = \sqrt{2} / 2 \end{gathered} $$ From the right triangle $OO_1N$, we find that $$ OO_1 = \sqrt{ON^2 - ON^2} = \sqrt{5 - 1/2} = 3 / \sqrt{2} \text{.} $$ Therefore, $$ V(OABC) = \frac{1}{3} s \cdot OO_1 = \frac{1}{3} 2\sqrt{2} \cdot 3 / \sqrt{2} = 2 $$ ## Answer 2." cfd2f58a7142,"## Task B-3.7. Determine all integers $a$ for which the solutions of the equation $x^{2}-(3+a) x+3 a+3=0$ are also integers. What are these numbers?","-1$ and $a=7$, the solutions of the equation $x^{2}-(3+a) x+3 a+3=0$ are also integers.",medium,"## First Solution. By applying Viète's formulas to the solutions $x_{1}, x_{2}$ of the given equation, we get: $$ \left\{\begin{array}{l} x_{1}+x_{2}=3+a \\ x_{1} \cdot x_{2}=3 a+3 \end{array}\right. $$ If we express, for example, $x_{1}$ in terms of $x_{2}$ from equation $(\star)$, we obtain $x_{1}=\frac{3 x_{2}-6}{x_{2}-3}$. Now, $x_{1}$ can be written in the form $$ x_{1}=\frac{3 x_{2}-6}{x_{2}-3}=\frac{3\left(x_{2}-3\right)+3}{x_{2}-3}=3+\frac{3}{x_{2}-3} $$ Since $x_{1}$ is an integer, from equation ($* \star$) we conclude that $x_{2}-3 \in\{1,3,-1,-3\}$. Thus, we have the following possibilities: $$ \begin{gathered} x_{2}-3=1 \Rightarrow x_{2}=4 \Rightarrow x_{1}=6 \\ x_{2}-3=3 \Rightarrow x_{2}=6 \Rightarrow x_{1}=4 \\ x_{2}-3=-1 \Rightarrow x_{2}=2 \Rightarrow x_{1}=0 \\ x_{2}-3=-3 \Rightarrow x_{2}=0 \Rightarrow x_{1}=2 \end{gathered} $$ From $x_{1}=6, x_{2}=4$ and equation $(\star)$, we get $a=7$, similarly from $x_{1}=4$ and $x_{2}=6$. From $x_{1}=0, x_{2}=2$ and equation $(\star)$, we get $a=-1$, similarly from $x_{1}=2$ and $x_{2}=0$. Therefore, for the integers $a=-1$ and $a=7$, the solutions of the equation $x^{2}-(3+a) x+3 a+3=0$ are also integers." 1accab01fa42,"61. Given $a \neq 0, S_{1}=a, S_{2}=\frac{2}{S_{1}}, \cdots, S_{2018}=\frac{2}{S_{2017}}$, express $S_{2018}=$ in terms of $a$",$2 / a$,easy,Reference answer: $2 / a$ 6ca7f46eaf92,"14. Insert parentheses in the equation: $1: 2: 3: 4: 5: 6: 7: 8: 9: 10=7$, to make it true.",$1: 2: 3: 4: 5:(6: 7: 8: 9: 10)=7$,easy,14. Answer: $1: 2: 3: 4: 5:(6: 7: 8: 9: 10)=7$. eaa952215004,"16. Fnargs are either red or blue and have 2,3 or 4 heads. A group of six Fnargs consisting of one of each possible form is made to line up such that no immediate neighbours are the same colour nor have the same number of heads. How many ways are there of lining them up from left to right? A 12 B 24 С 60 D 120 E 720",See reasoning trace,medium,"Solution A We let R2, R3, R4 be red Fnargs with 2, 3 and 4 heads, respectively, and B2, B3, B4 be blue Fnargs with 2,3 and 4 heads, respectively. We need to count the number of ways of lining up R2, R3, R4, B2, B3, B4 so that no two Fnargs that are next to each other have the same colour or the same number of heads. Suppose that a row of these six Fnargs begins with R2 at the left hand end. The second Fnarg in the row must be blue and have 3 or 4 heads. So the row either begins R2, B3 or R2, B4. If the row begins R2, B3 the third Fnarg must be red and cannot have 3 heads. Since R2 is already in the line up, the third Fnarg must be R4. The fourth Fnarg must be blue and cannot have 4 heads. Since B3 is already in the row, this fourth Fnarg must be B2. This leaves just R3 and B4 to be placed. So, as the colours must alternate, the fifth and sixth Fnargs in the row must be R3 and B4 in this order from left to right. Hence the line up must be R2, B3, R4, B2, R3, B4. A similar argument shows that if the row begins R2, B4, then the line up must be as above but with 3 and 4 interchanged, that is, R2, B4, R3, B2, R4, B3. So there are just two ways to complete a row that begins with R2. A similar argument shows that whichever Fnarg begins a row, there are just two ways to complete the row. Since the row may begin with any of the six Fnargs, the total number of ways to line them up is $6 \times 2=12$. FOR INVESTIGATION 16.1 As a result of a genetic modification there are now Fnargs with 2, 3, 4 or 5 heads, but still only red or blue. In how many ways can we line up eight Fnargs consisting of one of each possible form so that two adjacent Fnargs have neither the same colour nor the same number of heads? 16.2 As a result of a further genetic modification there are now red, blue and green Fnargs, each with $2,3,4$ or 5 heads. In how many ways can we line up twelve Fnargs consisting of one of each possible form so that two adjacent Fnargs have neither the same colour nor the same number of heads?" 617bff6b4ad0,"2. Two bottles of equal volume are filled with a mixture of water and juice. In the first bottle, the ratio of the quantities of water and juice is $2: 1$, and in the second bottle, it is $4: 1$. If we pour the contents of both bottles into a third bottle, what will be the ratio of the quantities of water and juice in it?",See reasoning trace,medium,"First method: Let $v_{1}$ and $s_{1}$ be the quantities of water and juice in the first barrel, and $v_{2}$ and $s_{2}$ be the quantities of water and juice in the second barrel, with $V$ being the volume of each of the two barrels. Then we have: $$ v_{1}: s_{1}=2: 1 \text{ and } v_{2}: s_{2}=4: 1, $$ which means $$ \begin{aligned} & v_{1}=\frac{2}{3} V \text{ and } s_{1}=\frac{1}{3} V \\ & v_{2}=\frac{4}{5} V \text{ and } s_{2}=\frac{1}{5} V \end{aligned} $$ The total quantity of water from both barrels is $$ v_{3}=v_{1}+v_{2}=\frac{22}{15} V $$ 1 POINT and the total quantity of juice from both barrels is $$ s_{3}=s_{1}+s_{2}=\frac{8}{15} V $$ The ratio of the quantities of water and juice in the third barrel is $v_{3}: s_{3}=\frac{22}{15} V: \frac{8}{15} V=11: 4$. ##" 3ead58484177,"10. (15 points) As shown in Figure 3, three circles intersect to form seven regions. Fill in the numbers $1, 2, 2, 4, 4, 8, 8$ into these seven regions so that the product of the four numbers inside each circle is equal (denoted as $P$). The filling method shown in Figure 4 satisfies the condition, with $P=64$ at this time. For all filling methods that meet the above requirements, find the maximum and minimum values of $P$.",See reasoning trace,medium,"10. Let $a \sim g$ represent a permutation of $1,2,2,4,4,8,8$ that satisfies the conditions of the filling method in Figure 7. Then $P=a b c e = a b d f = a c d g$. Note that, $$ \text { abcdefg }=1 \times 2 \times 2 \times 4 \times 4 \times 8 \times 8=2^{12} \text {. } $$ Thus, $P^{3}=a^{3} b^{2} c^{2} d^{2} efg =2^{12} a^{2} b c d$. Since $P^{3}$ is a cube number, $a^{2} b c d$ must also be a cube number. And $a^{2} b c d$ is a positive integer power of 2, with $$ \begin{array}{l} 2^{4}=1^{2} \times 2 \times 2 \times 4 \leqslant a^{2} b c d \\ \leqslant 8^{2} \times 8 \times 4 \times 4=2^{13}, \end{array} $$ Therefore, $2^{6} \leqslant a^{2} b c d \leqslant 2^{12}$. Figures 4 and 8 show the filling methods that make $a^{2} b c d=2^{6}$ and $2^{12}$, respectively. Thus, $P_{\max }=\sqrt[3]{2^{12} \times 2^{12}}=256$, $$ P_{\min }=\sqrt[3]{2^{12} \times 2^{6}}=64 . $$" 43c659112b0a,"(a) Determine the point of intersection of the lines with equations $y=4 x-32$ and $y=-6 x+8$. (b) Suppose that $a$ is an integer. Determine the point of intersection of the lines with equations $y=-x+3$ and $y=2 x-3 a^{2}$. (The coordinates of this point will be in terms of $a$.) (c) Suppose that $c$ is an integer. Show that the lines with equations $y=-c^{2} x+3$ and $y=x-3 c^{2}$ intersect at a point with integer coordinates. (d) Determine the four integers $d$ for which the lines with equations $y=d x+4$ and $y=2 d x+2$ intersect at a point with integer coordinates.","-2$ : The lines with equations $y=-2 x+4$ and $y=-4 x+2$ intersect at $(-1,6)$.",medium,"(a) To determine the point of intersection, we equate the two expressions for $y$ to successively obtain: $$ \begin{aligned} 4 x-32 & =-6 x+8 \\ 10 x & =40 \\ x & =4 \end{aligned} $$ When $x=4$, using the equation $y=4 x-32$, we obtain $y=4 \cdot 4-32=-16$. Therefore, the lines intersect at $(4,-16)$. (b) To determine the point of intersection, we equate the two expressions for $y$ to successively obtain: $$ \begin{aligned} -x+3 & =2 x-3 a^{2} \\ 3+3 a^{2} & =3 x \\ x & =1+a^{2} \end{aligned} $$ When $x=1+a^{2}$, using the equation $y=-x+3$, we obtain $y=-\left(1+a^{2}\right)+3=2-a^{2}$. Therefore, the lines intersect at $\left(1+a^{2}, 2-a^{2}\right)$. (c) Since $c$ is an integer, then $-c^{2}$ is a integer that is less than or equal to 0 . The two lines have slopes $-c^{2}$ and 1 . Since $-c^{2} \leq 0$, these slopes are different, which means that the lines are not parallel, which means that they intersect. To determine the point of intersection, we equate the two expressions for $y$ to successively obtain: $$ \begin{aligned} -c^{2} x+3 & =x-3 c^{2} \\ 3+3 c^{2} & =x+c^{2} x \\ 3+3 c^{2} & =x\left(1+c^{2}\right) \end{aligned} $$ Since $c^{2} \geq 0$, then $1+c^{2} \geq 1$, which means that we can divide both sides by $1+c^{2}$ to obtain $x=\frac{3+3 c^{2}}{1+c^{2}}=3$. In particular, this means that the $x$-coordinate of the point of intersection is an integer. When $x=3$, using the equation $y=-c^{2} x+3$, we obtain $y=-c^{2} \cdot 3+3=3-3 c^{2}$. Since $c$ is an integer, then $y=3-3 c^{2}$ is an integer. Therefore, the lines intersect at a point whose coordinates are integers. (d) To determine the point of intersection in terms of $d$, we equate the two expressions for $y$ to successively obtain: $$ \begin{aligned} d x+4 & =2 d x+2 \\ 2 & =d x \end{aligned} $$ For the value of $x$ to be an integer, we need $d \neq 0$ and $\frac{2}{d}$ to be an integer. Since $d$ is itself an integer, then $d$ is a divisor of 2 , which means that $d$ equals one of 1 , $-1,2,-2$. We still need to confirm that, for each of these values of $d$, the coordinate $y$ is also an integer. When $x=\frac{2}{d}$, using the equation $y=d x+4$, we obtain $y=d \cdot\left(\frac{2}{d}\right)+4=2+4=6$. Therefore, when $d=1,-1,2,-2$, the lines intersect at a point with integer coordinates. We can verify this in each case: - $d=1$ : The lines with equations $y=x+4$ and $y=2 x+2$ intersect at $(2,6)$. - $d=-1$ : The lines with equations $y=-x+4$ and $y=-2 x+2$ intersect at $(-2,6)$. - $d=2$ : The lines with equations $y=2 x+4$ and $y=4 x+2$ intersect at $(1,6)$. - $d=-2$ : The lines with equations $y=-2 x+4$ and $y=-4 x+2$ intersect at $(-1,6)$." 03e7e510f3bd,"## Task 4 - 140624 A cyclist rode with constant speed on a road from $A$ to $B$. He started in $A$ at 6:00 AM and covered $14 \mathrm{~km}$ every hour. A second cyclist rode on the same road with constant speed from $B$ to $A$. He started on the same day at 8:00 AM in $B$ and covered $21 \mathrm{~km}$ every hour. Both cyclists met exactly at the midpoint of the road from $A$ to $B$. At what time did they meet? How long is the road from $A$ to $B$?",See reasoning trace,medium,"The first cyclist had been cycling for exactly two hours by 8:00 AM, covering a distance of 28 km. From this point on, the second cyclist covered exactly 7 km more per hour than the first. Since they met exactly at the midpoint of the route from A to B, and because $28 \div 7 = 4$, this happened exactly 4 hours after the second cyclist started, which was at 12:00 PM. By this time, because $6 \times 14 = 84$ and $4 \times 21 = 84$, each of them had covered exactly 84 km. The length of the route from A to B is therefore $2 \times 84 = 168$ km, because $2 \times 84 = 168$. Solutions of the second round 1974 taken from [5] ### 3.16 XV. Olympiad 1975 ### 3.16.1 First Round 1975, Class 6" 43ba98ffff6a,"13. (15 points) As shown in the figure, ten points are evenly marked on the circumference of a circle. Place the ten natural numbers $1 \sim 10$ on these ten points. Rotate a line passing through the center of the circle. When the line does not pass through any marked points, it divides the numbers 1 to 10 into two groups. For each arrangement, there are five ways to divide the numbers as the line rotates. For each division, there is a product of the sums of the two groups, and the smallest of these five products is denoted as $K$. What is the maximum value of $K$ among all possible arrangements?","In all the arrangements, $K$ is maximized at 756",easy,"【Analysis】No matter how these 10 numbers are placed, their sum remains unchanged, $1+2+3+4+5+6+7+8+9+10=55$, divided into two groups, the closer the sums of the two groups, the larger the required product will be. 【Solution】 $$ \begin{array}{l} 1+2+3+4+5+6+7+8+9+10=55 \\ 55=27+28 \end{array} $$ Therefore, as long as the 10 numbers are divided into two parts with sums of 27 and 28, the product at this time is $27 \times 28=756$. Answer: In all the arrangements, $K$ is maximized at 756." e0775010c9de,"3. (BUL 1) Find all polynomials $f(x)$ with real coefficients for which $$ f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right) $$",0 \). We now look for nonzero solutions. We note that plugging in \( x = 0 \) we get \( f(0)^2 = f(0,medium,"3. An obvious solution is \( f(x) = 0 \). We now look for nonzero solutions. We note that plugging in \( x = 0 \) we get \( f(0)^2 = f(0) \); hence \( f(0) = 0 \) or \( f(0) = 1 \). If \( f(0) = 0 \), then \( f \) is of the form \( f(x) = x^k g(x) \), where \( g(0) \neq 0 \). Plugging this formula into \( f(x) f(2x^2) = f(2x^3 + x) \) we get \( 2^k x^{2k} g(x) g(2x^2) = (2x^2 + 1)^k g(2x^3 + x) \). Plugging in \( x = 0 \) gives us \( g(0) = 0 \), which is a contradiction. Hence \( f(0) = 1 \). For an arbitrary root \( \alpha \) of the polynomial \( f \), \( 2\alpha^3 + \alpha \) must also be a root. Let \( \alpha \) be a root of the largest modulus. If \( |\alpha| > 1 \) then \( |2\alpha^3 + \alpha| > 2|\alpha|^3 - |\alpha| > |\alpha| \), which is impossible. It follows that \( |\alpha| \leq 1 \) and hence all roots of \( f \) have modules less than or equal to 1. But the product of all roots of \( f \) is \( |f(0)| = 1 \), which implies that all the roots have modulus 1. Consequently, for a root \( \alpha \) it holds that \( |\alpha| = |2\alpha^3 - \alpha| = 1 \). This is possible only if \( \alpha = \pm i \). Since the coefficients of \( f \) are real it follows that \( f \) must be of the form \( f(x) = (x^2 + 1)^k \) where \( k \in \mathbb{N}_0 \). These polynomials satisfy the original formula. Hence, the solutions for \( f \) are \( f(x) = 0 \) and \( f(x) = (x^2 + 1)^k, k \in \mathbb{N}_0 \)." 82d48a9cde31,"Task A-1.7. (10 points) Ana has four times as many years as Peter had when Ana was as old as Peter is now. When Peter is as old as Ana is now, together they will have 95 years. How old is Ana, and how old is Peter?",See reasoning trace,medium,"## Solution. Let the number of Ana's years be $A$, and the number of Petar's years be $P$. Ana had $P$ years (how old Petar is now) $A-P$ years ago. 1 point $A-P$ years ago, Petar had $P-(A-P)=2P-A$ years, 1 point so according to the problem, Ana is now $4(2P-A)$ years old, hence $A=4(2P-A)$. 1 point Petar will have $A$ years (how old Ana is now) in $A-P$ years, 1 point and Ana will then have $A+(A-P)=2A-P$ years 1 point so according to the problem, $(2A-P)+A=95$, i.e., $3A-P=95$. 1 point By solving the obtained system $A=4(2P-A), 3A-P=95$ 1 point we get $A=40, P=25$. Ana is 40 years old, and Petar is 25." af725d4cd9f2,"Example 4. Find the locus of the midpoint of the chord passing through the imaginary vertex $B(0,-b)$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,",See reasoning trace,medium,"Let $M (x, y)$ be the midpoint of the chord that meets the condition, then we have $$ k_{B \boldsymbol{\mu}}=\frac{y+b}{x}, k_{\circ}=\frac{y}{x} .(x \neq 0) $$ By $k_{B M} \cdot k_{o_{\mathrm{M}}}=\frac{b^{2}}{a^{2}}$, i.e., $\frac{y+b}{x} \cdot \frac{y}{x}$ $$ =\frac{b^{2}}{a^{2}}, $$ we get $\quad b^{2} x^{2}-a^{2} y^{2}-a^{2} b y=0$. Therefore, $$ \frac{\left(y+\frac{b}{2}\right)^{2}}{\frac{b^{2}}{4}}-\frac{x^{2}}{\frac{a^{2}}{4}}=1 . $$ This is a hyperbola with the center at $\left(0,-\frac{b}{2}\right)$ and foci on the $y$-axis. (Note: The problem should also discuss the singularity of the trajectory, which is omitted here)" a6eaaf320c6b,"Example 12. What is the minimum number of trials that need to be conducted so that with a probability of at least $\alpha(0<\alpha<1)$, one can expect the event $A$ to occur at least once, if the probability of event $A$ in one trial is $p$?","0.02$ and $\alpha=0.98$, then by the last formula we get $n_{0}=80$.",medium,"Solution. We require that the probability of event $A$ occurring at least once in $n$ trials (see formula (3.1.7)) is not less than $\alpha$: $$ 1-q^{n} \geq \alpha, \text { or }\left(1-(1-p)^{n}\right) \geq \alpha \text {. } $$ Solving this inequality for $n$, we obtain the inequality $$ n \geq \frac{\lg (1-\alpha)}{\lg (1-p)} $$ From this, we conclude that the minimum number of trials $n_{0}$, satisfying the condition of the example, is determined by the formula $$ n_{0}=\left[\frac{\lg (1-\alpha)}{\lg (1-p)}\right]+1 $$ where the square brackets denote the integer part of the number. In particular, if $p=0.02$ and $\alpha=0.98$, then by the last formula we get $n_{0}=80$." fc46ac3ff0de,"Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo are students from different parts of Brazil who were chosen to represent their country in international olympiads. After several weeks of training, some friendships were formed. We then asked each of them how many friends they had made in the group. Arnaldo, Bernaldo, Cernaldo, and Dernaldo answered, respectively, that they had made 1, 2, 3, and 4 friends within the group. How many of the group members are friends with Ernaldo?",2,medium,"Solution Suppose Ernaldo has $x$ friends within the group. Since Dernaldo has 4 friends and the group has 5 members, everyone is a friend of Dernaldo. Let's remove Dernaldo from the group. Thus, Arnaldo, Bernaldo, Cernaldo, and Ernaldo now have $0, 1, 2$, and $x-1$ friends within the subgroup, respectively. Since Arnaldo no longer has any friends within the subgroup, we can ignore him. The following table shows the number of friends each member has within the subgroup. | Member | Number of friends | | :--- | :---: | | Bernaldo | 1 | | Cernaldo | 2 | | Ernaldo | $x-1$ | Obviously, the two friends of Cernaldo within the subgroup are Bernaldo and Ernaldo. Moreover, since Bernaldo has only one friend within the group, Cernaldo must be that friend. In particular, Ernaldo and Bernaldo are not friends. Therefore, Cernaldo is also the only friend of Ernaldo within the subgroup, i.e., $x-1=1$. We conclude that Cernaldo and Dernaldo are the only friends of Ernaldo. The answer is 2." e3f2c440d68f,"We are given 5771 weights weighing 1,2,3,...,5770,5771. We partition the weights into $n$ sets of equal weight. What is the maximal $n$ for which this is possible?",2886,medium,"1. First, we need to find the sum of the weights from 1 to 5771. This can be calculated using the formula for the sum of the first \( n \) natural numbers: \[ S = \sum_{k=1}^{5771} k = \frac{5771 \times 5772}{2} \] Calculating this, we get: \[ S = \frac{5771 \times 5772}{2} = \frac{33315412}{2} = 16657706 \] 2. We need to partition this sum \( S \) into \( n \) sets of equal weight. Let \( a \) be the weight of each set. Then: \[ S = a \times n \] Since \( a \) must be an integer and each set must have at least one weight, the minimum possible value for \( a \) is 5771 (the largest single weight). 3. To find the maximum \( n \), we need to determine the largest integer \( n \) such that: \[ a \geq 5771 \quad \text{and} \quad a \times n = 16657706 \] This implies: \[ n \leq \frac{16657706}{5771} \] Calculating this, we get: \[ n \leq \frac{16657706}{5771} \approx 2886 \] 4. To verify, we check if \( n = 2886 \) is possible. If \( n = 2886 \), then: \[ a = \frac{16657706}{2886} = 5771 \] This means each set must have a total weight of 5771. 5. We can construct such sets as follows: \[ \{ \{5771\}, \{1, 5770\}, \{2, 5769\}, \ldots, \{2885, 2886\} \} \] Each of these pairs sums to 5771, and there are 2886 such pairs. Thus, the maximal \( n \) for which the weights can be partitioned into sets of equal weight is \( \boxed{2886} \)." 3e7808be8b22,"1. In the Cartesian coordinate system $x O y$, points $A$ and $B$ lie on the parabola $y^{2}=2 x$, satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-1, F$ is the focus of the parabola. Then the minimum value of $S_{\triangle O F A}+S_{\triangle O F B}$ is $\qquad$",See reasoning trace,medium,"$-1 . \frac{\sqrt{2}}{2}$. It is known that $F\left(\frac{1}{2}, 0\right)$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $x_{1}=\frac{y_{1}^{2}}{2}, x_{2}=\frac{y_{2}^{2}}{2}$. Notice that, $$ \begin{array}{l} -1=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2} \\ =\frac{1}{4}\left(y_{1} y_{2}\right)^{2}+y_{1} y_{2} . \end{array} $$ Thus, $y_{1} y_{2}=-2$. Therefore, $S_{\triangle O F A}+S_{\triangle O F B}$ $$ \begin{array}{l} =\frac{1}{2}|O F|\left|y_{1}\right|+\frac{1}{2}|O F|\left|y_{2}\right| \\ =\frac{1}{4}\left(\left|y_{1}\right|+\left|y_{2}\right|\right) \geqslant \frac{1}{2} \sqrt{\left|y_{1} y_{2}\right|}=\frac{\sqrt{2}}{2}, \end{array} $$ where the equality holds if and only if $\left|y_{1}\right|=\left|y_{2}\right|=\sqrt{2}$, i.e., when point $A(1, \sqrt{2})$ and $B(1,-\sqrt{2})$ or $A(1,-\sqrt{2})$ and $B(1, \sqrt{2})$. Thus, $S_{\triangle O F A}+S_{\triangle O F B}$ achieves the minimum value $\frac{\sqrt{2}}{2}$." bd246de80902,"3. As shown in Figure 3, given that $M$ is a point inside rectangle $A B C D$, $A B=1, B C=2, t=$ $A M \cdot M C+B M \cdot M D$. Then the minimum value of $t$ is $\qquad$",See reasoning trace,medium,"3. 2 . As shown in Figure 9, draw perpendiculars from point $M$ to $A B$ and $B C$, with the feet of the perpendiculars being $P$ and $Q$ respectively. Let $A P=a$, $$ \begin{array}{l} B P=b, \\ B Q=c, Q C=d . \end{array} $$ Then $a+b=1, c+d=2$. Notice that, $$ \begin{array}{l} A M^{2} \cdot M C^{2}=\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right) \\ =a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2} \\ =(a d+b c)^{2}+(a b-c d)^{2} \geqslant(a d+b c)^{2} \end{array} $$ Therefore, $A M \cdot M C \geqslant a d+b c$, with equality holding if and only if $a b=c d$. Similarly, $B M \cdot M D \geqslant a c+b d$, with equality holding if and only if $a b=c d$. $$ \begin{array}{l} \text { Hence } t \geqslant(a d+b c)+(a c+b d) \\ =(a+b)(c+d)=2, \end{array} $$ with equality holding if and only if $a b=c d$. For example, when $a=0.5+0.1 \sqrt{6}, b=0.5-0.1 \sqrt{6}$, $c=0.1, d=1.9$, we have $a b=c d$. Therefore, the minimum value of $t$ is 2." 2d00597015ef,"Three, find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer. (Note: The square mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{i}$ is defined as $\left.\left[\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right]^{\frac{1}{2}}\right)$",$n=337$,medium,"$$ \begin{array}{c} \text { Three, let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \\ =\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \text { be a } \end{array} $$ perfect square $m^{2}$, then $6 m^{2}=(n+1)(2 n+1)$. Since $6 m^{2}$ is even, it must be odd. We can set $n=3 p+q$, where $q=-1, 1$ or 3. If $q=2$, then $6 m^{2}=(6 p+4)(12 p+7)$ $=72 p^{2}+90 p+28$. Since $6|72, 6| 90$, but $6 \nmid 28$, this case is impossible. If $q=-1$, then $6 m^{2}=(6 p)(12 p-1)$, i.e., $m^{2}=p(12 p-1)$. Since $(p, 12 p-1) = 1$, both $p$ and $12 p-1$ must be perfect squares. Let $p=s^{2}$, $12 p-1=t^{2}$, we get $t^{2}-12 s^{2}=-1$. Since $t^{2} \equiv 0$ (mod 3), this case is also impossible. If $q=1$, then $6 m^{2}=(6 p+2)(12 p+3)$, i.e., $m^{2}=(3 p+1)(4 p+1)$. Since $(3 p+1, 4 p+1)=1$, both $3 p+1=u^{2}$ and $4 p+1=v^{2}$, we get $3 v^{2}+1=4 u^{2}$. Clearly, $u=v=1$ is a solution, but it leads to $p=0, m=1, n=1$, which does not meet the requirement. To find a solution for $n>1$, by trying integer values of $u$ from 2 to 12, $v$ has no integer value, but when $u=13$, $v=15$. Therefore, the smallest solution is $p=56, m=195, n=337$. The answer is $n=337$." 90f07b163fac,11.5. Solve the equation in integers: $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3$.,". $(\mathrm{x}, \mathrm{y}, z)=(2,2,3),(1,3,8),(1,4,5),(-4,1,1)$ and all permutations of these triples",medium,"Answer. $(\mathrm{x}, \mathrm{y}, z)=(2,2,3),(1,3,8),(1,4,5),(-4,1,1)$ and all permutations of these triples. In total, there are 18 solutions. Solution. Without loss of generality, we can assume that $a \leq b \leq c$. Notice that if $x<0$, then $1+\frac{1}{x}<1$, and if $x \geq 1$, then $1+\frac{1}{x} \leq 2$. Therefore, among the numbers $a \leq b \leq c$, there is at most one negative number. 1) Suppose $a<00)$. By symmetry, we assume that the circle $\Omega$ is tangent to the $x$-axis at point $F$ above the $x$-axis. Then the equation of the circle $\Omega$ is $$ (x-1)^{2}+(y-r)^{2}=r^{2} \text {. } $$ Substituting $x=\frac{y^{2}}{4}$ into equation (1) and simplifying, we get $$ \left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}-2 r y=0 \text {. } $$ Clearly, $y>0$. Thus, $r=\frac{1}{2 y}\left(\left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}\right)=\frac{\left(y^{2}+4\right)^{2}}{32 y}$. According to the given conditions, we know that equation (2) has exactly one positive solution $y$, which corresponds to the unique common point of the circle $\Omega$ and the parabola $\Gamma$. Consider the minimum value of $f(y)=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$. By the AM-GM inequality, we have $y^{2}+4=y^{2}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3} \geqslant 4 \sqrt[4]{y^{2}\left(\frac{4}{3}\right)^{3}}$. Thus, $f(y) \geqslant \frac{1}{32 y} \cdot 16 \sqrt{y^{2}\left(\frac{4}{3}\right)^{3}}=\frac{4 \sqrt{3}}{9}$, with equality if and only if $y^{2}=\frac{4}{3}$, i.e., $y=\frac{2 \sqrt{3}}{3}$, at which $f(y)$ attains its minimum value $\frac{4 \sqrt{3}}{9}$. From the fact that equation (2) has a solution, we know that $r \geqslant \frac{4 \sqrt{3}}{9}$. Furthermore, if $r>\frac{4 \sqrt{3}}{9}$, noting that $f(y)$ varies continuously with $y$, and that $f(y)$ can be arbitrarily large as $y \rightarrow 0^{+}$ and $y \rightarrow+\infty$, it follows that equation (2) has solutions in both $\left(0, \frac{2 \sqrt{3}}{3}\right)$ and $\left(\frac{2 \sqrt{3}}{3},+\infty\right)$, contradicting the uniqueness of the solution. In summary, only $r=\frac{4 \sqrt{3}}{9}$ satisfies the conditions, and at this time, the unique common point of the circle $\Omega$ and the parabola $\Gamma$ is $\left(\frac{1}{3}, \frac{2 \sqrt{3}}{3}\right)$." bfedb81a5454,"A2. For every integer $n \geqslant 1$ consider the $n \times n$ table with entry $\left\lfloor\frac{i j}{n+1}\right\rfloor$ at the intersection of row $i$ and column $j$, for every $i=1, \ldots, n$ and $j=1, \ldots, n$. Determine all integers $n \geqslant 1$ for which the sum of the $n^{2}$ entries in the table is equal to $\frac{1}{4} n^{2}(n-1)$.",All integers $n$ for which $n+1$ is a prime,hard,"Answer: All integers $n$ for which $n+1$ is a prime. Solution 1. First, observe that every pair $x, y$ of real numbers for which the sum $x+y$ is integer satisfies $$ \lfloor x\rfloor+\lfloor y\rfloor \geqslant x+y-1 . $$ The inequality is strict if $x$ and $y$ are integers, and it holds with equality otherwise. We estimate the sum $S$ as follows. $$ \begin{array}{r} 2 S=\sum_{1 \leqslant i, j \leqslant n}\left(\left\lfloor\frac{i j}{n+1}\right\rfloor+\left\lfloor\frac{i j}{n+1}\right\rfloor\right)=\sum_{1 \leqslant i, j \leqslant n}\left(\left\lfloor\frac{i j}{n+1}\right\rfloor+\left\lfloor\frac{(n+1-i) j}{n+1}\right\rfloor\right) \\ \geqslant \sum_{11$, then $d|(4 n+5), d|(7 n+6)$. Thus $d \mid(7 n+6-(4 n+5))=3 n+1$, $$ \begin{array}{l} d \mid((4 n+5)-(3 n+1))=n+4, \\ d \mid((3 n+1)-2(n+4))=n-7, \\ d \mid((n+4)-(n-7))=11 . \end{array} $$ Since 11 is a prime number, then $d=11$. Let $n-7=11 k$, then $01$. Find all possible values of the number $a$.","2^{N}\left(a_{0}\right)^{N_{0}} \ldots\left(a_{q}\right)^{N_{q}}$, where $N>N_{0}+\ldots+N_{q}$. Sin",medium,"7. The values $a=2$ and $a=3$ are possible: for example, $2^{2}-1=2+1, (3-1)^{2}=3+1$. Suppose that for some $a>3$ the required equality holds: $$ A=\left(a^{m_{1}}-1\right) \ldots\left(a^{m_{n}}-1\right)=\left(a^{k_{1}}+1\right) \ldots\left(a^{k_{l}}+1\right) $$ Lemma 1. Any $m_{i}$ and $a-1$ are powers of two. Proof. Let $\gamma$ be an arbitrary odd divisor of the number $m_{i}$ (possibly $\gamma=1$), and let $p$ be any prime divisor of the number $a^{\gamma}-1$. Then $A$ is divisible by $p$ (see fact 8), which means that in the right product there is a factor $a^{k_{j}}+1$ that is divisible by $p$ (see fact 9). Note that $$ \left(\left(a^{k_{j}}\right)^{\gamma}+1\right)-\left(\left(a^{\gamma}\right)^{k_{j}}-1\right)=2 $$ On the other hand, $\left(a^{k_{j}}\right)^{\gamma}+1$ is divisible by $p$ since $a^{k_{j}}+1$ is divisible by $p$, and $\left(a^{\gamma}\right)^{k_{j}}-1$ is divisible by $p$ since $a^{\gamma}-1$ is divisible by $p$. Therefore, 2 is divisible by $p$, hence $p=2$. Thus, 2 is the only prime divisor of the number $a^{\gamma}-1$, meaning that $a^{\gamma}-1$ is a power of two (see fact 10), i.e., there exists some $K$ such that $2^{K}=a^{\gamma}-1=\left(a^{\gamma-1}+\ldots+a+1\right)(a-1)$. Since $\gamma$ is odd, the expression in the first parenthesis is an odd divisor of the number $2^{K}$, i.e., 1. Therefore, $\gamma=1$. Consequently, $a-1=2^{K}$. Moreover, we have proved that one is the only odd divisor of the number $m_{i}$, meaning that $m_{i}$ is a power of two. The lemma is proved. Since $a-1>2$, then $a-1$ is divisible by 4, so $a^{k}+1$ gives a remainder of 2 when divided by 4 for any $k$ (see fact 7). Let $a_{i}=\frac{a^{2^{i}}+1}{2}$. Then $a_{i}$ is an odd number. We will need the following formula: $$ \begin{aligned} a^{2^{d}}-1=(a-1)(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right) & \ldots\left(a^{2^{d-1}}+1\right)= \\ & =2^{K} \cdot 2^{d} \cdot a_{0} \ldots a_{d-1} . \end{aligned} $$ Lemma 2. All $a_{i}$ are pairwise coprime. Proof. Consider the numbers $a_{i}$ and $a_{j}$ and prove that they are coprime. We can assume that $i>j$. By (2), $a^{2^{i}}-1$ is divisible by $a_{j}$, and on the other hand, $a^{2^{i}}+1$ is divisible by $a_{i}$. Therefore, any common divisor of the numbers $a_{i}$ and $a_{j}$ is also a divisor of the number $$ 2=\left(a^{2^{i}}+1\right)-\left(a^{2^{i}}-1\right) . $$ To complete the proof, it remains to note that the numbers $a_{i}$ and $a_{j}$ are odd. The lemma is proved. Multiplying the representations (2) for $a^{2^{m_{i}}}-1$, we obtain that $A$ is represented as $A=2^{N}\left(a_{0}\right)^{N_{0}} \ldots\left(a_{q}\right)^{N_{q}}$, where $N>N_{0}+\ldots+N_{q}$. Since each of the numbers $a^{k_{j}}+1$ gives a remainder of 2 when divided by 4, then $l=N$ (see (1)). But each of $a^{k_{j}}+1$ is divisible by one of the numbers $a_{i}$ (indeed, if $k_{j}=2^{r} s$, where $s$ is odd, then $a^{k_{j}}+1$ is divisible by $a_{r}$). Then, since $l>N_{0}+\ldots+N_{q}$, more than $N_{i}$ numbers $a^{k_{j}}+1$ are divisible by some $a_{i}$, and therefore, $A$ is divisible by $a_{i}^{N_{i}+1}$. This contradicts the fact that $a_{i}$ are odd and pairwise coprime." 5f3642b9d7d2,"A2. Let $\alpha$ and $\beta$ be acute angles in a right triangle that is not isosceles. Then $\sin (\alpha+\beta)$ is equal to: (A) $-\cos 2 \alpha$ (B) $2 \cos ^{2} \alpha$ (C) $2 \sin ^{2} \beta$ (D) $\sin ^{2} \alpha+\cos ^{2} \beta$ (E) 1","\frac{\pi}{2}$, then $\sin (\alpha+\beta)=1$.",easy,"A2. Given $\alpha+\beta=\frac{\pi}{2}$, then $\sin (\alpha+\beta)=1$." b3a96af9bddd,"Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite. The sum of the coordinates of vertex $S$ is: $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$",\textbf{(E),easy,"Graph the three points on the coordinate grid. Noting that the opposite sides of a [parallelogram](https://artofproblemsolving.com/wiki/index.php/Parallelogram) are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$. The sum of the x-coordinates and y-coordinates is $9$, so the answer is $\boxed{\textbf{(E)}}$." f06166b1d966,"20. Let $n(n \geqslant 3)$ be a positive integer, and let positive numbers $\alpha$ and $\beta$ satisfy $\frac{\alpha}{\beta}=\frac{n-1}{n-2}$. Also, $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{\alpha}=1$, find the minimum value of $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{\alpha}}$.",\left(\frac{1}{n-1}\right)^{\frac{1}{n-1}}\left(\frac{n-1}{n}\right)^{\frac{n}{n-1}}=(n-1) n^{-\frac,medium,"20. Let $y_{i}=x_{i}^{\sigma-\beta}\left(1-x_{i}^{a}\right)$. By the weighted arithmetic-geometric mean inequality, we have $y_{i}^{\alpha}=\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left(\frac{\alpha}{\alpha-\beta} x_{i}^{\alpha}\right)^{\alpha \beta}\left(1-x_{i}^{\alpha}\right)^{\circ}$ $$ \leqslant\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left[\frac{(\alpha-\beta) \cdot \frac{\alpha}{\alpha-\beta} x_{i}^{o}+\alpha\left(1-x_{i}^{q}\right)}{(\alpha-\beta)+\alpha}\right]^{2-\beta}=\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left(\frac{\alpha}{2 \alpha-\beta}\right)^{2 \alpha-\beta} \text {. } $$ Then $y_{i} \leqslant\left(\frac{\alpha-\beta}{\alpha}\right)^{\frac{2-\beta}{\alpha}}\left(\frac{\alpha}{2 \alpha-\beta}\right)^{\frac{2 \alpha-\beta}{\alpha}}=\left(\frac{1}{n-1}\right)^{\frac{1}{n-1}}\left(\frac{n-1}{n}\right)^{\frac{n}{n-1}}=(n-1) n^{-\frac{n}{n-1}}$, hence $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{a}}=\sum_{i=1}^{n} \frac{x_{i}^{a}}{x_{i}^{-\beta}\left(1-x_{i}^{a}\right)} \geqslant(n-1)^{-1} \cdot n_{n-1}^{n-1} \cdot \sum_{i=1}^{n} x_{i}^{a}=\frac{n}{n-1} \cdot \sqrt[n-1]{n}$. The equality holds when $\frac{\alpha}{\alpha-\beta} x_{i}^{o}=1-x_{i}^{\alpha}$, i.e., $x_{i}^{\alpha}=\frac{\alpha-\beta}{2 \alpha-\beta}=\frac{1}{n}$, which means $x_{i}=\left(\frac{1}{n}\right)^{\frac{1}{\alpha}}$. Therefore, the minimum value of $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{\beta}}$ is $\frac{n}{n-1} \cdot \sqrt[n-1]{n}$." f8a019f290ba,"7. In a free-throw test, a person only needs to make 3 shots to pass and does not need to continue shooting, but each person can shoot at most 5 times. The probability that a player with a shooting accuracy of $\frac{2}{3}$ passes the test is . $\qquad$",\frac{8}{27}$; The probability of making 3 out of 4 shots is $\frac{2}{3} \times\left(\frac{2}{3}\ri,medium,"7. $\frac{64}{81}$. Analysis: The probability of making 3 out of 3 shots is $\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$; The probability of making 3 out of 4 shots is $\frac{2}{3} \times\left(\frac{2}{3}\right)^{2} \times \frac{1}{3} \mathrm{C}_{3}^{2}=$ $\frac{8}{27}$; The probability of making 3 out of 5 shots is $\frac{2}{3} \times\left(\frac{2}{3}\right)^{2} \times$ $\left(\frac{1}{3}\right)^{2} \mathrm{C}_{4}^{2}=\frac{16}{81}$. The probability of passing the test is $\frac{8}{27}+\frac{8}{27}+\frac{16}{81}=$ $\frac{64}{81}$." 732260397057,Example 4 Find the integer solution of the equation $\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\left[\frac{x}{3!}\right]+\cdots+$ $\left[\frac{x}{10!}\right]=1001$.,584$ is the unique solution of the equation.,easy,"Solution: $\because$ the solution $x$ of the equation is a natural number, and $xx+\frac{x}{2}+\frac{x}{6}+\frac{x}{24}+\frac{x}{120}-4 \\ =\frac{103}{60} x-4, \therefore 583 1. By (3), the k denominations not in box A must be in every other box. Similarly for the other boxes. So each box contains the 4k denominations that are excluded from one of the other boxes. So the total number of coins must be 5k. So n must be a multiple of 5 and at least 10. Now for n = 10k, we can take the first 2k columns below to give the coins excluded from each box. A: 1, 10; 11, 20; 21, 30; 31, 40; ... B: 2, 9; 12, 19; 22, 29; 32, 39; ... C: 3, 8; 13, 18; 23, 28; 33, 38; ... D: 4, 7; 14, 17; 24, 27; 34, 37; ... E: 5, 6; 15, 16; 25, 26; 35, 36; ... Similarly, for n = 10k + 5, we can take the first 2k+1 columns below to give the coins excluded from each box. A: 1, 8, 15; 16, 25; 26, 35; ... B: 2, 9, 13; 17, 24; 27, 34; ... C: 3, 10, 11; 18, 23; 28, 33; ... D: 4, 6, 14; 19, 22; 29, 32; ... E: 5, 7, 12; 20, 21; 30, 31; ... For example, for n = 15, we take A = (2,3,4,5,6,7,9,10,11,12,13,14) B = (1,3,4,5,6,7,8,10,11,12,14,15) C = (1,2,4,5,6,7,8,9,12,13,14,15) D = (1,2,3,5,7,8,9,10,11,12,13,15) E = (1,2,3,4,6,8,9,10,11,13,14,15). 15th Mexican 2001 © John Scholes jscholes@kalva.demon.co.uk 9 March 2004 Last corrected/updated 9 Mar 04" 58444c02afd2,"Mrs. Siroka was expecting guests in the evening. First, she prepared 25 sandwiches for them. Then she calculated that each guest could take two, but three would not be enough for everyone. She thought that if she made another 10 sandwiches, each guest could take three, but not four. This still seemed too little to her. In the end, she prepared a total of 52 sandwiches. Each guest could then take four sandwiches, but five would not be enough for everyone. How many guests was Mrs. Siroka expecting? She herself is on a diet and never eats in the evening.",See reasoning trace,medium,"First, let's work with the part of the problem where 25 sandwiches are considered. According to this, Mrs. Siroka expected no more than 12 guests, because $25: 2=12$, remainder 1, which means that 12 people could take two sandwiches each, but only one would be left. Here we also find out that Mrs. Siroka expected more than 8 guests, because $25: 3=8$, remainder 1, which means that with 8 guests, everyone could take three sandwiches. So far, it is plausible that she expected 9, 10, 11, or 12 guests. Now let's consider only the part of the problem that talks about 35 sandwiches. We determine that Mrs. Siroka expected at most 11 guests, since $35: 3=11$, remainder 2, and more than 8 guests, since $35: 4=8$, remainder 3. Therefore, Mrs. Siroka could have expected 9, 10, or 11 guests. Next, let's work only with the consideration of 52 sandwiches. According to this, Mrs. Siroka expected at most 13 guests, because $52: 4=13$, and more than 10 guests, because $52: 5=10$, remainder 2. She therefore expected 11, 12, or 13 guests. We see that the only number of guests that fits all the data in the problem is 11. Another solution. Just as in the first paragraph of the previous solution, we determine that Mrs. Siroka could have expected 9, 10, 11, or 12 guests. For each number, we will check whether it corresponds to the other data in the problem. 9 guests: With 35 sandwiches, everyone could take three sandwiches, not four, because $9 \cdot 3 < 35$. With 52 sandwiches, everyone could take four sandwiches, but even five, because $9 \cdot 4 < 52$. With 52 sandwiches, everyone could take four sandwiches, but even five, because $10 \cdot 4 < 52$. With 52 sandwiches, everyone could take four sandwiches, but not five, because $11 \cdot 4 \leq 52$. This number of guests fits the entire problem. 12 guests: With 35 sandwiches, everyone could not take three sandwiches, because $12 \cdot 3 > 35$. This number of guests is rejected. Mrs. Siroka expected 11 guests." 0b98272376cf,"4. From the 2022 natural numbers 1 to 2022, select several distinct numbers such that the sum of any 2 of them is not a multiple of 5. The maximum number of numbers that can be selected is $\qquad$.",See reasoning trace,easy,$811$ 2c4c4b6b7a6a,17. Find the largest natural number that cannot be represented as the sum of two composite numbers. (Recall that a natural number is called composite if it is divisible by some natural number other than itself and one.),11,easy,"17. Answer: 11. All large numbers can be represented in the required way: odd numbers as ""9 + an even number greater than two"", and even numbers as ""8 + an even number greater than two""." df85b8aa46cb,"1. Given two quadratic trinomials $f(x)$ and $g(x)$ with leading coefficients of 1, each having two real roots, and $f(1)=g(2)$, $g(1)=f(2)$. Find the sum of the four roots of these two quadratic trinomials.","0$, and $-c$ is the sum of the roots of the equation $g(x)=0$. Therefore, the sum of the four roots ",easy,"1.6. $$ \text { Let } f(x)=x^{2}+a x+b, g(x)=x^{2}+c x+d \text {. } $$ According to the conditions in the problem, we have $$ \begin{array}{l} 1+a+b=4+2 c+d, \\ 4+2 a+b=1+c+d . \end{array} $$ (1) - (2) gives $$ -3-a=3+c \Rightarrow-a-c=6 \text {. } $$ According to Vieta's formulas, $-a$ is the sum of the roots of the equation $f(x)=0$, and $-c$ is the sum of the roots of the equation $g(x)=0$. Therefore, the sum of the four roots of these two quadratic trinomials is 6." 21cffdab3b1a,"1. As shown in Figure 1, in the Cartesian coordinate system, the graph of the quadratic function $y=a x^{2}+m c(a \neq$ $0)$ passes through three vertices $A$, $B$, and $C$ of the square $A B O C$, and $a c=-2$. Then the value of $m$ is ( ). (A) 1 (B) -1 (C) 2 (D) -2","-2$, so $m=1$.",easy,"-1.A. From the problem, we know $A(0, m c), O A=m c=B C$. Also, $B$ and $C$ are symmetric with respect to the $y$-axis, so $C\left(\frac{1}{2} m c, \frac{1}{2} m c\right)$. Thus, $\frac{1}{2} m c=a\left(\frac{1}{2} m c\right)^{2}+m c$, which gives $a m c=-2$. Also, $a c=-2$, so $m=1$." c5d876d77664,"8. (10 points) If the expression $\frac{1}{1 \times 2}-\frac{1}{3 \times 4}+\frac{1}{5 \times 6}-\frac{1}{7 \times 8}+\cdots+\frac{1}{2007 \times 2008}$ is converted to a decimal, then the first digit after the decimal point is $\qquad$ .",: 4,medium,"【Analysis】Based on the periodicity of addition and subtraction of the fractional sequence operation symbols, group the fractional sequence to find approximate values, and perform estimation. 【Solution】Solution: $\frac{1}{1 \times 2}-\frac{1}{3 \times 4} \approx 0.416$ $$ \begin{array}{l} \frac{1}{5 \times 6}-\frac{1}{7 \times 8} \approx 0.01548 \\ \frac{1}{9 \times 10}-\frac{1}{11 \times 12} \approx 0.003 \\ \frac{1}{13 \times 14}-\frac{1}{15 \times 16} \approx 0.00133 \\ \frac{1}{17 \times 18}-\frac{1}{19 \times 20} \approx 0.00063 \end{array} $$ $\cdots$ Reasoning that the difference between every two fractions is closer to 0, and it is a finite sum, so the first digit after the decimal point is 4. Therefore, the answer is: 4." 534e48b1fa5d,"Example 12 Find all polynomials $P(x)$ that satisfy $x P(x-1) \equiv (x-2) P(x) \, (x \in \mathbf{R})$.",See reasoning trace,medium,"Solving: Substituting 0,2 into the identity in the problem, we know that the polynomial has roots 0 and 1, i.e., it is divisible by the polynomial $x^{2}-x$. Next, substituting $P(x)=\left(x^{2}-x\right) Q(x)$ into the identity, we know that the polynomial $Q(x)$ satisfies the identity $Q(x)=$ $Q(x-1)$, from which we get $Q(0)=Q(1)=Q(2)=\cdots$ Therefore, $Q(x)=a$ (a constant), so the polynomial we are looking for is $P(x)=a\left(x^{2}-x\right)$. Conversely, it is easy to verify that all such polynomials also satisfy the identity in the problem." c043b1c2ad65,"Nathan has discovered a new way to construct chocolate bars, but it’s expensive! He starts with a single $1\times1$ square of chocolate and then adds more rows and columns from there. If his current bar has dimensions $w\times h$ ($w$ columns and $h$ rows), then it costs $w^2$ dollars to add another row and $h^2$ dollars to add another column. What is the minimum cost to get his chocolate bar to size $20\times20$?",5339,medium,"To find the minimum cost to construct a $20 \times 20$ chocolate bar, we need to consider the cost of adding rows and columns. The cost to add a row is $w^2$ dollars, and the cost to add a column is $h^2$ dollars, where $w$ is the current width and $h$ is the current height of the chocolate bar. We will compare two strategies: 1. Adding all rows first and then all columns. 2. Adding rows and columns in an alternating fashion. ### Strategy 1: Adding all rows first, then all columns 1. Start with a $1 \times 1$ chocolate bar. 2. Add 19 rows to make it a $1 \times 20$ chocolate bar. - The cost to add each row is $1^2 = 1$ dollar. - Total cost for adding 19 rows: $19 \times 1 = 19$ dollars. 3. Add 19 columns to make it a $20 \times 20$ chocolate bar. - The cost to add each column is $20^2 = 400$ dollars. - Total cost for adding 19 columns: $19 \times 400 = 7600$ dollars. 4. Total cost for this strategy: $19 + 7600 = 7619$ dollars. ### Strategy 2: Adding rows and columns in an alternating fashion 1. Start with a $1 \times 1$ chocolate bar. 2. Add a row and a column alternately until reaching $20 \times 20$. - The cost sequence will be: $1^2, 2^2, 2^2, 3^2, 3^2, \ldots, 20^2$. 3. Calculate the total cost: - Sum of squares of the first 20 natural numbers: $1^2 + 2^2 + 3^2 + \ldots + 20^2$. - The formula for the sum of squares of the first $n$ natural numbers is: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] - For $n = 20$: \[ \sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870 \] - Since we are adding rows and columns alternately, each term appears twice except the first and last terms: \[ 2 \left( \sum_{k=1}^{20} k^2 \right) - 1^2 - 20^2 = 2 \cdot 2870 - 1 - 400 = 5740 - 401 = 5339 \] Thus, the minimum cost to construct a $20 \times 20$ chocolate bar is $5339$ dollars. The final answer is $\boxed{5339}$." 1e182c75fd7d,"1.6. Find the distance between two intersecting medians of the faces of a regular tetrahedron with edge 1. (Consider all possible arrangements of the medians.) ## § 2. Angles between lines and planes",See reasoning trace,medium,"1.6. Let $A B C D$ be a given regular tetrahedron, $K$ - the midpoint of $A B$, $M$ - the midpoint of $A C$. Consider the projection onto the plane $\mathrm{b}$, perpendicular to the face $A B C$ and passing through the line $A B$. Let $D_{1}$ be the projection of vertex $D$, $M_{1}$ - the projection of point $M$, i.e., the midpoint of segment $A K$. The distance between the lines $C K$ and $D M$ is equal to the distance from point $K$ to the line $D_{1} M_{1}$. In the right-angled triangle $D_{1} M_{1} K$, the leg $K M_{1}$ is equal to 1/4, and the leg $D_{1} M_{1}$ is equal to the height of the tetrahedron $A B C D$, i.e., it is equal to $\sqrt{2 / 3}$. Therefore, the hypotenuse is $\sqrt{35 / 48}$, and thus the desired distance is $\sqrt{2 / 35}$. If $N$ is the midpoint of edge $C D$, then to find the distance between the medians $C K$ and $B N$, we can consider the projection onto the same plane as in the previous case. Let $N_{1}$ be the projection of point $N$, i.e., the midpoint of segment $D_{1} K$. In the right-angled triangle $B N_{1} K$, the leg $K B$ is equal to $1 / 2$, and the leg $K N_{1}$ is equal to $\sqrt{1 / 6}$. Therefore, the hypotenuse is $\sqrt{5 / 12}$, and the desired distance is $\sqrt{1 / 10}$." 2da913175d8e,"13. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $P\left(\frac{\sqrt{6}}{2}, \frac{1}{2}\right)$, and the eccentricity is $\frac{\sqrt{2}}{2}$, with a moving point $M(2, t)(t>0)$. (1) Find the standard equation of the ellipse; (2) Find the equation of the circle with $O M$ as its diameter and which is intersected by the line $3 x-4 y-5=0$ to form a chord of length 2; (3) Let $F$ be the right focus of the ellipse. Draw a perpendicular from point $F$ to $O M$ which intersects the circle with $O M$ as its diameter at point $N$. Prove that the length of segment $O N$ is a constant, and find this constant.",x_{N}^{2}+y_{N}^{2}=\left(-\frac{t}{2} y_{N}+1\right)^{2}+y_{N}^{2}=\left(\frac{t^{2}}{4}+1\right) y,medium,"(1) $\frac{3}{2 a^{2}}+\frac{1}{4 b^{2}}=1 \Rightarrow \frac{3 b^{2}}{2 a^{2}}=b^{2}-\frac{1}{4} \Rightarrow \frac{3}{2}\left(1-e^{2}\right)=\frac{3}{4}=b^{2}-\frac{1}{4} \Rightarrow b^{2}=1, a^{2}=2$, So the standard equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$. (2) Let the equation of the circle be $x^{2}+y^{2}-2 x-t y=0$, the center of the circle is $\left(1, \frac{t}{2}\right)$, and the radius $r=\sqrt{1+\frac{t^{2}}{4}}$, then the distance from the center of the circle to the line is $d=\frac{|3-2 t-5|}{5}=\frac{t}{2} \Rightarrow t=4$. So the equation of the circle is $x^{2}+y^{2}-2 x-4 y=0$. (3) The equation of the line perpendicular to point $F$ is $y=-\frac{2}{t}(x-1)$, Combining $\left\{\begin{array}{l}x=-\frac{t}{2} y+1, \\ x^{2}+y^{2}-2 x-t y=0\end{array} \Rightarrow\left(\frac{t^{2}}{4}+1\right) y^{2}-t y-1=0\right.$. Then $|O N|^{2}=x_{N}^{2}+y_{N}^{2}=\left(-\frac{t}{2} y_{N}+1\right)^{2}+y_{N}^{2}=\left(\frac{t^{2}}{4}+1\right) y_{N}^{2}-t y_{N}+1=2$, so the length of segment $O N$ is a constant $\sqrt{2}$." 098cf757626d,"2. Given the quadratic trinomial in $x$ with integer coefficients $a x^{2}+b x+c$, when $x$ takes the values $1, 3, 6, 8$, a student recorded the values of this quadratic trinomial as $1, 5, 25, 50$. Upon checking, only one of these results is incorrect. The incorrect result is ( ). (A) When $x=1$, $a x^{2}+b x+c=1$ (B) When $x=3$, $a x^{2}+b x+c=5$ (C) When $x=6$, $a x^{2}+b x+c=25$ (D) When $x=8$, $a x^{2}+b x+c=50$",See reasoning trace,easy,"2. (C). Notice that $(n-m) \mid\left[\left(a n^{2}+b n+c\right)-\left(a m^{2}+b m+c\right)\right]$. From $(6-1) \mid (25-1), (8-6) \mid (50-25)$, but $(81) \mid (50-1)$, this leads to the result being $\mathbf{C}$." 7bedf5b3e702,"4. (10 points) Solve the equation $$ \sqrt{7-x^{2}+6 x}+\sqrt{6 x-x^{2}}=7+\sqrt{x(3-x)} $$",3,medium,"# Solution: The domain of the variable x in our problem is the interval [0;3]. By completing the square in the expressions under the square roots on the left side of the equation or by plotting the graphs, we notice that the values of the first root do not exceed 4, and the second does not exceed 3, with the minimum values being reached at $\mathrm{x}=3$. The value of the left side of the equation does not exceed 7, while the right side is no less than 7. Therefore, equality is possible only when both sides of the equation are simultaneously equal to 7, which is achieved at $x=3$. Answer: 3" 9ddb61a7c688,"## Task 1 - 260721 Anne, Bernd, and Peter help in the garden during the apple harvest. All three use baskets of the same size. Anne needs 10 minutes to fill a basket, Bernd takes 15 minutes, and little Peter even 30 minutes. How long would it take for the three children to fill a basket together? We assume that the picking speed does not change for any of the three helpers.",See reasoning trace,easy,"Anne has filled $\frac{1}{10}$ of her basket in one minute, Bernd $\frac{1}{15}$, and Peter $\frac{1}{30}$. Together, they have filled $$ \frac{1}{10}+\frac{1}{15}+\frac{1}{30}=\frac{1}{5} $$ of a basket after one minute. Thus, they need 5 minutes together to fill one basket." 9bd5e66ea5ec,"9. (15 points) As shown in Figure 2, circle $\odot M$ is externally tangent to circle $\odot O$ at point $C$, with radii $r$ and $R$ for $\odot M$ and $\odot O$ respectively. Line $TPQ$ is tangent to $\odot M$ at point $T$, and intersects $\odot O$ at points $P$ and $Q$. Find the value of $\frac{CQ - CP}{PQ}$.",See reasoning trace,medium,"II. 9. As shown in Figure 6, extend $PC$ and $QC$ to intersect $\odot M$ at points $D$ and $E$ respectively. Since $M, C, O$ are collinear, $\Rightarrow$ Isosceles $\triangle MCD \sim$ Isosceles $\triangle OCP$ $\Rightarrow \frac{CD}{CP}=\frac{r}{R} \Rightarrow \frac{PD}{CP}=\frac{R+r}{R}$ $\Rightarrow \frac{PD \cdot PC}{PC^2}=\frac{R+r}{R}$. Also, $PD \cdot PC=PT^2$, hence $\frac{PT^2}{PC^2}=\frac{R+r}{R}$ $\Rightarrow \frac{PT}{PC}=\sqrt{\frac{R+r}{R}}$. Similarly, $\frac{QT}{QC}=\sqrt{\frac{R+r}{R}}$. From equations (1) and (2), $$ \begin{array}{l} \frac{QT-PT}{QC-PC}=\sqrt{\frac{R+r}{R}} \Rightarrow \frac{PQ}{QC-PC}=\sqrt{\frac{R+r}{R}} \\ \Rightarrow \frac{QC-PC}{PQ}=\sqrt{\frac{R}{R+r}} . \end{array} $$" 4371f73eeebb,"1. Brothers Anže and Uroš have a total of 312 euros. If Anže gave Uroš $4 \%$ of his amount, Uroš would have 4 times as much as Anže. Calculate the amounts that Anže and Uroš have.","Anže had 65 euros, and Uroš had 247 euros",medium,"1. We consider that the brothers together have $x+y=312$ euros. From the problem statement, we write the equation $\left(x-\frac{4}{100} x\right) \cdot 4=y+\frac{4}{100} x$. In the equation, we remove the parentheses $4 x-\frac{4}{25} x=312-x+\frac{1}{25}$ and rearrange it $96 x+24 x=7800$. We calculate $x=65$. Then we calculate $y=247$. We write the answer: Anže had 65 euros, and Uroš had 247 euros. We write the relationship $x+y=312$ or $x=312-y \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ ![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-06.jpg?height=57&width=1636&top_left_y=1822&top_left_x=210) ![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-06.jpg?height=54&width=1636&top_left_y=1869&top_left_x=207) ![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-06.jpg?height=57&width=1636&top_left_y=1919&top_left_x=210) Answer . ........................................................................................................." f2b95e76d7c1,"7、A and B read a 120-page book, starting on October 1st. A reads 8 pages every day; B reads 13 pages every day, but he takes a break every two days. By the end of the long holiday on October 7th, A and B $\qquad$ compared $\qquad$ who read more, and by how many pages.",See reasoning trace,easy,"【Analysis】Two people read books together for 7 days, A read $7 \times 8=56$ pages. In 7 days, since $7 \div 3=2 \cdots 1$, B rested for 2 days, read books for 5 days, and read $5 \times 13=65$ pages. Therefore, B read more than A, 9 pages more." 93388ace7b2e,"2. When $\theta$ takes all real numbers, the area of the figure enclosed by the line $$ x \cos \theta+y \sin \theta=4+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) $$ is ( ). (A) $\pi$ (B) $4 \pi$ (C) $9 \pi$ (D) $16 \pi$",See reasoning trace,easy,"2.1). The equation of the line becomes $(x-1) \cos \theta+(y-1) \sin \theta=4$. Therefore, the distance from point $A(1,1)$ to the line is $$ d=\frac{4}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=4 \text {. } $$ Thus, when $\theta \in \mathbf{R}$, the line $$ x \cos \theta+y \sin \theta=4+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) $$ encloses a figure which is a circle with center at point $A$ and radius 4." 7d4c6d9b4921,"## Task 3 - 110613 Four faces of a wooden cube with an edge length of $3 \mathrm{~cm}$ are painted red, the other two remain unpainted. Afterwards, the cube is cut into exactly 27 smaller cubes, each with an edge length of $1 \mathrm{~cm}$. Determine the number of these small cubes that have no red-painted face, exactly one red-painted face, exactly two red-painted faces, exactly three red-painted faces! Distinguish the following cases: a) The unpainted faces do not share an edge. b) The unpainted faces share an edge. It is sufficient to provide the numbers without justification.",See reasoning trace,easy,"a) Exactly 3 small cubes have no red-painted surface. Exactly 12 small cubes have exactly one red-painted surface. Exactly 12 small cubes have exactly two red-painted surfaces. None of the cubes have three red-painted surfaces. b) The numbers in this case, in the above order, are: $4 ; 12 ; 9 ; 2$." 26aa1a98746a,"Solve the system of equations $$ \begin{aligned} & \cos x+\cos y+\cos z=\frac{3 \sqrt{3}}{2} \\ & \sin x+\sin y+\sin z=\frac{3}{2} \end{aligned} $$",See reasoning trace,medium,"I. solution. We square both equations and then add the resulting equations. Taking into account the identity $\cos ^{2} \alpha+\sin ^{2} \alpha=1$, we obtain: $$ \begin{gathered} 3+2(\cos x \cos y+\cos x \cos z+\cos y \cos z)+ \\ +2(\sin x \sin y+\sin x \sin z+\sin y \sin z)=\frac{27}{4}+\frac{9}{4}=\frac{36}{4}=9 \end{gathered} $$ from which: $$ (\cos x \cos y+\sin x \sin y)+(\cos x \cos z+\sin x \sin z)+(\cos y \cos z+\sin y \sin z)=3 $$ Applying the addition formula $\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)$ to all three terms: $\cos (x-y)+\cos (x-z)+$ $\cos (y-z)=3$. Since the value of the cosine function cannot be greater than 1, this equation can only be satisfied if $\cos (x-y)=\cos (x-z)=\cos (y-z)=1$, that is, $(x-y),(x-z),(y-z)$ are integer multiples of $2 \pi$: $$ y=x+2 k_{2} \pi ; \quad z=x+2 k_{3} \pi $$ where $k_{2}, k_{3}$ are arbitrary integers. Then $\cos x=\cos y=\cos z$ and $\sin x=\sin y=\sin z$; substituting these into the original equations: $$ \cos x+\cos x+\cos x=\frac{3 \sqrt{3}}{2}, \quad \sin x+\sin x+\sin x=\frac{3}{2} $$ in other words $$ \cos x=\frac{\sqrt{3}}{2}, \quad \sin x=\frac{1}{2} $$ from which: $$ x=\frac{\pi}{6}+2 k_{1} \pi $$ where $k_{1}$ is an arbitrary integer. Therefore, the solution is: $$ x=\frac{\pi}{6}+2 k_{1} \pi, \quad y=\frac{\pi}{6}+2 k_{2} \pi, \quad z=\frac{\pi}{6}+2 k_{3} \pi $$ where $k_{1}, k_{2}, k_{3}$ are arbitrary integers. II. solution. Imagine the problem as if we had to represent the vector $\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$ as the sum of three unit vectors in the complex plane. By introducing the unit-length complex unknowns $s=\cos x+i \sin x, t=\cos y+i \sin y$, $u=\cos z+i \sin z$, the equation $s+t+u=\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$ in the complex numbers precisely means that the real part of the left side, $\cos x+\cos y+\cos z$, is equal to the real part of the right side, $\frac{3 \sqrt{3}}{2}$, and the imaginary part of the left side, $\sin x+\sin y+\sin z$, is equal to the imaginary part of the right side, $\frac{3}{2}$; these are precisely the equations of the problem. Write the triangle inequality for the sum $s+t+u=\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$: $$ 3=|s|+|t|+|u| \geq|s+t+u|=\left|\frac{3 \sqrt{3}}{2}+i \frac{3}{2}\right|=\sqrt{\left(\frac{3 \sqrt{3}}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}}=3 $$ The obtained equality precisely means that the vectors $s, t, u$ are parallel and equally oriented, so $|s|=|t|=|u|=1$ implies $s=t=u$, hence $$ s=t=u=\frac{1}{3}\left(\frac{3 \sqrt{3}}{2}+i \frac{3}{2}\right)=\frac{\sqrt{3}}{2}+i \frac{1}{2} $$ Therefore, $\cos x=\cos y=\cos z=\frac{\sqrt{3}}{2}, \sin x=\sin y=\sin z=\frac{1}{2}$, so the solution is: $$ x=\frac{\pi}{6}+2 k \pi, \quad y=\frac{\pi}{6}+2 m \pi, \quad z=\frac{\pi}{6}+2 n \pi $$ where $k, m, n$ are arbitrary integers." ec5c82120797,"Booin d.A. Given two sequences: $2,4,8,16,14,10,2$ and 3, 6, 12. In each of them, each number is obtained from the previous one according to the same rule. a) Find this rule. b) Find all natural numbers that transform into themselves (according to this rule). c) Prove that the number $2^{1991}$ will become a single-digit number after several transformations.",a) Double the sum of the digits; b) 18,medium,"a) The law can be guessed, noticing, for example, that while the number is a single digit, it doubles, and then - apparently not. And the fact that 10 turns into 2 suggests that it is not the number itself that doubles, but the sum of its digits. Thus, the sought law has been discovered: ""Double the sum of the digits."" Of course, this is not a proof in the strict mathematical sense of the word. For example, this way one can ""prove"" that the number sixty is divisible by all numbers. Indeed, 60 is divisible by 1, by 2, by 3, by 4, by 5, by 6... However, for solving the problem, it is only necessary to find a ""sufficiently simple"" rule, following which one can obtain such a sequence. The ability to see, to feel the pattern (which was required in this problem) is no less important for a mathematician than the ability to reason strictly! If you find some other (but also ""sufficiently simple"") law that gives the sequences 2, 4, 8, 16, 14, 10, 2 and 3, 6, 12, please write to us (and on the Olympiad, such a solution would also be accepted!). b) See the solution to problem 5 for grades 5-6. c) Notice that if a number is at least three digits, then the sum of its digits is less than the number itself. Therefore, the number will decrease until it becomes a two-digit or a single-digit number. The only remaining danger is to get into a ""fixed point"" - 18. But this is impossible in our case, since the initial number was not divisible by 9 (prove strictly the following property of our law of obtaining one number from another: if some number is divisible by 9, then the number from which it is obtained is also divisible by 9). ## Answer a) Double the sum of the digits; b) 18." ad6bdc97e9ee,"9.4. (Jury, SRV, 79). Find all triples of numbers $a, b, c \in \mathbf{N}$, which are the lengths of the sides of a triangle with the diameter of the circumscribed circle equal to 6.25.","100-c^{2}$, from which $c=6$. Therefore, the sought triplet can only consist of the numbers $5,5,6$,",medium,"9.4. Let $a, b, c \in \mathbf{N}$ be the lengths of the sides of a triangle with a diameter $2 R=6.25$ of the circumscribed circle, area $S$, and semiperimeter $p=(a+b+c) / 2$. Since $a, b, c \leqslant 2 R$, we have $$ a, b, c \in\{1 ; 2 ; 3 ; 4 ; 5 ; 6\} $$ Further, we have the equalities $$ (a b c)^{2}=(4 S \cdot R)^{2}=p(p-a)(p-b)(p-c)(4 R)^{2} $$ from which we obtain $$ 64 a^{2} b^{2} c^{2}=625(a+b+c)(c+b-a)(a+c-b)(a+b-c) $$ Thus, the number $64 a^{2} b^{2} c^{2}$ is divisible by 625, which means that at least two of the three numbers $a, b, c$ are equal to 5. Let, for example, $a=b=5$, then $$ 64 c^{2}=(10+c) c^{2}(10-c) $$ i.e., $64=100-c^{2}$, from which $c=6$. Therefore, the sought triplet can only consist of the numbers $5,5,6$, which, as verification shows, satisfy the condition of the problem." e2d4aaf0fe27,"Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $$ a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. } $$ (1) Find the general term formula of the sequence $\left\{a_{n}\right\}$; (2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n}}\left(n \in \mathbf{N}_{+}\right)$, and for any positive integer $n(n \geqslant 2)$, the inequality $$ \sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24} $$ always holds, find the maximum value of the integer $m$.",See reasoning trace,medium,"(1) From $$ \begin{array}{l} a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\ \Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\ \text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) . \end{array} $$ Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a geometric sequence with the first term 3 and common ratio 3. Thus, $\frac{1}{a_{n}}+1=3^{n} \Rightarrow a_{n}=\frac{1}{3^{n}-1}\left(n \in \mathbf{N}_{+}\right)$. (2) From (1), we get $b_{n}=1+\frac{1}{a_{n}}=3^{n}$. Therefore, $\log _{3} b_{k}=k(k=1,2, \cdots)$. Hence, $\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}$ $\Leftrightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}>\frac{m}{24}$. Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$. Then $$ \begin{array}{l} f(n+1)-f(n) \\ =\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\ =\frac{1}{2 n+1}-\frac{1}{2 n+2}>0 . \end{array} $$ Therefore, $f(n)$ is monotonically increasing. Hence, $f(n)_{\min }=f(2)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$. Thus, $\frac{7}{12}>\frac{m}{24} \Rightarrow m<14$. Therefore, the maximum integer value of $m$ is 13." ed365ad57752,"An ant starts at vertex $A$ in equilateral triangle $\triangle ABC$ and walks around the perimeter of the triangle from $A$ to $B$ to $C$ and back to $A$. When the ant is $42$ percent of its way around the triangle, it stops for a rest. Find the percent of the way from $B$ to $C$ the ant is at that point",26\%,medium,"1. First, we need to understand the total distance the ant travels around the perimeter of the equilateral triangle $\triangle ABC$. Since the ant starts at vertex $A$ and walks around the perimeter back to $A$, the total distance is the perimeter of the triangle. 2. Let the side length of the equilateral triangle be $s$. The perimeter of the triangle is $3s$. 3. The ant travels $42\%$ of the total perimeter. Therefore, the distance traveled by the ant is: \[ 0.42 \times 3s = 1.26s \] 4. Each side of the triangle represents $\frac{1}{3}$ or approximately $33.33\%$ of the total perimeter. Therefore, the ant travels one full side from $A$ to $B$ and part of the next side from $B$ to $C$. 5. The distance from $A$ to $B$ is $s$, which is $33.33\%$ of the total perimeter. The remaining distance the ant travels is: \[ 1.26s - s = 0.26s \] 6. The remaining distance $0.26s$ is the distance the ant travels from $B$ towards $C$. To find the percentage of the way from $B$ to $C$, we calculate: \[ \frac{0.26s}{s} \times 100\% = 26\% \] Therefore, the ant is $26\%$ of the way from $B$ to $C$ when it stops for a rest. The final answer is $\boxed{26\%}$" 12ae77403417,"2. If $a_{1}, a_{2}, \cdots, a_{n}$ is a set of real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=k$. Find the minimum value of $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$.","x^{2}$ is a convex function on $(-\infty,+\infty)$, so $\frac{1}{n}\left[f\left(a_{1}\right)+\cdots+",easy,"2. $f(x)=x^{2}$ is a convex function on $(-\infty,+\infty)$, so $\frac{1}{n}\left[f\left(a_{1}\right)+\cdots+f\left(a_{n}\right)\right] \geqslant f\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)$. Therefore, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n \cdot\left(\frac{k}{n}\right)^{2}=\frac{k^{2}}{n}$." bcaf22db44cd,"1. Solve the inequality $$ (2+\sqrt{3})^{x}+2<3(\sqrt{2-\sqrt{3}})^{2 x} $$ In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-20 ; 53)$.",The solution to the inequality is $(-\infty ; 0)$,easy,Answer: The solution to the inequality is $(-\infty ; 0)$. The desired sum $-19-18-\cdots-2-1=-190$. We write -190 as the answer. 87e6507bfd66,"6. Let real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=1$, and let the maximum and minimum values of $a b+b c+c a$ be $M$ and $m$, respectively, then $M-m=$",$\frac{3}{2}$,easy,"Answer $\frac{3}{2}$. Analysis According to the Cauchy-Schwarz inequality $(a b+b c+c a)^{2} \leq\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+a^{2}\right)=1$, so $M=1$, or according to, $$ 2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0 $$ we can also get $M=1$, by $$ (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 0 $$ we can get $a b+b c+c a \leq-\frac{1}{2}$, i.e., $m=-\frac{1}{2}$, so $M-m=\frac{3}{2}$" 498df8c00862,"For a sequence $ a_i \in \mathbb{R}, i \in \{0, 1, 2, \ldots\}$ we have $ a_0 \equal{} 1$ and \[ a_{n\plus{}1} \equal{} a_n \plus{} \sqrt{a_{n\plus{}1} \plus{} a_n} \quad \forall n \in \mathbb{N}.\] Prove that this sequence is unique and find an explicit formula for this recursively defined sequence.",a_n = \frac{n(n + 3),medium,"1. **Initial Condition and Recurrence Relation:** We start with the initial condition \( a_0 = 1 \) and the recurrence relation: \[ a_{n+1} = a_n + \sqrt{a_{n+1} + a_n} \quad \forall n \in \mathbb{N}. \] 2. **Isolating \( a_{n+1} \):** To isolate \( a_{n+1} \), we let \( x = a_{n+1} \). Then the recurrence relation becomes: \[ x = a_n + \sqrt{x + a_n}. \] Squaring both sides to eliminate the square root, we get: \[ x^2 = (a_n + \sqrt{x + a_n})^2. \] Expanding the right-hand side: \[ x^2 = a_n^2 + 2a_n\sqrt{x + a_n} + (x + a_n). \] Simplifying, we have: \[ x^2 = a_n^2 + x + 3a_n + 2a_n\sqrt{x + a_n}. \] Rearranging terms: \[ x^2 - x - a_n^2 - 3a_n = 2a_n\sqrt{x + a_n}. \] Squaring both sides again to eliminate the square root: \[ (x^2 - x - a_n^2 - 3a_n)^2 = 4a_n^2(x + a_n). \] 3. **Finding the Explicit Formula:** We hypothesize that the sequence \( a_n \) can be expressed in a closed form. Let's assume: \[ a_n = \frac{n(n + 3)}{2} + 1. \] We will verify this hypothesis by induction. 4. **Base Case:** For \( n = 0 \): \[ a_0 = \frac{0(0 + 3)}{2} + 1 = 1, \] which matches the initial condition. 5. **Inductive Step:** Assume \( a_n = \frac{n(n + 3)}{2} + 1 \) holds for some \( n \). We need to show it holds for \( n + 1 \): \[ a_{n+1} = a_n + \sqrt{a_{n+1} + a_n}. \] Using the hypothesis: \[ a_n = \frac{n(n + 3)}{2} + 1. \] We need to find \( a_{n+1} \): \[ a_{n+1} = \frac{n(n + 3)}{2} + 1 + \sqrt{a_{n+1} + \frac{n(n + 3)}{2} + 1}. \] Let \( a_{n+1} = \frac{(n+1)((n+1) + 3)}{2} + 1 \): \[ a_{n+1} = \frac{(n+1)(n + 4)}{2} + 1 = \frac{n^2 + 5n + 4}{2} + 1 = \frac{n^2 + 5n + 6}{2}. \] We need to verify: \[ \frac{n^2 + 5n + 6}{2} = \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{n^2 + 5n + 6}{2} + \frac{n(n + 3)}{2} + 1}. \] Simplifying the right-hand side: \[ \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{n^2 + 5n + 6 + n^2 + 3n + 2}{2}} = \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{2n^2 + 8n + 8}{2}} = \frac{n(n + 3)}{2} + 1 + \sqrt{n^2 + 4n + 4} = \frac{n(n + 3)}{2} + 1 + (n + 2). \] Simplifying further: \[ \frac{n(n + 3)}{2} + 1 + n + 2 = \frac{n^2 + 3n}{2} + n + 3 = \frac{n^2 + 5n + 6}{2}. \] This confirms our hypothesis. 6. **Conclusion:** The sequence \( a_n \) is unique and given by: \[ a_n = \frac{n(n + 3)}{2} + 1. \] The final answer is \( \boxed{ a_n = \frac{n(n + 3)}{2} + 1 } \)." ec08e6eb761b,"$20 \cdot 25$ As shown, $AB$ is tangent to $\odot O$ at $A$, and point $D$ is inside the circle, with $DB$ intersecting the circle at $C$. If $BC=DC=3, OD=2, AB=6$, then the radius of the circle is (A) $3+\sqrt{3}$. (B) $15 / \pi$. (C) $9 / 2$. (D) $2 \sqrt{6}$. (E) $\sqrt{22}$. (27th American High School Mathematics Examination, 1976)",$(E)$,easy,"[Solution] Let $E$ be the intersection of the circle and the extension of $BD$, $FG$ be the diameter through $D$, and $r$ be the radius of the circle. Then $BC \cdot BE = AB^2$, which means $3 \cdot (DE + 6) = 36$, giving $DE = 6$. Similarly, from $DE \cdot DC = DF \cdot DG$, we have $18 = r^2 - 4$, yielding $r = \sqrt{22}$. Therefore, the answer is $(E)$." 10051e186d8c,"2. First method: Find the digits $a, b, c$ such that the number $\overline{387 a b c}$ is divisible by 5, 6, and 7. Any even number divisible by 5 ends in 0, so $c=0$. For divisibility by 3, it is necessary that $3+8+7+a+b=18+a+b$ is divisible by 3. Since 18 is divisible by 3, $a+b$ must be divisible by 3. Let's take the smallest number divisible by 6 and 5, which is 387030. Direct verification shows that this number is also divisible by 7, and therefore is the desired number. If we add $210=5 \cdot 6 \cdot 7$ to this number, its divisibility by 5, 6, and 7 will not change. Therefore, 387240, 387450, 387660, 387870 are also solutions. Second method: A number divisible by 5, 6, and 7 is divisible by $210=5 \cdot 6 \cdot 7$ and has the form $210k$. We need to find $k$ such that $387000 \leq 210k \leq 387999$. We get $k=1843, \ldots 1847$.","387030, 387240, 387450, 387660, 387870",easy,"Answer: 387030, 387240, 387450, 387660, 387870. Evaluation recommendations: Students may find only one or a few solutions. If only one solution is found, the task is scored 3 points; if several but not all solutions are found - 5 points. Full solution - 7 points." b54743580306,"Let the notation $|X|$ represent the number of elements in the set $X$. Suppose $|X|=n\left(n \in \mathbf{N}_{+}\right)$, and all subsets of $X$ are $\left\{A_{1}, A_{2}, \cdots, A_{m}\right\}\left(m=2^{n}\right)$. It is known that the sum of the number of elements in the intersections of these subsets is $$ S=\sum_{i=1}^{m}\left(\sum_{j=1}^{m}\left|A_{i} \cap A_{j}\right|\right)=n \cdot 2^{2(n-1)} . $$ Try to find the sum $$ T=\sum_{i=1}^{m-1}\left(\sum_{j=i+1}^{m}\left|A_{i} \cap A_{j}\right|\right) . $$ (2002, Zhejiang Province Mathematical Summer Camp)",See reasoning trace,medium,"Consider the following $m \times m$ matrix $$ \left(\begin{array}{cccc} \left|A_{1} \cap A_{1}\right| & \left|A_{1} \cap A_{2}\right| & \cdots & \left|A_{1} \cap A_{m}\right| \\ \left|A_{2} \cap A_{1}\right| & \left|A_{2} \cap A_{2}\right| & \cdots & \left|A_{2} \cap A_{m}\right| \\ \vdots & \vdots & \vdots & \vdots \\ \left|A_{m} \cap A_{1}\right| & \left|A_{m} \cap A_{2}\right| & \cdots & \left|A_{m} \cap A_{m}\right| \end{array}\right) $$ Let the diagonal from the top left to the bottom right be denoted as $l$. Since the intersection operation of sets is commutative, the matrix is symmetric about $l$. Therefore, the sum of the numbers on $l$ is $$ R=\sum_{i=1}^{m}\left|A_{i} \cap A_{i}\right|=\sum_{i=1}^{m}\left|A_{i}\right| . $$ Then $T=\frac{1}{2}(S-R)$. It is easy to see that $R=\mathrm{C}_{n}^{1}+2 \mathrm{C}_{n}^{2}+\cdots+n \mathrm{C}_{n}^{n}=n \cdot 2^{n-1}$. Thus, $T=\frac{1}{2}(S-R)=\frac{1}{2}\left[n \cdot 2^{\left.\alpha_{n-1}\right)}-n \cdot 2^{n-1}\right]$ $$ =\frac{n}{2}\left(4^{n-1}-2^{n-1}\right) \text {. } $$" 1bc35e9a9c13,"An isosceles triangle has angles of $50^\circ,x^\circ,$ and $y^\circ$. Find the maximum possible value of $x-y$. [i]Proposed by Nathan Ramesh",30^\circ,medium,"1. **Identify the properties of the isosceles triangle:** - An isosceles triangle has two equal angles. - The sum of the angles in any triangle is $180^\circ$. 2. **Set up the equations based on the given angles:** - Let the angles of the isosceles triangle be $50^\circ$, $x^\circ$, and $y^\circ$. - Since it is an isosceles triangle, two of these angles must be equal. 3. **Consider the possible cases for the angles:** - Case 1: $50^\circ$ is one of the equal angles. - If $50^\circ$ is one of the equal angles, then the other equal angle must also be $50^\circ$. - Therefore, the third angle must be $180^\circ - 50^\circ - 50^\circ = 80^\circ$. - This gives us the angle sets $(50^\circ, 50^\circ, 80^\circ)$. - Case 2: $50^\circ$ is the unique angle. - If $50^\circ$ is the unique angle, then the other two angles must be equal. - Let these equal angles be $x^\circ$. - Therefore, $50^\circ + x^\circ + x^\circ = 180^\circ$. - Solving for $x$, we get $2x = 130^\circ \implies x = 65^\circ$. - This gives us the angle sets $(50^\circ, 65^\circ, 65^\circ)$. 4. **Calculate $x - y$ for each case:** - For the angle set $(50^\circ, 50^\circ, 80^\circ)$: - If $x = 80^\circ$ and $y = 50^\circ$, then $x - y = 80^\circ - 50^\circ = 30^\circ$. - If $x = 50^\circ$ and $y = 80^\circ$, then $x - y = 50^\circ - 80^\circ = -30^\circ$. - For the angle set $(50^\circ, 65^\circ, 65^\circ)$: - If $x = 65^\circ$ and $y = 65^\circ$, then $x - y = 65^\circ - 65^\circ = 0^\circ$. 5. **Determine the maximum possible value of $x - y$:** - From the calculations above, the maximum possible value of $x - y$ is $30^\circ$. The final answer is $\boxed{30^\circ}$" 79deb1414d80,44th Putnam 1983,"- n/(2r + 1). Integrating it over the adjacent interval (2n/(2r+2), 2n/(2r+1) ) gives + (r + 1) (2n/",medium,": ln(4/π). Note that on the interval (2n/2n, 2n/(2n-1) ) the integrand is n - n/x; on (2n/(2n-1), 2n/(2n-2) ) it is n/x - (n- 1); on (2n/(2n-2), 2n/(2n-3) ) it is (n-1) - n/x; ... ; on (2n/5, 2n/4) it is n/x - 2; on (2n/4, 2n/3) it is 2 - n/x; and on (2n/3, 2n/2) it is n/x - 1. Integrating the constant term over the interval (2n/(2r+1), 2n/2r) gives - r (2n/2r - 2n/(2r+1) ) = - n/(2r + 1). Integrating it over the adjacent interval (2n/(2r+2), 2n/(2r+1) ) gives + (r + 1) (2n/(2r+1) - 2n/(2r+2) ) = n/(2r + 1). So over the entire interval (1, n) the constant terms integrate to zero. Integrating the 1/x term over (2n/(2r+1), 2n/2r) gives n ln( (2r+1)/2r ), whilst over (2n/2r, 2n/(2r-1) ) gives - n ln( 2r/(2r-1) ). So over the combined interval (2n/(2r+1), 2n/(2r-1) ) we get - n ln ( 2r/(2r-1) 2r/(2r+1) ). Summing we get that integral over (2n/(2n-1), 2n/3) is - n ln P n , where P n = ∏ 2 n-1 2r/(2r-1) 2r/(2r+1). That leaves the intervals at each end which integrate to n ln 3/2 and - n ln(2n/(2n-1) ). So multiplying by 1/n and taking the limit we get - ln(3/4 π/2) + ln 3/2 = ln(4/π). 44th Putnam 1983 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001" cefc7c652637,"## Task A-1.2. In an isosceles trapezoid, the midline is of length $l$, and the diagonals are mutually perpendicular. Determine the area of the trapezoid.",See reasoning trace,medium,"## First Solution. ![](https://cdn.mathpix.com/cropped/2024_05_30_a9c12f6adaba3761a080g-02.jpg?height=446&width=788&top_left_y=591&top_left_x=631) Let: $|A B|=a, |C D|=c, |B D|=d$, and the height of the trapezoid be $v$. Notice that in an isosceles trapezoid, the diagonals are congruent, so $|A C|=|B D|=d$. According to the problem, $l=\frac{a+c}{2}$. Extend the line segment $\overline{A B}$ to point $E$ such that $|B E|=|D C|$. Then $B E C D$ is a parallelogram, and $|C E|=|D B|$, and $\overline{C E} \| \overline{D B}$, so $|C E|=d$. Triangle $A E C$ is an isosceles right triangle with legs of length $d$, because $|A C|=|C E|=d$ and $A C \perp C E (C E \| B D, B D \perp A C)$. Therefore, the height $v$ is equal to half the length of the hypotenuse of the right triangle $A C E$, i.e., $v=\frac{1}{2}|A E|=\frac{a+c}{2}=l$. (5 points) Finally, $$ P(A B C D)=\frac{a+c}{2} \cdot v=l \cdot v=l^{2} $$" a75b074adcde,"[ Arithmetic operations. Numerical identities] Using five threes, arithmetic operations, and exponentiation, form the numbers from 1 to 10. #",See reasoning trace,medium,"First solution: $1=(3: 3)^{333}$, $2=(3-3: 3) \times 3: 3$, $3=3 \times(3: 3) \times(3: 3)$ $4=(3+3: 3) \times(3: 3)$ $5=3+3: 3+3: 3$ $6=(3+3) \times(3: 3)^{3}$, $7=(3+3)+(3: 3)^{3}$, $8=(3 \cdot 3)-(3: 3)^{3}$ $9=(3 \cdot 3) \times(3: 3)^{3}$, $10=(3 \cdot 3)+(3: 3)^{3}$ Second solution. $1=(3 \times 3+3): 3-3$ $2=(3+3+3-3): 3$ $3=(3 \times 3+3-3): 3$ $4=(3+3+3+3): 3$ $5=(3 \times 3+3+3): 3$ $6=3 \times 3+3-3-3$ $7=3: 3+(3 \times 3-3)$ $8=(3 \times 3 \times 3-3): 3$, $9=3+3+3+3-3$ $10=3: 3+3+3+3$ (Solution by 7th grade students of the ""Traditional Gymnasium"" in Moscow) Send a comment" 3434e606b22b,"I4.1 Given that $x$ and $y$ are real numbers such that $|x|+x+y=10$ and $|y|+x-y=10$. If $P=x+y$, find the value of $P$.",See reasoning trace,medium,"If $x \geq 0, y \geq 0 ;\left\{\begin{aligned} 2 x+y & =10 \\ x & =10\end{aligned}\right.$, solving give $x=10, y=-10$ (reject) If $x \geq 0, y<0,\left\{\begin{array}{l}2 x+y=10 \\ x-2 y=10\end{array}\right.$, solving give $x=6, y=-2$ If $x<0, y \geq 0,\left\{\begin{array}{l}y=10 \\ x=10\end{array}\right.$, reject If $x<0, y<0 ;\left\{\begin{aligned} y & =10 \\ x-2 y & =10\end{aligned}\right.$, solving give $x=30, y=10$ (reject) $$ \begin{array}{l} \therefore x=6, y=-2 \\ P=x+y=6-2=4 \end{array} $$" 2bdc1e015a2b,"The letter $A$ has a vertical line of symmetry and the letter $B$ does not. How many of the letters H L O R X D P E have a vertical line of symmetry? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5",(C),easy,"If a letter is folded along its vertical line of symmetry, both halves of the letter would match exactly. Three of the given letters, $\mathrm{H}, \mathrm{O}$ and $\mathrm{X}$, have a vertical line of symmetry, as shown. $$ \begin{array}{llllllll} H & L & \phi & R & X & D & P & E \end{array} $$ ANSWER: (C)" 7dcf60b3a376,"3. Solve the equation $$ 2 x+2+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}+\operatorname{arctg}(x+2) \cdot \sqrt{x^{2}+4 x+5}=0 $$",-1,medium,"Answer: -1. Solution. Let $f(x)=x+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the sum of increasing functions). Due to its oddness, it is increasing on the entire real line. Further, we have $$ f(x)=-f(x+2) ; \quad f(x)=f(-x-2) $$ Since an increasing function takes each of its values exactly once, it follows that $x=-x-2$, from which $x=-1$. Remark. The monotonicity of the left side of the original equation could also have been established using the derivative. Evaluation. 12 points for a correct solution. If the answer is guessed but not proven that there are no other solutions, 3 points." 9a18fe118a40,"Find the real numbers $x, y$ and $z$ such that $$ \left\{\begin{array}{l} x+y+z=3 \\ x^{2}+y^{2}+z^{2}=3 \\ x^{3}+y^{3}+z^{3}=3 \end{array}\right. $$",y=z=1$.,medium,"First, by trial and error, we find that $x=y=z=1$ works. We will now prove that this is the only solution. To do this, 1 the idea is to obtain several polynomials of the degrees that interest us, in order to get expressions that we can use. For example, $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)$, so $$ x y+y z+z x=\frac{(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)}{2}=\frac{3^{2}-3}{2}=3 $$ Similarly, $$ \begin{aligned} & (x+y+z)^{3}=\left(x^{3}+y^{3}+z^{3}\right)+3\left(x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y\right)+6 x y z \text { and } \\ & (x+y+z)\left(x^{2}+y^{2}+z^{2}\right)=\left(x^{3}+y^{3}+z^{3}\right)+\left(x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y\right) . \end{aligned} $$ We deduce that $$ \begin{aligned} x y z & =\frac{(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)-3(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)+3\left(x^{3}+y^{3}+z^{3}\right)}{6} \\ & =\frac{3^{3}-3-3^{3}+3^{2}}{6} \\ & =1 \end{aligned} $$ We conclude that $$ (t-x)(t-y)(t-z)=t^{3}-3 t^{2}+3 t-1=(t-1)^{3} $$ and therefore that $x=y=z=1$." 9bf088226da1,24th Putnam 1963,"x n+1 y (n+1) and that is easily proved by induction. So we seek a solution to x n+1 y (n+1) = f(x),",medium,"The first step is to try to simplify Ey. Experimenting with small n, we soon suspect that Ey = x n+1 y (n+1) and that is easily proved by induction. So we seek a solution to x n+1 y (n+1) = f(x), subject to y(1) = y'(1) = ... = y (n) (1) = 0. The trick is to use the Taylor expansion: y(x) = y(1) + (x-1)y'(1) + (x-1) 2 /2! y''(1) + ... + (x-1) n y (n) (1) + ∫ 1 x g(t) dt, where g(t) = (x - t) n y n+1 (t)/n! = (x - t) n f(t)/(n! t n+1 ). 24th Putnam 1963 © John Scholes jscholes@kalva.demon.co.uk 5 Feb 2002" 98b73171a15d,"Example 5 Let the set $M=\{1,2,3, \cdots, 1000\}$. For any non-empty subset $X$ of $M$, let $\alpha_{X}$ denote the sum of the largest and smallest numbers in $X$. Then the arithmetic mean of all such $\alpha_{X}$ is $\qquad$ (1991, National High School Mathematics Competition)",See reasoning trace,medium,"Solution: Construct the subset $X^{\prime}=\{1001-x \mid x \in X\}$, then all non-empty subsets can be divided into two categories: $X^{\prime}=X$ and $X^{\prime} \neq X$. When $X^{\prime}=X$, it must be that $X^{\prime}=X=M$, thus, $\alpha_{X}=1001$. When $X^{\prime} \neq X$, let $x$ and $y$ be the maximum and minimum numbers in $X$, respectively, then $1001-x$ and $1001-y$ are the minimum and maximum numbers in $X^{\prime}$, respectively. Therefore, $a_{X}=x+y$, $\alpha_{X^{\prime}}=2002-x-y$. Thus, $\frac{a_{X}}{2} \frac{\alpha_{X^{\prime}}}{2}=1001$. Hence, the arithmetic mean of the required $\alpha_{x}$ is 1001. 4 Solution using the principle of inclusion-exclusion Using the principle of inclusion-exclusion to solve certain subset competition problems can directly calculate the result of the problem." 09c22ac2b036,"10,11 Find the lateral surface area of a regular triangular pyramid if its height is 4 and the apothem is 8.",288,medium,"Let $M$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, and $K$ be the midpoint of $BC$. Then $DM$ is the height of the pyramid, $DK$ is the apothem of the pyramid, and $AK$ is the height of the equilateral triangle $ABC$, with $AK$ passing through point $M$. Denote $AB = BC = AC = a$. Then $$ AK = \frac{a \sqrt{3}}{2}, \quad KM = \frac{1}{3} AK = \frac{a \sqrt{3}}{6}. $$ On the other hand, from the right triangle $DKM$, we find that $$ KM = \sqrt{DK^2 - DM^2} = \sqrt{64 - 16} = 4 \sqrt{3}. $$ From the equality $\frac{a \sqrt{3}}{6} = 4 \sqrt{3}$, we find that $a = 24$. If $S$ is the lateral surface area of the pyramid, then $$ S = 3 S_{\triangle BDC} = 3 \cdot \frac{1}{2} AB \cdot DK = 3 \cdot \frac{1}{2} a \cdot DK = 3 \cdot \frac{1}{2} \cdot 24 \cdot 8 = 288 $$ ## Answer 288.00 ## Problem" 78119337dd15,"Through the vertex of a regular quadrilateral pyramid and the midpoints of two adjacent sides of the base, a plane is drawn. Find the area of the resulting section if the side of the base of the pyramid is $a$, and the lateral edge is $2a$. #",$\frac{a^2 \sqrt{29}}{8}$,medium,"Let $M$ and $N$ be the midpoints of sides $AB$ and $BC$ of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$, respectively, with $AB = a$ and $AP = 2a$. Let $O$ be the center of the square $ABCD$, and $K$ be the intersection point of $MN$ and $BD$. Then $OK$ is the orthogonal projection of the inclined line $PK$ on the plane $ABCD$, and $OK \perp MN$. By the theorem of three perpendiculars, $PK \perp MN$, i.e., $PK$ is the height of triangle $PMN$. Since $MN$ is the midline of triangle $ABC$, $$ MN = \frac{1}{2} AC = \frac{1}{2} a \sqrt{2}, \quad OK = \frac{1}{2} OB = \frac{1}{4} BD = \frac{1}{4} a \sqrt{2} $$ By the Pythagorean theorem in right triangles $AOP$ and $ROK$, we find that $$ OP = \sqrt{AP^2 - AO^2} = \sqrt{4a^2 - \frac{a^2}{2}} = \frac{1}{2} a \sqrt{14} $$ $$ PK = \sqrt{OP^2 + OK^2} = \sqrt{\frac{7a^2}{2} + \frac{a^2}{8}} = \frac{1}{4} a \sqrt{58} $$ Therefore, $$ S_{\triangle PMN} = \frac{1}{2} MN \cdot PK = \frac{1}{2} \cdot \frac{1}{2} a \sqrt{2} \cdot \frac{1}{4} a \sqrt{58} = \frac{1}{8} a^2 \sqrt{29} $$ ## Answer $\frac{a^2 \sqrt{29}}{8}$." fcc57471c445,"Rectangles that measure $4 \times 2$ are positioned in a pattern in which the top left vertex of each rectangle (after the top one) is placed at the midpoint of the bottom edge of the rectangle above it, as shown. When a total of ten rectangles are arranged in this pattern, what is the perimeter of the figure? (A) 48 (B) 64 (C) 90 (D) 84 (E) 100 ![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-030.jpg?height=409&width=339&top_left_y=1124&top_left_x=1294)",(D),medium,"Solution 1 The first and tenth rectangles each contribute an equal amount to the perimeter. They each contribute two vertical sides (each of length 2), one full side of length 4 (the top side for the first rectangle and the bottom side for the tenth rectangle), and one half of the length of the opposite side. That is, the first and tenth rectangles each contribute $2+2+4+2=10$ to the perimeter. Rectangles two through nine each contribute an equal amount to the perimeter. They each contribute two vertical sides (each of length 2), one half of a side of length 4, and one half of the length of the opposite side (which also has length 4). That is, rectangles two through nine each contribute $2+2+2+2=8$ to the perimeter. Therefore, the total perimeter of the given figure is $(2 \times 10)+(8 \times 8)=20+64=84$. ## Solution 2 One method for determining the perimeter of the given figure is to consider vertical lengths and horizontal lengths. Each of the ten rectangles has two vertical sides (a left side and a right side) which contribute to the perimeter. These 20 sides each have length 2 , and thus contribute $20 \times 2=40$ to the perimeter of the figure. Since these are the only vertical lengths contributing to the perimeter, we now determine the sum of the horizontal lengths. There are two types of horizontal lengths which contribute to the perimeter: the bottom side of a rectangle, and the top side of a rectangle. The bottom side of each of the first nine rectangles contributes one half of its length to the perimeter. That is, the bottom sides of the first nine rectangles contribute $\frac{1}{2} \times 4 \times 9=18$ to the perimeter. The entire bottom side of the tenth rectangle is included in the perimeter and thus contributes a length of 4 . Similarly, the top sides of the second rectangle through to the tenth rectangle contribute one half of their length to the perimeter. That is, the top sides of rectangles two through ten contribute $\frac{1}{2} \times 4 \times 9=18$ to the perimeter. The entire top side of the first rectangle is included in the perimeter and thus contributes a length of 4. In total, the horizontal lengths included in the perimeter sum to $18+4+18+4=44$. Since there are no additional lengths which contribute to the perimeter of the given figure, the total perimeter is $40+44=84$. ## Solution 3 Before they were positioned to form the given figure, each of the ten rectangles had a perimeter of $2 \times(2+4)=12$. When the figure was formed, some length of each of the ten rectangles' perimeter was ""lost"" (and thus is not included) in the perimeter of the given figure. These lengths that were lost occur where the rectangles touch one another. There are nine such locations where two rectangles touch one another (between the first and second rectangle, between the second and third rectangle, and so on). In these locations, each of the two rectangles has one half of a side of length 4 which is not included in the perimeter of the given figure. That is, the portion of the total perimeter of the ten rectangles that is not included in the perimeter of the figure is $9 \times(2+2)=36$. Since the total perimeter of the ten rectangles before they were positioned into the given figure is $10 \times 12=120$, then the perimeter of the given figure is $120-36=84$. ANSWER: (D)" 089631043433,"Cátia leaves school every day at the same time and returns home by bicycle. When she cycles at $20 \mathrm{~km} / \mathrm{h}$, she arrives home at 4:30 PM. If she cycles at $10 \mathrm{~km} / \mathrm{h}$, she arrives home at 5:15 PM. At what speed should she cycle to arrive home at 5:00 PM?",See reasoning trace,medium,"Let $t$ be the time she spends cycling at $20 \mathrm{~km} / \mathrm{h}$. Cycling at $10 \mathrm{~km} / \mathrm{h}$, she takes twice the time she spends cycling at $20 \mathrm{~km} / \mathrm{h}$, which is $2 t$. However, since she takes 45 minutes longer, we have: $$ 2 t-t=45 \Longrightarrow t=45 \mathrm{~min} $$ Therefore, she leaves school daily at $$ 4: 30 \mathrm{~h}-45 \mathrm{~min}=3: 45 \mathrm{~h} $$ and the distance to her home is $$ 45 \mathrm{~min} \times 20 \mathrm{~km} / \mathrm{h}=\frac{3}{4} \times 20=15 \mathrm{~km} $$ To cover $15 \mathrm{~km}$ in $5: 00 \mathrm{~h}-3: 45 \mathrm{~h}=1: 15 \mathrm{~h}=\frac{5}{4} \mathrm{~h}$, she needs to maintain a speed of $$ \frac{15 \mathrm{~km}}{\frac{5}{4} \mathrm{~h}}=12 \mathrm{~km} / \mathrm{h} $$" 188956a6400e,"How many ways are there to choose distinct positive integers $a, b, c, d$ dividing $15^6$ such that none of $a, b, c,$ or $d$ divide each other? (Order does not matter.) [i]Proposed by Miles Yamner and Andrew Wu[/i] (Note: wording changed from original to clarify)",1225,medium,"1. **Prime Factorization and Representation**: We start by noting that \( 15^6 = 3^6 \cdot 5^6 \). Each divisor of \( 15^6 \) can be written in the form \( 3^x \cdot 5^y \) where \( 0 \leq x \leq 6 \) and \( 0 \leq y \leq 6 \). 2. **Distinct Exponents**: We need to choose four distinct positive integers \( a, b, c, d \) such that none of them divide each other. This implies that the exponents \( x_i \) and \( y_i \) in the representations \( a = 3^{x_1}5^{y_1} \), \( b = 3^{x_2}5^{y_2} \), \( c = 3^{x_3}5^{y_3} \), and \( d = 3^{x_4}5^{y_4} \) must all be distinct. If any two \( x_i \) or \( y_i \) are equal, one number would divide the other, violating the condition. 3. **Choosing Distinct Exponents**: We need to choose 4 distinct values for \( x \) and 4 distinct values for \( y \) from the set \(\{0, 1, 2, 3, 4, 5, 6\}\). The number of ways to choose 4 distinct values from 7 is given by the binomial coefficient: \[ \binom{7}{4} \] Since we need to choose 4 distinct values for both \( x \) and \( y \), the total number of ways to choose these values is: \[ \left( \binom{7}{4} \right)^2 \] 4. **Ensuring No Divisibility**: Given the chosen values, we need to ensure that none of the numbers divide each other. To do this, we can assume without loss of generality (WLOG) that \( x_1 < x_2 < x_3 < x_4 \). This ordering of \( x \)-values forces the \( y \)-values to be in strictly decreasing order \( y_1 > y_2 > y_3 > y_4 \) to ensure that none of the numbers divide each other. 5. **Calculating the Number of Ways**: The number of ways to choose 4 distinct values from 7 is: \[ \binom{7}{4} = 35 \] Therefore, the total number of ways to choose the \( x \)-values and \( y \)-values is: \[ \left( \binom{7}{4} \right)^2 = 35^2 = 1225 \] 6. **Conclusion**: Since the ordering of \( x \)-values uniquely determines the ordering of \( y \)-values, there is exactly 1 way to arrange the \( y \)-values for each choice of \( x \)-values. Thus, the total number of ways to choose the distinct positive integers \( a, b, c, d \) such that none of them divide each other is: \[ \boxed{1225} \]" a989598d7370,14.5.13 ** Find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.,"(n+1,1)=1$, and $2 n+1$ is odd, we have $\left\{\begin{array}{l}n+1=2 u^{2}, \\ 2 n+1=3 v^{2}\end{ar",medium,"Let $\frac{1}{n}\left(1^{2}+2^{2}+\cdots+n^{2}\right)$ be a square number $m^{2}$, then $6 m^{2}-(n+1)(2 n+1)$. Since $(n+1,2 n+1)=(n+1,1)=1$, and $2 n+1$ is odd, we have $\left\{\begin{array}{l}n+1=2 u^{2}, \\ 2 n+1=3 v^{2}\end{array}\right.$ or $\left\{\begin{array}{l}n+1=6 u^{2}, \\ 2 n+1=v^{2} .\end{array}\right.$ The latter gives $12 u^{2}-1=v^{2}$. Thus, $v^{2} \equiv-1(\bmod 4)$, which is impossible. From the former, we get $4 u^{2}-3 v^{2}=1$. By testing $u$ from 1 to 13, we find the solutions $u=1, v=1$ (which does not meet the problem's requirements) and $u=13, v=$ 15. This yields the smallest positive integer $n=337$." 2d610ea1702f,"In the sequence $32,8, \longrightarrow, \longrightarrow x$, each term after the second is the average of the two terms immediately before it. The value of $x$ is (A) 17 (B) 20 (C) 44 (D) 24 (E) 14 ## Part B: Each correct answer is worth 6.",(A),easy,"The third term in the sequence is, by definition, the average of the first two terms, namely 32 and 8 , or $\frac{1}{2}(32+8)=\frac{1}{2}(40)=20$. The fourth term is the average of the second and third terms, or $\frac{1}{2}(8+20)=14$. The fifth term is the average of the third and fourth terms, or $\frac{1}{2}(20+14)=17$. Therefore, $x=17$. ANSWER: (A)" 41cb26cb2374,"1. Determine the integer solutions of the equation $p(x+y)=xy$, where $p$ is a given prime number.","p+1, y=p(p+1)$ and $x=p-1, y=p(1-p)$.",medium,"Solution. Note that the pair $(0,0)$ is a solution to the problem and that the pair $(x, y)$ in which exactly one of the numbers is equal to 0 is not a solution to the problem. Let $p(x+y)=x y \neq 0$. Then $p \mid x y$, and since $p$ is a prime number, it follows that $p \mid x$ or $p \mid y$. Let, for example, $x=m p$, where $m$ is an integer and $m \neq 0$. It is easy to see that in this case $m$ cannot be $1$. Further, $p(m p+y)=m p y$, so $y=\frac{m p}{m-1}$. 1) If $m-1=1$, then $x=2 p, y=2 p$ is a solution to the given equation. 2) If $m-1=-1$, then $m=0$, which contradicts the assumption. 3) If $m-1=p$, then $x=p(p+1), y=p+1$ is a solution to the given equation. 4) If $m-1=-p$, then $x=p(1-p), y=p-1$ is a solution to the given equation. 5) If $m-1 \notin\{-1,1,-p, p\}$, then $y=\frac{m p}{m-1}$ is not an integer. Similarly, the case when $y$ is divisible by $p$ is considered, and in this case, we obtain two more solutions: $x=p+1, y=p(p+1)$ and $x=p-1, y=p(1-p)$." 1b2cd9ec6d09,"2. Given $\sin 2 x=\frac{\sin \theta+\cos \theta}{2}$, $\cos ^{2} x=\sin \theta \cdot \cos \theta$. Then, the value of $\cos 2 x$ is ( ). (A) $\frac{-1 \pm \sqrt{33}}{8}$ (B) $\frac{-1+\sqrt{33}}{8}$ (C) $\frac{-1-\sqrt{33}}{8}$ (D) 0",\frac{-1-\sqrt{33}}{8}$.,easy,"2. C. Notice $$ \begin{array}{l} \sin ^{2} 2 x=\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}=\frac{1+2 \sin \theta \cdot \cos \theta}{4} \\ =\frac{1+2 \cos ^{2} x}{4}, \end{array} $$ then $1-\cos ^{2} 2 x=\frac{2+\cos 2 x}{4}$. Thus, $\cos 2 x=\frac{-1 \pm \sqrt{33}}{8}$. $$ \begin{array}{l} \text { Also, } \cos 2 x=2 \cos ^{2} x-1=2 \sin \theta \cdot \cos \theta-1 \\ =\sin 2 \theta-1<0, \end{array} $$ Therefore, $\cos 2 x=\frac{-1-\sqrt{33}}{8}$." a10490c47663,"15. Around a $10 \sqrt{2}$ meters $\times 10$ meters rectangular building, there is a large, lush grassland. At a point $5 \sqrt{2}$ meters in front of the midpoint of the longer wall of the building, a wooden stake is driven into the ground. A 20-meter-long rope is then used to tie a goat to the stake. The area of the grassland that the goat can reach (in square meters) is ( ). (A) $162.5 \pi+25$ (B) $250 \pi+25$ (C) $300 \pi+50$ (D) $325 \pi+50$ (E) None of the above answers are correct",10=$ $B F$. The area where the goat can eat grass includes: (1) $\triangle P A B$. (2) A quarter cir,medium,"15. (D) Let $A B C D$ be a building, and the stake $P$ is located outside the long side $A B$. Extend $P A$ to $E$ and $P B$ to $F$, it is easy to see that $A E=10=$ $B F$. The area where the goat can eat grass includes: (1) $\triangle P A B$. (2) A quarter circle with $P$ as the center, between the radii (length 20) $P E, P F$. (3) An eighth circle with $A$ as the center, between the radii (length 10) $A D, A E$. (4) An eighth circle with $B$ as the center, between the radii (length 10) $B C, B F$. Therefore, the total area is $50+300 \pi+\frac{25 \pi}{2}+\frac{25 \pi}{2}$." 84fa7500812e,"4. (5 points) When Xiaohong was 3 years old, her mother's age was the same as Xiaohong's age this year; when her mother is 78 years old, Xiaohong's age will be the same as her mother's age this year. How old is Xiaohong's mother this year.",Mom is 53 years old this year,medium,"53 【Analysis】Let the age difference between mom and Xiaohong be $x$ years, then according to “When Xiaohong was 3 years old, mom's age was the same as Xiaohong's age this year;” we get Xiaohong's age this year is: $x+3$ years; According to “When mom is 78 years old, Xiaohong's age is the same as mom's age this year,” we get Xiaohong's current age is: $78-x$ years; According to Xiaohong's age + age difference = mom's age, we can set up the equation to solve the problem. 【Solution】Solution: Let the age difference between mom and Xiaohong be $x$ years, then Xiaohong's current age is $x+3$ years, and mom's current age is $78-x$ years. According to the problem, we can set up the equation: $$ \begin{aligned} x+3+x & =78-x \\ 2 x+3 & =78-x \\ 2 x+x & =78-3 \\ 3 x & =75 \\ x & =25 \\ 78-25 & =53 \text { (years) } \end{aligned} $$ Answer: Mom is 53 years old this year. The answer is: 53." 92f25b56b031,"204. Divide the number 10 into 2 parts, the difference of which is 5.",10 ; 2 x=5$ and $x=2 \frac{1}{2}$.,easy,"204. Let the smaller part be $x$, the larger $5+x$. Then, by the condition $2 x+5=10 ; 2 x=5$ and $x=2 \frac{1}{2}$." bad11a1b6708,"8. Adam the Ant started at the left-hand end of a pole and crawled $\frac{2}{3}$ of its length. Benny the Beetle started at the right-hand end of the same pole and crawled $\frac{3}{4}$ of its length. What fraction of the length of the pole are Adam and Benny now apart? A $\frac{3}{8}$ B $\frac{1}{12}$ C $\frac{5}{7}$ D $\frac{1}{2}$ E $\frac{5}{12}$",\frac{5}{12}$.,easy,8. E Adam the Ant has crawled $\frac{2}{3}$ of the length of the pole and so is $\frac{1}{3}$ of the length of the pole from the right-hand end. Benny the Beetle has crawled $\frac{3}{4}$ of the length of the pole and so is $\frac{1}{4}$ of the length of the pole from the left-hand end. Hence the fraction of the length of the pole that Adam and Benny are apart is $\left(1-\frac{1}{3}-\frac{1}{4}\right)=\frac{5}{12}$. c28bdc1a8467,"Let $x0$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & y^{\prime}=\left(2 \arcsin \frac{2}{3 x+1}+\sqrt{9 x^{2}+6 x-3}\right)^{\prime}= \\ & =2 \cdot \frac{1}{\sqrt{1-\left(\frac{2}{3 x+1}\right)^{2}}} \cdot\left(-\frac{2}{(3 x+1)^{2}} \cdot 3\right)+\frac{1}{2 \sqrt{9 x^{2}+6 x-3}} \cdot(18 x+6)= \\ & =-\frac{2(3 x+1)}{\sqrt{(3 x+1)^{2}-2^{2}}} \cdot \frac{6}{(3 x+1)^{2}}+\frac{9 x+3}{\sqrt{9 x^{2}+6 x-3}}= \\ & =-\frac{12}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}+\frac{3(3 x+1)}{\sqrt{9 x^{2}+6 x-3}}= \\ & =\frac{-12+3(3 x+1)^{2}}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}=\frac{-12+27 x^{2}+18 x+3}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}= \\ & =\frac{27 x^{2}+18 x-9}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}=\frac{3\left(9 x^{2}+6 x-3\right)}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}= \\ & =\frac{3 \sqrt{9 x^{2}+6 x-3}}{3 x+1} \end{aligned} $$ Problem Kuznetsov Differentiation 14-15" 851930219ada,"3. Find the number of four-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3.",See reasoning trace,medium,"Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be 1, 4, 7; if the first is 4, the last can be 2, 5, 8; if the first is 6, the last can be 0, 3, (6 does not fit), 9; if the first is 8, the last can be 1, 4, 7. In total, there are 3 + 3 + 3 + 3 = 12 options for the first and last digits. For each of these options, there are 8 * 7 ways to choose the two middle digits. In total, 56 * 12 = 672 ways. ## Criteria. Errors. Numbers of the form 6 ** 6 are incorrectly counted - two points are deducted. Without explanation, it is stated that each first digit corresponds to three variants of the last - one point is deducted (this is not obvious, as it is only due to the impossibility of 6). Numbers starting with zero are incorrectly counted - two points are deducted. Arithmetic errors in the correct solution method - one point is deducted for each error. Advancements. If it is written that the number of pairs (second digit, third digit) is not 7 * 8, but 9 * 9 or 8 * 8, etc., the problem is considered unsolved and no more than 2 points are given." 42cbea3e268c,"5. Given the inequality $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant 9$ holds for any positive real numbers $x, y$. Then the minimum value of the positive real number $a$ is ( ). (A) 2 (B) 4 (C) 6 (D) 8",See reasoning trace,easy,"5.B. $$ (x+y)\left(\frac{1}{x}+\frac{a}{y}\right)=1+a+\frac{y}{x}+\frac{a x}{y} \text {. } $$ When $\frac{y}{x}=2$, $1+a+\frac{y}{x}+\frac{a x}{y}=3+\frac{3}{2} a$. From $3+\frac{3}{2} a \geqslant 9$, we get $a \geqslant 4$. $$ \text { Also, } 1+4+\frac{y}{x}+\frac{4 x}{y} \geqslant 5+2 \sqrt{\frac{y}{x} \cdot \frac{4 x}{y}}=9 \text {, } $$ Thus, the minimum value of $a$ is 4." bce25bb74569,"10.2. Determine all values of the real parameter $a$ for which the equation $$ \left|a x^{2}-6\right|=|2 a x|+|3 a| \text { has at least one real solution. } $$",$a \in\left(-\infty ;-\frac{3}{2}\right] \cup(0 ;+\infty)$,medium,"Solution: 1. Let $a=0$, then the equation has no solutions. 2. Let $a \neq 0$, then the equation is equivalent to $\left|x^{2}-\frac{6}{a}\right|=2|x|+3 \Leftrightarrow\left[\begin{array}{c}x^{2}-\frac{6}{a}=2|x|+3, \\ \frac{6}{a}-x^{2}=2|x|+3\end{array} \Leftrightarrow\left[\begin{array}{l}(|x|-1)^{2}=4+\frac{6}{a}, \\ (|x|+1)^{2}=\frac{6}{a}-2 .\end{array}\right.\right.$ We observe that $(|x|-1)^{2} \geq 0, \forall x \in R$, and $\operatorname{Im}(|x|-1)^{2}=[0 ;+\infty)$. We also observe that $(|x|+1)^{2} \geq 1, \forall x \in R$, and $\operatorname{Im}(|x|+1)^{2}=[1 ;+\infty)$. The given equation will have at least one real solution if the following conditions are satisfied: $\left[\begin{array}{l}4+\frac{6}{a} \geq 0, \\ \frac{6}{a}-2 \geq 1\end{array} \Leftrightarrow\left[\begin{array}{l}a \in\left(-\infty ;-\frac{3}{2}\right.\end{array}\right] \cup(0 ;+\infty), \Leftrightarrow a \in\left(-\infty ;-\frac{3}{2}\right] \cup(0 ;+\infty)\right.$. Answer: $a \in\left(-\infty ;-\frac{3}{2}\right] \cup(0 ;+\infty)$" 7c654bffaae3,"7. If $A$ is the sum of the absolute values of all roots of the equation $$ x=\sqrt{19}+\frac{91}{\sqrt{19}+\cdots \sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}} $$ Find $A^{2}$.","\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the onl",easy,"7. The fraction to the right of the equation can be simplified to $\frac{a x+b}{c x+d}$ $(a, b, c, d$ are real numbers), then the equation is a quadratic equation, with at most two roots. H $x=\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the only two roots of the original equation. Therefore, $A=\sqrt{383}$, i.e., $A^{2}=383$." 0913d968519b,"8. Given $a>0, b>0, a^{3}+b^{3}=1$. Then the range of $a+b$ is . $\qquad$",See reasoning trace,easy,"8. $(1, \sqrt[3]{4}]$. Let $u=a+b$. Then $u>0$, and $a=u-b(00 \end{array} \Leftrightarrow 1 \frac{999}{1000} $$ Substituting the expressions for $T_{n}$ and $T_{k}$, we get: $$ \frac{n \cdot \frac{r^{2}}{2} \sin \frac{2 \pi}{n}}{r^{2} \pi} > \frac{999}{1000} $$ Simplifying, we have: $$ \frac{n}{2 \pi} \sin \frac{2 \pi}{n} > \frac{999}{1000} $$ or $$ n \cdot \sin \frac{2 \pi}{n} > \frac{999}{500} \pi \approx 6.276717 $$ If $n=81$, then (1) cannot hold because $81 \cdot \sin \frac{2 \pi}{81}$ is an integer. Let $x=\frac{2 \pi}{n}$. It suffices to show that $\frac{\sin x}{x}$ is monotonically increasing as $x$ decreases (Fig. 2). This is because $\sin x < x$, and the graph of $\sin x$ is concave from below. Therefore, the line $O B_{2}$ intersects the segment $A_{1} B_{1}$ at an internal point $P$. Thus, if $x_{1} < x_{2}$, then: $$ \frac{A_{1} P}{O A_{1}} = \frac{A_{2} B_{2}}{O A_{2}} = \frac{\sin x_{2}}{x_{2}} $$ which is what we wanted to prove. Therefore, regular polygons with at least 82 sides approximate the area of their circumscribed circle with an error less than one thousandth. Based on the work of Kiss-Tóth Christian (Fazekas M. Főv. Gyak. Gimn., 8th grade) Note. It is not entirely obvious how to find the value $n=81$. With a calculator and some patience, we can reach this value relatively quickly, but with some background knowledge, we can find the threshold more quickly. It is known that if $x$ is a real number, then the sine of the angle whose radian measure is $x$ is the sum of the following infinite series: $$ x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots + \ldots $$ This is not really necessary here, but it helps to obtain upper and lower bounds (which become increasingly accurate) by including alternating terms. For all positive $x$, the following inequality holds: $$ \sin x > x - \frac{x^{3}}{3!} $$ Dividing by $x > 0$, we get: $$ \frac{\sin x}{x} > 1 - \frac{x^{2}}{6} $$ If we rewrite the inequality $n \cdot \sin \frac{2 \pi}{n} > \frac{999}{500} \pi$ as: $$ \frac{\sin \frac{2 \pi}{n}}{\frac{2 \pi}{n}} > 0.999 $$ then it certainly holds if: $$ 1 - \frac{\left(\frac{2 \pi}{n}\right)^{2}}{6} > 0.999 $$ Rearranging, we get: $$ n > \frac{2 \pi}{\sqrt{0.006}} \approx 81.12 $$ Therefore, (1) certainly holds if $n \geq 82$, and it can be quickly determined that (1) does not hold for $n=81$, either by using a calculator or by considering the error in the inequality (*), which is smaller than the absolute value of the next term, $\frac{x^{5}}{5!}$. ![](https://cdn.mathpix.com/cropped/2024_05_02_61aa8ebde2ec7a9dd7d9g-2.jpg?height=1290&width=1110&top_left_y=270&top_left_x=496)" 55f6a6c8da26,Joana wrote the numbers from 1 to 10000 on the blackboard and then erased all multiples of 7 and 11. What number remained in the 2008th position?,"27$ numbers in the $34^{\text {th }}$ line. Therefore, after erasing the multiples of 7 and 11 in th",medium,"Initially observe that, from 1 to 77, Joana erased 11 multiples of 7 and 7 multiples of 11. Since 77 is a multiple of 7 and 11, she erased $11+7-1=17$ numbers, leaving $77-17=60$ numbers. Now, grouping the first 10000 numbers into groups of 77 consecutive numbers, this reasoning applies to each of the lines below, that is, in each line 60 numbers remained. $$ \begin{array}{ccccc} \text { 1st line: } & 1, & 2, & \ldots, & 77 \\ 2^{\text {nd }} \text { line: } & 78, & 79, & \ldots, & 154 \\ 3^{\text {rd }} \text { line: } & 155, & 158, & \ldots, & 231 \end{array} $$ Since $2008=33 \times 60+28$, we know that among the first $33 \times 77=2541$ numbers, $33 \times 60=1980$ numbers remained without being erased. $$ \text { 33rd line: } \quad \ldots, \quad \ldots, \quad \ldots, \quad 2541 $$ We still need to count 28 numbers. Let's then examine the 34th line, which starts with 2542. Since the erased numbers are in columns $7,11,14,21,22,28,33,35$, etc., and up to the $35^{\text {th }}$ column, eight numbers have been erased, there remain $35-8=27$ numbers in the $34^{\text {th }}$ line. Therefore, after erasing the multiples of 7 and 11 in this line, the 28th number is 2577. Thus, the number in the 2008th position is 2577." 50beaa5f7eb0,"61.If a cyclist rides at a speed of 10 km/h, he will be 1 hour late. If he rides at a speed of 15 km/h, he will arrive 1 hour early. At what speed should he ride to arrive on time?",12 km/h,medium,"61.If there were two cyclists, with the first one's speed being $10 \mathrm{km} / \mathrm{h}$ and the second one's speed being $15 \mathrm{km} / \mathrm{h}$. Then, according to the problem, if the first cyclist started 2 hours earlier than the second, they would arrive at the destination simultaneously. In this case, in 2 hours, the first cyclist would cover 20 km, and the second cyclist would be able to catch up in 4 hours - already at the final destination. Therefore, the entire distance is 60 km. The question of the problem can be rephrased as: ""At what speed should a cyclist travel to cover the entire distance in 5 hours?"". Answer: 12 km/h." db54d1c8a98c,"10. One hundred rectangles are arranged edge-to-edge in a continuation of the pattern shown. Each rectangle measures $3 \mathrm{~cm}$ by $1 \mathrm{~cm}$. What is the perimeter, in $\mathrm{cm}$, of the completed shape? A 800 B 700 С 602 D 600 E 502","800$. When two rectangles are placed together at a ""join"" in the pattern, there is an overlap of len",easy,"10. C The perimeter of each rectangle is $2 \times(3+1) \mathrm{cm}=8 \mathrm{~cm}$. The total perimeter, in $\mathrm{cm}$, of the separate rectangles is therefore $100 \times 8=800$. When two rectangles are placed together at a ""join"" in the pattern, there is an overlap of length $1 \mathrm{~cm}$ and the total perimeter is reduced by $2 \mathrm{~cm}$. In the complete pattern there are 99 joins and hence its perimeter, in cm, is $800-99 \times 2=800-198=602$." b39e400f9ce1,"Let $ABC$ be a triangle with $AB = 26$, $BC = 51$, and $CA = 73$, and let $O$ be an arbitrary point in the interior of $\vartriangle ABC$. Lines $\ell_1$, $\ell_2$, and $\ell_3$ pass through $O$ and are parallel to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, respectively. The intersections of $\ell_1$, $\ell_2$, and $\ell_3$ and the sides of $\vartriangle ABC$ form a hexagon whose area is $A$. Compute the minimum value of $A$.",280,medium,"1. **Identify the problem and given values:** - We are given a triangle \( \triangle ABC \) with sides \( AB = 26 \), \( BC = 51 \), and \( CA = 73 \). - An arbitrary point \( O \) inside the triangle is chosen, and lines \( \ell_1 \), \( \ell_2 \), and \( \ell_3 \) pass through \( O \) and are parallel to \( \overline{AB} \), \( \overline{BC} \), and \( \overline{CA} \), respectively. - These lines intersect the sides of \( \triangle ABC \) forming a hexagon whose area we need to minimize. 2. **Determine the area of \( \triangle ABC \) using Heron's formula:** - First, calculate the semi-perimeter \( s \) of \( \triangle ABC \): \[ s = \frac{AB + BC + CA}{2} = \frac{26 + 51 + 73}{2} = 75 \] - Using Heron's formula, the area \( A_{ABC} \) is: \[ A_{ABC} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{75 \cdot (75-26) \cdot (75-51) \cdot (75-73)} \] \[ = \sqrt{75 \cdot 49 \cdot 24 \cdot 2} = \sqrt{176400} = 420 \] 3. **Analyze the hexagon formed by the parallel lines:** - Let \( \ell_1 \) intersect \( AC \) at \( D \) and \( BC \) at \( G \). - Let \( \ell_2 \) intersect \( AB \) at \( F \) and \( AC \) at \( I \). - Let \( \ell_3 \) intersect \( AB \) at \( E \) and \( BC \) at \( H \). - The hexagon \( DEFGHI \) is formed by these intersections. 4. **Use properties of similar triangles and parallel lines:** - The triangles \( \triangle EFO \), \( \triangle OGH \), and \( \triangle DOI \) are similar to \( \triangle ABC \) and their areas are proportional to the square of the ratios of corresponding sides. - By the properties of similar triangles and parallel lines, the sum of the areas of these smaller triangles is minimized when \( O \) is the centroid of \( \triangle ABC \). 5. **Calculate the minimum area of the hexagon:** - When \( O \) is the centroid, the area of the hexagon \( DEFGHI \) is \( \frac{2}{3} \) of the area of \( \triangle ABC \). - Therefore, the minimum area \( A \) of the hexagon is: \[ A = \frac{2}{3} \times 420 = 280 \] The final answer is \( \boxed{280} \)." 00ec211cdbeb,"$[$ A row of soldiers is called incorrect if no three consecutive soldiers stand in order of height (neither in ascending nor in descending order). How many incorrect rows can be formed from $n$ soldiers of different heights if a) $n=4$; b) $n=5$?",See reasoning trace,medium,"Let's place a ""+"" between neighboring soldiers if the right one is taller than the left one, and a ""-"" otherwise. According to the condition, the plus and minus signs in the incorrect rows should alternate. It is clear that the number of rows starting with a plus is equal to the number of rows starting with a minus. a) Let's find the number of rows of the type *+*_*+*. Denote the soldiers by the letters $A, B, C$ and $D$ in decreasing order of height. In the row of the specified type, $A$ can stand in the second or fourth position (after a plus). In the row ${ }^{*} A^{*+*}$, soldier $B$ can stand in the first or fourth position. In the first case, we have one row - $B A D C$, in the second case - two: $C$ and $D$ can be placed in the remaining positions arbitrarily. In the row *+*_*A, soldier $B$ can stand only in the second position. There are again two such rows. In total, we have 5 rows of the type *+*_*+*, and all incorrect rows are twice as many. b) Let's find the number of rows of the type *+*_*+*_*. Denote the soldiers by the letters $A, B, C, D$ and $E$ in decreasing order of height. Here, $A$ can stand in the second or fourth position. But these cases are symmetric, so it is enough to consider only the first: * $A^{*+*}-*$. In this case, $B$ can stand in the first or fourth position. In the first case ( $B A^{*+*}{ }^{*}$ ), $C$ can stand only in the fourth position, and $D$ and $E$ can be placed in the remaining positions arbitrarily (2 rows). In the second case $\left({ }^{*} A^{*} B^{*}\right)$, we have 6 rows: $C, D$ and $E$ can be placed arbitrarily. In total, we have 8 rows of the type * $A^{*+*-*}$, and all incorrect rows are four times as many. ## Answer a) 10 rows; b) 32 rows." 477c280084f1,"10. From any point $P$ on the parabola $y^{2}=2 x$, draw a perpendicular to its directrix $l$, with the foot of the perpendicular being $Q$. The line connecting the vertex $O$ and $P$ and the line connecting the focus $F$ and $Q$ intersect at point $R$. Then the equation of the locus of point $R$ is $\qquad$.","-2 x^{2}+x$. Hint: Let $P\left(x_{1}, y_{1}\right), R(x, y)$, then $Q\left(-\frac{1}{2}, y_{1}\right",easy,"10. $y^{2}=-2 x^{2}+x$. Hint: Let $P\left(x_{1}, y_{1}\right), R(x, y)$, then $Q\left(-\frac{1}{2}, y_{1}\right), F\left(\frac{1}{2}, 0\right)$, then the equation of $O P$ is: $y=\frac{y_{1}}{x_{1}} x, F Q$'s equation is: $y=-y_{1}\left(x-\frac{1}{2}\right)$, eliminate $x_{1}, y_{1}$ (substitution method) and you are done." 54399439896e,"6. The lengths of the sides of a triangle are expressed in meters by whole numbers. Determine them, if the triangle has a perimeter of $72 m$ and the longest side of the triangle is divided by the point of tangency of the inscribed circle in the ratio $3: 4$.",See reasoning trace,medium,"SOLUTION. We will use the general fact that the points of tangency of the inscribed circle divide the perimeter of the triangle into six segments, and that any two of them that emanate from the same vertex of the triangle are congruent. (Tangents from a given point to a given circle are symmetric with respect to the line connecting the given point to the center of the given circle.) In our problem, the longest side of the triangle is divided into segments whose lengths we denote by $3x$ and $4x$, while the lengths of the segments from the vertex opposite the longest side are denoted by $y$ (see Fig. 1). The sides of the triangle thus have lengths $7x$, $4x+y$, and $3x+y$, where $x$ and $y$ are unknown positive numbers (we take lengths without units). For $7x$ to be the length of the longest side, it must be true that $7x > 4x + y$ or $3x > y$. Let us emphasize that the sought numbers $x$ and $y$ do not necessarily have to be integers, but according to the problem statement, this is true for the numbers $7x$, $4x+y$, and $3x+y$. ![](https://cdn.mathpix.com/cropped/2024_04_17_93ab0272c174b6cbb42cg-2.jpg?height=457&width=688&top_left_y=982&top_left_x=684) Fig. 1 The information about the perimeter of the triangle can be written as the equation $$ 72 = 7x + (3x + y) + (4x + y) \text{ or } 36 = 7x + y $$ Since $7x$ is an integer, $y = 36 - 7x$ is also an integer; and since according to the problem statement, the numbers $4x + y$ and $3x + y$ are also integers, $x = (4x + y) - (3x + y)$ is an integer as well. Therefore, from this point on, we are looking for pairs of positive integers $x$ and $y$ for which $$ 3x > y \quad \text{and} \quad 7x + y = 36. $$ From this, it follows that $7x < 36 < 7x + 3x = 10x$, so $x \leq 5$ and simultaneously $x \geq 4$. For $x = 4$, we have $y = 8$ and $(7x, 4x + y, 3x + y) = (28, 24, 20)$; for $x = 5$, we have $y = 1$ and $(7x, 4x + y, 3x + y) = (35, 21, 16)$. The sides of the triangle are therefore $(28, 24, 20)$ or $(35, 21, 16)$. (The triangle inequalities are clearly satisfied.) GUIDING AND SUPPLEMENTARY PROBLEMS:" 2a285e8f66ae,"25. For $n \in \mathbf{N}^{*}$, let $f(n)$ denote the smallest positive integer such that: $n \mid \sum_{k=1}^{f(n)} k$. Find all $n \in \mathbf{N}^{*}$ such that $f(n)=2 n-1$.",2 n-1 \) if and only if \( n \) is a power of 2.,medium,"25. First, we prove: If \( n=2^{m}, m \in \mathbf{N} \), then \( f(n)=2 n-1 \). In fact, on one hand, \[ \sum_{k=1}^{2 n-1} k=(2 n-1) n=\left(2^{m+1}-1\right) 2^{m} \] is divisible by \( n \). On the other hand, if \( l \leqslant 2 n-2 \), then \[ \sum_{k=1}^{l} k=\frac{1}{2} l(l+1) \] Since one of \( l \) and \( l+1 \) is odd, and \[ l+1 \leqslant 2 n-1=2^{m+1}-1, \] the above sum cannot be divisible by \( 2^{m} \) (because \( 2^{m+1} \times l(l+1) \)). Next, we prove: When \( n \) is not a power of 2, \( f(n) < 2 n-1 \). Let \( n=2^{m} p \), where \( p \) is an odd number. We show that there exists \( l < 2 n-1 \) such that \( 2^{m+1} \mid l \) and \( p \mid (l+1) \) (in this case, of course, \( 2^{m} p \left\lvert\, \frac{l(l+1)}{2}\right. \), thus \( f(n) < 2 n-1 \)). Since \( \left(2^{m+1}, p\right)=1 \), by the Chinese Remainder Theorem, \[ l \equiv 0\left(\bmod 2^{m+1}\right), \quad l \equiv p-1(\bmod p), \] has a solution \( l \equiv x_{0}\left(\bmod 2^{m+1} p\right) \). Therefore, there exists \( l_{0}, 0 < l_{0} \leqslant 2^{m+1} p \) satisfying the above system of congruences. Note that \( 2 n-1 \neq 0\left(\bmod 2^{m+1}\right) \), and \( 2 n+1 \not \equiv 0(\bmod p) \), so \( 2 n-1 \) and \( 2 n \) are not solutions to this system of congruences, hence \( 0 < l_{0} < 2 n-1 \), i.e., \( f(n) < 2 n-1 \). In summary, \( f(n) = 2 n-1 \) if and only if \( n \) is a power of 2." 31683b61e2a1,"5. How many pairs of positive integers $(a, b)$ are there, both not exceeding 10 , such that $a \leq \operatorname{gcd}(a, b) \leq b$ and $a \leq \operatorname{lcm}(a, b) \leq b$ ?",$10+5+3+2+2+1+1+1+1+1=27$,easy,"Answer: 27 Solution: $\operatorname{gcd}(a, b) \geq a \Longrightarrow \operatorname{gcd}(a, b)=a$. Likewise, $\operatorname{lcm}(a, b)=b$. This can only happen if $b$ is a multiple of $a$. Hence the answer is $10+5+3+2+2+1+1+1+1+1=27$." bdefd2ffd8c5,"2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$.",See reasoning trace,medium,"2.1. Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$. Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the perpendiculars, respectively. Therefore, $$ \begin{array}{l} 2|M F|-|M P|=|M A|-|M P| \\ \leqslant|M A|-|M B|=|A B|, \end{array} $$ where $|A B|$ is the distance from point $P$ to the right directrix, which equals 1. Thus, when the y-coordinate of point $M$ is 1, the maximum value of $2|M F|-|M P|$ is 1." 02bded19fab6,"Positive integers m and n are both greater than 50, have a least common multiple equal to 480, and have a greatest common divisor equal to 12. Find m + n.",156,medium,"1. We start with the given information: - Least Common Multiple (LCM) of \( m \) and \( n \) is 480. - Greatest Common Divisor (GCD) of \( m \) and \( n \) is 12. - Both \( m \) and \( n \) are greater than 50. 2. We use the relationship between LCM and GCD: \[ \text{LCM}(m, n) \times \text{GCD}(m, n) = m \times n \] Substituting the given values: \[ 480 \times 12 = m \times n \] \[ 5760 = m \times n \] 3. We need to find pairs \((m, n)\) such that \( m \times n = 5760 \) and both \( m \) and \( n \) are greater than 50. Additionally, \( \text{GCD}(m, n) = 12 \). 4. We express \( m \) and \( n \) in terms of their GCD: \[ m = 12a \quad \text{and} \quad n = 12b \] where \( \text{GCD}(a, b) = 1 \). Substituting these into the product equation: \[ (12a) \times (12b) = 5760 \] \[ 144ab = 5760 \] \[ ab = \frac{5760}{144} \] \[ ab = 40 \] 5. We need to find pairs \((a, b)\) such that \( ab = 40 \) and \( \text{GCD}(a, b) = 1 \). The pairs \((a, b)\) that satisfy this are: \[ (1, 40), (2, 20), (4, 10), (5, 8) \] 6. We now convert these pairs back to \( m \) and \( n \): \[ (m, n) = (12a, 12b) \] - For \((1, 40)\): \((m, n) = (12 \times 1, 12 \times 40) = (12, 480)\) (not valid since both must be > 50) - For \((2, 20)\): \((m, n) = (12 \times 2, 12 \times 20) = (24, 240)\) (not valid since both must be > 50) - For \((4, 10)\): \((m, n) = (12 \times 4, 12 \times 10) = (48, 120)\) (not valid since both must be > 50) - For \((5, 8)\): \((m, n) = (12 \times 5, 12 \times 8) = (60, 96)\) (valid since both are > 50) 7. The only valid pair is \((60, 96)\). Therefore, the sum \( m + n \) is: \[ 60 + 96 = 156 \] The final answer is \(\boxed{156}\)." d10737062e12,"1. Find the sum of the numbers: $$ 6+66+666+6666+\cdots+\underbrace{66666 \ldots 6}_{2018 \text { of them }} $$",$\frac{2\left(10^{2019}-18172\right)}{27}$,medium,"Answer: $\frac{2\left(10^{2019}-18172\right)}{27}$. ## Solution: $$ \begin{gathered} 6+66+666+6666+\cdots+\underbrace{66666 \ldots 6}_{2018 \text { items }}=\frac{2}{3}(9+99+999+9999+\cdots+\underbrace{99999 \ldots 9}_{2018 \text { items }})= \\ =\frac{2}{3}\left(10-1+10^{2}-1+10^{3}-1+\cdots+10^{2018}-1\right)=\frac{2}{3}\left(\frac{10\left(10^{2018}-1\right)}{10-1}-2018\right)= \\ =\frac{2}{3}\left(\frac{10^{2019}-10-2018 \cdot 9}{9}\right)=\frac{2\left(10^{2019}-18172\right)}{27} \end{gathered} $$" 93b29e2ce33d,12. (2001 National High School Competition Question) The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is,"\frac{y^{2}-2}{2 y-3}$. It is easy to see that $x \geqslant 2$, so $x^{2}-3 x+2 \geqslant 0$ and $y=",medium,"12. Fill $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$. Reason: From $y=x+\sqrt{x-3 x+2}$, we have $\sqrt{x^{2}-3 x+2}=y-x \geqslant 0$. Squaring both sides of this equation gives $(2 y-3) x=y^{2}-2$. Thus, $y \neq \frac{3}{2}$ and $x=\frac{y^{2}-2}{2 y-3}$. From $y-x=y-\frac{y^{2}-2}{2 y-3} \geqslant 0$, we get $\frac{y^{2}-3 y+2}{2 y-3} \geqslant 0$, which implies $1 \leqslant y<\frac{3}{2}$ or $y \geqslant 2$. For any $y \geqslant 2$, let $x=\frac{y^{2}-2}{2 y-3}$. It is easy to see that $x \geqslant 2$, so $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$. For any $1 \leqslant y<\frac{3}{2}$, similarly let $x=\frac{y^{2}-2}{2 y-3}$. It is easy to see that $x \leqslant 1$. Thus, $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$. Therefore, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$." 225c4fa927b3,7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$.,0,medium,"Solution. Domain: $x^{3}+2 x+1>0$. From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as $\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$. Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{3}{40}$ (no solutions), or $5^{x}=5^{-1} \Leftrightarrow x_{1}=-1$ (does not satisfy the domain). From the second equation, we have $x^{3}+2 x+1=1 \Leftrightarrow x^{3}+2 x=0 \Leftrightarrow x\left(x^{2}+2\right)=0, x_{3}=0, x^{2}+2 \neq 0$. Answer: 0." ee976fc4af0f,$d(n)$ shows the number of positive integer divisors of positive integer $n$. For which positive integers $n$ one cannot find a positive integer $k$ such that $\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )$ is a perfect square.,n,medium,"To solve the problem, we need to determine for which positive integers \( n \) one cannot find a positive integer \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square. 1. **Understanding the function \( d(n) \)**: - The function \( d(n) \) represents the number of positive divisors of \( n \). - For example, \( d(6) = 4 \) because the divisors of 6 are 1, 2, 3, and 6. 2. **Initial observations**: - If \( n \) is a prime number \( p \), then \( d(p) = 2 \) because the only divisors are 1 and \( p \). - If \( n \) is a power of a prime, say \( n = p^k \), then \( d(p^k) = k + 1 \). 3. **Key insight**: - We need to find \( n \) such that no matter how many times we apply \( d \), we never get a perfect square. - A perfect square is an integer that can be expressed as \( m^2 \) for some integer \( m \). 4. **Inductive step**: - Suppose for \( n = 1, 2, \ldots, m-1 \), we can find a \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square. - We need to prove this for \( n = m \). 5. **Case analysis**: - If \( d(m) \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers, then we need to check if \( d(m) \) can be a perfect square. - If \( q > 2 \), then \( q-1 \) is even, and \( p^{q-1} \) is a perfect square. - If \( q = 2 \), then \( d(m) = p \), and we need to check if \( m \) can be written as \( r_1^{p-1} \) where \( r_1 \) and \( p \) are primes. 6. **Conclusion**: - If \( n \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers, then \( d(n) = q \), \( d(q) = 2 \), and \( d(2) = 2 \), which is not a perfect square. - Therefore, for such \( n \), we cannot find a \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square. The final answer is \( \boxed{ n } \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers." f9f50400f04f,"7. Given real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=3$. Then the maximum value of $a b c$ is $\qquad$ .",f\left(-\frac{1}{3}\right)=\frac{5}{27}$.,medium,"7. $\frac{5}{27}$. Notice that among $a^{2}, b^{2}, c^{2}$, there is always one that does not exceed 1, let's assume $c^{2} \leqslant 1$. Then $c \in[-1,1]$. $$ \begin{array}{l} \text { Also } a b c=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] c \\ =\frac{1}{2}\left[(1-c)^{2}-\left(3-c^{2}\right)\right] c=c^{3}-c^{2}-c . \\ \text { Let } f(x)=x^{3}-x^{2}-x(x \in[-1,1]) . \end{array} $$ Then by $f^{\prime}(x)=3 x^{2}-2 x-1=0$, we get $$ x_{1}=-\frac{1}{3}, x_{2}=1 \text {. } $$ It is easy to see that $f(x)_{\max }=f\left(-\frac{1}{3}\right)=\frac{5}{27}$." 85e1679fe25b,"33. Calculate the determinants of the third order: a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$ b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$",See reasoning trace,easy,"Solution. a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3 \cdot 5 \cdot 3+2 \cdot 3 \cdot 3+2 \cdot 4 \cdot 1-1 \cdot 5 \cdot 3-2 \cdot 2 \cdot 3-$ $-3 \cdot 3 \cdot 4=45+18+8-15-12-36=71-63=8$; b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=a c b+b a c+c b a-c \cdot c \cdot c-b \cdot b \cdot b-a \cdot a \cdot a=$ $=3 a b c-a^{3}-b^{3}-c^{3}$. 34-39. Calculate the determinants:" 505a7f66b506,"10. The volumes of seven cubes of different sizes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$, respectively. They are stacked vertically to form a tower, with the volume of the cubes decreasing from bottom to top. Except for the bottom cube, the bottom surface of each cube is completely on top of the cube below it. The surface area of the tower (including the bottom) is ( ). (A) 644 (B) 658 (C) 664 (D) 720 (E) 749",See reasoning trace,medium,"10. B. From the problem, we know that the edge lengths of each cube are 1, $2, 3, 4, 5, 6, 7$. Therefore, the required surface area is $$ 6 \sum_{n=1}^{7} n^{2}-2 \sum_{n=1}^{6} n^{2}=840-182=658 \text {. } $$" 8b3188cd2357,"8. Given a non-empty set $M$ satisfying $$ M \subseteq\{0,1, \cdots, n\}\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right) \text {. } $$ If there exists a non-negative integer $k(k \leqslant n)$, such that for any $a \in M$, we have $2 k-a \in M$, then the set $M$ is said to have property $P$. Let the number of sets $M$ with property $P$ be $f(n)$. Then the value of $f(9)-f(8)$ is $\qquad$",See reasoning trace,medium,"8. 31 . $$ \begin{array}{l} \text { When } n=2 \text {, } \\ M=\{0\},\{1\},\{2\},\{0,2\},\{0,1,2\} \end{array} $$ have property $P$, and the corresponding $k$ values are $0, 1, 2, 1, 1$. Thus, $f(2)=5$. When $n=k$, the number of sets $M$ with property $P$ is $f(t)$. Then, when $n=k+1$, we have $$ f(t+1)=f(t)+g(t+1), $$ where $g(t+1)$ represents the number of sets $M$ that have property $P$ and include $t+1$. Next, we calculate the expression for $g(t+1)$ in terms of $t$. At this point, $2 k \geqslant t+1$, i.e., $k \geqslant \frac{t+1}{2}$, so we discuss $n=t$ in odd and even cases. (1) When $t$ is even, $t+1$ is odd, at this time, $$ k \geqslant \frac{t+2}{2} \text {. } $$ Thus, for each $k$, $t+1$ and $2 k-t-1$ must belong to the set $M$, and $t$ and $2 k-t \cdots \cdots k$ and $k$ have $t+1-k$ pairs of numbers, each pair of numbers must either both belong to or not belong to the set $M$. Therefore, for each $k$, the number of sets $M$ with property $P$ is $$ \mathrm{C}_{t+1-k}^{0}+\mathrm{C}_{t+1-k}^{1}+\cdots+\mathrm{C}_{t+1-k}^{t+1-k}=2^{t+1-k} . $$ Then $g(t+1)=2^{\frac{1}{2}}+2^{\frac{t-2}{2}}+\cdots+2+1$ $$ =2 \times 2^{\frac{1}{2}}-1 \text {. } $$ (2) When $t$ is odd, $t+1$ is even, at this time, $$ k \geqslant \frac{t+1}{2} \text {. } $$ Similarly, $$ \begin{array}{l} g(t+1)=2^{\frac{t+1}{2}}+2^{\frac{t-1}{2}}+\cdots+2+1 \\ =2 \sqrt{2} \times 2^{\frac{t}{2}}-1 . \end{array} $$ Thus $f(t+1)=\left\{\begin{array}{cc}f(t)+2 \times 2^{\frac{1}{2}}-1, & t \text { is even; } \\ f(t)+2 \sqrt{2} \times 2^{\frac{1}{2}}-1, & t \text { is odd. }\end{array}\right.$ By the summation method, we get $$ f(n)=\left\{\begin{array}{ll} 6 \times 2^{\frac{n}{2}}-n-5, & n \text { is even; } \\ 4 \times 2^{\frac{n+1}{2}}-n-5, & n \text { is odd. } \end{array}\right. $$ Thus $f(9)-f(8)$ $$ =4 \times 2^{5}-9-5-\left(6 \times 2^{4}-8-5\right)=31 \text {. } $$" 5d70ea6b8be8,"4. Given real numbers $a, b$ satisfy $$ |a-1|+\sqrt{(a-2)(b-2)^{2}}+\left|b^{2}+1\right|=a \text {. } $$ Then $a^{b}=(\quad$. (A) $\frac{1}{4}$ (B) $\frac{1}{2}$ (C) 1 (D) 2","2, b=0$, the equality holds. Therefore, $a^{b}=1$.",easy,"4. C. From the given, we know $$ 1=|a-1|-(a-1)+\sqrt{(a-2)(b-2)^{2}}+\left|b^{2}+1\right| \geqslant 1 \text {, } $$ when $a=2, b=0$, the equality holds. Therefore, $a^{b}=1$." f20006cdd169,"Let $n$ be an integer bigger than $0$. Let $\mathbb{A}= ( a_1,a_2,...,a_n )$ be a set of real numbers. Find the number of functions $f:A \rightarrow A$ such that $f(f(x))-f(f(y)) \ge x-y$ for any $x,y \in \mathbb{A}$, with $x>y$.",\sum_{k=0,medium,"1. **Define the function \( g \):** Let \( g \) be the function defined by \( g(x) = f(f(x)) \). The given condition is \( f(f(x)) - f(f(y)) \ge x - y \) for any \( x, y \in \mathbb{A} \) with \( x > y \). 2. **Analyze the condition:** Since \( x > y \), we have \( x - y > 0 \). Therefore, \( f(f(x)) - f(f(y)) \ge x - y \) implies that \( g(x) = f(f(x)) \) is a strictly increasing function. 3. **Properties of \( g \):** Because \( \mathbb{A} \) is a finite set of real numbers, and \( g \) is strictly increasing, \( g \) must be a bijective function. This means \( g \) is a permutation of \( \mathbb{A} \). 4. **Identity function:** Since \( g \) is a bijective function and strictly increasing, and \( \mathbb{A} \) is finite, the only strictly increasing bijective function on a finite set is the identity function. Therefore, \( g(x) = x \) for all \( x \in \mathbb{A} \). 5. **Implication for \( f \):** If \( g(x) = x \), then \( f(f(x)) = x \) for all \( x \in \mathbb{A} \). This means \( f \) must be an involution, i.e., \( f \) is a permutation of \( \mathbb{A} \) with order 2 (since applying \( f \) twice returns the original element). 6. **Counting such permutations:** We need to count the number of permutations of \( \mathbb{A} \) that are involutions. An involution is a permutation where every element is either a fixed point or part of a 2-cycle. The number of such permutations can be calculated using the formula: \[ \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{2^k k! (n-2k)!} \] Here, \( k \) represents the number of 2-cycles, and \( \lfloor n/2 \rfloor \) is the greatest integer less than or equal to \( n/2 \). The final answer is \( \boxed{ \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{2^k k! (n-2k)!} } \)" f01f81094b1e,"3. Master Jure is installing central heating, and he charges by taking half the cost of the materials used and adding 170 kuna per day of work. How much did Mr. Jurić pay Master Jure, and how much was the material if the total bill was 3348 kuna, and the job was completed in 3 days?",See reasoning trace,medium,"3. The master worked for 3 days, and the value of his daily wages is 510 kuna, while the material and his work excluding daily wages are worth 2838 kuna: 2 points If the master's work is worth half the price of the material, then the price of the material is twice the price of the master. Let the master Jure earn $\square$ (or $x$) kuna. Then the price of the material is $\square \square$ (or $2x$) kuna, and together this amounts to $\square \square \square$ (or 3x) kuna. By dividing 2838:3, we get that $\square$ (or $x$) is 946 kuna. 4 points Including the daily wages, the master Jure earned $510+946=1456$ kuna. 2 points The price of the material is 2838 - 946 or 946 $\cdot$ 2, i.e., 1892 kuna. 2 points .$T O T A L 10$ POINTS" c08e43ab8af4,"Which are the two consecutive odd numbers in the number sequence, the sum of whose squares is of the form $\frac{n(n+1)}{2}$, where $n$ is a natural number.",See reasoning trace,medium,"Let the two adjacent odd numbers be $2k-1$ and $2k+1$, then $$ (2k-1)^{2}+(2k+1)^{2}=\frac{n(n+1)}{2} $$ which means $$ 16k^{2}+4=n(n+1) $$ Multiplying this by 4 and adding 1, $$ 64k^{2}+17=(2n+1)^{2} $$ Denoting $(2n+1)$ by $m$, we get $$ m^{2}-64k^{2}=(m+8k)(m-8k)=17 $$ Here, by assumption, $m$ is positive, so the first factor on the left side is also positive, and thus the second factor must also be positive and smaller than the first factor. Since 17 is a prime number, the only possible factorization is $$ m+8k=17, \quad m-8k=1, \quad \text{and thus} \quad 16k=17-1=16 $$ From this, $$ k=1, \quad m=2n+1=9, \quad \text{and thus} \quad n=4 $$ Indeed, $$ 1^{2}+3^{2}=\frac{4 \cdot 5}{2} $$" 9cb59283fa43,"## Task 2 864 Komsomol members came to the GDR. Half of them participated in a Subbotnik for Chilean patriots. The other half helped the FDJ members in setting up playgrounds and green spaces. The next day, the Soviet friends traveled in three equally strong groups to different districts of the GDR. a) How many Komsomol members participated in the Subbotnik? b) How many Komsomol members were in each of the three groups?",See reasoning trace,easy,"a) $864: 2=432$. At the Subbotnik, 432 Komsomol members participated. b) $864: 3=288$. In each of the three groups, 288 Komsomol members belonged. ### 13.5 14th Olympiad 1976 ### 13.5.1 1st Round 1976, Grade 4" 6e94c5ef6d8d,"### 6.345 Find the coefficients $p$ and $q$ of the equation $x^{4}-10 x^{3}+37 x^{2}+p x+q=0$, if among its roots there are two pairs of equal numbers.","$p=-60, q=36$",medium,"Solution. Let $x_{1}=x_{2}, x_{3}=x_{4}$ be the roots of the original equation. By Vieta's theorem, we have: $\left\{\begin{array}{l}x_{1}+x_{3}=5, \\ x_{1}^{2}+4 x_{1} x_{3}+x_{3}^{2}=37, \\ 2 x_{1} x_{3}\left(x_{1}+x_{3}\right)=-\dot{p}, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}+x_{3}=5, \\ \left(x_{1}+x_{3}\right)^{2}+2 x_{1} x_{3}=37, \\ 2 x_{1} x_{3}\left(x_{1}+x_{3}\right)=-p, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1} x_{3}=6, \\ 10 x_{1} x_{3}=-p, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Rightarrow\right.\right.\right.$ $\Rightarrow p=-60, q=36$. Answer: $p=-60, q=36$." 1e3b1c28f180,"1. When $2 x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by 10, and when $5 x^{2}$ was subtracted from it, its maximum value decreased by $\frac{15}{2}$. How will the maximum value of $f(x)$ change if $3 x^{2}$ is added to it?",. It will increase by $\frac{45}{2}$,medium,"Answer. It will increase by $\frac{45}{2}$. Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its maximum value, its leading coefficient is negative, and the maximum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \cdot \frac{b^{2}}{4 a^{2}}-\frac{b^{2}}{2 a}+c=-\frac{b^{2}}{4 a}+c$. If we add $2 x^{2}$ to $f(x)$, we get the quadratic trinomial $(a+2) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a+8}+c$. If we subtract $5 x^{2}$ from $f(x)$, we get the quadratic trinomial $(a-5) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a-20}+c$. From this, we get the equations $$ \left\{\begin{array} { l } { - \frac { b ^ { 2 } } { 4 a + 8 } + c = - \frac { b ^ { 2 } } { 4 a } + c + 1 0 , } \\ { - \frac { b ^ { 2 } } { 4 a - 2 0 } + c = - \frac { b ^ { 2 } } { 4 a } + c - \frac { 1 5 } { 2 } } \end{array} \Leftrightarrow \left\{\begin{array} { l } { \frac { b ^ { 2 } } { 4 a } - \frac { b ^ { 2 } } { 4 a + 8 } = 1 0 , } \\ { \frac { b ^ { 2 } } { 4 a - 2 0 } - \frac { b ^ { 2 } } { 4 a } = \frac { 1 5 } { 2 } } \end{array} \Leftrightarrow \left\{\begin{array}{l} \frac{2 b^{2}}{4 a(a+2)}=10 \\ \frac{5 b^{2}}{4 a(a-5)}=\frac{15}{2} \end{array}\right.\right.\right. $$ By dividing one equation of the last system by the other, we get $\frac{2(a-5)}{5(a+2)}=\frac{20}{15}$, from which $a=-5$. Then $b^{2}=300$, and the maximum value of $f(x)$ is $-\frac{300}{-20}+c=15+c$. If we add $3 x^{2}$ to the quadratic trinomial $f(x)$, we get the function $(a+3) x^{2}+b x+c$, the maximum value of which is $-\frac{b^{2}}{4 a+12}+c=-\frac{300}{-8}+c=\frac{75}{2}+c$, which is $\frac{45}{2}$ more than the maximum of the original function." 68ecdac82d11,"There are $64$ booths around a circular table and on each one there is a chip. The chips and the corresponding booths are numbered $1$ to $64$ in this order. At the center of the table there are $1996$ light bulbs which are all turned off. Every minute the chips move simultaneously in a circular way (following the numbering sense) as follows: chip $1$ moves one booth, chip $2$ moves two booths, etc., so that more than one chip can be in the same booth. At any minute, for each chip sharing a booth with chip $1$ a bulb is lit. Where is chip $1$ on the first minute in which all bulbs are lit?",64,medium,"1. **Determine the position of each chip at minute \( m \):** - Each chip \( x \) moves \( x \) booths every minute. Therefore, at minute \( m \), chip \( x \) will be at booth \( mx \mod 64 \). - Specifically, chip 1 will be at booth \( m \mod 64 \). 2. **Condition for lighting bulbs:** - A bulb is lit if chip 1 shares a booth with any other chip. Therefore, we need to find \( m \) such that for every chip \( x \) (where \( x \neq 1 \)), there exists a booth \( k \) such that \( mx \equiv m \mod 64 \). 3. **Simplify the condition:** - The condition \( mx \equiv m \mod 64 \) can be rewritten as \( m(x - 1) \equiv 0 \mod 64 \). - This implies that \( m \) must be a multiple of \( \frac{64}{\gcd(x-1, 64)} \) for all \( x \). 4. **Finding the smallest \( m \):** - We need \( m \) to be a common multiple of \( \frac{64}{\gcd(x-1, 64)} \) for all \( x \) from 2 to 64. - The smallest \( m \) that satisfies this condition is the least common multiple (LCM) of these values. 5. **Calculate the LCM:** - We need to find the LCM of \( \frac{64}{\gcd(x-1, 64)} \) for \( x = 2, 3, \ldots, 64 \). - Notice that \( \gcd(x-1, 64) \) can take values that are divisors of 64. The possible values of \( \frac{64}{\gcd(x-1, 64)} \) are the divisors of 64. 6. **Determine the LCM of the divisors of 64:** - The divisors of 64 are \( 1, 2, 4, 8, 16, 32, 64 \). - The LCM of these divisors is 64 itself. 7. **Verify the solution:** - Since \( m = 64 \) is a multiple of all \( \frac{64}{\gcd(x-1, 64)} \), it satisfies the condition \( m(x-1) \equiv 0 \mod 64 \) for all \( x \). - Therefore, at minute 64, chip 1 will be at booth \( 64 \mod 64 = 0 \), which is equivalent to booth 64. 8. **Conclusion:** - Chip 1 will be at booth 64 at the first minute in which all bulbs are lit. The final answer is \( \boxed{64} \)." c84b69b2a2d3,"8. Let the arithmetic sequence $\left\{a_{n}\right\}$ have all terms as integers, with the first term $a_{1}=2019$, and for any positive integer $n$, there always exists a positive integer $m$ such that $a_{1}+a_{2}+\cdots+a_{n}=a_{m}$. The number of such sequences $\left\{a_{n}\right\}$ is $\qquad$","2019 = 3 \times 673 \Rightarrow k-2 = -1, 1, 3, 673, 2019$, i.e., $d = -2019, 1, 3, 673, 2019$. Henc",medium,"Let $\{a_n\}$ be an arithmetic sequence with a common difference $d$, and according to the problem, $d \in \mathbf{Z}$. There exists a positive integer $k$ such that $a_1 + a_2 = a_k \Rightarrow 2a_1 + d = a_1 + (k-1)d \Rightarrow d = \frac{a_1}{k-2}$. Thus, $a_1 + a_2 + \cdots + a_n = n a_1 + \frac{n(n-1)}{2} d = a_1 + (n-1) a_1 + \frac{n(n-1)}{2} d$ $= a_1 + (n-1)(k-2) d + \frac{n(n-1)}{2} d = a_1 + \left[(n-1)(k-2) + \frac{n(n-1)}{2}\right] d$, which implies there exists a positive integer $m = (n-1)(k-2) + \frac{n(n-1)}{2} + 1$ that satisfies the condition. Therefore, we only need $(k-2) \mid a_1 = 2019 = 3 \times 673 \Rightarrow k-2 = -1, 1, 3, 673, 2019$, i.e., $d = -2019, 1, 3, 673, 2019$. Hence, there are 5 sequences $\{a_n\}$ that satisfy the condition." 6b0f6b9a667c,"In the given operation, the letters $a, b$, and $c$ represent distinct digits and are different from 1. Determine the values of $a, b$, and $c$. $$ \begin{array}{r} a b b \\ \times \quad c \\ \hline b c b 1 \end{array} $$","5, b=3$ and $c=7$.",medium,"Since the product of $b$ by $c$ ends in 1, then $b \times c$ can be 21 or 81, and therefore $3 \times 7$ or $9 \times 9$. The only possibility to write the product of two distinct numbers less than 10 is $21=3 \times 7$. Thus, we have only two possible cases. 1st Case: If $b=7$ and $c=3$, we should have $$ \begin{array}{r} a 77 \\ \times \quad 3 \\ \hline 7371 \end{array} $$ but this is impossible, since $\frac{7371}{3}=2457$ has four digits. 2nd Case: If $b=3$ and $c=7$, we have and, since $\frac{3731}{7}=533$, necessarily $a=5$. Therefore, the only possibility is $a=5, b=3$ and $c=7$." 6209141519d7,"Which of the following integers cannot be written as the sum of four consecutive odd integers? $\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$",\textbf{(D),medium,"Solution 1 Let our $4$ numbers be $n, n+2, n+4, n+6$, where $n$ is odd. Then, our sum is $4n+12$. The only answer choice that cannot be written as $4n+12$, where $n$ is odd, is $\boxed{\textbf{(D)}\text{ 100}}$. Solution 2 If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$; then, the sum is $8n$. All the integers are divisible by $8$ except $\boxed{\textbf{(D)}~100}$. Solution 3 If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$, the sum is $4a+12$, and $4a+12$ divided by $4$ gives $a+3$. This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$, so the answer is $\boxed{\textbf{(D)}~100}$. Solution 4 From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$. Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$. Since $n \equiv 1 \pmod{2}$, we can express it as the equation $n = 2a + 1$ for some integer $a$. Multiplying 4 to each side of the equation yields $4n = 8a + 4$, and taking mod 8 gets us $4n \equiv 4 \pmod{8}$, so $b = 0$. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\boxed{\textbf{(D)}~100}$. Solution 5 Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \[16=1+3+5+7\] \[40=7+9+11+13\] \[72=15+17+19+21\] \[200=47+49+51+53\] All of the answer choices can be a sum of consecutive odd numbers except $100$, so the answer is $\boxed{\textbf{(D)} 100}$." 3186d8920cf4,"## Task Condition Find the derivative. $y=\operatorname{arctg} \frac{\operatorname{tg} \frac{x}{2}+1}{2}$",See reasoning trace,medium,"## Solution $y^{\prime}=\left(\operatorname{arctg} \frac{\operatorname{tg} \frac{x}{2}+1}{2}\right)^{\prime}=\frac{1}{1+\left(\frac{\operatorname{tg}(x / 2)+1}{2}\right)^{2}} \cdot\left(\frac{\operatorname{tg} \frac{x}{2}+1}{2}\right)^{\prime}=$ $=\frac{4}{\operatorname{tg}^{2} \frac{x}{2}+2 \operatorname{tg} \frac{x}{2}+5} \cdot \frac{1}{2} \cdot \frac{1}{\cos ^{2} \frac{x}{2}} \cdot \frac{1}{2}=\frac{1}{\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}+5 \cos ^{2} \frac{x}{2}}=$ $=\frac{1}{1+\sin x+4 \cos ^{2} \frac{x}{2}}=\frac{1}{1+\sin x+2 \cos x+2}=\frac{1}{\sin x+2 \cos x+3}$ ## Problem Kuznetsov Differentiation 10-31" ac8112875977,"Example 5 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=a_{n}+$ $\sqrt{a_{n}}+\frac{1}{4}$. Then $a_{99}=(\quad$. (A) $2550 \frac{1}{4}$ (B) 2500 (C) $2450 \frac{1}{4}$ (D) 2401","\left(\frac{99+1}{2}\right)^{2}=2500$. Hence, the correct choice is (B).",easy,"Solution: From the recursive formula, we have $a_{n+1}=\left(\sqrt{a_{n}}+\frac{1}{2}\right)^{2}$, which means $\sqrt{a_{n+1}}=\sqrt{a_{n}}+\frac{1}{2}$. Therefore, $\left\{\sqrt{a_{n}}\right\}$ is an arithmetic sequence. Thus, we have $$ \sqrt{a_{n}}=\sqrt{a_{1}}+(n-1) \times \frac{1}{2}=\frac{n+1}{2} \text {. } $$ Therefore, $a_{99}=\left(\frac{99+1}{2}\right)^{2}=2500$. Hence, the correct choice is (B)." a577fcd049b3,"## Task 1 - 221231 All non-negative integer solutions of the system of equations $$ \begin{aligned} x_{1}+11 x_{2}+21 x_{3}+31 x_{4}+41 x_{5} & =55 \\ 2 x_{1}+12 x_{2}+22 x_{3}+32 x_{4}+42 x_{5} & =60 \\ 3 x_{1}+13 x_{2}+23 x_{3}+33 x_{4}+43 x_{5} & =65 \\ 4 x_{1}+14 x_{2}+24 x_{3}+34 x_{4}+44 x_{5} & =70 \\ 5 x_{1}+15 x_{2}+25 x_{3}+35 x_{4}+45 x_{5} & =75 \end{aligned} $$ are to be determined.",See reasoning trace,medium,"The above system of equations is obviously equivalent to the simpler one $$ \begin{aligned} x_{1}+11 x_{2}+21 x_{3}+31 x_{4}+41 x_{5} & =55 \\ x_{1}+x_{2}+x_{3}+x_{4}+x_{5} & =5 \end{aligned} $$ since (2) can be derived by simple subtraction from the first two equations, and conversely, all equations of the original system can be obtained by adding a suitable multiple of (2) to (1). Subtracting (2) from (1) and dividing by 10, we immediately obtain the additional equation $$ x_{2}+2 x_{3}+3 x_{4}+4 x_{5}=5 $$ We now distinguish the following three mutually exclusive cases: 1. Case: $x_{5}=1$. From (3), it immediately follows that $$ x_{2}=1, x_{3}=x_{4}=0 $$ which, together with (2), yields $x_{1}=3$, which is indeed a solution to our system of equations. 2. Case: $x_{5}=0, x_{4}=1$ Here, (3) gives rise to the two subcases: a) $x_{2}=0, x_{3}=1$, and $x_{1}=3$ from (2) b) $x_{2}=2, x_{3}=0$, and $x_{1}=2$ from (2) both of which also solve (1), and are thus solutions. 3. Case: $x_{4}=x_{5}=0$ Here, using (3) again, we get three subcases, namely a) $x_{2}=1, x_{3}=2$, and $x_{1}=2$ from (2) b) $x_{2}=3, x_{3}=1$, and $x_{1}=1$ from (2) c) $x_{2}=5, x_{3}=0$, and $x_{1}=0$ from (2) all of which also solve (1) and are thus solutions. In summary, the following 6 solutions $$ \left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \in\{(0,5,0,0,0),(1,3,1,0,0),(2,1,2,0,0),(2,2,0,1,0),(3,0,1,1,0),(3,1,0,0,1)\} $$ of the given system of equations have been obtained." 68c07e04f548,"3. Let the function $y=(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right)(x \in[0,1])$, then the minimum value of $y$ is $\qquad$","\frac{\sqrt{2}}{2}$, $y_{\text {min }}=2\left(\frac{\sqrt{2}}{2}+1\right)=\sqrt{2}+2$.",medium,"Let $x=\sin \theta, \theta \in\left[0, \frac{\pi}{2}\right]$, then $\frac{\theta}{2} \in\left[0, \frac{\pi}{4}\right] \Rightarrow \cos \frac{\theta}{2}>\sin \frac{\theta}{2}$. Thus, $y=(\sqrt{1+\sin \theta}+\sqrt{1-\sin \theta}+2)\left(\sqrt{1-\sin ^{2} \theta}+1\right)$ $=\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}+\cos \frac{\theta}{2}-\sin \frac{\theta}{2}+2\right)(\cos \theta+1)=4 \cos ^{2} \frac{\theta}{2}\left(\cos \frac{\theta}{2}+1\right)$. When $\cos \frac{\theta}{2}=\frac{\sqrt{2}}{2}$, $y_{\text {min }}=2\left(\frac{\sqrt{2}}{2}+1\right)=\sqrt{2}+2$." f15dd645e4a2,"Let $x$ and $y$ be real numbers such that $\frac{\sin{x}}{\sin{y}} = 3$ and $\frac{\cos{x}}{\cos{y}} = \frac{1}{2}$. The value of $\frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.",107,medium,"1. Given the equations: \[ \frac{\sin{x}}{\sin{y}} = 3 \quad \text{and} \quad \frac{\cos{x}}{\cos{y}} = \frac{1}{2} \] We can express \(\sin{x}\) and \(\cos{x}\) in terms of \(\sin{y}\) and \(\cos{y}\): \[ \sin{x} = 3 \sin{y} \quad \text{and} \quad \cos{x} = \frac{1}{2} \cos{y} \] 2. We need to find the value of \(\frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}}\). Using the double-angle identities: \[ \sin{2x} = 2 \sin{x} \cos{x} \quad \text{and} \quad \sin{2y} = 2 \sin{y} \cos{y} \] \[ \frac{\sin{2x}}{\sin{2y}} = \frac{2 \sin{x} \cos{x}}{2 \sin{y} \cos{y}} = \frac{\sin{x} \cos{x}}{\sin{y} \cos{y}} \] Substituting \(\sin{x} = 3 \sin{y}\) and \(\cos{x} = \frac{1}{2} \cos{y}\): \[ \frac{\sin{2x}}{\sin{2y}} = \frac{3 \sin{y} \cdot \frac{1}{2} \cos{y}}{\sin{y} \cos{y}} = \frac{3}{2} \] 3. Next, we use the double-angle identity for cosine: \[ \cos{2x} = 2 \cos^2{x} - 1 \quad \text{and} \quad \cos{2y} = 2 \cos^2{y} - 1 \] We need to find \(\frac{\cos{2x}}{\cos{2y}}\): \[ \frac{\cos{2x}}{\cos{2y}} = \frac{2 \cos^2{x} - 1}{2 \cos^2{y} - 1} \] Substituting \(\cos{x} = \frac{1}{2} \cos{y}\): \[ \cos^2{x} = \left(\frac{1}{2} \cos{y}\right)^2 = \frac{1}{4} \cos^2{y} \] Therefore: \[ \cos{2x} = 2 \left(\frac{1}{4} \cos^2{y}\right) - 1 = \frac{1}{2} \cos^2{y} - 1 \] \[ \cos{2y} = 2 \cos^2{y} - 1 \] Thus: \[ \frac{\cos{2x}}{\cos{2y}} = \frac{\frac{1}{2} \cos^2{y} - 1}{2 \cos^2{y} - 1} \] 4. To simplify \(\frac{\cos{2x}}{\cos{2y}}\), we need to find \(\cos^2{y}\). Using the Pythagorean identity: \[ \sin^2{x} + \cos^2{x} = 1 \] Substituting \(\sin{x} = 3 \sin{y}\) and \(\cos{x} = \frac{1}{2} \cos{y}\): \[ (3 \sin{y})^2 + \left(\frac{1}{2} \cos{y}\right)^2 = 1 \] \[ 9 \sin^2{y} + \frac{1}{4} \cos^2{y} = 1 \] Using \(\sin^2{y} + \cos^2{y} = 1\): \[ 9 (1 - \cos^2{y}) + \frac{1}{4} \cos^2{y} = 1 \] \[ 9 - 9 \cos^2{y} + \frac{1}{4} \cos^2{y} = 1 \] \[ 9 - 1 = 9 \cos^2{y} - \frac{1}{4} \cos^2{y} \] \[ 8 = \frac{35}{4} \cos^2{y} \] \[ \cos^2{y} = \frac{32}{35} \] 5. Substituting \(\cos^2{y} = \frac{32}{35}\) into the expression for \(\frac{\cos{2x}}{\cos{2y}}\): \[ \cos{2x} = \frac{1}{2} \cdot \frac{32}{35} - 1 = \frac{16}{35} - 1 = \frac{16 - 35}{35} = \frac{-19}{35} \] \[ \cos{2y} = 2 \cdot \frac{32}{35} - 1 = \frac{64}{35} - 1 = \frac{64 - 35}{35} = \frac{29}{35} \] \[ \frac{\cos{2x}}{\cos{2y}} = \frac{\frac{-19}{35}}{\frac{29}{35}} = \frac{-19}{29} \] 6. Finally, we combine the results: \[ \frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}} = \frac{3}{2} + \frac{-19}{29} \] Converting to a common denominator: \[ \frac{3}{2} = \frac{3 \cdot 29}{2 \cdot 29} = \frac{87}{58} \] \[ \frac{-19}{29} = \frac{-19 \cdot 2}{29 \cdot 2} = \frac{-38}{58} \] \[ \frac{87}{58} + \frac{-38}{58} = \frac{87 - 38}{58} = \frac{49}{58} \] The final answer is \(49 + 58 = \boxed{107}\)" 3d5261fa9895,"Example 1 Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people.",See reasoning trace,medium,"Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people. From the equality of the total number of soldiers, we get the equation $$ 1001 x + 1 = 1002 y + 4 \text{. } $$ Rearranging gives $$ 1001(x - y) - 3 = y \text{. } $$ Since $x$ and $y$ are both positive integers, the smallest value for $x - y$ is 1, at which point, $y = 998$. Thus, the minimum number of soldiers in this troop count is $$ 998 \times 1002 + 4 = 1000000 \text{ (people). } $$" 5fdfa62ca169,"4. Find all roots of the equation $1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$ $+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$ In the Answer, indicate the sum of the found roots.",$\{21\}$,easy,"4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$" 25b1ef025757,"1. Given the polynomial $$ P(x)=x^{2019}+2018 x^{2017}+2016 x+2015 $$ Find all integers $n$ such that $$ P(n) \mid P(P(n)+1) $$","-1-2018-2016+2015=-2020 \nmid 6050, P(0)=2015 \nmid 6050$ and $P(1)=6050$. Therefore, the only solut",medium,"1. For every integer $n$ we have: $$ P(P(n)+1) \equiv P(1)=1+2018+2016+2015=6050 \quad(\bmod P(n)) $$ Therefore, to satisfy the given condition, it must hold that $P(n) \mid 6050$. However, note that for $n \geqslant 2$ we have $P(n)>2^{2019}>6050$, and for $n \leqslant-2$ we have $P(n)<-2^{2019}<-6050$. Thus, the only possibilities are $n \in\{-1,0,1\}$. We directly compute $P(-1)=-1-2018-2016+2015=-2020 \nmid 6050, P(0)=2015 \nmid 6050$ and $P(1)=6050$. Therefore, the only solution is $n=1$." 8484e7c95ffd,"C4. On a flat plane in Camelot, King Arthur builds a labyrinth $\mathfrak{L}$ consisting of $n$ walls, each of which is an infinite straight line. No two walls are parallel, and no three walls have a common point. Merlin then paints one side of each wall entirely red and the other side entirely blue. At the intersection of two walls there are four comers: two diagonally opposite corners where a red side and a blue side meet, one corner where two red sides meet, and one corner where two blue sides meet. At each such intersection, there is a two-way door connecting the two diagonally opposite corners at which sides of different colours meet. After Merlin paints the walls, Morgana then places some knights in the labyrinth. The knights can walk through doors, but cannot walk through walls. Let $k(\mathfrak{L})$ be the largest number $k$ such that, no matter how Merlin paints the labyrinth $\mathfrak{L}$, Morgana can always place at least $k$ knights such that no two of them can ever meet. For each $n$, what are all possible values for $k(\mathfrak{L})$, where $\mathfrak{L}$ is a labyrinth with $n$ walls?","The only possible value of $k$ is $k=n+1$, no matter what shape the labyrinth is",medium,"Answer: The only possible value of $k$ is $k=n+1$, no matter what shape the labyrinth is. Solution 1. First we show by induction that the $n$ walls divide the plane into $\binom{n+1}{2}+1$ regions. The claim is true for $n=0$ as, when there are no walls, the plane forms a single region. When placing the $n^{\text {th }}$ wall, it intersects each of the $n-1$ other walls exactly once and hence splits each of $n$ of the regions formed by those other walls into two regions. By the induction hypothesis, this yields $\left(\binom{n}{2}+1\right)+n-\binom{n+1}{2}+1$ regions, proving the claim. Now let $G$ be the graph with vertices given by the $\binom{n+1}{2}+1$ regions, and with two regions connected by an edge if there is a door between them. We now show that no matter how Merlin paints the $n$ walls, Morgana can place at least $n+1$ knights. No matter how the walls are painted, there are exactly $\binom{n}{2}$ intersection points, each of which corresponds to a single edge in $G$. Consider adding the edges of $G$ sequentially and note that each edge reduces the number of connected components by at most one. Therefore the number of connected components of $\mathrm{G}$ is at least $\binom{n+1}{2}+1-\binom{n}{2}-n+1$. If Morgana places a knight in regions corresponding to different connected components of $G$, then no two knights can ever meet. Now we give a construction showing that, no matter what shape the labyrinth is, Merlin can colour it such that there are exactly $n+1$ connected components, allowing Morgana to place at most $n+1$ knights. First, we choose a coordinate system on the labyrinth so that none of the walls run due north-south, or due east-west. We then have Merlin paint the west face of each wall red, and the east face of each wall blue. We label the regions according to how many walls the region is on the east side of: the labels are integers between 0 and $n$. We claim that, for each $i$, the regions labelled $i$ are connected by doors. First, we note that for each $i$ with $0 \leqslant i \leqslant n$ there is a unique region labelled $i$ which is unbounded to the north. Now, consider a knight placed in some region with label $i$, and ask them to walk north (moving east or west by following the walls on the northern sides of regions, as needed). This knight will never get stuck: each region is convex, and so, if it is bounded to the north, it has a single northernmost vertex with a door northwards to another region with label $i$. Eventually it will reach a region which is unbounded to the north, which will be the unique such region with label $i$. Hence every region with label $i$ is connected to this particular region, and so all regions with label $i$ are connected to each other. As a result, there are exactly $n+1$ connected components, and Morgana can place at most $n+1$ knights. Comment. Variations on this argument exist: some of them capture more information, and some of them capture less information, about the connected components according to this system of numbering. For example, it can be shown that the unbounded regions are numbered $0,1, \ldots, n-1, n, n-1, \ldots, 1$ as one cycles around them, that the regions labelled 0 and $n$ are the only regions in their connected components, and that each other connected component forms a single chain running between the two unbounded ones. It is also possible to argue that the regions are acyclic without revealing much about their structure. Solution 2. We give another description of a strategy for Merlin to paint the walls so that Morgana can place no more than $n+1$ knights. Merlin starts by building a labyrinth of $n$ walls of his own design. He places walls in turn with increasing positive gradients, placing each so far to the right that all intersection points of previously-placed lines lie to the left of it. He paints each in such a way that blue is on the left and red is on the right. For example, here is a possible sequence of four such lines $\ell_{1}, \ell_{2}, \ell_{3}, \ell_{4}$ : We say that a region is ""on the right"" if it has $x$-coordinate unbounded above (note that if we only have one wall, then both regions are on the right). We claim inductively that, after placing $n$ lines, there are $n+1$ connected components in the resulting labyrinth, each of which contains exactly one region on the right. This is certainly true after placing 0 lines, as then there is only one region (and hence one connected component) and it is on the right. When placing the $n^{\text {th }}$ line, it then cuts every one of the $n-1$ previously placed lines, and since it is to the right of all intersection points, the regions it cuts are exactly the $n$ regions on the right. The addition of this line leaves all previous connected components with exactly one region on the right, and creates a new connected component containing exactly one region, and that region is also on the right. As a result, by induction, this particular labyrinth " b8ac251e19ac,"Example 2 (2000 National High School League Question) The minimum value of the distance from a lattice point (a point with integer coordinates) to the line $y=$ $\frac{5}{3} x+\frac{4}{5}$ is ( ). A. $\frac{\sqrt{34}}{170}$ B. $\frac{\sqrt{34}}{85}$ C. $\frac{1}{20}$ D. $\frac{1}{30}$","-1$, $y_{0}=-1$, $\left|25 x_{0}-15 y_{0}+12\right|=2$, hence the minimum value of the required $d$ ",medium,"Solution: Choose B. Reason: Let the integer point be $\left(x_{0}, y_{0}\right)$, then the distance from it to the line $y=\frac{5}{3} x+\frac{4}{5}$ is $$ d=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{\sqrt{25^{2}+15^{2}}}=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{5 \sqrt{34}} . $$ Since $x_{0}, y_{0} \in \mathbf{Z}$, $25 x_{0}+15 y_{0}$ is a multiple of 5, so $\left|25 x_{0}-15 y_{0}+12\right| \geqslant 2$. And when $x_{0}=-1$, $y_{0}=-1$, $\left|25 x_{0}-15 y_{0}+12\right|=2$, hence the minimum value of the required $d$ is $\frac{2}{5 \sqrt{34}}=\frac{\sqrt{34}}{85}$." 6ff8f4067040,"Example 4 Let $D=\{1,2, \cdots, 10\}, f(x)$ be a one-to-one mapping from $D$ to $D$, and let $f_{n+1}(x)=$ $f\left(f_{n}(x)\right), n \in \mathbf{N}_{+} ; f_{1}(x)=f(x)$. Try to find a permutation $x_{1}, x_{2}, \cdots, x_{10}$ of $D$, such that $\sum_{i=1}^{10} x_{i} f_{2520}(i)=220$.","1,2, \cdots, 10), f_{0}(i)=i, f_{1}(i), f_{2}(i), \cdots, f_{10}(i)$, at least two of them are equal",medium,"For any $i(i=1,2, \cdots, 10), f_{0}(i)=i, f_{1}(i), f_{2}(i), \cdots, f_{10}(i)$, at least two of them are equal. Therefore, from $f$ being a one-to-one mapping, it follows that there exists $1 \leqslant r_{i} \leqslant 10$, such that $f_{r_{i}}(i)=i$. Since $2520=2^{3} \times 3^{2} \times 5 \times 7$ is the least common multiple of $1, 2, 3, \cdots, 10$, $r_{i}$ divides $2520 (i=1,2, \cdots, 10)$, which leads to $f_{2520}(i)=i (i=1,2, \cdots, 10)$. According to the sorting principle, $\sum_{i=1}^{10} x_{i} \cdot i \geqslant 10 \times 1 + 9 \times 2 + 8 \times 3 + \cdots + 1 \times 10 = 220$, and the equality holds if and only if $x_{1}, x_{2}, \cdots, x_{9}, x_{10}$ are $10, 9, 8, \cdots, 2, 1$ respectively. Therefore, the required permutation is $10,9,8,7,6,5,4,3,2,1$." 4907298572ae,"2. (10 points) A team of 8 people completed $\frac{1}{3}$ of a project in 30 days. Then, 4 more people were added to complete the remaining part of the project. Therefore, the total time to complete the project was $\qquad$ days. The team of 8 people completed $\frac{1}{3}$ of a project in 30 days. Then, 4 more people were added to complete the remaining part of the project. Therefore, the total time to complete the project was $\qquad$ days.",It took a total of 70 days,medium,"【Analysis】Regarding this project as a unit ""1"", use “$\frac{1}{3} \div 30 \div 8=\frac{1}{720}$” to find the work efficiency of 1 person in 1 day, then the total work efficiency of 12 people is $\frac{1}{720} \times 12=\frac{1}{60}$. Calculate the remaining work volume, and then use ""work volume ÷ work efficiency = work time"" to find the time used later, and thus determine the total time used to complete this project. 【Solution】Solution: The work efficiency of one person is $\frac{1}{3} \div 30 \div 8=\frac{1}{720}$, The total work efficiency of 12 people is $\frac{1}{720} \times 12=\frac{1}{60}$, $$ \begin{array}{l} =40+30 \\ =70 \text { (days) } \end{array} $$ Answer: It took a total of 70 days. Therefore, the answer is: 70." 2387816d1c93,"2. Find the smallest natural number that can be written in the form $3 a^{2}-a b^{2}-2 b-4$, where $a$ and $b$ are natural numbers.",2,medium,"III/2. The answer is 2. If we take $a=4$ and $b=3$, we get $3 a^{2}-a b^{2}-2 b-4=2$. It is sufficient to show that the equation $3 a^{2}-a b^{2}-2 b-4=1$ has no solutions in natural numbers. We rearrange the equation to $3 a^{2}-a b^{2}=2 b+5$. Since the right side is odd, the left side must also be odd, so $a$ must be an odd number and $b$ must be an even number. The number $b^{2}$ is therefore divisible by 4, and the remainder of $a^{2}$ when divided by 4 is 1. The left side thus has a remainder of 3 when divided by 4, while the right side has a remainder of 1, since $2 b$ is divisible by 4. We have reached a contradiction, so the equation has no solutions in natural numbers. Answer that the number 2 can be obtained. ..................................................................... Answer that the number 2 is obtained when $a=4$ and $b=3$. ................................................. Explanation that we need to show that the number 1 cannot be obtained. .................... 1 point ![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-15.jpg?height=52&width=1633&top_left_y=745&top_left_x=220) Observation that $b$ must be an even number. ............................................... 1 point Justification that even with the above assumption, we reach a contradiction. ........... 2 points" a5a87a2e1dac,"Example 25 Let $x, y, z \in(0,1)$, satisfying: $$\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2,$$ Find the maximum value of $x y z$.",See reasoning trace,medium,"Let $u=\sqrt[6]{x y z}$, then by the condition and the AM-GM inequality, we have $$\begin{aligned} 2 u^{3}= & 2 \sqrt{x y z}=\frac{1}{\sqrt{3}} \sum \sqrt{x(3-3 x)} \\ \leqslant & \frac{1}{\sqrt{3}} \sum \frac{x+(3-3 x)}{2} \\ & =\frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \\ & \leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} \cdot \sqrt[3]{x y z} \\ & =\frac{3 \sqrt{3}}{2}-\sqrt{3} u^{2} \end{aligned}$$ Thus, $$4 u^{3}+2 \sqrt{3} u^{2}-3 \sqrt{3} \leqslant 0$$ That is, $$(2 u-\sqrt{3})\left(2 u^{2}+2 \sqrt{3} u+3\right) \leqslant 0$$ Therefore, $u \leqslant \frac{\sqrt{3}}{2}$. From this, we know that $x y z \leqslant \frac{27}{64}$, with equality when $$x=y=z=\frac{3}{4}$$ Thus, the maximum value sought is $\frac{27}{64}$." 432db21cd7f5,"3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $A$. Segment $O_{2} A$ intersects circle $\omega_{1}$ again at point $K$, and segment $O_{1} A$ intersects circle $\omega_{2}$ again at point $L$. The line passing through point $A$ parallel to $K L$ intersects circles $\omega_{1}$ and $\omega_{2}$ again at points $C$ and $D$ respectively. Segments $C K$ and $D L$ intersect at point $N$. Find the angle between the lines $O_{1} A$ and $O_{2} N$.",$90^{\circ}$,medium,"Answer: $90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-19.jpg?height=468&width=736&top_left_y=1999&top_left_x=680) Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. Therefore, their vertex angles are equal. Let the common value of these angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle A D L = \angle A C K = \alpha$. Since $K L \| C D$, we also get $\angle K L N = \angle L K N = \alpha$. Then $$ \angle L N K = 180^{\circ} - 2 \angle K L N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle A O_{2} L = 180^{\circ} - \angle K O_{2} L $$ Therefore, the quadrilateral $L N K O_{2}$ is cyclic, from which we get $$ \angle K O_{2} N = \angle K L N = \alpha = \frac{1}{2} \angle A O_{2} L $$ This means that the line $O_{2} N$ is the angle bisector of the vertex angle in the isosceles triangle $A O_{2} L$, and therefore also the altitude. Thus, the lines $O_{2} N$ and $O_{1} A$ are perpendicular." 1c51a58ec2d1,1. Determine which number is greater $\log _{3} 4$ or $\log _{4} 5$.,See reasoning trace,easy,"Solution. From the inequality of means, it follows that $$ \log _{4} 5 \cdot \log _{4} 3 \leq\left(\frac{\log _{4} 5+\log _{4} 3}{2}\right)^{2}=\left(\frac{\log _{4} 15}{2}\right)^{2}<1 $$ therefore $$ \left.\log _{4} 5<\frac{1}{\log _{4} 3}\right] \log _{3} 4 $$" fd4507600d39,Example 2 Solve the equation $x^{3}+(1+\sqrt{2}) x^{2}-2=0$.,See reasoning trace,medium,"【Analysis】This is a cubic equation, which cannot be solved by conventional methods. Therefore, we adopt the idea of ""reversing the roles"", treating the constant $\sqrt{2}$ as an ""unknown"", while regarding $x$ as a constant. Let $\sqrt{2}=t$, then the original equation becomes $$ \begin{array}{l} x^{3}+(1+t) x^{2}-t^{2}=0 \\ \Rightarrow t^{2}-x^{2} t-\left(x^{3}+x^{2}\right)=0 \\ \Rightarrow t_{1}=-x, t_{2}=x^{2}+x \\ \Rightarrow \sqrt{2}=-x, \sqrt{2}=x^{2}+x \\ \Rightarrow x=-\sqrt{2}, \frac{-1 \pm \sqrt{1+4 \sqrt{2}}}{2} . \end{array} $$" 2bab2384bde8,21st Putnam 1960,"n(a + b) implies a(b - n) = nb > 0, so a and b must both exceed n. Let b = n + d, then a = n(n+d)/d ",easy,"ab = n(a + b) implies a(b - n) = nb > 0, so a and b must both exceed n. Let b = n + d, then a = n(n+d)/d = n + n 2 /d. This is a solution iff d divides n 2 . So f(n) = the number of divisors of n 2 . 21st Putnam 1960 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002" bd275911a8cc,"10. Let $M$ be the set of all $\triangle ABC$ satisfying the conditions $BC=1, \angle A=2 \angle B$. If $\triangle ABC \in M$, and the angle bisector $CD$ of $\angle C$ divides the opposite side $AB$ into two segments in the ratio $AD:DB=t$, then $\triangle ABC$ is associated with the real number $t$. Determine the set of all such real numbers $t$.","2 \angle B, \therefore 0^{\circ}<3 \angle B<180^{\circ}$, i.e., $0^{\circ}<\angle B<60^{\circ}$. The",medium,"10. Let $\triangle A B C \in M$. Draw the angle bisector $A E$ of $\angle A$, then!! According to the given, $\angle 1=\angle 2=\angle B$. Therefore, $\triangle A E C \sim \triangle B A C$. Combining with the properties of the angle bisector of a triangle, we get $t=\frac{A D}{D B}=\frac{A C}{D C}=\frac{E C}{A C}=\frac{B E}{A B}$. Since $\triangle A E B$ is an isosceles triangle, we have $\cos B=\frac{\frac{1}{2} \cdot A B}{B E}=\frac{A B}{2 \overline{B E}}$, thus $t=\frac{1}{2 \cos B}$. On the other hand, $\because 0^{\circ}<\angle B+\angle A<180^{\circ}$, $\angle A=2 \angle B, \therefore 0^{\circ}<3 \angle B<180^{\circ}$, i.e., $0^{\circ}<\angle B<60^{\circ}$. Therefore, the set of all such $t$ is $\left\{t \left\lvert\, t=\frac{1}{2 \cos \bar{B}}\right., \quad 0^{\circ}<\angle B<60^{\circ}\right\}$ $=\left(\frac{1}{2}, 1\right)$." 8b3d51d8bc70,"3 . The two real roots of the equation $2 x^{2}-2 \sqrt{2} x+\frac{1}{2}=0$ are $\frac{1+\sqrt{2}}{2}$ and $\frac{\sqrt{2}-1}{2}$, then the factorization of $2 x^{2}-2 \sqrt{2} x+\frac{1}{2}$ in the real number range is ( ). (A) $\left(x-\frac{1+\sqrt{2}}{2}\right)\left(x-\frac{\sqrt{2-1}}{2}\right)$; (B) $2\left(x+\frac{\sqrt{2}+1}{2}\right)\left(x+\frac{\sqrt{2}-1}{2}\right)$ (C) $(2 x-\sqrt{2}-1)(2 x-\sqrt{2}+1)$, (D) $2\left(x-\frac{\sqrt{2}+1}{2}\right)\left(x-\frac{\sqrt{2}-1}{2}\right)$.",See reasoning trace,easy,3) (D) 33f232c082e1,"1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)",27,medium,"Answer: 27. Solution. If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads than red ones. Therefore, there are no fewer than $\frac{80}{3}=26 \frac{2}{3}$ green beads, which means there are at least 27. An example of a necklace containing exactly 27 green beads is as follows: GKR 3 GKR ... ; GKR (26 times); GK." 2328ff95a593,"2. Given that $A, B, C, D$ are four non-coplanar points on a sphere, and $A B=B C=A D=1, B D=A C=\sqrt{2}, B C \perp$ $A D$, then the volume of the sphere is $\qquad$ .",See reasoning trace,easy,"As shown in the figure, by supplementing the tetrahedron $D-ABC$ to form a cube, we can see that the diameter of its circumscribed sphere is $\sqrt{3}$, so the volume of this sphere is $$ V=\frac{4}{3} \pi \cdot\left(\frac{\sqrt{3}}{2}\right)^{3}=\frac{\sqrt{3}}{2} \pi $$" c058520091bf,A 25-meter long coiled cable is cut into pieces of 2 and 3 meters. In how many ways can this be done if the order of the different sized pieces also matters?,See reasoning trace,medium,"Let the number of 2-meter pieces be $x$, and the number of 3-meter pieces be $y$. First, we need to solve the following Diophantine equation: $$ 2 x+3 y=25 $$ From the equation, $$ x=\frac{25-3 y}{2} $$ Since we are looking for integer solutions, $25-3 y$ must be even. This means that $3 y$ and thus $y$ must be odd. Furthermore, since $x>0$, $3 y<25$. Therefore, $y$ must be a positive odd number less than 8; its possible values are: $1, 3, 5, 7$. The corresponding values of $x$ are: $\quad 11, 8, 5, 2$. In the first case, we have a total of 12 pieces of wire (1 piece of 3 meters and 11 pieces of 2 meters), and the number of possible arrangements is $\frac{12!}{1!\cdot 11!}=12$. For the other cases, similarly: $\frac{11!}{3!\cdot 8!}=165$, $\frac{10!}{5!\cdot 5!}=252$, and $\frac{9!}{2!\cdot 7!}=36$. Thus, the total number of possible cases is: $165+252+36+12=465$. Krisztian Balog (Miskolc, Foldes F. Gymnasium, 12th grade)" 8bb961ed1f8b,"28 Three sides $O A B, O A C$ and $O B C$ of a tetrahedron $O A B C$ are right-angled triangles, i.e. $\angle A O B=\angle A O C=\angle B O C=90^{\circ}$. Given that $O A=7, O B=2$ and $O C=6$, find the value of (Area of $\triangle O A B)^{2}+(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2}$.",(1052) } \\ \text { Note that }(\text { Area of } \triangle O A B)^{2}=(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2} \text {,medium,$\begin{array}{l}28 \text { Answer: (1052) } \\ \text { Note that }(\text { Area of } \triangle O A B)^{2}=(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2} \text {. } \\ \text { So (Area of } \triangle O A B)^{2}+(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2} \\ =2 \times\left\{(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2}\right\} \\ =2 \times\left((1 / 2 \times 7 \times 6)^{2}+(1 / 2 \times 7 \times 2)^{2}+(1 / 2 \times 2 \times 6)^{2}\right\}=1052 \text {. }\end{array}$ f30f43b21e29,"26. In a competition of fun and ingenuity, 9 points were awarded for each correctly completed task, and 5 points were deducted for each uncompleted or incorrectly completed task. It is known that the team was offered no more than 15 tasks and scored 57 points. How many tasks did the team complete correctly?","57$, from which $x=\frac{57+5 y}{9}$. Since $x$ is a natural number, $(57+5 y)$ must be divisible by",easy,"26. Let $x$ be the number of tasks completed correctly, and $y$ be the number of tasks completed incorrectly or not at all. Then we have $9 x-5 y=57$, from which $x=\frac{57+5 y}{9}$. Since $x$ is a natural number, $(57+5 y)$ must be divisible by 9. The smallest value of $y$ for which this condition is met is 3. In this case, $x=8$. Since $8+3=11$ and $11<15$, the found value of $x$ satisfies the condition." 3efe14f8e63e,"Four brothers inherited a circular meadow. Three of them have one goat each. One of the brothers suggests that they tie the 3 goats with equally long chains and secure the ends of the chains to suitable points on the perimeter of the meadow, so that the goats can graze one-quarter of the meadow each, as allowed by the chain, and finally the fourth brother will mow what the goats cannot reach. However, another brother is concerned that the areas accessible to the individual goats will overlap. a) Show that this situation can be avoided. b) Determine by trial and error what percentage of the meadow's radius the chains should be. (It is sufficient to determine 2 significant figures.)",See reasoning trace,medium,"a) The sections of the meadow that can be grazed by individual goats do not overlap if, for example, the three tethering points of the chain are chosen as the vertices of an equilateral triangle \(ABC\) inscribed in a circle \(k\) that encloses the meadow, and of course, the length of the chains is such that each goat can only reach \(1/4\) of the meadow. We will show that circles centered at \(A\), \(B\), and \(C\) that are tangent to each other and have the same radius - or \(AB/2\) radius - cover more than \(1/4\) of the area of \(k\). Knowing this, it is obvious that by keeping the points \(A\), \(B\), and \(C\) unchanged and reducing the radius to less than \(AB/2\), a radius \(\varrho\) can be found such that the circle centered at \(A\) (and similarly at \(B\) and \(C\)) with radius \(\varrho\) covers \(1/4\) of the meadow, and there is no overlap. ![](https://cdn.mathpix.com/cropped/2024_05_02_bd6b240496d6b0a63396g-1.jpg?height=589&width=512&top_left_y=459&top_left_x=793) Figure 1 Let the circle centered at \(A\) with radius \(AB/2\) intersect \(k\) at point \(D\) on the shorter arc \(AB\). This point is not on the chord \(AB\), so it lies on the side of the perpendicular bisector of \(AB\) that contains \(A\). Therefore, if \(F\) is the midpoint of the arc \(AB\) (Figure 1), \[ \angle DAB > \angle FAB = 30^\circ \] Thus, the central angle corresponding to the arc of \(k_1\) that lies inside \(k\) is greater than \(120^\circ\), so the area \(t\) of the part of \(k_1\) that lies inside \(k\) is greater than one-third of the area of \(k_1\): \[ t > \frac{1}{3} \pi \left(\frac{AB}{2}\right)^2 \] The right-hand side is exactly one-fourth of the area of \(k\): \[ \frac{\pi}{12} AB^2 = \frac{\pi}{12} \cdot 3R^2 = \frac{\pi R^2}{4} \] where \(R\) is the radius of \(k\); indeed, by a simple calculation, \(AB = R \sqrt{3}\). b) Let the length of the chain in question be \(\varrho\), the two intersection points of the circle centered at \(A\) with radius \(\varrho\) and \(k\) be \(M\) and \(N\), the center of \(k\) be \(O\), the midpoint of \(MN\) be \(G\), and \(\angle AOM = x\) (measured in degrees; Figure 2), which we will approximate. ![](https://cdn.mathpix.com/cropped/2024_05_02_bd6b240496d6b0a63396g-1.jpg?height=502&width=702&top_left_y=1904&top_left_x=687) Figure 2 The part of the meadow that the goat can reach is divided into the segments cut off by \(AM\) and \(AN\) from \(k\) and the sector \(AMN\), and the former are obtained by removing the area of the kite \(OMAN\) from the sector \(OMN\). Clearly, \(\angle MAN = 180^\circ - x\), so the requirement, using symmetry, is \[ \varrho^2 \pi \cdot \frac{180^\circ - x}{360^\circ} + R^2 \pi \cdot \frac{2x}{360^\circ} - OA \cdot MG = \frac{1}{4} R^2 \pi \] where \(MG = R \sin x\), and \(\varrho = 2R \sin \frac{x}{2}\). Simplification in the first term is given by the following transformation using the right triangle \(AMG\): \[ \varrho^2 = 2R \varrho \sin \frac{x}{2} = 2R \cdot AG = 2R(OA - OG) = 2R^2(1 - \cos x) \] since by the inscribed angle theorem, \(\angle AMG = \angle AMN = \angle AON / 2\). Our equation thus becomes: \[ \left(1 - \frac{x}{180^\circ}\right) \cos x + \frac{\sin x}{\pi} = \frac{3}{4} \] and since \(\varrho < R \sqrt{3} / 2\), therefore \[ \sin \frac{x}{2} = \frac{\varrho}{2R} < \frac{\sqrt{3}}{4} < 0.4331, \quad \frac{x}{2} < 25^\circ 40', \quad x < 51^\circ 20' \] The value of the left-hand side at \(x = 45^\circ\) is 0.7554, which is greater than the right-hand side, and at \(x = 46^\circ\) it is 0.7461, which is already smaller. The sought value of \(x\) is expected around \(45^\circ 35'\) based on the excess of 0.0054 and the deficit of 0.0039. The values of \(\varrho / R\) corresponding to the chosen angles are 0.7654 and 0.7814, so only 77% and 78% are feasible. At \(x = 45^\circ 35'\), the left-hand side of the equation is 0.7500, so this is the sought value of the angle, and to this \(\varrho / R = 0.7748\), so to three significant figures, \(\varrho\) is 77.5% of \(R\). János Kuhár (Budapest, Berzsenyi D. Gymnasium, 1st year)" 8c39d0b36f68,"9. Let $\triangle A B C$ have internal angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and $\angle A - \angle C = \frac{\pi}{2}, a, b, c$ form an arithmetic sequence. Then the value of $\cos B$ is . $\qquad$",See reasoning trace,easy,"II. $\frac{3}{4}$. It is easy to know, $$ \begin{array}{l} 2 b=a+c \\ \Leftrightarrow 2 \sin (A+C)=\sin A+\sin C \\ \Leftrightarrow 2 \sin \left(2 C+\frac{\pi}{2}\right)=\sin \left(C+\frac{\pi}{2}\right)+\sin C \\ \Leftrightarrow 2 \cos 2 C=\cos C+\sin C \\ \Rightarrow \cos C-\sin C=\frac{1}{2} . \end{array} $$ Therefore, $\cos B=-\cos (A+C)$ $$ =-\cos \left(2 C+\frac{\pi}{2}\right)=\sin 2 C=\frac{3}{4} \text {. } $$" ced04438053d,"A right-angled triangle has side lengths that are integers. What could be the last digit of the area's measure, if the length of the hypotenuse is not divisible by 5?","k\left(u^{2}-v^{2}\right)$, $b=2 k \cdot u v$, and $c=k\left(u^{2}+v^{2}\right)$, where $k$ is any n",medium,"Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to the Pythagorean theorem, we then have $$ a^{2}+b^{2}=c^{2} $$ A square number can give a remainder of 0, 1, or 4 when divided by 5. If neither $a$ nor $b$ is divisible by 5, then the left side of (1) can give a remainder of 2 (in the form $1+1$), 0 (in the form $1+4$), or 3 (in the form $4+4$) when divided by 5. Since 2 and 3 cannot be the remainder of a square number $-c^{2}$ when divided by 5, and 0 is excluded by the condition, if the length of the hypotenuse is not divisible by 5, then the length of one of the legs will be divisible by 5. If both legs are even, then their product is divisible by 4. We claim that this is also true if there is an odd leg. First, note that both legs cannot be odd. In this case, in (1), $a^{2}+b^{2}$ would be even, and thus $c^{2}$, as an even square number, would be divisible by 4, while $a^{2}$ and $b^{2}$, when divided by 4, would each give a remainder of 1, making their sum 2. In this case, however, equality could not hold in (1). If now one of the legs, say $a$, is odd, then $b$ is even, and $c$ is odd. From (1), $$ b^{2}=(c+a)(c-a) $$ Here, both factors on the right side are even. If neither is divisible by 4, then they each give a remainder of 2 when divided by 4, so their sum, $2c$, is divisible by 4. This would, however, make $c$ even, which is not true. Thus, in (2), the right side is also divisible by $2^{3}=8$. We know that a square number, in this case $b^{2}$, has every prime factor in its prime factorization with an even exponent. This means that $b^{2}$ is divisible by $2^{4}$, and thus $b$ is divisible by 4. This proves that if (1) holds, then $ab$ is divisible by 4. The area of the triangle, $ab/2$, is thus an even number. Since there is a leg divisible by 5, the area is also divisible by $2 \cdot 5=10$, so the last digit of its measure is 0. Remark. The statement - that on the one hand, in every Pythagorean triple there is a number divisible by 5, and on the other hand, the product of the legs is divisible by 4 - can also be easily shown based on the known characterization of Pythagorean triples. According to this characterization, Pythagorean triples can be written in the form $a=k\left(u^{2}-v^{2}\right)$, $b=2 k \cdot u v$, and $c=k\left(u^{2}+v^{2}\right)$, where $k$ is any natural number, and $u$ and $v$ are relatively prime numbers of different parities, with $u>v$. See, for example, Rademacher and Toeplitz's book *Numbers and Figures*." c3ff7dde8b90,"The diagonals of convex quadrilateral $BSCT$ meet at the midpoint $M$ of $\overline{ST}$. Lines $BT$ and $SC$ meet at $A$, and $AB = 91$, $BC = 98$, $CA = 105$. Given that $\overline{AM} \perp \overline{BC}$, find the positive difference between the areas of $\triangle SMC$ and $\triangle BMT$. [i]Proposed by Evan Chen[/i]",336,medium,"1. **Identify the given information and setup the problem:** - The diagonals of convex quadrilateral \( BSCT \) meet at the midpoint \( M \) of \( \overline{ST} \). - Lines \( BT \) and \( SC \) meet at \( A \). - Given lengths: \( AB = 91 \), \( BC = 98 \), \( CA = 105 \). - \( \overline{AM} \perp \overline{BC} \). 2. **Determine the lengths \( BM \) and \( MC \):** - Since \( M \) is the midpoint of \( \overline{ST} \) and \( \overline{AM} \perp \overline{BC} \), we can use the fact that \( \triangle ABM \) and \( \triangle AMC \) are right triangles. - Using the Pythagorean theorem in \( \triangle ABM \): \[ AB^2 = AM^2 + BM^2 \implies 91^2 = AM^2 + BM^2 \] - Using the Pythagorean theorem in \( \triangle AMC \): \[ AC^2 = AM^2 + MC^2 \implies 105^2 = AM^2 + MC^2 \] - Subtract the first equation from the second: \[ 105^2 - 91^2 = MC^2 - BM^2 \] - Simplify: \[ (105 - 91)(105 + 91) = MC^2 - BM^2 \implies 14 \times 196 = MC^2 - BM^2 \implies 2744 = MC^2 - BM^2 \] - Let \( BM = x \) and \( MC = y \). Then: \[ y^2 - x^2 = 2744 \] - Also, from the Pythagorean theorem: \[ 91^2 = AM^2 + x^2 \quad \text{and} \quad 105^2 = AM^2 + y^2 \] - Subtracting these: \[ 105^2 - 91^2 = y^2 - x^2 \implies 2744 = y^2 - x^2 \] - Solving for \( x \) and \( y \): \[ x = 35 \quad \text{and} \quad y = 63 \] 3. **Construct parallelogram \( BSXT \):** - Since \( BX \) bisects \( ST \), \( X \) is on \( BC \) and \( MX = 35 \), implying \( XC = 28 \). 4. **Calculate the areas of \( \triangle SMC \) and \( \triangle BMT \):** - Notice that \( [SMC] - [BMT] = [SXC] + [SMX] - [BMT] = [SXC] \), because \( BSXT \) is a parallelogram. - Since \( BT \parallel AB \) and \( BT \parallel SX \), \( AB \parallel SX \) and \( \triangle ABC \sim \triangle SXC \). - Therefore, \( SX = 26 \) and \( SC = 30 \). 5. **Calculate the area of \( \triangle SXC \):** - Using the similarity ratio: \[ \frac{SX}{AB} = \frac{SC}{AC} \implies \frac{26}{91} = \frac{30}{105} \] - The area of \( \triangle ABC \) is: \[ \text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC) \] - Since \( \triangle ABC \sim \triangle SXC \), the area ratio is: \[ \left(\frac{26}{91}\right)^2 = \frac{[SXC]}{[ABC]} \] - Calculate the area of \( \triangle ABC \): \[ [ABC] = \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC) \] - Using the similarity ratio: \[ [SXC] = \left(\frac{26}{91}\right)^2 \times [ABC] = \left(\frac{26}{91}\right)^2 \times \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC) \] - Simplify: \[ [SXC] = \left(\frac{26}{91}\right)^2 \times 91 \times 98 \times \sin(\angle ABC) = 336 \] The final answer is \( \boxed{336} \)." af8a3352421c,"A-3. For a given positive integer $m$, find all positive integer triples $(n, x, y)$, where $m, n$ are coprime, and satisfy $$ \left(x^{2}+y^{2}\right)^{m}=(x y)^{n} . $$",See reasoning trace,medium,"When $m$ is odd, the equation has no solution. When $m$ is even, the only solution is $(n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right)$. We will prove this below. If $(n, x, y)$ is a solution to the equation, then by the arithmetic-geometric mean inequality, we have $(x y)^{n}=\left(x^{2}+y^{2}\right)^{n} \geqslant(2 x y)^{n}$. Therefore, $n>m$. Let $p$ be a prime number, and let $a, b$ be the highest powers of $p$ in $x, y$ respectively. Then $(x y)^{n}$ contains the highest power of $p$ as $(a+b)^{n}$. If $a1$ be a fixed positive integer. For a nonempty string of base-ten digits $S$, let $c(S)$ be the number of ways to split $S$ into contiguous nonempty strings of digits such that the base-ten number represented by each string is divisible by $m$. These strings are allowed to have leading zeroes. In terms of $m$, what are the possible values that $c(S)$ can take? For example, if $m=2$, then $c(1234)=2$ as the splits 1234 and $12 \mid 34$ are valid, while the other six splits are invalid.",0$ and $c(00 \ldots 0)=2^{n-1}$ if there are $n$ zeroes in the string. Now we show that these are th,medium,"Answer: 0 and $2^{n}$ for all nonnegative integer $n$ First, we note that $c(1)=0$ and $c(00 \ldots 0)=2^{n-1}$ if there are $n$ zeroes in the string. Now we show that these are the only possibilities. Note that a split can be added if and only if the string before this split (ignoring all other splits) represent a multiple of $m$ (if there is a split before it, then removing the digits before this preceding split is equivalent to subtracting the removed number times a power of 10 , which will also be a multiple of $m$, so the remaining number between the two splits remain a multiple of $m$ ). Thus, whether we can add a split or not depends only on the string itself and no other splits, so $c(S)$ is 0 if the number is not divisible by $m$ and a power of two otherwise." 12527edeca54,"## a) Compute $$ \lim _{n \rightarrow \infty} n \int_{0}^{1}\left(\frac{1-x}{1+x}\right)^{n} d x $$ b) Let $k \geq 1$ be an integer. Compute $$ \lim _{n \rightarrow \infty} n^{k+1} \int_{0}^{1}\left(\frac{1-x}{1+x}\right)^{n} x^{k} d x $$",See reasoning trace,medium,"Solution. $a$ ) The limit equals $\frac{1}{2}$. The result follows immediately from $b$ ) for $k=0$. b) The limit equals $\frac{k!}{2^{k+1}}$. We have, by the substitution $\frac{1-x}{1+x}=y$, that $$ \begin{aligned} n^{k+1} \int_{0}^{1}\left(\frac{1-x}{1+x}\right)^{n} x^{k} d x & =2 n^{k+1} \int_{0}^{1} y^{n}(1-y)^{k} \frac{d y}{(1+y)^{k+2}} \\ & =2 n^{k+1} \int_{0}^{1} y^{n} f(y) d y \end{aligned} $$ where $$ f(y)=\frac{(1-y)^{k}}{(1+y)^{k+2}} $$ We observe that $$ f(1)=f^{\prime}(1)=\cdots=f^{(k-1)}(1)=0 $$ We integrate $k$ times by parts $\int_{0}^{1} y^{n} f(y) d y$, and by (7) we get $$ \int_{0}^{1} y^{n} f(y) d y=\frac{(-1)^{k}}{(n+1)(n+2) \ldots(n+k)} \int_{0}^{1} y^{n+k} f^{(k)}(y) d y $$ One more integration implies that $$ \begin{aligned} \int_{0}^{1} y^{n} f(y) d y= & \frac{(-1)^{k}}{(n+1)(n+2) \ldots(n+k)(n+k+1)} \\ & \times\left(\left.f^{(k)}(y) y^{n+k+1}\right|_{0} ^{1}-\int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y\right) \\ = & \frac{(-1)^{k} f^{(k)}(1)}{(n+1)(n+2) \ldots(n+k+1)} \\ & +\frac{(-1)^{k+1}}{(n+1)(n+2) \ldots(n+k+1)} \int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y \end{aligned} $$ It follows that $$ \lim _{n \rightarrow \infty} 2 n^{k+1} \int_{0}^{1} y^{n} f(y) d y=2(-1)^{k} f^{(k)}(1) $$ since $$ \lim _{n \rightarrow \infty} \int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y=0 $$ $f^{(k+1)}$ being continuous and hence bounded. Using Leibniz's formula we get that $$ f^{(k)}(1)=(-1)^{k} \frac{k!}{2^{k+2}} $$ and the problem is solved." 8364b4d3f3ea,4.50. Calculate the solid angle of a cone with an angle of $2 \alpha$ at the vertex.,OM - OH = 1 - \cos \alpha$. The solid angle of the cone is equal to the area of the spherical segmen,medium,"4.50. Let $O$ be the vertex of the cone, $OH$ its height. Construct a sphere of radius 1 with center $O$ and consider its section by a plane passing through the line $OH$. Let $A$ and $B$ be points on the cone lying on the sphere; $M$ be the point of intersection of the ray $OH$ with the sphere (Fig. 40). Then $HM = OM - OH = 1 - \cos \alpha$. The solid angle of the cone is equal to the area of the spherical segment cut off by the base of the cone. According to problem 4.24, this area is $2 \pi R h = 2 \pi (1 - \cos \alpha)$." 987858bc91cd,8. Convert the parametric equations $\left\{\begin{array}{l}x=\frac{\cos \theta}{1+\cos \theta} \\ y=\frac{\sin \theta}{1+\cos \theta}\end{array}\right.$ to a Cartesian equation is $\qquad$ .,"-2\left(x-\frac{1}{2}\right)$ Hint: You can express $\sin \theta, \cos \theta$ in terms of $x, y$, a",easy,"8. $y^{2}=-2\left(x-\frac{1}{2}\right)$ Hint: You can express $\sin \theta, \cos \theta$ in terms of $x, y$, and then use the square relationship to eliminate $\theta$." 5d010d82ba9a,"## 9. Megacube From identical small cubes, one large cube is assembled. The length of the edge of the large cube is a hundred times larger than the length of the edge of each small cube. The large cube is placed on a table, and all its sides, except the one that touches the table, are painted red. How many small cubes have exactly two red sides? Result: $\quad 788$",385&width=529&top_left_y=2060&top_left_x=1248),easy,"## Solution. Along each of the 4 edges of the top face of the cube, there are 98 cubes $s$ with two red sides. Along each of the 4 side edges of the cube, there are 99 cubes with two red sides. The number of cubes with exactly two red sides is $4 \cdot 98 + 4 \cdot 99 = 392 + 396 = 788$. ![](https://cdn.mathpix.com/cropped/2024_05_30_44689c7bad1ce05a818cg-5.jpg?height=385&width=529&top_left_y=2060&top_left_x=1248)" 3c494cbaae03,"Let $a,b,c$ be positive real numbers. Find all real solutions $(x,y,z)$ of the system: \[ ax+by=(x-y)^{2} \\ by+cz=(y-z)^{2} \\ cz+ax=(z-x)^{2}\]","(0, 0, 0), (0, 0, c), (a, 0, 0), (0, b, 0)",easy,"\( (0, 0, c) \). The final answer is \( \boxed{ (0, 0, 0), (0, 0, c), (a, 0, 0), (0, b, 0) } \)." 267909ab091c,"7. Given that $z$ is a complex number, and $|z|=1$. When $\mid 1+z+$ $3 z^{2}+z^{3}+z^{4}$ | takes the minimum value, the complex number $z=$ $\qquad$ or . $\qquad$",-\frac{1}{4} \pm \frac{\sqrt{15}}{4}$ i.,easy,"7. $-\frac{1}{4} \pm \frac{\sqrt{15}}{4}$ i. Notice, $$ \begin{array}{l} \left|1+z+3 z^{2}+z^{3}+z^{4}\right| \\ =\left|\frac{1}{z^{2}}+\frac{1}{z}+3+z+z^{2}\right| \\ =\left|\left(z+\frac{1}{z}\right)^{2}+z+\frac{1}{z}+1\right| \\ =\left|(z+\bar{z})^{2}+z+\bar{z}+1\right| \\ =\left|4 \operatorname{Re}^{2} z+2 \operatorname{Re} z+1\right| \\ \geqslant \frac{16-4}{16}=\frac{3}{4}, \end{array} $$ Here, $\operatorname{Re} z=-\frac{1}{4}$. Thus, $z=-\frac{1}{4} \pm \frac{\sqrt{15}}{4}$ i." 0d294d5c3fab,"Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and \[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$. Prove that they are converges and find their limits.",\lim_{n \to \infty,medium,"1. **Base Case Verification:** We start by verifying the base case for \( n = 1 \). Given \( x_1 = 1 \) and \( y_1 = \sqrt{3} \), we need to check if \( x_1^2 + y_1^2 = 4 \): \[ x_1^2 + y_1^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \] Thus, the base case holds. 2. **Inductive Step:** Assume that for some \( n = t \), the relation \( x_t^2 + y_t^2 = 4 \) holds. We need to show that \( x_{t+1}^2 + y_{t+1}^2 = 4 \). 3. **Using the Recurrence Relations:** From the given recurrence relations: \[ x_{n+1} y_{n+1} = x_n \quad \text{and} \quad x_{n+1}^2 + y_n = 2 \] We can express \( x_{t+1} \) and \( y_{t+1} \) in terms of \( x_t \) and \( y_t \): \[ x_{t+1}^2 = 2 - y_t \] \[ y_{t+1} = \frac{x_t}{x_{t+1}} \] 4. **Substituting \( y_{t+1} \):** Using the inductive hypothesis \( x_t^2 + y_t^2 = 4 \), we substitute \( y_{t+1} \): \[ y_{t+1}^2 = \left( \frac{x_t}{x_{t+1}} \right)^2 = \frac{x_t^2}{x_{t+1}^2} \] Substituting \( x_{t+1}^2 = 2 - y_t \): \[ y_{t+1}^2 = \frac{x_t^2}{2 - y_t} \] 5. **Simplifying \( y_{t+1}^2 \):** Using the inductive hypothesis \( x_t^2 = 4 - y_t^2 \): \[ y_{t+1}^2 = \frac{4 - y_t^2}{2 - y_t} \] 6. **Summing \( x_{t+1}^2 \) and \( y_{t+1}^2 \):** \[ x_{t+1}^2 + y_{t+1}^2 = (2 - y_t) + \frac{4 - y_t^2}{2 - y_t} \] Simplify the second term: \[ \frac{4 - y_t^2}{2 - y_t} = \frac{(2 - y_t)(2 + y_t)}{2 - y_t} = 2 + y_t \] Therefore: \[ x_{t+1}^2 + y_{t+1}^2 = (2 - y_t) + (2 + y_t) = 4 \] This completes the inductive step. 7. **Conclusion:** By induction, \( x_n^2 + y_n^2 = 4 \) for all \( n \). 8. **Finding the Limits:** Let \( x_n = 2 \sin a_n \) and \( y_n = 2 \cos a_n \) where \( a_n \in (0, \pi/2) \). From the recurrence relations: \[ x_{n+1}^2 = 4 \sin^2 a_{n+1} = 2(1 - \cos a_n) \] \[ 2 \sin^2 a_{n+1} = 1 - \cos a_n \] Since the sequences are positive, we have: \[ a_{n+1} = \frac{a_n}{2} \] As \( n \to \infty \), \( a_n \to 0 \), so \( x_n \to 0 \) and \( y_n \to 2 \). The final answer is \( \boxed{ \lim_{n \to \infty} x_n = 0 } \) and \( \lim_{n \to \infty} y_n = 2 \)." 6a490ba3860d,"2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya: - Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades. The best method for Vasya is the one that gives him the highest grade for the half-year. ![](https://cdn.mathpix.com/cropped/2024_05_06_be2097b0254fa1f5b26ag-10.jpg?height=328&width=534&top_left_y=1912&top_left_x=1366) a) (from 6th grade. 1 point). Which method is the best for Vasya? b) (from 6th grade. 2 points). Then the teacher thought and added: - Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade. Prove that under these conditions, Method A is not the best for Vasya.",See reasoning trace,easy,"# Solution. a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A. b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now gives less than Method B, so Method A is not the best." 446809f166ad,"4. The curve represented by the equation $\frac{x^{2}}{\sin \left(19^{2 \infty}\right)^{\circ}}+\frac{y^{2}}{\cos \left(19^{2 \infty}\right)^{\circ}}=1$ is ( ). (A) Hyperbola (B) Ellipse with foci on the $x$-axis (C) Ellipse with foci on the $y$-axis (D) None of the above answers is correct",(C),easy,"4.C. $$ \begin{array}{l} 19^{2007}=19 \times\left(19^{2}\right)^{1003}=19 \times(360+1)^{1003} \\ =19 \times(360 n+1)\left(n \in \mathbf{N}_{+}\right) . \end{array} $$ Therefore, $\sin \left(19^{2007}\right)^{\circ}=\sin (360 \times 19 n+19)^{\circ}=\sin 19^{\circ}$. Similarly, $\cos \left(19^{2007}\right)^{\circ}=\cos 19^{\circ}$. Since $\cos 19^{\circ}>\sin 19^{\circ}>0$, the answer is (C)." 410a63716b59,"6. (10 points) As shown in the figure, the area of square $A B C D$ is 840 square centimeters, $A E=E B, B F=2 F C, D F$ intersects $E C$ at $G$. Then the area of quadrilateral $A E G D$ is square centimeters.",See reasoning trace,medium,"510 【Analysis】This figure easily leads one to think about finding the areas of triangles $BCE$, $CDF$, and $CGF$. The most challenging part is finding the area of $\triangle CGF$. According to the given conditions, we should draw $GM \perp BC$ with the intersection point at $M$. This forms two sets of similar triangles corresponding to $\triangle BCE$ and $\triangle CDF$. By using the property that the ratio of the areas of similar triangles is equal to the square of the similarity ratio, we can find the area of $\triangle CGF$. Finally, based on the figure, we can calculate the area of the required shape. 【Solution】Draw $GM \perp BC$ with the intersection point at $M$, $$ \begin{array}{l} \therefore \triangle FMG \sim \triangle FCD \Rightarrow FM: FC = MG: CD, \\ \because BF = 2FC \Rightarrow BC = 3FC, \\ \therefore MG = 3FM, \\ \because \triangle CGM: \triangle CEB \Rightarrow CM: CB = GM: BE, \\ BC = 2BE, \\ \therefore GM = \frac{1}{2} CM = 3FM \Rightarrow CM = 6FM, \\ \therefore FM: FC = 1: 7, \\ CM: CB = 2: 7, \\ S \triangle BCE = \frac{1}{4} \times \square ABCD = 210, \\ S \triangle CGM = 4 \div 49 \times 210 = 840 \div 49 = 120 \div 7, \\ S \triangle CDF = S \square ABCD \div 6 = 140, \\ S \triangle MGF = 140 \times 1 \div 49 = 140 \div 49 = 20 \div 7, \\ S \triangle CGM + \triangle MGF = 120 \div 7 + 20 \div 7 = 20, \\ 840 - 210 - 140 + 20 = 510 \text{ (square centimeters). } \end{array} $$ Therefore, the area of quadrilateral $AEGD$ is 510 square centimeters." 3c71d460e377,"3. In $\triangle A B C$, $\lg \operatorname{tg} A+\lg \operatorname{tg} C=$ $2 \lg \operatorname{tg} B$. Then the range of $\angle B$ is () . (A) $0<\angle B \leqslant \frac{\pi}{3}$ (B) $\frac{\pi}{3} \leqslant \angle B<\frac{\pi}{2}$ (C) $0<\angle B \leqslant \frac{\pi}{6}$ (D) $\frac{\pi}{6} \leqslant \angle B<\frac{\pi}{2}$",\operatorname{tg} A \operatorname{tg} C \leqslant\left(\frac{\operatorname{tg} A+\operatorname{tg} C,easy,"3. (B). From the given, we have $\operatorname{tg}^{2} B=\operatorname{tg} A \operatorname{tg} C \leqslant\left(\frac{\operatorname{tg} A+\operatorname{tg} C}{2}\right)^{2}$. Therefore, $2 \operatorname{tg} B \leqslant \operatorname{tg} A+\operatorname{tg} C=\operatorname{tg}(A+C)(1-\operatorname{tg} A \operatorname{tg} C)=$ $-\operatorname{tg} B \cdot\left(1-\operatorname{tg}^{2} B\right)$. Then $\operatorname{tg}^{2} B \geqslant 3, \operatorname{tg} B \geqslant \sqrt{3}$." a67ce229aa95,"2. The students went to see an exhibition and had to pay 120 EUR for the bus fare. Since three students fell ill, each of the remaining students had to pay an additional 2 EUR for the transport. How many students signed up for the exhibition?",.,medium,"2. From the text, we write the equations $n \cdot c=120$ and $(n-3)(c+2)=120$. Solving the resulting system of equations, we get the quadratic equation $c^{2}+2 c-80=0$. Considering the positive solution $c=8$ and calculating $n=15$. ![](https://cdn.mathpix.com/cropped/2024_06_07_93a0a0489a8a740fd9e9g-11.jpg?height=60&width=1639&top_left_y=1009&top_left_x=206) ![](https://cdn.mathpix.com/cropped/2024_06_07_93a0a0489a8a740fd9e9g-11.jpg?height=57&width=1639&top_left_y=1062&top_left_x=206) Solving the system ................................................................................................................................. ![](https://cdn.mathpix.com/cropped/2024_06_07_93a0a0489a8a740fd9e9g-11.jpg?height=57&width=1639&top_left_y=1162&top_left_x=206) ![](https://cdn.mathpix.com/cropped/2024_06_07_93a0a0489a8a740fd9e9g-11.jpg?height=57&width=1642&top_left_y=1211&top_left_x=207) Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point" ce901c1e63cd,"1. If 2009 numbers $a_{1}, a_{2}, \cdots, a_{2009}$ satisfy: $a_{1}$ $=3, a_{n}^{2}-\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) a_{n}+\frac{1}{2009}=0$ (where $n=$ $2,3,4, \cdots, 2009$ ), then the maximum value that $a_{2009}$ can reach is","3\left(\frac{1}{2009}\right)^{2007}, \therefore\left(a_{2009}\right)_{\max }=\frac{1}{3} \cdot 2009^",medium,"1. $\left(a_{2009}\right)_{\max }=\frac{1}{3} \cdot 2009^{2007}$. Solution: $a_{n}=$ $$ \begin{array}{l} \frac{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) \pm \sqrt{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right)^{2}-\frac{4}{2009}}}{2} \\ =\frac{\left(\frac{a_{n-1}}{2009}+\frac{1}{a_{n-1}}\right) \pm\left|\frac{a_{n-1}}{2009}-\frac{1}{a_{n-1}}\right|}{2}, \end{array} $$ Therefore, $a_{n}=\frac{a_{n-1}}{2009}\left(a_{n-1}>\sqrt{2009}\right)$ or $a_{n}=\frac{1}{a_{n-1}}\left(a_{n-1}\right.$ $<\sqrt{2009})$, Hence $\left(a_{2008}\right)_{\min }=3\left(\frac{1}{2009}\right)^{2007}, \therefore\left(a_{2009}\right)_{\max }=\frac{1}{3} \cdot 2009^{2007}$." 07baeab52c2d,"Example 5 In space, there are four spheres with radii of $2$, $2$, $3$, and $3$. Each sphere is externally tangent to the other three spheres. There is another smaller sphere that is externally tangent to these four spheres. Find the radius of the smaller sphere.",See reasoning trace,easy,"Let the radius of the required sphere be $r=\frac{1}{x}$. Then, by Descartes' theorem, we have $$ \begin{array}{l} \left(\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+x\right)^{2} \\ =3\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{3^{2}}+x^{2}\right) \\ \Rightarrow\left(\frac{5}{3}+x\right)^{2}=3\left(\frac{13}{18}+x^{2}\right) \\ \Rightarrow x=\frac{11}{6} \Rightarrow r=\frac{6}{11} . \end{array} $$" c97df889ebc5,"6. Given positive integers $m, n$ satisfying $m+n=19$. Then the maximum number of solutions to the equation $\cos m x=\cos n x$ in the interval $[0, \pi]$ is","\frac{k_{2}}{m-n}$, i.e., $19 k_{2}=(m-n) k_{1}$, thus $k_{1}$ is a multiple of 19. But $k_{1} \leqs",medium,"6.18. Since $m+n=19$ is odd, then $m \neq n$. Without loss of generality, assume $m>n$. The solutions to the equation $\cos m x=\cos n x$ are $x=\frac{2 k_{1} \pi}{m+n}, x=\frac{2 k_{2} \pi}{m-n}\left(k_{1}, k_{2} \in \mathbb{Z}\right)$. From $0 \leqslant \frac{2 k_{1} \pi}{m+n} \leqslant \pi, 0 \leqslant \frac{2 k_{2} \pi}{m-n} \leqslant \pi$, we get $k_{1}=0,1,2, \cdots, \frac{m+n-1}{2}$, $k_{2}=0,1,2, \cdots, \frac{m-n-1}{2}$. Therefore, the number of roots of the equation is no more than $$ \left(\frac{m+n-1}{2}+1\right)+\left(\frac{m-n-1}{2}+1\right)=m+1 \text {. } $$ Since $n \geqslant 1$, then $m \leqslant 18$, so the number of roots is no more than 19. However, there cannot be 19 roots, because this would include the root when $k_{1}=k_{2}=0$, thus the number of roots is no more than 18. Next, we prove that the number of roots of the equation is exactly 18. For this, we only need to prove that when $k_{1}, k_{2}$ are not equal to 0, $\frac{2 k_{1} \pi}{m+n} \neq \frac{2 k_{2} \pi}{m-n}$. If not, then $\frac{k_{1}}{19}=\frac{k_{2}}{m-n}$, i.e., $19 k_{2}=(m-n) k_{1}$, thus $k_{1}$ is a multiple of 19. But $k_{1} \leqslant \frac{m+n}{2}=\frac{19}{2}$, so this is impossible." c18818a5d78f,"1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+2 \operatorname{tg} \alpha-4 \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=\frac{1}{3}$. Find $\operatorname{ctg} \alpha$.",2 or $\frac{1}{3}$,medium,"Answer: 2 or $\frac{1}{3}$. Solution. $\operatorname{tg} 2 \gamma=\frac{2 \operatorname{tg} \gamma}{1-\operatorname{tg}^{2} \gamma}=\frac{2 / 3}{1-1 / 9}=\frac{3}{4}$. Then the given equality can be transformed as follows: $$ \frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \operatorname{tg} 2 \gamma}+2 \operatorname{tg} \alpha-3=0 \Leftrightarrow \frac{\operatorname{tg} \alpha+\frac{3}{4}}{1-\frac{3}{4} \operatorname{tg} \alpha}+2 \operatorname{tg} \alpha-3=0 $$ By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$, from which $\operatorname{tg} \alpha=3$ or $\operatorname{tg} \alpha=\frac{1}{2}$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \alpha=\frac{1}{3}$ or $\operatorname{ctg} \alpha=2$." 86d1da6d258a,"11.231 The base of the pyramid $S A B C$ is a right isosceles triangle $A B C$, with the length of the hypotenuse $A B=4 \sqrt{2}$. The lateral edge $S C$ of the pyramid is perpendicular to the plane of the base, and its length is 2. Find the measure of the angle and the distance between the lines, one of which passes through point $S$ and the midpoint of edge $A C$, and the other through point $C$ and the midpoint of edge $A B$.",$\frac{\pi}{3} ; \frac{2\sqrt{3}}{3}$,medium,"Solution. ![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-279.jpg?height=289&width=471&top_left_y=1210&top_left_x=190) Let $K$ and $M$ be the midpoints of edges $AC$ and $AB$, and $N$ be the midpoint of segment $AM$. $CM \| KN \Rightarrow CM \| SKN \Rightarrow$ the distance $x$ between $CM$ and $SK$ is equal to the distance between $CM$ and the plane $SKN$, or, in other words, it is equal to the height of the pyramid $MSKN$ dropped from $M$ to $SKN$. Additionally, the angle between $CM$ and $SK$ is equal to $\angle SKN$ (or $180^\circ - \angle SKN$). $CM$ is the median of the right $\triangle ABC \Rightarrow KN = \frac{CM}{2} = \frac{AB}{4} = \sqrt{2}$, $SK = \sqrt{KC^2 + SC^2} = 2\sqrt{2} \quad \left(AC = \frac{AB}{\sqrt{2}} = 4\right), \quad SM = \sqrt{CM^2 + SC^2} = 2\sqrt{3}$, $SN = \sqrt{MN^2 + SM^2} = \sqrt{14}$. By the cosine theorem for $\triangle SKN$, $\cos \angle SKN = \frac{KN^2 + SK^2 - SN^2}{2 \cdot KN \cdot SK} = -\frac{1}{2}, \angle SKN = 120^\circ \Rightarrow$ $\Rightarrow S_{SKN} = \frac{KN \cdot SK \cdot \sin 120^\circ}{2} = \sqrt{3}, V_{MSKN} = \frac{x}{3} \cdot S_{SKN} = \frac{x}{\sqrt{3}}$. On the other hand, $V_{MSKN} = \frac{1}{3} SC \cdot S_{KMN} = \frac{2}{3} \cdot \frac{S_{ABC}}{8} = \frac{2}{3} \Rightarrow x = \frac{2\sqrt{3}}{3}$. Answer: $\frac{\pi}{3} ; \frac{2\sqrt{3}}{3}$." 51b5f6434e3e,"4. (10 points) On Beta Planet, there are seven countries, each of which has exactly four friendly countries and two enemy countries. There are no three countries that are all enemies with each other. For such a planetary situation, a total of $\qquad$ three-country alliances, where all countries are friends with each other, can be formed.",: 7,medium,"【Analysis】Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ to represent seven countries, and use dashed lines to connect to indicate enemy relations, and solid lines to connect to indicate friendly relations. Then each country has 2 dashed lines and 4 solid lines. There are a total of $7 \times 2 \div 2=7$ (lines) dashed lines, and the rest are solid lines. First, it must be explained that these 7 points must be connected by 7 dashed lines to form a closed loop. $A_{2}$ must be connected to two points with dashed lines, let's denote them as $A_{1}, A_{3}$, and $A_{3}$ must be connected to another point with a dashed line, denoted as $A_{4} ; A_{4}$ is connected to $A_{5}$ with a dashed line, otherwise the remaining 3 points would have mutual enemy relations; $A_{5}$ is connected to $A_{6}$ with a dashed line, otherwise the remaining two points cannot be connected by 2 dashed lines; $A_{6}$ is connected to $A_{7}$ with a dashed line, and finally $A_{7}$ can only be connected to $A_{1}$ with a dashed line. The final connection diagram is as follows. As long as the three selected points do not have any two adjacent, the condition is met, there are $135,136,146,246,247,257,357$, a total of 7 kinds. (For clarity, we use $1,2,3,4,5,6,7$ to represent $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ respectively)【Solution】Solution: Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ to represent seven countries, and use dashed lines to connect to indicate enemy relations, and solid lines to connect to indicate friendly relations. Selecting three points such that no two are adjacent meets the condition, there are $135,136,146,246,247,257,357$, a total of 7 kinds. (For clarity, we use $1,2,3,4,5,6,7$ to represent $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ respectively), so for such a planetary situation, a total of 7 alliances of three countries that are all friendly can be formed. Therefore, the answer is: 7." 417a66047c3f,"10.024. The lengths of the parallel sides of the trapezoid are 25 and 4 cm, and the lengths of the non-parallel sides are 20 and 13 cm. Find the height of the trapezoid.",12 cm,easy,"## Solution. Given $B C=4 \text{ cm}, A D=25 \text{ cm}, A B=20 \text{ cm}, C D=13 \text{ cm}$ (Fig. 10.24). Draw $B E \perp A D$ and $C F \perp A D$. Let $B E=C F=h, A E=x$, $F D=y$. Then from $\triangle A B E$ and $\triangle C F D$ we find $h^{2}=20^{2}-x^{2}=13^{2}-y^{2}$. Considering that $y=25-4-x=21-x$, we have $20^{2}-x^{2}=13^{2}-(21-x)^{2}$ or $42 x=672$, from which $x=16$ (cm). Therefore, $h=\sqrt{20^{2}-16^{2}}=12$ (cm). Answer: 12 cm." c8b48379f5e8,"Kanel-Belov A.Y. A unit square is divided into a finite number of smaller squares (the sizes of which may vary). Can the sum of the perimeters of the squares intersecting the main diagonal be greater than 1993? (If a square intersects the diagonal at a single point, this is also considered an intersection.)",Yes,medium,"Let's divide the unit square into four equal smaller squares. The squares that intersect the diagonal only at a vertex will be called squares of the first level. Each of the remaining squares will be divided into four smaller squares. The squares that intersect the diagonal at a vertex will be called squares of the second level. We will again divide each of the remaining squares into four smaller squares, and so on (see figure). ![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-31.jpg?height=380&width=466&top_left_y=1365&top_left_x=796) We will perform this operation 500 times. We will get $2^{k}$ squares of the $k$-th level. The side length of each of them is $2^{-k}$. Therefore, their total perimeter is 4. Consequently, the total perimeter of all squares of all levels is $4 \cdot 500 > 1993$. ## Answer Yes." 6420a9f93622,"The hyperbola $3 x^{2}-y^{2}+24 x+36=0$ has a focus $F$ and a right directrix $l$. The ellipse $C$ has $F$ and $l$ as its corresponding focus and directrix. A line parallel to $y=x$ is drawn through $F$, intersecting the ellipse $C$ at points $A$ and $B$. It is known that the center of $C$ is inside the circle with $A B$ as its diameter. Find the range of the eccentricity of the ellipse $C$.",See reasoning trace,medium,"14. Transform the hyperbola equation to $\frac{(x-4)^{2}}{4} \cdots \frac{y^{2}}{12}=1$, where the coordinates of $F$ are $(8,0)$, and the equation of the right directrix $l$ is $x=5$. Using $F$ as the pole and $F x$ as the polar axis, let $\mathrm{e}$ be the eccentricity of the ellipse $C$, then the equation of $C$ is $\rho = \frac{3 e}{1 \cdots \cos \theta}$. Let $\theta=\frac{\pi}{4}$ or $\frac{5 \pi}{4}$, we have $|F A|=\frac{3 \mathrm{e}}{1-\frac{\sqrt{2}}{2} \mathrm{e}}, |F B|=\frac{3 \mathrm{e}}{1+\frac{\sqrt{2}}{2} \mathrm{e}}$. On the other hand, let the center of $C$ be $I$, then $|F I|=c$, and $\frac{a^{2}}{c}-c=3$ (where $a, b, c$ are the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse $C$, respectively). Using the cosine rule, we know that $|A I|^{2}=|F A|^{2}+|F I|^{2}-\sqrt{2}|F A| \cdot|F I|, |B I|^{2}=|F B|^{2}+|F I|^{2}+\sqrt{2}|F B| \cdot|F I|, |A B|^{2}=|F A|^{2}+|F B|^{2}+2|F A| \cdot|F B|$. According to the problem, we have: $|A I|^{2} + |B I|^{2} > |A B|^{2}$. Therefore, we should have $2 c^{2}+\sqrt{2}(|F B| \cdots|F A|) c > 2|F A| \cdot|F B|$, i.e., $2 c^{2}+$ Since $\frac{a^{2}}{c}-c=3$, we have $c=\frac{3 \mathrm{e}^{2}}{1-\mathrm{e}^{2}}$. Substituting this into the above equation, we get $$ \frac{\mathrm{e}^{2}}{\left(1-\mathrm{e}^{2}\right)^{2}}-\frac{\mathrm{e}^{2}}{\left(1-\mathrm{e}^{2}\right)\left(1-\frac{1}{2} \mathrm{e}^{2}\right)}>\frac{1}{1-\frac{1}{2} \mathrm{e}^{2}} \text{. } $$ Solving this, we get $0<\mathrm{e}<\sqrt{2-\sqrt{2}}$." e718a3e46011,"6. If from the set $S=\{1,2, \cdots, 20\}$, we take a three-element subset $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, such that it simultaneously satisfies: $a_{2}-a_{1} \geqslant 5,4 \leqslant a_{3}-a_{2} \leqslant 9$, then the number of all such subsets $A$ is $\qquad$ (answer with a specific number).",See reasoning trace,medium,"6.251. $a_{2}-a_{1} \geqslant 5 \Leftrightarrow\left(a_{2}-4\right)-a_{1} \geqslant 1$, $4 \leqslant a_{3}-a_{2} \leqslant 9 \Leftrightarrow 1 \leqslant\left(a_{3}-7\right)-\left(a_{2}-4\right) \leqslant 6$. Let $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-4, a_{3}^{\prime}=a_{3}-7$, then $a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{Z}$, satisfying $1 \leqslant a_{1}^{\prime}0)$ and the line $x=1$ enclose a closed figure with an area of $\frac{4}{3}$. Then the coefficient of the $x^{-18}$ term in the expansion of $\left(x+\frac{a}{x}\right)^{20}$ is $\qquad$,"\frac{4}{3} \Rightarrow a=1$. Therefore, the term containing $x^{-18}$ is $\mathrm{C}_{20}^{19} x\le",medium,"14. 20 . According to the problem, we know $2 \int_{0}^{1} \sqrt{a x} \mathrm{~d} x=\frac{4}{3} \Rightarrow a=1$. Therefore, the term containing $x^{-18}$ is $\mathrm{C}_{20}^{19} x\left(\frac{1}{x}\right)^{19}$." b23a51c38e9f,"Let $x$ be a real number and $n$ a natural number greater than or equal to 1. Calculate: $$ 1+x+x^{2}+\cdots+x^{n} $$",See reasoning trace,medium,"A case distinction must be made according to $x=1$ and $x \neq 1$ and we find: $$ 1+x+x^{2}+\cdots+x^{n}= \begin{cases}\frac{1-x^{n+1}}{1-x} & \text { if } x \neq 1 \\ n+1 & \text { if } x=1\end{cases} $$ ## - Inequalities - ## Basic Manipulations ## Theorem 1.1.5. Let $a, b, c, d, \lambda$ be real numbers. — Link between strict and non-strict: If $a0$, then $G O=\sqrt{100 r}=\sqrt{100} \times \sqrt{r}=10 \sqrt{r}$. Thus, $S V=G O=10 \sqrt{r}$. Using a similar analysis, $$ H O^{2}=Q O^{2}-Q H^{2}=(16+r)^{2}-(16-r)^{2}=\left(256+32 r+r^{2}\right)-\left(256-32 r+r^{2}\right)=64 r $$ Since $H O>0$, then $H O=\sqrt{64 r}=\sqrt{64} \times \sqrt{r}=8 \sqrt{r}$. Thus, $T V=H O=8 \sqrt{r}$. From (b), $S T=40$ and so $S V+T V=40$. Therefore, $10 \sqrt{r}+8 \sqrt{r}=40$ which gives $18 \sqrt{r}=40$ and so $\sqrt{r}=\frac{40}{18}=\frac{20}{9}$. Finally, $r=\left(\frac{20}{9}\right)^{2}=\frac{400}{81}$." 61fc11c1608e,"In the diagram, the radius of the larger circle is 3 times the radius of the smaller circle. What fraction of the area of the larger circle is not shaded? (A) $\frac{8}{9}$ (B) $\frac{2}{3}$ (D) $\frac{7}{9}$ (E) $\frac{1}{3}$ (C) $\frac{5}{6}$ ![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-012.jpg?height=247&width=244&top_left_y=2267&top_left_x=1339)",(A),medium,"The fraction of the area of the larger circle that is not shaded does not depend on the actual radius of either circle, and so we begin by letting the radius of the smaller circle be 1 and thus the radius of the larger circle is 3 . In this case, the area of the smaller circle is $\pi(1)^{2}=\pi$. The area of the larger circle is $\pi(3)^{2}=9 \pi$. The area of the larger circle that is not shaded is $9 \pi-\pi=8 \pi$. Therefore, the fraction of the area of the larger circle that is not shaded is $\frac{8 \pi}{9 \pi}=\frac{8}{9}$. (Alternately, we could note that the fraction of the area of the larger circle that is shaded is $\frac{\pi}{9 \pi}=\frac{1}{9}$ and so the fraction of the area of the larger circle that is not shaded is $1-\frac{1}{9}=\frac{8}{9}$.) ANswer: (A)" b5d78ca72d7c,"4. For which integer values of $a$ and $b$ does the system $$ \left\{\begin{array}{l} m^{2}+n^{2}=b \\ \frac{m^{n}-1}{m^{n}+1}=a \end{array}\right. $$ have a solution (for $m$ and $n$) in the set $\mathbb{Z}$ of integers.",See reasoning trace,medium,"Solution. From the first equation, we get $m^{n}=\frac{1+a}{1-a}=1+\frac{2 a}{1-a}$, which means that the number $\frac{2 a}{1-a}$ must be an integer. Since $a$ and $a-1$ are coprime for $a \neq 0$, the number $\frac{2 a}{1-a}$ is an integer if and only if $a=-1,2,3$. For $a=-1$ we have $m^{n}=0$, so we get $m=0$ and $n$ is any integer different from 0. For $a=2$ we get $m^{n}=-3$, so $m=-3, n=1$. For $a=3$, we get $m^{n}=-2$, so $m=-2, n=1$. On the other hand, for $a=0$ we have $m^{n}=1$, from which we get $n=0, m$ is any integer different from zero or $m=1$ and $n$ is any integer. Therefore, the system has a solution in the set of integers for: $a=-1, b=n^{2}, n \neq 0, a=0, b=1+n^{2} ; a=2, b=10 ; a=3, b=5$. ## 4th year" 41d38aa13212,"## Task B-3.1. The intersection of a right circular cone by a plane containing its axis of symmetry is a right triangle. Calculate the surface area of the sphere inscribed in the cone if the height of the cone is $2 \mathrm{~cm}$.","R+R \sqrt{2}$ (2 points), from which $R=\frac{h}{1+\sqrt{2}}=\frac{2}{1+\sqrt{2}}(1$ point).",medium,"## Solution. ![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-17.jpg?height=522&width=1402&top_left_y=961&top_left_x=248) The cross-section of the cone and the inscribed sphere is an isosceles right triangle with an inscribed circle of radius $R$. The generators of the cone are the legs, and the diameter of the base of the cone is the hypotenuse of this right triangle. (sketch) The radius $R$ of the inscribed circle in a right triangle is calculated using the formula $R=\frac{1}{2}(a+b-c)$, where $a$ and $b$ are the legs, and $c$ is the hypotenuse (or using the area: $\left.P=R \cdot \frac{1}{2}(a+b+c)=\frac{1}{2} c h\right)$. Since $h=2 \mathrm{~cm}$, it follows that $a=b=s=2 \sqrt{2} \mathrm{~cm}$ and $c=2 r=4 \mathrm{~cm}$, so $$ R=\frac{1}{2}(2 \sqrt{2}+2 \sqrt{2}-4)=2(\sqrt{2}-1) \mathrm{cm} $$ Substituting $R$ into the formula for the surface area of the sphere, $O=4 r^{2} \pi$, we get $$ O=16 \pi(3-2 \sqrt{2}) \mathrm{cm}^{2} $$ Note: The student can also find the radius of the sphere using the equality $h=R+R \sqrt{2}$ (2 points), from which $R=\frac{h}{1+\sqrt{2}}=\frac{2}{1+\sqrt{2}}(1$ point)." 32c9e5c33baa,"A softball team played ten games, scoring $1,2,3,4,5,6,7,8,9$, and $10$ runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? $\textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55$","45$, which is $C$",easy,"To score twice as many runs as their opponent, the softball team must have scored an even number. Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much. Therefore, the total runs by the opponent is $(2+4+6+8+10)+(1+2+3+4+5) = 45$, which is $C$" c1a53c65a73d,"Task 1. Determine all pairs $(a, b)$ of positive integers for which $$ a^{2}+b \mid a^{2} b+a \quad \text { and } \quad b^{2}-a \mid a b^{2}+b $$","a^{2}+a+1$ and $a^{2} b+a=a^{2}(a+1)+a=$ $a^{3}+a^{2}+a=a\left(a^{2}+a+1\right)$, so the first divis",medium,"Solution. From $a^{2}+b \mid a^{2} b+a$ it follows that $$ a^{2}+b \mid\left(a^{2} b+a\right)-b\left(a^{2}+b\right)=a-b^{2} . $$ From $b^{2}-a \mid a b^{2}+b$ it follows that $$ b^{2}-a \mid\left(a b^{2}+b\right)-a\left(b^{2}-a\right)=b+a^{2} . $$ We see that $a^{2}+b\left|a-b^{2}\right| a^{2}+b$. This means that $a^{2}+b$ is equal to $a-b^{2}$ up to a sign. Therefore, we have two cases: $a^{2}+b=b^{2}-a$ and $a^{2}+b=a-b^{2}$. In the second case, $a^{2}+b^{2}=a-b$. But $a^{2} \geq a$ and $b^{2} \geq b > -b$, so this is impossible. Therefore, we must have the first case: $a^{2}+b=b^{2}-a$. This gives $a^{2}-b^{2}=-a-b$, so $(a+b)(a-b)=-(a+b)$. Since $a+b$ is positive, we can divide by it and get $a-b=-1$, so $b=a+1$. All pairs that can satisfy the conditions are thus of the form $(a, a+1)$ for a positive integer $a$. We check these pairs. We have $a^{2}+b=a^{2}+a+1$ and $a^{2} b+a=a^{2}(a+1)+a=$ $a^{3}+a^{2}+a=a\left(a^{2}+a+1\right)$, so the first divisibility relation is satisfied. Furthermore, $b^{2}-a=(a+1)^{2}-a=a^{2}+a+1$ and $a b^{2}+b=a(a+1)^{2}+(a+1)=a^{3}+2 a^{2}+2 a+1=$ $a\left(a^{2}+a+1\right)+a^{2}+a+1=(a+1)\left(a^{2}+a+1\right)$, so the second divisibility relation is also satisfied. The pairs $(a, a+1)$ thus satisfy the conditions and are therefore the exact solutions." b70422f2aac5,"3) What is the mean (arithmetic) of the numbers 1, 2, 3, . ., 1999, 2000? (A) 999 (B) 999.5 (C) 1000 (D) 1000.5 (E) 1001.","2001 \times 1000$. Therefore, the arithmetic mean is $S / 2000=2001 / 2=$ 1000.5.",easy,"3) The answer is $(D)$ Just pair the first number with the last, the second with the penultimate, and so on, until you get 1000 pairs. Since the sum of two numbers in a pair is 2001, we will have that the sum of all the numbers is $S=2001 \times 1000$. Therefore, the arithmetic mean is $S / 2000=2001 / 2=$ 1000.5." 603fe7af16c1,"Example 8. Find the differential equation for which the function $y=C_{1} x+C_{2}$, depending on two arbitrary constants, is the general solution.",0$ satisfies the condition of the problem.,easy,"Solution. Differentiating the given function twice, we eliminate the parameters $C_{1}$ and $C_{2}$: $$ y^{\prime}=C_{1}, y^{\prime \prime}=0 $$ The equation $y^{\prime \prime}=0$ satisfies the condition of the problem." 3348f1162614,"Let $M$ and $m$ be the largest and the smallest values of $x$, respectively, which satisfy $4x(x - 5) \le 375$. Find $M - m$.",20,medium,"1. Start with the given inequality: \[ 4x(x - 5) \le 375 \] 2. Expand and rearrange the inequality to standard quadratic form: \[ 4x^2 - 20x \le 375 \] \[ 4x^2 - 20x - 375 \le 0 \] 3. Factor the quadratic expression: \[ 4x^2 - 20x - 375 = 0 \] To factor this, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -20\), and \(c = -375\): \[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 4 \cdot (-375)}}{2 \cdot 4} \] \[ x = \frac{20 \pm \sqrt{400 + 6000}}{8} \] \[ x = \frac{20 \pm \sqrt{6400}}{8} \] \[ x = \frac{20 \pm 80}{8} \] \[ x = \frac{100}{8} \quad \text{or} \quad x = \frac{-60}{8} \] \[ x = \frac{25}{2} \quad \text{or} \quad x = -\frac{15}{2} \] 4. The quadratic expression factors as: \[ 4x^2 - 20x - 375 = 4(x - \frac{25}{2})(x + \frac{15}{2}) \] 5. Determine the intervals where the inequality holds: \[ 4(x - \frac{25}{2})(x + \frac{15}{2}) \le 0 \] The critical points are \(x = \frac{25}{2}\) and \(x = -\frac{15}{2}\). We test the intervals \((-\infty, -\frac{15}{2})\), \((- \frac{15}{2}, \frac{25}{2})\), and \((\frac{25}{2}, \infty)\). 6. Testing the intervals: - For \(x \in (-\infty, -\frac{15}{2})\), both factors \( (x - \frac{25}{2}) \) and \( (x + \frac{15}{2}) \) are negative, so the product is positive. - For \(x \in (-\frac{15}{2}, \frac{25}{2})\), one factor is negative and the other is positive, so the product is negative. - For \(x \in (\frac{25}{2}, \infty)\), both factors are positive, so the product is positive. Therefore, the inequality \(4(x - \frac{25}{2})(x + \frac{15}{2}) \le 0\) holds for: \[ x \in \left[-\frac{15}{2}, \frac{25}{2}\right] \] 7. Identify the largest and smallest values of \(x\): \[ M = \frac{25}{2}, \quad m = -\frac{15}{2} \] 8. Calculate \(M - m\): \[ M - m = \frac{25}{2} - \left(-\frac{15}{2}\right) = \frac{25}{2} + \frac{15}{2} = \frac{40}{2} = 20 \] The final answer is \(\boxed{20}\)" e7c2aadf9e85,"[ Inequality A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? #",16 bugs,medium,"Let there be $x$ bugs in the first jar, then in the second jar - no less than $x+1$ bug, in the third jar - no less than $x+2$ bugs, ..., in the tenth jar no less than $x+9$ bugs. Therefore, the total number of bugs is no less than $10 x+45$. Considering that a total of 150 bugs were distributed, we get $x \leq 10$. On the other hand, in the tenth jar there should be no more than $2 x$ bugs, in the ninth - no more than $2 x-1$ bugs, and so on. This means that in the first jar - no more than $2 x-9$ bugs, and the total number of bugs - no more than $20 x-45$. Since a total of 150 bugs were distributed, then $x \geq 10$. Thus, there are exactly 10 bugs in the first jar, and 19 or 20 in the last jar. Let's find the sum of eleven consecutive numbers starting from ten: $10+11+\ldots+19+20=165$. Since there should be a total of 150 bugs, the jar with 15 bugs is missing. Therefore, the distribution is uniquely determined: 10, 11, 12, 13, 14, 16, 17, 18, 19, and 20 bugs from the first to the tenth jar respectively. Thus, there are 16 bugs in the sixth jar. ## Answer 16 bugs. ## Problem" fb0d09ba1595,"## Task 1 - 220921 Determine all natural numbers $n$ that satisfy the following conditions (1) and (2): (1) $n-9$ is a prime number. (2) $n^{2}-1$ is divisible by 10.","2$ and thus $n=11$. Indeed, $11^{2}-1=120$ is also divisible by 10.",easy,"Due to (2), $n^{2}-1$ is even, so $n^{2}$ is odd, so $n$ is odd, so $n-9$ is even, so due to (1) $n-9=2$ and thus $n=11$. Indeed, $11^{2}-1=120$ is also divisible by 10." 9e0fd9da55e1,"A right truncated pyramid is inscribed around a sphere of radius $r$, with its base being a square. What is the volume of this truncated pyramid, if the diagonal of its base is $4 r$?",See reasoning trace,medium,"Let $A E F C$ be a section through the center $O$ of the sphere, perpendicular to two opposite sides of the base and the top of the pyramid; $D B$ is the height of the pyramid. Since $\angle C O D = \angle O A B$, therefore $\triangle C D O \sim \triangle O B A$, and thus $$ A B: r = r: C D $$ but $$ A B = \frac{A E}{2} = r \sqrt{2} $$ and thus $$ C D = \frac{C F}{2} = \frac{r}{2} \sqrt{2} $$ from which $$ C F = r \sqrt{2} $$ The volume of the pyramid is: $$ K = \frac{2 r}{3}\left(\overline{A E}^{2} + \overline{C F}^{2} + A E \cdot C F\right) $$ into which the values of $A B$ and $C F$ are substituted: $$ K = \frac{2 r}{3}\left(8 r^{2} + 2 r^{2} + 4 r^{2}\right) $$ or $$ K = \frac{28 r^{3}}{3} $$ (András Hrivnák.) Number of solutions: 39." 6308c95c3629,"$4 \cdot 231$ Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfying $$ \sum_{k=1}^{5} k x_{k}=a, \sum_{k=1}^{5} k^{3} \cdot x_{k}=a^{2}, \sum_{k=1}^{5} k^{5} x_{k}=a^{3} . $$","j^{2}$, and $x_{j}=j$, with the rest of the numbers being 0. Since $j$ can be $1,2,3,4,5$, the only ",medium,"[Solution] Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ that satisfy the conditions. Then, by the Cauchy-Schwarz inequality, $$ \begin{array}{l} a^{4}=\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2}=\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} x_{k}^{\frac{1}{2}}\right)\left(k^{\frac{5}{2}} x_{k}^{\frac{1}{2}}\right)\right]^{2} \\ \leqslant\left(\sum_{k=1}^{5} k x_{k}\right)\left(\sum_{k=1}^{5} k^{5} x_{k}\right)=a \cdot a^{3}=a^{4}, \\ \frac{k^{\frac{5}{2}} \cdot x_{k}^{\frac{1}{2}}}{k^{\frac{1}{2}} \cdot x_{k}^{\frac{1}{2}}}=\frac{k^{2} \cdot x_{k}^{\frac{1}{2}}}{x_{k}^{\frac{1}{2}}}=\text { constant. }(k=1,2,3,4,5) \end{array} $$ Thus, $\frac{k^{\frac{5}{2}} \cdot x_{k}^{\frac{1}{2}}}{k^{\frac{1}{2}} \cdot x_{k}^{\frac{1}{2}}}=\frac{k^{2} \cdot x_{k}^{\frac{1}{2}}}{x^{\frac{1}{2}}}=$ constant. $(k=1,2,3,4,5)$ If $x_{1}=x_{2}=\cdots=x_{5}=0$, then $a=0$. If $x_{1}, x_{2}, \cdots, x_{5}$ have two or more non-zero values, then $\circledast$ cannot hold. Therefore, $x_{1}, x_{2}, \cdots, x_{5}$ can only have one non-zero value. Suppose $x_{j} \neq 0$, then, $$ j x_{j}=a, j^{3} x_{j}=a^{2}, j^{5} x_{j}=a^{3}, $$ From the first two equations, we get $a=j^{2}$, and $x_{j}=j$, with the rest of the numbers being 0. Since $j$ can be $1,2,3,4,5$, the only values for $a$ that satisfy the conditions are $a=0,1,4,9,16,25$." bc7b6ab95a45,"12. (12 points) Given $$ \begin{array}{l} f(x, y) \\ =x^{3}+y^{3}+x^{2} y+x y^{2}-3\left(x^{2}+y^{2}+x y\right)+3(x+y), \end{array} $$ and $x, y \geqslant \frac{1}{2}$. Find the minimum value of $f(x, y)$.",See reasoning trace,medium,"12. When $x \neq y$, multiplying both sides of the function by $x-y$ gives $$ \begin{array}{l} (x-y) f(x, y) \\ =\left(x^{4}-y^{4}\right)-3\left(x^{3}-y^{3}\right)+3\left(x^{2}-y^{2}\right) . \\ \text { Let } g(x)=x^{4}-3 x^{3}+3 x^{2} . \end{array} $$ Then $f(x, y)=\frac{g(x)-g(y)}{x-y}$ is the slope of the line segment connecting two points on the graph of $g(x)$. When $x=y$, $f(x, y)=4 x^{3}-9 x^{2}+6 x$. Thus, we only need to find the minimum value of the derivative of $g(x)$ for $x \geqslant \frac{1}{2}$, $$ h(x)=4 x^{3}-9 x^{2}+6 x $$ It is easy to find that when $x \geqslant \frac{1}{2}$, $$ h(x)=4 x^{3}-9 x^{2}+6 x $$ the minimum value is 1." ac7e8a21be39,"2. For numbers $a$ and $b$ such that $a + 10b = 1$, what is the maximum possible value of the product $a \cdot b$? Determine the values of $a$ and $b$ for which this value is achieved.",See reasoning trace,medium,"2. For numbers $a$ and $b$ such that $a+10 b=1$, what is the maximum possible value of the product $a \cdot b$? Determine the values of $a$ and $b$ for which this value is achieved. ## Solution. $$ \begin{aligned} a b & =(1-10 b) b=b-10 b^{2} \\ & =-10 b^{2}+b=-10\left(b^{2}-\frac{1}{10} b\right) \\ & =-10\left(b^{2}-\frac{1}{10} b+\frac{1}{400}-\frac{1}{400}\right) \\ & =-10\left(b^{2}-\frac{1}{10} b+\frac{1}{400}\right)+\frac{1}{40} \\ & =\frac{1}{40}-10\left(b-\frac{1}{20}\right)^{2} . \end{aligned} $$ The square of any number is a non-negative number, and so is its tenfold. Therefore, it follows that $\frac{1}{40}-10\left(b-\frac{1}{20}\right)^{2} \leq \frac{1}{40}$. If $a b=\frac{1}{40}$, then $10\left(b-\frac{1}{20}\right)^{2}=0$ $b-\frac{1}{20}=0$ $b=\frac{1}{20}$. Finally, we have $a=1-10 b=1-10 \cdot \frac{1}{20}=\frac{1}{2}$. Thus, the maximum possible value of the product $a \cdot b$ is $\frac{1}{40}$." c32c49408499,3. Given that $a \cos x + b \cos 2x \geqslant -1 (x \in \mathbf{R})$ always holds. Then the range of $a + b$ is $\qquad$,"-1, b=0$, $a \cos x+b \cos 2 x=-\cos x \geqslant-1(x \in \mathbf{R})$. Therefore, the range of $a+b$",easy,"3. $[-1,2]$. Let $\cos x=-\frac{1}{2} \Rightarrow a+b \leqslant 2$. Also, when $a=\frac{4}{3}, b=\frac{2}{3}$, $2 b \cos ^{2} x+a \cos x-b+1$ $=\frac{1}{3}(2 \cos x+1)^{2} \geqslant 0$. Let $\cos x=1 \Rightarrow a+b \geqslant-1$. Also, when $a=-1, b=0$, $a \cos x+b \cos 2 x=-\cos x \geqslant-1(x \in \mathbf{R})$. Therefore, the range of $a+b$ is $[-1,2]$." 0f3eb9da597d,"4. On each side of the cube, there is a natural number written. Each vertex of the cube is associated with the product of the three numbers that are on the faces containing that vertex. The sum of all eight numbers associated with the vertices is 455. What is the sum of all the numbers written on the sides of the cube?",See reasoning trace,medium,"4. On each side of the cube, a natural number is written. Each vertex of the cube is associated with the product of the three numbers that are on the faces that determine that vertex. The sum of all eight of these numbers is 455. What is the sum of all the numbers written on the sides of the cube? ## Solution. Let the vertices of the cube be standardly labeled as $A, B, \ldots, H$. Let the number written on the bottom face be $x$, and on the top face $a$. Let the number written on the front face be $y$, and on the back face $b$. Let the number written on the left face be $z$, and on the right face $c$. ![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-17.jpg?height=432&width=1558&top_left_y=2047&top_left_x=250) Then, the number associated with vertex $A$ is $x y z$, with vertex $B$ is $x y c$, with vertex $C$ is $x b c$, with vertex $D$ is $x z b$, with vertex $E$ is $y z a$, with vertex $F$ is $y a c$, with vertex $G$ is $a b c$, and with vertex $H$ is $z a b$. We have: $$ \begin{gathered} x y z + x y c + x b c + x z b + y z a + y a c + a b c + z a b = 455 \\ x \cdot (y z + y c + b c + z b) + a \cdot (y z + y c + b c + z b) = 455 \\ (y z + y c + b c + z b) \cdot (x + a) = 455 \\ {[y \cdot (z + c) + b \cdot (c + z)] \cdot (x + a) = 455} \\ (z + c) \cdot (y + b) \cdot (x + a) = 455 \end{gathered} $$ The prime factorization of 455 is: $$ 455 = 5 \cdot 7 \cdot 13 $$ Clearly, the sums of the numbers on opposite faces of the cube are 5, 7, and 13 (in any order). Then, the sum of all the numbers written on the cube is $$ x + y + z + a + b + c = (z + c) + (y + b) + (z + c) = 5 + 7 + 13 = 25 $$" 26dc23837f31,"14) Emanuele has made a long journey and cannot sleep. After returning to Italy, at exactly 11:11 Italian time, he states: ""I haven't slept for 53 hours and 53 minutes."" At what time did he last wake up, knowing that at that moment he was in Korea and now he is in Italy (remember that the time difference between Italy and Korea is 7 hours ahead)? (A) 12:04 (B) $12: 18$ (C) $12: 58$ (D) 13:04 (E) $13: 18$.",See reasoning trace,medium,"14) The answer is (B) Let's reason according to Italian time first. A period of 53 hours and 53 minutes is equivalent to 2 days, 5 hours, and 53 minutes. Therefore, to calculate when Emanuele woke up, we need to subtract 5 hours and 53 minutes from 11:11. The most convenient way to perform this calculation is to subtract 6 hours and then add 7 minutes. This easily gives us that the required time is 5:18 (Italian time). A period of 53 hours and 53 minutes is equivalent to 2 days, 5 hours, and 53 minutes. Therefore, to calculate when Emanuele woke up, we need to subtract 5 hours and 53 minutes from 11:11. The most convenient way to perform this calculation is to subtract 6 hours and then add 7 minutes. This easily gives us that the required time is 5:18. To get the Korean time, we just need to add 7 hours to the Italian time. Therefore, Emanuele woke up at 12:18 (Korean time)." e9fb69147a33,"1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 3375. The answer should be presented as an integer.",1680,medium,"Answer: 1680. Solution. Since $3375=3^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three threes, three fives, and two ones, or (b) one three, one nine, three fives, and three ones. We will calculate the number of variants in each case. (a) First, we choose three places out of eight for the threes $\left(C_{8}^{3}=\frac{8!}{3!5!}\right.$ ways), then three places out of the remaining five for the fives ( $C_{5}^{3}=\frac{5!}{3!2!}$ ways). Finally, the remaining places are occupied by ones. By the multiplication rule, we get $C_{8}^{3} \cdot C_{5}^{3}=\frac{8!}{3!3!2!}=560$ ways. (b) Reasoning similarly, we find that the number of ways in this case is $\frac{8!}{3!3!}=1120$. Finally, we get $560+1120=1680$ ways." 1e3bf941c6eb,1. The range of the function $f(x)=\sin ^{4} x \tan x+\cos ^{4} x \cot x$ is,See reasoning trace,medium,"Solution: $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2},+\infty\right)$ Notice that $$ f(x)=\sin ^{4} x \frac{\sin x}{\cos x}+\cos ^{4} x \frac{\cos x}{\sin x}=\frac{\sin ^{6} x+\cos ^{6} x}{\sin x \cos x}=\frac{2-\frac{3}{2} \sin ^{2} 2 x}{\sin 2 x} $$ Let $t=\sin 2 x \in[-1,0) \cup(0,1]$, then $$ f(x)=\frac{2-\frac{3}{2} t^{2}}{t}=\frac{2}{t}-\frac{3}{2} t $$ It is easy to see that $g(t)=\frac{2}{t}-\frac{3}{2} t$ is a decreasing function on the intervals $[-1,0)$ and $(0,1]$. Therefore, the range of $f(x)$ (i.e., $g(t)$) is $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2},+\infty\right)$." ec0fd2225ed8,"1. Function $$ f(x)=\sin ^{4} x+\sin x \cdot \cos x+\cos ^{4} x $$ The maximum value is $\qquad$.","\frac{1}{2}$, $f(x)$ reaches its maximum value $\frac{9}{8}$.",easy,"1. $\frac{9}{8}$. Notice that, $$ \begin{array}{l} f(x)=\sin ^{4} x+\sin x \cdot \cos x+\cos ^{4} x \\ =\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cdot \cos ^{2} x+ \\ \quad \sin x \cdot \cos x \\ =1-\frac{1}{2} \sin ^{2} 2 x+\frac{1}{2} \sin 2 x \\ =-\frac{1}{2}\left(\sin 2 x-\frac{1}{2}\right)^{2}+\frac{9}{8} . \end{array} $$ Therefore, when $\sin 2 x=\frac{1}{2}$, $f(x)$ reaches its maximum value $\frac{9}{8}$." d8c4bd54b928,For which integers $n \geq 3$ does there exist a regular $n$-gon in the plane such that all its vertices have integer coordinates in a rectangular coordinate system?,4,medium,"1. **Assume there exists an $n$-gon all of whose vertices are lattice points.** - We will work in the complex plane and label the vertices counterclockwise by $p_1, p_2, \ldots, p_n$. - The center of the $n$-gon, call it $q$, is the centroid of the vertices: \[ q = \frac{p_1 + p_2 + \cdots + p_n}{n} \] - Since all vertices are lattice points, $q$ must also have rational real and imaginary components. 2. **Consider the rotation of vertices around the center $q$.** - Let $\omega = e^{2\pi i / n}$ be the primitive $n$-th root of unity. Then, the vertices can be expressed as: \[ p_k = q + \omega^{k-1}(p_1 - q) \quad \text{for} \quad k = 1, 2, \ldots, n \] - Specifically, for $k=2$, we have: \[ p_2 = q + \omega(p_1 - q) \] - Solving for $\omega$, we get: \[ \omega = \frac{p_2 - q}{p_1 - q} \] - Since $p_1, p_2, q$ are lattice points, the right-hand side (RHS) has both rational real and imaginary components. 3. **Analyze the properties of $\omega$.** - We know that: \[ \omega = \cos\left(\frac{2\pi}{n}\right) + i \sin\left(\frac{2\pi}{n}\right) \] - For $\omega$ to have rational real and imaginary components, both $\cos\left(\frac{2\pi}{n}\right)$ and $\sin\left(\frac{2\pi}{n}\right)$ must be rational. 4. **Apply Niven's Theorem.** - Niven's Theorem states that the only rational values of $\cos(\theta)$ for $\theta$ in degrees are $\cos(0^\circ) = 1$, $\cos(60^\circ) = \frac{1}{2}$, $\cos(90^\circ) = 0$, $\cos(120^\circ) = -\frac{1}{2}$, and $\cos(180^\circ) = -1$. - For $\cos\left(\frac{2\pi}{n}\right)$ to be rational, $\frac{2\pi}{n}$ must correspond to one of these angles. 5. **Determine the possible values of $n$.** - The angles $\frac{2\pi}{n}$ corresponding to rational $\cos$ values are: \[ \frac{2\pi}{n} = 0, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \pi \] - Solving for $n$, we get: \[ n = 1, 6, 4, 3, 2 \] - Since $n \geq 3$, the only possible values are $n = 3$ and $n = 4$. 6. **Verify the existence of such $n$-gons.** - For $n = 3$ (triangle), it is impossible to have all vertices as lattice points in a regular triangle. - For $n = 4$ (square), it is possible to have all vertices as lattice points (e.g., vertices at $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$). Conclusion: The only integer $n \geq 3$ for which there exists a regular $n$-gon with all vertices having integer coordinates is $n = 4$. The final answer is $\boxed{4}$." ef78efeee2a2,"Zaslavsky A.A. Let $O, I$ be the centers of the circumcircle and incircle of a right triangle; $R, r$ be the radii of these circles; $J$ be the point symmetric to the right-angle vertex with respect to $I$. Find $O J$.",$R-2r$,medium,"Let $ABC$ be a given right triangle, $\angle C=90^{\circ}$. It is obvious that the circle with center $J$ and radius $2r$ touches $AC$ and $BC$. We need to prove that it also touches the circumcircle $\Omega$ of triangle $ABC$; from this it will follow that $OJ=R-2r$. Consider the circle $\omega$ that touches $AC$, $BC$ at points $P$, $Q$ respectively, and touches $\Omega$ internally at point $T$; we need to prove that $J$ is the center of $\omega$. Since $T$ is the center of homothety of $\omega$ and $\Omega$, the lines $TP$, $TQ$ intersect the circumcircle again at points where the tangents are parallel to $AC$ and $BC$, that is, at the midpoints $B'$, $A'$ of the arcs $AC$ and $C$. Therefore, the lines $AA'$ and $BB'$ intersect at point $I$. By Pascal's theorem (see problem 57105), applied to the broken line $C A A' T B' B$, the points $P$, $I$, $Q$ lie on the same line. Since the line $PQ$ is perpendicular to the angle bisector of $\angle C$, $P$, $Q$ are the projections of $J$ onto $AC$ and $BC$, which means that $J$ is the center of $\omega$. ![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-35.jpg?height=1418&width=1532&top_left_y=800&top_left_x=252) Answer $R-2r$." 3b2a701f0ba6,"Let $ABC$ be a right triangle with a right angle at $C.$ Two lines, one parallel to $AC$ and the other parallel to $BC,$ intersect on the hypotenuse $AB.$ The lines split the triangle into two triangles and a rectangle. The two triangles have areas $512$ and $32.$ What is the area of the rectangle? [i]Author: Ray Li[/i]",256,medium,"1. Let $ABC$ be a right triangle with a right angle at $C$. Let $D$ lie on $AC$, $E$ on $AB$, and $F$ on $BC$. The lines $DE$ and $EF$ are parallel to $BC$ and $AC$ respectively, forming a rectangle $DEFG$ inside the triangle. 2. Given that the areas of the two triangles formed are $512$ and $32$, we need to find the area of the rectangle $DEFG$. 3. Since $\triangle ADE \sim \triangle EFB$, the ratio of their areas is the square of the ratio of their corresponding sides. Let $AC = a$ and $BC = b$. Let $AD = a - l$ and $DF = l$, where $l$ is the length of the rectangle. 4. By the property of similar triangles, we have: \[ \left(\frac{a - l}{l}\right)^2 = \frac{32}{512} = \frac{1}{16} \] Solving for $\frac{a - l}{l}$, we get: \[ \frac{a - l}{l} = \frac{1}{4} \] Therefore: \[ a - l = \frac{l}{4} \] Solving for $a$, we get: \[ a = \frac{5l}{4} \] 5. Now, we need to find the area of the rectangle. The area of $\triangle ADE$ is given by: \[ \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times DE = 32 \] Since $AD = \frac{l}{4}$ and $DE = l$, we have: \[ \frac{1}{2} \times \frac{l}{4} \times l = 32 \] Simplifying, we get: \[ \frac{l^2}{8} = 32 \] Solving for $l^2$, we get: \[ l^2 = 256 \] 6. The area of the rectangle $DEFG$ is: \[ \text{Area of } DEFG = l \times w = l \times l = l^2 = 256 \] The final answer is $\boxed{256}$" 7b4a620bf7a8,"31. In one month, three Wednesdays fell on even dates. What date will the second Sunday of the month be?",See reasoning trace,easy,"31. If in one month three Wednesdays fall on even dates, this is possible under the condition that the month is not February and starts on a Tuesday. Then Wednesday will be on the 2nd, 16th, and 30th, and the second Sunday will be on the 13th." 5c4f8be3b4e8,"Example 1 The graph of the function $y=f(x)$ is symmetric to the graph of the function $g(x)=\log _{2} x(x>0)$ with respect to the origin, then the expression for $f(x)$ is (). (A) $f(x)=\frac{1}{\log _{2} x}(x>0)$ (B) $f(x)=\log _{2}(-x)(x>0)$ (C) $f(x)=-\log _{2} x(x<0)$ (D) $f(x)=-\log _{2}(-x)(x<0)$","\log _{2} x(x>$ $0) \Rightarrow f(x)=-\log _{2}(-x)(x<0)$, choose D.",easy,"The symmetric point of $(x, y)$ with respect to the origin is $(-x,-y)$, so $g(x)=\log _{2} x(x>$ $0) \Rightarrow f(x)=-\log _{2}(-x)(x<0)$, choose D." de5b34f7d9b1,"## Task A-4.5. A cube is formed from 27 identical white cubes, and all its outer faces are painted black. (a) One of these cubes is randomly selected and placed on a table on a randomly chosen face. What is the probability that all five visible faces of the cube will be white? (b) A cube is on the table with all five visible faces being white. What is the probability that the sixth face of this cube is also white?",See reasoning trace,medium,"## Solution. After painting the outer sides of the large cube black, 8 small cubes have 3 white faces, 12 small cubes have 4 white faces, 6 small cubes have 5 white faces, and 1 small cube has 6 white faces. Each of the 27 small cubes can be placed on the table in 6 different ways (depending on which face is on the table), so the total number of events is $27 \cdot 6$. Only the cubes with 5 or 6 white faces can be placed on the table such that all 5 visible faces are white. The cube with 6 white faces (there is only one such cube) can be placed in 6 ways, while any of the 6 cubes with 5 white faces can be placed on the table in exactly one way. Therefore, the probability that all five visible faces of the cube are white is $$ \frac{6 \cdot 1 + 1 \cdot 6}{27 \cdot 6} = \frac{2}{27} $$ Let $S$ be the event that the selected cube has 6 white faces, and let $H$ be the event that the selected cube is placed on the table such that all 5 visible faces are white. We calculate the conditional probability of event $S$ given that event $H$ has occurred, i.e., $$ P(S \mid H) = \frac{P(S \cap H)}{P(H)} $$ Note that the probability $P(H)$ was calculated in part (a) of the problem. The probability $P(S \cap H)$ is equal to the probability of event $S$, because if event $S$ occurs, then event $H$ must also occur. This probability is $\frac{1}{27}$ (there is only one completely white cube on the table). Finally, we have $$ P(S \mid H) = \frac{P(S \cap H)}{P(H)} = \frac{P(S)}{P(H)} = \frac{\frac{1}{27}}{\frac{2}{27}} = \frac{1}{2} $$" 8bcfb6e55a1a,"In a certain sequence of 80 numbers, any term, except for the two endpoints, is equal to the product of its neighboring terms. The product of the first 40 terms of the sequence is 8 and the product of all the terms is also 8. Determine the terms of the sequence.",See reasoning trace,medium,"Let $a_{1}, a_{2}, \ldots, a_{80}$ be the numbers of this sequence. For each $i \geq 1$, we have $$ \left\{\begin{array}{l} a_{i+1}=a_{i} \cdot a_{i+2} \\ a_{i+2}=a_{i+1} \cdot a_{i+3} \end{array}\right. $$ Consequently, $a_{i+1}=a_{i} \cdot a_{i+1} \cdot a_{i+3}$ and, since $a_{i+1} \neq 0$, as the product of the terms of the sequence is $8 \neq 0$, it follows that $a_{i} \cdot a_{i+3}=1$. Any two numbers in the sequence, whose indices differ by 3, have a product equal to 1. Therefore, the product of any six consecutive numbers in this sequence is always equal to 1. Given that the product of the first 40 terms of the sequence is 8, we conclude that the product of the first four terms is also 8, because the remaining 36 terms form six groups of six consecutive terms of the sequence, and in each of these groups, the product is equal to 1. That is, $a_{1} a_{2} a_{3} a_{4}=8$. Since $a_{i} \cdot a_{i+3}=1$, it follows that $a_{1} a_{4}=1$ and, therefore, $a_{2} a_{3}=8$. We also have the hypothesis that the product of the 80 terms of the sequence is 8, from which we can conclude that $a_{1} a_{2}=8$, since the last 78 terms can be grouped into 13 groups of six consecutive terms, each with a product equal to 1, as we have already seen. Thus, from $a_{2} a_{3}=8$, $a_{1} a_{2}=8$, and $a_{2}=a_{1} a_{3}$, it follows that $a_{1} a_{2}^{2} a_{3}=64$ and $a_{2}^{3}=64$. Therefore, $$ a_{1}=2, a_{2}=4 \text{ and } a_{3}=2 $$ Observe, also, that the entire sequence is now determined. Indeed, we have $$ 2,4,2, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}, 2,4,2, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}, \ldots $$ where the first six terms repeat, always in the same order." e225cb359ea7,Example 2 Find the positive integer solutions of the equation $3^{x}-5^{y}=z^{2}$.,See reasoning trace,medium,"Taking modulo 2, we get $$ 0 \equiv z^{2}(\bmod 2) \Rightarrow 21 z \text {. } $$ Taking modulo 4, we get $$ (-1)^{x}-1 \equiv 0(\bmod 4) \Rightarrow 2 \mid x \text {. } $$ Let $x=2 x_{1}$. Then $\left(3^{x_{1}}+z\right)\left(3^{x_{1}}-z\right)=5^{y}$. Let $\left\{\begin{array}{l}3^{x_{1}}+z=5^{\alpha}, \\ 3^{x_{1}}-z=5^{\beta},\end{array}\right.$ where $\alpha+\beta=y$, and $\alpha>\beta$. Adding the two equations, we get $$ 2 \times 3^{x_{1}}=5^{\beta}\left(5^{\alpha-\beta}+1\right) \text {. } $$ Therefore, $\beta=0$. Thus, $\alpha=y$. (1) If $x_{1}=1$, then $5^{y}+1=6$. Hence $y=1, x=2 x_{1}=2, z=5^{y}-3^{x_{1}}=2$. Thus, $(x, y, z)=(2,1,2)$. (2) If $x_{1} \geqslant 2$, then taking modulo 9, we get $$ 5^{y}+1 \equiv 0(\bmod 9), $$ i.e., $5^{y} \equiv-1(\bmod 9)$. However, for any $y \in \mathbf{N}_{+}$, we have $$ 5^{y} \equiv 5,7,-1,4,2,1(\bmod 9) \text {. } $$ Thus, $y \equiv 3(\bmod 6)$. Let $y=6 y_{1}+3$. Then $$ 5^{6 y_{1}+3}+1=2 \times 3^{x_{1}} \text {. } $$ And $5^{6 y_{1}+3}+1=\left(5^{3}\right)^{2 y_{1}+1}+1$ has a factor $$ 5^{3}+1=2 \times 3^{2} \times 7 \text {, } $$ but $7 \times 2 \times 3^{x_{1}}$, so there is no positive integer solution in this case. In summary, the positive integer solution to the original equation is $$ (x, y, z)=(2,1,2) . $$" 31b79ec30195,"11. (20 points) Let point $Z$ move on the circle $|z|=3$ in the complex plane, and let $w=\frac{1}{2}\left(z+\frac{1}{z}\right)$. The trajectory of $w$ is the curve $\Gamma$. A line $l$ passes through the point $P(1,0)$ and intersects the curve $\Gamma$ at points $A$ and $B$, and intersects the imaginary axis at point $M$. If $\overrightarrow{M A}=t \overrightarrow{A P}$ and $\overrightarrow{M B}=s \overrightarrow{B P}$, find the value of $t+s$.",See reasoning trace,medium,"11. Given $|z|=3$, let $z=3 e^{\theta \mathrm{i}}$. Thus, $w=\frac{1}{2}\left(3 \mathrm{e}^{\theta \mathrm{i}}+\frac{1}{3} \mathrm{e}^{-\theta \mathrm{i}}\right)$ $=\frac{5}{3} \cos \theta+\frac{4}{3} \mathrm{i} \sin \theta$. Eliminating the parameter $\theta$ yields $\Gamma: \frac{x^{2}}{\frac{25}{9}}+\frac{y^{2}}{\frac{16}{9}}=1$. Therefore, $P(1,0)$ is the right focus of $\Gamma$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(0, y_{0}\right)$. Then $$ \begin{array}{l} \left(x_{1}, y_{1}-y_{0}\right)=t\left(1-x_{1},-y_{1}\right) \\ \Rightarrow x_{1}=\frac{t}{1+t}, y_{1}=\frac{y_{0}}{1+t} \\ \Rightarrow \frac{\left(\frac{t}{1+t}\right)^{2}}{\frac{25}{9}}+\frac{\left(\frac{y_{0}}{1+t}\right)^{2}}{\frac{16}{9}}=1 \\ \Rightarrow \frac{16}{25} t^{2}+2 t+1-\frac{9}{16} y_{0}^{2}=0 . \end{array} $$ Similarly, $\frac{16}{25} s^{2}+2 s+1-\frac{9}{16} y_{0}^{2}=0$. Since $s t<0$, $s$ and $t$ are the roots of the equation $$ \frac{16}{25} x^{2}+2 x+1-\frac{9}{16} y_{0}^{2}=0 $$ By Vieta's formulas, we have $$ t+s=\frac{-2}{\frac{16}{25}}=-\frac{25}{8} . $$" c465e4abc686,"If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$, and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$, then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is $\textbf{(A) }6\qquad \textbf{(B) }2\sqrt{6}\qquad \textbf{(C) }\frac{8\sqrt{3}}{3}\qquad \textbf{(D) }3\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$",\textbf{(D),easy,"Solution by e_power_pi_times_i Notice that quadrilateral $Q_1Q_2Q_3Q_4$ consists of $3$ equilateral triangles with side length $2$. Thus the area of the quadrilateral is $3\cdot(\frac{2^2\cdot\sqrt{3}}{4}) = 3\cdot\sqrt{3} = \boxed{\textbf{(D) } 3\sqrt{3}}$." 411a66b2c442,"4. From the vertex of the obtuse angle $A$ of triangle $A B C$, the altitude $A D$ is dropped. A circle with center $D$ and radius $D A$ is drawn, intersecting sides $A B$ and $A C$ again at points $M$ and $N$ respectively. Find $A C$, if $A B=c, A M=m$ and $A N=n$.",$\frac{m c}{n}$,medium,"Answer: $\frac{m c}{n}$. Solution. We will prove that $A M \cdot A B = A N \cdot A C$. This can be done in different ways. First method. In the right triangles $A D B$ and $A D C$, draw the altitudes $D P$ and $D Q$ respectively (see Fig. 10.4a). Then $A P \cdot A B = A D^{2} = A Q \cdot A C$. Since triangles $A D M$ and $A D N$ are isosceles, we have $A P = \frac{1}{2} A M$ and $A Q = \frac{1}{2} A N$. Substituting $A P$ and $A Q$ in the equation $A P \cdot A B = A Q \cdot A C$, we get the required result. Second method. We will prove that the quadrilateral $B M N C$ is cyclic, then the required equality will follow from the theorem of the segments of secants applied to point $A$ and the circumcircle of quadrilateral $B M N C$ (see Fig. 10.4b). Let $\angle A N M = \alpha$, then $\angle A O M = 2 \alpha$ (inscribed and central angles subtending the same arc). Additionally, from the isosceles triangle $A D M: \angle M A D = 90^{\circ} - \alpha$, so $\angle A B C = \alpha$. From the equality $\angle A B C = \angle A N M$, it follows that $B M N C$ is cyclic. After proving the required equality, it is sufficient to substitute the given data from the problem and obtain the answer. Third method. Let the given circle intersect segments $B D$ and $C D$ at points $K$ and $L$ respectively, and let its radius be $R$ (see Fig. 10.4c). Then, by the theorem of the segments of secants: $B A \cdot B M = B L \cdot B K$, that is, $c(c - m) = B K(B K + 2 R)$. From triangle $A B D$ by the Pythagorean theorem: $c^{2} = (B K + R)^{2} + R^{2} = 2 R^{2} + B K^{2} + 2 B K \cdot R$. Therefore, $c(c - m) = c^{2} - 2 R^{2}$, from which $m = 2 R^{2}$. By conducting a similar reasoning for side $A C$, we get $A C \cdot n = 2 R^{2}$. Then $A C = \frac{m c}{n}$. Note that in this solution method, instead of the Pythagorean theorem, the cosine theorem can be applied to triangle $B A K$. ![](https://cdn.mathpix.com/cropped/2024_05_06_ca89b09c7a9d835f2b71g-2.jpg?height=474&width=734&top_left_y=183&top_left_x=1186) Grading criteria. ![](https://cdn.mathpix.com/cropped/2024_05_06_ca89b09c7a9d835f2b71g-2.jpg?height=1036&width=752&top_left_y=732&top_left_x=1188) ""+"" - a complete and justified solution is provided “士” - a generally correct reasoning is provided, containing minor gaps or inaccuracies (e.g., $m$ and $n$ are confused) “耳” - the plan of the solution is correct and the correct answer is obtained, but some of the used facts are not proven (e.g., it is used but not proven that quadrilateral $BMNC$ is cyclic) ""干"" - the plan of the solution is correct, but the solution itself contains errors or is not completed “Ғ” - there is no clear plan of the solution, but some significant facts are justified, from which the solution can be obtained “-"" - only the answer is provided “-"" - the problem is not solved or solved incorrectly" 1da599452248,"## Task $7 / 76$ The smallest natural number $n$ is sought, with the property that both the cross sum $Q(n)$ of the number $n$ and the cross sum $Q(n+1)$ of the successor of $n$ are divisible by 5.",49999$. It is $Q(n)=40$ and $Q(n+1)=5$.,medium,"The number $n$ must end in the digit 9; for with any other digit, $Q(n+1)=Q(n)+1$, and it would not be possible for both $Q(n)$ and $Q(n+1)$ to be divisible by 5. Let $n=10a+9$, where $a$ is a natural number with $a>0$. It is easy to see that the number $a$ must also end in the digit 9; otherwise, $Q(n+1)=Q(a)+1=Q(n)-8$ because $Q(n)=Q(a)+9$. Therefore, $a=10b+9$ with $b \neq 0$, of course, so $n=100b+99$. Similarly, one can see that $b$ must end in the digit 9, otherwise $Q(n+1)=Q(b)+1=Q(n)-17$ would result. Thus, $b=10c+9$ and hence $n=1000c+999$. Continuing this reasoning leads to $c=10d+9$ with $n=10000d+9999$ and $Q(n+1)=Q(d)+1=Q(n)-35$. Here the chain breaks off. One recognizes that $Q(n)$ and $Q(n+1)$ are always both either divisible by 5 or not divisible by 5 if and only if the last four digits of the number $n$ are all 9. The smallest natural number with the required property is thus $n=49999$. It is $Q(n)=40$ and $Q(n+1)=5$." 8574768b7ae8,"Example 2 Find all solutions to the indeterminate equations $$x^{2}-8 y^{2}=-1$$ and $$x^{2}-8 y^{2}=1$$","1,2, \cdots$ one by one, instead of finding the periodic continued fraction of $\sqrt{d}$. From the ",medium,"From Example 4 in §3, we know that $\sqrt{8}=\langle 2, \overline{1,4}\rangle$. The period is 2. Therefore, by Theorem 1 and Theorem 3, the indeterminate equation (21) has no solution, and the smallest positive solution of the indeterminate equation (22) is $x=h_{1}, y=k_{1}$. It is easy to find that $$h_{1} / k_{1}=\langle 2,1\rangle=3 / 1$$ Thus, by Corollary 4, the complete solution of the indeterminate equation (22) is $$x+y \sqrt{8}= \pm(3 \pm \sqrt{8})^{j}, \quad j=0,1,2,3, \cdots$$ When $d$ is relatively small or a special number, we can find the smallest positive solution of $$x^{2}-d y^{2}=-1 \quad \text { or } x^{2}-d y^{2}=1$$ by trying $y=1,2, \cdots$ one by one, instead of finding the periodic continued fraction of $\sqrt{d}$. From the above discussion, when the first solution found is the smallest positive solution of $x^{2}-d y^{2}=-1$, we can find all solutions of these two equations according to Corollary 4 (ii); when it is the smallest positive solution of $x^{2}-d y^{2}=1$, $x^{2}-d y^{2}=-1$ has no solution, and we can find all solutions of $x^{2}-d y^{2}=1$ according to Corollary 4 (i). For example, in the case of Example 2, when $y=1$, $x=3, y=1$ is a solution of the indeterminate equation (22), so the indeterminate equation (21) has no solution. This immediately gives us all solutions of the indeterminate equation (22). Here is another example." 917ae61812c6,"8. [30] Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N$ ). Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number, at which point that player loses. For how many values of $N$ between 2 and 100 inclusive does player 1 have a winning strategy?",47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2,medium,"Answer: 47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2 . First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$ (such as $a$ ), and you will have a smaller odd number. Eventually you will get to an odd prime and lose. Now consider even numbers that aren't powers of 2 . As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win. Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2. Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win. Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 ." c708235a65ee,"If $4 x+12=48$, the value of $x$ is (A) 12 (B) 32 (C) 15 (D) 6 (E) 9",(E),easy,"If $4 x+12=48$, then $4 x=48-12$ or $4 x=36$, and so $x=\frac{36}{4}=9$. ANsWER: (E)" 061b00bbb3ce,"XLVII OM - I - A palindromic number is defined as a natural number whose decimal representation read from left to right is the same as when read from right to left. Let $ (x_n) $ be the increasing sequence of all palindromic numbers. Determine all prime numbers that are divisors of at least one of the differences $ x_{n+1}-x_n $.",See reasoning trace,medium,"The decimal representation of the palindromic number $ x_n $ (respectively: ($ 2m $)-digit or ($ 2m-1 $)-digit) looks like this: or ($ c_i \in \{ 0,1,2,3,4,5,6,7,8,9\} $ for $ i = 1, \ldots ,m;\ c_1 \ne 0 $). There are three possible situations: Case I. $ c_m \ne 9 $. If $ x_n $ has the form (1), then $ x_{n+1} $ (the smallest palindromic number greater than $ x_n $) has the decimal representation where $ d_m = c_m + 1 $. Justification: Example (3) shows that there are palindromic numbers greater than $ x_n $ and having the same initial digits $ c_1,c_2,\ldots,c_{m-1} $ as $ x_n $; thus, $ x_{n+1} $ must be one of these numbers. Its next ($ m $-th) digit cannot be $ c_m $, because then the number would be identical to $ x_n $ (formula (1)). By taking $ d_m= c_m +1 $ as the next digit, we get the only palindromic number with the initial digits $ c_1,c_2,\ldots,c_{m-1},d_m $ (its further digits are already uniquely determined; see formula (3)); and by writing any digit greater than $ d_m $ on the $ m $-th place, we get even larger numbers; therefore, formula (3) indeed represents the number $ x_{n+1} $. Analogous reasoning shows that if $ x_n $ has the form (2), then $ x_{n+1} $ has the decimal representation where $ d_m = c_m + 1 $. We calculate the difference $ x_{n+1} -x_n $ using the usual subtraction algorithm (that is, by writing the digits of the number $ x_n $ under the corresponding digits of the number $ x_{n+1} $ and performing ""written subtraction""). The result of the subtraction: Case II. Not all digits of the number $ x_n $ (given by formula (1) or (2)) are nines, but the digit $ c_m $ is a nine: $ c_m = 9 $. Let $ c_k $ be the last digit different from $ 9 $ in the sequence $ c_1,c_2,\ldots,c_{m-1},c_m $. Then where $ l = 2m-2k $ or $ l = 2m-1-2k $, depending on whether the number $ x_n $ is expressed by formula (1) or (2). In each of these cases, the smallest palindromic number greater than $ x_n $ has the representation where $ d_k = c_k + 1 $. (Detailed justification - very similar to the one conducted in Case I.) The subtraction algorithm now gives the result Case III. All digits of the number $ x_n $ are nines: Then $ x_{n+1} = \overline{1 \underbrace{00\ldots 0}_{p-1}1} $, so Cases I, II, III exhaust all possibilities. The results obtained in formulas (5), (6), (7) show that the only prime divisors of the differences $ x_{n+1}-x_n $ are the numbers $ 2 $, $ 5 $, and $ 11 $." db832faad7a9,"## Calculate the indefinite integral: $$ \int \frac{3 x^{3}-x^{2}-12 x-2}{x(x+1)(x-2)} d x $$",See reasoning trace,medium,"## Solution $$ \int \frac{3 x^{3}-x^{2}-12 x-2}{x(x+1)(x-2)} d x=\int \frac{3 x^{3}-x^{2}-12 x-2}{x^{3}-x^{2}-2 x} d x= $$ Under the integral, we have an improper fraction. Let's separate the integer part: $$ \begin{array}{ll} 3 x^{3}-x^{2}-12 x-2 & \mid x^{3}-x^{2}-2 x \\ \frac{3 x^{3}-3 x^{2}-6 x}{2 x^{2}-6 x-2} & 3 \end{array} $$ We get: $$ =\int\left(3+\frac{2 x^{2}-6 x-2}{x^{3}-x^{2}-2 x}\right) d x=\int 3 d x+\int \frac{2 x^{2}-6 x-2}{x(x+1)(x-2)} d x= $$ Let's decompose the proper rational fraction into partial fractions using the method of undetermined coefficients: $$ \begin{aligned} & \frac{2 x^{2}-6 x-2}{x(x+1)(x-2)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-2}= \\ & =\frac{A(x+1)(x-2)+B x(x-2)+C x(x+1)}{x(x+1)(x-2)}= \\ & =\frac{A\left(x^{2}-x-2\right)+B\left(x^{2}-2 x\right)+C\left(x^{2}+x\right)}{x(x+1)(x-2)}= \\ & =\frac{(A+B+C) x^{2}+(-A-2 B+C) x-2 A}{x(x+1)(x-2)}= \\ & \left\{\begin{array}{l} A+B+C=2 \\ -A-2 B+C=-6 \\ -2 A=-2 \end{array}\right. \end{aligned} $$ $$ \left\{\begin{array}{l} B+C=1 \\ -2 B+C=-5 \\ A=1 \end{array}\right. $$ Add to the second row the first row multiplied by 2: $$ \begin{aligned} & \left\{\begin{array}{l} B+C=1 \\ 3 C=-3 \\ A=1 \end{array}\right. \\ & \left\{\begin{array}{l} B=2 \\ C=-1 \\ A=1 \end{array}\right. \\ & \frac{2 x^{2}-6 x-2}{x(x+1)(x-2)}=\frac{1}{x}+\frac{2}{x+1}-\frac{1}{x-2} \end{aligned} $$ Then we get: $$ \begin{aligned} & =\int 3 d x+\int\left(\frac{1}{x}+\frac{2}{x+1}-\frac{1}{x-2}\right) d x= \\ & =3 x+\ln |x|+2 \cdot \ln |x+1|-\ln |x-2|+C \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-28$ » Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals Ukrainian Banner Network - Last modified: 11:38, 7 April 2009. - Content is available under CC-BY-SA 3.0." 1aafa29609df,"When the integer $300^{8}$ is written out, it has $d$ digits. What is the value of $d$ ?",See reasoning trace,easy,"We note that $300^{8}=3^{8} \cdot 100^{8}=3^{8} \cdot\left(10^{2}\right)^{8}=6561 \cdot 10^{16}$. Multiplying 6561 by $10^{16}$ is equivalent to appending 16 zeroes to the right end of 6561 , creating an integer with 20 digits." ab631bf0e6df,315. The numbers $2 \overline{a c}+1$ and $3 \overline{a c}+1$ are perfect squares. Find $\overline{a c}$.,"9, z=11$, hence $x=40$.",medium,"315. First method. By setting $\overline{a c}=x$, we obtain the system $\left\{\begin{array}{l}2 x+1=y^{2} \\ 3 x+1=z^{2}\end{array}\right.$, i.e., $\frac{y^{2}-1}{2}$ and $\frac{z^{2}-1}{3}$ represent equal integers. It is easy to find $x=2 t^{2}+2 t$ and $x=3 t_{1}^{2} \pm 2 t_{1}$. According to the problem's condition, we get $\left\{\begin{array}{l}2 t^{2}+2 t>9 \\ 2 t^{2}+2 t<100\end{array}\right.$ and $\left\{\begin{array}{l}3 t_{1}^{2} \pm 2 t_{1} \geq 9 \\ 3 t_{1}^{2} \pm 2 t_{1}<100\end{array}\right.$, from which we find $x=12,24,40,60,84 ; x=21,40,65$, $96,16,33,56,85$. From this, we conclude that $\overline{a c}=40$. Second method. Excluding $x$ from the system $\left\{\begin{array}{l}2 x+1=y^{2} \\ 3 x+1=z^{2}\end{array}\right.$, we find the equation $3 y^{2}-2 z^{2}=1$. Solving this equation using the method indicated in № 266, we get $y=9, z=11$, hence $x=40$." b55eb456b498,"## Calculate the limit of the function: $\lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}$",See reasoning trace,medium,"## Solution Substitution: $x=y+\pi \Rightarrow y=x-\pi$ $x \rightarrow \pi \Rightarrow y \rightarrow 0$ We get: $$ \begin{aligned} & \lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}=\lim _{y \rightarrow 0} \frac{1+\cos 3(y+\pi)}{\sin ^{2} 7(y+\pi)}= \\ & =\lim _{y \rightarrow 0} \frac{1+\cos (3 y+3 \pi)}{\sin ^{2}(7 y+7 \pi)}=\lim _{y \rightarrow 0} \frac{1+\cos (3 y+\pi)}{\sin ^{2}(7 y+\pi)}= \\ & =\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{(-\sin 7 y)^{2}}=\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{\sin ^{2} 7 y}= \end{aligned} $$ Using the substitution of equivalent infinitesimals: $\sin 7 y \sim 7 y$, as $y \rightarrow 0(7 y \rightarrow 0)$ $1-\cos 3 y \sim \frac{(3 y)^{2}}{2}$, as $y \rightarrow 0$ We get: $=\lim _{y \rightarrow 0} \frac{\frac{(3 y)^{2}}{2}}{(7 y)^{2}}=\lim _{y \rightarrow 0} \frac{9 y^{2}}{2 \cdot 49 y^{2}}=\lim _{y \rightarrow 0} \frac{9}{2 \cdot 49}=\frac{9}{98}$ ## Problem Kuznetsov Limits 13-3" e1fa27641313,"79. A store sells a certain product and makes a profit of $m$ yuan per item, with a profit margin of $20\%$ (profit margin $=\frac{\text{selling price - purchase price}}{\text{purchase price}} \times 100\%$). If the purchase price of this product increases by $25\%$, and the store raises the selling price so that it still makes a profit of $m$ yuan per item, what is the profit margin after the price increase? A. $25\%$ B. $20\%$ C. $16\%$ D. $12.5\%$",C,easy,Answer: C 1d78e5c8fad3,"## Task 3 Calculate in writing: $438+8006+98+728$",9270$,easy,$438+8006+98+728=9270$ e16eae0a273a,"In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, then $m\angle ADB=$ $\textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ$",$\textbf{(C)},easy,"Draw the diagram using the information above. In triangle $DCB,$ note that $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, so $m\angle CDB=40^\circ.$ Because $AB \parallel CD,$ we have $m\angle CDB= m\angle DBA = 40^\circ.$ Triangle $DAB$ is isosceles, so $m\angle ADB = 180 - 2(40) = 100^\circ.$ The answer is $\textbf{(C)}.$ -edited by coolmath34" 0e0f3bf94219,"A5. Dion and Jaap participated in a running race. The number of runners who finished before Dion is equal to the number of runners who finished after him. The number of runners who finished before Jaap is three times the number of runners who finished after him. In the final ranking, there are 10 other participants between Dion and Jaap. No runners finished at the same time, and everyone finished the race. How many runners participated in this race? A) 22 B) 23 C) 41 D) 43 E) 45","11$, so $n=4 \times 11+1=45$. The number of participants was therefore 45.",medium,"A5. E) 45 Let the number of participants be $n$. The number of participants who finished before Dion is $\frac{1}{2}(n-1)$ (half of all participants except Dion). The number of participants who finished before Jaap is $\frac{3}{4}(n-1)$. Since exactly 10 participants finished between Dion and Jaap, it follows that $\frac{3}{4}(n-1)-\frac{1}{2}(n-1)$ is equal to 11 (Dion himself and the 10 participants between Dion and Jaap). It follows that $\frac{1}{4}(n-1)=11$, so $n=4 \times 11+1=45$. The number of participants was therefore 45." a9129096c3cf,What is the sum of the first $n$ even and the first $n$ odd numbers?,See reasoning trace,medium,"The sum of the first $2,3,4, \ldots, n$ even numbers: $$ \begin{gathered} 2+4=6=2^{2}+2 \\ 2+4+6=12=3^{2}+3 \\ 2+4+6+8=20=4^{2}+4 \text { and so } \\ 2+4+6+\ldots+2 n=n^{2}+n=n(n+1) \end{gathered} $$ The sum of the first $2,3,4, \ldots, n$ odd numbers: $$ \begin{gathered} 1+3=4=2^{2} \\ 1+3+5=9=3^{2} \\ 1+3+5+7=16=4^{2} \quad \text { and so } \\ 1+3+5+7+\ldots+(2 n-1)=n^{2} \end{gathered} $$ (László Vámossy, Budapest.) Note. Using the sum formula for an arithmetic progression $S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)$, in the first case $$ S_{n}=\frac{n}{2}(2+2 n)=n(n+1) $$ and in the second case $$ S_{n}=\frac{n}{2}(1+2 n-1)=\frac{n}{2} \cdot 2 n=n^{2} $$" f42eb5ab8d8a,"1. A snowdrift 468 cm high decreased in height by 6 cm in the first hour, by 12 cm in the second hour, ..., by $6 k$ cm in the -th hour. After some time $T$, the snowdrift melted completely. What fraction of the snowdrift's height melted in the time $\frac{T}{2}$?",$\frac{7}{26}$,medium,"Answer: $\frac{7}{26}$. Solution: From the equation $6+12+\cdots+6n=468$, we find $6 \cdot \frac{n(n+1)}{2}=468$ or $n^{2}+n-$ $156=0$. From this, $n=12$, meaning the snowdrift melted in 12 hours. In 6 hours, it melted by $6 \cdot \frac{6 \cdot 7}{2}=126$ cm in height, which is $\frac{126}{468}=\frac{7}{26}$. Comment: Correct solution - 20 points. Height of 126 cm found, but answer not obtained or incorrectly calculated - 18 points. Only $T=12$ found - 10 points. Correct approach but error in formula - 5 points. Solution started but progress insignificant - 1 point. Only correct answer without solution - 0 points." bd27851c4e34,"14. How many different cubes are there with three faces coloured red and three faces coloured blue? A 1 B 2 C 3 D 4 E 5",See reasoning trace,medium,"14. B Consider the case where two opposite faces are coloured red. Whichever of the four remaining faces is also coloured red, the resulting arrangement is equivalent under rotation to a cube with top, bottom and front faces coloured red. Hence, there is only one distinct colouring of a cube consisting of three red and three blue faces with two opposite faces coloured red. Now consider the case where no two opposite faces are coloured red. This is only possible when the three red faces share a common vertex and, however these faces are arranged, the resulting arrangement is equivalent under rotation to a cube with top, front and right-hand faces coloured red. Hence there is also only one distinct colouring of a cube consisting of three red and three blue faces in which no two opposite faces are coloured red. Therefore there are exactly two different colourings of the cube as described in the question." 79ddcdf70418,"3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel.","If the original speed was maintained, it would take 180 days",medium,"3. (6 points) A construction company builds a tunnel. When $\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \%$ and reducing the daily working hours to $80 \%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\qquad$ 180 days. 【Solution】Solution: $\left(1-\frac{1}{3}\right) \div[(1+20 \%) \times 80 \%]$ $$ \begin{array}{l} =\frac{2}{3} \div[120 \% \times 80 \%], \\ =\frac{2}{3} \div \frac{24}{25} \\ =\frac{25}{36} ; \\ 185 \div\left(\frac{1}{3}+\frac{25}{36}\right) \\ =185 \div \frac{37}{36}, \\ =180 \text { (days). } \end{array} $$ Answer: If the original speed was maintained, it would take 180 days. Therefore, the answer is: 180." 20b994785ecd,"11. (12 points) There is a cylinder, the height of which is 3 times the radius of the base. It is divided into two cylinders, a larger one and a smaller one, as shown in the figure. The surface area of the larger cylinder is 3 times that of the smaller cylinder. Therefore, the volume of the larger cylinder is $\qquad$ times that of the smaller cylinder.",: 11,medium,"【Solution】Solution: Let the radius of the base of this cylinder be $r$, then the height is $3 r$, the height of the small cylinder is $h$, so the height of the large cylinder is $(3 r-h)$; Since the surface area of the large cylinder is 3 times that of the small cylinder, thus $h=\frac{\mathrm{r}}{4}$, then the height of the large cylinder is $\frac{11}{4} r$; and since the base areas of the two cylinders are the same, $$ \frac{11}{4} r \div \frac{r}{4}=11 \text {, } $$ so the volume of the large cylinder is also 11 times that of the small cylinder. Therefore, the answer is: 11." fc87488c9b63,"Let $ f(x) = x^3 + ax + b $, with $ a \ne b $, and suppose the tangent lines to the graph of $f$ at $x=a$ and $x=b$ are parallel. Find $f(1)$.",1,medium,"1. Given the function \( f(x) = x^3 + ax + b \), we need to find \( f(1) \) under the condition that the tangent lines to the graph of \( f \) at \( x = a \) and \( x = b \) are parallel. 2. The condition for the tangent lines to be parallel is that their slopes must be equal. The slope of the tangent line to the graph of \( f \) at any point \( x \) is given by the derivative \( f'(x) \). 3. Compute the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx}(x^3 + ax + b) = 3x^2 + a \] 4. Set the slopes at \( x = a \) and \( x = b \) equal to each other: \[ f'(a) = f'(b) \] Substituting the derivative, we get: \[ 3a^2 + a = 3b^2 + a \] 5. Simplify the equation: \[ 3a^2 + a - 3b^2 - a = 0 \] \[ 3a^2 = 3b^2 \] \[ a^2 = b^2 \] 6. Since \( a \neq b \), the only solution is \( a = -b \). 7. Now, we need to find \( f(1) \): \[ f(1) = 1^3 + a \cdot 1 + b = 1 + a + b \] 8. Substitute \( b = -a \) into the expression: \[ f(1) = 1 + a - a = 1 \] The final answer is \( \boxed{1} \)." 08a76b4df4d3,"1. Find the numbers of the form $\overline{a b c d}$, written in base $10, \overline{c d}<\overline{a b}$, knowing that the product of the number formed by the first two digits of the number and the number formed by the last two digits of it is equal to 300, and $\frac{(\overline{a b}, \overline{c d})}{[\overline{a b}, \overline{c d}]}=0.08(3)$; where $(\overline{a b}, \overline{c d})$ and $[\overline{a b}, \overline{c d}]$ denote, respectively, the greatest common divisor and the least common multiple of the numbers $\overline{a b}$ and $\overline{c d}$. Prof. Gheorghe Iacoviță",See reasoning trace,medium,"Solution: Let $\overline{ab}=m, \overline{cd}=n$. We can write $m>n, mn=300, \frac{(m, n)}{[m, n]}=0.08(3)$. $\frac{(m, n)}{[m, n]}=0.08(3) \Leftrightarrow \frac{(m, n)}{[m, n]}=\frac{1}{12} \Leftrightarrow 12(m, n)=[m, n] ;$ from which $12(m, n)[m, n]=[m, n]^{2} ;$ and since $(m, m)[m, n]=mn,$ we have: $12 \cdot 300=[m, n]^{2} \Rightarrow [m, n]=60,$ thus $(m, n)=5$. Then $m=5x, n=5y, (x, y)=1$; in conclusion, we will have: $m \cdot n=300 \Rightarrow 5x \cdot 5y=300 \Rightarrow xy=12$. Since $m>n \Rightarrow x>y$, we have the following cases: 1) $x=4, y=3 \Rightarrow m=20, n=15$; i.e., $\overline{ab}=20, \overline{cd}=15$, so $\overline{abcd}=2015$. 2) $x=6, y=2,$ but $(x, y) \neq 1$. 3) $x=12, y=1 \Rightarrow m=60, n=5$, which are not two-digit numbers. In conclusion, $\overline{abcd}=2015$. Scoring: | Let $\overline{ab}=m, \overline{cd}=n \cdot \frac{(m, n)}{[m, n]}=0.08(3) \Leftrightarrow \frac{(m, n)}{[m, n]}=\frac{1}{12} \Leftrightarrow 12(m, n)=[m, n]$ | $\mathbf{1 p}$ | | :--- | :--- | | $12(m, m)[m, n]=[m, n]^{2}$ | $\mathbf{1 p}$ | | Find $[m, n]=60$ and $(m, n)=5$ | $\mathbf{1 p}$ | | $m=5x, n=5y, (x, y)=1 ;$ from which $m \cdot n=300 \Rightarrow 5x \cdot 5y=300 \Rightarrow xy=12$. | $\mathbf{1 p}$ | | Discuss the cases. | $\mathbf{2 p}$ | | Finalization. | $\mathbf{1 p}$ |" b230620d05d4,"8.70 In chess, the king traverses an $8 \times 8$ chessboard, visiting each square exactly once and returning to the starting point (the king's moves follow the usual rules of chess), and when the centers of the squares the king has visited are connected by line segments in sequence, the resulting path is a non-intersecting closed polyline. (1) Provide an example showing that the king can move a total of exactly 28 steps along horizontal and vertical lines. (2) Prove that the total number of steps the king makes along horizontal and vertical directions is at least 28. (3) If the side length of each square is 1, find the maximum and minimum possible lengths of the king's path.",See reasoning trace,medium,"[Solution] (1) See the figure (a) below. (2) Consider the boundary of the chessboard composed of 28 edge squares. Clearly, the King passes through each of these boundary squares exactly once as it traverses the entire chessboard. Number these edge squares in the order the King passes through them as $1, 2, \cdots, 28$, and divide the King's entire path into 28 segments: from square 1 to square 2, from square 2 to square 3, ..., from square 28 to square 1. According to the given conditions, the King's path does not intersect itself, so square $i$ and $i+1$ must be adjacent, for $i=1, 2, \cdots, 28$ (square 29 is square 1). Otherwise, if square 1 and 2 are not adjacent, they would divide the boundary into two parts. When the King moves from a square in one part to a square in the other part, the path taken must intersect the path from square 1 to square 2, which is a contradiction. Since squares $i$ and $i+1$ are adjacent, they must be of different colors. Therefore, when the King moves from square $i$ to square $i+1$ after several steps, at least one of these steps must be from one color to another, and this step must be either along a horizontal or vertical line. This means that the total number of steps the King takes along horizontal and vertical lines is at least 28. (3) Clearly, the total length of the King's path is no less than 64, and there exists a path of length 64 (as shown in figure (b)). On the other hand, from (1) and (2), we know that at least 28 steps of the King's entire path are along horizontal or vertical lines. Therefore, the maximum possible length of the path is $28 + 36\sqrt{2}$." 003f70267c95,"## SUBJECT IV Let the natural number $n=\overline{11 \ldots 1}+\overline{22 \ldots 2}+\ldots+\overline{88 \ldots 8}+\overline{99 \ldots 9}$, where each number of the form $\overline{a a \ldots a}$ contains 2015 digits of $a$. Determine how many digits of 9 the number $n$ contains. ## Note: All subjects are mandatory; Each subject is graded out of 7 points; No points are awarded by default; The actual working time is 2 hours from the moment the subject is received. \section*{MATHEMATICS OLYMPIAD LOCAL STAGE 14.02.2015 GRADE 5 Solutions and grading criteria} untranslated section title: \section*{OLIMPIADA DE MATEMATICĂ ETAPA LOCALÄ 14.02.2015 CLASA a V-a Soluții şi barem de corectare} (Note: The section title is untranslated as it appears to be specific to the original document's context and structure.)",See reasoning trace,easy,"## SUBJECT IV We have $n=\overline{11 \ldots 1}+\overline{22 \ldots 2}+\ldots+\overline{88 \ldots 8}+\overline{99 \ldots 9}=(1+2+\ldots+8+9) \cdot \overline{11 \ldots 1}$ $\qquad$ 1 point $1+2+\ldots+8+9=45$ $\qquad$ 2 points . We obtain $n=45 \cdot \overline{11 \ldots 1}$, $\qquad$ which leads to $n=\overline{499 \ldots .95}$, a number with 2016 digits $\qquad$ 1 point . Finalization, $n$ contains 2014 nines 1 point . Grading: $7 p$" 87110c2b1162,"$10 \cdot 25$ is the sum of an arithmetic sequence with an even number of terms. The sums of the odd-numbered terms and the even-numbered terms are 24 and 30, respectively. If the last term exceeds the first term by 10.5, then the number of terms in the arithmetic sequence is (A) 20. (B) 18. (C) 12. (D) 10. (E) 8. (24th American High School Mathematics Examination, 1973)",$(E)$,medium,"[Solution] Let $a, d$ and $2 n$ represent the first term, common difference, and number of terms of the sequence, respectively. If $S_{1}$ and $S_{2}$ represent the sums of all odd-numbered terms and all even-numbered terms, respectively, then $$ \begin{array}{l} S_{1}=\frac{n}{2}[2 a+(n-1) \cdot 2 d]=24, \\ S_{2}=\frac{n}{2}[2(a+d)+(n-1) \cdot 2 d]=30 . \end{array} $$ The difference $S_{2}-S_{1}=\frac{n}{2} \cdot 2 d=6$, i.e., $n d=6$. If $l$ represents the last term, then $\quad l-a=a+(2 n-1) d-a=10.5$. $\therefore 2 n d-d=10.5$, i.e., $12-d=10.5$, yielding $d=1.5$. $\therefore n=\frac{6}{d}=\frac{6}{1.5}=4$, or $2 n=8$. Therefore, the answer is $(E)$." 64a947817f75,"One, (20 points) A toy factory has a total of 273 labor hours and 243 units of raw materials for producing a batch of teddy bears and kittens. Producing a teddy bear requires 9 labor hours and 12 units of raw materials, with a profit of 144 yuan; producing a kitten requires 6 labor hours and 3 units of raw materials, with a profit of 81 yuan. Under the constraints of labor and raw materials, how should the production of teddy bears and kittens be reasonably arranged to maximize the total profit from producing teddy bears and kittens?",See reasoning trace,medium,"Let the number of bears and cats produced be $x$ and $y$, respectively, with the total profit being $z$ yuan, requiring $m$ labor hours and $n$ units of raw materials. Then, $$ \left\{\begin{array}{l} m \leqslant 273, n \leqslant 243, \\ 9 x+6 y=m, \\ 12 x+3 y=n . \end{array}\right. $$ From equations (1) and (2), we get $x=\frac{2 n-m}{15}, y=\frac{4 m-3 n}{15}$. From equations (1) and (2), it is known that $31 m, 3 \mid n$. Let $m=3 k, n=3 t\left(k, t \in \mathbf{N}_{+}, k \leqslant 91, t \leqslant\right.$ 81). Then $$ \begin{array}{l} z=144 x+81 y \\ =144 \times \frac{2 n-m}{15}+81 \times \frac{4 m-3 n}{15} \\ =12 m+3 n=36 k+9 t . \end{array} $$ Since $36>9,91>81$, the value of $z$ is maximized when $k$ is as large as possible and $t$ follows suit, while ensuring that $x$ and $y$ are positive integers. When $k=91$, we have $$ \begin{array}{l} x=\frac{2 \times 3 t-3 k}{15}=\frac{2 t-k}{5}=\frac{2 t-91}{5} \\ =\frac{2 t-1}{5}-18 . \end{array} $$ Thus, $2 t-1$ is a multiple of 5. Since $2 t-1$ is also an odd number, let $2 t-1=5\left(2 t_{1}+1\right)$. Then $t=5 t_{1}+3 \leqslant 81$. Solving, we get $t_{1} \leqslant 15 \frac{3}{5}$. Therefore, the maximum value of $t_{1}$ is 15. Hence, the maximum value of $t$ is 78. At this point, $x=13, y=26$. Thus, the maximum value of $z$ is $$ 36 \times 91+9 \times 78=3978 \text { (yuan) } \text {. } $$ Therefore, by producing 13 bears and 26 cats, the total profit can reach its maximum." e58ddc616b8b,"12. In the tetrahedron $ABCD$, there is a point $O$ inside such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{OA_1}=\frac{BO}{OB_1}=\frac{CO}{OC_1}=\frac{DO}{OD_1}=k$. Then $k=$ $\qquad$ .","1$, hence $k=3$.",easy,"12. $3 \frac{V_{A B C D}}{V_{O B C D}}=\frac{A A_{1}}{O A_{1}}=\frac{A O}{O A_{1}}+\frac{O A_{1}}{O A_{1}}=1+k$, thus $\frac{V_{O B C D}}{V_{A B C D}}=\frac{1}{1+k}$. Similarly, $\frac{V_{(X \cdot D A}}{V_{A B C D}}=\frac{V_{O D A B}}{V_{A B C D}}=\frac{V_{O A B C}}{V_{A B C D}}=\frac{1}{1+k}$. Since $V_{O B C D}+V_{\triangle C D A}+V_{O D A B}+V_{O A B C}=V_{A B C D}$, we have $\frac{4}{1+k}=1$, hence $k=3$." 67f88d8ecaa7,"Find those triples of integers, the product of which is four times their sum, and one of which is twice the sum of the other two.","0$, that is, $b=-a$, then $c=0$, and these triplets $(a,-a, 0)$ - for any integer $a$ - are also sol",medium,"Solution. For the triplet of numbers $a, b, c$, according to the conditions, we can write the following equations: $a b c=4(a+b+c)$, and for example, $c=2(a+b)$. Substituting this: $$ a b \cdot 2(a+b)=4(a+b+2(a+b)) $$ Rearranging and factoring out $(a+b)$: $$ 2 a b(a+b)=12(a+b) $$ If $a+b \neq 0$, we arrive at the equation $a b=6$. If $a= \pm 1$, then $b= \pm 6$ and $c= \pm 14$ (the signs of $a, b$, and $c$ are of course the same). In this case, $$ a b c= \pm 84 \quad \text { and } \quad a+b+c= \pm 21 $$ thus the requirements of the problem are satisfied. If $a= \pm 2$, then $b= \pm 3$ and $c= \pm 10$. Now, $$ a b c= \pm 60 \quad \text { and } \quad a+b+c= \pm 15 $$ and this also meets the requirements of the problem. If $a= \pm 3$ or $\pm 6$, we get the same solutions as above, only the roles of $a$ and $b$ are swapped. Finally, if $a+b=0$, that is, $b=-a$, then $c=0$, and these triplets $(a,-a, 0)$ - for any integer $a$ - are also solutions to the problem." 24b25dad06c7,"$30 \cdot 28$ If $x$ is a natural number, let $y=x^{4}+2 x^{3}+2 x^{2}+2 x+1$, then (A) $y$ must be a perfect square. (B) There are a finite number of $x$ such that $y$ is a perfect square. (C) $y$ must not be a perfect square. (D) There are infinitely many $x$ such that $y$ is a perfect square. (China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1990)",$(C)$,easy,"[Solution] Let $y=\left(x^{4}+2 x^{2}+1\right)+\left(2 x^{3}+2 x\right)$ $=\left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right)$ $=\left(x^{2}+1\right)(x+1)^{2}$, Since $(x+1)^{2}$ is a perfect square, but $x^{2}+1$ is not a perfect square, then $\left(x^{2}+1\right)(x+1)^{2}$ is not a perfect square. Therefore, the answer is $(C)$." e98ea665b108,"12. All 45 students in Class 3(1) have signed up for the school sports meet. Apart from the tug-of-war competition where the entire class participates, each student must participate in at least one of the other three events. It is known that 39 students in the class have signed up for the kick shuttlecock competition, and 28 students have signed up for the basketball shooting. How many students have signed up for all three events?","】Solution: According to the problem, all students in the class participate in the tug-of-war competition, so we only need to consider the number of students who sign up for all three events, which is the number of students who sign up for both skipping and basketball shooting",easy,"【Answer】Solution: According to the problem, all students in the class participate in the tug-of-war competition, so we only need to consider the number of students who sign up for all three events, which is the number of students who sign up for both skipping and basketball shooting. By the principle of inclusion-exclusion, the number of students is $39+28-45=22$. Answer: 22 students have signed up for all three events." dbe8d4e63a27,"Find the height of a triangular pyramid, the lateral edges of which are pairwise perpendicular and equal to 2, 3, and 4. #",$\frac{12}{\sqrt{61}}$,medium,"Let the lateral edges $DA, DB$, and $DC$ of the triangular pyramid $ABCD$ be pairwise perpendicular, and $DA=2, DB=3$, $DC=4$. Consider the triangular pyramid $ABCD$ with vertex $C$ and base $ABD$. Its lateral edge $CD$ is perpendicular to the two intersecting lines $BD$ and $AD$ in the plane $ABD$. Therefore, $CD$ is perpendicular to this plane. Thus, $CD=4$ is the height of the pyramid $ABCD$. The base of this pyramid is a right triangle $ABD$ with legs $AD=2$ and $BD=3$. Therefore, $$ V_{ABCD}=\frac{1}{3} S_{\triangle ABD} \cdot CD=\frac{1}{3} \cdot \frac{1}{2} AD \cdot BD \cdot CD=\frac{1}{6} \cdot 2 \cdot 3 \cdot 4=4. $$ Let $DK$ be the desired height of the pyramid $ABCD$. Then $$ V_{ABCD}=\frac{1}{3} S_{\triangle ABC} \cdot DK $$ From this, we find that $$ DK=\frac{3 V_{ABCD}}{S_{\triangle ABC}}=\frac{12}{S_{\triangle ABC}}. $$ It remains to find the area of triangle $ABC$. Let the lines $AK$ and $BC$ intersect at point $M$. Then the line $BC$ is perpendicular to the two intersecting lines $DK$ and $AD$ in the plane $AMD$. Therefore, $BC \perp DM$, i.e., $DM$ is the height of the right triangle $BCD$ drawn from the vertex of the right angle, and $AM$ is the height of triangle $ABC$. Further, we have: $$ \begin{gathered} DM=\frac{CD \cdot BD}{BC}=\frac{4 \cdot 3}{\sqrt{4^2+3^2}}=\frac{12}{5} \\ AM=\sqrt{DM^2 + AD^2}=\sqrt{\left(\frac{12}{5}\right)^2 + 4}=\frac{2\sqrt{61}}{5} \\ S_{\triangle ABC}=\frac{1}{2} BC \cdot AM=\frac{1}{2} \cdot 5 \cdot \frac{2\sqrt{61}}{5}=\sqrt{61} \end{gathered} $$ Therefore, $$ DK=\frac{12}{S_{\triangle ABC}}=\frac{12}{\sqrt{61}}. $$ ## Answer $\frac{12}{\sqrt{61}}$." 6687308accf5,"5. For which integers $a$ and $b$ does the system of equations $$ \left\{\begin{array}{c} |a x-2 y+b|=3 \\ (3 x-4 y-1)(4 x-y-10)(x+5 y+6)=0 \end{array}\right. $$ have an infinite number of solutions?","$a=8, b=-17 ; a=8 ; b=-23$",easy,"Answer: $a=8, b=-17 ; a=8 ; b=-23$." b460fb7e601d,"2. In a computer game, a turtle moves across a grid on the computer screen, which contains 11 columns and 5 rows. Initially, it is located at the bottom-left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for example, taking one step up from cell $(3,4)$, the turtle will end up in cell $(3,0)$. Where will the turtle be after executing the following program: 1) 1 step down; 2) 2 steps to the right; 3) 3 steps up; 4) 4 steps to the left; 5) 5 steps down; 6) 6 steps to the right; .. ; 2016) 2016 steps to the left; 2017) 2017 steps down?","$(4 ; 1)\|(4,1)\| 4 ; 1 \| 4,1$",easy,"Answer: $(4 ; 1)\|(4,1)\| 4 ; 1 \| 4,1$" d230de21343d,The positive integers $ a$ and $ b$ are such that the numbers $ 15a \plus{} 16b$ and $ 16a \minus{} 15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?,481^2,medium,"1. **Define the problem in terms of equations:** Let \( 15a + 16b = m^2 \) and \( 16a - 15b = n^2 \) for some integers \( m \) and \( n \). 2. **Solve for \( a \) and \( b \) in terms of \( m \) and \( n \):** \[ 15a + 16b = m^2 \quad \text{(1)} \] \[ 16a - 15b = n^2 \quad \text{(2)} \] To solve for \( a \) and \( b \), we can use the method of elimination. Multiply equation (1) by 16 and equation (2) by 15: \[ 16(15a + 16b) = 16m^2 \quad \Rightarrow \quad 240a + 256b = 16m^2 \quad \text{(3)} \] \[ 15(16a - 15b) = 15n^2 \quad \Rightarrow \quad 240a - 225b = 15n^2 \quad \text{(4)} \] Subtract equation (4) from equation (3): \[ (240a + 256b) - (240a - 225b) = 16m^2 - 15n^2 \] \[ 481b = 16m^2 - 15n^2 \] \[ b = \frac{16m^2 - 15n^2}{481} \quad \text{(5)} \] Similarly, add equation (3) and equation (4): \[ (240a + 256b) + (240a - 225b) = 16m^2 + 15n^2 \] \[ 480a + 31b = 16m^2 + 15n^2 \] Substitute \( b \) from equation (5): \[ 480a + 31\left(\frac{16m^2 - 15n^2}{481}\right) = 16m^2 + 15n^2 \] \[ 480a = 16m^2 + 15n^2 - \frac{31(16m^2 - 15n^2)}{481} \] \[ a = \frac{481(16m^2 + 15n^2) - 31(16m^2 - 15n^2)}{480 \cdot 481} \] \[ a = \frac{481 \cdot 16m^2 + 481 \cdot 15n^2 - 31 \cdot 16m^2 + 31 \cdot 15n^2}{480 \cdot 481} \] \[ a = \frac{(481 \cdot 16 - 31 \cdot 16)m^2 + (481 \cdot 15 + 31 \cdot 15)n^2}{480 \cdot 481} \] \[ a = \frac{450m^2 + 512n^2}{480 \cdot 481} \] 3. **Check the integrality conditions:** For \( a \) and \( b \) to be integers, the numerators of both expressions must be divisible by 481. This implies: \[ 16m^2 \equiv 15n^2 \pmod{481} \] This congruence must hold for \( m \) and \( n \). 4. **Find the smallest \( m \) and \( n \) that satisfy the conditions:** By trial and error or systematic search, we find that \( m = 481 \) and \( n = 481 \) satisfy the conditions. Substituting these values: \[ a = \frac{450 \cdot 481^2 + 512 \cdot 481^2}{480 \cdot 481} = \frac{(450 + 512) \cdot 481^2}{480 \cdot 481} = \frac{962 \cdot 481}{480} \] \[ b = \frac{16 \cdot 481^2 - 15 \cdot 481^2}{481} = \frac{481^2}{481} = 481 \] 5. **Verify the solution:** \[ 15a + 16b = 15 \cdot \frac{962 \cdot 481}{480} + 16 \cdot 481 = 481^2 \] \[ 16a - 15b = 16 \cdot \frac{962 \cdot 481}{480} - 15 \cdot 481 = 481^2 \] Thus, the smallest possible value of the smaller square is \( 481^2 \). The final answer is \( \boxed{481^2} \)." 1793bd227101,Example 7 The range of the function $y=\frac{x^{2}-1}{x^{2}+2}$ is $\qquad$ .,See reasoning trace,medium,"From the given, we have $x^{2}=\frac{2 y+1}{1-y} \geqslant 0$, solving this yields $-\frac{1}{2} \leqslant y<1$, therefore, the range of the function $y=\frac{x^{2}-1}{x^{2}+2}$ is $\left[-\frac{1}{2}, 1\right)$. Solution Guidance Construct an inequality that the dependent variable satisfies, and obtain the range of the function by solving the inequality. This is another important method for studying the range of a function, or rather, this is an application of the ""construction method"" in solving problems related to the range of a function." fd9115a5a898,"Let $f: [0, 1] \to \mathbb{R}$ be a continuous strictly increasing function such that \[ \lim_{x \to 0^+} \frac{f(x)}{x}=1. \] (a) Prove that the sequence $(x_n)_{n \ge 1}$ defined by \[ x_n=f \left(\frac{1}{1} \right)+f \left(\frac{1}{2} \right)+\cdots+f \left(\frac{1}{n} \right)-\int_1^n f \left(\frac{1}{x} \right) \mathrm dx \] is convergent. (b) Find the limit of the sequence $(y_n)_{n \ge 1}$ defined by \[ y_n=f \left(\frac{1}{n+1} \right)+f \left(\frac{1}{n+2} \right)+\cdots+f \left(\frac{1}{2021n} \right). \]",\ln 2021,hard,"### Part (a) 1. **Define the sequence and integral:** \[ x_n = \sum_{k=1}^n f\left(\frac{1}{k}\right) - \int_1^n f\left(\frac{1}{x}\right) \, dx \] 2. **Analyze the behavior of \( f \left( \frac{1}{x} \right) \):** Since \( f \) is continuous and strictly increasing, and given that \( \lim_{x \to 0^+} \frac{f(x)}{x} = 1 \), we have: \[ f\left(\frac{1}{x}\right) \sim \frac{1}{x} \quad \text{as} \quad x \to \infty \] 3. **Use the Riemann sum approximation:** The integral \( \int_1^n f\left(\frac{1}{x}\right) \, dx \) can be approximated by the Riemann sum: \[ \int_1^n f\left(\frac{1}{x}\right) \, dx \approx \sum_{k=1}^n f\left(\frac{1}{k}\right) - f\left(\frac{1}{n}\right) \] This implies: \[ f\left(\frac{1}{n}\right) < x_n < f(1) \] 4. **Bound the sequence \( x_n \):** Since \( f \left( \frac{1}{n} \right) \to 0 \) as \( n \to \infty \) and \( f(1) \) is a finite constant, \( x_n \) is bounded. 5. **Show that \( x_n \) is monotonic:** \[ x_{n+1} = \sum_{k=1}^{n+1} f\left(\frac{1}{k}\right) - \int_1^{n+1} f\left(\frac{1}{x}\right) \, dx \] \[ x_{n+1} = x_n + f\left(\frac{1}{n+1}\right) - \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx \] Since \( f \left( \frac{1}{n+1} \right) < \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx < f \left( \frac{1}{n} \right) \), we have: \[ f\left(\frac{1}{n+1}\right) - \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx < 0 \] Thus, \( x_{n+1} < x_n \), meaning \( x_n \) is strictly decreasing. 6. **Conclude convergence:** Since \( x_n \) is bounded and strictly decreasing, by the Monotone Convergence Theorem, \( x_n \) converges. ### Part (b) 1. **Define the sequence \( y_n \):** \[ y_n = f\left(\frac{1}{n+1}\right) + f\left(\frac{1}{n+2}\right) + \cdots + f\left(\frac{1}{2021n}\right) \] 2. **Rewrite \( y_n \) using \( x_n \):** \[ y_n = \sum_{k=n+1}^{2021n} f\left(\frac{1}{k}\right) \] \[ y_n = \sum_{k=1}^{2021n} f\left(\frac{1}{k}\right) - \sum_{k=1}^n f\left(\frac{1}{k}\right) \] \[ y_n = x_{2021n} - x_n + \int_1^{2021n} f\left(\frac{1}{x}\right) \, dx - \int_1^n f\left(\frac{1}{x}\right) \, dx \] \[ y_n = x_{2021n} - x_n + \int_n^{2021n} f\left(\frac{1}{x}\right) \, dx \] 3. **Evaluate the integral:** \[ \int_n^{2021n} f\left(\frac{1}{x}\right) \, dx = \int_1^{2021} n f\left(\frac{1}{nu}\right) \, du \] Since \( f\left(\frac{1}{nu}\right) \sim \frac{1}{nu} \) for large \( n \): \[ \int_1^{2021} n f\left(\frac{1}{nu}\right) \, du \sim \int_1^{2021} \frac{1}{u} \, du = \ln 2021 \] 4. **Conclude the limit:** \[ \lim_{n \to \infty} y_n = \lim_{n \to \infty} \left( x_{2021n} - x_n \right) + \ln 2021 \] Since \( x_n \) converges to a constant, \( x_{2021n} - x_n \to 0 \): \[ \lim_{n \to \infty} y_n = \ln 2021 \] The final answer is \( \boxed{ \ln 2021 } \)" ac7d28e7e959,"1. As shown in Figure 14, quadrilateral $A B C D$ is a square, and an equilateral triangle $\triangle A B E$ is constructed outward from side $A B$. $C E$ intersects $B D$ at point $F$. Then $\angle A F D=$ $\qquad$ degrees.","\angle C E B=$ $15^{\circ}$, and by the axial symmetry of the square, we get $\angle F A B=$ $\angle",easy,"(Given that $\angle E C B=\angle C E B=$ $15^{\circ}$, and by the axial symmetry of the square, we get $\angle F A B=$ $\angle E C B=15^{\circ}$. Therefore, $\angle A F D=60^{\circ}$.)" aa7225dbf6cf,4. Given the functions $f(x)=x^{2}+4 x+3$ and $g(x)=x^{2}+2 x-1$. Find all integer solutions to the equation $f(g(f(x)))=g(f(g(x)))$.,. $x=-2$,medium,"4. Answer. $x=-2$. Solution. Let's represent the functions as $f(x)=x^{2}+4 x+3=(x+2)^{2}-1$ and $g(x)=x^{2}+2 x-1=(x+1)^{2}-2 . \quad$ Then $\quad f(g(x))=\left((x+1)^{2}-2+2\right)^{2}-1=(x+1)^{4}-1$, $g(f(x))=\left((x+2)^{2}-1+1\right)^{2}-2=(x+2)^{4}-2$. Performing similar operations, we get $g(f(g(x)))=(x+1)^{8}-2$ and $f(g(f(x)))=(x+2)^{8}-1$. Thus, the equation is $$ \begin{gathered} (x+1)^{8}-(x+2)^{8}=1 \\ \left((x+1)^{4}-(x+2)^{4}\right)\left((x+1)^{4}+(x+2)^{4}\right)=1 \end{gathered} $$ Obviously, the second bracket can only be equal to 1, which means the first bracket is also equal to 1. This system has a unique solution $x=-2$. Grading criteria. Correct solution - 7 points, all function compositions are correctly written but the solution is incomplete or incorrect - 2 points, the solution is incorrect or only the answer - $\mathbf{0}$ points." e9c2f41e70c1,"8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2018}$ is divided by 7 is $\qquad$",See reasoning trace,easy,"8. 4 . Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, thus, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Therefore, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0. Since $2018=336 \times 6+2, 336 \times 10=3360$, the remainder 2 corresponds to a term whose units digit is 4. Hence, $$ \begin{array}{l} b_{2018}=a_{3364}=3364^{3}-3364 \\ \equiv 4^{3}-4 \equiv 4(\bmod 7) . \end{array} $$" 37af4752f53b,"Antal and Béla set off from Hazul to Cegléd on their motorcycles. From the one-fifth point of the journey, Antal turns back for some reason, which causes him to ""switch gears,"" and he manages to increase his speed by a quarter. He immediately sets off again from Hazul. Béla, traveling alone, reduces his speed by a quarter. They eventually meet again and travel the last part of the journey together at a speed of $48 \mathrm{~km} /$ hour, arriving 10 minutes later than planned. What can we calculate from this?",See reasoning trace,medium,"I. Solution: Due to the turnaround, by the time Béla caught up, Antal Béla had traveled 2/5 more of the entire distance. From the turnaround to the catch-up, the two travelers moved at 5/4 and 3/4 of their original speeds, respectively, or in other words, their speeds, measured in a suitable unit, were 5 and 3, respectively. Thus, from the turnaround to the catch-up, Antal traveled 2/5 more of the distance he himself had covered, compared to Béla. Since this difference is also 2/5 of the distance from home to Cegléd, Antal traveled the same distance from the separation to the meeting as the distance from home to Cegléd. One-fifth of this was the return journey, so the catch-up occurred at 4/5 of the entire distance; and at the time they were originally scheduled to arrive in Cegléd, because in this time Béla would have traveled 4/5 instead of 3/5 of the distance at their original speed, or the entire remaining distance at the moment of their separation. Therefore, during the 10-minute, 1/6-hour delay, they traveled the last 1/5 of the distance, so its length is 48/6 = 8 km, and the total distance is 5 * 8 = 40 km, which is how far they live from Cegléd. Antal turned back after traveling the first 8 km. A natural question would be what their original speed was, and what speed they exerted when traveling alone, as well as how long they spent on the journey, and how much later Antal set out from home the second time compared to the first. We cannot calculate this because the above reasoning is valid for any original speed up to the point of catch-up. At most, we can say about Antal's second departure that it happened when they should have already traveled 8 + 8 * 4/5 = 14.4 km. Imre Soós (Budapest, Rákóczi F. g. II. o. t.) ![](https://cdn.mathpix.com/cropped/2024_05_02_92fa52488c391b7db95eg-1.jpg?height=881&width=485&top_left_y=814&top_left_x=798) II. Solution: Let the distance from Cegléd to the travelers' home be $s \mathrm{~km}$, the planned speed $v \mathrm{~km} /$ hour, and the planned travel time $t = s / v$ hours. Furthermore, let Béla's distance traveled alone be $s_{1} \mathrm{~km}$, then Antal's distance traveled alone is $s_{1} + 2s / 5$. They traveled these distances at 3v/4 and 5v/4 speeds, respectively, in the same amount of time, so $$ \frac{s_{1}}{\frac{3 v}{4}} = \frac{s_{1} + \frac{2 s}{5}}{\frac{5 v}{4}} $$ from which $$ s_{1} = 3s / 5 $$ and the time spent traveling alone is $s_{1} : 3v / 4 = 4t / 5$ hours. Therefore, the catch-up occurred at a distance of $s / 5 + s_{1} = 4s / 5 \mathrm{~km}$ from home, $t / 5 + 4t / 5 = t$ hours after (the first) departure - that is, exactly when they were supposed to arrive - and at this point, $s / 5 \mathrm{~km}$ of the journey still lay ahead of the travelers. They covered this in the remaining 1/6 hour from the actual travel time of $t + 1/6$ hours. $$ \frac{s}{5} = 48 \frac{\mathrm{km}}{\text{hour}} \cdot \frac{1}{6} \text{ hour, } $$ and from this, $s = 40 \mathrm{~km}$. Zoltán Endreffy (Budapest, I. István Gymn. II. o. t.) Remarks. 1. The speeds could not be determined because the only data given in absolute units (distance) is for the last segment of the journey. The motion graph cannot connect the segments before and after the catch-up. 2. The second solution is not substantially different from the first. Readers should compare the two to see how the need for setting up equations can be eliminated." aab165cc7360,"3. Let $n$ and $m$ be positive integers of different parity, and $n > m$. Find all integers $x$ such that $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}$ is a perfect square. (Pan Chengdu)","m}^{n-1}\left(\frac{x^{2^{1}}+1}{2}\right)$. And $\left(\frac{x^{2^{4}}+1}{2}, \frac{x^{2^{j}}+1}{2}",medium,"3. Let $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}=A^{2}$, where $A \in \mathbf{N}$, then $$ A^{2}=\prod_{i=m}^{n-1}\left(x^{2^{4}}+1\right) \text {. } $$ Notice that for $i \neq j,\left(x^{2^{2}}+1, x^{2^{2}}+1\right)=\left\{\begin{array}{ll}1, & 2 \mid x, \\ 2, & 2 \nmid x .\end{array}\right.$ Case 1: If $2 \mid x$, then $\forall m \leqslant i \leqslant n-1$, the number $x^{2^{i}}+1$ is a perfect square, so it can only be $x=0$; Case 2: If $2 \nmid x$, then $2^{n-m} \prod_{i=m}^{n-1}\left(\frac{x^{2^{1}}+1}{2}\right)$. And $\left(\frac{x^{2^{4}}+1}{2}, \frac{x^{2^{j}}+1}{2}\right)=1$, so it must be that $2 \mid (n-m)$, which is a contradiction. In summary, the only integer $x$ that satisfies the condition is $x=0$." 3a77e13d7d85,"Find the greatest real number $C$ such that, for all real numbers $x$ and $y \neq x$ with $xy = 2$ it holds that \[\frac{((x + y)^2 - 6)((x - y)^2 + 8)}{(x-y)^2}\geq C.\] When does equality occur?",18,medium,"1. Let \( a = (x - y)^2 \). Note that \( (x + y)^2 = (x - y)^2 + 4xy \). Given \( xy = 2 \), we have: \[ (x + y)^2 = a + 8 \] 2. Substitute these expressions into the given inequality: \[ \frac{((x + y)^2 - 6)((x - y)^2 + 8)}{(x - y)^2} = \frac{(a + 2)(a + 8)}{a} \] 3. Simplify the expression: \[ \frac{(a + 2)(a + 8)}{a} = \frac{a^2 + 10a + 16}{a} = a + 10 + \frac{16}{a} \] 4. To find the minimum value of \( a + 10 + \frac{16}{a} \), we use the AM-GM inequality: \[ a + \frac{16}{a} \geq 2\sqrt{a \cdot \frac{16}{a}} = 2\sqrt{16} = 8 \] 5. Therefore: \[ a + 10 + \frac{16}{a} \geq 10 + 8 = 18 \] 6. Equality in the AM-GM inequality holds when \( a = \frac{16}{a} \), which gives \( a^2 = 16 \) or \( a = 4 \). 7. When \( a = 4 \), we have: \[ (x - y)^2 = 4 \implies x - y = \pm 2 \] 8. Also, since \( (x + y)^2 = a + 8 \): \[ (x + y)^2 = 4 + 8 = 12 \implies x + y = \pm 2\sqrt{3} \] 9. Solving the system of equations \( x + y = 2\sqrt{3} \) and \( x - y = 2 \), we get: \[ x = \sqrt{3} + 1, \quad y = \sqrt{3} - 1 \] 10. Similarly, solving \( x + y = -2\sqrt{3} \) and \( x - y = 2 \), we get: \[ x = -1 - \sqrt{3}, \quad y = 1 - \sqrt{3} \] Therefore, the greatest real number \( C \) is \( 18 \), and equality occurs when \( \{x, y\} = \{\sqrt{3} + 1, \sqrt{3} - 1\} \) or \( \{x, y\} = \{-1 - \sqrt{3}, 1 - \sqrt{3}\} \). The final answer is \( \boxed{18} \)." 2a1c7eb9e1bc,"$ABC$ be atriangle with sides $AB=20$ , $AC=21$ and $BC=29$. Let $D$ and $E$ be points on the side $BC$ such that $BD=8$ and $EC=9$. Find the angle $\angle DAE$.",45^\circ,medium,"1. **Identify the given information and apply the cosine rule to find $\angle BAC$:** - Given: $AB = 20$, $AC = 21$, $BC = 29$. - Using the cosine rule in $\triangle ABC$: \[ c^2 = a^2 + b^2 - 2ab \cos(C) \] where $a = 20$, $b = 21$, and $c = 29$. \[ 29^2 = 20^2 + 21^2 - 2 \cdot 20 \cdot 21 \cos(\angle BAC) \] \[ 841 = 400 + 441 - 840 \cos(\angle BAC) \] \[ 841 = 841 - 840 \cos(\angle BAC) \] \[ 0 = -840 \cos(\angle BAC) \] \[ \cos(\angle BAC) = 0 \] Therefore, $\angle BAC = 90^\circ$. 2. **Determine the positions of points $D$ and $E$ on $BC$:** - Given: $BD = 8$ and $EC = 9$. - Since $BC = 29$, we have $DE = BC - BD - EC = 29 - 8 - 9 = 12$. 3. **Analyze the triangle properties:** - Since $\angle BAC = 90^\circ$, $\triangle ABC$ is a right triangle with $A$ as the right angle. - By the Pythagorean theorem, $AB = BE$ and $AC = DC$ because $AB$ and $AC$ are the legs of the right triangle. 4. **Calculate $\angle DAE$:** - Since $AB = BE$ and $AC = DC$, triangles $\triangle ABD$ and $\triangle AEC$ are congruent by the hypotenuse-leg (HL) theorem. - Therefore, $\angle BAD = \angle EAC$. 5. **Use the angle sum property:** - $\angle DAE = \angle BAD + \angle EAC$. - Since $\angle BAC = 90^\circ$, we have: \[ \angle DAE = \frac{\angle BAC}{2} = \frac{90^\circ}{2} = 45^\circ \] The final answer is $\boxed{45^\circ}$." 1d8704154f0b,"10. (20 points) In the Cartesian coordinate system $x O y$, the vertices of $\triangle A B C$ are $B(-1,0)$, $C(1,0)$, and $A$ is a moving point. $H$ and $G$ are the orthocenter and centroid of $\triangle A B C$, respectively. If the midpoint $K$ of segment $H G$ lies on the line $B C$: (1) Find the equation of the trajectory $E$ of point $A$; (2) A line passing through point $C$ with a slope of 1 intersects the curve $E$ at point $D$. If there exist two points $P$ and $Q$ on the curve $E$ such that points $P$, $C$, $Q$, and $D$ are concyclic, find the minimum value of the area of the circumcircle of quadrilateral $P C Q D$.","-\frac{1}{2}$, i.e., $l_{P Q}: y=-x-\frac{1}{2}$, the area of the circle reaches its minimum value $",medium,"10. (1) Let point $A(x, y)(y \neq 0)$. Then the orthocenter $H\left(x_{H}, y_{H}\right)$ of $\triangle A B C$ satisfies the system of equations \[ \left\{\begin{array}{l} x_{H}=x, \\ \left(x_{H}+1\right)(1-x)-y_{H} y=0 \end{array}\right. \] \[ \Rightarrow x_{H}=x, y_{H}=\frac{1-x^{2}}{y}. \] The centroid $G\left(\frac{x}{3}, \frac{y}{3}\right)$, thus the midpoint $K\left(\frac{2 x}{3}, \frac{3-3 x^{2}+y^{2}}{6 y}\right)$ of segment $H G$. Since point $K$ lies on the $x$-axis, we have $3-3 x^{2}+y^{2}=0$, which means the locus equation of point $A$ is $x^{2}-\frac{y^{2}}{3}=1(y \neq 0)$. (2) As shown in Figure 2, let $C D$ and $P Q$ intersect at point $M\left(x_{0}, y_{0}\right)$. Then the parametric equation of line $C D$ is \[ \left\{\begin{array}{l} x=x_{0}+\frac{\sqrt{2}}{2} t, \\ y=y_{0}+\frac{\sqrt{2}}{2} t \end{array} \text { ( } t \text { is a parameter). }\right. \] Substituting into the equation of curve $E$ gives \[ t^{2}+\left(3 x_{0}-2 y_{0}\right) t+3 x_{0}^{2}-y_{0}^{2}-3=0. \] Thus, $|M C||M D|=-t_{1} t_{2}=-\left(3 x_{0}^{2}-y_{0}^{2}-3\right)$. Let the inclination angle of line $P Q$ be $\alpha$, and the parametric equation of line $P Q$ be \[ \left\{\begin{array}{l} x=x_{0}+t \cos \alpha, \\ y=y_{0}+t \sin \alpha \end{array} \text { ( } t \text { is a parameter). }\right. \] Substituting into the equation of curve $E$ gives \[ \left(3-4 \sin ^{2} \alpha\right) t^{2}+2\left(3 x_{0} \cos \alpha-y_{0} \sin \alpha\right) t+3 x_{0}^{2}-y_{0}^{2}-3=0. \] Then $|M P||M Q|=-t_{3} t_{4}=-\frac{3 x_{0}^{2}-y_{0}^{2}-3}{3-4 \sin ^{2} \alpha}$. Since points $P, C, Q, D$ are concyclic, \[ \Rightarrow|M C||M D|=|M P||M Q| \] \[ \Rightarrow-\left(3 x_{0}^{2}-y_{0}^{2}-3\right)=-\frac{3 x_{0}^{2}-y_{0}^{2}-3}{3-4 \sin ^{2} \alpha} \] \[ \Rightarrow 3-4 \sin ^{2} \alpha=1 \Rightarrow \sin ^{2} \alpha=\frac{1}{2}. \] Since $\alpha \in[0, \pi)$ and $C D$ intersects $P Q$, we have $\alpha=\frac{3 \pi}{4}, k_{P Q}=-1$. By the problem, we can set $l_{P Q}: y=-x+m$, and combining $l_{c D}$: $y=x-1$, the equation of the quadratic curve system passing through points $P, C, Q, D$ is \[ \begin{array}{l} (x-y-1)(x+y-m)+\lambda\left(3 x^{2}-y^{2}-3\right)=0 \\ \Rightarrow(1+3 \lambda) x^{2}-(1+\lambda) y^{2}-(1+m) x+ \\ \quad(m-1) y+m-3 \lambda=0 . \end{array} \] Since equation (1) represents a circle, we have \[ 1+3 \lambda=-(1+\lambda) \Rightarrow \lambda=-\frac{1}{2}. \] At this point, equation (1) becomes \[ x^{2}+y^{2}+2(m+1) x-2(m-1) y-2 m-3=0. \] Let the radius of the corresponding circle be $R$, then \[ R^{2}=2 m^{2}+2 m+5 \geqslant \frac{9}{2}. \] When $m=-\frac{1}{2}$, i.e., $l_{P Q}: y=-x-\frac{1}{2}$, the area of the circle reaches its minimum value $\frac{9 \pi}{2}$." d914d8b36488,"A three-digit $\overline{abc}$ number is called [i]Ecuadorian [/i] if it meets the following conditions: $\bullet$ $\overline{abc}$ does not end in $0$. $\bullet$ $\overline{abc}$ is a multiple of $36$. $\bullet$ $\overline{abc} - \overline{cba}$ is positive and a multiple of $36$. Determine all the Ecuadorian numbers.",864,medium,"To determine all the Ecuadorian numbers, we need to satisfy the given conditions. Let's break down the problem step by step. 1. **Condition 1: $\overline{abc}$ does not end in $0$.** - This implies that $c \neq 0$. 2. **Condition 2: $\overline{abc}$ is a multiple of $36$.** - Since $36 = 4 \times 9$, $\overline{abc}$ must be divisible by both $4$ and $9$. - For divisibility by $4$, the last two digits $\overline{bc}$ must form a number divisible by $4$. - For divisibility by $9$, the sum of the digits $a + b + c$ must be divisible by $9$. 3. **Condition 3: $\overline{abc} - \overline{cba}$ is positive and a multiple of $36$.** - $\overline{abc} - \overline{cba} = 100a + 10b + c - (100c + 10b + a) = 99a - 99c = 99(a - c)$. - For $99(a - c)$ to be a multiple of $36$, $a - c$ must be a multiple of $4$ (since $99$ is already a multiple of $9$). 4. **Range of $\overline{abc}$:** - Since $\overline{abc}$ is a three-digit number, $100 \leq \overline{abc} \leq 999$. - Given that $\overline{abc}$ is a multiple of $36$, we can write $\overline{abc} = 36n$ for some integer $n$. - Therefore, $100 \leq 36n \leq 999$ implies $3 \leq n \leq 27$. 5. **Finding valid $n$:** - We need to check each $n$ in the range $3 \leq n \leq 27$ to see if $\overline{abc} = 36n$ satisfies all conditions. Let's check each $n$: - For $n = 3$: $\overline{abc} = 36 \times 3 = 108$ (fails as $108 - 801$ is negative) - For $n = 4$: $\overline{abc} = 36 \times 4 = 144$ (fails as $144 - 441$ is negative) - For $n = 5$: $\overline{abc} = 36 \times 5 = 180$ (fails as $180 - 081$ is negative) - For $n = 6$: $\overline{abc} = 36 \times 6 = 216$ (fails as $216 - 612$ is negative) - For $n = 7$: $\overline{abc} = 36 \times 7 = 252$ (fails as $252 - 252$ is zero) - For $n = 8$: $\overline{abc} = 36 \times 8 = 288$ (fails as $288 - 882$ is negative) - For $n = 9$: $\overline{abc} = 36 \times 9 = 324$ (fails as $324 - 423$ is negative) - For $n = 10$: $\overline{abc} = 36 \times 10 = 360$ (fails as $360 - 063$ is negative) - For $n = 11$: $\overline{abc} = 36 \times 11 = 396$ (fails as $396 - 693$ is negative) - For $n = 12$: $\overline{abc} = 36 \times 12 = 432$ (fails as $432 - 234$ is negative) - For $n = 13$: $\overline{abc} = 36 \times 13 = 468$ (fails as $468 - 864$ is negative) - For $n = 14$: $\overline{abc} = 36 \times 14 = 504$ (fails as $504 - 405$ is negative) - For $n = 15$: $\overline{abc} = 36 \times 15 = 540$ (fails as $540 - 045$ is negative) - For $n = 16$: $\overline{abc} = 36 \times 16 = 576$ (fails as $576 - 675$ is negative) - For $n = 17$: $\overline{abc} = 36 \times 17 = 612$ (fails as $612 - 216$ is positive but not a multiple of $36$) - For $n = 18$: $\overline{abc} = 36 \times 18 = 648$ (fails as $648 - 846$ is negative) - For $n = 19$: $\overline{abc} = 36 \times 19 = 684$ (fails as $684 - 486$ is positive but not a multiple of $36$) - For $n = 20$: $\overline{abc} = 36 \times 20 = 720$ (fails as $720 - 027$ is negative) - For $n = 21$: $\overline{abc} = 36 \times 21 = 756$ (fails as $756 - 657$ is positive but not a multiple of $36$) - For $n = 22$: $\overline{abc} = 36 \times 22 = 792$ (fails as $792 - 297$ is positive but not a multiple of $36$) - For $n = 23$: $\overline{abc} = 36 \times 23 = 828$ (fails as $828 - 828$ is zero) - For $n = 24$: $\overline{abc} = 36 \times 24 = 864$ (satisfies all conditions) - For $n = 25$: $\overline{abc} = 36 \times 25 = 900$ (fails as it ends in $0$) - For $n = 26$: $\overline{abc} = 36 \times 26 = 936$ (fails as $936 - 639$ is positive but not a multiple of $36$) - For $n = 27$: $\overline{abc} = 36 \times 27 = 972$ (fails as $972 - 279$ is positive but not a multiple of $36$) Thus, the only number that satisfies all conditions is $864$. The final answer is $\boxed{864}$." 01302e5e70c0,"Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$. ![](https://cdn.mathpix.com/cropped/2024_05_06_2fdcf97aa9799d0d4cd6g-32.jpg?height=327&width=495&top_left_y=1316&top_left_x=479)",14,medium,"Answer: 14. Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=x$. In the right-angled triangle $C K H$ we have $\angle C K H=90^{\circ}-\angle C=\angle A=30^{\circ}$, so $K C=2 \cdot C H=2 \cdot(C M+M H)=$ $2 \cdot(3+x)=6+2 x$. In the right-angled triangle $A B C$ we have $\angle A=30^{\circ}$, so $B C=2 \cdot A C$. Setting up and solving the corresponding equation $31+2 x+3=2 \cdot(4+6+2 x)$, we find $x=7$. Then $L M=2 x=2 \cdot 7=14$. ![](https://cdn.mathpix.com/cropped/2024_05_06_2fdcf97aa9799d0d4cd6g-33.jpg?height=414&width=703&top_left_y=96&top_left_x=379) Fig. 7: to the solution of problem 9.6" 7665cfe351fb,"The lateral sides of the trapezoid are 7 and 11, and the bases are 5 and 15. A line drawn through the vertex of the smaller base parallel to the larger lateral side cuts off a triangle from the trapezoid. Find the sides of this triangle. #","$7,10,11$",easy,"The specified straight line divides the trapezoid into a triangle and a parallelogram. ## Solution Let $A D$ and $B C$ be the bases of trapezoid $A B C D$, where $$ A B=7, B C=5, C D=11, A D=15 \text {. } $$ Draw a line through vertex $C$ parallel to the side $A B$. Let this line intersect the base $A D$ at point $K$. Then $A B C K$ is a parallelogram. Therefore, $$ C K=A B=7, D K=A D-A K=A D-B C=15-5=10 . $$ ## Answer $7,10,11$." 11487da1c3f8,"## Task 1 - 191231 Let $n$ and $m$ be natural numbers with $n \geq 1, m \geq 1$; $N$ be the set of natural numbers from 1 to $n$ and $M$ the set of natural numbers from 1 to $m$. Determine the number of all those subsets of $N$ that have common elements with $M$.",See reasoning trace,medium,"A subset of $N=\{1,2, \ldots, n\}$ has obviously no common elements with $M=\{1,2, \ldots, m\}$ if and only if it is also a subset of $M \backslash N$. By subtracting the number of these from the total number of subsets of $N$, we then obtain for the number $A_{m, n}$ of all subsets of $N$ that have common elements with $M$, $$ A_{m, n}= \begin{cases}2^{n}-1, & \text { if } n \leq m \\ 2^{n-m}\left(2^{m}-1\right), & \text { if } n>m\end{cases} $$ or, summarized, $$ A_{m, n}=2^{n}-2^{\max (n-m, 0)} $$" 1f68097e2bcf,"## Task B-1.3. Solve the equation $$ \frac{2 x}{x-1}-\frac{x}{x+2}-\frac{5 x}{(x-1)(x+2)}=\frac{4}{x^{2}+x-2}-\frac{2}{x-1} $$",0$.,easy,"## Solution. Let's write the trinomial $x^{2}+x-2$ in the form of a product: $$ x^{2}+x-2=(x+2)(x-1) $$ Then the given equation becomes $$ \frac{2 x(x+2)-x(x-1)-5 x}{(x-1)(x+2)}=\frac{4-2(x+2)}{(x-1)(x+2)}, \quad x \neq 1, x \neq-2 $$ Then we have $$ \begin{array}{cc} 2 x^{2}+4 x-x^{2}+x-5 x=4-2 x-4 & 1 \text { point } \\ x^{2}+2 x=0 & \\ x(x+2)=0 . & 1 \text { point } \end{array} $$ Since $x \neq-2$, the only solution is $x=0$." 7ea0b9dacd8f,"Example 5: On the ground, there are 10 birds pecking, and among any 5 birds, at least 4 birds are on the same circumference. What is the maximum number of birds on the circumference that has the most birds?",See reasoning trace,medium,"Analysis and Solution: Use points to represent birds. Let the circle with the most points have $r$ points. Clearly, $r \geqslant 4$. Can $r=4$? Let's first explore the case where $r=4$. If $r=4$, i.e., each circle has at most 4 points, but we are considering circles that pass through at least 4 points, so each ""4-point circle"" has exactly 4 points. Step 1: Calculate the number of ""4-point circles"". Note the condition: any 5-point set has 4 points on a circle, i.e., each ""5-point set"" corresponds to a ""4-point circle"". Thus, we can use a mapping to count the number of circles. In fact, each ""5-point set"" corresponds to one ""4-point circle"", so there are $C_{10}^{5}=252$ ""4-point circles"". However, each ""4-point circle"" can belong to 6 different ""5-point sets"", and is counted 6 times, so the number of ""4-point circles"" is $\frac{252}{6}=42$. These ""4-point circles"" are distinct. If not, there would be two different 4-point sets $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ on the same circle, but $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ would have at least 5 distinct points, and these 5 points would be on the same circle, contradicting $r=4$. Step 2: Observe the partition and use a middle quantity to count twice. The 42 different 4-point circles can be seen as 42 different (possibly intersecting) subsets, denoted as: $M_{1}, M_{2}, \cdots, M_{42}$, which satisfy: $\left|M_{i}\right|=4$, $\left|M_{i} \cap M_{j}\right| \leqslant 2$. Using the ""addition technique"", we can calculate the number of triangles $S$. On one hand, $S=C_{10}^{3}=120$. On the other hand, $S \geqslant 42 \mathrm{C}_{4}^{3}=168$, which is a contradiction. Therefore, $r>4$. Let $M$ be the circle with the most points, and by $r>4$, $M$ has at least 5 points $A, B, C, D, E$. We will prove: the other points (except possibly one point) are on this circle. (*) In fact, suppose there are two points $P, Q$ not on circle $M$, then among the 5 points $P, Q, A, B, C$, 4 points are on a circle. But $P, Q$ are not on circle $A B C$, so $P, Q$ must be on a circle with two of $A, B, C$, say $P, Q, A, B$ are on circle $M_{1}$. Similarly, consider the 5 points $P, Q, C, D, E$, there must be 4 points among $P, Q$ and $C, D, E$ on a circle, say $P, Q, C, D$ are on circle $M_{2}$. Now consider the 5 points $P, Q, A, C, E$, there must be 4 points among $P, Q$ and $A, C, E$ on a circle. (i) If $P, Q, A, C$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{1}$, so $P Q A B C$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction. (ii) If $P, Q, A, E$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{1}$, so $P Q A B E$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction. (iii) If $P, Q, C, E$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{2}$, so $P Q C D E$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction. In summary, conclusion $(*)$ holds, i.e., $r \geqslant 9$. Finally, $r=9$ is possible, i.e., 9 points are on the same circle, and the other point is outside the circle, which clearly meets the condition. In conclusion, the minimum value of $r$ is 9." cbbeb14fe022,"What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2, -10, 5)$ and the other on the sphere of radius 87 with center $(12, 8, -16)$?",137,medium,"1. **Identify the centers and radii of the spheres:** - Sphere 1: Center at $(-2, -10, 5)$ with radius $19$. - Sphere 2: Center at $(12, 8, -16)$ with radius $87$. 2. **Calculate the distance between the centers of the two spheres using the distance formula:** \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Substituting the given coordinates: \[ d = \sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} \] Simplify the expressions inside the square root: \[ d = \sqrt{(12 + 2)^2 + (8 + 10)^2 + (-16 - 5)^2} \] \[ d = \sqrt{14^2 + 18^2 + (-21)^2} \] Calculate the squares: \[ d = \sqrt{196 + 324 + 441} \] Sum the values inside the square root: \[ d = \sqrt{961} \] Simplify the square root: \[ d = 31 \] 3. **Determine the largest possible distance between a point on each sphere:** - The farthest distance occurs when the points are on the line connecting the centers and are on the outermost points of each sphere. - This distance is the sum of the distance between the centers and the radii of both spheres: \[ \text{Maximum distance} = d + r_1 + r_2 \] Substituting the values: \[ \text{Maximum distance} = 31 + 19 + 87 \] Calculate the sum: \[ \text{Maximum distance} = 137 \] The final answer is $\boxed{137}$" 4cc8843ba89d,"## Find the point of intersection of the line and the plane. $\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}$ $2 x-3 y-5 z-7=0$",See reasoning trace,medium,"## Solution Let's write the parametric equations of the line. $\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}=t \Rightarrow$ $\left\{\begin{array}{l}x=-2+t \\ y=2 \\ z=-3\end{array}\right.$ Substitute into the equation of the plane: $2(-2+t)-3 \cdot 2-5 \cdot(-3)-7=0$ $-4+2 t-6+15-7=0$ $2 t-2=0$ $t=1$ Find the coordinates of the intersection point of the line and the plane: $$ \left\{\begin{array}{l} x=-2+1=-1 \\ y=2 \\ z=-3 \end{array}\right. $$ We get: $(-1 ; 2 ;-3)$ Problem Kuznetsov Analytic Geometry 14-10" 4e36c7529a94,"$a, b, c$ are positive real numbers such that $$(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)=1$$ At most, how many of the numbers: $$a-\frac {b}{c}, a-\frac {c}{b}, b-\frac {a}{c}, b-\frac {c}{a}, c-\frac {a}{b}, c-\frac {b}{a}$$ can be bigger than $1$?",4,medium,"1. **Given Condition:** We start with the given condition: \[ (\sqrt{ab} - 1)(\sqrt{bc} - 1)(\sqrt{ca} - 1) = 1 \] where \(a, b, c\) are positive real numbers. 2. **Claim:** We need to determine how many of the numbers: \[ a - \frac{b}{c}, \quad a - \frac{c}{b}, \quad b - \frac{a}{c}, \quad b - \frac{c}{a}, \quad c - \frac{a}{b}, \quad c - \frac{b}{a} \] can be greater than 1. 3. **Example to Achieve 4:** Consider the specific values \(b = c = \frac{3}{2}\) and \(a = 2 + \frac{4\sqrt{2}}{3}\). We need to check if these values satisfy the given condition and if they allow four of the expressions to be greater than 1. 4. **Assumption and Simplification:** Assume without loss of generality that \(a \geq b \geq c\). We analyze the case when \(c \geq 2\): - For \(a = b = c = 2\), all expressions are equal to 1. - For \(a > 2\), the product \((\sqrt{ab} - 1)(\sqrt{ac} - 1)(\sqrt{bc} - 1)\) becomes greater than 1, which contradicts the given condition. Thus, \(c < 2\). 5. **Case Analysis:** - If \(c \leq 1\), then \(c - \frac{b}{a}\) and \(c - \frac{a}{b}\) are both less than 1. Additionally, \(b - \frac{a}{c} \leq b - a < 0\), so at most three expressions can be greater than 1. - If \(1 < c < 2\), we analyze further: - \(c - \frac{a}{b} < 1\), so at most five expressions can be greater than 1. 6. **Checking for 5 Expressions:** Suppose five expressions can be greater than 1. This implies: - \(c - \frac{b}{a} > 1\) - \(b - \frac{a}{c} > 1\) - \(a - \frac{b}{c} > 1\) These yield: \[ bc > a + c \quad \text{and} \quad ac > a + b \] 7. **Further Implications:** - Since \(b > 2\) (otherwise \(b - \frac{a}{c} > 1\) would yield a contradiction), we have \(a > b > 2 > c\). - From \(c - \frac{b}{a} > 1\), we get \(ac > a + b\). - From \(b - \frac{a}{c} > 1\), we get \(bc > a + c\). 8. **Contradiction:** Combining these, we get: \[ ac > a + b > 2\sqrt{ab} \quad \text{and} \quad bc > a + c > 2\sqrt{ac} \] This implies: \[ abc^2 > 4a\sqrt{bc} \quad \text{or} \quad bc^3 > 16 \] Since \(c^2 < 4\), we get \(bc > 4\). Finally, \(ab > ac > bc > 4\), which contradicts the given condition \((\sqrt{ab} - 1)(\sqrt{bc} - 1)(\sqrt{ca} - 1) = 1\). 9. **Conclusion:** Therefore, at most four of the given expressions can be greater than 1. The final answer is \( \boxed{ 4 } \)." 930957bb1294,"$A$, $B$, $C$, and $D$ are points on a circle, and segments $\overline{AC}$ and $\overline{BD}$ intersect at $P$, such that $AP=8$, $PC=1$, and $BD=6$. Find $BP$, given that $BP1, x>0$, and $(2 x)^{\log _{b} 2}-(3 x)^{\log _{b} 3}=0$, then $x$ is (A) $\frac{1}{216}$. (B) $\frac{1}{6}$. (C) 1 . (D) 6 . (E) cannot be uniquely determined. (32nd American High School Mathematics Examination, 1981)",$(B)$,medium,"[Solution] The given equation is equivalent to the following equation: $$ (2 x)^{\log _{b} 2}=(3 x)^{\log _{b} 3} \text {, } $$ or $\frac{2^{\log _{b} 2}}{3^{\log _{b} 3}}=\frac{x^{\log _{b} 3}}{x^{\log _{b} 2}}$, which simplifies to $\frac{2^{\log _{b} 2}}{3^{\log _{b} 3}}=x^{\log _{b} 3-\log _{b} 2}$. Taking the logarithm with base $b$ on both sides of the above equation, we get $$ \begin{array}{l} \log _{b} 2^{\log _{b} 2}-\log _{b} 3^{\log _{b} 3}=\left(\log _{b} 3-\log _{b} 2\right) \log _{b} x, \\ \left(\log _{b} 2\right)^{2}-\left(\log _{b} 3\right)^{2}=\left(\log _{b} 3-\log _{b} 2\right) \log _{b} x . \end{array} $$ Thus, $$ -\left(\log _{b} 2+\log _{b} 3\right)=\log _{b} x, $$ or $$ -\log _{b}(2 \cdot 3)=\log _{b} x, $$ Therefore, $\log _{b} \frac{1}{6}=\log _{b} x$, which gives $x=\frac{1}{6}$. Hence, the answer is $(B)$." 272fe8d77c31,"3. We are given the values of the differentiable real functions $f, g, h$, as well as the derivatives of their pairwise products, at $x=0$ : $$ f(0)=1 ; g(0)=2 ; h(0)=3 ;(g h)^{\prime}(0)=4 ;(h f)^{\prime}(0)=5 ;(f g)^{\prime}(0)=6 . $$ Find the value of $(f g h)^{\prime}(0)$.",f^{\prime} g h+f g^{\prime} h+f g h^{\prime}=\left((f g)^{\prime} h+(g h)^{\prime} f+\right.$ $\left,easy,"Solution: 16 By the product rule, $(f g h)^{\prime}=f^{\prime} g h+f g^{\prime} h+f g h^{\prime}=\left((f g)^{\prime} h+(g h)^{\prime} f+\right.$ $\left.(h f)^{\prime} g\right) / 2$. Evaluated at 0 , this gives 16 ." d64ebdbe9629,"38. Find the largest number by which each of the fractions $\frac{154}{195}$, $\frac{385}{156}$, and $\frac{231}{130}$ can be divided to yield natural numbers.","2 \cdot 7 \cdot 11, 385=5 \cdot 7 \cdot 11$ and $231=3 \cdot 7 \cdot 11$, the desired numerator is 7",medium,"38. The numerator of the fraction of the desired number should be the greatest common divisor of the numerators of the given fractions, and the denominator should be the least common multiple of the denominators of these fractions. Since $154=2 \cdot 7 \cdot 11, 385=5 \cdot 7 \cdot 11$ and $231=3 \cdot 7 \cdot 11$, the desired numerator is 77. Since $195=3 \cdot 5 \cdot 13, 156=2^{2} \cdot 3 \cdot 13$ and $130=2 \cdot 5 \cdot 13$, the desired denominator is $2^{2} \cdot 3 \cdot 5 \cdot 13=780$. Therefore, the desired number is $\frac{77}{780}$." a55f18f1d97b,![](https://cdn.mathpix.com/cropped/2024_05_06_1ea0b100610baa73554bg-02.jpg?height=300&width=491&top_left_y=613&top_left_x=481),48,easy,"Answer: 48. Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$." f014b709f964,"$a$ is irrational , but $a$ and $a^3-6a$ are roots of square polynomial with integer coefficients.Find $a$",a \in \left\{-1 - \sqrt{2,medium,"Given that \(a\) is irrational and both \(a\) and \(a^3 - 6a\) are roots of a polynomial with integer coefficients, we need to find the value of \(a\). 1. Let \(a\) and \(a^3 - 6a\) be roots of a polynomial with integer coefficients. This implies that there exist integers \(s\) and \(p\) such that: \[ a + (a^3 - 6a) = s \quad \text{and} \quad a(a^3 - 6a) = p \] Simplifying these, we get: \[ a^3 - 5a = s \quad \text{(Equation 1)} \] \[ a^4 - 6a^2 = p \quad \text{(Equation 2)} \] 2. Multiply Equation 1 by \(a^2 + sa - 1\): \[ (a^3 - 5a)(a^2 + sa - 1) = 0 \] Expanding this, we get: \[ a^5 + sa^4 - a^3 - 5a^3 - 5sa^2 + 5a = 0 \] Simplifying, we have: \[ a^5 + sa^4 - 6a^3 - 5sa^2 + 5a = 0 \quad \text{(Equation 3)} \] 3. Multiply Equation 2 by \(a + s\): \[ (a^4 - 6a^2)(a + s) = 0 \] Expanding this, we get: \[ a^5 + sa^4 - 6a^3 - 6sa^2 = 0 \quad \text{(Equation 4)} \] 4. Subtract Equation 4 from Equation 3: \[ (a^5 + sa^4 - 6a^3 - 5sa^2 + 5a) - (a^5 + sa^4 - 6a^3 - 6sa^2) = 0 \] Simplifying, we get: \[ (5 - s^2 + p)a + s + ps = 0 \] Since \(a\) is irrational, the coefficient of \(a\) must be zero: \[ 5 - s^2 + p = 0 \quad \text{and} \quad s + ps = 0 \] 5. Solving \(s + ps = 0\): \[ s(1 + p) = 0 \] This gives us two cases: - Case 1: \(s = 0\) - Case 2: \(1 + p = 0\) 6. For Case 1 (\(s = 0\)): \[ 5 - s^2 + p = 0 \implies 5 + p = 0 \implies p = -5 \] The polynomial becomes: \[ x^2 - 5 = 0 \implies x = \pm \sqrt{5} \] 7. For Case 2 (\(1 + p = 0\)): \[ p = -1 \] \[ 5 - s^2 - 1 = 0 \implies s^2 = 4 \implies s = \pm 2 \] The polynomial becomes: \[ x^2 \pm 2x - 1 = 0 \] Solving these quadratics: \[ x^2 + 2x - 1 = 0 \implies x = -1 \pm \sqrt{2} \] \[ x^2 - 2x - 1 = 0 \implies x = 1 \pm \sqrt{2} \] Thus, the possible values of \(a\) are: \[ \boxed{a \in \left\{-1 - \sqrt{2}, -\sqrt{5}, 1 - \sqrt{2}, -1 + \sqrt{2}, \sqrt{5}, 1 + \sqrt{2}\right\}} \]" 18a7f052c782,"3. Before the trial run of one of the units of the hydropower station under construction, it was found that a fishing net is located $S$ km upstream from the dam. The river current speed is $v$ km/h. The hydropower station workers decided to go there by boat. Removing the net will take 5 minutes. What should be the boat's own speed so that the entire trip (including the time required to remove the net) takes no more than 45 minutes?",. $x \geqslant \frac{3 S+\sqrt{9 S^{2}+4 v^{2}}}{2}$,easy,"Answer. $x \geqslant \frac{3 S+\sqrt{9 S^{2}+4 v^{2}}}{2}$. Criteria. «+.» Answer $x=\ldots$ instead of $x \geqslant \ldots ;$ «士» answer found in different units of measurement; «干» in the correct solution, the comparison of the roots of the quadratic equation with zero was not performed, leading to an additional piece of the solution $\left(0 ; \frac{3 S-\sqrt{9 S^{2}+4 v^{2}}}{2}\right)$." 4e887c662bcb,"4. (8 points) There is a magical tree with 123 fruits on it. On the first day, 1 fruit will fall from the tree. Starting from the second day, the number of fruits that fall each day is 1 more than the previous day. However, if the number of fruits on the tree is less than the number that should fall on a certain day, then on that day it will start over by dropping 1 fruit, and continue according to the rule, and so on. So, on which day will all the fruits on the tree have fallen?","】Solution: Since $1+2+3+4+5+6+7+8+9+10+11+12+13+14+15=120$, by the sixteenth day, there are not enough 16 to start over",easy,"17 【Answer】Solution: Since $1+2+3+4+5+6+7+8+9+10+11+12+13+14+15=120$, by the sixteenth day, there are not enough 16 to start over. $1+2=3$ that is $1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+1+2=123$ (items). Therefore, the answer is: 17 days." c0d7aeabdbf9,At most how many numbers can we select from the first 1983 positive integers so that the product of any two selected numbers does not appear among the selected numbers?,See reasoning trace,medium,"Clearly, we get an appropriate set of numbers if the product of any two of the selected numbers is greater than 1983. Since $1983 < 45 \cdot 46$, selecting the numbers 45, 46, ..., 1982, 1983 results in an appropriate set of numbers. This means that we can select 1939 numbers that meet the conditions. If $a \cdot b \leq 1983$, then at most two of $a$, $b$, and $a b$ can be among the selected numbers. Let's try the products of the form $(44-i)(45+i)=1980-i-i^{2}$! Denote the factors of this product by $a_{i}, b_{i}$, and the product itself by $c_{i}$ for $i=1, \ldots, 42$. It immediately follows that $c_{i+1}p)$. It is easy to see that $\frac{n}{p q} \in \mathbf{Z}$. Then $f(n)=1+p+q \equiv 1+p(\bmod q)$, $g(n)=n+\frac{n}{p}=q(1+p) \frac{n}{p q}(\bmod q)$. Hence $(1+p)^{k} \equiv f^{k}(n)=g(n) \equiv 0(\bmod q)$. Since $q$ is a prime, we have $q \mid(1+p)$. Thus, $pn$. Then $r=\frac{m+n-30}{2}$ and $y=\frac{m-n}{2}$. The table shows all possible values solving the equation in positive integers and the respective values for $r, y$ and $b$. \begin{tabular}{c|c|c|c|c} $m$ & $n$ & $r$ & $y$ & $b$ \\ 225 & 1 & 98 & 112 & 128 \\ 75 & 3 & 24 & 36 & 54 \\ 45 & 5 & 10 & 20 & 40 \\ 25 & 9 & 2 & 8 & 32 \\ 15 & 15 & 0 & 0 & 0 \end{tabular} The final line of the table is not relevant as we do not have $m>n$ which leads to the ratio of red to yellow being indeterminate. However, it is useful to check we have considered all pairs of numbers multiplying to 225 . Hence the possible numbers of red beads are $2,10,24$ or 98 ." 6fc0a1f5c50d,"5. Let the function $f(x)=x^{2}+6 x+8$. If $f(b x+c)=4 x^{2}+16 x+15$, then $c-2 b=(\quad)$. (A) 3 (B) 7 (C) -3 (D) -7",See reasoning trace,easy,"5. C. Take $x=-2$, we have $f(c-2b)=16-16 \times 2+15=-1$. And when $x^{2}+6x+8=-1$, we have $x=-3$. Therefore, $$ c-2b=-3 \text{. } $$" cca44737f3ce,"12.129. Find the volume of a regular quadrilateral pyramid if the side of its base is equal to $a$, and the dihedral angle at the base is $\alpha$.",$\frac{a^{3}}{6} \operatorname{tg} \alpha$,easy,"Solution. The volume of the pyramid $V=\frac{1}{3} S H$ (Fig. 12.129). The area of the base: $S=a^{2}$. The height of the pyramid from $\triangle S O E \quad S O=H=\frac{a}{2} \operatorname{tg} \alpha$. Therefore, $V=\frac{1}{3} \frac{a^{3}}{2} \operatorname{tg} \alpha=\frac{a^{3}}{6} \operatorname{tg} \alpha$. Answer: $\frac{a^{3}}{6} \operatorname{tg} \alpha$." 45074fae0df6,"The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$. The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at the same place will come to $\textbf{(A)}\ $1.70 \qquad \textbf{(B)}\ $1.65 \qquad \textbf{(C)}\ $1.20 \qquad \textbf{(D)}\ $1.05 \qquad \textbf{(E)}\ $0.95$",\textbf{(D),medium,"Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie. With the information, \[3s+7c+p=3.15\] \[4s+10c+p=4.20\] Subtract the first equation from the second to get \[s+3c=1.05\] That means $s=1.05-3c$. Substituting it back in the second equation results in. \[4.20-12c+10c+p=4.20\] Solving for $p$ yields $p=2c$. With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \boxed{\textbf{(D)}\ $1.05}$." c96f5f577ad5,"Suppose that $m$ and $n$ are relatively prime positive integers with $A = \tfrac mn$, where \[ A = \frac{2+4+6+\dots+2014}{1+3+5+\dots+2013} - \frac{1+3+5+\dots+2013}{2+4+6+\dots+2014}. \] Find $m$. In other words, find the numerator of $A$ when $A$ is written as a fraction in simplest form. [i]Proposed by Evan Chen[/i]",2015,medium,"1. First, we need to find the sum of the even numbers from 2 to 2014. This sequence is an arithmetic series with the first term \(a = 2\), the common difference \(d = 2\), and the last term \(l = 2014\). The number of terms \(n\) in this series can be found using the formula for the \(n\)-th term of an arithmetic sequence: \[ l = a + (n-1)d \implies 2014 = 2 + (n-1) \cdot 2 \implies 2014 = 2n \implies n = 1007 \] The sum \(S\) of the first \(n\) terms of an arithmetic series is given by: \[ S = \frac{n}{2} (a + l) = \frac{1007}{2} (2 + 2014) = \frac{1007 \cdot 2016}{2} = 1007 \cdot 1008 \] 2. Next, we need to find the sum of the odd numbers from 1 to 2013. This sequence is also an arithmetic series with the first term \(a = 1\), the common difference \(d = 2\), and the last term \(l = 2013\). The number of terms \(n\) in this series can be found similarly: \[ l = a + (n-1)d \implies 2013 = 1 + (n-1) \cdot 2 \implies 2013 = 2n - 1 \implies 2014 = 2n \implies n = 1007 \] The sum \(S\) of the first \(n\) terms of this arithmetic series is: \[ S = \frac{n}{2} (a + l) = \frac{1007}{2} (1 + 2013) = \frac{1007 \cdot 2014}{2} = 1007^2 \] 3. Now, we can express \(A\) as follows: \[ A = \frac{2 + 4 + 6 + \dots + 2014}{1 + 3 + 5 + \dots + 2013} - \frac{1 + 3 + 5 + \dots + 2013}{2 + 4 + 6 + \dots + 2014} \] Substituting the sums we found: \[ A = \frac{1007 \cdot 1008}{1007^2} - \frac{1007^2}{1007 \cdot 1008} \] 4. Simplify each term: \[ \frac{1007 \cdot 1008}{1007^2} = \frac{1008}{1007} \] \[ \frac{1007^2}{1007 \cdot 1008} = \frac{1007}{1008} \] 5. Combine the terms: \[ A = \frac{1008}{1007} - \frac{1007}{1008} \] 6. Find a common denominator and simplify: \[ A = \frac{1008^2 - 1007^2}{1007 \cdot 1008} = \frac{(1008 - 1007)(1008 + 1007)}{1007 \cdot 1008} = \frac{1 \cdot 2015}{1007 \cdot 1008} \] 7. Since 2015 is relatively prime to \(1007 \cdot 1008\), the fraction is already in its simplest form. Therefore, the numerator \(m\) is: \[ \boxed{2015} \]" a8fdcc2bb31e,"7. Given $\alpha, \beta \in[0, \pi]$, then the maximum value of $(\sin \alpha+\sin (\alpha+\beta)) \cdot \sin \beta$ is",\frac{\cos \alpha}{\sin \beta}$ and $\sin ^{2} \frac{\beta}{2} = \frac{1}{2} \cos ^{2} \frac{\beta}{,medium,"7. $\frac{8 \sqrt{3}}{9}$. According to the Cauchy-Schwarz inequality, we have $$ \begin{aligned} (\sin \alpha + \sin (\alpha + \beta))^{2} & = (\sin \alpha + \sin \alpha \cos \beta + \cos \alpha \sin \beta)^{2} = (\sin \alpha (1 + \cos \beta) + \cos \alpha \sin \beta)^{2} \\ & \leqslant \left(\sin ^{2} \alpha + \cos ^{2} \alpha\right)\left((1 + \cos \beta)^{2} + \sin ^{2} \beta\right) = 2 + 2 \cos \beta = 4 \cos ^{2} \frac{\beta}{2}. \end{aligned} $$ Given $\beta \in [0, \pi]$, we have $\frac{\beta}{2} \in \left[0, \frac{\pi}{2}\right]$, thus $$ \begin{aligned} (\sin \alpha + \sin (\alpha + \beta)) \cdot \sin \beta & \leqslant 2 \sin \beta \cos \frac{\beta}{2} = 4 \sin \frac{\beta}{2} \cos ^{2} \frac{\beta}{2} = 8 \sqrt{\sin ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2}} \\ & \leqslant 8 \sqrt{\left(\frac{\sin ^{2} \frac{\beta}{2} + \frac{1}{2} \cos ^{2} \frac{\beta}{2} + \frac{1}{2} \cos ^{2} \frac{\beta}{2}}{3}\right)^{3}} = \frac{8 \sqrt{3}}{9}. \end{aligned} $$ In the above two inequalities, equality holds if and only if $\frac{\sin \alpha}{1 + \cos \beta} = \frac{\cos \alpha}{\sin \beta}$ and $\sin ^{2} \frac{\beta}{2} = \frac{1}{2} \cos ^{2} \frac{\beta}{2}$. In this case, $\beta = 2 \arctan \frac{\sqrt{2}}{2}$, $\alpha = \arctan \sqrt{2}$. Therefore, the maximum value of $(\sin \alpha + \sin (\alpha + \beta)) \cdot \sin \beta$ is $\frac{8 \sqrt{3}}{9}$." b6f58345408a,,See reasoning trace,medium,"Solution. $(0, b)$ and $(a, 0)$ are solutions for any integers $a, b$ .... 1 point $(a, b)$ is a solution if and only if $(a, -b)$ is a solution, reduction to the case $a b > 0$ 1 point $a^{2} + 2 b^{2} + 2 a + 1 \leq 2 a b$ 1 point $(a - 2 b)^{2} + (a + 2)^{2} \leq 2$ 1 point $|a + 2| \leq \sqrt{2}, |a - 2 b| \leq \sqrt{2}, a \in \{-3, -2, -1\}$, $(a, b) \in \{(-3, -2); (-3, -1); (-1, -1)\}$ (after verification) 2 points $S_{1} = \{(-3, -2); (-3, -1); (-1, -1); (-3, 2); (-3, 1); (-1, 1)\}$, $S_{2} = \{(a, 0) \mid a \in \mathbb{Z}\} \cup \{(0, b) \mid b \in \mathbb{Z}\}, S = S_{1} \cup S_{2}$. 1 point" 1da7e9ace6f7,"Example 3 Let $a, b, c$ be positive numbers, and $abc=1$, find the minimum value of $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$.",See reasoning trace,medium,"Solution: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}^{+}\right)$, then $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1}=\frac{y}{y+2 x}+\frac{z}{z+2 y}+$ $\frac{x}{x+2 z}$. By the Cauchy-Schwarz inequality, we have $$ \begin{array}{c} {[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\ \left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \geqslant(x+y+z)^{2} . \end{array} $$ Thus, $\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} \geqslant$ $$ \frac{(x+y+z)^{2}}{[y(y+2 x)+z(z+2 y)+x(x+2 z)]}=1, $$ which means $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$. Equality holds if and only if $a=b=c=1$. Therefore, the minimum value sought is 1." 5338043ae7ab,"1. Given the sets $U=\{1,2,3,4,5,6,7,8\}, A=\{1,2,3,4,5\}, B=\{4,5,6,7,8\}$, the number of subsets of set $U$ that are not subsets of set $A$ and are not subsets of set $B$ is",196$.,easy,"The number of subsets of set $U$ is $2^{8}=256$, the number of subsets of set $A$ is $2^{5}=32$, and the number of subsets of set $B$ is $2^{5}=32$, among which the sets that are subsets of both $A$ and $B$ are $2^{2}=4$. Therefore, the number of sets that are neither subsets of set $A$ nor subsets of set $B$ is $256-32-32+4=196$." 279bad827ad4,"Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$. A fence is located at the horizontal line $y = 0$. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$. Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.",273,medium,"Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$. Also note that $E(0)=1+\frac{2E(0)+E(1)}{3}$. This gives $E(0)=E(1)+3$. Plugging this into the equation for $E(1)$ gives that \[E(1)=1+\frac{E(2)+3E(1)+3}{4},\] or $E(1)=E(2)+7$. Iteratively plugging this in gives that $E(n)=E(n+1)+4n+3$. Thus $E(23)=E(24)+95$, $E(22)=E(23)+91=E(24)+186$, and $E(21)=E(22)+87=E(24)+273=\boxed{273}$." b16c99827758,"3.20. Calculate $$ \int_{1}^{i} \frac{\ln ^{2} z}{z} d z $$",See reasoning trace,medium,"Solution. Let's take some region $D$ that does not contain 0. In this region, as it is not difficult to verify, the function $$ f(z)=\frac{\ln ^{2} z}{z} $$ is analytic, and therefore, by the corollary of Cauchy's theorem, $$ \begin{gathered} \int_{1}^{i} \frac{\ln ^{2} z}{z} d z=\int_{1}^{i} \ln ^{2} z d \ln z=\left.\frac{\ln ^{3} z}{3}\right|_{1} ^{i}=\frac{\ln ^{3} i-\ln ^{3} 1}{3}= \\ =\frac{\ln ^{3} i}{3}=\frac{(\ln |i|+i \arg i)^{3}}{3}=\frac{1}{3}\left(i \frac{\pi}{2}\right)^{3}=\frac{-i \pi^{3}}{24} \end{gathered} $$ ### 3.12 . CAUCHY INTEGRAL FORMULA ### 3.12.1. CAUCHY INTEGRAL FORMULA Let $f(z)$ be a single-valued analytic function in a simply connected domain $D$ and $L \subset D$ be an arbitrary piecewise continuous closed curve bounding a region $G \subset D$. Then for any point $z_{0} \in G$, the Cauchy integral formula holds: $$ f\left(z_{0}\right)=\frac{1}{2 \pi i} \int_{L} \frac{f(z)}{z-z_{0}} d z $$ The Cauchy formula allows us to determine the value of the analytic function $f(z)$ at any point inside a certain region of analyticity based on its values on a closed contour bounding this region. 3.12.2. FORMULA FOR DERIVATIVES For any point $z_{0} \in G \subset D$, the derivative $f^{\prime}\left(z_{0}\right)$ can be found by differentiating the right-hand side of (3.43) under the integral sign, leading to the following expression for $f^{\prime}\left(z_{0}\right):$ $$ f^{\prime}\left(z_{0}\right)=\frac{1}{2 \pi i} \int \frac{f(z)}{\left(z-z_{0}\right)^{2}} d z $$ By applying repeated differentiation to (3.46), it can be shown by induction that the formula for the $n$-th order derivative of the analytic function $f(z)$ is: $$ f^{(n)}\left(z_{0}\right)=\frac{n!}{2 \pi i} \int \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z $$ ## Examples." de6626b79a1f,"7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box?",20,medium,"Answer: 20. Solution: Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $190, y>0, z>0)$, we have $$ \left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) \leq 22 $$ From AM-GM we have $$ \frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6, \quad \frac{4 z}{y}+\frac{9 y}{z} \geq 12 $$ Therefore $$ 22 \leq\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) $$ Now from (1) and (3) we get $$ \left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)=22 $$ which means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$. Therefore $\frac{4 x}{y}=\frac{y}{x}, \frac{z}{x}=\frac{9 x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2 x, z=3 x$. Finally if we substitute for $y$ and $z$, in $x+y+z=12$, we get $x=2$, therefore $y=2 \cdot 2=4$ and $z=3 \cdot 2=6$. Thus the unique solution is $(x, y, z)=(2,4,6)$." 191fd25428de,11.4. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge length a. The ends of the segment intersecting edge $C_{1} D_{1}$ lie on the lines $A A_{1}$ and $B C$. What is the minimum length that this segment can have,"A M > a$ and $y = B N > a$. Considering projections onto the planes $A A_{1} B$ and $A B C$, we get ",medium,"11.4. Let points $M$ and $N$ lie on the lines $A A_{1}$ and $B C$ respectively, and the segment $M N$ intersects the line $C_{1} D_{1}$ at point $L$. Then points $M$ and $N$ lie on the rays $A A_{1}$ and $B C$, with $x = A M > a$ and $y = B N > a$. Considering projections onto the planes $A A_{1} B$ and $A B C$, we get $C_{1} L: L D_{1} = a: (x - a)$ and $C_{1} L: L D_{1} = (y - a): a$. Therefore, $(x - a)(y - a) = a^{2}$, i.e., $x y = (x + y) a$, and thus, $(x y)^{2} = (x + y)^{2} a^{2} \geq 4 x y a^{2}$, i.e., $x y \geq 4 a^{2}$. Consequently, $M N^{2} = x^{2} + y^{2} + a^{2} = (x + y)^{2} - 2 x y + a^{2} = (x y)^{2} / a^{2} - 2 x y + a^{2} = (x y - a^{2})^{2} / a^{2} \geq 9 a^{2}$. The minimum value of the length of segment $M N$ is $3 a$; it is achieved when $A M = B N = 2 a$." 99af8d139b2a,"## Task 18/77 Given is a cylindrical beaker of height $H$. The (homogeneous) mantle has the mass $M$, the mass of the bottom is negligible. The glass is filled with a (homogeneous) liquid of mass $m_{H}$ to the brim, but the liquid drips out through an opening in the bottom until it is completely empty. Required is a function that gives the position of the center of mass as a function of the fill height $h$, as well as the smallest distance of the center of mass from the bottom.",f\left(h_{E}\right)<0.5 H=f(0)=f(H)$ and b) $f(h)$ is continuous.,medium,"Due to the homogeneity and symmetry, the center of gravity lies on the axis of the cylinder, so that its position is completely determined by the height $s$ above the ground. Also due to the homogeneity, the center of gravity of the mantle is at a height of 0.5H, and the center of gravity of the liquid is at a height of $0.5 h$ above the ground (where $0 \leq h \leq H$ applies). The mass of the liquid at the fill height $h$ is $m=\frac{h}{H} m_{H}$. For the two moments $M_{G}$ and $M_{F}$ (referenced to the ground), the following applies: $$ M_{G}=M \cdot 0.5 H=0.5 H M \quad ; \quad M_{F}=m \cdot 0.5 h=0.5 \frac{h^{2}}{H} \cdot m_{H} $$ The total moment then results in $$ M_{G}+M_{F}=0.5 H M+0.5 \frac{h^{2}}{H} m_{H}=\frac{1}{2 H}\left(H^{2} M+h^{2} m_{H}\right) $$ If the total moment is divided by the total mass, the distance $s$ of the center of gravity from the ground is obtained: $$ s=f(h)=\frac{\frac{1}{2 H}\left(H^{2} M+h^{2} m_{H}\right)}{M+\frac{h}{H} m_{H}}=0.5 \frac{H^{2} M+h^{2} m_{H}}{H M+h m_{H}} $$ For $h=0$ and for $h=H$, this yields (as expected due to symmetry) $$ s_{0}=s_{H}=f(0)=f(H)=0.5 H $$ Possible extreme value points of the function are obtained by setting the first derivative to zero, where it suffices to set the numerator to zero: $$ \begin{gathered} s^{\prime}=f^{\prime}(h)=\frac{m_{H}}{2} \cdot \frac{h^{2} m_{H}+2 h H M-H^{2} M}{\left(H M+h m_{H}\right)^{2}} \\ h^{2} m_{H}+2 h H M-H^{2} M=0 \rightarrow h_{E}=\frac{H}{m_{H}}\left(\sqrt{M\left(M+m_{H}\right)}-M\right) \end{gathered} $$ (since a negative fill height is meaningless, the negative root value is discarded). From this it follows $$ s_{E}=f\left(h_{E}\right)=\frac{H M}{m_{H}}\left(\sqrt{1+\frac{m_{H}}{M}}-1\right) $$ That this value is a minimum can be confirmed in various ways: 1. by checking the second derivative; it is $f^{\prime \prime}\left(h_{E}\right)>0$; 2. by the observations that a) $s_{E}=f\left(h_{E}\right)<0.5 H=f(0)=f(H)$ and b) $f(h)$ is continuous." 9193b5226c2c,"Example 8. Find the directional derivative along the radius vector $\mathbf{r}$ for the function $u=\sin r$, where $r=|\mathbf{r}|$.",See reasoning trace,medium,"Solution. According to formula (2), the derivative of the function with respect to the direction of the radius vector $\mathbf{r}$ is $$ \frac{\partial u}{\partial r}=\left(\operatorname{grad} \sin r, r^{0}\right) $$ We find the gradient of this function: $$ \begin{aligned} \text { grad } \sin r & =\frac{\partial(\sin r)}{\partial x} \mathbf{i}+\frac{\partial(\sin r)}{\partial y} \mathbf{j}+\frac{\partial(\sin r)}{\partial z} \mathbf{k}= \\ & =\frac{\partial(\sin r)}{\partial r} \frac{\partial r}{\partial x} \mathbf{i}+\frac{\partial(\sin r)}{\partial r} \frac{\partial r}{\partial y} \mathbf{j}+\frac{\partial(\sin r)}{\partial r} \frac{\partial r}{\partial z} \mathbf{k}= \\ & =\left(\frac{\partial r}{\partial x} \mathbf{i}+\frac{\partial r}{\partial y} \mathbf{j}+\frac{\partial r}{\partial z} \mathbf{k}\right) \cos r=r^{0} \cos r \end{aligned} $$ Substituting (4) into (3), we get $$ \frac{\partial u}{\partial r}=\left(\mathbf{r}^{0} \cos r, \mathbf{r}^{0}\right)=\left(\mathbf{r}^{0}, \mathbf{r}^{0}\right) \cos r=\cos r $$" f2a13be5df7c,"What magnitude force does Jonathan need to exert on the physics book to keep the rope from slipping? (A) $Mg$ (B) $\mu_k Mg$ (C) $\mu_k Mg/\mu_s$ (D) $(\mu_s + \mu_k)Mg$ (E) $Mg/\mu_s$",\frac{Mg,medium,"1. **Identify the forces acting on the book:** - The gravitational force acting downward: \( F_g = Mg \) - The normal force exerted by the surface on the book: \( N \) - The static friction force that prevents the book from slipping: \( F_s \) 2. **Express the static friction force:** The maximum static friction force is given by: \[ F_s = \mu_s N \] where \( \mu_s \) is the coefficient of static friction. 3. **Set up the inequality for static friction:** To prevent the book from slipping, the static friction force must be at least equal to the gravitational force: \[ \mu_s N \ge Mg \] 4. **Solve for the normal force \( N \):** Rearrange the inequality to solve for \( N \): \[ N \ge \frac{Mg}{\mu_s} \] 5. **Interpret the result:** The normal force \( N \) represents the force Jonathan needs to exert to keep the book from slipping. Therefore, the required force is: \[ N \ge \frac{Mg}{\mu_s} \] 6. **Match the result with the given choices:** The correct choice is (E) \( \frac{Mg}{\mu_s} \). The final answer is \( \boxed{ \frac{Mg}{\mu_s} } \)" 62ae4a13db49,"(7) There are 5 pairs of shoes with different sizes. If 4 shoes are taken out, the probability of being able to form exactly one pair is $\qquad$",\frac{\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{10}^{,easy,(7) $\frac{4}{7}$ Hint: $P=\frac{\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{10}^{4}}=\frac{120}{210}=\frac{4}{7}$. 30e522957d45,12th Swedish 1972,"1, we have x = 1, y = 1. Suppose a > 1. Then a(4y+1) = 1-3y, so y = -(a-1)/(4a+3) -1. Contradiction",easy,"1 Solution For a = 1, we have x = 1, y = 1. Suppose a > 1. Then a(4y+1) = 1-3y, so y = -(a-1)/(4a+3) -1. Contradiction. So there are no solutions with a > 1. Thanks to Suat Namli 12th Swedish 1972 © John Scholes jscholes@kalva.demon.co.uk 12 February 2004 Last corrected/updated 12 Feb 04" cfa0d0c04b1a,"9. The sum of the first $n$ terms of an arithmetic sequence is 2000, the common difference is 2, the first term is an integer, and $n>1$, the sum of all possible values of $n$ is $\qquad$ .",See reasoning trace,easy,"9. 4835 . $$ \begin{array}{l} S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}=n\left(a_{1}+n-1\right)=2000=2^{4} \times 5^{3} \text {, so the required value is }(1+5+25+125) \times \\ (1+2+4+8+16)-1=4835 . \end{array} $$" 32b243c88d63,"\left.\begin{array}{l}{[\quad \text { Pythagorean Theorem (direct and inverse). }} \\ \text { [ The ratio in which the bisector divides the side. }]\end{array}\right] The height of a triangle, equal to 2, divides the angle of the triangle in the ratio 2:1, and the base of the triangle - into parts, the smaller of which is equal to 1. Find the area of the triangle.",$\frac{11}{3}$,medium,"## First Method. Let $BD=2$ be the height of triangle $ABC$, and $AD=1$. In the right triangles $ABD$ and $CBD$, the common leg is $BD$, and according to the problem, the leg $AD$ of the first triangle is less than the leg $CD$ of the second. Therefore, $\angle ABD < \angle CBD$. Let $\angle ABD = \alpha$. Then $\angle CBD = 2\alpha$. From the right triangle $ABD$, we find that $$ \tan \alpha = \frac{AD}{BD} = \frac{1}{2} $$ Then $$ \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{4}{3} $$ Then, in the right triangle $CBD$, $$ CD = BD \cdot \tan \angle CBD = 2 \cdot \tan 2\alpha = 2 \cdot \frac{4}{3} = \frac{8}{3} $$ Therefore, $$ S_{\triangle ABC} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \left(1 + \frac{8}{3}\right) \cdot 2 = \frac{11}{3} $$ ## Second Method. Let $BD=2$ be the height of triangle $ABC$, and $AD=1$. In the right triangles $ABD$ and $CBD$, the common leg is $BD$, and according to the problem, the leg $AD$ of the first triangle is less than the leg $CD$ of the second. Therefore, $\angle ABD < \angle CBD$. Let $\angle ABD = \alpha$. Then $\angle CBD = 2\alpha$. Suppose the angle bisector of $\angle DBC$ intersects segment $CD$ at point $E$. Then triangle $ABE$ is isosceles, because its height $BD$ is also the angle bisector. Therefore, $DE = AD = 1$. Since $BD = 2DE$, by the property of the angle bisector of a triangle, $BC = 2CE$. Let $CE = x$. By the Pythagorean theorem, $$ BC^2 = BD^2 + CD^2, \text{ or } 4x^2 = 4 + (1 + x)^2 $$ From this equation, we find that $x = \frac{5}{3}$. Therefore, $$ S_{\triangle ABC} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \left(1 + 1 + \frac{5}{3}\right) \cdot 2 = \frac{11}{3} $$ ## Answer $\frac{11}{3}$" 0d8b926255d3,"4. A sequence of 300 natural numbers is written in a row. Each number, starting from the third, is equal to the product of the two preceding numbers. How many perfect squares can there be among these numbers? (Provide all answers and prove that there are no others.)","0, 100 or 300",easy,"Answer: 0, 100 or 300. If after number a, number $b$ is written, then the numbers аb and $а b^{2}$ follow them. The number $a b^{2}$ is a perfect square if and only if a is a perfect square. Among the first three numbers, squares can be all three, one, or none (examples of all three situations are trivial). Hence, the answer follows immediately." 1844c7a3de9b,"Example 2 Given $A=\left\{x \mid x^{2}+p x+q=0, x \in \mathrm{R}\right\}, B=\left\{x \mid x^{2}-3 x+2\right.$ $=0, x \in \mathrm{R}\}$, and $A \cap B=A$, find the values or relationship of $p$ and $q$.","\{1,2\}$, the equation $x^{2}+p x+q=0$ has two roots 1 and 2, then $p=$ $-3, q=2$.",medium,"Solution: From the problem, we know that $B=\left\{x \mid x^{2}-3 x+2=0\right\}=\{1,2\}$. From $A \cap B=A$, we get $A \subseteq B$, so $A=\varnothing,\{1\},\{2\}$, or $\{1,2\}$. When $A=\varnothing$, the equation $x^{2}+p x+q=0$ has no real roots, then $\Delta=p^{2}-4 q<0$, i.e., $p^{2}<4 q$; When $A=\{1\}$, the equation $x^{2}+p x+q=0$ has a repeated root 1, then $p=-2, q=1$; When $A=\{2\}$, the equation $x^{2}+p x+q=0$ has a repeated root 2, then $p=-4, q=4$; When $A=\{1,2\}$, the equation $x^{2}+p x+q=0$ has two roots 1 and 2, then $p=$ $-3, q=2$." f1d9a4a19732,"On an island there are two types of inhabitants: Heroes who always tell the truth and Villains who always lie. Four inhabitants are seated around a table. When each is asked ""Are you a Hero or a Villain?"", all four reply ""Hero"". When asked ""Is the person on your right a Hero or a Villain?"", all four reply ""Villain"". How many Heroes are present? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4","""Hero"" and Villains will lie and answer ""Hero""",medium,"First, we note that no matter how many Heroes are present, all four would always reply ""Hero"" when asked ""Are you a Hero or a Villain?"". (This is because Heroes will tell the truth and answer ""Hero"" and Villains will lie and answer ""Hero"".) When each is asked ""Is the person on your right a Hero or a Villain?"", all four reply ""Villain"", so any Hero that is at the table must have a Villain on his right (or he would have answered ""Hero"") and any Villain at the table must have a Hero on his right (or he would have had a Villain on his right and answered ""Hero""). In other words, Heroes and Villains must alternate around the table, so there are 2 Heroes and 2 Villains. (It is worth checking that when 2 Heroes and 2 Villains sit in alternate seats, the answers are indeed as claimed.) ANSWER: (C)" b3b9db690645,"10. (14 points) Given the set $A$ consisting of the positive integers $1,2, \cdots, 1000$, first randomly select a number $a$ from set $A$, and after selecting, return $a$ to set $A$; then randomly select another number $b$ from set $A$. Find the probability that $\frac{a}{b}>\frac{1}{3}$.",See reasoning trace,medium,"10. Solution 1 Note that $$ P\left(\frac{a}{b}>\frac{1}{3}\right)=1-P\left(\frac{a}{b} \leqslant \frac{1}{3}\right) \text {. } $$ By $\frac{a}{b} \leqslant \frac{1}{3} \Leftrightarrow a \leqslant \frac{1}{3} b \Leftrightarrow a \leqslant\left[\frac{1}{3} b\right]$. Then $P\left(\frac{a}{b} \leqslant \frac{1}{3}\right)=\frac{\sum_{b=1}^{1000}\left[\frac{1}{3} b\right]}{1000^{2}}$ $$ =\frac{\sum_{k=1}^{332} 3 k+333+333}{1000^{2}}=\frac{333}{2000} \text {. } $$ Therefore, $P\left(\frac{a}{b}>\frac{1}{3}\right)=1-\frac{333}{2000}=\frac{1667}{2000}$. Solution 2 When $a=1$, $b=1$ or 2, there are 2 ways; When $a=2$, $b$ increases to $3, 4, 5$, there are $2+3$ ways; When $a=3$, $b$ increases to $6, 7, 8$, there are $2+2 \times 3$ ways; When $a=333$, $b$ has $2+332 \times 3$ ways; when $334 \leqslant a \leqslant 1000$, $b$ has 1000 ways. Thus, $P\left(\frac{a}{b}>\frac{1}{3}\right)$ $$ \begin{array}{l} =\frac{2+(2+3)+(2+2 \times 3)+\cdots+(2+332 \times 3)+667 \times 1000}{1000^{2}} \\ =\frac{333(2+166 \times 3)+667 \times 1000}{1000^{2}} \\ =\frac{1667}{2000} . \end{array} $$" 2b92c3616957,"A cube-shaped container with an edge length of $12 \mathrm{~cm}$ was filled to $\frac{5}{8}$ of its capacity with a liquid, and then it was slightly tilted along one of its edges. The diagram shows the cross-section of the container with the horizontal level of the liquid inside. We know that the length of segment $L C$ is exactly twice the length of segment $K B$. Determine the length of segment $L C$. ![](https://cdn.mathpix.com/cropped/2024_05_02_2ef57db81a3ef472104fg-1.jpg?height=265&width=271&top_left_y=331&top_left_x=922)",5$. The length of the segment $L C$ is $10 \mathrm{~cm}$.,easy,"Solution. Before the impact, the fluid level was at $\frac{5}{8}$ of the 12 cm height, i.e., at $7.5 \mathrm{~cm}$. The area of this rectangle's cross-section is $12 \cdot 7.5=90 \mathrm{~cm}^{2}$. After the impact, the area of the trapezoid $B C L K$ will be $90 \mathrm{~cm}^{2}$. Let $L C=2 x$, then according to the text, $K B=x$, so we can write the area of the trapezoid as: $$ \frac{(2 x+x) \cdot 12}{2}=90 $$ from which $x=5$. The length of the segment $L C$ is $10 \mathrm{~cm}$." 826c7275421f,"18. (12 points) Given vectors $\boldsymbol{m}=(\sin A, \cos A)$, $\boldsymbol{n}=(\cos B, \sin B)$, $\boldsymbol{m} \cdot \boldsymbol{n}=\sin 2 C$, and $\angle A$, $\angle B$, $\angle C$ are the angles opposite to sides $a$, $b$, $c$ of $\triangle ABC$ respectively. (1) Find the size of $\angle C$; (2) If $\sin A$, $\sin C$, $\sin B$ form a geometric sequence, and $\overrightarrow{C A} \cdot \overrightarrow{C B}=18$, find the value of $c$.",18 \Rightarrow c^2 = a b = 36 \Rightarrow c = 6$.,medium,"18. (1) From the given information, $$ \begin{array}{l} \sin A \cdot \cos B + \cos A \cdot \sin B = \sin 2 C \\ \Rightarrow \sin C = \sin 2 C \Rightarrow \cos C = \frac{1}{2}. \end{array} $$ Since $\angle C$ is an interior angle of $\triangle ABC$, we have $\angle C = \frac{\pi}{3}$. (2) From the given information, $$ \begin{array}{l} \sin^2 C = \sin A \cdot \sin B \Rightarrow c^2 = a b. \\ \text{Also, } \overrightarrow{C A} \cdot \overrightarrow{C B} = 18 \text{, then } \end{array} $$ $a b \cos C = 18 \Rightarrow c^2 = a b = 36 \Rightarrow c = 6$." 49d35c21c3fd,"1. Given $a+b+c=0, a>b>c$. Then the range of $\frac{c}{a}$ is $\qquad$ .",See reasoning trace,easy,"$$ \text{2.1. }-2b>c, \text{ get } a>0, cb>c, \text{ get} $$ $$ 1>\frac{b}{a}>\frac{c}{a} \text{. } $$ Substitute equation (1) into equation (2) to get $$ 1>-1-\frac{c}{a}>\frac{c}{a} \text{. } $$ Solving, we get $-2<\frac{c}{a}<-\frac{1}{2}$." 0e1c8704cb40,1. Is the sum $1^{2008}+2^{2008}+3^{2008}+4^{2008}+5^{2008}+6^{2008}$ divisible by 5? Explain your answer.,"1$. Since $2^{1}=2$, $2^{2}=4,2^{3}=8,2^{4}=16,2^{5}=32, \ldots$, we conclude that the powers of the",medium,"Solution. First, let's note that $1^{2008}=1$. Since $2^{1}=2$, $2^{2}=4,2^{3}=8,2^{4}=16,2^{5}=32, \ldots$, we conclude that the powers of the number 2 end in $2,4,8$ and 6 and repeat in this order depending on whether the exponent modulo 4 gives a remainder of $1,2,3$ or is divisible by 4. Since 2008 is divisible by 4, we have that $2^{2008}$ ends in 6. Similarly, $3^{2008}$ ends in $1,4^{2008}$ ends in $6,5^{2008}$ ends in 5 and $6^{2008}$ ends in 6. Since $1+6+1+6+5+6=25$, which ends in 5, it follows that the sum $1^{2008}+2^{2008}+3^{2008}+4^{2008}+5^{2008}+6^{2008}$ also ends in 5, i.e., is divisible by 5." e75b673080eb,"2. Positive integers $x, y$, for which $\gcd(x, y)=3$, are the coordinates of a vertex of a square with its center at the origin and an area of $20 \cdot \text{lcm}(x, y)$. Find the perimeter of the square.",$p=24 \sqrt{5}$,easy,Answer: $p=24 \sqrt{5}$. 064dbed86b5e,"3. When dividing the number $a$ by 7, we get a remainder of 3, and when dividing the number $b$ by 7, we get a remainder of 4. What is the remainder when the square of the sum of the numbers $a$ and $b$ is divided by 7? Justify your answer.",See reasoning trace,easy,"3. Write: $a=7 \cdot x+3$ ..... 1 point Write $b=7 \cdot y+4$ ..... 1 point $(a+b)^{2}=(7 x+7 y+7)^{2}=$ ..... 1 point $=(7 \cdot(x+y+1))^{2}=$ ..... 1 point $=7 \cdot\left(7 \cdot(x+y+1)^{2}\right)$ ..... 1 point Written conclusion: The remainder is 0, since $(a+b)^{2}$ is a multiple of the number 7. ..... 1 point Note: A solution obtained by trial and error is graded with two points." baa44113c360,"[ Fermat's Little Theorem ] Find the remainder when $2^{100}$ is divided by 101. #",1,easy,"According to Fermat's little theorem, it is equal to 1. ## Answer 1." 6dba6a16b978,"## Calculate the limit of the function: $\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}$",See reasoning trace,medium,"## Solution Substitution: $$ \begin{aligned} & x=y+3 \Rightarrow y=x-3 \\ & x \rightarrow 3 \Rightarrow y \rightarrow 0 \end{aligned} $$ ## We get: $\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{5+(y+3)}-2}{\sin \pi(y+3)}=$ $=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{\sin (\pi y+3 \pi)}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{\sin (\pi y+\pi)}=$ $=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{-\sin \pi y}=$ Using the substitution of equivalent infinitesimals: $\sin \pi y \sim \pi y, \text{ as } y \rightarrow 0 (\pi y \rightarrow 0)$ We get: $$ \begin{aligned} & =\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{-\pi y}= \\ & =\lim _{y \rightarrow 0} \frac{(\sqrt[3]{y+8}-2)\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\ & =\lim _{y \rightarrow 0} \frac{y+8-8}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\ & =\lim _{y \rightarrow 0} \frac{y}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\ & =\lim _{y \rightarrow 0} \frac{1}{-\pi\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\ & =\frac{1}{-\pi\left(\sqrt[3]{(0+8)^{2}}+2 \sqrt[3]{0+8}+4\right)}=\frac{1}{-\pi\left(2^{2}+2 \cdot 2+4\right)}= \\ & =-\frac{1}{12 \pi} \end{aligned} $$ Problem Kuznetsov Limits 16-23" a526e2b71a48,"2. (9th Canadian Mathematical Competition) $N$ is an integer, its representation in base $b$ is 777. Find the smallest positive integer $b$ such that $N$ is a fourth power of an integer in decimal notation.","x^{4}$, (1) has an integer solution for $x$. Since 7 is a prime number, it follows from equation (1)",medium,"2. This problem is equivalent to finding the smallest positive integer $b$, such that the equation $7 b^{2}+7 b+7=x^{4}$, (1) has an integer solution for $x$. Since 7 is a prime number, it follows from equation (1) that 7 is a divisor of $x$. Therefore, let $x=7 k$, then equation (1) becomes $b^{2}+b+1=7^{3} k^{4}$. The smallest $b$ occurs when $k$ is at its minimum. Taking $k=1$, we then have $b^{2}+b+1=343, b^{2}+b-342=0$, which is $(b-18)(b+19)=0$, yielding the positive integer solution $b=18$. Thus, we have $(777)_{18}=\left(7^{4}\right)_{10}$." 8b484530ab41,10. Evaluate $\int_{1}^{\infty}\left(\frac{\ln x}{x}\right)^{2011} d x$.,"$\frac{2011!}{2010^{2012}}$ By the chain rule, $\frac{d}{d x}(\ln x)^{n}=\frac{n \ln ^{n-1} x}{x}$",medium,"Answer: $\frac{2011!}{2010^{2012}}$ By the chain rule, $\frac{d}{d x}(\ln x)^{n}=\frac{n \ln ^{n-1} x}{x}$. We calculate the definite integral using integration by parts: $$ \int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\left[\frac{(\ln x)^{n}}{-2010 x^{2010}}\right]_{x=1}^{x=\infty}-\int_{x=1}^{\infty} \frac{n(\ln x)^{n-1}}{-2010 x^{2011}} d x $$ But $\ln (1)=0$, and $\lim _{x \rightarrow \infty} \frac{(\ln x)^{n}}{x^{2010}}=0$ for all $n>0$. So $$ \int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\int_{x=1}^{\infty} \frac{n(\ln x)^{n-1}}{2010 x^{2011}} d x $$ It follows that $$ \int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\frac{n!}{2010^{n}} \int_{x=1}^{\infty} \frac{1}{x^{2011}} d x=\frac{n!}{2010^{n+1}} $$ So the answer is $\frac{2011!}{2010^{2012}}$." b81778ad3e93,"6. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}, F_{2}$, and $P$ is any point on the ellipse that does not coincide with the left or right vertices. Points $I, G$ are the incenter and centroid of $\triangle P F_{1} F_{2}$, respectively. Given that for any point $\boldsymbol{P}, \boldsymbol{I} \boldsymbol{G}$ is always perpendicular to the $\boldsymbol{x}$-axis, the eccentricity of the ellipse is $\qquad$","3 c$, thus $e=\frac{1}{3}$;",easy,"6. $\frac{1}{3}$ Analysis: Let $\left|P F_{1}\right|=n,\left|P F_{2}\right|=m$, then $n+m=2 a$; it is easy to know $$ \operatorname{uri}_{P G}=\frac{1}{3}\left(\operatorname{urf}_{P F_{1}}+P F_{2}\right), P I=\frac{\operatorname{un}_{1}+n P F_{2}}{m+n+2 c}=\frac{1}{2} g \frac{\operatorname{un}_{1}+n P F_{2}}{a+c} $$ Since $I G \perp F_{1} F_{2}$, we have $a=3 c$, thus $e=\frac{1}{3}$;" fe71e657fe44,"$\left[\begin{array}{l}{[\text { Ratio of areas of triangles with a common angle }} \\ {\left[\begin{array}{l}\text { Law of Cosines }\end{array}\right]}\end{array}\right.$ In triangle $A B C$, angle $A$ is $60^{\circ} ; A B: A C=3: 2$. Points $M$ and $N$ are located on sides $A B$ and $A C$ respectively, such that $B M=M N=N C$. Find the ratio of the area of triangle $A M N$ to the area of triangle $A B C$.",$\frac{4}{25}$,medium,"Let $AB = 3t, AC = 2t, BM = MN = NC = x$. By the Law of Cosines, $$ MN^2 = AM^2 + AN^2 - 2 \cdot AM \cdot AN \cdot \cos 60^{\circ}, \quad x^2 = (3t - x)^2 + (2t - x)^2 - (3t - x)(2t - x) $$ from which $x = \frac{7}{5} t$. Then $$ AM = 3t - x = 3t - \frac{7}{5} t = \frac{8}{5} t, \quad AN = 2t - x = 2t - \frac{7}{5} t = \frac{3}{5} t, $$ therefore, $$ \frac{S_{\triangle AMN}}{S_{\triangle ABC}} = \frac{AM}{AB} \cdot \frac{AN}{AC} = \frac{8/5 t}{3t} \cdot \frac{3/5 t}{2t} = \frac{4}{25} $$ ## Answer $\frac{4}{25}$" 14b17da97f2a,"Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$, and the sum of the second series is $r_2$. What is $r_1 + r_2$? $\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$",1,medium,"Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$. Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$. This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$. As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$. Using [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas) we get that the sum of these two roots is $\boxed{1}$." 04a9d6f01d03,3. Solve the equation $\log _{4}\left(1+\log _{4}\left(3^{x}-\sqrt{\left(5^{0}+4^{2}\right)^{2}}\right)\right)=e^{0}$,\sqrt{(17)^{2}}=17 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldo,medium,"3. We find that $e^{0}=1$. We also simplify the radicand $\sqrt{\left(5^{0}+4^{2}\right)^{2}}=\sqrt{(17)^{2}}=17$. Using the relation $1=\log _{4} 4$ we get $1+\log _{4}\left(3^{x}-17\right)=4$. Rearranging, $\log _{4}\left(3^{x}-17\right)=3$. Using the definition of logarithm, $64=3^{x}-17$. Simplifying, $3^{x}=81$. Solving and calculating, $x=4$. ![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-11.jpg?height=57&width=1642&top_left_y=1933&top_left_x=207) Calculation $\sqrt{\left(5^{0}+4^{2}\right)^{2}}=\sqrt{(17)^{2}}=17 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldot" 24d88b17898c,"3.2.8* The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{2}=2, a_{n+2}+a_{n+1}+a_{n}=0$. Find the general term formula $a_{n}$.",See reasoning trace,medium,"The characteristic equation is $x^{2}+x+1=0$. The characteristic roots are solved as $x_{1}=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}, x_{2}=$ $\cos \frac{2 \pi}{3}-\mathrm{i} \sin \frac{2 \pi}{3}$. Therefore, $a_{n}=A_{1}\left(\cos \frac{2 n \pi}{3}+i \sin \frac{2 n \pi}{3}\right)+A_{2}\left(\cos \frac{2 n \pi}{3}-\mathrm{i} \sin \frac{2 n \pi}{3}\right)$. Substituting the initial conditions $a_{1}=1, a_{2}=2$, we get $$ \left\{\begin{array}{l} \Lambda_{1}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)+\Lambda_{2}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)=1, \\ \Lambda_{1}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\Lambda_{2}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=2 . \end{array}\right. $$ Solving, we get $A_{1}=\frac{9+\sqrt{3} \mathrm{i}}{6}, A_{2}=\frac{-9-\sqrt{3} \mathrm{i}}{6}$. Therefore, $$ \begin{aligned} a_{n} & =\frac{-9+\sqrt{3} i}{6}\left(\cos \frac{2 n \pi}{3}+\mathrm{i} \sin \frac{2 n \pi}{3}\right)+\frac{-9-\sqrt{3} i}{6}\left(\cos \frac{2 n \pi}{3}-\mathrm{i} \sin \frac{2 n \pi}{3}\right) \\ & =-3 \cos \frac{2 n \pi}{3}-\frac{\sqrt{3}}{3} \sin \frac{2 n \pi}{3} . \end{aligned} $$" d6015c620b64,Example 1. Find the zeros of the function $f(z)=1+\cos z$ and determine their orders.,"(2 n+1) \pi(n=0, \pm 1, \pm 2, \ldots)$ are zeros of the second order of the given function.",easy,"Solution. Setting $f(z)$ to zero, we get $\cos z=-1$, from which $z_{n}=$ $(2 n+1) \pi(n=0, \pm 1, \pm 2, \ldots)$ - these are the zeros of the given function. Next, $$ \begin{aligned} & f^{\prime}[(2 n+1) \pi]=-\sin (2 n+1) \pi=0 \\ & f^{\prime \prime}[(2 n+1) \pi]=-\cos (2 n+1) \pi=1 \neq 0 \end{aligned} $$ Therefore, the points $z_{n}=(2 n+1) \pi(n=0, \pm 1, \pm 2, \ldots)$ are zeros of the second order of the given function." 719c12063f1b,"Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients. Find all functions $f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ satisfying the following conditions: [list] [*] $f$ maps the zero polynomial to itself, [*] for any non-zero polynomial $P \in \mathbb{R}[x]$, $\text{deg} \, f(P) \le 1+ \text{deg} \, P$, and [*] for any two polynomials $P, Q \in \mathbb{R}[x]$, the polynomials $P-f(Q)$ and $Q-f(P)$ have the same set of real roots. [/list] [i]Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha[/i]",f(Q) = Q \text{ or,medium,"1. **Initial Conditions and Definitions:** - Let \( f: \mathbb{R}[x] \rightarrow \mathbb{R}[x] \) be a function satisfying: - \( f(0) = 0 \) - For any non-zero polynomial \( P \in \mathbb{R}[x] \), \( \deg f(P) \leq 1 + \deg P \) - For any two polynomials \( P, Q \in \mathbb{R}[x] \), the polynomials \( P - f(Q) \) and \( Q - f(P) \) have the same set of real roots. 2. **Involution Property:** - From \( B(f(Q), Q) \), we have that \( 0 \) and \( Q - f(f(Q)) \) have the same set of real roots. This implies \( f(f(Q)) = Q \), so \( f \) is an involution. 3. **Roots Preservation:** - From \( B(0, Q) \), we have that \( f(Q) \) and \( Q \) have the same real roots. 4. **Degree Analysis for Linear Polynomials:** - Suppose \( \deg R = 1 \). Then \( A(R) \) implies \( \deg f(R) \in \{0, 1, 2\} \). - Since \( f \) is an involution and preserves roots, \( f(R) \) must be of the form \( cR \) or \( cR^2 \) or a constant polynomial \( C \). 5. **Claim 1: \( \deg f(R) \neq 2 \):** - If \( f(R) = cR^2 \), then \( cR^2 - f(d) \) and \( R - d \) must have the same set of real roots for some \( d \). Choosing \( d \) such that \( f(d) \) has a different sign from \( c \), \( cR^2 - f(d) \) will not have any real roots, leading to a contradiction. Hence, \( f(R) \neq cR^2 \). 6. **Claim 2: \( \deg f(R) \neq 0 \):** - Suppose \( f(R_1) = C \) for some \( R_1 \). Then \( A(C) \) implies \( \deg f(C) \in \{0, 1\} \). - If \( \deg f(C) = 1 \), then \( B(C, 0) \) implies \( f(C) = 0 \), which is a contradiction. Hence, \( \deg f(C) = 0 \). - \( B(R_1, C) \) implies \( R_1 - f(C) \) and \( 0 \) have the same set of real roots, so \( R_1 = f(C) \), leading to a contradiction. 7. **Consistency for Different Roots:** - Let \( R, R' \) have different roots. From \( B(R, R') \), \( cR - R' \) and \( R - dR' \) must have the same set of roots, implying \( mc = 1 \) and \( -m = -d \). This gives \( cd = 1 \). - Choosing another polynomial \( R_2 \) with different roots from both, and letting \( f(R_2) = kR_2 \), we get \( cd = dk = ck = 1 \), so \( c = d = \pm 1 \). 8. **Final Cases:** - If \( f(T) = T \) for all \( T \in \mathbb{R}[x] \), then \( B(R, Q) \) implies \( f(Q) - R \) and \( Q - R \) have the same set of real roots, leading to \( f(Q) = Q \). - If \( f(T) = -T \) for all \( T \in \mathbb{R}[x] \), then \( B(R, Q) \) implies \( f(Q) - R \) and \( Q + R \) have the same set of real roots, leading to \( f(Q) = -Q \). Conclusion: \[ \boxed{f(Q) = Q \text{ or } f(Q) = -Q \text{ for all } Q \in \mathbb{R}[x]} \]" 612d8955bff5,6. The maximum value of the volume $V$ of a tetrahedron inscribed in a sphere with radius $R$ is $\qquad$ .,See reasoning trace,medium,"6. $\frac{8 \sqrt{3}}{27} R^{3}$. Let the tetrahedron be $P-ABC$, and the circumradius of $\triangle ABC$ be $r$. Then $$ S_{\triangle ABC}=2 r^{2} \sin A \cdot \sin B \cdot \sin C \leqslant \frac{3 \sqrt{3}}{4} r^{2} . $$ The equality holds if and only if $\angle A=\angle B=\angle C=60^{\circ}$. If the distance from the center of the sphere $O$ to the plane $ABC$ is $h$, then $$ \begin{array}{l} V \leqslant \frac{1}{3} S_{\triangle ABC}(R+h) \leqslant \frac{\sqrt{3}}{4} r^{2}(R+h) \\ =\frac{\sqrt{3}}{4}\left(R^{2}-h^{2}\right)(R+h) \\ =\frac{\sqrt{3}}{8}(R+h)(R+h)(2 R-2 h) \\ \leqslant \frac{\sqrt{3}}{8}\left(\frac{R+h+R+h+2 R-2 h}{3}\right)^{3} \\ =\frac{8 \sqrt{3}}{27} R^{3} . \end{array} $$ The equality holds if and only if the tetrahedron $P-ABC$ is a regular tetrahedron." 9d70ca312a18,11. Find the sum of the values of the polynomial \(x^{5}-1.7 \cdot x^{3}+2.5\) at \(x=19.1\) and \(x=-19.1\).,See reasoning trace,easy,"11. Solution. If the sign of \(x\) is changed to the opposite, the given polynomial will take the form: \(-x^{5}+1.7 x^{3}+2.5\). The sum of the given polynomial and the obtained one is 5." 2544916ff117,"## Calculate the indefinite integral: $$ \int(7 x-10) \sin 4 x \, d x $$",See reasoning trace,medium,"## Solution $$ \int(7 x-10) \sin 4 x d x= $$ Let: $$ \begin{aligned} u=7 x-10 ; d u & =7 d x \\ d v=\sin 4 x d x ; v & =-\frac{1}{4} \cos 4 x \end{aligned} $$ Using the integration by parts formula $\int u d v=u v-\int v d u$. We get: $$ \begin{aligned} & =(7 x-10) \cdot\left(-\frac{1}{4} \cos 4 x\right)-\int\left(-\frac{1}{4} \cos 4 x\right) \cdot 7 d x= \\ & =\frac{1}{4} \cdot(10-7 x) \cos 4 x+\frac{7}{4} \int \cos 4 x d x=\frac{1}{4} \cdot(10-7 x) \cos 4 x+\frac{7}{4} \cdot \frac{1}{4} \cdot \sin 4 x+C= \\ & =\frac{1}{4} \cdot(10-7 x) \cos 4 x+\frac{7}{16} \cdot \sin 4 x+C \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-26 » Categories: Kuznetsov Problem Book Integrals Problem 1 | Integrals Ukrainian Banner Network - Last modified: 19:52, 3 March 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals $1-27$ ## Material from PlusPi" 4ec9087933e3,"The height of a regular quadrilateral pyramid $S A B C D$ ($S$ - the vertex) is $\sqrt{3}$ times the length of the base edge. Point $E$ is the midpoint of the apothem lying in the face $A S B$. Find the angle between the line $D E$ and the plane $A S C$. #",$45^{\circ}$,medium,"Let $S H$ be the height of the pyramid, $M$ and $N$ be the midpoints of edges $A B$ and $S A$ respectively (Fig.1). Let $A B=a, S H=a \sqrt{3}$. On the extension of edge $C D$ beyond point $D$, lay off the segment $D P=\frac{1}{4} C D=\frac{1}{4} a$. Since $E N$ is the midline of triangle $A S M$, then $N E\|A M\| D P$ and $N E=\frac{1}{2} A M=\frac{1}{4} a=D P$, hence, quadrilateral $D P N E$ is a parallelogram, so $P N=D E$ and $P N \| D E$. Therefore, the angle $\phi$ between the plane $A S C$ and the line $D E$ is equal to the angle between this plane and the line $P N$. The plane $A S C$ passes through the line $S H$, which is perpendicular to the plane $A B C D$, so the planes $A S C$ and $A B C D$ are perpendicular, meaning that the perpendicular $P Q$, dropped from point $P$ to the line $A C$, is a perpendicular to the plane $A S C$. Therefore, the distance from point $P$ to the plane $A S C$ is equal to the length of the segment $P Q$. From the isosceles right triangle $C P Q$, we find that $$ P Q=\frac{C P}{\sqrt{2}}=\frac{\operatorname{Su}^{2}}{\sqrt{2}}=\frac{5 n}{4 \sqrt{2}} $$ Let $F$ be the orthogonal projection of point $E$ onto the base plane of the pyramid, $K$ be the midpoint of $C D$. Then $F$ is the midpoint of $4 M$, so $$ \begin{gathered} F K=\frac{3}{4} a, E F=\frac{1}{2} S H=\frac{a \sqrt{3}}{2} \\ P N=D E=\sqrt{D K^{2}+K F^{2}+E F^{2}}=\sqrt{\frac{\alpha^{2}}{4}+\frac{9}{16} a^{2}+\frac{3}{4} a^{2}}=\frac{5}{4} a \end{gathered} $$ From the right triangle $P Q N$ (Fig.2), we find that $$ \sin \phi=\sin \angle P N Q=\frac{P Q}{P N}=\frac{\frac{5 s}{\sqrt{2}}}{\frac{\sqrt{2}}{\sqrt{a}}}=\frac{1}{\sqrt{2}} $$ Therefore, $\phi=45^{\circ}$. ## Answer $45^{\circ}$." 8e1397d7a828,"$12 \cdot 6$ Let non-zero complex numbers $x, y$ satisfy $x^{2}+x y+y^{2}=0$, then the value of the algebraic expression $\left(\frac{x}{x+y}\right)^{1990}+\left(\frac{y}{x+y}\right)^{1990}$ is (A) $2^{-1989}$. (B) -1 . (C) 1 . (D) None of the above answers is correct. (China High School Mathematics League, 1990)",$(B)$,medium,"[Solution] Let $y=\omega x, \omega \neq 1$, substituting into the given condition yields $1+\omega+\omega^{2}=0$. Therefore, $(1-\omega)\left(1+\omega+\omega^{2}\right)=0$, which implies $\omega^{3}=1$. $$ \begin{aligned} \therefore \quad \text { the original expression } & =\frac{1}{(1+\omega)^{1990}}+\frac{\omega^{1990}}{(1+\omega)^{1990}}=\frac{1+\omega^{1990}}{\left(-\omega^{2}\right)^{1990}} \\ & =\frac{1+\omega^{3 \times 663+1}}{\omega^{2 \times(3 \times 663+1)}}=\frac{1+\omega}{\omega^{2}}=-1 . \end{aligned} $$ Therefore, the answer is $(B)$." 8d5e77df0376,3.303. $1-\sin ^{2} \alpha-\sin ^{2} \beta+2 \sin \alpha \sin \beta \cos (\alpha-\beta)$.,$\cos ^{2}(\alpha-\beta)$,medium,"## Solution. $$ \begin{aligned} & 1-\sin ^{2} \alpha-\sin ^{2} \beta+2 \sin \alpha \sin \beta \cos (\alpha-\beta)= \\ & =1-\frac{1-\cos 2 \alpha}{2}-\frac{1-\cos 2 \beta}{2}+2 \sin \alpha \sin \beta \cos (\alpha-\beta)= \\ & =\frac{1}{2}(\cos 2 \alpha+\cos 2 \beta)+2 \sin \alpha \sin \beta \cos (\alpha-\beta)= \\ & =\left[\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right]= \\ & =\frac{1}{2} \cdot 2 \cos (\alpha+\beta) \cos (\alpha-\beta)+2 \sin \alpha \sin \beta \cos (\alpha-\beta)= \\ & =\cos (\alpha-\beta)(\cos (\alpha+\beta)+2 \sin \alpha \sin \beta)= \\ & =\left[\sin x \sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))\right]= \\ & =\cos (\alpha-\beta)\left(\cos (\alpha+\beta)+2 \cdot \frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))\right)= \\ & =\cos (\alpha-\beta)(\cos (\alpha+\beta)+\cos (\alpha-\beta)-\cos (\alpha+\beta))= \\ & =\cos (\alpha-\beta) \cos (\alpha-\beta)=\cos ^{2}(\alpha-\beta) . \end{aligned} $$ Answer: $\cos ^{2}(\alpha-\beta)$." 396741f841b8,"5. Let $a, b$ be real numbers, and the function $f(x)=a x+b$ satisfies: for any $x \in[0,1]$, we have $|f(x)| \leq 1$. Then the maximum value of $a b$ is $\qquad$ .",$\frac{1}{4}$,easy,"Answer $\frac{1}{4}$. Solution It is easy to know that $a=f(1)-f(0), b=f(0)$, then $$ a b=f(0) \cdot(f(1)-f(0))=-\left(f(0)-\frac{1}{2} f(1)\right)^{2}+\frac{1}{4}(f(1))^{2} \leq \frac{1}{4}(f(1))^{2} \leq \frac{1}{4} . $$ When $2 f(0)=f(1)= \pm 1$, i.e., $a=b= \pm \frac{1}{2}$, $a b=\frac{1}{4}$. Therefore, the maximum value of $a b$ is $\frac{1}{4}$." 6ec889d703b9,"What is the largest possible area of a quadrilateral with sides $1,4,7,8$ ? $ \textbf{(A)}\ 7\sqrt 2 \qquad\textbf{(B)}\ 10\sqrt 3 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 12\sqrt 3 \qquad\textbf{(E)}\ 9\sqrt 5 $",18,medium,"1. To maximize the area of a quadrilateral with given side lengths, the quadrilateral should be cyclic. This is because a cyclic quadrilateral has the maximum possible area for given side lengths. 2. We use Brahmagupta's formula to find the area of a cyclic quadrilateral. Brahmagupta's formula states that the area \( A \) of a cyclic quadrilateral with sides \( a, b, c, d \) is given by: \[ A = \sqrt{(s-a)(s-b)(s-c)(s-d)} \] where \( s \) is the semiperimeter of the quadrilateral, defined as: \[ s = \frac{a + b + c + d}{2} \] 3. For the given sides \( a = 1 \), \( b = 4 \), \( c = 7 \), and \( d = 8 \), we first calculate the semiperimeter \( s \): \[ s = \frac{1 + 4 + 7 + 8}{2} = \frac{20}{2} = 10 \] 4. Next, we substitute the values into Brahmagupta's formula: \[ A = \sqrt{(10-1)(10-4)(10-7)(10-8)} \] 5. Simplify the expression inside the square root: \[ A = \sqrt{(9)(6)(3)(2)} \] 6. Calculate the product inside the square root: \[ 9 \times 6 = 54 \] \[ 54 \times 3 = 162 \] \[ 162 \times 2 = 324 \] 7. Finally, take the square root of 324: \[ A = \sqrt{324} = 18 \] The final answer is \(\boxed{18}\)" 6d7efa68ffa1,"Mr. Ambulando is at the intersection of $5^{\text{th}}$ and $\text{A St}$, and needs to walk to the intersection of $1^{\text{st}}$ and $\text{F St}$. There's an accident at the intersection of $4^{\text{th}}$ and $\text{B St}$, which he'd like to avoid. [center][/center] Given that Mr. Ambulando wants to walk the shortest distance possible, how many different routes through downtown can he take?",56,medium,"1. **Identify the problem and constraints:** - Mr. Ambulando starts at the intersection of $5^{\text{th}}$ and $\text{A St}$. - He needs to reach the intersection of $1^{\text{st}}$ and $\text{F St}$. - He must avoid the intersection of $4^{\text{th}}$ and $\text{B St}$. - He wants to take the shortest path possible. 2. **Determine the grid dimensions and movements:** - The grid is a $4 \times 5$ rectangle (4 blocks south and 5 blocks east). - The total number of movements required is $4$ south and $5$ east. 3. **Calculate the total number of paths without any constraints:** - The total number of paths from $(5^{\text{th}}, \text{A St})$ to $(1^{\text{st}}, \text{F St})$ is given by the binomial coefficient: \[ \binom{9}{4} = \frac{9!}{4!5!} \] - Calculating this: \[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \] 4. **Account for the constraint (avoiding $4^{\text{th}}$ and $\text{B St}$):** - We need to subtract the paths that pass through $(4^{\text{th}}, \text{B St})$. - Calculate the number of paths from $(5^{\text{th}}, \text{A St})$ to $(4^{\text{th}}, \text{B St})$: \[ \binom{2}{1} = \frac{2!}{1!1!} = 2 \] - Calculate the number of paths from $(4^{\text{th}}, \text{B St})$ to $(1^{\text{st}}, \text{F St})$: \[ \binom{6}{3} = \frac{6!}{3!3!} = 20 \] - The total number of paths passing through $(4^{\text{th}}, \text{B St})$ is: \[ 2 \times 20 = 40 \] 5. **Subtract the invalid paths from the total paths:** - The number of valid paths is: \[ 126 - 40 = 86 \] 6. **Verify the solution:** - The initial solution provided states the answer is $56$. Let's re-evaluate the steps to ensure accuracy. 7. **Re-evaluate the paths avoiding the accident:** - Consider the paths that avoid $(4^{\text{th}}, \text{B St})$ by passing through either $(5^{\text{th}}, \text{C St})$ or $(3^{\text{rd}}, \text{A St})$. 8. **Calculate paths through $(5^{\text{th}}, \text{C St})$:** - Paths from $(5^{\text{th}}, \text{A St})$ to $(5^{\text{th}}, \text{C St})$: \[ \binom{2}{2} = 1 \] - Paths from $(5^{\text{th}}, \text{C St})$ to $(1^{\text{st}}, \text{F St})$: \[ \binom{6}{3} = 20 \] - Total paths through $(5^{\text{th}}, \text{C St})$: \[ 1 \times 20 = 20 \] 9. **Calculate paths through $(3^{\text{rd}}, \text{A St})$:** - Paths from $(5^{\text{th}}, \text{A St})$ to $(3^{\text{rd}}, \text{A St})$: \[ \binom{2}{2} = 1 \] - Paths from $(3^{\text{rd}}, \text{A St})$ to $(1^{\text{st}}, \text{F St})$: \[ \binom{6}{3} = 20 \] - Total paths through $(3^{\text{rd}}, \text{A St})$: \[ 1 \times 20 = 20 \] 10. **Sum the valid paths:** - Total valid paths: \[ 20 + 20 = 40 \] The final answer is $\boxed{56}$." 2223d86db39c,14. (3 points) The unit digit of $1^{8}+2^{8}+3^{8}+4^{8}+5^{8}+6^{8}+7^{8}+8^{8}+9^{8}+10^{8}+11^{8}$ is,: 4,medium,"14. (3 points) The unit digit of $1^{8}+2^{8}+3^{8}+4^{8}+5^{8}+6^{8}+7^{8}+8^{8}+9^{8}+10^{8}+11^{8}$ is $\qquad$ 4. 【Solution】Solution: Since any power of 1 is 1, the unit digit of $1^{8}$ is 1, $2^{8}=(2 \times 2 \times 2 \times 2) \times(2 \times 2 \times 2 \times 2)=16^{2}$, the unit digit is 6, $3^{8}=(3 \times 3 \times 3 \times 3) \times(3 \times 3 \times 3 \times 3)=81^{2}$, the unit digit is 1. Similarly: The unit digit of $4^{8}$ is 6, the unit digit of $5^{8}$ is 5, the unit digit of $6^{8}$ is 6, the unit digit of $7^{8}$ is 1, the unit digit of $8^{8}$ is 6, the unit digit of $9^{8}$ is 1, the unit digit of $10^{8}$ is 0, and the unit digit of $11^{8}$ is 1, $$ 1+6+1+6+5+6+1+6+1+0+1=34 \text {, } $$ Therefore, the unit digit of $1^{8}+2^{8}+3^{8}+4^{8}+5^{8}+6^{8}+7^{8}+8^{8}+9^{8}+10^{8}+11^{8}$ is 4; Hence the answer is: 4" e612cb8b2349,"An integer from 10 to 99 inclusive is randomly chosen so that each such integer is equally likely to be chosen. The probability that at least one digit of the chosen integer is a 6 is (A) $\frac{1}{5}$ (B) $\frac{1}{10}$ (C) $\frac{1}{9}$ (D) $\frac{19}{90}$ (E) $\frac{19}{89}$",(A),medium,"From 10 to 99 inclusive, there is a total of 90 integers. (Note that $90=99-10+1$.) If an integer in this range includes the digit 6, this digit is either the ones (units) digit or the tens digit. The integers in this range with a ones (units) digit of 6 are 16, 26, 36, 46, 56, 66, 76, 86, 96 . The integers in this range with a tens digit of 6 are $60,61,62,63,64,65,66,67,68,69$. In total, there are 18 such integers. (Notice that 66 is in both lists and $9+10-1=18$.) Therefore, the probability that a randomly chosen integer from 10 to 99 inclusive includes the digit 6 is $\frac{18}{90}=\frac{1}{5}$. ANSWER: (A)" a754c9121a2d,"Task B-1.6. How many ordered pairs of integers $x, y, 00, x_{1}, x_{2}, \cdots, x_{n} \in[0, a](n \geqslant 2)$ and satisfy $$x_{1} x_{2} \cdots x_{n}=\left(a-x_{1}\right)^{2}\left(a-x_{2}\right)^{2} \cdots\left(a-x_{n}\right)^{2}$$ Find the maximum value of $x_{1} x_{2} \cdots x_{n}$.",\left[\left(a-x_{1}\right)\left(a-x_{2}\right) \cdots(a-\right.$ $\left.\left.x_{n}\right)\right]^{\,medium,"16. By the inequality of arithmetic and geometric means, we have $\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{1}{2 n}}=\left[\left(a-x_{1}\right)\left(a-x_{2}\right) \cdots(a-\right.$ $\left.\left.x_{n}\right)\right]^{\frac{1}{n}} \leqslant a-\frac{x_{1}+\cdots+x_{n}}{n} \leqslant a-\left(x_{1} \cdots x_{n}\right)^{\frac{1}{n}}$. Let $y=\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{1}{2 n}} \geqslant 0$, then we have $y \leqslant a-y^{2}$, i.e., $y^{2}+y-a \leqslant 0$. Solving the inequality gives $0 \leqslant y \leqslant \frac{-1+\sqrt{4 a+1}}{2}$, hence the maximum value of $x_{1} x_{2} \cdots x_{n}$ is $\left(\frac{-1+\sqrt{4 a+1}}{2}\right)^{2 n}$." 396124d0b88c,164. Find a perfect number of the form $p^{2} q$.,. 28,easy,"164. From the equation $\left(p^{2}+p+1\right)(q+1)=2 p^{2} q$, we find $q=1+\frac{2(p+1)}{p^{2}-p-1}$, from which we obtain: 1) $p^{2}-p-1=1, p=2, q=7$ 2) $p^{2}-p-1=2$, no solutions. Moreover, the numbers $p+1$ and $p^{2}-p-1$ have no common divisors, since $\gcd \left(p+1, p^{2}-p-1\right)=\gcd[p(p+1), p^{2}-p-1]=\gcd(2 p+1, p+1)=\gcd(p, p+1)=1$. Answer. 28." 6d1df86c7efb,"5. In a regular tetrahedron $P-ABC$, $AB=1, AP=2$, a plane $\alpha$ passing through $AB$ bisects its volume. Then the cosine of the angle formed by edge $PC$ and plane $\alpha$ is $\qquad$.",See reasoning trace,medium,"5. $\frac{3 \sqrt{5}}{10}$. Let the midpoints of $AB$ and $PC$ be $K$ and $M$, respectively. Then it is easy to prove that the plane $ABM$ is the plane $\alpha$. By the median length formula, we have $$ \begin{array}{l} A M^{2}=\frac{1}{2}\left(A P^{2}+A C^{2}\right)-\frac{1}{4} P C^{2} \\ =\frac{1}{2}\left(2^{2}+1^{2}\right)-\frac{1}{4} \times 2^{2}=\frac{3}{2} \\ \Rightarrow K M=\sqrt{A M^{2}-A K^{2}}=\frac{\sqrt{5}}{2} . \end{array} $$ Since the projection of line $PC$ on plane $\alpha$ is line $MK$, and given $CM=1, KC=\frac{\sqrt{3}}{2}$, we get $$ \cos \angle K M C=\frac{K M^{2}+M C^{2}-K C^{2}}{2 K M \cdot M C}=\frac{3 \sqrt{5}}{10} . $$ Therefore, the cosine of the angle formed by the edge $PC$ and the plane $\alpha$ is $$ \text { } \frac{3 \sqrt{5}}{10} \text { . } $$" 18db36bce7d0,"Find the real number $k$ such that $a$, $b$, $c$, and $d$ are real numbers that satisfy the system of equations \begin{align*} abcd &= 2007,\\ a &= \sqrt{55 + \sqrt{k+a}},\\ b &= \sqrt{55 - \sqrt{k+b}},\\ c &= \sqrt{55 + \sqrt{k-c}},\\ d &= \sqrt{55 - \sqrt{k-d}}. \end{align*}",1018,medium,"1. **Substitute \( c' = -c \) and \( d' = -d \):** Let \( c' = -c \) and \( d' = -d \). Then, for \( x \in \{a, b\} \), we have \( x = \sqrt{55 \pm \sqrt{k+x}} \). For \( x \in \{c', d'\} \), we have \( -x = \sqrt{55 \pm \sqrt{k+x}} \). In either case, \( a, b, c', d' \) satisfy \( x^2 = 55 \pm \sqrt{k+x} \). 2. **Square both sides of the equation:** \[ x^2 = 55 \pm \sqrt{k+x} \] Squaring both sides, we get: \[ (x^2 - 55)^2 = (\sqrt{k+x})^2 \] Simplifying, we obtain: \[ (x^2 - 55)^2 = k + x \] 3. **Expand and rearrange the equation:** \[ (x^2 - 55)^2 = x^4 - 110x^2 + 3025 \] Therefore, the equation becomes: \[ x^4 - 110x^2 + 3025 = k + x \] Rearranging, we get: \[ x^4 - 110x^2 - x + (3025 - k) = 0 \] 4. **Determine the product of the roots:** The product of the roots of the polynomial \( x^4 - 110x^2 - x + (3025 - k) = 0 \) is given by the constant term divided by the leading coefficient (which is 1 in this case). Thus, the product of the roots is: \[ 3025 - k = ab c' d' = ab cd = 2007 \] 5. **Solve for \( k \):** \[ 3025 - k = 2007 \] Solving for \( k \), we get: \[ k = 3025 - 2007 = 1018 \] The final answer is \( \boxed{1018} \)." 402f0110c390,"By which natural numbers can the fraction $\frac{3 m-n}{5 n+2 m}$ be reduced, given that it is reducible and that the numbers $m$ and $n$ are coprime.",Divisible by 17,easy,"If $3 m-n$ and $5 n+2 m$ are divisible by $d$, then the numbers $17 m=5(3 m-n)+5 n+2 m$ and $17 n=3(5 n+2 m)-2(3 m-n)$ are also divisible by $d$. But $\operatorname{HOK}(17 m, 17 n)=17$. Therefore, $d=17$. This is possible, for example, when $m=1, n=3$ (or when $m=$ $6, n=1)$. ## Answer Divisible by 17." 57c3f4625324,"Marvin had a birthday on Tuesday, May 27 in the leap year $2008$. In what year will his birthday next fall on a Saturday? $\mathrm{(A)}\ 2011 \qquad \mathrm{(B)}\ 2012 \qquad \mathrm{(C)}\ 2013 \qquad \mathrm{(D)}\ 2015 \qquad \mathrm{(E)}\ 2017$",(E),medium,"$\boxed{(E)}$ $2017$ There are $365$ days in a non-leap year. There are $7$ days in a week. Since $365 = 52 \cdot 7 + 1$ (or $365\equiv 1 \pmod{ 7}$), the same date (after February) moves ""forward"" one day in the subsequent year, if that year is not a leap year. For example: $5/27/08$ Tue $5/27/09$ Wed However, a leap year has $366$ days, and $366 = 52 \cdot 7 + 2$ . So the same date (after February) moves ""forward"" two days in the subsequent year, if that year is a leap year. For example: $5/27/11$ Fri $5/27/12$ Sun You can keep counting forward to find that the first time this date falls on a Saturday is in $2017$: $5/27/13$ Mon $5/27/14$ Tue $5/27/15$ Wed $5/27/16$ Fri $5/27/17$ Sat" 8193a120adf7,"3. Let $l$ and $m$ be two skew lines, and on $L$ there are three points $A$, $B$, and $C$, with $A B = B C$. Through $A$, $B$, and $C$, perpendiculars $A D$, $B E$, and $C F$ are drawn to $m$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $A D = \sqrt{15}$, $B E = \frac{7}{2}$, and $C F = \sqrt{10}$, find the distance between $l$ and $m$.","\sqrt{15 - x^{2}} - \sqrt{10 - x^{2}}$, so $A, B, C$ must be on the same side of point $L$.",medium,"3. Let $L M$ be the line determined by lines $l$ and $m$. Suppose plane $P$ is parallel to line $l$, and through points $A, B, C$, perpendiculars $A G, B H, C K$ are drawn to plane $P$, with feet of the perpendiculars being $G, H, K$, respectively, as shown in Figure 15-12. Then points $G, H, K$ lie on a line $L^{\prime}$ parallel to $l$, and lines $C D, H E, K F$ are connected. Since $C K \perp$ plane $P, B H \perp$ plane $P, A G \perp$ plane $P, A B = B C$, it follows that $H K = G H$. Also, since $C F, B E, A D$ are all perpendicular to $m$, then $F K, E H, D G$ are also perpendicular to $m$. Thus, $K F \parallel E H \parallel G D$, and we have $2 E H = F K + D G$. Let the common perpendicular distance be $x$, then $L M = C K = B H = A G = x$, $E H = \sqrt{\frac{49}{4} - x^{2}}, F K = \sqrt{10 - x^{2}}, D G = \sqrt{15 - x^{2}}$. Therefore, we get $2 \sqrt{\frac{49}{4} - x^{2}} = \sqrt{10 - x^{2}} + \sqrt{15 - x^{2}}$, solving this yields $x = \pm \sqrt{6}$ (the negative root is discarded). Hence, the distance between $l$ and $m$ is $\sqrt{6}$. Note: When $A, B, C$ are not all on the same side of point $L$, then $2 \sqrt{\frac{49}{4} - x^{2}} = \sqrt{15 - x^{2}} - \sqrt{10 - x^{2}}$, so $A, B, C$ must be on the same side of point $L$." 88774b1bdce8,9.039. $\frac{\log _{5}\left(x^{2}+3\right)}{4 x^{2}-16 x}<0$.,$\quad x \in(0 ; 4)$,easy,"## Solution. $\log _{5}\left(x^{2}+3\right)>0$ for $x \in R$, therefore $4 x^{2}-16 x<0, x(x-4)<0$. Answer: $\quad x \in(0 ; 4)$." 536ea7138ab8,"178. Strange Multiplication. I was often asked to explain the following fact, which will undoubtedly interest many readers who were not previously aware of it. If someone correctly performs addition but cannot multiply or divide by numbers greater than 2, it turns out that they can obtain the product of any two numbers using the following strange method. For example, suppose we need to multiply 97 by 23. We form two columns of numbers: | 97 | 23 | | ---: | :---: | | 48 | (46) | | 24 | (92) | | 12 | (184) | | 6 | (368) | | 3 | 736 | | 1 | 1472 | | | 2231 | We successively divide the numbers in the first column by 2, discarding the remainder, until we get 1, and multiply the numbers in the second column by 2 the same number of times. If we cross out the products that stand opposite even numbers in the left column (we have enclosed them in parentheses) and add the remaining ones, we get the correct answer: 2231 Why?",is obtained: we simply act in the binary system,easy,"178. Let's write down in a row the remainders from dividing the numbers in the first column by 2. We get 1000011, or, if written in reverse order, 1100001. But the last number is 97 in the binary system, that is, \(1+2^{5}+2^{6}\). By adding the numbers in the second column opposite the remainders equal to 1, we get \(23 \times 1+23 \times 2^{5}+23 \times 2^{6}=2231\). Now it is clear why the correct answer is obtained: we simply act in the binary system." ec73d38ab5f8,"Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ that satisfy the equation $$ f^{a b c-a}(a b c)+f^{a b c-b}(a b c)+f^{a b c-c}(a b c)=a+b+c $$ for all $a, b, c \geq 2$. (Here $f^{k}$ means $f$ applied $k$ times.)",$f(n)=n-1$ for $n \geq 3$ with $f(1)$ and $f(2)$ arbitrary; check these work,medium,"The answer is $f(n)=n-1$ for $n \geq 3$ with $f(1)$ and $f(2)$ arbitrary; check these work. Lemma We have $f^{t^{2}-t}\left(t^{2}\right)=t$ for all $t$. Proof. We say $1 \leq k \leq 8$ is good if $f^{t^{9}-t^{k}}\left(t^{9}\right)=t^{k}$ for all $t$. First, we observe that $$ f^{t^{9}-t^{3}}\left(t^{9}\right)=t^{3} \quad \text { and } \quad f^{t^{3}-t}\left(t^{3}\right)=t \Longrightarrow f^{t^{9}-t}\left(t^{9}\right)=t . $$ so $k=1$ and $k=3$ are good. Then taking $(a, b, c)=\left(t, t^{4}, t^{4}\right),(a, b, c)=\left(t^{2}, t^{3}, t^{4}\right)$ gives that $k=4$ and $k=2$ are good, respectively. The lemma follows from this $k=1$ and $k=2$ being good. Now, letting $t=a b c$ we combine $$ \begin{aligned} f^{t-a}(a)+f^{t-b}(b)+f^{t-c}(c) & =a+b+c \\ f^{t^{2}-a b}\left(t^{2}\right)+f^{t^{2}-t}\left(t^{2}\right)+f^{t^{2}-c}\left(t^{2}\right) & =a b+t+c \\ \Longrightarrow\left[f^{t-a}(t)-a\right]+\left[f^{t-b}(t)-b\right] & =\left[f^{t-a b}(t)-a b\right] \end{aligned} $$ by subtracting and applying the lemma repeatedly. In other words, we have proven the second lemma: Lemma Let $t$ be fixed, and define $g_{t}(n)=f^{t-n}(t)-n$ for $nq>\max \{a, b\}$ be primes. Suppose $s=a^{p} b^{q}$ and $t=s^{2} ;$ then $$ p g_{t}(a)+q g_{t}(b)=g_{t}\left(a^{p} b^{q}\right)=g_{t}(s)=f^{s^{2}-s}(s)-s=0 . $$ Now $$ q \mid g_{t}(a)>-a \text { and } p \mid g_{t}(b)>-b \Longrightarrow g_{t}(a)=g_{t}(b)=0 . $$ and so we conclude $f^{t-a}(t)=a$ and $f^{t-b}(t)=b$ for $a, b \geq 2$. In particular, if $a=n$ and $b=n+1$ then we deduce $f(n+1)=n$ for all $n \geq 2$, as desired. Remark. If you let $c=(a b)^{2}$ after the first lemma, you recover the 2-variable version!" 211760a04863,"A2. We throw two dice at once multiple times and each time we add the numbers of dots on the top faces of both. What is the maximum number of different sums of dots we can get? (A) 10 (B) 11 (C) 12 (D) 18 (E) 20","1+1,3=1+2=2+1$, $4=1+3=2+2=3+1,5=1+4=2+3=3+2=4+1,6=1+5=$ $2+4=3+3=4+2=5+1,7=1+6=2+5=3+4=4+3=5+2=6+1$",easy,"A2. Possible sums of the dots on the upper faces of the dice are: $2=1+1,3=1+2=2+1$, $4=1+3=2+2=3+1,5=1+4=2+3=3+2=4+1,6=1+5=$ $2+4=3+3=4+2=5+1,7=1+6=2+5=3+4=4+3=5+2=6+1$, $8=2+6=3+5=4+4=5+3=6+2,9=3+6=4+5=5+4=6+3$, $10=4+6=5+5=6+4,11=5+6=6+5$ and $12=6+6$, a total of 11 different sums." cd71baf3b9f1,"[The product of the lengths of the segments of chords and the lengths of the segments of secants] [Properties and characteristics of an isosceles triangle.] [Thales' theorem and the theorem of proportional segments] On the extension of side $A D$ of rhombus $A B C D$ beyond point $D$, point $K$ is taken. Lines $A C$ and $B K$ intersect at point $Q$. It is known that $A K=14$ and that points $A, B$ and $Q$ lie on a circle with radius 6, the center of which lies on segment $A A$. Find $B K$.",7,medium,"Let $O$ be the center of the circle. Use the similarity of triangles $K Q O$ and $K B A$. ## Solution Let $O$ be the center of the circle, $F$ be the second intersection point of the circle with the line $A$. Since triangle $A O Q$ is isosceles $(O A=O Q=6)$, then $\angle A Q O=\angle O A Q=\angle B A Q$. Therefore, $O Q \| A B$. Hence, $K Q: K B=K O: K A=4: 7,4 / 7 K B^{2}=K Q \cdot K B=K F \cdot K A=$ 28, from which $B K=7$. ![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-37.jpg?height=432&width=783&top_left_y=1&top_left_x=643) ## Answer 7." 5f98fb29105c,"1. If the complex number $z$ satisfies $|z-1|+|z-3-2 \mathrm{i}|=2 \sqrt{2}$, then the minimum value of $|z|$ is",1,easy,"【Analysis】Answer: 1. Let $z=x+y \mathrm{i}$, then the geometric meaning of $|z-1|+|z-3-2 \mathrm{i}|=2 \sqrt{2}$ is that the sum of the distances from point $P(x, y)$ to points $A(1,0), B(3,2)$ is $2 \sqrt{2}$. Since $|A B|=2 \sqrt{2}$, point $P$ lies on the line segment $A B$, thus: $|O P| \geq 1$. That is, when $z=1$, the minimum value of $|z|$ is $1$." b1d950d6ed7b,"2. (6 points) Arrange some small circles with the same radius according to the pattern shown in the figure: the 1st figure has 6 small circles, the 2nd figure has 10 small circles, the 3rd figure has 16 small circles, the 4th figure has 24 small circles, $\cdots$, following this pattern, the 6th figure has $\qquad$ small circles.",The number of small dots in the 6th shape is 46,medium,"【Analysis】According to the problem, the number of small dots at the four corners of each shape is 4. In the 1st shape, the number of small circles is 6, which can be written as $6=1 \times(1+1)+4$; in the 2nd shape, the number of small circles is 10, which can be written as $10=2 \times(2+1)+4$; in the 3rd shape, the number of small circles is 16, which can be written as $16=3 \times(3+1)+4$; in the 4th shape, the number of small circles is 24, which can be written as $24=4 \times(4+1)+4 ; \cdots$ Therefore, for the $n$th shape, the number of small dots can be written as: $n \times(n+1)+4$, from which the problem can be solved. 【Solution】According to the analysis of the problem, the number of small circles in the $n$th shape is $n \times(n+1)+4$. When $n=6$, the number of small circles in the shape is: $6 \times 7+4=46$ (circles). Answer: The number of small dots in the 6th shape is 46. Therefore, the answer is: 46." e8cfb17fb26e,"3. In the regular triangular frustum $A^{\prime} B^{\prime} C^{\prime}-A B C$, $A^{\prime} B^{\prime} : A B = 5 : 7$, the sections $A B C^{\prime}$ and $A^{\prime} B C^{\prime}$ divide the frustum into three pyramids $C^{\prime}-A B C$, $B-A^{\prime} B^{\prime} C^{\prime}$, and $C^{\prime}-A A B A^{\prime}$. The volume ratio of $V_{1}, V_{2}, V_{3}$ is $\qquad$",See reasoning trace,easy,3. $49: 25: 35$ e6566284559b,"Given $0 \leqslant x, y, z \leqslant 1$, solve the equation: $$\frac{x}{1+y+z x}+\frac{y}{1+z+x y}+\frac{z}{1+x+y z}=\frac{3}{x+y+z} .$$",y=z=1$.,easy,"3. It is not difficult to prove: $\frac{x}{1+y+z x} \leqslant \frac{1}{x+y+z}$, $\frac{y}{1+z+x y} \leqslant \frac{1}{x+y+z}$, $\frac{z}{1+x+y z} \leqslant \frac{1}{x+y+z}$. Therefore, if the equality holds, it is easy to get $x=y=z=1$. Note: This problem can also be solved as follows: First prove $\frac{x}{1+y+z x} \leqslant \frac{x}{x+y+z}$, etc., so we can get $x+y+z \geqslant 3$, hence $x=y=z=1$." d12d8f4f29c6,"1.57. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distance from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm.",15 and 20 cm,medium,"1.57. Since $A C$ is the diameter of the circle (Fig. 1.44), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5 \quad$ (cm). In $\triangle B O F$ we have $O F=$ $=\sqrt{O B^{2}-B F^{2}}=\sqrt{12.5^{2}-12^{2}}=3.5$ (cm); hence, $M F=12.5-3.5=9$ (cm), $M K=12.5+3.5=16$ (cm). From $\triangle M B F$ and $\triangle M A K$ we find the required distances: $\quad M B=\sqrt{M F^{2}+B F^{2}}=\sqrt{9^{2}+12^{2}}=15(\mathrm{~cm})$, $M A=\sqrt{M K^{2}+K A^{2}}=\sqrt{16^{2}+12^{2}}=20(\mathrm{~cm})$. Answer: 15 and 20 cm." e31ae5cdd5bc,"A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$. Which of the following numbers cannot equal $A+P$? $\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$",B,medium,"Let the rectangle's length be $a$ and its width be $b$. Its area is $ab$ and the perimeter is $2a+2b$. Then $A + P = ab + 2a + 2b$. Factoring, we have $(a + 2)(b + 2) - 4$. The only one of the answer choices that cannot be expressed in this form is $102$, as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$, keeping $a$ and $b$ as positive integers. Our answer is then $\boxed{B}$. Note: The original problem only stated that $A$ and $P$ were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice." d694be60abec,"3. From point $M$, lying inside triangle $A B C$, perpendiculars are drawn to the sides $B C, A C, A B$, with lengths $k, l$, and $m$ respectively. Find the area of triangle $A B C$, if $\angle C A B=\alpha$ and $\angle A B C=\beta$. If the answer is not an integer, round it to the nearest integer. $$ \alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}, k=3, l=2, m=4 $$",67,medium,"Answer: 67. Solution. Denoting the sides of the triangle by $a, b, c$, using the sine theorem we get $S=\frac{k a+l b+m c}{2}=R(k \sin \alpha+l \sin \beta+m \sin \gamma)$. Since, in addition, $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, we can express $R=\frac{k \sin \alpha+l \sin \beta+m \sin \gamma}{2 \sin \alpha \sin \beta \sin \gamma}$. Therefore, $S=\frac{(k \sin \alpha+l \sin \beta+m \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}$. After substituting the numerical values of $k, l, m, \alpha, \beta, \gamma$ we get $$ S=\frac{(3 \sin \alpha+2 \sin \beta+4 \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}=\frac{(3+4 \sqrt{2}+2 \sqrt{6})^{2}}{\sqrt{3}+1} \approx 67 $$" ef98013a43fd,"24. (5 points) Yunbao and Rourou each solve a division",See reasoning trace,easy,$23$ 7cb14ccb73d2,"4. In a football tournament where each team played against each other once, teams A, B, C, D, E, and F participated. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\mathrm{E}$ could have?",See reasoning trace,medium,"4. In a match where one of the teams won, the teams together score 3 points; in a match that ended in a draw, - 2 points. Since 7 is not divisible by 3, a team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. In total, as is easy to check, 15 matches were played. Therefore, all teams together scored no more than $2 \cdot 3 + 3 \cdot 12 = 42$ points. Of these, 35 points were scored by teams A, B, V, G, and D. Therefore, team $\mathrm{E}$ scored no more than $42 - 35 = 7$ points. How it could have scored exactly 7 points is shown in the table. | | A | B | V | G | D | E | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | A | X | 3 | 3 | 1 | 0 | 0 | | B | 0 | X | 3 | 3 | 1 | 0 | | V | 0 | 0 | X | 3 | 3 | 1 | | G | 1 | 0 | 0 | X | 3 | 3 | | D | 3 | 1 | 0 | 0 | X | 3 | | E | 3 | 3 | 1 | 0 | 0 | X |" b925f8789c57,"9. (10 points) Given that 7 red balls and 5 white balls weigh 43 grams, and 5 red balls and 7 white balls weigh 47 grams, then 4 red balls and 8 white balls weigh $\qquad$ grams.",4 red balls and 8 white balls together weigh 49 g,easy,"【Solution】Solution: $(47-43) \div(7-5)$ $=4 \div 2$ $=2$ (g) $47+2=49$ (g) Answer: 4 red balls and 8 white balls together weigh 49 g. Therefore, the answer is: 49." e75d7537f8ba,"Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$. The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",076,medium,"We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram. Let $x$ be the number of men with all three risk factors. Since ""the probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$,"" we can tell that $x = \frac{1}{3}(x+14)$, since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$. Let $y$ be the number of men with no risk factors. It now follows that \[y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.\] The number of men with risk factor A is $10+2 \cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$. So the answer is $21+55=\boxed{076}$." b8092fc20852,"8. If $\arcsin x < \arccos x$, then the range of values for $x$ is","\arcsin x$ and $y=\arccos x$. From the graphs, it can be determined that $\arcsin x<\arccos x$ if an",easy,"8. $-1 \leqslant x<\frac{\sqrt{2}}{2}$. Draw the graphs of $y=\arcsin x$ and $y=\arccos x$. From the graphs, it can be determined that $\arcsin x<\arccos x$ if and only if $-1 \leqslant x<\frac{\sqrt{2}}{2}$." c034f3fea99b,"10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the shortest distance from $P$ along the side surface of the prism through the edge $C C_{1}$ to point $M$ is $3 \sqrt{2}$. Let this shortest distance intersect $C C_{1}$ at point $N$. (1) Prove: $A_{1} B / /$ plane $M N P$; (2) Find the tangent value of the dihedral angle (acute angle) formed by plane $M N P$ and plane $A B C$.",See reasoning trace,medium,"10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles. As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position of $P_{1}$, and connect $M P_{1}$. Then $M P_{1}$ is the shortest path from point $P$ along the lateral surface of the prism through $C C_{1}$ to point $M$. Let $P C=x$. Then $P_{1} C=x$. In the right triangle $\triangle M A P_{1}$, note that, $$ \begin{array}{l} (2+x)^{2}+3^{2}=18 \\ \Rightarrow x=1 . \end{array} $$ Thus, $P$ is the midpoint of $B C$. Therefore, $N C=1$. Let $A_{1} C$ intersect $M N$ at point $Q$. Then $Q$ is the midpoint of $A_{1} C$. So, $A_{1} B / / P Q$. Hence, $A_{1} B / /$ plane $M N P$. (2) As shown in Figure 3, connect $C H$ and $P P_{1}$. Then $P P_{1}$ is the intersection line of plane $M N P$ and plane $A B C$. Draw $N H \perp P P_{1}$ at point $H$. Since $C C_{1} \perp$ plane $A B C$, it follows that $C H \perp P P_{1}$. Therefore, $\angle N H C$ is the plane angle of the dihedral angle formed by plane $N M P$ and plane $A B C$. In the right triangle $\triangle P H C$, we have $$ \angle P C H=\frac{1}{2} \angle P C P_{1}=60^{\circ} \text {, } $$ Then $C H=\frac{1}{2}$. In the right triangle $\triangle N C H$, $$ \tan \angle N H C=\frac{N C}{C H}=2 . $$" 5b9f5c74eb59,"6. Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}+z_{2}\right|=20,\left|z_{1}^{2}+z_{2}^{2}\right|=16$, then the minimum value of $\left|z_{1}^{3}+z_{2}^{3}\right|$ is","20, \\ z_{1}^{2}+z_{2}^{2}=16\end{array} \Rightarrow z_{1} z_{2}=192\right.$, then $z_{1}, z_{2}$ ar",medium,"$$ \begin{array}{l} \left|z_{1}^{3}+z_{2}^{3}\right|=\left|\left(z_{1}+z_{2}\right)\left(z_{1}^{2}+z_{2}^{2}-z_{1} z_{2}\right)\right|=10\left|3\left(z_{1}^{2}+z_{2}^{2}\right)-\left(z_{1}+z_{2}\right)^{2}\right| \\ \geqslant 10\left(\left|z_{1}+z_{2}\right|^{2}-3\left|z_{1}^{2}+z_{2}^{2}\right|\right)=10(400-3 \cdot 16)=3520, \end{array} $$ The equality holds when the vectors corresponding to $\left(z_{1}+z_{2}\right)^{2}$ and $z_{1}^{2}+z_{2}^{2}$ are in the same direction. For example, if $\left\{\begin{array}{l}z_{1}+z_{2}=20, \\ z_{1}^{2}+z_{2}^{2}=16\end{array} \Rightarrow z_{1} z_{2}=192\right.$, then $z_{1}, z_{2}$ are the roots of the equation $x^{2}-20 x+192=0$, solving which gives $z_{1}=10+2 \sqrt{23} \mathrm{i}, z_{2}=10-2 \sqrt{23} \mathrm{i}$." c55941e76f1c,"10. The prime factor decomposition of 2021 is $43 \times 47$. What is the value of $53 \times 57$ ? A 2221 B 2521 C 2921 D 3021 E 3031",(43+10)(47+10)=43 \times 47+43 \times 10+10 \times 47+10 \times 10$ $=2021+(43+47+10) \times 10=2021,easy,10. D Note that $53 \times 57=(43+10)(47+10)=43 \times 47+43 \times 10+10 \times 47+10 \times 10$ $=2021+(43+47+10) \times 10=2021+100 \times 10=2021+1000=3021$. 2024b3f89753,"We received a cube with the length of its edge expressed in centimeters as a whole number. We painted all its sides red and then cut it into smaller cubes with an edge of $1 \mathrm{~cm}$. - Lukáš claims that the number of smaller cubes with two painted sides is ten times greater than those with three painted sides. - Martina says that the number of smaller cubes with two painted sides is fifteen times greater than those with three painted sides. However, only one of them is right — who is it? And what was the length of the edge of the original cube? (L. Simünek)","10$ small cubes with two colored faces. However, the edge of the original cube also included two sma",medium,"From the formulation of the task, it follows that the edge of the original cube measured at least $2 \mathrm{~cm}$. A small cube has three colored faces if its vertex was originally a vertex of the large cube. Therefore, there are as many such small cubes as there are vertices of the cube, which is 8. A small cube has exactly two colored faces if one of its edges originally formed an edge of the large cube and none of its vertices were originally vertices of the large cube. Since the large cube had 12 edges, the number of small cubes with exactly two colored faces is a multiple of twelve. According to Lukáš, there are $10 \cdot 8=80$ such small cubes, which is not possible because 80 is not a multiple of twelve. Martina is right, who claims that there are $15 \cdot 8=120$ such small cubes. On each edge of the large cube, we obtained 120 : $12=10$ small cubes with two colored faces. However, the edge of the original cube also included two small cubes with three colored faces, so the length of the edge corresponded to twelve small cubes. The edge of the original cube measured $12 \mathrm{~cm}$." 1849b5d54a96,"Sabrina has a fair tetrahedral die whose faces are numbered 1, 2, 3, and 4, respectively. She creates a sequence by rolling the die and recording the number on its bottom face. However, she discards (without recording) any roll such that appending its number to the sequence would result in two consecutive terms that sum to 5. Sabrina stops the moment that all four numbers appear in the sequence. Find the expected (average) number of terms in Sabrina's sequence.",10,medium,"1. Define \( e_i \) as the expected number of moves to get all four numbers in the sequence given that we have currently seen \( i \) distinct values. We need to find \( e_0 \). 2. We start with the following equations: \[ e_0 = e_1 + 1 \] \[ e_1 = \frac{1}{3}e_1 + \frac{2}{3}e_2 + 1 \] \[ e_2 = \frac{2}{3}e_2 + \frac{1}{3}e_3 + 1 \] 3. To solve for \( e_3 \), we consider the two sub-states: - The last roll was a 1. - The last roll was either 2 or 3. Let \( a \) and \( b \) represent the expected number of moves needed to finish for these sub-states, respectively. We have: \[ a = \frac{1}{3}a + \frac{2}{3}b + 1 \] \[ b = \frac{1}{3}a + \frac{1}{3}b + \frac{1}{3}e_4 + 1 \] 4. Since \( e_4 = 0 \) (as all four numbers have been seen), we can simplify the equations: \[ a = \frac{1}{3}a + \frac{2}{3}b + 1 \implies a - \frac{1}{3}a = \frac{2}{3}b + 1 \implies \frac{2}{3}a = \frac{2}{3}b + 1 \implies a = b + \frac{3}{2} \] \[ b = \frac{1}{3}a + \frac{1}{3}b + 1 \implies 3b = a + b + 3 \implies 2b = a + 3 \implies a = 2b - 3 \] 5. Solving the system of equations \( a = b + \frac{3}{2} \) and \( a = 2b - 3 \): \[ b + \frac{3}{2} = 2b - 3 \implies \frac{3}{2} + 3 = b \implies b = \frac{9}{2} \] \[ a = b + \frac{3}{2} = \frac{9}{2} + \frac{3}{2} = 6 \] 6. Since we must be in state \( b \) when we get to \( e_3 \), we have \( e_3 = b = \frac{9}{2} \). 7. Plugging \( e_3 = \frac{9}{2} \) back into the equations for \( e_2 \) and \( e_1 \): \[ e_2 = \frac{2}{3}e_2 + \frac{1}{3} \cdot \frac{9}{2} + 1 \implies 3e_2 = 2e_2 + \frac{9}{2} + 3 \implies e_2 = \frac{15}{2} \] \[ e_1 = \frac{1}{3}e_1 + \frac{2}{3} \cdot \frac{15}{2} + 1 \implies 3e_1 = e_1 + 15 + 3 \implies 2e_1 = 18 \implies e_1 = 9 \] 8. Finally, substituting \( e_1 = 9 \) into the equation for \( e_0 \): \[ e_0 = e_1 + 1 = 9 + 1 = 10 \] The final answer is \( \boxed{10} \)." b759686dde9c,"8 The line $\sqrt{3} x-y-\sqrt{3}=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$ ($A$ is above the x-axis), and intersects the x-axis at point $F$. If $\overrightarrow{O F}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}$, then $\mu^{2}-$ $\lambda^{2}=$ $\qquad$ .","0$ passes through the focus $(1, 0)$ of the parabola $y^{2}=4 x$. Substituting $y=\sqrt{3} x-\sqrt{3",medium,"$8 \frac{1}{2}$ Hint: The line $\sqrt{3} x-y-\sqrt{3}=0$ passes through the focus $(1, 0)$ of the parabola $y^{2}=4 x$. Substituting $y=\sqrt{3} x-\sqrt{3}$ into $y^{2}=4 x$, we get $3 x^{2}-10 x+3=0$. Therefore, $A(3, 2 \sqrt{3})$ and $B\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$. $\overrightarrow{O F}=(1,0), \overrightarrow{O A}=(3,2 \sqrt{3}), \overrightarrow{O B}=\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$. Since $\overrightarrow{O F}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B},(1,0)=\lambda(3,2 \sqrt{3})+$ $\mu\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$, solving gives $\lambda=\frac{1}{4}, \mu=\frac{3}{4}$, so $\mu^{2}-\lambda^{2}=\frac{1}{2}$." 02ad172bf7be,"$9 \cdot 6$ For $a>b>c>0, \quad m>n>0(m, n$ are integers), which of the following relations is correct? (A) $a^{m} b^{n}>b^{m} c^{n}>c^{n} a^{m}$. (B) $a^{m} b^{n}>c^{n} a^{m}>b^{n} c^{m}$. (C) $a^{m} c^{n}>a^{m} b^{n}>b^{n} c^{m}$. (D) $b^{n} c^{m}>c^{n} a^{m}>a^{m} b^{n}$. (E) $a^{m} c^{n}>b^{n} c^{m}>a^{m} b^{n}$. (China Beijing Junior High School Mathematics Competition, 1983)",$(B)$,easy,"[Solution] Given $a>b>c>0, m>n>0$, we have $$ a^{m} c^{n}b>1=c$, then $$ b^{m} c^{n}=b^{m} \cdot 1^{n}=b^{m}b>c>0, \text{ when } m>n>0$, we have $$ a^{m} b^{n}>a^{m} c^{n}>b^{m} c^{n}>b^{n} c^{m} \text {. } $$ Therefore, the answer is $(B)$." 0dcab06a65a7,74. Find the minimum value of the algebraic expression $\sqrt{x^{2}-8 x+41}+\sqrt{x^{2}-4 x+13}$.,$\quad 2 \sqrt{17}$,easy,Answer: $\quad 2 \sqrt{17}$ d21bc41905f3,"21. (1999 National Team Selection Test) For non-negative real numbers $x_{1}, x_{2}$, $\cdots, x_{n}$ satisfying the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.",See reasoning trace,medium,"21. Use the adjustment method to explore the maximum value of the sum in the problem. First, for $x, y>0$, we compare $(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}$ with $x^{4}-x^{5}+y^{4}-y^{5}$, $$ \begin{aligned} & (x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\ = & x y\left(4 x^{2}+6 x y+4 y^{2}\right)-x y\left(5 x^{3}+10 x^{2} y+10 x y^{2}+5 y^{3}\right) \\ \geqslant & \frac{7}{2} x y\left(x^{2}+2 x y+y^{2}\right)-5 x y\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right) \\ = & \frac{1}{2} x y(x+y)^{2}[7-10(x+y)] . \end{aligned} $$ As long as $x, y>0, x+y>0$, then, among them, there must be two whose sum $\leqslant \frac{2}{3}0, x+y=1$. For this case, $x^{4}-x^{3}+y^{4}-y^{6}=x^{4}(1-x)+y^{4}(1-y)=x y\left(x^{3}+y^{3}\right)=x y\left[(x+y)^{3}-3 . x y(x+\right.$ $y)]=x y(1-3 x y)=\frac{1}{3}(3 x y)(1-3 x y)$. When $3 x y=\frac{1}{2}$, the above expression reaches its maximum value, $x^{4}-x^{5}+y^{4}-y^{5}=\frac{1}{6}\left(1-\frac{1}{2}\right)=\frac{1}{12}$. This is the maximum value sought in the problem, and the $x_{1}, x_{2}, \cdots, x_{n}$ that can achieve this maximum value are such that only two are not equal to 0, represented by $x$ and $y$, then $x, y>0, x+y=1, x y=\frac{1}{6}$. Solving the quadratic equation $\lambda^{2}-\lambda+\frac{1}{6}=0$ gives $x=\frac{3+\sqrt{3}}{6}, y=\frac{3-\sqrt{3}}{6}$. Of course, it can also be $x=\frac{3-\sqrt{3}}{6}, y=\frac{3+\sqrt{3}}{6}$. Verification shows that if $x_{1}, x_{2}, \cdots, x_{n}$ have only these two non-zero numbers, then the sum in the problem indeed reaches the maximum value $\frac{1}{12}$." 2be6c2d2ef31,Determine all positive integers $n$ such that $n$ divides $\phi(n)^{d(n)}+1$ but $d(n)^5$ does not divide $n^{\phi(n)}-1$.,n = 2,medium,"1. **Initial Assumptions and Simplifications:** - We need to find all positive integers \( n \) such that \( n \) divides \( \phi(n)^{d(n)} + 1 \) but \( d(n)^5 \) does not divide \( n^{\phi(n)} - 1 \). - The only solution is \( n = 2 \), which clearly works since \( \phi(2) = 1 \) and \( d(2) = 2 \). Thus, \( 2 \mid 1^2 + 1 = 2 \) and \( 2^5 = 32 \) does not divide \( 2^1 - 1 = 1 \). 2. **Assume \( n > 2 \):** - If \( n \) is not squarefree, then \( \gcd(n, \phi(n)) > 1 \), which contradicts the first condition. Therefore, \( n \) must be squarefree. - Since \( \phi(n) \) is even, \( n \) must be odd. Let \( n = p_1 p_2 \cdots p_k \) for distinct odd primes \( p_1, \ldots, p_k \). 3. **Divisor Function and Euler's Totient Function:** - We have \( d(n) = 2^k \) because \( n \) is a product of \( k \) distinct primes. - Therefore, \( n \mid \phi(n)^{2^k} + 1 \) and \( \nu_2(n^{\phi(n)} - 1) < 5k \). 4. **Valuation of \( \phi(n) \):** - Claim: \( \nu_2(\phi(n)) \ge k^2 + k \). - Proof: For each \( p_i \), \( \phi(n)^{2^{k+1}} \equiv 1 \pmod{p_i} \) and \( \phi(n)^{2^k} \equiv -1 \pmod{p_i} \). Thus, \( \mathrm{ord}_{p_i}(\phi(n)) = 2^{k+1} \), implying \( \nu_2(p_i - 1) \ge k + 1 \). - Hence, \( \nu_2(\phi(n)) = \nu_2\left(\prod_{i=1}^k (p_i - 1)\right) \ge k(k + 1) = k^2 + k \). 5. **Valuation of \( n^{\phi(n)} - 1 \):** - By the Lifting The Exponent (LTE) lemma: \[ \nu_2(n^{\phi(n)} - 1) = \nu_2(n - 1) + \nu_2(n + 1) + \nu_2(\phi(n)) - 1 \] - Since \( n = p_1 p_2 \cdots p_k \) and each \( p_i \equiv 1 \pmod{2^{k+1}} \), we have: \[ \nu_2(n - 1) \ge k + 1, \quad \nu_2(n + 1) \ge k + 1 \] - Therefore: \[ \nu_2(n^{\phi(n)} - 1) \ge (k + 1) + (k + 1) + (k^2 + k) - 1 = k^2 + 2k + 1 \] - This value must be less than \( 5k \), so \( k \le 2 \). 6. **Case Analysis:** - If \( k = 1 \), then \( n = p \) for some odd prime \( p \). We need \( p \mid \phi(p)^{2} + 1 \), which simplifies to \( p \mid (p-1)^2 + 1 \). This is impossible for \( p > 2 \). - If \( k = 2 \), let \( n = pq \) for distinct odd primes \( p \) and \( q \). We have \( p \equiv q \equiv 1 \pmod{8} \). 7. **Further Analysis for \( k = 2 \):** - Claim: \( p, q \equiv 9 \pmod{16} \). - Proof: Suppose not. Without loss of generality, assume \( p \equiv 1 \pmod{16} \). Then: \[ \nu_2(n^{\phi(n)} - 1) = \nu_2(n^{(p-1)(q-1)} - 1) = \nu_2(n - 1) + \nu_2(n + 1) + \nu_2((p-1)(q-1)) - 1 \] - Since \( p \equiv 1 \pmod{16} \), \( \nu_2(p - 1) \ge 4 \). Similarly, \( \nu_2(q - 1) \ge 4 \). Thus: \[ \nu_2(n - 1) \ge 3, \quad \nu_2(n + 1) \ge 1, \quad \nu_2((p-1)(q-1)) \ge 7 \] - Therefore: \[ \nu_2(n^{\phi(n)} - 1) \ge 3 + 1 + 7 - 1 = 10 \] - This is impossible since it contradicts \( \nu_2(n^{\phi(n)} - 1) < 5k = 10 \). 8. **Conclusion:** - There are no solutions for \( n > 2 \). The final answer is \( \boxed{ n = 2 } \)." 1f5fcf5e6e40,"Let the unit area rectangle $ABCD$ have the incenter of the triangle determined by vertices $A, B, C$ as $O$. Consider the rectangle with $O$ and $D$ as opposite vertices, and the other two vertices on the perimeter of the original rectangle. What is the area of this rectangle?",See reasoning trace,medium,"We will show that the rectangle with vertex $O$ is cut by the diagonal $AC$ into a triangle - which is shaded in the diagram - whose area is equal to the part of the $\triangle ACD$ that protrudes from the rectangle. (In the diagram, the two dotted right-angled triangles are shown.) This implies that the areas of the two figures, the rectangle with vertex $O$ and the $\triangle ACD$, are equal, so the area of the rectangle in question is $1/2$. # $1986-12-447-1$. eps Indeed, the radius of the inscribed circle of $\triangle ABC$ perpendicular to the diagonal $AC$ cuts the shaded triangle into two right-angled triangles, each with a leg of length $r$. Due to the equal vertex angles, one of these is congruent to the triangle at vertex $C$, and the other is congruent to the triangle at vertex $A$, both with a leg of length $r$." e8cd3f509acf,"Let $a_1=2021$ and for $n \ge 1$ let $a_{n+1}=\sqrt{4+a_n}$. Then $a_5$ can be written as $$\sqrt{\frac{m+\sqrt{n}}{2}}+\sqrt{\frac{m-\sqrt{n}}{2}},$$ where $m$ and $n$ are positive integers. Find $10m+n$.",45,medium,"1. We start with the given sequence \(a_1 = 2021\) and the recursive formula \(a_{n+1} = \sqrt{4 + a_n}\). 2. Calculate the first few terms of the sequence: \[ a_2 = \sqrt{4 + a_1} = \sqrt{4 + 2021} = \sqrt{2025} = 45 \] \[ a_3 = \sqrt{4 + a_2} = \sqrt{4 + 45} = \sqrt{49} = 7 \] \[ a_4 = \sqrt{4 + a_3} = \sqrt{4 + 7} = \sqrt{11} \] \[ a_5 = \sqrt{4 + a_4} = \sqrt{4 + \sqrt{11}} \] 3. We are given that \(a_5\) can be written in the form: \[ a_5 = \sqrt{\frac{m + \sqrt{n}}{2}} + \sqrt{\frac{m - \sqrt{n}}{2}} \] 4. Let us denote: \[ x = \sqrt{\frac{m + \sqrt{n}}{2}} \quad \text{and} \quad y = \sqrt{\frac{m - \sqrt{n}}{2}} \] Then: \[ a_5 = x + y \] Squaring both sides: \[ a_5^2 = (x + y)^2 = x^2 + y^2 + 2xy \] 5. We know: \[ x^2 = \frac{m + \sqrt{n}}{2} \quad \text{and} \quad y^2 = \frac{m - \sqrt{n}}{2} \] Adding these: \[ x^2 + y^2 = \frac{m + \sqrt{n}}{2} + \frac{m - \sqrt{n}}{2} = \frac{2m}{2} = m \] 6. Also: \[ 2xy = 2 \sqrt{\left(\frac{m + \sqrt{n}}{2}\right) \left(\frac{m - \sqrt{n}}{2}\right)} = 2 \sqrt{\frac{(m + \sqrt{n})(m - \sqrt{n})}{4}} = 2 \sqrt{\frac{m^2 - n}{4}} = \sqrt{m^2 - n} \] 7. Therefore: \[ a_5^2 = m + \sqrt{m^2 - n} \] 8. We know \(a_5 = \sqrt{4 + \sqrt{11}}\), so: \[ a_5^2 = 4 + \sqrt{11} \] Thus: \[ 4 + \sqrt{11} = m + \sqrt{m^2 - n} \] 9. By comparing both sides, we can deduce: \[ m = 4 \quad \text{and} \quad \sqrt{m^2 - n} = \sqrt{11} \] Squaring the second equation: \[ m^2 - n = 11 \implies 4^2 - n = 11 \implies 16 - n = 11 \implies n = 5 \] 10. Finally, we calculate \(10m + n\): \[ 10m + n = 10 \cdot 4 + 5 = 40 + 5 = 45 \] The final answer is \(\boxed{45}\)." 2a3ebfc75e98,"8. On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5(x-(1 / 2))^{2}$ pass, the angle between which is $45^{\circ}$.",$M(0 ; 0)$ or $M(0 ;-3)$,medium,"# Solution: On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5 \cdot\left(x-\frac{1}{2}\right)^{2}$ pass, with the angle between them being $45^{\circ}$. Solution (without using derivatives). $y=0.5 \cdot(x-1 / 2)^{2}, M\left(0 ; y_{0}\right) \cdot$ The equation $0.5 \cdot(x-1 / 2)^{2}=y_{0}+k x$, or $x^{2}-2\left(k+\frac{1}{2}\right) x-2 y_{0}+\frac{1}{4}=0$, has a unique solution if $\frac{D}{4}=k^{2}+k+2 y_{0}=0$. The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=-1$ (1), $k_{1} \cdot k_{2}=2 y_{0}$ (2). From the condition $\alpha_{2}-\alpha_{1}=45^{\circ}, \operatorname{tg}\left(\alpha_{2}-\alpha_{1}\right)=1$ it follows that $\frac{\operatorname{tg} \alpha_{2}-\operatorname{tg} \alpha_{1}}{1+\operatorname{tg} \alpha_{2} \cdot \operatorname{tg} \alpha_{1}}=1$, or $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=1$. Hence, $k_{2}-k_{1}=1+k_{2} k_{1}, k_{2}-k_{1}=1+2 y_{0}$, (3). Solving the system $\left\{\begin{array}{c}k_{2}-k_{1}=1+2 y_{0}, \\ k_{1}+k_{2}=-1,\end{array}\right.$ we find $\left\{\begin{array}{c}k_{2}=y_{0}, \\ k_{1}=-y_{0}-1\end{array}\right.$ and $k_{1} k_{2}=-y_{0}^{2}-y_{0}$. Considering (2), we get the condition $2 y_{0}=-y_{0}^{2}-y_{0}$, or $y_{0}^{2}+3 y_{0}=0$. There are two possible solutions. 1) $y_{0}=0, M(0 ; 0), k_{1}=-1, k_{2}=0$. Equations of the tangents: $y_{1}=-x ; y_{2}=0$. 2) $y_{0}=-3, M(0 ;-3), k_{1}=2, k_{2}=-3$. Equations of the tangents: $y_{1}=2 x-3 ; y_{2}=-3 x-3$. Answer: $M(0 ; 0)$ or $M(0 ;-3)$." 95a9d66fb11c,"Example 2 Let $x \in \mathbf{R}$, try to find the minimum value of the function $f(x)=\left(x^{2}+4 x+5\right)\left(x^{2}+4 x+2\right)+2 x^{2}+8 x+1$.","-\frac{9}{2}$, $g(u)$ reaches its minimum value, and when $u>-\frac{9}{2}$, $g(u)$ is a strictly inc",easy,"Let $u=x^{2}+4 x$, then $$ f(x)=g(u)=(u+5)(u+2)+2 u+1=u^{2}+9 u+11 \text {. } $$ By the properties of quadratic functions, when $u=-\frac{9}{2}$, $g(u)$ reaches its minimum value, and when $u>-\frac{9}{2}$, $g(u)$ is a strictly increasing function of $u$. Since $u=x^{2}+4 x=(x+2)^{2}-4$, we know that $u \geqslant-4$, and when $x=-2$, $u=-4$. Therefore, when $x=-2$, $f(x)$ reaches its minimum value, which is -9." c5147848ac2a,"1. (16 points) The dividend is five times larger than the divisor, and the divisor is four times larger than the quotient. Find the dividend. #",See reasoning trace,easy,"# Answer: 100 Solution. From the condition of the problem, it follows that the quotient is 5. Then the divisor is 20, and the dividend is 100." ba98074f773a,"Matilda has a summer job delivering newspapers. She earns $\$ 6.00$ an hour plus $\$ 0.25$ per newspaper delivered. Matilda delivers 30 newspapers per hour. How much money will she earn during a 3 hour shift? (A) $\$ 40.50$ (B) $\$ 18.75$ (C) $\$ 13.50$ (D) $\$ 25.50$ (E) $\$ 28.50$",(A),easy,"During a 3 hour shift, Matilda will deliver $3 \times 30=90$ newspapers. Therefore, she earns a total of $3 \times \$ 6.00+90 \times \$ 0.25=\$ 18.00+\$ 22.50=\$ 40.50$ during her 3 hour shift. ANSWER: (A)" 9101143d1bbf,(7) The sum of all positive integers $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ is $\qquad$,See reasoning trace,medium,"(7) 33 Hint: By the convexity of the sine function, when $x \in\left(0, \frac{\pi}{6}\right)$, $\frac{3}{\pi} x$ $\sin x\frac{3}{\pi} \times \frac{\pi}{12}=\frac{1}{4}, \\ \sin \frac{\pi}{10}\frac{3}{\pi} \times \frac{\pi}{9}=\frac{1}{3} . \end{array} $ Therefore, $\sin \frac{\pi}{13}<\frac{1}{4}<\sin \frac{\pi}{12}<\sin \frac{\pi}{11}<\sin \frac{\pi}{10}<\frac{1}{3}<\sin \frac{\pi}{9}$. Hence, the positive integer values of $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ are 10, 11, and 12, and their sum is 33." d3e4e8116cd1,"3. Given the circle $C: x^{2}+y^{2}=24$, the line $l: \frac{x}{12}+\frac{y}{8}=1$, and point $P$ on $l$, the ray $O P$ intersects the circle at point $R$. Point $Q$ is on $O P$ and satisfies: $|O Q| \cdot|O P|=|O R|^{2}$. When point $P$ moves along $l$, find the equation of the trajectory of point $Q$ and describe what kind of curve the trajectory is.","\frac{x}{12}+\frac{y}{8}$, we rearrange to get $(x-1)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{4",easy,"3. From $\frac{x^{2}}{24}+\frac{y^{2}}{24}=\frac{x}{12}+\frac{y}{8}$, we rearrange to get $(x-1)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{4}$, which represents a circle with center $\left(1, \frac{3}{2}\right)$ and radius $\frac{\sqrt{13}}{2}$." ca127d6de693,"Find all real numbers $ a,b,c,d$ such that \[ \left\{\begin{array}{cc}a \plus{} b \plus{} c \plus{} d \equal{} 20, \\ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150. \end{array} \right.\]","(5, 5, 5, 5)",medium,"1. We start with the given system of equations: \[ \begin{cases} a + b + c + d = 20 \\ ab + ac + ad + bc + bd + cd = 150 \end{cases} \] 2. We know from algebraic identities that: \[ (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) \] 3. Substituting the given values into the identity: \[ 20^2 = a^2 + b^2 + c^2 + d^2 + 2 \cdot 150 \] 4. Simplifying the equation: \[ 400 = a^2 + b^2 + c^2 + d^2 + 300 \] 5. Solving for \(a^2 + b^2 + c^2 + d^2\): \[ a^2 + b^2 + c^2 + d^2 = 400 - 300 = 100 \] 6. Using the Cauchy-Schwarz inequality: \[ (a^2 + b^2 + c^2 + d^2)(1^2 + 1^2 + 1^2 + 1^2) \geq (a + b + c + d)^2 \] 7. Substituting the known values: \[ 4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2 \] 8. Simplifying with the given sums: \[ 4 \cdot 100 \geq 20^2 \] 9. This simplifies to: \[ 400 \geq 400 \] 10. Since equality holds in the Cauchy-Schwarz inequality, it implies that \(a = b = c = d\). 11. Given \(a + b + c + d = 20\) and \(a = b = c = d\), we have: \[ 4a = 20 \implies a = 5 \] 12. Therefore, \(a = b = c = d = 5\). 13. Verifying the second condition: \[ ab + ac + ad + bc + bd + cd = 6 \cdot 5^2 = 6 \cdot 25 = 150 \] 14. Both conditions are satisfied with \(a = b = c = d = 5\). The final answer is \( \boxed{ (5, 5, 5, 5) } \)" 966dcc64cedc,"4. Given $a_{0}=-1, a_{1}=1, a_{n}=2 a_{n-1}+3 a_{n-2}+3^{n}(n \geqslant 3)$, find $a_{n}$.",\frac{1}{16}\left[(4 n-3) \cdot 3^{n+1}-7(-1)^{n}\right]$ is the solution.,medium,"4. Let $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots$ $$ \begin{array}{l} -2 x \cdot f(x)=-2 a_{0} x-2 a_{1} x^{2}-\cdots-2 a_{n-1} x^{n}-\cdots, \\ -3 x^{2} \cdot f(x)=-3 a_{0} x^{2}-\cdots-3 a_{n-2} x^{n}-\cdots, \\ -\frac{1}{1-3 x}=-1-3 x-3^{2} x^{2}-\cdots-3^{n} x^{n}-\cdots . \end{array} $$ Adding the above four equations and using the given conditions, we get $\left(1-2 x-3 x^{2}\right) \cdot f(x)-\frac{1}{1-3 x}=-2$. Thus, $f(x)=\frac{6 x-1}{(1+x)(1-3 x)^{2}}=\frac{A}{1+x}+\frac{B}{(1-3 x)^{2}}+\frac{C}{1-3 x}$, then $A=\left.f(x)(1+x)\right|_{x=-1}=-\frac{7}{16}, B=\left.f(x)(1-3 x)\right|_{x=\frac{1}{3}}=\frac{3}{4}$, $0=\lim _{x \rightarrow \infty} x f(x)=\lim _{x \rightarrow \infty}\left(\frac{A x}{1+x}+\frac{B x}{(1-3 x)^{2}}+\frac{C x}{1-3 x}\right)=A-\frac{1}{3} C$. Therefore, $C=3 A=-\frac{21}{16}$. $$ \text { Hence } \begin{aligned} f(x) & =-\frac{7}{16(1+x)}+\frac{3}{4(1-3 x)^{2}}-\frac{21}{16(1-3 x)} \\ & =-\frac{7}{16} \sum_{k=0}^{\infty}(-1)^{k} \cdot x^{k}+\frac{3}{4} \sum_{k=0}^{\infty} C_{k+1}^{1} 3^{k} x^{k}-\frac{21}{16} \sum_{k=0}^{\infty} 3^{k} x^{k} \\ & =\sum_{k=0}^{\infty}\left[\frac{(4 k .-3) \cdot 3^{k+1}-7(-1)^{k}}{16}\right] x^{k} . \end{aligned} $$ Therefore, $a_{n}=\frac{1}{16}\left[(4 n-3) \cdot 3^{n+1}-7(-1)^{n}\right]$ is the solution." ced55ad62ebc,"11.1. At first, there were natural numbers, from 1 to 2021. And they were all white. Then the underachiever Borya painted every third number blue. Then the top student Vova came and painted every fifth number red. How many numbers remained white? (7 points) #",See reasoning trace,easy,"# Solution Borya repainted the numbers divisible by 3, a total of [2021:3]=673 numbers. Vova repainted [2021:5]=404 numbers. 673+404=1077. However, numbers divisible by 15 were counted twice. [2021:15]=134. Therefore, the number of white numbers remaining is 2021-1077+134=1078. Answer: 1078 white numbers | criteria | points | | :--- | :---: | | 1. correct solution | 7 | | 2. Not accounted for numbers divisible by 15 being counted twice. | 4 | | 3. For each arithmetic error | 1 |" 43e09d44ce64,"11. Given that the internal angles $A, B, C$ of $\triangle ABC$ have opposite sides $a, b, c$ respectively, and $\sqrt{3} b \cos \frac{A+B}{2}=c \sin B$. (1) Find the size of $\angle C$; (2) If $a+b=\sqrt{3} c$, find $\sin A$.",\frac{1}{2}$ or $\sin A=1$.,medium,"(1) $\sqrt{3} \sin B \cos \frac{A+B}{2}=\sin C \sin B \Rightarrow \sqrt{3} \sin \frac{C}{2}=\sin C \Rightarrow \cos \frac{C}{2}=\frac{\sqrt{3}}{2}$, so $\frac{C}{2}=30^{\circ} \Rightarrow C=60^{\circ}$. $$ \begin{array}{l} \text { (2) } \sin A+\sin B=\sqrt{3} \sin C=\frac{3}{2} \Rightarrow \sin A+\sin \left(120^{\circ}-A\right) \\ =\sin A+\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A=\frac{3}{2} \sin A+\frac{\sqrt{3}}{2} \cos A=\frac{3}{2} \\ \Rightarrow \frac{\sqrt{3}}{2} \sin A+\frac{1}{2} \cos A=\sin \left(A+30^{\circ}\right)=\frac{\sqrt{3}}{2} \Rightarrow A=30^{\circ} \text { or } A=90^{\circ} . \end{array} $$ So $\sin A=\frac{1}{2}$ or $\sin A=1$." 10984f03d7b5,"There is a regular $17$-gon $\mathcal{P}$ and its circumcircle $\mathcal{Y}$ on the plane. The vertices of $\mathcal{P}$ are coloured in such a way that $A,B \in \mathcal{P}$ are of diff erent colour, if the shorter arc connecting $A$ and $B$ on $\mathcal{Y}$ has $2^k+1$ vertices, for some $k \in \mathbb{N},$ including $A$ and $B.$ What is the least number of colours which suffices?",4,medium,"1. **Identify the relevant values of \(2^k + 1\):** Since the minor arc \(AB\) has at most 9 vertices on it, the relevant values of \(2^k + 1\) are \(3, 5,\) and \(9\). This is because: \[ 2^1 + 1 = 3, \quad 2^2 + 1 = 5, \quad 2^3 + 1 = 9 \] 2. **Enumerate the vertices:** Enumerate the vertices of the 17-gon as \(V_0, V_1, \ldots, V_{16}\). 3. **Determine the conditions for different colors:** We have that \(V_i\) and \(V_j\) are of different colors if \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\). This is derived from the relevant values of \(2^k + 1\). 4. **Establish a lower bound for the number of colors:** Since \(V_0, V_2, V_4\) must have pairwise distinct colors, we have \(n \geq 3\). 5. **Check if 4 colors suffice:** If there are 4 colors \(A, B, C,\) and \(D\), we can try to color the vertices such that the condition is satisfied. One possible coloring is: \[ \{ V_0, V_1, \cdots, V_{16} \} = \{ A, D, B, B, C, C, D, D, B, B, C, A, D, D, B, B, C \} \] This coloring ensures that no two vertices \(V_i\) and \(V_j\) with \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\) have the same color. Therefore, \(n \leq 4\). 6. **Prove that 3 colors are insufficient:** Assume there exists a solution with \(n = 3\). Without loss of generality, let \(V_0 = A\) and \(V_2 = B\). Then: \[ V_4 = C, \quad V_6 = A, \quad V_8 = B, \quad V_{10} = C, \quad V_{12} = A, \quad V_{14} = B, \quad V_{16} = C \] Continuing this pattern, we find that \(V_{13} = A\), which contradicts the assumption that \(n = 3\) is a solution because \(V_0\) and \(V_{13}\) would be the same color, violating the condition for different colors. Therefore, the least number of colors required is \(4\). The final answer is \(\boxed{4}\)." 804ec3c009e9,"10. (20 points) Find the number of all positive integer solutions $(x, y, z)$ to the equation $\arctan \frac{1}{x}+\arctan \frac{1}{y}+\arctan \frac{1}{z}=\frac{\pi}{4}$.",See reasoning trace,medium,"10. By symmetry, let $x \leqslant y \leqslant z$. Taking the tangent of both sides of the given equation, we get $\frac{\frac{1}{y}+\frac{1}{z}}{1-\frac{1}{y z}}=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$ $\Rightarrow \frac{y+z}{y z-1}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$. If $x \geqslant 5$, then $1-\frac{2}{x+1} \geqslant 1-\frac{2}{5+1}=\frac{2}{3} \Rightarrow \frac{y+z}{y z-1} \geqslant \frac{2}{3}$. However, when $z \geqslant y \geqslant x \geqslant 5$, $$ \frac{y+z}{y z-1} \leqslant \frac{5+5}{25-1}=\frac{5}{12}<\frac{2}{3}, $$ which is a contradiction. Therefore, $x=2,3,4$. When $x=2$, $$ \begin{array}{l} y z-1=3(y+z) \\ \Rightarrow(y-3)(z-3)=10 \\ \Rightarrow(y, z)=(4,13),(5,8) . \end{array} $$ When $x=3$, $$ \begin{array}{l} y z-1=2(y+z) \\ \Rightarrow(y-2)(z-2)=5 \\ \Rightarrow(y, z)=(3,7) . \end{array} $$ When $x=4$, $$ \begin{array}{l} 3 y z-3=5(y+z) \\ \Rightarrow 3=5\left(\frac{1}{y}+\frac{1}{z}\right)+\frac{3}{y z} \leqslant 5\left(\frac{1}{4}+\frac{1}{4}\right)+\frac{3}{16}, \end{array} $$ which is a contradiction. Thus, the ordered triples $(x, y, z)$ that satisfy $z \geqslant y \geqslant x$ are $$ (x, y, z)=(2,4,13),(2,5,8),(3,3,7) \text {. } $$ By permuting the order, we can obtain $6+6+3=15$ solutions. In conclusion, the number of ordered positive integer solutions to the equation is 15." 9a743c307da0,"I. Fill in the Blanks (8 points each, total 64 points) 1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$.",See reasoning trace,easy,"$-, 1.327$ $$ C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327 $$" 3ec16e9cb3a8,"## Find the numbers of the form $\overline{a b c}$ knowing that when dividing $\overline{a b c}$ by $\overline{b c}$, the quotient is 6 and the remainder is 5. Note - Effective working time 2 hours. - All - 7 points are awarded for each correctly solved ## National Mathematics Olympiad Local stage -14.02.2015 Grading criteria Grade 5",68&width=1545&top_left_y=676&top_left_x=274),medium,"## Problem 4 According to the natural number theorem, we have $\overline{a b c}=6 \cdot \overline{b c}+5$ $1 p$ $\overline{b c}=20 a-1$ 2p For $a=1=>\overline{b c}=19=>\overline{a b c}=119$ For $a=2=>\overline{b c}=39=>\overline{a b c}=239$ For $a=3=>\overline{b c}=59=>\overline{a b c}=359$ For $a=4=>\overline{b c}=79=>\overline{a b c}=479$ For $a=5=>\overline{b c}=99=>\overline{a b c}=599$ For $a \geq 6$ there are no solutions......................................................................................3p ![](https://cdn.mathpix.com/cropped/2024_06_07_dfc9162f0525ae4a7f10g-3.jpg?height=68&width=1545&top_left_y=676&top_left_x=274)" f2d35798e49a,,"100$. By solving the equation, we find that there are 27 students and 73 apples.",easy,"Solution. If the number of students is denoted by $x$, then the basket will contain $2 x+19$ apples. From the condition that the total number of students and apples is 100, we get $x+2 x+19=100$. By solving the equation, we find that there are 27 students and 73 apples." a78e2dcf98f3,"(BMO 2007) Find all real functions $f$ such that for all $x, y \in \mathbb{R}$, $$ f(f(x)+y)=f(f(x)-y)+4 f(x) y $$",g\left(c^{\prime}+2\left(g(a)+a^{2}-g(b)-b^{2}\right)\right)$ for all $c^{\prime} \in \mathbb{R}$. T,medium,"Let $g(x)=f(x)-x^{2}$. The equation becomes $$ g\left(g(x)+x^{2}+y\right)=g\left(g(x)+x^{2}-y\right) $$ For all real numbers $a, b, c$, by taking $x=a$ and $y=c-b^{2}-g(b)$, and $x=b, y=c-a^{2}-g(b)$, we obtain $$ g\left(g(a)+a^{2}-g(b)-b^{2}+c\right)=g\left(g(a)+a^{2}+g(b)+b^{2}-c\right)=g\left(g(b)+b^{2}-g(a)-a^{2}+c\right) $$ The equality between the first and the third member shows that $g\left(c^{\prime}\right)=g\left(c^{\prime}+2\left(g(a)+a^{2}-g(b)-b^{2}\right)\right)$ for all $c^{\prime} \in \mathbb{R}$. Thus, either $g(x)+x^{2}$ is a constant function, which implies that $f$ is constant, and then $f(x)=0$ for all real $x$ by returning to the original equation; or there exist $a \neq b$ such that $g(a)+a^{2} \neq g(b)+b^{2}$, and $g$ is periodic. In this case, let $T$ be its period. Taking $a=b+T$, we get $g\left(c^{\prime}\right)=g\left(c^{\prime}+2\left(2 y T+T^{2}\right)\right)$. Therefore, $g$ is $4 y T+2 T^{2}$-periodic. As $y$ ranges over $\mathbb{R}$, $4 y T+2 T^{2}$ also ranges over $\mathbb{R}$, so $g\left(c^{\prime}\right)=g\left(c^{\prime}+r\right)$ for all real $r$. Thus, $g$ is constant, and $f(x)=x^{2}+C$ where $C \in \mathbb{R}$. Conversely, if $g$ is constant, it clearly satisfies equation (VI.1) which is equivalent to the given equation, so these functions $f$ are solutions to the problem for any real $C$." 6f20ba4d7816,"A cube with $3$-inch edges is to be constructed from $27$ smaller cubes with $1$-inch edges. Twenty-one of the cubes are colored red and $6$ are colored white. If the $3$-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white? $\textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3}$",\textbf{(A),easy,"For the least possible surface area that is white, we should have 1 cube in the center, and the other 5 with only 1 face exposed. This gives 5 square inches of white surface area. Since the cube has a surface area of 54 square inches, our answer is $\boxed{\textbf{(A) }\frac{5}{54}}$." 6f0e86ffa25d,"4. The maximum value $M(a)$ of the function $f(x)=\left|x^{2}-a\right|$ on the interval $[-1,1]$ has its minimum value equal to $\qquad$ .","\frac{1}{2}$ when, $\dot{M}(a)$ has the minimum value of $\frac{1}{2}$.",medium,"4. $\frac{1}{2}$. Since $f(x)=\left|x^{2}-a\right|$ is an even function, $\therefore M(a)$ is the maximum value of $f(x)$ in the interval $0 \leqslant x \leqslant 1$. When $a \leqslant 0$, $f(x)=x^{2}-a$, it is easy to know that $M(a)=1-a$. When $a>0$, from the graph we can see that. If $\sqrt{2 a} \geqslant 1$, then $M(a)$ $=a$; If $\sqrt{2 a} \leqslant 1$, then $M(a)$ $=f(1)=1-a$. $$ \therefore M(a)=\left\{\begin{array}{r} 1-a,\left(a \leqslant \frac{1}{2}\right) \\ a .\left(a \geqslant \frac{1}{2}\right) \end{array}\right. $$ Since $a \leqslant \frac{1}{2}$ when $M(a) \backslash$, and $a \geqslant \frac{1}{2}$ when $M(a) \lambda$. $\therefore a=\frac{1}{2}$ when, $\dot{M}(a)$ has the minimum value of $\frac{1}{2}$." e23984a7d103,"5. Let $a$, $b$, $c$ ($a<5$) be positive real numbers, and satisfy $a^{2}-a-2 b-2 c=0$ and $a+2 b-2 c+3=0$. Then the size relationship between $a$, $b$, and $c$ is ( ). (A) $a0 \Rightarrow a>3 \text {. } $$ Also, $a>0 \\ \Rightarrow c>a, \\ 4(b-a)=a^{2}-6 a-3=(a-3)^{2}-12b . \end{array} $$ Therefore, $c>a>b$." 5b951404c283,"## Task A-2.2. Determine all triples $(a, b, c)$ of real numbers for which $$ a^{2}+b^{2}+c^{2}=1 \quad \text { and } \quad(2 b-2 a-c) a \geqslant \frac{1}{2} $$","\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$ and $(a, b, c)=\left(-\frac",medium,"## First Solution. Let $a, b$, and $c$ be such that the given conditions are satisfied. Then we have $$ (2 b-2 a-c) a \geqslant \frac{a^{2}+b^{2}+c^{2}}{2} $$ By multiplying the inequality by 2 and moving terms to one side, we get $$ 0 \geqslant a^{2}+b^{2}+c^{2}-4 a b+4 a^{2}+2 a c $$ It follows that $$ 0 \geqslant a^{2}+2 a c+c^{2}+b^{2}-4 a b+4 a^{2} $$ or $$ 0 \geqslant(a+c)^{2}+(b-2 a)^{2} $$ Since $(a+c)^{2} \geqslant 0$ and $(b-2 a)^{2} \geqslant 0$, we have $$ 0 \leqslant(a+c)^{2}+(b-2 a)^{2} $$ Thus, $(a+c)^{2}+(b-2 a)^{2}=0$, which implies $c=-a$ and $b=2 a$ From the condition $a^{2}+b^{2}+c^{2}=1$, we get $$ a^{2}+4 a^{2}+a^{2}=1, \quad \text { i.e. } \quad a^{2}=\frac{1}{6} $$ The desired triples are: $(a, b, c)=\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$ and $(a, b, c)=\left(-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$." 9fa1467da4e9,"37. Overtime Work. A certain company offered 350 of its employees to work overtime, with each man being offered a reward of 10 dollars, and each woman - 8 dollars and 15 cents. All women agreed to this offer, while some men refused. Upon calculation, it was found that the total reward amount does not depend on the number of male employees. What is the total reward amount paid to all women?",See reasoning trace,easy,"37. Let $m$ be the total number of men, and $x$ be the fraction of men who refused to work overtime. Then the total amount of compensation paid is $$ T=8.15(350-m)+10(1-x) m=2852.50+m(1.85-10 x) $$ which is independent of $m$ only if $x=0.185$. It is also known that $m<350$ and that both $m$ and $0.185 m$ are integers. Therefore, $m=200$. From this, it follows that 150 women received a total compensation of 1220 dollars and 50 cents. SOLUTIONS" 1a79ae3f48b2,"Carlson was given a bag of candies: chocolate and caramels. In the first 10 minutes, Carlson ate 20% of all the candies, and 25% of them were caramels. After that, Carlson ate three more chocolate candies, and the proportion of caramels among the candies Carlson had eaten decreased to 20%. How many candies were in the bag Carlson was given?",60 candies,easy,"In the first portion of candies eaten, there were 3 times more chocolate ones than caramels, and then their number became 4 times more. This means that the number of additional chocolate candies eaten equals the number of caramels. Thus, at the beginning, Karlson ate 3 caramels and 9 chocolate candies, and these 12 candies make up $20 \%$ of all the candies. Therefore, there were 60 candies in the bag in total. ## Answer 60 candies." 4da5d503e0b0,2. Suppose the function $f(x)-f(2 x)$ has derivative 5 at $x=1$ and derivative 7 at $x=2$. Find the derivative of $f(x)-f(4 x)$ at $x=1$.,"1$. This is $g^{\prime}(x)+2 g^{\prime}(2 x)$ at $x=1$, or $5+2 \cdot 7=19$.",easy,"Solution: 19 Let $g(x)=f(x)-f(2 x)$. Then we want the derivative of $$ f(x)-f(4 x)=(f(x)-f(2 x))+(f(2 x)-f(4 x))=g(x)+g(2 x) $$ at $x=1$. This is $g^{\prime}(x)+2 g^{\prime}(2 x)$ at $x=1$, or $5+2 \cdot 7=19$." 35387191a3cf,"## Task 1 - 320831 If $a, b, c$ are the hundreds, tens, and units digits of a three-digit natural number, then this number is briefly denoted by $\overline{a b c}$. Similarly, a two-digit number with tens and units digits $b$ and $c$ is denoted by $\overline{b c}$. Determine all $a, b, c$ for which $\overline{a b c}$ is a three-digit number and $\overline{b c}$ is a two-digit number, such that the equation $\overline{a b c}=(\overline{b c})^{b}$ holds!","6, b=2, c=5$ are exactly the digits of the required type.",medium,"I. If $a, b, c$ are digits of the required type, then: It cannot be $b=0$, because then $\overline{b c}$ would not be a two-digit number. It cannot be $b=1$, because then $(\overline{b c})^{b}=\overline{b c}$, which could not be equal to the three-digit number $\overline{a b c}$. It cannot be $b \geq 3$, because then $(\overline{b c})^{b} \geq 30^{3}$ would be at least a five-digit number, and thus could not be equal to the three-digit number $\overline{a b c}$. Therefore, $b=2$ must hold, and thus $\overline{a b c}=(\overline{b c})^{2}$ must hold. Therefore, $c$ must be the units digit of the number $c^{2}$. Since $2^{2}=4, 3^{2}=9, 4^{2}=16, 7^{2}=49, 8^{2}=64, 9^{2}=81$, this cannot be true for $c=2 ; 3 ; 4 ; 7 ; 8 ; 9$, and thus can only be true for $c=0 ; 1 ; 5 ; 6$. Furthermore, from the tens digits of $20^{2}=400, 21^{2}=441, 26^{2}=676$ it is clear: The number $(\overline{b c})^{2}$ with $b=2$ does not have the required tens digit 2 for $c=0 ; 1 ; 6$ as required by $\overline{a b c}=(\overline{b c})^{2}$; it can only have this (among the possibilities $c=0 ; 1 ; 5 ; 6$) for $c=5$. This finally leads to $25^{2}=625$, giving $a=6$. II. From $a=6, b=2, c=5$, the required equation $\overline{a b c}=(\overline{b c})^{b}$ is satisfied because $625=25^{2}$. With I. and II., it is shown: The digits $a=6, b=2, c=5$ are exactly the digits of the required type." 4215d6532a3e,"9. Let $N$ be the smallest positive integer such that $N / 15$ is a perfect square, $N / 10$ is a perfect cube, and $N / 6$ is a perfect fifth power. Find the number of positive divisors of $N / 30$.","2^{m} 3^{n} 5^{p}$ for some nonnegative integers $m, n, p$. Since $N / 15=$ $2^{m} 3^{n-1} 5^{p-1}$ ",medium,"Answer: 8400 Solution: $N$ must be of the form $N=2^{m} 3^{n} 5^{p}$ for some nonnegative integers $m, n, p$. Since $N / 15=$ $2^{m} 3^{n-1} 5^{p-1}$ is a perfect square, we have $m \equiv 0 \bmod 2$ and $n \equiv p \equiv 1 \bmod 2$. Since $N / 10=$ $2^{m-1} 3^{n} 5^{p-1}$ is a perfect cube, we have $n \equiv 0 \bmod 3$ and $m \equiv p \equiv 1 \bmod 3$. Since $N / 6=2^{m-1} 3^{n-1} 5^{p}$ is a perfect fifth power, we have $p \equiv 0 \bmod 5$ and $m \equiv n \equiv 1 \bmod 5$. By Chinese remainder theorem, we get $m \equiv 16 \bmod 30, n \equiv 21 \bmod 30$ and $p \equiv 25 \bmod 30$. Since $N$ is as small as possible, we take $m=16, n=21, p=25$, so $N=2^{16} 3^{21} 5^{25}$. Thus the number of positive divisors of $N / 30=2^{15} 3^{20} 5^{24}$ is $16 \cdot 21 \cdot 25=8400$." 7c2173b0c2db,"GS. 4 Let $x, y$ and $z$ be real numbers that satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1$ and $z+\frac{1}{x}=\frac{7}{3}$. Find the value of $x y z$. (Reference 2010 FG2.2)",1$,medium,"Method 1 From (1), $x=4-\frac{1}{y}=\frac{4 y-1}{y}$ $$ \Rightarrow \frac{1}{x}=\frac{y}{4 y-1} $$ Sub. (4) into (3): $z+\frac{y}{4 y-1}=\frac{7}{3}$ $$ z=\frac{7}{3}-\frac{y}{4 y-1} $$ From (2): $\frac{1}{z}=1-y$ $z=\frac{1}{1-y} \ldots \ldots$ (6) (5) $=$ (6): $\frac{1}{1-y}=\frac{7}{3}-\frac{y}{4 y-1}$ $\frac{1}{1-y}=\frac{28 y-7-3 y}{3(4 y-1)}$ $3(4 y-1)=(1-y)(25 y-7)$ $12 y-3=-25 y^{2}-7+32 y$ $25 y^{2}-20 y+4=0$ $(5 y-2)^{2}=0$ $$ y=\frac{2}{5} $$ Sub. $y=\frac{2}{5}$ into (6): $z=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$ Sub. $y=\frac{2}{5}$ into (1): $x+\frac{5}{2}=4 \Rightarrow x=\frac{3}{2}$ Method 2 $\left\{\begin{array}{l}x+\frac{1}{y}=4 \cdots \cdots(1) \\ y+\frac{1}{z}=1 \cdots \cdots(2) \\ z+\frac{1}{x}=\frac{7}{3} \cdots \cdots(3)\end{array}\right.$ (1) $\times(2): \quad x y+1+\frac{x}{z}+\frac{1}{y z}=4$ $x\left(y+\frac{1}{z}\right)+\frac{1}{y z}=3$ Sub. (2) into the eqt.: $x+\frac{x}{x y z}=3$ Let $a=x y z$, then $x+\frac{x}{a}=3 \cdots \cdots(4)$ (2) $\times(3): y\left(\frac{7}{3}\right)+\frac{y}{a}=\frac{4}{3} \Rightarrow y\left(\frac{7}{3}+\frac{1}{a}\right)=\frac{4}{3} \ldots \ldots$. (1) $\times(3): z(4)+\frac{z}{a}=\frac{25}{3} \Rightarrow z\left(4+\frac{1}{a}\right)=\frac{25}{3} \cdots \cdots$ (4) $\times(5) \times(6): a\left(1+\frac{1}{a}\right)\left(\frac{7}{3}+\frac{1}{a}\right)\left(4+\frac{1}{a}\right)=\frac{100}{3}$ $\frac{(a+1)(7 a+3)(4 a+1)}{3 a^{2}}=\frac{100}{3}$ which reduces to $28 a^{3}-53 a^{2}+22 a+3=0$ $\Rightarrow(a-1)^{2}(28 a+3)=0$ $\therefore a=1$ $x y z=\frac{2}{5} \times \frac{5}{3} \times \frac{3}{2}=1$ Method $3(1) \times(2) \times(3)-(1)-(2)-(3)$ : $x y z+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y z}-\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{28}{3}-\frac{22}{3} \Rightarrow x y z+\frac{1}{x y z}=2$ $x y z=1$" 76f8a400d304,"11. (12 points) If the sum of the digits of a natural number is the same as the sum of the digits of its triple, but different from the sum of the digits of its double, we call such a number a ""wonderful number"". Therefore, the smallest ""wonderful number"" is $\qquad$",: 144,medium,"11. (12 points) If the sum of the digits of a natural number is the same as the sum of the digits of its triple, but different from the sum of the digits of its double, we call such a number a “wonderful number”. Therefore, the smallest “wonderful number” is $\qquad$ . $\qquad$ 【Solution】Solution: According to the problem, let this number be $x$. Based on the rule of casting out nines, the sum of the digits of the sum = the sum of the digits of the addends - the number of carries $\times 9$. If $x$ is a single-digit number, then it can only be $9, 9+9=18$, which does not meet the requirement that the sum of the digits of $x+x$ is different from the sum of the digits of $x$; if $x$ is a two-digit number, and the sum of these two digits is 18, then it can only be 99, and $99+99=198$ does not meet the requirement that the sum of the digits of $x+x$ is different from the sum of the digits of $x$; If $x$ is a two-digit number, and the sum of these two digits is 9, then the sum of the digits of its double has one carry, and the final sum of the digits is still 9; if $x$ is a three-digit number, and the sum of the digits is 9, then we can let this three-digit number's double have no carry, the smallest is 144, which meets the requirement; Therefore, the answer is: 144" 5f0fc05a96b2,"If $N$ is the smallest positive integer whose digits have a product of 1728 , then the sum of the digits of $N$ is (A) 28 (B) 26 (C) 18 (D) 27 (E) 21",(A),medium,"In order to find $N$, which is the smallest possible integer whose digits have a fixed product, we must first find the minimum possible number of digits with this product. (This is because if the integer $a$ has more digits than the integer $b$, then $a>b$.) Once we have determined the digits that form $N$, then the integer $N$ itself is formed by writing the digits in increasing order. (Given a fixed set of digits, the leading digit of $N$ will contribute to the largest place value, and so should be the smallest digit. The next largest place value should get the next smallest digit, and so on.) Note that the digits of $N$ cannot include 0 , else the product of its digits would be 0 . Also, the digits of $N$ cannot include 1, otherwise we could remove the 1s and obtain an integer with fewer digits (thus, a smaller integer) with the same product of digits. Since the product of the digits of $N$ is 1728 , we find the prime factorization of 1728 to help us determine what the digits are: $$ 1728=9 \times 192=3^{2} \times 3 \times 64=3^{3} \times 2^{6} $$ We must try to find a combination of the smallest number of possible digits whose product is 1728 . Note that we cannot have 3 digits with a product of 1728 since the maximum possible product of 3 digits is $9 \times 9 \times 9=729$. Let us suppose that we can have 4 digits with a product of 1728 . In order for $N$ to be as small as possible, its leading digit (that is, its thousands digit) must be as small as possible. From above, this digit cannot be 1 . This digit also cannot be 2 , since otherwise the product of the remaining 3 digits would be 864 , which is larger than the product of 3 digits can be. Can the thousands digit be 3 ? If so, the remaining 3 digits have a product of 576 . Can 3 digits have a product of 576 ? If one of these 3 digits were 7 or less, then the product of the 3 digits would be at most $7 \times 9 \times 9=567$, which is too small. Therefore, if we have 3 digits with a product of 576 , then each digit is 8 or 9 . Since the product is even, then at least one of the digits would have to be 8 , leaving the remaining two digits to have a product of $576 \div 8=72$. These two digits would then have to be 8 and 9 . Thus, we can have 3 digits with a product of 576 , and so we can have 4 digits with a product of 1728 with smallest digit 3 . Therefore, the digits of $N$ must be $3,8,8,9$. The smallest possible number formed by these digits is when the digits are placed in increasing order, and so $N=3889$. The sum of the digits of $N$ is $3+8+8+9=28$. ANSWER: (A)" 885f98e53f6f,"5. In a regular quadrilateral pyramid $P-ABCD$, the side faces are equilateral triangles with a side length of 1. $M, N$ are the midpoints of edges $AB, BC$ respectively. The distance between the skew lines $MN$ and $PC$ is $\qquad$.",\frac{\sqrt{2}}{4}$.,easy,"Let $O$ be the midpoint of $A C$, $P O \perp$ base $A B C D \Rightarrow$ plane $P A C \perp$ base $A B C D$. Connect $O B$, intersecting $M N$ at $H$. Then $O B \perp$ plane $P A C \Rightarrow O H$ is the distance from line $M N$ to plane $P A C$, which is the distance between the skew lines $M N$ and $P C$. According to the problem, the side length of the base $A B C D$ is $1 \Rightarrow O H=\frac{\sqrt{2}}{4}$." df5b4da17d10,"\section*{ A regular tetrahedron with vertices \(A, B, C\), and \(D\) and edge length \(a\) is intersected by six pairwise distinct planes, each of which contains exactly one edge of the tetrahedron and the midpoint of the opposite edge. a) How many sub-bodies are created in total if all cuts are imagined to be performed simultaneously? b) Calculate the volumes of the individual sub-bodies using the edge length \(a\).",See reasoning trace,medium,"} Since \(A C=A D\), \(A\) lies on the perpendicular bisector plane \(\epsilon\) of \(C D\). Similarly, \(B\) lies on \(\epsilon\). Therefore, \(\epsilon\) is the plane among the intersection planes that contains \(A B\). Correspondingly, the other intersection planes coincide with the perpendicular bisector planes of the tetrahedron edges. As a regular tetrahedron, \(A B C D\) has a circumcenter \(S\) with \(S A=S B=S C=S D\), which therefore lies on each of the mentioned perpendicular bisector planes, i.e., on all intersection planes. Consequently, the plane through \(A, B, S\) is the intersection plane that contains \(A B\). Similarly, the other intersection planes coincide with the planes connecting one tetrahedron edge and \(S\). Each of the faces adjacent to \(S\) of the tetrahedron \(S B C S\) thus lies in one of the intersection planes. The same applies to the tetrahedra \(A B D S, A C D S, B C D S\). The desired decomposition of \(A B C D\) can therefore be obtained by further decomposing the four tetrahedra \(A B C S, A B D S, A C D S, B C D S\) (into which \(A B C D\) can initially be decomposed). For the further decomposition of \(A B C S\), only those cutting edges are relevant that (besides through \(S\)) pass through interior points of \(A B C S\). These are exactly those that pass through interior points of the triangle \(A B C\), i.e., those that pass through \(D\), one of the vertices \(A, B, C\), and a midpoint of an edge. They divide the area of the triangle \(A B C\) by its medians into 6 equal-area (even congruent) sub-areas. Thus, the tetrahedron \(A B C S\) is divided into exactly 6 equal-volume (even congruent) sub-bodies. The same applies to the tetrahedra \(A B D S, A C D S, B C D S\) which are congruent to \(A B C S\). Therefore, a total of 24 equal-volume tetrahedra are created. Each of them thus has the volume \(V_{K}=\frac{1}{24} V_{T}\), where \(V_{T}\) is the volume of the tetrahedron \(A B C D\). Since \(V_{T}=\frac{a^{3}}{12} \sqrt{2}\), it follows that \[ V_{K}=\frac{a^{3}}{288} \sqrt{2} \] Adapted from [5] \subsection*{7.16 XIV. Olympiad 1974} \subsection*{7.16.1 First Round 1974, Grade 10}" efbb99555726,"11. Given that $\theta$ is an acute angle, and $\frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $\frac{\sin 3 \theta}{\sin \theta}=$ $\qquad$",2+\frac{1}{3}=\frac{7}{3}$.,medium,"$=.11 \cdot \frac{7}{3}$. Solution 1: From the given and the triple angle cosine formula, we get $4 \cos ^{2} \theta-3=\frac{1}{3}$, which means $4 \cos ^{2} \theta=\frac{10}{3}$. Therefore, $\frac{\sin 3 \theta}{\sin \theta}=3-4 \sin ^{2} \theta=4 \cos ^{2} \theta-1=\frac{7}{3}$. Solution 2: Let $\frac{\sin 3 \theta}{\sin \theta}=x$, then $$ \begin{array}{l} x-\frac{1}{3}=\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta} \\ =\frac{\sin (3 \theta-\theta)}{\sin \theta \cdot \cos \theta}=\frac{\sin 2 \theta}{\frac{1}{2} \sin 2 \theta}=2 . \end{array} $$ Thus, $x=2+\frac{1}{3}=\frac{7}{3}$." 04780af92bc9,"The value of $\frac{3}{a+b}$ when $a=4$ and $b=-4$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless}$","0$, the value of the fraction $\frac{3}{4 - (-4)}$ is $\textbf{(E)}\ \text{meaningless}$",easy,"As the denominator is $4 - (-4) = 0$, the value of the fraction $\frac{3}{4 - (-4)}$ is $\textbf{(E)}\ \text{meaningless}$" 208de055fb9e,"10.5. On the board, all proper divisors of some composite natural number $n$, increased by 1, were written down. Find all such numbers $n$ for which the numbers on the board turn out to be all proper divisors of some natural number $m$. (Proper divisors of a natural number $a>1$ are all its natural divisors, different from $a$ and from 1.) (A. Khryabrov)",$n=4$ or $n=8$,medium,"Answer: $n=4$ or $n=8$. First solution. Note that the number 2 is not written on the board, because 1 is not a proper divisor of $n$; hence, $m$ is odd. Therefore, all the divisors of $m$ are odd, and thus all the divisors of $n$ are even. Thus, $n$ does not divide by odd prime numbers, meaning $n$ is a power of two (and all its divisors are also powers of two). If $n$ is divisible by 16, then 4 and 8 are its proper divisors, so 5 and 9 are written on the board. Therefore, $m$ is divisible by 45, and in particular, 15 is its proper divisor. But the number 15 could not have been written, since 14 is not a power of two. Therefore, $n$ cannot be divisible by 16. The remaining composite powers of two, $n=4$ and $n=8$, fit: for them, we can respectively set $m=9$ and $m=15$. Second solution. Let $a_{1}n / a_{2}>\ldots>$ $>n / a_{k}$ be the proper divisors of the number $n$. Then they correspond to $a_{k}, a_{k-1}, \ldots, a_{1}$, meaning $a_{i} a_{k+1-i}=n$ for all $i=1,2, \ldots, k$. A similar reasoning can be applied to the divisors of the number $m$. Suppose $k \geqslant 3$. Then $a_{1} a_{k}=a_{2} a_{k-1}=n$ and $\left(a_{1}+1\right)\left(a_{k}+1\right)=$ $=\left(a_{2}+1\right)\left(a_{k-1}+1\right)=m$, from which $a_{1}+a_{k}=a_{2}+a_{k-1}=m-$ $-n-1$. We see that the pairs of numbers $\left(a_{1}, a_{k}\right)$ and ( $a_{2}, a_{k-1}$ ) have the same sum and product; by Vieta's theorem, they are pairs of roots of the same quadratic equation, meaning these pairs must coincide - a contradiction. Thus, $k \leqslant 2$; this is possible if $n=p^{2}, n=p^{3}$, or $n=p q$, where $p$ and $q$ are prime numbers (in the last case, we assume $p ab - a - b$, then $$ \begin{array}{l} by = n - ax > ab - a - b - a(b-1) = -b \\ \Rightarrow y > -1 \Rightarrow y \geq 0. \end{array} $$ Thus, any integer greater than $ab - a - b$ can be expressed as $$ \begin{array}{l} ax + by \quad (x, y \in \mathbb{N}). \end{array} $$ If $ab - a - b = ax + by \quad (x, y \in \mathbb{N})$, then $$ \begin{array}{l} ab = a(x + 1) + b(y + 1) \\ \Rightarrow b \mid (x + 1) \Rightarrow x + 1 \geq b. \end{array} $$ Similarly, $y + 1 \geq a$. Then $ab = a(x + 1) + b(y + 1) \geq 2ab$, which is a contradiction. Thus, the lemma is proved. Returning to the original problem. By the lemma, we know that any integer greater than $5 \times 6 - 5 - 6 = 19$ can be expressed as $5x + 6y \quad (x, y \in \mathbb{N})$, except for $1, 5, 6$. Among the numbers less than 19 that can be written as $5x + 6y$, we have $$ \begin{aligned} 5 \times 2 & = 10, 5 + 6 = 11, 6 \times 2 = 12, \\ 5 \times 3 & = 15, 5 \times 2 + 6 = 16, \\ 5 & + 6 \times 2 = 17, 6 \times 3 = 18, \end{aligned} $$ leaving nine numbers, which are $$ 2, 3, 4, 7, 8, 9, 13, 14, 19. $$ Therefore, if Player A starts by writing 19, the remaining eight numbers can be divided into four pairs: $(2, 3), (4, 7), (8, 9), (13, 14)$. When Player B writes one of these numbers, Player A writes the other number in the pair. After all eight numbers are written, it will be Player B's turn to write 1, so Player A wins. Thus, Player A has a winning strategy." 29e6fb2628ec,"Let $p$ and $q$ be different prime numbers. How many divisors does the number have: a) $p q ;$ b) $p^{2} q ;$ c) $p^{2} q^{2}$; d) $p^{m} q^{n}$ ?",a) 4 divisors; b) 6; c) 9; d) $(m+1)(n+1)$ divisors,easy,"г) Each divisor also has the form $p^{k} q^{l}$, and the exponent $k$ can be chosen in $m+1$ ways (from 0 to $m$), and p ## Answer a) 4 divisors; b) 6; c) 9; d) $(m+1)(n+1)$ divisors. Submit a comment" d5d5bc016b6d,"317. Find $y^{\prime}$, if $y=\ln \sin ^{3} 5 x$.",See reasoning trace,easy,"Solution. We will use formulas VII, VIII, and XIII: $$ y^{\prime}=\frac{1}{\sin ^{3} 5 x} \cdot 3 \sin ^{2} 5 x \cos 5 x \cdot 5=\frac{15 \cos 5 x}{\sin 5 x}=15 \operatorname{ctg} 5 x $$ 318-323. Find the derivatives of the following functions:" d4dc89f7299f,211. $\int x^{2} \sin x d x$,See reasoning trace,medium,"Solution. We have $$ \int x^{2} \sin x d x=\left|\begin{array}{l} u=x^{2}, d v=\sin x d x \\ d u=2 x d x, v=-\cos x \end{array}\right|=-x^{2} \cos x+2 \int x \cos x d x $$ To find the integral obtained on the right side of the equation, we integrate by parts again: $$ \int x \cos x d x=x \sin x+\cos x+C $$ (see the solution of example 207). As a result, we obtain the final answer: $$ \int x^{2} \sin x d x=-x^{2} \cos x+2 x \sin x+2 \cos x+C $$" 9d8e800faf3d,"4. Let $\sigma=\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right)$ be a permutation of $(1,2,3, \ldots, n)$. A pair $\left(a_{i}, a_{j}\right)$ is said to correspond to an inversion of $\sigma$, if $ia_{j}$. (Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1)$, $(4,3),(4,1),(5,3),(5,1),(3,1)$.) How many permutations of $(1,2,3, \ldots n)$, $(n \geq 3)$, have exactly two inversions?",See reasoning trace,medium," Solution: In a permutation of $(1,2,3, \ldots, n)$, two inversions can occur in only one of the following two ways: (A) Two disjoint consecutive pairs are interchanged: $$ \begin{aligned} & (1,2,3, j-1, j, j+1, j+2 \ldots k-1, k, k+1, k+2, \ldots, n) \\ & \quad \longrightarrow(1,2, \ldots j-1, j+1, j, j+2, \ldots, k-1, k+1, k, k+2, \ldots, n) \end{aligned} $$ (B) Each block of three consecutive integers can be permuted in any of the following 2 ways; $$ \begin{aligned} & (1,2,3, \ldots k, k+1, k+2, \ldots, n) \longrightarrow(1,2, \ldots, k+2, k, k+1, \ldots, n) \\ & (1,2,3, \ldots k, k+1, k+2, \ldots, n) \longrightarrow(1,2, \ldots, k+1, k+2, k, \ldots, n) \end{aligned} $$ Consider case (A). For $j=1$, there are $n-3$ possible values of $k$; for $j=2$, there are $n-4$ possibilities for $k$ and so on. Thus the number of permutations with two inversions of this type is $$ 1+2+\cdots+(n-3)=\frac{(n-3)(n-2)}{2} $$ In case (B), we see that there are $n-2$ permutations of each type, since $k$ can take values from 1 to $n-2$. Hence we get $2(n-2)$ permutations of this type. Finally, the number of permutations with two inversions is $$ \frac{(n-3)(n-2)}{2}+2(n-2)=\frac{(n+1)(n-2)}{2} $$ " ee93b8032268,"107 The 100th digit after the decimal point of the number $(\sqrt{5}+2)^{1997}$ is A. 0 B. 1 C. 2 D. Greater than 2",See reasoning trace,easy,"107 A. $(\sqrt{5}+2)^{1997}-(\sqrt{5}-2)^{1997}$ is an integer. Therefore, the decimal parts of $(\sqrt{5}+2)^{1997}$ and $(\sqrt{5}-2)^{1997}$ are exactly the same. And $(\sqrt{5}-2)^{1997}<(\sqrt{5}-2)^{200}<0.3^{200}<0.1^{100}$. So the 100th digit after the decimal point of $(\sqrt{5}+2)^{1997}$ is 0." 567dc13370c0,"Example 4 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}$, $Q$ is a point inside the triangle, $\angle Q B A=$ $\angle Q C A=20^{\circ}$. Find the degree measure of $\angle Q A B$.","\angle A B Q$, making point $E$ the symmetric point of $A$ with respect to $B Q$. To find such a poi",medium,"Solution: As shown in Figure 4, let $B Q$ intersect $A C$ at $D$, and draw a perpendicular from $D$ to $B C$ intersecting $Q C$ at $E$. Connect $B E$. From $\angle Q 13 C=30^{\circ}=\angle A C B$, we know that $D E$ is the perpendicular bisector of $B C$. From $\angle Q C B=10^{\circ}$, we know $\angle E B C=10^{\circ}, \angle Q B E=20^{\circ}=\angle Q B A$. From $\angle E D B=60^{\circ}=\angle E D C$, we know $\angle B D A$ $=60^{\circ}=\angle B D E$. Thus, point $A$ is symmetric to point $E$ with respect to $B D$. $$ \begin{array}{l} \text { Then } \angle Q A B=\angle Q E B \\ =\angle E B C+\angle E C B=20^{\circ} . \end{array} $$ Here, we notice that $B Q$ is the angle bisector of $\angle A Q C$, so we consider taking a point $E$ on $Q C$ such that $\angle E B Q=\angle A B Q$, making point $E$ the symmetric point of $A$ with respect to $B Q$. To find such a point $E$, it is precisely the intersection of the perpendicular bisector of $B C$ with $Q C$. Furthermore, from $\angle Q B C=30^{\circ}=\angle A C B$, we think that the intersection point $D$ of $B Q$ and $A C$ should be another point on the perpendicular bisector of $B C$. Therefore, we chose the above method to find the symmetric point $E$ of $A$ with respect to $B Q$." 1e22b3e6fa90,"The value of $2^{5}-5^{2}$ is (A) 0 (B) -3 (C) -7 (D) 3 (E) 7",(E),easy,"Calculating, $2^{5}-5^{2}=32-25=7$. ANswer: (E)" 4f70bb4da7f8,"$34 \cdot 4$ For positive integers $n$ and $a$, define $n_{a}!$ as $$ n_{a}!=n(n-a)(n-2 a)(n-3 a) \cdots(n-k a), $$ where $k$ is the largest integer for which $n>k a$. Then the quotient of $72_{8}! \div 182!$ is (A) $4^{5}$. (B) $4^{6}$. (C) $4^{8}$. (D) $4^{9}$. (E) $4^{12}$. (24th American High School Mathematics Examination, 1973)",$(D)$,easy,"[Solution] $72_{8}!=72 \cdot 64 \cdot 56 \cdot 48 \cdot 40 \cdot 32 \cdot 24 \cdot 16 \cdot 8$ $$ =8^{9} \cdot 9!, $$ and $$ \begin{aligned} 182! & =18 \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2 \\ & =2^{9} \cdot 9!, \end{aligned} $$ then $72_{8}! \div 18_{2}!=8^{9} \div 2^{9}=4^{9}$. Therefore, the answer is $(D)$." 73d2b8a89ea5,5. Let $A B C$ be a triangle. Let $E$ be a point on the segment $B C$ such that $B E=2 E C$. Let $F$ be the mid-point of $A C$. Let $B F$ intersect $A E$ in $Q$. Determine $B Q / Q F$.,See reasoning trace,medium," Solution: Let $C Q$ and $E T$ meet $A B$ in $S$ ![](https://cdn.mathpix.com/cropped/2024_06_05_2df523a359bae2f5e442g-2.jpg?height=328&width=545&top_left_y=671&top_left_x=454) and $T$ respectively. We have $$ \frac{[S B C]}{[A S C]}=\frac{B S}{S A}=\frac{[S B Q]}{[A S Q]} $$ Using componendo by dividendo, we obtain $$ \frac{B S}{S A}=\frac{[S B C]-[S B Q]}{[A S C]-[A S Q]}=\frac{[B Q C]}{[A Q C]} $$ Similarly, We can prove $$ \frac{B E}{E C}=\frac{[B Q A]}{[C Q A]}, \quad \frac{C F}{F A}=\frac{[C Q B]}{[A Q B]} $$ But $B D=D E=E C$ implies that $B E / E C=2 ; C F=F A$ gives $C F / F A=1$. Thus $$ \frac{B S}{S A}=\frac{[B Q C]}{[A Q C]}=\frac{[B Q C] /[A Q B]}{[A Q C] /[A Q B]}=\frac{C F / F A}{E C / B E}=\frac{1}{1 / 2}=2 $$ Now $$ \frac{B Q}{Q F}=\frac{[B Q C]}{[F Q C]}=\frac{[B Q A]}{[F Q A]}=\frac{[B Q C]+[B Q A]}{[F Q C]+[F Q A]}=\frac{[B Q C]+[B Q A]}{[A Q C]} $$ This gives $$ \frac{B Q}{Q F}=\frac{[B Q C]+[B Q A]}{[A Q C]}=\frac{[B Q C]}{[A Q C]}+\frac{[B Q A]}{[A Q C]}=\frac{B S}{S A}+\frac{B E}{E C}=2+2=4 $$ (Note: $B S / S A$ can also be obtained using Ceva's theorem. One can also obtain the result by coordinate geometry.) " f10d2629c7a5,"Burrito Bear has a white unit square. She inscribes a circle inside of the square and paints it black. She then inscribes a square inside the black circle and paints it white. Burrito repeats this process indefinitely. The total black area can be expressed as $\frac{a\pi+b}{c}$. Find $a+b+c$. [i]2022 CCA Math Bonanza Individual Round #4[/i]",0,medium,"1. **Initial Setup:** - The area of the initial white unit square is \(1\). - The area of the inscribed circle is \(\pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}\). - The area of the inscribed square within the circle is \(\left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\). 2. **General Formulas:** - For the \(i\)-th iteration, the area of the square is \(\left(\frac{1}{2^{i-1}}\right)^2 = \frac{1}{2^{2(i-1)}} = \frac{1}{4^{i-1}}\). - The area of the circle inscribed in this square is \(\pi \left(\frac{1}{2^i}\right)^2 = \frac{\pi}{4^i}\). 3. **Total White Area:** - The total white area is the sum of the areas of all the white squares. - The area of the \(i\)-th white square is \(\frac{1}{4^{i-1}}\). - The total white area is: \[ \sum_{i=1}^{\infty} \frac{1}{4^{i-1}} = \sum_{i=0}^{\infty} \frac{1}{4^i} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \] 4. **Total Black Area:** - The total black area is the sum of the areas of all the black circles minus the areas of all the white squares. - The area of the \(i\)-th black circle is \(\frac{\pi}{4^i}\). - The total area of the black circles is: \[ \sum_{i=1}^{\infty} \frac{\pi}{4^i} = \frac{\pi}{4} \sum_{i=0}^{\infty} \frac{1}{4^i} = \frac{\pi}{4} \cdot \frac{4}{3} = \frac{\pi}{3} \] - The total black area is then: \[ \text{Total black area} = \text{Total area of black circles} - \text{Total white area} = \frac{\pi}{3} - \frac{4}{3} = \frac{\pi - 4}{3} \] 5. **Final Calculation:** - The total black area can be expressed as \(\frac{a\pi + b}{c}\) where \(a = 1\), \(b = -4\), and \(c = 3\). - Therefore, \(a + b + c = 1 - 4 + 3 = 0\). The final answer is \(\boxed{0}\)" 3302a005839a,"3. On each of the lines $x=2$ and $x=15$, there are 400 points with ordinates $1,2,3, \ldots, 400$. In how many ways can three points be chosen from the 800 marked points so that they form the vertices of a right triangle?",320868,medium,"Answer: 320868. Solution. There are two possibilities. 1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of segments of a right triangle, we get that $CH^2 = AH \cdot BH$, i.e., $AH \cdot BH = 169$. Since $AH$ and $BH$ are integers, the following cases are possible: $AH = BH = 13$, $AH = 169$ and $BH = 1$, $AH = 1$ and $BH = 169$. In the first of these cases, the hypotenuse $AB$, equal to 26, can be placed in $374 \cdot 2 = 748$ ways (400 - 26 ways of placement on each of the two given lines), and the position of vertex $C$ is uniquely determined. In the second and third cases, the length of the hypotenuse is 170, and it can be placed in $2(400 - 170) = 460$ ways. For each position of the hypotenuse, there are two ways to place the vertex - resulting in $2 \cdot 460 = 920$ ways. 2) One of the legs of the triangle (let's call it $BC$) is perpendicular to the given lines, and the second leg ($AC$) lies on one of the given lines. Then the position of the leg $BC$ can be chosen in 400 ways. For each variant of the placement of the leg $BC$, the vertex $A$ can be placed in 798 ways (all points except the already chosen $B$ and $C$ are suitable) - in total, we get $400 \cdot 798 = 319200$ ways. In total, we get $748 + 920 + 319200 = 320868$ ways." 2d1dee76b5fa,"$\left.\begin{array}{l}\text { [Inclusion-Exclusion Principle]} \\ {[\quad \text { Word In the garden, Anya and Vitya had 2006 rose bushes. Vitya watered half of all the bushes, and Anya watered half of all the bushes. It turned out that exactly three bushes, the most beautiful ones, were watered by both Anya and Vitya. How many rose bushes were left unwatered?",3 bushes,easy,"Vitya watered 1003 bushes, of which 1000 he watered alone, and three - together with Anya. Similarly, Anya watered 1003 bushes, of which 1000 she watered alone, and three - with Vitya. Therefore, together they watered $1000+1000+3=2003$ bushes. Thus, $2006-2003=3$ rose bushes remained unwatered. ## Answer 3 bushes." 1a31424f9664,"Example 8 Let $D$ be a point inside an acute-angled $\triangle ABC$, $$ \begin{array}{l} \angle A D B=\angle A C B+90^{\circ}, \text { and } A C \cdot B D=A D \cdot B C . \\ \text { Find the value of } \frac{A B \cdot C D}{A C \cdot B D} \text { . } \end{array} $$",See reasoning trace,medium,"【Analysis】With $C$ as the origin and $C A$ as the real axis, establish a complex plane, and let $|\overrightarrow{C A}|=r$. According to the geometric meaning of complex number multiplication and division, the vector $\overrightarrow{D B}$ can be regarded as $\overrightarrow{D A}$ after rotation and scaling. Therefore, $$ \begin{array}{l} \overrightarrow{D B}=\overrightarrow{D A} \cdot \frac{|\overrightarrow{D B}|}{|\overrightarrow{D A}|}\left(\cos \left(\angle A C B+90^{\circ}\right)+\right. \\ \left.\quad \mathrm{i} \sin \left(\angle A C B+90^{\circ}\right)\right) \\ =\frac{1}{r}\left(r-z_{D}\right) z_{B} \mathrm{i} . \\ \text { Also, } \overrightarrow{D B}=z_{B}-z_{D} \text {, then } \\ z_{D}=\frac{r z_{B}(1-\mathrm{i})}{r-z_{B} \mathrm{i}} \\ \Rightarrow \frac{A B \cdot C D}{A C \cdot B D}=\frac{\left|z_{B}-r\right|\left|z_{D}\right|}{|r|\left|z_{B}-z_{D}\right|}=|1-\mathrm{i}|=\sqrt{2} . \end{array} $$" 056fafe55c87,"## Task A-4.5. In a circle, a finite number of real numbers are arranged. Each number is colored red, white, or blue. Each red number is twice as small as the sum of its two neighboring numbers, each white number is equal to the sum of its two neighboring numbers, and each blue number is twice as large as the sum of its two neighboring numbers. Let $b$ be the sum of all white numbers, and $p$ be the sum of all blue numbers, where both sums are different from 0. Determine the ratio $\frac{b}{p}$.",-\frac{3}{2}$.,medium,"## Solution. Let $x_{1}, x_{2}, \ldots, x_{n}$ be the given numbers. For simplicity, denote $x_{n+1}=x_{1}$ and $x_{-1}=x_{n}$. Let $c$ be the sum of all red numbers, and $S=c+b+p$ be the sum of all numbers. If $x_{i}$ is a red number, then $x_{i-1}+x_{i+1}=2 x_{i}$. If $x_{i}$ is a white number, then $x_{i-1}+x_{i+1}=x_{i}$. If $x_{i}$ is a red number, then $x_{i-1}+x_{i+1}=\frac{1}{2} x_{i}$. Thus, we have written the conditions from the problem as equations that connect three consecutive numbers. 2 points Consider the sum of all these equations. If we sum all these equations (for $i=1,2, \ldots, n$) on the right side of the equation, we will get $$ 2 c+b+\frac{1}{2} p $$ while on the left side, we will get $$ \sum_{i=1}^{n} x_{i-1}+\sum_{i=1}^{n} x_{i+1}=2 S=2 c+2 b+2 p $$ Therefore, $2 c+2 b+2 p=2 c+b+\frac{1}{2} p$, 1 point which simplifies to $\frac{b}{p}=-\frac{3}{2}$." 4447fa3cee54,"Example 8. In a Go tournament, ten players participate. Each player competes against the other nine players once. All players have different scores. The first and second place players have never lost a game, and the total score of the first two players is 10 points more than the second place player. The fourth place player's score is the same as the total score of the last four players. Please answer: How many points did each of the first to sixth place players get? (The winner gets 1 point per game, the loser gets 0 points, and in a draw, both get 0.5 points)",See reasoning trace,medium,"Let the scores of the ten selected players be $x_{1}, x_{2}, \cdots, x_{10}$. (1) Obviously, the result of the match between the first and second place is a draw. Thus, $x_{1} \leqslant 8.5, x_{2} \leqslant 8$. (2) The last four players have a total of 6 matches among themselves, so their total score $x_{1} + x_{8} + x_{9} + x_{10} \geqslant 6$. (3) Since the score of the fourth place is equal to the total score of the last four players, $x_{4} \geqslant 6$, thus $x_{3} \geqslant 6.5$. (4) If $x_{3} \geqslant 7$, then $x_{1} + x_{2} = x_{3} + 10 \geqslant 17$, which contradicts the result in (1), so $x_{3} < 7$, hence $x_{3} = 6.5$. (5) Therefore, $x_{4} = 6, x_{1} + x_{2} = 16.5$, thus $x_{1} = 8.5, x_{2} = 8$. (6) All ten players participate in a total of 45 matches, scoring 45 points, so the total score of the last six players is $45 - (8.5 + 8 - 6.5 + 6) = 16$ points, and the total score of the last four players is 6 points, thus $x_{5} + x_{6} = 10$, hence $x_{5} = 5.5, x_{6} = 4.5$. In summary, the scores of the top six players are 8.5 points, 8 points, 6.5 points, 6 points, 5.5 points, and 4.5 points, respectively." db0c51625023,"# Task 4. Maximum 20 points ## Option 1 At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved | Name | Maximum number of stereometry | :--- | :---: | :---: | | Andrey | 7 | 7 | | Volodya | 6 | 3 | | Zhanna | 3 | 18 | | Petya | 12 | 3 | | Galina | 7 | 14 | Help Andrey decide which pair of students to take into his team if the team's only goal is to win the tournament.",- 2 points,medium,"# Solution: Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve. Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points for the team. Volodya, instead of solving 1 problem in planimetry, can solve 2 problems in stereometry. Since 1 problem in planimetry is less valuable than 2 problems in stereometry, he should specialize in stereometry problems, earning $12 * 6 = 72$ points for the team. Zhanna, instead of solving 1 problem in stereometry, can solve 6 problems in planimetry. Since 1 problem in stereometry is less valuable than 6 problems in planimetry, she should specialize in planimetry problems, earning $7 * 18 = 126$ points for the team. In total, the team of Andrey, Volodya, and Zhanna can earn a maximum of 84 + 72 + 126 = 282 points. Let's find out what the maximum result the team of Andrey, Petya, and Galina could achieve. Petya, instead of solving 1 problem in planimetry, can solve 4 problems in stereometry. Since 1 problem in planimetry is less valuable than 4 problems in stereometry, he should specialize in stereometry problems, earning $12 * 12 = 144$ points for the team. Galina, instead of solving 1 problem in stereometry, can solve 2 problems in planimetry. Since 1 problem in stereometry is less valuable than 2 problems in planimetry, she should specialize in planimetry problems, earning $7 * 14 = 98$ points for the team. In total, the team of Andrey, Petya, and Galina can earn a maximum of $84 + 144 + 98 = 326$ points. Thus, Andrey should invite Petya and Galina to his team. ## Grading Criteria: To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Volodya, Zhanna, Petya, and Galina. A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points. A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Andrey will invite to the team) - 2 points. A comparison of the benefits of each pair of participants and the correct answer - 2 points. Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point. Any arithmetic error that led to a significant distortion of the results is penalized by 5 points. ## Variant 2 A school is holding the remote stage of a team tournament in physics, where the results of the participants are evaluated based on the number of points earned for a fully solved problem. A fully solved problem in kinematics is worth 14 points, and a problem in thermodynamics is worth 24 points. The team that scores the most points wins the tournament. Volodya is forming his team of 3 people, in which he will be the captain. He is considering whether to invite Andrey and Tatyana, or Semyon and Maria. Therefore, he asked all the candidates to honestly indicate in the table their capabilities for solving problems during the allocated time for this stage of the tournament. It is known that the opportunity costs of each student for solving problems in kinematics and thermodynamics are always constant. | Name | Maximum number of kinematics problems if solving only them | Maximum number of thermodynamics problems if solving only them | | :--- | :---: | :---: | | Volodya | 7 | 7 | | Semyon | 4 | 8 | | Maria | 8 | 2 | | Andrey | 2 | 12 | | Tatyana | 2 | 1 | Help Volodya decide which pair of students to invite to his team if the team's only goal is to win the tournament. ## Solution: Let's find out what the maximum result the team of Volodya, Semyon, and Maria could achieve. Volodya, instead of solving 1 problem in thermodynamics, can solve 1 problem in kinematics. Since a problem in thermodynamics is more valuable, he should specialize in thermodynamics problems, earning $24 * 7 = 168$ points for the team. Semyon, instead of solving 1 problem in kinematics, can solve 2 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 2 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 8 = 192$ points for the team. Maria, instead of solving 1 problem in thermodynamics, can solve 4 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 4 problems in kinematics, she should specialize in kinematics problems, earning $14 * 8 = 112$ points for the team. In total, the team of Volodya, Semyon, and Maria can earn a maximum of $168 + 192 + 112 = 472$ points. Let's find out what the maximum result the team of Volodya, Andrey, and Tatyana could achieve. Andrey, instead of solving 1 problem in kinematics, can solve 6 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 6 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 12 = 288$ points" d3911ff65a04,"A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square? [asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2)); draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H);[/asy] $\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad \textbf{(E)}\ 2 - \sqrt{2}$",\mathrm{(A),medium,"Let's assume that the side length of the octagon is $x$. The area of the center square is just $x^2$. The triangles are all $45-45-90$ triangles, with a side length ratio of $1:1:\sqrt{2}$. The area of each of the $4$ identical triangles is $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$, so the total area of all of the triangles is also $x^2$. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is $x$ and the other side length is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$, so the area of all of the rectangles is $2x^2\sqrt{2}$. The ratio of the area of the square to the area of the octagon is $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$. Cancelling $x^2$ from the fraction, the ratio becomes $\dfrac{1}{2\sqrt2+2}$. Multiplying the numerator and the denominator each by $2\sqrt{2}-2$ will cancel out the radical, so the fraction is now $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$" f861a2767919,"## 133. Math Puzzle $6 / 76$ A catch device that secures the hoist basket in a shaft fails in at most one out of 1000 operational cases. Another independent safety device fails in at most one out of 100 cases where it is called upon. What is the probability that the occupants will be saved by the safety devices if the hoist system fails?",1)$ that the safety system will function.,easy,"The probability that the two (independent) safety devices of the hoist basket fail simultaneously is, according to the and-rule: $$ P_{\text {Failure }}=P_{1} \cdot P_{2}=\frac{1}{1000} \cdot \frac{1}{100}=10^{-5}=0.001 \% $$ Thus, the probability that the safety system functions is $$ P=1-P_{\text {Failure }}=1-10^{-5}=0.99999=99.999 \% $$ This means that it can almost be considered certain $(P=1)$ that the safety system will function." 305006475d94,"6. In the complex plane, the number of intersection points between the curve $z^{4}+z=1$ and the circle $|z|$ $=1$ is ( ). (A) 0 (B) 1 (C) 2 (D) 3","\frac{\pi}{3}$ or $\frac{5 \pi}{3}$ do not satisfy $1+\cos 3 \theta=\cos \theta$, so the two curves ",medium,"6.A. Since $|z|=1$, we can let $z=\cos \theta+\mathrm{i} \sin \theta(0 \leqslant \theta<2 \pi)$. From $z^{3}+1=\bar{z}$ $$ \begin{array}{l} \Rightarrow 1+\cos 3 \theta+\mathrm{i} \sin 3 \theta=\cos \theta-\mathrm{i} \sin \theta \\ \Rightarrow \cos 3 \theta=\cos \theta-1, \sin 3 \theta=-\sin \theta \\ \Rightarrow(\cos \theta-1)^{2}+(-\sin \theta)^{2}=\cos ^{2} 3 \theta+\sin ^{2} 3 \theta=1 \\ \Rightarrow(\cos \theta-1)^{2}+(-\sin \theta)^{2}=1 \Rightarrow 2 \cos \theta=1 \\ \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \text { or } \frac{5 \pi}{3} . \end{array} $$ We also know that $\theta=\frac{\pi}{3}$ or $\frac{5 \pi}{3}$ do not satisfy $1+\cos 3 \theta=\cos \theta$, so the two curves have no intersection points." 0bcc68056e5b,"A truck has had new brake linings fitted to all four of its wheels. A lining is considered completely worn out if it has traveled $15000 \mathrm{~km}$ on a rear wheel, or $25000 \mathrm{~km}$ on a front wheel. How far can the truck travel until all four linings are completely worn out, if at the appropriate time the front lining pair is swapped with the rear pair?","the question of the problem, it is not necessary to construct $B$ and $D$, but to carry out the tire swap, it is precisely this that needs to be known, at which position of the odometer the tire swap should take place",medium,"I. solution. Consider the state of the tires after, for example, $1000 \mathrm{~km}$ of travel. The rear tires have worn out by $11/5$ of their part, and the front tires by $1/25$ of their part. (We assume that the car travels on an average quality road, and the wear is directly proportional to the distance traveled.) If we now swap the tires, then after another $1000 \mathrm{~km}$ of travel, each of the four tires will have traveled an equal distance and can be considered worn out by $1/15 + 1/25 = 8/75$ of their part. The tires can travel $2000 \mathrm{~km}$ more times the ratio of the total wear to the wear that has already occurred, which is $1$ to the $8/75$ wear ratio. Therefore, the car can travel $2000 \cdot 75 / 8 = 18750 \mathrm{~km}$ before the tires are completely worn out, with half of this distance covered in the original arrangement and the other half with the swapped tires. László Jereb (Sopron, Széchenyi I. Gymnasium II. grade) Note. Similarly, it can be shown that if a tire wears out after $a \mathrm{~km}$ on one wheel and $b \mathrm{~km}$ on the other, then with the above swap, the distance that can be traveled with four tires is $$ \frac{1}{\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)} $$ This value is known as the harmonic mean of the two numbers. II. solution. The ratio of the allowed kilometers is $15000: 25000 = 3: 5$, so if a tire is moved from the rear wheel to the front wheel, the distance it can still travel increases by a factor of $5/3$. If it is moved from the front to the rear, it decreases by a factor of $3/5$. If the car has already traveled $x$ km with the tires, then the front tires could still travel $25000 - x$ km, but after the swap, they can only travel $\frac{3}{5}(25000 - x)$ km, while the rear tires, instead of the remaining $15000 - x$ km, can still be used for $\frac{5}{3}(15000 - x)$ km. We need to determine $x$ such that the distance the two types of tires can still travel is equal: $$ \frac{3}{5}(25000 - x) = \frac{5}{3}(15000 - x) $$ From this, $x = 9375 \mathrm{~km}$, and the distance traveled after the swap is $3 \cdot 15625 / 5 = 9375 \mathrm{~km}$, so the car can travel $18750 \mathrm{~km}$. Béla Pénzes (Esztergom, Temesvári Pelbárt Gymnasium II. grade) Note. We can also use the fact that the distance traveled before and after the swap is the same, otherwise, by the time the rear tire is completely worn out, the other tire would still be usable. This leads to the somewhat simpler equation $3(25000 - x) / 5 = x$. ![](https://cdn.mathpix.com/cropped/2024_05_02_c7e6978c848383369babg-1.jpg?height=686&width=900&top_left_y=1473&top_left_x=602) III. solution. Viewing a new tire as $100 \%$ and a completely worn-out tire as $0 \%$, we will illustrate how the value state of the tires changes with the distance traveled. The value graph for the two types of tires from the initial setup is obtained by connecting the point $A$ on the value axis marked $100 \%$ with a straight line to the point on the distance axis marked $15000 \mathrm{~km}$ or $25000 \mathrm{~km}$. When the tires are swapped, the slopes of their graphs also swap, and the four graph segments corresponding to the two distance segments form a parallelogram, which we want to construct. The two types of tires reach the $0 \%$ value state after traveling the same distance, so the two graphs intersect on the distance axis, which will be the point $C$ opposite $A$ in the parallelogram. The other two vertices $B$ and $D$ correspond to the tire swap, so the two vertices are above the same point on the distance axis, meaning we know the direction of the $BD$ diagonal: it is parallel to the value axis. From this, we can determine the position of the other diagonal. The segments of the lines parallel to the $BD$ diagonal that lie in the $BAD$ angle region are bisected by the $AC$ diagonal. Thus, taking any line parallel to the value axis, the segment $B^*D^*$ between the two graphs that is bisected by $F^*$ is on the $AC$ diagonal, so $AF^*$ intersects the distance axis at $C$, and the corresponding abscissa value is the sought distance. Notes. 1. To answer the question of the problem, it is not necessary to construct $B$ and $D$, but to carry out the tire swap, it is precisely this that needs to be known, at which position of the odometer the tire swap should take place. 2. This solution is essentially a graphical version of the first solution. The similarity would be even more evident if we completed the $D^*AB^*$ triangle to a parallelogram and drew the diagonal through the point $C^*$, which would correspond to the diagonal. 3. The two diagonals bisect each other, so their intersection point is on a line parallel to the distance axis at the height corresponding to $50 \%$. If we reflect one wheel's value graph over this line, the reflection intersects the other graph at the point corresponding to the tire swap. We can also arrive at this directly by reasoning, since after the swap, one tir" bf4d258c43c2,"Solve the following inequality on the set of real numbers: $$ \sqrt{x+6}>x-6 . $$","4$, i.e., $x=10$ is valid.",medium,"Let's plot the functions $$ f(x)=\sqrt{x+6} \quad(\text { Domain: } x \geq-6) \quad \text { and } \quad g(x)=x-6 \quad(\text { Domain: } x \in \mathbf{R}) $$ in the same coordinate system. ![](https://cdn.mathpix.com/cropped/2024_05_02_f8b7dd0c4231739d08aag-1.jpg?height=623&width=634&top_left_y=304&top_left_x=735) The elements of the solution set of the inequality can only be chosen from real numbers not less than -6. $f(-6)=0, g(-6)=-12$, so the inequality holds at $x=-6$. Both functions are strictly increasing, but except for an initial small segment, $f(x)$ grows more slowly than $g(x)$. Let's calculate where $g(x)$ ""catches up"" with $f(x)$. The $$ \sqrt{x+6}=x-6 \quad(x \geq-6) $$ equation is traditionally solved by squaring both sides, which leads to a consequential equation. The $$ x+6=(x-6)^{2} $$ equation will contain the root(s) of the original equation, but false roots may also appear. (Squaring is not an equivalent transformation.) The resulting $$ x^{2}-13 x+30=0 $$ quadratic equation has roots: $x_{1}=10, x_{2}=3$. The latter is a false root. The original equation is satisfied by $x_{1}=10$, but not by $x_{2}=3$. Based on all this, the solution to the inequality is: $$ -6 \leq x<10 $$ Hegedús Dalma (Siófok, Perczel Mór Gymn., I. o.t.) Note. Many fell into the trap because they did not check the roots, so they did not realize that 3 is a false root. Perhaps the false root is more easily revealed if the $\sqrt{x+6}=x-6$ equation is written in the form $$ (x+6)-\sqrt{x-6}-12=0 $$ This is a quadratic equation in $\sqrt{x+6}$. According to the quadratic formula, $$ \sqrt{x+6}=\frac{1 \pm \sqrt{1+48}}{2}=\left\{\begin{array}{c} 4 \\ -3 \end{array}\right. $$ Since $\sqrt{x+6} \geq 0$, only $\sqrt{x+6}=4$, i.e., $x=10$ is valid." 2d38fc6cb43f,"2. In the Cartesian coordinate system, the coordinates of the three vertices of the right triangle $Rt \triangle OAB$ are $O(0,0), A(1,0)$, and $B(1,2)$. Take any point $C(x, 2x) (0 0, x_{2} > 0$, the expression in the square brackets is clearly greater ",medium,"(1) Proof by contradiction. Assume that the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ are on the same branch, such as $C_{1}$, with coordinates $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right), \left(x_{3}, y_{3}\right)$, respectively. Without loss of generality, assume $0 < y_{1} < y_{2} < y_{3} < 0$. Thus, $$ \begin{aligned} & |P Q|^{2}+|Q R|^{2}-|P R|^{2} \\ = & {\left[\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}-\left(x_{1}-x_{3}\right)^{2}\right] } \\ & +\left[\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}-\left(y_{1}-y_{3}\right)^{2}\right] \\ = & \left(2 x_{2}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}+2 x_{1} x_{3}\right) \\ & +\left(2 y_{2}^{2}-2 y_{1} y_{2}-2 y_{2} y_{3}+2 y_{1} y_{3}\right) \\ = & 2\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)+2\left(y_{2}-y_{1}\right)\left(y_{2}-y_{3}\right) < 0. \end{aligned} $$ Since $x_{1} x_{2} x_{3} = 1$ and $x_{1} x_{2} x_{3} > 0, x_{2} > 0$, the expression in the square brackets is clearly greater than 0, so $1 - x_{1} x_{2} = 0$. Hence, $x_{1} x_{2} = 1$. Therefore, the coordinates of $Q$ are $\left(\frac{1}{x_{2}}, x_{2}\right)$, and the coordinates of $R$ are $\left(x_{2}, \frac{1}{x_{2}}\right)$, which means $Q$ and $R$ are symmetric with respect to the line $y = x$. The lines $PQ$ and $PR$ are two mutually symmetric lines passing through point $P$ and forming a $30^{\circ}$ angle with $y = x$. It is easy to see that their slopes are $75^{\circ}$ and $15^{\circ}$, respectively. Without loss of generality, assume the slope of $PQ$ is $75^{\circ}$, so its equation is $y + 1 = \tan 75^{\circ} \cdot (x + 1)$. That is, $y + 1 = (2 + \sqrt{3})(x + 1)$. Substituting this into the hyperbola equation $xy = 1$, we find that the coordinates of $Q$ are $(2 - \sqrt{3}, 2 + \sqrt{3})$, and by symmetry, the coordinates of $R$ are $(2 + \sqrt{3}, 2 - \sqrt{3})$." 410086a577e4,"2. A production team in a factory is tasked with producing a batch of parts. When each worker works at their original position, it takes 9 hours to complete the production task. If the positions of workers A and B are swapped, with the production efficiency of other workers remaining unchanged, the task can be completed one hour earlier; if the positions of workers C and D are swapped, with the production efficiency of other workers remaining unchanged, the task can also be completed one hour earlier. Question: If the positions of A and B, as well as C and D, are swapped simultaneously, with the production efficiency of other workers remaining unchanged, how many minutes earlier can the production task be completed?","10$ (parts) per hour, meaning each part takes $\frac{60}{10}=6$ (minutes) to complete. Therefore, fo",medium,"2.【Solution】Divide the total task into 72 parts. Originally, $\overline{9}=8$ parts were completed per hour, and each part took $\overline{8}=$ 7.5 minutes. After A and B are swapped, $\frac{72}{8}=9$ parts are completed per hour, which is 1 part more than before. Since the efficiency of other workers remains unchanged, this additional part is the extra work done by A and B. Similarly, after C and D are swapped, they also complete 1 additional part per hour. After both swaps, A and B, as well as C and D, each complete 1 additional part per hour, so the entire group completes $8+1+1=10$ (parts) per hour, meaning each part takes $\frac{60}{10}=6$ (minutes) to complete. Therefore, for each part of the task, the time is reduced by $7.5-6=1.5$ (minutes). For 72 parts, the total time saved is $72 \times 1.5=108$ (minutes)." 178ad563b39f,"In base-$2$ notation, digits are $0$ and $1$ only and the places go up in powers of $-2$. For example, $11011$ stands for $(-2)^4+(-2)^3+(-2)^1+(-2)^0$ and equals number $7$ in base $10$. If the decimal number $2019$ is expressed in base $-2$ how many non-zero digits does it contain ?",6,medium,"To convert the decimal number \(2019\) into base \(-2\), we need to follow a systematic approach. Let's break down the steps: 1. **Identify the largest power of \(-2\) less than or equal to \(2019\):** \[ (-2)^{12} = 4096 \] Since \(4096\) is greater than \(2019\), we use the next lower power: \[ (-2)^{11} = -2048 \] Since \(-2048\) is negative and we need a positive number, we use: \[ (-2)^{11} + (-2)^{12} = 2048 - 4096 = -2048 \] This is not helpful, so we use: \[ 2^{11} = 2048 \] 2. **Subtract \(2048\) from \(2019\):** \[ 2019 - 2048 = -29 \] 3. **Identify the largest power of \(-2\) less than or equal to \(-29\):** \[ (-2)^5 = -32 \] Since \(-32\) is less than \(-29\), we use: \[ -29 + 32 = 3 \] 4. **Identify the largest power of \(-2\) less than or equal to \(3\):** \[ (-2)^2 = 4 \] Since \(4\) is greater than \(3\), we use the next lower power: \[ (-2)^1 = -2 \] Since \(-2\) is less than \(3\), we use: \[ 3 + 2 = 5 \] 5. **Identify the largest power of \(-2\) less than or equal to \(5\):** \[ (-2)^2 = 4 \] Since \(4\) is less than \(5\), we use: \[ 5 - 4 = 1 \] 6. **Identify the largest power of \(-2\) less than or equal to \(1\):** \[ (-2)^0 = 1 \] Since \(1\) is equal to \(1\), we use: \[ 1 - 1 = 0 \] 7. **Combine all the powers of \(-2\) used:** \[ 2019 = 2^{11} - 2^5 + 2^2 + 2^1 + 2^0 \] 8. **Convert these powers into base \(-2\):** \[ 2019 = (-2)^{12} + (-2)^{11} + (-2)^5 + (-2)^2 + (-2)^1 + (-2)^0 \] 9. **Write the number in base \(-2\):** \[ 2019_{10} = 1100000100111_{-2} \] 10. **Count the non-zero digits:** The number \(1100000100111\) has \(6\) non-zero digits. The final answer is \(\boxed{6}\)." 8526ec4e005e,"3.48. The height of the cone is $H$, the angle between the slant height and the height is $\alpha$. A smaller cone is inscribed in this cone such that the vertex of the smaller cone coincides with the center of the base of the larger cone, and the corresponding slant heights of both cones are perpendicular to each other. Find the volume of the inscribed cone.",See reasoning trace,medium,"3.48. According to the condition, $S O=H, O C \perp A S$, $O D \perp B S, \angle O S B=\alpha$ (Fig. 3.50); it is required to find the volume of the cone $O D C$. Obviously, $\angle O D O_{1}=\angle O S B$ as acute angles with mutually perpendicular sides. From $\triangle S O D$ we find $O D=H \sin \alpha$; further, from $\triangle O O_{1} D$ we get $$ \begin{gathered} O O_{1}=O D \sin \alpha=H \sin ^{2} \alpha \\ O_{1} D=O D \cos \alpha=H \sin \alpha \cos \alpha \end{gathered} $$ Finally, we have $$ \begin{gathered} V=\frac{1}{3} \pi O_{1} D^{2} \cdot O O_{1}=\frac{1}{3} \pi H^{2} \sin ^{2} \alpha \cos ^{2} \alpha \cdot H \sin ^{2} \alpha= \\ =\frac{1}{3} \pi H^{3} \sin ^{4} \alpha \cos ^{2} \alpha \end{gathered} $$" de7534ed673b,"3. In the afternoon, 5 classes need to be scheduled: Physics, Chemistry, Biology, and two self-study periods. If the first class cannot be Biology and the last class cannot be Physics, then the number of different scheduling methods is ( ) kinds. (A) 36 (B) 39 (C) 60 (D) 78",39$ (ways).,easy,"3. B. The total number of unrestricted arrangements is $\frac{A_{5}^{5}}{2}=60$; the number of arrangements with biology in the first period or physics in the last period is $\frac{\mathrm{A}_{4}^{4}}{2}=12$ each, and among these, the number of arrangements with biology in the first period and physics in the last period is $\frac{A_{3}^{3}}{2}=3$. Therefore, the number of arrangements that meet the requirements is $60-12 \times 2+3=39$ (ways)." 873f3cbf9a70,"7. The equation $\left\{\begin{array}{l}x=\frac{1}{2}\left(e^{t}+e^{-t}\right) \cos \theta, \\ y=\frac{1}{2}\left(e^{t}-e^{-t}\right) \sin \theta\end{array}\right.$ where $\theta$ is a constant $\left(\theta \neq \frac{n}{2} \pi, n \in \mathbf{Z}\right), t$ is a parameter, then the shape of the figure it represents is . $\qquad$",1$.,easy,"7. Hyperbola Hint, converting the equation to a standard form is $\frac{x^{2}}{\cos ^{2} \theta}-\frac{y^{2}}{\sin ^{2} \theta}=1$." caa2c2889811,"For example, in the sequence $x_{1}, x_{2}, \cdots, x_{n}, \cdots$, the sum of any three consecutive terms is 20, and $x_{1}=9, x_{12}=7$. Find the value of $x_{2000}$.","x_{n}(n \geqslant 1)$. Therefore, $\left\{x_{n}\right\}$ is a periodic sequence with a period of 3, ",easy,"Given that $x_{n+1}+x_{n+2}+x_{n+3}=20$ $$ x_{n}+x_{n+1}+x_{n+2}=20 $$ Subtracting the two equations, we get $x_{n+3}=x_{n}(n \geqslant 1)$. Therefore, $\left\{x_{n}\right\}$ is a periodic sequence with a period of 3, so $x_{4}=x_{1}=9, x_{12}=$ $x_{3}=7$. Since $x_{1}+x_{2}+x_{3}=20$, we have $x_{2}=4$. Hence, $x_{2000}=x_{2}=4$" 0895280ab3c8,"Someone draws a 2400 franc annuity 8 times at the end of every third year, what is the present value of the annuity at $4 \%$?",See reasoning trace,easy,"The current value of the annuity $$ t=\frac{2400}{e^{3}}+\frac{2400}{e^{6}}+\ldots+\frac{2400}{e^{24}} $$ which simplifies to $$ t=\frac{2400}{e^{24}} \cdot \frac{e^{24-1}}{e^{3}-1} $$ Upon calculation, $t=11723$ forints. (Miksa Roth.) The problem was solved by: Barna D., Beck F., Bojedain F., Détshy K., Fekete J., Kallos M., Schiffer H., Szabó I., Weisz J." fb43a4dd9d9c,#,. $(\sqrt{6} - 1)$,medium,"# Solution. When ten identical spheres are arranged in this way, the centers $A, B, C, D$ of four of them are located at the vertices of a regular tetrahedron, and the points of contact are located on the edges of this tetrahedron. Therefore, the edge of the tetrahedron is equal to four radii of these spheres, the radius of the outer sphere is greater than the radius of the sphere circumscribed around the tetrahedron by a quarter of the length of the tetrahedron's edge, and the radius of the inner sphere is smaller than the distance from the center of the tetrahedron to its face by the same amount. Consider the section of the tetrahedron by the plane $A B M$ (Fig. 3). ![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-23.jpg?height=651&width=662&top_left_y=1907&top_left_x=366) Fig. 2 ![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-23.jpg?height=691&width=759&top_left_y=1867&top_left_x=1114) Fig. 3 Let the length of the edge of the tetrahedron be $a$, the radius of the sphere circumscribed around the pyramid of spheres be $R$, and the radius of the sphere inscribed in the center of the pyramid of spheres be $r$. In triangle $\triangle A B M: A M = B M = \frac{a \sqrt{3}}{2} ; M E = M H = \frac{1}{3} A M = \frac{a \sqrt{3}}{6}$; $A H = B E = \frac{2}{3} A M = \frac{a \sqrt{3}}{3}$ $A E = B H = \sqrt{A M^{2} - M E^{2}} = \frac{2 a}{\sqrt{6}} = \frac{a \sqrt{6}}{3}$. From the similarity of triangles $\triangle A E M$ and $\triangle A H O$ we have $\frac{A O}{A M} = \frac{A H}{A E} = \frac{a \sqrt{3}}{a \sqrt{6}} = \frac{\sqrt{2}}{2}$; $A O = B O = \frac{\sqrt{2}}{2} A M = \frac{a \sqrt{6}}{4}$. In triangle $\triangle A B O: S_{A B O} = \frac{A H \cdot B O}{2} = \frac{A B \cdot F O}{2} \Rightarrow F O = \frac{A H \cdot B O}{A B} = \frac{a^{2} \sqrt{18}}{12 a} = \frac{a \sqrt{2}}{4}$. Then $R = A O + A L = \frac{a \sqrt{6}}{4} + \frac{a}{4} = \frac{a(\sqrt{6} + 1)}{4} ;$ and $r = F O - F K = \frac{a \sqrt{2}}{4} - \frac{a}{4} = \frac{a(\sqrt{2} - 1)}{4}$. $$ \frac{r}{R} = \frac{(\sqrt{2} - 1)}{(\sqrt{6} + 1)} = \frac{(\sqrt{6} - 1)(\sqrt{2} - 1)^{4}}{5} ; r = \frac{(\sqrt{6} - 1)(\sqrt{2} - 1)}{5} R = (\sqrt{6} - 1) $$ Answer. $(\sqrt{6} - 1)$. ## Variant-31" 335215189dce,"## Task 2 Calculate the sum and the difference of 56 and 23.",79 ; \quad 56-23=33$,easy,$56+23=79 ; \quad 56-23=33$ 691104dffb28,"1643. Given two independent random variables: $X$ - the number of heads appearing in 2 tosses of a five-kopeck coin, and $Y$ - the number of points that appear when rolling a die. Find the distribution law of the difference of these two random variables $X-Y$, the mathematical expectations and variances of the random variables $X, Y, X-Y$, and verify that $E(X-Y)=E X-E Y$ and $D(X-Y)=D X+D Y$.",D X+D Y$. It also holds: $3.417=0.5+2.917=3.417$.,medium,"Solution. Since the random variable $X$ takes the value $m$ with probability $P_{m 2}=C_{2}^{m} p^{m} q^{2-m}$ (Bernoulli's formula), where $m=0,1,2$ and $p=0.5$, the distribution table of this random variable is | $x$ | 0 | 1 | 2 | | :---: | :---: | :---: | :---: | | $p(x)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ | From this, $$ E X=0 \cdot \frac{1}{4}+1 \cdot \frac{1}{2}+2 \cdot \frac{1}{4}=1 $$ $D X=E(X-E X)^{2}=E(X-1)^{2}=(0-1)^{2} \cdot \frac{1}{4}+(1-1)^{2} \cdot \frac{1}{2}+$ $+(2-1)^{2} \cdot \frac{1}{4}=0.5$ The distribution table of the random variable $Y$ is: | $y$ | 1 | 2 | 3 | 4 | 5 | 6 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $p(y)$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | Then $$ \begin{aligned} & E Y=1 \cdot \frac{1}{6}+2 \cdot \frac{1}{6}+3 \cdot \frac{1}{6}+4 \cdot \frac{1}{6}+5 \cdot \frac{1}{6}+6 \cdot \frac{1}{6}=3.500 \\ & D Y=(1-3.5)^{2} \cdot \frac{1}{6}+(2-3.5)^{2} \cdot \frac{1}{6}+(3-3.5)^{2} \cdot \frac{1}{6}+(4- \\ & \quad-3.5)^{2} \cdot \frac{1}{6}+(5-3.5)^{2} \cdot \frac{1}{6}+(6-3.5)^{2} \cdot \frac{1}{6}=2.917 \end{aligned} $$ To obtain the distribution of $X-Y$, we will construct the following table, which will give the various possible cases: $X=$ $=x_{i}, Y=y_{j}$, the probabilities of these cases: $P\left(X=x_{i}, Y=y_{j}\right)=$ $=P\left(X=x_{i}\right) P\left(Y=y_{j}\right)$ (the random variables are independent) and the values of $X-Y$ in each of them: | Various possible cases | $X=x_{i}$ | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | | $Y=y_{j}$ | 1 | 2 | 3 | 4 | 5 | 6 | 1 | 2 | 3 | | Probabilities of these cases | $P\left(X=x_{i}, Y=y_{j}\right)$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | | Values of $X-Y$ | $x_{i}-y_{j}$ | -1 | -2 | -3 | -4 | -5 | -6 | 0 | -1 | -2 | Continuation | Various possible cases | $X=x_{i}$ | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 2 | 2 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | | $Y=y_{j}$ | 4 | 5 | 6 | 1 | 2 | 3 | 4 | 5 | 6 | | Probabilities of these cases | $P\left(X=x_{i}, Y=y_{j}\right)$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | $\frac{1}{24}$ | | Values of $X-Y$ | $x_{l}-y_{j}$ | -3 | -4 | -5 | 1 | 0 | -1 | -2 | -3 | -4 | We notice that $X-Y$ sometimes takes the same value in different cases. Using the addition theorem of probabilities, we combine these cases into one (the random variable $X-Y$ takes a given value). Then we get the following distribution table of the random variable $X-Y$: | -6 | -5 | -4 | -3 | | :---: | :---: | :---: | :---: | :---: | | $\frac{1}{24}$ | $\frac{1}{24}+\frac{1}{12}$ | $\frac{1}{24}+\frac{1}{24}+\frac{1}{12}$ | $\frac{1}{24}+\frac{1}{12}+\frac{1}{24}$ | | | | | | | -2 | -1 | 0 | 1 | | $\frac{1}{24}+\frac{1}{12}+\frac{1}{24}$ | $\frac{1}{24}+\frac{1}{12}+\frac{1}{24}$ | $\frac{1}{12}+\frac{1}{24}$ | $\frac{1}{24}$ | or | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\frac{1}{24}$ | $\frac{1}{8}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{8}$ | $\frac{1}{24}$ | From this, $$ \begin{aligned} E(X-Y)= & -6 \cdot \frac{1}{24}+(-5) \cdot \frac{1}{8}-4 \cdot \frac{1}{6}-3 \cdot \frac{1}{6}-2 \cdot \frac{1}{6}- \\ & -1 \cdot \frac{1}{6}+0 \cdot \frac{1}{8}+1 \cdot \frac{1}{24}=-2.5 \end{aligned} $$ We check property 3 of the mathematical expectation: $E(X-Y)=$ $=E X-E Y$. It holds: $-2.5=1-3.5=-2.5$. Then $$ \begin{aligned} & D(X-Y)=(-6+2.5)^{2} \cdot \frac{1}{24}+(-5+2.5)^{2} \cdot \frac{1}{8}+(-4+ \\ & +2.5)^{2} \frac{1}{6}+(-3+2.5)^{2} \cdot \frac{1}{6}+(-2+2.5)^{2} \cdot \frac{1}{6}+(-1+ \\ & \quad+2.5)^{2} \cdot \frac{1}{6}+(0+2.5)^{2} \cdot \frac{1}{8}+(1+2.5)^{2} \cdot \frac{1}{24}=3.417 . \end{aligned} $$ We check property 3 of the variance: $D(X-Y)=D X+D Y$. It also holds: $3.417=0.5+2.917=3.417$." d955270d03aa,"17. Last year's match at Wimbledon between John Isner and Nicolas Malut, which lasted 11 hours and 5 minutes, set a record for the longest match in tennis history. The fifth set of the match lasted 8 hours and 11 minutes. Approximately what fraction of the whole match was taken up by the fifth set? A $\frac{1}{5}$ B $\frac{2}{5}$ C $\frac{3}{5}$ D $\frac{3}{4}$ E $\frac{9}{10}$",491$ minutes and 11 hours and 5 minutes is $11 \times 60+5=665$ minutes. Therefore the exact fractio,medium,"Solution: D We have that 8 hours and 11 minutes is $8 \times 60+11=491$ minutes and 11 hours and 5 minutes is $11 \times 60+5=665$ minutes. Therefore the exact fraction taken by the last set is $\frac{491}{665}$. So we need to decide which of the given fractions this is closest to. This can be done in more than one way. For example, as 491 is just below 500 and 665 is almost $\frac{1}{3}$ rds of 2000 , we have that $\frac{491}{665} \sim \frac{500}{\frac{1}{3}(2000)}=\frac{1}{\frac{1}{3}(4)}=\frac{3}{4}$. (Here "" $\sim$ "" means ""is approximately equal to"".) Another method is to say that $\frac{491}{665} \sim \frac{490}{665}=\frac{98}{135}=\frac{14}{19} \sim \frac{15}{20}=\frac{3}{4}$." cc2a6ca03d9c,"2. In the set of real numbers, solve the equation $$ (5-2 x)^{2}+\left(\frac{1}{2-x}-1\right)^{2}=9 $$",z$ and obtain the equation $z^{2}+2 z-3=0$ whose solutions are 1 and -3. From $2 t-\frac{1}{t}=1$ we,medium,"Solution. Clearly, $x \neq 2$. We introduce the substitution $2-x=t$. Now $5-2 x=2 t+1$, so we sequentially obtain the equations $$ \begin{aligned} & (1+2 t)^{2}+\left(\frac{1}{t}-1\right)^{2}=9 \\ & 4 t^{2}+4 t+1+\frac{1}{t^{2}}-\frac{2}{t}+1=9 \end{aligned} $$ $$ \begin{aligned} & 4 t^{2}-4+\frac{1}{t^{2}}+4 t-\frac{2}{t}=3 \\ & \left(2 t-\frac{1}{t}\right)^{2}+2\left(2 t-\frac{1}{t}\right)=3 \end{aligned} $$ We introduce the substitution $2 t-\frac{1}{t}=z$ and obtain the equation $z^{2}+2 z-3=0$ whose solutions are 1 and -3. From $2 t-\frac{1}{t}=1$ we get $2 t^{2}-t-1=0$ and obtain the solutions $t_{1}=1$ and $t_{2}=-\frac{1}{2}$. From $2 t-\frac{1}{t}=-3$ we get $2 t^{2}+3 t-1=0$ and obtain the solutions $t_{3 / 4}=\frac{-3 \pm \sqrt{17}}{4}$. Finally, by substituting into $x=2-t$ we obtain the solutions $x_{1}=1$, $x_{2}=\frac{5}{2}$ and $x_{3 / 4}=\frac{11 \pm \sqrt{17}}{4}$." 3228e6ab296a,"4. Given real numbers $a$ and $b$ satisfy $ab = 2a + 2b - 3$. Then the minimum value of $a^{2} + b^{2}$ is ( ). (A) 1 (B) 2 (C) 3 (D) 4","2$, i.e., $y=1, a=b=1$, the equality holds.",easy,"4. B. Let $x=a+b, y=a b$. Then $y=2 x-3$. Therefore, $$ \begin{array}{l} a^{2}+b^{2}=(a+b)^{2}-2 a b \\ =x^{2}-2 y=x^{2}-2(2 x-3) \\ =x^{2}-4 x+6=(x-2)^{2}+2 \geqslant 2 . \end{array} $$ When $x=2$, i.e., $y=1, a=b=1$, the equality holds." f97fbb97ec96,"2. Let the real-coefficient quadratic equation $x^{2}+a x+2 b-$ $2=0$ have two distinct real roots, one of which lies in the interval $(0,1)$, and the other in the interval $(1,2)$. Then the range of $\frac{b-4}{a-1}$ is . $\qquad$",See reasoning trace,easy,"2. $\left(\frac{1}{2}, \frac{3}{2}\right)$. Solution: According to the problem, let the two distinct real roots be $x_{1}, x_{2}$, and $0 < x_{1} < 1 < x_{2} < 2$, then $1 < x_{1} + x_{2} = -a < 3, 0 < x_{1} x_{2} = 2b - 2 < 2$. Thus, $-3 < a < -1, 1 < b < 2$, which means $-\frac{1}{2} < \frac{1}{a-1} < -\frac{1}{4}, -3 < b-4 < -2$. Therefore, $\frac{1}{2} < \frac{b-4}{a-1} < \frac{3}{2}$, so the range of values is $\left(\frac{1}{2}, \frac{3}{2}\right)$." 3ff22b1ceac6,"A standard die is a cube whose faces are labelled 1, 2, 3, 4, 5, 6. Raffi rolls a standard die three times obtaining the results $p, q$ and $r$ on the first, second and third rolls, respectively. What is the probability that $p0$ $$ \Rightarrow a<-2-\sqrt{3} \text { or }-2+\sqrt{3}4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, the equation $12=k!+l!$ gives one more solution $k=l=3$." 2160a51080eb,"The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle? $\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}$",\textbf{(C),medium,"Let $a$ be the short side of the rectangle, and $b$ be the long side of the rectangle. The diagonal, therefore, is $\sqrt{a^2 + b^2}$. We can get the equation $\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}$. Cross-multiplying, we get $a\sqrt{a^2 + b^2} = b^2$. Squaring both sides of the equation, we get $a^2 (a^2 + b^2) = b^4$, which simplifies to $a^4 + a^2b^2 - b^4 = 0$. Solving for a quadratic in $a^2$, using the quadratic formula we get $a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}$ which gives us $\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}$. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is $\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}$. Solution by: vedadehhc" 928085d74dae,"4. Given the complex sequence $\left\{a_{n}\right\}$ satisfies: $a_{n+1}^{2}-a_{n} a_{n+1}+$ $a_{n}^{2}=0(n \geqslant 0)$, and $a_{n+1} \neq a_{n-1}(n \geqslant 1), a_{1}=1$, then $\sum_{n=0}^{2006} a_{n}$ is $\qquad$.",See reasoning trace,medium,"4. 2 . Solution: According to the problem, when $n \geqslant 1$, $a_{n+1}^{2}-a_{n} \cdot a_{n+1}+a_{n}^{2}=$ $0, a_{n-1}^{2}-a_{n} \cdot a_{n-1}+a_{n}^{2}=0$, and $a_{n+1} \neq a_{n-1}$, so $a_{n-1}$, $a_{n+1}$ are the two roots of the quadratic equation $x^{2}-a_{n} \cdot x+a_{n}^{2}=0$, Therefore, $a_{n-1}+a_{n+1}=a_{n}$, thus $a_{n+3}=a_{n+2}-a_{n+1}=$ $\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n}, a_{n+6}=-a_{n+3}=a_{n}$, hence, $$ \begin{array}{l} \sum_{n=0}^{2006} a_{n}=\sum_{n=0}^{2006}\left(a_{n+1}-a_{n+2}\right)=a_{1}-a_{2008}=a_{1}-a_{4}=a_{1} \\ -\left(-a_{1}\right)=2 a_{1}=2 . \end{array} $$" 1c168cfce829,5. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive semi-prime numbers that can be semi-prime?,5,medium,"5. Note that an odd semiprime number can only be the sum of two and an odd prime number. Let's show that three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assume the opposite. Then we get that the numbers $2n-1, 2n+1, 2n+3$ are prime, and all of them are greater than 3. But one of these three numbers is divisible by 3. This leads to a contradiction. Therefore, three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Note that among any six consecutive natural numbers, there are three consecutive odd numbers, so there cannot be more than five consecutive semiprime numbers. Five consecutive numbers can be semiprimes; for example, $30=17+13, 31=29+2, 32=19+13, 33=31+2, 34=23+11$. There are other examples as well. Answer: 5." 100484cbb8f2,"11. Given $\alpha, \beta \in\left[0, \frac{\pi}{4}\right]$. Then the maximum value of $\sin (\alpha-\beta)+$ $2 \sin (\alpha+\beta)$ is $\qquad$ .",See reasoning trace,medium,"11. $\sqrt{5}$. $$ \begin{array}{l} \text { Since } \sin (\alpha-\beta)+2 \sin (\alpha+\beta) \\ =3 \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta, \end{array} $$ By the Cauchy-Schwarz inequality, we have $$ \begin{array}{l} \text { the above expression } \leqslant \sqrt{(3 \sin \alpha)^{2}+\cos ^{2} \alpha} \cdot \sqrt{\cos ^{2} \beta+\sin ^{2} \beta} \\ =\sqrt{8 \sin ^{2} \alpha+1} \leqslant \sqrt{5} . \end{array} $$ When $\alpha=\frac{\pi}{4}, \beta=\arcsin \frac{1}{\sqrt{10}}$, $$ \sin (\alpha-\beta)+2 \sin (\alpha+\beta) $$ achieves the maximum value $\sqrt{5}$." 6b46977ea397,"7.1. Pasha and Sasha made three identical toy cars. Sasha made a third of the first car, a fifth of the second car, and a fifteenth of the third car. What part of the entire work did Sasha do?",- 0 points,easy,"Answer: one fifth Solution 1: In total, Sasha made $1 / 3+1 / 5+1 / 15=3 / 5$ of a car. This constitutes $3 / 5: 3=1 / 5$ of three cars. Solution 2: Divide each car into 15 parts. In total, there are 45 parts. Sasha made five parts of the first car, 3 parts of the second car, and 1 part of the third car. That is, a total of 9 parts out of 45, in other words, one fifth. Criteria: Only the answer - 0 points. Partial solution of the problem - 0 points." d134f0862e1c,"16. Ali is arranging the books on his bookshelves. He puts half his books on the bottom shelf and two-thirds of what remains on the second shelf. Finally he splits the rest of his books over the other two shelves so that the third shelf contains four more books than the top shelf. There are three books on the top shelf. How many books are on the bottom shelf? A 60 B 50 C 40 D 30 E 20","\frac{1}{3}$ of his books on the second shelf, leaving $\left(1-\frac{1}{2}-\frac{1}{3}\right)=\frac",medium,"16. D Since Ali places half his books on the bottom shelf and $\frac{2}{3}$ of the remainder on the second shelf, he places $\frac{2}{3} \times \frac{1}{2}=\frac{1}{3}$ of his books on the second shelf, leaving $\left(1-\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{6}$ of his books for the top two shelves. There are three books on the top shelf and four more, so seven books, on the third shelf. Therefore these 10 books represent $\frac{1}{6}$ of the total number of books on the bookshelves. Hence there are 60 books on the bookshelves and half of these, or 30 books, on the bottom shelf." 5be18f25eff0,"Daddy decided to give his son Mojmir a monthly allowance. Mojmir received his first allowance in January. Daddy increased the allowance by 4 Kč every month. If Mojmir didn't spend any, he would have 900 Kč after the twelfth allowance before Christmas. How many Kč did Mojmir receive for his first allowance in January? (L. Hozová)",See reasoning trace,easy,"Let's denote the amount of Mojmír's pocket money in January in Kč as $x$. In February, Mojmír received $x+4$, in March $x+8$, in April $x+12, \ldots$, in December $x+44$. According to the problem, we know that $$ 12 x+(4+8+12+16+20+24+28+32+36+40+44)=900 . $$ After simplification, we get: $$ \begin{aligned} 12 x+264 & =900, \\ 12 x & =636, \\ x & =53 . \end{aligned} $$ Mojmír received 53 Kč in January." f32c9fb6db90,"The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is: $\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none of these}$",\textbf{(E),medium,"Let $\triangle ABC$ be an isosceles triangle with $AB=AC=12,$ and $BC = 6$. Draw altitude $\overline{AD}$. Since $A$ is the apex of the triangle, the altitude $\overline{AD}$ is also a median of the triangle. Therefore, $BD=CD=3$. Using the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $AD=\sqrt{12^2-3^2}=3\sqrt{15}$. The area of $\triangle ABC$ is $\frac12\cdot 6\cdot 3\sqrt{15}=9\sqrt{15}$. If $a,b,c$ are the sides of a triangle, and $A$ is its area, the circumradius of that triangle is $\frac{abc}{4A}$. Using this formula, we find the circumradius of $\triangle ABC$ to be $\frac{6\cdot 12\cdot 12}{4\cdot 9\sqrt{15}}=\frac{2\cdot 12}{\sqrt{15}}=\frac{24\sqrt{15}}{15}=\frac{8\sqrt{15}}{5}$. The answer is $\boxed{\textbf{(E)}\ \text{none of these}}$" 285125fce985,"16. Chloe chose a three-digit integer with all its digits different and wrote it on lots of pieces of paper. Peter picked some of the pieces of paper and added the three-digit integers on them. His answer was 2331. How many pieces of paper did Peter pick? A 2331 B 21 C 9 D 7 E 3",3 \times 3 \times 7 \times 37$. Note also that $3 \times 37=111$. Since Claire's three-digit integer,medium,"16. C First note that the factorisation of 2331 into prime numbers is $2331=3 \times 3 \times 7 \times 37$. Note also that $3 \times 37=111$. Since Claire's three-digit integer has all its digits different, and both $3 \times 111=333$ and $7 \times 111=777$ have repeated digits, the factors of Claire's integer do not include both 3 and 37 . However, since $3 \times 3 \times 7=63$, which is not a three-digit integer, the factors of Claire's integer must include 37. Therefore Claire's integer is $37 \times 7=259$ and hence the number of pieces of paper Peter picked is $2331 \div 259=9$." cc43f4965875,"# A person was building a house. Only the roof was left to be done. The builder wanted to achieve the goal that raindrops falling on the roof would slide off it as quickly as possible. Determine the angle of inclination of the roof necessary to achieve this goal. Neglect the friction of the raindrops on the roof. ![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-1.jpg?height=477&width=825&top_left_y=1258&top_left_x=610) Answer: $\alpha=45^{\circ}$ #","1$, i.e., $\alpha=45^{\circ}$.",easy,"# Solution and Evaluation Criteria: Newton's second law for a rolling drop: $\quad m g \sin \alpha=m a$. Distance traveled by the drop: $s=\frac{1}{2} \frac{x}{\cos \alpha}$, where $x-$ is the width of the house. (4 points) Time of rolling: $t=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 \cdot \frac{1}{2} \frac{x}{\cos \alpha}}{g \sin \alpha}}=\sqrt{\frac{2 x}{g \sin 2 \alpha}}$. The rolling time will be minimal if $\sin 2 \alpha=1$, i.e., $\alpha=45^{\circ}$." 1dfeccf757cc,"Example 18 Let $a_{n}$ be the number of $n$-digit numbers, formed by $1,2,3,4,5$, that have at least 3 different digits, where 1 and 2 are not adjacent. Find $a_{n}$ and $a_{5}$.","5, x_{2}=23, x_{3}=4 x_{2}+3 x_{1}=107, x_{4}=4 x_{3}+3 x_{2}=497, x_{5}=4 x_{4}+3 x_{3}=$ $2309, y_",medium,"Let $x_{n}$ be the number of $n$-digit numbers formed by $1,2,3,4,5$ where 1 and 2 are not adjacent, $b_{n}$ be the number of such $n$-digit numbers ending in 1 or 2, $c_{n}$ be the number of such $n$-digit numbers not ending in 1 or 2, and $y_{n}$ be the number of $n$-digit numbers with at most two different digits. Then, $$ a_{n}=x_{n}-y_{n}, x_{n}=b_{n}+c_{n} . $$ The $b_{n}$ $n$-digit numbers ending in 1 or 2 and where 1 and 2 are not adjacent can be divided into two categories: those ending in 1 or 2 with the second last digit not being 1 or 2, which are $2 c_{n-1}$ in number, and those ending in 11 or 22, which are $b_{n-1}$ in number. Thus, $$ b_{n}=b_{n-1}+2 c_{n-1} \text {. } $$ The $c_{n}$ $n$-digit numbers ending in 3, 4, or 5 and where 1 and 2 are not adjacent have the first $n-1$ digits forming an $n-1$-digit number where 1 and 2 are not adjacent, so $$ c_{n}=3 x_{n-1} \text {. } $$ By eliminating $b_{n}$ and $c_{n}$ from (1), (2), and (3), we get $x_{n}-4 x_{n-1}-3 x_{n-2}=0$. The characteristic equation is $r^{2}-4 r-3=0$, with roots $r_{1,2}=2 \pm \sqrt{7}$, hence $$ x_{n}=k_{1}(2+\sqrt{7})^{n}+k_{2}(2-\sqrt{7})^{n} \text {. } $$ Define $x_{0}$ such that $x_{2}-4 x_{1}-3 x_{0}=0$. Given $x_{1}=5, x_{2}=2^{5}-2=23$, we get $x_{0}=1$, thus $$ \left\{\begin{array}{l} k_{1}+k_{2}=1, \\ k_{1}(2+\sqrt{7})+k_{2}(2-\sqrt{7})=5, \end{array}\right. $$ Solving, we get $k_{1}=\frac{3+\sqrt{7}}{2 \sqrt{7}}, k_{2}=-\frac{3-\sqrt{7}}{2 \sqrt{7}}$, so $$ x_{n}=\frac{3+\sqrt{7}}{2 \sqrt{7}}(2+\sqrt{7})^{n}-\frac{3-\sqrt{7}}{2 \sqrt{7}}(2-\sqrt{7})^{n}. $$ There are $C_{5}^{1}=5$ $n$-digit numbers with all digits the same, and $n$-digit numbers with exactly two different digits and 1 and 2 adjacent are $\left(C_{5}^{2}-1\right)\left(2^{n}-2\right)=9 \cdot 2^{n}-18$ in number, so $y_{n}=5+9 \cdot 2^{n}-18=9 \cdot 2^{n}-13$, thus $$ a_{n}=x_{n}-y_{n}=\frac{3+\sqrt{7}}{2 \sqrt{7}}(2+\sqrt{7})^{n}-\frac{3-\sqrt{7}}{2 \sqrt{7}}(2-\sqrt{7})^{n}-9 \cdot 2^{n}-13 . $$ Given $x_{1}=5, x_{2}=23, x_{3}=4 x_{2}+3 x_{1}=107, x_{4}=4 x_{3}+3 x_{2}=497, x_{5}=4 x_{4}+3 x_{3}=$ $2309, y_{5}=9 \cdot 2^{5}-13=275$, we have $a_{5}=x_{5}-y_{5}=2309-275=2034$." be58b46aae17,"6. $P$ is the midpoint of the height $V O$ of the regular quadrilateral pyramid $V-A B C D$. If the distance from point $P$ to the side face is 3, and the distance from point $P$ to the base is 5, then the volume of the regular quadrilateral pyramid is",\frac{1}{3} \cdot 15^{2} \cdot 10=750$.,easy,"$$ \begin{array}{l} \text { As shown, } V P=5, P H=3 \Rightarrow V H=4, \\ \triangle V P H \sim \triangle V E O \Rightarrow \frac{V H}{P H}=\frac{V O}{E O} \\ \Rightarrow O E=\frac{15}{2} \Rightarrow A B=15 . \end{array} $$ So $V_{V-A B C D}=\frac{1}{3} \cdot 15^{2} \cdot 10=750$." 1ccea09a9c1e,"## Calculate the definite integral: $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin x \, dx}{2+\sin x} $$",See reasoning trace,medium,"## Solution We will use the universal substitution: $$ t=\operatorname{tg} \frac{x}{2} $$ From which: $$ \begin{aligned} & \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\ & x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\ & x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1 \end{aligned} $$ Substitute: $$ \begin{aligned} & \int_{0}^{\frac{\pi}{2}} \frac{\sin x d x}{2+\sin x}=\int_{0}^{1} \frac{\frac{2 t}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{2+\frac{2 t}{1+t^{2}}}=\int_{0}^{1} \frac{4 t \cdot d t}{\left(1+t^{2}\right)^{2} \cdot \frac{2+2 t^{2}+2 t}{1+t^{2}}}= \\ & =\int_{0}^{1} \frac{2 t \cdot d t}{\left(1+t^{2}\right) \cdot\left(1+t+t^{2}\right)}= \end{aligned} $$ Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients: $$ \begin{aligned} & \frac{2 t}{\left(1+t^{2}\right) \cdot\left(1+t+t^{2}\right)}=\frac{A t+B}{t^{2}+1}+\frac{C t+D}{t^{2}+t+1}= \\ & =\frac{(A t+B)\left(t^{2}+t+1\right)+(C t+D)\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(t^{2}+t+1\right)}= \\ & =\frac{A t^{3}+A t^{2}+A t+B t^{2}+B t+B+C t^{3}+C t+D t^{2}+D}{\left(t^{2}+1\right)\left(t^{2}+t+1\right)}= \end{aligned} $$ $$ \begin{aligned} & =\frac{(A+C) t^{3}+(A+B+D) t^{2}+(A+B+C) t+(B+D)}{\left(t^{2}+1\right)\left(t^{2}+t+1\right)} \\ & \left\{\begin{array}{l} A+C=0 \\ A+B+D=0 \\ A+B+C=2 \\ B+D=0 \end{array}\right. \end{aligned} $$ Subtract the first equation from the third: $$ \begin{aligned} & \left\{\begin{array}{l} A+C=0 \\ A+B+D=0 \\ B=2 \\ B+D=0 \end{array}\right. \\ & \left\{\begin{array}{l} A+C=0 \\ A+D=-2 \\ B=2 \\ D=-2 \end{array}\right. \\ & \left\{\begin{array}{l} A+C=0 \\ A=0 \\ B=2 \\ D=-2 \end{array}\right. \\ & \left\{\begin{array}{l} C=0 \\ A=0 \\ B=2 \\ D=-2 \end{array}\right. \\ & \frac{2 t}{\left(1+t^{2}\right) \cdot\left(1+t+t^{2}\right)}=\frac{2}{t^{2}+1}-\frac{2}{t^{2}+t+1} \end{aligned} $$ Then: $$ \begin{aligned} & =\int_{0}^{1}\left(\frac{2}{t^{2}+1}-\frac{2}{t^{2}+t+1}\right) d t=\int_{0}^{1}\left(\frac{2}{t^{2}+1}-\frac{2 d\left(t+\frac{1}{2}\right)}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right)= \\ & =\left.2 \cdot \operatorname{arctg} t\right|_{0} ^{1}-\left.2 \cdot \frac{1}{\sqrt{\frac{3}{4}}} \operatorname{arctg} \frac{t+\frac{1}{2}}{\sqrt{\frac{3}{4}}}\right|_{0} ^{1}=\left.2 \cdot \operatorname{arctg} t\right|_{0} ^{1}-\left.\frac{4}{\sqrt{3}} \operatorname{arctg} \frac{2 t+1}{\sqrt{3}}\right|_{0} ^{1}= \\ & =2 \cdot \operatorname{arctg} 1-2 \cdot \operatorname{arctg} 0-\frac{4}{\sqrt{3}} \operatorname{arctg} \frac{2 \cdot 1+1}{\sqrt{3}}+\frac{4}{\sqrt{3}} \operatorname{arctg} \frac{2 \cdot 0+1}{\sqrt{3}}= \\ & =2 \cdot \frac{\pi}{4}-0-\frac{4}{\sqrt{3}} \operatorname{arctg} \sqrt{3}+\frac{4}{\sqrt{3}} \operatorname{arctg} \frac{1}{\sqrt{3}}= \end{aligned} $$ $$ =\frac{\pi}{2}-\frac{4}{\sqrt{3}} \cdot \frac{\pi}{3}+\frac{4}{\sqrt{3}} \cdot \frac{\pi}{6}=\frac{\pi}{2}-\frac{4 \pi}{3 \sqrt{3}}+\frac{2 \pi}{3 \sqrt{3}}=\frac{\pi}{2}-\frac{2 \pi}{3 \sqrt{3}} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals - Last edited: 19:53, 2 May 2009. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 8-30 ## Material from PlusPi" 2f96f5e1a4bc,"13.014. Find three numbers if the first is $80 \%$ of the second, the second is related to the third as $0.5: 9 / 20$, and the sum of the first and third is 70 more than the second number.","$80,100,90$",easy,"Solution. Let $x$ be the first number, $y$ be the second number, and $z$ be the third number. Then, according to the condition, $\left\{\begin{array}{l}x=0.8 y, \\ \frac{y}{z}=\frac{0.5}{9 / 20}, \\ x+z=y+70 .\end{array}\right.$ Solving the system, we get $x=80 ; y=100 ; z=90$. Answer: $80,100,90$." e6217ebfd19d,"5. (10 points) Among the divisors of a four-digit palindrome, exactly 3 are prime, and 39 are not prime. The value of the four-digit palindrome is",: 6336,medium,"6336 【Solution】According to the rule that the sum of the digits in the odd and even positions of a double palindrome is equal, it must be a multiple of 11, and the number of factors is $3+39=42$. A four-digit number contains 3 prime numbers, so 42 needs to be decomposed into the product of 3 numbers. $42=2 \times 3 \times 7$. Therefore, the double palindrome can be written as $a \times b^{2} \times c^{6}$. Let's see what the smallest case is when the prime numbers are the smallest. $11 \times 3^{2} \times 2^{6}$ $=6336$. When the prime numbers are a bit larger, such as $b=5$ and $c=2$, $11 \times 5^{2} \times 2^{6}=17600$ (which does not meet the condition of being a four-digit number). Therefore, the answer is: 6336." 2f62826ee0d2,"7. If the expression $\frac{1}{1 \times 2}-\frac{1}{3 \times 4}+\frac{1}{5 \times 6}-\frac{1}{7 \times 8}+\cdots-\frac{1}{2007 \times 2008}+\frac{1}{2009 \times 2010}$ is converted to a decimal, then the first digit after the decimal point is $\qquad$ .",4,easy,Answer: 4 6a2309470749,"4. Given the three sides of $\triangle A B C$ are $A B=$ $2 \sqrt{a^{2}+576}, B C=\sqrt{a^{2}+14 a+625}, A C=$ $\sqrt{a^{2}-14 a+625}$, where $a>7$. Then the area of $\triangle A B C$ is $\qquad$",S_{\triangle A B D}-S_{\triangle C B D}-S_{\triangle A C D}=168$.,medium,"4.168. Notice $$ \begin{array}{l} A B^{2}=(2 a)^{2}+48^{2}, \\ B C^{2}=(a+7)^{2}+24^{2}, \\ A C^{2}=(a-7)^{2}+24^{2} . \end{array} $$ As shown in Figure 6, with $A B$ as the hypotenuse, construct a right triangle $\triangle A B D$ on one side of $\triangle A B C$, such that $$ \begin{array}{l} B D=2 a, A D=48, \\ \angle A D B=90^{\circ} . \end{array} $$ Take a point $E$ on $B D$ such that $B E=a+7, E D=a-7$, and take the midpoint $F$ of $A D$, construct rectangle $E D F C_{1}$. Since $B C_{1}^{2}=B E^{2}+E C_{1}^{2}=(a+7)^{2}+24^{2}=B C^{2}$, $$ A C_{1}^{2}=C_{1} F^{2}+A F^{2}=(a-7)^{2}+24^{2}=A C^{2} \text {, } $$ thus point $C$ coincides with point $C_{1}$. And $S_{\triangle A B D}=48 a, S_{\triangle C B D}=24 a, S_{\triangle A C D}=24(a-7)$, then $S_{\triangle B B C}=S_{\triangle A B D}-S_{\triangle C B D}-S_{\triangle A C D}=168$." b311f00c450b,"How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? $\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$",\textbf{(E),easy,"We compute $2^8+1=257$. We're all familiar with what $6^3$ is, namely $216$, which is too small. The smallest cube greater than it is $7^3=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$, which therefore will clearly be the largest cube less than $2^{18}+1$. So, the required number of cubes is $64-7+1= \boxed{\textbf{(E) }58}$." 349d25e6f888,"6. The positive integer solution of the system of equations $\left\{\begin{array}{l}\frac{z^{y}}{y}=y^{2 x}, \\ 2^{z}=2 \cdot 4^{x}, \\ x+y+z=12\end{array}\right.$ is $\qquad$","1, \\ y=8\end{array}\right.$ or $\left\{\begin{array}{l}x=2, \\ y=5\end{array}\right.$ or $\left\{\b",easy,"$6 . x=2, y=5, z=5$. From $2^{z}=2^{2 x+1} \Rightarrow z=2 x+1$. Substituting into $x+y+z=12 \Rightarrow 3 x+y=11$. $\because x, y$ are positive integers then $\left\{\begin{array}{l}x=1, \\ y=8\end{array}\right.$ or $\left\{\begin{array}{l}x=2, \\ y=5\end{array}\right.$ or $\left\{\begin{array}{l}x=3, \\ y=2 .\end{array}\right.$ Substituting into $z^{y}=y^{2 x+1}$, we get $z=5$. That is, $x=2, y=5, z=5$ is the solution." 2953e880ab1a,2. Solve the equation $\left(\frac{3 x}{2}\right)^{\log _{3}(8 x)}=\frac{x^{7}}{8}$.,"$x=\frac{729}{8}, x=2$",medium,"Answer: $x=\frac{729}{8}, x=2$. Solution. By taking the logarithm to base 3, we obtain an equation equivalent to the original: $$ \log _{3}(8 x) \cdot \log _{3}\left(\frac{3 x}{2}\right)=\log _{3}\left(\frac{x^{7}}{8}\right) \Leftrightarrow\left(\log _{3} x+3 \log _{3} 2\right)\left(1+\log _{3} x-\log _{3} 2\right)=7 \log _{3} x-3 \log _{3} 2 $$ Let $\log _{3} x=y, \log _{3} 2=a$. After expanding the brackets and combining like terms, the equation becomes $y^{2}+(2 a-6) y-3 a^{2}+6 a=0$. We solve the quadratic equation for $y$: $$ \frac{D}{4}=(a-3)^{2}+\left(3 a^{2}-6 a\right)=4 a^{2}-12 a+9=(2 a-3)^{2}, y=3-a \pm(2 a-3) \Leftrightarrow\left[\begin{array}{l} y=a \\ y=6-3 a \end{array}\right. $$ We find $x$: if $y=a$, then $\log _{3} x=\log _{3} 2$ and $x=2$; if $y=6-3 a$, then $\log _{3} x=6-3 \log _{3} 2$ and $x=\frac{729}{8}$." 4a7acdb701c3,"There are 20 people - 10 boys and 10 girls. How many ways are there to form a company where the number of boys and girls is equal? #",$C_{20}^{10}=184756$ ways,medium,"For each such company, associate a set of 10 people, which includes all the girls who joined the company and all the boys who did not join it. ## Solution Let there be some company consisting of $k$ boys and $k$ girls. We will associate it with a set of 10 people, which includes $k$ girls who joined the company and $10-k$ boys who did not join it. The established correspondence is clearly one-to-one. Therefore, the desired number is equal to the number of ways to choose 10 people out of 20. ## Answer $C_{20}^{10}=184756$ ways." 0e5ceab1f9b5,"Let $w_1, w_2, \dots, w_n$ be [complex numbers]. A line $L$ in the [complex plane] is called a mean [line] for the [points] $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$, $w_2 = - 7 + 64i$, $w_3 = - 9 + 200i$, $w_4 = 1 + 27i$, and $w_5 = - 14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the [slope] of this mean line.",163,medium,"Solution 1 $\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0$ $\sum_{k=1}^5 z_k = 3 + 504i$ Each $z_k = x_k + y_ki$ lies on the complex line $y = mx + 3$, so we can rewrite this as $\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^5 y_ki$ $3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)$ Matching the real parts and the imaginary parts, we get that $\sum_{k=1}^5 x_k = 3$ and $\sum_{k=1}^5 (mx_k + 3) = 504$. Simplifying the second summation, we find that $m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489$, and substituting, the answer is $m \cdot 3 = 489 \Longrightarrow m = 163$. Solution 2 We know that $\sum_{k=1}^5 w_k = 3 + 504i$ And because the sum of the 5 $z$'s must cancel this out, $\sum_{k=1}^5 z_k = 3 + 504i$ We write the numbers in the form $a + bi$ and we know that $\sum_{k=1}^5 a_k = 3$ and $\sum_{k=1}^5 b_k = 504$ The line is of equation $y=mx+3$. Substituting in the polar coordinates, we have $b_k = ma_k + 3$. Summing all 5 of the equations given for each $k$, we get $504 = 3m + 15$ Solving for $m$, the slope, we get $\boxed{163}$ Solution 3 The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$. Since we now have two points, namely that one and $(0, 3i)$, we can simply find the slope between them, which is $\boxed{163}$ by the good ol' slope formula." 91fcaef2c867,"1. $\sqrt{ }(\lg 3)^{2}-\lg 9+1$ equals $(\quad)$. (A) $\pm(\lg 3-1)$. (B) $\lg 3-1$ (C) $1-\lg 3$. (D) 1.",See reasoning trace,easy,"C Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 6b3768b988fd,"2 Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number. Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number.","|(2 x+3)(2 x-9)|$, we know that only when $|2 x+3|=1$ or $|2 x-9|=1$, the number $\left|4 x^{2}-12 x",easy,"2. From $\left|4 x^{2}-12 x-27\right|=|(2 x+3)(2 x-9)|$, we know that only when $|2 x+3|=1$ or $|2 x-9|=1$, the number $\left|4 x^{2}-12 x-27\right|$ can possibly be a prime number. According to this, the required $x=-2,-1,4$ or 5, and the corresponding $\left|4 x^{2}-12 x-27\right|$ are 13, 11, 11 or 13, all of which are prime numbers." bfe0c383dadc,"(7) Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{20} x^{20}$, then $a_{0}+$ $a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=$ $\qquad$",See reasoning trace,easy,"(7) -9 Hint: Let $x=0$, we get $a_{0}=1$. Differentiate both sides $$ \begin{array}{c} 10\left(1+x-x^{2}\right)^{9}(1-2 x)=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+20 a_{20} x^{19} . \\ \text { Let } x=1, a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-10 \text {, thus } \\ a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-9 . \end{array} $$" a03f35a3b68a,"Mary is $20\%$ older than Sally, and Sally is $40\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday? $\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11$",$\mathrm{(B)}$,easy,"Let $m$ be Mary's age, let $s$ be Sally's age, and let $d$ be Danielle's age. We have $s=.6d$, and $m=1.2s=1.2(.6d)=.72d$. The sum of their ages is $m+s+d=.72d+.6d+d=2.32d$. Therefore, $2.32d=23.2$, and $d=10$. Then $m=.72(10)=7.2$. Mary will be $8$ on her next birthday. The answer is $\mathrm{(B)}$." 6ac221f0872e,"Solve the following equation: $$ x-2=\sqrt{4-3 \sqrt{4-3 \sqrt{10-3 x}}} $$",See reasoning trace,medium,"Solution. Let's substitute $\sqrt{1} x$ with $(a+2)-\mathrm{t}$: $$ a=\sqrt{4-3 \sqrt{4-3 \sqrt{10-3 a-6}}}=\sqrt{4-3 \sqrt{4-3 \sqrt{4-3 a}}} $$ Let $f(x)=\sqrt{4-3 x}$. We are looking for the value of $a$ such that $a=f(f(f(a)))$. If $a=f(a)$, then this is clearly true. $a=f(a)$ holds if, under the conditions $a \geq 0, a \leq \frac{4}{3}$, $a^{2}+3 a-4=0$, from which $a=1$, so $x=3$. We need to check if there is any other $a$ that satisfies the equation $a=f(f(f(a)))$. $a$ cannot be negative because it is an element of the range of $f$, so it is sufficient to look for further solutions of $a$ in the intervals $[0 ; 1)$ and $\left(1 ; \frac{4}{3}\right]$. $f$ maps elements from the interval $[0 ; 1)$ to the interval $(1 ; 2]$, and elements from the interval $\left(1 ; \frac{4}{3}\right]$ to the interval $[0 ; 1)$. Therefore, if $a \in[0 ; 1)$, then $f(f(f(a))) \in(1 ; 2]$; if $a \in(1 ; 2]$, then $f(f(f(a))) \in[0 ; 1)$ (assuming the expression is meaningful). This means that the equation has no other solutions. Remark. The reasoning is incomplete if it claims that $$ a=f(f(\ldots(f(a) \ldots))) $$ is only possible if $a=f(a)$. Several submissions simplified the solution of the problem in this way, which is incorrect. This is not necessarily true even for monotonic functions. The function $f(x)=-x$ is monotonic, yet it is not true that $f(f(x))=x$ only holds if $x=f(x)$ (since $f(f(x))=x$ for all $x$). Similarly, it is conceivable that $a \neq f(a), a \neq f(f(a))$, but $a=f(f(f(a)))$. This is about functions that are not monotonic, as discussed in problem B. 3541 (KöMaL 2003/2, p. 89), where the function $f(x)=\frac{x-3}{x-2}$ is mentioned in the remark, and a method is shown for constructing such first-degree rational functions. The function $f(x)=-\frac{2 x+7}{x+3}$, examined in problem B. 3481, is also such, and the solution of the problem relies precisely on this property (KöMaL 2002/1, p. 40).[^0] [^0]: ${ }^{1}$ With this substitution, we obtained problem B. 3579, whose solution can be found on p. 153 of the 2003/3 issue of KöMaL." f3d55276de79,"5. (10 points) As shown in the figure, a rectangular piece of paper is 20 cm long and 16 cm wide. If a small rectangle 8 cm long and 4 cm wide is cut from this paper, and at least one side of the small rectangle is on the edge of the original rectangle, then the maximum perimeter of the remaining paper is ( ) cm. A. 72 B. 80 C. 88 D. 96",The maximum perimeter of the remaining paper piece is 88 cm,medium,"【Analysis】As shown in Figure (1): A rectangle 8 cm long and 4 cm wide is cut from one corner, and the perimeter of the remaining paper piece is equal to the original rectangle. According to the perimeter formula of a rectangle: $c=(a+b) \times 2$, substitute the data into the formula to solve. As shown in Figure (2), if a rectangle 8 cm long is cut along the original edge with a width of 4 cm, then the perimeter of the remaining paper piece will increase by $8+8=16$ cm. Solve accordingly. 【Solution】Solution: (1) $(20+16) \times 2$, $$ \begin{array}{l} =36 \times 2, \\ =72 \text { (cm), } \end{array} $$ (2) $(20+16) \times 2+8 \times 2$, $$ \begin{array}{l} =36 \times 2+16, \\ =72+16, \\ =88 \text { (cm), } \end{array} $$ Answer: The maximum perimeter of the remaining paper piece is 88 cm. Therefore, the correct choice is: $C$." 92605dce086a,"6. $a, b, c, d, e$ are 5 elements randomly selected from the set $\{1,2,3,4,5\}$ (repetition allowed). Then the probability that $a b c d + e$ is odd is $\qquad$",See reasoning trace,easy,"$a b c d+e=$ odd $\Leftrightarrow a b c d=$ odd, $e=$ even; or $a b c d=$ even, $e=$ odd. $a b c d$ being odd has $3^{4}=81$ possibilities, $a b c d$ being even has $5^4-3^{4}=544$ possibilities. Therefore, the required probability is $$ P=\frac{81 \times 2+544 \times 3}{5^{5}}=\frac{1794}{3125} . $$" c9a8cf784c95,"2. Find all values of the variable $x$, for each of which both expressions $f(x)=\sqrt{16-x^{2}+6 x}$ and $g(x)=|x-3|$ are defined, and $\min (f(x) ; g(x))>\frac{5-x}{2}$.",$x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 8\right]$,medium,"Answer: $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 8\right]$. Solution. Both functions are defined when $16-x^{2}+6 x \geqslant 0 \Leftrightarrow -2 \leqslant x \leqslant 8$. Note that the following two statements are equivalent: ""the smaller of two numbers is greater than $A$"" and ""both numbers are greater than $A$"". Therefore, the given inequality is equivalent to the system of inequalities $$ \left\{\begin{array}{l} f(x)>\frac{5-x}{2} \\ g(x)>\frac{5-x}{2} \end{array}\right. $$ Since the functions $f(x)$ and $g(x)$ are non-negative on the domain of definition, both inequalities of the system are satisfied when $\frac{5-x}{2} < 0$, and considering the domain of definition - when $x \in(5 ; 8]$. If $x \leqslant 5$, then both the left and right sides of both inequalities are non-negative, and squaring them is an equivalent transformation. We get $\left\{\begin{array}{l}16-x^{2}+6 x>\frac{x^{2}-10 x+25}{4} \\ x^{2}-6 x+9>\frac{x^{2}-10 x+25}{4}\end{array} \Leftrightarrow\left\{\begin{array}{l}5 x^{2}-34 x-39<0 \\ 3 x^{2}-14 x+11>0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \in\left(-1 ; \frac{39}{5}\right), \\ x \in(-\infty ; 1) \cup\left(\frac{11}{3} ;+\infty\right)\end{array} \Leftrightarrow x \in(-1 ; 1) \cup\left(\frac{11}{3} ; \frac{39}{5}\right)\right.\right.\right.$. Considering the interval $x \in(-\infty ; 5]$, we get $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 5\right]$. Combining the results, we finally find that $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 8\right]$." 198f1455342f,"13.421 If a two-digit number is divided by a certain integer, the quotient is 3 and the remainder is 8. If the digits in the dividend are swapped while the divisor remains the same, the quotient becomes 2 and the remainder is 5. Find the original value of the dividend.",53,easy,"Solution. Let $\overline{x y}$ be the desired number, $z$ be the divisor. Then, according to the condition, \[ \left\{\begin{array}{l} 10 x+y=3 z+8 \\ 10 y+x=2 z+5 \end{array}\right. \] By eliminating $z$ from these equations, we have: $17 x-28 y=1 \Leftrightarrow 28(x-y)=1+11 x$. But $12 \leq 1+11 x \leq 100$. In this range, only the numbers $28, 56, 84$ are divisible by 28. We directly verify that only $56=1+11 x \Rightarrow x=5 \Rightarrow y=3$. Answer: 53." b367be126394,"79. In a triangle, three lines parallel to its sides and tangent to the inscribed circle have been drawn. They cut off three triangles from the given one. The radii of the circumcircles of these triangles are $R_{1}, R_{2}, R_{3}$. Find the radius of the circumcircle of the given triangle.",$R=R_{1}+R_{2}+R_{3}$,medium,"79. Prove that if $a, b, c$ are the lengths of the sides of a triangle, then the perimeters of the cut-off triangles will be $2(p-a), 2(p-b)$, $2(p-c)$. Consequently, if $R$ is the radius of the circumscribed circle, then $R_{1}+R_{2}+R_{3}=\left(\frac{p-a}{p}+\frac{p-b}{p}+\frac{p-c}{p}\right) R=R$. Answer: $R=R_{1}+R_{2}+R_{3}$." ac66f577db49,"Two positive integers $a$ and $b$ are prime-related if $a = pb$ or $b = pa$ for some prime $p$. Find all positive integers $n$, such that $n$ has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related. Note that $1$ and $n$ are included as divisors.",n,medium,"To solve the problem, we need to find all positive integers \( n \) such that \( n \) has at least three divisors, and all the divisors can be arranged in a circle so that any two adjacent divisors are prime-related. 1. **Prime Power Case:** - If \( n \) is a prime power, say \( n = p^k \) where \( p \) is a prime and \( k \geq 1 \), then the divisors of \( n \) are \( 1, p, p^2, \ldots, p^k \). - For \( n \) to have at least three divisors, \( k \geq 2 \). - However, in this case, \( 1 \) and \( p \) must be adjacent, and \( p \) and \( p^2 \) must be adjacent. But \( 1 \) and \( p^2 \) are not prime-related, so \( n \) cannot be a prime power. 2. **Perfect Square Case:** - If \( n \) is a perfect square, say \( n = m^2 \), then \( n \) has an odd number of divisors. - Consider the sum of the exponents of the primes in the prime factorization of a number. Each move (from one divisor to the next) changes the parity of this sum. - Starting from \( 1 \) and ending at \( 1 \) after an odd number of moves implies that the parity of the sum of the exponents changes an odd number of times, which is a contradiction. - Therefore, \( n \) cannot be a perfect square. 3. **General Case:** - We claim that integers \( n \) with at least three prime factors work. - Consider \( n = p^a q^b \) where \( p \) and \( q \) are primes and \( a, b \geq 1 \). - Assume \( b \) is odd. We can construct a sequence of divisors as follows: \[ 1, q, pq, p^2q, \ldots, p^{a-1}q, q^2p^{a-1}, q^2p^{a-2}, \ldots, pq^2, q^2, q^3, q^3p, \ldots, q^b, q^bp, q^bp^2, \ldots, q^bp^{a-1} \] - This sequence maintains the condition that any two adjacent divisors are prime-related. - The remaining divisors are of the form \( p^i \) and \( p^a q^j \) for \( 1 \leq j \leq b \). We attach the following string: \[ p^a q^b, p^a q^{b-1}, \ldots, p^a q, p^a, p^{a-1}, p^{a-2}, \ldots, p \] - This construction works for \( n = p^a q^b \). 4. **Inductive Step:** - Assume that for \( n = \prod_{i=1}^{k-1} p_i^{a_i} \), there is a valid construction. - For \( n' = \prod_{i=1}^{k+1} p_i^{a_i} \), consider the sequence: \[ b_1, b_2, \ldots, b_l, p b_l, p, b_{l-1}, \ldots, p b_3, p b_2, p^2 b_2, p^2 b_3, \ldots, p^2 b_l, \ldots \] - If \( a_{k+1} \) is even, the sequence ends with \( p^{a_{k+1}} b_l \). If \( a_{k+1} \) is odd, it ends with \( p^{a_{k+1}} b_2 \). - Add the following string: \[ p^{a_{k+1}} b_1, p^{a_{k+1}-1} b_1, \ldots, p^2 b_1, p b_1 \] - This construction works, and every number which is not a square or a prime power works. \(\blacksquare\) The final answer is all positive integers \( \boxed{ n } \) that are not prime powers or perfect squares." c7e14bd2120a,"## Task A-4.6. Let $M$ and $N$ be the feet of the altitudes from vertices $A$ and $B$ of an acute-angled triangle $ABC$. Let $Q$ be the midpoint of segment $\overline{M N}$, and $P$ be the midpoint of side $\overline{A B}$. If $|M N|=10$ and $|A B|=26$, determine the length of $|P Q|$.",\sqrt{|P M|^{2}-|Q M|^{2}}=\sqrt{13^{2}-5^{2}}=12$.,medium,"## Solution. Since $\varangle B M A=\varangle B N A=90^{\circ}$, quadrilateral $A B M N$ is cyclic. The center of the circle circumscribed around quadrilateral $A B C D$ is point $P$. Segments $\overline{P N}$ and $\overline{P M}$ are radii of this circle, so $|P N|=|P M|=|A P|=13$. ![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-28.jpg?height=677&width=877&top_left_y=1706&top_left_x=521) Since $Q$ is the midpoint of segment $\overline{M N}$, it follows that $|Q N|=|Q M|=5$. 1 point Triangle $P N M$ is isosceles, so triangles $N P Q$ and $M P Q$ are right triangles. 1 point By the Pythagorean theorem, we have $|P Q|=\sqrt{|P M|^{2}-|Q M|^{2}}=\sqrt{13^{2}-5^{2}}=12$." 19e67e23268b,,See reasoning trace,easy," Solution: It is obvious that $x>y$. On the other hand $xy^{3}+2 y^{2}+1 \Longleftrightarrow y(y+3)>0$. Therefore if $y>0$ or $y<-3$ the problem has no solution. Direct verification yields all pairs $(x, y)$ which satisfy the equality $x^{3}=y^{3}+2 y^{2}+1$, namely $(-2,-3),(1,-2)$ and $(1,0)$. " 5b3729c05064,"4. Given that $A_{1}, B_{1}, C_{1}, D_{1}$ are points on the four sides of square $A B C D$. The following are two propositions about it: (1) If $A_{1} B_{1} C_{1} D_{1}$ is a rectangle with unequal adjacent sides, then the rectangle $A_{1} B_{1} C_{1} D_{1}$ must have a side parallel to one of the diagonals of the square $A B C D$; (2) If $A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid, then the isosceles trapezoid $A_{1} B_{1} C_{1} D_{1}$ must have a base parallel to one of the diagonals of the square $A B C D$. Then, ( ). (A) (1) and (2) are both true propositions (B) (1) and (2) are both false propositions (C) (1) is a false proposition, (2) is a true proposition (D) (1) is a true proposition, (2) is a false proposition",See reasoning trace,medium,"4. (D). (1) As shown in the figure, it is easy to know that $\angle A D_{1} A_{1} + \angle D D_{1} C_{1} = 90^{\circ}, \angle D D_{1} C_{1} + \angle D C_{1} D_{1} = 90^{\circ}, \angle A D_{1} A_{1} = \angle D C_{1} D_{1}$, thus $\triangle A D_{1} A_{1} \sim \triangle D C_{1} D_{1}$. Also, Rt $\triangle A D_{1} A_{1} \cong \operatorname{Rt} \triangle C B_{1} C_{1}, A D_{1} = C B_{1}, A A_{1} = C C_{1}$, hence $D D_{1} = B B_{1}, B A_{1} = D C_{1}$. Since $A_{1} D_{1} \neq D_{1} C_{1}$, $\triangle A D_{1} A_{1}$ and $\triangle D C_{1} D_{1}$ are not congruent. Therefore, $A A_{1} \neq D D_{1}$. Without loss of generality, assume $D D_{1} > A A_{1}$. Now assume $A_{1} D_{1} X B D$, then $\angle A A_{1} D_{1} \neq \angle A D_{1} A_{1}$. Without loss of generality, assume $\angle A A_{1} D_{1} > \angle A D_{1} A_{1}$. Thus, it is easy to know that $D D_{1} > D C_{1}$, so $D D_{1} - A D_{1} > |D C_{1} - A A_{1}|$, i.e., $D D_{1} - C B_{1} > |D C_{1} - A A_{1}|$. Draw $D_{1} E \perp B C$ at $E$, $A_{1} F \perp D C$ at $F$, connect $A_{1} C_{1}$, $B_{1} D_{1}$, then $B_{1} E = B B_{1} - A D_{1}, C_{1} F = |D C_{1} - A A_{1}|, A_{1} F = D_{1} E$. We have $B_{1} D_{1} = \sqrt{D_{1} E^{2} + B_{1} E^{2}} > \sqrt{A_{1} F^{2} + C_{1} F^{2}} = A_{1} C_{1}$. This contradicts the fact that the diagonals of a rectangle are equal. Therefore, $A_{1} D_{1} \parallel B D$. (2) As shown in the figure, through a non-midpoint $P$ on the diagonal $A C$, we can draw $B_{1} D_{1}$ intersecting $B C$ and $A D$ at points $B_{1}$ and $D_{1}$ respectively, such that $\angle A P D_{1} \neq 45^{\circ}$. Draw $A_{1} C_{1} \perp B_{1} D_{1}$ through $P$, intersecting $A B$ and $C D$ at points $A_{1}$ and $C_{1}$ respectively. Therefore, $A, A_{1}, P, D_{1}$ are concyclic, and $C, C_{1}, P, B_{1}$ are concyclic, $\angle A_{1} D_{1} P = \angle A_{1} A P = 45^{\circ} = \angle P A D_{1} = \angle D_{1} A_{1} P$, $\angle P B_{1} C_{1} = \angle P C C_{1} = 45^{\circ} = \angle P C B_{1} = \angle P C_{1} B_{1}$, thus $\angle A_{1} D_{1} P = \angle P B_{1} C_{1}, A_{1} D_{1} \parallel B_{1} C_{1}, A_{1} P = P D_{1}, P C_{1} = B_{1} P, A_{1} C_{1} = B_{1} D_{1}$, and it is easy to know that $A_{1} D_{1} \neq B_{1} C_{1}$. Therefore, $A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid. Clearly, its bases are not parallel to the diagonals of the square $A B C D$. Counterexamples do exist." 52f2085fa4ea,3. Given the ellipse $x^{2}+4(y-a)^{2}=4$ and the parabola $x^{2}=2 y$ have common points. Then the range of values for $a$ is $\qquad$ .,See reasoning trace,medium,"3. $\left[-1, \frac{17}{8}\right]$. Notice that, the parametric equation of the ellipse is $$ \left\{\begin{array}{l} x=2 \cos \theta, \\ y=a+\sin \theta \end{array}(\theta \in \mathbf{R})\right. \text {. } $$ Substituting into the parabola equation, we get $$ \begin{array}{l} 4 \cos ^{2} \theta=2(a+\sin \theta) \\ \Rightarrow a=2 \cos ^{2} \theta-\sin \theta \\ =-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8} . \end{array} $$ Since $\sin \theta \in[-1,1]$, then $a \in\left[-1, \frac{17}{8}\right]$." de4613e148cc,BMO 1966,"s > √(3/2). Note that 3 - √3 = 1.27, √(3/2) = 1.22, so the range is fairly narrow. The square must b",medium,": 3 - √3 >= s > √(3/2). Note that 3 - √3 = 1.27, √(3/2) = 1.22, so the range is fairly narrow. The square must be small enough to fit into the hole and big enough to stick. The sticking condition is easy. Inscribe a circle in the hexagon. If the square will fit into the circle, then it obviously does not stick. So we require the diagonal of the square to be > √3, or s > √(3/2). On the other hand if the diagonal exceeds √3, then it obviously does stick because the parallel opposite sides of the hexagon are only a distance √3 apart. It seems fairly clear that the maximum square has sides parallel to a pair of opposite sides and each vertex a distance x from on endpoint of those sides. For the square to have equal sides, we then need √3 (1 - x) = 1 + x, or x = 2 - √3, and hence s = 3 - √3. But how do we prove it? 2nd BMO 1966 © John Scholes jscholes@kalva.demon.co.uk 30 Aug 2000" bcac7c491ecf,"3. Let $AB$ be a chord of the unit circle $\odot O$. If the area of $\odot O$ is exactly equal to the area of the square with side $AB$, then $\angle AOB=$ $\qquad$ (to 0.001 degree).",See reasoning trace,easy,3. $124.806^{\circ}$ 4345cbf6cf6c,"$\left[\begin{array}{l}{[\text { Prime numbers and their properties }} \\ {[\underline{\text { Arithmetic of residues (other) })]}}\end{array}\right.$ a) $p, p+10, p+14$ - prime numbers. Find $p$. b) $p, 2 p+1,4 p+1$ - prime numbers. Find $p$.",$p=3$,easy,"Consider the remainders when dividing by 3. One of these numbers is divisible by 3. ## Answer $p=3$." b37c82928f5c,"7. Let $x, y$ be real numbers, and satisfy $$ \left\{\begin{array}{l} (x-1)^{3}+1998(x-1)=\cdots 1, \\ (y-1)^{3}+1998(y-1)=1 . \end{array}\right. $$ Then $x+y=(\quad)$. (A) 1 (B) -1 (C) 2 (D) -2",See reasoning trace,easy,"7.C Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. However, it seems there was a misunderstanding in your request. The text ""7.C"" does not require translation as it is already in a form that is the same in both Chinese and English. If you have more text to translate or need further assistance, feel free to let me know!" 19fb9e199600,"1. Given in $\triangle A B C$, $\angle A=90^{\circ}, M$ is the midpoint of side $B C$, and point $N$ on the extension of $B C$ satisfies $A M \perp A N$. The incircle of $\triangle A B C$ touches sides $A B$ and $A C$ at points $E$ and $F$, respectively. $E F$ intersects $A N$ and $B C$ at points $P$ and $Q$. Then $P N - Q N$ ( ). (A) greater than 0 (B) equal to 0 (C) less than 0 (D) uncertain",\angle N Q P \Rightarrow P N=Q N$.,easy,"-1.B. As shown in Figure 5. From $A M \perp A N$, we know $$ \begin{aligned} & \angle A N M=90^{\circ}-\angle A M N=90^{\circ}-\angle A M C \\ & =90^{\circ}-\left(180^{\circ}-2 \angle A C M\right) \\ & =2\left(\angle A C M-45^{\circ}\right) \\ = & 2(\angle A C M-\angle A F E) \\ =2( & \angle A C M-\angle C F Q)=2 \angle C Q F . \end{aligned} $$ Therefore, $\angle N P Q=\angle N Q P \Rightarrow P N=Q N$." 03ff646fc550,"4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters? ![](https://cdn.mathpix.com/cropped/2024_05_06_535cfcd84333341737d0g-01.jpg?height=281&width=374&top_left_y=676&top_left_x=844)",120,easy,"Answer: 120. Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm." 256d3c908012,"4. 122 Solve the system of equations $\left\{\begin{array}{l}\sqrt{x-1}+\sqrt{y-3}=\sqrt{x+y}, \\ \lg (x-10)+\lg (y-6)=1 .\end{array}\right.$",See reasoning trace,medium,"[Solution] The original system of equations can be transformed into the system of equations $$\left\{\begin{array}{l} x y-3 x-y-1=0 \\ x y-6 x-10 y+50=0 \end{array}\right.$$ Subtracting (4) from (3) and simplifying, we get $$x=17-3 y .$$ Substituting (5) into (3) and rearranging, we get $$3 y^{2}-25 y+52=0$$ Thus $\square$ $$y_{1}=4, y_{2}=\frac{13}{3}$$ Substituting the values of $y$ into (5), we get $x_{1}=5, x_{2}=4$, so the solutions to the system of equations are $$\left\{\begin{array} { l } { x _ { 1 } = 5 , } \\ { y _ { 1 } = 4 ; } \end{array} \quad \left\{\begin{array}{l} x_{2}=4 \\ y_{2}=\frac{13}{3} \end{array}\right.\right.$$ Upon verification, neither of these pairs satisfies equation (2), so the original system of equations has no solution." be6d88124ad3,"$19 \cdot 27$ The sum of the $n$ interior angles and one exterior angle of a convex $n$-sided polygon is $1350^{\circ}$, then $n$ equals (A) 6. (B) 7. (C) 8. (D) 9. (E) 10. (China Beijing Junior High School Mathematics Competition, 1984)",$(D)$,easy,"[Solution] The sum of the interior angles of a convex $n$-sided polygon is $(n-2) \cdot 180^{\circ}$. Let one of the exterior angles be $\alpha$, then $$ (n-2) \cdot 180^{\circ} + \alpha = 1350^{\circ} \text{, and } 0^{\circ} < \alpha < 180^{\circ} \text{. } $$ By the property of integer divisibility, $90 \mid \alpha$, so $\alpha = 90^{\circ}$, and solving for $n$ gives $n = 9$. Therefore, the answer is $(D)$." ac60636ca00f,"Let's find such numbers that when multiplied by 12345679, we get numbers consisting of 9 identical digits.",See reasoning trace,easy,"A 9-digit number consisting of equal digits has the general form 111111111a, where $a$ can take any value from 1 to 9. Therefore, $$ 12345679 x = 111111111 a $$ from which $$ x = 9 a $$ the sought numbers are thus: $$ 9, 18, 27, 36, 45, 54, 63, 72, 81 $$ (Raab Rezsó, Györ.) Number of solutions: 22." c1c5f762681e,"2. List the natural numbers $1, 2, \cdots, k^{2}$ in a square table (as shown in Table 1), then select any one number from the table, and subsequently remove the row and column containing that number. Then, perform the same operation on the remaining $(k-1)^{2}$ numbers in the square table, and continue this process $k$ times. The sum of the $k$ selected numbers is 1 \begin{tabular}{|c|c|c|c|} \hline 1 & 2 & $\cdots$ & $k$ \\ \hline$k+1$ & $k+2$ & $\cdots$ & $2 k$ \\ \hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\ \hline$(k-1) k+1$ & $(k-1) k+2$ & $\cdots$ & $k^{2}$ \\ \hline \end{tabular}",See reasoning trace,medium,"2. $\frac{1}{2} k\left(k^{2}+1\right)$. Divide Table 1 into the following two number tables: It is easy to see that each number in Table 1 equals the sum of the two numbers in the same position in the divided Table 2 and Table 3. Therefore, the sum of the $k$ selected numbers, which are neither in the same row nor in the same column, is exactly $$ \begin{array}{l} S=(1+2+\cdots+k)+[0+k+\cdots+(k-1) k] \\ =\frac{1}{2} k(k+1)+\frac{1}{2} k^{2}(k-1)=\frac{1}{2} k\left(k^{2}+1\right) . \end{array} $$" 5799f74be733,"4. Given: $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)$ ( $n$ is a natural number). Then the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is (A) $1991{ }^{-1}$; (B) $-1991^{-1}$; (C) $(-1)^{n} 1991$; (D) $(-1)^{n} 1991^{-1}$.",See reasoning trace,easy,"4. (D) $$ \text{Solution} \begin{aligned} 1+x^{2} & =1+\frac{1}{4}\left(1991^{\frac{2}{n}}-2+1991^{-\frac{2}{n}}\right) \\ & =\left[\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{2}, \end{aligned} $$ Therefore, the original expression $=\left(-1991^{-\frac{1}{n}}\right)^{n}$ $$ =(-1)^{n} 1991^{-1} \text{. } $$" 2214aea35b75,"$$ \left\{\begin{array}{l} (x-1)(y-1)(z-1)=x y z-1 \\ (x-2)(y-2)(z-2)=x y z-2 \end{array}\right. $$ (Vladimir Bragin) Answer: $x=1, y=1, z=1$.",See reasoning trace,medium," Solution 1. By expanding the parentheses and reducing common terms we obtain $$ \left\{\begin{array}{l} -(x y+y z+z x)+(x+y+z)=0 \\ -2(x y+y z+z x)+4(x+y+z)=6 \end{array}\right. $$ From the first equation we can conclude that $x y+y z+z x=x+y+z$. By substituting this into the second equation, we obtain that $x+y+z=3$. We now have to solve the system $$ \left\{\begin{array}{l} x+y+z=3 \\ x y+y z+z x=3 \end{array}\right. $$ If we square the first equation, we get $x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=9$. Hence $x^{2}+y^{2}+z^{2}=3=x y+y z+z x$. We will prove that if $x^{2}+y^{2}+z^{2}=x y+y z+z x$, than $x=y=z$ : $$ \begin{gathered} x^{2}+y^{2}+z^{2}=x y+y z+z x \quad \Longleftrightarrow \\ 2 x^{2}+2 y^{2}+2 z^{2}=2 x y+2 y z+2 z x \quad \Longleftrightarrow \\ x^{2}-2 x y+y^{2}+x^{2}-2 x z+z^{2}+y^{2}-2 y z+z^{2}=0 \\ (x-y)^{2}+(x-z)^{2}+(y-z)^{2}=0 \end{gathered} $$ The sum of three squares is 0 , so all of them are zeroes, which implies $x=y=z$. That means $x=y=z=1$. Solution 1'. We will show one more way to solve the system (1). Express $z=3-x-y$ from first equation and substitute it into the second one: $$ \begin{array}{cc} x y+(y+x)(3-x-y)=3 \quad \Longleftrightarrow \\ x y+3 x+3 y-2 x y-x^{2}-y^{2}=3 & \Longleftrightarrow \\ x^{2}+y^{2}+x y-3 x-3 y+3=0 & \Longleftrightarrow \\ x^{2}+x(y-3)+y^{2}-3 y+3=0 \end{array} $$ Let us solve it as a quadratic equation over variable $x$ : $$ \begin{aligned} x & =\frac{(3-y) \pm \sqrt{(y-3)^{2}-4\left(y^{2}-3 y+3\right)}}{2}= \\ & =\frac{(3-y) \pm \sqrt{y^{2}-6 y+9-4 y^{2}+12 y-12}}{2}= \\ & =\frac{(3-y) \pm \sqrt{-3 y^{2}+6 y-3}}{2}= \\ & =\frac{(3-y) \pm \sqrt{-3(y-1)^{2}}}{2} \end{aligned} $$ We can conclude that $y=1$, because otherwise the square root wouldn't exist. It follows that $x=\frac{3-1 \pm 0}{2}=1$, and then $z=1$. " ca72234bcf77,"## Task A-2.4. Determine all triples of natural numbers ( $a, b, c$ ) for which $$ 2^{a} \cdot 5^{b}-1=11 \cdot 3^{c} $$","(2, 2, 2)$.",medium,"## Solution. Let's look at the remainders that the expressions on the left and right sides of the given equation give when divided by 5. The first addend on the left side is divisible by 5, so the left side gives a remainder of 4 when divided by 5. On the other hand, the number 11 gives a remainder of 1 when divided by 5. Finally, the expression $3^{c}$, depending on the value of the number $c$, gives remainders $3, 4, 2, 1$, which then cyclically repeat with a period of 4. We conclude that, for the right side to also give a remainder of 4 when divided by 5, the number $c$ must be of the form $4c_{1} + 2$, for some non-negative integer $c_{1}$. In particular, $c$ is an even number. Now let's look at the remainders that the same expressions give when divided by 8. Since $c$ is an even number, the number $11 \cdot 3^{c} = 11 \cdot 9^{2c_{1} + 1}$ gives a remainder of 3 when divided by 8. Therefore, the number $2^{a} \cdot 5^{b} = 1 + 11 \cdot 3^{c}$ must give a remainder of 4 when divided by 8. This is only possible if $a = 2$ (if $a = 1$, the left side gives a remainder of 2; if $a \geq 3$, the left side is divisible by 8). Finally, let's look at the remainders that these expressions give when divided by 3. The right side is divisible by 3, so the number $2^{a} \cdot 5^{b} = 4 \cdot 5^{b}$ must give a remainder of 1. Since $5^{b}$ gives periodic remainders 2 and 1, we conclude that $b$ is even, i.e., $b = 2b_{1}$, for some natural number $b_{1}$. The left side can now be factored using the difference of squares: $$ \left(2 \cdot 5^{b_{1}} - 1\right)\left(2 \cdot 5^{b_{1}} + 1\right) = 11 \cdot 3^{c} $$ On the left side of the equation, we have the product of two consecutive odd numbers, and therefore they are relatively prime. Neither factor can be equal to 1, since they are at least $2 \cdot 5^{1} - 1 = 9$. Therefore, one factor must be equal to 11, and the other to $3^{c}$. In the first case, we have $2 \cdot 5^{b_{1}} - 1 = 11$, which has no solution. In the second case, we have $2 \cdot 5^{b_{1}} + 1 = 11$, from which $b_{1} = 1$, i.e., $b = 2$. Then $2 \cdot 5^{b_{1}} - 1 = 9 = 3^{c}$, i.e., $c = 2$. Thus, the only solution to the equation is $(a, b, c) = (2, 2, 2)$." c59f06d6bb72,"For all integers $n\geq 1$ we define $x_{n+1}=x_1^2+x_2^2+\cdots +x_n^2$, where $x_1$ is a positive integer. Find the least $x_1$ such that 2006 divides $x_{2006}$.",531,medium,"1. **Define the sequence and initial conditions:** Given the sequence defined by \( x_{n+1} = x_1^2 + x_2^2 + \cdots + x_n^2 \) for \( n \geq 1 \) and \( x_1 \) is a positive integer. We need to find the smallest \( x_1 \) such that \( 2006 \) divides \( x_{2006} \). 2. **Establish the recurrence relation:** From the problem, we have: \[ x_2 = x_1^2 \] For \( n \geq 3 \), we can write: \[ x_{n+1} = x_1^2 + x_2^2 + \cdots + x_n^2 \] This can be rewritten as: \[ x_{n+1} = x_n + x_n^2 \] This implies \( x_{n+1} = x_n (1 + x_n) \). 3. **Analyze the divisibility by 2006:** Since \( 2006 = 2 \cdot 17 \cdot 59 \), we need \( x_{2006} \) to be divisible by 2, 17, and 59. 4. **Check divisibility by 2:** Notice that \( x_n \) is even for all \( n \geq 3 \) because \( x_2 = x_1^2 \) is even if \( x_1 \) is even. Hence, \( x_{2006} \) is even. 5. **Check divisibility by 17 and 59:** We need to ensure that \( x_{2006} \) is divisible by both 17 and 59. We will use the properties of quadratic residues and the Chinese Remainder Theorem. 6. **Divisibility by 59:** If \( x_{2006} \) is divisible by 59, then there must be some \( x_i \) such that \( x_i \equiv 0 \pmod{59} \) or \( x_i + 1 \equiv 0 \pmod{59} \). This implies: \[ x_i \equiv -1 \pmod{59} \] We need to check if \( x_1^2 \equiv -1 \pmod{59} \) has a solution. Since -1 is not a quadratic residue modulo 59, \( x_1 \) must be divisible by 59. 7. **Divisibility by 17:** Similarly, for 17, we need: \[ x_1^2 \equiv -1 \pmod{17} \] We find that -1 is a quadratic residue modulo 17, and the solutions are: \[ x_1 \equiv \pm 4 \pmod{17} \] 8. **Combine results using the Chinese Remainder Theorem:** We have: \[ x_1 \equiv 0 \pmod{59} \] \[ x_1 \equiv 4 \pmod{17} \quad \text{or} \quad x_1 \equiv -4 \pmod{17} \] Solving these congruences using the Chinese Remainder Theorem, we get: \[ x_1 = 59k \quad \text{and} \quad 59k \equiv 4 \pmod{17} \] Since \( 59 \equiv 8 \pmod{17} \), we have: \[ 8k \equiv 4 \pmod{17} \] Solving for \( k \): \[ k \equiv 8^{-1} \cdot 4 \pmod{17} \] The modular inverse of 8 modulo 17 is 15 (since \( 8 \cdot 15 \equiv 1 \pmod{17} \)), so: \[ k \equiv 15 \cdot 4 \equiv 60 \equiv 9 \pmod{17} \] Thus, \( k = 17m + 9 \) for some integer \( m \), and: \[ x_1 = 59(17m + 9) = 1003m + 531 \] The smallest positive \( x_1 \) is when \( m = 0 \), giving: \[ x_1 = 531 \] The final answer is \( \boxed{531} \)" 23677db12a6d,"How many squares with sides along the grid lines can be drawn on an $8 \times 8$ board? #",In Middle Boltai,medium,"Squares of $1 \times 1$ can be drawn 64. Squares of $2 \times 2$ can be drawn 49 (take the square in the top right corner and first slide it along the top strip - this gives seven squares, and then all these squares can be moved down seven times). Squares of $3 \times 3$ can be drawn 36, and so on. The total number of squares is $64+49+36+25+16+9+4+1=204$. In Old Kalitva, there are 50 schoolchildren, and in Middle Boltai - 100 schoolchildren. Where should a school be built to minimize the total distance traveled by all schoolchildren? ## Solution Let's take 50 schoolchildren from Old Kalitva and 50 schoolchildren from Middle Boltai and pair them up, with one schoolchild from each city in each pair. For each such pair, the distance they travel to school does not depend on the location of the school (of course, it should be built somewhere on the road between the cities), but there are still 50 schoolchildren left in Middle Boltai. Therefore, to minimize the total distance traveled by all schoolchildren, the school should be built in Middle Boltai. ## Answer In Middle Boltai." abb6d0c5dace,"Let $a, b, c, d$ be a permutation of the numbers $1, 9, 8,4$ and let $n = (10a + b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n.$",\frac{5,medium,"To solve this problem, we need to determine the probability that \(1984!\) is divisible by \(n\), where \(n = (10a + b)^{10c + d}\) and \(a, b, c, d\) are permutations of the numbers \(1, 9, 8, 4\). 1. **Identify the possible values of \(10a + b\):** Since \(a\) and \(b\) are permutations of \(1, 9, 8, 4\), the possible values of \(10a + b\) are: \[ 10 \cdot 1 + 9 = 19, \quad 10 \cdot 1 + 8 = 18, \quad 10 \cdot 1 + 4 = 14, \quad 10 \cdot 9 + 1 = 91, \quad 10 \cdot 9 + 8 = 98, \quad 10 \cdot 9 + 4 = 94, \] \[ 10 \cdot 8 + 1 = 81, \quad 10 \cdot 8 + 9 = 89, \quad 10 \cdot 8 + 4 = 84, \quad 10 \cdot 4 + 1 = 41, \quad 10 \cdot 4 + 9 = 49, \quad 10 \cdot 4 + 8 = 48. \] 2. **Identify the possible values of \(10c + d\):** Similarly, since \(c\) and \(d\) are permutations of \(1, 9, 8, 4\), the possible values of \(10c + d\) are the same as above: \[ 19, 18, 14, 91, 98, 94, 81, 89, 84, 41, 49, 48. \] 3. **Determine the divisibility of \(1984!\) by \(n\):** We need to check if \(1984!\) is divisible by \(n = (10a + b)^{10c + d}\). For \(1984!\) to be divisible by \(n\), \(10a + b\) must be a prime number, and \(10c + d\) must be such that \((10a + b)^{10c + d}\) divides \(1984!\). 4. **Check the prime numbers among \(10a + b\):** The prime numbers among the possible values of \(10a + b\) are: \[ 19, 41, 89. \] 5. **Count the valid permutations:** - For \(10a + b = 19\), the possible values of \(10c + d\) are \(18, 14, 91, 98, 94, 81, 89, 84, 41, 49, 48\). Since \(19\) is a prime number, \(19^{10c + d}\) will divide \(1984!\) for any \(10c + d\). - For \(10a + b = 41\), the possible values of \(10c + d\) are the same as above. Since \(41\) is a prime number, \(41^{10c + d}\) will divide \(1984!\) for any \(10c + d\). - For \(10a + b = 89\), the possible values of \(10c + d\) are the same as above. Since \(89\) is a prime number, \(89^{10c + d}\) will divide \(1984!\) for any \(10c + d\). Therefore, there are \(3\) valid values for \(10a + b\) and \(4!\) permutations of \(c\) and \(d\) for each valid value of \(10a + b\). 6. **Calculate the total number of valid permutations:** \[ 3 \times 4! = 3 \times 24 = 72. \] 7. **Calculate the total number of permutations:** Since \(a, b, c, d\) are permutations of \(1, 9, 8, 4\), there are \(4!\) total permutations: \[ 4! = 24. \] 8. **Calculate the probability:** The probability that \(1984!\) is divisible by \(n\) is: \[ \frac{72}{24} = 3. \] The final answer is \(\boxed{\frac{5}{6}}\)." 6890fc4b69d3,"## Find all continuous functions $f:[1,8] \rightarrow R$, such that $$ \int_{1}^{2} f^{2}\left(t^{3}\right) d t+2 \int_{1}^{2} f\left(t^{3}\right) d t=\frac{2}{3} \int_{1}^{8} f(t) d t-\int_{1}^{2}\left(t^{2}-1\right)^{2} d t $$","x^{2 / 3}-1$ for every $x \in[1,8] \backslash C$ satisfies the conditions.",medium,"Solution. Using the substitution $t=u^{3}$ we get $$ \frac{2}{3} \int_{1}^{8} f(t) d t=2 \int_{1}^{2} u^{2} f\left(u^{3}\right) d u=2 \int_{1}^{2} t^{2} f\left(t^{3}\right) d u $$ Hence, by the assumptions, $$ \int_{1}^{2}\left(f^{2}\left(t^{3}\right)+\left(t^{2}-1\right)^{2}+2 f\left(t^{3}\right)-2 t^{2} f\left(t^{3}\right)\right) d t=0 $$ Since $f^{2}\left(t^{3}\right)+\left(t^{2}-1\right)^{2}+2 f\left(t^{3}\right)-2 t^{2} f\left(t^{3}\right)=\left(f\left(t^{3}\right)\right)^{2}+\left(1-t^{2}\right)^{2}+2\left(1-t^{2}\right) f\left(t^{3}\right)=\left(f\left(t^{3}\right)+1-t^{2}\right)^{2} \geq$ 0 , we get $$ \int_{1}^{2}\left(f\left(t^{3}\right)+1-t^{2}\right)^{2} d t=0 $$ The continuity of $f$ implies that $f\left(t^{3}\right)=t^{2}-1,1 \leq t \leq 2$, thus, $f(x)=x^{2 / 3}-1,1 \leq x \leq 8$. Remark. If the continuity assumption for $f$ is replaced by Riemann integrability then infinitely many $f$ 's would satisfy the given equality. For example if $C$ is any closed nowhere dense and of measure zero subset of $[1,8]$ (for example a finite set or an appropriate Cantor type set) then any function $f$ such that $f(x)=x^{2 / 3}-1$ for every $x \in[1,8] \backslash C$ satisfies the conditions." 4b1893a2d230,"4. Tomi dropped a rubber ball from a certain height onto a hard floor. After each bounce, it reached $\frac{3}{4}$ of the previous height. The ball bounced four times from the ground. At the moment it fell for the fifth time, his neighbor Jani caught it. From what height did Tomi drop the ball if it traveled a total distance of $653 \mathrm{~cm}$? Write down your answer.",63&width=1636&top_left_y=2264&top_left_x=210),medium,"4. Let the height from which the ball is dropped to the ground be $x$. After the first bounce from the ground, the ball reached $\frac{3}{4} x$ of the initial height, just as much when dropped from that height. After the second bounce, the ball reached $\frac{3}{4}\left(\frac{3}{4} x\right)=\frac{9}{16} x$ of the initial height, just as much when dropped. After the third bounce and drop, the ball traveled a distance of $\frac{27}{64} x \cdot 2$. After the fourth bounce and drop, the ball traveled a distance of $\frac{81}{256} x \cdot 2$. Thus, the total distance traveled by the ball is $x+2 x\left(\frac{3}{4}+\frac{9}{16}+\frac{27}{64}+\frac{81}{256}\right)=653$. We rearrange the equation to $653=\frac{653}{128} x$ and solve for $x=128 \mathrm{~cm}$. ![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-11.jpg?height=74&width=1642&top_left_y=2056&top_left_x=207) ![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-11.jpg?height=71&width=1636&top_left_y=2112&top_left_x=210) ![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-11.jpg?height=71&width=1636&top_left_y=2163&top_left_x=210) ![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-11.jpg?height=69&width=1636&top_left_y=2213&top_left_x=207) ![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-11.jpg?height=63&width=1636&top_left_y=2264&top_left_x=210)" e00b9e972527,"3. let $n$ be a natural number. Determine the number of pairs $(a, b)$ of natural numbers for which the following equation is satisfied: $$ (4 a-b)(4 b-a)=2010^{n} $$ ## Solution",See reasoning trace,medium,"Set $x=4 a-b$ and $y=4 b-a$. Because $x y>0$ and $x+y=3(a+b)>0$, $x$ and $y$ are natural numbers. Furthermore $$ a=\frac{4 x+y}{15}, \quad b=\frac{4 y+x}{15} $$ so $4 x+y$ and $4 y+x$ must both be divisible by 15. However, because of $15 \mid x y$, at least one of the two numbers $x, y$ is divisible by 3 and at least one of the two is divisible by 5. Thus, both must be divisible by 15 and in this case $a, b$ are actually whole. These considerations provide a bijection between the set of all pairs $(a, b)$ that fulfill the above equation and the set $S$ of all pairs $(x, y)$ of natural numbers with the properties - $15 \mid x$ and $15 \mid y$. - $x y=2010^{n}$. Because $2010=2 \cdot 3 \cdot 5 \cdot 67$, $x$ is of the form $x=2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 67^{d}$ with $0 \leq a, d \leq n$ and $1 \leq b, c \leq n-1$ and $y=2010^{n} / x$. So in total there are $$ (n+1)^{2}(n-1)^{2}=\left(n^{2}-1\right)^{2} $$ Solutions to this equation." 07b8c88a90c5,3. Let's call a natural number special if one of its digits can be replaced by another digit so that all digits in the resulting number are distinct. Numbers in which all digits are already distinct are also considered special. How many special ten-digit numbers exist? (20 points),$414 \cdot 9!$,medium,"Solution: Let's divide the numbers into groups: 1) Numbers in which all digits are different - there are a total of $9 \cdot 9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9 \cdot 9$ !. 2) Numbers in which two non-zero digits are the same and do not stand in the highest place. There are 9 such digits, pairs of positions where these digits can stand - $\frac{9 \cdot 8}{2}=36$, the remaining digits can be arranged in the remaining 8 positions in $8 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=8 \cdot 8$ ! ways. In total, there are such numbers $36 \cdot 8 \cdot 9!$. 3) Numbers in which two zero digits are the same. There are pairs of positions where these digits can stand $-\frac{9 \cdot 8}{2}=36$, the remaining digits can be arranged in the remaining 8 positions in $9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9$ ! ways. In total, there are such numbers $36 \cdot 9$ !. 4) Numbers in which two non-zero digits are the same, and one of them stands in the highest place. There are 9 such digits, pairs of positions where these digits can stand - 9, the remaining digits can be arranged in the remaining 8 positions in $9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9$ ! ways. In total, there are such numbers $81 \cdot 9$ !. In total, there are $(9+36 \cdot 8+36+81) \cdot 9!=414 \cdot 9!$ special numbers. Answer: $414 \cdot 9!$" ee9e2c56e2a7,"Let $\mathcal{P}$ be a set of monic polynomials with integer coefficients of the least degree, with root $k \cdot \cos\left(\frac{4\pi}{7}\right)$, as $k$ spans over the positive integers. Let $P(x) \in \mathcal{P}$ be the polynomial so that $|P(1)|$ is minimized. Find the remainder when $P(2017)$ is divided by $1000$. [i]Proposed by [b] eisirrational [/b][/i]",167,medium,"1. **Identify the polynomial \( P(x) \) with the root \( k \cdot \cos\left(\frac{4\pi}{7}\right) \):** Given that \( k \) spans over the positive integers, we need to find the monic polynomial with integer coefficients of the least degree that has \( k \cdot \cos\left(\frac{4\pi}{7}\right) \) as a root. 2. **Consider the polynomial \( x^7 - k^7 \):** Since \( k \cdot \cos\left(\frac{4\pi}{7}\right) \) is a root, the polynomial \( P(x) \) must divide \( x^7 - k^7 \). We can factor \( x^7 - k^7 \) as: \[ x^7 - k^7 = (x - k)(x^6 + kx^5 + k^2x^4 + k^3x^3 + k^4x^2 + k^5x + k^6) \] 3. **Choose \( k = 1 \) to minimize \( |P(1)| \):** For \( k = 1 \), the polynomial simplifies to: \[ x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) \] The polynomial \( P(x) \) is then: \[ P(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \] 4. **Evaluate \( P(1) \):** \[ P(1) = 1^6 + 1^5 + 1^4 + 1^3 + 1^2 + 1 + 1 = 7 \] This confirms that \( |P(1)| = 7 \), which is minimized for \( k = 1 \). 5. **Find the remainder when \( P(2017) \) is divided by 1000:** \[ P(2017) = 2017^6 + 2017^5 + 2017^4 + 2017^3 + 2017^2 + 2017 + 1 \] We need to compute this modulo 1000. First, reduce 2017 modulo 1000: \[ 2017 \equiv 17 \pmod{1000} \] Therefore: \[ P(2017) \equiv 17^6 + 17^5 + 17^4 + 17^3 + 17^2 + 17 + 1 \pmod{1000} \] 6. **Compute powers of 17 modulo 1000:** \[ 17^2 = 289 \] \[ 17^3 = 17 \cdot 289 = 4913 \equiv 913 \pmod{1000} \] \[ 17^4 = 17 \cdot 913 = 15521 \equiv 521 \pmod{1000} \] \[ 17^5 = 17 \cdot 521 = 8857 \equiv 857 \pmod{1000} \] \[ 17^6 = 17 \cdot 857 = 14569 \equiv 569 \pmod{1000} \] 7. **Sum the results modulo 1000:** \[ P(2017) \equiv 569 + 857 + 521 + 913 + 289 + 17 + 1 \pmod{1000} \] \[ P(2017) \equiv 3167 \pmod{1000} \] \[ P(2017) \equiv 167 \pmod{1000} \] The final answer is \( \boxed{167} \)." 48aedff783f0,Let $ ABC$ be a triangle such that \[ \frac{BC}{AB \minus{} BC}\equal{}\frac{AB \plus{} BC}{AC}\] Determine the ratio $ \angle A : \angle C$.,1:2,medium,"1. Given the equation: \[ \frac{BC}{AB - BC} = \frac{AB + BC}{AC} \] Let's denote the sides of the triangle as follows: \( BC = a \), \( AC = b \), and \( AB = c \). Substituting these into the given equation, we get: \[ \frac{a}{c - a} = \frac{c + a}{b} \] 2. Cross-multiplying to eliminate the fractions, we obtain: \[ a \cdot b = (c - a)(c + a) \] Simplifying the right-hand side using the difference of squares: \[ ab = c^2 - a^2 \] 3. Rearranging the equation to isolate \( c^2 \): \[ c^2 = ab + a^2 \] 4. Using the Law of Cosines in triangle \( ABC \): \[ c^2 = a^2 + b^2 - 2ab \cos C \] Equating the two expressions for \( c^2 \): \[ ab + a^2 = a^2 + b^2 - 2ab \cos C \] 5. Simplifying the equation: \[ ab = b^2 - 2ab \cos C \] Rearranging to solve for \( \cos C \): \[ ab + 2ab \cos C = b^2 \] \[ ab (1 + 2 \cos C) = b^2 \] \[ 1 + 2 \cos C = \frac{b}{a} \] 6. Using the Law of Sines, we know: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Since \( C = 2A \), we can use the double-angle identity for sine: \[ \sin 2A = 2 \sin A \cos A \] 7. Substituting \( C = 2A \) into the Law of Sines: \[ \frac{a}{\sin A} = \frac{b}{\sin 2A} = \frac{b}{2 \sin A \cos A} \] Simplifying: \[ \frac{a}{\sin A} = \frac{b}{2 \sin A \cos A} \] \[ a = \frac{b}{2 \cos A} \] 8. Since \( \cos A \) must be a positive value less than or equal to 1, we conclude that: \[ \cos A = \frac{b}{2a} \] 9. Given that \( C = 2A \), the ratio of the angles is: \[ \angle A : \angle C = 1 : 2 \] The final answer is \(\boxed{1:2}\)" c126dea1e8ff,"Find all primes of the form $a^{2}-1$, with $a \geqslant 2$ natural.","(a-1)(a+1)$, it is divisible by $a-1$ and $a+1$. Therefore, these must be 1 and $p$. Since they are ",easy,"3 is the only prime of this form. Indeed, if a prime $p$ is of the form $a^{2}-1=(a-1)(a+1)$, it is divisible by $a-1$ and $a+1$. Therefore, these must be 1 and $p$. Since they are distinct and $a-1$ is the smallest of them, $a-1=1$ and $p=a^{2}-1=2^{2}-1=3$. Conversely, $2^{2}-1=3$ is indeed prime." c5dab61079ee,"4. An $8 \times 6$ grid is placed in the first quadrant with its edges along the axes, as shown. A total of 32 of the squares in the grid are shaded. A line is drawn through $(0,0)$ and $(8, c)$ cutting the shaded region into two equal areas. What is the value of $c$ ?",$c=4$,medium,"4. Solution 1 Since 32 of the $1 \times 1$ squares are shaded, then the total shaded area is 32 . We want to draw a line through $(0,0)$ and $(8, c)$ that divides the shaded region into two pieces, each with area $\frac{32}{2}=16$. As long as the slope of the line segment through $(0,0)$ and $(8, c)$ is not too large, the bottom piece will be a triangle with vertices $(0,0),(8,0)$, and $(8, c)$. This triangle is right-angled at $(8,0)$ and so has base of length 8 and height $c$. For the area of this triangle to be 16 , we need $\frac{1}{2}(8)(c)=16$ or $4 c=16$. Thus, $c=4$. (Note that if $c=4$, the line segment passes through $(0,0)$ and $(8,4)$. This line segment has slope $\frac{4-0}{8-0}=\frac{1}{2}$ and so passes through the points $(2,1)$ and $(4,2)$. This confirms that the line segment does not pass through any unshaded regions.) Solution 2 We move the upper right shaded $2 \times 2$ square to complete a $8 \times 4$ shaded rectangle, as shown. The shaded area is now a rectangle with bottom left vertex $(0,0)$ and top right vertex $(8,4)$. The area of the rectangle is cut in half by its diagonal passing through $(0,0)$ and $(8,4)$. Note the region affected by this move lies above this diagonal. (This is because the diagonal has slope $\frac{4-0}{8-0}=\frac{1}{2}$ and so passes through the points $(2,1)$ and $(4,2)$.) This means that the areas of the regions below and above the line were unaffected by the move. Thus, reversing the move does not change the areas of the regions below and above the line. Therefore, the line drawn through $(0,0)$ and $(8,4)$ cuts the original region into two pieces of equal area. Thus, $c=4$. ANSWER: $c=4$" b3a645db7dc2,"4. On the plane, there are points $A, B, C$ and $D$ such that $|A B|=|B C|=$ $|A C|=|C D|=10 \mathrm{~cm}$ and $|A D|=17 \mathrm{~cm}$. What is the measure of angle $\Varangle A D B$?",See reasoning trace,medium,"II/4. Since point $C$ is equally distant from points $A, B$, and $D$, points $A, B$, and $D$ lie on a circle $\mathrm{s}$ with center $C$ and radius $10 \mathrm{~cm}$. Because $|A B|=10 \mathrm{~cm}$, triangle $A B C$ is equilateral and $\angle A C B=60^{\circ}$. Since $|A D|>|A B|$, point $D$ lies on the same side of line $A B$ as point $C$. Therefore, $\angle A D B$ and $\angle A C B$ are inscribed and central angles over the same arc, and $\angle A D B=30^{\circ}$. (In the figure, one of the two possible positions of point $D$ is drawn.) ![](https://cdn.mathpix.com/cropped/2024_06_07_af92d6d33b11d7748e87g-08.jpg?height=434&width=417&top_left_y=1913&top_left_x=1413) Cyclic nature of points $A, B, D: \mathbf{2}$ points. Central angle $60^{\circ}: \mathbf{1}$ point. Inscribed angle $30^{\circ}: \mathbf{1}$ point. Justification that because $|A D|>|A B|$, points $C$ and $D$ lie on the same side of line $A B$: 3 points. If the contestant (incorrectly!) claims that the position of point $D$ is uniquely determined by the conditions of the problem: at most 6 points." b41078e3a86d,"8. Let $a>0, b>0$. Then the inequality that does not always hold is ( ). (A) $\frac{2}{\frac{1}{a}+\frac{1}{b}} \geqslant \sqrt{a b}$ (B) $\frac{1}{a}+\frac{1}{b} \geqslant \frac{4}{a+b}$ (C) $\sqrt{|a-b|} \geqslant \sqrt{a}-\sqrt{b}$ (D) $a^{2}+b^{2}+1>a+b$",See reasoning trace,easy,"8. A. $$ \text { From } \frac{1}{a}+\frac{1}{b} \geqslant 2 \sqrt{\frac{1}{a b}} \Rightarrow \frac{2}{\frac{1}{a}+\frac{1}{b}} \leqslant \sqrt{a b} \text {. } $$ Therefore, option (A) does not always hold." 45e871dc5d84,"20 Find all integer pairs $(x, y)$, such that $x^{2}+3 y^{2}=1998 x$.","0,5$, $37^{2}-48 r^{2}$ is a perfect square, so $(x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540",medium,"20. Let $(x, y)$ be the integer pairs that satisfy the equation, it is known that $3 \mid x^{2}$, hence $3 \mid x$, set then $$\begin{array}{l} x=3 x_{1} \\ 3 x_{1}^{2}+y^{2}=1998 x_{1} \end{array}$$ Thus, $$3 \mid y^{2},$$ hence $$3 \mid y \text {. }$$ Set again $$y=3 y_{1},$$ then $$x_{1}^{2}+3 y_{1}^{2}=666 x_{1}$$ Proceeding similarly, we can set $$x=27 m, y=27 n$$ we get $$m^{2}+3 n^{2}=74 m$$ Rearranging and completing the square, we get $$(m-37)^{2}+3 n^{2}=37^{2}$$ Analyzing the parity of both sides of this equation, we find that $m$ and $n$ are both even, thus $$3 n^{2}=37^{2}-(m-37)^{2} \equiv 0(\bmod 8)$$ Hence set then thus hence $$\begin{array}{c} n=4 r \\ 48 r^{2} \leqslant 37^{2} \\ r^{2} \leqslant 28 \\ |r| \leqslant 5 \end{array}$$ For each $$|r|=0,1, \cdots, 5$$ calculate the value of $37^{2}-48 r^{2}$, we find that only when $|r|=0,5$, $37^{2}-48 r^{2}$ is a perfect square, so $(x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540)$" 5dd748bd60d9,"Task 3. (15 points) Laboratory engineer Sergei received an object for research consisting of about 200 monoliths (a container designed for 200 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (statistical probability) that a randomly selected monolith will be sandy loam is $\frac{1}{9}$. The relative frequency that a randomly selected monolith will be marine clayey loam is $\frac{11}{18}$. How many monoliths of lake-glacial genesis does the object contain, if there are no marine sandy loams among the sandy loams?",. 77,medium,"# Solution. Let's determine the exact number of monoliths. It is known that the probability of a monolith being loamy sand is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, there are 198 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(198: 9=22)$ and part of the clay loams. Let's find out how many monoliths of lacustrine-glacial origin are among the clay loams: The total number of clay loams is 198:9$\cdot$8=176. Among them, marine clay loams constitute $\frac{11}{18}$ of the total, which is 121. Therefore, the number of clay loams of lacustrine-glacial origin is 55. Thus, the total number of monoliths of lacustrine-glacial origin is: $22+55=77$. Answer. 77." b1a01aafd793,"Triangle $ABC$ has $AB = 13, BC = 14$, and $AC = 15$. The points $D, E$, and $F$ are the midpoints of $\overline{AB}, \overline{BC}$, and $\overline{AC}$ respectively. Let $X \neq E$ be the intersection of the circumcircles of $\triangle BDE$ and $\triangle CEF$. What is $XA + XB + XC$? $\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}$",3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}$,medium,"Let us also consider the circumcircle of $\triangle ADF$. Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of $\triangle ABC$ which is $P$, Also, since $m\angle ADP = m\angle AFP = 90^\circ$. $ADPF$ is cyclic, similarly, $BDPE$ and $CEPF$ are also cyclic. With this, we know that the circumcircles of $\triangle ADF$, $\triangle BDE$ and $\triangle CEF$ all intersect at $P$, so $P$ is $X$. The question now becomes calculating the sum of the distance from each vertex to the circumcenter. We can calculate the distances with coordinate geometry. (Note that $XA = XB = XC$ because $X$ is the circumcenter.) Let $A = (5,12)$, $B = (0,0)$, $C = (14, 0)$, $X= (x_0, y_0)$ Then $X$ is on the line $x = 7$ and also the line with slope $-\frac{5}{12}$ that passes through $(2.5, 6)$. $y_0 = 6-\frac{45}{24} = \frac{33}{8}$ So $X = (7, \frac{33}{8})$ and $XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}$" acba4e42a8fb,"1. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha$ $>\beta$. Without solving the equation, use the relationship between roots and coefficients to find the value of $\frac{2}{\alpha}+3 \beta^{2}$. (8th ""Zu Chongzhi Cup"" Mathematics Invitational Competition)",$\frac{1}{8}(403-85 \sqrt{17})$ ),easy,(Tip: Construct a conjugate. Answer: $\frac{1}{8}(403-85 \sqrt{17})$ ) 720642abbd99,"## Task B-4.6. Let $z=(1-i \sqrt{3})^{p}$ and $w=(1+i)^{q}$. Determine the smallest natural numbers $p$ and $q$ such that the complex numbers $z$ and $w$ are equal.","12$, $q=24$.",medium,"## First Solution. Let's write the complex numbers $1-i \sqrt{3}, 1+i$ in trigonometric form: $$ \begin{aligned} 1-i \sqrt{3} & =2\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right) \\ 1+i & =\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \end{aligned} $$ Then we have $$ \begin{aligned} & z=2^{p}\left(\cos \frac{5 p \pi}{3}+i \sin \frac{5 p \pi}{3}\right) \\ & w=\sqrt{2}^{q}\left(\cos \frac{q \pi}{4}+i \sin \frac{q \pi}{4}\right) \end{aligned} $$ If $z=w$, then $2^{p}=\sqrt{2}^{q}$ and $\frac{5 p \pi}{3}=\frac{q \pi}{4}+2 k \pi, k \in \mathbb{Z}$. From $2^{p}=\sqrt{2}^{q}$, or $2^{p}=2^{\frac{q}{2}}$, it follows that $q=2 p$. $$ \begin{gathered} \frac{5 p \pi}{3}-\frac{q \pi}{4}=2 k \pi, k \in \mathbb{Z} \\ \frac{5 p \pi}{3}-\frac{2 p \pi}{4}=2 k \pi, k \in \mathbb{Z} \\ \frac{7 p \pi}{6}=2 k \pi, k \in \mathbb{Z} \\ p=\frac{12}{7} k, k \in \mathbb{Z} \end{gathered} $$ Thus, the smallest value of $k$ for which $p$ is a natural number is 7. Then $p=12, q=24$. ## Second Solution. Notice that from $|z|=|w|$ it follows that $2^{p}=(\sqrt{2})^{q}$. Then $2 p=q$. We are looking for $p$ such that $(1-i \sqrt{3})^{p}=(1+i)^{2 p}$, or, $(1-i \sqrt{3})^{p}=(2 i)^{p},\left(\frac{1-i \sqrt{3}}{2}\right)^{p}=i^{p}$. The left side can have only 6 values: $1, \frac{1}{2}-i \frac{\sqrt{3}}{2},-\frac{1}{2}-i \frac{\sqrt{3}}{2},-1,-\frac{1}{2}+i \frac{\sqrt{3}}{2}, \frac{1}{2}+i \frac{\sqrt{3}}{2}$, (for $p \in\{6 m, 6 m+1,6 m+2,6 m+3,6 m+4,6 m+5\}, m \in \mathbb{N}$ ). The right side can have 4 values $1, i,-1,-i$ (for $p \in\{4 l, 4 l+1,4 l+2,4 l+3\}, l \in$ $\mathbb{N}$ ). Since the left and right sides must be equal, $p$ is such that both are equal to 1 or both are equal to -1. In the first case, $p=6 m=4 l$, so $p$ is divisible by 3 and 4, hence $p=12 k, k \in \mathbb{N}$. In the second case, $p=6 m+3=4 l+2$ which is impossible because $p$ would be both even and odd at the same time. Finally, the solution consists of all numbers $p$ divisible by 12. The smallest sought numbers are $p=12$, $q=24$." 77a7ef8bee32,"3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ again at point $X$ and circle $\omega_{2}$ again at point $Y$. Point $M$ is the midpoint of segment $XY$. Find the angle $OMX$.",$90^{\circ}$,medium,"Answer: $90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-33.jpg?height=497&width=691&top_left_y=1970&top_left_x=702) Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A C$, i.e., $\angle A X C=\angle B A C=\alpha$. Similarly, we get $\angle A Y B=\alpha$. Extend quadrilateral $C X Y B$ to form triangle $X Y Z$ (see figure). Note that $$ \angle B O C=2 \angle B A C=2 \alpha=\angle Z X Y+\angle Z Y X=180^{\circ}-\angle X Z Y $$ which means quadrilateral $O B Z C$ is cyclic. Then $$ \angle O Z B=\angle O C B=\angle O B C=\angle O Z C $$ Therefore, ray $Z O$ is the angle bisector of isosceles triangle $X Z Y$, which is also the median and altitude. Thus, point $M$ lies on ray $Z O$ and $\angle O M X=90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-34.jpg?height=500&width=691&top_left_y=818&top_left_x=702) Solution 2. Let line $X Y$ intersect circle $\omega$ again at point $K$. Since line $A C$ is tangent to circle $\omega_{2}$, $$ \angle B Y K=\angle B Y A=\angle B A C $$ Angle $Y B A$ subtends arc $Y A$ of circle $\omega_{2}$, not containing point $B$. Therefore, it is equal to the angle between chord $Y A$ and the tangent to $\omega_{2}$ at point $A$, i.e., angle $X A C$. Considering the cyclic quadrilateral $A B C K$, we get $$ \angle Y B A=\angle X A C=\angle K A C=\angle K B C $$ which implies $\angle Y B K=\angle A B C$. Therefore, triangles $Y B K$ and $A B C$ are similar by two angles. Then $$ \frac{Y K}{A C}=\frac{B K}{B C} \Longleftrightarrow Y K=\frac{A C \cdot B K}{B C} $$ Similarly, it can be verified that triangles $C A X$ and $C B K$ are similar and $\frac{A X}{B K}=\frac{A C}{B C}$. Thus, $$ A X=\frac{A C \cdot B K}{B C}=Y K $$ From this, it follows that point $M$ is the midpoint of segment $A K$. Note that point $O$ is the circumcenter of triangle $A K B$ and, therefore, lies on the perpendicular bisector of side $A K$. Hence, $\angle O M X=\angle O M K=90^{\circ}$." 7142726c5da3,"2. Around a table are sitting 4 people, each of whom can either be a knight (who always tells the truth) or a knave (who always lies). Each of those present states: ""Of the other three people sitting at this table with me, the knaves are exactly two."" What is the total number of knaves sitting at the table? (A) none (B) definitely 2 (C) definitely all 4 (D) definitely 1 (E) the information provided is not sufficient to determine it",is $(E)$,easy,"(2) The correct answer is $(E)$. If at the table there are two scoundrels and two knights, each knight actually tells the truth, because they see two scoundrels, and each scoundrel actually lies, because they see only one scoundrel. Similarly, if at the table there are only scoundrels, each sees three scoundrels, and thus actually lies. The provided elements therefore do not allow concluding whether there are two or four scoundrels at the table. Problem proposed by Matteo Protopapa." 0b2c3d5ad87d,"On the paper, there were several consecutive positive multiples of a certain natural number greater than one. Radek pointed to one of the written numbers: when he multiplied it by the number that was adjacent to it on the left, he got a product that was 216 less than when he multiplied it by the number that was adjacent to it on the right. Which number could Radek have pointed to? Find all possible options. (L. Šimünek)",See reasoning trace,medium,"Let the natural number whose multiples were written on the paper be denoted by $n$; according to the problem, $n>1$. The numbers on the paper can thus be denoted as $$ (k-1) n, \quad k n, \quad(k+1) n, $$ where the unknown $k$ is a natural number; for all three expressions to be positive, $k>1$. Radek thus received the products $(k-1) k n^{2}$ and $(k+1) k n^{2}$, whose difference is $2 k n^{2}$. It is given that $2 k n^{2}=216$, which simplifies to $$ k n^{2}=108=2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \text {. } $$ Respecting the conditions $n>1$ and $k>1$, we get three different solutions: | $n$ | 2 | 3 | 6 | | :---: | :---: | :---: | :---: | | $k$ | 27 | 12 | 3 | | $k n$ | 54 | 36 | 18 | There are three possible numbers that Radek could have pointed to: 54, 36, or 18. Scoring suggestion. 2 points for the equation $k n^{2}=108=2^{2} \cdot 3^{3}$ or its equivalent; 1 point for each solution that meets the problem's conditions; 1 point for a correctly formulated conclusion. Listing the three terms of the sequence is not a necessary part of the solution. For clarity, we list them: a) $52,54,56$; b) $33,36,39$; c) $12,18,24$." 3be50c66f17f,"One, (40 points) Find the smallest positive integer $m$ such that the equation $$ 2^{x}+3^{y}-5^{z}=2 m $$ has no positive integer solutions for $x, y, z$.",See reasoning trace,medium,"$$ \begin{array}{l} 2=2^{2}+3-5,4=2+3^{3}-5^{2}, \\ 6=2^{3}+3-5,8=2^{2}+3^{2}-5, \\ 10=2^{3}+3^{3}-5^{2}, 12=2^{3}+3^{2}-5, \\ 14=2^{4}+3-5,16=2^{5}+3^{2}-5^{2}, \\ 18=2^{4}+3^{3}-5^{2}, 20=2^{4}+3^{2}-5 . \end{array} $$ Below we prove that $2^{x}+3^{y}-5^{z}=22$ has no solution. If $x=1$, then $3^{y}=20+5^{z}$. Clearly, this equation has no solution. When $x \geqslant 2$, $2^{x} \equiv 0(\bmod 4)$, then $(-1)^{y} \equiv 3(\bmod 4)$. Thus, $y$ is odd. $$ \text { Also, } 2^{x}+3^{y}-5^{z} \equiv(-1)^{x}-(-1)^{z} \equiv 1(\bmod 3) \text {, } $$ Therefore, $x$ is odd and $z$ is even. Let $x=2 \alpha+1, y=2 \beta+1\left(\beta \in \mathbf{N}, \alpha \in \mathbf{N}_{+}\right)$. Then $2^{x}+3^{y}-5^{z}=2^{2 \alpha+1}+3^{2 \beta+1}-5^{z}$ $=2 \times 4^{\alpha}+3 \times 9^{\beta}-5^{z}$ $\equiv 2(-1)^{\alpha}+3(-1)^{\beta}$ $\equiv 0$ or 1 or $-1(\bmod 5)$. But $22 \equiv 2(\bmod 5)$, so this equation has no integer solution. Therefore, the smallest positive integer $m$ that satisfies the condition is 11." a2d6807df97f,"18. If the three sides of a right-angled triangle are $a, b, c, \angle B=90^{\circ}$, then, the nature of the roots of the equation $$ a\left(x^{2}-1\right)-2 c x+b\left(x^{2}+1\right)=0 $$ with respect to $x$ is ( ). (A) It has two equal real roots (B) It has two unequal real roots (C) It has no real roots (D) Cannot be determined",See reasoning trace,easy,"18. A. $$ \begin{array}{l} \Delta=(2 c)^{2}-4(a+b)(b-a) \\ =4\left(c^{2}+a^{2}-b^{2}\right)=0 . \end{array} $$" 65ba4758ab10,"4. Let $S_{n}=1+\frac{1}{1+\frac{1}{3}}+\frac{1}{1+\frac{1}{3}+\frac{1}{6}}+\cdots+$ $\frac{1}{1+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{1}{k_{n}}}$, where $k_{n}=\frac{n(n+1)}{2}$ $\left(n \in \mathbf{N}_{+}\right)$, and let $T_{0}$ be the largest integer $T$ that satisfies the inequality $S_{2006}>T$. Among the following 4 numbers, which one is closest to $T_{0}$? (A) 2006 (B) 2005 (C) 1006 (D) 1005",See reasoning trace,medium,"4.C. $$ \begin{array}{l} \text { Given } \frac{1}{a_{n}}=1+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{2}{n(n+1)} \\ =2\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{n(n+1)}\right] \\ =2\left(1-\frac{1}{n+1}\right)=\frac{2 n}{n+1}, \end{array} $$ we know that $a_{n}=\frac{n+1}{2 n}=\frac{1}{2}\left(1+\frac{1}{n}\right)$. Therefore, $$ \begin{array}{l} S_{2000}=\sum_{k=1}^{2000} a_{k}=\frac{1}{2} \sum_{k=1}^{2000}\left(1+\frac{1}{k}\right) \\ =\frac{1}{2} \times 2006+\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006}\right) . \end{array} $$ $$ \begin{array}{l} \text { Also, } 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006} \\ >1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+ \\ \cdots+(\underbrace{\frac{1}{1024}+\frac{1}{1024}+\cdots+\frac{1}{1024}}_{\text {12 terms }})+ \\ (\underbrace{\frac{1}{2048}+\frac{1}{2048}+\cdots+\frac{1}{2048}}_{\text {22 terms }}) \\ >1+\frac{10}{2}=6, \end{array} $$ then $S_{2000}>1003+3=1006$. And $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006}$ $$ \begin{array}{l} <1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)+ \\ \cdots+(\underbrace{\frac{1}{512}+\frac{1}{512}+\cdots+\frac{1}{512}}_{512 \text{ terms }})+ \\ (\underbrace{\frac{1}{1024}+\frac{1}{1024}+\cdots+\frac{1}{1024}}_{1024 \text{ terms }}) \\ <10+1=11, \end{array} $$ then $S_{2000}<1003+5.5=1008.5$. Thus, $1006b>0)$. If there exists a point $P$ on the ellipse, different from $A_{1}$ and $A_{2}$, such that $\overrightarrow{P O} \cdot \overrightarrow{P A_{2}}=0$, where $O$ is the origin, then the range of the eccentricity $e$ of the ellipse is ( ). (A) $\left(0, \frac{1}{2}\right)$ (B) $\left(0, \frac{\sqrt{2}}{2}\right)$ (C) $\left(\frac{1}{2}, 1\right)$ (D) $\left(\frac{\sqrt{2}}{2}, 1\right)$",See reasoning trace,medium,"7. D. From the given, we know $\angle O P A_{2}=90^{\circ}$. Let $P(x, y)(x>0)$. The equation of the circle with $\mathrm{OA}_{2}$ as its diameter is $$ \left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4} \text {, } $$ Combining this with the ellipse equation, we get $$ \left(1-\frac{b^{2}}{a^{2}}\right) x^{2}-a x+b^{2}=0 \text {. } $$ From the given, we know that this equation has real roots in $(0, a)$. This leads to $0\frac{1}{2} \text {. } $$ Therefore, the range of $e$ is $\left(\frac{\sqrt{2}}{2}, 1\right)$." bbcc2c861860,"The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ is equal to: $\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$",\textbf{(A),easy,"Simplifying $\sqrt{\dfrac{4}{3}}$ yields $\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$. Simplifying $\sqrt{\dfrac{3}{4}}$ yields $\dfrac{\sqrt{3}}{2}$. $\dfrac{2\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}=\dfrac{4\sqrt{3}}{6}-\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{6}$. Since we cannot simplify further, the correct answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}$" e46149427b8f,"The points $A(-6,-1), B(2,3), C(-1,4)$ are given in a Cartesian coordinate system. Determine the point $D$ such that the quadrilateral $A B C D$ is an isosceles trapezoid $(A B \| C D)$.","-5, v=2$.",medium,"I. Solution: By construction, the point $D$ is to be established as the intersection of the line parallel to $AB$ (briefly $a$) passing through $C$ and the circle centered at $A$ with radius $BC$; let us follow this line of thought using the methods of coordinate geometry. The slope of $a$ from the given data is $\frac{3+1}{2+6}=\frac{1}{2}$, so the equation of $c$ is: $$ y-4=\frac{1}{2}(x+1) $$ On the other hand, the square of the length of the side $BC$ is $(4-3)^{2}=(-1-2)^{2}=10$, so the equation of the circle mentioned is: $$ (x+6)^{2}+(y+1)^{2}=10 $$ Now, the coordinates of $D$ can be found from the system of equations formed by (1) and (2): $D_{1}(-5,2)$ and $D_{2}(-9,0)$. Of these, $D_{1}$ forms a ""conventional"" trapezoid, while $D_{2}$ completes $ABC$ into a parallelogram. László Gyuris (Gyöngyös, Vak Bottyán g. III. o. t.) II. Solution: Utilize the axial symmetry of the isosceles trapezoid, and construct $D$ as the reflection of $C$ over the axis, which is the perpendicular bisector of the parallel sides. The points on the axis $t$, being equidistant from $A$ and $B$, satisfy the equation $$ (x+6)^{2}+(y+1)^{2}=(x-2)^{2}+(y-3)^{2} $$ which, when rearranged in the usual form, is $$ 2 x+y+3=0 $$ The line through $C$ and perpendicular to $t$ is identical to $c$, so the midpoint $F$ of side $CD$, which we wish to reflect $C$ over, is given by the solution of the system of equations formed by (1) and (3): $F(-3,3)$. Now, denoting the coordinates of $D$ by $(u, v)$, from the fact that $F$ bisects $CD$: $$ \frac{-1+u}{2}=-3 \quad \text { and } \quad \frac{4+v}{2}=3 $$ and from this, $u=-5, v=2$." 8a8b8d28d6de,"45. a) What values can the following quantities take: 1) angle $A$ of triangle $A B C$; 2) angle $B$ 3) angle $C$, if $A \leqslant B \leqslant C$? b) The area of a triangle with sides $a, b$, and $c$, where $a \leqslant b \leqslant c$, is 1. What values can 1) side $a$ of the triangle; 2) side $b$ 3) side $c$ take?",See reasoning trace,medium,"45. a) It is clear that 1) $09\)? (1) \(-3-3\) (3) \(x-3\) (5) \(x>1\) or \(x<-3\) )",See reasoning trace,easy,"} The following inequalities are equivalent to the given one \[ \begin{aligned} x^{2}+2 x-3 & >0 \\ (x+1)^{2} & >4 \\ (x+1)^{2}-2^{2} & >0 \\ (x-1)(x+3) & >0 \end{aligned} \] By case analysis, we obtain that the last condition is equivalent to: \(x>1\) or \(x<-3\). Thus, it is shown that among the conditions (1) to (5), only condition (5) is equivalent to the given one. Solutions of the second round 1969 adopted from [5] \subsection*{7.11.3 Third Round 1969, Class 10}" aa556e3b9b3a,"5. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1$, and subsequent terms are given by the formula $a_{n}=a_{n-1}+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$, then $a_{4}=$ A. $\frac{7}{4}$ B. $-\frac{7}{4}$ C. $\frac{4}{7}$ D. $-\frac{4}{7}$",$A$,easy,"$$ a_{n}=a_{n-1}+\frac{1}{n-1}-\frac{1}{n} \Rightarrow a_{n}+\frac{1}{n}=a_{n-1}+\frac{1}{n-1}=\cdots=a_{1}+1=2 \Rightarrow a_{n}=2-\frac{1}{n} \text {, } $$ Thus $a_{4}=2-\frac{1}{4}=\frac{7}{4}$. Therefore, the answer is $A$." 51d5fb188d40,"452. Natural numbers $a$ and $b$, where $a>b$, are coprime. Find all values of the greatest common divisor of the numbers $a+b$ and $a-b$.",1 or 2,medium,"$\triangle$ Let $$ \text { GCD }(a+b, a-b)=d $$ Then $$ (a+b): d, \quad(a-b): d $$ Therefore, the sum and difference of the numbers $a+b$ and $a-b$ are divisible by $d$. The sum of these numbers is $2 a$, and the difference is $-2 b$. We get: $$ 2 a: d, \quad 2 b: d . $$ But the numbers $a$ and $b$ are coprime by condition, so 2 is divisible by $d: 2: d$. Hence $d=1$ or $d=2$. Are both of these cases possible? Both: $d=1$, if the numbers $a$ and $b$ have different parity, and $d=2$, if they are odd. Answer: 1 or 2." e481961153f2,"## Calculate the definite integral: $$ \int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x \, d x $$",See reasoning trace,medium,"## Solution $$ \int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x d x= $$ Let's denote: $$ \begin{aligned} & u=x^{2}-3 x ; d u=(2 x-3) d x \\ & d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x \end{aligned} $$ Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get: $$ \begin{aligned} & =\left.\left(x^{2}-3 x\right) \cdot\left(-\frac{1}{2} \cos 2 x\right)\right|_{0} ^{3}-\int_{0}^{3}\left(-\frac{1}{2} \cos 2 x\right) \cdot(2 x-3) \cdot d x= \\ & =\left(3^{2}-3 \cdot 3\right) \cdot\left(-\frac{1}{2} \cos (2 \cdot 3)\right)-\left(0^{2}-3 \cdot 0\right) \cdot\left(-\frac{1}{2} \cos (2 \cdot 0)\right)+\frac{1}{2} \int_{0}^{3} \cos 2 x \cdot(2 x-3) \cdot d x= \\ & =0 \cdot\left(-\frac{1}{2}\right) \cdot \cos 6-0 \cdot\left(-\frac{1}{2}\right) \cos 0+\frac{1}{2} \int_{0}^{3} \cos 2 x \cdot(2 x-3) \cdot d x=\frac{1}{2} \int_{0}^{3} \cos 2 x \cdot(2 x-3) \cdot d x= \end{aligned} $$ Let's denote: $$ \begin{aligned} & u=2 x-3 ; d u=2 d x \\ & d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x \end{aligned} $$ Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get: $$ \begin{aligned} & =\frac{1}{2} \cdot\left(\left.(2 x-3) \cdot \frac{1}{2} \sin 2 x\right|_{0} ^{3}-\int_{0}^{3} \frac{1}{2} \sin 2 x \cdot 2 d x\right)= \\ & =\frac{1}{2} \cdot\left((2 \cdot 3-3) \cdot \frac{1}{2} \sin (2 \cdot 3)-(2 \cdot 0-3) \cdot \frac{1}{2} \sin (2 \cdot 0)-\int_{0}^{3} \sin 2 x d x\right)= \end{aligned} $$ $$ \begin{aligned} & =\frac{1}{2} \cdot\left(3 \cdot \frac{1}{2} \cdot \sin 6-(-3) \cdot \frac{1}{2} \cdot 0-\left.\left(-\frac{1}{2} \cos 2 x\right)\right|_{0} ^{3}\right)= \\ & =\frac{1}{2} \cdot\left(\frac{3}{2} \sin 6+\frac{1}{2} \cos (2 \cdot 3)-\frac{1}{2} \cos (2 \cdot 0)\right)= \\ & =\frac{1}{2} \cdot\left(\frac{3}{2} \sin 6+\frac{1}{2} \cos 6-\frac{1}{2}\right)=\frac{3}{4} \sin 6+\frac{1}{4} \cos 6-\frac{1}{4}=\frac{3 \sin 6+\cos 6-1}{4} \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 2-14 » Categories: Kuznetsov's Problem Book Integrals Problem 2 | Integrals Ukrainian Banner Network - Last modified: 21:12, 26 February 2009. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 2-15 ## Material from PlusPi" 63ace64e1cf8,"Let $P$ be a cubic polynomial with $P(0) = k$, $P(1) = 2k$, and $P(-1) = 3k$. What is $P(2) + P(-2)$ ? $\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$",\textbf{(E),easy,"Let $P(x) = Ax^3+Bx^2 + Cx+D$. Plugging in $0$ for $x$, we find $D=k$, and plugging in $1$ and $-1$ for $x$, we obtain the following equations: \[A+B+C+k=2k\] \[-A+B-C+k=3k\] Adding these two equations together, we get \[2B=3k\] If we plug in $2$ and $-2$ in for $x$, we find that \[P(2)+P(-2) = 8A+4B+2C+k+(-8A+4B-2C+k)=8B+2k\] Multiplying the third equation by $4$ and adding $2k$ gives us our desired result, so \[P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}\]" 0e643b0aec11,"The height drawn to the base of an isosceles triangle is $h$ and is twice as large as its projection on the lateral side. Find the area of the triangle. #",$h^{2} \sqrt{3}$,medium,"Find the angle between the given height and the lateral side of the isosceles triangle. ## Solution Let $B P$ be the projection of the height $B K$ of the isosceles triangle $A B C (A B = B C)$ on the lateral side $B C$. Since $B P = \frac{1}{2} B K$, then $\angle K B P = 60^{\circ}$. Therefore, $$ C K = B K \operatorname{tg} 60^{\circ} = h \sqrt{3}, A C = 2 C K = 2 h \sqrt{3} . $$ Thus, $$ S_{\triangle \mathrm{ABC}} = \frac{1}{2} A C \cdot B K = h^{2} \sqrt{3} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-24.jpg?height=320&width=1106&top_left_y=1455&top_left_x=476) Answer $h^{2} \sqrt{3}$." 8b069948b087,"5. Find all values of $a$, for each of which the system $$ \left\{\begin{array}{l} |y|+|y-x| \leqslant a-|x-1| \\ (y-4)(y+3) \geqslant(4-x)(3+x) \end{array}\right. $$ has exactly two solutions.",when $a=7$,medium,"# Problem 5. Answer: when $a=7$. Solution. By the triangle inequality: $$ |y|+|y-x|+|x-1| \geqslant|y-(y-x)-(x-1)|=1 $$ therefore, the first inequality can have solutions only when $a \geqslant 1$. In this case, it is equivalent to the system of inequalities $$ -\frac{a-1}{2} \leqslant y \leqslant \frac{a+1}{2}, \quad-\frac{a-1}{2} \leqslant x \leqslant \frac{a+1}{2}, \quad x-\frac{a+1}{2} \leqslant y \leqslant x+\frac{a-1}{2} $$ which defines a hexagon on the coordinate plane if $a>1$, and a triangle if $a=1$. The second inequality of the system is equivalent to the inequality $(x-0.5)^{2}+(y-0.5)^{2} \geqslant 49 / 2$ and thus defines the complement to the entire plane of the interior of a circle with center $(1 / 2,1 / 2)$ and radius $7 / \sqrt{2}$. If $a=1$, the system has no solutions. Since for $a>1$ the vertices $\left(-\frac{a-1}{2},-\frac{a-1}{2}\right)$ and $\left(\frac{a+1}{2}, \frac{a+1}{2}\right)$ of the larger diagonal of the hexagon are the farthest from the point $(1 / 2,1 / 2)$ among all points of the hexagon, the intersection of the considered sets will consist of exactly two points if and only if the circle of the second set passes through these vertices. The point $(1 / 2,1 / 2)$ is the midpoint of the larger diagonal, and the length of this diagonal is $a \sqrt{2}$. Therefore, $a \sqrt{2}=14 / \sqrt{2}$, i.e., $a=7$." 6ddd45227e79,"$$ \begin{aligned} & \text { [Isosceles, inscribed, and circumscribed trapezoids] } \\ & \text { [ Right triangle with an angle of \$30^\circ\$ ] } \\ & \text { [ Pythagorean theorem (direct and inverse). ] } \\ & \text { [ Equilateral (equiangular) triangle ] } \end{aligned} $$ Quadrilateral $A B C D$ is inscribed in a circle with center $O, \angle B O A=\angle C O D=60^{\circ}$. The perpendicular $B K$, dropped to side $A D$, is equal to $6 ; A D=3 B C$. Find the area of triangle $C O D$.",$\frac{63 \sqrt{3}}{4}$,medium,"$A B C D$ - is an isosceles trapezoid. ## Solution Since $-A B=-C D$, then $B C \| A D$. Therefore, $A B C D$ - is an isosceles trapezoid. $\angle B D A=1 / 2 \angle A O B=30^{\circ}$, hence, $K D=B K \operatorname{ctg} 30^{\circ}=6 \sqrt{3}$, On the other hand, $K D=1 / 2(A D+B C)=2 B C$, from which $A K=B C=3 \sqrt{3}, A B^{2}=$ $A K^{2}+B K^{2}=63$. Triangle $A O B$ is equilateral, therefore, $S_{C O D}=S_{A O B}=\frac{\sqrt{3}}{4} A B^{2}=\frac{63 \sqrt{3}}{4}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_82cd6cdb4fc29ee56532g-86.jpg?height=280&width=483&top_left_y=1029&top_left_x=793) ## Answer $\frac{63 \sqrt{3}}{4}$." f4cf6f52148e,"14. Two spheres $O_{1}$ and $O_{2}$, both with a radius of 1, are tangent to each other and are also tangent to the two half-planes of the right dihedral angle $\alpha-l-\beta$. There is another smaller sphere $O$ with radius $r (r<1)$ that is also tangent to the two half-planes of the dihedral angle, and is externally tangent to both spheres $O_{1}$ and $O_{2}$. Then the value of $r$ is $\qquad$",3-\sqrt{7}$.,medium,"$14 \cdot 3-\sqrt{7}$. It is known that the centers of the three spheres $O, O_{1}, O_{2}$ are all on the bisecting plane $\gamma$ of the right dihedral angle $\alpha-l-\beta$, as shown in Figure 8. In the plane $\gamma$, draw $O_{1} A \perp l, O_{2} B \perp l, O C \perp l$, with the feet of the perpendiculars being $A, B, C$ respectively. Let the spheres $O_{1}$ and $O_{2}$ be tangent at point $T$. Then $$ \begin{array}{l} O O_{1}=O O_{2}=r+1, O_{1} A=O_{2} B=\sqrt{2}, \\ O C=\sqrt{2} r, O_{1} T=O_{2} T=1 . \end{array} $$ Thus, $O T=\sqrt{2}-\sqrt{2} r$. In the right triangle $\triangle O T O_{1}$, by $O O_{1}^{2}=O T^{2}+O_{1} T^{2}$, we get $(r+1)^{2}=(\sqrt{2}-\sqrt{2} r)^{2}+1$. Noting that $r<1$, solving yields $r=3-\sqrt{7}$." f350fca07d40,"Find consecutive integers bounding the expression \[\frac{1}{x_1 + 1}+\frac{1}{x_2 + 1}+\frac{1}{x_3 + 1}+... +\frac{1}{x_{2001} + 1}+\frac{1}{x_{2002} + 1}\] where $x_1 = 1/3$ and $x_{n+1} = x_n^2 + x_n.$",2 \text{ and,medium,"1. Given the sequence \( x_1 = \frac{1}{3} \) and \( x_{n+1} = x_n^2 + x_n \), we need to find consecutive integers bounding the sum: \[ S = \sum_{k=1}^{2002} \frac{1}{x_k + 1} \] 2. Using the hint, we can rewrite \( x_{k+1} \) as: \[ x_{k+1} = x_k^2 + x_k = x_k(1 + x_k) \] This implies: \[ \frac{1}{1 + x_k} = \frac{1}{x_k} - \frac{1}{x_{k+1}} \] 3. Therefore, the sum \( S \) can be expressed as a telescoping series: \[ S = \sum_{k=1}^{2002} \left( \frac{1}{x_k} - \frac{1}{x_{k+1}} \right) \] 4. Notice that in a telescoping series, most terms cancel out: \[ S = \left( \frac{1}{x_1} - \frac{1}{x_2} \right) + \left( \frac{1}{x_2} - \frac{1}{x_3} \right) + \cdots + \left( \frac{1}{x_{2001}} - \frac{1}{x_{2002}} \right) + \left( \frac{1}{x_{2002}} - \frac{1}{x_{2003}} \right) \] This simplifies to: \[ S = \frac{1}{x_1} - \frac{1}{x_{2003}} \] 5. Given \( x_1 = \frac{1}{3} \), we have: \[ \frac{1}{x_1} = 3 \] 6. To find \( \frac{1}{x_{2003}} \), we observe that \( x_n \) grows very quickly. For large \( n \), \( x_n \) becomes very large, making \( \frac{1}{x_{2003}} \) very small. Thus, \( \frac{1}{x_{2003}} \approx 0 \). 7. Therefore, the sum \( S \) is approximately: \[ S \approx 3 \] 8. Since \( \frac{1}{x_{2003}} \) is positive but very small, \( S \) is slightly less than 3. Hence, the consecutive integers bounding \( S \) are 2 and 3. The final answer is \( \boxed{2 \text{ and } 3} \)" f11be3cc86e8,"1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 14, and 14, respectively. Find the greatest possible value of $f(x)$.",15,medium,"Answer: 15. Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text{v}}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n-1.5)^{2}+c$. Since $f(n)=6, f(n+1)=14$, we get $\frac{9}{4} a+c=6, \frac{a}{4}+c=14$, from which $a=-4, c=15$. But $c=f\left(x_{\text{v}}\right)$ and is the maximum value of the function." 5b1ac8f72ff6,"[b]p1A[/b] Positive reals $x$, $y$, and $z$ are such that $x/y +y/x = 7$ and $y/z +z/y = 7$. There are two possible values for $z/x + x/z;$ find the greater value. [b]p1B[/b] Real values $x$ and $y$ are such that $x+y = 2$ and $x^3+y^3 = 3$. Find $x^2+y^2$. [b]p2[/b] Set $A = \{5, 6, 8, 13, 20, 22, 33, 42\}$. Let $\sum S$ denote the sum of the members of $S$; then $\sum A = 149$. Find the number of (not necessarily proper) subsets $B$ of $A$ for which $\sum B \ge 75$. [b]p3[/b] $99$ dots are evenly spaced around a circle. Call two of these dots ”close” if they have $0$, $1$, or $2$ dots between them on the circle. We wish to color all $99$ dots so that any two dots which are close are colored differently. How many such colorings are possible using no more than $4$ different colors? [b]p4[/b] Given a $9 \times 9$ grid of points, count the number of nondegenerate squares that can be drawn whose vertices are in the grid and whose center is the middle point of the grid. PS. You had better use hide for answers. Collected [url=https://artof",\frac{7,medium,"Given the equations: \[ x + y = 2 \] \[ x^3 + y^3 = 3 \] We start by using the identity for the sum of cubes: \[ (x + y)^3 = x^3 + y^3 + 3xy(x + y) \] Substituting the given values: \[ 2^3 = 3 + 3xy \cdot 2 \] \[ 8 = 3 + 6xy \] Solving for \(xy\): \[ 8 - 3 = 6xy \] \[ 5 = 6xy \] \[ xy = \frac{5}{6} \] Next, we use the identity for the sum of squares: \[ x^2 + y^2 = (x + y)^2 - 2xy \] Substituting the known values: \[ x^2 + y^2 = 2^2 - 2 \cdot \frac{5}{6} \] \[ x^2 + y^2 = 4 - \frac{10}{6} \] \[ x^2 + y^2 = 4 - \frac{5}{3} \] \[ x^2 + y^2 = \frac{12}{3} - \frac{5}{3} \] \[ x^2 + y^2 = \frac{7}{3} \] The final answer is \(\boxed{\frac{7}{3}}\)." 3c737ecae347,"Shapovalov A.V. At a round table, 12 places were prepared for the jury with each place labeled with a name. Nikolai Nikolaevich, who arrived first, absent-mindedly sat not in his own place, but in the next one clockwise. Each subsequent jury member, approaching the table, would take their own place or, if it was already occupied, walk around the table clockwise and sit in the first available place. The resulting seating arrangement of the jury members depends on the order in which they approached the table. How many different seating arrangements can arise?",1024 ways,medium,"Let's consider a certain way of seating the jury members. We will call a jury member lucky if they are sitting in their own seat. The first of the unlucky ones (excluding Nikolai Nikolaevich) to approach the table is the one whose seat is taken by Nikolai Nikolaevich (another unlucky one would sit in their still free seat, which contradicts their unluckiness). He takes the seat of the next (clockwise) unlucky jury member. The second of the unlucky ones to approach is the one whose seat is taken by the first unlucky one (for the same reason), and so on. Thus, each unlucky one sits in the next ""unlucky"" seat after their own. Therefore, the seating arrangement is uniquely determined by the division of the jury into lucky and unlucky members. Nikolai Nikolaevich and the one whose seat he took are always unlucky. Any subset of the jury members, not including these two, can be the set of lucky ones. Such a seating arrangement can be implemented, for example, as follows: after Nikolai Nikolaevich, all those we have chosen as lucky (in any order) enter, and then all the others in the order of their seating around the table clockwise. Therefore, the number of seating arrangements is equal to the number of subsets of a set of 10 people, which is \(2^{10}=1024\). ## Answer 1024 ways." ea71e69641af,"We folded a rectangular piece of paper along one of its diagonals. After the fold, the four vertices ended up at the four vertices of a trapezoid, three sides of which are of equal length. What is the length of the shorter side of the original rectangle if the longer side is 12 cm?",See reasoning trace,medium,"Solution. Let the rectangle be $ABCD (BC < AB), AB = 12 \, \text{cm}$, and $AC$ be the diagonal along which the rectangle is folded, meaning that when the $\triangle ABC$ is reflected over $AC$, we get the $\triangle AB'C$. In the trapezoid $ACB'D$, $AD = DB' = B'C$ (note that $AD \neq AC$, because $AD < AC$, since in the right triangle $\triangle DAC$, the leg is smaller than the hypotenuse). ![](https://cdn.mathpix.com/cropped/2024_05_02_51ce44eec2c64e8a38e9g-1.jpg?height=512&width=507&top_left_y=351&top_left_x=798) Consider the circle $k$ with $AC$ as its diameter, and its center at $O$. $\angle ADC = 90^\circ$ (one of the angles of the given rectangle), by the converse of Thales' theorem, point $D$ lies on circle $k$. Since $\angle ABC = 90^\circ$, by reflection, $\angle AB'C = 90^\circ$. Therefore, points $B$ and $B'$ also lie on circle $k$. Based on the above and the folding of the rectangle, $A, D, B', C$ lie on the semicircle $k$ with diameter $AC$. We know that $AD = DB' = B'C$, therefore $$ \angle B'OC = \frac{\angle AOC}{3} = \frac{180^\circ}{3} = 60^\circ $$ (in a circle, equal chords subtend equal central angles). By the theorem of inscribed and central angles: $$ \angle B'AC = \frac{\angle B'OC}{2}, \quad \text{so} \quad \angle B'AC = \frac{60^\circ}{2} = 30^\circ, $$ and thus $\angle CAB = 30^\circ$. If one of the acute angles in a right triangle is $30^\circ$, then the opposite leg is half the hypotenuse. Let $BC = b$, then $AC = 2b$, and in the right triangle $\triangle ABC$, by the Pythagorean theorem: $(2b)^2 = 12^2 + b^2$, from which $b = 4\sqrt{3} \approx 6.93$. Therefore, the shorter side of the given rectangle is approximately $6.93 \, \text{cm}$." f48e2fe1c3bd,"Vassilyev N.B. In a round glass, the axial cross-section of which is the graph of the function $y=x^{4}$, a cherry - a sphere of radius $r$ - is dropped. For what largest $r$ will the sphere touch the lowest point of the bottom? (In other words, what is the maximum radius $r$ of a circle lying in the region $y \geq x^{4}$ and containing the origin?)",to the original problem,medium,"Let's first solve another problem: construct a circle with its center on the $y$-axis, which touches the $x$-axis, and determine the smallest radius $r$ for which it has a common point with the curve $y=x^{4}$, different from the origin (see the figure). In other words, for what smallest $r$ does the system of equations $$ y=x^{4}, \quad x^{2}+(y-r)^{2}=r^{2} $$ have a non-zero solution. Intuitively, it is clear that these problems are equivalent, and we will later prove this rigorously. Substitute $y=x^{4}$ into the equation of the circle. Combine like terms and divide by $x^{2} (x>0)$: $$ x^{6}-2 r x^{2}+1=0 $$ Express $r$ in terms of $x$: $$ r(x)=\frac{1}{2}\left(x^{4}+\frac{1}{x^{2}}\right). $$ The desired number $r_{0}$ is the minimum of this function. The derivative of the function is $$ r^{\prime}(x)=2 x^{3}-\frac{1}{x^{3}} $$ For $x>0$, this derivative behaves as follows: it is negative for $xx_{0}$. This means that $r(x)$ decreases for $0x_{0}$. Thus, the smallest $r$ for which the circle has a common point with the curve $y=x^{4}$ is $$ r_{0}=r\left(x_{0}\right)=\frac{3 \sqrt[3]{2}}{4} $$ It remains to show that this $r_{0}$ gives the answer to the original problem. First, let's show that the corresponding cherry is entirely contained within the glass. Indeed, for any $x \neq 0$, we have $r_{0} \leq r(x)$. Substituting this inequality into the expression for $r(x)$, we get that for any $x$, $$ x^{6}-2 r_{0} x^{2}+1 \geq 0 $$ Multiplying both sides by $x^{2}$ and substituting $y=x^{4}$, we obtain $$ x^{2}+\left(y-r_{0}\right)^{2} \geq r_{0}^{2} $$ for all $x, y=x^{4}$. But this means that the cherry is inside the glass. It remains to show that if $r>r_{0}$, the cherry will not touch the origin. Indeed, in this case, $$ x_{0}^{6}-2 r x_{0}^{2}+1<0 $$ This means that the equation $x^{6}-2 r x^{2}+1=0$ has a non-zero solution, and the situation is the same as when $a=4$; b) when $a=2$, we get $x=0$ and $r=1 / 2$ - the radius of curvature of the graph at the point $(0 ; 0)$, the maximum cherry touches the parabola at a single point c) when $04$. Then $$ \begin{array}{l} \left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=|I|-\left|\bigcup_{k=1}^{6} A_{k}\right| \\ =|I|-\sum_{k=1}^{6}\left|A_{k}\right|+\sum_{j4, y_{j}>4, y_{k}>4$ of $\sum_{k=1}^{6} y_{k}=15$ in positive integers, then $$ \begin{array}{l} \left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=\mathrm{C}_{14}^{5}-6 \mathrm{C}_{10}^{5}+\mathrm{C}_{5}^{2} \mathrm{C}_{6}^{5} \\ =2002-1512+90=580 . \end{array} $$ Since the 5 new weapons are all different, swapping their positions results in different arrangements, so the number of ways to equip the new weapons is $$ 580 \times 5!=69600 \text {. } $$ (Provided by Kong Yangxin)" 552d67e2c060,"In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.",is $10$,medium,"We can set $AE=ED=m$. Set $BD=k$, therefore $CD=3k, AC=4k$. Thereafter, by Stewart's Theorem on $\triangle ACD$ and cevian $CE$, we get $2m^2+14=25k^2$. Also apply Stewart's Theorem on $\triangle CEB$ with cevian $DE$. After simplification, $2m^2=17-6k^2$. Therefore, $k=1, m=\frac{\sqrt{22}}{2}$. Finally, note that (using [] for area) $[CED]=[CAE]=3[EDB]=3[AEB]=\frac{3}{8}[ABC]$, because of base-ratios. Using Heron's Formula on $\triangle EDB$, as it is simplest, we see that $[ABC]=3\sqrt{7}$, so your answer is $10$." 2abbf55c6e56,"7. Cut a 7-meter long iron wire into two segments (it can also be directly cut into two segments), so that the difference in length between these two segments does not exceed 1 meter. If these two segments are used to form two circles respectively, then the maximum value of the sum of the areas of these two circles is $\qquad$ square meters.","\frac{x^2 + y^2}{4 \pi}$. Viewing this as the equation of a circle, $S$ reaches its maximum value of",medium,"7. $\frac{25}{4 \pi}$. Let the lengths of the two segments be $x$ meters and $y$ meters, respectively. Then $x$ and $y$ satisfy the following relationships: $$ \left\{\begin{array}{l} x>0, \\ y>0, \\ x+y \leqslant 7, \\ |x-y| \leqslant 1 . \end{array}\right. $$ The planar region is the shaded part shown in Figure 4. The sum of the areas of the two circles is $S = \frac{x^2 + y^2}{4 \pi}$. Viewing this as the equation of a circle, $S$ reaches its maximum value of $\frac{25}{4 \pi}$ square meters when the circle passes through point $A(4,3)$ or $B(3,4)$." 00fdfd9307ae,"5. Six equal circles $\mathrm{O}, \mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{O}_{4}$, $0_{5}$, on the surface of a sphere with radius 1, $\odot O$ is tangent to the other five circles and $\odot \mathrm{O}_{1}$ is tangent to $\odot \mathrm{O}_{2}$, $\odot \mathrm{O}_{2}$ is tangent to $\odot \mathrm{O}_{3}$, $\odot O_{3}$ is tangent to $\odot O_{4}$, $\odot O_{4}$ is tangent to $\odot O_{5}$, $\odot O_{5}$ is tangent to $\odot O_{1}$, find the radius of these circles.",See reasoning trace,medium,"Let the center of the sphere be $K$, and $\odot O$ and $\odot O_{1}$ touch at $A$. The common tangent line through $A$ is $l$. Then, $K A, O A, O_{1} A$ are all in the plane perpendicular to $l$ through $A$. In this plane, $K A=1$, $O A=O_{1} A=r$. By plane geometry (see Fig. (1)), $K O=K O_{1}=\sqrt{1-r}$, $$ \begin{array}{l} x\left(=O B=\frac{1}{2} O O_{1}\right)=\frac{r \sqrt{1-r^{2}}}{1} \\ =r \sqrt{1-r^{2}}, \\ y(=A B)=\frac{r^{2}}{1}=r^{*} \text {. } \\ \frac{1}{2} O_{1} C \times O K=S_{\triangle O O_{1 K}}=S_{\text {quadrilateral } O_{1} A \cup K} \\ -S_{\triangle O 14 O} \\ =r \sqrt{1-r^{2}}-x y=r \sqrt{1 \cdots r^{2}} \\ -r^{3} \sqrt{1-r}=r \sqrt{1-r^{2}}\left(1-r^{2}\right) \text {, } \\ z\left(=O_{1} C\right)=\approx r\left(1-r^{2}\right) \\ \frac{1}{2} h_{r}=S_{\angle O 1 A O}=r_{\}_{j}}, \quad h=\frac{2 \times y}{r} \\ =2 r \sqrt{1-r^{2}} \text {. } \\ \end{array} $$ Let the plane of $\odot O$ be $Q$, then plane 2 is perpendicular to $K O$, and the distances from $O, O, O, O, O_{5}$ to plane 2 are all $h=2 r \sqrt{1-r}$, so $O_{1}, O_{2}, O_{3}, O_{4}, O_{6}$ are on the same plane $R$, and $K O \perp$ plane $R$, with the foot of the perpendicular being $C$ (see Fig. (2)). The intersection of plane $R$ with the sphere is a circle (with center $C$), and it is easy to see that the pentagon $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4} \mathrm{O}_{0}$ is an inscribed regular pentagon of this circle, and $$ C O_{1}=z, \quad O . O_{-}=2 x, \angle O_{1} C O_{\angle}=\frac{2 \pi}{5} . $$ Therefore, $$ \begin{array}{l} \frac{1}{\sin ^{2} \frac{\pi}{5}}=\left(\frac{z}{x}\right)^{2}=\left(\frac{2\left(1-r^{2}\right)}{\sqrt{1-r^{2}}}\right)^{2} \\ =4\left(1-r^{2}\right), \end{array} $$ which implies $$ 4 r^{2}=4-\frac{1}{\sin ^{2} \frac{\pi}{5}}=3-\operatorname{ctg} \frac{2 \pi}{5} . $$ Thus, $$ r=\frac{1}{2} \sqrt{3-\operatorname{ctg}^{2} \frac{\pi}{5}} . $$" 0d509569ee0f,"In the drawing below, $C$ is the intersection point of $A E$ and $B F$, $A B=B C$, and $C E=C F$. If $\angle C E F=50^{\circ}$, determine the angle $\angle A B C$. ![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-10.jpg?height=552&width=885&top_left_y=1363&top_left_x=491)",See reasoning trace,medium,"Solution Since triangle $CEF$ is isosceles, we have $\angle CEF = \angle CFE = 50^{\circ}$ and $$ \begin{aligned} \angle FCE & =180^{\circ}-\angle CEF-\angle CFE \\ & =180^{\circ}-50^{\circ}-50^{\circ} \\ & =80^{\circ} \end{aligned} $$ We also have $\angle BCA = \angle FCE$, as they are vertically opposite angles. Furthermore, since $ABC$ is isosceles, we have $\angle BAC = \angle BCA = 80^{\circ}$. Finally, we can conclude that $$ \begin{aligned} \angle ABC & =180^{\circ}-\angle BAC-\angle BCA \\ & =180^{\circ}-80^{\circ}-80^{\circ} \\ & =20^{\circ} \end{aligned} $$" 56d3b233e216,"Example 7 The roots of the equation $x^{2}+p x+q=0$ are both positive integers, and $p+q=1992$. Then the ratio of the larger root to the smaller root is $\qquad$.",See reasoning trace,easy,"Given: $\because x_{1}+x_{2}=-p, x_{1} x_{2}=q$, $$ \begin{array}{l} \therefore x_{1} x_{2}-x_{1}-x_{2}=q+p=1992, \\ \left(x_{1}-1\right)\left(x_{2}-1\right)=1993 . \end{array} $$ $\because 1993$ is a prime number, $$ \therefore\left\{\begin{array}{l} x_{1}-1=1, \\ x_{2}-1=1993 . \end{array}\right. $$ Solving, we get $\left\{\begin{array}{l}x_{1}=2, \\ x_{2}=1994 .\end{array}\right.$ $$ \text { Therefore, } \frac{x_{2}}{x_{1}}=997 \text {. } $$" e7f68ea8801b,"Example 2. Consider the sequence $\left\{a_{n}\right\}$ defined by $a_{1}=2, a_{n+1}=\frac{a_{n}}{a_{n}+3}$ $(n \geqslant 1)$. Solve the following (1) If $b_{n}=\frac{1}{a_{n}}$, find the relationship between $b_{n+1}$ and $b_{n}$; (2) Find the general term of the sequence $\left\{a_{n}\right\}$.",\frac{2}{2 \cdot 3^{n-1}-1}$.,medium,"Solution: (1) From the given, we have $$ a_{n}=-\frac{1}{b_{n}}, \quad a_{n+1}=\frac{1}{b_{n+1}} . $$ Substituting the above into the known equation $$ a_{n+1}=\frac{a_{n}}{a_{n}+3} $$ we get $$ 3_{n} \frac{1}{b_{n}+3}=\frac{\frac{1}{b_{n}}}{b_{n}} . $$ From this, we obtain $b_{n+1}=3 b_{n}+1$. (2) From the result in (1), we have $b_{n+1}-b_{n}$ $=3\left(b_{n}-b_{n-1}\right)$. Let $c_{n}=b_{n+1}-\dot{b}_{n}$, then $c_{n+1}=3 c_{n}$. Since $\quad b_{1}=\frac{1}{2}, \quad a_{2}=\frac{2}{2+3}=\frac{2}{5}$, $\therefore b_{2}=\frac{5}{2}$. $\therefore \quad c_{1}=b_{2}-b_{1}=\frac{5}{2}-\frac{1}{2}=2$. When $n \geqslant 2$, $b_{n}=\frac{1}{2}+\sum_{k=1}^{n} 2 \cdot 3^{k-1}=2 \cdot 3^{n-1}-1$ The above formula still holds when $n=1$. Thus, $b_{n}=\frac{2 \cdot 3^{n-1}-1}{2}$, Therefore, $a_{n}=\frac{2}{2 \cdot 3^{n-1}-1}$." 525dc3970e28,"Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $AF$.",336,medium,"By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle. Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and $DB=\frac{840-x}{2}\cdot\sqrt3.$ Let the brackets denote areas. We have \[[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}\] and \[[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).\] We set up and solve an equation for $x:$ \begin{align*} \frac{[AFG]}{[BED]}&=\frac89 \\ \frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\ \frac{2x^2}{(840-x)^2}&=\frac89 \\ \frac{x^2}{(840-x)^2}&=\frac49. \end{align*} Since $00.$ Therefore, we take the positive square root for both sides: \begin{align*} \frac{x}{840-x}&=\frac23 \\ 3x&=1680-2x \\ 5x&=1680 \\ x&=\boxed{336}. \end{align*} ~MRENTHUSIASM" 1e61f6f3247d,There exist two positive numbers $ x$ such that $ \sin(\arccos(\tan(\arcsin x)))\equal{}x$. Find the product of the two possible $ x$.,1,medium,"1. Let $\arcsin x = \theta$. Then, $x = \sin \theta$ and $\theta \in [0, \frac{\pi}{2}]$ since $x$ is a positive number. 2. We need to find $\tan(\arcsin x)$. Using the right triangle representation, we have: \[ \sin \theta = x \quad \text{and} \quad \cos \theta = \sqrt{1 - x^2} \] Therefore, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}} \] 3. Let $\beta = \arccos \left( \frac{x}{\sqrt{1 - x^2}} \right)$. Then, $\cos \beta = \frac{x}{\sqrt{1 - x^2}}$. 4. We need to find $\sin \beta$. Using the right triangle representation again, we have: \[ \cos \beta = \frac{x}{\sqrt{1 - x^2}} \quad \text{and} \quad \sin \beta = \sqrt{1 - \cos^2 \beta} \] Therefore, \[ \sin \beta = \sqrt{1 - \left( \frac{x}{\sqrt{1 - x^2}} \right)^2} = \sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - x^2 - x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}} \] 5. Given that $\sin(\arccos(\tan(\arcsin x))) = x$, we have: \[ \sqrt{\frac{1 - 2x^2}{1 - x^2}} = x \] 6. Squaring both sides, we get: \[ \left( \sqrt{\frac{1 - 2x^2}{1 - x^2}} \right)^2 = x^2 \implies \frac{1 - 2x^2}{1 - x^2} = x^2 \] 7. Multiplying both sides by $1 - x^2$, we obtain: \[ 1 - 2x^2 = x^2 (1 - x^2) \implies 1 - 2x^2 = x^2 - x^4 \] 8. Rearranging terms, we get: \[ x^4 - 3x^2 + 1 = 0 \] 9. Let $y = x^2$. Then, the equation becomes: \[ y^2 - 3y + 1 = 0 \] 10. Solving this quadratic equation using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get: \[ y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \] 11. Therefore, the solutions for $x^2$ are: \[ x^2 = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x^2 = \frac{3 - \sqrt{5}}{2} \] 12. Since $x$ is positive, we take the positive square roots: \[ x = \sqrt{\frac{3 + \sqrt{5}}{2}} \quad \text{and} \quad x = \sqrt{\frac{3 - \sqrt{5}}{2}} \] 13. We are asked to find the product of the two possible values of $x$. Therefore, we calculate: \[ \left( \sqrt{\frac{3 + \sqrt{5}}{2}} \right) \left( \sqrt{\frac{3 - \sqrt{5}}{2}} \right) = \sqrt{\left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right)} \] 14. Simplifying the product inside the square root: \[ \left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right) = \frac{(3 + \sqrt{5})(3 - \sqrt{5})}{4} = \frac{9 - 5}{4} = \frac{4}{4} = 1 \] 15. Therefore, the product of the two possible values of $x$ is: \[ \sqrt{1} = 1 \] The final answer is $\boxed{1}$" ebbc68614e4f,"# The commander of a tank battalion, in celebration of being awarded a new military rank, decided to invite soldiers to a tank festival, where the main delicacy is buckwheat porridge. The commander discovered that if the soldiers are lined up by height, there is a certain pattern in the change of their demand functions. The demand for buckwheat porridge of the shortest soldier is given by $Q d=510-5.1 P$, the next tallest soldier's demand is $Q d=520-5.2 P$, the next one's is $Q d=530-5.3 P$, and so on. The commander's individual demand for buckwheat porridge is $Q d=500-5 P$. The battalion commander always tries to feed as many soldiers as possible. Apart from the commander and the invited soldiers, no one else demands buckwheat porridge. At the festival, 25 perfectly competitive firms offer buckwheat porridge, with each firm's supply function being $Q s=302 P$. The battalion commander, after consulting with his family, decided to make a surprise - initially, the guests and the commander will make decisions about consuming buckwheat porridge, assuming that everyone will pay for it themselves, but at the end of the celebration, the commander's family will pay the total bill from their savings, which amount to 2,525,000 monetary units. If this amount is insufficient, the commander will ask all guests to equally share the remaining bill. It is known that the equilibrium price is set at 20 monetary units. Assume that if guests, like the commander, have to make a certain fixed payment, this payment does not affect their demand for buckwheat porridge. (a) How is the individual demand of the 60th soldier invited to the celebration described? How will the aggregate demand for buckwheat porridge be described if the commander plans to invite 40 soldiers and does not refuse the porridge himself? Explain the construction of the demand function in this case. (b) How many soldiers did the commander invite to the celebration? Did the guests have to pay part of the bill? If so, how much did each of them pay?","$Q_{d}=1100-11 P, Q_{d}=28700-287 P$",medium,"# Solution (a) The commander tries to feed as many soldiers as possible, which means he will invite relatively short soldiers first - all other things being equal, their consumption of porridge is less. Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n- (5+0.1 n) P$, where $n$ is the ordinal number of the soldier. Also note that all consumers are willing to purchase the product at $P \in [0 ; 100]$, so to find the market demand, it is sufficient to sum the individual demand functions. The individual demand of the 60th soldier is $Q_{d}=1100-11 P$. To find the demand of 40 soldiers, we can use the formula for the arithmetic progression: $Q_{d}=\frac{510-5.1 P+500+10 n-(5+0.1 n) P)}{2} \cdot n$ $Q_{d}=\frac{1410-14.1 P}{2} \cdot 40$ $Q_{d}=28200-282 P$ Adding the commander's demand, we get the market demand: $Q_{d}=28700-287 P$. Answer: $Q_{d}=1100-11 P, Q_{d}=28700-287 P$. (b) Knowing the individual supply of one firm, we can find the market supply: $Q_{s}=302 P \cdot 25=7550 P$ It is known that $Q_{s}(20)=Q_{d}(20)$, so $Q_{d}=7550 \cdot 20=151000$. From the previous part, we know that $Q_{d}=5 n^{2}+505 n+500-\left(0.05 n^{2}+5.05 n+5\right) P$ $Q_{d}(20)=4 n^{2}+404 n+400=151000$ $(4 n+1004)(n-150)=0$ $n_{1}=-251, n_{2}=150$. Obviously, $n>0$. The commander invited 150 guests. $T R=151000 \cdot 20=3020000>2525000$ - the expenses for buckwheat porridge exceed the family budget, which means the soldiers will help pay part of the bill. Each soldier will contribute an amount of $\frac{495000}{150}=3300$ monetary units. Answer: the commander invited 150 guests, each of whom paid 3300 monetary units. ## Grading Criteria (a) Individual demand is formulated - (3p), market demand is formulated - (5p). (b) Market supply function is formulated - (2p), equilibrium quantity is determined - (2p), market demand function in terms of the number of guests is formulated - (2p), maximum number of guests is determined - (3p), behavior of guests and their payment of the bill is determined - (3p). A penalty of 1 point for arithmetic errors that did not lead to significant distortion of the results. Alternative solutions may be awarded full points if they contain a correct and justified sequence of actions." 95c6e546369c,![](https://cdn.mathpix.com/cropped/2024_05_06_1f1bf0225c3b69484645g-36.jpg?height=561&width=616&top_left_y=89&top_left_x=419),67,easy,"Answer: 67. Solution. Since inscribed angles subtended by the same arc are equal, then $\angle A C F=$ $\angle A B F=81^{\circ}$ and $\angle E C G=\angle E D G=76^{\circ}$. Since a right angle is subtended by the diameter, $$ \angle F C G=\angle A C F+\angle E C G-\angle A C E=81^{\circ}+76^{\circ}-90^{\circ}=67^{\circ} $$" 66b2ae876f7c,"10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that $$ a_{k}=\left\{\begin{array}{ll} k, & S_{k-1}0$ that $f^{k}(0)=0$. Since $f(m)=m+a$ or $f(m)=m-b$, it follows that $k$ can be written as $k=r+s$, where $r a-s b=0$. Since $a$ and $b$ are relatively prime, it follows that $k \geq a+b$. Let us now prove that $f^{a+b}(0)=0$. In this case $a+b=r+s$ and hence $f^{a+b}(0)=(a+b-s) a-s b=(a+b)(a-s)$. Since $a+b \mid f^{a+b}(0)$ and $f^{a+b}(0) \in S$, it follows that $f^{a+b}(0)=0$. Thus for $(a, b)=1$ it follows that $k=a+b$. For other $a$ and $b$ we have $k=\frac{a+b}{(a, b)}$." 62b1ee09caa3,"Karolína wrote all three-digit numbers formed by the digits 1, 2, and 3, in which no digit was repeated and 2 was in the tens place. Nikola wrote all three-digit numbers formed by the digits 4, 5, and 6, in which no digit was repeated. Kuba chose one number from Karolína and one number from Nikola so that the sum of these two numbers was even. What was the digit in the units place of the product of the numbers chosen by Kuba? Determine all possibilities. (L. Hozová)",See reasoning trace,medium,"Karolína wrote the numbers $$ 123, \quad 321 . $$ Nikola wrote the numbers $$ 456, \quad 465, \quad 546, \quad 564, \quad 645, \quad 654 . $$ Both of Karolína's numbers are odd. For an even sum, Kuba had to choose an odd number from Nikola. The even sums are given by the following cases: $$ 123+465, \quad 123+645, \quad 321+465, \quad 321+645 $$ In all cases, the digit in the units place of the product is determined by an odd multiple of 5, which can only be 5. Evaluation. 1 point for Karolína's and Nikola's numbers; 2 points for Kuba's choice of an even sum; 2 points for determining the last digit of the product. Note. It is not necessary to list the products to reach the correct conclusion. However, for the quartet of possibilities in $(*)$, these products are, in order, $$ 57195, \quad 79335, \quad 149265, \quad 207045 . $$" 38b6d7ba8928,"7. Suppose ten people each holding a bucket come to a water tap to fill their buckets. Let the time required to fill the bucket of the $i$-th person $(i=1,2, \cdots, 10)$ be $T_{i}$ minutes. Assume these $T_{i}$ are all different. When only one water tap is available, how should the order of these ten people be arranged to minimize their total waiting time (including the time each person spends filling their own bucket)? What is this minimum total time? (You must prove your argument).",See reasoning trace,medium,"7. If the buckets are filled in the order of $i_{1}, i_{2}, \cdots, i_{10}$, then the total time required is $$ f(t)=T_{i_{1}}+\left(T_{i_{1}}+T_{i_{2}}\right)+\cdots+\left(T_{i_{1}}+T_{i_{2}}+\cdots+T_{i_{10}}\right)=10 T_{i_{1}}+9 T_{i_{2}}+\cdots+2 T_{i_{9}}+T_{i_{10}} \text {. } $$ Assuming $T_{1}1,7558$ so $3 \lg 56+\lg 57>7,0001$.) In the third case, it does not hold, so our problem can only have two solutions: $$ \begin{array}{lll} n_{1}=3, & A_{1}=5, & B_{1}=6 ; \\ n_{2}=4, & A_{2}=2, & B_{2}=5 . \end{array} $$ Indeed, $$ \begin{gathered} \frac{1}{56^{3}}=0.00000569 \ldots \quad \text { and } \\ \frac{1}{25^{4}}=0.04^{4}=0.2^{8}=0.00000256 \end{gathered} $$ Imre Major (Budapest, I. István Gymn., II. o. t. ) Béla Tarsó (Veszprém, Lovassy L. Gymn., II. o. t. )" 711f31e09e6f,"Two rectangles $P Q U V$ and $W S T V$ overlap as shown. What is the area of $P Q R S T V$ ? (A) 35 (B) 24 (C) 25 (D) 17 (E) 23 ![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-075.jpg?height=498&width=412&top_left_y=171&top_left_x=1301)",14+15-6=23$.,medium,"Solution 1 Since $P Q U V$ and $W S T V$ are rectangles that share a common right angle at $V$, then $P Q, W S$ and $V T$ are parallel, as are $P V, Q U$, and $S T$. This tells us that all of the angles in the diagram are right angles. Since $P Q U V$ is a rectangle, then $V U=P Q=2$. Since $V T=5$ and $V U=2$, then $U T=V T-V U=5-2=3$. Note that $R S T U$ is a rectangle, since it has four right angles. Therefore, the area of $P Q R S T V$ equals the sum of the areas of rectangles $P Q U V$ and $R S T U$, or $2 \times 7+3 \times 3=23$. (We could also consider the area of $P Q R S T V$ to be the sum of the areas of rectangle $P Q R W$ and rectangle $W S T V$.) ## Solution 2 Since $P Q U V$ and $W S T V$ are rectangles that share a common right angle at $V$, then $P Q, W S$ and $V T$ are parallel, as are $P V, Q U$, and $S T$. This tells us that all of the angles in the diagram are right angles. We can consider $P Q R S T V$ to be a large rectangle $P X T V$ with a smaller rectangle $Q X S R$ removed. ![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-127.jpg?height=506&width=415&top_left_y=636&top_left_x=909) The area of rectangle $P X T V$ is $7 \times 5=35$. Since $P Q U V$ is a rectangle, then $Q U=P V=7$. Since $P V$ is parallel to $Q U$ and $S T$, then $R U=S T=3$. Thus, $Q R=Q U-R U=7-3=4$. Since $W S T V$ is a rectangle, then $W S=V T=5$. Since $V T$ is parallel to $W S$ and $P Q$, then $W R=P Q=2$. Thus, $R S=W S-W R=5-2=3$. Therefore, rectangle $Q X S R$ is 4 by 3, and so has area 12 . Therefore, the area of $P Q R S T V$ is $35-12=23$. ## Solution 3 Since $P Q U V$ and $W S T V$ are rectangles that share a common right angle at $V$, then $P Q, W S$ and $V T$ are parallel, as are $P V, Q U$, and $S T$. This tells us that all of the angles in the diagram are right angles. If we add up the areas of rectangle $P Q U V$ and $W S T V$, we get exactly the region $P Q R S T V$, but have added the area of $W R U V$ twice. Thus, the area of $P Q R S T V$ equals the area of $P Q U V$ plus the area of $W S T V$ minus the area of $W R U V$. We note that rectangle $P Q U V$ is 2 by 7 , rectangle $W S T V$ is 3 by 5 , and rectangle $W R U V$ is 2 by 3 (since $W R=P Q=2$ and $R U=S T=3)$. Therefore, the area of $P Q R S T V$ equals $2 \times 7+3 \times 5-2 \times 3=14+15-6=23$." 037ace9ff192,"Kayla went to the fair with $\$ 100$. She spent $\frac{1}{4}$ of her $\$ 100$ on rides and $\frac{1}{10}$ of her $\$ 100$ on food. How much money did she spend? (A) $\$ 65$ (B) $\$ 32.50$ (C) $\$ 2.50$ (D) $\$ 50$ (E) $\$ 35$",(E),easy,"Kayla spent $\frac{1}{4} \times \$ 100=\$ 25$ on rides and $\frac{1}{10} \times \$ 100=\$ 10$ on food. The total that she spent was $\$ 35$. ANSWER: (E)" ec9757450e88,10.3. Find all values of the parameter $a$ for which the equation $a x^{2}+\sin ^{2} x=a^{2}-a$ has a unique solution.,$a=1$,easy,"Answer: $a=1$. Solution. Note that only $x=0$ can be the unique root of the equation, since due to the evenness of the functions involved, for any solution $x_{0} \neq 0$, $(-x_{0})$ will also be a solution. Therefore, we necessarily get $a^{2}-a=0 \Leftrightarrow a=0$ or $a=1$. Let's check these values. When $a=0$, we have the equation $\sin ^{2} x=0$, which has an infinite set of solutions $x=k \pi(k \in Z)$. If $a=1$, then the equation $x^{2}+\sin ^{2} x=0$ has a unique solution $x=0$." 6010b18b42b3,"3. In the acute triangle $\triangle A B C$, $\angle A: \angle B, \angle C$ are its interior angles. Let $T=\operatorname{ctg} 2 A+\operatorname{ctg} 2 B+\operatorname{ctg} 2 C$. Then it must be ( ). (A.) T $>0$ (B) $\cdots \geqslant 0$ (c) $T<0$ (D) $T \leqslant 0$",See reasoning trace,medium,"3. (C). From $\angle A + \angle B > \frac{\pi}{2}$, we get $\angle A > \frac{\pi}{2} - \angle B$, which means $\operatorname{tg} A > \operatorname{tg}\left(\frac{\pi}{2} - \angle B\right) = \operatorname{ctg} B$. Similarly, $\operatorname{tg} B > \operatorname{ctg} C, \operatorname{tg} C > \operatorname{ctg} A$. Therefore, $$ \begin{array}{l} \operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C > \operatorname{ctg} A + \operatorname{ctg} B + \operatorname{ctg} C \\ \operatorname{ctg} 2 A + \operatorname{ctg} 2 B + \operatorname{ctg} 2 C \\ = \frac{\cos^2 A - \sin^2 A}{2 \sin A \cos A} + \operatorname{ctg} 2 B + \operatorname{ctg} 2 C \\ = \frac{1}{2} (\operatorname{ctg} A - \operatorname{tg} A + \operatorname{ctg} B - \operatorname{tg} B + \operatorname{ctg} C - \operatorname{tg} C) \\ < 0 \end{array} $$" fb8de6e1580d,"5. The teacher wrote thirteen natural numbers on the blackboard and asked Xiaoming to calculate the average (to two decimal places). Xiaoming's answer was 12.43, but the teacher said the last digit was wrong, while the other digits were correct. What should the correct answer be?",See reasoning trace,easy,5. 12.46 c9c0032d4497,"7. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catches two fish, he will say he caught six fish. In response to the question: ""How many fish did your left neighbor catch?"", the answers (not necessarily in the order in which the fishermen stand) were $12,12,20,24,32,42$ and 56. In response to the question: ""How many fish did your right neighbor catch?"", six fishermen answered $12,14,18$, $32,48,70$. What did the seventh one answer?",ed $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cdot 14 \cdot 18 \cdot 32 \cdot 48 \cdot 70}=16$,medium,"Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cdot 14 \cdot 18 \cdot 32 \cdot 48 \cdot 70}=16$. Note that such a situation could indeed arise if the fishermen caught $2,7,3,8,4,2,2$ fish, and their measure of lying was $10,6,6,8,7,8,6$, respectively." c1bc1b407ffd,"13.266. Two athletes start running simultaneously - the first from $A$ to $B$, the second from $B$ to $A$. They run at different but constant speeds and meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the length of $A B$.",500 m,medium,"Solution. Consider the graphs of the runs of two athletes (Fig. 13.9). Let $AB = x$ m, $C$ and $D$ be the points of the first and second meetings, $v_{1}$ and $v_{2}$ be the speeds of the first and second athlete. Then $\frac{300}{v_{1}}=\frac{x-300}{v_{2}}\left({ }^{*}\right)$. During the time between the meetings $(KL)$, the athletes ran: the first $(x-300)+400=x+100$ m; the second $300+(x-400)=x-100$ m. Thus, $\frac{x+100}{v_{1}}=\frac{x-100}{v_{2}}(* *)$. From $\left({ }^{*}\right)$ and (**) we get $\frac{x+100}{300}=\frac{x-100}{x-300} \Rightarrow x=500 \mathrm{~m}$. Answer: 500 m." d84973711dc9,"From a survey at our school, it was found that - all children who like mathematics solve the Mathematical Olympiad, - $90\%$ of children who do not like mathematics do not solve the Mathematical Olympiad, - $46\%$ of children solve the Mathematical Olympiad. What percentage of children in our school like mathematics? (L. Hozová)",See reasoning trace,medium,"The Mathematical Olympiad (MO) is solved by all children who like mathematics, but only one-tenth of those who do not like mathematics. Let's say $x \%$ of children like mathematics, and thus solve MO. Then $(46-x) \%$ of children do not like mathematics and also solve MO. The total number of children who do not like mathematics is ten times more, i.e., $10(46-x)\%$. Altogether, we have $$ x + 10(46 - x) = 100 $$ from which simple manipulations yield $x = 40$. The number of children who like mathematics is $40 \%$. Notes. Different notation leads to different manipulations. Suppose $y \%$ of children do not like mathematics. Then $\frac{9}{10} y \%$ of children do not solve MO. This corresponds to $54 \%$ of children $(100 - 46 = 54)$. Altogether, we have $$ \frac{9}{10} y = 54 $$ from which simple manipulations yield $y = 60$. The given relationships can be illustrated using a Venn diagram as follows: ![](https://cdn.mathpix.com/cropped/2024_04_17_5ebc08916757a9a5715ag-2.jpg?height=345&width=668&top_left_y=1792&top_left_x=700) Evaluation. 2 points for auxiliary notation and partial observations; 2 points for the result; 2 points for the quality of the commentary." 621e3338362f,"1. The complex number $z$ satisfies $|z|=5$, and $(3+4 \mathrm{i}) z$ is a pure imaginary number. Then $\bar{z}=$ $\qquad$",See reasoning trace,easy,"$$ -1 \cdot \bar{z}= \pm(4-3 \mathrm{i}) $$ From the given information, we have $$ \begin{array}{l} |(3+4 \mathrm{i}) z|=|3+4 \mathrm{i}||z|=25 \\ \Rightarrow(3+4 \mathrm{i}) z= \pm 25 \mathrm{i} \\ \Rightarrow z=\frac{ \pm 25 \mathrm{i}}{3+4 \mathrm{i}}= \pm(4+3 \mathrm{i}) \\ \Rightarrow \bar{z}= \pm(4-3 \mathrm{i}) . \end{array} $$" ed30b46d6d88,"Let $1,7,19,\ldots$ be the sequence of numbers such that for all integers $n\ge 1$, the average of the first $n$ terms is equal to the $n$th perfect square. Compute the last three digits of the $2021$st term in the sequence. [i]Proposed by Nathan Xiong[/i]",261,medium,"1. Let \( x_n \) be the \( n \)-th term and \( s_n \) be the sum of the first \( n \) terms for all \( n \in \mathbb{Z}_{+} \). Since the average of the first \( n \) terms is \( n^2 \), we have: \[ \frac{s_n}{n} = n^2 \implies s_n = n^3 \] This implies that the sum of the first \( n \) terms is \( n^3 \). 2. To find the \( n \)-th term \( x_n \), we use the relationship: \[ x_n = s_n - s_{n-1} \] where \( s_n = n^3 \) and \( s_{n-1} = (n-1)^3 \). Therefore: \[ x_n = n^3 - (n-1)^3 \] 3. We expand \( n^3 - (n-1)^3 \) using the binomial theorem: \[ n^3 - (n-1)^3 = n^3 - (n^3 - 3n^2 + 3n - 1) = 3n^2 - 3n + 1 \] 4. We need to compute the last three digits of the \( 2021 \)-st term in the sequence, \( x_{2021} \): \[ x_{2021} = 2021^3 - 2020^3 \] 5. To simplify the calculation modulo 1000, we use: \[ 2021 \equiv 21 \pmod{1000} \quad \text{and} \quad 2020 \equiv 20 \pmod{1000} \] Therefore: \[ x_{2021} \equiv 21^3 - 20^3 \pmod{1000} \] 6. Calculate \( 21^3 \) and \( 20^3 \): \[ 21^3 = 21 \times 21 \times 21 = 441 \times 21 = 9261 \] \[ 20^3 = 20 \times 20 \times 20 = 400 \times 20 = 8000 \] 7. Subtract the results modulo 1000: \[ 21^3 - 20^3 \equiv 9261 - 8000 \equiv 1261 - 1000 \equiv 261 \pmod{1000} \] 8. Therefore, the last three digits of the \( 2021 \)-st term are: \[ \boxed{261} \]" ed8c5dbdaa60,"Let $\mathbb{N}$ denote the set of positive integers, and let $S$ be a set. There exists a function $f :\mathbb{N} \rightarrow S$ such that if $x$ and $y$ are a pair of positive integers with their difference being a prime number, then $f(x) \neq f(y)$. Determine the minimum number of elements in $S$.",4,medium,"To determine the minimum number of elements in \( S \) such that there exists a function \( f: \mathbb{N} \rightarrow S \) with the property that if \( x \) and \( y \) are positive integers whose difference is a prime number, then \( f(x) \neq f(y) \), we will proceed as follows: 1. **Construct a function \( f \) with \( |S| = 4 \):** Define the function \( f \) as follows: \[ f(4k) = 4, \quad f(4k+1) = 1, \quad f(4k+2) = 2, \quad f(4k+3) = 3 \] for \( k \in \mathbb{N} \). 2. **Verify that this construction works:** We need to check that for any \( x \) and \( y \) such that \( |x - y| \) is a prime number, \( f(x) \neq f(y) \). - If \( x \equiv 0 \pmod{4} \), then \( f(x) = 4 \). The possible values of \( y \) such that \( |x - y| \) is a prime number are \( y \equiv 1, 2, 3 \pmod{4} \). For these values, \( f(y) = 1, 2, 3 \) respectively, which are all different from 4. - If \( x \equiv 1 \pmod{4} \), then \( f(x) = 1 \). The possible values of \( y \) such that \( |x - y| \) is a prime number are \( y \equiv 0, 2, 3 \pmod{4} \). For these values, \( f(y) = 4, 2, 3 \) respectively, which are all different from 1. - If \( x \equiv 2 \pmod{4} \), then \( f(x) = 2 \). The possible values of \( y \) such that \( |x - y| \) is a prime number are \( y \equiv 0, 1, 3 \pmod{4} \). For these values, \( f(y) = 4, 1, 3 \) respectively, which are all different from 2. - If \( x \equiv 3 \pmod{4} \), then \( f(x) = 3 \). The possible values of \( y \) such that \( |x - y| \) is a prime number are \( y \equiv 0, 1, 2 \pmod{4} \). For these values, \( f(y) = 4, 1, 2 \) respectively, which are all different from 3. Thus, the function \( f \) defined above satisfies the required condition. 3. **Show that \( |S| < 4 \) does not work:** Suppose \( |S| < 4 \). Then \( |S| \) can be at most 3. Consider the values \( f(1), f(3), f(6), f(8) \). The differences between these values are: \[ |1 - 3| = 2, \quad |3 - 6| = 3, \quad |6 - 8| = 2, \quad |1 - 6| = 5, \quad |1 - 8| = 7, \quad |3 - 8| = 5 \] All these differences are prime numbers. Therefore, \( f(1), f(3), f(6), f(8) \) must all be different. However, if \( |S| < 4 \), it is impossible to assign different values to these four numbers, leading to a contradiction. Therefore, the minimum number of elements in \( S \) is 4. The final answer is \( \boxed{4} \)." e57038fec844,"A circle $\omega$ and a point $P$ not lying on it are given. Let $ABC$ be an arbitrary equilateral triangle inscribed into $\omega$ and $A', B', C'$ be the projections of $P$ to $BC, CA, AB$. Find the locus of centroids of triangles $A' B'C'$.",\text{The midpoint of,medium,"1. **Given**: A circle $\omega$ with center $O$ and a point $P$ not lying on $\omega$. An equilateral triangle $ABC$ is inscribed in $\omega$. Let $A', B', C'$ be the projections of $P$ onto the sides $BC, CA, AB$ respectively. 2. **Claim**: The locus of the centroids of triangles $A'B'C'$ is the midpoint of $OP$. 3. **Fixing $ABC$ and varying $P$**: Consider $P$ moving along a line perpendicular to $BC$. The projections $A'$ and $C'$ of $P$ onto $BC$ and $AB$ respectively will move with the same speed and in the same direction parallel to $BC$. 4. **Centroid of $A'B'C'$**: The centroid $G$ of triangle $A'B'C'$ is given by: \[ G = \frac{A' + B' + C'}{3} \] 5. **Behavior of projections**: When $P$ is on $BC$, the projections $A', B', C'$ are such that $A'$ lies on $BC$, $B'$ lies on $CA$, and $C'$ lies on $AB$. The centroid $G$ of $A'B'C'$ will be influenced by the positions of these projections. 6. **Distance from perpendicular bisector**: The distance of the centroid $G$ from the perpendicular bisector of $BC$ is half the distance from $P$ to the perpendicular bisector of $BC$. This is because the projections $A'$ and $C'$ move symmetrically with respect to the perpendicular bisector of $BC$. 7. **Generalization to all sides**: By symmetry, the same argument applies to the projections onto the other sides $CA$ and $AB$. Therefore, the centroid $G$ of $A'B'C'$ will always be at the midpoint of the segment $OP$. 8. **Conclusion**: The locus of the centroids of triangles $A'B'C'$ is the midpoint of $OP$. \[ \boxed{\text{The midpoint of } OP} \]" 7456afde6a38,"9. (16 points) Given that the odd function $f(x)$ is increasing on the interval $(-\infty, 0)$, and $f(-2)=-1, f(1)=0$, when $x_{1}>0, x_{2}>0$, we have $f\left(x_{1} x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$, find the solution set of the inequality $\log _{2}|f(x)+1|<0$.",See reasoning trace,medium,"9. From $\log _{2}|f(x)+1|<0$, we get $0<|f(x)+1|<1$, so $-11$ (C) $-2 a + 1 , } \\ { 1 - a ^ { 2 } \leqslant a - 1 } \end{array} \text { or } \left\{\begin{array}{l} a^{2}+1 \geqslant a+1, \\ 1-a^{2}0, \\ (a-1)(a+2) \geqslant 0\end{array}\right.\text{ or } \left\{\begin{array}{l}a(a-1) \geqslant 0, \\ (a-1)(a+2)>0 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}a>1 \text { or } a1 .\end{array}\right.$ Therefore, $a>1$ or $a \leqslant-2$. However, the condition in the original problem is “$A$ is not a proper subset of $B$”, so we have $-21)$. Let the slopes of the lines $A P$, $B P$, $A Q$, and $B Q$ be $k_{1}$, $k_{2}$, $k_{3}$, and $k_{4}$, respectively. (1) Prove: $k_{1}+k_{2}+k_{3}+k_{4}=0$; (2) Let $F_{1}$ and $F_{2}$ be the right foci of the ellipse and hyperbola, respectively. If $P F_{2} \parallel Q F_{1}$, find the value of $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}$.",See reasoning trace,medium,"(1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and we know $A(-a, 0), B(a, 0)$, then we have $$ \begin{array}{c} k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \frac{x_{1}}{y_{1}}, \\ k_{3}+k_{4}=\frac{y_{2}}{x_{2}+a}+\frac{y_{2}}{x_{2}-a}=\frac{2 x_{2} y_{2}}{x_{2}^{2}-a^{2}}=-\frac{2 b^{2}}{a^{2}} \frac{x_{2}}{y_{2}} . \end{array} $$ Let $O$ be the origin, then $2 \overrightarrow{O P}=\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})=2 \lambda \cdot \overrightarrow{O Q}$, so $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$, hence $O, P, Q$ are collinear, thus $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$, from (1) and (2) we can get $k_{1}+k_{2}+k_{3}+k_{4}=0$. (2) Since point $Q$ is on the ellipse, we have $\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1$, and from (1) we know $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$, so $\left(x_{1}, y_{1}\right)=$ $\lambda\left(x_{2}, y_{2}\right)$, thus $x_{2}=\frac{1}{\lambda} x_{1}, y_{2}=\frac{1}{\lambda} y_{1}$, hence $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=\lambda^{2}$, (3) Since point $P$ is on the hyperbola, we have $\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1$, (4) From (3) and (4) we can get $x_{1}^{2}=\frac{\lambda^{2}+1}{2} a^{2}, y_{1}^{2}=\frac{\lambda^{2}-1}{2} b^{2}$. Since $P F_{2} \parallel Q F_{1}$, we have $\frac{\left|O F_{2}\right|}{\left|O F_{1}\right|}=\frac{|O P|}{|O Q|}=\lambda$, thus $\lambda^{2}=\frac{\left|O F_{2}\right|^{2}}{\left|O F_{1}\right|^{2}}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$, hence $$ \frac{x_{1}^{2}}{y_{1}^{2}}=\frac{\lambda^{2}+1}{\lambda^{2}-1} \cdot \frac{a^{2}}{b^{2}}=\frac{a^{4}}{b^{4}} . $$ From (1) we get $\left(k_{1}+k_{2}\right)^{2}-\frac{4 b^{4}}{a^{4}} \frac{x_{1}^{2}}{y_{1}^{2}}-4$, similarly $\left(k_{3}+k_{4}\right)^{2}=4$. On the other hand, $k_{1} k_{2}=$ $\frac{y_{1}}{x_{1}+a} \cdot \frac{y_{1}}{x_{1}-a}=\frac{y_{1}^{2}}{x_{1}^{2}-a^{2}}=\frac{b^{2}}{a^{2}}, k_{3} k_{4}=\frac{y_{2}}{x_{2}+a} \cdot \frac{y_{2}}{x_{2}-a}=\frac{y_{2}^{2}}{x_{2}^{2}-a^{2}}=-\frac{b^{2}}{a^{2}}$, thus $$ k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}=\left(k_{1}+k_{2}\right)^{2}+\left(k_{3}+k_{4}\right)^{2}-2\left(k_{1} k_{2}+k_{3} k_{4}\right)=8 . $$" 1ee851037949,"11. An infinite geometric series has sum 2020. If the first term, the third term, and the fourth term form an arithmetic sequence, find the first term.","2020$, or $a=2020(1-r)$. We also have $a r^{2}-a=a r^{3}-a r^{2}$. Since the sum of the geometric se",medium,"Answer: $1010(1+\sqrt{5})$ Solution: Let $a$ be the first term and $r$ be the common ratio. Thus, $\frac{a}{1-r}=2020$, or $a=2020(1-r)$. We also have $a r^{2}-a=a r^{3}-a r^{2}$. Since the sum of the geometric series is nonzero, $a \neq 0$, and so we have $r^{2}-1=r^{3}-r^{2}$, or $r^{3}-2 r^{2}+1=0$. Since the sum of the geometric series is finite, $r$ cannot be 1 , so $r^{3}-2 r^{2}+1=(r-1)\left(r^{2}-r-1\right)=0$ implies $r^{2}-r-1=0$. Solving this equation gives $r=\frac{1-\sqrt{5}}{2}$ (since $|r|<1$ ). This gives us $a=2020\left(\frac{1+\sqrt{5}}{2}\right)=1010(1+\sqrt{5})$." e873915d86bf,"B1. Find all integers $z$ for which $\frac{5 z^{2}+3}{z-1}$ is also an integer. (20 points)",60&width=1579&top_left_y=1329&top_left_x=273),medium,"B1. The expression $\frac{5 z^{2}+3}{z-1}$ is first transformed as follows: $$ \frac{5 z^{2}+3}{z-1}=5 z+\frac{5 z+3}{z-1}=5 z+5+\frac{8}{z-1} $$ Since $z$ is an integer, $\frac{8}{z-1}$ must also be an integer. This means that $z-1$ must be one of the divisors of the number 8. The divisors of 8 are $-8,-4,-2,-1,1,2,4$ and 8, so $z \in\{-7,-3,-1,0,2,3,5,9\}$. ![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=68&width=1579&top_left_y=1071&top_left_x=273) ![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=65&width=1576&top_left_y=1121&top_left_x=274) The realization that $\frac{8}{z-1}$ must be an integer ......................................................... ![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=55&width=1576&top_left_y=1229&top_left_x=274) ![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=60&width=1579&top_left_y=1278&top_left_x=273) ![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=60&width=1579&top_left_y=1329&top_left_x=273)" 9ecef50aa252,"\section*{ Given a triangle \(A B C\) with side length \(B C=a\) and height \(A D=h_{a}\). The line \(g\) is the parallel to \(B C\) through \(A\). Calculate the volume \(V\) of the body that is formed by rotating the triangular area \(A B C\) around \(g\), in terms of \(a\) and \(h_{a}\)!",V_{Z}-V_{K_{1}}-V_{K_{2}}=\pi \cdot h_{a}^{2} \cdot a-\frac{1}{3} \pi \cdot h_{a}^{2} \cdot x-\frac{,medium,"} ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1363.jpg?height=397&width=345&top_left_y=607&top_left_x=273) If the triangle is rotated around the line \(g\), a cylindrical body is formed from which a cone is cut out on the sides of the base and top faces. The cylinder \(Z\) and both cones \(K_{1}, K_{2}\) have the height \(h_{a}\) as their radius. The height of the cylinder is \(s\), and the heights of the two cones are \(x\) and \(a-x\). For the volume of the desired body, we then get: \(V=V_{Z}-V_{K_{1}}-V_{K_{2}}=\pi \cdot h_{a}^{2} \cdot a-\frac{1}{3} \pi \cdot h_{a}^{2} \cdot x-\frac{1}{3} \pi \cdot h_{a}^{2} \cdot(a-x)=\frac{2}{3} \pi \cdot h_{a}^{2} \cdot a\)" e52e63f105b7,"6・3 If $a=(-3)^{-4}, b=-3^{4}, c=-3^{-4}$, then the incorrect conclusion among the following is (A) $\log _{3} a=-4$. (B) $\lg b c=0$. (C) $a, c$ are opposite numbers. (D) $a, b$ are reciprocals. (China Sichuan Province Junior High School Mathematics League, 1989)",$(D)$,easy,"[Solution] Given $a=(-3)^{-4}=\frac{1}{(-3)^{4}}=\frac{1}{3^{4}}>0$, and $$ c=-3^{-4}=-\frac{1}{3^{4}}<0, $$ then $a, c$ are opposite numbers, and $\log _{3} a=-4, \lg b c=\lg 1=0$. Therefore, the answer is $(D)$." 6bb8f077fdd2,"12. As shown in Figure 1, there are five cities $A, B, C, D, E$, connected by routes $\widetilde{A B}, \widetilde{A D}, \widetilde{A E}, \widetilde{B C}$, $\widetilde{B D}, \widetilde{C D}$, and $\widetilde{D E}$. Then, from city $A$ to $B$, there are ( ) different paths, requiring each path to be taken once, where each path can pass through a city multiple times. (A) 7 (B) 9 (C) 12 (D) 16 (E) 18",16$ different schemes in total.,medium,"12. D. Notice that, there is only one road passing through cities $C$ and $E$. Therefore, cities $C$ and $E$ can be merged into city $D$, making Graph 1 equivalent to Graph 2. Discussing the two scenarios based on the first direction chosen from city $A$. (1) $A \rightarrow D$. $$ A \rightarrow D \rightarrow A \rightarrow B \rightarrow D \rightarrow B, $$ There are $2 \times 1 \times 2=4$ different schemes; $$ A \rightarrow D \rightarrow B \rightarrow D \rightarrow A \rightarrow B $$ or $A \rightarrow D \rightarrow B \rightarrow A \rightarrow D \rightarrow B$, There are $2 \times 2 \times 2=8$ different schemes. Thus, there are $4+8=12$ different schemes in total. (2) $A \rightarrow B$. $$ A \rightarrow B \rightarrow D \rightarrow A \rightarrow D \rightarrow B, $$ There are $2 \times 2=4$ different schemes. By the principle of addition, there are $12+4=16$ different schemes in total." 0b4b83b1d5ad,"Which of the following equations could represent the relationship between the variables $x$ and $y$ in the table? (A) $y=x+0.5$ (B) $y=2 x-0.5$ (C) $y=0.5 x+1$ (D) $y=1.5 x$ (E) $y=x^{2}+0.5$ | $x$ | $y$ | | :---: | :---: | | 1 | 1.5 | | 2 | 3 | | 3 | 4.5 | | 4 | 6 |",1.5 x$.,easy,"A3 All points from the table: $(1,1.5),(2,3),(3,4.5)$ and $(4,6)$ belong only to the line $y=1.5 x$." 414c74d31f1f,"【Question 18】 Two natural numbers are divided, the quotient is 15, and the remainder is 5. It is known that the sum of the dividend, divisor, quotient, and remainder is 2169. Find the dividend.",2015$.,easy,"【Analysis and Solution】 Let the divisor be $x$, then the dividend is $15 x+5$; $$ (15 x+5)+x+15+5=2169 \text {; } $$ Solving for $x$ gives $x=134$; The dividend is $15 \times 134+5=2015$." aec5bec5871e,"2. Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations $$ \begin{array}{l} x^{2}+a x+b=0, \\ x^{2}+b x+c=0, \\ x^{2}+c x+a=0 \end{array} $$ have exactly one common root, find the value of $a^{2}+b^{2}+c^{2}$.",See reasoning trace,medium,"2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$. Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively. Thus, $p q r=-1$. If any two of $p, q, r$ are equal, assume without loss of generality that $p=q$, then the three equations have a common root $p$. Therefore, $$ p=r \Rightarrow p=r=q=-1 \Rightarrow a=b=c, $$ which is a contradiction. Now assume $p, q, r$ are distinct. Then the three equations have the form $$ \begin{array}{l} (x-p)(x-r)=0, \\ (x-p)(x-q)=0, \\ (x-q)(x-r)=0 . \end{array} $$ Thus, $a=-p-r=q r, b=-q-p=r p$, $c=-r-q=p q$. Hence, $-2(p+q+r)=p q+q r+r p$, $$ \begin{array}{l} -1=(q+1)(p+1)(r+1) \\ =p q+q r+r p+p+q+r . \end{array} $$ Therefore, $p+q+r=1, p q+q r+r p=-2$. Then $a^{2}+b^{2}+c^{2}$ $$ =2\left(p^{2}+q^{2}+r^{2}+p q+q r+r p\right)=6 \text {. } $$" 65ad769a2f90,"1. Find the number of divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3. Find the sum of such divisors.","2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3, the number 2 can be chosen with exponents ",medium,"1. Solution. Among the divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3, the number 2 can be chosen with exponents $0,1,2,3$ (4 options), the number 3 - with exponents 1,2 (2 options), and the number 5 - with exponents $0,1,2$ (3 options). Therefore, the total number of divisors is $4 \cdot 2 \cdot 3=24$. The sum of the divisors of the number $b=2^{3} \cdot 5^{2}$ is $(1+2+4+8)(1+5+25)=465$. Then the sum of the divisors of the number $a$, which are divisible by 3, is $\left(3+3^{2}\right) \cdot 465=12 \cdot 465=5580$." 67660528e1f5,"5. Fyodor starts writing pairs of natural numbers $(a, b)$ on the board, where $a2$, line $l_{1}: y=\frac{m-2}{m} x+2$, line $l_{2}: y=-x+2 m$ and the $y$-axis form a triangle with an area of 30. Then the value of $m$ is $(\quad)$. (A) 6 (B) 12 (C) $\frac{1+\sqrt{61}}{2}$ (D) $1+\sqrt{61}$",30 \Rightarrow m=6$.,easy,"-1. A. Notice that, line $l_{1}$ intersects the $y$-axis at point $(0,2)$, line $l_{2}$ intersects the $y$-axis at point $(0,2 m)$, and the intersection point of lines $l_{1}$ and $l_{2}$ is $(m, m)$. Then $\frac{1}{2} m(2 m-2)=30 \Rightarrow m=6$." d3424a13bb63,"9. If $\tan x+\tan y=5$ and $\tan (x+y)=10$, find $\cot ^{2} x+\cot ^{2} y$. Answer: 96 Solution: We know that $$ \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=10 $$",See reasoning trace,easy,"This leads to $$ \tan x \tan y=\frac{1}{2} . $$ Hence, $$ \begin{aligned} \cot ^{2} x+\cot ^{2} y & =\frac{1}{\tan ^{2} x}+\frac{1}{\tan ^{2} y} \\ & =\frac{\tan ^{2} y+\tan ^{2} x}{\tan ^{2} x \tan ^{2} y} \\ & =\frac{(\tan x+\tan y)^{2}-2 \tan x \tan y}{(\tan x \tan y)^{2}} \\ & =\frac{5^{2}-1}{\frac{1}{4}} \\ & =96 . \end{aligned} $$" 0e57070d85ce,"The first question: Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy $$ \begin{array}{l} 9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015) \text {. Find } \\ \quad\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right) \end{array} $$ the maximum value.",See reasoning trace,medium,"Solution 1 First prove a lemma Lemma $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in \mathbf{R}_{+}$, $n \geqslant 2, a_{i}>b_{i}(1 \leqslant i \leqslant n)$. Then $\prod_{i=1}^{n}\left(a_{i}-b_{i}\right) \leqslant\left(\sqrt[n]{\prod_{i=1}^{n} a_{i}}-\sqrt[n]{\prod_{i=1}^{n} b_{i}}\right)^{n}$. Proof (1) Prove the case when $n=2^{k}\left(k \in \mathbf{Z}_{+}\right)$. Use mathematical induction on $k$. When $k=1$, $\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \leqslant\left(\sqrt{a_{1} a_{2}}-\sqrt{b_{1} b_{2}}\right)^{2}$ $\Leftrightarrow a_{1} b_{2}+a_{2} b_{1}-2 \sqrt{a_{1} b_{2} a_{2} b_{1}} \geqslant 0$ $\Leftrightarrow\left(\sqrt{a_{1} b_{2}}-\sqrt{a_{2} b_{1}}\right)^{2} \geqslant 0$. The proposition holds. Assume the proposition holds when $k=s$, Then when $k=s+1$, $\prod_{i=1}^{2^{s+1}}\left(a_{i}-b_{i}\right)$ $=\left[\prod_{i=1}^{2^{s}}\left(a_{i}-b_{i}\right)\right]\left[\prod_{i=1}^{2^{s}}\left(a_{i+2^{s}}-b_{i+2^{s}}\right)\right]$ $\leqslant\left[\left(\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} a_{i}}-\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} b_{i}}\right)\right.$ $\left.\left(\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} a_{i+2^{s}}}-\sqrt[2^{s}]{\prod_{i=1}^{2^{s}} b_{i+2^{s}}}\right)\right]^{2^{s}}$ $\leqslant\left(\sqrt{\prod_{i=1}^{2^{s+1}} a_{i}}-\sqrt{\prod_{i=1}^{2^{s+1}} b_{i}}\right)^{2^{s+1}}$. Thus, the proposition holds in this case. (2) Prove: If the proposition holds when $n=r+1$, then the proposition also holds when $n=r$ $\left(r \geqslant 2, r \in \mathbf{Z}_{+}\right)$. Let $A=\sqrt[r]{\prod_{i=1}^{r} a_{i}}, B=\sqrt[r]{\prod_{i=1}^{r} b_{i}}$, $a_{i}^{*}=a_{i}, b_{i}^{*}=b_{i}(1 \leqslant i \leqslant r)$, $a_{r+1}=A, b_{r+1}=B$. Then $A>B$. By the induction hypothesis, $\prod_{i=1}^{r+1}\left(a_{i}^{*}-b_{i}^{*}\right)$ $\leqslant\left(\sqrt{\prod_{i=1}^{r+1} a_{i}^{*}}-\sqrt{\prod_{i=1}^{r+1} b_{i}^{*}}\right)^{r+1}$ $=(A-B)^{r+1}$ $\Rightarrow \prod_{i=1}^{r}\left(a_{i}-b_{i}\right)=\frac{\prod_{i=1}^{r+1}\left(a_{i}^{*}-b_{i}^{*}\right)}{A-B} \leqslant(A-B)^{r}$. In summary, the lemma is proved. For any $i(1 \leqslant i \leqslant 2015)$, by the given conditions, $9 a_{i}>11 a_{i+1}^{2} \geqslant 9 a_{i+1}^{2} \geqslant 0 \Rightarrow a_{i}-a_{i+1}^{2}>0$ If $a_{2016}-a_{1}^{2} \leqslant 0$, then $M=\prod_{i=2010}^{2016}\left(a_{i}-a_{i+1}^{2}\right) \leqslant 0\left(a_{2017}=a_{1}\right)$. Therefore, we only need to consider the case $a_{2016}-a_{1}^{2}>0$, in which case, $\prod_{i=1}^{2016} a_{i}>\prod_{i=1}^{2016} a_{i+1}^{2} \Rightarrow S=\sqrt{\prod_{i=1}^{2016} a_{i}}$. If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$. Below, consider the case $a_{2016}-a_{1}^{2}>0$. Let $m_{i}=\frac{a_{i}}{a_{i+1}^{2}}(i=1,2, \cdots, 2015)$, $m_{2016}=\frac{a_{2016}}{a_{1}^{2}}$, Then, $\prod_{i=1}^{2016} m_{i}=\prod_{i=1}^{2016} \frac{1}{a_{i}}$. Also, $m_{i}>\frac{11}{9}(i=1,2, \cdots, 2015)$, so $P=\prod_{i=1}^{2016}\left[\left(m_{i}-1\right) a_{i}^{2}\right]$ $=\left[\prod_{i=1}^{2016}\left(m_{i}-1\right)\right]\left(\prod_{i=1}^{2016} a_{i}^{2}\right)$ $=\left[\prod_{i=1}^{2016}\left(m_{i}-1\right)\right]\left(\prod_{i=1}^{2016} \frac{1}{m_{i}^{2}}\right)$ $=\prod_{i=1}^{2016} \frac{m_{i}-1}{m_{i}^{2}}$. Notice that, $\frac{m_{i}-1}{m_{i}^{2}}=-\left(\frac{1}{m_{i}}-\frac{1}{2}\right)^{2}+\frac{1}{4} \leqslant \frac{1}{4}$. Thus, $P=\prod_{i=1}^{2016} \frac{m_{i}-1}{m_{i}^{2}} \leqslant \frac{1}{4^{2016}}$. The equality holds if and only if $m_{i}=2(i=1,2, \cdots, 2016)$, which means $a_{i}=2 a_{i+1}^{2}(i=1,2, \cdots, 2015), a_{2016}=2 a_{1}^{2}$. Solving this, we get $a_{i}=\frac{1}{2}(i=1,2, \cdots, 2016)$. Upon verification, this satisfies the given inequalities, indicating that the equality can hold. In summary, the maximum value sought is $\frac{1}{4^{2016}}$." 5baacbece00f,"For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n!$ If $\frac{2017!}{2^{n}}$ is an integer, then the maximum value of $n$ is $\qquad$ .",See reasoning trace,medium,"II. 1.2 010. From $2^{10}=1024<2017<2048=2^{11}$, we know that the number of factor 2s in 2017! is $$ \sum_{k=1}^{10}\left[\frac{2017}{2^{k}}\right]=2010 . $$" f2dda331ba1c,"Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n = 1,2,...)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^n$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$), but have not taken exactly the same route within that time. Determine all possible values of $q$. Proposed by Jeremy King, UK",q = 1,medium,"1. **Initial Setup and Assumptions:** Let \( q \) be a positive rational number. Two ants start at the same point \( X \) in the plane. In the \( n \)-th minute (\( n = 1, 2, \ldots \)), each ant chooses to walk due north, east, south, or west, covering a distance of \( q^n \) meters. After a certain number of minutes, they meet at the same point in the plane but have not taken exactly the same route. 2. **Representation in the Complex Plane:** Consider the ants' movements in the complex plane. Let the \( i \)-th move of ant 1 be denoted by \( \varepsilon_i \) and the \( i \)-th move of ant 2 be denoted by \( \varepsilon'_i \). Clearly, \( \varepsilon_i, \varepsilon'_i \in \{\pm 1, \pm i\} \). 3. **Equating the Paths:** If the ants meet at the \( n \)-th minute, we have: \[ \sum_{i=1}^n \varepsilon_i q^i = \sum_{i=1}^n \varepsilon_i' q^i \] Dividing both sides by \( q \), we get: \[ \sum_{i=1}^n \varepsilon_i q^{i-1} = \sum_{i=1}^n \varepsilon_i' q^{i-1} \] Rearranging, we obtain: \[ \sum_{i=1}^n (\varepsilon_i - \varepsilon_i') q^{i-1} = 0 \] 4. **Conjugation and Coefficients:** Conjugating the equation, we note that for every \( \varepsilon_i \), \( 1 \) and \( -1 \) remain unchanged, while \( i \mapsto -i \) and \( -i \mapsto i \). Adding the conjugated equation to the original, we find that every coefficient of \( q^k \) is either \( 0, \pm 2 \), or \( \pm 4 \). 5. **Application of the Rational Root Theorem:** Since \( q \) is a positive rational number, by the Rational Root Theorem, the possible values of \( q \) are \( \{1, 2, 2^{-1}\} \). 6. **Verification of \( q = 1 \):** Clearly, \( q = 1 \) works because the ants can take different paths and still meet at the same point after any number of minutes. 7. **Verification of \( q = 2 \):** For \( q = 2 \), consider the left-right moves as \( i \) moves and the up-down moves as \( j \) moves. Since their paths are not exactly the same, at least one of the \( i \) or \( j \) moves must be different for the two ants. Suppose the \( i \) moves are different. We must then have: \[ \pm 2^{x_1} \pm \cdots \pm 2^{x_r} = \pm 2^{y_1} \pm \cdots \pm 2^{y_s} \] for some \( r, s \) with \( \{x_i\} \ne \{y_i\} \) and \( x_1 < \cdots < x_r \) and \( y_1 < \cdots < y_s \). Taking the positive terms on one side and the negative on the other, we get a contradiction as two different binary numbers cannot be equal. 8. **Verification of \( q = 2^{-1} \):** For \( q = 2^{-1} \), we move in the reverse direction, i.e., start at the endpoint and move towards the starting point. This case reduces to the case \( q = 2 \), which we have shown is not possible. Therefore, the only possible value for \( q \) is \( 1 \). The final answer is \( \boxed{ q = 1 } \)." 80b40b579bf3,"## Task 4 - 250514 In each of the figures a), b), c), the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are to be entered into the circles. Each of these numbers should (in each such entry) occur exactly once. For some circles, the number to be entered is already specified. Furthermore, for each entry, the following should apply: If you add the four numbers on each side of the triangle, the sum should be the same for each of the three sides. a) Find an entry in Figure a) where the sum for each of the three sides is 17! b) Find as many entries as possible in Figure b) where the sum for each of the three sides is 21! c) Find as many entries as possible in Figure c) where the sum for each of the three sides is the same! For each of these entries, give this value! a) ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=314&width=369&top_left_y=865&top_left_x=455) b) ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=320&width=357&top_left_y=862&top_left_x=878) c) ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=323&width=351&top_left_y=861&top_left_x=1292)",See reasoning trace,medium,"Figure a shows a solution to task a); the figures b 1, 2, 3, 4 show four solutions to task b); the figures c 1, 2 show two solutions to task c), in both of these solutions, the sum for each side of the triangle is 20. ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=306&width=348&top_left_y=1463&top_left_x=294) $a$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=303&width=331&top_left_y=1459&top_left_x=680) $b 1$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=299&width=348&top_left_y=1461&top_left_x=1045) $b 2$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=303&width=343&top_left_y=1459&top_left_x=1416) $b 3$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=314&width=352&top_left_y=1836&top_left_x=475) $b 4$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=302&width=351&top_left_y=1845&top_left_x=858) $c 1$ ![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0108.jpg?height=309&width=354&top_left_y=1844&top_left_x=1225) $c 2$ Solutions from the 1st Round 1985 adopted from [5] ### 2.26.2 2nd Round 1985, Class 5" 097588d6db98,"1. Given the function $f(x)=2 x+3$, then the inverse function of $f^{-1}(x+1)$ is A. $y=\frac{x-5}{2}$ B. $y=\frac{x+5}{2}$ C. $y=2 x+5$ D. $y=2 x+2$","\frac{x-3}{2}, f^{-1}(x+1)=\frac{x-2}{2}$.",easy,"1. $\mathrm{D} \quad f^{-1}(x)=\frac{x-3}{2}, f^{-1}(x+1)=\frac{x-2}{2}$." 48ce26810b7c,"29. Choose any three numbers from $a, b, c, d, e$ and find their sum, exactly obtaining the ten different numbers $7,11,13,14,19,21,22,25,26,28$. Then $a+b+c+d+e=$ ( i). MATHEMATICE YoUTH CLUE A. 25 B. 31 C. 37 D. 43",B,easy,Reference answer: B e1686e7d2446,"In a triangle, the length of side $AB$ is $10 \mathrm{~cm}$, the length of side $AC$ is $5.1 \mathrm{~cm}$, and $\angle CAB = 58^{\circ}$. Determine $\angle BCA$ to the nearest hundredth of a degree.",246&width=499&top_left_y=1991&top_left_x=802),medium,"Solution. In a triangle, using the usual notation, we write the cosine rule for side $a$: $$ a^{2}=b^{2}+c^{2}-2 b c \cdot \cos \alpha $$ By repeatedly applying the cosine rule, we can write: $$ c^{2}=a^{2}+b^{2}-2 a b \cdot \cos \gamma $$ From this, $$ \cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b} $$ Substitute the expression for $a^{2}$ from (1) into (2): $$ \cos \gamma=\frac{b^{2}+c^{2}+b^{2}-2 b c \cdot \cos \alpha-c^{2}}{2 b \sqrt{b^{2}+c^{2}-2 b c \cdot \cos \alpha}} $$ Combine and simplify the fraction by $2 b \neq 0$. We get: $$ \cos \gamma=\frac{b-c \cdot \cos \alpha}{\sqrt{b^{2}+c^{2}-2 b c \cdot \cos \alpha}} $$ Then substitute the given values and perform the indicated operations: $\cos \gamma \approx -0.02346$. The cosine value is negative, so the angle we are looking for is greater than $90^{\circ}$. The value of $\gamma$ is $180^{\circ}-88.6557^{\circ} \approx 91.35^{\circ}$. Juhász Anikó (Eger, Gárdonyi G. Gymnasium, 12th grade) Note. Many incorrect and incomplete solutions were received for this problem. If we first calculate the length of side $a$ using the cosine rule: $$ a^{2}=b^{2}+c^{2}-2 b c \cdot \cos \alpha=5.1^{2}+10^{2}-2 \cdot 5.1 \cdot 10 \cdot \cos 58^{\circ} $$ then $a^{2} \approx 71.958$ and from this $a \approx 8.48 \, \text{cm}$. If we then write the sine rule, we get: $$ \sin \gamma=\frac{\sin 58^{\circ}}{a} \cdot c $$ The first surprise comes after substitution: if we substitute the approximate value of $a$, then $\sin \gamma \approx 1.000056$ is obtained, which is not possible. Before we think that the triangle does not exist, let's consider that the triangle can be constructed without any further issues. The problem lies elsewhere. When we rounded $a$ to two decimal places, 8.48 is smaller than the actual value, thus reducing the denominator in the sine rule and increasing the value of the fraction and thus $\sin \gamma$. The consequence of rounding is usually the inaccuracy of the final result, but in this case, $\gamma$ is in a critical range: approximately $90^{\circ}$, with a sine value close to 1. The rounding, therefore, changes not the value of the result but its nature. It also becomes clear that $\sin \gamma$ is approximately 1, indicating that the triangle is nearly a half-cut equilateral triangle, with an angle of about $90^{\circ}$ at vertex $C$. Let's calculate more precisely. With four decimal places, $a \approx 8.4828$ and then $\sin \gamma \approx 0.9997$. If someone gets this far, they only need to determine whether $\gamma$ is an acute or obtuse angle. $\gamma$ is the largest angle in the triangle, but nothing more can be deduced from the given data. This can be clarified using the cosine rule. ![](https://cdn.mathpix.com/cropped/2024_05_02_d93feee26c4d303e08c4g-1.jpg?height=246&width=499&top_left_y=1991&top_left_x=802)" b792f73f7438,"Given a square $ABCD$ with side length $6$. We draw line segments from the midpoints of each side to the vertices on the opposite side. For example, we draw line segments from the midpoint of side $AB$ to vertices $C$ and $D$. The eight resulting line segments together bound an octagon inside the square. What is the area of this octagon?",6,medium,"1. **Define the vertices and midpoints:** Given the square \(ABCD\) with side length \(6\), we can place the vertices at: \[ A(0,0), \quad B(6,0), \quad C(6,6), \quad D(0,6) \] The midpoints of each side are: \[ E(3,0), \quad F(6,3), \quad G(3,6), \quad H(0,3) \] 2. **Find the intersection points:** We draw line segments from the midpoints to the opposite vertices. Specifically: - From \(E\) to \(C\) and \(D\) - From \(F\) to \(A\) and \(D\) - From \(G\) to \(A\) and \(B\) - From \(H\) to \(B\) and \(C\) We need to find the intersection points of these lines inside the square. Let's denote these intersection points as follows: - \(AF \cap BH = J\) - \(BH \cap DE = I\) - \(AF \cap CE = K\) - \(DE \cap CF = L\) 3. **Calculate the coordinates of the intersection points:** - For \(J\), the line \(AF\) has the equation \(y = \frac{1}{2}x\) and the line \(BH\) has the equation \(y = -\frac{1}{2}x + 3\). Solving these equations: \[ \frac{1}{2}x = -\frac{1}{2}x + 3 \implies x = 3, \quad y = \frac{1}{2}(3) = \frac{3}{2} \] So, \(J(3, \frac{3}{2})\). - For \(I\), the line \(BH\) has the equation \(y = -\frac{1}{2}x + 3\) and the line \(DE\) has the equation \(y = x - 3\). Solving these equations: \[ -\frac{1}{2}x + 3 = x - 3 \implies 3 = \frac{3}{2}x - 3 \implies 6 = \frac{3}{2}x \implies x = 4, \quad y = 4 - 3 = 1 \] So, \(I(4, 1)\). - For \(K\), the line \(AF\) has the equation \(y = \frac{1}{2}x\) and the line \(CE\) has the equation \(y = -\frac{1}{2}x + 6\). Solving these equations: \[ \frac{1}{2}x = -\frac{1}{2}x + 6 \implies x = 6, \quad y = \frac{1}{2}(6) = 3 \] So, \(K(6, 3)\). - For \(L\), the line \(DE\) has the equation \(y = x - 3\) and the line \(CF\) has the equation \(y = -\frac{1}{2}x + 6\). Solving these equations: \[ x - 3 = -\frac{1}{2}x + 6 \implies x + \frac{1}{2}x = 9 \implies \frac{3}{2}x = 9 \implies x = 6, \quad y = 6 - 3 = 3 \] So, \(L(6, 3)\). 4. **Calculate the area of the octagon:** The octagon is symmetric and can be divided into four congruent rhombuses. Each rhombus has vertices at the center of the square and two intersection points. For example, consider the rhombus \(IJKO\) with: - \(I(4, 1)\) - \(J(3, \frac{3}{2})\) - \(K(6, 3)\) - \(O(3, 3)\) The area of one rhombus can be calculated using the formula for the area of a rhombus: \[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \] where \(d_1\) and \(d_2\) are the lengths of the diagonals. For rhombus \(IJKO\): - \(d_1 = \text{distance between } I \text{ and } K = 2\) - \(d_2 = \text{distance between } J \text{ and } O = \frac{3}{2}\) Therefore, the area of one rhombus is: \[ \text{Area} = \frac{1}{2} \times 2 \times \frac{3}{2} = \frac{3}{2} \] Since there are four such rhombuses in the octagon, the total area of the octagon is: \[ 4 \times \frac{3}{2} = 6 \] The final answer is \(\boxed{6}\)." 669da9258ada,"Question 17 Given a sequence $a_{1}, a_{2}, a_{3}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1$ holds for all natural numbers $n$. Is it possible to choose $a_{1}$ such that the first 100000 terms of this sequence are all even, and the 100001st term is odd? (Federal Republic of Germany 1981 Mathematics Competition)",See reasoning trace,medium,"We can determine $a_{1}$ such that the first 100000 terms of this sequence are all even, and the 100001st term is odd. For example, we can choose $$ \begin{array}{l} a_{1}=2^{100000}-2, \\ a_{2}=\left[\frac{3}{2} a_{1}\right]+1=3 \times 2^{99999}-3+1=3 \times 2^{99999}-2, \\ a_{3}=\left[\frac{3}{2} a_{2}\right]+1=3^{2} \times 2^{99998}-3+1=3^{2} \times 2^{99998}-2, \\ \ldots \ldots \\ a_{99999}=\left[\frac{3}{2} a_{99998}\right]+1=3^{99998} \times 2^{2}-3+1=3^{99998} \times 2^{2}-2, \\ a_{100000}=\left[\frac{3}{2} a_{99999}\right]+1=3^{99999} \times 2-3+1, \\ a_{100001}=\left[\frac{3}{2} a_{100000}\right]+1=3^{100000}-3+1=3^{100000}-2. \end{array} $$ Therefore, $a_{1}, a_{2}, \cdots, a_{99999}, a_{100000}$ are all even, and $a_{100001}$ is odd." 8ecc54dde36a,"Three different numbers from the list 2, 3, 4, 6 have a sum of 11 . What is the product of these numbers? (A) 24 (B) 72 (C) 36 (D) 48 (E) 32",(C),easy,"The sum of 2,3 and 6 is $2+3+6=11$. Their product is $2 \cdot 3 \cdot 6=36$. ANswer: (C)" 11cf982644a4,"6.22 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)$ ( $n$ is a natural number). Then the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is (A) $1991^{-1}$. (B) $-1991^{-1}$. (C) $(-1)^{n} 1991$. (D) $(-1)^{n} 1991^{-1}$. (China Junior High School Mathematics Competition, 1991)",$(D)$,medium,"[Solution] Since $1+x^{2}=1+\frac{1}{4}\left(1991 \frac{2}{n}-2+1991^{-\frac{2}{n}}\right)$ $$ \begin{aligned} & =\frac{1}{4}\left(1991^{\frac{2}{n}}+2+1991^{-\frac{2}{n}}\right) \\ & =\left[\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{2} . \\ \therefore \quad \text { the original expression }= & {\left[\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)-\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{n} } \\ = & \left(-1991^{-\frac{1}{n}}\right)^{n}=(-1)^{n} 1991^{-1} . \end{aligned} $$ Therefore, the answer is $(D)$." 22b7bdb5e76c,13. Determine the maximum positive integer $k$ such that $k^{2}$ divides $\frac{n !}{(n-6) !}$ for every $n>6$.,12,medium,"13. The answer is 12. Let $N=\frac{n !}{(n-6) !}=(n-5)(n-4)(n-3)(n-2)(n-1) n$. First, let's find the maximum $q$ such that $q$ divides $N, \forall n$. Among six consecutive integers, three are divisible by 2, and of these three, one is certainly divisible by 4. Therefore, $2^{4}$ divides $N$. Among six consecutive integers, there are two integers divisible by 3. Therefore, $3^{2}$ divides $N$. Among six consecutive integers, only one is divisible by 5. Therefore, 5 divides $N$. Thus, $$ q \geq 2^{4} \cdot 3^{2} \cdot 5 $$ Since we are looking for the maximum $q$ such that $q$ divides $N, \forall n$, consider $n_{1}=7, n_{2}=13$. Certainly, $q$ divides both $N_{1}$ and $N_{2}$, so we have that $$ q \text { divides } \operatorname{GCD}\left(N_{1}, N_{2}\right) $$ We find that $N_{1}=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7, N_{2}=8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13$. Therefore, $\operatorname{GCD}\left(N_{1}, N_{2}\right)=2^{4} \cdot 3^{2} \cdot 5$. Now, from 1 and 2, it follows that $q=2^{4} \cdot 3^{2} \cdot 5$. To extract the maximum square from $q$, we find that $k^{2}=2^{4} \cdot 3^{2}$, which means $k=12$." d9f1685125aa,"The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy \[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\] where $a,b$, and $c$ are (not necessarily distinct) digits. Find the three-digit number $abc$.",447,medium,"1. **Convert the repeating decimals to fractions:** Let \( x = 0.abab\overline{ab} \) and \( y = 0.abcabc\overline{abc} \). For \( x \): \[ x = 0.abab\overline{ab} = \frac{abab}{9999} \] where \( abab \) is the four-digit number formed by repeating \( ab \). For \( y \): \[ y = 0.abcabc\overline{abc} = \frac{abcabc}{999999} \] where \( abcabc \) is the six-digit number formed by repeating \( abc \). 2. **Set up the equation given in the problem:** \[ x + y = \frac{33}{37} \] 3. **Express \( x \) and \( y \) in terms of \( a, b, \) and \( c \):** \[ x = \frac{100a + 10b + a + b}{9999} = \frac{101a + 11b}{9999} \] \[ y = \frac{100000a + 10000b + 1000c + 100a + 10b + c}{999999} = \frac{100100a + 10010b + 1001c}{999999} \] 4. **Combine the fractions:** \[ \frac{101a + 11b}{9999} + \frac{100100a + 10010b + 1001c}{999999} = \frac{33}{37} \] 5. **Find a common denominator:** \[ \frac{(101a + 11b) \cdot 999999 + (100100a + 10010b + 1001c) \cdot 9999}{9999 \cdot 999999} = \frac{33}{37} \] 6. **Simplify the numerator:** \[ (101a + 11b) \cdot 999999 + (100100a + 10010b + 1001c) \cdot 9999 = 33 \cdot 9999 \cdot 999999 \] 7. **Solve for \( a, b, \) and \( c \):** \[ 101a \cdot 999999 + 11b \cdot 999999 + 100100a \cdot 9999 + 10010b \cdot 9999 + 1001c \cdot 9999 = 33 \cdot 9999 \cdot 999999 \] Simplifying the left-hand side: \[ 101a \cdot 999999 + 11b \cdot 999999 + 100100a \cdot 9999 + 10010b \cdot 9999 + 1001c \cdot 9999 \] Grouping terms: \[ (101 \cdot 999999 + 100100 \cdot 9999)a + (11 \cdot 999999 + 10010 \cdot 9999)b + 1001 \cdot 9999c = 33 \cdot 9999 \cdot 999999 \] 8. **Solve for \( a, b, \) and \( c \):** \[ 101 \cdot 999999 + 100100 \cdot 9999 = 101000000000 - 101 + 1001000000 - 1001 = 101000000000 - 1102 \] \[ 11 \cdot 999999 + 10010 \cdot 9999 = 11000000000 - 11 + 100100000 - 10010 = 11000000000 - 10021 \] \[ 1001 \cdot 9999 = 10000000 - 1001 = 9998999 \] Simplifying: \[ 101000000000 - 1102 + 11000000000 - 10021 + 9998999 = 33 \cdot 9999 \cdot 999999 \] Solving for \( a, b, \) and \( c \): \[ a = 4, b = 4, c = 7 \] The final answer is \( \boxed{447} \)." fd6c42371b67,"4. The last digit of $1+2^{2}+3^{3}+4^{4}+\cdots+1992^{1992}$ is ( ). A. 8 B. 2 C. 4 D. 0",See reasoning trace,easy,"4. D. For the last digit of $a_{n}\left(n \in \mathbf{N}_{+}\right)$, it can be proven that it cycles with a period of 4 (or 2 or 1), and it only depends on the last digit of $a$. It can be verified that the sum of the last digits of $a^{n}$ for any 10 consecutive $n$ is 7. Therefore, the last digit of the original expression is the last digit of $7 \times 199 + 1 + 6$, which is 0." dd80adc43f78,"## Task 5 - 210835 Someone withdraws a certain amount of money from their savings account. They receive this amount paid out in a total of 29 banknotes, exclusively in 10-mark notes, 20-mark notes, and 50-mark notes. The number of 10-mark notes is 1 less than the number of 20-mark notes. The number of 50-mark notes is greater than twice but less than three times the number of 20-mark notes. Determine the amount of money withdrawn!",See reasoning trace,easy,"Assume that $x$ twenty-mark bills and $y$ fifty-mark bills were issued. Then $(x-1)$ ten-mark bills were issued, and it holds that: $$ (x-1)+x+y=29, \quad \text { thus } \quad 2 x+y=30 \quad \text { and } \quad 2 xy \text{ follows } 5 x>30, \text{ so } x>6. (2) From (1) and (2) it follows that $x=7$, thus $y=16$. Therefore, 6 ten-mark bills, 7 twenty-mark bills, and 16 fifty-mark bills, amounting to $1000 \mathrm{M}$, were issued." cd20d9fc787d,"7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $\sqrt{6}+\sqrt{2}$. The distances from the center of the inscribed circle of triangle $D E F$ to points $A$ and $C$ are 2 and $2 \sqrt{2}$, respectively. Find the radius of the circumscribed circle around triangle $D E F$. (16 points)",1,medium,"Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=2, C O=2 \sqrt{2}$. Let $O E=x, A E=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=4, \\ (\sqrt{2}+\sqrt{6}-y)^{2}+x^{2}=8 .\end{array}\right.$ Solving the system, we get $y=\sqrt{2}, x=\sqrt{2}$. Then $\quad \angle D A C=\angle B C A=45^{\circ}, \quad B C=2 \sqrt{3}$, ![](https://cdn.mathpix.com/cropped/2024_05_06_d66465e084edd0e7e96fg-11.jpg?height=414&width=620&top_left_y=564&top_left_x=1369) $\angle F C A=\operatorname{arctg} \frac{1}{\sqrt{3}}=30^{\circ}, \quad \angle A B E=30^{\circ}, \quad A B=2 \sqrt{2}$. $\angle D F E=90^{\circ}, D E=A B \cos 45^{\circ}=2 . \quad R_{o n .}=D E / 2=1$. ## Answer: 1." 482d72f438f4,"4. As shown in Figure 3, given that the area of the equilateral $\triangle A B C$ is $S$, $M$ and $N$ are the midpoints of sides $A B$ and $A C$ respectively, $M E \perp B C$ at $E$, and $E F \perp A C$ at $F$. Let the areas of $\triangle A M N$, $\triangle B E M$, $\triangle E F C$, and quadrilateral $M E F N$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively. Then $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=$ $\qquad$ .","S-\left(S_{1}+S_{2}+S_{3}\right)=\frac{11}{32} S$. Hence, $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=\frac{11}",medium,"4. $\frac{11}{21}$. As shown in Figure 6, connect $B N$, and draw the altitude $A H$. Since $\triangle A B C$ is an equilateral triangle, we know that $B N \perp A C$. From $\triangle M B E \backsim \triangle A B H$, we get $$ \frac{S_{2}}{S}=\frac{1}{8}, $$ which means $S_{2}=\frac{1}{8} S$. Since $\frac{B E}{B H}=\frac{1}{2}$, we have $\frac{C E}{B C}=\frac{3}{4}$. Also, since $\triangle E C F \backsim \triangle B C N$, we get $\frac{S_{3}}{S}=\frac{9}{32}$, which means $S_{3}=\frac{9}{32} S$. From $\triangle A M N \backsim \triangle A B C$, we have $\frac{S_{1}}{S}=\frac{1}{4}$, which means $S_{1}=\frac{1}{4} S$. Therefore, $S_{4}=S-\left(S_{1}+S_{2}+S_{3}\right)=\frac{11}{32} S$. Hence, $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=\frac{11}{21}$." 5403036ef121,"$1 \cdot 32$ Find the smallest positive integer $n$, such that the last three digits of its cube are 888. (6th American Invitational Mathematics Examination, 1988)",2 k=192$.,medium,"[Solution 1] If the cube of a positive integer ends in 8, then this number itself must end in 2, i.e., it can be written as $$ n=10 k+2 \text { ( } k \text { is a non-negative integer) } $$ Thus, $$ \begin{aligned} n^{3} & =(10 k+2)^{3} \\ & =1000 k^{3}+600 k^{2}+120 k+8 . \end{aligned} $$ Here, $120 k$ determines the tens digit of $n^{3}$. Since it is required that the tens digit of $n^{3}$ is 8, then the units digit of $12 k$ should be 8, i.e., the units digit of $k$ is 4 or 9, so we can set $$ k=5 m+4 \text { ( } m \text { is a non-negative integer). } $$ At this time, $n^{3}=[10(5 m+4)+2]^{3}$ $$ =125000 m^{3}+315000 m^{2}+264600 m+74088 \text {. } $$ To make the hundreds digit of $n^{3}$ be 8, it is necessary that the units digit of $2646 m$ is 8, the smallest $m=3$. At this time $$ \begin{array}{c} k=5 m+4=19, \\ n=10 k+2=192 . \end{array} $$ We can find that $n^{3}=7077888$, its last three digits are 888. Therefore, the smallest $n$ is 192. [Solution 2] By the problem, we have $$ 1000 \mid n^{3}-888 $$ So $$ \begin{array}{c} 1000 \mid n^{3}+112-1000, \\ 1000 \mid n^{3}+112 . \end{array} $$ So $n$ is even, let $n=2 k$, then $$ 1000 \mid 8 k^{3}+112=8\left(k^{3}+14\right) \text {. } $$ This requires $$ 125 \mid k^{3}+14 \text {. } $$ First, it must be $5 \mid k^{3}+14$, so the units digit of $k$ is 6 or 1. Thus, we can set $k=5 m+1$ ( $m$ is a non-negative integer) $$ \begin{aligned} k^{3}+14 & =(5 m+1)^{3}+14 \\ & =125 m^{3}+75 m^{2}+15 m+15 \\ & =5\left(25 m^{3}+15 m^{2}+3 m+3\right) . \end{aligned} $$ To make $125 \mid k^{3}+14$, it is only necessary that $$ 25 \mid 15 m^{2}+3 m+3 \text {. } $$ Thus, the units digit of $15 m^{2}+3 m+3$ is 0 or 5, since $$ 15 m^{2}+3 m+3=3 m(5 m+1)+3 \text {. } $$ $m(5 m+1)$ is definitely even, so the units digit of $3 m(5 m+1)$ is definitely 2. Thus, the units digit of $m(5 m+1)$ must be 4, at this time the units digit of $m$ must be 4 or 9. For this, set $m=5 t+4,(t$ is a non-negative integer $)$. $$ \begin{aligned} & 15 m^{2}+3 m+3 \\ = & 15(5 t+4)^{2}+3(5 t+4)+3 \\ = & 15 \cdot 25 t^{2}+40 \cdot 15 t+16 \cdot 15+15 t+15 \\ = & 15 \cdot 25 t^{2}+24 \cdot 25 t+15 t+15 \cdot 17 \end{aligned} $$ To make $25 \mid 15 m^{2}+3 m+3$, it is only necessary that i.e., $$ \begin{array}{c} 25 \mid 15 t+15 \cdot 17, \\ 5 \mid 3 t+51 . \end{array} $$ Take the smallest $t=3$. Then $\quad m=5 t+4=19, k=5 m+1=96$. Thus $\quad n=2 k=192$." 1034364d68b5,A rectangle has sides of length $\sin x$ and $\cos x$ for some $x$. What is the largest possible area of such a rectangle?,"\frac{1}{2} \sin 2 x$. But $\sin 2 x \leq 1$, with equality holding for $x=\pi / 4$, so the maximum ",easy,"Solution: We wish to maximize $\sin x \cdot \cos x=\frac{1}{2} \sin 2 x$. But $\sin 2 x \leq 1$, with equality holding for $x=\pi / 4$, so the maximum is $\frac{1}{2}$." 75ed5517c0f8,"The height $BL$ of the rhombus $ABCD$, dropped to the side $AD$, intersects the diagonal $AC$ at point $E$. Find $AE$, if $BL=$ $8, AL: LD=3: 2$. #",$3 \sqrt{5}$,easy,"$A E$ is the bisector of triangle $B A L$. ## Solution Since $A E$ is the bisector of triangle $B A L$, then $B E: E L=A B: A L=A D: A L=5: 3$. From the right triangle, it is clear that $A L: B L=3: 4$, which means $A L=6$. Additionally, $E L=3 / 8 B L=3$. Therefore, $A E=$ $\sqrt{A L^{2}+E L^{2}}=3 \sqrt{5}$ ![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-31.jpg?height=435&width=834&top_left_y=383&top_left_x=612) Answer $3 \sqrt{5}$." 36b04a0669f1,Find all primes that can be written both as a sum and as a difference of two primes (note that $ 1$ is not a prime).,5,medium,"1. **Consider the problem statement**: We need to find all primes \( p \) that can be written both as a sum and as a difference of two primes. 2. **Set up the equations**: Let \( p \) be a prime such that: \[ p = x + y \quad \text{and} \quad p = z - w \] where \( x, y, z, w \) are primes. 3. **Consider the case \( p > 2 \)**: Since \( p \) is a prime greater than 2, it must be odd. This is because the only even prime is 2, and 2 cannot be written as a difference of two primes (since the difference of two primes would be at least 1). 4. **Analyze modulo 2**: Since \( p \) is odd, one of \( x \) or \( y \) must be 2 (the only even prime), and the other must be odd. Similarly, one of \( z \) or \( w \) must be 2, and the other must be odd. 5. **Assign values**: Without loss of generality, assume \( x = 2 \). Then: \[ p = 2 + y \implies y = p - 2 \] Since \( y \) must be a prime, \( p - 2 \) must also be a prime. 6. **Consider the difference equation**: Similarly, assume \( w = 2 \). Then: \[ p = z - 2 \implies z = p + 2 \] Since \( z \) must be a prime, \( p + 2 \) must also be a prime. 7. **Check for small primes**: We need to find primes \( p \) such that both \( p - 2 \) and \( p + 2 \) are primes. - For \( p = 3 \): \[ p - 2 = 1 \quad (\text{not a prime}) \] - For \( p = 5 \): \[ p - 2 = 3 \quad (\text{prime}), \quad p + 2 = 7 \quad (\text{prime}) \] - For \( p = 7 \): \[ p - 2 = 5 \quad (\text{prime}), \quad p + 2 = 9 \quad (\text{not a prime}) \] 8. **Verify the solution**: The only prime \( p \) that satisfies both conditions is \( p = 5 \). The final answer is \(\boxed{5}\)." c3501c093195,26th Putnam 1965,"(f n + x)/(f n + 1) (*). So if f n (x) tends to a limit k(x), then k(x) = ( k(x) + x) /( k(x) + 1), ",medium,"It is almost obvious that f n+1 = (f n + x)/(f n + 1) (*). So if f n (x) tends to a limit k(x), then k(x) = ( k(x) + x) /( k(x) + 1), and hence k(x) = √x. Obviously, f n (0) = 1/n, so k(0) = 0. We notice also that f n (x) = (√x) N/D, where N = (1 + √x) n + (1 - √x) n and D = (1 + √x) n - (1 - √x) n . Suppose 0 1, put y = (√x - 1)/(1 + √x) and we again get k(x) = √x. It is clear from (*), that for x < 0, f n (x) does not tend to a limit. 26th Putnam 1965 © John Scholes jscholes@kalva.demon.co.uk 25 Jan 2002" c3e31e9df6d8,"The real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are the consecutive terms of an arithmetic sequence. If $$ \frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\frac{x_{4}}{x_{3}+x_{5}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957 $$ what is the value of $n$ ?",3916,medium,"Since the numbers $x_{1}, x_{2}, \ldots, x_{n}$ form an arithmetic sequence, then for each integer $k$ with $2 \leq k \leq n-1$, we have $x_{k}-x_{k-1}=x_{k+1}-x_{k}$. Rearranging, we obtain $2 x_{k}=x_{k-1}+x_{k+1}$ and so $\frac{x_{k}}{x_{k-1}+x_{k+1}}=\frac{1}{2}$ for each integer $k$ with $2 \leq k \leq n-1$. We note that there are $(n-1)-2+1=n-2$ integers $k$ in this range. Therefore, starting with the given equation $$ \frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957 $$ we obtain $(n-2) \cdot \frac{1}{2}=1957$ which gives $n-2=3914$ and so $n=3916$. ANSWER: 3916" 0ab3d138676f,"Petra has an equilateral triangle $ABC$. First, she folded the triangle so that point $A$ coincided with point $C$. Then she folded the resulting shape so that point $B$ coincided with point $C$. She then drew the outline of this shape on paper and found that its area is $12 \, \text{cm}^2$. Determine the area of the original triangle. (E. Novotná)",See reasoning trace,medium,"After the first fold, point $A$ coincided with point $C$. The line along which the fold was made is the axis of the segment $A C$, which in the equilateral triangle $A B C$ passes through point $B$. After this fold, Petra obtained the triangle $B E C$, which is exactly half of the triangle $A B C$ (point $E$ denotes the midpoint of the segment $A C$, see the figure). ![](https://cdn.mathpix.com/cropped/2024_04_17_52f9a35295cef28fb34dg-1.jpg?height=486&width=579&top_left_y=2170&top_left_x=767) After the second fold, point $B$ also coincided with point $C$. The line along which the fold was made is the axis of the segment $B C$, which in the original triangle passes through point $A$. After this fold, Petra obtained the quadrilateral $O E C D$ ( $D$ denotes the midpoint of the segment $B C$ and $O$ is the intersection of the axes). ![](https://cdn.mathpix.com/cropped/2024_04_17_52f9a35295cef28fb34dg-2.jpg?height=517&width=591&top_left_y=621&top_left_x=755) The axes of the sides in the equilateral triangle $A B C$ are axes of symmetry of this triangle. It follows that the quadrilaterals $O E C D, O D B F$, and $O F A E$ are congruent to each other and each has an area of $12 \mathrm{~cm}^{2}$ ( $F$ denotes the midpoint of the segment $A B$ ). The area of the triangle $A B C$ is equal to the sum of the areas of these three quadrilaterals: $$ S_{A B C}=3 \cdot 12=36\left(\mathrm{~cm}^{2}\right) . $$ Evaluation. 2 points for the correct interpretation of the first fold (triangle $B E C$); 2 points for the correct interpretation of the second fold (quadrilateral $O E C D$); 2 points for expressing the area of the triangle $A B C$." 3817df6b329a,"## Task B-4.1. Determine all natural numbers $n$ for which $$ \frac{(9!n!)^{2}-(8!(n+1)!)^{2}}{(9!n!)^{2}-18 \cdot(8!)^{2} n!(n+1)!+(8!(n+1)!)^{2}}>0 $$",See reasoning trace,medium,"## Solution. Let's write the left side of the given inequality in the form $$ \frac{(9 \cdot 8!)^{2}(n!)^{2}-(8!)^{2}((n+1) \cdot n!)^{2}}{(9 \cdot 8!)^{2}(n!)^{2}-18 \cdot(8!)^{2} n!\cdot(n+1) \cdot n!+(8!)^{2}((n+1) \cdot n!)^{2}} $$ If we divide both the numerator and the denominator by $(8!n!)^{2}$, we get $$ \frac{81-(n+1)^{2}}{81-18(n+1)+(n+1)^{2}} $$ Notice the difference of squares in the numerator and the square of a difference in the denominator. $$ \frac{(9-(n+1))(9+(n+1))}{(9-(n+1))^{2}}=\frac{(8-n)(10+n)}{(8-n)^{2}} $$ After canceling out $8-n$ (with the condition $n \neq 8$), we get $$ \frac{10+n}{8-n}>0 $$ which means $8-n>0$ (the numerator is always positive). Thus, $n<8$, so the final solution is $n \in\{1,2,3,4,5,6,7\}$." c90b12c41454,4. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face?,16,medium,"Answer: 16. Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9). Consider a face containing the vertex where the number 6 is placed. Since the sum of the numbers in the other three vertices is not less than 10, the sum of all numbers in the vertices of this face is not less than 16. ![](https://cdn.mathpix.com/cropped/2024_05_06_73b5fdf426b8c910f128g-1.jpg?height=432&width=465&top_left_y=2371&top_left_x=1435) An example of an arrangement where the smallest sum of numbers in the vertices of one face is 16 is shown in the figure: the sum of the numbers in the front face is $2+3+5+6=16$." 31dfe94ea50f,"Find all positive integers $(m, n)$ that satisfy $$m^2 =\sqrt{n} +\sqrt{2n + 1}.$$","(13, 4900)",medium,"We are given the Diophantine equation: \[ m^2 = \sqrt{n} + \sqrt{2n + 1} \] where \( m \) and \( n \) are positive integers. 1. **Square both sides of the equation:** \[ m^4 = (\sqrt{n} + \sqrt{2n + 1})^2 \] Expanding the right-hand side: \[ m^4 = n + 2\sqrt{n(2n + 1)} + (2n + 1) \] Simplifying: \[ m^4 = 3n + 1 + 2\sqrt{n(2n + 1)} \] 2. **Isolate the square root term:** \[ 2\sqrt{n(2n + 1)} = m^4 - 3n - 1 \] 3. **Square both sides again to eliminate the square root:** \[ 4n(2n + 1) = (m^4 - 3n - 1)^2 \] Expanding the right-hand side: \[ 4n(2n + 1) = m^8 - 6m^4n - 2m^4 + 9n^2 + 6n + 1 \] 4. **Consider the fact that \( \gcd(n, 2n + 1) = 1 \):** This implies that \( n \) and \( 2n + 1 \) are coprime. Therefore, we can write: \[ n = x^2 \quad \text{and} \quad 2n + 1 = y^2 \] where \( x \) and \( y \) are coprime positive integers. 5. **Substitute \( n = x^2 \) and \( 2n + 1 = y^2 \) into the original equation:** \[ x + y = m^2 \] and \[ 2x^2 + 1 = y^2 \] 6. **Solve the system of equations:** \[ 2x^2 + 1 = (m^2 - x)^2 \] Expanding and simplifying: \[ 2x^2 + 1 = m^4 - 2m^2x + x^2 \] Rearranging terms: \[ x^2 - 2m^2x + m^4 - 2x^2 - 1 = 0 \] \[ -x^2 - 2m^2x + m^4 - 1 = 0 \] \[ (x + m^2)^2 - 2m^4 = -1 \] 7. **Solve the Diophantine equation \( a^2 - 2b^4 = 1 \):** The only positive integer solutions to this equation are \( (a, b) = (1, 1) \) and \( (239, 13) \). 8. **Since \( m > 1 \), the only solution is \( (x + m^2, m) = (239, 13) \):** \[ x = 239 - m^2 = 239 - 13^2 = 239 - 169 = 70 \] 9. **Calculate \( n \):** \[ n = x^2 = 70^2 = 4900 \] Conclusion: The only solution of the equation in positive integers is \( (m, n) = (13, 4900) \). The final answer is \( \boxed{ (13, 4900) } \)" abf7ad5831fb,"Example 6 In trapezoid $A B C D$, $A D / / B C, B C=$ $B D=1, A B=A C, C D<1, \angle B A C+\angle B D C$ $=180^{\circ}$ Find the value of $C D$.",\sqrt{2} - 1$.,medium,"Solution: As shown in Figure 7, construct: point $D$ as the symmetric point of $\mathrm{K} B C$, and connect $A E$, $B E$, and $C E$. Let $A E^{1} \cdot \mathrm{j}$ $B C$ intersect at point $F$. If $A I) B C$, then points $A$ and $E$ are equidistant from $B C$, so, $$ A F = F E. $$ Let $C I) = C E = x$, $A F' = F E = m$. Since $\angle B A C + \angle B D C = 180^{\circ}$, we have $\angle B A C + \angle B E C = 180^{\circ}$. Therefore, $A, B, E, C$ are concyclic. By $A B = A C$, we get $\angle A B C = \angle A C B$. Thus, $\angle 1 = \angle A C B = \angle A B C = \angle 2$. Also, $\angle E B F = \angle E A C$, so, $\triangle B F E \sim \triangle A C E$. Therefore, $\frac{B E}{F E} = \frac{A E}{C E}$. Thus, $2 m^{2} = A E \cdot F E = B E \cdot C E = x$. By the angle bisector property, $$ \frac{B F}{C F} = \frac{B E}{C E} = \frac{1}{x}. $$ Since $B F + C F = 1$, we have, $$ B F = \frac{1}{x+1}, \quad C F = \frac{x}{x+1}. $$ From equation (2) and the intersecting chords theorem, we get $$ m^{2} = A F \cdot F' E' = B F \cdot F' C = \frac{x}{(x+1)^{2}}. $$ Substituting equation (3) into equation (1), we get $$ \frac{2 x}{(x+1)^{2}} = x. $$ Solving for $x$, we get $x = \sqrt{2} - 1$. Therefore, $C D = \sqrt{2} - 1$." a7f57ccddb46,"25. Arrange the numbers $0,0,1,1,2,2,4,4$ in a row (0 can be placed first), so that no four consecutive numbers form 2014. There are $\qquad$ ways.",2401$.,easy,"【Analysis】There are $C_{8}^{2} \times C_{6}^{2} \times C_{4}^{2} \times C_{2}^{2}=2520$ ways to arrange. If 2014 appears, then 2014 has 5 possible positions. The remaining digits can be arranged arbitrarily, resulting in $5 \times P_{4}^{4}=120$ ways. Among these, 20142014 is counted once in the first position and once in the fifth position, which is a repetition. Therefore, there are 119 ways in which 2014 appears. Thus, the number of arrangements that meet the requirements is $2520-119=2401$." 1fc2aaede7be,"Example 3 If $p$ and $p+2$ are both prime numbers, then these two prime numbers are called ""twin primes"". Consider the following two sequences. Fibonacci sequence: $1,1,2,3,5,8, \cdots$ (the sequence satisfying $F_{1}=1, F_{2}=1$, $F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots$). Twin prime sequence: $3,5,7,11,13,17,19, \cdots$ (the sequence formed by writing all twin prime pairs in ascending order). Question: Which positive integers appear in both of the sequences above?",See reasoning trace,medium,"This is a problem related to the properties of the Fibonacci sequence. A certain term $F_{n}$ appears in the twin prime sequence if and only if $F_{n}-2$ and $F_{n}$ are both prime, or $F_{n}$ and $F_{n}+2$ are both prime. Therefore, to negate that $F_{n}$ appears in the twin prime sequence, we need to prove that $F_{n}$ is composite, or that $F_{n}-2$ and $F_{n}+2$ are both composite. Notice that, $F_{4}=3, F_{5}=5, F_{7}=13$ all appear in the twin prime sequence. Below, we prove that when $n \geqslant 8$, $F_{n}$ does not appear in the twin prime sequence. Using the recursive formula of the Fibonacci sequence, we have: $$\begin{aligned} F_{n+m+1} & =F_{1} F_{n+m-1}+F_{2} F_{n+m} \\ & =F_{1} F_{n+m-1}+F_{2}\left(F_{n-m-1}+F_{n+m-2}\right) \\ & =F_{2} F_{n+m-2}+F_{3} F_{n+m-1} \\ & =\cdots \\ & =F_{n-1} F_{m+1}+F_{n} F_{m+2} \end{aligned}$$ In (2), let $m=n-1$, then we have $F_{2 n}=F_{n}\left(F_{n+1}+F_{n-1}\right)$. Since $F_{2}=1$, when $n \geqslant 3$, $F_{2 n}$ is composite. On the other hand, it is well-known that: $F_{n}^{2}=F_{n-1} F_{n+1}+(-1)^{n}, n=1,2, \cdots$ (this conclusion is easily proven by mathematical induction). Combining with (2), we have $$\begin{aligned} F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2} \\ & =F_{2 n}^{2}+\left(F_{2 n-1}+F_{2 n}\right)^{2} \\ & =2 F_{2 n}^{2}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1} \\ & =2 F_{2 n+1} F_{2 n-1}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1}+2 \end{aligned}$$ Therefore, $$\begin{aligned} F_{4 n+1}-2 & =F_{2 n-1}\left(2 F_{2 n+1}+2 F_{2 n}+F_{2 n-1}\right) \\ & =F_{2 n-1}\left(F_{2 n+1}+F_{2 n+2}+F_{2 n+1}\right) \\ & =F_{2 n-1}\left(F_{2 n+3}+F_{2 n+1}\right) \end{aligned}$$ Thus, when $n \geqslant 2$, $F_{4 n+1}-2$ is composite. Furthermore, we also have $$\begin{aligned} F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2}=\left(F_{2 n+2}-F_{2 n+1}\right)^{2}+F_{2 n+1}^{2} \\ & =2 F_{2 n+1}^{2}+F_{2 n+2}\left(F_{2 n+2}-2 F_{2 n+1}\right) \\ & =2 F_{2 n+1}^{2}-F_{2 n+2} F_{2 n-1} \\ & =2 F_{2 n} F_{2 n+2}-F_{2 n+2} F_{2 n-1}-2 \\ & =F_{2 n+2}\left(2 F_{2 n}-F_{2 n-1}\right)-2 \\ & =F_{2 n+2}\left(F_{2 n}+F_{2 n-2}\right)-2 \end{aligned}$$ Therefore, when $n \geqslant 2$, $F_{4 n+1}+2$ is also composite. Similarly, it can be proven that when $n \geqslant 2$, the numbers $F_{4 n+3}-2$ and $F_{4 n+3}+2$ are also composite. In summary, only the numbers $3,5,13$ appear in both of the above sequences." c0a88cf726a9,"In the following figure, $\angle C A B=2 \cdot \angle C B A, A D$ is an altitude and $M$ is the midpoint of $A B$. If $A C=2 \text{ cm}$, find the length of the segment $D M$. ![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-27.jpg?height=365&width=700&top_left_y=510&top_left_x=755) #",365&width=698&top_left_y=1599&top_left_x=759),medium,"Solution Let $K$ be the midpoint of $A C$. Since the circle with center $K$ and diameter $A C$ passes through $D$, it follows that $C K=A K=D K=1 \mathrm{~cm}$. From this last equality, it follows that triangle $A D K$ is isosceles and thus $\angle K D A=\angle K A D$. The segment $K M$ is the midline of triangle $A B C$, implying that $K M \| B C$ and thus $\angle K M D=\angle C B A$. By the Exterior Angle Theorem, $$ \angle D K M=\angle A D K-\angle D M K=\angle D M K $$ Finally, since the last equalities show that triangle $D K M$ is isosceles, it follows that $D M=D K=1 \mathrm{~cm}$. ![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-27.jpg?height=365&width=698&top_left_y=1599&top_left_x=759)" 05604e6f627a,"3. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{7}{6}$, and for any positive integer $n$, the quadratic equation $a_{n} x^{2}-$ $a_{n+1} x+1=0$ has two roots $\alpha_{n}, \beta_{n}$ that satisfy $6 \alpha_{n}-2 \alpha_{n} \beta_{n}+6 \beta_{n}=3$. Then the general term formula for $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$","\frac{1}{2}$, we have $a_{n}-\frac{2}{3}=\left(\frac{1}{2}\right)^{n}$, i.e., $a_{n}=\frac{1}{2^{n}}",medium,"3. $\frac{1}{2^{n}}+\frac{2}{3}$. By Vieta's formulas, we have $$ \alpha_{n}+\beta_{n}=\frac{a_{n+1}}{a_{n}}, \quad \alpha_{n} \beta_{n}=\frac{1}{a_{n}}, $$ thus $$ 3=6\left(\alpha_{n}+\beta_{n}\right)-2 \alpha_{n} \beta_{n}=\frac{6 a_{n+1}-2}{a_{n}}, $$ rearranging, we get $$ a_{n+1}=\frac{1}{2} a_{n}+\frac{1}{3}, $$ which means $$ a_{n+1}-\frac{2}{3}=\frac{1}{2}\left(a_{n}-\frac{2}{3}\right). $$ Since $a_{1}-\frac{2}{3}=\frac{1}{2}$, we have $a_{n}-\frac{2}{3}=\left(\frac{1}{2}\right)^{n}$, i.e., $a_{n}=\frac{1}{2^{n}}+\frac{2}{3}$." 59bec400f451,"The cells of a $100 \times 100$ table are colored white. In one move, it is allowed to select some $99$ cells from the same row or column and recolor each of them with the opposite color. What is the smallest number of moves needed to get a table with a chessboard coloring? [i]S. Berlov[/i]",100,medium,"1. **Proving the minimality:** - To achieve a chessboard coloring, we need to ensure that the cells alternate in color such that no two adjacent cells (horizontally or vertically) share the same color. - Consider the set \( S \) of cells on the main diagonal, i.e., cells \((i, i)\) for \( i = 1, 2, \ldots, 100 \). In a chessboard coloring, these cells must all be of the same color (either all black or all white). - Each cell in \( S \) lies in a different row and column from every other cell in \( S \). Therefore, to change the color of each cell in \( S \), we need at least one move per cell, resulting in a minimum of \( |S| = 100 \) moves. 2. **Construction for 100 moves:** - We need to find a sequence of moves that will achieve the desired chessboard coloring in exactly 100 moves. - Consider the cells \((2i, 2(51-i))\) for \( i = 1, 2, \ldots, 50 \). These cells are chosen such that each cell is in a different row and column. - For each cell \((2i, 2(51-i))\), perform the following: - Select the 99 other cells in the same row and recolor them. - Select the 99 other cells in the same column and recolor them. - This special move for each cell \((2i, 2(51-i))\) requires 2 moves (one for the row and one for the column). - Since there are 50 such cells, the total number of moves is \( 50 \times 2 = 100 \). Thus, the smallest number of moves needed to achieve a chessboard coloring is 100. The final answer is \(\boxed{100}\)." 531f93437971,"4. (15 points) A ball was thrown vertically upwards from a balcony. It is known that it hit the ground after 6 seconds. Given that the initial speed of the ball is 20 m/s, determine the height of the balcony. The acceleration due to gravity is 10 m/s².",60 m,easy,"Answer: 60 m. Solution. The equation of motion of the ball: $$ y=0=h_{0}+v_{0} t-\frac{g t^{2}}{2}=h_{0}+20 \cdot 6-\frac{10 \cdot 6^{2}}{2} $$ We get $h_{0}=60 m$ m." 96af4074415e,"$7 \cdot 74$ There are several locks, and now six people each hold a part of the keys. It is known that any two people trying to open the locks together will have exactly one lock that they cannot open, while any three people can open all the locks. How many locks are there at least? --- Note: The translation keeps the original format and structure of the text, including the mathematical notation and the",15$ is the minimum number of locks.,medium,"[Solution 1] Since every pair of people has one lock they cannot open, let the total number of these locks be $k$, then $$ k \leqslant C_{6}^{2}=15. $$ But $k k$ 4) $b_{k}$ and each city $a_{i}$, where $i > k$ 5) $b_{k}$ and each city $b_{i}$, where $i < k$. Thus, cities $a_{i}$ and $a_{j}$ are connected by a flight of the airline with number $\max \left(a_{i}, a_{j}\right)$, cities $b_{i}$ and $b_{j}$ by the airline with number $\max \left(b_{i}, b_{j}\right)$, and cities $a_{i}$ and $b_{j}$ by the airline with number $\min \left(b_{i}, b_{j}\right)$." 0c6fb98c7c95,"## Calculate the areas of figures bounded by lines given in polar coordinates. $$ r=\frac{5}{2} \sin \phi, r=\frac{3}{2} \sin \phi $$",See reasoning trace,medium,"## Solution $$ S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi $$ Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, and then we get: ![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-38.jpg?height=1313&width=1060&top_left_y=1000&top_left_x=886) $$ \begin{aligned} & S=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}\left(\frac{25}{4} \sin ^{2} \varphi-\frac{9}{4} \sin ^{2} \varphi\right) d \varphi= \\ & =2 \int_{-\pi / 2}^{\pi / 2} \sin ^{2} \varphi d \varphi= \end{aligned} $$ $$ \begin{aligned} & =2 \int_{-\pi / 2}^{\pi / 2} \frac{1-\cos 2 \varphi}{2} d \varphi= \\ & =\left.\left(\varphi-\frac{1}{2} \sin 2 \varphi\right)\right|_{-\pi / 2} ^{\pi / 2}= \\ & =\left(\frac{\pi}{2}-\frac{1}{2} \sin \pi\right)-\left(-\frac{\pi}{2}-\frac{1}{2} \sin (-\pi)\right)=\frac{\pi}{2}+\frac{\pi}{2}=\pi \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-20 »$ Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals - Last edited: 11:23, 6 July 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 16-21 ## Material from PlusPi" db71176e7fa5,"Let $y=f(x)$ be a function of the graph of broken line connected by points $(-1,\ 0),\ (0,\ 1),\ (1,\ 4)$ in the $x$ -$y$ plane. Find the minimum value of $\int_{-1}^1 \{f(x)-(a|x|+b)\}^2dx.$ [i]2010 Tohoku University entrance exam/Economics, 2nd exam[/i]",\frac{8,medium,"To find the minimum value of the integral \(\int_{-1}^1 \{f(x)-(a|x|+b)\}^2 \, dx\), we need to first express the function \(f(x)\) and then set up the integral. 1. **Define the function \(f(x)\):** The function \(f(x)\) is given by the piecewise linear function connecting the points \((-1, 0)\), \((0, 1)\), and \((1, 4)\). Therefore, we can write: \[ f(x) = \begin{cases} 1 + x & \text{for } -1 \leq x < 0, \\ 1 + 3x & \text{for } 0 \leq x \leq 1. \end{cases} \] 2. **Set up the integral:** We need to minimize the integral: \[ \int_{-1}^1 \{f(x) - (a|x| + b)\}^2 \, dx. \] This can be split into two integrals: \[ \int_{-1}^0 \{(1 + x) - (a(-x) + b)\}^2 \, dx + \int_0^1 \{(1 + 3x) - (ax + b)\}^2 \, dx. \] 3. **Simplify the integrands:** For \(x \in [-1, 0)\): \[ (1 + x) - (a(-x) + b) = 1 + x - (-ax + b) = 1 + x + ax - b = (1 - b) + (1 + a)x. \] For \(x \in [0, 1]\): \[ (1 + 3x) - (ax + b) = 1 + 3x - (ax + b) = 1 - b + (3 - a)x. \] 4. **Set up the integrals with the simplified expressions:** \[ \int_{-1}^0 \{(1 - b) + (1 + a)x\}^2 \, dx + \int_0^1 \{(1 - b) + (3 - a)x\}^2 \, dx. \] 5. **Expand the integrands:** \[ \int_{-1}^0 \{(1 - b)^2 + 2(1 - b)(1 + a)x + (1 + a)^2 x^2\} \, dx + \int_0^1 \{(1 - b)^2 + 2(1 - b)(3 - a)x + (3 - a)^2 x^2\} \, dx. \] 6. **Integrate each term:** \[ \int_{-1}^0 (1 - b)^2 \, dx + \int_{-1}^0 2(1 - b)(1 + a)x \, dx + \int_{-1}^0 (1 + a)^2 x^2 \, dx + \int_0^1 (1 - b)^2 \, dx + \int_0^1 2(1 - b)(3 - a)x \, dx + \int_0^1 (3 - a)^2 x^2 \, dx. \] 7. **Evaluate the integrals:** \[ (1 - b)^2 \int_{-1}^0 dx + 2(1 - b)(1 + a) \int_{-1}^0 x \, dx + (1 + a)^2 \int_{-1}^0 x^2 \, dx + (1 - b)^2 \int_0^1 dx + 2(1 - b)(3 - a) \int_0^1 x \, dx + (3 - a)^2 \int_0^1 x^2 \, dx. \] \[ (1 - b)^2 [x]_{-1}^0 + 2(1 - b)(1 + a) \left[\frac{x^2}{2}\right]_{-1}^0 + (1 + a)^2 \left[\frac{x^3}{3}\right]_{-1}^0 + (1 - b)^2 [x]_0^1 + 2(1 - b)(3 - a) \left[\frac{x^2}{2}\right]_0^1 + (3 - a)^2 \left[\frac{x^3}{3}\right]_0^1. \] \[ (1 - b)^2 (0 - (-1)) + 2(1 - b)(1 + a) \left(0 - \frac{1}{2}\right) + (1 + a)^2 \left(0 - \left(-\frac{1}{3}\right)\right) + (1 - b)^2 (1 - 0) + 2(1 - b)(3 - a) \left(\frac{1}{2} - 0\right) + (3 - a)^2 \left(\frac{1}{3} - 0\right). \] \[ (1 - b)^2 + 2(1 - b)(1 + a) \left(-\frac{1}{2}\right) + (1 + a)^2 \left(\frac{1}{3}\right) + (1 - b)^2 + 2(1 - b)(3 - a) \left(\frac{1}{2}\right) + (3 - a)^2 \left(\frac{1}{3}\right). \] 8. **Combine like terms and simplify:** \[ 2(1 - b)^2 - (1 - b)(1 + a) + \frac{(1 + a)^2}{3} + (1 - b)(3 - a) + \frac{(3 - a)^2}{3}. \] 9. **Differentiate with respect to \(a\) and \(b\) and set to zero to find critical points:** \[ \frac{\partial}{\partial a} \left(2(1 - b)^2 - (1 - b)(1 + a) + \frac{(1 + a)^2}{3} + (1 - b)(3 - a) + \frac{(3 - a)^2}{3}\right) = 0, \] \[ \frac{\partial}{\partial b} \left(2(1 - b)^2 - (1 - b)(1 + a) + \frac{(1 + a)^2}{3} + (1 - b)(3 - a) + \frac{(3 - a)^2}{3}\right) = 0. \] Solving these equations, we get: \[ 2a + 3b = 5, \] \[ a + 2b = 3. \] 10. **Solve the system of equations:** \[ \begin{cases} 2a + 3b = 5, \\ a + 2b = 3. \end{cases} \] Multiply the second equation by 2: \[ 2a + 4b = 6. \] Subtract the first equation from this result: \[ (2a + 4b) - (2a + 3b) = 6 - 5, \] \[ b = 1. \] Substitute \(b = 1\) into the second equation: \[ a + 2(1) = 3, \] \[ a = 1. \] 11. **Substitute \(a = 1\) and \(b = 1\) back into the integral:** \[ \int_{-1}^0 (1 + x - (1(-x) + 1))^2 \, dx + \int_0^1 (1 + 3x - (1x + 1))^2 \, dx, \] \[ \int_{-1}^0 (x)^2 \, dx + \int_0^1 (2x)^2 \, dx, \] \[ \int_{-1}^0 x^2 \, dx + \int_0^1 4x^2 \, dx, \] \[ \left[\frac{x^3}{3}\right]_{-1}^0 + 4\left[\frac{x^3}{3}\right]_0^1, \] \[ \left(0 - \left(-\frac{1}{3}\right)\right) + 4\left(\frac{1}{3} - 0\right), \] \[ \frac{1}{3} + \frac{4}{3} = \frac{5}{3}. \] The final answer is \(\boxed{\frac{8}{3}}\)." 3b060af2df9a,"Alice wanders in the Forest of Forgetfulness where she cannot remember the day of the week. She meets the lion and the unicorn. The lion lies on Monday, Tuesday, and Wednesday and tells the truth the other days, while the unicorn lies only on Thursday, Friday, and Saturday. ""Yesterday was a day when I lied,"" says the lion. ""Yesterday was a day when I lied,"" says the unicorn. Question: What day is it today?",See reasoning trace,medium,"The sought day cannot be Monday, Tuesday, or Wednesday. Let's consider the case of Monday (the other two days are eliminated in a similar manner): if it is Monday, then the unicorn is telling the truth and the lion is lying. So the unicorn is right in saying that the previous day, it was lying. However, the day before Monday is Sunday, and on Sunday, the unicorn tells the truth, so the sought day cannot be Monday (this contradicts the fact that the unicorn is right!). We can eliminate Friday, Saturday, and Sunday in the same way (by analyzing what the lion says this time) so we are left with Thursday. We can also verify the consistency of the lion's and unicorn's statements on Thursday to ensure our reasoning is correct!" 21fba2bea7de,"1. Compute the sum: $$ \sqrt{2018-\sqrt{2019 \cdot 2017}}+\sqrt{2016-\sqrt{2017 \cdot 2015}}+\cdots+\sqrt{2-\sqrt{3 \cdot 1}} $$",$\frac{\sqrt{2019}-1}{\sqrt{2}}$,medium,"Answer: $\frac{\sqrt{2019}-1}{\sqrt{2}}$. Solution: Using the known formula $\sqrt{a-\sqrt{(a+1)(a-1)}}=\frac{\sqrt{a+1}-\sqrt{a-1}}{\sqrt{2}}$, valid for $a \geq 1$, we get that each of the roots can be represented as: $$ \begin{gathered} \sqrt{2018-\sqrt{2019 \cdot 2017}}=\frac{\sqrt{2019}-\sqrt{2017}}{\sqrt{2}} \\ \sqrt{2016-\sqrt{2017 \cdot 2015}}=\frac{\sqrt{2017}-\sqrt{2015}}{\sqrt{2}} \\ \cdots \\ \sqrt{2-\sqrt{3 \cdot 1}}=\frac{\sqrt{3}-\sqrt{1}}{\sqrt{2}} \end{gathered} $$ Adding all these equalities, we have $$ \sqrt{2018-\sqrt{2019 \cdot 2017}}+\sqrt{2016-\sqrt{2017 \cdot 2015}}+\cdots+\sqrt{2-\sqrt{3 \cdot 1}}=\frac{\sqrt{2019}-\sqrt{1}}{\sqrt{2}} $$" dc5a3270ffac,"3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12 \%$. However, this was not enough, so it was decided to reduce the length by $20 \%$ and the width by $10 \%$. By what percentage did the perimeter of the rectangular flower bed decrease from the original version?",18,medium,"Answer: 18. Solution. Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $2(0.9 x + 0.8 y) = 0.88 \cdot 2(x + y)$ or $x = 4 y$. The original perimeter: $10 y$. After the second reduction: $0.8 x = 3.2 y$ - length of the flower bed, $0.9 y$ - width of the flower bed, $2(3.2 y + 0.9 y) = 8.2 y$ - perimeter. The perimeter decreased by $1.8 y$ or by $18\%$." 3f4a7cff3fd7,"What is the value of the sum $A+B$, if $$ \frac{1}{A}=1-\frac{1-x}{y}, \quad \frac{1}{B}=1-\frac{y}{1-x}, \quad x=\frac{1-a}{1-\frac{1}{a}}, \quad y=1-\frac{1}{x} $$ and $a \neq \pm 1$.","1$ if $a \neq-1,0,1$.",medium,"According to the third relationship, $x$ depends on $a$, and according to the fourth relationship, $y$ depends on $x$, so ultimately the value of $A+B$ is determined by the value of $a$ according to the first two relationships: $$ \frac{1}{A}=\frac{x+y-1}{y}, \quad \frac{1}{B}=\frac{1-x-y}{1-x}=\frac{x+y-1}{x-1} $$ The numerators of the two expressions are identical. Thus, $$ A+B=\frac{y+(x-1)}{x+y-1}=1 $$ However, we must determine the conditions under which our result is valid. $A$, and $B$ are defined if $$ \begin{gathered} y \neq 0, \text { and } \frac{1}{A} \neq 0 \text {, that is, } \frac{1-x}{y} \neq 1 \text {, or } 1-x \neq y, \quad x+y-1 \neq 0 \\ 1-x \neq 0 \text {, and } \frac{1}{B} \neq 0 \text {, that is, } \frac{y}{1-x} \neq 1 \text {, or } x \neq 1 \text {, and } x+y-1 \neq 0 \end{gathered} $$ It will suffice to examine the last two conditions, because if these are satisfied, the division in (1) can also be performed, and from the fourth relationship, $y=0$ cannot result. $y$ is not defined in the case of $x=0$, but this is excluded by $a \neq 1$. $x$ is not defined, beyond the case of $a=1$, also in the case of $a=0$, which we must exclude. $x=1$ would only occur in the case of $a=1 / a$, which is excluded by $a \neq \pm 1$. Finally, the condition $x+y-1 \neq 0$ transforms, considering the fourth relationship, then the third relationship, and the previous considerations, as follows: $$ x-\frac{1}{x} \neq 0, \quad x^{2} \neq 1, \quad x \neq-1, \quad a-1 \neq 1-\frac{1}{a} $$ $a^{2}-2 a+1=(a-1)^{2} \neq 0$, which is again satisfied. According to all this, $A+B=1$ if $a \neq-1,0,1$." c274a7d2d616,"2) The city of Mystery is $500 \mathrm{~km}$ from Mouseville and $1200 \mathrm{~km}$ from Duckburg. What is the minimum possible value for the distance between Mouseville and Duckburg? (A) $500 \mathrm{~km}$ (B) $700 \mathrm{~km}$ (C) $1200 \mathrm{~km}$ (D) $1300 \mathrm{~km}$ (E) $1700 \mathrm{~km}$.",$(B)$,easy,"2) The answer is $(B)$. In any triangle $A B C$, the length of $A B$ is greater than or equal to the length of $B C$ minus the length of $A C$ (assuming $B C \geq A C$), and it is equal to this difference only if $A B C$ are collinear and $A$ is between $B$ and $C$." 4671d3d34080,"8.6. Ostap Bender put new tires on the car ""Gnu Antelope"". It is known that the front tires of the car wear out after 25,000 km, while the rear tires wear out after 15,000 km (the tires are the same both in the front and in the rear, but the rear ones wear out more). After how many kilometers should Ostap Bender swap these tires to ensure that the ""Gnu Antelope"" travels the maximum possible distance? What is this distance?","The tires should be swapped after 9375 km, then a total of 18750 km can be traveled",medium,"Solution. Let Ostap Bender swap the tires after x kilometers. Then the rear tires have used up [x/15000] of their resource, and the front tires [x/25000]. After the swap, they can work for another $$ 25000 \cdot\left(1-\frac{x}{15000}\right) \text { and } 15000 \cdot\left(1-\frac{x}{25000}\right) $$ kilometers, respectively. Thus, the total distance that can be traveled is no more than $$ x+25000\left(1-\frac{x}{15000}\right)=25000-\frac{2}{3} x $$ and no more than $$ x+15000\left(1-\frac{x}{25000}\right)=15000+\frac{2}{5} x $$ The maximum distance can be traveled if these expressions are equal (otherwise, either the first or the second set of tires will wear out earlier, since when the first expression increases, the second decreases and vice versa). Thus, $$ 25000-\frac{2}{3} x=15000+\frac{2}{5} x $$ from which $10000=\frac{16}{15} x$, or $x=9375$. Answer: The tires should be swapped after 9375 km, then a total of 18750 km can be traveled." acf0fe2ff10a,"Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}b_{2}>\cdots>b_{100}$). Try to find $$ \left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{100}-b_{100}\right| $$ the value of the expression.",See reasoning trace,medium,"First, prove: For any term in the algebraic expression $$ \left|a_{k}-b_{k}\right|(k=1,2, \cdots, 100) $$ the larger number among $a_{k}$ and $b_{k}$ must be greater than 100, and the smaller number must not exceed 100. (1) If $a_{k} \leqslant 100$ and $b_{k} \leqslant 100$, then by $$ a_{1}b_{k+1}>\cdots>b_{100} $$ we get that $a_{1}, a_{2}, \cdots, a_{k}, b_{k}, b_{k+1}, \cdots, b_{100}$ are 101 numbers, all of which do not exceed 100, which is impossible; (2) If $a_{k}>100$ and $b_{k}>100$, then by $$ a_{100}>a_{99}>\cdots>a_{k+1}>a_{k}>100 $$ and $b_{1}>b_{2}>\cdots>b_{k}>100$ we get that $b_{1}, b_{2}, \cdots, b_{k}, a_{k}, a_{k+1}, \cdots, a_{100}$ are 101 numbers, all of which are greater than 100, which is also impossible. Thus, in the 100 absolute values of the algebraic expression $$ \left|a_{1}-b_{1}\right|,\left|a_{2}-b_{2}\right|, \cdots,\left|a_{100}-b_{100}\right| $$ the smaller numbers are $1,2, \cdots, 100$, and the larger numbers are 101, $102, \cdots, 200$. Therefore, the original expression is $$ \begin{array}{l} =(101+102+\cdots+200)-(1+2+\cdots+100) \\ =10000 . \end{array} $$" bb2b1fc1a1b8,"4. Let's write the natural number $a$ in its canonical form: $$ a=p_{1}^{s_{1}} \cdot p_{2}^{s_{2}} \cdot \ldots \cdot p_{n}^{s_{n}} $$ where $p_{1}, p_{2}, \ldots, p_{n}$ are distinct prime numbers, and $s_{1}, s_{2}, \ldots, s_{n}$ are natural numbers. It is known that the number of natural divisors of $a$, including 1 and $a$, is equal to $\left(s_{1}+1\right)\left(s_{2}+1\right) \cdot \ldots \cdot\left(s_{n}+1\right)$. According to the Let's calculate the sum of the divisors: $$ \Sigma_{d}=1+p+p^{2}+\ldots+p^{100}=\frac{p^{101}-1}{p-1}=\frac{a \sqrt[100]{a}-1}{\sqrt[100]{a}-1} $$ Let's calculate the product of the divisors: $$ \Pi_{d}=p^{1+2+\ldots+100}=p^{50 \cdot 101}=\sqrt{a^{101}} $$",1) $\Sigma_{d}=\frac{a_{100}^{a}-1}{\sqrt[100]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{101}}$,easy,Answer: 1) $\Sigma_{d}=\frac{a_{100}^{a}-1}{\sqrt[100]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{101}}$. 1dc9225d8208,"9. Let $a+b=1, b>0, a \neq 0$, then the minimum value of $\frac{1}{|a|}+\frac{2|a|}{b}$ is Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.","\sqrt{2}-1, b=\sqrt{2} a=2-\sqrt{2}$; when $a<0$, $\frac{1}{|a|}+\frac{2|a|}{b} \geq 2 \sqrt{2}-1$, ",medium,"Solve $2 \sqrt{2}-1$. Analysis $\frac{1}{|a|}+\frac{2|a|}{b}=\frac{a+b}{|a|}+\frac{2|a|}{b}=\frac{a}{|a|}+\left(\frac{b}{|a|}+\frac{2|a|}{b}\right) \geq \frac{a}{|a|}+2 \sqrt{\frac{b}{|a|} \cdot \frac{2|a|}{b}}=2 \sqrt{2}+\frac{a}{|a|}$, where the equality holds when $\frac{b}{|a|}=\frac{2|a|}{b}$, i.e., $b^{2}=2 a^{2}$. When $a>0$, $\frac{1}{|a|}+\frac{2|a|}{b} \geq 2 \sqrt{2}+1$, the minimum value is reached when $a=\sqrt{2}-1, b=\sqrt{2} a=2-\sqrt{2}$; when $a<0$, $\frac{1}{|a|}+\frac{2|a|}{b} \geq 2 \sqrt{2}-1$, the minimum value $2 \sqrt{2}-1$ is reached when $a=-(\sqrt{2}+1), b=-\sqrt{2} a$." d9b22f8fcac0,"A7. The odometer of a car indicates that the car has traveled $2010 \mathrm{~km}$. It is an odometer with six wheels and no digit after the decimal point, so the reading of the meter is 002010. However, this odometer skips the digit 4 and jumps directly from 3 to 5, for each of the wheels. How many kilometers has the car actually traveled? A) 1409 B) 1467 C) 1647 D) 1787 E) 1809","729$ kilometers, the fourth wheel from the right jumps. So in the position 002010, the number of kil",easy,"A7. B) 1467 At 9 kilometers, the second wheel from the right jumps. After it has jumped nine times, that is, after 9.9 kilometers, the third wheel from the right jumps. After $9 \cdot 9 \cdot 9=729$ kilometers, the fourth wheel from the right jumps. So in the position 002010, the number of kilometers traveled is equal to $2 \cdot 729+1 \cdot 9=1467$." 6ce7cc74d01b,"4. In triangle $ABC$, side $AB$ is equal to $\sqrt{11}$, and the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are in the ratio $9: 25$. a) Find the ratio of the distances from point $A$ to the lines $\ell$ and $BC$. b) Find the distance from point $C$ to the line $\ell$ and the radius of the circle $\Omega$.",". a) $9: 16$, b) $d_{C}=\frac{275}{54}, R=3$",medium,"Answer. a) $9: 16$, b) $d_{C}=\frac{275}{54}, R=3$. Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\circ}$). By the theorem on the angle between a tangent and a chord, the angle between the tangent and $A B$ is $\gamma$, and the angle between the tangent and $B C$ is $2 \gamma$. Therefore, the distance $d_{A}$ from point $A$ to the tangent is $A B \sin \gamma = 2 R \sin ^{2} \gamma$. Similarly, the distance $d_{C}$ from point $C$ to the tangent is $B C \sin 2 \gamma = 2 R \sin ^{2} 2 \gamma$. Thus, $\frac{25}{9}=\frac{d_{C}}{d_{A}}=4 \cos ^{2} \gamma$, from which $\cos ^{2} \gamma=\frac{25}{36}$. Then $\sin ^{2} \gamma=\frac{11}{36}, R=\frac{A B}{\sin 2 \gamma}=3$. Therefore, $d_{A}=\frac{11}{6}, d_{C}=\frac{275}{54}$. The distance from point $A$ to line $B C$ is $$ A B \cdot \sin \angle A B C=A B \sin 3 \gamma=A B\left(3 \sin \gamma-4 \sin ^{3} \gamma\right)=\frac{88}{27} $$ The ratio of the distances from point $A$ to lines $\ell$ and $B C$ is $\frac{11}{6}: \frac{88}{27}=\frac{9}{16}$." fecd00962563,"* Given the quadratic equation in $x$: $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$ has real roots, find the value of the real number $a$.","1$, substituting into $a x_{0}^{2}+x_{0}+a^{2}=0$ gives $a^{2}+a+1=0$. Since $a \in \mathbf{R}, a^{2",medium,"Suppose the original equation has a real root $x_{0}$, substituting into the original equation gives $a(1+\mathrm{i}) x_{0}^{2}+\left(1+a^{2} \mathrm{i}\right) x_{0}+a^{2}+\mathrm{i}=0$, which is $\left(a x_{0}^{2}+x_{0}+a^{2}\right)+\mathrm{i}\left(a x_{0}^{2}+a^{2} x_{0}+1\right)=0$. According to the necessary and sufficient condition for the equality of complex numbers, we have $\left\{\begin{array}{l}a x_{0}^{2}+x_{0}+a^{2}=0 \\ a x_{0}^{2}+a^{2} x_{0}+1=0,\end{array}\right.$ solving this gives $x_{0}=1$ or $a=\pm 1$. When $x_{0}=1$, substituting into $a x_{0}^{2}+x_{0}+a^{2}=0$ gives $a^{2}+a+1=0$. Since $a \in \mathbf{R}, a^{2}+a+1=\left(a+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, so it does not hold. Similarly, when $a=+1$, it does not hold either. When $a=-1$, the equation becomes $x_{0}^{2}-x_{0}-1=0$, so $x_{0}=\frac{1 \pm \sqrt{5}}{2}$ satisfies the condition, hence $a=-1$." 75956004e196,33. Find the largest positive integer $n$ such that $\sqrt{n-100}+\sqrt{n+100}$ is a rational number.,\sqrt{n+100}-\sqrt{n-100}$ should be as small as possible. We rule out $\ell-k=1$ since it leads to ,medium,"33. Answer: 2501 First we note that if $\sqrt{n-100}+\sqrt{n+100}$ is a rational number, then both $\sqrt{n-100}$ and $\sqrt{n+100}$ are rational. Since $n$ is a positive integer, this implies that $\sqrt{n-100}$ and $\sqrt{n+100}$ are integers. Let $\sqrt{n-100}=k$ and $\sqrt{n+100}=\ell$. Squaring both sides of the equations, we obtain $n-100=k^{2}$ and $n+100=\ell^{2}$. This gives $$ 200=\ell^{2}-k^{2}=(\ell-k)(\ell+k) . $$ Now for $n$ to be the largest positive integer for which $\sqrt{n-100}+\sqrt{n+100}$ is rational, $\ell-k=\sqrt{n+100}-\sqrt{n-100}$ should be as small as possible. We rule out $\ell-k=1$ since it leads to $\ell$ and $k$ being non-integers from equation (1). Thus we have $\ell-k=2$, and it follows from equation (1) that $\ell+k=100$. Hence $\ell=51$ and $k=49$, and $n=k^{2}+100=49^{2}+100=2501$." 382968882724,"11. (IND $\left.3^{\prime}\right)^{\mathrm{IMO}}$ Given a circle with two chords $A B, C D$ that meet at $E$, let $M$ be a point of chord $A B$ other than $E$. Draw the circle through $D, E$, and $M$. The tangent line to the circle $D E M$ at $E$ meets the lines $B C, A C$ at $F, G$, respectively. Given $\frac{A M}{A B}=\lambda$, find $\frac{G E}{E F}$.",\frac{\lambda}{1-\lambda}$.,medium,"11. Assume $\mathcal{B}(A, E, M, B)$. Since $A, B, C, D$ lie on a circle, we have $\angle G C E = \angle M B D$ and $\angle M A D = \angle F C E$. Since $F D$ is tangent to the circle around $\triangle E M D$ at $E$, we have $\angle M D E = \angle F E B = \angle A E G$. Consequently, $\angle C E F = 180^{\circ} - \angle C E A - \angle F E B = 180^{\circ} - \angle M E D - \angle M D E = \angle E M D$ and $\angle C E G = 180^{\circ} - \angle C E F = 180^{\circ} - \angle E M D = \angle D M B$. It follows that $\triangle C E F \sim \triangle A M D$ and $\triangle C E G \sim \triangle B M D$. From the first similarity we obtain $C E \cdot M D = A M \cdot E F$, and from the second we obtain $C E \cdot M D = B M \cdot E G$. Hence $$ \begin{gathered} A M \cdot E F = B M \cdot E G \Longrightarrow \\ \frac{G E}{E F} = \frac{A M}{B M} = \frac{\lambda}{1-\lambda} . \end{gathered} $$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-543.jpg?height=371&width=353&top_left_y=1023&top_left_x=894) If $\mathcal{B}(A, M, E, B)$, interchanging the roles of $A$ and $B$ we similarly obtain $\frac{G E}{E F} = \frac{\lambda}{1-\lambda}$." 0cb373a5a9a8,"6 Consider a rectangular array of single digits $d_{i, j}$ with 10 rows and 7 columns, such that $d_{i+1, j}-d_{i, j}$ is always 1 or -9 for all $1 \leq i \leq 9$ and all $1 \leq j \leq 7$, as in the example below. For $1 \leq i \leq 10$, let $m_{i}$ be the median of $d_{i, 1}, \ldots, d_{i, 7}$. Determine the least and greatest possible values of the mean of $m_{1}, m_{2}, \ldots, m_{10}$. Example: \begin{tabular}{lcccccccc} & $d_{i, 1}$ & $d_{i, 2}$ & $d_{i, 3}$ & $d_{i, 4}$ & $d_{i, 5}$ & $d_{i, 6}$ & $d_{i, 7}$ & $m_{i}$ \\ $i=1$ & 2 & 7 & 5 & 9 & 5 & 8 & 6 & median is 6 \\ $i=2$ & 3 & 8 & 6 & 0 & 6 & 9 & 7 & median is 6 \\ $i=3$ & 4 & 9 & 7 & 1 & 7 & 0 & 8 & median is 7 \\ $i=4$ & 5 & 0 & 8 & 2 & 8 & 1 & 9 & median is 5 \\ $i=5$ & 6 & 1 & 9 & 3 & 9 & 2 & 0 & median is 3 \\ $i=6$ & 7 & 2 & 0 & 4 & 0 & 3 & 1 & median is 2 \\ $i=7$ & 8 & 3 & 1 & 5 & 1 & 4 & 2 & median is 3 \\ $i=8$ & 9 & 4 & 2 & 6 & 2 & 5 & 3 & median is 4 \\ $i=9$ & 0 & 5 & 3 & 7 & 3 & 6 & 4 & median is 4 \\ $i=10$ & 1 & 6 & 4 & 8 & 4 & 7 & 5 & median is 5 \end{tabular}","a$ if and only if $M_{i+10-a, 8-j}=9-a$. Summing over $i$ (and noting that $i+10-a$ varies over $0,1",medium,"Solution 1: Note that rearranging the columns does not change the medians, hence we may sort the first row, so that $d_{1,1} \leq d_{1,2} \leq \ldots \leq d_{1,7}$. The calculations are much simplified if we subtract $i-1$ from each row. In other words, we put $D_{i, j}=d_{i, j}-(i-1)$. This subtracts $i-1$ from the median $m_{i}$ as well - that is if $M_{i}$ is the median of $D_{i, j} \mathrm{~s}$, then $M_{i}=m_{i}-(i-1)$. Thus the sum of the $M_{i} \mathrm{~s}$ is equal to the sum of the $m_{i} \mathrm{~s}$ minus $0+1+2+\ldots+9=45$. We shall show that sum of $M_{i}$ 's is 0 , so that the sum of the $m_{i} \mathrm{~s}$ is 45 and the average is always 4.5 . Note that since $D_{1,1} \leq D_{1,2} \leq \ldots \leq D_{1,7}$ the entry $D_{1,4}$ is a median. The fourth column will continue to contain a median until $d_{i, 7}=0$ (at which point the third column will contain a median), that is $10-D_{1,7}$ times (note that $\left.d_{1,7}=D_{1,7}\right)$. The sum of those medians is then equal $D_{1,4}\left(10-D_{1,7}\right)$. After that, median moves to the third column and stays there until $d_{i, 6}=0$ (this may be no time at all, if $d_{1,6}=d_{1,7}$, but that will not affect the calculation). The contribution of those medians is $D_{1,3}\left(D_{1,7}-D_{1,6}\right)$. Continuing this way we see that the medians in the second column contribute $D_{1,2}\left(D_{1,6}-D_{1,5}\right)$ and ones in the first column $D_{1,1}\left(D_{1,5}-D_{1,4}\right)$. A median then moves to the seventh column, but by that point its value has dropped, $D_{i, 7}=D_{1,7}-10$. The contribution of those medians is then $\left(D_{1,7}-10\right)\left(D_{1,4}-D_{1,3}\right)$. Similarly for those in sixth and fifth columns we get $\left(D_{1,6}-10\right)\left(D_{1,3}-D_{1,2}\right)$ and $\left(D_{1,5}-10\right)\left(D_{1,2}-D_{1,1}\right)$. Finally the median moves to the fourth column again, staying there remaining $D_{1,1}$ times, contributing $\left(D_{1,4}-10\right) D_{1,1}$. Overall, the sum of all medians is thus $$ \begin{array}{r} D_{1,4}\left(10-D_{1,7}\right)+D_{1,3}\left(D_{1,7}-D_{1,6}\right)+D_{1,2}\left(D_{1,6}-D_{1,5}\right)+ \\ D_{1,1}\left(D_{1,5}-D_{1,4}\right)+\left(D_{1,7}-10\right)\left(D_{1,4}-D_{1,3}\right)+\left(D_{1,6}-10\right)\left(D_{1,3}-D_{1,2}\right) \\ +\left(D_{1,5}-10\right)\left(D_{1,2}-D_{1,1}\right)+\left(D_{1,4}-10\right) D_{1,1} . \end{array} $$ It is fairly easy to see that this expression is in fact equal to 0 (for example, by considering the linear and quadratic terms separately). This means that the sum of new medians $M_{i}$ is zero, and the sum of the original $m_{i}$ 's is 45 , as wanted, Solution 2: We will prove a stronger claim: for all $a$, the number of $m_{i}$ 's equal to $a$ equals the number of $m_{i}$ 's equal to $9-a$. (By a pairing argument, this implies that the average of the $m_{i}^{\prime}$ 's is $9 / 2$.) Indeed, for $1 \leq j \leq 10$ let $M_{i, j}$ denote the $j$ th smallest entry in row $i$ of the table (so that $m_{i}=M_{i, 4}$ ); we will show that for all $a$ and $j$, the number of $M_{i, j}$ 's equal to $a$ equals the number of $M_{i, 8-j}$ 's equal to $9-a$. Henceforth, all row-indices are to be interpreted modulo 10 , and ""between"" is meant in the inclusive sense. It follows from the defining property of the table that for all $i$ between 1 and 10 , all $a$ between 0 and 9 , and all $k$ between 0 and $a$, the number of $k$ 's in row $i$ equals the number of $k+a$ 's in row $i+a$. Replacing $a$ by $9-a$, and summing over all $k$ between 0 and $a$, we find that the number of entries between 0 and $a$ in row $i$ equals the number of entries between $9-a$ and 9 in row $i+9-a$. Hence for all $j$, the number of entries between 0 and $a$ in row $i$ is greater than or equal to $j$ if and only if the number of entries between $9-a$ and 9 in row $i+9-a$ is greater than or equal to $j$. But this means that the $j$ smallest entries in row $i$ are all between 0 and $a$ if and only if the $j$ largest entries in row $i+9-a$ are all between $9-a$ and 9 . That is, $M_{i, j} \leq a$ if and only if $M_{i+9-a, 8-j} \geq 9-a$. Replacing $a$ by $a-1$, we see also that $M_{i, j} \leq a-1$ if and only if $M_{i+10-a, 8-j} \geq 10-a$. Combining the last two facts, we conclude that $M_{i, j}=a$ if and only if $M_{i+10-a, 8-j}=9-a$. Summing over $i$ (and noting that $i+10-a$ varies over $0,1, \ldots, 9 \bmod 10$ as $i$ does), we see that the number of $i$ 's with $M_{i, j}=a$ equals the number of $i$ 's with $M_{i, 8-j}=9-a$, as was claimed above." 52e064584be2,"Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $200$ distinct points. (Note that for $3$ distinct points, the minimum number of lines is $3$ and for $4$ distinct points, the minimum is $4$)",21,medium,"1. **Understanding the Problem:** We need to find the minimum number of lines that can be drawn on a plane such that they intersect in exactly 200 distinct points. 2. **General Formula for Intersections:** For \( n \) lines, the maximum number of intersection points is given by the combination formula: \[ \binom{n}{2} = \frac{n(n-1)}{2} \] This formula counts the number of ways to choose 2 lines out of \( n \) lines, each pair of which intersects at a unique point. 3. **Finding the Minimum \( n \):** We need to find the smallest \( n \) such that: \[ \frac{n(n-1)}{2} \geq 200 \] Solving for \( n \): \[ n(n-1) \geq 400 \] \[ n^2 - n - 400 \geq 0 \] Solving the quadratic equation \( n^2 - n - 400 = 0 \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ n = \frac{1 \pm \sqrt{1 + 1600}}{2} = \frac{1 \pm \sqrt{1601}}{2} \] Approximating \( \sqrt{1601} \approx 40.01 \): \[ n = \frac{1 + 40.01}{2} \approx 20.505 \] Since \( n \) must be an integer, we round up to the next whole number: \[ n = 21 \] 4. **Verification:** Check if \( n = 21 \) satisfies the condition: \[ \frac{21 \times 20}{2} = 210 \] Since 210 is greater than 200, \( n = 21 \) is a valid solution. 5. **Configuration:** The solution provided suggests using 10 pairs of parallel lines and 1 additional line that intersects all other lines. However, this configuration does not match the requirement of exactly 200 distinct intersection points. Instead, we should use the general formula directly. 6. **Conclusion:** The minimum number of lines required to achieve exactly 200 distinct intersection points is 21. The final answer is \( \boxed{ 21 } \)" dfb26ee0bd38,"## Task A-1.1. Determine all ordered pairs of integers $(x, y)$ for which $$ x^{2} y+4 x^{2}-3 y=51 $$",See reasoning trace,medium,"## Solution. Let's rewrite the initial equality in a different way: $$ \begin{aligned} x^{2} y+4 x^{2}-3 y & =51 & & \\ x^{2}(y+4)-3 y & =51 & & \\ x^{2}(y+4)-3(y+4)+12 & =51 & & 1 \text { point } \\ \left(x^{2}-3\right)(y+4) & =39 . & & 2 \text { points } \end{aligned} $$ The factors \(x^{2}-3\) and \(y+4\) on the left side are divisors of the number 39, and their product is 39. We conclude that $$ x^{2}-3 \in\{-39,-13,-3,-1,1,3,13,39\} $$ which means \(x^{2} \in\{-36,-10,0,2,4,6,16,42\}\). Of these numbers, only 0, 4, and 16 are perfect squares. For \(y\), we have \(y+4=\frac{39}{x^{2}-3}\). If \(x^{2}=0\), i.e., \(x=0\), then \(y=-17\). 1 point If \(x^{2}=4\), i.e., \(x= \pm 2\), then \(y=35\). 1 point If \(x^{2}=16\), i.e., \(x= \pm 4\), then \(y=-1\). We conclude that all solutions to the initial equation are $$ (x, y)=(0,-17),(2,35),(-2,35),(4,-1),(-4,-1) $$" bad2215f161c,"4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-5 ; 5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$.","(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expressi",easy,"Answer: 110 Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-11 \leqslant y-1 \leqslant 9$ and $-8 \leqslant 2-x \leqslant 12$. Therefore, $(y-1)(2-x)+2 \leqslant 9 \cdot 12+2=110$. The maximum value of the expression is achieved when $a=c=-5, b=d=5$." 6f1321a75f68,"$3 \cdot 3$ Label the endpoints of a diameter of a circle with the number 1, then equally divide each of the resulting semicircles, and write the sum of the numbers at the endpoints of the two semicircles at their division points (first step). Then equally divide the 4 arcs obtained, and write the sum of the numbers at the endpoints of these arcs at their division points (second step). Find the sum of all numbers marked on the circle after performing this process $n$ times.",See reasoning trace,easy,"[Solution] Let the sum of all numbers after the $k$-th step be denoted as $S_{k}$, then $S_{0}=2, S_{1}=$ 6. During the $(k+1)$-th step, since each number in $S_{k}$ is at the endpoint of exactly two arcs, each number is used twice, so we have $S_{k+1}=S_{k}+2 S_{k}=3 S_{k}$. By this recurrence relation, we get $$ S_{n}=3 S_{n-1}=3^{n} S_{0}=2 \times 3^{n}, $$ which means after $n$ steps, the sum of all numbers marked on the circle is $2 \times 3^{n}$." 98c3b1197634,"7. (6 points) $n$ is a positive integer, and a prime number of the form $2^{n}-1$ is called a Mersenne number, for example: $2^{2}-1=3,2^{3}-1=7$ are Mersenne numbers. Recently, American scholars have refreshed the largest Mersenne number, $n=74207281$, and this Mersenne number is also the largest known prime number so far, its unit digit is . $\qquad$",: 1,medium,"【Analysis】According to the problem, this Mersenne number is $2^{n}-1=2^{74207281}-1$, to find the unit digit of the Mersenne number, we only need to find the unit digit of $2^{74207281}$, and the unit digit of $2^{74207281}$ can be determined according to the periodic rule. 【Solution】Solution: According to the analysis, this Mersenne number is $2^{n}-1=2^{74207281}-1$, $\because 2^{1}=2 ; 2^{2}=4 ; 2^{3}=8 ; 2^{4}=16 ; 2^{5}=32$; $2^{6}=64 ; 2^{7}=128 ; 2^{8}=256 ; 2^{9}=512 ; 2^{10}=1024 \cdots$ From this, we can see that the unit digit of $2^{n}$ is: $2, 4, 8, 6, 2, 4, 8, 6, 2 \cdots$ That is, when $n=1,5,9, \cdots$, the unit digit is $2 ; n=2,6,10, \cdots$, the unit digit is 4 ; $n=3,7,11, \cdots$, the unit digit is $8 ; n=4,8,12, \cdots$, the unit digit is 6 ; In summary, the unit digit of $2^{n}$ appears in a cycle, with a period of 4, and $74207281=4 \times 18551820+1$, so the unit digit of $2^{74207281}$ is the same as that of $2^{1}$, it can be concluded that the unit digit of $2^{74207281}$ is 2, and the unit digit of $2^{74207281}-1$ is: $2-1=1$. Therefore, the answer is: 1." c6515c36b1b3,"Let $N$ and $M$ be positive integers, and $p$ and $q$ be distinct prime numbers. Suppose that $N+M$ is a five-digit number, $N$ is divisible by $p$, and the number of its divisors is $q$, while $M$ is divisible by $q$, and the number of its divisors is $p$. Determine the possible values of $N$ and $M$.",See reasoning trace,medium,"To determine the number of divisors of a number, we multiply the numbers that are one greater than the exponents in its prime factorization. In our case, this product must be a prime number. If more than one prime factor were present in the prime factorization, the number of divisors would not be a prime number; thus, both numbers must be prime powers. From this, it follows that if the number $N$ has $q$ divisors, then its exponent is $(q-1)$. Since only one prime appears in the prime factorization of $N$ and $N$ is divisible by $p$, this prime must be $p$. Therefore, $N=p^{q-1}$. Similarly, we get $M=q^{p-1}$. We need to examine when the sum $N+M=p^{q-1}+q^{p-1}$ is a five-digit number. The right side of the equation remains the same if $p$ and $q$ are swapped, so for simplicity, we assume $p \leq q$. If $p^{q-1}$ is at least a six-digit number, then the sum $p^{q-1}+q^{p-1}$ is also at least a six-digit number. Since $2^{18}$, $3^{12}$, $5^{10}$, and $7^{6}$ are all at least six-digit numbers, we only need to check the following 13 cases (the corresponding $p^{q-1}+q^{p-1}$ values are listed in the table). | $p q q$ | 2 | 3 | 5 | 7 | 11 | 13 | 17 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 2 | 4 | 7 | 21 | 71 | 1035 | 4109 | **65553** | | 3 | | 18 | 106 | 778 | **59170** | | | | 5 | | | 1250 | **18026** | | | | In these cases, the possible values of $N$ and $M$ (considering that they can be swapped) are: | $p$ | $q$ | $N$ | $M$ | | :---: | :---: | :---: | :---: | | 2 | 17 | 65536 | 17 | | 17 | 2 | 17 | 65536 | | 3 | 11 | 59049 | 121 | | 11 | 3 | 121 | 59049 | | 5 | 7 | 15625 | 2401 | | 7 | 5 | 2401 | 15625 | Note. Unfortunately, during the grading, relatively few solutions received full marks. Many provided the primes but not the numbers $N$ and $M$, which was a result of inattention. Many also provided only three solutions instead of six, and did not realize that $N$ and $M$ can be swapped. Many did not provide sufficiently detailed reasoning, resulting in lost points (usually either they did not derive the expressions for $N$ and $M$ in terms of $p$ and $q$, or they only listed the correct primes, showed that they were indeed correct, but did not explain why other solutions were not possible). Those who provided only the final result without any reasoning received 0 points, as the competition rules state that points cannot be awarded for just the result. However, there were also some excellent solutions." 80f66803a0a6,"Find all increasing sequences $a_1,a_2,a_3,...$ of natural numbers such that for each $i,j\in \mathbb N$, number of the divisors of $i+j$ and $a_i+a_j$ is equal. (an increasing sequence is a sequence that if $i\le j$, then $a_i\le a_j$.)",a_n = n,medium,"1. **Claim 1:** If \( p \) is a prime, then \( a_{2^{p-2}} = 2^{p-2} \). **Proof:** From \( P(2^{p-2}, 2^{p-2}) \), we have: \[ \tau(2^{p-1}) = \tau(2a_{2^{p-2}}) \] Since \( \tau(2^{p-1}) = p \) (because \( 2^{p-1} \) has \( p \) divisors), we get: \[ \tau(2a_{2^{p-2}}) = p \] Since \( p \) is prime, the only integers \( n \) for which \( \tau(n) = p \) are of the form \( q^{p-1} \) for some prime \( q \). Therefore: \[ 2a_{2^{p-2}} = 2^{p-1} \implies a_{2^{p-2}} = 2^{p-2} \] which proves the claim. \(\blacksquare\) 2. **Claim 2:** If \( n \) is a positive integer and \( p \) is prime such that \( p > n \ge 2 \), then: \[ a_{p-n} + a_n = p \] **Proof:** From \( P(p-n, n) \), we have: \[ \tau(p) = \tau(a_{p-n} + a_n) \] Since \( \tau(p) = 2 \) (because \( p \) is prime), we get: \[ \tau(a_{p-n} + a_n) = 2 \] This implies that \( a_{p-n} + a_n \) is a prime number. 3. **Subclaim 2.1:** If \( n \) is a positive integer and \( p \) and \( q \) are primes such that \( p > q > n \ge 2 \), then: \[ a_{p-n} \ne a_{q-n} \] **Proof:** Suppose there exists \( n, p, q \) such that \( a_{p-n} = a_{q-n} \). Then this means \( a_{p-n} + a_n = a_{q-n} + a_n = r \) for some prime \( r \). Since \( p > q \ge 3 \), this means there is a positive integer \( k \) which lies strictly between \( p \) and \( q \), where \( k \) is composite. Since \( a_1, a_2, a_3, \ldots \) is an increasing sequence, this means that \( a_{k-n} + a_n = r \), which implies that \( a_{k-n} + a_n \) is prime. However, from \( P(k-n, n) \): \[ \tau(k) = \tau(a_{k-n} + a_n) \implies \tau(k) = 2 \] which implies that \( k \) is prime, which is a contradiction. \(\blacksquare\) 4. Now, we prove Claim 2. Let \( m \) be the smallest prime such that \( 2^{m-2} > p \), and let \( S \) be the set of all positive integers less than \( 2^{m-2} \) which are of the form \( q-n \) for some prime \( q \), and \( T \) be the set \( \{s_k \mid k \in S\} \). Since \( s_k < s_{2^{m-2}} = 2^{m-2} \), this means all the elements of \( T \) must be less than \( 2^{m-2} \) as well. We know that all of the elements from \( T \) must be of the form \( p-n \) as well, which implies that \( T \subseteq S \). Furthermore, according to Subclaim 2.1, we find that \( a_{p-n} \ne a_{q-n} \) for any primes \( p \) and \( q \) satisfying \( p > q > n \ge 2 \). This implies that all the elements of \( T \) are distinct. This shows that \( |S| = |T| \), which implies that \( S = T \). Therefore, \( a_{p-n} = p-n \) for all \( n \) and prime \( p \) satisfying \( p > n \ge 2 \). \(\blacksquare\) 5. Finally, we use induction to complete the proof that \( a_n = n \) for all positive integers \( n \). The base cases are when \( n = 1 \) and \( n = 2 \), which easily follows from Claim 1. For the inductive step, suppose that we know that \( a_n = n \) for all \( n = 1, 2, 3, \ldots, k \). According to Bertrand's Postulate, there exists a prime \( p \) such that \( k < p < 2k \). Also, \( p - k < 2k - k = k \). From our induction hypothesis, this implies that \( a_{p-k} = p - k \). Hence, from Claim 2: \[ a_k + a_{p-k} = p \implies a_k + (p - k) = p \implies a_k = k \] which completes the proof. \(\blacksquare\) The final answer is \( \boxed{ a_n = n } \) for all positive integers \( n \)." f92e6270f1ca,"7. The function $$ f(x)=x+\cos x-\sqrt{3} \sin x $$ passes through a point on its graph where the slope of the tangent line is $k$. Then the range of values for $k$ is $\qquad$ .","1-\sin x-\sqrt{3} \cos x \\ =1-2 \sin \left(x+\frac{\pi}{3}\right) \in[-1,3]\end{array}$",easy,"$\begin{array}{l}\text { 7. }[-1,3] \text {. } \\ f^{\prime}(x)=1-\sin x-\sqrt{3} \cos x \\ =1-2 \sin \left(x+\frac{\pi}{3}\right) \in[-1,3]\end{array}$" 05325beabee5,2.199. $\left(\frac{\sqrt[3]{m n^{2}}+\sqrt[3]{m^{2} n}}{\sqrt[3]{m^{2}}+2 \sqrt[3]{m n}+\sqrt[3]{n^{2}}}-2 \sqrt[3]{n}+\frac{m-n}{\sqrt[3]{m^{2}}-\sqrt[3]{n^{2}}}\right):(\sqrt[6]{m}+\sqrt[6]{n})$.,$\sqrt[6]{m}-\sqrt[6]{n}$,medium,"## Solution. Domain of definition: $\left\{\begin{array}{l}m \neq n, \\ m \geq 0, \\ n \geq 0 .\end{array}\right.$ $$ \begin{aligned} & \left(\frac{\sqrt[3]{m n^{2}}+\sqrt[3]{m^{2} n}}{\sqrt[3]{m^{2}}+2 \sqrt[3]{m n}+\sqrt[3]{n^{2}}}-2 \sqrt[3]{n}+\frac{m-n}{\sqrt[3]{m^{2}}-\sqrt[3]{n^{2}}}\right):(\sqrt[6]{m}+\sqrt[6]{n})= \\ & =\left(\frac{\sqrt[3]{m n}(\sqrt[3]{m}+\sqrt[3]{n})}{(\sqrt[3]{m}+\sqrt[3]{n})^{2}}-2 \sqrt[3]{n}+\frac{(\sqrt[3]{m}-\sqrt[3]{n})\left(\sqrt[3]{m^{2}}+\sqrt[3]{m n}+\sqrt[3]{n^{2}}\right)}{(\sqrt[3]{m}-\sqrt[3]{n})(\sqrt[3]{m}+\sqrt[3]{n})}\right) \times \end{aligned} $$ $$ \begin{aligned} & \times \frac{1}{\sqrt[6]{m}+\sqrt[6]{n}}=\left(\frac{\sqrt[3]{m n}}{\sqrt[3]{m}+\sqrt[3]{n}}-2 \sqrt[3]{n}+\frac{\sqrt[3]{m^{2}}+\sqrt[3]{m n}+\sqrt[3]{n^{2}}}{\sqrt[3]{m}+\sqrt[3]{n}}\right) \cdot \frac{1}{\sqrt[6]{m}+\sqrt[6]{n}}= \\ & =\frac{\sqrt[3]{m n}-2 \sqrt[3]{n}(\sqrt[3]{m}+\sqrt[3]{n})+\sqrt[3]{m}+\sqrt[3]{m n}+\sqrt[3]{n^{2}}}{\sqrt[3]{m}+\sqrt[3]{n}} \cdot \frac{1}{\sqrt[6]{m}+\sqrt[6]{n}}= \end{aligned} $$ $$ \begin{aligned} & =\frac{\sqrt[3]{m^{2}}-\sqrt[3]{n^{2}}}{\sqrt[3]{m}+\sqrt[3]{n}} \cdot \frac{1}{\sqrt[6]{m}+\sqrt[6]{n}}=\frac{(\sqrt[3]{m}-\sqrt[3]{n})(\sqrt[3]{m}+\sqrt[3]{n})}{\sqrt[3]{m}+\sqrt[3]{n}} \cdot \frac{1}{\sqrt[6]{m}+\sqrt[6]{n}}= \\ & =\frac{\sqrt[3]{m}-\sqrt[3]{n}}{\sqrt[6]{m}+\sqrt[6]{n}}=\frac{(\sqrt[6]{m})^{2}-(\sqrt[6]{n})^{2}}{\sqrt[6]{m}+\sqrt[6]{n}}=\frac{(\sqrt[6]{m}+\sqrt[6]{n})(\sqrt[6]{m}-\sqrt[6]{n})}{\sqrt[6]{m}+\sqrt[6]{n}}=\sqrt[6]{m}-\sqrt[6]{n} . \end{aligned} $$ Answer: $\sqrt[6]{m}-\sqrt[6]{n}$." 40193bb76d79,"$1 \cdot 24$ Given $\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}=k$, then the value of $k$ is (A) 2 or -1. (B) 2. (C) -1. (D) 2 or 1. (Anhui Province, Anqing and other counties and cities junior high school mathematics competition, 1991)",$(A)$,easy,"［Solution］From the given and the property of proportional division, we have $$ \begin{array}{c} \frac{(y+z)+(z+x)+(x+y)}{x+y+z}=k, \\ \frac{2(x+y+z)}{x+y+z}=k . \end{array} $$ If $x+y+z \neq 0$, then $k=2$; If $x+y+z=0$, substituting $y+z=-x$ into $\frac{y+z}{x}=k$, we get $k=-1$. Therefore, $k=2$ or -1. Hence, the answer is $(A)$." 4e7bef2710e4,Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digits of the number 14641. Determine all solutions of the equation $x^{4}=n$ in the set $\mathbb{C}$.,See reasoning trace,medium,"## Solution. If between every two digits of the number 14641 we write 2013 zeros, we get a number of the form $1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1$, and we can express it using powers with base 10. Then, $$ \begin{aligned} n & =1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1 \\ & =10^{4 \cdot 2013+4}+4 \cdot 10^{3 \cdot 2013+3}+6 \cdot 10^{2 \cdot 2013+2}+4 \cdot 10^{2013+1}+1= \\ & =\left(10^{2013+1}\right)^{4}+4 \cdot\left(10^{2013+1}\right)^{3}+6 \cdot\left(10^{2013+1}\right)^{2}+4 \cdot 10^{2013+1}+1= \\ & =\binom{4}{0}\left(10^{2014}\right)^{4}+\binom{4}{1}\left(10^{2014}\right)^{3} \cdot 1+\binom{4}{2}\left(10^{2014}\right)^{2} \cdot 1^{2}+\binom{4}{3} 10^{2014} \cdot 1^{3}+\binom{4}{4} \cdot 1^{4} \end{aligned} $$ Then the solutions to the equation $x^{4}=n$, or $x^{4}=\left(10^{2014}+1\right)^{4}$, are the numbers $$ x_{1,2}= \pm\left(10^{2014}+1\right), \quad x_{3,4}= \pm\left(10^{2014}+1\right) \cdot i $$" 1f2c839de7f6,"2. If the line $y=k x+1$ intersects the circle $x^{2}+y^{2}+k x+$ $m y-4=0$ at points $P$ and $Q$, and points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, then the area of the plane region represented by the system of inequalities $$ \left\{\begin{array}{l} k x-y+1 \geqslant 0, \\ k x-m y \leqslant 0, \\ y \geqslant 0 \end{array}\right. $$ is ( ). (A) 2 (B) 1 (C) $\frac{1}{2}$ (D) $\frac{1}{4}$","0$, we know that the line $y=kx+1$ is perpendicular to $x+y=0$, and the center $\left(-\frac{k}{2},-",medium,"2.D. From the fact that points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, we know that the line $y=kx+1$ is perpendicular to $x+y=0$, and the center $\left(-\frac{k}{2},-\frac{m}{2}\right)$ of the circle lies on the line $x+y=0$, which gives $k=1$, and $\left(-\frac{k}{2}\right)+\left(-\frac{m}{2}\right)=0$. Solving these equations simultaneously, we get $k=1, m=-1$. Therefore, the system of inequalities $\left\{\begin{array}{l}x-y+1 \geqslant 0, \\ x+y \leqslant 0, \\ y \geqslant 0\end{array}\right.$ represents the plane region of the isosceles right triangle $\triangle ABO$ in Figure 3, with an area of $S=\frac{1}{4}$." 14409452357a,"[ The Law of Sines [ Area of a Triangle (using two sides and the included angle).] On a line passing through the center $O$ of a circle with radius 12, points $A$ and $B$ are taken such that $O A=15, A B=5$, and $A$ lies between $O$ and $B$. Tangents are drawn from points $A$ and $B$ to the circle, with the points of tangency lying on the same side of the line $O B$. Find the area of triangle $A B C$, where $C$ is the point of intersection of these tangents.",$\frac{150}{7}$,medium,"Apply the Law of Sines to triangle $ABC$. ## Solution Let $M$ and $N$ be the points of tangency of the circle with the lines passing through points $A$ and $B$ respectively, $\angle O A M=\alpha, \angle O B N=\beta$. Then $$ \frac{O M}{O A}=\sin \alpha, \frac{O N}{O B}=\sin \beta $$ Therefore, $$ \begin{gathered} \sin \alpha=\frac{4}{5}, \sin \beta=\frac{3}{5}, \cos \alpha=\frac{3}{5}, \cos \beta=\frac{4}{5} \\ B C=\frac{A B \sin \alpha}{\sin (\alpha-\beta)}=\frac{A B \sin \alpha}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{100}{7} \end{gathered} $$ Thus, $$ S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \sin \angle A B C=\frac{1}{2} \cdot 5 \cdot \frac{100}{7} \cdot \frac{3}{5}=\frac{150}{7} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-46.jpg?height=378&width=738&top_left_y=820&top_left_x=660) ## Answer $\frac{150}{7}$." feccb5a00080,"5. It is known that after 100 years, cadmium retains $95.76 \%$ of its original mass. Let the remaining mass of a mass of 1 after $x$ years be $y$, then the functional relationship between $x$ and $y$ is ( ). (A) $y=(0.9576)^{\frac{x}{100}}$ (B) $y=(0.9576)^{100 x}$ (C) $y=\left(\frac{0.9576}{100}\right)^{x}$ (D) $y=1-(0.0424)^{\frac{x}{100}}$",See reasoning trace,easy,5. A 691b562c8586,"3A. Determine all pairs of integers \(x\) and \(y\) that are solutions to the equation \[ (x+y+1993)^{2}=x^{2}+y^{2}+1993^{2} \]",See reasoning trace,medium,"Solution. We have $$ \begin{aligned} & (x+y+1993)^{2}=x^{2}+y^{2}+1993^{2} \\ & x^{2}+y^{2}+1993^{2}+2 x y+2 \cdot 1993 x+2 \cdot 1993 y=x^{2}+y^{2}+1993^{2} \\ & x y+1993 x+1993 y=0 \\ & (x+1993)(y+1993)=1993^{2} \end{aligned} $$ Given that 1993 is a prime number, and $x$ and $y$ are integers, the last equation yields the following six systems of linear equations: $$ \begin{aligned} & \left\{\begin{array}{l} x+1993=1 \\ y+1993=1993^{2} \end{array}\right. \\ & \left\{\begin{array}{l} x+1993=1993^{2} \\ y+1993=1 \end{array}\right. \\ & \left\{\begin{array}{l} x+1993=1993 \\ y+1993=1993 \end{array}\right. \\ & \left\{\begin{array}{l} x+1993=-1 \\ y+1993=-1993^{2} \end{array}\right. \\ & \left\{\begin{array}{l} x+1993=-1993^{2} \\ y+1993=-1 \end{array}\right. \\ & \left\{\begin{array}{l} x+1993=-1993 \\ y+1993=-1993 \end{array}\right. \end{aligned} $$ whose solutions are the pairs: $$ \begin{aligned} & (-1992,3970056),(3970056,-1992),(0,0), \\ & (1994,-3974042),(-3974042,-1994),(-3986,3986) \end{aligned} $$" 60b833a4e4e4,"3. Given the function $f(x)=\frac{2+x}{1+x}$. Let $$ \begin{array}{l} f(1)+f(2)+\cdots+f(1000)=m, \\ f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{1000}\right)=n . \end{array} $$ Then the value of $m+n$ is $\qquad$",3 \times 999+\frac{3}{2}=2998.5$.,easy,"3. 2998.5 It is clear that $x \neq -1$. Since $$ f(x)+f\left(\frac{1}{x}\right)=\frac{2+x}{1+x}+\frac{2+\frac{1}{x}}{1+\frac{1}{x}}=3, f(1)=\frac{3}{2} \text{, } $$ therefore, $m+n=3 \times 999+\frac{3}{2}=2998.5$." 4683691dd839,"What is the sum of the two smallest prime factors of $250$? $\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12$",\text{(C),easy,The prime factorization of $250$ is $2 \cdot 5^3$. The smallest two are $2$ and $5$. $2+5 = \boxed{\text{(C) }7}$. 50afce382b60,3. Determine the largest prime number whose all digits are distinct such that any permutation of its digits yields another prime number.,See reasoning trace,medium,"3. If the desired number is at least two digits, then in its representation, digits from the set $\{0,2,4,5,6,8\}$ must not appear (since by moving such a digit to the end of the number, we would obtain a number that is not prime). Thus, only four possible digits remain: $\{1,3,7,9\}$, so the desired number can be at most four digits. If it were four digits, then each of the digits $1,3,7$, and 9 would appear exactly once, but then by permuting the digits, we could obtain the number 1397, which is divisible by 11 and thus not prime. Similarly, for any possible choice of three out of these four digits, we can find a permutation where we do not get a prime number: $7|371,11| 319,7|791,7| 973$. Therefore, the desired number is at most two digits, and among them, we easily find the largest: it is 97 (it is prime, and by permuting its digits, we obtain the number 79, which is also prime)." 9e5637a8b088,"21. Let $p$ be a prime number such that the next larger number is a perfect square. Find the sum of all such prime numbers. (For example, if you think that 11 and 13 are two such prime numbers, then the sum is 24 .)","n^{2}, p=(n+1)(n-1)$. Since $p$ is prime, $n=2$. Hence $p=3$ is unique.",easy,"21 Ans: 3 . If $p+1=n^{2}, p=(n+1)(n-1)$. Since $p$ is prime, $n=2$. Hence $p=3$ is unique." 32ab60a514da,"4. On a circle, 1974 children are arranged and are playing a game of elimination as follows: the first child remains in the circle, the second child is eliminated, the third child remains, the fourth child is eliminated, and so on until only one child remains in the circle. Determine which child remains.",See reasoning trace,medium,"Solution. In the first round, children with even numbers are eliminated - only those with numbers of the form $2k+1$ remain, the smallest of which is 1 and the largest is 1973. In the second round, children with numbers remain such that the difference between adjacent numbers is equal to 4; the smallest of these numbers is 1, and the largest is 1973. We continue with this reasoning and obtain the table: | Round | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | | Smallest Num. | 1 | 1 | 5 | 13 | 13 | 45 | 109 | 109 | 365 | 877 | | Difference | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024 | | Largest Num. | 1973 | 1973 | 1973 | 1965 | 1965 | 1965 | 1901 | 1901 | 1901 | 1901 | In the last, eleventh round, the child with the number 877 is eliminated, and the child with the number 1901 remains. ## Second Year" bedea56b0c9a,"Given the complex number $z$ has a modulus of 1. Find $$ u=\frac{(z+4)^{2}-(\bar{z}+4)^{2}}{4 i} \text {. } $$ the maximum value.",See reasoning trace,medium,"$$ \begin{array}{l} \text { Given }|z|=1 \text {, we can set } z=\cos x+i \sin x \text {. Then, } \\ u=(4+\cos x) \sin x \text {. } \\ \text { By } u^{2}=(4+\cos x)^{2} \sin ^{2} x \\ =\frac{(4+\cos x)(4+\cos x)(\sqrt{6}+1)(1+\cos x) \cdot(\sqrt{6}+3)(1-\cos x)}{(\sqrt{6}+1)(\sqrt{6}+3)} \\ \leqslant \frac{1}{9+4 \sqrt{6}} \\ \cdot\left[\frac{(1+\cos x)+(1+\cos x)+(\sqrt{6}+1)(1+\cos x)+(\sqrt{6}+3)(1-\cos x)}{4}\right]^{4} \\ =\frac{1}{9+4 \sqrt{6}}\left(\frac{6+\sqrt{6}}{2}\right)^{4} \\ =\frac{9+24 \sqrt{6}}{4} \text {. } \\ \end{array} $$ When $4+\cos x=$ ( $\sqrt{6}+1)(1+\cos x)$ $$ =(\sqrt{6}+3)(1-\cos x) \text {, } $$ i.e., $\cos x=\frac{\sqrt{6}-2}{2}$, the maximum value is $$ u_{\max }=\frac{\sqrt{9+24 \sqrt{6}}}{2} $$ The coefficients $\sqrt{6}+1, \sqrt{6}+3$ are determined by the method of undetermined coefficients." 31dccb4b37b3,"At least how many passengers must occupy a seat in a metro car so that no more can sit down? (Assumptions: a) The ""width"" of each passenger is 1/6 of the seat width. b) Seated passengers cannot be forced to change their position.)",See reasoning trace,medium,"Two passengers cannot yet occupy the six-seat bench, because no matter how they sit, each of them can occupy at most 2 seats, and the remaining two seats will definitely accommodate another passenger. ![](https://cdn.mathpix.com/cropped/2024_05_02_1fc51a85693eabf5347ag-1.jpg?height=206&width=507&top_left_y=274&top_left_x=798) However, three passengers are sufficient to occupy the entire bench. Arrange the passengers so that the width of the spaces between them and at the ends of the bench is 1/8 of the width of the six-seat bench. Thus, the total width of all the spaces considered is $$ 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{8} = 1 $$ which is indeed the total width of the bench, and since $\frac{1}{8} < \frac{1}{6}$, more passengers will not fit." a19c6ad32ecc,"5. On the table, there are empty glasses lined up. Petya and Vasya take turns (starting with Petya) to fill them with drinks: Petya - lemonade, Vasya - compote. In one move, a player can fill one empty glass of their choice so that after their move, no two adjacent glasses contain the same drink. If all the glasses are filled as a result of the players' actions, the game ends in a draw. Otherwise, the player who cannot make a move loses. For which p will Vasya win regardless of Petya's actions?","$n \notin\{1,2,4,6\}$",medium,"Answer: $n \notin\{1,2,4,6\}$. Solution. Number the glasses from left to right with numbers from 1 to $n$. For $n=1$ and $n=2$, a draw is obvious. If $n$ is 4 or 6, Petya first fills the first glass with lemonade. For $n=4$, Petya can then fill one of the two last glasses and, thus, will not lose. For $n=6$, Petya fills the last glass (if it is empty) on his second move, and then one of the two middle glasses; if the last glass was filled by Vasya, Petya fills the third glass, and then the fifth. In any case, Petya will not lose. Now consider $n \notin\{1,2,4,6\}$. We will prove two statements. 1) If Vasya fills the glass with number 1 on his first move, he will always have a move. Let's call a segment a set of glasses standing between two consecutive glasses with lemonade. By the condition, each segment is non-empty, and any glass in the segment is either empty or contains compote. Let $k \geqslant 2$. After Petya's $k$-th move, $k-1$ segments are formed. By this point, Vasya has filled $k-1$ glasses, including the first one, which does not belong to any segment. Therefore, there will be a segment where all the glasses are empty. Vasya can then fill one of them with compote. 2) If Vasya fills the glass with number 1 on his first move, he cannot lose. Indeed, by 1), Vasya will always have a move and, thus, will achieve at least a draw. Let's describe Vasya's winning strategy. On his first move, Vasya always fills the edge glass (let's assume it is the first one; otherwise, we can renumber the glasses in reverse order). By 2), it is sufficient to show that Vasya can avoid a draw. Consider two cases. a) $n$ is odd. Vasya can play arbitrarily. A draw is impossible because in this case, the last glass would have been filled by Petya, which contradicts 1). b) $n=2 m n p u ~ m \geqslant 4$. On his second move, Vasya must fill a glass with an even number greater than 2. This is possible because there are at least three glasses with such numbers, and thus one of them is empty. Vasya can then play arbitrarily. Suppose the game ends in a draw. Then $m$ glasses with lemonade are placed in positions $2,3, \ldots, 2 m$. They must be on even positions, as no two glasses stand next to each other. But this is impossible, as there are only $m$ even positions, and one of them has already been used by Vasya." 0074dc07531c,"3. In the Cartesian coordinate system $x O y$, two point sets $M, N$ satisfy $M=\left\{(x, y) \mid x^{2}+y^{2}=1\right\}$, $N=\{(x, y)|| x-1|+| y-1 \mid=a\}$. If $M \bigcap N=\varnothing$, then the range of the real number $a$ is $\qquad$",See reasoning trace,easy,"Solution: As shown in the figure, set $N$ is the square $A B C D$, where $C D: x+y+a-2=0$. Therefore, $d=\frac{|a-2|}{\sqrt{2}}>1$ $\Rightarrow a>2+\sqrt{2}$ or $a<2-\sqrt{2}$. Thus, the range of the real number $a$ is $(-\infty, 2-\sqrt{2}) \cup(2+\sqrt{2},+\infty)$." 3b21fe66c47f,"# In trapezoid $A B C D$, the lateral side $B C$ is equal to the diagonal $B D$. On the smaller arc $A B$ of the circumscribed circle of triangle $A B C$, a point $E$ is chosen such that $B C=B E$. Find the angle $\angle A E D$.",$90^{\circ}$,medium,"Answer: $90^{\circ}$. ## Solution: From the isosceles triangle $B C D$ and the parallelism of $A B$ and $C D$, we get $\angle B C D = \angle B D C = \angle D B A = \alpha$. Let the line $C D$ intersect the circumcircle of triangle $A B C$ at point $F$. Then $B C F A$ is an inscribed, i.e., isosceles, trapezoid, from which the arcs $B C, B E$, and $A F$ are equal. Hence, $\angle B C E = \angle A C F = \beta$. $\angle E B A = \angle E C A = \gamma$, since these angles subtend the same arc. $\angle B C A = \angle B C E + \angle E C A = \beta + \gamma$. $\angle E B D = \angle E B A + \angle D B A = \gamma + \alpha$. Therefore, in the isosceles triangle $E B D$, the equality $\angle B D E = \angle B E D = \frac{180^{\circ} - \alpha - \gamma}{2}$ holds. Moreover, $\alpha = \angle B C D = \angle B C F = \angle B C A + \angle A C F = 2 \beta + \gamma$. $\angle A E D = \angle B E A - \angle B E D = (180^{\circ} - \angle B C A) - \frac{180^{\circ} - \alpha - \gamma}{2} = 180^{\circ} - \beta - \gamma - 90^{\circ} + \frac{2 \beta + 2 \gamma}{2} = 90^{\circ}$" 3ab1a4ad346a,"13.175. A motorboat left the pier simultaneously with a raft and traveled downstream $40 / 3$ km. Without stopping, it turned around and went upstream. After traveling $28 / 3$ km, it met the raft. If the speed of the river current is 4 km/h, what is the motorboat's own speed",$68 / 3$ km $/$ h,easy,"Solution. The boat and the raft were on the way $\frac{\frac{40}{3}-\frac{28}{3}}{4}=1$ hour. Let $v$ km $/ h$ be the boat's own speed. Downstream, the boat traveled $\frac{\frac{40}{3}}{4+v}$ hours, and upstream $\frac{28}{3(v-4)}$ hours. According to the condition, $\frac{40}{3(v+4)}+\frac{28}{3(v-4)}=1$, from which $v=\frac{68}{3}$ km $/$ h. Answer: $68 / 3$ km $/$ h." 605c19c65174,"1. Given real numbers $x, y$ satisfy $$ 17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. } $$ then the maximum value of $f(x, y)=\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is . $\qquad$","-5$, $f(x, y)$ reaches its maximum value of 7.",medium,"$-1.7$. From $17\left(x^{2}+y^{2}\right)-30 x y-16=0$, we get $$ \begin{array}{l} (x+y)^{2}+16(x-y)^{2}=16 . \\ \text { Let }\left\{\begin{array}{l} x+y=4 \cos \theta, \\ x-y=\sin \theta \end{array}(\theta \in \mathbf{R})\right. \\ \Rightarrow\left\{\begin{array}{l} x=2 \cos \theta+\frac{1}{2} \sin \theta, \\ y=2 \cos \theta-\frac{1}{2} \sin \theta . \end{array}\right. \end{array} $$ Then $f(x, y)$ $$ =\sqrt{(3 \sin \theta+4 \cos \theta)^{2}-3(3 \sin \theta+4 \cos \theta)+9} \text {. } $$ Therefore, when $3 \sin \theta+4 \cos \theta=-5$, $f(x, y)$ reaches its maximum value of 7." ba930dc569d5,"9.1. Find the smallest six-digit number that is a multiple of 11, where the sum of the first and fourth digits is equal to the sum of the second and fifth digits and is equal to the sum of the third and sixth digits.",. 100122,easy,"Solution. Let the desired number have the form $\overline{1000 x y}$, where $x, y$ are some digits. Then, by the condition, the sum of the first and fourth digits is 1, from which $x=y=1$. But the number 100011 is not divisible by 11. Therefore, we will look for a number of the form $\overline{1001 x y}$. Then $x=y=2$, and the number 100122 is divisible by 11. Answer. 100122." a7c81a3791f6,"Bootin D.A. Find in the sequence $2,6,12,20,30, \ldots$ the number standing a) at the 6th; b) at the 1994th position. Explain your answer.",a) 42; b) $1994 \cdot 1995=3978030$,medium,"$2=1 \cdot 2 ; 6=2 \cdot 3$ ## Solution We can notice that $2=1 \cdot 2, 6=2 \cdot 3, 12=3 \cdot 4$, and assume that the $n$-th term of the sequence is $n \cdot (n+1)$. Checking on the 4th ($20=4 \cdot 5$) and 5th ($30=5 \cdot 6$) terms of the sequence shows that we are correct. Therefore, the number at the 6th position is $6 \cdot 7=42$, and at the 1994th position, it is $1994 \cdot 1995=3978030$. Of course, this is not a strict mathematical proof. For example, this method could ""prove"" that the number sixty is divisible by all numbers. Indeed, 60 is divisible by 1, by 2, by 3, by 4, by 5, by 6, etc. However, for solving the problem, it is only necessary to find a sufficiently simple rule that can generate such a sequence. The ability to see and feel the pattern (which was required in this problem) is just as important for a mathematician as the ability to reason strictly! If you find any other (but also ""sufficiently simple"") rule that gives the sequence $2, 6, 12, 20, 30$, please write to us (and such a solution would also be accepted on the olympiad!). ## Answer a) 42; b) $1994 \cdot 1995=3978030$. ## Problem" 5284bb59ab6f,10th Brazil 1988,"0 for all n Solution f(30) = f(2) + f(3) + f(5) = 0 and f(n) is non-negative, so f(2) = f(3) = f(5) ",medium,"f(n) = 0 for all n Solution f(30) = f(2) + f(3) + f(5) = 0 and f(n) is non-negative, so f(2) = f(3) = f(5) = 0. For any positive integer n not divisible by 2 or 5 we can find a positive integer m such that mn = 7 mod 10 (because 1·7 = 3·9 = 7 mod 10, so if n = 1 mod 10 we take any m = 7 mod 10 and vice versa, whilst if n = 3 mod 10 we take any m = 9 mod 10 and vice versa). But then f(mn) = 0, so f(n) = 0. It is a trivial induction that f(2 a 5 b n) = f(5 b n) = f(n), so f is identically zero. Thanks to Suat Namli . 10th Brazil 1988 © John Scholes jscholes@kalva.demon.co.uk 5 February 2004 Last corrected/updated 5 Feb 04" b4960cb75693,"4. The sequence $\left\{x_{n}\right\}$ is defined as follows: $$ x_{1}=\frac{2}{3}, x_{n+1}=\frac{x_{n}}{2(2 n+1) x_{n}+1}\left(n \in \mathbf{Z}_{+}\right) \text {. } $$ Then $x_{1}+x_{2}+\cdots+x_{2014}=$ $\qquad$",1-\frac{1}{4029}=\frac{4028}{4029}$.,medium,"4. $\frac{4028}{4029}$. By definition, $\frac{1}{x_{n+1}}=2(2 n+1)+\frac{1}{x_{n}}$. Let $y_{n}=\frac{1}{x_{n}}$. Thus, $y_{1}=\frac{3}{2}$, and $y_{n+1}-y_{n}=2(2 n+1)$. Therefore, $y_{n}=\left(y_{n}-y_{n-1}\right)+\left(y_{n-1}-y_{n-2}\right)+\cdots+$ $\left(y_{2}-y_{1}\right)+y_{1}$ $=2[(2 n-1)+(2 n-3)+\cdots+3]+\frac{3}{2}$ $=2\left(n^{2}-1\right)+\frac{3}{2}=\frac{1}{2}(2 n+1)(2 n-1)$, $\sum_{i=1}^{2014} x_{i}=\sum_{i=1}^{2014} \frac{1}{y_{i}}=\sum_{i=1}^{2014} \frac{2}{(2 i+1)(2 i-1)}$ $=\sum_{i=1}^{2014}\left(\frac{1}{2 i-1}-\frac{1}{2 i+1}\right)$ $=1-\frac{1}{4029}=\frac{4028}{4029}$." 7eb12054cda6,2.259. $\left(\sqrt[3]{\frac{8 z^{3}+24 z^{2}+18 z}{2 z-3}}-\sqrt[3]{\frac{8 z^{2}-24 z^{2}+18 z}{2 z+3}}\right)-\left(\frac{1}{2} \sqrt[3]{\frac{2 z}{27}-\frac{1}{6 z}}\right)^{-1}$.,0,medium,"Solution. Domain of definition: $z \neq \pm \frac{3}{2}, z \neq 0$. $$ \begin{aligned} & \left(\sqrt[3]{\frac{8 z^{3}+24 z^{2}+18 z}{2 z-3}}-\sqrt[3]{\frac{8 z^{2}-24 z^{2}+18 z}{2 z+3}}\right)-\left(\frac{1}{2} \sqrt[3]{\frac{2 z}{27}-\frac{1}{6 z}}\right)^{-1}= \\ & =\sqrt[3]{\frac{2 z\left(4 z^{2}+12 z+9\right)}{2 z-3}}-\sqrt[3]{\frac{2 z\left(4 z^{2}-12 z+9\right)}{2 z+3}}-\left(\frac{1}{2} \sqrt[3]{\frac{4 z^{2}-9}{54 z}}\right)^{-1}= \\ & =\sqrt[3]{\frac{2 z(2 z+3)^{2}}{2 z-3}}-\sqrt[3]{\frac{2 z(2 z-3)^{2}}{2 z+3}}-2 \sqrt[3]{\frac{54 z}{4 z^{2}-9}}= \\ & =\frac{\sqrt[3]{2 z(2 z+3)^{2}}}{\sqrt[3]{2 z-3}}-\frac{\sqrt[3]{2 z(2 z-3)^{2}}}{\sqrt[3]{2 z+3}}-\frac{2 \sqrt[3]{54 z}}{\sqrt[3]{(2 z-3)(2 z+3)}}= \end{aligned} $$ $$ \begin{aligned} & =\frac{\sqrt[3]{2 z(2 z+3)^{3}}-\sqrt[3]{2 z(2 z-3)^{3}}-2 \sqrt[3]{54 z}}{\sqrt[3]{(2 z-3)(2 z+3)}}= \\ & =\frac{\sqrt[3]{2 z}(2 z+3-2 z+3-6)}{\sqrt[3]{4 z^{2}-9}}=\frac{\sqrt[3]{2 z} \cdot 0}{\sqrt[3]{4 z^{2}-9}}=0 \end{aligned} $$ Answer: 0." 930e34371940,"7. Given positive real numbers $a, b$ satisfy $a b(a+b)=4$, then the minimum value of $2 a+b$ is $\qquad$ Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.",See reasoning trace,easy,"Given $a b(a+b)=4 \Rightarrow a^{2}+a b=\frac{4}{b}$, then $(2 a+b)^{2}=4 a^{2}+4 a b+b^{2}=\frac{16}{b}+b^{2}$ $=\frac{8}{b}+\frac{8}{b}+b^{2} \geqslant 3 \sqrt[3]{64}=12$, equality holds when $b=2, a=\sqrt{3}-1$. Therefore, the minimum value of $2 a+b$ is $2 \sqrt{3}$." c505e3f97b42,"On the coordinate plane, let $C$ be a circle centered $P(0,\ 1)$ with radius 1. let $a$ be a real number $a$ satisfying $0100 a^{2}$, hence $\sqrt{N}>10 a$. By the condition, the number $\sqrt{N}$ is an integer. Therefore, $\sqrt{N} \geq 10 a+1$, which means $N \geq (10 a+1)^{2}=100 a^{2}+20 a+1$. On the other hand, $N<1700$, but the first two digits must form the number $a^{2}=16$. Thus, the largest possible value of $N$ is 1681. ## Answer 1681.00" 42041cb1cf9e,"8. The calculation result of $1.322 - \frac{a}{99}$, rounded to two decimal places using the round half up method, is 1.10. The natural number $a$ is . $\qquad$",See reasoning trace,easy,"Reference answer: 22 Exam point: Calculation -- approximate number" 832a404de873,"4.1. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 2021 less than the originally thought number. What is N?",See reasoning trace,easy,"Answer: 2022 Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a number that is $\mathrm{N}-1$ less than the original number." ef23c54293eb,". With inspiration drawn from the rectilinear network of streets in New York, the Manhattan distance between two points $(a, b)$ and $(c, d)$ in the plane is defined to be $$ |a-c|+|b-d| \text {. } $$ Suppose only two distinct Manhattan distances occur between all pairs of distinct points of some point set. What is the maximal number of points in such a set?",nine,medium,"Answer: nine. Let $$ \left\{\left(x_{1}, y_{1}\right), \ldots,\left(x_{m}, y_{m}\right)\right\}, \quad \text { where } \quad x_{1} \leq \cdots \leq x_{m} $$ be the set, and suppose $m \geq 10$. A special case of the Erdős-Szekeres Theorem asserts that a real sequence of length $n^{2}+1$ contains a monotonic subsequence of length $n+1$. (Proof: Given a sequence $a_{1} \ldots, a_{n^{2}+1}$, let $p_{i}$ denote the length of the longest increasing subsequence ending with $a_{i}$, and $q_{i}$ the length of the longest decreasing subsequence ending with $a_{i}$. If $i5$.,. $x \in[3 ; 35) \cup(120 ;+\infty)$,medium,"Answer. $x \in[3 ; 35) \cup(120 ;+\infty)$. Solution. Note that the second radicand can be written as $(\sqrt{x+1}-9)^{2}$. The inequality takes the form $\sqrt{\sqrt{x+1}-2}+|\sqrt{x+1}-9|>5$. Let $\sqrt{\sqrt{x+1}-2}=t$. Then we get $$ t+\left|t^{2}-7\right|>5 \Leftrightarrow\left[\begin{array} { l } { t ^ { 2 } - 7 > 5 - t } \\ { t ^ { 2 } - 7 0 , } \\ { t ^ { 2 } - t - 2 < 0 } \end{array} \Leftrightarrow \left[\begin{array}{l} t \in(-\infty ;-4) \cup(3 ;+\infty) \\ t \in(-1 ; 2) \end{array}\right.\right.\right. $$ Returning to the variable $x$, we get the system ![](https://cdn.mathpix.com/cropped/2024_05_06_6022a90abea870d9ae11g-01.jpg?height=290&width=1474&top_left_y=1594&top_left_x=291)" 77808e3f472d,"1. $a, b$ are positive integer divisors of 1995, then the number of ordered pairs $(a, b)$ is (). (A) 16 (B) 32 (C) 64 (D) 256","3 \cdot 5 \cdot 7 \cdot 19$, the number of positive divisors is 24 $=16$. For the pair $(a, b)$, $a$",easy,"1. D $1995=3 \cdot 5 \cdot 7 \cdot 19$, the number of positive divisors is 24 $=16$. For the pair $(a, b)$, $a$ has 16 possible values, and for each $a$, $b$ also has 16 possible values. Therefore, the number of pairs $(a, b)$ is $16 \times 16=256$ (pairs)." 39c2a7203df0,"Let $\mathbb{N}$ denote the set of all positive integers. An ordered pair $(a;b)$ of numbers $a,b\in\mathbb{N}$ is called [i]interesting[/i], if for any $n\in\mathbb{N}$ there exists $k\in\mathbb{N}$ such that the number $a^k+b$ is divisible by $2^n$. Find all [i]interesting[/i] ordered pairs of numbers.","(2^k l + 1, 2^k q - 1)",medium,"1. **Define the valuation function**: For any positive integer \( n \), let \( v(n) \) be the largest integer such that \( 2^{v(n)} \mid n \). 2. **Lemma**: If \( b \equiv 1 \pmod{2^{v(a-1)}} \), then there exists \( k \in \mathbb{N} \) such that \( a^k \equiv b \pmod{2^n} \) for any \( n \in \mathbb{N} \). 3. **Proof of Lemma**: - Let \( v(a-1) = p \), \( a = 2^p l + 1 \) where \( l \) is odd, and \( b = 2^p m + 1 \). - For \( n \le p \), the statement is obvious (just take \( k = 1 \)). - Suppose \( n > p \). Note that: \[ a^{2^{n-p}} - 1 = \sum_{i=1}^{2^{n-p}} (2^p l)^i \binom{2^{n-p}}{i} \] - Note that \( v((2^p l)^i \binom{2^{n-p}}{i}) \ge pi + n - p - v(i) > n \). So \( a^{2^{n-p}} \equiv 1 \pmod{2^n} \). - If \( a, a^2, \ldots, a^{2^{n-p}} \) have distinct values modulo \( 2^n \), then they take all values of the form \( 2^p m + 1 \), so \( a^k \equiv b \pmod{2^n} \) for some \( k \). - Otherwise, the order of \( a \) modulo \( 2^n \) is less than \( 2^{n-p} \), thus we must have \( a^{2^{n-p-1}} \equiv 1 \pmod{2^n} \). - But we have: \[ a^{2^{n-p-1}} - 1 = \sum_{i=1}^{2^{n-p-1}} (2^p l)^i \binom{2^{n-p-1}}{i} \] - For \( i \ge 2 \), we have \( v((2^p l)^i \binom{2^{n-p-1}}{i}) = pi + n - p - 1 - v(i) \ge n \), and for \( i = 1 \), we have \( v((2^p l)^i \binom{2^{n-p-1}}{i}) = v(2^p l 2^{n-p-1}) = n - 1 \). So \( 2^n \nmid a^{2^{n-p-1}} - 1 \), a contradiction. Our lemma is proved. 4. **Parity Consideration**: - Since \( a^k + b \) is even, then \( a \) and \( b \) have the same parity. - Suppose they are both even. For \( n > v(b) \), we have \( a^k \equiv -b \pmod{2^n} \). So we must have \( k v(a) = v(b) \), or \( k = \frac{v(a)}{v(b)} \). Thus \( a^{\frac{v(a)}{v(b)}} + b \) is divisible by \( 2^n \) for all \( n > v(b) \), which is impossible. Therefore \( a \) and \( b \) are both odd. 5. **Divisibility Consideration**: - For \( n = v(a^2 - 1) \), we have \( a^k \equiv -b \pmod{2^n} \). Note that either \( a^2 - 1 \mid a^k - 1 \) or \( a^2 - 1 \mid a^k - a \). Thus \( -b - 1 \) or \( -b - a \) is divisible by \( 2^{v(a^2 - 1)} \), equivalently, \( 2^{v(a^2 - 1)} \) divides \( b + 1 \) or \( a + b \). 6. **Case Analysis**: - **Case (i)**: \( a \equiv 1 \pmod{4} \) - Let \( a = 2^k l + 1 \) where \( k \ge 2 \) and \( l \) is odd. Then \( a^2 - 1 = 4^k l^2 + 2^{k+1} l \), so \( v(a^2 - 1) = k + 1 \). - So \( b = 2^{k+1} m - 1 \) or \( b = 2^{k+1} m - 2^k l - 1 = 2^k (2m - l) - 1 = 2^k p - 1 \). Either way, we have \( b = 2^k q - 1 \) for some positive integer \( q \). - From our lemma, we know that there exists \( x \in \mathbb{N} \) such that \( a^x \equiv -b \pmod{2^n} \), so \( (2^k l + 1, 2^k q - 1) \) is an interesting pair for any \( k \ge 2 \) and odd \( l \). - **Case (ii)**: \( a \equiv -1 \pmod{4} \) - Let \( a = 2^k l - 1 \) where \( k \ge 2 \) and \( l \) is odd. Then \( a^2 - 1 = 4^k l^2 - 2^{k+1} l \), so \( v(a^2 - 1) = k + 1 \). - Then \( b = 2^{k+1} m - 1 \) or \( b = 2^{k+1} m - 2^k l + 1 = 2^k q + 1 \) where \( q \) is odd. - **Subcase (ii.i)**: \( b = 2^{k+1} m - 1 \) - Note that \( v(a^2 - 1) = v(4^k l^2 - 2^{k+1} l) = k + 1 \). Since \( -b \equiv 1 \pmod{2^{k+1}} \), then it follows from our lemma that there exists \( x \) such that \( (a^2)^x \equiv -b \pmod{2^n} \). So \( (2^k l - 1, 2^{k+1} m - 1) \) is interesting for \( k \ge 2 \) and odd \( l \). - **Subcase (ii.ii)**: \( b = 2^k q + 1 \) - Consider a sufficiently large \( n \). It is easy to see that the inverse of \( a \) modulo \( 2^n \) is of the form \( a^* \equiv 2^k r - 1 \) for odd \( r \). Since \( -b a^* \equiv 1 \pmod{2^{k+1}} \), from our lemma we know that there exists \( x \) such that \( (a^2)^x \equiv -b a^* \pmod{2^n} \), which gives \( a^{2x+1} \equiv -b \pmod{2^n} \). Therefore \( (2^k l - 1, 2^k q + 1) \) is interesting for \( k \ge 2 \) and odd \( l, q \). The final answer is \( \boxed{ (2^k l + 1, 2^k q - 1) } \) and \((2^k l - 1, 2^k q + 1)\) for \( k \ge 2 \) and odd \( l, q \)." bccc96ffafbf,,See reasoning trace,medium,"Solution. If a natural number decomposes into $p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{n}^{a_{n}}$, where $p_{1}, p_{2}, \ldots p_{n}$ are different prime numbers, then the number of its natural divisors is equal to $\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{n}+1\right)$. Since $2015=1 \cdot 2015=5 \cdot 403=13 \cdot 155=31 \cdot 65=5 \cdot 13 \cdot 31$, it follows that the sought number has one of the forms $a=2^{2014}, b=2^{402} \cdot 3^{4}, c=2^{154} \cdot 3^{12}, d=2^{64} \cdot 3^{30}$ or $e=2^{30} \cdot 3^{12} \cdot 5^{4}$, because to obtain the smallest numbers, the largest exponents must correspond to the smallest prime numbers $4 p$ We have $a>e$ since from $2^{2014}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{1984}>3^{12} \cdot 5^{4}$ which is obviously true. We have $b>e$ since from $2^{402} \cdot 3^{4}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{372}>3^{8} \cdot 5^{4}$ which is obviously true. We have $c>e$ since from $2^{154} \cdot 3^{12}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{124}>5^{4}$ which is obviously true. We have $d>e$ since from $2^{64} \cdot 3^{30}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{34} \cdot 3^{18}>5^{4}$ which is obviously true. Therefore, the smallest natural number with exactly 2015 divisors is $2^{30} \cdot 3^{12} \cdot 5^{4} \ldots$" adb211e34308,"In an acute-angled triangle, two altitudes are equal to 3 and $2 \sqrt{2}$, and their point of intersection divides the third altitude in the ratio 5:1, counting from the vertex of the triangle. Find the area of the triangle.",6,medium,"Let $\alpha$ and $\beta$ be the acute angles of a triangle, from the vertices of which the given altitudes are drawn. Formulate a system of trigonometric equations in terms of $\alpha$ and $\beta$. ## Solution Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes of triangle $ABC$, and $H$ be their point of intersection; $A A_{1}=3, B B_{1}=2 \sqrt{2}$. Denote $$ \angle C A B=\alpha, \angle A B C=\beta, H C_{1}=x $$ Then $$ C C_{1}=6 x, \angle A H C_{1}=\beta, \angle B H C_{1}=\alpha . $$ From the right triangles $A A_{1} B$ and $B B_{1} A$, we find that $$ A B=\frac{3}{\sin \beta}=\frac{2 \sqrt{2}}{\sin \alpha} $$ Therefore, $\sin \beta=\frac{3 \sin \alpha}{2 \sqrt{2}}$. From the right triangles $B C_{1} C$ and $B C_{1} H$, we find that $$ B C_{1}=\frac{6 x}{\tan \beta}=x \tan \alpha . $$ Therefore $$ \tan \alpha \tan \beta=6 \Rightarrow \sin \alpha \sin \beta=6 \cos \alpha \cos \beta \Rightarrow \sin ^{2} \alpha \sin ^{2} \beta=36 \cos ^{2} \alpha \cos ^{2} \beta $$ Substitute into the last equation $$ \sin ^{2} \beta=\frac{9 \sin ^{2} \alpha}{8}, \cos ^{2} \beta=1-\sin ^{2} \beta=1-\frac{9 \sin ^{2} \alpha}{8}, \cos ^{2} \alpha=1-\sin ^{2} \alpha $$ We get the equation $$ 35 \sin ^{4} \alpha-68 \sin ^{2} \alpha+32=0 $$ from which we find that $\sin \alpha=\frac{2}{\sqrt{5}}$ ( $\alpha$ and $\beta$ are less than 90° ). Therefore, $$ \cos \alpha=\frac{1}{\sqrt{5}}, \sin \beta=\frac{3}{\sqrt{10}}, \cos \beta=\frac{1}{\sqrt{10}}, A B=\frac{3}{\sin \beta}=\sqrt{10} $$ From triangle $ABC$, we find by the sine rule: $$ A C=\frac{A B \sin \beta}{\sin (\alpha+\beta)}=3 \sqrt{2} $$ Therefore, $$ S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \alpha=\frac{1}{2} \cdot \sqrt{10} \cdot 3 \sqrt{2} \cdot \frac{2}{\sqrt{5}}=6 $$ ## Answer 6." 32a4bc6e79a9,"$\overline{H K}=\overline{A K}-\overline{A H}=\frac{1}{2}(\overline{A Q}+\overline{A N})-\frac{1}{2}(\overline{A M}+\overline{A P})=$ $=\frac{1}{2}\left(\frac{1}{2}(\overline{A E}+\overline{A D})+\frac{1}{2}(\overline{A B}+\overline{A C})\right)-$ $-\frac{1}{2}\left(\frac{1}{2} \overline{A B}+\frac{1}{2}(\overline{A C}+\overline{A D})\right)=\frac{1}{4} \overline{A E}$. Therefore, $H K=\frac{1}{4} A E=1.75$.",1,easy,"Answer: 1.75. ![](https://cdn.mathpix.com/cropped/2024_05_06_1015c2cf6918084885d1g-1.jpg?height=540&width=691&top_left_y=1712&top_left_x=1231)" e5c69208010a,"2. Given $\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta=\frac{3}{4}$, $\sin (\alpha-\beta) \neq \frac{1}{2}$. Then $\sin (\alpha+\beta)=$ $\qquad$ .",See reasoning trace,medium,"$$ \begin{array}{l} \text { 2. } \frac{1}{2} \text {. } \\ 0=\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta-\frac{3}{4} \\ =\frac{1-\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}- \\ \frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]-\frac{3}{4} \\ =\frac{1}{4}-\frac{1}{2}(\cos 2 \alpha-\cos 2 \beta)- \\ \frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ =\frac{1}{4}+\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)- \\ \frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ =\left[\frac{1}{2}-\sin (\alpha+\beta)\right]\left[\frac{1}{2}-\sin (\alpha-\beta)\right] \text {. } \\ \end{array} $$ Since $\sin (\alpha-\beta) \neq \frac{1}{2}$, then, $$ \sin (\alpha+\beta)=\frac{1}{2} . $$" 48d0e495d132,"377. The density function $f(x)$ of a random variable $X$, whose possible values are contained in the interval $(a, b)$, is given. Find the density function of the random variable $Y=3X$.","3 x$, then $3 a99,2-$ is not a two-digit number.",medium,"5. Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x=a+b+c$. According to the condition, the numbers $a, b, c$ and $x$ satisfy the equation $10^{4} a+100 b+c=x^{3}$ (see fact 11). Therefore, $$ x^{3}=10^{4} a+100 b+c44$; 2) $x=45(x-1=44), 45^{3}=91125, a=9, b=11, c=25$; 3) $x=54(x+1=55), 54^{3}=157464,15+74+64>54$; 4) $x=55(x-1=54), 55^{3}=166375,16+63+75>55$; 5) $x=89(x-1=88, x+1=90), 89^{3}=704969,70+49+$ $+69>89$ 6) $x=98(x+1=99), 98^{3}=941192,94+11+92>98$; 7) $x=99,99^{3}=970299,97+2+99>99,2-$ is not a two-digit number." 418d442338cc,"## Subject 1 a) The 20th term is 77 .................................................. $1 p$ $S=1+5+9+\ldots+77=780$ $.2 p$ b) The $n$-th term of the sequence is $t_{n}=4(n-1)+1=4 n-3, n \in \mathbf{N}^{*} \ldots . . .1 p$ $t_{2013}=8049$................................................................................... $1 p$ c) We check if the equation $4 n-3=2013$ has solutions in $\mathbf{N}$.......................1p $n=504$, so 2013 is a term of the sequence ......................................... $1 p$",See reasoning trace,easy,"## Subject 1 Consider the sequence: $1,5,9,13, \ldots$. a) Calculate the sum of the first 20 terms of the sequence; b) Determine the 2013th term of the sequence; c) Verify if 2013 is a term of the sequence. Prof. Ecaterina Botan, prof. Contanu Mihai" 34b8cfff279e,,See reasoning trace,medium," Solution. Let $R$ be the circumradius of $\triangle A B C$ and $\Delta$ its area. We have $O D=R \cos A$ and $D C=\frac{a}{2}$, so $$ [O D C]=\frac{1}{2} \cdot O D \cdot D C=\frac{1}{2} \cdot R \cos A \cdot R \sin A=\frac{1}{2} R^{2} \sin A \cos A $$ Again $H E=2 R \cos C \cos A$ and $E A=c \cos A$. Hence $$ [H E A]=\frac{1}{2} \cdot H E \cdot E A=\frac{1}{2} \cdot 2 R \cos C \cos A \cdot c \cos A=2 R^{2} \sin C \cos C \cos ^{2} A $$ Further $$ [G F B]=\frac{\Delta}{6}=\frac{1}{6} \cdot 2 R^{2} \sin A \sin B \sin C=\frac{1}{3} R^{2} \sin A \sin B \sin C $$ Equating (1) and (2) we get $\tan A=4 \sin C \cos C$. And equating (1) and (3), and using this relation we get $$ \begin{aligned} 3 \cos A & =2 \sin B \sin C=2 \sin (C+A) \sin C \\ & =2(\sin C+\cos C \tan A) \sin C \cos A \\ & =2 \sin ^{2} C\left(1+4 \cos ^{2} C\right) \cos A \end{aligned} $$ Since $\cos A \neq 0$ we get $3=2 t(-4 t+5)$ where $t=\sin ^{2} C$. This implies $(4 t-3)(2 t-1)=0$ and therefore, since $\sin C>0$, we get $\sin C=\sqrt{3} / 2$ or $\sin C=1 / \sqrt{2}$. Because $\triangle A B C$ is acute, it follows that $\widehat{C}=\pi / 3$ or $\pi / 4$. We observe that the given conditions are satisfied in an equilateral triangle, so $\widehat{C}=\pi / 3$ is a possibility. Also, the conditions are satisfied in a triangle where $\widehat{C}=\pi / 4, \widehat{A}=\tan ^{-1} 2$ and $\widehat{B}=\tan ^{-1} 3$. Therefore $\widehat{C}=\pi / 4$ is also a possibility. Thus the two possible values of $\widehat{C}$ are $\pi / 3$ and $\pi / 4$. " 64e2e2c1234d,"10.083. The sum of the lengths of the diagonals of a rhombus is $m$, and its area is $S$. Find the side of the rhombus.",See reasoning trace,medium,"Solution. $A C+B D=m ; A C=x, B D=y . S=\frac{1}{2} d_{1} d_{2}$ (Fig. 10.081). Solving the system $\left\{\begin{array}{l}x+y=m, \\ 2 S=x y,\end{array}\right.$ we get $x_{1}=\frac{m+\sqrt{m^{2}-8 S}}{2}, y_{1}=\frac{m-\sqrt{m^{2}-8 S}}{2}$ or $x_{2}=\frac{m-\sqrt{m^{2}-8 S}}{2}, y_{2}=\frac{m+\sqrt{m^{2}-8 S}}{2}$. From this: $B C^{2}=\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{2}\right)^{2}=\left(\frac{m+\sqrt{m^{2}-8 S}}{2 \cdot 2}\right)^{2}+\left(\frac{m-\sqrt{m^{2}-8 S}}{2 \cdot 2}\right)^{2}=\frac{m^{2}-4 S}{4}$. $B C=\frac{\sqrt{m^{2}-4 S}}{2}$. Answer: $\quad \frac{\sqrt{m^{2}-4 S}}{2}$ ![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-596.jpg?height=402&width=500&top_left_y=91&top_left_x=110) Fig. 10.81 ![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-596.jpg?height=402&width=564&top_left_y=91&top_left_x=688) Fig. 10.82" d08655920d92,5th Chinese 1990,2u. We can rearrange the second equation slightly as: u(a 4 + b 4 + c 4 ) + v(a 2 b 2 + b 2 c 2 + c ,medium,": there is a solution iff v = 2u. We can rearrange the second equation slightly as: u(a 4 + b 4 + c 4 ) + v(a 2 b 2 + b 2 c 2 + c 2 a 2 ) = (2u + v)/4 (13n) 4 . But a 2 b 2 + b 2 c 2 + c 2 a 2 = (a 2 + b 2 + c 2 ) 2 /2 - (a 4 + b 4 + c 4 )/2, so the equation becomes: (u - v/2) (a 4 + b 4 + c 4 ) + v(a 2 + b 2 + c 2 ) 2 /2 = (2u + v)/4 (13n) 4 . Using the other equation this gives: (2u - v)(a 4 + b 4 + c 4 )/2 = (2u - v)/4 (13n) 4 . So if 2u - v is non-zero we can derive the integer equation: 2(a 4 + b 4 + c 4 ) = (13n) 4 . It is unlikely that this has solutions. Indeed, we can use the standard descent argument. If it has a solution in positive integers, then take the solution with smallest n. The lhs is even, so n must be even. Hence a 4 + b 4 + c 4 = 0 mod 8. But fourth powers are all 0 or 1 mod 8, so a, b, and c must all be even. But that means a/2, b/2, c/2, n/2 is a solution with smaller n. Contradiction. On the other hand, if 2u - v is zero, then the second equation adds nothing. Indeed, dividing by u it becomes simply (a 2 + b 2 + c 2 ) 2 = (13n) 4 , which is just the square of the first equation. We can easily satisfy the first equation by taking a = 12n, b = 4n, c = 3n. 5th Chinese 1990 © John Scholes jscholes@kalva.demon.co.uk 1 May 2002" 94fe36f7dfba,"2.40 Form a second-degree equation, one of whose roots is equal to the sum, and the other is equal to the product of the roots of the equation $a x^{2}+b x+c=0$.",See reasoning trace,medium,"2.40 If the roots of the desired equation are denoted by $\bar{x}_{1}$ and $\bar{x}_{2}$, and those of the given equation by $x_{1}$ and $x_{2}$, then from the condition it follows that $\bar{x}_{1}=x_{1}+x_{2} ; \bar{x}_{2}=x_{1} x_{2}$. Using Vieta's theorem, we get $\bar{x}_{1}=-\frac{b}{a} ; \bar{x}_{2}=\frac{c}{a}$. Hence, $\bar{x}_{1}+\overline{x_{2}}=\frac{c-b}{a} ; \bar{x}_{1} \bar{x}_{2}=-\frac{b c}{a^{2}}$. Therefore, the desired equation will take the form $$ x^{2}+\frac{b-c}{a} x-\frac{b c}{a^{2}}=0, \text { or } a^{2} x^{2}+(a b-a c) x-b c=0 $$" 1b95fb642353,"8. Let $\mathcal{R}$ be the finite region in the plane bounded by the $x$-axis and the graph of the curve given by the equation $2 x^{2}+5 y=10$. Given three points in $\mathcal{R}$, which of the following statements is always true? (A) at least two of the three points have a distance $\geq \frac{\sqrt{5}}{2}$ (B) at least two of the three points have a distance $\leq 3$ (C) the sum of the distances between the points is $\leq \frac{9}{2} \sqrt{5}$ (D) the sum of the squares of the distances between the points is $\leq 38$ (E) the product of the distances between the points is $\leq 16 \sqrt{5}$",(B),medium,"8. The answer is (B). It is a segment of a parabola with the vertex $V=(0,2)$ and intersections with the axes $A=(-\sqrt{5}, 0)$ and $B=(\sqrt{5}, 0)$. Two of the three points must be in the same quadrant, and therefore their distance cannot exceed the length of the segment $A V$, which is 3 by the Pythagorean theorem (indeed, the part of the region in the first quadrant is entirely contained within the circle with diameter $B V$, and the part in the second quadrant is entirely contained within the circle with diameter $A V$). Note that all other answers are false: (A) at least two of the three points have a distance $\geq \frac{\sqrt{5}}{2}$: obviously the points can be taken as close as desired! (C) the sum of the distances between the points is $\leq \frac{9}{2} \sqrt{5}$: just take $A, B$, and $V$ to get $2 \sqrt{5}+6>\frac{9}{2} \sqrt{5}$. (D) the sum of the squares of the distances between the points is $\leq 38$: taking $A, B, B$ the sum of the squares of the distances is equal to 40 (if one seeks 3 distinct points with a sum of the squares of the distances greater than 38, just take $A, B, C$ with $C$ very close to $B$). (E) the product of the distances between the points is $\leq 16 \sqrt{5}$: taking $A, B$, and $V$ again gives $18 \sqrt{5}>16 \sqrt{5}$." ab48617afb6b,"10. (10 points) In 3 pencil cases, there are a total of 15 pencils and 14 pens. Each pencil case must contain at least 4 pencils and 2 pens. If the number of pencils in each pencil case is not less than the number of pens, then the pencil case with the most pens and pencils combined can have at most ( ) pens and pencils. A. 12 B. 14 C. 15 D. 16","】: Solution: Pencils are only 1 more than pens, so in each bag, either pencils $=$ pens, or pencils $=$ pens +1",medium,"【Answer】: Solution: Pencils are only 1 more than pens, so in each bag, either pencils $=$ pens, or pencils $=$ pens +1. The bag with the most should have the most, other bags should be as few as possible, the least can only be 4 pencils and 3 pens, the remaining: $15-4-4=7$ (pencils), $14-3-3=8$ (pens), which does not meet the requirement, so we take away one pen, the number of pens and pencils in the bags is as shown in the table below: \begin{tabular}{|l|l|l|l|} \hline & 4 & 4 & 7 \\ \hline Pencils & & & \\ \hline Pens & 3 & 4 & 7 \\ \hline \end{tabular} Obviously, the bag with the most pens and pencils has: $7+7=14$ pens and pencils. Therefore, the answer is: $B$." d304e1c7d625,"Determine all positive integers $n>1$ such that for any divisor $d$ of $n,$ the numbers $d^2-d+1$ and $d^2+d+1$ are prime. [i]Lucian Petrescu[/i]","n = 2, 3, 6",medium,"To determine all positive integers \( n > 1 \) such that for any divisor \( d \) of \( n \), the numbers \( d^2 - d + 1 \) and \( d^2 + d + 1 \) are prime, we proceed as follows: 1. **Consider the divisors of \( n \):** Let \( d \) be a divisor of \( n \). We need to check the primality of \( d^2 - d + 1 \) and \( d^2 + d + 1 \). 2. **Analyze the congruence conditions:** - If \( d \equiv 1 \pmod{3} \) and \( d > 1 \), then: \[ d^2 + d + 1 \equiv 1^2 + 1 + 1 \equiv 3 \equiv 0 \pmod{3} \] This implies \( d^2 + d + 1 \) is divisible by 3. Since \( d > 1 \), \( d^2 + d + 1 > 3 \), so \( d^2 + d + 1 \) cannot be prime. - If \( d \equiv 2 \pmod{3} \) and \( d > 2 \), then: \[ d^2 - d + 1 \equiv 2^2 - 2 + 1 \equiv 4 - 2 + 1 \equiv 3 \equiv 0 \pmod{3} \] This implies \( d^2 - d + 1 \) is divisible by 3. Since \( d > 2 \), \( d^2 - d + 1 > 3 \), so \( d^2 - d + 1 \) cannot be prime. 3. **Determine possible values of \( n \):** From the above analysis, \( d \) must be such that \( d \equiv 0 \pmod{3} \) or \( d = 1 \) or \( d = 2 \). - If \( n = 3^k \), then \( d = 3^i \) for \( i \leq k \). For \( d = 3 \): \[ 3^2 + 3 + 1 = 9 + 3 + 1 = 13 \quad (\text{prime}) \] \[ 3^2 - 3 + 1 = 9 - 3 + 1 = 7 \quad (\text{prime}) \] However, for \( k \geq 2 \), \( 9^2 + 9 + 1 = 91 = 7 \times 13 \) is not prime. Thus, \( k < 2 \). - If \( n = 2 \cdot 3^k \), then \( d = 1, 2, 3 \). For \( d = 2 \): \[ 2^2 + 2 + 1 = 4 + 2 + 1 = 7 \quad (\text{prime}) \] \[ 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \quad (\text{prime}) \] For \( d = 3 \), as shown above, both expressions are prime. However, for \( k \geq 2 \), the same issue with \( 9 \) arises. 4. **Conclusion:** The possible values of \( n \) are \( 2, 3, \) and \( 6 \). The final answer is \( \boxed{ n = 2, 3, 6 } \)." bfa095dc53e7,"Task B-2.3. (20 points) In a right triangle $A B C$, point $D$ is the foot of the altitude dropped from vertex $C$ to the hypotenuse $\overline{A B}$, point $E$ is the midpoint of segment $\overline{C D}$, and point $F$ is the intersection of lines $A E$ and $B C$. If $|A D|=4$ and $|B D|=9$, determine the length of segment $\overline{A F}$.","\frac{|AB|}{|DB|}$, i.e., $\frac{5 + x}{2x} = \frac{13}{9}$, from which we find $x = \frac{45}{17}$.",medium,"Solution. From the equality $|CD|^2 = |AD| \cdot |BD|$ for the right triangle $ABC$, we get $|CD| = 6$. Since $E$ is the midpoint of segment $\overline{CD}$, we have $|DE| = 3$. From the right triangle $AED$, we have $|AE| = 5$. ![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-11.jpg?height=409&width=524&top_left_y=1107&top_left_x=651) Let $G$ be a point on side $\overline{BC}$ such that $DG \parallel AF$. Since $\overline{EF}$ is the midline of triangle $CDG$, with $|EF| = x$, we have $|DG| = 2x$. From the similarity of triangles $AFB$ and $DGB$, we get the equality $\frac{|AF|}{|DG|} = \frac{|AB|}{|DB|}$, i.e., $\frac{5 + x}{2x} = \frac{13}{9}$, from which we find $x = \frac{45}{17}$. Finally, $|AF| = \frac{130}{17}$." c2a1cf64b8f7,2. The chords $AB$ and $CD$ of a circle intersect at a right angle. The lengths of the chords $AD=60$ and $BC=25$ are known. Calculate the radius of the circle.,"\varphi$ and $\measuredangle DCA = \psi$ (note that $\varphi + \psi = 90^{\circ}$, according to the ",medium,"2. Let $O$ be the center of the circle, and $E$ a point on the circle such that $AE$ is a diameter. Denote $\measuredangle BAC = \varphi$ and $\measuredangle DCA = \psi$ (note that $\varphi + \psi = 90^{\circ}$, according to the problem statement). Then we have $\measuredangle AOD = 2\psi$ and thus $\measuredangle EAD = \measuredangle OAD = \frac{180^{\circ} - 2\psi}{2} = 90^{\circ} - \psi = \varphi$. Therefore, since the peripheral angles over the chords $BC$ and $ED$ are both equal to $\varphi$, it follows that $ED = BC = 25$. Now, using the Pythagorean theorem, we have $(2r)^2 = AE^2 = AD^2 + ED^2 = 60^2 + 25^2 = 3600 + 625 = 4225$, so we get $r = \frac{\sqrt{4225}}{2} = \frac{65}{2}$." 5fe121336240,"3. A pentagram has five vertices. Draw straight lines through these five vertices so that each line must pass through two of them. Then, the total number of lines that can be drawn is ( ). (A) 5 (B) 10 (C) 15 (D) 20",See reasoning trace,easy,"3. B Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 6778294a4021,"a) Two heavy bodies start moving on two inclined planes without initial velocity. The inclined planes form angles $\alpha$ and $\beta$ with the horizontal plane, respectively. One body falls from point $A$, the other from point $B$. $O A=a, O B=b$. After how many seconds will the line $M N$, which connects the two moving bodies, be horizontal? Under what conditions is this ![](https://cdn.mathpix.com/cropped/2024_05_02_cedbfe4ef8ea5fcedd1ag-1.jpg?height=262&width=409&top_left_y=281&top_left_x=847) b) If $\alpha$ and $\beta$ are not given, but we know the angle $\varphi$ between the two planes, calculate the angles $\alpha$ and $\beta$, under the condition that the heavy bodies starting from points $A$ and $B$ without initial velocity arrive at $O$ simultaneously.",See reasoning trace,medium,"I.) The line $M N$ is horizontal if $$ N O \sin \beta = M O \sin \alpha $$ If $t$ denotes the time it takes for the bodies to reach $M$ and $N$, respectively, then: $$ \left(b - \frac{1}{2} g t^{2} \sin \beta\right) \sin \beta = \left(a - \frac{1}{2} g t^{2} \sin \alpha\right) \sin \alpha $$ From this equation: $$ t = \sqrt{\frac{2(a \sin \alpha - b \sin \beta)}{g(\sin^2 \alpha - \sin^2 \beta)}} $$ Furthermore, $$ O M = \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \sin \beta \text{ and } O N = \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \sin \alpha $$ The problem is only possible if $t$ has a real value and $$ O M \geq 0, O N \geq 0 $$ These conditions are satisfied if: $$ \frac{a \sin \alpha - b \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \geq 0, \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \geq 0 $$ First, assume that $\alpha > \beta$, then it must be: $$ a \sin \alpha - b \sin \beta \geq 0, \text{ and } b \sin \alpha - a \sin \beta \geq 0 $$ From this, $$ \frac{\sin \alpha}{\sin \beta} \geq \frac{a}{b} \geq \frac{\sin \beta}{\sin \alpha} $$ If $\frac{a}{b} = \frac{\sin \alpha}{\sin \beta}$, then $O M = O N = 0$, meaning the two moving points arrive at $O$ simultaneously; if $\frac{a}{b} = \frac{\sin \beta}{\sin \alpha}$, then $t = 0$, meaning $A B$ is a horizontal line. If $\alpha = \beta$, then $a = b$ and $t = \frac{0}{0}$, which means $M N$ is horizontal throughout the entire motion. II.) Since in this case $$ a = \frac{1}{2} g t^{2} \sin \alpha \text{ and } b = \frac{1}{2} g t^{2} \sin \beta $$ we get: $$ \frac{1}{2} g t^{2} = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} $$ From this, $$ \frac{a + b}{a - b} = \frac{\tan \frac{\alpha + \beta}{2}}{\tan \frac{\alpha - \beta}{2}} $$ From this equation, $$ \tan \frac{\alpha - \beta}{2} = \frac{a - b}{a + b} \tan \frac{\alpha + \beta}{2} $$ But $\alpha + \beta + \gamma = 180^\circ$ so $$ \tan \frac{\alpha + \beta}{2} = \cot \frac{\gamma}{2} $$ Thus, $$ \tan \frac{\alpha - \beta}{2} = \frac{a - b}{a + b} \cot \frac{\gamma}{2} $$ From this, we can calculate $\alpha - \beta$; we know $\alpha + \beta$, so we can determine $\alpha$ and $\beta$. The problem was solved by: Friedmann Bernát, Goldstein Zsigmond, Goldziher Károly, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes, Szabó István." 49b61417b224,![](https://cdn.mathpix.com/cropped/2024_05_06_1ea0b100610baa73554bg-10.jpg?height=412&width=579&top_left_y=309&top_left_x=437),9,medium,"Answer: 9.5. Solution. Extend rays $AB$ and $DC$ until they intersect at point $X$ (see figure). ![](https://cdn.mathpix.com/cropped/2024_05_06_1ea0b100610baa73554bg-10.jpg?height=505&width=505&top_left_y=920&top_left_x=466) Figure for the solution to problem 8.7.1 In the isosceles triangle $HCD$, we have $\angle CHD = \angle CDH$. In the right triangle $XHD$, we have $\angle HXD = 90^{\circ} - \angle XDH = 90^{\circ} - \angle CHD = \angle XHC$. Therefore, triangle $XHC$ is isosceles, and $CX = CH = CD$. Then, in triangle $AXD$, segment $BC$ is the midline (since $BC \| AD$, and point $C$ is the midpoint of segment $XD$), so $BC = \frac{1}{2} AD = 9.5$." 2a85d00259f9,"Kanel-Belov A.Y. A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In some cube, the number 10 is written. Through this cube, three layers $1 \times 20 \times 20$, parallel to the faces of the cube, pass. Find the sum of all numbers outside these layers.",333,medium,"Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with G (there are 361 such cubes). These cubes are completely covered by 19 columns lying in G. The sum of all numbers in these columns (which is 19) exceeds $S$ by the sum of 19 numbers lying in a perpendicular column containing K. The latter sum is clearly $1-10=-9$. Therefore, $S=19-(-9)=28$. Finally, we have: $361-28=333$. ## Answer 333." fabb4ccf0979,"4. For the geometric sequence $a_{1}, a_{2}, a_{3}, a_{4}$, it satisfies $a_{1} \in(0,1), a_{2} \in(1,2), a_{3} \in(2,3)$, then the range of $a_{4}$ is $\qquad$",See reasoning trace,easy,"4. $(2 \sqrt{2}, 9)$ Analysis: According to the problem, we have: $\left\{\begin{array}{c}0 1 \). This is because \( d \) should increase more than \( v_2\left(\varphi\left(\frac{n}{d}\right)\right) \) decreases. Consider the bound: \[ v_2\left(\varphi\left(\frac{n}{d}\right)\right) + v_2(\varphi(d)) = v_2\left(\varphi\left(\frac{n}{d}\right) \varphi(d)\right) \geq v_2(\varphi(n)) - 1 \] This follows by observing individual prime powers dividing \( \frac{n}{d} \) and \( d \). Then we have: \[ v_2\left(\varphi\left(\frac{n}{d}\right)\right) + d \geq v_2(\varphi(n)) + 1 + d - v_2(\varphi(d)) - 2 \geq v_2(\varphi(n)) + 1 \] because \( d - v_2(\varphi(d)) - 2 = 0 \) for \( d = 2, 3 \) and \( d \geq \log_2(d) + 2 \geq v_2(\varphi(d)) + 2 \) for \( d \geq 4 \). 5. **Implication of the claim:** This claim implies: \[ v_2(S) \geq v_2(\varphi(n)) + 1 \geq v_2(n) \] This is because \( \varphi \) can only knock off one 2 from \( n \), but every odd prime dividing \( n \) contributes to \( v_2(\varphi(n)) \). In particular, if the inequality is strict, then the number of orbits \( \frac{S}{n} \) is even. Thus, we must have equality in both inequalities, i.e., \( n \) is a power of 2 and \( n \in \{2, 3, 4\} \). 6. **Manual verification:** Manually checking, \( B_2 \) has 3 orbits, and \( n = 2 \) is the only \( n \) that works. The final answer is \( \boxed{ n = 2 } \)." 6d9dbfa8a46b,"3. Boy Zhenya lives in a building with a total of 100 floors. The elevator takes 1 second to travel between adjacent floors. Unfortunately, only two buttons in the elevator work: “+13 floors” and “-7 floors”. The buttons are pressed instantly. How many seconds will it take Zhenya to get home from the 1st floor to the 24th floor? (The elevator can only travel to existing floors)",See reasoning trace,easy,"Answer: 107 Examples of writing answers: 45" 039855a46a1c,"Solve the following system of equations: $$ \begin{aligned} & 1-\frac{12}{3 x+y}=\frac{2}{\sqrt{x}} \\ & 1+\frac{12}{3 x+y}=\frac{6}{\sqrt{y}} \end{aligned} $$","4+2 \sqrt{3} \approx 7.464, y=12+6 \sqrt{3} \approx 22.392$.",medium,"Solution. Assuming $x, y>0$, by adding the two equations and dividing both sides of the obtained equation by 2, we get: $$ 1=\frac{3}{\sqrt{x}}+\frac{1}{\sqrt{y}} $$ By subtracting the first equation from the second and dividing both sides by 2: $$ \frac{12}{3 x+y}=\frac{3}{\sqrt{y}}-\frac{1}{\sqrt{x}} $$ Multiplying equations (1) and (2): $$ \frac{12}{3 x+y}=\frac{9}{y}-\frac{1}{x}=\frac{9 x-y}{x y} $$ After bringing to a common denominator: $$ 12 x y=(9 x-y)(3 x+y)=27 x^{2}+6 x y-y^{2} $$ Factoring the quadratic equation: $(3 x-y) \cdot(9 x+y)=0$. Since $x, y>0$, only $3 x-y=0$ is possible, so $x=\frac{y}{3}$. Substituting this into (1): $$ 1=\frac{\sqrt{3}}{\sqrt{y}}+\frac{3}{\sqrt{y}} $$ From this, $\sqrt{y}=3+\sqrt{3} \Rightarrow y=12+6 \sqrt{3}$ and $x=4+2 \sqrt{3}$. Therefore, the solution to the system of equations is: $x=4+2 \sqrt{3} \approx 7.464, y=12+6 \sqrt{3} \approx 22.392$." 02ec4d8883d5,"Find all functions $f\colon \mathbb{R}\to\mathbb{R}$ that satisfy $f(x+y)-f(x-y)=2y(3x^2+y^2)$ for all $x,y{\in}R$ ______________________________________ Azerbaijan Land of the Fire :lol:",f(x) = x^3 + a,medium,"1. Let \( P(x, y) \) be the assertion \( f(x+y) - f(x-y) = 2y(3x^2 + y^2) \). 2. First, we test the function \( f(x) = x^3 + a \) to see if it satisfies the given functional equation. Assume \( f(x) = x^3 + a \). 3. Substitute \( f(x) = x^3 + a \) into the functional equation: \[ f(x+y) = (x+y)^3 + a \] \[ f(x-y) = (x-y)^3 + a \] 4. Calculate \( f(x+y) - f(x-y) \): \[ f(x+y) - f(x-y) = (x+y)^3 + a - ((x-y)^3 + a) \] \[ = (x+y)^3 - (x-y)^3 \] 5. Expand the cubes: \[ (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \] \[ (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 \] 6. Subtract the expanded forms: \[ (x+y)^3 - (x-y)^3 = (x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) \] \[ = x^3 + 3x^2y + 3xy^2 + y^3 - x^3 + 3x^2y - 3xy^2 + y^3 \] \[ = 6x^2y + 2y^3 \] 7. Compare this with the right-hand side of the original equation: \[ 2y(3x^2 + y^2) \] \[ = 6x^2y + 2y^3 \] 8. Since both sides match, \( f(x) = x^3 + a \) is indeed a solution. 9. To show that this is the only solution, consider the general form of the functional equation. Suppose \( f \) is any function satisfying the given equation. We need to show that \( f(x) \) must be of the form \( x^3 + a \). 10. Let \( x = 0 \) in the original equation: \[ f(y) - f(-y) = 2y(y^2) \] \[ f(y) - f(-y) = 2y^3 \] 11. This implies that \( f(y) = f(-y) + 2y^3 \). Let \( f(y) = g(y) + y^3 \), where \( g(y) \) is an even function (i.e., \( g(y) = g(-y) \)). 12. Substitute \( f(x) = g(x) + x^3 \) into the original equation: \[ g(x+y) + (x+y)^3 - g(x-y) - (x-y)^3 = 2y(3x^2 + y^2) \] \[ g(x+y) - g(x-y) + (x+y)^3 - (x-y)^3 = 2y(3x^2 + y^2) \] 13. From the previous steps, we know that \( (x+y)^3 - (x-y)^3 = 6x^2y + 2y^3 \). Therefore: \[ g(x+y) - g(x-y) + 6x^2y + 2y^3 = 2y(3x^2 + y^2) \] \[ g(x+y) - g(x-y) = 0 \] 14. Since \( g(x+y) = g(x-y) \) for all \( x \) and \( y \), and \( g \) is an even function, it must be constant. Let \( g(x) = c \). 15. Therefore, \( f(x) = x^3 + c \). 16. Let \( c = a \), where \( a \) is a constant. Thus, the general solution is \( f(x) = x^3 + a \). \(\blacksquare\) The final answer is \( \boxed{ f(x) = x^3 + a } \) where \( a \in \mathbb{R} \)." ece74075f537,"8,9 | | A circle with its center on the diagonal $A C$ of parallelogram $A B C D$ touches the line $A B$ and passes through points $C$ and $D$. Find the sides of the parallelogram if its area $S=\sqrt{2}$, and $\angle B A C=\arcsin \frac{1}{3}$.","$\sqrt{2}, \sqrt{3}$",medium,"Let $O$ be the center of the circle, $F$ the point of tangency of the circle with the line $AB$. Denote $OC = OD = OF = R$, $AB = CD = x$, $AD = BC = y$, $\angle ACD = \angle BAC = \alpha$. Then $$ \sin \alpha = \frac{1}{3}, \cos \alpha = \frac{2 \sqrt{2}}{3} $$ Extend the radius $OF$ to intersect the chord $CD$ at point $M$. Since $OF \perp AB$ and $CD \parallel AB$, then $OM \perp CD$, so $M$ is the midpoint of $CD$. From the right triangle $OMC$, we find that $$ R = OC = \frac{MC}{\cos \alpha} = \frac{x}{2 \cos \alpha} = \frac{3x}{4 \sqrt{2}}, OM = OC \sin \alpha = \frac{1}{3} R = \frac{x}{4 \sqrt{2}} $$ Then $$ FM = OF + OM = R + \frac{1}{3} R = \frac{4}{3} R = \frac{4}{3} \cdot \frac{3x}{4 \sqrt{2}} = \frac{x}{\sqrt{2}}, $$ and since $MF$ is the height of the parallelogram $ABCD$, then $S = CD \cdot MF$, or $\sqrt{2} = x \cdot \sqrt{\sqrt{2}}$, from which $AB = x = \sqrt{2}$. Let $P$ be the projection of vertex $C$ onto the line $AB$. Then $CP = MF = \sqrt{2}$. From the right triangle $APC$, we find that $$ AC = \frac{CP}{\sin \angle PAC} = \frac{CP}{\sin \alpha} = \frac{\sqrt{2}}{\frac{1}{3}} = \frac{3x}{\sqrt{2}} = 3 $$ By the cosine rule $$ y^2 = BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos \angle BAC = 2 + 9 - 2 \cdot \sqrt{2} \cdot 3 \cdot \frac{2 \sqrt{2}}{3} = 11 - 8 = 3 $$ Therefore, $BC = \sqrt{3}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-17.jpg?height=1146&width=1835&top_left_y=170&top_left_x=154) ## Answer $\sqrt{2}, \sqrt{3}$." 0221d5502524,"$11 \cdot 22$ Find 8 positive integers $n_{1}, n_{2}, \cdots, n_{8}$, such that they have the following property: For each integer $k, -1985 \leqslant k \leqslant 1985$, there are 8 integers $a_{1}, a_{2}, \cdots, a_{8}$, each $a_{i}$ belonging to the set $\{-1,0,1\}$, such that $k=\sum_{i=1}^{8} a_{i} n_{i}$.","3^{i-1}, \quad i=1,2, \cdots, 8$.",medium,"［Solution］Consider the ternary representation of 1985 $$ 1985=2 \cdot 3^{6}+2 \cdot 3^{5}+3^{3}+3^{2}+3+2 \text {. } $$ However, by the problem's condition, the coefficients in the representation cannot be 2, and since 2=3-1, then 1985 can be expressed as $$ 1985=3^{7}-3^{5}+3^{4}-3^{3}-3^{2}-3-1 $$ Similarly, we have $\quad 1984=3^{7}-3^{5}+3^{3}+3^{2}+3+1$. Now starting from 1984, subtract 1 successively, two cases may arise: One is like 1984, where the last few terms are positive, and can be reduced by 1, without increasing the number of terms; the other is like 1985, where the last few terms are negative, and subtracting 1 will result in $-2 \cdot 3^{i} \quad(i=0,1,2,3)$, and since $$ -2 \cdot 3^{i}=-3^{i+1}+3^{i} $$ replacing $-2 \cdot 3^{i}$ with $-3^{i+1}+3^{i}$, although it adds one term, since the expression for 1985 has only 7 terms, adding one term will not exceed 8 terms, and when $i+1=4$ or 7, it will exactly cancel out with the positive terms in front, thus keeping the number of terms unchanged. Therefore, for $k \in[0,1985]$, all numbers can be represented as $$ k=\sum_{i=1}^{8} a_{i} 3^{i-1}, \quad a_{i} \in\{-1,0,1\} \text { . } $$ If the $a_{i}$ in the formula are replaced with their opposites, then for $k \in[-1985,0]$, the above formula can also be used. Thus, the required $n_{i}=3^{i-1}, \quad i=1,2, \cdots, 8$." e67c8bb8a280,"$4 \cdot 103$ real numbers $\alpha, \beta$ satisfy the system of equations $$ \left\{\begin{array}{l} \alpha^{3}-3 \alpha^{2}+5 \alpha-17=0 \\ \beta^{3}-3 \beta^{2}+5 \beta+11=0 . \end{array}\right. $$ Find $\alpha+\beta$.",2$.,medium,"[Solution] The system of equations is transformed into $$ \begin{array}{l} (\alpha-1)^{3}+2(\alpha-1)-14=0, \\ (\beta-1)^{3}+2(\beta-1)+14=0 . \end{array} $$ Adding the two equations, we get $(\alpha-1)^{3}+(\beta-1)^{3}+2(\alpha+\beta-2)=0$, which simplifies to $(\alpha+\beta-2)\left[(\alpha-1)^{2}+(\beta-1)^{2}-(\alpha-1)(\beta-1)+2\right]=0$. However, when $x, y$ are both real numbers, the equation $$ x^{2}+y^{2}-x y+2=0 $$ does not hold, otherwise $x=\frac{y \pm \sqrt{-3 y^{2}-8}}{2}$ would contradict the fact that $x$ is real. Therefore, $\alpha+\beta-2=0$, which means $\alpha+\beta=2$." b6863a573c51,"(Infinite Solitaire)(^) You all know the rules of solitaire: there are marbles on a board, and you eliminate marbles by jumping over them with another marble, etc. Now consider the plane and place marbles on all integer points of the negative half-plane. The goal of the game is to place a marble as high as possible in a finite number of moves. The drawing shows how to place a marble at height 1. What is the maximum height that can be reached? ![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-019.jpg?height=369&width=1459&top_left_y=1873&top_left_x=214) - Solutions to the exercises -",See reasoning trace,medium,"It is easy to raise a ball to height 1, 2, and 3, and with a bit of patience, one can raise a ball to height 4, but that is the limit. Let $\varphi$ be the golden ratio, which satisfies $\varphi^{2}=\varphi+1$: $$ \varphi=\frac{1+\sqrt{5}}{2} $$ Then, we need to sum over all points with coordinates $(x, y)$ on which there are balls of $\varphi^{y-|x|}$. I leave it to the reader to verify that this quantity remains constant or decreases when the balls are moved, and that in the initial position, it is equal to $\varphi^{5}$. This proves that it is impossible to raise a ball to height 5 or more. ## 2 Course/TD on Basic Strategies ## - Recurrence - The principle of recurrence is fundamentally linked to the notion of integers: every time one needs to prove that a result is true for all integers from a certain rank, whether in arithmetic, geometry, algebra... one can be led to reason by recurrence. The proof by recurrence is based on the fact that every integer $n$ has a successor $n+1$, and that if one ""climbs step by step"" the set $\mathbb{N}$ of natural numbers, starting from 0, going from one integer to the next, and repeating indefinitely, one thus fully traverses $\mathbb{N}$. We will therefore prove that a property is true for the integer $n=0$ (initialization), and then that if it is true for any integer $n \geqslant 0$, it is still true for the next integer $n+1$ (induction): we deduce from this that the property is true for all natural numbers $n$. This is the reasoning by recurrence. It should be noted that both parts of the proof are important, even if the first is often quite obvious. One must not forget to initialize the recurrence, that is, to prove that the property is true for $n=0$. One can start the recurrence at a value other than 0, for example, at $1, 2, \cdots$ or $K$; it will then be necessary to prove (initialization) that the property is true for $n=K$, and then (induction) that for $n \geqslant K$, if it is true for $n$, it is still true for $n+1$. It should also be noted that induction has some variants. For example, it sometimes happens that one has to prove that if the property is true for all integers $k \leqslant n$, then it is true for $n+1$." 0f16505bcea6,"11. As shown in the figure, there are 10 cards, each card has a number from $0 \sim 9$, and there are no duplicates. Xiaoyu and Xiaomu each take 5 cards and start playing a card game. The rules of the game are as follows: (1) Both players calculate the sum of the numbers on all 5 cards they have as their initial score. (2) Each turn, both players play 1 card from their hand. Each player will add the number on the card they played to their current score, and then subtract the number on the card played by the opponent. After all cards are played, Xiaomu's score is twice that of Xiaoyu's. What is Xiaoyu's initial score? $\qquad$ .",See reasoning trace,easy,$20$ 14659f7bd786,"12. What digits do the decimal representations of the following numbers end with: 1) $135^{x}+31^{y}+56^{x+y}$, if $x \in N, y \in N$ 2) $142+142^{2}+142^{3}+\ldots+142^{20}$ 3) $34^{x}+34^{x+1}+34^{2 x}$, if $x \in N$.",See reasoning trace,easy,"12. 3) $\mathrm{Solution.} \mathrm{If} \mathrm{a} \mathrm{number} \mathrm{ends} \mathrm{in} \mathrm{four,} \mathrm{then}$ even powers of it end in 6, and odd powers end in 4. Therefore, one of the first two terms ends in four, and the other ends in six. The third term ends in six, so the decimal representation of the sum ends in six." a557d8704699,6. [25] Into how many regions can a circle be cut by 10 parabolas?,"201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$",medium,"Answer: 201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$. We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, the number of regions is given in terms of the number of intersections of the parabolas that occur within the circle, since every time two parabolas cross a new region is created. Since two pairs of parallel lines intersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always make all 4 points of intersection lie inside the circle, the parallel lines case is the best we can do. In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines. We can compute the answer for pairs of parallel lines as follows - when we add the $k$ th set of parallel lines, there are already $2 k-2$ lines that the two new lines can intersect, meaning that each of the lines adds $2 k-1$ new regions ${ }^{1}$. This means that we add $4 k-2$ regions when adding the $k$ th set of lines, making the answer $1+2+6+10+14+\cdots+(4 n-2)=1+2(1+3+5+7+\cdots+(2 n-1))=1+2 \cdot n^{2}=2 n^{2}+1$. Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there are $n$ parabolas inside the circle, and the they intersect in $K$ points total, then we claim that the number of regions the circle is divided into will be at most $K+n+r+1$, where $r$ is the number of parabolas that intersect the circle itself in exactly four points. We will prove this by induction. In the base case of $n=0$, we are just saying that the circle itself consists of exactly one region. To prove the inductive step, suppose that we have $n$ parabolas with $K$ points of intersection. We want to show that if we add an additional parabola, and this parabola intersects the other parabolas in $p$ points, then this new parabola adds either $p+1$ or $p+2$ regions to the circle, and that we get $p+2$ regions if and only if the parabola intersects the circle in exactly four points. We will do this by considering how many regions the parabola cuts through, following its path from when it initially enters the circle to when it exits the circle for the last time. When it initially enters the circle, it cuts through one region, thereby increasing the number of regions by one ${ }^{2}$. Then, for each other parabola that this parabola crosses, we cut through one additional region. It is also possible for the parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects ${ }^{1}$ This is the maximum possible number of new regions, but it's not too hard to see that this is always attainable. ${ }^{2}$ While the fact that a curve going through a region splits it into two new regions is intuitively obvious, it is actually very difficult to prove. The proof relies on some deep results from algebraic topology and is known as the Jordan Curve Theorem. If you are interested in learning more about this, see http://en.wikipedia.org/wiki/Jordan_curve_theorem the circle in four points, and also adds one additional region. Therefore, the number of regions is either $p+1$ or $p+2$, and it is $p+2$ if and only if the parabola intersects the circle in four points. This completes the induction and proves the claim. So, we are left with trying to maximize $K+n+r+1$. Since a pair of parabolas intersects in at most 4 points, and there are $\binom{n}{2}$ pairs of parabolas, we have $K \leq 4\binom{n}{2}=2 n^{2}-2 n$. Also, $r \leq n$, so $K+n+r+1 \leq 2 n^{2}+1$. On the other hand, as explained in the paragraphs giving the intuition, we can attain $2 n^{2}+1$ by making the parabolas sufficiently steep that they act like pairs of parallel lines. Therefore, the answer is $2 n^{2}+1$, as claimed." cc5eac6e9186,"## Calculate the limit of the numerical sequence: $\lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\ & =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \cdot\left(n^{2}+2 n+1+n^{2}-2 n+1\right)}{(n+1)^{3}+(n-1)^{3}}= \\ & =\lim _{n \rightarrow \infty} \frac{4 n\left(2 n^{2}+2\right)}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}} 8 n\left(n^{2}+1\right)}{n^{3}\left((n+1)^{3}+(n-1)^{3}\right)}= \\ & =\lim _{n \rightarrow \infty} \frac{8\left(1+\frac{1}{n^{2}}\right)}{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}=\frac{8 \cdot 1}{1^{3}+1^{3}}=4 \end{aligned} $$ ## Problem Kuznetsov Limits 3-24" a7d09e396a6a,b) Find how many 4-digit numbers written in base 10 are written in base 2 using exactly one digit 0.,See reasoning trace,medium,"Solution. a) Writing in base 10 a number written with $n$ digits of 1 in base 2, we have: $\overline{11 \ldots 1}_{(2)}=2^{n-1}+2^{n-2}+\ldots+2^{2}+2+1=2^{n}-1$ $1 p$ Numbers of four digits of the form $2^{n}-1$ are $2^{10}-1=1023,2^{11}-1=2047,2^{12}-1=$ ![](https://cdn.mathpix.com/cropped/2024_06_07_3dd00a25927e70aa5fc0g-2.jpg?height=62&width=1589&top_left_y=1666&top_left_x=230) b) Numbers that can be written in base 2 using only one digit 0 can be expressed as the difference between two numbers, where the minuend $D$ is written in base 2 only with the digit 1, and the subtrahend $S$ is written in base 2 using only one digit 1, followed by some zeros: ![](https://cdn.mathpix.com/cropped/2024_06_07_3dd00a25927e70aa5fc0g-2.jpg?height=108&width=612&top_left_y=1949&top_left_x=716) $1 p$ Converting to base 10, the minuend $D$ is a number of the form $2^{n}-1$, and the subtrahend $S$ is of the form $2^{p}$, where $0 \leq p \leq n-2$, since for $p=n-1$ the difference is equal to $2^{n-1}-1$, which is written only with 1s. Therefore, the difference $D-S$ is between $2^{n}-1-2^{n-2}$ and $2^{n}-2$, and since $D-S$ is a four-digit number, we get $2^{n}-2 \geq$ 1000 and $2^{n}-1-2^{n-2} \leq 9999$, from which $10 \leq n \leq 13$ $1 p$ If the minuend $D$ is $2^{10}-1=1023$, for the difference to have four digits, the subtrahend $S$ can only take the values $1,2,4,8$ or 16 (5 possibilities). If $D=2^{11}-1$, then the subtrahend $S$ can be $1,2,2^{2}, \ldots, 2^{9}$ (10 possibilities), for $D=2^{12}-1$ there are 11 ways to choose $S$, and for $D=2^{13}-1$, the subtrahend can be chosen in 12 ways. Thus, there are $5+10+11+12=38$ four-digit numbers that can be written in base 2 with only one digit 0 $3 p$" 48ee855fc2cc,"## Find the derivative $y_{x}^{\prime}$. \[ \left\{\begin{array}{l} x=\arccos \frac{1}{t} \\ y=\sqrt{t^{2}-1}+\arcsin \frac{1}{t} \end{array}\right. \]",See reasoning trace,medium,"## Solution $x_{t}^{\prime}=\left(\arccos \frac{1}{t}\right)^{\prime}=\frac{-1}{\sqrt{1-\left(\frac{1}{t}\right)^{2}}} \cdot\left(-\frac{1}{t^{2}}\right)=\frac{1}{t \cdot \sqrt{t^{2}-1}}$ $y_{t}^{\prime}=\left(\sqrt{t^{2}-1}+\arcsin \frac{1}{t}\right)^{\prime}=\frac{1}{2 \sqrt{t^{2}-1}} \cdot 2 t+\frac{1}{\sqrt{1-\left(\frac{1}{t}\right)^{2}}} \cdot\left(-\frac{1}{t^{2}}\right)=$ $=\frac{t^{2}}{t \cdot \sqrt{t^{2}-1}}-\frac{1}{t \cdot \sqrt{t^{2}-1}}=\frac{t^{2}-1}{t \cdot \sqrt{t^{2}-1}}=\frac{\sqrt{t^{2}-1}}{t}$ We obtain: $y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{\sqrt{t^{2}-1}}{t}\right) /\left(\frac{1}{t \cdot \sqrt{t^{2}-1}}\right)=\frac{\sqrt{t^{2}-1} \cdot t \cdot \sqrt{t^{2}-1}}{t}=$ $=t^{2}-1$ ## Problem Kuznetsov Differentiation 16-17" 6bfe0296eb8d,"7. Let $A B C D E F$ be a regular hexagon of area 1 . Let $M$ be the midpoint of $D E$. Let $X$ be the intersection of $A C$ and $B M$, let $Y$ be the intersection of $B F$ and $A M$, and let $Z$ be the intersection of $A C$ and $B F$. If $[P]$ denotes the area of polygon $P$ for any polygon $P$ in the plane, evaluate $[B X C]+[A Y F]+[A B Z]-[M X Z Y]$.",zero,easy,"Answer: 0 Let $O$ be the center of the hexagon. The desired area is $[A B C D E F]-[A C D M]-[B F E M]$. Note that $[A D M]=[A D E] / 2=[O D E]=[A B C]$, where the last equation holds because $\sin 60^{\circ}=\sin 120^{\circ}$. Thus, $[A C D M]=[A C D]+[A D M]=[A C D]+[A B C]=[A B C D]$, but the area of $A B C D$ is half the area of the hexagon. Similarly, the area of $[B F E M]$ is half the area of the hexagon, so the answer is zero." 8624c7291b25,"## Calculate the indefinite integral: $$ \int \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}} d x $$",See reasoning trace,medium,"## Solution $$ \int \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}} d x= $$ We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients: $$ \begin{aligned} & \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= \\ & =\frac{A(x+2)^{3}+B_{1}(x-2)(x+2)^{2}+B_{2}(x-2)(x+2)+B_{3}(x-2)}{(x-2)(x+2)^{3}}= \\ & =\frac{A\left(x^{3}+6 x^{2}+12 x+8\right)+B_{1}\left(x^{3}+2 x^{2}-4 x-8\right)+B_{2}\left(x^{2}-4\right)+B_{3}(x-2)}{(x-2)(x+2)^{3}}= \\ & =\frac{\left(A+B_{1}\right) x^{3}+\left(6 A+2 B_{1}+B_{2}\right) x^{2}+\left(12 A-4 B_{1}+B_{3}\right) x+\left(8 A-8 B_{1}-4 B_{2}-2 B_{3}\right)}{(x-2)(x+2)^{3}} \\ & \left\{\begin{array}{l} A+B_{1}=1 \\ 6 A+2 B_{1}+B_{2}=6 \\ 12 A-4 B_{1}+B_{3}=13 \\ 8 A-8 B_{1}-4 B_{2}-2 B_{3}=6 \end{array}\right. \end{aligned} $$ Add the third equation multiplied by 2 to the fourth equation: $$ \left\{\begin{array}{l} A+B_{1}=1 \\ 6 A+2 B_{1}+B_{2}=6 \\ 12 A-4 B_{1}+B_{3}=13 \\ 32 A-16 B_{1}-4 B_{2}=32 \end{array}\right. $$ Add the second equation multiplied by 4 to the fourth equation: $$ \begin{aligned} & \left\{\begin{array}{l} A+B_{1}=1 \\ 6 A+2 B_{1}+B_{2}=6 \\ 12 A-4 B_{1}+B_{3}=13 \\ 56 A-8 B_{1}=56 \end{array}\right. \\ & \left\{\begin{array}{l} A+B_{1}=1 \\ 6 A+2 B_{1}+B_{2}=6 \\ 12 A-4 B_{1}+B_{3}=13 \\ 7 A-B_{1}=7 \end{array}\right. \end{aligned} $$ Add the first equation to the fourth equation: $$ \begin{aligned} & \left\{\begin{array}{l} A+B_{1}=1 \\ 6 A+2 B_{1}+B_{2}=6 \\ 12 A-4 B_{1}+B_{3}=13 \\ 8 A=8 \end{array}\right. \\ & \left\{\begin{array}{l} B_{1}=0 \\ 2 B_{1}+B_{2}=0 \\ -4 B_{1}+B_{3}=1 \\ A=1 \end{array}\right. \\ & \left\{\begin{array}{l} B_{1}=0 \\ B_{2}=0 \\ B_{3}=1 \\ A=1 \end{array}\right. \\ & \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}}=\frac{1}{x-2}+\frac{1}{(x+2)^{3}} \end{aligned} $$ Then: $$ \int \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}} d x=\int\left(\frac{1}{x-2}+\frac{1}{(x+2)^{3}}\right) d x=\ln |x-2|-\frac{1}{2(x+2)^{2}}+C $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+6-31 »$ Categories: Kuznetsov's Integral Problems Problem 6 | Integrals - Last edited: 09:11, April 10, 2009. - Content is available under CC-BY-SA 3.0." dfc8c34a3f04,"1. In $\triangle A B C$, $|\overrightarrow{A B}|=2,|\overrightarrow{A C}|=3, \overrightarrow{A B} \cdot \overrightarrow{C A}>0$, and the area of $\triangle A B C$ is $\frac{3}{2}$, then $\angle B A C$ equals A. $150^{\circ}$ B. $30^{\circ}$ C. $120^{\circ}$ D. $60^{\circ}$","Since $\sin \angle B A C=\frac{1}{2}$ and $\angle B A C$ is an obtuse angle $\Rightarrow \angle B A C=150^{\circ}$, so the answer is $A$",easy,"Answer Since $\sin \angle B A C=\frac{1}{2}$ and $\angle B A C$ is an obtuse angle $\Rightarrow \angle B A C=150^{\circ}$, so the answer is $A$." cf496ada4e25,"Initially 255, set 2009 can be decomposed into the sum of squares of four positive integers, among which, the ratio of two numbers is $\frac{5}{14}$, and the ratio of the other two numbers is $\frac{1}{2}$. Write down this decomposition.",See reasoning trace,medium,"Solution: Let the required decomposition be $$ 2009=(14 x)^{2}+(5 x)^{2}+(2 y)^{2}+y^{2} \text {. } $$ Obviously, $221 x^{2}+5 y^{2}=2009$, hence $$ x^{2}=\frac{2009-5 y^{2}}{221}<10 \text {. } $$ Thus, $x=1, 2$ or 3. (1) If $x=1, y^{2}=\frac{2009-221 x^{2}}{5}=\frac{1788}{5}$, no solution; (2) If $x=2, y^{2}=\frac{2009-221 x^{2}}{5}=225$, hence $y=15$; (3) If $x=3, y^{2}=\frac{2009-221 x^{2}}{5}=4$, hence $y=2$. In summary, the required decompositions are $$ \begin{array}{l} 2009=30^{2}+28^{2}+15^{2}+10^{2}, \\ 2009=42^{2}+15^{2}+4^{2}+2^{2} . \end{array} $$ (Tian Yonghai, Suihua Educational Institute, Heilongjiang Province, 152054)" a8db3afae10b,"Example 15 (1998 National High School Competition Question) Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$. For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$ and prove your conclusion.","3$, so $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$, i.e., $l(a)=\frac{-8-\sqrt{64-3",medium,"Solution: $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$, $\therefore \max _{x \in \mathbf{R}} f(x)=3-\frac{16}{a}$. (1) $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a}$, and $f(0)=3$, so $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$, i.e., $l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=$ $\frac{2}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2}$, with equality holding if and only if $a=$ -8. Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, $l(a)$ attains its maximum value $\frac{\sqrt{5}+1}{2}$ if and only if $a=-8$." 700157d48c96,"6. Cut a square with a side of 4 into rectangles, the sum of the perimeters of which is 25.",283&width=283&top_left_y=1526&top_left_x=181),easy,"For example, two rectangles $2 \times 0.5$ and one rectangle $3.5 \times 4-$ cm. The total perimeter is $2 * 2 * (2 + 0.5) + 2 * (3.5 + 4) = 25$. ![](https://cdn.mathpix.com/cropped/2024_05_06_2bec25c7ce83461908d3g-2.jpg?height=283&width=283&top_left_y=1526&top_left_x=181)" 99a5753854fb,"## Calculate the lengths of the arcs of the curves given by the equations in polar coordinates. $$ \rho=1-\sin \varphi, -\frac{\pi}{2} \leq \varphi \leq -\frac{\pi}{6} $$",See reasoning trace,medium,"## Solution The length of the arc of a curve given by an equation in polar coordinates is determined by the formula $$ L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi $$ Let's find $\frac{d \rho}{d \phi}$: $$ \frac{d \rho}{d \phi}=(-\cos \phi) $$ We get: $$ \begin{aligned} L & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{(1-\sin \phi)^{2}+(-\cos \phi)^{2}} d \phi= \\ & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{1-2 \sin \phi+\sin ^{2} \phi+\cos ^{2} \phi} d \phi= \\ & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{2-2 \sin \phi} d \phi= \\ & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{2-2 \cos \left(\frac{\pi}{2}-\phi\right)} d \phi= \\ & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{4 \sin ^{2}\left(\frac{\pi}{4}-\frac{\phi}{2}\right)} d \phi= \\ & =\int_{-\pi / 2}^{-\pi / 6} 2 \sin \left(\frac{\pi}{4}-\frac{\phi}{2}\right) d \phi=\left.4 \cdot \cos \left(\frac{\pi}{4}-\frac{\phi}{2}\right)\right|_{-\pi / 2} ^{-\pi / 6}= \\ & =4\left(\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{-\pi}{6}\right)-\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{-\pi}{2}\right)\right)=4\left(\cos \left(\frac{\pi}{3}\right)-\cos \left(\frac{\pi}{2}\right)\right)=4\left(\frac{1}{2}-0\right)=2 \end{aligned} $$ Categories: Kuznetsov Problem Book Integrals Problem 19 | Integrals Ukrainian Banner Network - Last modified: 13:05, 27 May 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 19-12 ## Material from PlusPi" 8a19f2706862,"9.6. First, Betya writes the numbers $1, 2$ in her notebook, and Nik writes $3, 4$ in his notebook. Then, at the beginning of each minute, Nik and Betya each write down a quadratic polynomial with real coefficients, whose roots are the two numbers in their notebooks, denoted as $f(x)$ and $g(x)$, respectively. If the equation $f(x)=g(x)$ has two distinct real roots $x_{1}$ and $x_{2}$, one of them immediately changes the numbers in their notebook to $x_{1}$ and $x_{2}$; if the equation $f(x)=g(x)$ does not have two distinct real roots $x_{1}$ and $x_{2}$, they do nothing and wait for the next minute to start. After some time, one of the numbers in Betya's notebook is 5. Find all possible values of the other number in her notebook.",See reasoning trace,easy,"9.6. It is not hard to see that the two numbers in each person's exercise book are always the roots of the equation $$ \alpha(x-1)(x-2)+\beta(x-3)(x-4)=0 $$ where $\alpha, \beta$ are parameters that depend on time. At a certain moment, 5 is a root of this equation. Substituting 5, we get $6 \alpha+\beta=0$. Thus, $\alpha(x-1)(x-2)+\beta(x-3)(x-4)$ $=\alpha\left(-5 x^{2}+39 x-70\right)$ $=-\alpha(x-5)(5 x-14)$. The other root is $\frac{14}{5}$." cde0be4a1e0f,"14. $\log _{\frac{\sqrt{3}}{3}}\left(\log _{8} \frac{\sqrt{2}}{2}-\log _{3} \frac{\sqrt{3}}{3}\right)$. In № $15-27$ solve the equations:",2,easy,"14. We should switch to the logarithm with base $\frac{\sqrt{3}}{3}$. Then replace $\frac{1}{3}$ with $\left(\frac{\sqrt{3}}{3}\right)^{2}$. Answer: 2." beb53692b8d1,"If $x * y=x+y^{2}$, then $2 * 3$ equals (A) 8 (B) 25 (C) 11 (D) 13 (E) 7",(C),easy,"If $x * y=x+y^{2}$, then $2 * 3$ equals (A) 8 (B) 25 (C) 11 (D) 13 (E) 7 ## Solution $2 * 3^{2}=2+3^{2}=11$ ANSwer: (C)" d2bf73c28c52,"A3. When we throw 3 fair dice of different colors. In how many cases can we get a sum of 10 dots? (A) 30 (B) 27 (C) 10 (D) 6 (E) 36",is $B$,easy,"A3. The sum of 10 points can be obtained in six possible outcomes, namely, in the outcomes $\{1,3,6\},\{1,4,5\}$ and $\{2,3,5\}$ each $3!=6$, and in the outcomes $\{2,2,6\},\{3,3,4\}$ and $\{4,4,2\}$ each $\frac{3!}{2!}=3$. If we add up all the possibilities $6 \cdot 3+3 \cdot 3$ we get 27 possibilities. The correct answer is $B$." a31ba6b907aa,"Ya Chenenko I.V. In the store, they sell DVD discs - individually and in two types of packages (packages of different types differ in quantity and cost). Vasya calculated how much money is needed to buy $N$ discs (if it is more profitable to buy more discs than needed - Vasya does so): | $N$ | 1 | 2 | 3 | 4 | 5 | $6-10$ | 11 | 12 | 13 | 14 | 15 | $16-20$ | 21 | 22 | $23-25$ | 26 | 27 | 28 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | c. | 20 | 40 | 60 | 80 | 100 | 111 | 131 | 151 | 171 | 191 | 211 | 222 | 242 | 262 | 265 | 285 | 305 | 325 | How many discs were in the packages and at what price were the packages sold? How much money does Vasya need to buy at least 29 discs?",For sale are: individual disks at a price of 20 rubles; packages of 10 disks at a price of 111 rubles per package; packages of 25 disks at a price of 265 rubles per package,medium,"If the most profitable purchase consists of several blocks (disks individually and packages), then the disks contained in any block or any set of these blocks were purchased in a way that cannot be ""improved"" (made more profitable) - otherwise, by ""improving"" the payment method for part of the disks, we would be able to make the entire purchase more profitable. From this, it follows that if we used a certain package for the purchase, then to buy the number of disks contained in this package at the minimum price, we can buy just this package. From the table, it can be seen that disks in quantities of 1, 2, 3, 4, 5 were purchased individually at a price of 20 rubles per disk. Purchasing no less than 6 disks at this price does not correspond, so in this case, at least one package was purchased. But nothing else, besides this package, could have been purchased, as it would have resulted in this package optimizing the purchase of fewer than 6 disks. But then we would have used such a method to purchase such a (less than 6) quantity. When purchasing 7 disks, the same package should have been used. The same sequentially (using the previously obtained information) we establish for 8, 9, and 10 disks. The price of 11 disks is higher than 10. Therefore, this package contains only 10 disks. Thus, we have learned the parameters of one of the packages: it contains 10 disks and costs 111 rubles. Now notice that the price of purchasing 24 and 25 disks is the same, while for purchasing 26, it is different. When purchasing 25 disks, we could not use individual disks (since otherwise, by refusing one individual disk, we could have bought 24 disks 20 rubles cheaper than 25 disks, while the table shows the same price for these quantities). And it is impossible to get 25 disks with packages of 10 disks. Therefore, we have established another type of package: it contains 25 disks and costs 265 rubles. This package cannot contain more than 25 disks, as then 26 disks would cost the same. Thus, we have established 2 types of packages: 1) 111 rubles for 10 disks (11.1 rubles per disk); 2) 265 rubles for 25 disks (10.6 rubles per disk). Since the condition states that there should be only two types of packages, we have determined all possible options. Let's determine the optimal price for purchasing 29 disks. We can buy three packages of 10 disks, which will cost 333 rubles. Purchasing individual disks is irrational: as soon as we buy one individual disk (for 20 rubles), we will need to buy 28 disks, the optimal price of which, according to the condition, is 325 rubles. In total, it will be 345 rubles. If we buy a package of 25 disks, we will need to buy 4 more disks. From the table, it can be seen that it is cheapest to buy them individually. But, as we just found out, the price will be no less than 345 rubles. Therefore, the initially considered option (3 packages of 10 disks) is the most rational. ## Answer For sale are: individual disks at a price of 20 rubles; packages of 10 disks at a price of 111 rubles per package; packages of 25 disks at a price of 265 rubles per package. To purchase no less than 29 disks, 333 rubles are required." fac6e424b700,"Determine all functions $f : \mathbb R \to \mathbb R$ satisfying the following two conditions: (a) $f(x + y) + f(x - y) = 2f(x)f(y)$ for all $x, y \in \mathbb R$, and (b) $\lim_{x\to \infty} f(x) = 0$.",f(x) = 0,medium,"1. Let \( P(x, y) \) denote the assertion \( f(x + y) + f(x - y) = 2f(x)f(y) \). 2. Let \( Q(x) \) denote the condition \( \lim_{x \to \infty} f(x) = 0 \). ### Case 1: \( f(0) = 0 \) 3. From \( P(0, 0) \), we have: \[ f(0 + 0) + f(0 - 0) = 2f(0)f(0) \implies 2f(0) = 2f(0)^2 \implies f(0)(1 - f(0)) = 0 \] Thus, \( f(0) = 0 \) or \( f(0) = 1 \). 4. Assume \( f(0) = 0 \). Then, from \( P(x, 0) \): \[ f(x + 0) + f(x - 0) = 2f(x)f(0) \implies 2f(x) = 0 \implies f(x) = 0 \quad \forall x \in \mathbb{R} \] This is a valid solution. ### Case 2: \( f(0) = 1 \) 5. Assume \( f(0) = 1 \). Then, from \( P(0, x) \): \[ f(0 + x) + f(0 - x) = 2f(0)f(x) \implies f(x) + f(-x) = 2f(x) \implies f(-x) = f(x) \] This shows that \( f \) is an even function. 6. From \( P(x, -x) \): \[ f(x - x) + f(x + x) = 2f(x)f(-x) \implies f(0) + f(2x) = 2f(x)^2 \implies 1 + f(2x) = 2f(x)^2 \] 7. Using \( Q(2x) \), we have: \[ \lim_{x \to \infty} f(2x) = 0 \] Therefore: \[ \lim_{x \to \infty} (1 + f(2x)) = 1 \quad \text{and} \quad \lim_{x \to \infty} 2f(x)^2 = 0 \] This implies: \[ 1 = 0 \quad \text{(contradiction)} \] Since the assumption \( f(0) = 1 \) leads to a contradiction, the only valid solution is \( f(0) = 0 \), which implies \( f(x) = 0 \) for all \( x \in \mathbb{R} \). The final answer is \( \boxed{ f(x) = 0 } \) for all \( x \in \mathbb{R} \)." 29263a07b786,B4. Een parallellogram heeft twee zijden van lengte 4 en twee zijden van lengte 7. Ook heeft een van de diagonalen lengte 7. (Let op: het plaatje hiernaast is niet op schaal.) Hoe lang is de andere diagonaal?,See reasoning trace,medium," B4. 9 We tekenen enkele hulplijnen, zoals in de figuur. Het parallellogram is $A B C D$ met $|A B|=4$ en $|A D|=|B C|=7$. We zijn op zoek naar $|A C|$. Aangezien $|A D|=|B D|$, is driehoek $\triangle A B D$ een gelijkbenige driehoek. Laat $E$ het punt zijn midden tussen $A$ en $B$. Dan zijn de driehoeken $\triangle A E D$ en $\triangle B E D$ congruent en hoek $\angle A E D$ is een rechte hoek. Vanuit het punt $C$ laten we een loodlijn neer op het verlengde van $A B$, noem het snijpunt $F$. Dan is $\angle D A E=\angle C B F$ vanwege $\mathrm{F}$-hoeken, en $\angle A E D=\angle B F C=90^{\circ}$ vanwege de loodlijn. Verder geldt $|A D|=|B C|$, ![](https://cdn.mathpix.com/cropped/2024_04_17_848643ef5340edd7a266g-2.jpg?height=434&width=363&top_left_y=1485&top_left_x=1486) dus we hebben de congruente driehoeken $\triangle A E D \cong \triangle B F C$. Hieruit volgt dat $|B F|=2$. We passen nu de stelling van Pythagoras toe in driehoek $\triangle B F C$. Dit geeft dat voor de hoogte $h$ van deze driehoek geldt dat $2^{2}+h^{2}=7^{2}$, dus $h^{2}=45$. Nu passen we de stelling van Pythagoras toe in driehoek $\triangle A F C$ om de diagonaal $d=|A C|$ te vinden: $h^{2}+6^{2}=d^{2}$, oftewel $45+36=d^{2}$, dus $d=9$. " 05c69a4647b1,"13. Given that $p$ and $q$ are both prime numbers, and that the quadratic equation in $x$, $x^{2}-(8 p-10 q) x+5 p q=0$, has at least one positive integer root. Find all prime number pairs $(p, q)$.",See reasoning trace,medium,"13. From the sum of the two roots of the equation being $8 p-10 q$, we know that if one root is an integer, the other root is also an integer. From the product of the two roots being $5 p q$, we know that the other root is also a positive integer. Let the two positive integer roots of the equation be $x_{1}$ and $x_{2} \left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have $$ \begin{array}{l} x_{1}+x_{2}=8 p-10 q, \\ x_{1} x_{2}=5 p q . \end{array} $$ From equation (2), we get the following possible cases for $x_{1}$ and $x_{2}$: $$ \left\{\begin{array}{l} x_{1}=1,5, p, q, 5 p, 5 q, \\ x_{2}=5 p q, p q, 5 q, 5 p, q, p, \end{array}\right. $$ Substitute $x_{1}+x_{2}=1+5 p q, 5+p q, p+5 q, q+5 p$ into equation (1). When $x_{1}+x_{2}=1+5 p q$, $1+5 p q=8 p-10 q$, but $1+5 p q>10 p>8 p-10 q$, so there is no solution in this case. When $x_{1}+x_{2}=5+p q$, $5+p q=8 p-10 q$, then we have $(p+10)(q-8)=-85$. Since $p$ and $q$ are both primes, it is only possible that $$ \left\{\begin{array}{l} q-8=-5,-1, \\ p+10=17,85 . \end{array}\right. $$ Therefore, $(p, q)=(7,3)$. When $x_{1}+x_{2}=p+5 q$, $p+5 q=8 p-10 q$, so $7 p=15 q$, which is impossible. When $x_{1}+x_{2}=q+5 p$, $q+5 p=8 p-10 q$, so $3 p=11 q$, thus, we have $(p, q)=(11,3)$. In summary, the prime pairs $(p, q)$ that satisfy the conditions are $(7,3)$ or $(11,3)$." c3d29eeb51ae,"4. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$. When $n \geqslant 2$, $a_{n}, S_{n}, S_{n}-\frac{1}{2}\left(S_{n}\right.$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$) form a geometric progression, then $\lim _{n \rightarrow \infty} n^{2} a_{n}=$ $\qquad$",See reasoning trace,medium,"4. $-\frac{1}{2}$. From the condition, when $n \geqslant 2$, $$ S_{n}^{2}=a_{n}\left(S_{n}-\frac{1}{2}\right)=\left(S_{n}-S_{n-1}\right)\left(S_{n}-\frac{1}{2}\right), $$ Thus, $$ \frac{1}{S_{n}}-\frac{1}{S_{n-1}}=2 $$ Therefore, $$ \frac{1}{S_{n}}=\frac{1}{S_{1}}+2(n-1)=2 n-1, $$ So, $S_{n}=\frac{1}{2 n-1}$. Hence, $$ a_{n}=S_{n}-S_{n-1}=\frac{1}{2 n-1}-\frac{1}{2 n-3}=-\frac{2}{(2 n-1)(2 n-3)} . $$ Therefore, $$ \lim _{n \rightarrow \infty} n^{2} a_{n}=\lim _{n \rightarrow \infty}-\frac{2 n^{2}}{(2 n-1)(2 n-3)}=\lim _{n \rightarrow \infty}-\frac{2}{\left(2-\frac{1}{n}\right)\left(2-\frac{3}{n}\right)}=-\frac{1}{2} $$" ee41e58cee63,"54. A deck of playing cards has 52 cards. By convention, $A, J, Q, K$ are valued as 1 point, 11 points, 12 points, 13 points respectively. If you draw several cards at random, disregarding suits, to ensure that there are three cards with the same point value, you need to draw at least $\qquad$ cards; to ensure that there are two cards whose point values add up to 15, you need to draw at least $\qquad$ cards.","27,29",easy,"Reference answer: 27,29" da4ad0038a24,"Example 1 Given that $AB$ is a chord of $\odot O$ with radius 1, and $AB=a<1$. Using $AB$ as one side, construct a regular $\triangle ABC$ inside $\odot O$, and let $D$ be a point on $\odot O$ different from point $A$, such that $DB=AB=a$, and the extension of $DC$ intersects $\odot O$ at point $E$. Then the length of $AE$ is ( ). (A) $\frac{\sqrt{5}}{2} a$ (B) 1 (C) $\frac{\sqrt{3}}{2}$ (D) $a$",B,easy,"Solve As shown in Figure 2, connect $A D, O E, O A$. Notice that $$ D B=A B=B C \text {. } $$ By Judgment 1, $B$ is the circumcenter of $\triangle A C D$. Then, by Property 2(1), $\angle A D C=\frac{1}{2} \angle A B C$. Thus, $O$ is the circumcenter of $\triangle A D E$. Therefore, $\angle A D E=\frac{1}{2} \angle A O E$. Hence, $\angle A O E=\angle A B C=60^{\circ}$. Also, since $O E=O A$, $\triangle O A E$ is an equilateral triangle. So, $A E=O A=1$. Therefore, the answer is B." 472b30e524b6,"Anna thinks of an integer. - It is not a multiple of three. - It is not a perfect square. - The sum of its digits is a prime number. The integer that Anna is thinking of could be (A) 12 (B) 14 (C) 16 (D) 21 (E) 26",not (A) or (D),easy,"12 and 21 are multiples of $3(12=4 \times 3$ and $21=7 \times 3)$ so the answer is not (A) or (D). 16 is a perfect square $(16=4 \times 4)$ so the answer is not $(\mathrm{C})$. The sum of the digits of 26 is 8 , which is not a prime number, so the answer is not (E). Since 14 is not a multiple of a three, 14 is not a perfect square, and the sum of the digits of 14 is $1+4=5$ which is prime, then the answer is 14 , which is choice (B). ANSWER: (B)" b1f25511deb8,"2. $36 S$ is a subset of $\{1,2, \cdots, 1989\}$, and the difference between any two numbers in $S$ cannot be 4 or 7. How many elements can $S$ have at most?",905$ elements.,medium,"[Solution] First, we prove: In any 11 consecutive integers in the set $\{1,2, \cdots, 1989\}$, at most 5 can be elements of $S$. In fact, let $T=\{1,2, \cdots, 11\}$. We examine the subsets $\{1,5\},\{2,9\}$, $\{3,7\},\{4,11\},\{6,10\},\{8\}$ of $T$. Clearly, each of these subsets can have at most one element in $S$. If $T$ has 6 elements in $S$, then $$ 8 \in S \Rightarrow 1 \notin S \Rightarrow 5 \in S \Rightarrow 9 \notin S \Rightarrow 2 \in S \Rightarrow 6 \notin S \Rightarrow 10 \in S \Rightarrow 3 $$ $\notin S \Rightarrow 7 \in S \Rightarrow 11 \notin S \Rightarrow 4 \in S \Rightarrow 8 \notin S$. This leads to a contradiction. Therefore, $T$ can have at most 5 elements in $S$. A subset of $T$ with 5 elements is possible, for example: $$ T^{\prime}=\{1,3,4,6,9\}, $$ Let $A=\{k+1, k+2, \cdots, k+11\}$ be a subset of $\{1,2, \cdots, 1989\}$. Then, by analogy with the above proof, $A$ can have at most 5 elements in $S$. Since $S^{\prime}=\left\{k+11 n \mid k \in T^{\prime}, n \in I, k+11 n \leqslant 1989.\right\}$ and $1989=180 \times 11+9$. Therefore, $S$ can have at most $181 \times 5=905$ elements." e8d277975e7e,"$$ \text { Three. (25 points) In } \triangle A B C \text {, } A B=8+2 \sqrt{6} \text {, } $$ $B C=7+2 \sqrt{6}, C A=3+2 \sqrt{6}, O, I$ are the circumcenter and incenter of $\triangle A B C$, respectively. Is the length of segment $O I$ a rational number or an irrational number? Make a judgment and explain your reasoning.",See reasoning trace,medium,"Three, OI=2.5 is a rational number. Let $a, b, c, p, S, R, r$ represent the three sides, the semi-perimeter, the area, the circumradius, and the inradius of $\triangle ABC$, respectively. Thus, $$ \begin{array}{l} p=\frac{1}{2}(a+b+c)=9+3 \sqrt{6}, \\ p-a=2+\sqrt{6}, p-b=6+\sqrt{6}, \\ p-c=1+\sqrt{6}. \end{array} $$ Then $a b c=(7+2 \sqrt{6})(3+2 \sqrt{6})(8+2 \sqrt{6})$ $$ \begin{array}{l} =50(12+5 \sqrt{6}), \\ S=\sqrt{p(p-a)(p-b)(p-c)} \\ =\sqrt{(9+3 \sqrt{6})(2+\sqrt{6})(6+\sqrt{6})(1+\sqrt{6})} \\ =\sqrt{3(3+\sqrt{6})(1+\sqrt{6})(2+\sqrt{6})(6+\sqrt{6})} \\ =\sqrt{3(9+4 \sqrt{6})(18+8 \sqrt{6})} \\ =\sqrt{6}(9+4 \sqrt{6})=3(8+3 \sqrt{6}), \\ R=\frac{a b c}{4 S}=\frac{50(12+5 \sqrt{6})}{4 \times 3(8+3 \sqrt{6})} \\ =\frac{5}{6}(3+2 \sqrt{6}), \\ r=\frac{S}{p}=\frac{3(8+3 \sqrt{6})}{9+3 \sqrt{6}}=\frac{6+\sqrt{6}}{3}. \end{array} $$ Also, $O I^{2}=R^{2}-2 R r$ $=\left[\frac{5}{6}(3+2 \sqrt{6})\right]^{2}-2 \times \frac{5}{6}(3+2 \sqrt{6}) \times$ $$ \begin{aligned} & \frac{6+\sqrt{6}}{3} \\ = & \frac{25(33+12 \sqrt{6})}{36}-\frac{5}{9}(30+15 \sqrt{6}) \\ = & \frac{25(33+12 \sqrt{6})-20(30+15 \sqrt{6})}{36} \\ = & \frac{825-600}{36}=\frac{225}{36}, \end{aligned} $$ Thus, $O I=\sqrt{\frac{225}{36}}=\frac{15}{6}=2.5$. This is a rational number." 65530d37f9a4,"3. In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the maximum value of $n$? Prove your conclusion. In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the maximum value of $n$? Prove your conclusion.",See reasoning trace,medium,"3. If a $9 \times 9$ grid is divided into 9 $3 \times 3$ grids, and each small cell in the same $3 \times 3$ grid is filled with the same number, and the numbers in any two different $3 \times 3$ grids are different, then each row and each column will have exactly three different numbers. Therefore, the maximum value of $n$ is no more than 3. The following proof shows that as long as each row and each column have at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least 3 times in a row and at least 3 times in a column. When a number appears no less than 3 times in a row, mark the cells containing this number in that row with a symbol (1). Since each row has at most three different numbers, there are at most four cells in the same row that are not marked with a symbol (1), and at least five cells are marked with a symbol (1). Therefore, at least $5 \times 9$ cells in the entire grid are marked with a symbol (1). Similarly, when a number appears no less than 3 times in a column, mark the cells containing this number in that column with a symbol (2). By the same reasoning, at least $5 \times 9$ cells in the entire grid are marked with a symbol (2). Since $5 \times 9 + 5 \times 9 > 9 \times 9$, there is at least one cell that is marked with both symbol (1) and symbol (2). Clearly, the number in this cell appears at least 3 times in the row it is in and at least 3 times in the column it is in. In conclusion, the maximum value of $n$ is 3." 550db32099d4,"Nils is playing a game with a bag originally containing $n$ red and one black marble. He begins with a fortune equal to $1$. In each move he picks a real number $x$ with $0 \le x \le y$, where his present fortune is $y$. Then he draws a marble from the bag. If the marble is red, his fortune increases by $x$, but if it is black, it decreases by $x$. The game is over after $n$ moves when there is only a single marble left. In each move Nils chooses $x$ so that he ensures a final fortune greater or equal to $Y$ . What is the largest possible value of $Y$?",\frac{2^n,medium,"To solve this problem, we need to determine the largest possible value of \( Y \) that Nils can ensure as his final fortune after \( n \) moves. We will use induction to prove that \( Y(n) = \frac{2^n}{n+1} \). 1. **Base Case:** - When \( n = 1 \), there is one red marble and one black marble. Nils should bet \( x = 0 \) to ensure his fortune does not decrease, as betting any positive amount could result in a loss if the black marble is drawn. Therefore, \( Y(1) = 1 \). 2. **Inductive Step:** - Assume that for \( k \) red marbles, the largest possible \( Y \) is \( Y(k) = \frac{2^k}{k+1} \). - We need to show that for \( n = k+1 \) red marbles, the largest possible \( Y \) is \( Y(k+1) = \frac{2^{k+1}}{(k+1)+1} = \frac{2^{k+1}}{k+2} \). 3. **Inductive Hypothesis:** - Suppose Nils bets \( x \) when there are \( n \) red marbles and one black marble. The fortune after drawing a red marble will be \( y + x \), and after drawing a black marble, it will be \( y - x \). - The expected fortune after one move should be equal to the largest possible \( Y \) for \( n-1 \) red marbles, which is \( Y(n-1) \). 4. **Expected Value Calculation:** - The expected value of Nils' fortune after one move is given by: \[ \frac{n}{n+1}(y + x) + \frac{1}{n+1}(y - x) \] - Simplifying this, we get: \[ \frac{n(y + x) + (y - x)}{n+1} = \frac{ny + nx + y - x}{n+1} = \frac{(n+1)y + (n-1)x}{n+1} \] 5. **Ensuring Final Fortune \( Y \):** - To ensure a final fortune of \( Y \), we need: \[ \frac{(n+1)y + (n-1)x}{n+1} \geq Y(n-1) \] - Solving for \( x \), we get: \[ (n+1)y + (n-1)x \geq (n+1)Y(n-1) \] \[ (n-1)x \geq (n+1)Y(n-1) - (n+1)y \] \[ x \geq \frac{(n+1)Y(n-1) - (n+1)y}{n-1} \] 6. **Substituting \( Y(n-1) \):** - Using the inductive hypothesis \( Y(n-1) = \frac{2^{n-1}}{n} \), we get: \[ x \geq \frac{(n+1)\frac{2^{n-1}}{n} - (n+1)y}{n-1} \] \[ x \geq \frac{(n+1)2^{n-1} - n(n+1)y}{n(n-1)} \] 7. **Final Calculation:** - To ensure the largest possible \( Y \), we need to maximize \( x \). Setting \( y = \frac{2^n}{n+1} \), we get: \[ x = \frac{2^{n-1} - \frac{2^n}{n+1}}{2^{n-1} + \frac{2^n}{n+1}} \] \[ x = \frac{2^{n-1}(n+1) - 2^n}{2^{n-1}(n+1) + 2^n} \] \[ x = \frac{2^{n-1}n + 2^{n-1} - 2^n}{2^{n-1}n + 2^{n-1} + 2^n} \] \[ x = \frac{2^{n-1}n - 2^{n-1}}{2^{n-1}n + 2^{n-1} + 2^n} \] \[ x = \frac{2^{n-1}(n - 1)}{2^{n-1}(n + 1) + 2^n} \] \[ x = \frac{n - 1}{n + 1 + 2} \] \[ x = \frac{n - 1}{n + 3} \] Thus, the largest possible value of \( Y \) that Nils can ensure is \( \frac{2^n}{n+1} \). The final answer is \( \boxed{ \frac{2^n}{n+1} } \)." a4cecf5c1f02,"(4) If there exists an obtuse angle $\alpha$, such that $\sin \alpha-\sqrt{3} \cos \alpha=\log _{2}\left(x^{2}-x+2\right)$ holds, then the range of real number $x$ is ( ). (A) $\{x \mid-1 \leqslant x<0$ or $12 n$, which is impossible. Therefore, the number of selected numbers cannot exceed $n$." 71c7adef4624,"40. Let $n \geqslant 2$ be a positive integer. Find the maximum value of the constant $C(n)$ such that for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in(0, 1)$ $(i=1,2, \cdots, n)$, and $\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant ib>0)$ with left and right foci $F_{1}, F_{2}$, and the right vertex $A$, and $P$ is any point on the ellipse $C$. It is known that the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is 3, and the minimum value is 2. (1) Find the equation of the ellipse $C$; (2) If the line $l: y=k x+m$ intersects the ellipse $C$ at points $M, N$ (where $M, N$ are not the left or right vertices), and the circle with diameter $M N$ passes through point $A$. Prove that the line $l$ passes through a fixed point, and find the coordinates of this fixed point.",See reasoning trace,medium,"(1) Let $P(a \cos \theta, b \sin \theta), F_{1}(-c, 0), F_{2}(c, 0)$, then $\overrightarrow{P F_{1}}=(-c-a \cos \theta,-b \sin \theta)$, $\overrightarrow{P F_{2}}=(c-a \cos \theta,-b \sin \theta)$, thus $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=a^{2} \cos ^{2} \theta-c^{2}+b^{2} \sin ^{2} \theta=c^{2} \cos ^{2} \theta+b^{2}-c^{2}$, hence $\left\{\begin{array}{l}a^{2}-c^{2}=3, \\ b^{2}-c^{2}=2\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=4, \\ b^{2}=3 .\end{array}\right.\right.$ Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. (2) Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), A(2,0)$, solving the system of equations gives $$ \left\{\begin{array}{l} y=k x+m, \\ 3 x^{2}+4 y^{2}=12 \end{array} \Rightarrow\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-12=0 \Rightarrow x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}, x_{1} x_{2}=\frac{4 m^{2}-12}{3+4 k^{2}}\right. \text {. } $$ Thus, $\overrightarrow{A M} \cdot \overrightarrow{A N}=\left(x_{1}-2\right)\left(x_{2}-2\right)+\left(k x_{1}+m\right)\left(k x_{2}+m\right)$ $$ \begin{array}{l} =\left(1+k^{2}\right) x_{1} x_{2}+(k m-2)\left(x_{1}+x_{2}\right)+m^{2}+4=0 \\ \Rightarrow\left(1+k^{2}\right)\left(4 m^{2}-12\right)-8 k m(k m-2)+\left(m^{2}+4\right)\left(3+4 k^{2}\right)=0 \\ \Rightarrow 7 m^{2}+16 k m+4 k^{2}=(7 m+2 k)(m+2 k)=0 . \end{array} $$ When $7 m+2 k=0$, $m=-\frac{2}{7} k \Rightarrow y=k x-\frac{2}{7} k=k\left(x-\frac{2}{7}\right)$, at this time, the line $y=k x+m$ passes through the fixed point $\left(\frac{2}{7}, 0\right)$; When $m+2 k=0$, $m=-2 k \Rightarrow y=k x-2 k=k(x-2)$, at this time, the line $y=k x+m$ passes through the fixed point $A(2,0)$, which does not meet the requirement, so it is discarded. In summary, the line $l$ passes through a fixed point, and the coordinates of the fixed point are $\left(\frac{2}{7}, 0\right)$." a3e82ae21273,How many integers $a$ divide $2^{4} \times 3^{2} \times 5$?,2^{n}$ possibilities.,medium,"It should be noted here that a divisor $n$ of $2^{4} \times 3^{2} \times 5$ is of the form $n=2^{x} \times 3^{y} \times 5^{z}$. Indeed, if a prime number other than 2, 3, or 5 divided $n$, then it would also divide $2^{4} \times 3^{2} \times 5$, which is not possible. It is also necessary that $x \leqslant 4, y \leqslant 2$, and $z \leqslant 1$, and this will give us all possible divisors. In summary, to create a divisor, we need to 1. Choose $x$ between 0 and 4, so 5 possibilities, 2. Choose $y$ between 0 and 2, so 3 possibilities, 3. Choose $z$ between 0 and 1, so 2 possibilities, This gives a total of $5 \times 3 \times 2=30$ possible divisors. ## Proposition 2. For a set of size $n$, there are $2^{n}$ ""subsets"". Proof. The reasoning is as follows: consider the first element, you have the choice to keep it or leave it out, that is, 2 possibilities. The same for the second element, either you keep it or you leave it out. etc... In total, this gives you $2 \times 2 \times \cdots \times 2=2^{n}$ possibilities." 88e06f9aabc5,"1. Given $P=\{x \mid x=3 k, k \in \mathbf{Z}\}, Q=\{x \mid x=3 k+1, k \in \mathbf{Z}\}, S$ $\{x \mid x=3 k-1, k \in \mathbf{Z}\}$. If $a \in P, b \in Q, c \in S$, then ( ). A. $a+b-c \in P$ B. $a+b-c \in Q$ C. $a+b-c \in S$ D. $a+b-c \in P \cup Q$",See reasoning trace,easy,"1. C. $$ a+b-c=3k+(3m+1)-(3n-1)=3(k+m-n)+2=3(k+m-n+1)-1 \in S . $$" d0d94bc33544,"## Task 4 What is the smallest two-digit number?",See reasoning trace,easy,"The smallest two-digit number is 10. ### 11.6.2 2nd Round 1969, Class 2" 4fccb911679d,"(Application of the algorithm) Find the GCD $d$ of the integers 663 and 182 and integers $u, v \in \mathbb{Z}$ such that $d=663 u+182 v$.","13$ and the solution pair $(u, v)=(11,-3)$.",medium,"Let's apply the 2-step algorithm. First, we set $a=182$ (the smaller of the two numbers) and $b=663$ (the larger). Step $0: a_{0}=b=663, a_{1}=a=182, u_{0}=v_{1}=0$ and $v_{0}=u_{1}=1$ Step 1: $a_{0}=663=3 \times 182+117$ so we set $a_{2}=117$. With $q=3$, we apply the formula of the algorithm to find $u_{2}=u_{0}-q u_{1}=-3$ and $v_{2}=v_{0}-q v_{1}=1$. Step 2: $a_{1}=182=1 \times 117+65$. We set $a_{3}=65, q=1, u_{3}=u_{1}-q u_{2}=4$ and $v_{3}=v_{1}-q v_{2}=-1$. Step $3: a_{2}=117=1 \times 65+52$. We set $a_{4}=52, q=1, u_{4}=u_{2}-q u_{3}=-7$ and $v_{4}=v_{2}-q v_{3}=2$. Step $4: a_{3}=65=1 \times 52+13$. We set $a_{5}=13, q=1, u_{5}=u_{3}-q u_{4}=11$ and $v_{5}=v_{3}-q v_{4}=-3$. Step 5: $a_{4}=52=4 \times 13+0$. Since $a_{6}=0$, we have found the GCD $d=a_{5}=13$ and we know that it can be written as $$ 13=d=a_{5}=a u_{5}+b v_{5}=11 \times 182-3 \times 663 $$ Thus, $d=13$ and the solution pair $(u, v)=(11,-3)$." 48420e5abb0c,"573. Do there exist natural numbers $a, b$, and $c$ such that the decimal representation of the sum $$ a^{60}+b^{60}+c^{60} $$ ends in the digit 9? 574. Prove that the square of a two-digit number $\overline{a b}(b \neq 0)$ ends in $\overline{a b}$ only when $\overline{a b}$ is 25 or 76.",See reasoning trace,medium,"$\triangle$ Numbers $x=\overline{a b}$ and $x^{2}=\overline{a b^{2}}$ end with the same two digits if and only if their difference is divisible by 100: $$ \left(x^{2}-x\right): 100, \quad x(x-1) \vdots 100 $$ Two consecutive natural numbers $x-1$ and $x$ are coprime (see the supporting problem 454 from § 11), so their product is divisible by 100 only when one of them is divisible by 25 and the other by 4. (It is necessary to consider that these factors are two-digit numbers, so none of them can be divisible by 100.) Consider two cases. 1) Let the number $x$ be divisible by 25, and $x-1$ by 4. From the first condition, we find that $x \in\{25 ; 50 ; 75\}$. Of these numbers, only 25 satisfies the second condition. 2) Let $x$ be divisible by 4, and $x-1$ by 25. From the second condition, we get that $(x-1) \in\{25 ; 50 ; 75\}$, i.e., $x \in\{26 ; 51 ; 76\}$. The first condition is satisfied only by 76." fe1cf071d5aa,"10. The solution set of the inequality $\sqrt{a x+b}>x \quad(a>0, b>0)$ is",See reasoning trace,easy,"10. $\left[-\frac{b}{a}, \frac{1}{2}\left(a+\sqrt{a^{2}+4 b}\right)\right)$ From the graph, the solution set is $\left[-\frac{b}{a}, \frac{a+\sqrt{a^{2}+4 b}}{2}\right)$." 49884b857a79,"3. Let the function be $$ \begin{array}{l} f(x, y) \\ =\sqrt{x^{2}+y^{2}-6 y+9}+\sqrt{x^{2}+y^{2}+2 \sqrt{3} x+3}+ \\ \quad \sqrt{x^{2}+y^{2}-2 \sqrt{3} x+3} . \end{array} $$ Then the minimum value of $f(x, y)$ is ( ).",See reasoning trace,easy,"3. C. Let $A(0,3), B(-\sqrt{3}, 0), C(\sqrt{3}, 0)$, $D(0,1), P(x, y)$. Then $f(x, y)=|P A|+|P B|+|P C|$. Notice that, $$ \angle A D B=\angle B D C=\angle C D A=120^{\circ} \text {. } $$ Thus, $D$ is the Fermat point of $\triangle A B C$. Therefore, the minimum value of $f(x, y)$ is achieved when point $P$ is $D$, and the minimum value is $$ |D A|+|D B|+|D C|=6 \text {. } $$" fe1bce3bdd6a,"15. As shown in the figure, the area of square $\mathrm{ABCD}$ is 196 square centimeters, and it contains two partially overlapping smaller squares. The larger of the two smaller squares has an area that is 4 times the area of the smaller one, and the overlapping area of the two squares is 1 square centimeter. Therefore, the area of the shaded part is $\qquad$ square centimeters.","36$, so the area of the two smaller rectangles is $36 \times 2=72\left(\mathrm{~cm}^{2}\right)$.",medium,"【Answer】 72 【Analysis】Key point: Skillful area calculation The area of the square is 196, so the side length is 14. The overlapping area is 1, so the side length is 1; the area of the larger square is 4 times that of the smaller square, so the side length of the larger square is twice that of the smaller square, and the sum of the side lengths of the larger and smaller squares is $14+1=15$. Therefore, the side length of the smaller square is $15 \div 3=5$, and the side length of the larger square is $5 \times 2=10$. The area of the smaller rectangle is $(5-1) \times(10-1)=36$, so the area of the two smaller rectangles is $36 \times 2=72\left(\mathrm{~cm}^{2}\right)$." 14726a734a89,"Through each face of the cube, a plane was drawn. Into how many parts will these planes divide the space?",27\) parts.,easy,"The planes divide the space into 27 parts. Indeed, let's first draw two planes through opposite faces. They will divide the space into three ""layers"". Now, let's draw the remaining four planes. They will cut out nine sections in each layer. (It might be helpful here to look at what happens if you draw lines through the sides of a square on a plane.) In total, we have \(3 \times 9 = 27\) parts." fd4b3ec150ad,"12. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{0}=1, a_{1}=2, a_{n+2}=a_{n}+a_{n+1}^{2}$, then $a_{2006} \equiv$ $\qquad$ $(\bmod 7)$",See reasoning trace,easy,"12. 6 Hint: $a_{0} \equiv 1(\bmod 7), a_{1} \equiv 2(\bmod 7)$, $$ \begin{array}{l} a_{2} \equiv 2^{2}+1 \equiv 5(\bmod 7), a_{3} \equiv 6(\bmod 7), \\ a_{4} \equiv 6(\bmod 7), a_{5} \equiv 0(\bmod 7), \\ a_{6} \equiv 6(\bmod 7), a_{7} \equiv 1(\bmod 7), \\ a_{8} \equiv 2(\bmod 7) . \end{array} $$ Therefore, $a_{n+7} \equiv a_{n}(\bmod 7)$, so $a_{2006} \equiv a_{7 \times 286+4} \equiv a_{4} \equiv 6(\bmod 7)$." e8b8709106fd,"321. $\sqrt{2 x+5}+\sqrt{x-1}=8$. Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. 321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.",10$.,medium,"Solution. Isolate one of the radicals and square both sides of the equation: $$ (\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1) $$ Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side: $$ 16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x $$ Square both sides of the equation again: $(16 \sqrt{x-1})^{2}=(58-x)^{2} ; 256 x-256=3364-116 x+x^{2} ; x^{2}-372 x+3620=0$. Solve the resulting quadratic equation. Since $a=1, b=-372$, $c=3620, \quad D=b^{2}-4 a c=(-372)^{2}-4 \cdot 1 \cdot 3620=138384-14480=123904$; $\sqrt{D}=352$, then $$ x_{1}=\frac{-b-\sqrt{D}}{2 a}=\frac{372-352}{2}=10 ; \quad x_{2}=\frac{-b+\sqrt{D}}{2 a}=\frac{372+352}{2}=362 $$ Check both values $x_{1}=10 ; x_{2}=362$. If $x=10$, then $\sqrt{2 \cdot 10+5}+\sqrt{10-1}=8 ; \sqrt{25}+\sqrt{9}=8 ; 5+3=8$; $8=8$ If $x=362$, then $\sqrt{2 \cdot 362+5}+\sqrt{362-1}=8 ; \sqrt{729}+\sqrt{361}=8 ; 27+$ $+19 \neq 8$; therefore, $x=362$ is an extraneous root. The root of the given equation is $x=10$." 72978342835e,"A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle? $\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$",\textbf{(D),medium,"Solution 1 In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the [Pythagorean theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_theorem) we have $AC=4$. Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\overline{AC}$. Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$. We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$. Hence we get the following two equations: \begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*} Simplifying both, we get \begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*} As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$. Now there are two possibilities: either $\frac{4-s}s=-2$, or $\frac{4-s}s=2$. In the first case clearly $s<0$, which puts the center on the wrong side of $A$, so this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$, this circle turns out to have the same radius as circle $B$, with center directly left of center $B$, and tangent to $B$ directly above center $A$.) The second case solves to $s=\frac 43$. We then have $4r = s^2 = \frac {16}9$, hence $r = \boxed{\frac 49}$. More generally, for two large circles of radius $a$ and $b$, the radius $c$ of the small circle is $c = \frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2} = \frac{1}{\left(1/\sqrt{a}+1/\sqrt{b}\right)^2}$. Equivalently, we have that $1/\sqrt{c} = 1/\sqrt{a} + 1/\sqrt{b}$. Solution 2 The horizontal line is the equivalent of a circle of curvature $0$, thus we can apply [Descartes' Circle Formula](https://artofproblemsolving.com/wiki/index.php/Descartes%27_Circle_Formula). The four circles have curvatures $0, 1, \frac 14$, and $\frac 1r$. We have $2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2$ Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$ \[\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0\] \[\frac{16}{r^2}-\frac{40}{r}+9=0\] \[\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0\] Obviously $r$ cannot equal $4$, therefore $r = \boxed{\frac 49}$. Solution 3 (Basically 1 but less complicated) As in solution 1, in triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem or pythagorean triples in general, we have $AC=4$. Let $r$ be the radius. Let $s$ be the perpendicular intersecting point $S$ and line $BC$. $AC=s$ because $s,$ both perpendicular radii, and $AC$ form a rectangle. We just have to find $AC$ in terms of $r$ and solve for $r$ now. From the Pythagorean theorem and subtracting to get lengths, we get $AC=s=4=\sqrt{(r+1)^2 - (1-r)^2} + \sqrt{(r+4)^2 - (4-r)^2}$, which is simply $4=\sqrt{4r}+\sqrt{16r} \implies \sqrt{r}=\frac{2}{3} \implies r= \boxed{\textbf{(D) } \frac{4}{9}}.$ ~Wezzerwez7254 Video Solution [https://youtu.be/zOwYoFOUg2U](https://artofproblemsolving.comhttps://youtu.be/zOwYoFOUg2U)" c01cb1c3fdb1,,"6$ teachers, and only tea is drunk by $28-23=5$ teachers.",easy,"Solution. Since 16 teachers do not drink either coffee or tea, we get that $50-16=34$ teachers drink either coffee or tea. However, 28 teachers drink tea, and 29 drink coffee, so $(28+29)-34=57-34=23$ teachers drink ![](https://cdn.mathpix.com/cropped/2024_06_05_0c08f972e294b2535969g-05.jpg?height=264&width=397&top_left_y=827&top_left_x=1056) both coffee and tea. Only coffee is drunk by $29-23=6$ teachers, and only tea is drunk by $28-23=5$ teachers." b85fa2e7e60a,Find the largest $n$ such that the last nonzero digit of $n!$ is $1$.,1,medium,"1. We start by noting that the factorial of a number \( n \), denoted \( n! \), is the product of all positive integers up to \( n \). Specifically, \( n! = n \times (n-1) \times (n-2) \times \cdots \times 1 \). 2. The last nonzero digit of \( n! \) is influenced by the factors of 2 and 5 in the factorial. This is because every pair of factors 2 and 5 contributes a trailing zero in the decimal representation of \( n! \). 3. For \( n = 0 \), we have \( 0! = 1 \). The last nonzero digit is 1. 4. For \( n = 1 \), we have \( 1! = 1 \). The last nonzero digit is 1. 5. For \( n \geq 2 \), \( n! \) will always include at least one factor of 2. Since \( n! \) is the product of all integers from 1 to \( n \), and 2 is included in this product, \( n! \) will be even for \( n \geq 2 \). 6. To determine the last nonzero digit of \( n! \) for \( n \geq 2 \), we need to consider the factors of 2 and 5. The presence of these factors will introduce trailing zeros, and we need to look at the remaining product after removing these trailing zeros. 7. For \( n = 2 \), we have \( 2! = 2 \). The last nonzero digit is 2. 8. For \( n = 3 \), we have \( 3! = 6 \). The last nonzero digit is 6. 9. For \( n = 4 \), we have \( 4! = 24 \). The last nonzero digit is 4. 10. For \( n = 5 \), we have \( 5! = 120 \). The last nonzero digit is 2 (after removing the trailing zero). 11. Continuing this process, we observe that for \( n \geq 2 \), the last nonzero digit of \( n! \) is not 1. 12. Therefore, the largest \( n \) such that the last nonzero digit of \( n! \) is 1 is \( n = 1 \). The final answer is \( \boxed{1} \)." 93243eefb0e6,"1. One car covers a distance of 120 km 18 minutes faster than the other. If the first car reduced its speed by 12 km/h, and the second car increased its speed by $10 \%$, they would spend the same amount of time on the same distance. Find the speeds of the cars.","$100 \mathrm{km} / \mathrm{h}, 80 \mathrm{km} / \mathrm{h}$",easy,"Solution: Let $x$ be the speed of the first car, $y$ be the speed of the second car $\frac{120}{x}+\frac{18}{60}=\frac{120}{y}+5, \quad x-12=1.1 y, \quad \frac{40}{x}+\frac{1}{10}=\frac{44}{x-12}, \quad x^{2}-52 x-4800=0, \quad x=100, \quad y=80$ Answer: $100 \mathrm{km} / \mathrm{h}, 80 \mathrm{km} / \mathrm{h}$" adcf1abd0072,"If $\log_{10}2=a$ and $\log_{10}3=b$, then $\log_{5}12=$? $\textbf{(A)}\ \frac{a+b}{a+1}\qquad \textbf{(B)}\ \frac{2a+b}{a+1}\qquad \textbf{(C)}\ \frac{a+2b}{1+a}\qquad \textbf{(D)}\ \frac{2a+b}{1-a}\qquad \textbf{(E)}\ \frac{a+2b}{1-a}$",\textbf{(D),easy,"From the [change of base formula](https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula), $\log_{5}12 = \frac{\log_{10}12}{\log_{10}5}$. For the numerator, $2 \cdot 2 \cdot 3 = 12$, so $\log_{10}12 = \log_{10}2 + \log_{10}2 + \log_{10}3 = 2a+b$. For the denominator, note that $\log_{10}10 = 1$. Thus, $\log_{10}5 = \log_{10}10 - \log_{10}2 = 1 - a$. Thus, the fraction is $\frac{2a+b}{1-a}$, so the answer is $\boxed{\textbf{(D)}}$." ad25be8ed4a5,"5. Positive integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ form a strictly increasing geometric sequence, and satisfy $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=211$, then $a_{1}=$ $\qquad$ , common ratio $q=$ $\qquad$ .","a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>5 a_{1} \Rightarrow a_{1}<42$. Let $q=\frac{m}{n},(m, n)=1$, then $\le",medium,"Since 211 is a prime number $\Rightarrow q$ is not an integer, and $211=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>5 a_{1} \Rightarrow a_{1}<42$. Let $q=\frac{m}{n},(m, n)=1$, then $\left.a_{5}=a_{1} q^{4}=\frac{a_{1} m^{4}}{n^{4}} \Rightarrow n^{4} \right\rvert\, a_{1} \Rightarrow n=2, a_{1}=16$. When $a_{1}=16, n=2$, $a_{2} \geqslant 24, a_{3} \geqslant 36, a_{4} \geqslant 54, a_{5} \geqslant 81 \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \geqslant 211$, equality holds when $q=\frac{3}{2}$, which satisfies the problem." 637306af075b,"5) Simone writes the number 3 on the blackboard, then erases it and replaces it with its square, 9, then erases the 9 and replaces it with its square, 81. She repeats this operation a total of 2006 times: each time she replaces the written number with its square. What is the unit digit of the last number written? (A) 1 , (B) $3, \quad$ (C) 5 , (D) $7, \quad$ (E) 9 .",(A),easy,"5. The answer is (A). The first three terms of the sequence being constructed are 3, 9, 81. We have reached a term whose units digit is 1, so every time we replace the number with its square, the units digit will not change anymore: $1 \times 1=1$. All terms from the third onward will have 1 as the units digit." 42f1b0bca969,"2. In an acute triangle, the interior angles $A, B$ satisfy $\tan A-\frac{1}{\sin 2 A}=\tan B$ and $\cos ^{2} \frac{B}{2}=\frac{\sqrt{6}}{3}$, then $\sin 2 A=$",See reasoning trace,easy,"$$ \begin{array}{l} \tan A-\frac{1}{\sin 2 A}=\tan A-\frac{\tan ^{2} A+1}{2 \tan A}=\frac{\tan ^{2} A-1}{2 \tan A}=-\frac{1}{\tan 2 A}=\tan B \Rightarrow \tan 2 A=-\frac{1}{\tan B} \\ \Rightarrow 2 A=90^{\circ}+B, \text{ so } \sin 2 A=\cos B=2 \cos ^{2} \frac{B}{2}-1=\frac{2 \sqrt{6}-3}{3} . \end{array} $$" 0df5b79c798b,"4. A school's chess team consists of 2 boys and 3 girls. A photographer is taking a group photo, requiring the 5 people to stand in a row, with girls in the middle and boys on the sides. The number of such arrangements is ( ). (A) 2 (B) 4 (C) 5 (D) 6 (E) 12",12 .\end{array},easy,\begin{array}{l}\text { 4. } E \text {. } \\ 2(3!)=12 .\end{array} 0c1070e05b68,"1. Five brothers were dividing their father's inheritance equally. The inheritance included three houses. Since the houses couldn't be divided, the three older brothers took them, and the younger brothers were given money: each of the three older brothers paid $2000. How much did one house cost? ## Solution:",$5000,easy,"Answer: $5000. Since each of the brothers who received money got $3000, the total inheritance was estimated at $15000. Therefore, each house was worth a third of this amount." 034ccd9efab4,Determine the value of $2^{4}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}\right)$.,31,easy,"Solution 1 Multiplying through, we obtain $$ 2^{4}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}\right)=16+\frac{16}{2}+\frac{16}{4}+\frac{16}{8}+\frac{16}{16}=16+8+4+2+1=31 $$ ## Solution 2 Using a common denominator inside the parentheses, we obtain $$ 2^{4}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}\right)=16\left(\frac{16}{16}+\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=16\left(\frac{31}{16}\right)=31 $$ ANSWER: 31" 4cc8070c9098,"1. Find all pairs of primes $p, q$ for which there exists a positive integer a such that $$ \frac{p q}{p+q}=\frac{a^{2}+1}{a+1} $$ (Ján Mazák, Róbert Tóth)",See reasoning trace,medium," Solution. First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $p q$ and $p+q$ are relatively prime: the product $p q$ is divisible by two primes only (namely $p$ and $q$ ), while the sum $p+q$ is divisible by neither of these primes. We will look for a positive integer $r$ which can be a common divisor of both $a+1$ and $a^{2}+1$. If $r \mid a+1$ and, at the same time, $r \mid a^{2}+1$, then $r \mid(a+1)(a-1)$ and also $r \mid\left(a^{2}+1\right)-\left(a^{2}-1\right)=2$, so $r$ can only be one of the numbers 1 and 2 . Thus the fraction $\frac{a^{2}+1}{a+1}$ either is in lowest terms, or will be in lowest terms when reduced by two, depending on whether the integer $a$ is even, or odd. If $a$ is even, we must have $$ p q=a^{2}+1 \quad \text { and } \quad p+q=a+1 $$ The numbers $p, q$ are thus the roots of the quadratic equation $x^{2}-(a+1) x+a^{2}+1=0$, whose discriminant $$ (a+1)^{2}-4\left(a^{2}+1\right)=-3 a^{2}+2 a-3=-2 a^{2}-(a-1)^{2}-2 $$ is apparently negative, so the equation has no solution in the real numbers. If $a$ is odd, we must have (taking into account the reduction by two) $$ 2 p q=a^{2}+1 \quad \text { and } \quad 2(p+q)=a+1 $$ The numbers $p, q$ are thus the roots of the quadratic equation $2 x^{2}-(a+1) x+a^{2}+1=0$, whose discriminant is negative as well. Therefore, there is no pair of distinct primes $p, q$ satisfying the conditions. It remains to analyze the case of $p=q$. Then, so we must have $$ \frac{p \cdot q}{p+q}=\frac{p \cdot p}{p+p}=\frac{p}{2} $$ $$ p=\frac{2\left(a^{2}+1\right)}{a+1}=2 a-2+\frac{4}{a+1} ; $$ this is an integer if and only if $a+1 \mid 4$, i. e. $a \in\{1,3\}$, so $p=2$ or $p=5$. To summarize, there are exactly two pairs of primes satisfying the conditions, namely $p=q=2$ and $p=q=5$. " 1c544170152e,3. A rectangular photograph with dimensions of 30 cm and 4 dm was enlarged several times to create a rectangular billboard. The area of the billboard is 48 m². What are its length and width?,"the width of the billboard is 6 m, the length is 8 m",medium,"3. 4) $30 \times 40=1200\left(\mathrm{~cm}^{2}\right)$ - the area of the photograph; 2) $480000: 1200=400$ (times) - the area of the billboard is 400 times larger than the area of the photograph. Since when each side of a rectangle is increased by $k$ times, its area increases by $k \times k$ times, in our case, each side increases by 20 times $(20 \times 20=400)$ 3) $30 \times 20=600$ (cm) - the width of the billboard; 4) $40 \times 20=800$ (cm) - the length of the billboard. Answer: the width of the billboard is 6 m, the length is 8 m." f1eee241fc65,"a) The measurements of the sides of a rectangle are integers, and the measurement of its area is twice the measurement of its perimeter. What could be the lengths of the sides of the rectangle? - b) The measurements of the edges of a cuboid are integers, two faces of the body are squares, and the measurement of its volume is twice the measurement of its surface area. What could be the lengths of the edges of the cuboid?","2 c, f=d$ for (b).",medium,"a) Denoting the sides of the rectangle by $c$ and $d$ as required, $$ c d=4(c+d), \quad \text { from which } \quad(c-4)(d-4)=16 $$ thus $(c-4)$ and $(d-4)$ have the same sign. However, they cannot be negative, otherwise $c, d \geq 1$, and $(c-4),(d-4) \geq-3$ would imply that their product is at most 9. Therefore, $c-4$ is a positive divisor of 16, and $$ d=4+\frac{16}{c-4} $$ thus $$ \begin{aligned} & c-4=16, \quad 8, \quad 4, \quad 2, \quad 1, \\ & c=20, \quad 12, \quad 8, \quad 6, \quad 5, \\ & d=5, \quad 6, \quad 8, \quad 12, \quad 20 \end{aligned} $$ The last two pairs of $c, d$ values are already obtained in the reverse order of $d, c$, so there are three essentially different pairs of side lengths. b) Since two faces of the solid are squares, one of the three edges meeting at each vertex is equal; denote this by $e$, and the third edge by $f$. Thus, according to the requirement, $$ e^{2} f=4\left(e^{2}+e f+e f\right)=4 e(e+2 f) $$ from which, given $e \neq 0$, similarly to the previous, $$ e f=4 e+8 f, \quad(e-8)(f-4)=32 $$ where $(e-8)$ and $(f-4)$ are both positive, since otherwise $e-8 \geq-7$ and $f-4 \geq-3$ would imply that their product is at most 21. Therefore, $(e-8)$ is a positive divisor of 32, $$ f=4+\frac{32}{e-8} $$ possible value pairs: $$ \begin{array}{rrrrrr} e-8=32, & 16, & 8, & 4, & 2, & 1 \\ e=40, & 24, & 16, & 12, & 10, & 9 \\ f=5, & 6, & 8, & 12, & 20, & 36 \end{array} $$ We found that there are 6 different edge ratios for the rectangular prism, and each corresponds to a unit of length, with one case specifically yielding a cube. (The unit of measurement is significant; the edge pairs $(e, f)$ cannot be simplified, as the two sides of (2) represent quantities of different dimensions, just as the two sides of (1) do - only the numerical values are equal.) Remark. Dividing (1) by $4 c d$ and (2) by $4 e^{2} f$, $$ \frac{1}{c}+\frac{1}{d}=\frac{1}{4}, \quad \text { or } \quad \frac{2}{e}+\frac{1}{f}=\frac{1}{4} $$ From these, we can also arrive at the solutions, and we see that every pair of $c, d$ values for (a) has a corresponding solution $e=2 c, f=d$ for (b)." 40ccd60a8bbc,"24. In square $R S T U$ a quarter-circle arc with centre $S$ is drawn from $T$ to $R$. A point $P$ on this arc is 1 unit from $T U$ and 8 units from $R U$. What is the length of the side of square RSTU? A 9 B 10 C 11 D 12 E 13","5$ or $x=13$, but $x>8$ so the length of the side of the square RSTU is 13 .",medium,"24. E In the diagram $V$ is the point where the perpendicular from $P$ meets TS. Let the side of the square RSTU be $x$. So the radius of the arc from $R$ to $T$ is $x$. Therefore $S P$ has length $x, P V$ has length $x-8$ and $V S$ has length $x-1$. Applying Pythagoras' Theorem to triangle $P V S$ : $$ (x-8)^{2}+(x-1)^{2}=x^{2} \text {. So } x^{2}-16 x+64+x^{2}-2 x+1=x^{2} \text {. } $$ Therefore $x^{2}-18 x+65=0$, so $(x-5)(x-13)=0$. Hence $x=5$ or $x=13$, but $x>8$ so the length of the side of the square RSTU is 13 ." 237ebcfd8fcf,"Let $p$ and $q$ be two given positive integers. A set of $p+q$ real numbers $a_1 90^{\circ}$, which is a contradiction! If $P_{2}$ is inside the convex hull $\Lambda$, let the line $P_{1} P_{2}$ intersect a face $\pi$ of $\Lambda$ at point $P_{2}^{\prime}$, and let the projection of $P_{1}$ on the plane $\pi$ be $H$. Since $P_{2}^{\prime}$ is inside the plane $\pi$, there exists a vertex $P_{i}$ on the plane $\pi$ such that $\angle P_{i} P_{2}^{\prime} H \geqslant 90^{\circ}$. Therefore, $$ P_{1} P_{2}^{\prime 2} + P_{i} P_{2}^{\prime 2} = P_{1} H^{2} + H P_{2}^{\prime 2} + P_{i} P_{2}^{\prime 2} \leqslant P_{1} H^{2} + P_{i} H^{2} = P_{1} P_{i}^{2}, $$ so $\angle P_{1} P_{2} P_{i} > \angle P_{1} P_{2}^{\prime} P_{i} \geqslant 90^{\circ}$, which is a contradiction! Therefore, (a) holds. For (b), consider the plane $p_{ij}$ passing through $Q_{ij} = \overrightarrow{O P}_{i} + \overrightarrow{O P}_{j}$ and perpendicular to the line segment $P_{i}P_{j}$. All points of $\Lambda_{i}$ are on the same side of $p_{ij}$ as $Q_{i}$ (otherwise, there would be an obtuse triangle), and all points of $\Lambda_{j}$ are also on the same side of $p_{ij}$ as $Q_{i}$. Therefore, (b) holds. By (a) and (b), the volume of $\bigcup_{i=1}^{n} \Lambda_{i}$ (which is no greater than the volume of $\Gamma$) is the sum of the volumes of all $\Lambda_{i}(1 \leqslant i \leqslant n)$, which is $n$ times the volume of $\Lambda$. Therefore, $n \leqslant 8$ holds. If $P_{1}, P_{2}, \cdots, P_{n}$ are coplanar, the same argument shows that $n \leqslant 4$ holds. Therefore, $n \leqslant 8$. Hence, the maximum value of $n$ is 8." 26b5e09abb1b,"21. Given a square with side length $x$ inscribed in a right-angled triangle with side lengths $3, 4, 5$, and one vertex of the square coincides with the right-angle vertex of the triangle; a square with side length $y$ is also inscribed in the right-angled triangle with side lengths $3, 4, 5$, and one side of the square lies on the hypotenuse of the right-angled triangle. Then the value of $\frac{x}{y}$ is ( ). (A) $\frac{12}{13}$ (B) $\frac{35}{37}$ (C) 1 (D) $\frac{37}{35}$ (E) $\frac{13}{12}$",See reasoning trace,medium,"21. D. As shown in Figure 4, let $A D=x$, and the square $A F E D$ is inscribed in the right triangle $\triangle A B C$. Then $$ \begin{array}{l} \triangle A B C \backsim \triangle D E C \Rightarrow \frac{D E}{A B}=\frac{C D}{A C} \\ \Rightarrow \frac{x}{4}=\frac{3-x}{3} \Rightarrow x=\frac{12}{7} . \end{array} $$ As shown in Figure 5, let $Q R=y$, and the square $S T Q R$ is inscribed in the right triangle $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then $$ \begin{array}{l} \triangle A^{\prime} B^{\prime} C^{\prime} \backsim \triangle R B^{\prime} Q \backsim \triangle S T C^{\prime} \\ \Rightarrow \frac{4}{3} y+y+\frac{3}{4} y=5 \Rightarrow y=\frac{60}{37} . \\ \text { Hence } \frac{x}{y}=\frac{37}{35} . \end{array} $$" 641789c52edb,"Find the functions $f$ : $\mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$, we have $$ f(x-f(y))=1-x-y $$","f(y)$, we find $f(y)=1-f(0)-y$. Let $c=1-f(0)$, in the initial equation, we have $1-x-y=f(x+y-c)=-(x",easy,"By setting $x=f(y)$, we find $f(y)=1-f(0)-y$. Let $c=1-f(0)$, in the initial equation, we have $1-x-y=f(x+y-c)=-(x+y-c)+c=2 c-x-y$ thus $c=\frac{1}{2}$. The only possible solution is therefore $x \mapsto \frac{1}{2}-x$, which indeed satisfies the initial equation." 11b6127cd788,"Is there a rational pair of numbers $a, b$ for which the following equality holds? $$ \frac{a+b}{a}+\frac{a}{a+b}=b $$",to the question is affirmative,medium,"I. solution. If there exists a suitable pair of numbers $a, b$, then the quotient $a / b=r$ is also a rational number. (It cannot be $b=0$, because then (1) would not be satisfied, as its left side could only be 2.) Substituting $a=b r$ into (1), we get: $$ \frac{r+1}{r}+\frac{r}{r+1}=b $$ From this, we immediately see that every rational number, except for $r=0$ and $r=-1$, gives a suitable pair of numbers $b, a$: $$ b=\frac{r+1}{r}+\frac{r}{r+1}, \quad a=r+1+\frac{r^{2}}{r+1} $$ Pintz János (Budapest, Fazekas M. High School) Remark. The following is even simpler than the above solution. The two terms on the left side are rational numbers and reciprocals of each other (so neither can be 0). Denoting the first by $r^{\prime}$, $$ \begin{aligned} & b=r^{\prime}+\frac{1}{r^{\prime}}, \quad \text { where } \frac{a+b}{a}=r^{\prime}, \text { so } \\ & a=\frac{1}{r^{\prime}-1} \cdot b=\frac{1}{r^{\prime}-1}\left(r^{\prime}+\frac{1}{r^{\prime}}\right), \text { if } r^{\prime} \neq 1 \end{aligned} $$ If $r^{\prime}$ is a rational number different from 0 and 1, then (2) and (3) give rational $b$ and $a$. (It can also be seen that we have obtained all rational solutions.) II. solution. Removing the fractions and rearranging (1) in terms of $a$, we get: $$ (2-b) a^{2}+b(2-b) a+b^{2}=0 $$ For $b$, only values can be used for which the discriminant $r_{1}$ has a rational square root: $$ D=b^{2}(2-b)^{2}-4(2-b) b^{2}=b^{2}\left(b^{2}-4\right)=r_{1}^{2} $$ that is, $$ \begin{aligned} & b^{2}-4=r_{2}^{2} \\ & b^{2}=r_{2}^{2}+4 \end{aligned} $$ where $r_{2}$ is also a rational number. Looking for solutions in integers, only $r_{2}=0, b^{2}=4, b= \pm 2$ are possible. However, $b=2$ leads to a contradiction with (4); for $b=-2$, $a=1$ is the only root, so the pair of numbers $a, b=1,-2$ is a solution, and we can already state that the answer to the question is affirmative. Takács László (Sopron, Széchenyi I. High School) Remarks. 1. Generally, looking for rational $b$ and $r_{2}$ in the form $B / N$, and $R / N$ respectively, where $B, R, N$ are integers, (5) transforms into: $$ R^{2}+(2 N)^{2}=B^{2} $$ According to this, every Pythagorean triple provides a rational solution: $$ b=\frac{u^{2}+v^{2}}{u v},\left(r_{2}=\frac{u^{2}-v^{2}}{u v}\right), a=\ldots=\frac{u^{2}+v^{2}}{v(v-u)} $$ where $u, v$ are integers different from each other and from 0. Takács László 2. However, a single pair of numbers $a, b$ could have been more easily obtained from the rearrangement of (4) in terms of $b$: $$ (1-a) b^{2}+a(2-a) b+2 a^{2}=0 $$ indeed, this equation simplifies to a first-degree equation for $a=1$, and thus its solution is rational: $b=-2$." aaa25416ec9c,"2. Find all real numbers $x$ and $y$ such that $$ x^{2}+2 y^{2}+\frac{1}{2} \leq x(2 y+1) $$",See reasoning trace,easy," Solution: We write the inequality in the form $$ 2 x^{2}+4 y^{2}+1-4 x y-2 x \leq 0 $$ Thus $\left(x^{2}-4 x y+4 y^{2}\right)+\left(x^{2}-2 x+1\right) \leq 0$. Hence $$ (x-2 y)^{2}+(x-1)^{2} \leq 0 $$ Since $x, y$ are real, we know that $(x-2 y)^{2} \geq 0$ and $(x-1)^{2} \geq 0$. Hence it follows that $(x-2 y)^{2}=0$ and $(x-1)^{2}=0$. Therefore $x=1$ and $y=1 / 2$. " bee801537344,"Example 7 As shown in the figure, in the right triangle $\triangle ABC$, it is known that $BC=a$. If the line segment $PQ$ of length $2a$ has point $A$ as its midpoint, for what value of the angle between $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is the value of $\overrightarrow{B P} \cdot \overrightarrow{C Q}$ maximized? Also, find this maximum value.","1$, i.e., $\theta=0$ (which means $\overrightarrow{P Q}$ is in the same direction as $\overrightarro",medium,"Analyze: Combining the diagram, according to the rules of vector operations, transform the problem into solving using function knowledge. Solution: Since $\overrightarrow{A B} \perp \overrightarrow{A C}$, it follows that $\overrightarrow{A B} \cdot \overrightarrow{A C}=0$. Since $\overrightarrow{A P}=-\overrightarrow{A Q}, \overrightarrow{B P}=\overrightarrow{A P}-\overrightarrow{A B}, \overrightarrow{C Q}=\overrightarrow{A Q}-\overrightarrow{A C}$, it follows that $\overrightarrow{B P} \cdot \overrightarrow{C Q}=(\overrightarrow{A P}-\overrightarrow{A B}) \cdot(\overrightarrow{A Q}-\overrightarrow{A C})$ $$ \begin{array}{l} =\overrightarrow{A P} \cdot \overrightarrow{A Q}-\overrightarrow{A P} \cdot \overrightarrow{A C}-\overrightarrow{A B} \cdot \overrightarrow{A Q}+\overrightarrow{A B} \cdot \overrightarrow{A C} \\ =-a^{2}-\overrightarrow{A P} \cdot \overrightarrow{A C}+\overrightarrow{A B} \cdot \overrightarrow{A P} \\ =-a^{2}+\overrightarrow{A P}(\overrightarrow{A B}-\overrightarrow{A C}) \\ =-a^{2}+\frac{1}{2} \overrightarrow{P Q} \cdot \overrightarrow{B C} \\ =-a^{2}+a^{2} \cdot \cos \theta \end{array} $$ Therefore, when $\cos \theta=1$, i.e., $\theta=0$ (which means $\overrightarrow{P Q}$ is in the same direction as $\overrightarrow{B C}$), $\overrightarrow{B P} \cdot \overrightarrow{C Q}$ is maximized, and its maximum value is 0." e51446c2f790,"8. Ryan's car averages 40 miles per gallon of gasoline, Tom's car averages 10 miles per gallon, Ryan and Tom drive their respective cars the same distance. Considering both cars together, on average, they can travel ( ) miles per gallon of gasoline. (A) 10 (B) 16 (C) 25 (D) 30 (E) 40",See reasoning trace,easy,"8. B. $$ \frac{2}{\frac{1}{40}+\frac{1}{10}}=16 $$" 18f6ffcd8383,"2. Given that $p$ and $q$ are both prime numbers, and satisfy $5 p^{2} + 3 q = 59$. Then the triangle with side lengths $p+3, 1-p+q, 2 p+q-4$ is ( ). (A) acute triangle (B) right triangle (C) obtuse triangle (D) isosceles triangle","13^{2}$, the triangle with side lengths $5,12,13$ is a right triangle.",easy,"2.B. Since $5 p^{2}+3 q$ is odd, $p$ and $q$ must be one odd and one even. Since $p$ and $q$ are both prime numbers, one of $p$ and $q$ must be 2. If $q=2$, then $p^{2}=\frac{53}{5}$, which does not satisfy the condition, so we discard it; If $p=2$, then $q=13$. In this case, $$ p+3=5,1-p+q=12,2 p+q-4=13 \text {. } $$ Since $5^{2}+12^{2}=13^{2}$, the triangle with side lengths $5,12,13$ is a right triangle." da87e5c0b004,"1. On a very long road, a race was organized. 20 runners started at different times, each running at a constant speed. The race continued until each runner had overtaken all the slower ones. The speeds of the runners who started first and last were the same, and the speeds of the others were different from theirs and distinct from each other. What could be the number of overtakes, if each involved exactly two people? In your answer, indicate the largest and smallest possible numbers in any order, separated by a semicolon. Example of answer format: $10 ; 20$",notation: $10 ; 20$,easy,"Answer: $18 ; 171$ || $171 ; 18$ || 18,171 || 171,18 || $18 ; 171 ;|| 171 ; 18$ Examples of answer notation: $10 ; 20$" 492f46f64852,"4. Given that $x$ and $y$ are real numbers, $x^{2}-x y+y^{2}=1$. Then the maximum value of $x+2 y$ is $\qquad$ .",-(5 x-4 y)^{2} \leqslant 0$ $\Rightarrow x+2 y \leqslant \frac{2 \sqrt{21}}{3}$ (the coefficients 3 ,easy,"4. $\frac{2 \sqrt{21}}{3}$. Notice that, $3(x+2 y)^{2}-28\left(x^{2}-x y+y^{2}\right)=-(5 x-4 y)^{2} \leqslant 0$ $\Rightarrow x+2 y \leqslant \frac{2 \sqrt{21}}{3}$ (the coefficients 3 and 28 are obtained by the method of undetermined coefficients, which involves treating $x$ as the main variable, and ensuring that the discriminant of $(x+2 y)^{2}-m\left(x^{2}-x y+y^{2}\right)$ is a perfect square)." e401eac7ffa1,"5. One of the mascots for the 2012 Olympic Games is called 'Wenlock' because the town of Wenlock in Shropshire first held the Wenlock Olympian Games in 1850. How many years ago was that? A 62 B 152 C 158 D 162 E 172",See reasoning trace,easy,"Solution: D All you have to do is a subtraction: $2012-1850=162$. For more information about the Wenlock Olympian Games, go to their website: http://www.wenlock-olympian-society.org.uk/" 27ada2e4da29,"A $3$ by $3$ determinant has three entries equal to $2$, three entries equal to $5$, and three entries equal to $8$. Find the maximum possible value of the determinant.",405,medium,"1. To find the maximum possible value of the determinant of a \(3 \times 3\) matrix with three entries equal to \(2\), three entries equal to \(5\), and three entries equal to \(8\), we need to consider the arrangement of these numbers in the matrix. The determinant of a \(3 \times 3\) matrix \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) is given by: \[ \det(A) = aei + bfg + cdh - ceg - bdi - afh \] 2. We need to find an arrangement of the numbers \(2, 5, 8\) that maximizes this determinant. One possible arrangement is: \[ A = \begin{pmatrix} 8 & 5 & 2 \\ 2 & 8 & 5 \\ 5 & 2 & 8 \end{pmatrix} \] 3. Calculate the determinant of this matrix: \[ \det(A) = \begin{vmatrix} 8 & 5 & 2 \\ 2 & 8 & 5 \\ 5 & 2 & 8 \end{vmatrix} \] 4. Using the formula for the determinant of a \(3 \times 3\) matrix: \[ \det(A) = 8 \cdot (8 \cdot 8 - 5 \cdot 2) + 5 \cdot (5 \cdot 5 - 2 \cdot 8) + 2 \cdot (2 \cdot 2 - 8 \cdot 5) \] 5. Simplify each term: \[ 8 \cdot (64 - 10) + 5 \cdot (25 - 16) + 2 \cdot (4 - 40) \] \[ = 8 \cdot 54 + 5 \cdot 9 + 2 \cdot (-36) \] \[ = 432 + 45 - 72 \] \[ = 405 \] 6. Therefore, the maximum possible value of the determinant is: \[ \boxed{405} \]" 84daf82b40b3,"## Task 3 - 241213 Determine all real numbers $x$ for which $2 x-3,5 x-14$ and $\frac{2 x-3}{5 x-14}$ are integers.",See reasoning trace,medium,"For a real number $x$, if the numbers $$ g=2 x-3 \quad ; \quad h=5 x-14 \quad \text { and } \quad k=\frac{2 x-3}{5 x-14}=\frac{g}{h} $$ are integers, then from (1) and (2), $$ 5 g=10 x-15 \quad ; \quad 2 h=10 x-28 $$ and therefore $$ 5 g-2 h=13 $$ Considering the equation $g=h k$ derived from (3), we get $$ (5 k-2) h=13 $$ Thus, $5 k-2$ is a divisor of 13, and therefore one of the numbers $1, -1, 13, -13$. However, only 13 has the form $5 k-2$ with an integer $k$, and thus from (4) we further get $h=1$, and from (2) $5 x=h+14=15, x=3$. Therefore, only $x=3$ can have the required property. Indeed, $2 \cdot 3-3=3, 5 \cdot 3-14=1$, and $\frac{2 \cdot 3-3}{5 \cdot 3-14}=3$ are integers. Thus, only $x=3$ has the required property. Adapted from $[5]$" ad9fea65e403,"Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.",96,medium,"1. **Identify the given information and setup the problem:** - Square \(ABCD\) with side length \(AB = 12\). - Point \(P\) lies on the diagonal \(AC\) such that \(AP > CP\). - \(O_1\) and \(O_2\) are the circumcenters of \(\triangle ABP\) and \(\triangle CDP\) respectively. - \(\angle O_1 P O_2 = 120^\circ\). 2. **Determine the coordinates of the vertices of the square:** - Place the square in the coordinate plane with \(A = (0, 0)\), \(B = (12, 0)\), \(C = (12, 12)\), and \(D = (0, 12)\). 3. **Find the coordinates of point \(P\) on the diagonal \(AC\):** - The equation of diagonal \(AC\) is \(y = x\). - Let \(P = (x, x)\) where \(0 < x < 12\). 4. **Calculate the circumcenters \(O_1\) and \(O_2\):** - The circumcenter of a right triangle is the midpoint of the hypotenuse. - For \(\triangle ABP\), the hypotenuse is \(AB\), so \(O_1\) is the midpoint of \(AB\): \[ O_1 = \left( \frac{0 + 12}{2}, \frac{0 + 0}{2} \right) = (6, 0) \] - For \(\triangle CDP\), the hypotenuse is \(CD\), so \(O_2\) is the midpoint of \(CD\): \[ O_2 = \left( \frac{12 + 0}{2}, \frac{12 + 12}{2} \right) = (6, 12) \] 5. **Use the given angle \(\angle O_1 P O_2 = 120^\circ\) to find \(x\):** - The coordinates of \(O_1\) and \(O_2\) are \((6, 0)\) and \((6, 12)\) respectively. - The coordinates of \(P\) are \((x, x)\). - Calculate the vectors \(\overrightarrow{O_1P}\) and \(\overrightarrow{O_2P}\): \[ \overrightarrow{O_1P} = (x - 6, x), \quad \overrightarrow{O_2P} = (x - 6, x - 12) \] - Use the dot product to find the angle between the vectors: \[ \overrightarrow{O_1P} \cdot \overrightarrow{O_2P} = (x - 6)(x - 6) + x(x - 12) \] \[ = (x - 6)^2 + x(x - 12) \] \[ = x^2 - 12x + 36 + x^2 - 12x \] \[ = 2x^2 - 24x + 36 \] - The magnitudes of the vectors are: \[ |\overrightarrow{O_1P}| = \sqrt{(x - 6)^2 + x^2}, \quad |\overrightarrow{O_2P}| = \sqrt{(x - 6)^2 + (x - 12)^2} \] \[ |\overrightarrow{O_1P}| = \sqrt{2x^2 - 12x + 36}, \quad |\overrightarrow{O_2P}| = \sqrt{2x^2 - 24x + 144} \] - Using the cosine formula for the angle between vectors: \[ \cos(120^\circ) = -\frac{1}{2} \] \[ \frac{2x^2 - 24x + 36}{\sqrt{(2x^2 - 12x + 36)(2x^2 - 24x + 144)}} = -\frac{1}{2} \] - Simplify and solve for \(x\): \[ 2(2x^2 - 24x + 36) = -(2x^2 - 12x + 36)(2x^2 - 24x + 144) \] \[ 4x^2 - 48x + 72 = -(4x^4 - 72x^3 + 432x^2 - 864x + 5184) \] \[ 4x^2 - 48x + 72 = -4x^4 + 72x^3 - 432x^2 + 864x - 5184 \] \[ 4x^4 - 72x^3 + 436x^2 - 912x + 5256 = 0 \] - Solving this quartic equation is complex, but we can use the given form \(AP = \sqrt{a} + \sqrt{b}\) to find \(a\) and \(b\). 6. **Find \(a\) and \(b\) from the given form \(AP = \sqrt{a} + \sqrt{b}\):** - Given \(AP = 6\sqrt{2} + 2\sqrt{6}\), we identify \(a = 72\) and \(b = 24\). The final answer is \(a + b = 72 + 24 = \boxed{96}\)." 0fc9c41f2bf3,"5. A boy wrote the first twenty natural numbers on a piece of paper. He did not like how one of them was written and crossed it out. It turned out that among the 19 remaining numbers, there is a number equal to the arithmetic mean of these 19 numbers. Which number did he cross out? If the",21,medium,"# Solution: The sum of the numbers on the sheet, initially equal to $1+2+3+\ldots+20=210$ and reduced by the crossed-out number, is within the range from 210-20=190 to 210-1=209. Moreover, it is a multiple of 19, as it is 19 times one of the addends. Since among the numbers $190,191,192, \ldots 209$ only 190 and 209 are multiples of 19, either the number $20=210-190$ or $1=210-209$ was erased. In both cases, the arithmetic mean of the numbers remaining on the sheet does not match the erased number. Answer: 21." fdc8ccf95284,"1. Vernonia High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniors playing all three sports, given that twice this number are members of the lacrosse team.",(74+26+2 n)-$ $(17+18+13)+(n)=100+2 n-48+n=52+3 n$. It is easily seen that $n=11$.,easy,"Answer: 11 Solution: Suppose that $n$ seniors play all three sports and that $2 n$ are on the lacrosse team. Then, by the principle of inclusion-exclusion, $85=(74+26+2 n)-$ $(17+18+13)+(n)=100+2 n-48+n=52+3 n$. It is easily seen that $n=11$." a4f18add8ccf,"6. The general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=\tan n \cdot \tan (n-1)$, and for any positive integer $n$, there is $a_{1}+a_{2}+\cdots+a_{n}=A \tan n+B n$ holds, then $A=$ $\qquad$ $. B=$ $\qquad$ .","\frac{1}{\tan 1}, B=-1$ satisfy the requirements of the problem.",medium,"6. $\frac{1}{\tan 1},-1$ $$ \begin{aligned} a_{1}+a_{2}+\cdots+a_{n} & =\sum_{k=1}^{n} \tan k \tan (k-1)=\sum_{k=1}^{n}\left(\frac{\tan k-\tan (k-1)}{\tan 1}-1\right) \\ & =\sum_{k=1}^{n} \frac{\tan k}{\tan 1}-\sum_{k=1}^{n-1} \frac{\tan k}{\tan 1}-n=\frac{\tan n}{\tan 1}-n . \end{aligned} $$ Thus, $A=\frac{1}{\tan 1}, B=-1$ satisfy the requirements of the problem." 74bbe07028e5,40th Putnam 1979,βx + α(1 - x). Then the integral becomes ∫ α β t λ dt/(β - α) = 1/(1 + λ) (β λ+1 - a λ+1 )/(β - α). ,medium,": β β/(β - α) /( e α α/(β - α) ) Let t = βx + α(1 - x). Then the integral becomes ∫ α β t λ dt/(β - α) = 1/(1 + λ) (β λ+1 - a λ+1 )/(β - α). We evaluate the limits of 1/(1 + λ) λ and (β λ+1 - a λ+1 ) λ /(β - α) λ separately. Put k = 1/λ. The first limit is lim 1/(1 + 1/k) k = 1/e. To evaluate the second, note that β x = e x ln β = (1 + x ln β + O(x 2 ) ), so the expression is (1 + 1/k (β ln β - α ln α)/(β - α) + O(1/k 2 ) ) k , which tends to the limit exp( (β ln β - α ln α)/(β - α) ) as k tends to infinity, in other words, β β/(β - α) / α α/(β - α) . 40th Putnam 1979 © John Scholes jscholes@kalva.demon.co.uk 4 Dec 1999" ffd9a83cbdd4,"3 The number of different planes passing through any three vertices of a parallelepiped $V$ is $m$, among which the number of planes that can bisect the volume of $V$ is $n$, then $\frac{n}{m}$ equals A. $\frac{1}{5}$ B. $\frac{1}{4}$ C. $\frac{1}{3}$ D. $\frac{3}{10}$","6$, i.e., $n=6$. The total number of different planes through $V$'s three vertices is $6+3 \times 2+",easy,"3 D. Through $V$'s three vertices and bisecting $V$'s volume, the number of different planes is $3 \times 2=6$, i.e., $n=6$. The total number of different planes through $V$'s three vertices is $6+3 \times 2+8 \times 1=20$, i.e., $m=20$, hence $\frac{n}{m}=\frac{6}{20}=\frac{3}{10}$." a39d4d0080f4,"Let $a_{i}$ and $b_{i}$, where $i \in \{1,2, \dots, 2005 \}$, be real numbers such that the inequality $(a_{i}x-b_{i})^{2} \ge \sum_{j=1, j \not= i}^{2005} (a_{j}x-b_{j})$ holds for all $x \in \mathbb{R}$ and all $i \in \{1,2, \dots, 2005 \}$. Find the maximum possible number of positive numbers amongst $a_{i}$ and $b_{i}$, $i \in \{1,2, \dots, 2005 \}$.",4009,medium,"1. **Define Variables and Rewrite the Inequality:** Let \( n = 2005 \), \( A = \sum_{i=1}^n a_i \), and \( B = \sum_{i=1}^n b_i \). The given inequality is: \[ (a_i x - b_i)^2 \ge \sum_{j \neq i} (a_j x - b_j) \] This can be rewritten as: \[ (a_i x - b_i)^2 \ge \sum_{j \neq i} (a_j x - b_j) \] Expanding and simplifying, we get: \[ a_i^2 x^2 - 2a_i b_i x + b_i^2 \ge (A - a_i)x - (B - b_i) \] Rearranging terms, we obtain: \[ a_i^2 x^2 - (2a_i b_i + A - a_i)x + (b_i^2 + B - b_i) \ge 0 \quad \forall x \in \mathbb{R} \] 2. **Case When \( a_i = 0 \):** If \( a_i = 0 \) for some \( i \), the inequality becomes: \[ b_i^2 \ge \sum_{j \neq i} b_j \] This implies: \[ b_i^2 \ge B - b_i \] Solving for \( b_i \), we get: \[ b_i^2 + b_i - B \ge 0 \] This quadratic inequality holds if \( b_i \) is sufficiently large. However, if \( a_i = 0 \), then \( A = 0 \) must hold, and \( b_i \) must be large enough to satisfy the inequality. 3. **Case When \( a_i \neq 0 \):** If \( a_i \neq 0 \), the inequality is a quadratic in \( x \) with leading coefficient \( a_i^2 > 0 \). For the quadratic to be non-negative for all \( x \), its discriminant must be non-positive: \[ (2a_i b_i + A - a_i)^2 \le 4a_i^2 (b_i^2 + B - b_i) \] Expanding and simplifying, we get: \[ 4a_i^2 b_i^2 + 4a_i b_i (A - a_i) + (A - a_i)^2 \le 4a_i^2 b_i^2 - 4a_i^2 b_i + 4a_i^2 B \] Simplifying further, we obtain: \[ 4a_i A b_i \le 4a_i^2 B - (A - a_i)^2 \] 4. **Summing Up the Inequalities:** If all \( a_i \) are positive, we can divide by \( 4a_i A \) and sum over all \( i \): \[ b_i \le \frac{a_i}{A} B - \frac{(A - a_i)^2}{4a_i A} \] Summing these inequalities for \( i = 1, 2, \ldots, n \), we get: \[ B \le B - \sum_{i=1}^n \frac{(A - a_i)^2}{4a_i A} \] This implies: \[ \sum_{i=1}^n \frac{(A - a_i)^2}{4a_i A} \le 0 \] Since the sum of non-negative terms is zero, each term must be zero, implying \( A = a_i \) for all \( i \). This leads to a contradiction unless \( a_i = 0 \) for all \( i \). 5. **Conclusion:** Therefore, we can have at most one \( a_i \) that is non-positive, and the rest can be positive. This means we can have at most \( 2n - 1 = 4009 \) positive numbers among \( a_i \) and \( b_i \). The final answer is \( \boxed{ 4009 } \)." 1825f24300d7,"6. (20 points) Each of the two baskets contains white and black balls, and the total number of balls in both baskets is 25. A ball is randomly drawn from each basket. It is known that the probability that both drawn balls will be white is 0.54. Find the probability that both drawn balls will be black.",0,medium,"Answer: 0.04. Solution. Let in the $i$-th basket there be $n_{i}$ balls, among which $k_{i}$ are white, $i=1,2$. Then $\frac{k_{1}}{n_{1}} \cdot \frac{k_{2}}{n_{2}}=0.54=\frac{27}{50}$. Therefore, for some natural number $m$, the equalities $k_{1} \cdot k_{2}=27 m, n_{1} \cdot n_{2}=50 m$ hold. One of the numbers $n_{i}$ is divisible by 5, then the other one is also divisible by 5 (since $n_{1}+n_{2}=25$). Let $n_{1} \leq n_{2}$. There are two possible cases: 1) $n_{1}=5, n_{2}=20$. Then $m=2, k_{1} \cdot k_{2}=54$, and $k_{1} \leq 5, k_{2} \leq 20$, so $k_{1}=3, k_{2}=18$. 2) $n_{1}=10, n_{2}=15$. Then $m=3, k_{1} \cdot k_{2}=81$, and $k_{1} \leq 10, k_{2} \leq 15$, so $k_{1}=9, k_{2}=9$. In both cases, the probability of drawing two black balls is $\left(1-\frac{k_{1}}{n_{1}}\right) \cdot\left(1-\frac{k_{2}}{n_{2}}\right)=0.04$. | Points | Grading Criteria | | :---: | :--- | | 20 | The correct answer is obtained with a justified solution. | | 10 | Not all possible cases of ball distribution in the baskets are considered in the solution, or there is no justification that all possible cases have been considered. | | 0 | The solution does not meet any of the above conditions. |" 4b79d2680b32,"3 Draw a line through the focus $F$ of the parabola $y^{2}=8(x+2)$ with an inclination angle of $60^{\circ}$. If this line intersects the parabola at points $A, B$, the perpendicular bisector of chord $A B$ intersects the x-axis at point $P$, then the length of segment $P F$ is $\mathrm{J}($. (A) $\frac{16}{3}$ (B) $\frac{8}{3}$ (C) $\frac{16 \sqrt{3}}{3}$ (I)) $8 \sqrt{3}$",(A),easy,"From the given conditions, we know that $F$ is the origin of coordinates, so the equation of the line on which the chord $AB$ lies is $y=\sqrt{3} x$. Substituting into the equation of the parabola, we get $$ 3 x^{2}-8 x-16=0 $$ Let the midpoint of $AB$ be $E$. Then the x-coordinate of $E$ is $\frac{4}{3}$. Given that the inclination angle of line $AB$ is $60^{\circ}$, we know $|FE|=2 \times \frac{4}{3}=\frac{8}{3},|PF|=2|FE|=\frac{16}{3}$, so the answer is (A)." 7d86eb5354f1,"Example 4.20. Find the particular solution of the equation $y^{\prime \prime}-3 y^{\prime}+2 y=0$, satisfying the given initial conditions $y(0)=1$, $y^{\prime}(0)=-1$.",See reasoning trace,medium,"Solution. Let's write down the characteristic equation $$ r^{2}-3 r+2=0 $$ its roots are $r_{1}=1, r_{2}=2$. Therefore, the general solution has the form: $$ y=C_{1} e^{x}+C_{2} e^{2 x} $$ Next, using the initial conditions, we determine the values of the constants $C_{1}$ and $C_{2}$. For this, we substitute the given values $x=0, y=1$ into the general solution; as a result, we obtain one of the equations connecting $C_{1}$ and $C_{2}$: $$ 1=C_{1}+C_{2} $$ The second equation for $C_{1}$ and $C_{2}$ is obtained as follows. We differentiate the general solution: $$ y^{\prime}=C_{1} e^{x}+2 C_{2} e^{2 x} $$ and substitute the given values $x=0$, $y^{\prime}=-1$ into the obtained expression: $$ -1=C_{1}+2 C_{2} $$ From the system $$ \begin{aligned} & C_{1}+C_{2}=1, \\ & C_{1}+2 C_{2}=-1 \end{aligned} $$ we find $C_{1}=3, C_{2}=-2$. Therefore, the required particular solution has the form $$ y=3 e^{x}-2 e^{2 x} $$ ## 4.8. Linear Non-Homogeneous Differential Equations of the Second Order with Constant Coefficients A linear non-homogeneous differential equation of the second order with constant coefficients has the form: $$ y^{\prime \prime}+p y^{\prime}+q y=f(x) . $$ It differs from the corresponding linear homogeneous equation $$ y^{\prime \prime}+p y^{\prime}+q y=0 . $$ by the presence of some function $f(x)$ on the right-hand side. To find the general solution of equation (1), we first need to find the general solution $\bar{y}$ of equation (2), and then find some particular solution $y^{*}$ of equation (1). Their sum is the general solution of the given non-homogeneous equation (1): $$ y=\bar{y}+y^{*} $$ We will provide the rule for finding a particular solution $y^{*}$ of equation (1) in the following two cases: the right-hand side $f(x)$ has the form $$ f(x)=e^{k x} P_{n}(x) $$ where $P_{n}(x)$ is a polynomial of degree $n$; the right-hand side $f(x)$ has the form $$ f(x)=a \cos \lambda x+b \sin \lambda x $$ Let's consider each of these cases separately. I. Suppose the right-hand side of equation (1) has the form $$ f(x)=e^{k x} P_{n}(x) $$ and the number $k$ is not a root of the characteristic equation $$ r^{2}+p r+q=0 $$ corresponding to the homogeneous equation (2). Then a particular solution of equation (1) should be sought in the form $$ y^{*}=e^{k x} Q_{n}(x), $$ where $Q_{n}(x)$ is some polynomial of the same degree $n$ with undetermined coefficients. If, however, the number $k$ is a root of the characteristic equation (5), then a particular solution of equation (1) should be sought in the form $$ y^{*}=x^{m} e^{k x} Q_{n}(x) $$ where $m$ is the multiplicity of the root $k$ (i.e., $m=1$ if $k$ is a simple root, and $m=2$ if $k$ is a double root). II. Suppose now the right-hand side of equation (1) has the form: $$ f(x)=a \cos \lambda x+b \sin \lambda x $$ and the numbers $\pm \lambda i$ are not roots of the characteristic equation (5). Then a particular solution of equation (1) should be sought in the form $$ y^{*}=A \cos \lambda x+B \sin \lambda x, $$ where $A$ and $B$ are undetermined coefficients. If, however, the complex numbers $\pm \lambda i$ are roots of the characteristic equation (5), then a particular solution of equation (1) should be sought in the form $$ y^{*}=x(A \cos \lambda x+B \sin \lambda x) . $$" 917fd32ff8e5,Evaluate $ \int_0^1 (1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{n \minus{} 1})\{1 \plus{} 3x \plus{} 5x^2 \plus{} \cdots \plus{} (2n \minus{} 3)x^{n \minus{} 2} \plus{} (2n \minus{} 1)x^{n \minus{} 1}\}\ dx.$,n^2,medium,"1. Define the functions \( a_n \) and \( b_n \) as follows: \[ a_n = 1 + x + x^2 + \cdots + x^{n-1} \] \[ b_n = 1 + 3x + 5x^2 + \cdots + (2n-3)x^{n-2} + (2n-1)x^{n-1} \] 2. We need to evaluate the integral: \[ f(n) = \int_0^1 a_n b_n \, dx \] 3. Observe that: \[ a_{n+1} = a_n + x^n \] \[ b_{n+1} = b_n + (2n+1)x^n \] 4. Therefore: \[ a_{n+1} b_{n+1} = (a_n + x^n)(b_n + (2n+1)x^n) \] \[ = a_n b_n + (2n+1)x^{2n} + (2n+1)a_n x^n + b_n x^n \] 5. Simplify the expression: \[ a_{n+1} b_{n+1} = a_n b_n + (2n+1)x^{2n} + [(2n+1)a_n + b_n]x^n \] 6. Notice that: \[ [(2n+1)a_n + b_n]x^n = 2[(n+1) + (n+2)x + \cdots + 2nx^{n-1}]x^n \] 7. Thus: \[ a_{n+1} b_{n+1} - a_n b_n = (x^{2n+1})' + 2(x^{n+1} + x^{n+2} + \cdots + x^{2n})' \] 8. Integrate both sides from 0 to 1: \[ \int_0^1 (a_{n+1} b_{n+1} - a_n b_n) \, dx = \int_0^1 (x^{2n+1})' \, dx + 2 \int_0^1 (x^{n+1} + x^{n+2} + \cdots + x^{2n})' \, dx \] 9. Evaluate the integrals: \[ \int_0^1 (x^{2n+1})' \, dx = x^{2n+1} \Big|_0^1 = 1 \] \[ 2 \int_0^1 (x^{n+1} + x^{n+2} + \cdots + x^{2n})' \, dx = 2 \left( \sum_{k=n+1}^{2n} \int_0^1 x^k \, dx \right) \] \[ = 2 \left( \sum_{k=n+1}^{2n} \frac{1}{k+1} \right) \] 10. Combine the results: \[ f(n+1) - f(n) = 2n + 1 \] 11. Given that \( f(1) = 1 \), we can solve the recurrence relation: \[ f(n) = n^2 \] The final answer is \( \boxed{n^2} \)." d66e088967cd,B3. Determine the domain of the function $f(x)=\ln \frac{1}{\cos x}$ and write down the points where the function $f(x)$ has tangents parallel to the angle bisectors of the even quadrants.,-1$. It follows that $-1=\frac{1}{\frac{1}{\cos x}} \cdot \frac{-1}{\cos ^{2} x}(-\sin x)=\operatorn,medium,"B3. For $f(x)=\ln \frac{1}{\cos x}$ to be defined, it must hold that $\frac{1}{\cos x}>0$ or $\cos x>0$. The domain is thus the union of intervals of the form $\left(-\frac{\pi}{2}+2 k \pi, \frac{\pi}{2}+2 k \pi\right)$ for all integers $k$. The tangent will be parallel to the bisectors of the even quadrants if its slope is -1, i.e., $f^{\prime}(x)=-1$. It follows that $-1=\frac{1}{\frac{1}{\cos x}} \cdot \frac{-1}{\cos ^{2} x}(-\sin x)=\operatorname{tg} x$, so the valid $x$ are those from the domain for which $x \stackrel{\pi}{=}-\frac{\pi}{4}+2 k \pi$. For these $x$, it also holds that $f(x)=\ln \left(\frac{1}{\cos \left(\frac{\pi}{4}\right)}\right)=\frac{1}{2} \ln 2$. We get the points $T\left(-\frac{\pi}{4}+2 k \pi, \frac{1}{2} \ln 2\right)$." 5eb09e07aa47,"2. If $a^{2}+a+1=0$, determine the value of the expression $$ a^{1987}+\frac{1}{a^{1987}} $$",See reasoning trace,easy,"Solution. If $a^{2}+a+1=0$, then $a \neq 1$, so we multiply the given equation by $a-1$ to get $(a-1)\left(a^{2}+a+1\right)=0$, i.e., $a^{3}-1=0$, which means $a^{3}=1$. Therefore, $a^{1987}=\left(a^{3}\right)^{662} a=1^{662} a=1 \cdot a=a$, so $$ a^{1987}+\frac{1}{a^{1987}}=a+\frac{1}{a} $$ From the condition $a^{2}+a+1=0$, we sequentially obtain $a^{2}+1=-a$ and if we divide by $a$, we find $a+\frac{1}{a}=-1$. Finally, $$ a^{1987}+\frac{1}{a^{1987}}=a+\frac{1}{a}=-1 $$" 8ebb28dcfe84,"8. (10 points) Given that the product of 4 prime numbers is 11 times their sum, then their sum is ( ) A. 46 B. 47 C. 48 D. There are no numbers that meet the condition",: D,medium,"8. (10 points) Given that the product of 4 prime numbers is 11 times their sum, then their sum is ( ) A. 46 B. 47 C. 48 D. No numbers meet the condition 【Solution】Let the four prime numbers be $a, b, c, d$. According to the problem: $a b c d$ is a multiple of 11, so one of these 4 prime numbers must be 11. Let $d$ be 11. $a b c d=11(a+b+c+d)$, rearranging gives $a b c=a+b+c+11$ If $a, b, c$ are odd, then $a b c$ is odd, and $a+b+c+11$ is even, which is a contradiction. Therefore, among $a, b, c$, there must be an even prime number 2, let $c=2$. Thus, $2 a b=a+b+2+11$ Since $2 a b$ is even, $a+b+2+11$ must also be even. Therefore, only one of $a, b$ can be odd. So we let $b=2$. $$ \begin{array}{l} 4 a=a+2+2+11 \\ \therefore a=5 \\ a+b+c+d=5+2+2+11=20 \end{array} $$ Therefore, the answer is: D." 49eda30b8a0a,"Sprilkov N.P. Nезнayka does not know about the existence of multiplication and exponentiation operations. However, he has mastered addition, subtraction, division, and square root extraction, and he also knows how to use parentheses. Practicing, Nезнayka chose three numbers 20, 2, and 2 and formed the expression $\sqrt{(2+20) : 2}$. Can he, using exactly the same three numbers 20, 2, and 2, form an expression whose value is greater than 30?",It can,easy,"$\frac{20}{2-\sqrt{2}}=\frac{20(2+\sqrt{2})}{2}=20+10 \sqrt{2}>20+10$. There are other solutions. ## Answer It can." ebd33dccf960,"10. If the function $f(x)=-\frac{1}{2} x^{2}+\frac{13}{2}$ has a minimum value of $2a$ and a maximum value of $2b$ on the interval $[a, b]$, find $[a, b]$.","[1, 3]$ or $\left[-2 - \sqrt{17}, \frac{13}{4}\right]$.",medium,"10. (1) If $0 \leqslant a < b$, then $f(x)$ attains its minimum value $2a$ at $x=a$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}$, yielding $a = -2 - \sqrt{17}$; (3) When $a < b \leqslant 0$, $f(x)$ is monotonically increasing on $[a, b]$, so $f(a) = 2a, f(b) = 2b$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}, 2b = -\frac{1}{2}b^2 + \frac{13}{2}$. However, the roots of the equation $\frac{1}{2}x^2 + 2x - \frac{13}{2} = 0$ have opposite signs, so there is no interval satisfying $a < b < 0$. In summary, $[a, b] = [1, 3]$ or $\left[-2 - \sqrt{17}, \frac{13}{4}\right]$." fbfa513193dc,"The digits from 1 to 9 are each used exactly once to write three one-digit integers and three two-digit integers. The one-digit integers are equal to the length, width and height of a rectangular prism. The two-digit integers are equal to the areas of the faces of the same prism. What is the surface area of the rectangular prism? (A) 176 (B) 184 (C) 186 (D) 198 (E) 212 ## Grade 7",See reasoning trace,medium,"Suppose that the length, or the width, or the height of the rectangular prism is equal to 5 . The product of 5 with any of the remaining digits has a units (ones) digit that is equal to 5 or it is equal to 0 . This means that if the length, or the width, or the height of the rectangular prism is equal to 5 , then at least one of the two-digit integers (the area of a face) has a units digit that is equal to 5 or 0. However, 0 is not a digit that can be used, and each digit from 1 to 9 is used exactly once (that is, 5 cannot be used twice), and so it is not possible for one of the dimensions of the rectangular prism to equal 5 . Thus, the digit 5 occurs in one of the two-digit integers (the area of a face). The digit 5 cannot be the units digit of the area of a face, since this would require that one of the dimensions be 5 . Therefore, one of the areas of a face has a tens digit that is equal to 5 . The two-digit integers with tens digit 5 that are equal to the product of two different one-digit integers (not equal to 5 ) are $54=6 \times 9$ and $56=7 \times 8$. Suppose that two of the dimensions of the prism are 7 and 8 , and so one of the areas is 56 . In this case, the digits 5, 6,7, and 8 have been used, and so the digits 1,2,3,4, and 9 remain. Which of these digits is equal to the remaining dimension of the prism? It cannot be 1 since the product of 1 and 7 does not give a two-digit area, nor does the product of 1 and 8 . It cannot be 2 since the product of 2 and 8 is 16 and the digit 6 has already been used. It cannot be 3 since $3 \times 7=21$ and $3 \times 8=24$, and so the areas of two faces share the digit 2 . It cannot be 4 since $4 \times 7=28$ and the digit 8 has already been used. Finally, it cannot be 9 since $9 \times 7=63$ and the digit 6 has already been used. Therefore, it is not possible for 7 and 8 to be the dimensions of the prism, and thus 6 and 9 must be two of the three dimensions. Using a similar systematic check of the remaining digits, we determine that 3 is the third dimension of the prism. That is, when the dimensions of the prism are 3,6 and 9 , the areas of the faces are $3 \times 6=18$, $3 \times 9=27$, and $6 \times 9=54$, and we may confirm that each of the digits from 1 to 9 has been used exactly once. Since the areas of the faces are 18, 27 and 54, the surface area of the rectangular prism is $2 \times(18+27+54)$ or $2 \times 99=198$. ANSWER: (D) #" 9ce7538d0e21,"4. Let $a, b, c$ be the lengths of the sides of any triangle, and let $s=a+b+c, t=ab+bc+ca$. Then $(\quad)$. (A) $tn$ of positive integers such that $$ 1=\operatorname{gcd}(n+1, m+1)=\operatorname{gcd}(n+2, m+2)=\cdots=\operatorname{gcd}(m, 2 m-n) $$",See reasoning trace,easy," Solution: Observe that $1=\operatorname{gcd}(n+r, m+r)=\operatorname{gcd}(n+r, m-n)$. Thus each of the $m-n$ consecutive positive integers $n+1, n+2, \ldots, m$ is coprime to $m-n$. Since one of these is necessarily a multiple of $m-n$, this is possible only when $m-n=1$. Hence each pair is of the form $(n, n+1)$, where $n \in \mathbb{N}$. " 9944a3e05c11,"8. Given $00)$ intersects the $O x$ axis at point $A$, and the $O y$ axis at point $B$ (see the figure). Find the smallest possible value of the area of triangle $A B O$.",2,medium,"Answer: 2. Solution. The abscissa of point $A$ of intersection with the $O x$ axis: $0=k x+k+1 ; x=$ ![](https://cdn.mathpix.com/cropped/2024_05_06_74e32b870097d181f24bg-1.jpg?height=271&width=231&top_left_y=1966&top_left_x=1729) $-\left(1+\frac{1}{k}\right)$. The ordinate of point $B$ of intersection with the $O y$ axis: $y=k \cdot 0+k+1 ; y=k+1$. Therefore, $S_{A B O}=\frac{1}{2} \cdot O A \cdot O B=\frac{1}{2}(k+1)\left(1+\frac{1}{k}\right)=\frac{1}{2}\left(2+k+\frac{1}{k}\right)$. By the inequality between the arithmetic mean and the geometric mean, $k+\frac{1}{k} \geq 2$, and equality is achieved when $k=1$, so the minimum value of $S_{A B O}$ is also achieved when $k=1$. Thus, the smallest possible area of triangle $A B O$ is 2. Comment. The correct answer is found based on the consideration of an example, and the minimality is not proven - 1 point. The expression for the area of the triangle in terms of $k$ is found - 5 points. If the logic is correct but errors in transformations prevent obtaining the correct answer - 4 points. The statement that the minimum value of the expression $k+\frac{1}{k}$ is achieved when $k=1$ can be used without proof." cb249a6b10a1,"## Task Condition Find the derivative. $y=\ln \cos \frac{2 x+3}{2 x+1}$",See reasoning trace,medium,"## Solution $y^{\prime}=\left(\ln \cos \frac{2 x+3}{2 x+1}\right)^{\prime}=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(\cos \frac{2 x+3}{2 x+1}\right)^{\prime}=$ $=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(-\sin \frac{2 x+3}{2 x+1}\right) \cdot\left(\frac{2 x+3}{2 x+1}\right)^{\prime}=$ $=-\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot\left(1+\frac{2}{2 x+1}\right)^{\prime}=-\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot\left(-\frac{2}{(2 x+1)^{2}} \cdot 2\right)=$ $=\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot \frac{4}{(2 x+1)^{2}}=\frac{4 \operatorname{tg} \frac{2 x+3}{2 x+1}}{(2 x+1)^{2}}$ ## Problem Kuznetsov Differentiation 8-17" b1426b23f332,"## Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$. $$ y=\arcsin \frac{x}{5}, y=\arcsin x, y=\frac{\pi}{2} $$",See reasoning trace,medium,"## Solution Since the $O y$ axis is the axis of rotation, the volume is found using the formula: ![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-36.jpg?height=825&width=1356&top_left_y=1044&top_left_x=561) $$ V=\pi \cdot \int_{a}^{b} x^{2} d y $$ Express $x$ in terms of $y$ and find the limits of integration: $$ \begin{aligned} & y=\arcsin \frac{x}{5} \Rightarrow x=5 \sin y \\ & y=\arcsin x \Rightarrow x=\sin y \end{aligned} $$ From the problem statement, we already have: $y_{2}=\frac{\pi}{2}$. Now, find the lower limit: $$ 5 \sin y=\sin y \Rightarrow y_{1}=0 $$ Find the volume of the solid as the difference in volumes of two solids of revolution: $$ \begin{aligned} & V=\pi \cdot \int_{0}^{\pi / 2}(5 \sin y)^{2} d y-\pi \cdot \int_{0}^{\pi / 2} \sin ^{2} y d y= \\ & =\pi \cdot \int_{0}^{\pi / 2}\left(25 \sin ^{2} y-\sin y\right) d y= \\ & =12 \pi \cdot \int_{0}^{\pi / 2} 2 \sin ^{2} y d y= \\ & =12 \pi \cdot \int_{0}^{\pi / 2}\left(2 \sin ^{2} y-1+1\right) d y= \\ & =12 \pi \cdot \int_{0}^{\pi / 2}(-\cos 2 y+1) d y= \\ & =\left.12 \pi \cdot\left(y-\frac{\sin 2 y}{2}\right)\right|_{0} ^{\pi / 2}=12 \pi \cdot\left(\frac{\pi}{2}-0\right)-(0-0)=6 \pi^{2} \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-18$ "" Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals | Problems for Checking - Last edited: 05:25, June 11, 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 21-19 ## Material from Plusi" b824cca76173,"Let $ABC$ be a triangle and $I$ its incenter. Suppose $AI=\sqrt{2}$, $BI=\sqrt{5}$, $CI=\sqrt{10}$ and the inradius is $1$. Let $A'$ be the reflection of $I$ across $BC$, $B'$ the reflection across $AC$, and $C'$ the reflection across $AB$. Compute the area of triangle $A'B'C'$.",\frac{24,medium,"1. **Identify the points of tangency and reflections:** Let $D, E, F$ be the points of tangency of the incircle with sides $BC, AC,$ and $AB$ respectively. The points $A', B', C'$ are the reflections of the incenter $I$ across the sides $BC, AC,$ and $AB$ respectively. 2. **Similarity of triangles:** Note that triangle $A'B'C'$ is similar to triangle $DEF$ with a ratio of $2:1$. This is because the reflections of the incenter $I$ across the sides of the triangle form a triangle that is homothetic to the contact triangle $DEF$ with a homothety ratio of $2$. 3. **Area of triangle $DEF$:** To find the area of triangle $DEF$, we first need to compute the area of the smaller triangles $EFI, DFI,$ and $DEI$. 4. **Area of triangle $EFI$:** By the Pythagorean theorem, we have: \[ AF = AE = \sqrt{AI^2 - FI^2} = \sqrt{2 - 1} = 1 \] Let $G$ be the intersection of $AI$ and $EF$. Since triangle $FGI$ is similar to triangle $AFI$, we have: \[ \frac{FG}{AF} = \frac{FI}{AI} \implies \frac{FG}{1} = \frac{1}{\sqrt{2}} \implies FG = \frac{1}{\sqrt{2}} \] Similarly, \[ \frac{GI}{FI} = \frac{FI}{AI} \implies \frac{GI}{1} = \frac{1}{\sqrt{2}} \implies GI = \frac{1}{\sqrt{2}} \] Therefore, the area of triangle $EFI$ is: \[ \text{Area of } EFI = \frac{1}{2} \cdot EF \cdot GI = \frac{1}{2} \cdot \frac{2}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \] 5. **Area of triangles $DFI$ and $DEI$:** By similar calculations, we find the areas of triangles $DFI$ and $DEI$: \[ \text{Area of } DFI = \frac{2}{5}, \quad \text{Area of } DEI = \frac{3}{10} \] 6. **Total area of triangle $DEF$:** Summing the areas of the three smaller triangles, we get: \[ \text{Area of } DEF = \frac{1}{2} + \frac{2}{5} + \frac{3}{10} = \frac{6}{5} \] 7. **Area of triangle $A'B'C'$:** Since triangle $A'B'C'$ is similar to triangle $DEF$ with a ratio of $2:1$, the area of $A'B'C'$ is: \[ \text{Area of } A'B'C' = 4 \times \text{Area of } DEF = 4 \times \frac{6}{5} = \frac{24}{5} \] The final answer is $\boxed{\frac{24}{5}}$" 1fa2d8a159bd,"11. (15th ""Hope Cup"" Senior High School Competition) Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{n+1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}$ $\left(n \in \mathbf{N}^{*}\right)$, then $a_{2004}=$ $\qquad$ .",See reasoning trace,medium,"11. Given $a_{1}=1, a_{n+1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}$, we have $$ \begin{array}{l} a_{2}=\frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{-2 \sqrt{3}+4}{2}=2-\sqrt{3}, \\ a_{3}=\frac{\sqrt{3}(2-\sqrt{3})-1}{(2-\sqrt{3})+\sqrt{3}}=\frac{2 \sqrt{3}-4}{2}=-2+\sqrt{3}, \\ a_{1}=\frac{\sqrt{3}(-2+\sqrt{3})-1}{(-2+\sqrt{3})+\sqrt{3}}=\frac{-2 \sqrt{3}+2}{2 \sqrt{3}-2}=-1, \\ a_{5}=\frac{-\sqrt{3}-1}{-1+\sqrt{3}}=-\frac{2 \sqrt{3}+4}{2}=-2-\sqrt{3}, \\ a_{6}=\frac{\sqrt{3}(-2-\sqrt{3})-1}{(-2-\sqrt{3})+\sqrt{3}}=\frac{-2 \sqrt{3}-4}{-2}=2+\sqrt{3}, \\ a_{7}=\frac{\sqrt{3}(2+\sqrt{3})-1}{(2+\sqrt{3})+\sqrt{3}}=\frac{2 \sqrt{3}+2}{2 \sqrt{3}+2}=1 . \end{array} $$ Thus, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with period $T=6$, and $2004 \div 6=334$, so $a_{2004}=a_{6}=2+\sqrt{3}$. Another solution: Let $a_{n}=\tan \theta_{n},-\frac{\pi}{2}<\theta_{n}<\frac{\pi}{2}(n=1.2,3, \cdots)$, using trigonometric formulas, we get $$ \tan \theta_{n+1}=u_{n-1}=\frac{\sqrt{3} a_{n}-1}{a_{n}+\sqrt{3}}=\frac{a_{n}-\frac{1}{\sqrt{3}}}{1+\frac{a_{n}}{\sqrt{3}}}=\frac{\tan \theta_{n}-\tan 30^{\circ}}{1+\tan \theta_{n} \cdot \tan 30^{\circ}}=\tan \left(\theta_{n}-30^{\circ}\right), $$ and $-\frac{\pi}{2}<\theta_{w}<\frac{\pi}{2}$, so $\theta_{n+1}=\theta_{n}-30^{\circ}, n=1,2,3, \cdots$ Thus, $\left\{\theta_{\infty}\right\}$ is an arithmetic sequence with a common difference of $-30^{\circ}$, where $\theta_{1}=\arctan 1=45^{\circ}$, Therefore, $\theta_{n}=45^{\circ}-(n-1) \times 30^{\circ}, n=1,2,3, \cdots$ Hence, $\theta_{2001}=45^{\circ}-2003 \times 30^{\circ}=45^{\circ}-167 \times 360^{\circ}+30^{\circ}=75^{\circ}+167 \times 360^{\circ}$, $$ a_{2004}=\tan \theta_{2004}=\tan 75^{\circ}=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=2+\sqrt{3} . $$" 84aa62cf9360,Oleg Mushkarov,\Varangle B_{1} L B$. We get from $\triangle K B C$ that $B L=\frac{1}{\sqrt{5}}$ and therefore $\ta,medium,"12.4. We may assume that the edges of the cube have length 1 . Denote $M=A_{1} B \cap K B_{1}$ and set $K B=x$. Since $\triangle K B M \sim \triangle A_{1} M B_{1}$ it follows that $\frac{M B}{A_{1} M}=x$. Now using the identity $A_{1} M+M B=\sqrt{2}$, we get $M B=\frac{x \sqrt{2}}{x+1}$. Denote by $V$ the volume of the tetrahedron $K B C B_{1}$. Then $$ V=\frac{1}{3} B B_{1} \cdot S_{K B C}=\frac{x}{6} $$ On the other hand, the altitude of $K B C B_{1}$ ![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-155.jpg?height=630&width=596&top_left_y=141&top_left_x=1115) This and (1) imply that $$ \frac{x}{6}=\frac{x \sqrt{6\left(2 x^{2}+1\right)}}{12(x+1)} $$ and we obtain easily that $x=\frac{1}{2}$. Denote by $L$ the foot of the perpendicular from $B$ to $K C$. Then $K C \perp B L$ and $K C \perp B B_{1}$ which shows that $K C \perp B_{1} L$. Hence $\alpha=\Varangle B_{1} L B$. We get from $\triangle K B C$ that $B L=\frac{1}{\sqrt{5}}$ and therefore $\tan \alpha=\sqrt{5}$." da7c1524cb8b,"Example 3. Find all roots of the equation $$ 2\left|x^{2}+2 x-5\right|=x-1 $$ satisfying the inequality $x<\sqrt{2}$.",See reasoning trace,medium,"Solution. The given equation is equivalent to the following system of equations ![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-037.jpg?height=134&width=1133&top_left_y=624&top_left_x=470) which on the set $(-\infty ; \sqrt{2})$ is equivalent to the following system $$ \left\{\begin{array} { l } { 2 ( x ^ { 2 } + 2 x - 5 ) = x - 1 , } \\ { 1 \leqslant x < \sqrt{2} } \end{array}\right. $$ For $x > 1$, we have the following chain of numerical inequalities: $\sqrt{113} / 4 > 9 / 4 \Leftrightarrow \sqrt{113} > 9 \Leftrightarrow 113 > 81$. The last numerical inequality is true, and therefore the original inequality is also true. From the inequality $\sqrt{113} / 4 - 5 / 4 \geqslant \sqrt{2}$, we have $\sqrt{113} > 4 \sqrt{2} + 5 \Leftrightarrow 113 \geqslant 32 + 40 \sqrt{2} + 25 \Leftrightarrow 56 \geqslant 40 \sqrt{2} \Leftrightarrow 7 \geqslant 5 \sqrt{2} \Leftrightarrow 49 \geqslant 50$. The last inequality is false, and therefore the original inequality is also false; hence, $\frac{\sqrt{113} - 5}{4}$ is less than $\sqrt{2}$. Thus, the condition $1 \leqslant \frac{\sqrt{113} - 5}{4} < \sqrt{2}$ is satisfied, and therefore the number $\frac{\sqrt{113} - 5}{4}$ is a solution to the second system. The number $\frac{-5 - \sqrt{113}}{4}$ is negative, and therefore it does not belong to the interval $[1 ; \sqrt{2})$ and is not a solution to the system. Thus, the second system of the combined system has a unique root $\frac{\sqrt{113} - 5}{4}$. Therefore, the combined system of the two systems, and consequently the original equation, has a unique root, the number $\frac{\sqrt{113} - 5}{4}$." f3cc283ce08b,"2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds $$ f(x-y)=f(x)+f(y)-2xy $$",4052169,easy,"Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$. Answer: 4052169. (C) ![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-02.jpg?height=67&width=1157&top_left_y=1003&top_left_x=498)" 1e7b1cfec48e,"[ Invariants $]$ On the board, the numbers $1,2, \ldots, 20$ are written. It is allowed to erase any two numbers $a$ and $b$ and replace them with the number $a b + a + b$. What number can remain on the board after 19 such operations?",$21!-1$,easy,"Note that $a b+a+b+1=(a+1)(b+1)$. This means that if all numbers are increased by 1, the product of the new numbers under the specified operation does not change. At the beginning (and therefore at the end), it is equal to $21!$. ## Answer $21!-1$." 78364ce51aaa,"8. As shown in Figure 1, let $P$ be a point in the plane of $\triangle A B C$, and $$ A P=\frac{1}{5} A B+\frac{2}{5} A C \text {. } $$ Then the ratio of the area of $\triangle A B P$ to the area of $\triangle A B C$ is ( ). (A) $\frac{1}{5}$ (B) $\frac{1}{2}$ (C) $\frac{2}{5}$ (D) $\frac{2}{3}$",See reasoning trace,medium,"8. C. As shown in Figure 4, extend $A P$ to $E$ such that $A P = \frac{1}{5} A E$. Connect $B E$, and draw $E D$ $/ / B A$ intersecting the extension of $A C$ at point $D$. From $A P = \frac{1}{5} A B + \frac{2}{5} A C^{\circ}$, we get $A C = C D$. Therefore, quadrilateral $A B E D$ is a parallelogram. Thus, $\frac{S_{\triangle A B P}}{S_{\triangle B E}} = \frac{1}{5}$. $$ \text{Also, } \frac{S_{\triangle A B E}}{S_{\triangle A B C}} = \frac{\frac{1}{2} S_{\square A B S D}}{\frac{1}{4} S_{\square \triangle A B E D}} = 2 \text{, so } \frac{S_{\triangle A B P}}{S_{\triangle B C}} = \frac{2}{5} \text{. } $$" 75a39d2d48cc,5. [4] Call a positive integer $n$ weird if $n$ does not divide $(n-2)$ !. Determine the number of weird numbers between 2 and 100 inclusive.,26$ weird numbers.,medium,"Answer: 26 We claim that all the weird numbers are all the prime numbers and 4 . Since no numbers between 1 and $p-2$ divide prime $p,(p-2)$ ! will not be divisible by $p$. We also have $2!=2$ not being a multiple of 4 . Now we show that all other numbers are not weird. If $n=p q$ where $p \neq q$ and $p, q \geq 2$, then since $p$ and $q$ both appear in $1,2, \ldots, n-2$ and are distinct, we have $p q \mid(n-2)!$. This leaves the only case of $n=p^{2}$ for prime $p \geq 3$. In this case, we can note that $p$ and $2 p$ are both less than $p^{2}-2$, so $2 p^{2} \mid(n-2)$ ! and we are similarly done. Since there are 25 prime numbers not exceeding 100 , there are $25+1=26$ weird numbers." cb06a8f5e9db,"2. A cruciform labyrinth represents the intersection of two segments, each 2 kilometers long, with a common midpoint. A nearsighted policeman on a defective motorcycle is chasing a gangster. The policeman's speed is 20 times that of the gangster, but he can only see 1 meter (in all directions), and his motorcycle can turn (including making a U-turn) no more than 60 times. Will the policeman be able to catch the gangster?","1000$, and $\frac{6}{n-7}<\frac{1}{2}$. This means that in 10 such operations, we will either catch ",medium,"2. Solution. Let's conduct the reasoning in the general case. Let the lengths of all corridors be $a$, and the speed of the policeman be $n$ times the speed of the cyclist. We will drive around all the corridors (this will take five turns), and return to the center. If the gangster is not caught, it means he has moved from one corridor to another, and consequently, he has moved away from the center by no more than $\frac{6 a}{n}$. Now, let's choose $x$ such that by driving around distances $x$ from the center, we would definitely catch the gangster if he does not pass through the center. For this, it is sufficient to satisfy the inequality $\frac{6 a}{n}+\frac{7 x}{n} \leqslant x$, from which $x \geqslant \frac{6 a}{n-7}$. Thus, each time the length of the corridor that needs to be driven around is multiplied by $\frac{6}{n-7}$. In our case, $a=1000$, and $\frac{6}{n-7}<\frac{1}{2}$. This means that in 10 such operations, we will either catch the gangster or he will be at a distance from the center not exceeding $1000 \cdot \frac{1}{2^{1} 0}<1$. Therefore, being in the center, the policeman will see him." eee5ac2cd290,"Solve the following equation: $$ \sqrt{2 x+8}-\sqrt{x+5}=7 . $$",See reasoning trace,medium,"$$ \sqrt{2 x+8}=7+\sqrt{x+5}: $$ Squaring and simplifying $$ x-46=14 \sqrt{x+5} $$ Squaring again $$ x^{2}-92 x+2116=196 x+980 $$ which is $$ x^{2}-288 x+1136=0 $$ from which $$ x_{1}=284\left[x_{2}=4\right] $$ Since the roots of the equation obtained by squaring may include those that do not satisfy the original equation (it is only true that every root of the original equation satisfies the equation obtained by squaring, but not vice versa), it must be checked by substitution which roots satisfy the original equation as well. In this case, we find that 284 satisfies $(\sqrt{576}-\sqrt{289}=24-17=7)$, while 4 does not satisfy our equation. $$ (\sqrt{16}-\sqrt{9}=4-3 \neq 7 . \quad \text { However, } 4 \text { would satisfy the } \sqrt{2 x+8}+\sqrt{x+5}=7 $$ equation, which 284 does not satisfy.) Magyar Károly (Debrecen, 3rd grade chemical engineering technology II. class)" 44598df45f6b,"Example 8 As shown in the figure, given $A(0, a), B(0, b), (00$, then $\overrightarrow{C A}=(0, a)-(x, 0)=(-x, a), \overrightarrow{C B}=(0, b)-(x, 0)=(-x, b)$. So $\overrightarrow{C A} \cdot \overrightarrow{C B}=x^{2}+a b$ $$ |\overrightarrow{C A}|=\sqrt{x^{2}+a^{2}}, |\overrightarrow{C B}|=\sqrt{x^{2}+b^{2}} $$ Therefore, $\cos \angle A C B=\frac{\overrightarrow{C A} \cdot \overrightarrow{C B}}{|\overrightarrow{C A}||\overrightarrow{C B}|}=\frac{x^{2}+a b}{\sqrt{x^{2}+a^{2}} \cdot \sqrt{x^{2}+b^{2}}}$ Let $t=x^{2}+a b$, then $x^{2}=t-a b$ Thus, $\cos \angle A C B=\frac{t}{\sqrt{\left(t-a b+a^{2}\right)\left(t-a b+b^{2}\right)}}$ $$ \begin{array}{l} =\frac{t}{\sqrt{t^{2}+(a-b)^{2} t-a b(a-b)^{2}}} \\ =\frac{1}{\sqrt{-a b(a-b)^{2} \cdot \frac{1}{t^{2}}+(a-b)^{2} \cdot \frac{1}{t}+1}} \end{array} $$ When $\frac{1}{t}=\frac{1}{2 a b}$, i.e., $t=2 a b$, $\cos \angle A C B$ achieves its minimum value $\frac{2 \sqrt{a b}}{a+b}$. That is, when $x=\sqrt{a b}$, the maximum value of $\angle A C B$ is $\arccos \frac{2 \sqrt{a b}}{a+b}$." 06363a649c1f,"35. On the coordinate plane, there is a line $l, O \perp l$. Consider a vector $\overrightarrow{O P}=\vec{p}$, where $P \in l$; then $p=|\vec{p}|$ is the distance from the line to the origin. Denote by $\varphi$ the angle that $\vec{e}=\frac{\vec{p}}{p}$, the unit vector of $\vec{p}$, forms with the unit vector $\vec{i}$. Let $\vec{r}=\overrightarrow{O M}$ be the variable radius vector of an arbitrary point $M \in l$. Taking into account that the vector $\overrightarrow{P M}=\vec{r}-\vec{p}$ is perpendicular to the vector $\vec{p}$, derive the vector equation of the line $l$ in the form $\vec{r} \cdot \vec{e}=\vec{p} \cdot \vec{e}$ and then from this, obtain the normal equation of the line $l$ in coordinates: $x \cos \varphi + y \sin \varphi = p$. Discuss the cases of different positions of the line relative to the coordinate axes.","0$, it means that the line passes through the origin, forming an angle of $\varphi+90^{\circ}$ with ",medium,"35. Since vectors $\vec{p}$ and $\vec{r}-\vec{p}$ are perpendicular, we have: $(\vec{r}-\vec{p}) \vec{p}=0$. From this, given that $\vec{p}=\overrightarrow{p e}$, we obtain: $$ \overrightarrow{r e}=p $$ Here $\vec{r}=x \vec{i}+y \vec{j}, \vec{e}=\vec{i} \cos \varphi+\vec{j} \sin \varphi$. Multiplying, we get: $$ x \cos \varphi+y \sin \varphi=p $$ If in this equation $p=0$, it means that the line passes through the origin, forming an angle of $\varphi+90^{\circ}$ with the $O x$ axis. If $p \neq 0, \varphi=0$, the equation takes the form: $x=p$ - this is the equation of a line parallel to the $O y$ axis and passing at a distance $p$ from the origin. For $\varphi=90^{\circ}$, we have: $y=p$ - the equation of a line parallel to the $O x$ axis and passing at a distance $p$ from the origin. Finally, the equations $y=0$ and $x=0$ are the equations of the $O x$ and $O y$ axes." ee2c4bd68532,"4. We call a set of professors and committees on which they serve a university if (1) given two distinct professors there is one and only one committee on which they both serve, (2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and (3) there are at least two professors on each committee; there are at least two committees. What is the smallest number of committees a university can have?",6$ committees. This minimum is attainable - just take four professors and let any two professors for,medium,"Solution: Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least $2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee." aaa9182b8920,"## Task B-2.1. Mara has decided to reshape her rectangular flower bed into a square flower bed. If she chooses one side of the rectangular flower bed as the side of the square flower bed, the double area of the resulting square will be $12 \mathrm{~m}^{2}$ larger than the area of the rectangle. If she chooses the other side of the rectangular flower bed as the side of the square flower bed, the sum of the area of the square and twice the area of the rectangle will be $16 \mathrm{~m}^{2}$. What are the dimensions of Mara's flower bed?",See reasoning trace,medium,"## First solution. Let $x$ and $y$ be the lengths of the sides of the rectangle. Then we have $$ \begin{aligned} & 2 x^{2}=x y+12 \\ & y^{2}+2 x y=16 \end{aligned} $$ If we express $y=2 x-\frac{12}{x}$ from the first equation and substitute it into the second, we get the equation $$ \left(2 x-\frac{12}{x}\right)^{2}+2 x\left(2 x-\frac{12}{x}\right)=16 $$ which simplifies to $$ 4 x^{2}-48+\frac{144}{x^{2}}+4 x^{2}-24-16=0 $$ After rearranging, we obtain the biquadratic equation $$ x^{4}-11 x^{2}+18=0 $$ Then $x_{1}^{2}=9, x_{2}^{2}=2$. If $x=\sqrt{2}$, then $y=-4 \sqrt{2}$, which is impossible. Therefore, $x=3$, and $y=2$. The flower bed has one side of length $3 \mathrm{~m}$, and the other $2 \mathrm{~m}$." 9727454ab220,9.4. How many three-digit natural numbers $n$ exist for which the number $n^{3}-n^{2}$ is a perfect square,22,easy,"Answer: 22. Solution. Let $n^{3}-n^{2}=m^{2}$ for some natural number $m$. Then $n^{2}(n-1)=m^{2}$, and therefore $n-1$ must also be a perfect square: $n-1=a^{2}$. Thus, $n=a^{2}+1$ and $m=\left(a^{2}+1\right) a$. Therefore, we need to find all three-digit numbers $n$ that are one more than perfect squares. Such numbers are $101=10^{2}+1, 122=11^{2}+1, \ldots, 962=31^{2}+1$ (the next square $32^{2}=1024-$ is already too large). ## 9th grade" 414178436d76,"## Zadatak A-4.2. (4 boda) Dana je elipsa čija je jednadžba $x^{2}+4 y^{2}=36$. Kružnica $k$ ima središte u točki $(0,3)$ i prolazi žarištima dane elipse. Odredi sva sjecišta kružnice $k$ s elipsom.",See reasoning trace,medium,"## Rješenje. ![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-24.jpg?height=805&width=939&top_left_y=543&top_left_x=570) Iz jednadžbe je vidljivo da su poluosi elipse $a=6, b=3$. Budući da je točka $(0,3)$ na elipsi, zbroj udaljenosti te točke od žarišta je $2 a=12$. Odavde slijedi da je polumjer kružnice $k$ jednak 6 . Jednadžba kružnice $k$ je $x^{2}+(y-3)^{2}=36$. Tražimo rješenja sustava $$ \begin{array}{r} x^{2}+4 y^{2}=36 \\ x^{2}+(y-3)^{2}=36 \end{array} $$ Oduzmemo li te dvije jednadžbe dobivamo $4 y^{2}-(y-3)^{2}=0$. Odavde dobivamo $4 y^{2}=(y-3)^{2}$, tj. $2 y= \pm(y-3)$. Imamo dva rješenja $y_{1}=-3, y_{2}=1$. Iz jednadžbe elipse dobivamo $x_{1}=0$ i $x_{2,3}= \pm \sqrt{36-4}= \pm 4 \sqrt{2}$. Tražene točke su $(0,-3),(4 \sqrt{2}, 1),(-4 \sqrt{2}, 1)$. Varijacija. (Drugi dokaz da polumjer kružnice iznosi 6.) Budući da je $\sqrt{a^{2}-b^{2}}=\sqrt{27}=3 \sqrt{3}$, zaključujemo da se žarišta elipse nalaze u točkama $(3 \sqrt{3}, 0)$ i $(-3 \sqrt{3}, 0)$. Odavde slijedi da je polumjer kružnice $k$ jednak $\sqrt{27+9}=6$. " 7979db7db870,"## Task A-1.1. Determine all pairs of natural numbers $(m, n)$ that satisfy the equation $$ m(m-n)^{2}(m+n)=m^{4}+m n^{3}-99 n $$",See reasoning trace,medium,"## First Solution. By simplifying the expression, we have in sequence: $$ \begin{aligned} m\left(m^{2}-2 m n+n^{2}\right)(m+n) & =m\left(m^{3}+n^{3}\right)-99 n, & & \\ m\left(m^{2}-2 m n+n^{2}\right)(m+n) & =m(m+n)\left(m^{2}-m n+n^{2}\right)-99 n, & & 3 \text { points } \\ 99 n & =m(m+n)(m n), & & \\ 99 & =m^{2}(m+n) . & & 4 \text { points } \end{aligned} $$ The prime factorization of the number 99 is $99=3 \cdot 3 \cdot 11$. The only divisors of 99 that are also squares are 1 and 9. Therefore, we have two cases: if $m=1$, then $m+n=99$, i.e., $n=98$; and if $m=3$, then $m+n=11$, i.e., $n=8$. Thus, the solutions for $(m, n)$ are $(1,98)$ and $(3,8)$." cec0c3305ba1,"The sides of a parallelogram are 8 and 3; the bisectors of two angles of the parallelogram, adjacent to the longer side, divide the opposite side into three parts. Find each of them.","$3,2,3$",easy,"Each of the specified bisectors cuts off an isosceles triangle from the parallelogram. ## Solution Let $B K$ and $C M$ be the bisectors of angles $B$ and $C$ of parallelogram $A B C D$, where $A D=B C=8, A B=C D=3$. Since $\angle A B K=\angle C B K=\angle A K B$, triangle $A B K$ is isosceles. Therefore, $A K=A B=3$. Similarly, $M D=D C=3$. ## Answer $3,2,3$." 8b0d8514b925,"1. Oleg has 550 rubles, and he wants to give his mother tulips for March 8, and there must be an odd number of them, and no color shade should be repeated. In the store where Oleg came, one tulip costs 49 rubles, and there are eleven shades of flowers available. How many ways are there for Oleg to give his mother flowers?",1024,medium,"Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 11. 1st method Using the property of binomial coefficients $$ \mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1} $$ and also considering their combinatorial meaning, we get that the number of ways to form a bouquet from an odd number of flowers of no more than 11 shades (with the condition that no shade should be repeated) is: $$ \mathrm{C}_{11}^{1}+\mathrm{C}_{11}^{3}+\mathrm{C}_{11}^{5}+\cdots+\mathrm{C}_{11}^{11}=2^{10}=1024 $$ 2nd method Consider 10 flowers of 10 different shades. A bouquet can be assembled from these flowers without considering the parity in $2^{10}$ ways. If the bouquet has an odd number of flowers, we keep it; if it has an even number, we add the unused eleventh flower. Thus, the total number of ways to assemble a bouquet is $2^{10}$. Answer: 1024." 866b232811cb,1. (5 points) Calculate: $928+217+64+8=$,See reasoning trace,easy,"【Solution】Solution: $928+217+64+8$, $$ \begin{array}{l} =928+217+72, \\ =928+72+217, \\ =1000+217, \\ =1217 . \end{array} $$" 9b7c0993b171,"Henry starts with a list of the first 1000 positive integers, and performs a series of steps on the list. At each step, he erases any nonpositive integers or any integers that have a repeated digit, and then decreases everything in the list by 1. How many steps does it take for Henry's list to be empty? [i]Proposed by Michael Ren[/i]",11,medium,"1. **Initial List and Conditions**: - Henry starts with the list of the first 1000 positive integers: \( \{1, 2, 3, \ldots, 1000\} \). - At each step, he erases any nonpositive integers or any integers that have a repeated digit. - After erasing, he decreases every remaining integer in the list by 1. 2. **Understanding the Erasure Condition**: - Nonpositive integers are erased immediately. - Integers with repeated digits are also erased. For example, numbers like 11, 22, 101, etc., will be erased. 3. **Effect of Decreasing by 1**: - After each step, every integer in the list is decreased by 1. This means that the list will eventually contain nonpositive integers, which will be erased. 4. **Key Observation**: - The critical observation is to determine the maximum number of steps it takes for any integer to be erased. - Consider the number 10. It will be decreased to 9, 8, 7, ..., 1, 0, and then -1. This process takes exactly 11 steps. 5. **Erasure of Numbers with Repeated Digits**: - Numbers like 11, 22, 33, ..., 99, 101, 111, etc., will be erased as soon as they appear in the list. - For example, 11 will be erased in the first step, 22 in the second step, and so on. 6. **General Case**: - For any number \( n \) in the list, if it does not have repeated digits, it will be decreased step by step until it becomes nonpositive. - The maximum number of steps required for any number to become nonpositive is determined by the number 10, which takes 11 steps. 7. **Conclusion**: - Since the number 10 takes exactly 11 steps to be erased, and all other numbers will be erased in fewer or equal steps, the entire list will be empty after 11 steps. \[ \boxed{11} \]" 0a8847d8c956,"5. As shown in Figure 3, the radii of the three semicircles are all $R$, and their centers $C_{1}, C_{2}, C_{3}$ are on the same straight line, with each center lying on the circumference of the other semicircle. $\odot C_{4}$ is tangent to these three semicircles. Let $r$ represent the radius of $\odot C_{4}$. Then, $R: r$ equals ( ). (A) $15: 4$ (B) $11: 3$ (C) $4: 1$ (D) $3: 1$",4: 1$.,easy,"5.C. As shown in Figure 7, connect $C_{1} C_{4}$ and $C_{2} C_{4}$. Then $C_{2} C_{4}=R-r$, $$ \begin{array}{r} C_{1} C_{4}=R+r, C_{1} C_{2}=R . \text { Therefore, } \\ (R-r)^{2}+R^{2}=(R+r)^{2} . \end{array} $$ Solving this, we get $R^{2}=4R$. Thus, $R: r=4: 1$." 9973c8f30db5,"One. (20 points) Let $x, y$ be non-zero real numbers, and satisfy $\frac{x \sin \frac{\pi}{5} + y \cos \frac{\pi}{5}}{x \cos \frac{\pi}{5} - y \sin \frac{\pi}{5}} = \tan \frac{9 \pi}{20}$. (1) Find the value of $\frac{y}{x}$; (2) In $\triangle ABC$, if $\tan C = \frac{y}{x}$, find the maximum value of $\sin 2A + 2 \cos B$.","\frac{1}{2}$, i.e., $\angle B=\frac{\pi}{3}$, $\sin 2 A+$ $2 \cos B$ achieves its maximum value of $",medium,"(1) Let $\frac{y}{x}=\tan \theta$. Then from the given information, $$ \begin{array}{c} \frac{\tan \frac{\pi}{5}+\tan \theta}{1-\tan \theta \cdot \tan \frac{\pi}{5}}=\tan \frac{9 \pi}{20} \\ \Rightarrow \tan \left(\theta+\frac{\pi}{5}\right)=\tan \frac{9 \pi}{20} \\ \Rightarrow \theta+\frac{\pi}{5}=k \pi+\frac{9 \pi}{20} \\ \Rightarrow \theta=k \pi+\frac{\pi}{4}(k \in \mathbf{Z}) . \end{array} $$ $$ \text { Hence } \frac{y}{x}=\tan \theta=\tan \left(k \pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4}=1 \text {. } $$ (2) From (1), we get $\tan C=1$. Since $0<\angle C<\pi$, therefore, $\angle C=\frac{\pi}{4}$. Then $\angle A+\angle B=\frac{3 \pi}{4} \Rightarrow 2 \angle A=\frac{3 \pi}{2}-2 \angle B$. Thus, $\sin 2 A+2 \cos B=\sin \left(\frac{3 \pi}{2}-2 B\right)+2 \cos B$ $$ \begin{array}{l} =-\cos 2 B+2 \cos B=-2 \cos ^{2} B+2 \cos B+1 \\ =-2\left(\cos B-\frac{1}{2}\right)^{2}+\frac{3}{2} . \end{array} $$ Therefore, when $\cos B=\frac{1}{2}$, i.e., $\angle B=\frac{\pi}{3}$, $\sin 2 A+$ $2 \cos B$ achieves its maximum value of $\frac{3}{2}$." 025a3ef988b6,"Find all triples $(x,y,z)$ of positive integers, with $z>1$, satisfying simultaneously that \[x\text{ divides }y+1,\quad y\text{ divides }z-1,\quad z\text{ divides }x^2+1.\]","(1, 1, 2)",medium,"We are given the following conditions for positive integers \(x, y, z\) with \(z > 1\): 1. \(x \text{ divides } y + 1\) 2. \(y \text{ divides } z - 1\) 3. \(z \text{ divides } x^2 + 1\) We need to find all such triples \((x, y, z)\). 1. **Express \(y\) in terms of \(x\):** Since \(x \mid (y + 1)\), we can write: \[ y + 1 = kx \quad \text{for some integer } k \] Therefore, \[ y = kx - 1 \] 2. **Express \(z\) in terms of \(y\):** Since \(y \mid (z - 1)\), we can write: \[ z - 1 = my \quad \text{for some integer } m \] Therefore, \[ z = my + 1 \] Substituting \(y = kx - 1\) into the equation for \(z\), we get: \[ z = m(kx - 1) + 1 = mkx - m + 1 \] 3. **Substitute \(z\) into the third condition:** Since \(z \mid (x^2 + 1)\), we have: \[ mkx - m + 1 \mid x^2 + 1 \] Let \(d = mkx - m + 1\). We need \(d \mid (x^2 + 1)\). This implies: \[ x^2 + 1 = c(mkx - m + 1) \quad \text{for some integer } c \] Expanding and rearranging, we get: \[ x^2 + 1 = cmkx - cm + c \] \[ x^2 - cmkx + cm + 1 - c = 0 \] 4. **Analyze the quadratic equation:** The quadratic equation in \(x\) is: \[ x^2 - cmkx + (cm + 1 - c) = 0 \] For \(x\) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by: \[ \Delta = (cmk)^2 - 4(cm + 1 - c) \] \[ \Delta = c^2m^2k^2 - 4cm - 4 + 4c \] \[ \Delta = c^2m^2k^2 - 4cm + 4c - 4 \] 5. **Check for perfect square solutions:** We need \(\Delta\) to be a perfect square. Let's test some small values for \(c, m, k\): - For \(c = 1\), \(m = 1\), \(k = 1\): \[ \Delta = 1^2 \cdot 1^2 \cdot 1^2 - 4 \cdot 1 \cdot 1 + 4 \cdot 1 - 4 = 1 - 4 + 4 - 4 = -3 \quad (\text{not a perfect square}) \] - For \(c = 1\), \(m = 1\), \(k = 2\): \[ \Delta = 1^2 \cdot 1^2 \cdot 2^2 - 4 \cdot 1 \cdot 1 + 4 \cdot 1 - 4 = 4 - 4 + 4 - 4 = 0 \quad (\text{perfect square}) \] This gives us a valid solution. Let's find \(x, y, z\): \[ y = 2x - 1 \] \[ z = 1(2x - 1) + 1 = 2x - 1 + 1 = 2x \] Therefore, the triple \((x, y, z) = (x, 2x - 1, 2x)\) is a solution. 6. **Verify the solution:** - \(x \mid (2x - 1 + 1) \Rightarrow x \mid 2x \Rightarrow \text{True}\) - \(2x - 1 \mid (2x - 1) \Rightarrow \text{True}\) - \(2x \mid (x^2 + 1)\) For \(2x \mid (x^2 + 1)\), we need: \[ x^2 + 1 = 2xk \quad \text{for some integer } k \] \[ x^2 + 1 = 2xk \] \[ x^2 - 2xk + 1 = 0 \] The discriminant of this quadratic equation is: \[ \Delta = (2k)^2 - 4 \cdot 1 \cdot 1 = 4k^2 - 4 \] \[ \Delta = 4(k^2 - 1) \] For \(\Delta\) to be a perfect square, \(k^2 - 1\) must be a perfect square. This implies \(k = 1\), giving us: \[ x^2 - 2x + 1 = 0 \] \[ (x - 1)^2 = 0 \] \[ x = 1 \] Therefore, the only solution is \(x = 1\), \(y = 1\), \(z = 2\). The final answer is \( \boxed{ (1, 1, 2) } \)." b2cec2101186,"Petr said to Pavel: ""Write a two-digit natural number that has the property that when you subtract the same two-digit natural number just written in reverse, you get a difference of 63."" Which number could Pavel have written? Determine all possibilities. (L. Hozová) Hint. What is the difference between the digits of Pavel's number?",7$.,medium,"The task can be solved as an alphametic: $$ \begin{array}{rr} a & b \\ -b & a \\ \hline 6 & 3 \end{array} $$ Since the difference is positive, it must be that $a > b$. Additionally, since the difference in the units place is 3, the calculation must involve a carry-over from the tens place. Since the difference in the tens place is 6, it must be that $a - b = 7$. Furthermore, since both numbers are two-digit numbers, it must be that $b > 0$. Altogether, this gives us two possibilities: $$ \begin{array}{r} 81 \\ -18 \\ \hline 63 \end{array} \quad \begin{array}{r} 92 \\ -29 \\ \hline 63 \end{array} $$ The number that Pavel could have written was 81 or 92. Note. A two-digit number written as $\overline{a b}$ can be expressed as $10a + b$. The previous notation is therefore equivalent to the equation $$ (10a + b) - (10b + a) = 63 $$ which, after rearrangement, leads to $a - b = 7$." 8153aeb2dd4f,"11. Given $(3 x+1)^{5}=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f$, then $a-b+c-d+e-f=$ $\qquad$",32,easy,answer: 32 ff714ce81513,"Find the area of the equilateral triangle that includes vertices at $\left(-3,5\right)$ and $\left(-5,9\right)$. $\text{(A) }3\sqrt{3}\qquad\text{(B) }10\sqrt{3}\qquad\text{(C) }\sqrt{30}\qquad\text{(D) }2\sqrt{15}\qquad\text{(E) }5\sqrt{3}$",5\sqrt{3,medium,"1. **Calculate the length of the side of the equilateral triangle:** The vertices given are \((-3, 5)\) and \((-5, 9)\). The distance \(s\) between these two points can be calculated using the distance formula: \[ s = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the given points: \[ s = \sqrt{(-5 - (-3))^2 + (9 - 5)^2} = \sqrt{(-5 + 3)^2 + (9 - 5)^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \] Therefore, the side length \(s\) of the equilateral triangle is: \[ s = \sqrt{20} \] 2. **Calculate the area of the equilateral triangle:** The formula for the area \(A\) of an equilateral triangle with side length \(s\) is: \[ A = \frac{s^2 \sqrt{3}}{4} \] Substituting \(s = \sqrt{20}\): \[ A = \frac{(\sqrt{20})^2 \sqrt{3}}{4} = \frac{20 \sqrt{3}}{4} = 5 \sqrt{3} \] 3. **Conclusion:** The area of the equilateral triangle is: \[ \boxed{5\sqrt{3}} \]" 6cb629861a5f,"In the coordinate plane, suppose that the parabola $C: y=-\frac{p}{2}x^2+q\ (p>0,\ q>0)$ touches the circle with radius 1 centered on the origin at distinct two points. Find the minimum area of the figure enclosed by the part of $y\geq 0$ of $C$ and the $x$-axis.",\sqrt{3,medium,"1. **System of Equations:** We start with the system of equations representing the parabola and the circle: \[ \begin{cases} y = -\frac{p}{2}x^2 + q \\ x^2 + y^2 = 1 \end{cases} \] Substituting \( y = -\frac{p}{2}x^2 + q \) into the circle equation \( x^2 + y^2 = 1 \), we get: \[ x^2 + \left(-\frac{p}{2}x^2 + q\right)^2 = 1 \] 2. **Simplifying the Equation:** Expanding and simplifying the equation: \[ x^2 + \left(-\frac{p}{2}x^2 + q\right)^2 = 1 \] \[ x^2 + \left(\frac{p^2}{4}x^4 - pqx^2 + q^2\right) = 1 \] \[ x^2 + \frac{p^2}{4}x^4 - pqx^2 + q^2 = 1 \] \[ \frac{p^2}{4}x^4 + x^2(1 - pq) + q^2 - 1 = 0 \] 3. **Condition for Tangency:** For the parabola to touch the circle at two distinct points, the quadratic equation in \( x^2 \) must have a double root. This implies the discriminant must be zero: \[ (1 - pq)^2 - 4 \cdot \frac{p^2}{4} \cdot (q^2 - 1) = 0 \] Simplifying the discriminant condition: \[ (1 - pq)^2 - p^2(q^2 - 1) = 0 \] \[ 1 - 2pq + p^2q^2 - p^2q^2 + p^2 = 0 \] \[ 1 - 2pq + p^2 = 0 \] Solving for \( q \): \[ q = \frac{p^2 + 1}{2p} \] 4. **Intersection Points:** The parabola \( y = -\frac{p}{2}x^2 + \frac{p^2 + 1}{2p} \) intersects the x-axis at: \[ -\frac{p}{2}x^2 + \frac{p^2 + 1}{2p} = 0 \] \[ -px^2 + p^2 + 1 = 0 \] \[ x^2 = \frac{p^2 + 1}{p^2} \] \[ x = \pm \frac{\sqrt{p^2 + 1}}{p} \] 5. **Area Calculation:** The area \( S(p) \) enclosed by the part of the parabola \( y \geq 0 \) and the x-axis is given by: \[ S(p) = 2 \int_{0}^{\frac{\sqrt{p^2 + 1}}{p}} \left( -\frac{p}{2}x^2 + \frac{p^2 + 1}{2p} \right) dx \] Evaluating the integral: \[ S(p) = 2 \left[ -\frac{p}{2} \int_{0}^{\frac{\sqrt{p^2 + 1}}{p}} x^2 \, dx + \frac{p^2 + 1}{2p} \int_{0}^{\frac{\sqrt{p^2 + 1}}{p}} dx \right] \] \[ S(p) = 2 \left[ -\frac{p}{2} \cdot \frac{x^3}{3} \Bigg|_{0}^{\frac{\sqrt{p^2 + 1}}{p}} + \frac{p^2 + 1}{2p} \cdot x \Bigg|_{0}^{\frac{\sqrt{p^2 + 1}}{p}} \right] \] \[ S(p) = 2 \left[ -\frac{p}{2} \cdot \frac{(\frac{\sqrt{p^2 + 1}}{p})^3}{3} + \frac{p^2 + 1}{2p} \cdot \frac{\sqrt{p^2 + 1}}{p} \right] \] \[ S(p) = 2 \left[ -\frac{p}{2} \cdot \frac{(p^2 + 1)^{3/2}}{3p^3} + \frac{(p^2 + 1)^{3/2}}{2p^2} \right] \] \[ S(p) = 2 \left[ -\frac{(p^2 + 1)^{3/2}}{6p^2} + \frac{(p^2 + 1)^{3/2}}{2p^2} \right] \] \[ S(p) = 2 \left[ \frac{(p^2 + 1)^{3/2}}{2p^2} - \frac{(p^2 + 1)^{3/2}}{6p^2} \right] \] \[ S(p) = 2 \left[ \frac{3(p^2 + 1)^{3/2} - (p^2 + 1)^{3/2}}{6p^2} \right] \] \[ S(p) = 2 \left[ \frac{2(p^2 + 1)^{3/2}}{6p^2} \right] \] \[ S(p) = \frac{2(p^2 + 1)^{3/2}}{3p^2} \] 6. **Minimizing the Area:** To find the minimum area, we take the derivative of \( S(p) \) with respect to \( p \) and set it to zero: \[ S'(p) = \frac{2}{3} \cdot \frac{3p^2(p^2 + 1)^{1/2} \cdot p - (p^2 + 1)^{3/2} \cdot 2p}{p^4} = 0 \] Simplifying: \[ S'(p) = \frac{2}{3} \cdot \frac{3p^3(p^2 + 1)^{1/2} - 2p(p^2 + 1)^{3/2}}{p^4} = 0 \] \[ 3p^3(p^2 + 1)^{1/2} = 2p(p^2 + 1)^{3/2} \] Dividing both sides by \( p(p^2 + 1)^{1/2} \): \[ 3p^2 = 2(p^2 + 1) \] \[ 3p^2 = 2p^2 + 2 \] \[ p^2 = 2 \] \[ p = \sqrt{2} \] 7. **Calculating the Minimum Area:** Substituting \( p = \sqrt{2} \) into \( S(p) \): \[ S(\sqrt{2}) = \frac{2((\sqrt{2})^2 + 1)^{3/2}}{3(\sqrt{2})^2} \] \[ S(\sqrt{2}) = \frac{2(2 + 1)^{3/2}}{3 \cdot 2} \] \[ S(\sqrt{2}) = \frac{2 \cdot 3^{3/2}}{6} \] \[ S(\sqrt{2}) = \frac{2 \cdot \sqrt{27}}{6} \] \[ S(\sqrt{2}) = \frac{2 \cdot 3\sqrt{3}}{6} \] \[ S(\sqrt{2}) = \sqrt{3} \] The final answer is \( \boxed{ \sqrt{3} } \)" ecb2c1736571,"Example 10 Let $a, b, c$ be positive real numbers, find the minimum value of $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$.",See reasoning trace,medium,"Let \[ \left\{ \begin{array}{l} x = a + 2b + c, \\ y = a + b + 2c, \\ z = a + b + 3c. \end{array} \right. \] Then we have \( x - y = b - c \) and \( z - y = c \), which leads to \[ \left\{ \begin{array}{l} a + 3c = 2y - x, \\ b = z + x - 2y, \\ c = z - y. \end{array} \right. \] Thus, \[ \begin{array}{l} \frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} = \\ \frac{2y - x}{x} + \frac{4(z + x - 2y)}{y} - \frac{8(z - y)}{z} = \\ -17 + 2\left(\frac{y}{x} + \frac{2x}{y}\right) + 4\left(\frac{z}{y} + \frac{2y}{z}\right) \geq \\ -17 + 4 \sqrt{\frac{y}{x} \cdot \frac{2x}{y}} + 8 \sqrt{\frac{z}{y} \cdot \frac{2y}{z}} = -17 + 12 \sqrt{2} \end{array} \] The equality in the above expression can be achieved. In fact, from the derivation process, the equality holds if and only if the equality in the AM-GM inequality holds, which is equivalent to \[ \left\{ \begin{array}{l} \frac{y}{x} = \frac{2x}{y}, \\ \frac{z}{y} = \frac{2y}{z}. \end{array} \right. \] This is equivalent to \[ \left\{ \begin{array}{l} y^2 = 2x^2, \\ z^2 = 2y^2, \end{array} \right. \] which is equivalent to \[ \left\{ \begin{array}{l} y = \sqrt{2}x, \\ z = \sqrt{2}y. \end{array} \right. \] Thus, \[ \left\{ \begin{array}{l} a + b + 2c = \sqrt{2}(a + 2b + c), \\ a + b + 3c = \sqrt{2}(a + b + 2c). \end{array} \right. \] Solving this system of equations, we get \[ \left\{ \begin{array}{l} b = (1 + \sqrt{2})a, \\ c = (4 + 3\sqrt{2})a. \end{array} \right. \] It is not difficult to verify that for any real number \(a\), if \(b = (1 + \sqrt{2})a\) and \(c = (4 + 3\sqrt{2})a\), then \[ \frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} = -17 + 12\sqrt{2}. \] Therefore, the minimum value is \(-17 + 12\sqrt{2}\)." e4698cc9c96f,"8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\circ}, 30^{\circ}, 60^{\circ}$, and $30^{\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$.",to the problem,medium,"Solution: Let's formulate several auxiliary statements. 1) Let the angular measure of the arc $AB$ (Fig.1) be $\varphi$. (This means that $\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord ![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=311&width=323&top_left_y=907&top_left_x=1569) $AB=2R\sin(\varphi/2)$. Here $R$ is the radius of the circle. 2) Let two chords $AA_1$ and $BB_1$ intersect at point $T$ (Fig.2). The angular measures of the arcs $AB$ and $A_1B_1$ are $\varphi$ and $v$. Triangles $ATB$ and $A_1TB_1$ are similar by two angles (equal angles are marked). The similarity ratio $$ k=AB / A_1B_1=\sin(\varphi/2) / \sin(v/2) $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=254&width=220&top_left_y=1643&top_left_x=193) Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=411&width=411&top_left_y=1551&top_left_x=480) Fig. 2 ![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=480&width=488&top_left_y=1505&top_left_x=955) Fig. 3 3) Referring to Fig. 3. Three chords intersect at one point, denoted as $T$. The angular measures of the six resulting arcs are marked on the figure. From the similarity of triangles $ATB$ and $A_1TB_1$ it follows (see point 2) the equality $AT / B_1T = \sin(\varphi/2) / \sin(v/2)$. Similarly, $\triangle B T C \sim \Delta B_1 T C_1 \Rightarrow B_1 T / C T = \sin(\psi/2) / \sin(u/2), \Delta C T A_1 \sim \Delta A T C_1 \Rightarrow C T / A T = \sin(\theta/2) / \sin(w/2)$. Multiplying the last three equalities, we get: $1 = AT / B_1T \cdot B_1T / CT \cdot CT / AT = \sin(\varphi/2) / \sin(v/2) \cdot \sin(\psi/2) / \sin(u/2) \cdot \sin(\theta/2) / \sin(w/2)$. Thus, the necessary (and in fact sufficient) condition for three chords to intersect at one point is the equality: $$ \sin(\varphi/2) \sin(\theta/2) \sin(\psi/2) = \sin(u/2) \sin(v/2) \sin(w/2) $$ Now it is not difficult to obtain the answer to the problem. Substituting the given data into this relation $w=150^\circ, \varphi=30^\circ, \theta=60^\circ, v=30^\circ$, and expressing $u$ from the equality $\varphi + u + \theta + v + \psi + w = 360^\circ$, we get the equation for determining the desired angle $\psi$: $$ \sin 15^\circ \sin(\psi/2) \sin 30^\circ = \sin 15^\circ \sin \left(\left(90^\circ - \psi\right)/2\right) \sin 75^\circ $$ From this, it is not difficult to obtain that $\psi = 60^\circ$. Answer: $60^\circ$." 77e6d9edf606,10. It is known that the 3 sides of a triangle are consecutive positive integers and the largest angle is twice the smallest angle. Find the perimeter of this triangle.,"A D+B D$, we have $x(x-1) /(x+1)+(x-1)^{2} /(x+1)=x+1$. Solving this, the only positive solution is ",easy,"10. Ans: 15 Let $\angle C=2 \angle A$ and $C D$ the bisector of $\angle C$. Let $B C=x-1, C A=x$ and $A B=x+1$. Then $\triangle A B C$ is similar to $\triangle C B D$. Thus $B D / B C=B C / A B$ so that $B D=(x-1)^{2} /(x+1)$. Also $C D / A C=C B / A B$ so that $A D=C D=$ $x(x-1) /(x+1)$. As $A B=A D+B D$, we have $x(x-1) /(x+1)+(x-1)^{2} /(x+1)=x+1$. Solving this, the only positive solution is $x=5$. Thus the perimeter of the triangle is $4+5+6=15$" d8f246138d26,"2. Given $f(x)=\frac{a}{a^{2}-2} \cdot\left(a^{x}-a^{-x}\right)(a>0$ and $a \neq 1)$ is an increasing function on $\mathbf{R}$. Then the range of the real number $a$ is ( ). (A) $(0,1)$ (B) $(0,1) \cup(\sqrt{2},+\infty)$ (C) $(\sqrt{2},+\infty)$ (D) $(0,1) \cup[\sqrt{2},+\infty)$",See reasoning trace,easy,"2.B. Let $x_{1}a^{x_{2}}, a^{-x_{1}}1$ when $a^{x_{1}}0$, then $a>\sqrt{2}$, which meets the conditions. In summary, the range of values for $a$ is $(0,1) \cup(\sqrt{2},+\infty)$." e5a54196cb18,"3. Given a convex quadrilateral $ABCD$ with area $P$. We extend side $AB$ beyond $B$ to $A_1$ such that $\overline{AB}=\overline{BA_1}$, then $BC$ beyond $C$ to $B_1$ such that $\overline{BC}=\overline{CB_1}$, then $CD$ beyond $D$ to $C_1$ so that $\overline{CD}=\overline{DC_1}$, and $DA$ beyond $A$ to $D_1$ such that $\overline{DA}=\overline{AD_1}$. What is the area of quadrilateral $A_1B_1C_1D_1$?",See reasoning trace,medium,"Solution. We connect $A$ with $C_{1}$, $B$ with $D_{1}$, $C$ with $A_{1}$, and $D$ with $B_{1}$. Then, $$ \begin{aligned} & P_{\triangle A B C}=P_{I}=P_{I I} \\ & P_{\triangle B C D}=P_{I I I}=P_{I V} \\ & P_{\triangle C D A}=P_{V}=P_{V I} \\ & P_{\triangle D A B}=P_{V I I}=P_{I I I} \end{aligned} $$ where $$ \begin{array}{ll} P_{I}=P_{\Delta A_{1} C}, & P_{I I}=P_{\Delta C A_{1} B_{1}} \\ P_{I I I}=P_{\Delta D C B_{1}}, & P_{I V}=P_{\Delta C_{1} D B_{1}} \\ P_{V}=P_{\Delta C_{1} D A}, & P_{V I}=P_{\Delta C_{1} D_{1} A} \\ P_{I I I}=P_{\Delta A D_{1} B}, & P_{V I I I}=P_{\Delta B D_{1} A_{1}} \end{array} $$ ![](https://cdn.mathpix.com/cropped/2024_06_05_ea01241e897ed9d0948dg-02.jpg?height=507&width=641&top_left_y=708&top_left_x=838) We have: $$ \begin{aligned} P_{\Delta A B C}+P_{\Delta B C D}+P_{\Delta C D A}+P_{\Delta D A B} & =P_{I}+P_{I I I}+P_{V}+P_{I I}= \\ & =P_{I I}+P_{I V}+P_{V I}+P_{V I I I}=2 P \\ P_{1}=P+\left(P_{I}+P_{I I}+P_{I I I}+P_{I V}+P_{V}+P_{I I}\right. & \left.+P_{V I I}+P_{I I I}\right)=P+(2 P+2 P)=5 P \end{aligned} $$" 7cd0fd46ba73,"16. If a positive integer cannot be written as the difference of two square numbers, then the integer is called a ""cute"" integer. For example, 1,2 and 4 are the first three ""cute"" integers. Find the $2010^{\text {th }}$ ""cute"" integer. (Note: A square number is the square of a positive integer. As an illustration, 1,4,9 and 16 are the first four square numbers.)",8030$.,medium,"16. Answer: 8030 Any odd number greater than 1 can be written as $2 k+1$, where $2 k+1=(k+1)^{2}-k^{2}$. Hence all odd integers greater than 1 are not ""cute"" integers. Also, since $4 m=(m+1)^{2}-(m-1)^{2}$, so that all integers of the form $4 m$, where $m>1$, are not ""cute"". We claim that all integers of the form $4 m+2$ are ""cute"". Suppose $4 m+2$ (for $m \geq 1$ ) is not ""cute"", then $$ 4 m+2=x^{2}-y^{2}=(x-y)(x+y) $$ for some integers positive integers $x$ and $y$. However, $x+y$ and $x-y$ have the same parity, so that $x-y$ and $x+y$ must all be even since $4 m+2$ is even. Hence $4 m+2$ is divisible by 4 , which is absurd. Thus, $4 m+2$ is ""cute"". The first few ""cute"" integers are $1,2,4,6,10 \cdots$. For $n>3$, the $n^{\text {th }}$ ""cute"" integer is $4 n-10$. Thus, the $2010^{\text {th }}$ ""cute"" integer is $4(2010)-10=8030$." 6f7c352ec28a,"1A. 2B. Determine $\log _{a} b, \log _{a b} b, \log _{a b^{2}} b$ and $\log _{a b^{3}} b$, if $$ \log _{a} b-\log _{a b} b=\log _{a b^{2}} b-\log _{a b^{3}} b $$",See reasoning trace,medium,"Solution. For $b=1$, the equality is true for any $a$, because all logarithms are equal to zero. For $b>0, b \neq 1$, the given condition can be written in the form $$ \frac{1}{\log _{b} a}-\frac{1}{\log _{b} a+1}=\frac{1}{\log _{b} a+2}-\frac{1}{\log _{b} a+3} $$ Let $\log _{b} a=x$. Then from the last equality, we have $$ \frac{1}{x}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+3} $$ from which we find $x=-\frac{3}{2}$. Therefore, the required values are $$ \log _{a} b=-\frac{2}{3}, \log _{a b} b=-2, \log _{a b^{2}} b=2, \log _{a b} 3 b=\frac{2}{3} $$" 70f828454ad6,"Example 1 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that both $a_{2}-$ $a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied, then the total number of different ways to select the numbers is $\qquad$. (1989 National High School League Question)","C_{10}^{3}$ methods, which is the total number of different selection methods.",medium,"Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$. $T=\left\{\left(a_{1}, a_{2}, a_{3}\right) \mid a_{1}, a_{2}, a_{3} \in S, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3\right\}$, $T^{\prime}=\left\{\left(a_{1}^{\prime}, a_{2}^{\prime}, a^{\prime}{ }_{3}\right) \mid a^{\prime}, a^{\prime}{ }_{2}, a_{3}^{\prime} \in S^{\prime}, a_{1}0, b>0) $$ $F_{1} 、 F_{2}$ are the two foci of the hyperbola, and $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ $=0$. Then the ratio of the inradius $r$ to the circumradius $R$ of $\triangle P F_{1} F_{2}$ is",See reasoning trace,medium,"4. $\frac{\sqrt{6}}{2}-1$. From $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=0 \Rightarrow \angle F_{1} P F_{2}=90^{\circ}$. Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n$. Also, $\left|F_{1} F_{2}\right|=2 c$, then $$ \begin{array}{l} R=c, r=\frac{1}{2}(m+n-2 c), \\ m^{2}+n^{2}=4 c^{2},|m-n|=2 a . \end{array} $$ Let $\frac{r}{R}=k$. Then $$ \begin{array}{l} r=k R=k c=\frac{1}{2}(m+n-2 c) \\ \Rightarrow m+n=(2 k+2) c . \end{array} $$ From conclusion (1) we get $$ \begin{array}{l} (2 k+2)^{2} c^{2}+4 a^{2}=8 c^{2} \\ \Rightarrow(k+1)^{2}=2-\frac{1}{e^{2}}=\frac{3}{2} \\ \Rightarrow k=\frac{\sqrt{6}}{2}-1 . \end{array} $$" 16951ba02008,"Example 25 A middle school is preparing to form an 18-person soccer team, consisting of students from 10 classes in the first year of high school, with at least one student from each class. The number of allocation schemes is $\qquad$ kinds.",24310$.,medium,"Solve: Fill in 24310. Reason: Construct a partition model, take 18 identical balls and arrange them in a row, use 9 partitions to divide the 18 balls into 10 intervals, the number of balls in the $i$-th interval $(1 \leqslant i \leqslant 10)$ corresponds to the number of student quotas for the $i$-th class. Therefore, the number of quota allocation schemes is equal to the number of ways to insert the partitions. Since the number of ways to insert partitions is $C_{17}^{9}$, the total number of quota allocation schemes is $C_{17}^{9}=24310$." 3a307dc14c00,"### 3.476 Find the maximum value of the expression $A=\sin ^{2}\left(\frac{15 \pi}{8}-4 \alpha\right)-\sin ^{2}\left(\frac{17 \pi}{8}-4 \alpha\right)$ for $0 \leq \alpha \leq \frac{\pi}{8}$.",$\frac{1}{\sqrt{2}}$ when $\alpha=\frac{\pi}{16}$,medium,"Solution. $A=\frac{1-\cos \left(\frac{15 \pi}{4}-8 \alpha\right)}{2}-\frac{1-\cos \left(\frac{17 \pi}{4}-8 \alpha\right)}{2}=\frac{1}{2}\left(\cos \left(\frac{\pi}{4}-8 \alpha\right)-\cos \left(\frac{\pi}{4}+8 \alpha\right)\right)=$ $=\sin \frac{\pi}{4} \cdot \sin 8 \alpha=\frac{\sin 8 \alpha}{\sqrt{2}}$. Since $0 \leq \alpha \leq \frac{\pi}{8}$, the maximum value of $A$ is achieved when $\sin 8 \alpha=1$, i.e., when $\alpha=\frac{\pi}{16}$. Answer: $\frac{1}{\sqrt{2}}$ when $\alpha=\frac{\pi}{16}$." d1daa698df15,"7. Let an ""operation"" be the act of randomly changing a known positive integer $n$ to a smaller non-negative integer (each number has the same probability). Then, the probability that the numbers $10$, $100$, and $1000$ all appear during the process of performing several operations to reduce 2019 to 0 is $\qquad$",See reasoning trace,medium,"7. $\frac{1}{1112111}$. Let the operation process produce $a_{1}, a_{2}, \cdots, a_{n}$ in addition to $2019, 10, 100, 1000, 0$. $$ \begin{array}{l} \text { Then the probability is } \frac{1}{2019} \times \frac{1}{10} \times \frac{1}{100} \times \frac{1}{1000} \prod_{k=1}^{n} \frac{1}{a_{k}} . \\ \text { Therefore, } p=\sum \frac{1}{2019} \times \frac{1}{10} \times \frac{1}{100} \times \frac{1}{1000} \prod_{k=1}^{n} \frac{1}{a_{k}} \\ =\frac{1}{2019} \times \frac{1}{1000} \times \frac{1}{100} \times \frac{1}{10} \prod_{k=10,100,1000}^{2018}\left(1+\frac{1}{k}\right) \\ =\frac{1}{1001 \times 101 \times 11}=\frac{1}{1112111} . \end{array} $$" ce48aaac94d5,"Question 24: If $f(x)$ is an odd function with a period of 2, and when $x \in [0,1), f(x)=2^{x}-1$, then $f\left(\log _{2} \frac{1}{24}\right)=$ _.",See reasoning trace,easy,"Question 24, Solution: Since $\log _{2} \frac{1}{24}=-\log _{2} 24=-(3+\log _{2} 3)$, according to the condition, we have: $$ \begin{array}{l} f\left(\log _{2} \frac{1}{24}\right)=f\left(-(3+\log _{2} 3)\right)=-f\left(3+\log _{2} 3\right)=-f\left(\log _{2} 3-1\right) \\ =-f\left(\log _{2} \frac{3}{2}\right)=-\left(2^{\log _{2} \frac{3}{2}}-1\right)=-\frac{1}{2} \end{array} $$" a844d520def3,"A5. What is $a$, if the equation $x^{2}=3(a x-3)$ has exactly one real solution? (A) $a=2$ or $a=-2$ (B) $a$ can only be 2 (C) $a=4$ (D) $a=6$ or $a=-6$ (E) $a=\frac{2}{3}$",0$. Its discriminant is $D=9 a^{2}-36$ and is equal to zero if $a=2$ or $a=-2$.,easy,A5. The equation is transformed into $x^{2}-3 a x+9=0$. Its discriminant is $D=9 a^{2}-36$ and is equal to zero if $a=2$ or $a=-2$. fe70f4e3b793,"45. The sum of the interior angles of a convex $n$-sided polygon is less than $2021^{\circ}$, then the maximum value of $n$ is",13,easy,Answer: 13 105ceda2d42a,"Hammie is in $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds? $\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5}\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$",\textbf{(E),easy,"Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$, we see that the median weight is 6 pounds. The average weight of the five kids is $\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26$. Hence,\[26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.\]" 0b94cbb308af,"In a triangle, the base is $4 \mathrm{~cm}$ larger than the corresponding height. If we increase both the base and the height by $12 \mathrm{~cm}$, we get a triangle whose area is five times the area of the original triangle. What are the base and height of the original triangle?",See reasoning trace,medium,"Let the base of the triangle be $a$, then its height is $m=a-4$. According to the problem, $$ (a+12)(a+8)=5 a(a-4) $$ or $$ a^{2}-10 a-24=0 $$ from which the positive root is: $a=12 \text{ cm}$, so $m=8 \text{ cm}$. (Leó Spitzer, Budapest.) The problem was also solved by: Bauer E., Breuer P., Dénes M., Döri V., Engler J., Erdős V., Fried E., Grün E., Grünwald M., Klein G., Koffer B., Köhler I., Lengyel K., Lengyel P., Neumann L., Rosenthal M., Rosinger J., Schnabel L., Sichermann F., Steiner L., Stolzer I., Szántó L., Szilárd V., Vilcsek A., Viola R., Wellis E." 0f44b128a3ff,"(4) Let $(1+x)^{16}=\sum_{i=0}^{16} a_{i} x^{i}$, then $\sum_{i=1}^{8} i a_{i}=$ $\qquad$",See reasoning trace,medium,"(4) 262144 Hint: Given: $a_{i}=\mathrm{C}_{16}^{i}(i=0,1, \cdots, 16)$, and $i a_{i}=$ $i \mathrm{C}_{16}^{i}=16 \mathrm{C}_{15}^{i-1}$, so $$ \sum_{i=1}^{8} i a_{i}=\sum_{i=1}^{8} 16 \mathrm{C}_{15}^{i-1}=16 \sum_{k=0}^{7} \mathrm{C}_{15}^{k}=16 \times 2^{14}=2^{18}=262144 . $$" 24ac2ca5bd1b,"Determine all pairs $(x, y)$ of natural integers such that $2^{2 x+1}+2^{x}+1=y^{2}$","4=2^{2}$ and $2^{9}+2^{4}+1=512+16+1=529=23^{2}$, $(4,23)$ and $(0,2)$ are the two solution pairs.",medium,"We notice that $x=0, y=2$ is a solution. From now on, assume $x \geqslant 1$. We start by factoring: $2^{2 x+1}+2^{x}=2^{x}\left(1+2^{x+1}\right)=y^{2}-1=(y-1)(y+1)$. Since $x \geqslant 1, (y+1)(y-1)$ is even, so $y$ is odd. Thus, both $y-1$ and $y+1$ are even, but since their difference is 2, only one is divisible by 4. In particular, either $y+1$ or $y-1$ is divisible by $2^{x-1}$. - If $2^{x-1}$ divides $y-1$, we set $y=1+2^{x-1} \times k$ for $k$ a natural number. We substitute back into the initial equation: $2^{2 x+1}+2^{x}+1=y^{2}=1+2^{x} k+2^{2(x-1)} k^{2}$. We get $2^{x+1}+1=k+k^{2} 2^{x-2}$, which simplifies to $1-k=2^{x-2}\left(k^{2}-8\right)$. Looking modulo 2, we see that $k$ is odd. If $k \geqslant 1$, $1-k$ is negative, so $k^{2}-8 \leqslant 0$ i.e., $0 \leqslant k \leqslant 2$. By parity, $k=1$, which contradicts $1-k=2^{x-2}\left(k^{2}-8\right)$. - If $2^{x-1}$ divides $y+1$, we set $y=-1+2^{x-1} \times k$ for $k$ a natural number. We substitute back into the initial equation: $2^{2 x+1}+2^{x}+1=y^{2}=1-2^{x} k+2^{2(x-1)} k^{2}$. We get $2^{x+1}+1=-k+k^{2} 2^{x-2}$, which simplifies to $1+k=2^{x-2}\left(k^{2}-8\right)$. Looking modulo 2, we see that $k$ is odd. Since $1+k$ is positive, $k^{2}-8$ is also positive, so $k \geqslant 3$. We have $1+k=2^{x-2}\left(k^{2}-8\right) \geqslant 2 k^{2}-16$, which rearranges to $2 k^{2}-k-17 \leqslant 0$. For $k \geqslant 4, 2 k^{2}-k-17=k(2 k-1)-17 \geqslant 4 \times 7-17=11>0$, so $3 \leqslant k \leqslant 4$. Since $k$ is odd, $k=3$, thus $4=2^{x-2}(9-8)=2^{x-2}$, so $x=4$. We also have $y=-1+3 \times 2^{4-1}=23$. Thus, we have found that all solutions are of the form $(4,23)$ and $(0,2)$. It remains to verify that these two pairs are indeed solutions. Conversely, since $2^{1}+2^{0}+1=4=2^{2}$ and $2^{9}+2^{4}+1=512+16+1=529=23^{2}$, $(4,23)$ and $(0,2)$ are the two solution pairs." f43123f950f5,"Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides; each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees, where $k$ is an integer greater than $1$. The number of possibilities for the pair $(x, k)$ is: $\textbf{(A)}\ \infty\qquad \textbf{(B)}\ \text{finite, but greater than 2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$",\textbf{(D),medium,"Each angle in each equiangular polygon is $\frac{180(n-2)}{n} = 180 - \frac{360}{n}$, where $n$ is the number of sides. As $n$ gets larger, each angle in the equiangular polygon approaches (but does not reach) $180^{\circ}$. That means $kx < 180$, so $x < \frac{180}{k}$. Recall that each angle in an equiangular triangle has $60^{\circ}$, each angle in an equiangular quadrilateral has $90^{\circ}$, and so on. If $k = 2$, then $x < 90$. That means the only option available is $x = 60$ because an equiangular triangle is the only equiangular polygon to have acute angles. If $k = 3$, then $x < 60$, and there are no values of $x$ that satisfy. As $k$ increases, $\frac{180}{k}$ decreases even further, so we can for sure say that there is only 1 possibility for the pair $(x, k)$, which is answer choice $\boxed{\textbf{(D)}}$." 97b1ca633861,"10. On a plane, 2011 points are marked. We will call a pair of marked points $A$ and $B$ isolated if all other points are strictly outside the circle constructed on $A B$ as its diameter. What is the smallest number of isolated pairs that can exist?",See reasoning trace,medium,"Solution. Connect points $A$ and $B$ with a segment if the pair $AB$ is isolated, and consider the resulting graph. Notice that it is connected. Indeed, suppose it splits into several connected components; then we find a pair of points from different components with the smallest distance between them. It is easy to see that such a pair is isolated, as any point that falls within their circle and is closer to both of them than they are to each other cannot lie in the same component with both of them. In a connected graph with 2011 vertices, there are at least 2010 edges, thus we have shown that there are at least 2010 isolated pairs. An example is 2011 points lying on a straight line (or on a semicircle)." a4b658ac7acd,"## Calculate the area of the figure bounded by the graphs of the functions: $$ y=\cos ^{5} x \cdot \sin 2 x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right) $$",See reasoning trace,medium,"## Solution Integrals $14-22$ ![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-42.jpg?height=691&width=1059&top_left_y=1094&top_left_x=904) $$ \begin{aligned} & S=\int_{0}^{\frac{\pi}{2}} \cos ^{5} x \cdot \sin 2 x d x=\int_{0}^{\frac{\pi}{2}} \cos ^{5} x \cdot 2 \sin x \cos x d x= \\ & =-2 \int_{0}^{\frac{\pi}{2}} \cos ^{6} x d(\cos x)=2 \int_{\frac{\pi}{2}}^{0} \cos ^{6} x d(\cos x)=\left.\frac{2 \cos ^{7} x}{7}\right|_{\frac{\pi}{2}} ^{0}=\frac{2}{7}-0=\frac{2}{7} \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-22$ » Categories: Kuznetsov Problem Book Integrals Problem 14 | Integrals Ukrainian Banner Network - Last edited on this page: 08:14, 11 June 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 14-23 ## Material from Plusi" 7a08c30caac5,"1665. It is known that the density of the distribution of a random variable $X$ has the form $$ p(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{x^{2}}{2 \sigma^{2}}} $$ Find the density of the distribution of the random variable $Y=X^{2}$.",See reasoning trace,medium,"Solution. First, let's find the distribution function of the random variable $Y$. Since it does not take negative values, then $F(y)=P(Y0$. Then $$ \begin{aligned} F(y) & =P(Y0 \end{aligned} $$ Therefore, the desired density function is given by the formula $$ \begin{aligned} \varphi(y)=F^{\prime}(y) & =\left\{\begin{array}{cc} 0 & \text { for } y \leqslant 0, \\ \frac{2}{\sigma \sqrt{2 \pi}} \cdot e^{-\frac{y}{2 \sigma^{2}}} \cdot \frac{1}{2 \sqrt{y}} \text { for } y>0 \end{array}=\right. \\ & =\left\{\begin{array}{cc} 0 \quad \text { for } y \leqslant 0, \\ \frac{1}{\sigma \sqrt{2 \pi y}} \cdot e^{-\frac{y}{2 \sigma^{2}}} & \text { for } y>0 . \end{array}\right. \end{aligned} $$" ac4afde2e098,"9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is",See reasoning trace,medium,"9.4. Let the circumcenter of $\triangle ABC$ be $O$. It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$. Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and $$ \begin{aligned} & \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\ & \text { Let }|\overrightarrow{O P}|=x . \\ & \text { Then } \sqrt{2} \leqslant x \leqslant 2 \sqrt{2} \text {, and } \\ & \overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M}) \cdot(\overrightarrow{O N}-\overrightarrow{O P}) \\ & =-\overrightarrow{O P}{ }^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P}-\overrightarrow{O M} \cdot \overrightarrow{O N} \\ = & -x^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\ \leqslant & -x^{2}+|\overrightarrow{O M}+\overrightarrow{O N}||\overrightarrow{O P}| \\ = & -x^{2}+4 x=4-(x-2)^{2} \leqslant 4, \end{aligned} $$ The equality holds if and only if $x=2$ and $\overrightarrow{O P}$ is in the same direction as $\overrightarrow{O M}+\overrightarrow{O N}$. Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4." 731dbd652cf1,"10.354. Point $M$ lies inside an equilateral triangle $A B C$. Calculate the area of this triangle, given that $A M=B M=$ $=2 \text{ cm}, \text{a} C M=1 \text{ cm}$.",$\frac{9 \sqrt{3}+3 \sqrt{15}}{8} \approx 3,medium,"Solution. Let the side of the triangle be $a$ and draw $M D \perp A B$ (Fig. 10.135). Since $A M=B M$, points $C, M, D$ lie on the height $C D$. In $\triangle A C D$ and $\triangle A M D$, we have $(1+M D)^{2}=a^{2}-\frac{a^{2}}{4}, M D^{2}=4-\frac{a^{2}}{4}$. Then $a^{2}=4\left(4-M D^{2}\right)$ and we get the quadratic equation $(1+M D)^{2}=$ $=3\left(4-M D^{2}\right)$ or $4 M D^{2}+2 M D-11=0$, from which $M D=\frac{3 \sqrt{5}-1}{4}$ (the second root is not suitable). Next, we find $a^{2}=4\left(4-M D^{2}\right)=16-\frac{46-6 \sqrt{5}}{4}=$ $=\frac{9+3 \sqrt{5}}{2}$ and, therefore, ![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0812.jpg?height=518&width=567&top_left_y=76&top_left_x=112) Fig. 10.136 .1 ![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0812.jpg?height=526&width=586&top_left_y=68&top_left_x=674) Fig. 10.136 .2 $$ S=\frac{a^{2} \sqrt{3}}{4}=\frac{(9+3 \sqrt{5}) \sqrt{3}}{2}=\frac{9 \sqrt{3}+3 \sqrt{15}}{8} \approx 3.4 \text{ cm}^{2} $$ Answer: $\frac{9 \sqrt{3}+3 \sqrt{15}}{8} \approx 3.4 \text{ cm}^{2}$." a2cc24610ca6,"Let $\triangle ABC$ be a triangle with $AB = 7$, $AC = 8$, and $BC = 3$. Let $P_1$ and $P_2$ be two distinct points on line $AC$ ($A, P_1, C, P_2$ appear in that order on the line) and $Q_1$ and $Q_2$ be two distinct points on line $AB$ ($A, Q_1, B, Q_2$ appear in that order on the line) such that $BQ_1 = P_1Q_1 = P_1C$ and $BQ_2 = P_2Q_2 = P_2C$. Find the distance between the circumcenters of $BP_1P_2$ and $CQ_1Q_2$.",3,medium,"To find the distance between the circumcenters of $\triangle BP_1P_2$ and $\triangle CQ_1Q_2$, we will follow the steps outlined in the solution sketch and provide detailed calculations and justifications. 1. **Identify the points and their properties:** - Given $AB = 7$, $AC = 8$, and $BC = 3$. - Points $P_1$ and $P_2$ lie on line $AC$ such that $A, P_1, C, P_2$ appear in that order. - Points $Q_1$ and $Q_2$ lie on line $AB$ such that $A, Q_1, B, Q_2$ appear in that order. - $BQ_1 = P_1Q_1 = P_1C$ and $BQ_2 = P_2Q_2 = P_2C$. 2. **Determine the circumcenters $O_1$ and $O_2$:** - Let $O_1$ be the circumcenter of $\triangle BP_1P_2$. - Let $O_2$ be the circumcenter of $\triangle CQ_1Q_2$. 3. **Show that $O_1$ lies on $\omega_2$ and $O_2$ lies on $\omega_1$:** - Since $P_1Q_1 = P_1C$ and $P_2Q_2 = P_2C$, triangles $BP_1P_2$ and $CQ_1Q_2$ are isosceles. - The circumcenters $O_1$ and $O_2$ are equidistant from the vertices of these triangles. 4. **Identify the midpoint $M$ of arc $\widehat{BAC}$ of $(ABC)$:** - $M$ is the center of a rotation that maps $Q_1BQ_2$ to $P_1CP_2$ and $\omega_2$ to $\omega_1$. 5. **Determine the midpoint $N$ of $BC$:** - $N$ has equal powers with respect to $\omega_1$ and $\omega_2$. - The perpendicular bisector $MN$ of $BC$ is the radical axis of $\omega_1$ and $\omega_2$. 6. **Reflect the circles over $MN$:** - Since $\omega_1$ and $\omega_2$ are reflections over $MN$, the distance between $O_1$ and $O_2$ is twice the distance from $N$ to $O_1$ or $O_2$. 7. **Calculate the distance between $O_1$ and $O_2$:** - Since $O_1$ lies on $MC$ and $O_2$ lies on $MB$, we need to find the distance between these points in $\triangle MBC$. - Using the properties of the circumcenters and the given distances, we can determine the exact distance. By symmetry and the properties of the circumcenters, the distance between $O_1$ and $O_2$ is equal to the length of $BC$, which is $3$. The final answer is $\boxed{3}$." 64b5d9e064d1,"2. Given $0$ $\frac{7}{2}, \therefore \cos \theta_{1} \leqslant \cos \theta_{2}$ $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n-1}=r$, contradiction! $\therefore n-1$ is the smallest upper bound of the left side of inequality (1). When $n \geqslant 3$, the smallest positive number $\lambda=n-1$, in summary, the smallest positive number $\lambda=\left\{\begin{array}{l}\frac{\sqrt{3} n}{3}(n=1,2) \\ n-1(n \geqslant 3)\end{array}\right.$." 04dd08a42dce,"Heather and Kyle need to mow a lawn and paint a room. If Heather does both jobs by herself, it will take her a total of nine hours. If Heather mows the lawn and, after she finishes, Kyle paints the room, it will take them a total of eight hours. If Kyle mows the lawn and, after he finishes, Heather paints the room, it will take them a total of seven hours. If Kyle does both jobs by himself, it will take him a total of six hours. It takes Kyle twice as long to paint the room as it does for him to mow the lawn. The number of hours it would take the two of them to complete the two tasks if they worked together to mow the lawn and then worked together to paint the room is a fraction $\tfrac{m}{n}$where $m$ and $n$ are relatively prime positive integers. Find $m + n$.",41,medium,"1. Let \( H_m \) and \( H_p \) be the hours Heather takes to mow the lawn and paint the room, respectively. Similarly, let \( K_m \) and \( K_p \) be the hours Kyle takes to mow the lawn and paint the room, respectively. 2. From the problem, we have the following information: - Heather does both jobs in 9 hours: \( H_m + H_p = 9 \) - Heather mows the lawn and Kyle paints the room in 8 hours: \( H_m + K_p = 8 \) - Kyle mows the lawn and Heather paints the room in 7 hours: \( K_m + H_p = 7 \) - Kyle does both jobs in 6 hours: \( K_m + K_p = 6 \) - Kyle takes twice as long to paint the room as to mow the lawn: \( K_p = 2K_m \) 3. Using \( K_p = 2K_m \) in \( K_m + K_p = 6 \): \[ K_m + 2K_m = 6 \implies 3K_m = 6 \implies K_m = 2 \] \[ K_p = 2K_m = 2 \times 2 = 4 \] 4. Substitute \( K_m = 2 \) and \( K_p = 4 \) into the equations: - \( H_m + K_p = 8 \): \[ H_m + 4 = 8 \implies H_m = 4 \] - \( K_m + H_p = 7 \): \[ 2 + H_p = 7 \implies H_p = 5 \] 5. Verify the values: - \( H_m + H_p = 4 + 5 = 9 \) (correct) - \( K_m + K_p = 2 + 4 = 6 \) (correct) 6. Calculate the combined work rates: - Heather and Kyle together mow the lawn: \[ \text{Rate of Heather mowing} = \frac{1}{H_m} = \frac{1}{4} \] \[ \text{Rate of Kyle mowing} = \frac{1}{K_m} = \frac{1}{2} \] \[ \text{Combined rate} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \] \[ \text{Time to mow together} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \text{ hours} \] - Heather and Kyle together paint the room: \[ \text{Rate of Heather painting} = \frac{1}{H_p} = \frac{1}{5} \] \[ \text{Rate of Kyle painting} = \frac{1}{K_p} = \frac{1}{4} \] \[ \text{Combined rate} = \frac{1}{5} + \frac{1}{4} = \frac{4}{20} + \frac{5}{20} = \frac{9}{20} \] \[ \text{Time to paint together} = \frac{1}{\frac{9}{20}} = \frac{20}{9} \text{ hours} \] 7. Total time to complete both tasks together: \[ \text{Total time} = \frac{4}{3} + \frac{20}{9} = \frac{12}{9} + \frac{20}{9} = \frac{32}{9} \text{ hours} \] 8. The fraction \(\frac{32}{9}\) is in simplest form, so \( m = 32 \) and \( n = 9 \). The final answer is \( 32 + 9 = \boxed{41} \)" 5b9f9da9c7db,"17. Let $S_{k}$ denote the sum of the $k$-th powers of the roots of the polynomial $x^{3}-5 x^{2}+8 x-13$, $S_{0}=3, S_{1}=5, S_{2}=9$. Let $a, b, c \in \mathbf{R}$, and $$ S_{k+1}=a S_{k}+b S_{k-1}+c S_{k-2}(k=2,3, \cdots) \text {. } $$ Then $a+b+c=(\quad$. (A) -6 (B) 0 (C) 6 (D) 10 (E) 26",See reasoning trace,medium,"By Vieta's formulas, we know $$ p+q+r=5, p q+q r+r p=8, p q r=13 \text {. } $$ Thus, $5 S_{k}+13 S_{k-2}$ $$ \begin{aligned} = & (p+q+r)\left(p^{k}+q^{k}+r^{k}\right)+ \\ & p q r\left(p^{k-2}+q^{k-2}+r^{k-2}\right) \\ = & p^{k+1}+q^{k+1}+r^{k+1}+p^{k} q+p^{k} r+q^{k} p+ \\ & q^{k} r+r^{k} p+r^{k} q+p^{k-1} q r+q^{k-1} p r+r^{k-1} p q \\ = & p^{k+1}+q^{k+1}+r^{k+1}+ \\ & \left(p^{k-1}+q^{k-1}+r^{k-1}\right)(p q+q r+r p) \\ = & S_{k+1}+8 S_{k-1} \\ \Rightarrow & S_{k+1}=5 S_{k}-8 S_{k-1}+13 S_{k-2} \\ & =a S_{k}+b S_{k-1}+c S_{k-2} \\ \Rightarrow & a=5, b=-8, c=13 \\ \Rightarrow & a+b+c=10 . \end{aligned} $$ 17. D. Let $p, q, r$ be the three roots of the polynomial $x^{3}-5 x^{2}+8 x-13$." 1e12bc3178e6,"10.1. Having found some polynomial of the sixth degree $x^{6}+a_{1} x^{5}+\ldots+a_{5} x+a_{6}$ with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt[3]{5}$, write in the answer the sum of its coefficients $a_{1}+a_{2}+\ldots+a_{6}$.",-47,medium,"Answer: -47. Solution. $x=\sqrt{2}+\sqrt[3]{5} \Rightarrow (x-\sqrt{2})^{3}=5 \Rightarrow x^{3}-3 x^{2} \sqrt{2}+3 x \cdot 2-2 \sqrt{2}=5$ $\Rightarrow x^{3}+6 x-5=(3 x^{2}+2) \cdot \sqrt{2} \Rightarrow (x^{3}+6 x-5)^{2}=(3 x^{2}+2)^{2} \cdot 2$ $\Rightarrow P_{6}(x)=(x^{3}+6 x-5)^{2}-2(3 x^{2}+2)^{2}$. In this case, the polynomial itself does not need to be explicitly obtained (for reference, note that this is the polynomial $x^{6}-6 x^{4}-10 x^{3}+12 x^{2}-60 x+17$), since we need the sum: $a_{1}+\ldots+a_{6}=P_{6}(1)-1=2^{2}-2 \cdot 5^{2}-1=-47$." 376f89ded075,"18. Ivana has two identical dice and on the faces of each are the numbers $-3,-2,-1,0,1,2$. If she throws her dice and multiplies the results, what is the probability that their product is negative? A $\frac{1}{4}$ B $\frac{11}{36}$ C $\frac{1}{3}$ D $\frac{13}{36}$ E $\frac{1}{2}$","\frac{6}{36}$, or if the first die is positive and the second is negative, with probability $\frac{2",easy,"18. C We can get a negative product if the first die is negative and the second positive, with probability $\frac{3}{6} \times \frac{2}{6}=\frac{6}{36}$, or if the first die is positive and the second is negative, with probability $\frac{2}{6} \times \frac{3}{6}=\frac{6}{36}$. Together this gives a probability of $\frac{12}{36}=\frac{1}{3}$." e6533ac67ea5,"A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player? $\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$",\mathrm{(E),easy,"The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{\mathrm{(E)}\ 6}$." 08f9a6476c50,"## Task $1 / 84$ Solve the equation $1984^{\lg x}=2 \cdot 1984^{2}-x^{\lg 1984}$ in real numbers $x$.",100$ as the unique solution independent of the parameter a.,easy,"The given equation is a special case of the equation $$ a^{\lg x}=2 a^{2}-x^{\lg a} $$ It holds that $$ \lg a^{\lg x}=\lg x \cdot \lg a=\lg x^{\lg a} $$ Thus, due to the uniqueness of the lg function, $$ a^{\lg x}=x^{\lg a} $$ Therefore, (1) is equivalent to the equation $$ a^{\lg x}=2 a^{2}-a^{\lg x} $$ By further equivalent transformation, we get $$ 2 a^{\lg x}=2 a^{2} \quad ; \quad \lg x=2 $$ and thus $x=100$ as the unique solution independent of the parameter a." da26d28bbcfd,5. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.,"5026$, at this point $n=7$.)",easy,"(Hint: Using the inequality $\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x}$, solve for $x=5026$, at this point $n=7$.)" 1320a2d73503,"Some unit-radius circles are drawn on the plane, and we color the center of each circle blue. We mark some points on the circumference of the circles in red such that exactly 2 red points lie on each circumference. What is the maximum number of blue points if there are a total of 25 colored points?",See reasoning trace,medium,"Translate the given text into English, preserving the original line breaks and formatting: --- Translate the text: Move a unit circle four times in the same direction by a distance of 0.0002. ![](https://cdn.mathpix.com/cropped/2024_05_02_9b69772ab09b921b6263g-1.jpg?height=228&width=336&top_left_y=222&top_left_x=884) The centers of the five resulting unit circles lie on a segment of length 0.0008; therefore, any two of them have two common points, and no three of them have a common point. To prove the latter property, assume that some three circles have a common point $Q$. Then the centers of these three circles would lie on the circumference of a unit circle centered at $Q$, which is impossible since the centers of the five circles are collinear. Now color the five centers red and the intersection points blue. Any two circles have two blue (intersection) points, which uniquely determine the two circles, so the number of blue points is $2 \cdot \binom{5}{2} = 20$. The conditions of the problem are thus satisfied by 20 unit circles whose centers are the blue points. Therefore, the number of blue points can be 20. We will show that more than 20 is not possible. With more than 20 blue points, there could be at most 4 red points. However, only points that are the intersection points of the unit circles drawn around the red points can be blue, since otherwise they could not be at a unit distance from at least two red points. However, four circles can have at most $2 \cdot \binom{4}{2} = 12$ intersection points (any two circles can have at most two). Therefore, with at most 4 red points, there can be at most 12 blue points, which is clearly a contradiction. With this, we are done: we have shown that there cannot be more than 20 blue points, and we have provided an example with 20. Hegedüs Dániel (Budapesti Fazekas M. Gyak. Ált. Isk. és Gimn., 10th grade) ---" 53cda03a76d1,"20. Ruth and Sarah decide to have a race. Ruth runs around the perimeter of the pool shown in the diagram while Sarah swims lengths of the pool. Ruth runs three times as fast as Sarah swims. Sarah swims six lengths of the pool in the same time Ruth runs around the pool five times. How wide is the pool? A $25 \mathrm{~m}$ B $40 \mathrm{~m}$ C $50 \mathrm{~m}$ D $80 \mathrm{~m}$ E $180 \mathrm{~m}$",(500+10 x) \mathrm{m}$. The total distance Sarah swims is $6 \times 50 \mathrm{~m}=300 \mathrm{~m}$.,easy,"20. B Let the width of the pool be $x \mathrm{~m}$. Therefore the total distance Ruth runs is $5(2 \times 50+2 x) \mathrm{m}=(500+10 x) \mathrm{m}$. The total distance Sarah swims is $6 \times 50 \mathrm{~m}=300 \mathrm{~m}$. Since Ruth runs three times as fast as Sarah swims, $500+10 x=3 \times 300$. Therefore $10 x=400$ and hence $x=40$." ceca1e8884a3,"$1 \cdot 38$ If $\left(r+\frac{1}{r}\right)^{2}=3$, then $r^{3}+\frac{1}{r^{3}}$ equals (A) 1 . (B) 2 . (C) 0 . (D) 3 . (E) 6 . (3rd American High School Mathematics Examination, 1952)",$(C)$,easy,"[Solution] From the assumption $\left(r+\frac{1}{r}\right)^{2}=3$, we know $r^{2}-1+\frac{1}{r^{2}}=0$, thus $r^{3}+\frac{1}{r^{3}}=\left(r+\frac{1}{r}\right)\left(r^{2}-1+\frac{1}{r^{2}}\right)=0$. Therefore, the answer is $(C)$." 5fc3c7399e40,"Example 11 Find the minimum value of the bivariate function $f(x, y)=(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}$.","(x-y)^{2}+\left[(x+1)-\left(-\frac{1}{y}\right)\right]^{2}$, we consider the square of the distance ",medium,"Solution: Since $(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}$ $$ =(x-y)^{2}+\left[(-x-1)-\frac{1}{y}\right]^{2}, $$ the above expression can be seen as the square of the distance between a point $(x,-x-1)$ on the line $y=-x-1$ and a point $\left(y, \frac{1}{y}\right)$ on the hyperbola $x y=1$, as shown in Figure 4-1. Therefore, the minimum value sought is the square of the shortest distance between the line and the hyperbola. From the graph of the function, it is known that the coordinates of points $A, B, C$ are $(1,1)$, $(-1,1)$, and $\left(-\frac{1}{2},-\frac{1}{2}\right)$, respectively. Thus, the required minimum value is $|B C|$. Also, $|B C|=\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{3}}{2}$, and $|B C|^{2}=\frac{1}{2}$. Hence, $f(x, y)_{\text {min }}=\frac{1}{2}$ is the answer. Note: By $(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}=(x-y)^{2}+\left[(x+1)-\left(-\frac{1}{y}\right)\right]^{2}$, we consider the square of the distance between a point $\left(y,-\frac{1}{y}\right)$ on the hyperbola $x y=-1$ and the line $y=x+1$." 21aa2402056f,"15. Let $\boldsymbol{i}, \boldsymbol{j}$ be the unit vectors in the positive directions of the $x, y$ axes in a Cartesian coordinate system. If the vectors $\boldsymbol{a}=(x+2) \boldsymbol{i}+y \boldsymbol{j}$, $\boldsymbol{b}=(x-2) \boldsymbol{i}+y \boldsymbol{j}$, and $|\boldsymbol{a}|-|\boldsymbol{b}|=2$. (1) Find the equation of the locus of point $P(x, y)$ that satisfies the above conditions; (2) Let $A(-1,0), F(2,0)$. Does there exist a constant $\lambda(\lambda>0)$ such that $\angle P F A=\lambda \angle P A F$ always holds? Prove your conclusion.","2\), such that \(\angle P F A=2 \angle P A F\) always holds.",medium,"(1) According to the definition of a hyperbola, we have \(a=1, c=2, b=\sqrt{3}\). The trajectory equation of point \(P(x, y)\) is \(x^{2}-\frac{y^{2}}{3}=1(x>0)\). (2) Let \(P(\sec \theta, \sqrt{3} \tan \theta)\), then \(\tan \angle P A F=\frac{\sqrt{3} \tan \theta}{\sec \theta+1}\) \[ \begin{array}{l} =\frac{\sqrt{3} \sin \theta}{\cos \theta+1}, \tan \angle P F A=\frac{\sqrt{3} \tan \theta}{2-\sec \theta}=\frac{\sqrt{3} \sin \theta}{2 \cos \theta-1}. \\ \text { Since } \tan 2 \angle P A F=\frac{\frac{2 \sqrt{3} \sin \theta}{\cos \theta+1}}{1-\left(\frac{\sqrt{3} \sin \theta}{\cos \theta+1}\right)^{2}} \\ =\frac{2 \sqrt{3} \sin \theta(\cos \theta+1)}{(\cos \theta+1)^{2}-3 \sin ^{2} \theta}=\frac{\sqrt{3} \sin \theta(\cos \theta+1)}{2 \cos ^{2} \theta+\cos \theta-1}=\frac{\sqrt{3} \sin \theta}{2 \cos \theta-1}=\tan \angle P F A, \end{array} \] Therefore, there exists a constant \(\lambda=2\), such that \(\angle P F A=2 \angle P A F\) always holds." 6c5381f784ea,"19. Given $m \in\{11,13,15,17,19\}$, $n \in\{1999,2000, \cdots, 2018\}$. Then the probability that the unit digit of $m^{n}$ is 1 is ( ). (A) $\frac{1}{5}$ (B) $\frac{1}{4}$ (C) $\frac{3}{10}$ (D) $\frac{7}{20}$ (E) $\frac{2}{5}$",\frac{2}{5}$.,medium,"19. E. Consider different cases. When $m=11$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 20 ways to choose; When $m=13$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose; When $m=15$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 5; When $m=17$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose; When $m=19$, $n \in\{2000,2002, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 10 ways to choose. In summary, the required result is $\frac{20+5+5+10}{5 \times 20}=\frac{2}{5}$." b9a7476b3657,"Example $1$, $, $ Given line segments $b, c$. Find the line segment $a$ such that $$ \frac{1}{a}=\frac{1}{b}+\frac{1}{c} . $$",See reasoning trace,medium,"Analysis: Consider $\mathrm{a}$ as $\mathrm{f}$, $\mathrm{b}$ as $\mathrm{u}$, and $\mathrm{c}$ as $\mathrm{v}$. Using the above physical construction method, we can construct $\mathrm{a}$. Construction: (1) Draw a line 1, take a point 0 on 1, and on both sides of 0 on 1, mark $O D=b$ and $O E=c$; (2) Draw perpendiculars DM, OT, and EN from D, O, and E to 1, and take an appropriate length for DM; (3) Extend MO to intersect EN at N; (4) Draw MT $\downarrow \mathrm{TO}$ at T; (5) Connect $\mathbf{T N}$ to intersect 1 at $\mathrm{F}$. Then OF is the desired length (Figure 2). This construction method is more complex than the usual algebraic method, which is to construct $\mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}$. However, it is more intuitive and does not require algebraic manipulation. Is there a simpler construction method? Yes, but we need to prove a proposition first: If points $\mathrm{C}$ and $\mathrm{B}$ are collinear, then $\frac{1}{\mathrm{OC}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}$. Proof: Extend $\mathrm{BO}$ to $\mathrm{D}$ such that $\mathrm{OD}=\mathrm{OA}$, and connect $\mathrm{DA}$. Since $\angle \mathrm{AOD}=180^{\circ}-\angle \mathrm{AOB}=60^{\circ}$ and $\mathrm{OD}=\mathrm{OA}$, $\triangle O A D$ is an equilateral triangle. - Therefore, $\angle \mathrm{ADO}=60^{\circ}=\frac{1}{2} \angle \mathrm{AOB}=\angle \mathrm{CUB}$, so OC is parallel to AD. Thus, $\triangle B C O \sim \triangle B B$, $$ \begin{array}{l} \frac{O B}{B D}=\frac{O C}{A D}, \quad \frac{O B}{O B+O A}=\frac{O C}{O A}, \\ O C=\frac{O A \cdot O B}{O A+O B} . \\ \end{array} $$ Therefore, $\frac{1}{O C}=\frac{1}{O A}+\frac{1}{O B}$. According to this proposition, the construction for Example 1 is easily solved." 30a7e0357237,"8. (4 points) Pleasant Goat and Lazy Goat have a total of 70 stamps. Pleasant Goat has 5 more than 4 times the number of stamps Lazy Goat has. Pleasant Goat has $\qquad$ stamps, and Lazy Goat has $\qquad$ stamps.","Happy Sheep has 57, Lazy Sheep has 13",easy,"【Solution】Solution: Let Lazy Sheep have $x$ tickets, then Happy Sheep has $(4 x+5)$ stamps, $$ \begin{array}{c} x+(4 x+5)=70 \\ 5 x+5=70 \\ 5 x=65 \\ x=13 \\ 13 \times 4+5=57 \text { (pieces) } \end{array} $$ Answer: Happy Sheep has 57, Lazy Sheep has 13. Therefore, the answer is: $57 ; 13$." 43aef52df66b,"9.11. The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that the sum of any seven consecutive numbers is negative, and the sum of any eleven consecutive numbers is positive. For what largest $n$ is this possible? 118 Chapter 9. Computation of Sums and Products $$ \text { 9.3. Sums } S_{k}(n)=1^{k}+2^{k}+\ldots+n^{k} $$ The sum $1+2+3+\ldots+n$ can be computed as follows. Add the equations $(k+1)^{2}=k^{2}+2 k+1$ for $k=1,2, \ldots, n$. After simplification, we get $(n+1)^{2}=1+2 S_{1}(n)+n$, where $S_{1}(n)$ is the desired sum. Therefore, $S_{1}(n)=\frac{n(n+1)}{2}$.","16$, such a sequence exists: $5,5,-13,5$, $5,5,-13,5,5,-13,5,5,5,-13,5,5$.",easy,"9.11. According to problem $9.10\left(a_{1}+a_{2}+\ldots+a_{7}\right)+\left(a_{2}+\ldots+a_{8}\right)+\ldots$ $\ldots+\left(a_{11}+\ldots+a_{17}\right)=\left(a_{1}+a_{2}+\ldots+a_{11}\right)+\left(a_{2}+\ldots+a_{12}\right)+\ldots+$ $+\left(a_{7}+\ldots+a_{17}\right)$. Therefore, $n<17$. For $n=16$, such a sequence exists: $5,5,-13,5$, $5,5,-13,5,5,-13,5,5,5,-13,5,5$." 580dfd49169d,"4. According to the 2002 March 5th Ninth National People's Congress Fifth Session 《Government Work Report》: ""In 2001, the Gross Domestic Product (GDP) reached 9593.3 billion yuan, an increase of $7.3 \%$ over the previous year. If the annual GDP growth rate remains the same during the 'Tenth Five-Year Plan' period (2001-2005), then by the end of the 'Tenth Five-Year Plan', China's annual GDP will be approximately $\qquad$ .""",See reasoning trace,easy,4. 127165 Prompt: $95933 \times(1+7.3 \%)^{4}$ bd1d099bb06d,"In the drawing below, points $E$ and $F$ belong to the sides $A B$ and $B D$ of triangle $\triangle A B D$ in such a way that $A E=A C$ and $C D=F D$. If $\angle A B D=60^{\circ}$, determine the measure of the angle $\angle E C F$. ![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-06.jpg?height=417&width=499&top_left_y=591&top_left_x=687)",\angle E A C$ and $2 \beta=\angle F D C$. Since triangles $\triangle E A C$ and $\triangle F D C$ ar,medium,"Solution Let $2 \alpha=\angle E A C$ and $2 \beta=\angle F D C$. Since triangles $\triangle E A C$ and $\triangle F D C$ are isosceles, it follows that $\angle A C E=\angle A E C=90^{\circ}-\alpha$ and $\angle D C F=\angle C F D=90^{\circ}-\beta$. Consequently, $\angle E C F=\alpha+\beta$. Now analyzing the sum of the angles of triangle $\triangle A B D$, we have $60^{\circ}+2 \alpha+2 \beta=180^{\circ}$, that is, $60^{\circ}=\alpha+\beta$. Since we already know that $\angle E C F=\alpha+\beta$, then $\angle E C F=60^{\circ}$." 9d0e726c9858,"3. Papa always picked up Misha from kindergarten at the same time and in the same car. But today, after firmly establishing that there would be pea porridge for dinner, Misha resolutely left the kindergarten and set off down the road to meet Papa. Having walked 5 km, Misha met Papa, got into the car, and they arrived home 10 minutes earlier than usual. At what speed does Misha's Papa drive?",- 7 points; only answer - 0 points,easy,"3. Speed: 60 km/h (or 1 km/min). Today, Misha and his dad saved 10 minutes that were previously spent driving 10 km (from the meeting point to the kindergarten and back). This means that in 1 minute, Misha's dad drives 1 km, i.e., his speed is 60 km/h. Criteria: Correct reasoning and correct answer - 7 points; only answer - 0 points." 4a306b9986ac,"Three generous friends, each with some money, redistribute the money as followed: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have? $\textbf{(A)}\ 108\qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 216\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 288$",D,medium,"If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \times 2 = 72$ dollars after the first step. Then Jan doubles his money, and Toy has $72 \times 2 = 144$ dollars after the second step. Then Toy doubles whatever Amy and Jan have. Since Toy ended up with $36$, he spent $144 - 36 = 108$ to double their money. Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total. And, just before this step, Toy had $144$ dollars. Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\boxed{D}$" 76a61d410157,"2.120. $\frac{5 \sqrt[3]{4 \sqrt[3]{192}}+7 \sqrt[3]{18 \sqrt[3]{81}}}{\sqrt[3]{12 \sqrt[3]{24}+6 \sqrt[3]{375}}}$. 2.120. $\frac{5 \sqrt[3]{4 \sqrt[3]{192}}+7 \sqrt[3]{18 \sqrt[3]{81}}}{\sqrt[3]{12 \sqrt[3]{24}+6 \sqrt[3]{375}}}$. The expression is already in a mathematical format and does not require translation. However, if you need an explanation or simplification of the expression, please let me know!",$\frac{31}{3}$,medium,"Solution. $$ \begin{aligned} & \frac{5 \sqrt[3]{4 \sqrt[3]{192}}+7 \sqrt[3]{18 \sqrt[3]{81}}}{\sqrt[3]{12 \sqrt[3]{24}+6 \sqrt[3]{375}}}=\frac{5 \sqrt[3]{4 \sqrt[3]{64 \cdot 3}}+7 \sqrt[3]{18 \sqrt[3]{27 \cdot 3}}}{\sqrt[3]{12 \sqrt[3]{8 \cdot 3}+6 \sqrt[3]{125 \cdot 3}}}= \\ & =\frac{5 \sqrt[3]{4 \cdot 4 \sqrt[3]{3}}+7 \sqrt[3]{18 \cdot 3 \sqrt[3]{3}}}{\sqrt[3]{12 \cdot 2 \sqrt[3]{3}+6 \cdot 5 \sqrt[3]{3}}}=\frac{5 \sqrt[3]{16 \sqrt[3]{3}}+7 \sqrt[3]{54 \sqrt[3]{3}}}{\sqrt[3]{24 \sqrt[3]{3}+30 \sqrt[3]{3}}}= \end{aligned} $$ $=\frac{5 \sqrt[3]{8 \cdot 2 \sqrt[3]{3}}+7 \sqrt[3]{27 \cdot 2 \sqrt[3]{3}}}{\sqrt[3]{54 \sqrt[3]{3}}}=\frac{5 \cdot 2 \sqrt[3]{2 \sqrt[3]{3}}+7 \cdot 3 \sqrt[3]{2 \sqrt[3]{3}}}{\sqrt[3]{27 \cdot 2 \sqrt[3]{3}}}=$ $=\frac{10 \sqrt[3]{2 \sqrt[3]{3}}+21 \sqrt[3]{2 \sqrt[3]{3}}}{3 \sqrt[3]{2 \sqrt[3]{3}}}=\frac{31 \sqrt[3]{2 \sqrt[3]{3}}}{3 \sqrt[3]{2 \sqrt[3]{3}}}=\frac{31}{3}$. Answer: $\frac{31}{3}$." c6ef0abf2b8e,"Condition of the Find the derivative. $$ y=\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}, x>0 $$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & y^{\prime}=\left(\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}\right)^{\prime}= \\ & =\frac{1}{24} \cdot 2 x \cdot \sqrt{x^{2}-4}+\frac{1}{24}\left(x^{2}+8\right) \cdot \frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x+\frac{2 x}{16} \cdot \arcsin \frac{2}{x}+\frac{x^{2}}{16} \cdot \frac{1}{\sqrt{1-\left(\frac{2}{x}\right)^{2}}} \cdot\left(-\frac{2}{x^{2}}\right)= \\ & =\frac{2 x\left(x^{2}-4\right)}{24 \sqrt{x^{2}-4}}+\frac{x\left(x^{2}+8\right)}{24 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x}-\frac{1}{8} \cdot \frac{x}{\sqrt{x^{2}-4}}= \\ & =\frac{3 x^{3}-3 x}{24 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x}=\frac{x^{3}-x}{8 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x} \end{aligned} $$ ## Problem Kuznetsov Differentiation 13-1" 5f02ff53b310,"There is a four-digit number $\overline{a b c d}, a+b+c+d=26, b \cdot d$'s tens digit equals $a+c$, and $b d-c^{2}$ is an integer power of 2. Try to find this four-digit number (explain the reasoning).","81$, so $1 \leqslant a+c \leqslant 8$. Thus $26=(b+d)+$ $(a+c) \leqslant(b+d)+8$, which means $b+d \",easy,"Parse Yan $0 \leqslant b \cdot d \leqslant 9 \cdot 9=81$, so $1 \leqslant a+c \leqslant 8$. Thus $26=(b+d)+$ $(a+c) \leqslant(b+d)+8$, which means $b+d \geqslant 18$. Therefore, $b=d=9, a+c=8$. Since $2^{m}=b d-c^{2}=$ $(9+c)(9-c)$, we have $c=7$, thus $a \doteq 1$. Hence, the required four-digit number is 1979." 0c261fcee13e,"9) To glue together the faces of two unit cubes, 0.25 grams of glue are required. How much glue is needed to build a $5 \times 5 \times 5$ cube from 125 unit cubes? (N.B. to ensure greater stability, all pairs of faces in contact are glued) (A) $180 \mathrm{~g}$ (B) $150 \mathrm{~g}$ (C) $90 \mathrm{~g}$ (D) $75 \mathrm{~g}$ (E) $125 \mathrm{~g}$.",See reasoning trace,easy,"9) The answer is $(D)$ The total number of faces is 900, and the non-glued faces are only the external ones, whose number is equal to the total surface area of the cube, which is 150. Therefore, there are 375 pairs of faces to glue, and thus 75 grams of glue are required." 1a98cf0dc3ce,". In the decimal writing of $A$, the digits appear in (strictly) increasing order from left to right. What is the sum of the digits of $9 A$?","10 A-A$ are $a_{1}, a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{k-1}-a_{k-2}, a_{k}-a_{k-1}-1,10-a_{k}$. Th",easy,". Let $A=\overline{a_{1} a_{2} \cdots a_{k}}$ be the decimal representation of $A$. By performing the subtraction $$ \begin{aligned} & \begin{array}{cccccc} a_{1} & a_{2} & a_{3} & \cdots & a_{k} & 0 \end{array} \\ & \begin{array}{llllll} - & a_{1} & a_{2} & \cdots & a_{k-1} & a_{k} \\ \hline \end{array} \end{aligned} $$ we find that the digits of $9 A=10 A-A$ are $a_{1}, a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{k-1}-a_{k-2}, a_{k}-a_{k-1}-1,10-a_{k}$. Their sum is $10-1=9$." 11e2b8145c48,"5. A knight is placed in each cell of a chessboard. What is the minimum number of knights that can be removed from the board so that no knight remains that attacks exactly four other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.)",8 knights,medium,"Answer: 8 knights. Solution. First, we will show that no fewer than 8 knights need to be removed. On the left diagram, all knights that attack exactly 4 squares of the board are marked (for convenience, they are highlighted in different colors). Let's call such knights bad. To stop a knight from attacking four others, one must remove either this knight or one of those it attacks. We will show that even to get rid of the bad black knights, one must free no fewer than 8 squares. On the middle diagram, the circles mark the squares under attack by the bad black knights. Three bad black knights in the upper left corner attack four squares marked with the symbol $\odot$. If only one of these squares is freed, one of the black knights will remain bad. Therefore, for this trio of knights, at least two squares need to be cleared. Thus, for all four trios, at least $4 \cdot 2 = 8$ squares need to be freed. Now, let's provide an example showing that 8 knights are sufficient. On the right diagram, the knights that need to be removed from the board are marked. ![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-07.jpg?height=304&width=1272&top_left_y=434&top_left_x=450)" 47a0fb147452,"Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 unit to the left, the second step is to jump 2 units to the right, the third step is to jump 3 units to the left, the fourth step is to jump 4 units to the right, and so on. (1) When it jumps to the tenth step, how far is the electronic flea from the origin? How many times has the red light been lit? (2) If the electronic flea needs to light up the red light ten times, what is the minimum number of jumps it needs to make? At this point, how far is it from the origin?",See reasoning trace,medium,"(1) Notice $$ \begin{array}{l} S=\mid-2-1+2-3+4-5+\cdots-9+101 \\ =1-2+1 \times 5 \mid=3 . \end{array} $$ Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times. (2) In fact, after the electronic flea jumps four times, it returns to the origin, and the red light has already flashed four times (including the initial jump). After that, the red light flashes once for each odd-numbered jump. Therefore, to make the red light flash ten times, the electronic flea needs to jump $4+$ $2 \times 6-1=15$ times. At this point, the distance of the electronic flea from the origin is $$ \begin{array}{l} S=|-2-1+2-3+4-5+\cdots-15| \\ =|-2+1 \times 7-15|=10 . \end{array} $$" 45bc10344b89,"What is the value of $2-(-2)^{-2}$? $\textbf{(A) } -2\qquad\textbf{(B) } \dfrac{1}{16}\qquad\textbf{(C) } \dfrac{7}{4}\qquad\textbf{(D) } \dfrac{9}{4}\qquad\textbf{(E) } 6$",\textbf{(C),easy,$2-(-2)^{-2}=2-\frac{1}{(-2)^2}=2-\frac{1}{4}=\frac{8}{4}-\frac{1}{4}=\boxed{\textbf{(C) }\frac{7}{4}}$ 0692363758c6,,"\frac{2}{3} \overline{A B}=\frac{2}{3}(6 ; \quad-3$; $-3)=(4 ;-2 ;-2)$. Therefore, $C(1+4 ;-1-2 ; 2-",easy,"Solution. Obviously, $\overline{A C}=\frac{2}{3} \overline{A B}=\frac{2}{3}(6 ; \quad-3$; $-3)=(4 ;-2 ;-2)$. Therefore, $C(1+4 ;-1-2 ; 2-2)$, i.e., $C(5 ;-3 ; 0)$." 53d16daaeac1,"Example 2 A bug crawls along a triangular iron ring, at each vertex the bug has an equal chance of crawling to one of the other two vertices. Then the probability that the bug is back at the starting point after $n$ moves is $\qquad$ [2]",See reasoning trace,medium,"【Analysis】Suppose the bug starts from vertex $A$ of $\triangle A B C$. Since there are two possible directions, clockwise or counterclockwise, at each vertex, there are $2^{n}$ possible ways of movement after $n$ steps. Let the number of ways the bug can end up at points $A, B, C$ after $n$ moves be denoted as $A_{n}, B_{n}, C_{n}$, respectively. Assume that a clockwise move from any vertex is denoted as “1” and a counterclockwise move as “-1”. Each walk of $n$ steps from point $A$ can correspond to an ordered array $$ \left(x_{1}, x_{2}, \cdots, x_{n}\right)\left(x_{i} \in\{1,-1\}, i=1,2, \cdots, n\right) $$ Obviously, there are $2^{n}$ distinct ordered arrays, i.e., $$ A_{n}+B_{n}+C_{n}=2^{n} \text {. } $$ Assume that in a certain walk, there are $k$ counterclockwise moves, and let $$ p(k)=\sum_{i=1}^{n} x_{i}=n-2 k(k=0,1, \cdots, n) \text {. } $$ Since $p(k) \equiv 0,1,2(\bmod 3)$ corresponds to the bug ending at points $A, C, B$ respectively, and the same $p(k)$ value corresponds to $\mathrm{C}_{n}^{k}$ different walks. The $n+1$ distinct $k$ values can be divided into three categories: $\left\{k_{i}^{(0)}\right\}, \left\{k_{i}^{(1)}\right\}, \left\{k_{i}^{(2)}\right\}$, satisfying $$ \begin{array}{l} p\left(k_{i}^{(0)}\right) \equiv 0(\bmod 3)(i=1,2, \cdots, s), \\ p\left(k_{i}^{(1)}\right) \equiv 1(\bmod 3)(i=1,2, \cdots, t), \\ p\left(k_{i}^{(2)}\right) \equiv 2(\bmod 3)(i=1,2, \cdots, r) . \\ \text { Clearly, } s + t + r = n + 1 . \\ \text { Since } A_{n}=\sum_{i=1}^{s} \mathrm{C}_{n}^{k_{n}^{(1)}}(i=1,2, \cdots, s), \\ C_{n}=\sum_{i=1}^{r} \mathrm{C}_{n}^{k_{n}^{(1)}}(i=1,2, \cdots, t), \\ B_{n}=\sum_{i=1}^{r} \mathrm{C}_{n}^{k_{n}^{(3)}}(i=1,2, \cdots, r), \end{array} $$ How can we find $A_{n}, B_{n}, C_{n}$? We need to find a special number that can easily yield $A_{n}, B_{n}, C_{n}$. $$ \begin{array}{l} \text { Note that, } \omega=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3} \text { (a cube root of 1). } \\ \text { Then } \omega^{p\left(k_{i}^{(0)}\right)}=\omega^{0}=1, \omega^{p\left(k_{k_{1}^{(1)}}\right)}=\omega^{1}=\omega, \\ \omega^{p\left(x_{i}^{(2)}\right)}=\omega^{2}=\bar{\omega} \text {. } \\ \text { Hence } \sum_{i=0}^{n} \mathrm{C}_{n}^{k} \omega^{n-2 k}=\frac{1}{\omega^{n}} \sum_{i=0}^{n} \mathrm{C}_{n}^{k} \omega^{2 n-2 k} \\ =\frac{1}{\omega^{n}} \sum_{i=0}^{n} \mathrm{C}_{n}^{k}\left(\omega^{2}\right)^{n-k}=\frac{1}{\omega^{n}}\left(1+\omega^{2}\right)^{n} \\ =\frac{1}{\omega^{n}}(-\omega)^{n}=(-1)^{n} \text {. } \\ \text { Also } \sum_{i=0}^{n} \mathrm{C}_{n}^{k} \omega^{n-2 k} \\ =\sum_{i=1}^{1} \mathrm{C}_{n}^{k^{(0)}} \omega^{p\left(k_{i}^{(0)}\right)}+\sum_{i=1}^{1} \mathrm{C}_{n}^{k_{k}^{(i)}} \omega^{p\left(k_{i}^{(j)}\right)}+\sum_{i=1}^{1} \mathrm{C}_{n}^{k^{(i)}} \omega^{p\left(k_{i}^{(2)}\right)} \\ =A_{n}+C_{n} \omega+B_{n} \bar{\omega} \\ =A_{n}-\frac{1}{2}\left(B_{n}+C_{n}\right)+\frac{\sqrt{3}}{2}\left(C_{n}-B_{n}\right) \mathrm{i} \text {, } \\ \end{array} $$ Combining with equation (2) we get $$ A_{n}-\frac{1}{2}\left(B_{n}+C_{n}\right)=(-1)^{n}, C_{n}-B_{n}=0 . $$ Combining with equation (1) we solve to get $A_{n}=\frac{2^{n}+(-1)^{n} \times 2}{3}$. Thus, the probability that the bug returns to the starting point after $n$ moves is $$ P=\frac{2^{n}+(-1)^{n} \times 2}{3 \times 2^{n}} \text {. } $$ From the above two examples, it can be seen that the periodicity of $\omega^{n}$ can provide a unique approach to solving problems." 6bb74514617f,"9.207. $\left\{\begin{array}{l}0,2^{\cos x} \leq 1, \\ \frac{x-1}{2-x}+\frac{1}{2}>0 .\end{array}\right.$",$x \in\left(0 ; \frac{\pi}{2}\right]$,medium,"Solution. Domain of definition: $x \neq 2$. Rewrite the given system of inequalities as $\left\{\begin{array}{l}0.2^{\cos x} \leq 0.2^{0}, \\ \frac{2 x-2+2-x}{2(2-x)}>0\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x \geq 0, \\ x(x-2)<0\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}-\frac{\pi}{2}+2 \pi n \leq x \leq \frac{\pi}{2}+2 \pi n, \\ 01$, in this case, the right side of equation (1) is even, and the left side is the sum of $x$ odd numbers, hence $x$ must be even. Thus, equation (1) can be rewritten as $$ \begin{array}{c} 7^{x-2}(7+1)+7^{x-4}(7+1)+\cdots+(7+1)=2^{y-1}, \\ (7+1)\left(7^{x-2}+7^{x-4}+\cdots+1\right)=2^{y-1}, \\ 7^{x-2}+7^{x-4}+\cdots+1=2^{y-4} . \end{array} $$ From this, we know that $$ y \geqslant 4 \text {. } $$ If $y=4$, then $x=2$, thus we obtain the second set of natural number solutions $(x, y)=$ $(2,4)$. If $y>4$, then the right side of equation (2) is even, and the left side is the sum of $\frac{x}{2}$ odd numbers, therefore, $x$ must be a multiple of 4. Thus, equation (2) can be rewritten as $$ \begin{array}{c} 7^{x-4}\left(7^{2}+1\right)+7^{x-8}\left(7^{2}+1\right)+\cdots+\left(7^{2}+1\right)=2^{y-4}, \\ \left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\cdots+1\right)=2^{y-4}, \end{array} $$ From (3), we see that $2^{y-4}$ must be divisible by $7^{2}+1=50$, which is impossible. Therefore, when $y>4$, the equation has no natural number solutions. In summary, the equation has only two sets of natural number solutions: $$ (x, y)=(1,1),(2,4) . $$" 0a5212a27a0e,"一、Fill in the Blanks (Total 3 questions, 10 points each) 1. In a 1000-meter race, when A reaches the finish line, B is 50 meters from the finish line; when B reaches the finish line, C is 100 meters from the finish line. So when A reaches the finish line, C is $\qquad$ meters from the finish line.","855$ (meters), so C is 1000 $-855=145$ (meters) away from the finish line.",easy,"1. 145 1.【Solution】When A runs 1000 meters, B runs 950 meters, and when B runs 1000 meters, C runs 900 meters. Therefore, when A runs 1000 meters, C runs $950 \times \frac{900}{1000}=855$ (meters), so C is 1000 $-855=145$ (meters) away from the finish line." 175042330263,"In a regular pentagon $A B C D E$, an equilateral triangle $A B M$ is contained. Determine the size of the angle $B C M$. (L. Hozová) Hint. What are the sizes of the interior angles of a regular pentagon?",597&width=631&top_left_y=1846&top_left_x=718),medium,"The size of the internal angles of an equilateral triangle is $60^{\circ}$, the size of the internal angles of a regular pentagon is $108^{\circ}$. At vertex $B$, we find that the size of angle $C B M$ is $108^{\circ}-60^{\circ}=48^{\circ}$. Segments $A B, B C$, and $B M$ are congruent, so triangle $C B M$ is isosceles with base $C M$. The internal angles at the base are congruent and their sum is a right angle. The size of angle $B C M$ is therefore $\frac{1}{2}\left(180^{\circ}-48^{\circ}\right)=66^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_e57040b3fafc3d67d9eag-4.jpg?height=603&width=629&top_left_y=792&top_left_x=722) Note. A general $n$-sided polygon can be divided into $n-2$ triangles (whose vertices are the vertices of the $n$-sided polygon), so the sum of the sizes of its internal angles is $(n-2) \cdot 180^{\circ}$. A regular $n$-sided polygon has all internal angles congruent, so the size of each is $\frac{n-2}{n} \cdot 180^{\circ}$. This explains the initial relationships for $n=3$ and $n=5$. A regular $n$-sided polygon can also be divided into $n$ congruent isosceles triangles with a common vertex at the center of the $n$-sided polygon. For $n=5$, we get that angle $A S B$ has a size of $\frac{1}{5} \cdot 360^{\circ}=72^{\circ}$, so angles $S A B, S B A$, etc., have a size of $\frac{1}{2}\left(180^{\circ}-72^{\circ}\right)=54^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_e57040b3fafc3d67d9eag-4.jpg?height=597&width=631&top_left_y=1846&top_left_x=718)" 28ac22c8c606,,4,medium,"Solution. The answer is 4. The polynomial $x(x-1 / 2)(x-1)(x-2)$, with roots 0, 1/2, 1, and 2, satisfies the given bound. Suppose there was a polynomial with at least 5 distinct roots. It is important to note that a polynomial has a finite number of roots, which allows us to take maximums and minimums. If a polynomial has 5 roots, at least 4 would be non-zero, and we distinguish two cases. 1. If -1 is a root, there is at least one root (in fact, at least two) with a non-zero absolute value different from 1. Let $A$ be a root with the largest absolute value among all roots and assume that $|A|>1$. Since -1 and $A$ are roots, $-A$ is a root; but then $-A \cdot A=-A^{2}$ would also be a root and $\left|A^{2}\right|>|A|$, which would contradict the maximality of the absolute value of $A$. If there are no roots with an absolute value $>1$, take $A$ as one of the roots with the smallest absolute value among all non-zero roots. Then $|A|<1$, and since -1 and $A$ are roots, $-A$ is a root and so is $-A \cdot A=-A^{2}$. But then, $\left|A^{2}\right|<|A|$, which would contradict the minimality of the absolute value of $A$ among the non-zero roots. 2. If -1 is not a root, there are at least three non-zero roots with a modulus different from 1. There are then two roots $A, B$ with the two largest absolute values (greater than 1) or two roots with the two smallest absolute values $A, B$ (less than 1 and non-zero). In either case, multiplying $A \cdot B$ leads to a contradiction with the maximality or minimality." 84e797624e53,"If $x=11, y=-8$, and $2 x-3 z=5 y$, what is the value of $z$ ? (A) -6 (B) 13 (C) 54 (D) $\frac{62}{3}$ (E) $-\frac{71}{3}$",(D),easy,"Since $x=11, y=-8$ and $2 x-3 z=5 y$, then $2 \times 11-3 z=5 \times(-8)$ or $22-3 z=-40$. Therefore, $3 z=22+40=62$ and so $z=\frac{62}{3}$. ANSWER: (D)" e4d06e8bc19d,"271. Find the differentials of the functions: 1) $y=x^{3}-3^{x}$; 2) $F(\varphi)=\cos \frac{\varphi}{3}+\sin \frac{3}{\varphi}$ 3) $z=\ln \left(1+e^{10 x}\right)+\operatorname{arcctg} e^{5 x} ;$ calculate $\left.d z\right|_{x=0 ; d x=0,1}$","0$ and $d x=0.1$, we get $d z=0.25$.",medium,"Solution. To find the derivative of the given function and, by multiplying it by the differential of the independent variable, we obtain[^8]the desired differential of the given function: 1) $d y=y^{\prime} d x=\left(x^{3}-3^{x}\right)^{\prime} d x=\left(3 x^{2}-3^{x} \ln 3\right) d x$; 2) $d F(\varphi)=d\left(\cos \frac{\varphi}{3}+\sin \frac{3}{\varphi}\right)=\left(\cos \frac{\varphi}{3}+\sin \frac{3}{\varphi}\right)^{\prime} d \varphi=$ $=\left[-\sin \frac{\varphi}{3} \cdot\left(\frac{\varphi}{3}\right)^{\prime}+\cos \frac{3}{\varphi} \cdot\left(\frac{3}{\varphi}\right)^{\prime}\right] d \varphi=-\left(\frac{1}{3} \sin \frac{\varphi}{3}+\frac{3}{\varphi^{2}} \cos \frac{3}{\varphi}\right) d \varphi ;$ 3) $d z=\left[\frac{\left(1+e^{10 x}\right)^{\prime}}{1+e^{10 x}}-\frac{\left(e^{6 x}\right)^{\prime}}{1+e^{10 x}}\right] d x=\left(\frac{10 e^{10 x}}{1+e^{10 x}}-\frac{5 e^{6 x}}{1+e^{10 x}}\right) d x=$ $$ =\frac{5 e^{5 x}\left(2 e^{5 x}-1\right)}{1+e^{10 x}} d x $$ Assuming $x=0$ and $d x=0.1$, we get $d z=0.25$." 9f417d3ba48a,"(7 Ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ has a focal distance of $2 c$, if the three numbers $a$, $b$, $c$ form a geometric sequence in order, then its eccentricity is $\qquad$ .",\frac{\sqrt{5}-1}{2}$.,easy,"(7) Since $a^{2}=b^{2}+c^{2}, b^{2}=a c$, then $a^{2}=a c+c^{2}$, and $e=\frac{c}{a}$, so $e^{2}+e-1=0$, its positive root $e=\frac{\sqrt{5}-1}{2}$." d0801bc128db,"6. Find the smallest positive integer $n$ such that $$ \begin{array}{l} \sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\ <\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} . \end{array} $$",See reasoning trace,medium,"6. From the known, we must have $n \geqslant 2$ 013. At this time, $$ \begin{array}{l} \sqrt{\frac{n-2011}{2012}}4023, \\ \sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\ \Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\ \Leftrightarrow n \geqslant 4024 . \end{array} $$ From equations (1) and (2), when $n \geqslant 4024$, $$ \begin{array}{l} \sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}}\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} . \end{array} $$ In summary, the smallest positive integer $n$ that satisfies the condition is 4024." 979c246e7751,"$4 \cdot 20$ Given 9 points in space, where no 4 points are coplanar. Connect several line segments between the 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?",See reasoning trace,medium,"[Solution] Divide 9 points into 3 groups: $\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}\right.$, $\left.A_{9}\right\}$, and connect a line between every two points that belong to different groups. Then these lines do not form a tetrahedron but do form 27 triangles. It is evident that the maximum number of triangles is no less than 27. The following proof shows that if the graph has 28 triangles, then there must be a tetrahedron. To prove this, we first prove the following lemma. Lemma: Given $n$ points in space, where no 4 points are coplanar. If several line segments are connected and there are no triangles in the graph, then the number of line segments is at most $\left[\frac{n^{2}}{4}\right]$. Proof of the Lemma: We use mathematical induction to prove that when $n$ points are connected by $\left[\frac{n^{2}}{4}\right]+1$ line segments, these line segments must form at least 1 triangle. When $n=3,4$, the proposition is obviously true. Assume the proposition holds for $n=2 k-1$. When $n=2 k$, $\left[\frac{n^{2}}{4}\right]+1=k^{2}+1$, i.e., there are $k^{2}+1$ line segments between $2 k$ points, which have $2 k^{2}+2$ endpoints. Since $$ \frac{2 k^{2}+2}{2 k}=k+\frac{1}{k}, $$ it follows that there must be a point $A$ that connects to at most $k$ line segments. Removing point $A$, the remaining $2 k-1$ points have at least $$ k(k-1)+1=\left[\frac{(2 k-1)^{2}}{4}\right]+1 $$ line segments. By the induction hypothesis, they must form at least 1 triangle, i.e., the proposition holds for $n=2 k$. When $n=2 k+1$, $\left[\frac{(2 k+1)^{2}}{4}\right]+1=k(k+1)+1$, i.e., there are $k(k+1)+1$ line segments between $2 k+1$ points, which have $2 k(k+1)+2$ endpoints. Since $$ \frac{2 k(k+1)+2}{2 k+1}=k+\frac{k+2}{2 k+1}, $$ it follows that there must be a point $A$ that connects to at most $k$ line segments. Removing point $A$, the remaining $2 k$ points have at least $k^{2}+1$ line segments. Thus, they must form at least 1 triangle, i.e., the proposition holds for $n=2 k+1$. Therefore, by induction, the proposition holds for all $n \geqslant 3$. Returning to the proof of the original problem. If there is no tetrahedron, let point $A_{1}$ be the point that connects to the most line segments. Since 28 triangles have 84 vertices, by the pigeonhole principle, there must be a point $A_{2}$ that is a common vertex of at least 10 triangles. If point $A_{2}$ connects to at most 6 line segments, then these lines have at most 6 other endpoints, and there are at least 10 line segments among these endpoints. By the lemma, there must be a triangle, and adding point $A_{2}$ forms a tetrahedron, which is a contradiction. Therefore, point $A_{2}$ must connect to at least 7 line segments, and point $A_{1}$ must connect to even more. (1) If point $A_{1}$ connects to 8 line segments, then since there is no tetrahedron, the subgraph formed by the remaining 8 points cannot have a triangle. By the lemma, there can be at most 16 line segments among these 8 points, so the graph can have at most 16 triangles, which is a contradiction. (2) If point $A_{1}$ connects to 7 line segments, then the subgraph formed by the other 7 endpoints of these 7 line segments cannot have a triangle. By the lemma, there can be at most 12 line segments among these 7 points. Thus, the graph can have at most 24 triangles, which is a contradiction. In conclusion, the graph can have at most 27 triangles." 8d17b167098e,"7 For non-negative real numbers $x_{i}(i=1,2, \cdots, n)$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.",1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$ is $\frac{1}{12}$.,medium,"7. When $n=1$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=0$. When $n=2$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=\left(x_{1}^{4}+x_{2}^{4}\right)-\left(x_{1}+x_{2}\right)\left(x_{1}^{4}-x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}-\right.$ $\left.x_{1} x_{2}^{3}+x_{2}^{4}\right)=x_{1}^{3} x_{2}-x_{1}^{2} x_{2}^{2}+x_{1} x_{2}^{3}=x_{1} x_{2}\left(1-3 x_{1} x_{2}\right)$. Since $x_{1} x_{2} \leqslant \frac{1}{4}$, the maximum value is $\frac{1}{12}$. When $n \geqslant 3$, then $$\begin{aligned} & -\left(x^{4}+y^{4}-x^{5}-y^{5}\right)+\left[(x+y)^{4}-(x+y)^{5}\right] \\ = & {\left[(x+y)^{4}-x^{4}-y^{4}\right]-\left[(x+y)^{5}-x^{5}-y^{5}\right] } \\ = & x y\left(4 x^{2}+4 y^{2}+6 x y\right)-x y\left(5 x^{3}+5 y^{3}+10 x y^{2}+10 x^{2} y\right) \\ \geqslant & x y\left(4 x^{2}+4 y^{2}+6 x y\right)-5 x y(x+y)^{3} \\ = & x y\left(\frac{7}{2} x^{2}+\frac{7}{2} y^{2}+\frac{x^{2}+y^{2}}{2}+6 x y\right)-5 x y(x+y)^{3} \\ \geqslant & x y\left(\frac{7}{2} x^{2}+\frac{7}{2} y^{2}+7 x y\right)-5 x y(x+y)^{3}>0 \end{aligned}$$ This is equivalent to $\frac{7}{2} x y(x+y)^{2} \cdot[7-10(x+y)]>0$, i.e., $x+y<\frac{7}{10}$. When $n \geqslant 3$, there are always 2 numbers whose sum $<\frac{2}{3}<\frac{7}{10}$. By continuously replacing $x$ and $y$ with $x+y$, it can eventually be reduced to the case when $n=2$. In summary, the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$ is $\frac{1}{12}$." afa5af6c27a2,"## Task 25/74 The dimensions of all cylinders are to be specified in which the measurement numbers of radius and height are natural numbers and in which the surface area and volume are numerically equal.",3; \quad r_{3} = 6; \quad V_{3} = 108 \pi = O_{3}$,medium,"If the measures of radius and height are denoted by $r$ and $h$ respectively, where $r, h > 0, r, h \in \mathbb{N}$, the task requires the validity of the equation $$ \pi r^{2} h = 2 \pi r(r + h) $$ From this, it immediately follows that $$ r h = 2 r + 2 h \rightarrow r = \frac{2 h}{h - 2} = 2 + \frac{4}{h - 2} $$ For $r$ to be a natural number, it is necessary and sufficient that $h$ takes one of the values 6, 4, or 3. Thus, there are three cylinders of the required type: a) $h_{1} = 6; \quad r_{1} = 3; \quad V_{1} = 54 \pi = O_{1}$ b) $h_{2} = 4; \quad r_{2} = 4; \quad V_{2} = 64 \pi = O_{2}$ c) $h_{3} = 3; \quad r_{3} = 6; \quad V_{3} = 108 \pi = O_{3}$" e89441e702d9,"3. Solve the system of equations: $$ \left\{\begin{aligned} x+y-2018 & =(y-2019) \cdot x \\ y+z-2017 & =(y-2019) \cdot z \\ x+z+5 & =x \cdot z \end{aligned}\right. $$","(3, 2021, 4), ( $-1,2019,-2)$",easy,"Answer: (3, 2021, 4), ( $-1,2019,-2)$. Solution: a similar solution to this problem is present in variant 1 under the same number. Note: 1 point for each correct solution found by trial and error." 18cd04a087cc,"4. In the Cartesian coordinate system $x O y$, the ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Chords $S T$ and $U V$ are parallel to the $x$-axis and $y$-axis, respectively, and intersect at point $P$. It is known that the lengths of segments $P U$, $P S$, $P V$, and $P T$ are $1$, $2$, $3$, and $6$, respectively. Then the area of $\triangle P F_{1} F_{2}$ is $\qquad$",See reasoning trace,medium,"4. $\sqrt{15}$. By symmetry, without loss of generality, let $P\left(x_{P}, y_{P}\right)$ be in the first quadrant. Then, by the given conditions, $$ \begin{array}{l} x_{P}=\frac{1}{2}(|P T|-|P S|)=2, \\ y_{P}=\frac{1}{2}(|P V|-|P U|)=1, \end{array} $$ which means point $P(2,1)$. Also, from $x_{P}=|P U|=1,|P S|=2$, we know $U(2,2), S(4,1)$. Substituting into the equation of the ellipse $C$ gives $$ \frac{4}{a^{2}}+\frac{4}{b^{2}}=\frac{16}{a^{2}}+\frac{1}{b^{2}}=1 \Rightarrow a^{2}=20, b^{2}=5 \text {. } $$ Therefore, $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{P}\right|$ $$ =\sqrt{a^{2}-b^{2}} y_{P}=\sqrt{15} \text {. } $$" 29db0f13436c,"7. Five contestants $A, B, C, D, E$ participate in the ""The Voice"" competition, and the five of them stand in a row for a group appearance. They each have a contestant number on their chest, and the sum of the five numbers is 35. It is known that the sum of the numbers of the contestants standing to the right of $\mathrm{E}$ is 13; the sum of the numbers of the contestants standing to the right of $D$ is 31; the sum of the numbers of the contestants standing to the right of $A$ is 21; the sum of the numbers of the contestants standing to the right of $C$ is 7. What is the sum of the numbers of the contestants standing at the far left and the far right? $\qquad$","35-31=4, B$ is on the far right, so $B=7,7+4=11$.",easy,"【Answer】 11 【Analysis】According to the problem, $D$ is on the far left, so $D=35-31=4, B$ is on the far right, so $B=7,7+4=11$." 6819d48767ef,"3. If $\left|3^{x}-\log _{3} x\right|=3^{x}+\log _{3} x=k$, then the value of $k$ is ( ). (A) 3, (B) 1. (C) 0. (D) Cannot be determined.",See reasoning trace,easy,"A Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." 61483f67e318,"5. Use square tiles of the same size to pave a square floor, with white tiles around the edges and black tiles in the middle (as shown in Fig. 1 and Fig. 2). When 80 white tiles are used around the square floor, the number of black tiles needed is $\qquad$. 保留源文本的换行和格式，直接输出翻译结果如下： 5. Use square tiles of the same size to pave a square floor, with white tiles around the edges and black tiles in the middle (as shown in Fig. 1 and Fig. 2). When 80 white tiles are used around the square floor, the number of black tiles needed is $\qquad$.","19$ pieces; therefore, the black tiles require $19 \times 19=361$ pieces.",easy,"【Answer】 361 pieces 【Key Point】Square Array Problem 【Analysis】 On a square floor paved with 80 white tiles, the black tiles inside each row have $(80-4) \div 4=19$ pieces; therefore, the black tiles require $19 \times 19=361$ pieces." 2e6f0448882c,"13. A homothety is defined by its center $S$ and coefficient $k$. Write the formula connecting $\vec{A}$ and $\vec{A}^{\prime}$, where $A^{\prime}$ is the image of point $A$ under this homothety. (Note that the answer does not depend on the choice of origin O.)",See reasoning trace,easy,"13. If a homothety with center $S$ and coefficient $k$ maps point $A$ to $A^{\prime}$, then $\left(\vec{A}^{\prime}-\vec{S}\right)=k(\vec{A}-\vec{S})$, from which: $$ \vec{A}^{\prime}=\overrightarrow{k A}+(1-k) \vec{S} $$" d1d9c164937e,"16. Let $a$ be the fractional part of $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}$, and $b$ be the fractional part of $\sqrt{6+3 \sqrt{3}}-\sqrt{6-3 \sqrt{3}}$, then the value of $\frac{2}{b}-\frac{1}{a}$ is $\qquad$",$\sqrt{6}-\sqrt{2}+1$,easy,Answer: $\sqrt{6}-\sqrt{2}+1$ 4467b23ec97a,"3. Given $A_{1}, A_{2}, \cdots ; B_{1}, B_{2}, \cdots$ are geometric sequences. Can $A_{1}+B_{1}, A_{2}+B_{2}, A_{3}+B_{3}$ and $A_{1}+B_{4}$ determine $A_{5}+B_{5}$?",See reasoning trace,medium,"3. Let $A_{1}=a p^{i-1}, B_{1}=b q^{i-1}(1-i-5)$, $S_{1}=A_{1}+B_{1}$, then $$ S_{i+1}(p+q)=S_{i+2}+S_{i} p q(1 \leqslant i \leqslant 3) . $$ If $S_{1} S_{3} \neq S_{2}^{2}$, then from $$ \left\{\begin{array}{l} S_{2}(p+q)-S_{1} p q=S_{3}, \\ S_{3}(p+q)-S_{2} p q=S_{4}, \end{array}\right. $$ $p+q$ and $p q$ can be uniquely determined, thus $S_{5}=S_{4}(p+q)-S_{3} p q$ is uniquely determined. If $S_{1} S_{3}=S_{2}^{2}$, i.e., $a b(p-q)^{2}=0$. If $a=0, b=0$ or $p=q$, then $S_{1}=b q^{1-1}, a p^{1-1}$ or $(a+b) p^{i-1}$. In this case, if $S_{1}=0$, then $S_{5}=0$; if $S_{1} \neq 0$, then $S_{5}=\frac{S_{3}^{2}}{S_{1}}$. In any case, $S_{5}$ can be uniquely determined from $S_{1}$ to $S_{4}$." 5c13ebf61b61,"2. Find all polynomials $p$ of odd degree $z$ with real coefficients, for which $$ p(p(x)) \leq (p(x))^{3} $$ holds for all $x \in \mathbb{R}$ and which have the coefficient of $x^{2}$ equal to 0.",See reasoning trace,medium,"III/2. Let's write $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$, where $a_{n} \neq 0$. The leading term of the polynomial $p(p(x))$ is $a_{n}\left(a_{n} x^{n}\right)^{n}=a_{n}^{n+1} x^{n^{2}}$, and the leading term of the polynomial $p(x)^{3}$ is $\left(a_{n} x^{n}\right)^{3}=a_{n}^{3} x^{3 n}$, both of which are of odd degree. Since the polynomial $p(x)^{3}-p(p(x))$ is non-negative everywhere, it must be of even degree, so the mentioned leading terms must cancel each other out. Therefore, $a_{n}^{n+1} x^{n^{2}}=a_{n}^{3} x^{3 n}$ or $n^{2}=3 n$ and $a_{n}^{n+1}=a_{n}^{3}$. It follows that $n=3$ and $a_{n}=1$, since $a_{n} \neq 0$, i.e., $p$ is a polynomial of degree 3 with a leading coefficient of 1. Since the coefficient of the polynomial $p$ at $x^{2}$ is 0, the polynomial $p$ is of the form $p(x)=x^{3}+a x+b$. When we consider this in the given inequality and simplify the inequality, we get $a x^{3}+a^{2} x+a b+b \leq 0$ for every $x \in \mathbb{R}$. This is only possible if $a=0$, since a polynomial of odd degree always takes positive values. The inequality simplifies to $b \leq 0$ in this case. The polynomials that satisfy the conditions of the problem are exactly the polynomials of the form $p(x)=x^{3}+b$, where $b \leq 0$. ![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=51&width=1633&top_left_y=1551&top_left_x=220) The degree of the polynomial $p^{3}(x)-p(p(x))$ must be even .............................. 1 point ![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=57&width=1633&top_left_y=1645&top_left_x=220) ![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=57&width=1633&top_left_y=1693&top_left_x=220) Substituting the assumption $p(x)=x^{3}+a x+b$ into the inequality .............................. 1 point ![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=49&width=1631&top_left_y=1803&top_left_x=218) ![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=60&width=1631&top_left_y=1849&top_left_x=224) ## Addendum: We can also determine that the constant term is less than or equal to 0 in another way: since $p$ is of odd degree, it has a real root $x_{0}$. Then $a_{0}=p(0)=p\left(p\left(x_{0}\right)\right) \leq p\left(x_{0}\right)^{3}=0 . \mathbf{1}$ point In addition, points can also be awarded for: Guessing at least one correct solution 1 point Excluding a polynomial of degree 1 in the analysis 1 point Correctly writing the general inequality $p(p(x)) \leq p(x)^{3}$" f79bad2b01f5,,1832,medium,"Answer: 1832. Solution. Since we need to find the nearest four-digit number, let's try to find it in the form $\overline{18 a b}$. Then the number $18 \cdot \overline{a b}=3^{2} \cdot(2 \cdot \overline{a b})$ must be a perfect square. From this, it follows that $2 \cdot \overline{a b}$ must also be a perfect square. Clearly, $\overline{a b} \in\{19,20,21 \ldots, 31\}$ do not meet this condition, but $\overline{a b}=32$ does. Therefore, the answer to the problem is the number 1832. Remark. The same answer could have been obtained by proving that $\overline{a b}=2 s^{2}$ for some natural number $s>3$." c96b4ab0e91b,The triangle $K_2$ has as its vertices the feet of the altitudes of a non-right triangle $K_1$. Find all possibilities for the sizes of the angles of $K_1$ for which the triangles $K_1$ and $K_2$ are similar.,"A = 60^\circ, B = 60^\circ, C = 60^\circ",medium,"To solve the problem, we need to determine the conditions under which the triangle \( K_1 \) and its orthic triangle \( K_2 \) are similar. The orthic triangle \( K_2 \) is formed by the feet of the altitudes of \( K_1 \). 1. **Identify the angles of the orthic triangle \( K_2 \):** - Let \( A, B, \) and \( C \) be the angles of triangle \( K_1 \). - The angles of the orthic triangle \( K_2 \) are given by \( 180^\circ - 2A \), \( 180^\circ - 2B \), and \( 180^\circ - 2C \). 2. **Set up the similarity condition:** - For \( K_1 \) and \( K_2 \) to be similar, their corresponding angles must be proportional. This means we need: \[ A = 180^\circ - 2A, \quad B = 180^\circ - 2B, \quad C = 180^\circ - 2C \] 3. **Solve the angle equations:** - Consider the equation \( A = 180^\circ - 2A \): \[ A + 2A = 180^\circ \] \[ 3A = 180^\circ \] \[ A = 60^\circ \] - Similarly, solving for \( B \) and \( C \): \[ B = 180^\circ - 2B \implies 3B = 180^\circ \implies B = 60^\circ \] \[ C = 180^\circ - 2C \implies 3C = 180^\circ \implies C = 60^\circ \] 4. **Verify the solution:** - If \( A = 60^\circ \), \( B = 60^\circ \), and \( C = 60^\circ \), then triangle \( K_1 \) is an equilateral triangle. - The orthic triangle of an equilateral triangle is also an equilateral triangle, hence \( K_1 \) and \( K_2 \) are similar. 5. **Consider other possibilities:** - Since the only solution to the equations \( A = 180^\circ - 2A \), \( B = 180^\circ - 2B \), and \( C = 180^\circ - 2C \) is \( A = B = C = 60^\circ \), there are no other possibilities. Conclusion: The only possibility for the angles of \( K_1 \) such that \( K_1 \) and \( K_2 \) are similar is when \( K_1 \) is an equilateral triangle. The final answer is \( \boxed{ A = 60^\circ, B = 60^\circ, C = 60^\circ } \)." 86c1dda9a37b,1. The smallest positive integer $n=$ that makes $3 n+1$ and $5 n+1$ both perfect squares,"k^{2}, 5 n+1=m^{2} \Rightarrow(m+k)(m-k)=2 n$ is even, and since $m+k, m-k$ have the same parity, $n",easy,"Given $3 n+1=k^{2}, 5 n+1=m^{2} \Rightarrow(m+k)(m-k)=2 n$ is even, and since $m+k, m-k$ have the same parity, $n$ must be even. Let $n=2 n_{1} \Rightarrow\left\{\begin{array}{l}m-k=2, \\ m+k=2 n_{1}\end{array} \Rightarrow k=n_{1}-1\right.$. Substituting into $6 n_{1}+1=\left(n_{1}-1\right)^{2} \Rightarrow n_{1}=8$, so the smallest positive integer $n=16$." 5b84f3c77e11,"Given the polynomial $$P(x) = 1+3x + 5x^2 + 7x^3 + ...+ 1001x^{500}.$$ Express the numerical value of its derivative of order $325$ for $x = 0$.",651 \cdot 325!,medium,"1. The given polynomial is: \[ P(x) = 1 + 3x + 5x^2 + 7x^3 + \cdots + 1001x^{500} \] This can be written in the general form: \[ P(x) = \sum_{k=0}^{500} (2k+1)x^k \] 2. We need to find the 325th derivative of \( P(x) \) evaluated at \( x = 0 \). The general term in the polynomial is \( (2k+1)x^k \). 3. The 325th derivative of \( (2k+1)x^k \) is non-zero only if \( k \geq 325 \). For \( k = 325 \), the term is \( (2 \cdot 325 + 1)x^{325} = 651x^{325} \). 4. The 325th derivative of \( 651x^{325} \) is: \[ \frac{d^{325}}{dx^{325}} (651x^{325}) = 651 \cdot 325! \] 5. For \( k > 325 \), the 325th derivative of \( (2k+1)x^k \) will involve higher powers of \( x \) and will be zero when evaluated at \( x = 0 \). 6. Therefore, the only term that contributes to the 325th derivative at \( x = 0 \) is \( 651x^{325} \). 7. Evaluating the 325th derivative at \( x = 0 \): \[ P^{(325)}(0) = 651 \cdot 325! \] The final answer is \( \boxed{651 \cdot 325!} \)" e6aa15c48885,"Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A$, and on $\overline{AB}$. $D$ is a point $4$ units from $B$, and on $\overline{AB}$. $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$: $\textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\\ \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \\ \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \\ \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \\ \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.$",choice $(B)$,medium,"If $P$ is on $A$, then the length is 10, eliminating answer choice $(B)$. If $P$ is equidistant from $C$ and $D$, the length is $2\sqrt{1^2+5^2}=2\sqrt{26}>10$, eliminating $(A)$ and $(C)$. If triangle $CDP$ is right, then angle $CDP$ is right or angle $DCP$ is right. Assume that angle $DCP$ is right. Triangle $APB$ is right, so $CP=\sqrt{4*6}=\sqrt{24}$. Then, $DP=\sqrt{28}$, so the length we are looking for is $\sqrt{24}+\sqrt{28}>10$, eliminating $(D)$. Thus, our answer is $(E)$. $\fbox{}$ Note: Say you are not convinced that $\sqrt{24}+\sqrt{28}>10$. We can prove this as follows. Start by simplifying the equation: $\sqrt{6}+\sqrt{7}>5$. Square both sides: $6+2\sqrt{42}+7>25$. Simplify: $\sqrt{42}>6$ Square both sides again: $42>36$. From here, we can just reverse our steps to get $\sqrt{24}+\sqrt{28}>10$." 8840818618db,"15. Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2)$, and $a_{0}=a_{1}=1$. Find the general term formula for $\left\{a_{n}\right\}$.",See reasoning trace,medium,"$$ \begin{array}{l} \sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1} \Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}-1=2 \sqrt{\frac{a_{n-1}}{a_{n-2}}} \\ \Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}+1=2\left(\sqrt{\frac{a_{n-1}}{a_{n-2}}}+1\right) \Rightarrow\left\{\sqrt{\frac{a_{n}}{a_{n-1}}}+1\right\} \text { is a geometric sequence with common ratio } 2 \\ \Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}+1=2^{n} \Rightarrow \frac{a_{n}}{a_{n-1}}=\left(2^{n}-1\right)^{2} \Rightarrow a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}(n \geqslant 2) . \\ \text { Therefore, } a_{n}=\left\{\begin{array}{l} 1, n=1, \\ \prod_{k=1}^{n}\left(2^{k}-1\right)^{2}, n \in \mathbf{N}, n \geqslant 2 . \end{array}\right. \end{array} $$" 941665dff5d8,"Prove the following inequality if we know that $a$ and $b$ are the legs of a right triangle , and $c$ is the length of the hypotenuse of this triangle: $$3a + 4b \le 5c.$$ When does equality holds?",3a + 4b \leq 5c,medium,"1. Given that \(a\) and \(b\) are the legs of a right triangle and \(c\) is the hypotenuse, we start with the Pythagorean theorem: \[ a^2 + b^2 = c^2 \] 2. Consider the expression \((4a - 3b)^2\). Since the square of any real number is non-negative, we have: \[ (4a - 3b)^2 \geq 0 \] 3. Expanding \((4a - 3b)^2\), we get: \[ (4a - 3b)^2 = 16a^2 - 24ab + 9b^2 \] Therefore, \[ 16a^2 - 24ab + 9b^2 \geq 0 \] 4. Using the Pythagorean identity \(a^2 + b^2 = c^2\), we can rewrite the inequality: \[ 16a^2 - 24ab + 9b^2 \geq 0 \] 5. Notice that \(25(a^2 + b^2) = 25c^2\). We can compare this with the expanded form: \[ 25(a^2 + b^2) = 25c^2 \] 6. Now, we need to show that: \[ 25c^2 \geq 16a^2 - 24ab + 9b^2 \] 7. Since \(a^2 + b^2 = c^2\), we can substitute \(c^2\) back into the inequality: \[ 25c^2 \geq 16a^2 - 24ab + 9b^2 \] 8. Dividing both sides by 25, we get: \[ c^2 \geq \frac{16a^2 - 24ab + 9b^2}{25} \] 9. To show that \(3a + 4b \leq 5c\), we can use the fact that: \[ (3a + 4b)^2 \leq (5c)^2 \] 10. Expanding both sides, we get: \[ 9a^2 + 24ab + 16b^2 \leq 25c^2 \] 11. Since \(a^2 + b^2 = c^2\), we can substitute \(c^2\) back into the inequality: \[ 9a^2 + 24ab + 16b^2 \leq 25(a^2 + b^2) \] 12. This simplifies to: \[ 9a^2 + 24ab + 16b^2 \leq 25a^2 + 25b^2 \] 13. Rearranging terms, we get: \[ 0 \leq 16a^2 - 24ab + 9b^2 \] 14. This is true since \((4a - 3b)^2 \geq 0\). 15. Therefore, we have shown that: \[ 3a + 4b \leq 5c \] 16. Equality holds when \((4a - 3b)^2 = 0\), which implies: \[ 4a = 3b \implies \frac{a}{b} = \frac{3}{4} \] 17. Using the Pythagorean theorem, if \(a = 3k\) and \(b = 4k\), then: \[ c = \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k \] 18. Thus, equality holds when \(a : b : c = 3 : 4 : 5\). The final answer is \( \boxed{ 3a + 4b \leq 5c } \). Equality holds when \(a : b : c = 3 : 4 : 5\)." b56ac68cacb6,"In triangle $ABC$, $\angle BAC = 45^{\circ}$. Let $P$ be the point on side $AC$ closer to $A$ that divides $AC$ into a ratio of $1:2$. We know that $\angle ABP = 15^{\circ}$. What is the measure of $\angle ACB$?",461&width=553&top_left_y=1938&top_left_x=775),medium,"I. Solution. Drop a perpendicular from point $C$ to segment $PB$, and let the foot of this perpendicular be $E$. In the right triangle $CEP$, $\angle CPE = \angle PAB + \angle PBA = 60^\circ$, thus $\angle PCE = 30^\circ$, and $PE = \frac{PC}{2}$. Since $PA$ is also equal to $\frac{PC}{2}$, it follows that $PE = PA$, so triangle $APE$ is isosceles. In this triangle, $$ \angle PAE = \angle PEA = \frac{\angle CPE}{2} = 30^\circ. $$ Therefore, $CE = EA$. Since $\angle EAB = \angle PAB - \angle PAE = 45^\circ - 30^\circ = 15^\circ$, and $\angle EBA = 15^\circ$, triangle $AEB$ is also isosceles, so $EA = EB$. We have thus found that $CE = EA = EB$, meaning triangle $CEB$ is also isosceles, and since there is a right angle at vertex $E$, $\angle CBE = \angle BCE = 45^\circ$. Based on this, $\angle ACB = \angle PCE + \angle BCE = 30^\circ + 45^\circ = 75^\circ$ (Figure 1). ![](https://cdn.mathpix.com/cropped/2024_05_02_0e0788b0e3b74d5445bbg-1.jpg?height=460&width=553&top_left_y=632&top_left_x=775) II. Solution. Let $x$ be one-third of side $b$, $T$ be the foot of the perpendicular from point $C$ to side $AB$, and $\delta_1 := \angle BCT$ (Figure 2). Then $\angle TCA$ is also $45^\circ$, and triangle $ATC$ is an isosceles right triangle, so $AT = TC = 3x \cdot \frac{\sqrt{2}}{2}$. Applying the Law of Sines in triangle $ABP$: $$ \begin{aligned} \frac{c}{x} & = \frac{\sin 120^\circ}{\sin (45^\circ - 30^\circ)} = \frac{\sin 120^\circ}{\sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ} = \\ & = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}} = \frac{2 \sqrt{3}}{\sqrt{6} - \sqrt{2}} = \frac{2 \sqrt{3} (\sqrt{6} + \sqrt{2})}{4} = \frac{\sqrt{18} + \sqrt{6}}{2} = \frac{3 \sqrt{2} + \sqrt{6}}{2} \end{aligned} $$ From this, $c = \frac{3 \sqrt{2} + \sqrt{6}}{2} x$, and $TB = AB - AT = \frac{3 \sqrt{2} + \sqrt{6}}{2} x - 3x \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{2} x$. Writing the tangent of $\delta_1$ in triangle $CTB$: $$ \tan \delta_1 = \frac{TB}{TC} = \frac{x \cdot \frac{\sqrt{6}}{2}}{3x \cdot \frac{\sqrt{2}}{2}} = \frac{\sqrt{3}}{3} $$ From this, $\delta_1 = 30^\circ$, so $\angle ABC = 45^\circ + 30^\circ = 75^\circ$. ![](https://cdn.mathpix.com/cropped/2024_05_02_0e0788b0e3b74d5445bbg-1.jpg?height=461&width=553&top_left_y=1938&top_left_x=775)" 22773288fa37,"6. (6 points) Two rectangles, each 4 cm long and 2 cm wide, are joined together (without overlapping) to form a new rectangle. The perimeter of the new rectangle is $\qquad$ cm, or $\qquad$ cm.",The perimeter of the new rectangle formed by combining is 20 cm or 16 cm,medium,"【Analysis】According to the method of combining two new rectangles into a larger rectangle, we can get: the length and width of the new rectangle are $4+4=8$ cm, 2 cm; or 4 cm, 4 cm, so the perimeter of the new rectangle is $(2+4+4) \times 2=20 \mathrm{~cm}$, or $4 \times 4$ $=16 \mathrm{~cm}$. 【Solution】Solution: $(4+4+2) \times 2$, $$ \begin{array}{l} =10 \times 2, \\ =20 \text { (cm), } \\ 4 \times 4=16 \text { (cm), } \end{array} $$ Answer: The perimeter of the new rectangle formed by combining is 20 cm or 16 cm. Therefore, the answer is: $20 ; 16$. 【Comment】The key is to know that there are two ways to combine two rectangles into one, and then use the perimeter formula for a rectangle $C=(a+b) \times 2$ to solve the problem." 608725758502,"1. If $q$ is a prime number, and $q+1$ is a perfect square, then $q$ is called a $P$-type prime. There are ( ) such $P$-type primes. (A) 0 (B) 1 (C) 2 (D) infinitely many","2$. In this case, $q=3$ meets the condition. Hence, there is only one P-type prime.",easy,"- 1. B. From the given, let $q+1=m^{2}\left(m \in \mathbf{Z}_{+}, m \geqslant 2\right)$. Then $q=m^{2}-1=(m+1)(m-1)$. Thus, when $m \geqslant 3$, $q$ must be a composite number, which contradicts the problem statement. Therefore, $m=2$. In this case, $q=3$ meets the condition. Hence, there is only one P-type prime." a4c816e487e2,"1. Given the complex number $z$ satisfies $|z|-z=3-\mathrm{i}$. Then $z=$ $\qquad$ .",z-\mathrm{i} \in \mathbf{R} \\ \Rightarrow z=a+\mathrm{i} \Rightarrow \sqrt{a^{2}+1}-3=a \\ \Rightar,easy,\begin{array}{l}\text {1. }-\frac{4}{3}+\mathrm{i} . \\ \text { From }|z|-3=z-\mathrm{i} \in \mathbf{R} \\ \Rightarrow z=a+\mathrm{i} \Rightarrow \sqrt{a^{2}+1}-3=a \\ \Rightarrow a=-\frac{4}{3} \Rightarrow z=-\frac{4}{3}+\mathrm{i} .\end{array} bdd110c02546,"4. The increasing sequence $1,3,4,9,10,12,13, \cdots$, consists of some positive integers, which are either powers of 3, or the sum of several different powers of 3. The 100th term of this sequence is (). A. 729 B. 972 C. 243 1). 981","63$ different sums of powers of 3, which are the first 63 terms of the sequence; the 64th term is $3",medium,"4. D Because the terms of this sequence are of the form: $k_{0} \cdot 3^{0}+k_{1} \cdot 3^{1}+k_{2} \cdot 3^{2}+\cdots+k_{m} \cdot 3^{m}$ (where each $k_{i} \in\{0,1\}$). The first 6 powers of 3, $3^{0}, 3^{1}, 3^{2}, \cdots, 3^{5}$, can form $2^{6}-1=63$ different sums of powers of 3, which are the first 63 terms of the sequence; the 64th term is $3^{6}=729$, starting from the 65th term, there are $C_{5}^{1}+C_{5}^{2}+\cdots+C_{5}^{5}=2^{5}-1=31$ terms that do not contain $3^{5}=243$, the 96th term should be $729+$ $243=972$, followed by $729+243+1,729+243+3,729+243+1+3$, thus the 100th term is $729+243+9=981$." 7b87adbc793f,"Example 4 If the expansion of $\left(1+x+x^{2}\right)^{1000}$ is $a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2000} x^{2000}$, then the value of $a_{0}+a_{3}+a_{6}+\cdots+a_{1998}$ is ( ). (A) $3^{333}$ (B) $3^{666}$ (C) $3^{999}$ (D) $3^{2001}$ (2001, National High School Mathematics Competition)",(C),easy,"Explanation: Let $f(x)=\left(1+x+x^{2}\right)^{1000}$ $$ \begin{array}{l} =a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2000} x^{2000}, \\ \omega=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}, \end{array} $$ $$ \begin{array}{l} \text { then } a_{0}+a_{3}+a_{6}+\cdots+a_{1998} \\ =\frac{f(1)+f(\omega)+f\left(\omega^{2}\right)}{3}=3^{999} . \end{array} $$ Therefore, the answer is (C)." 6d51bb54d157,"2. Condition A: $\sqrt{1+\sin \theta}=a$. Condition B: $\sin \frac{\theta}{2}+\cos \frac{\theta}{2}=a$. (A) A is a necessary and sufficient condition for B; (B) A is a necessary condition for B; (C) A is a sufficient condition for B; (D) A is neither a necessary condition nor a sufficient condition for B.",See reasoning trace,easy,"2. (D) In condition A, it is easy to see that $a \geqslant 0$, while in condition B, $a$ may take negative values. Therefore, A does not imply B, and B does not imply A, meaning A is neither a necessary condition nor a sufficient condition for B." b81dae7c8035,"3. As shown in Figure 1, in $\triangle A B C$, $A B=A C, \angle A=40^{\circ}$, extend $A C$ to $D$ such that $C D=B C$, point $P$ is the incenter of $\triangle A B D$. Then $\angle B P C=(\quad)$. (A) $145^{\circ}$ (B) $135^{\circ}$ (C) $120^{\circ}$ (D) $105^{\circ}$","35^{\circ}$, so, $\angle B P C=180^{\circ}-35^{\circ}=145^{\circ}$.",easy,"3. (A). Connect $P D$, by the property of the incenter we know $\angle B P D=90^{\circ}+\frac{1}{2} \angle A$. Also, $\angle B C D=90^{\circ}+\frac{1}{2} \angle A$, thus $\angle B P D=\angle B C D$. Therefore, $B, D, C, P$ are concyclic. It is easy to know $\angle D=35^{\circ}$, so, $\angle B P C=180^{\circ}-35^{\circ}=145^{\circ}$." 56f3157ffdfe,"2. In the horizontal equation $\overline{A B C} \times \overline{A B}+C \times D=2017$, the same letter represents the same digit, and different letters represent different digits. If the equation holds, then the two-digit number $\overline{\mathrm{AB}}$ is $\qquad$ .","】Solution: Since $0b>0)$ has its left focus at $F$. A line passing through point $F$ intersects the ellipse at points $A$ and $B$. When the line $AB$ passes through one of the vertices of the ellipse, its inclination angle is exactly $60^{\circ}$. （1）Find the eccentricity of the ellipse; （2）The midpoint of segment $AB$ is $G$, and the perpendicular bisector of $AB$ intersects the $x$-axis and $y$-axis at points $D$ and $E$, respectively. Let the area of $\triangle GDF$ be $S_{1}$, and the area of $\triangle OED$ (where $O$ is the origin) be $S_{2}$. Find the range of $\frac{S_{1}}{S_{2}}$.",See reasoning trace,medium,"Solution (1) According to the problem, when the line $AB$ passes through the vertex $(0, b)$ of the ellipse, its angle of inclination is $60^{\circ}$. Let $F(-c, 0)$, then $\frac{b}{c}=\tan 60^{\circ}=\sqrt{3}$, substituting $b=\sqrt{3} c$ into $a^{2}=b^{2}+c^{2}$, we get $a=2 c$, so the eccentricity of the ellipse is $$ e=\frac{c}{a}=\frac{1}{2} $$ 4 points (2) From (1), the equation of the ellipse can be set as $\frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{3 c^{2}}=1$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. According to the problem, the line $AB$ cannot be perpendicular to the $x, y$ axes, so we set the equation of the line $AB$ as $y=k(x+c)$, substituting it into $$ 3 x^{2}+4 y^{2}=12 c^{2} $$ we get $$ \left(4 k^{2}+3\right) x^{2}+8 c k^{2} x+4 k^{2} c^{2}-12 c^{2}=0 $$ then $x_{1}+x_{2}=\frac{-8 c k^{2}}{4 k^{2}+3}, y_{1}+y_{2}=\frac{6 c k}{4 k^{2}+3}$, so $G\left(\frac{-4 c k^{2}}{4 k^{2}+3}, \frac{3 c k}{4 k^{2}+3}\right)$. 9 points Since $GD \perp AB$, we have $\frac{\frac{3 c k}{4 k^{2}+3}}{\frac{-4 c k^{2}}{4 k^{2}+3}-x_{D}} \times k=-1, x_{D}=\frac{-c k^{2}}{4 k^{2}+3}$. 11 points Since $\triangle GFD \sim \triangle OED$, we have $$ \frac{S_{1}}{S_{2}}=\frac{|GD|^{2}}{|OD|^{2}}=\frac{\left(\frac{-4 c k^{2}}{4 k^{2}+3}-\frac{-c k^{2}}{4 k^{2}+3}\right)^{2}+\left(\frac{3 c k}{4 k^{2}+3}\right)^{2}}{\left(\frac{-c k^{2}}{4 k^{2}+3}\right)^{2}}=\frac{\left(3 c k^{2}\right)^{2}+(3 c k)^{2}}{\left(c k^{2}\right)^{2}}=\frac{9 c^{2} k^{4}+9 c^{2} k^{2}}{c^{2} k^{4}}=9+\frac{9}{k^{2}}>9 $$ Therefore, the range of $\frac{S_{1}}{S_{2}}$ is $(9,+\infty)$. 14 points" 1451848c4172,"Gavin has a collection of 50 songs that are each 3 minutes in length and 50 songs that are each 5 minutes in length. What is the maximum number of songs from his collection that he can play in 3 hours? (A) 100 (B) 36 (C) 56 (D) 60 (E) 45",(C),easy,"To maximize the number of songs used, Gavin should use as many of the shortest length songs as possible. (This is because he can always trade a longer song for a shorter song and shorten the total length used.) If Gavin uses all 50 songs of 3 minutes in length, this takes 150 minutes. There are $180-150=30$ minutes left, so he can play an additional $30 \div 5=6$ songs that are 5 minutes in length. In total, he plays $50+6=56$ songs. ANSWER: (C)" 2ec0b507edbf,"30. Find the number of subsets $\{a, b, c\}$ of $\{1,2,3,4, \ldots, 20\}$ such that $a 2^4, 6 \times 5 < 2$. Therefore, when $n \geqslant 5$, the sum of the numbers that appear after rolling the dice $n$ times cannot be greater than 2, which means this is an impossible event, and the probability of passing is 0. So, the maximum number of consecutive levels that can be passed is 4. (2) Let event $A_{n}$ be ""failing the $n$-th level"", then the complementary event $\overline{A_{n}}$ is ""passing the $n$-th level"". In the $n$-th level game, the total number of basic events is $6^n$. Level 1: The number of basic events contained in event $A_{1}$ is 2 (i.e., the cases where the number is 1 or 2), So the probability of passing this level is: $P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$ Level 2: The number of basic events contained in event $A_{2}$ is the sum of the number of positive integer solutions to the equation $x+y=a$ when $a$ takes the values $2,3,4$. That is, $C_{1}^{3}+C_{3}^{1}+C_{3}^{1}=1+2+3=6$ (cases). So the probability of passing this level is $P\left(\bar{A}_{2}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6}$ Level 3: The number of basic events contained in event $A_{3}$ is the sum of the number of positive integer solutions to the equation $x+y+z=a$ when $a$ takes the values $3,4,5,6,7,8$. That is, $C_{2}^{3}+C_{3}^{3}+C_{4}^{2}+C_{5}^{3}+C_{6}^{3}+C_{5}^{2}=1+3+6+10+15+21=56$ (cases). So the probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$ Therefore, the probability of passing the first 3 levels consecutively is: $P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$" 1f628059590d,"7. Write the equation of the line passing through the points: 1) \(A(3 ; 0)\) and \(B(3 ; 5)\); 3) \(E(0 ; 4)\) and \(K(2 ; 0)\) 2) \(C(2 ; 2)\) and \(D(-1 ;-1)\); 4) \(M(3 ; 2)\) and \(P(6 ; 3)\).",\frac{1}{3}\) and \(b=1\). Equation of the line: \(y=\frac{1}{3} x+1\).,medium,"7. 1) Equation of the line: \(x=3\). 2) \(y=x\). 3) S o l u t i o n. Equation of the line: \(y=k x+b\). We need to find \(k\) and \(b\). When \(x=0\) and \(y=4\), we get: \(4=0 \cdot x+b\), hence, \(b=4\). When \(x=2\) and \(y=0\), we have: \(0=k \cdot 2+4\), hence, \(k=-2\). Equation of the line: \(y=-2 x+4\). 4) S o l u t i o n. Let the equation of the desired line be: \(y=k x+b\). We need to find \(k\) and \(b\). Since the line passes through the point \(M(3 ; 2)\), then \(2=k \cdot 3+b\), and since, in addition, this line passes through the point \(P(6 ; 3)\), then \(3=k \cdot 6+b\). Therefore, solving the system of equations \[ \left\{\begin{array}{l} 3 k+b=2 \\ 6 k+b=3 \end{array}\right. \] we find \(k=\frac{1}{3}\) and \(b=1\). Equation of the line: \(y=\frac{1}{3} x+1\)." 7b17efd124fc,"Zkov G. A bank serves a million customers, the list of whom is known to Ostap Bender. Each has a six-digit PIN code, and different customers have different codes. In one move, Ostap Bender can choose any customer he has not yet chosen and peek at the digits of the code at any $N$ positions (he can choose different positions for different customers). Ostap wants to find out the code of the millionaire Koreiko. For what smallest $N$ can he guarantee to do this?",When $N=3$,medium,"It is not difficult to do when $N=3$. Since any combination of the first three digits occurs exactly 1000 times, by looking at these digits for everyone except Koreiko, Bender will know them for Koreiko as well. Then it is sufficient to look at the last three digits of Koreiko's code. We will prove that when $N<3$, it is impossible to reliably determine Koreiko's code $K$. For each of the 6 positions, choose a code that differs from $K$ only in that position. The corresponding digit in the chosen code will be called bad. Suppose Bender is unlucky, and on the first check of a non-Koreiko code, he gets the chosen code, but does not check the bad digit (there are four unchecked positions, so a code with a bad digit in one of these positions could have been selected). Suppose the same happens on the second and third checks of non-Koreiko codes (there are four unchecked positions, and of the four chosen codes with bad digits in these positions, no more than two have been previously checked, so such a code could have been selected). When all codes have been checked, we will see which $N<3$ positions have been checked in code $K$. Among the four unchecked positions, at least one coincides with the position of the bad digit in one of the three checked chosen codes - code $L$. But if $L$ and $K$ are swapped, the results of all checks will not change. Therefore, Bender will not be able to distinguish $L$ from $K$. ## Answer When $N=3$." c3f7f6304890,"6-114 Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=20$, find the maximum and minimum values of $x^{2}-x y+y^{2}$.","124, Q_{\text {minimum }}=3$.",medium,"[Solution] Squaring both sides of $|5 x+y|+|5 x-y|=20$, and simplifying, we get $25 x^{2}+y^{2}+\left|25 x^{2}-y^{2}\right|=200$. If $|5 x| \geqslant|y|$, then we have $$50 x^{2}=200, x= \pm 2,|y| \leqslant 10$$ If $|5 x| \leqslant|y|$, then we have $$2 y^{2}=200, y= \pm 10,|5 x| \leqslant 10,|x| \leqslant 2$$ Therefore, the image of the equation is a rectangle $A B C D$, where $$A(2,-10), B(2,10), C(-2,10), D(-2,-10)$$ By symmetry, we only need to consider the sides $A B$ and $B C$, and it is easy to see that on $A B$, $Q=x^{2}-x y+y^{2}=4-2 y+y^{2}$, $$3 \leqslant Q \leqslant 124$$ On $B C$, $84 \leqslant Q \leqslant 124$. $Q_{\text {maximum }}=124, Q_{\text {minimum }}=3$." 6c70dad58d6d,"Let $\alpha \neq 0$ be a real number. Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $$ f(f(x)+y)=\alpha x+\frac{1}{f\left(\frac{1}{y}\right)} $$ for all $x, y \in \mathbb{R}_{>0}$. (Walther Janous)","For $\alpha=1$ the only solution is $f(x)=x$, for other values of $\alpha$ there is no solution",medium,"Answer: For $\alpha=1$ the only solution is $f(x)=x$, for other values of $\alpha$ there is no solution. Of course, $\alpha>0$ must hold, otherwise the right side becomes negative for large $x$. By considering $x$ in the original equation, we see that $f$ is injective $\left(f\left(x_{1}\right)=f\left(x_{2}\right)\right.$ implies $\alpha x_{1}=\alpha x_{2}$ ). Similarly, by the arbitrary choice of $x$ on the right side, the function is surjective onto an interval $(a, \infty)$ with a suitable $a$ (e.g., $a=\frac{1}{f(1)}$ ) and thus, by choosing a small $x$ and a large function value $f\left(\frac{1}{y}\right)$ on the right side, it is generally surjective. We now replace $y$ with $f(y)$ and obtain $$ f(f(x)+f(y))=\alpha x+\frac{1}{f\left(\frac{1}{f(y)}\right)} $$ The left side is symmetric in $x$ and $y$, so it also holds that $$ \alpha x+\frac{1}{f\left(\frac{1}{f(y)}\right)}=\alpha y+\frac{1}{f\left(\frac{1}{f(x)}\right)} $$ Choosing any fixed value for $y$, we get $$ \frac{1}{f\left(\frac{1}{f(x)}\right)}=\alpha x+C $$ for a suitable $C$. Clearly, $C \geq 0$ must hold because the function only takes positive values. We substitute the obtained identity into equation (1) and get $$ f(f(x)+f(y))=\alpha x+\alpha y+C . $$ Due to injectivity, it follows that $$ f(x)+f(y)=f(z)+f(w) \text {, if } x+y=z+w \text {. } $$ In particular, for $x, y \geq 0$, it holds that $$ f(x+1)+f(y+1)=f(x+y+1)+f(1) . $$ With $g(x)=f(x+1)$, it follows for the function $g: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ that $$ g(x)+g(y)=g(x+y)+g(0) . $$ For $h(x)=g(x)-g(0)$, it holds that $h(x) \geq-g(0)$ and the Cauchy functional equation $$ h(x)+h(y)=h(x+y) . $$ If a function value $h(t)1$, then $f(x)=h(x-1)+g(0)=c x+d$ for suitable constants $c$ and $d$. This also holds for smaller $x$, as can be seen from equation (2) for example with the values $y=3$, $z=2$ and $w=x+1$. Since $f$ is surjective and $c$ must be positive, the constant term must of course be zero, otherwise small values would not be assumed or negative values would be assumed. Substituting $f(x)=c x$ into the original equation yields (by comparing coefficients) the conditions $c^{2}=\alpha$ and $c^{2}=1$. Since $c$ must be positive, we obtain the only solution $f(x)=x$, if $\alpha=1$. (Theresia Eisenkölbl)" 8a09b829e6ff,"Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together? $\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$",\textbf{(E),easy,"Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$. Solving for $x$, we get $x=\boxed{\textbf{(E)}\ 120}.$" b5a2ef69b23b,"7. (10 points) In a math competition, there are three","ed $A$ correctly is $3 b+a$, and the number of people who answered $A$ and other questions correctly is $3 b+a-5$, so we have: $3 b+a+3 b+a-5+3 b+2 a=39$, simplifying to $4 a+9 b=44$",medium,"【Analysis】From the problem, we get the following: The number of people who only answered $A$ correctly is $3 b+a$, and the number of people who answered $A$ and other questions correctly is $3 b+a-5$, so we have: $3 b+a+3 b+a-5+3 b+2 a=39$, simplifying to $4 a+9 b=44$. Then, we need to find the values of $a$ and $b$ by assigning values to $a$ and $b$, and find the maximum values of $a$ and $b$. Since the total number of people who answered $A$ correctly is $3 b+a+3 b+a-5=6 b+2 a-5$, we substitute the maximum values of $a$ and $b$ into $6 b+2 a-5$ to solve the problem. 【Solution】Solution: The number of people who only answered $A$ correctly is $3 b+a$, and the number of people who answered $A$ and other questions correctly is $3 b+a-5$, so we have: $3 b+a+3 b+a-5+3 b+2 a=39$, simplifying to: $4 a+9 b=44$, Since $a$ and $b$ are natural numbers, when $a=2$, $b=4$; when $a=11$, $b=0$, $$ \text { i.e. }\left\{\begin{array} { l } { a = 2 } \\ { b = 4 } \end{array} \text { or } \left\{\begin{array}{l} a=11 \\ b=0 \end{array}\right.\right. $$ The total number of people who answered $A$ correctly is $3 b+a+3 b+a-5=6 b+2 a-5$. Substituting the maximum values of $a$ and $b$ into $6 b+2 a-5$, we get the maximum number: $$ \begin{array}{l} 6 \times 4+2 \times 2-5 \\ =24+4-5 \\ =23 \text { (people) } \end{array} $$ Answer: The maximum number of people who answered $A$ correctly is 23. The answer is: 23." a33daa771527,2. The solution to the equation $1-\frac{2}{x}-\frac{1}{x^{2}}=2 x+x^{2}$ is,"0$. Let $x+\frac{1}{x}=t$, then we get $t^{2}+2 t-3=0$, which is $(t+3)(t-1)=0, t_{1}=1, t_{2}=-3$. ",easy,"2. $\frac{-3 \pm \sqrt{5}}{2}$. The original equation is $\left(x+\frac{1}{x}\right)^{2}+2\left(x+\frac{1}{x}\right)-3=0$. Let $x+\frac{1}{x}=t$, then we get $t^{2}+2 t-3=0$, which is $(t+3)(t-1)=0, t_{1}=1, t_{2}=-3$. When $t=1$, we get $x+\frac{1}{x}=1$, which has no solution; when $t=-3$, we get $x+\frac{1}{x}=-3, x=\frac{-3 \pm \sqrt{5}}{2}$. Upon verification, the solutions to the equation are $x=\frac{-3 \pm \sqrt{5}}{2}$." 27675a51ceb5,"[ Two tangents drawn from one point $]$ Given a circle of radius 1. From an external point $M$, two perpendicular tangents $MA$ and $MB$ are drawn to the circle. Between the points of tangency $A$ and $B$ on the smaller arc $AB$, a point $C$ is taken, and a third tangent $KL$ is drawn through it, forming a triangle $KLM$ with the tangents $MA$ and $MB$. Find the perimeter of this triangle.",2,easy,"Tangents drawn from a single point to a circle are equal to each other. ## Solution Since $KA = KC$ and $BL = LC$, then $$ \begin{gathered} ML + LK + KM = ML + (LC + CK) + KM = \\ = (ML + LC) + (CK + KM) = (ML + LB) + (AK + KM) = \\ = MB + AM = 1 + 1 = 2. \end{gathered} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-14.jpg?height=406&width=426&top_left_y=1509&top_left_x=816) ## Answer 2." d11b1d38eba7,"4. If there exists a real number $\theta$ that satisfies $\cos \theta-\sin ^{2} \theta=\alpha$, then the range of real number $\alpha$ is",See reasoning trace,easy,"4. $-\frac{5}{4} \leqslant \alpha \leqslant 1$. The above text has been translated into English, retaining the original text's line breaks and format." 7dd890e94993,"2. Given real numbers $a, b, x, y$ satisfy the system of equations: $$ \left\{\begin{array}{l} a x+b y=3, \\ a x^{2}+b y^{2}=7, \\ a x^{3}+b y^{3}=16, \\ a x^{4}+b y^{4}=42 . \end{array}\right. $$ Find the value of $a x^{5}+b y^{5}$.",20$.,easy,"提示 Set $$ c=x+y, d=-x y, a_{n}=a x^{n}+b y^{n} \text {. } $$ then $a_{n+1}=c a_{n}+d a_{n-1}$. $$ \text { Hence }\left\{\begin{array}{l} 7 c+3 d=16, \\ 16 c+7 d=42, \\ 42 c+16 d=a_{5} \end{array}\right. \text {. } $$ Solving yields $a_{5}=20$." 6babfb2bad37,"1. Find all real numbers $x$ and $y$ that satisfy the equation $$ \log _{2}\left(\cos ^{2}(x y)+\frac{1}{\cos ^{2}(x y)}\right)=\frac{1}{y^{2}-2 y+2} $$","\frac{1}{(y-1)^{2}+1} \leq 1$, so equality holds only if $\cos (x y)=\frac{1}{\cos (x y)}$ and $\fra",medium,"Solution. For the expression on the left side to be well-defined, we need $x y \neq \frac{\pi}{2}+k \pi, k \in \mathbb{Z}$. Since $\cos ^{2}(x y)+\frac{1}{\cos ^{2}(x y)} \geq 2$, it follows that $$ \log _{2}\left(\cos ^{2}(x y)+\frac{1}{\cos ^{2}(x y)}\right) \geq \log _{2} 2=1 $$ However, $\frac{1}{y^{2}-2 y+2}=\frac{1}{(y-1)^{2}+1} \leq 1$, so equality holds only if $\cos (x y)=\frac{1}{\cos (x y)}$ and $\frac{1}{(y-1)^{2}+1}=1$. From the second equation, we have $y=1$ and from the first equation, $x=k \pi, k \in \mathbb{Z}$. Therefore, the solutions are all pairs $(k \pi, 1)$, where $k \in \mathbb{Z}$." 43e7c13b9235,"4. Given an isosceles triangle $A B C$, where $\angle A=30^{\circ}, A B=A C$. Point $D$ is the midpoint of $B C$. On segment $A D$, point $P$ is chosen, and on side $A B$, point $Q$ is chosen such that $P B=P Q$. What is the measure of angle $P Q C ?$ (S. S. Korechkova)",$15^{\circ}$,medium,"Answer: $15^{\circ}$. Solution. Since $D$ is the midpoint of the base of the isosceles triangle, $A D$ is the median, bisector, and altitude of the triangle. Draw the segment $P C$. Since $\triangle P D B = \triangle P D C$ (by two sides and the right angle between them), $P C = P B = P Q$, meaning that all three triangles $\triangle P B C$, $\triangle P B Q$, and $\triangle P Q C$ are isosceles. Notice that $\angle A B D = \angle A C B = (180^{\circ} - 30^{\circ}) : 2 = 75^{\circ}$. Let $\angle P B D = \angle P C D = \alpha$, then $\angle P B Q = \angle P Q B = 75^{\circ} - \alpha$, and $\angle P Q C = \angle P C Q = \beta$. The sum of the angles in triangle $\triangle Q C B$ is $2 \alpha + 2 \cdot (75^{\circ} - \alpha) + 2 \beta = 180^{\circ}$, from which $\beta = 15^{\circ}$. Criteria. Correct answer without justification - 0 points. Proven that $P B = P C = P Q - 3$ points. ![](https://cdn.mathpix.com/cropped/2024_05_06_758bf2042e6fda9db2ecg-2.jpg?height=542&width=411&top_left_y=180&top_left_x=1508)" efac400a6d85,"5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard?",199,medium,"Answer: 199. Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If the guard demands 200, the outsider might refuse, as there is no difference in profit. If the guard demands more, it is more profitable for the outsider to refuse. The guard can ask for less, but the problem requires finding the largest amount. Thus, the answer is 199 coins." 2b1f242eb623,"13.155. In the first week of their vacation trip, the friends spent 60 rubles less than $2 / 5$ of the amount of money they brought with them; in the second week, $1 / 3$ of the remainder and another 12 rubles on theater tickets; in the third week, $3 / 5$ of the new remainder and another 31 rubles 20 kopecks on boat rides, after which they had 420 rubles left. How much money was spent over the three weeks of the trip?",2330 rubles,medium,"Solution. Let the initial amount be $x$ rubles. | Week | Spent | Balance | | :--- | :---: | :---: | | First | $\frac{2}{5} x-60$ (rubles) | $\frac{3}{5} x+60$ (rubles) | | Second | $\frac{1}{3}\left(\frac{3}{5} x+60\right)+12=\frac{1}{5} x+32$ (rubles) | $\frac{2}{5} x+28$ (rubles) | | Third | $\frac{3}{5}\left(\frac{2}{5} x+28\right)+31 \frac{1}{5} x=$ | $\frac{2}{5} x+28-\frac{6}{25} x-48=$ | | | $=\frac{6}{25} x+48$ (rubles) | $=\frac{4}{25} x-20$ (rubles) | According to the condition $\frac{4}{25} x-20=420$, from which $x=2750$ rubles. Over three weeks, $2750-420=2330$ (rubles) were spent. Answer: 2330 rubles." 2c81d8881958,"7. Let the function $f(x)=x^{3}+a x^{2}+b x+c\left(a, b, c\right.$ all be non-zero integers), if $f(a)=a^{3}, f(b)=b^{3}$, then the value of $c$ is A. -16 B. -4 C. 4 D. 16",See reasoning trace,easy,"$$ \begin{array}{l} \Rightarrow a|b \xrightarrow{b=k a} a+k a=-k \Rightarrow k(a+1)=-a \Rightarrow k| a \xrightarrow{a=m k} m k+1=-m \Rightarrow m(k+1)=-1 \\ \Rightarrow m=1, k=-2 \Rightarrow a=-2, b=4 \Rightarrow c=16, \text { so choose } D . \end{array} $$" 805dcb517882,$3.420 \sin 10^{\circ} \cdot \sin 20^{\circ} \cdot \sin 30^{\circ} \cdot \sin 40^{\circ} \cdot \sin 50^{\circ} \cdot \sin 60^{\circ} \cdot \sin 70^{\circ} \cdot \sin 80^{\circ}=\frac{3}{256} \cdot$,\frac{\sqrt{3}}{4} \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot\left(\sin 10^{\circ} \cdot \cos 10^{\,medium,"## Solution. $L=\frac{\sqrt{3}}{4} \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot\left(\sin 10^{\circ} \cdot \cos 10^{\circ} \cdot \cos 20^{\circ} \cdot \cos 40^{\circ}\right)=\frac{\sqrt{3}}{32} \sin 20^{\circ} \cdot \sin 40^{\circ} \times$ $\times \sin 80^{\circ}=\frac{\sqrt{3}}{64} \sin 20^{\circ} \cdot\left(\cos 40^{\circ}-\cos 120^{\circ}\right)=\frac{\sqrt{3}}{128}\left(\sin 20^{\circ}+2 \sin 20^{\circ} \cdot \cos 40^{\circ}\right)=$ $=\frac{\sqrt{3}}{128}\left(\sin 20^{\circ}-\sin 20^{\circ}+\sin 60^{\circ}\right)=\frac{3}{256}$." 787c0465576e,"3. On a glade, 25 gnomes gathered. It is known that 1) every gnome who put on a hat also put on shoes; 2) 12 gnomes came without a hat; 3) 5 gnomes came barefoot. Which gnomes are more and by how many: those who came in shoes but without a hat, or those who put on a hat?",. There are 6 more gnomes who put on a cap,medium,"Answer. There are 6 more gnomes who put on a cap. ## Solution. From condition 2, it follows that $25-12=13$ gnomes came in a cap. From condition 1, we get that exactly 13 gnomes came both in a cap and in shoes. From condition 3, it follows that a total of $25-5=20$ gnomes came in shoes. Thus, $20-13=7$ gnomes came in shoes but without a cap. Therefore, those who put on a cap (13 gnomes) are 6 more than those who came in shoes but without a cap (7 gnomes). ## Grading Criteria. - Correct solution - 7 points. - Correct answer with incomplete justification - 3-4 points. - Only the correct answer without justification - 2 points." 584ac454bec1,"11. (5 points) As shown in the figure, a cubic wooden block is placed on a table, with several points drawn on each face. The sum of the number of points on opposite faces is 13. Jingjing sees that the sum of the points on the front, left, and top faces is 16, while Qingqing sees that the sum of the points on the top, right, and back faces is 24. What is the number of points on the face touching the table? $\qquad$ .",: 6,medium,"【Solution】Solution: Since Jingjing sees the sum of the points on the front, left, and top faces is 16, and Qingqing sees the sum of the points on the top, right, and back faces is 24, Therefore, $(\text{front} + \text{back}) + (\text{left} + \text{right}) + \text{top} \times 2 = 16 + 24 = 40$, Also, because the sum of the points on opposite faces is 13, Then $13 + 13 + \text{top} \times 2 = 40$, \[ \begin{array}{l} 26 + \text{top} \times 2 = 40, \\ \text{top} \times 2 = 40 - 26, \\ \text{top} = 14 \div 2, \\ \text{top} = 7, \end{array} \] So the number of points on the face touching the table is $13 - 7 = 6$. Therefore, the answer is: 6." e48702d3f0ed,"7. The set $\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., x \in \mathbf{N}\right\}$ has $\qquad$ proper subsets.",See reasoning trace,easy,"$$ -1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2} \Rightarrow-2<\lg \frac{1}{x} \leqslant-1 \Rightarrow 1 \leqslant \lg x<2 \Rightarrow 10 \leqslant x \leqslant 99 . $$ Thus, the set contains 90 elements, so the number of its proper subsets is $2^{90}-1$." 05ecd8fe21bb,"8. Let $a>1$. If the equation $a^{x}=x$ has no real roots, then the range of the real number $a$ is $\qquad$ .",See reasoning trace,medium,"8. $a>\mathrm{e}^{\frac{1}{e}}$. From the graphs of the functions $y=a^{x}$ and $y=x$, we know that if $a>1$ and $a^{x}=x$ has no real roots, then $a^{x}>x$ always holds. Let $f(x)=a^{x}-x$. Then $f^{\prime}(x)=a^{x}(\ln a)-1>0 \Rightarrow x>-\log _{a}(\ln a)$. Therefore, $f(x)=a^{x}-x$ is decreasing on the interval $\left(-\infty,-\log _{a}(\ln a)\right)$ and increasing on the interval $\left(-\log _{a}(\ln a),+\infty\right)$. Thus, $f(x)$ achieves its minimum value at $x=-\log _{a}(\ln a)$, i.e., $$ \begin{array}{l} f(x)_{\min }=f\left(-\log _{a}(\ln a)\right) \\ =a^{-\log _{a}(\ln a)}-\left(-\log _{a}(\ln a)\right)>0 \\ \Rightarrow \frac{1}{\ln a}-\left(-\log _{a}(\ln a)\right)>0 . \\ \text { Also, } \frac{1}{\ln a}=\log _{a} \mathrm{e},-\log _{a}(\ln a)=\log _{a} \frac{1}{\ln a} \\ \Rightarrow \log _{a} \mathrm{e}>\log _{a} \frac{1}{\ln a} \Rightarrow \ln a>\frac{1}{\mathrm{e}} \Rightarrow a>\mathrm{e}^{\frac{1}{a}} . \end{array} $$" 9e05c5373711,A baker uses $6\tfrac{2}{3}$ cups of flour when she prepares $\tfrac{5}{3}$ recipes of rolls. She will use $9\tfrac{3}{4}$ cups of flour when she prepares $\tfrac{m}{n}$ recipes of rolls where m and n are relatively prime positive integers. Find $m + n.$,55,medium,"1. Convert the mixed numbers to improper fractions: \[ 6 \frac{2}{3} = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3} \] \[ 9 \frac{3}{4} = 9 + \frac{3}{4} = \frac{36}{4} + \frac{3}{4} = \frac{39}{4} \] 2. Find the ratio between the two amounts of flour: \[ \text{Ratio} = \frac{\frac{39}{4}}{\frac{20}{3}} = \frac{39}{4} \cdot \frac{3}{20} = \frac{39 \cdot 3}{4 \cdot 20} = \frac{117}{80} \] 3. Apply this ratio to the number of recipes: \[ \frac{5}{3} \cdot \frac{117}{80} = \frac{5 \cdot 117}{3 \cdot 80} = \frac{585}{240} \] 4. Simplify the fraction \(\frac{585}{240}\): \[ \frac{585}{240} = \frac{585 \div 15}{240 \div 15} = \frac{39}{16} \] 5. Identify \(m\) and \(n\) from the simplified fraction \(\frac{39}{16}\): \[ m = 39, \quad n = 16 \] 6. Add \(m\) and \(n\): \[ m + n = 39 + 16 = 55 \] The final answer is \(\boxed{55}\)." 500b9c5ab101,"3. Let $x=-\sqrt{2}+\sqrt{3}+\sqrt{5}, y=\sqrt{2}-\sqrt{3}+\sqrt{5}$, and $z=\sqrt{2}+\sqrt{3}-\sqrt{5}$. What is the value of the expression below? $$ \frac{x^{4}}{(x-y)(x-z)}+\frac{y^{4}}{(y-z)(y-x)}+\frac{z^{4}}{(z-x)(z-y)} $$",See reasoning trace,medium,"Answer: 20 Solution: Writing the expression as a single fraction, we have $$ \frac{x^{4} y-x y^{4}+y^{4} z-y z^{4}-z x^{4}+x z^{4}}{(x-y)(x-z)(y-z)} $$ Note that if $x=y, x=z$, or $y=z$, then the numerator of the expression above will be 0 . Thus, $(x-y)(x-z)(y-z) \mid\left(x^{4} y-x y^{4}+y^{4} z-y z^{4}-z x^{4}+x z^{4}\right)$. Moreover, the numerator can be factored as follows. $$ (x-y)(x-z)(y-z)\left(x^{2}+y^{2}+z^{2}+x y+y z+z x\right) $$ Hence, we are only evaluating $x^{2}+y^{2}+z^{2}+x y+y z+z x$, which is equal to $$ \begin{aligned} \frac{1}{2}\left[(x+y)^{2}+(y+z)^{2}+(z+x)^{2}\right] & =\frac{1}{2}\left[(2 \sqrt{5})^{2}+(2 \sqrt{2})^{2}+(2 \sqrt{3})^{2}\right] \\ & =2(5+2+3) \\ & =20 \end{aligned} $$" 5aef22ed1b7c,"## Task A-4.7. (20 points) The angle at the vertex of the axial section of a right circular cone is $2 \alpha$, and the radius of the base is $r$. A regular hexagonal prism is inscribed in the cone, with all its edges of equal length (one base of the prism lies in the plane of the base of the cone, and the remaining vertices are on the lateral surface of the cone). Calculate the surface area of the prism using $\alpha$ and $r$.",See reasoning trace,medium,"## Solution. Let all edges of the inscribed prism be of length $a$ and let $A B V$ be the axial section of the given cone, and $C D E F$ the section of the inscribed prism by the same plane (see figure). The shape $C D E F$ is a rectangle with side lengths $|D E|=a$ and $|E F|=2 a$ (since $\overline{E F}$ is the longer diagonal of the upper base of the prism, a regular hexagon). ![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-20.jpg?height=586&width=551&top_left_y=875&top_left_x=750) (5 points) Since $\varangle D E B = \varangle O V B = \alpha$ (angles with parallel legs), from the right triangle $D E B$ we get $|D B| = |D E| \operatorname{tg} \alpha$, or $r - a = a \operatorname{tg} \alpha$. From this, we express $a = \frac{r}{1 + \operatorname{tg} \alpha}$. (5 points) The surface area of the given prism is $O = 2 \cdot \frac{6 a^{2} \sqrt{3}}{4} + 6 a^{2} = a^{2} (3 \sqrt{3} + 6)$. (5 points) By substituting the expression for $a$, we finally get $O = \frac{r^{2} (3 \sqrt{3} + 6)}{(1 + \operatorname{tg} \alpha)^{2}}$. (5 points)" d8010caf4b4e,"The number $\sqrt {2}$ is equal to: $\textbf{(A)}\ \text{a rational fraction} \qquad \textbf{(B)}\ \text{a finite decimal} \qquad \textbf{(C)}\ 1.41421 \\ \textbf{(D)}\ \text{an infinite repeating decimal} \qquad \textbf{(E)}\ \text{an infinite non - repeating decimal}$",\textbf{(E),easy,"$2$ is not a perfect square, so $\sqrt{2}$ is irrational. The answer is $\boxed{\textbf{(E)}\ \text{an infinite non - repeating decimal}}.$ ~coolmath34" 72425e3941be,"【Question 7】 Teacher Li bought some fruit cakes at 4.8 yuan each and some chocolate cakes at 6.6 yuan each, spending a total of 167.4 yuan. It is known that the average price per cake is 6.2 yuan. Therefore, Teacher Li bought $\qquad$ fruit cakes and $\qquad$ chocolate cakes.",See reasoning trace,medium,"【Analysis and Solution】Average number problem. Hence, the ratio of the number of fruit cakes to chocolate cakes is $2: 7$; There are a total of $167.4 \div 6.2=27$ pieces; Teacher Li bought $27 \times \frac{2}{2+7}=6$ fruit cakes, and $27 \times \frac{7}{2+7}=21$ chocolate cakes. ( Method II ) Let Teacher Li bought $x$ fruit cakes and $y$ chocolate cakes; According to the problem, we have $\left\{\begin{array}{l}4.8 x+6.6 y=167.4 \\ 6.2(x+y)=167.4\end{array}\right.$; Solving, we get $\left\{\begin{array}{l}x=6 \\ y=21\end{array}\right.$; Teacher Li bought 6 fruit cakes and 21 chocolate cakes." 15bef450280c,"6.19 Solve the equation $1+2 x+4 x^{2}+\ldots+(2 x)^{n}+\ldots=3.4-1.2 x$, given that $|x|<0.5$.",\( x=\frac{1}{3} \),easy,"6.19 The left side of the equation is the sum of an infinite geometric progression, where \( b_{1}=1 \) and \( |q|=|2 x|<1 \), since \( |x|<0.5 \). According to formula (6.10), we have \[ \begin{aligned} & 1+2 x+4 x^{2}+\ldots=\frac{1}{1-2 x} \\ & \frac{1}{1-2 x}=3.4-1.2 x ;(3.4-1.2 x)(1-2 x)=1 \end{aligned} \] \[ 2.4 x^{2}-8 x+2.4=0 ; 3 x^{2}-10 x+3=0 \] from which \( x_{1}=3, x_{2}=\frac{1}{3} \). Only the root \( x=\frac{1}{3} \) satisfies the condition. Answer: \( x=\frac{1}{3} \)." ca0e758e4e6c,"25. A sequence $\left\{a_{n}\right\}$ of real numbers is defined by $a_{1}=1$ and for all integers $n \geq 1$, $$ a_{n+1}=\frac{a_{n} \sqrt{n^{2}+n}}{\sqrt{n^{2}+n+2 a_{n}^{2}}} . $$ Compute the sum of all positive integers $n<1000$ for which $a_{n}$ is a rational number.","1,9,121$ and the sum is $1+9+121=131$.",medium,"25. 131 Solution. First, note that for $k \geq 1$, $$ a_{k+1}^{2}=\frac{k(k+1) a_{k}^{2}}{k(k+1)+2 a_{k}^{2}} \Longleftrightarrow \frac{1}{a_{k+1}^{2}}-\frac{1}{a_{k}^{2}}=\frac{2}{k(k+1)}=\frac{2}{k}-\frac{2}{k+1} $$ and summing the second equation from $k=1$ to $k=n-1$ with $n \geq 2$, we get $$ \frac{1}{a_{n}^{2}}-\frac{1}{a_{1}^{2}}=\sum_{k=1}^{n-1}\left(\frac{1}{a_{k+1}^{2}}-\frac{1}{a_{k}^{2}}\right)=\sum_{k=1}^{n-1}\left(\frac{2}{k}-\frac{2}{k+1}\right)=2-\frac{2}{n} . $$ Since $a_{1}=1$, we see that $$ \frac{1}{a_{n}^{2}}=3-\frac{2}{n} \Longleftrightarrow a_{n}=\sqrt{\frac{n}{3 n-2}} $$ for all integers $n \geq 1$ and we wish to find the sum of all positive integers $n<1000$ such that $\frac{n}{3 n-2}$ is a square of some rational number. To help us look for such integers $n$, we use the following lemma that provides integer solutions to the generalized Pell equation. Lemma. ${ }^{1}$ Let $d$ be a squarefree positive integer, and let a and $b$ be positive integers such that $a^{2}-d b^{2}=1$. Set $u=a+b \sqrt{d}$. Then for each nonzero integer $n$, every solution of $x^{2}-d y^{2}=n$ is a power of $u$ times $x+y \sqrt{d}$ where $(x, y)$ is an integer solution of $x^{2}-d y^{2}=n$ with $|x| \leq \sqrt{|n|}(\sqrt{u}+1) / 2$ and $|y| \leq \sqrt{|n|}(\sqrt{u}+1) /(2 \sqrt{d})$. We now let $g=\operatorname{gcd}(n, 3 n-2)$. Then $g \in\{1,2\}$ since $g \mid 3 n-(3 n-2)=2$. We now consider the following cases: - Suppose $g=1$. Then $n=y^{2}$ and $3 n-2=x^{2}$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^{2}-3 y^{2}=-2$. Set $u=2+\sqrt{3}$, with $(2,1)$ being a solution of $x^{2}-3 y^{2}=1$ in positive integers. We now look for the positive integer solutions $(x, y)$ of $x^{2}-3 y^{2}=-2$ with $x \leq \sqrt{2}(\sqrt{u}+1) / 2 \approx 2.07$ and $y \leq \sqrt{2}(\sqrt{u}+1) /(2 \sqrt{3}) \approx 1.2$. We obtain $(x, y)=(1,1)$ as the only such integer solution, so by the above lemma, we see that all positive integer solutions $\left(x_{k}, y_{k}\right)$ of $x^{2}-3 y^{2}=-2$ are given by $x_{k}+y_{k} \sqrt{3}=(1+\sqrt{3})(2+\sqrt{3})^{k}$ for all integers $k \geq 0$. We now compute this product for small values of $k$ : As $n<1000$, we require that $y_{k} \leq 31$, so the positive integer solutions $\left(x_{k}, y_{k}\right)$ of $x^{2}-3 y^{2}=$ -2 with $y_{k} \leq 31$ are $\left(x_{k}, y_{k}\right)=(1,1),(5,3),(19,11)$ (which indeed have relatively prime coordinates). These correspond to the values of $n: n=1,9,121$. - Suppose $g=2$. Then $n=2 y^{2}$ and $3 n-2=2 x^{2}$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^{2}-3 y^{2}=-1$. Again, we set $u=2+\sqrt{3}$. We now look for the positive integer solutions $(x, y)$ of $x^{2}-3 y^{2}=-1$ with $x \leq(\sqrt{u}+1) / 2 \approx 1.47$ and $y \leq(\sqrt{u}+1) /(2 \sqrt{3}) \approx 0.85$. It turns out that there are no such solutions on these bounds, so by the above lemma, we conclude that $x^{2}-3 y^{2}=-1$ has no solutions in positive integers. Hence, the only positive integers $n<1000$ for which $a_{n}$ is a rational number are $n=1,9,121$ and the sum is $1+9+121=131$." d6b7581772d3,"What is the smallest perfect square larger than $1$ with a perfect square number of positive integer factors? [i]Ray Li[/i]",36,medium,"1. We need to find the smallest perfect square larger than $1$ that has a perfect square number of positive integer factors. 2. Let's list the perfect squares and count their factors: - $4 = 2^2$: The factors are $1, 2, 4$. There are $3$ factors. - $9 = 3^2$: The factors are $1, 3, 9$. There are $3$ factors. - $16 = 2^4$: The factors are $1, 2, 4, 8, 16$. There are $5$ factors. - $25 = 5^2$: The factors are $1, 5, 25$. There are $3$ factors. - $36 = 6^2$: The factors are $1, 2, 3, 6, 9, 12, 18, 36$. There are $9$ factors. 3. We observe that $36$ is the smallest perfect square larger than $1$ that has a perfect square number of factors ($9$ factors, which is $3^2$). Thus, the smallest perfect square larger than $1$ with a perfect square number of positive integer factors is $36$. The final answer is $\boxed{36}$" f10fb53fefc9,"25. Anna, Bridgit and Carol run in a $100 \mathrm{~m}$ race. When Anna finishes, Bridgit is $16 \mathrm{~m}$ behind her and when Bridgit finishes, Carol is $25 \mathrm{~m}$ behind her. The girls run at constant speeds throughout the race. How far behind was Carol when Anna finished? A $37 \mathrm{~m}$ B $41 \mathrm{~m}$ C $50 \mathrm{~m}$ D $55 \mathrm{~m}$ E $60 \mathrm{~m}$",63$ metres. Hence Carol finishes $(100-63)$ metres $=37$ metres behind Anna.,easy,"25. A Carol finishes 25 metres behind Bridgit, so she travels 75 metres while Bridgit runs 100 metres. Therefore she runs 3 metres for every 4 metres Bridgit runs. When Anna finishes, Bridgit has run 84 metres, so that at that time Carol has run $\frac{3}{4} \times 84$ metres $=63$ metres. Hence Carol finishes $(100-63)$ metres $=37$ metres behind Anna." f81a1e380833,"6. Draw a line through $P(3,0)$ such that the line segment $A B$ intercepted between the two lines $2 x-y-2=0$ and $x+y+3=0$ is exactly bisected by point $P$. Then the equation of this line is 保留源文本的换行和格式，直接输出翻译结果。","0$ Hint: Let $A(x, y)$, then $B(6-x,-y),\left\{\begin{array}{l}2 x-y-2=0, \\ (6-x)+(-y)+3=0,\end{arr",easy,"6. $8 x-y-24=0$ Hint: Let $A(x, y)$, then $B(6-x,-y),\left\{\begin{array}{l}2 x-y-2=0, \\ (6-x)+(-y)+3=0,\end{array}\left\{\begin{array}{l}x=\frac{11}{3}, \\ y=\frac{16}{3}\end{array}\right.\right.$" d4c0971518a0,"## Task 1 - 220831 On a day in 1981, Cathrin asks her grandfather about his birth year. The grandfather, a friend of puzzle questions, replied: ""I am older than 65 years, but younger than 100 years. The year of my birth is not divisible by 2, 3, or 5. The remainder when this year is divided by 60 is not a prime number."" Investigate whether these statements can all be true for a birth year and whether they uniquely determine the birth year! What is the birth year of the grandfather? Hint: The year should be given in full, e.g., not 11 but 1911.",See reasoning trace,medium,"I. If the information for a birth year applies, then: Since the grandfather was older than 65 years and younger than 100 years on a day in 1981, he was born before the corresponding date in 1916 and after the corresponding date in 1881. The year of his birth is therefore one of the numbers 1881, 1882, ..., 1916. Of these, only the following are not divisible by 2, 3, or 5: $$ 1883, 1889, 1891, 1897, 1901, 1903, 1907, 1909, 1913 $$ These numbers leave the following remainders when divided by 60: $23, 29, 31, 37, 41, 43, 47, 49, 53$. Of these, only 49 is not a prime number. Therefore, the information can only apply to the birth year 1909. II. This applies here; for if the grandfather was born in 1909, he was either 71 or 72 years old on a day in 1981, thus older than 65 and younger than 100 years; furthermore, 1909 is not divisible by 2, 3, or 5, and leaves a remainder of 49 when divided by 60, which is not a prime number. From I. and II. it follows: The information can apply in total, and it uniquely determines the birth year. It is 1909." 8d69be53ad3d,"1. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ the following holds $$ (x+1) f(x f(y))=x f(y(x+1)) $$",0$ and $f(x)=x$.,medium,"1. By substituting $x=0$ into the given functional equation (*), we obtain $f(0)=0$. If for some $a \neq 0$ it holds that $f(a)=0$, then for $y=a$ and any $x \neq 0$, equation (*) gives $f(a(x+1))=0$, hence $f \equiv 0$. Therefore, we further assume that $f(x) \neq 0$ for $x \neq 0$. Suppose that for some $y \neq 0$ it holds that $f(y) \neq y$. Choose $x$ such that $x f(y)=y(x+1)$ - that is, $x=\frac{y}{f(y)-y}$. Then from (*) it follows that $0=(x+1) f(x f(y)) - x f(y(x+1)) = f(y(x+1))$, hence $x=-1$. From $x f(y)=y(x+1)$ it follows that $f(y)=0$, i.e., $y=0$, contrary to the assumption. Therefore, the only solutions are the functions $f(x)=0$ and $f(x)=x$." 9a93c446009c,"2. If $\frac{x}{3 y}=\frac{y}{2 x-5 y}=\frac{6 x-15 y}{x}$, then the value of $\frac{4 x^{2}-5 x y+6 y^{2}}{x^{2}-2 x y+3 y^{2}}$ is ( ). (A) $\frac{9}{2}$. (B) $\frac{9}{4}$ (C) 5 (D) 6",See reasoning trace,medium,"2. (A). From the given conditions, we know that $x \neq 0, y \neq 0$. Transform the given equation and use the ratio to eliminate $y$, we get $$ \text { - } \begin{aligned} \frac{25 x}{75 y} & =\frac{15 y}{30 x-75 y}=\frac{6 x-15 y}{x} \\ & =\frac{25 x+15 y+16 x-15}{75}+\frac{1}{x} \\ & =\frac{31 x}{31 x}=1 . \end{aligned} $$ Thus, $x=3 y$. $$ \begin{array}{l} \text { Therefore, } \frac{4 x^{2}-5 x y+6 y^{2}}{x^{2}-2 x y+3 y^{2}}=\frac{36 y^{2}-15 y^{2}+6 y^{2}}{9 y^{2}-6 y^{2}+3 y^{2}} \\ =\frac{27 y^{2}}{6 y^{2}}=\frac{9}{2} . \end{array} $$" c73029c13a9b,"Example 7 Divide $n$ stones into two piles at random, and record the product of the number of stones in the two piles; then divide one of the piles into two piles, and record the product of the number of stones in these two piles; then divide one of the three piles into two piles, and record the product of the number of stones in these two piles; continue this process until the stones are divided into $n$ piles, each containing one stone. Find the sum of these products.",\frac{n(n-1)}{2}$ holds for all natural numbers $n$ greater than or equal to 2. This is the solution,medium,"When dividing into two piles, there is one product; when dividing into three piles, there are two products; $\cdots$; when dividing into $n$ piles, there are $n-1$ products. Now, we seek the sum of these $n-1$ products, denoted as $S(n)$. When $n=2$, $S(2)=1 \times 1=1=\frac{2 \times 1}{2}$ When $n=3$, $S(3)=1 \times 2+1 \times 1=3=\frac{3 \times 2}{2}$ When $n=4$, $S(4)=2 \times 2+1 \times 1+1 \times 1=6=\frac{4 \times 3}{2}$ $\qquad$ Based on this, we conjecture that $S(n)=\frac{n(n-1)}{2}$. We will prove this using mathematical induction. Assume that for $2 \leqslant n \leqslant k$, we have $S(n)=\frac{n(n-1)}{2}$. Then, when $n=k+1$, first divide $k+1$ stones into two piles with $m$ and $k+1-m$ stones, where $1 \leqslant m \leqslant k, m \in \mathbf{N}$. By symmetry, we can assume $1 \leqslant m \leqslant \frac{k+1}{2}$ If $m=1$, then $S(k+1)=1 \cdot k+S(k)=k+\frac{k(k-1)}{2}=\frac{(k+1) k}{2}$ If $m>1$, then $S(k+1)=m(k+1-m)+S(m)+S(k+1-m)$ $$ =m(k+1-m)+\frac{m(m-1)}{2}+\frac{(k+1-m)(k-m)}{2}=\frac{(k+1) k}{2} $$ Therefore, $S(n)=\frac{n(n-1)}{2}$ holds for all natural numbers $n$ greater than or equal to 2. This is the solution to the problem." d28ea06bb63d,"1.50 $\frac{\sqrt{x-4 \sqrt{x-4}}+2}{\sqrt{x+4 \sqrt{x-4}}-2}$ 1.51 $\left(\frac{|x-1|}{x-1} \cdot x^{2}-2 x \cdot \frac{|x+1|}{x+1}+2 x-4\right):|x-2|$.","for $x \in(4,8) \Rightarrow \frac{4}{\sqrt{x-4}}-1 ;$ for $x \in[8, \infty) \Rightarrow 1$",medium,"1.50 We have: 1) $\sqrt{x-4 \sqrt{x-4}}+2=\sqrt{x-4-4 \sqrt{x-4}+4}+2=$ $$ =\sqrt{(\sqrt{x-4}-2)^{2}}+2=|\sqrt{x-4}-2|+2 $$ 2) $\sqrt{x+4 \sqrt{x-4}}-2=\sqrt{(\sqrt{x-4}+2)^{2}}-2=$ $=\sqrt{x-4}+2-2=\sqrt{x-4}$ 3) $\frac{|\sqrt{x-4}-2|+2}{\sqrt{x-4}}=A$; If $\sqrt{x-4}-2<0$, then $\sqrt{x-4}<2$, i.e., $424$, this case is impossible. Therefore, $5+6+6+7=24$. This means the winning team scored 7 points." c37ef78123af,Let $a_n\ (n\geq 1)$ be the value for which $\int_x^{2x} e^{-t^n}dt\ (x\geq 0)$ is maximal. Find $\lim_{n\to\infty} \ln a_n.$,-\ln 2,medium,"1. Define the integral \( I_n(x) = \int_x^{2x} e^{-t^n} \, dt \). We need to find the value of \( x \) that maximizes \( I_n(x) \). 2. To find the maximum, we first compute the derivative of \( I_n(x) \) with respect to \( x \): \[ I_n'(x) = \frac{d}{dx} \left( \int_x^{2x} e^{-t^n} \, dt \right) \] Using the Leibniz rule for differentiation under the integral sign, we get: \[ I_n'(x) = e^{-(2x)^n} \cdot 2 - e^{-x^n} \] Simplifying, we have: \[ I_n'(x) = 2e^{-(2x)^n} - e^{-x^n} \] 3. Set the derivative equal to zero to find the critical points: \[ 2e^{-(2x)^n} - e^{-x^n} = 0 \] \[ 2e^{-(2x)^n} = e^{-x^n} \] Taking the natural logarithm on both sides: \[ \ln(2e^{-(2x)^n}) = \ln(e^{-x^n}) \] \[ \ln 2 - (2x)^n = -x^n \] \[ \ln 2 = (2^n - 1)x^n \] Solving for \( x^n \): \[ x^n = \frac{\ln 2}{2^n - 1} \] Therefore: \[ x = \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} = a_n \] 4. We need to find \( \lim_{n \to \infty} \ln a_n \): \[ \ln a_n = \ln \left( \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} \right) \] \[ \ln a_n = \frac{1}{n} \ln \left( \frac{\ln 2}{2^n - 1} \right) \] \[ \ln a_n = \frac{1}{n} \left( \ln (\ln 2) - \ln (2^n - 1) \right) \] 5. Evaluate the limit: \[ \lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n} \] Since \( \ln (\ln 2) \) is a constant, we focus on the term involving \( 2^n \): \[ \lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n} = \lim_{n \to \infty} \frac{-\ln (2^n - 1)}{n} \] Using the approximation \( \ln (2^n - 1) \approx \ln (2^n) = n \ln 2 \) for large \( n \): \[ \lim_{n \to \infty} \frac{-n \ln 2}{n} = -\ln 2 \] The final answer is \(\boxed{-\ln 2}\)." e8da83a42e62,"1. Find the smallest natural number, which when multiplied by 127008 gives a cube of a natural number. Explain your answer!","2^{5} \cdot 3^{4} \cdot 7^{2}$, it follows that the smallest natural number by which we need to mult",easy,"Solution. Since $127008=2^{5} \cdot 3^{4} \cdot 7^{2}$, it follows that the smallest natural number by which we need to multiply this number is the number $2 \cdot 3^{2} \cdot 7=126$." 680b54eaef9c,"Three numbers have sum $k$ (where $k\in \mathbb{R}$) such that the numbers are arethmetic progression.If First of two numbers remain the same and to the third number we add $\frac{k}{6}$ than we have geometry progression. Find those numbers?",\left(\frac{2k,medium,"1. Let the three numbers in arithmetic progression be \(a - d\), \(a\), and \(a + d\). Given that their sum is \(k\), we have: \[ (a - d) + a + (a + d) = k \] Simplifying, we get: \[ 3a = k \implies a = \frac{k}{3} \] Therefore, the three numbers are: \[ \frac{k}{3} - d, \frac{k}{3}, \frac{k}{3} + d \] 2. According to the problem, if we keep the first two numbers the same and add \(\frac{k}{6}\) to the third number, the sequence becomes a geometric progression. Thus, the new sequence is: \[ \frac{k}{3} - d, \frac{k}{3}, \frac{k}{3} + d + \frac{k}{6} \] Simplifying the third term: \[ \frac{k}{3} + d + \frac{k}{6} = \frac{k}{3} + \frac{k}{6} + d = \frac{k}{2} + d \] 3. For these numbers to be in geometric progression, the square of the middle term must equal the product of the first and third terms: \[ \left(\frac{k}{3}\right)^2 = \left(\frac{k}{3} - d\right) \left(\frac{k}{2} + d\right) \] Expanding the right-hand side: \[ \frac{k^2}{9} = \left(\frac{k}{3} - d\right) \left(\frac{k}{2} + d\right) \] \[ \frac{k^2}{9} = \frac{k}{3} \cdot \frac{k}{2} + \frac{k}{3} \cdot d - d \cdot \frac{k}{2} - d^2 \] \[ \frac{k^2}{9} = \frac{k^2}{6} + \frac{k}{6}d - d^2 \] 4. Rearranging the equation to form a quadratic in \(d\): \[ \frac{k^2}{9} = \frac{k^2}{6} + \frac{k}{6}d - d^2 \] \[ d^2 + \frac{k}{6}d - \frac{k^2}{18} = 0 \] 5. Solving this quadratic equation using the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = \frac{k}{6}\), and \(c = -\frac{k^2}{18}\): \[ d = \frac{-\frac{k}{6} \pm \sqrt{\left(\frac{k}{6}\right)^2 - 4 \cdot 1 \cdot \left(-\frac{k^2}{18}\right)}}{2 \cdot 1} \] \[ d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{4k^2}{18}}}{2} \] \[ d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{2k^2}{9}}}{2} \] \[ d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{8k^2}{36}}}{2} \] \[ d = \frac{-\frac{k}{6} \pm \sqrt{\frac{9k^2}{36}}}{2} \] \[ d = \frac{-\frac{k}{6} \pm \frac{3k}{6}}{2} \] \[ d = \frac{-\frac{k}{6} \pm \frac{k}{2}}{2} \] \[ d = \frac{-\frac{k}{6} + \frac{k}{2}}{2} \quad \text{or} \quad d = \frac{-\frac{k}{6} - \frac{k}{2}}{2} \] \[ d = \frac{\frac{k}{3}}{2} = \frac{k}{6} \quad \text{or} \quad d = \frac{-\frac{2k}{3}}{2} = -\frac{k}{3} \] 6. Therefore, the possible values for \(d\) are \(\frac{k}{6}\) and \(-\frac{k}{3}\). Substituting these back into the original arithmetic progression: - For \(d = \frac{k}{6}\): \[ \frac{k}{3} - \frac{k}{6}, \frac{k}{3}, \frac{k}{3} + \frac{k}{6} \implies \frac{k}{6}, \frac{k}{3}, \frac{k}{2} \] - For \(d = -\frac{k}{3}\): \[ \frac{k}{3} - \left(-\frac{k}{3}\right), \frac{k}{3}, \frac{k}{3} + \left(-\frac{k}{3}\right) \implies \frac{2k}{3}, \frac{k}{3}, 0 \] The final answer is \(\boxed{\left(\frac{k}{6}, \frac{k}{3}, \frac{k}{2}\right)}\) or \(\boxed{\left(\frac{2k}{3}, \frac{k}{3}, 0\right)}\)." fcc7cd951c4e,"41. Find all geometric progressions in which each term, starting from the third, is equal to the sum of the two preceding ones.",See reasoning trace,medium,"41. Let us consider such a geometric progression. Then for it, the equality holds $$ u_{1} q^{n+1}=u_{1} q^{n}+u_{1} q^{n-1} \text { for all } n \text {. } $$ From this, it follows that $$ \begin{gathered} q^{2}=q+1 \\ q=\frac{1 \pm \sqrt{5}}{2} \end{gathered} $$ Thus, the common ratio of a geometric progression, in which each term is equal to the sum of the two preceding terms, can only take the values $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$. It is clear that the converse is also true, i.e., that in any progression with one of these common ratios, each term is equal to the sum of the two preceding terms." a256d8150713,"12. The sum of two numbers $a$ and $b$ is 7 and the difference between them is 2 . What is the value of $a \times b$ ? A $8 \frac{1}{4}$ B $9 \frac{1}{4}$ C $10 \frac{1}{4}$ D $11 \frac{1}{4}$ E $12 \frac{1}{4}$",7$ and $a-b=2$. Adding these two equations together gives $2 a=9$. So $a=\frac{9}{2}$ and hence $b=7,easy,12. D Assume that $a>b$. Then $a+b=7$ and $a-b=2$. Adding these two equations together gives $2 a=9$. So $a=\frac{9}{2}$ and hence $b=7-\frac{9}{2}=\frac{14-9}{2}=\frac{5}{2}$. Therefore $a \times b=\frac{9}{2} \times \frac{5}{2}=\frac{45}{4}=11 \frac{1}{4}$. 0b63f27cba24,"8. Divide a circle into a group of $n$ equal parts and color each point either red or blue. Starting from any point, record the colors of $k(k \leqslant n)$ consecutive points in a counterclockwise direction, which is called a “$k$-order color sequence” of the circle. Two $k$-order color sequences are considered different if and only if the colors at corresponding positions are different in at least one place. If any two 3-order color sequences are different, then the maximum value of $n$ is . $\qquad$",See reasoning trace,easy,"8. 8 . In a 3rd-order color sequence, since each point has two color choices, there are $2 \times 2 \times 2=8$ kinds of 3rd-order color sequences. Given that $n$ points can form $n$ 3rd-order color sequences, we know $n \leqslant 8$. Thus, $n=8$ can be achieved. For example, determining the colors of eight points in a counterclockwise direction as “red, red, red, blue, blue, blue, red, blue” meets the conditions." a74306143706,"1. If $|a|=1,|b|=2, \boldsymbol{c}=a+b$, and $c \perp$ $a$, then the angle between vector $a$ and $b$ is degrees.",See reasoning trace,easy,"Given $c \perp a \Rightarrow c \cdot a=0 \Rightarrow(a+b) \cdot a=0$ $$ \Rightarrow|\boldsymbol{a}|^{2}+|\boldsymbol{a}| \cdot|\boldsymbol{b}| \cos \alpha=0 \Rightarrow \cos \alpha=-\frac{1}{2} \text {. } $$ Therefore, the angle $\alpha$ between vector $a$ and $b$ is $120^{\circ}$." cb7614634b09,"1. Let $p, q \in N$, and $1 \leqslant p0$) and $x_{2}=3$. $$ S_{K L M N}=x+(9-x)+(8-x)+10=3+6+5+10=24 $$ Answer: $S_{K L M N}=24 \mathrm{m}^{2}$. | Criteria | points | | :--- | :---: | | Complete solution of the problem. | 7 | | The area of one of the internal triangles is found, but the area of the entire quadrilateral is not calculated. | 5 | | The equation is justified but not solved. | 3 | | Incorrect solution. | 0 |" 860daeb3a565,"5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure. ![](https://cdn.mathpix.com/cropped/2024_05_06_9cd52224ba9edadf4a1fg-04.jpg?height=277&width=594&top_left_y=684&top_left_x=731)",500,easy,"Answer: 500. Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is $$ 3 \cdot 100 + 4 \cdot 50 = 500 $$" f78755894a3e,3. A regular pentagon and a regular icosagon are inscribed in the same circle. Which is greater: the sum of the squares of the lengths of all sides of the pentagon or the sum of the squares of the lengths of all sides of the icosagon?,the sum of the squares of the side lengths of a regular pentagon is greater,medium,"Answer: the sum of the squares of the side lengths of a regular pentagon is greater. Solution. Note that in a regular 20-sided polygon, the vertices taken every other one form a regular 10-sided polygon, and the vertices of this 10-sided polygon, taken every other one, form a regular 5-sided polygon. Therefore, it is sufficient to compare two quantities: $4 a_{20}^{2}$ and $a_{5}^{2}$, where $a_{20}$ and $a_{5}$ are the side lengths of the regular 20-sided and 5-sided polygons, respectively. Consider the corresponding fragment and introduce the vertex notations as shown in Fig. 11.3a, b. We can reason in different ways from here. First method. We use the fact that for $n>4$, the angle of any regular $n$-sided polygon is obtuse. Consider the triangle $A_{1} A_{2} A_{3}$ with the obtuse angle $A_{2}$ (see Fig. 11.3a). By the corollary of the cosine theorem, $2 a_{20}^{2}\frac{8}{3}\left(x^{2}+y^{2}\right)>4 a_{20}^{2}$, from which the answer follows. Third method. We use the formula for calculating the side lengths of regular $n$-sided polygons inscribed in a circle of radius $R: a_{n}=2 R \sin \frac{180^{\circ}}{n}$. Then $\frac{a_{5}^{2}}{4 a_{20}^{2}}=$ $\left(\frac{a_{5}}{2 a_{20}}\right)^{2}=\left(\frac{\sin 36^{\circ}}{2 \sin 9^{\circ}}\right)^{2}=\left(\frac{4 \cos 18^{\circ} \cos 9^{\circ} \sin 9^{\circ}}{2 \sin 9^{\circ}}\right)^{2}=$ ![](https://cdn.mathpix.com/cropped/2024_05_06_97c3a8739e7290d11eb7g-2.jpg?height=317&width=766&top_left_y=270&top_left_x=1156) $\left(2 \cos 18^{\circ} \cos 9^{\circ}\right)^{2}>2.25>1$, since $\cos 9^{\circ}>\cos 18^{\circ}>\cos 30^{\circ}=\frac{\sqrt{3}}{2}$. Therefore, $a_{5}^{2}>4 a_{20}^{2}$. ## Grading Criteria. “+"" - a complete and well-reasoned solution is provided “士” - a generally correct reasoning is provided, containing minor gaps or inaccuracies “-” - only the answer is provided “-"" - the problem is not solved or solved incorrectly" e79532e1d4d2,"4. Given real numbers $x, y, z$ satisfy $$ x^{2}+2 y^{2}+3 z^{2}=24 \text {. } $$ Then the minimum value of $x+2 y+3 z$ is $\qquad$",See reasoning trace,easy,"4. -12 . By Cauchy-Schwarz inequality, we have $$ \begin{array}{l} (x+2 y+3 z)^{2} \\ =(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\ \leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right) \end{array} $$ $$ =144 \text {. } $$ Therefore, $x+2 y+3 z \geqslant-12$, with equality holding if and only if $x=y=z=-2$. Hence, the minimum value of $x+2 y+3 z$ is -12." 8094792000e4,"Find all positive integers $m$ for which there exist three positive integers $a,b,c$ such that $abcm=1+a^2+b^2+c^2$.",m = 4,medium,"1. **Claim**: The only such \( m \) is \( m = 4 \), achieved by \((a, b, c) = (1, 1, 1)\). 2. **Proof**: Assume \( a, b, c \) are all positive integers such that \( abcm = 1 + a^2 + b^2 + c^2 \). 3. **Step 1**: Show that \( a, b, c \) are all odd. - If all of them are even, then the LHS is even but the RHS is odd, a contradiction. - If two of them are even, then the LHS is a multiple of \( 4 \) but \( a^2 + b^2 + c^2 + 1 \equiv 0 + 0 + 1 + 1 \equiv 2 \pmod{4} \), which is a contradiction since a multiple of \( 4 \) cannot be \( 2 \pmod{4} \). - If one of them is even, then the LHS is even and the RHS is odd again, so we have another contradiction. - Thus, \( a, b, c \) are all odd. 4. **Step 2**: Conclude that \( 4 \mid m \). - Since \( a, b, c \) are odd, \( a^2, b^2, c^2 \equiv 1 \pmod{4} \). - Therefore, \( a^2 + b^2 + c^2 + 1 \equiv 1 + 1 + 1 + 1 \equiv 0 \pmod{4} \). - Hence, \( 4 \mid abcm \implies 4 \mid m \). 5. **Step 3**: Assume for contradiction that there exists \( m \neq 4 \) and positive integers \( a, b, c \) such that the condition is satisfied. - Let \((A, B, C)\) be the triplet with \( A + B + C \) minimized for this particular \( m \). - Since \( 4 \mid m \), we must have \( m \geq 8 \). 6. **Step 4**: Assume WLOG that \( A \geq B \geq C \). - Note that \( m = \frac{1}{ABC} + \frac{A}{BC} + \frac{B}{CA} + \frac{C}{AB} \). - Since \( \frac{1}{ABC} \leq 1 \), \( B \leq A \leq CA \), and \( C \leq A \leq BA \), we have that \( \frac{1}{ABC}, \frac{B}{CA}, \frac{C}{AB} \) are all less than or equal to \( 1 \). - Therefore, \( \frac{A}{BC} \geq m - 3 \implies A \geq (m - 3)BC \). 7. **Step 5**: Rewrite the original equation as \( A^2 - (mBC)A + (B^2 + C^2 + 1) = 0 \). - By Vieta's formulas, \( (mBC - A, B, C) \) is also a solution for this particular \( m \). - We can verify that all entries are positive integers: \( mBC - A = \frac{B^2 + C^2 + 1}{A} \), so \( mBC - A > 0 \) (and clearly \( mBC - A \) is an integer). - Thus, \( (mBC - A, B, C) \) is indeed a positive integer solution for this particular \( m \). 8. **Step 6**: Since \( A + B + C \) is minimal, we must have \( (mBC - A) + B + C \geq A + B + C \implies mBC \geq 2A \). - Since \( A \geq (m - 3)BC \) as derived earlier, this means that \( mBC \geq 2A \geq (2m - 6)BC \implies m \geq 2m - 6 \implies 6 \geq m \). - But since \( m \geq 8 \), this is a contradiction! 9. **Conclusion**: Thus, \( m = 4 \) is indeed the only solution, as desired. \(\blacksquare\) The final answer is \( \boxed{ m = 4 } \)." a7ac2766c406,"The expression $\frac{a^{9} \times a^{15}}{a^{3}}$ is equal to (A) $a^{45}$ (B) $a^{8}$ (C) $a^{18}$ (D) $a^{14}$ (E) $a^{21}$",(E),easy,"The expression $\frac{a^{9} \times a^{15}}{a^{3}}$ is equal to (A) $a^{45}$ (B) $a^{8}$ (C) $a^{18}$ (D) $a^{14}$ (E) $a^{21}$ Solution $$ \begin{aligned} \frac{a^{9} \times a^{15}}{a^{3}} & =\frac{a^{24}}{a^{3}} \\ & =a^{21} \end{aligned} $$ ANSWER: (E)" b105f8f651c2,"## Task Condition Find the derivative. $$ y=\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5} $$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & y^{\prime}=\left(\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5}\right)^{\prime}= \\ & =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)\right)^{\prime}= \\ & =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x} \cdot \ln 5 \cdot(\sin 3 x \cdot \ln 5-3 \cos 3 x)+5^{x}(\cos 3 x \cdot 3 \cdot \ln 5-3(-\sin 3 x) \cdot 3)\right)= \\ & =\frac{5^{x}}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5 \cdot \sin 3 x-3 \ln 5 \cdot \cos 3 x+3 \ln 5 \cdot \cos 3 x+9 \sin 3 x\right)= \\ & =\frac{5^{x}}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5 \cdot \sin 3 x+9 \sin 3 x\right)=\frac{5^{x} \cdot \sin 3 x}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5+9\right)=5^{x} \cdot \sin 3 x \end{aligned} $$ ## Problem Kuznetsov Differentiation 15-23" 195e1f77a179,"5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$. points)",164,easy,"Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiplying them by 307 and considering linear combinations for integer $t$, we get values in natural numbers $\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t \in\{123,124,125,126\} \Rightarrow\right.$ (58,1), (41,6), $(24,11),(7,16)$ $59+47+35+23=164$ Answer: 164." f7b7d93a7750,"19. The square $A B C D$ has sides of length 105 . The point $M$ is the midpoint of side $B C$. The point $N$ is the midpoint of $B M$. The lines $B D$ and $A M$ meet at the point $P$. The lines $B D$ and $A N$ meet at the point $Q$. What is the area of triangle $A P Q$ ?",\frac{1}{2} \times(35 \sqrt{2}-21 \sqrt{2}) \times \frac{1}{2} \times$ $105 \sqrt{2}=\frac{1}{2} \ti,medium,"SolUTION 735 Let $V$ be the centre of the square $A B C D$. Let $W$ be the intersection between $B D$ and the line through $N$ parallel to $A B$. Triangles $A P B$ and $M P V$ are similar, with $B P: P V=A B$ : $M V=\frac{1}{2}$. Therefore, $B P=\frac{2}{3} \times B V=\frac{2}{3} \times \frac{1}{2} \times 105 \sqrt{2}=35 \sqrt{2}$. Similarly, triangles $A Q B$ and $N Q W$ are similar, with $B Q: Q W=$ $A B: N W=\frac{1}{4}$. Therefore, $B Q=\frac{4}{5} \times B W=\frac{4}{5} \times \frac{1}{4} \times 105 \sqrt{2}=$ $21 \sqrt{2}$ The area of $A P Q$ is $\frac{1}{2} \times Q P \times V A=\frac{1}{2} \times(35 \sqrt{2}-21 \sqrt{2}) \times \frac{1}{2} \times$ $105 \sqrt{2}=\frac{1}{2} \times 14 \sqrt{2} \times \frac{1}{2} \times 105 \sqrt{2}=735$." faba4d1847c1,,"\sqrt{2}+\sqrt{3}+\sqrt{5}$, and $C$ does not lie on $A B$. On the segment $A B$, we choose a point ",medium,"Solution: We will prove that it is possible to construct a segment of length 1. From a segment of length $a$, we can construct $A B=a(\sqrt{2}+\sqrt{3}+\sqrt{5})$ (by constructing a segment of length $a \sqrt{2}$ as the diagonal of a square with side $a$, then $a \sqrt{3}$ as the diagonal of a rectangle $a \times a \sqrt{2}$, and then $a \sqrt{5}$ as the diagonal of a rectangle $a \sqrt{2} \times a \sqrt{3}$). Let $A C=\sqrt{2}+\sqrt{3}+\sqrt{5}$, and $C$ does not lie on $A B$. On the segment $A B$, we choose a point $D$ such that $A D=a$. We construct a line $D E \| B C$ ($E$ lies on $A C$). From the similarity of triangles $A B C$ and $A D E$, we have that $\frac{A E}{A D}=\frac{A C}{A B}$, i.e., $A E=\frac{A C . A D}{A B}=1$." 4a59be805e3f,"[Decimal numeral system] Find all natural numbers that increase by 9 times when a zero is inserted between the units digit and the tens digit. #",45,easy,"Let's write our number in the form $10 a+b$, where $b$ is the units digit. We get the equation $100 a+b=9(10 a+b)$. From this, $10 a=8 b$, i.e., $5 a=4 b$. Thus, $b$ is divisible by 5. Considering the two cases $b=0, b=5$, we get the unique answer: 45." 66d06a33154c,"$9 \cdot 23$ The largest of the following four numbers is (A) $\operatorname{tg} 48^{\circ}+\operatorname{ctg} 48^{\circ}$. (B) $\sin 48^{\circ}+\cos 48^{\circ}$. (C) $\operatorname{tg} 48^{\circ}+\cos 48^{\circ}$. (D) $\operatorname{ctg} 48^{\circ}+\sin 48^{\circ}$. (Chinese Junior High School Mathematics League, 1988)",See reasoning trace,medium,"[Solution] Since $0(C)>(D)>(B)$, when $45^{\circ}(C)=(D)>(B)$, when $\alpha=45^{\circ}$; (A) $>(D)>(C)>(B)$, when $0^{\circ}<\alpha<45^{\circ}$, $D C, B B^{\prime}>O D$, we have $A A^{\prime}+B B^{\prime}>B B^{\prime}+D C>D C+O D$, and $A A^{\prime}+B B^{\prime}>A A^{\prime}+O D>D C+O D$. That is, $\operatorname{tg} \alpha+\operatorname{ctg} \alpha>\operatorname{ctg} \alpha+\sin \alpha>\sin \alpha+\cos \alpha$, and $\operatorname{tg} \alpha+\operatorname{ctg} \alpha>\operatorname{tg} \alpha+\cos \alpha>\sin \alpha+\cos \alpha$. Moreover, by $\triangle B^{\prime} E C \backsim \triangle C F A^{\prime}$, and $C B^{\prime}\operatorname{ctg} \alpha+\sin \alpha$. Thus, $(A)>(C)>(D)>(B)$." 3d75e0cabe35,"17. The sum of two integers $A$ and $B$ is 2010 . If the lowest common multiple of $A$ and $B$ is 14807 , write down the larger of the two integers $A$ or $B$.",$17 \times 67=1139$,medium,"17. Ans: 1139 . A direct way to solve the problem is to factor 14807 directly. Alternatively, one may hope for $A$ and $B$ to have common factors to simplify the problem. This is a good strategy because of the following fact: ""The greatest common divisor of $A$ and $B$, equals the greatest common divisor of $A+B$ and $\operatorname{lcm}(A, B)$."" 2010 is easily factored as $2 \times 3 \times 5 \times 67$. Checking that 67 is also a factor of 14807 , we can conclude that 67 is also a factor of $A$ and $B$. The problem is reduced to finding $a$ and $b$ such that $$ a+b=\frac{2010}{67}=30 \quad \text { and } \quad a b=\frac{14807}{67}=221 . $$ Since 221 can be factoredeasily, $a$ and $b$ must be 13 and 17 . So the answer is $17 \times 67=1139$." ee27d26d250c,"4. Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be non-negative real numbers whose sum is 1. Determine the maximum possible value of the expression $$ x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5} $$",See reasoning trace,medium,"Solution. Due to the non-negativity of the numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, the inequality holds $$ x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5} \leq\left(x_{2}+x_{4}\right)\left(x_{1}+x_{3}+x_{5}\right) $$ Considering the inequality $a b \leq \frac{1}{4}(a+b)^{2}$ for $a=x_{2}+x_{4}$ and $b=x_{1}+x_{3}+x_{5}$, we get $$ \left(x_{2}+x_{4}\right)\left(x_{1}+x_{3}+x_{5}\right) \leq \frac{1}{4}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=\frac{1}{4} $$ From (1) and (2), it is clear that the value of the given expression does not exceed $\frac{1}{4}$. The given expression for $x_{1}=x_{2}=\frac{1}{2}$ and $x_{3}=x_{4}=x_{5}=0$ is equal to $\frac{1}{4}$, so this is its maximum value (this value is also achieved for some other values of the variables). ## 4th year" 29a303d4680d,"If $S=6 \times 10000+5 \times 1000+4 \times 10+3 \times 1$, what is $S$ ? (A) 6543 (B) 65043 (C) 65431 (D) 65403 (E) 60541",(B),easy,"If $S=6 \times 10000+5 \times 1000+4 \times 10+3 \times 1$, what is $S$ ? (A) 6543 (B) 65043 (C) 65431 (D) 65403 (E) 60541 ## Solution $S=60000+5000+40+3$ $=65043$ ANSWER: (B)" 7b488ec559a0,"6. Two real numbers $a$ and $b$ are randomly chosen from $[-1,1]$. Then the probability that the equation $x^{2}+a x+b=0$ has real roots is $\qquad$",\frac{S}{4}=\frac{13}{24}$.,medium,"6. $\frac{13}{24}$. Notice, the equation $x^{2}+a x+b=0$ has real roots $$ \Leftrightarrow a^{2}-4 b \geqslant 0 \Leftrightarrow b \leqslant \frac{1}{4} a^{2} \text {. } $$ Thus, in the Cartesian coordinate system $a O b$, the set of real number pairs $(a, b)$ that satisfy the condition forms the shaded region in Figure 1, with an area of $$ S=2+2 \int_{0}^{1} \frac{1}{4} a^{2} \mathrm{~d} a=\frac{13}{6} . $$ Therefore, the required probability is $p=\frac{S}{4}=\frac{13}{24}$." fe65513b0768,"Joshua rolls two dice and records the product of the numbers face up. The probability that this product is composite can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $m+n$. [i]Proposed by Nathan Xiong[/i]",65,medium,"1. **Identify the total number of outcomes:** - When rolling two dice, each die has 6 faces, so the total number of outcomes is: \[ 6 \times 6 = 36 \] 2. **Identify non-composite products:** - A product is non-composite if it is either 1 or a prime number. - The product 1 can only occur with the pair \((1,1)\). - Prime products can occur with the following pairs: \[ (1,2), (2,1), (1,3), (3,1), (1,5), (5,1) \] - These pairs yield the prime products 2, 3, and 5. 3. **Count the non-composite outcomes:** - There is 1 way to get the product 1. - There are 6 ways to get a prime product (as listed above). 4. **Calculate the number of composite outcomes:** - The total number of outcomes is 36. - The number of non-composite outcomes is \(1 + 6 = 7\). - Therefore, the number of composite outcomes is: \[ 36 - 7 = 29 \] 5. **Calculate the probability of getting a composite product:** - The probability is the number of composite outcomes divided by the total number of outcomes: \[ \frac{29}{36} \] 6. **Express the probability in simplest form:** - The fraction \(\frac{29}{36}\) is already in simplest form because 29 and 36 are relatively prime. 7. **Compute \(m+n\):** - Here, \(m = 29\) and \(n = 36\). - Therefore, \(m+n = 29 + 36 = 65\). The final answer is \(\boxed{65}\)." 46f28dbcb19f,"4. Given real numbers $x, y$ satisfy $\frac{x^{2}}{3}+y^{2}=1$. Then $$ P=|2 x+y-4|+|4-x-2 y| $$ the range of values for $P$ is $\qquad$.",See reasoning trace,medium,"4. $[2,14]$. Let $x=\sqrt{3} \cos \theta, y=\sin \theta(\theta \in[0,2 \pi))$. Then $2 x+y=2 \sqrt{3} \cos \theta+\sin \theta$ $$ \begin{array}{l} =\sqrt{13} \sin (\theta+\alpha)<4, \\ x+2 y=\sqrt{3} \cos \theta+2 \sin \theta \\ =\sqrt{7} \sin (\theta+\beta)<4 . \end{array} $$ $$ \begin{array}{l} \text { Hence } P=|2 x+y-4|+|4-x-2 y| \\ =(4-2 x-y)+(4-x-2 y) \\ =8-3(x+y) \\ =8-3(\sqrt{3} \cos \theta+\sin \theta) \\ =8-6 \sin \left(\theta+\frac{\pi}{3}\right) . \end{array} $$ Since $\theta \in[0,2 \pi)$, we get $2 \leqslant P \leqslant 14$." 886c8b61c793,"23. Find the smallest positive integer $n(\geqslant 3)$, such that in any set of $n$ points in the plane with no three points collinear, there must be three points that are the vertices of a non-isosceles triangle.","\overparen{M A N} \cup \overparen{M B N} \), and \( l_{2} = \) line segment \( M N \). Take any \( P",medium,"23. First, when 6 points in a plane are the 5 vertices of a regular pentagon and its center, all triangles formed by these 6 points are isosceles triangles, so the required positive integer \( n \geqslant 7 \). Next, if there exist 7 points in a plane (where no three points are collinear) such that all triangles formed by these 7 points are isosceles triangles. Let the set of these 7 points be \( E \). Without loss of generality, assume the two farthest points in \( E \) are \( A \) and \( B \). Draw arcs centered at \( A \) and \( B \) with radius \( AB \) intersecting at points \( M \) and \( N \). Let \( l_{1} = \overparen{M A N} \cup \overparen{M B N} \), and \( l_{2} = \) line segment \( M N \). Take any \( P \in E, P \neq A \), \( P \neq B \), then \( \triangle P A B \) is an isosceles triangle. Thus, at least one of the following three cases holds: (1) \( P A = P B \leqslant A B \); (2) \( P A = A B \geqslant P B \); (3) \( P B = A B \geqslant P A \), so \( P \in l_{1} \cup l_{2} \). Since no three points in \( E \) are collinear, there are at most 2 points of \( E \) on \( l_{2} \), so at least \( 7 - 2 - 2 = 3 \) points of \( E \) are on \( \overparen{A M} \), \( \overparen{B M} \), \( \overparen{A N} \), \( \overparen{B N} \) (excluding endpoints). Without loss of generality, assume there is a point \( C \in E \) on \( \overparen{B N} \) (where \( C \neq B \), \( C \neq N \)), and \( \overparen{B C} \) contains no points of \( E \). Draw arcs centered at \( A \) and \( C \) with radius \( A C = A B \) intersecting at points \( M' \) and \( N' \). Let \( l_{3} = \overparen{M' A N'} \cup \overparen{M' C N'} \), and \( l_{4} = \) line segment \( M' N' \). Similarly, all points of \( E \) belong to \( l_{3} \cup l_{4} \). Therefore, \( E \subseteq (l_{1} \cup l_{2}) \cap (l_{3} \cup l_{4}) \), and \( \overparen{B C} \) contains no points of \( E \). So, \( E \subseteq \overparen{M' B} \cup \overparen{C N} \cup \{A\} \cup \{O\} \cup \{T\} \cup \{R\} \), where \( O \) is the intersection of \( l_{2} \) and \( l_{4} \), \( T \) is the intersection of \( l_{2} \) and \( \overparen{M' A} \), and \( R \) is the intersection of \( l_{4} \) and \( \overparen{A N} \). Let \( S_{1} = \overparen{M' B} \cup \{T\} \), \( S_{2} = \overparen{C N} \cup \{R\} \), then \( S_{1} \cup S_{2} \) contains at least \( 7 - 3 - 1 = 3 \) points of \( E \) different from \( A \), \( B \), and \( C \). So, \( S_{1} \) or \( S_{2} \) contains at least 2 points of \( E \) different from \( A \), \( B \), and \( C \). Without loss of generality, assume \( S_{2} \) contains 2 points \( P \) and \( Q \) of \( E \) different from \( A \), \( B \), and \( C \), then there are only two cases: (1) \( P \in \overparen{C N} \), \( Q \in \overparen{C N} \). Since \( \triangle P B C \) is an isosceles triangle and \( P B > B C \), we have \( P C = B C \). Similarly, from \( \triangle Q B C \) being an isosceles triangle, we get \( Q C = B C \). So, \( C P = C Q \), i.e., \( C \) lies on the perpendicular bisector of \( P Q \), but the perpendicular bisector of \( P Q \) only passes through \( A \) and not through \( C \), a contradiction. (2) \( P \in \overparen{C N} \), \( Q = R \). Similarly, we get \( P C = B C \). Since \( \triangle R B C \) is an isosceles triangle and \( B R = B N > B C \), we have \( R C = B C \), so \( R C = P C = B C \). In \( \triangle B C P \) and \( \triangle B C R \), \( B C = B C \), \( P C = R C \), \( \angle B C P \geqslant \angle B C N > \angle B C R \), so \( B P > B R \). This contradicts \( B P \leqslant B N = B R \). Therefore, there do not exist 7 points in a plane (where no three points are collinear) such that all triangles formed by these 7 points are isosceles triangles. In summary, the smallest positive integer \( n = 7 \)." 1b94ca16ad98,"4. Find all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$, such that for all integers $m, n$, we have $$ \begin{array}{r} f(f(m)+n)+f(m) \\ =f(n)+f(3 m)+2014 \end{array} $$",See reasoning trace,medium,"4. Let the function $f$ satisfy equation (1). Let $C=1007$, and define the function $g: \mathbf{Z} \rightarrow \mathbf{Z}$ such that for all $m \in \mathbf{Z}$, we have $$ g(m)=f(3 m)-f(m)+2 C . $$ In particular, $g(0)=2 C$. Thus, for all $m, n \in \mathbf{Z}$, we have $$ f(f(m)+n)=g(m)+f(n) \text {. } $$ By mathematical induction in two directions, we know that for all $m, n, t \in \mathbf{Z}$, we have $$ f(t f(m)+n)=t g(m)+f(n) \text {. } $$ Substituting $(r, 0, f(0))$ and $(0,0, f(r))$ for $(m, n, t)$, we get $$ \begin{array}{l} f(0) g(r)=f(f(r) f(0))-f(0) \\ =f(r) g(0) . \end{array} $$ If $f(0)=0$, then by $g(0)=2 C>0$, we have $$ f(r)=0 \text {, } $$ which contradicts equation (1). If $f(0) \neq 0$, then $g(r)=\alpha f(r)$, where $\alpha=\frac{g(0)}{f(0)}$ is a non-zero constant. By the definition of $g$, we have $$ f(3 m)=(1+\alpha) f(m)-2 C, $$ i.e., for all $m \in \mathbf{Z}$, we have $$ f(3 m)-\beta=(1+\alpha)(f(m)-\beta), $$ where $\beta=\frac{2 C}{\alpha}$. Using mathematical induction on $k$, we know that for all integers $k (k \geqslant 0), m$, we have $$ f\left(3^{k} m\right)-\beta=(1+\alpha)^{k}(f(m)-\beta) . $$ Since $3 \times 2014$, by equation (1), there exists $a \in \mathbf{Z}$ such that $d=f(a)$ is not divisible by 3. By equation (2), we have $$ f(n+t d)=f(n)+t g(a)=f(n)+\alpha t f(a), $$ i.e., for all $n, t \in \mathbf{Z}$, we have $$ f(n+t d)=f(n)+\alpha t d . $$ Since $(3, d)=1$, by Euler's theorem, there exists $k$ (we can take $k=\varphi(|d|)$) such that $$ d \mid\left(3^{k}-1\right) \text {. } $$ For each $m \in \mathbf{Z}$, by equation (5), we have $$ f\left(3^{k} m\right)=f(m)+\alpha\left(3^{k}-1\right) m \text {. } $$ Thus, equation (4) becomes $$ \left[(1+\alpha)^{k}-1\right](f(m)-\beta)=\alpha\left(3^{k}-1\right) m \text {. } $$ Since $\alpha \neq 0$, for $m \neq 0$, the right-hand side of equation (6) is not zero. Therefore, the first factor on the left-hand side of equation (6) is also not zero. Hence, $f(m)=\frac{\alpha\left(3^{k}-1\right) m}{(1+\alpha)^{k}-1}+\beta$. This indicates that $f$ is a linear function. Let $f(m)=A m+\beta (m \in \mathbf{Z}$, constant $A \in \mathbf{Q})$. Substituting this into equation (1), we know that for all $m \in \mathbf{Z}$, we have $$ \left(A^{2}-2 A\right) m+(A \beta-2 C)=0 $$ $\Leftrightarrow A^{2}=2 A$, and $A \beta=2 C$. The first equation is equivalent to $A \in\{0,2\}$. Since $C \neq 0$, by the second equation, we have $$ A=2, \beta=C \text {. } $$ Thus, the function that satisfies the conditions is $$ f(n)=2 n+1007 $$" d3a7b563c738,"$1 \cdot 68$ If $a, b$ and $c$ represent integers, the expressions $\sqrt{a+\frac{b}{c}}$ and $a \sqrt{\frac{b}{c}}$ are equal if and only if (A) $a=b=c=1$. (B) $a=b$ and $c=a=1$. (C) $c=\frac{b\left(a^{2}-1\right)}{a}$. (D) $a=b$ and $c$ is any value. (E) $a=b$ and $c=a-1$. (6th American High School Mathematics Examination, 1955)",$(C)$,easy,"[Solution] If $\sqrt{a+\frac{b}{c}}=a \sqrt{\frac{b}{c}}$, after simplification we have $(a c+b)=a^{2} b$, thus $$ c=\frac{b\left(a^{2}-1\right)}{a} \text {. } $$ Therefore, the answer is $(C)$." 7a06f499a553,"Let $ABC$ be an acute triangle with $AB = 3$ and $AC = 4$. Suppose that $AH,AO$ and $AM$ are the altitude, the bisector and the median derived from $A$, respectively. If $HO = 3 MO$, then the length of $BC$ is [img]https://cdn.artof A. $3$ B. $\frac72$ C. $4$ D. $\frac92$ E. $5$",\frac{7,medium,"1. **Define Variables and Relationships:** Let \( BC = a \), \( AB = c = 3 \), and \( AC = b = 4 \). We are given that \( HO = 3MO \). Let \( OM = x \), then \( HO = 3x \) and \( MO = x \). 2. **Use the Angle Bisector Theorem:** The angle bisector theorem states that the ratio of the two segments created by the angle bisector on the opposite side is equal to the ratio of the other two sides of the triangle. Therefore, we have: \[ \frac{AB}{AC} = \frac{y + x}{y - x} \] Given \( AB = 3 \) and \( AC = 4 \), we get: \[ \frac{3}{4} = \frac{y + x}{y - x} \] 3. **Solve the Ratio Equation:** Cross-multiplying gives: \[ 3(y - x) = 4(y + x) \] Simplifying, we get: \[ 3y - 3x = 4y + 4x \] \[ -7x = y \] Therefore, \( y = -7x \). 4. **Use the Median Relationship:** The median \( AM \) divides \( BC \) into two equal segments, so \( M \) is the midpoint of \( BC \). Therefore, \( BM = MC = \frac{a}{2} \). 5. **Use the Given Condition \( HO = 3MO \):** Since \( HO = 3x \) and \( MO = x \), we have: \[ H = 3M \] 6. **Calculate the Length of \( BC \):** Since \( y = -7x \), we substitute \( y \) into the equation for the angle bisector theorem: \[ \frac{3}{4} = \frac{-7x + x}{-7x - x} \] Simplifying, we get: \[ \frac{3}{4} = \frac{-6x}{-8x} \] \[ \frac{3}{4} = \frac{3}{4} \] This confirms our relationship. 7. **Determine the Length of \( BC \):** Since \( y = -7x \) and \( y = \frac{a}{2} \), we have: \[ -7x = \frac{a}{2} \] Solving for \( a \), we get: \[ a = -14x \] 8. **Final Calculation:** Since \( x \) is a positive length, we take the absolute value: \[ a = 14x \] Given the problem constraints and the relationships, we find that: \[ BC = \frac{7}{2} \] The final answer is \(\boxed{\frac{7}{2}}\)." cfb91ad1e736,"Let $S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$. Then $S$ equals: $\textbf{(A)}\ (x-2)^4 \qquad \textbf{(B)}\ (x-1)^4 \qquad \textbf{(C)}\ x^4 \qquad \textbf{(D)}\ (x+1)^4 \qquad \textbf{(E)}\ x^4+1$",\textbf{(C),easy,"Let $y = x-1$. Substitution results in \[S = y^4 + 4y^3 + 6y^2 + 4y + 1\] \[S = (y+1)^4\] Substituting back results in \[S = x^4\] The answer is $\boxed{\textbf{(C)}}$. This problem can also be solved by traditionally expanding and combining like terms (though it would take much longer)." ef362edbcac6,"24. Find all positive functions defined on $(0,+\infty)$ that satisfy $$ a b \leqslant \frac{1}{2}\left\{a f(a)+b f^{-1}(b)\right\} $$ where $a, b \in(0,+\infty)$.",See reasoning trace,medium,"24. (1) From the given conditions, it is easy to see that $f$ maps $(0,+\infty)$ one-to-one onto $(0,+\infty)$. (2) Let $x, y \in \mathbf{R}^{+}=(0,+\infty), z=f(y)$, then $$ 2 x z \leqslant x f(x)+z f^{-1}(z)=x f(x)+y f(y). $$ That is, $$ 2 x f(y) \leqslant x f(x)+y f(y), $$ and $$ 2 y f(x) \leqslant x f(y)+y f(x), $$ therefore thus $$ \begin{array}{c} x f(y)+y f(x) \leqslant x f(x)+y f(y), \\ (x-y)(f(x)-f(y)) \geqslant 0, \end{array} $$ This shows that $f(x)$ is an increasing function. (3) From (1) and (2), it is known that $f(x)$ is continuous; (4) From (1) and (2), we get $\frac{y-x}{y}-f(x) \leqslant f(y)-f(x) \leqslant \frac{y-x}{x} f(y)$. Therefore, $\frac{f(y)-f(x)}{y-x}$ lies between $\frac{f(x)}{y}$ and $\frac{f(y)}{x}$. Using the continuity of $f$, we have $f^{\prime}(x)=\frac{f(x)}{x}, x \in(0,+\infty)$, thus $f(x)=C x, C>0$. It is easy to verify that such a function also satisfies the given functional inequality. Therefore, the general form of the function is $$ f(x)=C x,(C>0 \text { is a constant }) \text {. } $$" 677029e759a8,"Example. On average, 6 letters are sent to the department per week, of which 2 are from abroad. Find the probability that 13 letters will be sent in 2 weeks, of which 5 are from abroad.",". $\mathrm{P}\left(\mu_{2}=8, \tilde{\mu}_{2}=5\right) \approx 0",medium,"SOLUTION. There are two independent simple streams of letters with intensities $\lambda_{1}=4$ (internal) and $\lambda_{2}=2$ (foreign). Let $\mu_{t}$ and $\tilde{\mu}_{t}$ be the number of events that occurred over time $t$ in the first and second streams, respectively. We are interested in the probability of the random event $\left\{\mu_{2}=8, \tilde{\mu}_{2}=5\right\}$. Using formula (1), we get $$ \begin{aligned} & \mathrm{P}\left(\mu_{2}=8\right)=\frac{8^{8}}{8!} e^{-8} \approx 0.1396 \\ & \mathrm{P}\left(\tilde{\mu}_{2}=5\right)=\frac{4^{5}}{5!} e^{-4} \approx 0.1563 \end{aligned} $$ (approximate values of the probabilities can be obtained from Table 5.3 in the book by L.N. Bolshev and N.V. Smirnov ""Tables of Mathematical Statistics"", Moscow: Nauka, 1983 or using the PEШЕБНИK.ВM package). Since the random events $\left\{\mu_{2}=8\right\}$ and $\left\{\tilde{\mu}_{2}=5\right\}$ are independent, we find the desired probability from relations (2) and (3) $$ \mathrm{P}\left(\mu_{2}=8, \tilde{\mu}_{2}=5\right)=\mathrm{P}\left(\mu_{2}=8\right) \mathrm{P}\left(\tilde{\mu}_{2}=5\right) \approx 0.1396 \cdot 0.1563 \approx 0.0218 $$ Answer. $\mathrm{P}\left(\mu_{2}=8, \tilde{\mu}_{2}=5\right) \approx 0.0218$. PROBLEM CONDITIONS. 1. In a certain city, the average number of road traffic accidents per day is 5. Find the probability that there will be 10 road traffic accidents over two days. 2. In Rutherford and Geiger's observations, a radioactive substance emitted on average $3.87 \alpha$-particles over a period of 7.5 seconds. Find the probability that this substance will emit at least one $\alpha$-particle in one second. 3. Calls are made to an information bureau on average 30 times per hour. Find the probability that the first call after the start of the shift will occur no later than 1 minute. 4. Calls are made to an information bureau on average 30 times per hour. Find the probability that the second call after the start of the shift will occur no earlier than 1 minute and no later than 2 minutes. 5. Buses on a certain route arrive at a specific stop on average every 15 minutes. Find the probability that the waiting time for a bus will be at least 20 minutes. 6. A certain amount of dough is sprinkled with 1000 raisins, after which everything is thoroughly mixed. From this dough, 100 buns are made. Find the probability that a specific bun will contain exactly 10 raisins. 7. The average density of pathogenic bacteria in one cubic meter of air is 80. A sample of 1 cubic decimeter of air is taken. Find the probability that at least one bacterium will be found in it. 8. A book of 500 pages contains 500 typos. Find the probability that a specific page will contain at least three typos. 9. Messages to an office arrive either by phone or by fax. On average, 20 messages arrive per hour, with phone calls being three times more frequent than faxes. What is the probability that the number of phone calls in an hour will be more than 15, and the number of faxes will be less than 5? 10. During an eight-hour workday, on average, 10 orders for the production of a certain product are received from Moscow and 5 orders from St. Petersburg. Find the probability that at least one order from St. Petersburg and no orders from Moscow will be received during an hour of work. Answers. 1. 0.1251. 2. 0.4031. 3. 0.3935. 4. 0.1740. 5. 0.2636. 6. 0.1251. 7. 0.0769. 8. 0.0803. 9. 0.1903. 10. 0.1331. ### 6.11. Formulas of Total Probability and Bayes PROBLEM STATEMENT. Consider a situation where a second random experiment is conducted following a certain random experiment, based on the results of the first. 1. Find the probability of a certain random event $A$ associated with the second experiment. 2. It is known that event $A$ occurred. Find the probability that the first experiment ended with a certain outcome. PLAN OF SOLUTION." 381bcfe29c3d,7.230. $\log _{x+1}(x-0.5)=\log _{x-0.5}(x+1)$.,1,medium,"Solution. Domain of definition: $\left\{\begin{array}{l}0 b\). From this rectangle, four congruent right triangles (one at each corner, with the legs lying on the sides of the rectangle) are to be cut off such that the remaining figure forms an octagon with equal sides. The side of the octagon is to be expressed in terms of \(a\) and \(b\) and constructed from \(a\) and \(b\). Additionally, the conditions under which the",See reasoning trace,medium,"} The lengths of the catheti of the cut-off right-angled triangular areas are denoted by \(x\) and \(y\), and the length of the hypotenuse, which is equal to the side length of the octagon to be calculated, is denoted by \(c\). Then we have: \[ \begin{aligned} & 2 x+c=2 a, \text { i.e., } \quad x=\frac{2 a-c}{2} \\ & 2 y+c=2 b, \text { i.e., } \quad y=\frac{2 b-c}{2} \\ & x^{2}+y^{2}=c^{2} \end{aligned} \] Thus, due to (1), (2), and (3), \[ \frac{(2 a-c)^{2}}{4}+\frac{(2 b-c)^{2}}{4}=c^{2} \Rightarrow c^{2}+2(a+b) c-2\left(a^{2}+b^{2}\right)=0 \] It follows that either \[ \begin{gathered} c=-(a+b)+\sqrt{(a+b)^{2}+2\left(a^{2}+b^{2}\right)} \text { or } \\ c=-(a+b)-\sqrt{(a+b)^{2}+2\left(a^{2}+b^{2}\right)} \end{gathered} \] must hold. Since \(a\), \(b\), and \(c\) are positive, \(c\) cannot satisfy relation (5). If the construction is possible at all, \(c\) must satisfy condition (4). Remark: The construction is possible if and only if \(c<2 \min (a, b)\), i.e., if \(\max (a, b)<\) \(3 \min (a, b)\). Taken from [2]" dcc8a1a3404b,,14,easy,"Answer: 14. Solution. Let there be $d$ girls at the disco, then there are twice as many boys, i.e., $d+d$. The number of girls, excluding Masha, is $d-1$ (all except herself). Since there are 8 more boys than the other girls, excluding Masha, then $d+d=(d-1)+8$. From this, it follows that $d=7$, so the total number of boys is $7+7=14$." 1b2622ee5cf9,"24. Molly, Dolly, Sally, Elly and Kelly are sitting on a park bench. Molly is not sitting on the far right and Dolly is not sitting on the far left. Sally is not sitting at either end. Kelly is not sitting next to Sally and Sally is not sitting next to Dolly. Elly is sitting to the right of Dolly but not necessarily next to her. Who is sitting at the far right end? A Molly B Dolly C Sally D Kelly E Elly",See reasoning trace,medium,"24. E The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions 2, 3 and 4 from the left-hand end. Were Sally to sit in place 2, neither Dolly nor Kelly could sit in places 1 or 3 as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places 4 and 5 which is impossible. Similarly, were Sally to sit in place 3, Dolly could not sit in place 2 or 4 and the question also tells us she cannot sit in place 1 so Dolly would have to sit in place 5 making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place 4, Dolly could sit in place 2, Kelly in place 1, Molly (who cannot sit in place 5) in place 3 leaving Elly to sit in place 5 at the right-hand end." c672da708565,"Let's write the following numerical expressions in a simpler form: (1) $$ \begin{gathered} \frac{1}{(1+\sqrt{1+\sqrt{8}})^{4}}+\frac{1}{(1-\sqrt{1+\sqrt{8}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{8}})^{3}}+\frac{2}{(1-\sqrt{1+\sqrt{8}})^{3}} \\ \frac{1}{(1+\sqrt{1+\sqrt{2}})^{4}}+\frac{1}{(1-\sqrt{1+\sqrt{2}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{2}})^{3}}+\frac{2}{(1-\sqrt{1+\sqrt{2}})^{3}} \\ \frac{1}{(1+\sqrt{1+\sqrt{1.6}})^{4}}+\frac{1}{\left(1-\sqrt{1+\sqrt{1.6}}\right)^{4}}+\frac{2}{\left(1+\sqrt{1+\sqrt{1.6}}\right)^{3}}+ \\ \\ +\frac{2}{\left(1-\sqrt{1+\sqrt{1.6}}\right)^{3}} \end{gathered} $$","8$, $x=2$, and $x=1.6$ are -1, -1, and 0, respectively.",medium,"The three expressions are of the following form: $$ \begin{aligned} \frac{1}{(1+\sqrt{1+\sqrt{x}})^{4}}+ & \frac{1}{(1-\sqrt{1+\sqrt{x}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{x}})^{3}}+ \\ & +\frac{2}{(1-\sqrt{1+\sqrt{x}})^{3}} \end{aligned} $$ By simplifying this, the desired forms can be obtained by substituting 8.2 and 1 for $x$. Moreover, this can be further simplified by introducing the notation $$ \sqrt{1+\sqrt{x}}=y $$ Let's bring the first two terms and the last two terms of (1) to a common denominator, choosing the product of the respective denominators as the common denominator, and immediately taking (2) into account. The common denominators are: $$ \begin{aligned} & (1+y)^{4}(1-y)^{4}=\left(1-y^{2}\right)^{4}=(-\sqrt{x})^{4}=x^{2} \\ & (1+y)^{3}(1-y)^{3}=\left(1-y^{2}\right)^{3}=(-\sqrt{x})^{3}=-x \sqrt{x} \end{aligned} $$ The sum of the two pairs of terms, after expanding the powers and taking (2) into account, is $$ \begin{gathered} \frac{1}{x^{2}}\left[(1-y)^{4}+(1+y)^{4}\right]=\frac{2}{x^{2}}\left(1+6 y^{2}+y^{4}\right)= \\ =\frac{2}{x^{2}}(1+6+6 \sqrt{x}+1+2 \sqrt{x}+x)=\frac{2}{x^{2}}(8+8 \sqrt{x}+x) \end{gathered} $$ and respectively $$ -\frac{2}{x \sqrt{x}}\left[(1-y)^{3}+(1+y)^{3}\right]=-\frac{4}{x \sqrt{x}}\left(1+3 y^{2}\right)=-\frac{4}{x \sqrt{x}}(4+3 \sqrt{x}) $$ Now, taking $x^{2}$ as the common denominator, the entire expression transforms into $$ \frac{2}{x^{2}}(8+8 \sqrt{x}+x-8 \sqrt{x}-6 x)=\frac{2}{x^{2}}(8-5 x) $$ Accordingly, the values of the given expressions for $x=8$, $x=2$, and $x=1.6$ are -1, -1, and 0, respectively." ff960e98101f,"1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds.",3 \cdot(0+2+4+\cdots+2998)+2 \cdot 1500=$ $3 \cdot \frac{2998 \cdot 1500}{2}+3000=6748500$. Then $f(,easy,"Answer: 6748501 Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 2998$. We get: $$ \begin{aligned} & f(2)-f(0)=3 \cdot 0+2 \\ & f(4)-f(2)=3 \cdot 2+2 \end{aligned} $$ $$ f(3000)-f(2998)=3 \cdot 2998+2 $$ Adding the equations, we get: $f(3000)-f(0)=3 \cdot(0+2+4+\cdots+2998)+2 \cdot 1500=$ $3 \cdot \frac{2998 \cdot 1500}{2}+3000=6748500$. Then $f(3000)=6748501$." 4098111ab29d,"The numerical value of $2^{-2} \times 2^{-1} \times 2^{0} \times 2^{1} \times 2^{2}$ is (A) 4 (B) 1 (C) 0 (D) $\frac{1}{4}$ (E) $\frac{1}{2}$",(B),easy,"Using exponent laws, $$ 2^{-2} \times 2^{-1} \times 2^{0} \times 2^{1} \times 2^{2}=2^{-2-1+0+1+2}=2^{0}=1 $$ Alternatively, $$ 2^{-2} \times 2^{-1} \times 2^{0} \times 2^{1} \times 2^{2}=\frac{1}{4} \times \frac{1}{2} \times 1 \times 2 \times 4=1 $$ ANSWER: (B)" affbaedd42df,"For each real number $m$, the function $f(x)=x^{2}-2 m x+8 m+4$ has a minimum value. What is the maximum of these minimum values?",20,medium,"A quadratic function of the form $f(x)=a x^{2}+b x+c$ with $a>0$ attains its minimum when $x=-\frac{b}{2 a}$. One way to see this is to complete the square to get $$ f(x)=a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a} $$ and note that since $a\left(x+\frac{b}{2 a}\right)^{2}$ is nonnegative, the minimum will occur when this quantity equals 0 . In a similar way, if $a<0$, the function attains its maximum when $x=-\frac{b}{2 a}$. For $f(x)=x^{2}-2 m x+8 m+4$, the minimum occurs when $x=-\frac{-2 m}{2}=m$. The minimum value of $f(x)$ is therefore $$ f(m)=m^{2}-2 m(m)+8 m+4=-m^{2}+8 m+4 $$ This is a quadratic function in the variable $m$ with a negative leading coefficient. Therefore, the maximum occurs when $m=-\frac{8}{2(-1)}=4$. The maximum of these minimum values is $-4^{2}+8(4)+4=20$. ANSWER: 20" f3ef1e31673e,"## Task A-2.4. Determine the largest natural number $n$ such that $n+10$ divides $n^{3}+100$.","n^{2}-10 n+100-\frac{900}{n+10}$ can also be achieved in some other way, for example, by polynomial ",medium,"## Solution. Notice that $n^{3}+1000$ is a sum of cubes, so it holds that $n^{3}+1000=(n+10)\left(n^{2}-10 n+100\right)$. Now we have $$ \begin{aligned} \frac{n^{3}+100}{n+10} & =\frac{n^{3}+1000-900}{n+10}=\frac{(n+10)\left(n^{2}-10 n+100\right)-900}{n+10} \\ & =n^{2}-10 n+100-\frac{900}{n+10} \end{aligned} $$ Therefore, if $n+10$ is a divisor of $n^{3}+100$, it must also be a divisor of 900, and vice versa. Since the greatest divisor of 900 is 900 itself, the largest natural number that satisfies the statement of the problem is $n=900-10=890$. Note: The equality $\frac{n^{3}+100}{n+10}=n^{2}-10 n+100-\frac{900}{n+10}$ can also be achieved in some other way, for example, by polynomial division. Any (correct) derivation is worth 3 points." 542b300de4df,A square number in the decimal system has the form: $\overline{a b 1 a b}$. Which is this square number?,See reasoning trace,medium,"I. Solution. Let the sought number be $2 x$, its square $4 x^{2}$, so according to our data $$ 4 x^{2}=10000 a+1000 b+100+a+b=1000(10 a+b)+100+10 a+b $$ thus $$ 4 x^{2}-100=1001(10 a+b) \ldots $$ Since $1001=7 \cdot 11 \cdot 13$ $$ 4\left(x^{2}-25\right)=7 \cdot 11 \cdot 13(10 a+b) $$ Therefore, $10 a+b$ must be a multiple of 4; let's replace it with $4 y$ and since $10 a+b<100$, then $y<25$. This results in $$ (x+5)(x-5)=7 \cdot 11 \cdot 13 y \ldots $$ This equation can be interpreted as the factors $7,11,13, y$ need to be grouped into two partial products such that the difference between these two products is 10, while also $y<25$. This cannot be achieved if one or three of the four factors are combined. We can only meet the requirement if we combine two and two factors, meaning we need to examine the possibilities of the following combinations: $$ 7 \cdot 11 \text { and } 13 y, 7 \cdot 13 \text { and } 11 y, \quad 11 \cdot 13 \text { and } 7 y $$ I. If one of the partial products is 77, then the other is 67 or 87. Neither of the latter is a multiple of 13 (13y). II. If one of the partial products is $7 \cdot 13=91$, then the other is 81 or 101. Neither of the latter is a multiple of 11 (11y). III. If one of the partial products is $11 \cdot 13=143$, then the other is 133 or 153. Of these, only 133 is a multiple of 7: $$ 7 y=133, \quad y=19<25 $$ Accordingly, there is only one solution, and it is: $$ \begin{gathered} x^{2}-25=7 \cdot 11 \cdot 13 \cdot 19=19019, \quad x^{2}=19044, \quad x=138 \\ 2 x=276 \quad \text { and } \quad 4 x^{2}=276^{2}=76176 \end{gathered} $$ $4 x^{2}$ is indeed a number of the form $a b 1 a b$! George Bizám (Bolyai g. VI. o. Bp. V.) II. Solution. In the square of an even number, the digit in the units place is $0,4,6$. If in the given case $b=0$, then $a$ must also be zero; in this case $00100=10^{2}$. The square of an even number is a multiple of 4; therefore, $10 a+4$ or $10+6$ must be a multiple of 4, and the following numbers are conceivable: $24124,44144,64164,84184$, $16116,36136,56156,76176,96196$ Among these, 44144, 36136, 96196 cannot be square numbers because 441, 361, 961 are square numbers. Among the rest, 76176 is a square number." 9830130373c5,"## Task 10/75 Given is a regular pyramid with a square base. Each side face forms an angle $\alpha$ with the base. Two spheres are inscribed in this pyramid such that one sphere touches all five faces of the pyramid, while the other touches the four side faces of the pyramid and the surface of the first sphere. In what ratio do the volumes of the two spheres stand to each other?",See reasoning trace,medium,"If the radius of the larger sphere is denoted by $\mathrm{R}$ and the radius of the smaller sphere by $r$, then the desired ratio is $$ V_{2}: V_{1}=\frac{4}{3} r^{3} \pi: \frac{4}{3} R^{3} \pi=r^{3}: R^{3} $$ ![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-261.jpg?height=668&width=511&top_left_y=237&top_left_x=287) In the figure (cross-sectional view), the following relationships are valid: $$ \frac{R}{h-R}=\frac{r}{h-2 R-r}=\cos \alpha $$ (it is $\triangle S M_{1} B \sim \triangle \quad S M_{2} C \sim \triangle \quad S A Z$ according to the main similarity theorem; each of these triangles contains the angle $Z S A$ and a right angle; this also results in the equality of the third angle $\alpha$. From these relationships, one obtains after a short calculation $$ R=h \frac{\cos \alpha}{1+\cos \alpha} \quad ; \quad r=h \frac{\cos \alpha(1-\cos \alpha)}{(1+\cos \alpha)^{2}} $$ Thus, the desired ratio is $$ \left(\frac{r}{R}\right)^{3}=\left(\frac{1-\cos \alpha}{1+\cos \alpha}\right)^{3}=\tan ^{6} \frac{\alpha}{2} $$" 89ae852f354a,"\section*{Task 1 - 081211} At the European Championships for women's rowing in August 1966, the GDR, as the most successful country, received 37 points, and the USSR received 36.5 points. Both countries received exactly one of the three medals awarded in each of the 5 disciplines: Single Sculls, Double Sculls, ""Coxed Four,"" Double Four, and Eight. How many gold medals, how many silver medals, and how many bronze medals did each of the two countries receive? The point system is based on the following table. \begin{tabular}{r|c|c|c} & Gold Medal & Silver Medal & Bronze Medal \\ \hline Single Sculls or Double Sculls & 6 & 5 & 4 \\ ""Coxed Four"" or Double Four & 9 & 7.5 & 6 \\ Eight & 12 & 10 & 8 \end{tabular} It is also known that the GDR performed better in the Double Sculls than in the Single Sculls and better in the Double Four than in the ""Coxed Four."" The USSR performed better in the Single Sculls than in the Double Sculls.",See reasoning trace,medium,"} The USSR has 36.5 points; it must have received a medal in at least one discipline with a non-integer point value. This is only the case for silver medals in the ""Coxed Fours"" and ""Double Fours."" Since there are only two such medals, the GDR can now have received a maximum of one of them. However, this would result in a non-integer total score, but the GDR has 36 points. In the disciplines ""Coxed Fours"" and ""Double Fours,"" the GDR can only receive a gold or bronze medal. Since it is known that the GDR performed better in the ""Double Fours"" than in the ""Coxed Fours,"" it follows that the GDR received the bronze medal in the ""Coxed Fours"" and the gold medal in the ""Double Fours."" Thus, the GDR received \(9 + 6 = 15\) points. Therefore, in the remaining three disciplines, the GDR received a total of \(36 - 15 = 21\) points. The maximum possible score here is 24 with three gold medals: \(12 + 6 + 6 = 24\). Since the GDR performed better in the ""Double Twos"" than in the ""Single,"" it could not have received a gold medal in the ""Single,"" but at most a silver medal, resulting in a maximum score of \(12 + 6 + 5 = 23\). Since the GDR only has 21 points in the three disciplines, it must have performed worse in (at least) one discipline than ""Eights"" gold, ""Double Twos"" gold, ""Single"" bronze. In the ""Eights,"" the point differences are two points, so exactly one point cannot be lost here. Therefore, it must be exactly one medal worse in ""Double Twos"" or ""Single."" If a silver medal had been achieved in the ""Double Twos,"" the GDR would have performed equally well in the ""Single,"" which is not the case. Therefore, a bronze medal must have been achieved in the ""Single."" Thus, for the GDR: Single: Bronze 4 Double Twos: Gold 6 Coxed Fours: Bronze 6 Double Fours: Gold 9 Eights: Gold 12 Total: 3 gold medals, 2 bronze medals, 37 points For the USSR, considering the impossibility of a gold medal in the disciplines where the GDR has already won, the maximum possible score is \(6 + 5 + 9 + 7.5 + 10 = 37.5\) points. Since it only has 36.5 points and the point difference of 1 only exists in the ""Single"" and ""Double Twos,"" it must have been exactly one medal worse in either the ""Single"" or the ""Double Twos."" If it had only the silver medal in the ""Single,"" it would have performed equally well in the ""Double Twos,"" which is not the case. Therefore, the USSR must have received the bronze medal in the ""Double Twos."" Thus, for the USSR: Single: Gold 6 Double Twos: Bronze 4 Coxed Fours: Gold 9 Double Fours: Silver 7.5 Eights: Silver 10 Total: 2 gold medals, 2 silver medals, 1 bronze medal, 36.5 points" 073d744ab51a,"2. Four functions $$ \begin{array}{l} y=\sin |x|, y=\cos |x|, \\ y=|\tan x|, y=-\ln |\sin x| \end{array} $$ with a period of $\pi$, which is monotonically decreasing on $\left(0, \frac{\pi}{2}\right)$ and is an even function is ( ). (A) $y=\sin |x|$ (B) $y=\cos |x|$ (C) $y=|\tan x|$ (D) $y=-\ln |\sin x|$",-\ln |\sin x|$ meets the requirements of the problem.,easy,"2. D. Since $y=\sin |x|$ is not a periodic function, option A is incorrect; Since $y=\cos |x|$ is a periodic function with the smallest positive period of $2 \pi$, option B is incorrect; Since $y=|\tan x|$ is monotonically increasing in $\left(0, \frac{\pi}{2}\right)$, option C is incorrect; The function $y=-\ln |\sin x|$ meets the requirements of the problem." cc6fbd5eeb17,"11.4. There is a $9 \times 2004$ grid, in which the positive integers from 1 to 2004 are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. Find the minimum possible value of the sum of the numbers in the first row.",See reasoning trace,medium,"11.4. $C_{2003}^{2}+1=2005004$. Consider a $9 \times n$ grid, where the integers from 1 to $n$ are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. We will prove by mathematical induction that the sum of the numbers in the first row is not less than $C_{n-1}^{2}+1$. When $n \leqslant 4$, the conclusion is obvious. Because each cell contains a number not less than 1, and when $n \leqslant 4$, we have $n \geqslant C_{n-1}^{2}+1$. Next, we perform the inductive step. If necessary, we can rearrange the columns so that the numbers in the first row are in non-decreasing order. Therefore, we can assume that the numbers in the first row are already in non-decreasing order. Let $S_{i}$ denote the number of numbers in the first row that are not less than $i$, and let $D_{i}=n-S_{i}$. Thus, $S_{1}=n, D_{1}=0$, and the sum of the numbers in the first row is $$ S=S_{1}+S_{2}+\cdots+S_{n}. $$ Rewriting the above, we get $$ \begin{array}{l} S=\left(n-D_{1}\right)+\left(n-D_{2}\right)+\cdots+\left(n-D_{n}\right) \\ \geqslant n(n-3)-\left(D_{1}+D_{2}+\cdots+D_{n-3}\right). \end{array} $$ If for any $i \leqslant n-3$, we have $D_{1} \leqslant i+1$, then $$ \begin{array}{l} S \geqslant n(n-3)-[0+3+4+\cdots+(n-2)] \\ =\frac{n^{2}-3 n+4}{2}. \end{array} $$ This is what we need to prove. Assume there exists some $k \leqslant n-3$ such that $D_{k} \geqslant k+2$. In this case, $k \geqslant 2$, so $k+2 \geqslant 4$. Since the first row contains at least $k+2$ numbers less than $k$, all such numbers are in the first $k+2$ columns. This means that all numbers in the first $k+2$ columns do not exceed $k+2$. Therefore, all numbers in the remaining columns are not less than $k+3$. Now, we divide the entire grid into two parts: the first part consists of the first $k+2$ columns, and the rest form the second part. Since $n-1 \geqslant k+2 \geqslant 4$, the sum of the numbers in the first row of the first part is not less than $\mathrm{C}_{k+1}^{2}+1$. If we subtract $k+2$ from each number in the second part, we get a grid with $n-(k+2) \geqslant 1$ columns that satisfies the problem's conditions. By the inductive hypothesis, the sum of the numbers in the first row of this part is not less than $(k+2)(n-k-2)+\dot{C}_{n-k-3}^{2}+1$. Adding these two estimates, we get $$ \begin{array}{l} S \geqslant C_{k+1}^{2}+1+(k+2)(n-k-2)+\mathrm{C}_{n-k-3}^{2}+1 \\ =\mathrm{C}_{n-1}^{2}+3. \end{array} $$ We also provide an example that achieves the minimum possible value (as shown in Table 1): Table 1 \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1 & 1 & 1 & 2 & 3 & 4 & $\cdots$ & $k$ & $\cdots$ & 1998 & 1999 & 2000 & 2001 & 2001 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2002 & 2002 & 2003 & 2004 \\ \hline 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ & $k+3$ & $\cdots$ & 2001 & 2002 & 2003 & 2003 & 2004 \\ \hline 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ & $k+3$ & $\cdots$ & 2001 & 2002 & 2003 & 2004 & 2004 \\ \hline \end{tabular}" 9909a4db254e,"In a hidden friend, suppose no one takes oneself. We say that the hidden friend has ""marmalade"" if there are two people $A$ and $ B$ such that A took $B$ and $ B $ took $A$. For each positive integer n, let $f (n)$ be the number of hidden friends with n people where there is no “marmalade”, i.e. $f (n)$ is equal to the number of permutations $\sigma$ of {$1, 2,. . . , n$} such that: *$\sigma (i) \neq i$ for all $i=1,2,...,n$ * there are no $ 1 \leq i 0 \implies y = \frac{1 \pm \sqrt{1 + 4k}}{2} \] Both values of \( y \) are positive, so each \( y \) gives two real solutions for \( x \) (i.e., \( x = \pm \sqrt{y} \)). Thus, there are four real solutions for \( x \). Hence, \( N(k) = 4 \). - **Case 4:** \( k = 0 \) \[ 1 + 4k = 1 \implies y = \frac{1 \pm 1}{2} = 1 \text{ or } 0 \] For \( y = 1 \), \( x = \pm 1 \) (two solutions). For \( y = 0 \), \( x = 0 \) (one solution). Thus, there are three real solutions for \( x \). Hence, \( N(k) = 3 \). - **Case 5:** \( k > 0 \) \[ 1 + 4k > 0 \implies y = \frac{1 \pm \sqrt{1 + 4k}}{2} \] Both values of \( y \) are positive, so each \( y \) gives two real solutions for \( x \) (i.e., \( x = \pm \sqrt{y} \)). Thus, there are four real solutions for \( x \). Hence, \( N(k) = 4 \). 4. Summarizing the values of \( N(k) \) as \( k \) ranges from \( -\infty \) to \( \infty \): \[ \boxed{(0, 2, 4, 2, 3)} \]" 4bbf592a69f4,"14. Given that $A B C-A_{1} B_{1} C_{1}$ is a regular triangular prism, $A B=B C$ $=C A=2, A A_{1}=\sqrt{2}, D$ and $E$ are the midpoints of $A C$ and $B C$ respectively. Then the angle formed by $A_{1} D$ and $C_{1} E$ is $\qquad$ .",See reasoning trace,easy,"$14.60^{\circ}$. Take the midpoint $F$ of $A_{1} B_{1}$, then $D E \Perp \frac{1}{2} A B \Perp A_{1} F$. Therefore, $E F \mathbb{\mathbb { I }} D A_{1}$. $\because E F=A_{1} D=C_{1} E=\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}$, and $C_{1} F=\frac{\sqrt{3}}{2} A_{1} B_{1}=\sqrt{3}$, $\therefore \triangle C_{1} E F_{1}$ is an equilateral triangle. Also, $\angle F E C_{1}$ is the angle formed by $A_{1} D$ and $C_{1} E$, thus we get." 9723faba8f13,"Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following ineqaulity $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ When does eqaulity holds?",\frac{3,hard,"### Part (a) We need to determine the value of the expression: \[ \frac{1+ab}{a-b} \cdot \frac{1+bc}{b-c} + \frac{1+bc}{b-c} \cdot \frac{1+ca}{c-a} + \frac{1+ca}{c-a} \cdot \frac{1+ab}{a-b} \] 1. **Generalization**: We generalize the problem by considering: \[ \frac{1+\lambda ab}{a-b} \cdot \frac{1+\lambda bc}{b-c} + \frac{1+\lambda bc}{b-c} \cdot \frac{1+\lambda ca}{c-a} + \frac{1+\lambda ca}{c-a} \cdot \frac{1+\lambda ab}{a-b} \] and prove that it equals \(\lambda\) for any \(\lambda \in \mathbb{R}\). 2. **Expression Simplification**: Let: \[ A = \frac{1+\lambda ab}{a-b}, \quad B = \frac{1+\lambda bc}{b-c}, \quad C = \frac{1+\lambda ca}{c-a} \] We need to show: \[ AB + BC + CA = \lambda \] 3. **Substitution and Simplification**: Substitute \(A\), \(B\), and \(C\) into the expression: \[ AB + BC + CA = \left(\frac{1+\lambda ab}{a-b}\right) \left(\frac{1+\lambda bc}{b-c}\right) + \left(\frac{1+\lambda bc}{b-c}\right) \left(\frac{1+\lambda ca}{c-a}\right) + \left(\frac{1+\lambda ca}{c-a}\right) \left(\frac{1+\lambda ab}{a-b}\right) \] 4. **Common Denominator**: Combine the terms over a common denominator: \[ AB + BC + CA = \frac{(1+\lambda ab)(1+\lambda bc)}{(a-b)(b-c)} + \frac{(1+\lambda bc)(1+\lambda ca)}{(b-c)(c-a)} + \frac{(1+\lambda ca)(1+\lambda ab)}{(c-a)(a-b)} \] 5. **Expand and Simplify**: Expand each term in the numerator: \[ (1+\lambda ab)(1+\lambda bc) = 1 + \lambda ab + \lambda bc + \lambda^2 ab \cdot bc \] Similarly for the other terms. After expansion, the terms involving \(\lambda^2\) will cancel out due to symmetry, and we are left with: \[ \frac{1 + \lambda(ab + bc + ca) + \lambda^2 abc}{(a-b)(b-c)(c-a)} \] 6. **Final Simplification**: The terms involving \(\lambda^2\) cancel out, and we are left with: \[ \frac{\lambda(ab + bc + ca)}{(a-b)(b-c)(c-a)} \] Since \(ab + bc + ca\) is symmetric and the denominator is the product of differences, the expression simplifies to \(\lambda\). Thus, we have: \[ \boxed{\lambda} \] ### Part (b) We need to determine the value of the expression: \[ \frac{1-ab}{a-b} \cdot \frac{1-bc}{b-c} + \frac{1-bc}{b-c} \cdot \frac{1-ca}{c-a} + \frac{1-ca}{c-a} \cdot \frac{1-ab}{a-b} \] 1. **Generalization**: We generalize the problem by considering: \[ \frac{1-\lambda ab}{a-b} \cdot \frac{1-\lambda bc}{b-c} + \frac{1-\lambda bc}{b-c} \cdot \frac{1-\lambda ca}{c-a} + \frac{1-\lambda ca}{c-a} \cdot \frac{1-\lambda ab}{a-b} \] and prove that it equals \(-\lambda\) for any \(\lambda \in \mathbb{R}\). 2. **Expression Simplification**: Let: \[ A = \frac{1-\lambda ab}{a-b}, \quad B = \frac{1-\lambda bc}{b-c}, \quad C = \frac{1-\lambda ca}{c-a} \] We need to show: \[ AB + BC + CA = -\lambda \] 3. **Substitution and Simplification**: Substitute \(A\), \(B\), and \(C\) into the expression: \[ AB + BC + CA = \left(\frac{1-\lambda ab}{a-b}\right) \left(\frac{1-\lambda bc}{b-c}\right) + \left(\frac{1-\lambda bc}{b-c}\right) \left(\frac{1-\lambda ca}{c-a}\right) + \left(\frac{1-\lambda ca}{c-a}\right) \left(\frac{1-\lambda ab}{a-b}\right) \] 4. **Common Denominator**: Combine the terms over a common denominator: \[ AB + BC + CA = \frac{(1-\lambda ab)(1-\lambda bc)}{(a-b)(b-c)} + \frac{(1-\lambda bc)(1-\lambda ca)}{(b-c)(c-a)} + \frac{(1-\lambda ca)(1-\lambda ab)}{(c-a)(a-b)} \] 5. **Expand and Simplify**: Expand each term in the numerator: \[ (1-\lambda ab)(1-\lambda bc) = 1 - \lambda ab - \lambda bc + \lambda^2 ab \cdot bc \] Similarly for the other terms. After expansion, the terms involving \(\lambda^2\) will cancel out due to symmetry, and we are left with: \[ \frac{1 - \lambda(ab + bc + ca) + \lambda^2 abc}{(a-b)(b-c)(c-a)} \] 6. **Final Simplification**: The terms involving \(\lambda^2\) cancel out, and we are left with: \[ \frac{-\lambda(ab + bc + ca)}{(a-b)(b-c)(c-a)} \] Since \(ab + bc + ca\) is symmetric and the denominator is the product of differences, the expression simplifies to \(-\lambda\). Thus, we have: \[ \boxed{-\lambda} \] ### Part (c) We need to prove the inequality: \[ \frac{1+a^2b^2}{(a-b)^2} + \frac{1+b^2c^2}{(b-c)^2} + \frac{1+c^2a^2}{(c-a)^2} \geq \frac{3}{2} \] 1. **Expression Analysis**: Consider the terms: \[ \frac{1+a^2b^2}{(a-b)^2}, \quad \frac{1+b^2c^2}{(b-c)^2}, \quad \frac{1+c^2a^2}{(c-a)^2} \] 2. **AM-GM Inequality**: By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have: \[ \frac{1+a^2b^2}{(a-b)^2} \geq \frac{2\sqrt{1 \cdot a^2b^2}}{(a-b)^2} = \frac{2ab}{(a-b)^2} \] Similarly for the other terms. 3. **Summation**: Summing these inequalities, we get: \[ \frac{1+a^2b^2}{(a-b)^2} + \frac{1+b^2c^2}{(b-c)^2} + \frac{1+c^2a^2}{(c-a)^2} \geq \frac{2ab}{(a-b)^2} + \frac{2bc}{(b-c)^2} + \frac{2ca}{(c-a)^2} \] 4. **Symmetry and Simplification**: By symmetry and the properties of the AM" af0940fd254e,"An ant starts from a vertex of a cube, walking only along the edges, until it returns to the initial vertex, without passing through any vertex twice. What is the longest walk that this ant can make?",See reasoning trace,medium,"In the figure, we have a path consisting of eight edges that the ant can take starting from the vertex identified as 1. ![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-084.jpg?height=314&width=309&top_left_y=1970&top_left_x=905) Would it be possible for her to make a path passing through nine edges? To make this path, she would have to pass through nine vertices, as the arrival vertex is the same as the starting vertex, since the little ant returns to the initial vertex. ![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-084.jpg?height=226&width=829&top_left_y=2491&top_left_x=702) Since the cube only has eight vertices, this walk is not possible. Therefore, the longest walk covers eight edges." 14ebacb02a52,Task B-2.6. Write some quadratic equation with real coefficients whose one solution is $\left(\frac{\sqrt{2}}{1-i}\right)^{2011}$.,See reasoning trace,medium,"## Solution. $$ \begin{aligned} & x_{1}=\left(\frac{\sqrt{2}}{1-i}\right)^{2011}=\left(\frac{\sqrt{2}}{1-i}\right)^{2010} \cdot \frac{\sqrt{2}}{1-i}=\left(\left(\frac{\sqrt{2}}{1-i}\right)^{2}\right)^{1005} \cdot \frac{\sqrt{2}}{1-i} \cdot \frac{1+i}{1+i}= \\ & =\left(\frac{2}{1-2 i-1}\right)^{1005} \cdot \frac{\sqrt{2}(1+i)}{2}=\left(\frac{1}{-i} \cdot \frac{i}{i}\right)^{1005} \cdot \frac{\sqrt{2}(1+i)}{2}=i^{1005} \cdot \frac{\sqrt{2}(1+i)}{2}= \\ & =i \cdot \frac{\sqrt{2}(1+i)}{2}=\frac{-\sqrt{2}+i \sqrt{2}}{2} \\ & x_{2}=\frac{-\sqrt{2}-i \sqrt{2}}{2} \\ & x_{1}+x_{2}=-\sqrt{2} \\ & x_{1} \cdot x_{2}=1 \\ & x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} \cdot x_{2}=0 \\ & x^{2}+\sqrt{2} \cdot x+1=0 \end{aligned} $$" 95417bc1c738,"(3) Does there exist a real number $a$ such that the function $f(x)=\log _{a}\left(a x^{2}-x\right)$ is increasing on the interval $[2,4]$? If it exists, find the allowed values of $a$; if not, explain the reason.",See reasoning trace,easy,"(3) $a \in(1,+\infty)$" 1f6580834f2b,"8. The domain of the function $f(x)$ is $x \in \mathbf{R}$, and $x \neq 1$, it is known that $f(x+1)$ is an odd function. When $x>1$, the decreasing interval of $f(x)$ is ( ). A. $\left[\frac{5}{4},+\infty\right)$ B. $\left(1, \frac{5}{4}\right)$ C. $\left[\frac{7}{4},+\infty\right)$ D. $\left(1, \frac{7}{4}\right]$","t$, then $-x+1=2-t, \therefore f(t)=-f(2-t)$, i.e., $f(x)=-f(2-x)$. When $x>1$, $2-x<1$, thus $f(x)=",easy,"8. C. Let $x+1=t$, then $-x+1=2-t, \therefore f(t)=-f(2-t)$, i.e., $f(x)=-f(2-x)$. When $x>1$, $2-x<1$, thus $f(x)=-f(2-x)=-\left[2\left(x-\frac{7}{4}\right)^{2}+\frac{7}{8}\right]$." 258fd12f696f,"Let's find the highest degree expression of $x$ such that $$ n x^{n+1}-(n+1) x^{n}+1 $$ and $x^{n}-n x+n-1$ are divisible.","1$, then (2) is identically 0, or $0 \cdot(x-1)^{2}$, and (1) is precisely $(x-1)^{2}$.",medium,"Add up the two polynomials, the sum is $$ n(x-1)\left(x^{n}-1\right) $$ Subtract the $n(x-1)$ multiple of (2) from (3), then we get $n^{2}(x-1)^{2}$. Therefore, the common factor can only be $(x-1)^{2}$. However, this is indeed a common factor, because (1) can be written as $$ \begin{aligned} & n x^{n+1}-(n+1) x^{n}+1=(x-1)\left[n x^{n}-\left(1+x+x^{2}+\ldots+x^{n-1}\right)\right]= \\ & \quad=(x-1)\left[\left(x^{n}-1\right)+\left(x^{n}-x\right)+\left(x^{n}-x^{2}\right)+\ldots+\left(x^{n}-x^{n-1}\right)\right] \end{aligned} $$ (2) is $$ \begin{gathered} x^{n}-n x+n-1=(x-1)\left[1+x+x^{2}+\ldots+x^{n-1}-n\right]= \\ \quad=(x-1)\left[(x-1)+\left(x^{2}-1\right)+\ldots+\left(x^{n-1}-1\right)\right] \end{gathered} $$ However, it is clear that $(x-1)$ can be factored out from both expressions in the square brackets if $n \geq 2$. If, however, $n=1$, then (2) is identically 0, or $0 \cdot(x-1)^{2}$, and (1) is precisely $(x-1)^{2}$." 1ae7abe44896,"Florián was thinking about what bouquet he would have tied for his mom for Mother's Day. In the florist's, according to the price list, he calculated that whether he buys 5 classic gerberas or 7 mini gerberas, the bouquet, when supplemented with a decorative ribbon, would cost the same, which is 295 crowns. However, if he bought only 2 mini gerberas and 1 classic gerbera without any additional items, he would pay 102 crowns. How much does one ribbon cost? (L. Šimůnek)","102 ; 2$ points for deriving the price of the ribbon $(d = 6, 295 - 5 \cdot 7 \cdot d = 85)$.",medium,"From the assignment, it follows that 5 classic gerberas cost the same as 7 minigerberas, meaning that the price of a classic gerbera and the price of a minigerbera are in the ratio of $7: 5$. If we represent the price of a classic gerbera with 7 equal parts, the price of a minigerbera will correspond to 5 such parts. The cost of purchasing 2 minigerberas and 1 classic gerbera represents $2 \cdot 5 + 7 = 17$ of these parts, so 1 part therefore corresponds to the amount 102 : 17 = 6 crowns. Five classic gerberas, or seven minigerberas, cost $5 \cdot 7 \cdot 6 = 210$ crowns. The price of the ribbon is $295 - 210 = 85$ crowns. Evaluation. 2 points for determining that the prices of a classic gerbera and a minigerbera are in the ratio of 7 : $5 ; 2$ points for determining the corresponding equation $17 d = 102 ; 2$ points for deriving the price of the ribbon $(d = 6, 295 - 5 \cdot 7 \cdot d = 85)$." 917adda4090d,"5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies $$ \begin{array}{l} f(x+3) \leqslant f(x)+3, \\ f(x+2) \geqslant f(x)+2, \end{array} $$ and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then $$ f(2015)= $$ $\qquad$",2016$.,easy,"5.2016. Notice, $$ \begin{array}{l} f(x)+3 \geqslant f(x+3) \\ =f(x+1+2) \geqslant f(x+1)+2 \\ \Rightarrow f(x)+1 \geqslant f(x+1) . \\ \text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\ =f(x+1+3) \leqslant f(x+1)+3 \\ \Rightarrow f(x+1) \geqslant f(x)+1 . \end{array} $$ Therefore, $f(x+1)=f(x)+1$. Thus, $f(2015)=2016$." d1ef30fc7184,"3. Given that $\triangle A B C$ is an equilateral triangle, the ellipse $\Gamma$ has one focus at $A$, and the other focus $F$ lies on the line segment $B C$. If the ellipse $\Gamma$ passes exactly through points $B$ and $C$, then its eccentricity is $\qquad$.","|C A|+|C F|=2 a \Rightarrow F$ is the midpoint of $B C$. Without loss of generality, assume the side",easy,"According to the problem, $|B A|+|B F|=|C A|+|C F|=2 a \Rightarrow F$ is the midpoint of $B C$. Without loss of generality, assume the side length of $\triangle A B C$ is 2, then $2 a=3, 2 c=\sqrt{3} \Rightarrow e=\frac{c}{a}=\frac{\sqrt{3}}{3}$." 4454154c83c2,"The incircle of $ABC$ touches the sides $BC,CA,AB$ at $A' ,B' ,C'$ respectively. The line $A' C'$ meets the angle bisector of $\angle A$ at $D$. Find $\angle ADC$.",90^\circ,medium,"1. **Identify the given elements and their properties:** - The incircle of $\triangle ABC$ touches the sides $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$ respectively. - The line $A'C'$ meets the angle bisector of $\angle A$ at $D$. - We need to find $\angle ADC$. 2. **Use the property of the incircle and cyclic quadrilaterals:** - Since $A'$, $B'$, and $C'$ are points where the incircle touches the sides, $A'$, $B'$, and $C'$ are known as the points of tangency. - The quadrilateral $BC'IA'$ is cyclic because the opposite angles sum up to $180^\circ$. 3. **Calculate the angles using the properties of cyclic quadrilaterals:** - Since $BC'IA'$ is cyclic, we have: \[ \angle C'A'I = \frac{\angle B}{2} \] - Notice that $\angle DA'C = 90^\circ + \frac{\angle B}{2}$ because $A'$ is the point of tangency and $D$ lies on the angle bisector of $\angle A$. 4. **Use the properties of the incenter and angle bisectors:** - The incenter $I$ is the intersection of the angle bisectors of $\triangle ABC$. - The angle $\angle BID$ is given by: \[ \angle BID = \frac{1}{2} (\angle A + \angle B) \] - The angle $\angle BIC$ is: \[ \angle BIC = 90^\circ + \frac{\angle A}{2} \] 5. **Calculate $\angle DIC$:** - Using the angles calculated above: \[ \angle DIC = \angle BIC - \angle BID = \left(90^\circ + \frac{\angle A}{2}\right) - \left(\frac{1}{2} (\angle A + \angle B)\right) \] Simplifying this: \[ \angle DIC = 90^\circ + \frac{\angle A}{2} - \frac{\angle A}{2} - \frac{\angle B}{2} = 90^\circ - \frac{\angle B}{2} \] 6. **Conclude that $A'DIC$ is cyclic:** - Since $\angle DIC = 90^\circ - \frac{\angle B}{2}$, we can conclude that $A'DIC$ is cyclic. 7. **Determine $\angle ADC$:** - In a cyclic quadrilateral, the opposite angles sum up to $180^\circ$. - Therefore, $\angle ADC = 180^\circ - \angle A'DI$. - Since $\angle A'DI = 90^\circ$, we have: \[ \angle ADC = 180^\circ - 90^\circ = 90^\circ \] The final answer is $\boxed{90^\circ}$" e42c5b263cef,"The cashier at the gallery sells tickets to visitors with a number according to the order in which they came that day. The first visitor gets a ticket with the number 1, the second with the number 2, etc. However, during the day, the yellow paper on which the tickets were printed ran out, so the cashier had to continue printing on red paper. Throughout the day, she sold as many yellow tickets as red ones. She found that the sum of the numbers on the yellow tickets was 1681 less than the sum of the numbers on the red tickets. How many tickets did she sell that day? (M. Mach) Idea. Notice by how much the numbers on the sold yellow and red tickets differ.",See reasoning trace,medium,"Let the number of yellow tickets be $n$. The first yellow ticket had the number 1, the second had 2, and so on, with the last yellow ticket having the number $n$. The first red ticket had the number $n+1$, the second red ticket had $n+2$, and so on, with the last red ticket having the number $2n$. Notice that the first red ticket has a number that is $n$ greater than the first yellow ticket. Similarly, the second red ticket has a number that is $n$ greater than the second yellow ticket; this applies to all such pairs of tickets, of which there are $n$ in total. Therefore, the sum of the numbers on the red tickets is $n^2$ greater than the sum of the numbers on the yellow tickets. From the problem, we know that $n^2 = 1681$, so $n = 41$. The cashier sold 41 yellow and 41 red tickets that day, for a total of 82 tickets." 459dbbf029b4,"$4 \cdot 70$ The sequence of functions $\left\{f_{n}(x)\right\}$ is recursively defined by: $$ \left\{\begin{array}{l} f_{1}(x)=\sqrt{x^{2}+48} \\ f_{n+1}(x)=\sqrt{x^{2}+6 f_{n}(x)} \end{array} \text { for each } n \geqslant 1 .\right. $$ Find all real solutions to the equation $f_{n}(x)=2 x$.","2$ has a unique solution $x=4$ on $(0,+\infty)$.",medium,"［Solution］From the given information, we have $$ f_{n}(x)>0 \quad(n=1,2, \cdots), $$ Therefore, the equation $f_{n}(x)=2 x$ has only positive solutions. First, we use mathematical induction to prove that $x=4$ is a solution to the equation $$ f_{n}(x)=2 x $$ In fact, when $n=1$, substituting $x=4$ into the left side of (1) yields $f_{1}(4)=$ $\sqrt{4^{2}+48}=8$; substituting into the right side of (1) yields $2 \times 4=8$. Therefore, $x=4$ is a solution to $f_{1}(x)=2 x$. Assume that $x=4$ is a solution to $f_{k}(x)=2 x$. That is, $f_{k}(4)=8$. Then $$ f_{k+1}(4)=\sqrt{4^{2}+6 f_{k}(4)}=\sqrt{4^{2}+48}=8=2 \times 4 . $$ Therefore, $x=4$ is also a solution to $f_{k+1}(x)=2 x$. By the principle of mathematical induction, for any natural number $n$, $x=4$ is a solution to equation (1). Next, we use mathematical induction to prove that the function $\frac{f_{n}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$. In fact, when $n=1$, $$ \frac{f_{1}(x)}{x}=\sqrt{1+\frac{48}{x^{2}}} $$ It is clearly a monotonically decreasing function on $(0,+\infty)$. Assume that $\frac{f_{k}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$. Then $$ \frac{f_{k+1}(x)}{x}=\sqrt{1+\frac{6}{x} \cdot \frac{f_{k}(x)}{x}} $$ is also a monotonically decreasing function on $(0,+\infty)$. By the principle of mathematical induction, for any natural number $n$, the function $\frac{f_{n}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$. Therefore, $\frac{f_{n}(x)}{x}=2$ has a unique solution $x=4$ on $(0,+\infty)$." c52f1b078310,"Task 2. (10 points) In triangle $ABC$ with sides $AB=9$, $AC=3$, $BC=8$, the bisector $AK$ is drawn, and a point $M$ is marked on side $AC$ such that $AM: CM=3: 1$. Point $N$ is the intersection of $AK$ and $BM$. Find $KN$. #",. $\frac{\sqrt{15}}{5}$,medium,"# Solution. Draw a line through point $A$ parallel to line $B C$ and intersecting line $B M$ at point $L$ (Fig. 1). ![](https://cdn.mathpix.com/cropped/2024_05_06_b2236406d6db6e8875c1g-17.jpg?height=366&width=1287&top_left_y=1462&top_left_x=453) Fig. 1. By the property of the angle bisector $\frac{B K}{C K}=\frac{A B}{A C}=\frac{9}{3}=\frac{3}{1} \Rightarrow \frac{B K}{8-B K}=\frac{3}{1} \Rightarrow B K=6$ Consider similar triangles $A M L$ and $C M B: \frac{A L}{B C}=\frac{A M}{M C}$. Therefore, $\frac{A L}{B C}=3, A L=3 B C=24$. Consider similar triangles $L A N$ and $B K N: \frac{A N}{N K}=\frac{A L}{B K}=\frac{24}{6}=\frac{4}{1}$. The length of the angle bisector is $A K=\sqrt{A B \cdot A C-B K \cdot C K}=\sqrt{9 \cdot 3-6 \cdot 2}=\sqrt{15}$. $K N=\frac{1}{5} A K=\frac{\sqrt{15}}{5}$ Answer. $\frac{\sqrt{15}}{5}$." 1b86034b4c4f,"6. Given three points $A, B, C$ on a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$, then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is equal to",See reasoning trace,medium,"6. -25 Hint: Since $|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}=|\overrightarrow{C A}|^{2}$, then $\angle B=90^{\circ}$. Thus, $\cos \angle A B C=0$, $$ \begin{array}{l} \cos \angle B A C=\frac{3}{5}, \cos \angle B C A=\frac{4}{5}, \text { so } \overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=3 \times 4 \times 0+4 \times 5 \times\left(-\frac{4}{5}\right)+3 \times \\ 5 \times\left(-\frac{3}{5}\right)=-25 \end{array} $$" bc9dc70273bf,"8. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B-b \cos A=\frac{4}{5} c$, then $\frac{\tan A}{\tan B}=$ $\qquad$ .",See reasoning trace,easy,"$$ \begin{array}{l} \sin A \cos B-\sin B \cos A=\frac{4}{5} \sin C=\frac{4}{5}(\sin A \cos B+\sin B \cos A) \\ \Rightarrow \frac{1}{5} \sin A \cos B=\frac{9}{5} \sin B \cos A \Rightarrow \frac{\sin A \cos B}{\sin B \cos A}=9 \Rightarrow \frac{\tan A}{\tan B}=9 \end{array} $$" e7c196439f4f,"9. In the arithmetic sequences $3,10,17, \cdots, 2005$ and $3,8, 13, \cdots, 2003$, the number of terms that have the same value is $\qquad$.",58$.,easy,"9.58. Subtract 3 from each term of the two sequences, transforming them into $0,7,14, \cdots$, 2002 and $0,5,10, \cdots, 2000$. The first sequence represents multiples of 7 not exceeding 2002, and the second sequence can be seen as multiples of 5 within the same range. Therefore, the common terms are multiples of 35. Hence $\left[\frac{2002}{35}\right]+1=58$." 658b10c234f2,"Let $A$, $B$, $C$, $D$ be four points on a line in this order. Suppose that $AC = 25$, $BD = 40$, and $AD = 57$. Compute $AB \cdot CD + AD \cdot BC$. [i]Proposed by Evan Chen[/i]",1000,medium,"1. Given the points \(A\), \(B\), \(C\), and \(D\) on a line in this order, we know the following distances: \[ AC = 25, \quad BD = 40, \quad AD = 57 \] 2. First, we calculate the distance \(CD\): \[ CD = AD - AC = 57 - 25 = 32 \] 3. Next, we calculate the distance \(BC\): \[ BC = BD - CD = 40 - 32 = 8 \] 4. Then, we calculate the distance \(AB\): \[ AB = AC - BC = 25 - 8 = 17 \] 5. Now, we need to compute \(AB \cdot CD + AD \cdot BC\): \[ AB \cdot CD + AD \cdot BC = 17 \cdot 32 + 57 \cdot 8 \] 6. We can simplify the expression step-by-step: \[ 17 \cdot 32 = 17 \cdot (30 + 2) = 17 \cdot 30 + 17 \cdot 2 = 510 + 34 = 544 \] \[ 57 \cdot 8 = 57 \cdot (10 - 2) = 57 \cdot 10 - 57 \cdot 2 = 570 - 114 = 456 \] 7. Adding these results together: \[ 544 + 456 = 1000 \] Thus, the value of \(AB \cdot CD + AD \cdot BC\) is \(\boxed{1000}\)." d36becc63495,"# A polynomial $P(x)$ with integer coefficients has the properties $$ P(1)=2019, \quad P(2019)=1, \quad P(k)=k, $$ where the number $k$ is an integer. Find this number $k$. #",$\quad k=1010$,easy,"# Solution. Since the polynomial $P(x)$ has integer coefficients, $P(a)-P(b)$ is divisible by $a-b$ for any integers $a$ and $b$. We get that $$ \begin{gathered} P(k)-P(1)=(k-2019) \text { is divisible by }(k-1), \\ P(k)-P(2019)=(k-1) \text { is divisible by }(k-2019) . \end{gathered} $$ This can only be true if $|k-1|=|k-2019|$. The solution to the obtained equation is $k=1010$. Answer: $\quad k=1010$." e147ef696f8e,"22. Consider a list of six numbers. When the largest number is removed from the list, the average is decreased by 1 . When the smallest number is removed, the average is increased by 1 . When both the largest and the smallest numbers are removed, the average of the remaining four numbers is 20 . Find the product of the largest and the smallest numbers.",375,easy,"22. Answer: 375 . Let $m$ and $M$ be the smallest and the largest numbers. Then $$ \frac{m+80}{5}+1=\frac{M+80}{5}-1=\frac{m+M+80}{6} $$ Solving the system, $m=15$ and $M=25$. Then $m M=375$." 6f8d6bf96214,,See reasoning trace,medium,"Solution. Let $a, b, c, d, e, f$ be the natural numbers written on the faces of the cube, such that $a$ and $b$ are numbers written on opposite faces, the numbers $c$ and $d$ are written on opposite faces, and the numbers $e$ and $f$ are written on opposite faces. At the vertices of the cube, the following numbers are written: $ace, acf, ade, adf, bce, bcf, bde, bdf$. From the condition of the problem, we get $$ a c e + a c f + a d e + a d f + b c e + b c f + b d e + b d f = (a + b)(c + d)(e + f) = 385 $$ The only way to write the number 385 as a product of three factors greater than 1 is $385 = 5 \cdot 7 \cdot 11$, from which it follows that $$ a + b + c + d + e + f = (a + b) + (c + d) + (e + f) = 5 + 7 + 11 = 23 $$ ## VIII Department" 5a760b99ed90,"1. Use $1,2,3,4,5$ to form a five-digit number, such that the difference between any two adjacent digits is at least 2. Then the number of such five-digit numbers is $\qquad$ .",See reasoning trace,medium,"2.1.14. Consider $\square$ classified by the middle number $a$: (1) If $a=1$, then 2 and 3 are on opposite sides of $a$, and 4 and 5 are on opposite sides of $a$, resulting in 4 permutations: $$ 24135,24153,35142,53142 \text {; } $$ (2) If $a=2$, then 4 and 5 are on opposite sides of $a$, and 1 and 3 are not adjacent to $a$, resulting in 2 permutations: 14253,35241 ; (3) If $a=3$, then 1 and 2 are on opposite sides of $a$, and 4 and 5 are on opposite sides of $a$, with 2 and 4 not adjacent to $a$, resulting in 2 permutations: 25314,41352 ; (4) By symmetry, $a=5$ is the same as (1), with 4 permutations; $a=4$ is the same as (2), with 2 permutations. Therefore, there are a total of 14 such five-digit numbers." 6654710b57f0,"6. Given that $P$ is a point on a sphere $O$ with radius $r$, and three mutually perpendicular moving chords $P A$, $P B$, and $P C$ are drawn through $P$. If the maximum distance from point $P$ to the plane $A B C$ is 1, then $r=$ $\qquad$",\frac{3}{2}$.,medium,"6. $\frac{3}{2}$. Let $P A=x, P B=y, P C=z$. Then $$ \begin{array}{l} A B^{2}=x^{2}+y^{2}, B C^{2}=y^{2}+z^{2}, C A^{2}=z^{2}+x^{2}, \\ S_{\triangle A B C}=\frac{1}{2} A B \cdot B C \sin \angle A B C \\ =\frac{1}{2} A B \cdot B C \sqrt{1-\cos ^{2} \angle A B C} \\ =\frac{1}{2} \sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}, \\ V_{\text {tetrahedron } P-A B C}=\frac{1}{6} x y z=\frac{1}{3} h S_{\triangle A B C} . \end{array} $$ Therefore, $h=\frac{x y z}{\sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}}=\frac{1}{\sqrt{\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}}}$ $$ \leqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{9}}=\sqrt{\frac{4 r^{2}}{9}} . $$ Thus, $r=\frac{3}{2}$." 827db0a12d37,"[ Touching circles ] [Common tangent to two circles] Two touching circles are inscribed in an angle of size $2 \alpha$. Find the ratio of the radius of the smaller circle to the radius of a third circle that touches the first two circles and one of the sides of the angle.",$\frac{2(1 + \cos \alpha)}{1 + \sin \alpha}$,medium,"The length of the segment of the common external tangent to two touching circles of radii $r$ and $R$, enclosed between the points of tangency, is $2 \sqrt{r R}$. ## Solution Let $r$ and $R$ be the radii of the given circles ($r < R$), $O$ and $Q$ their centers, $A$ and $B$ the points of tangency on one side of the angle, $x$ the radius of the third circle, and $C$ its point of tangency on the same side of the angle. Since $$ A B = A C + B C, \quad A B = 2 \sqrt{r R}, \quad A C = 2 \sqrt{r x}, \quad B C = 2 \sqrt{R x} $$ we have $$ 2 \sqrt{r x} + 2 \sqrt{R x} = 2 \sqrt{r R} $$ From this equation, we find that $$ x = \frac{r R}{(\sqrt{r} + \sqrt{R})^2} $$ Then $$ \frac{r}{x} = \frac{(\sqrt{r} + \sqrt{R})^2}{R} = \left(\frac{\sqrt{r} + \sqrt{R}}{\sqrt{R}}\right)^2 = \left(\sqrt{\frac{r}{R}} + 1\right)^2 $$ Dropping a perpendicular $OP$ from point $O$ to the radius $QB$, in the right triangle $OPQ$ $$ O Q = r + R, \quad P Q = R - r, \quad \angle P O Q = \alpha $$ Therefore, $$ R - r = (R + r) \sin \alpha \text{, or } 1 - \frac{r}{R} = \left(1 + \frac{r}{R}\right) \cdot \sin \alpha $$ From this, we find that $$ \frac{r}{R} = \frac{1 - \sin \alpha}{1 + \sin \alpha} = \frac{\cos^2 \alpha}{(1 + \sin \alpha)^2} $$ Thus, $$ \frac{r}{x} = \left(\frac{\cos \alpha}{1 + \sin \alpha} + 1\right)^2 = \frac{2(1 + \cos \alpha)}{1 + \sin \alpha} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-51.jpg?height=440&width=763&top_left_y=177&top_left_x=636) ## Answer $\frac{2(1 + \cos \alpha)}{1 + \sin \alpha}$." 18212c699577,"Find all triplets of natural numbers $(a, b, c)$ such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$.","(1,1,1)$ and $(a, b, c)=(3,2,1)$. We then verify that $1^{3}+1^{3}+1^{3}=3 \neq 1=(1 \times 1 \times",medium,"Suppose, without loss of generality, that $a \geqslant b \geqslant c \geqslant 1$. Since our equation involves terms of different degrees, we start by looking at what inequalities we might obtain. Here, we observe that: 口 $3 a^{3} \geqslant a^{3}+b^{3}+c^{3}=(a b c)^{2}$, and thus $3 a \geqslant(b c)^{2}$; $\Delta a^{2}$ divides $(a b c)^{2}-a^{3}=b^{3}+c^{3}$, and thus $a^{2} \leqslant b^{3}+c^{3} \leqslant 2 b^{3}$. Therefore, $b^{3} c^{5} \leqslant(b c)^{4} \leqslant 9 a^{2} \leqslant 18 b^{3}$, so $c^{5} \leqslant 18217$, and we conclude that 217 is prime, hence $a=1$; $\triangleright$ if $b=7$, then $a^{2}$ divides $b^{3}+1=344=2^{3} \times 43$, so $a$ divides 2; $\Delta$ if $b=8$, then $a^{2}$ divides $b^{3}+1=513$, which is not divisible by any prime $p \leqslant 19$; since $23^{2}=529>513$, we conclude that 513 is prime, hence $a=1$; $\Delta$ if $b=9$, then $a^{2}$ divides $b^{3}+1=730=2 \times 5 \times 73$, so $a=1$. The only two plausible triplets are $(a, b, c)=(1,1,1)$ and $(a, b, c)=(3,2,1)$. We then verify that $1^{3}+1^{3}+1^{3}=3 \neq 1=(1 \times 1 \times 1)^{2}$ and $3^{3}+2^{3}+1^{3}=26=(3 \times 2 \times 1)^{2}$, and we conclude that the only solutions are $(a, b, c)=(3,2,1)$ and its five permutations." 489137e29c9e,"8. A. As shown in Figure 3, in the rhombus $A B C D$ with side length 2, $B D=2, E$ and $F$ are moving points on $A D$ and $C D$ (including endpoints), and $A E+C F=2$. Then the range of the length of line segment $E F$ is $\qquad$",See reasoning trace,medium,"8. A. $\sqrt{3} \leqslant E F \leqslant 2$. From the given information, $\triangle A B D$ and $\triangle C B D$ are both equilateral triangles with side lengths of 2. Thus, $C F=2-A E=A D-A E=D E$ $\Rightarrow \triangle B D E \cong \triangle B C F$ $\Rightarrow \triangle B E F$ is an equilateral triangle. Therefore, we only need to consider the range of $B E$. When the moving point $E$ reaches point $D$ or $A$, the maximum value of $B E$ is 2; when $B E \perp A D$, i.e., $E$ is the midpoint of $A D$, the minimum value of $B E$ is $\sqrt{3}$. Therefore, $\sqrt{3} \leqslant E F \leqslant 2$." 6109482d6652,"A natural number is a [i]factorion[/i] if it is the sum of the factorials of each of its decimal digits. For example, $145$ is a factorion because $145 = 1! + 4! + 5!$. Find every 3-digit number which is a factorion.",145,medium,"To find every 3-digit number which is a factorion, we need to check if the number is equal to the sum of the factorials of its digits. Let's denote the 3-digit number as \( N = \overline{abc} \), where \( a, b, \) and \( c \) are its digits. 1. **Determine the range of digits:** - Since \( N \) is a 3-digit number, \( 100 \leq N < 1000 \). - The factorial of a digit \( d \) is denoted as \( d! \). - The maximum value for \( a, b, \) and \( c \) is 9, but we need to check the feasibility of each digit. 2. **Check the upper limit for factorial sums:** - The factorial of 9 is \( 9! = 362880 \). - For a 3-digit number, the sum of the factorials of its digits must be less than 1000. - Therefore, \( a, b, \) and \( c \) must be such that \( a! + b! + c! < 1000 \). 3. **Evaluate the possible values for \( a, b, \) and \( c \):** - Since \( 6! = 720 \) and \( 7! = 5040 \), any digit greater than 6 would make the sum exceed 1000. - Thus, \( a, b, \) and \( c \) must be \( \leq 5 \). 4. **Check each combination of digits:** - We need to check if \( N = 100a + 10b + c \) is equal to \( a! + b! + c! \). 5. **Case analysis:** - For \( a = 1 \): - \( N = 100 + 10b + c \) - Check if \( 100 + 10b + c = 1! + b! + c! \) - For \( b = 4 \) and \( c = 5 \): \[ 100 + 40 + 5 = 145 \] \[ 1! + 4! + 5! = 1 + 24 + 120 = 145 \] - Therefore, \( 145 \) is a factorion. 6. **Verify no other combinations:** - For \( a = 2 \): - \( N = 200 + 10b + c \) - Check if \( 200 + 10b + c = 2! + b! + c! \) - No valid combinations found. - For \( a = 3, 4, 5 \): - Similar checks show no valid combinations. Thus, the only 3-digit factorion is \( 145 \). The final answer is \( \boxed{145} \)." 0b7e69938325,"14. Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, define $$ S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {. } $$ (1) Find the minimum value of $S_{n}$; (2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$; (3) Under the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the minimum value of $S_{n}$, and provide a proof.","x_{2}=\cdots=x_{n}=\frac{1}{n}$, the minimum value $n$ is achieved.",medium,"14. (1) $S_{n} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{\frac{n-1}{n^{2}}}\right)^{2}$ $$ =4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=4 \times \frac{n-1}{n} \text {. } $$ When $x_{1}=x_{2}=\cdots=x_{n}=\frac{\sqrt{n-1}}{n}$, the minimum value $4 \times \frac{n-1}{n}$ is achieved. $$ \begin{array}{l} \text { (2) } S_{n}=\sum_{i=1}^{n}\left(x_{i}^{2}+2 \times \frac{n-1}{n^{2}}+\frac{(n-1)^{2}}{n^{4}} \cdot \frac{1}{x_{i}^{2}}\right) \\ =1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{4}} \sum_{i=1}^{n} \frac{1}{x_{i}^{2}} \\ \geqslant 1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{2}}=\left(1+\frac{n-1}{n}\right)^{2} . \end{array} $$ When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}$, the minimum value $$ \left(1+\frac{n-1}{n}\right)^{2} $$ is achieved. (3) Since $\left[\sum_{i=1}^{n} 1 \times\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2}$ $$ \leqslant\left(\sum_{i=1}^{n} 1^{2}\right) \cdot \sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {, } $$ Therefore, $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2}$ $$ \begin{array}{l} \geqslant \frac{1}{n}\left[\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2} \\ \geqslant \frac{1}{n}\left[1+\frac{n-1}{n^{2}} \cdot n^{2}\right]^{2}=n . \end{array} $$ When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, the minimum value $n$ is achieved." e8917df5255b,"In the quadrilateral $\mathrm{ABCD}$, the following angles are known: $B A C \angle=50^{\circ}, A B D \angle=60^{\circ}, D B C \angle=20^{\circ}, B D C \angle=30^{\circ}$. Is it true that the angle $C A D$ is $30^{\circ}$?","\angle ABC = 80^\circ$, which implies that $\angle DAC$ is indeed $30^\circ$.",medium,"Let's construct an equilateral triangle over the segment $AB$, and denote the vertex on the $BD$ segment as $G$. Draw a line through $D$ parallel to $AB$, intersecting $AG$ at $E$. $\angle EDG = \angle ABD = 60^\circ$ (so $E$ is above $C$), and $\angle DGE = \angle BGA = 60^\circ$, thus $\triangle DEG$ is equilateral, so $DE = EG$. According to the condition, $\angle GDC = \angle BDC = 30^\circ$, so $DC$ is the perpendicular bisector of segment $GE$. ![](https://cdn.mathpix.com/cropped/2024_05_02_84753f4460507e91dd1cg-1.jpg?height=729&width=588&top_left_y=346&top_left_x=758) In $\triangle ABC$, $\angle BCA = 180^\circ - (50^\circ + 60^\circ + 20^\circ) = 50^\circ$, so $\triangle ABC$ is isosceles with $BC = BA$, which is equal to $BG$ since $\triangle ABG$ is equilateral. Therefore, in $\triangle GBC$, $BG = BC$ and $\angle BCG = \angle BGC = \frac{180^\circ - 20^\circ}{2} = 80^\circ$, so $\angle ACG = \angle BCG - \angle BCA = 80^\circ - 50^\circ = 30^\circ$. In $\triangle GCE$, the perpendicular bisector of side $GE$ passes through the opposite vertex, so $CG = CE$. $\angle EGC = 180^\circ - (60^\circ + 80^\circ) = 40^\circ$, so $\angle ECG = 180^\circ - 2 \cdot 40^\circ = 100^\circ$, which implies $\angle ECG + \angle BCG = 100^\circ + 80^\circ = 180^\circ$, so points $E$, $C$, and $B$ are collinear. Therefore, quadrilateral $ABED$ is an isosceles trapezoid, with equal base angles, so $\angle DBA = \angle ABC = 80^\circ$, which implies that $\angle DAC$ is indeed $30^\circ$." 8e06365fad72,"## Find the derivative. $$ y=x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}+a^{\pi^{\sqrt{2}}} $$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & y^{\prime}=\left(x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}+a^{\pi^{\sqrt{2}}}\right)^{\prime}=\left(x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}\right)^{\prime}= \\ & =1+\frac{1}{\sqrt{2}} \cdot \frac{x+\sqrt{2}}{x-\sqrt{2}} \cdot\left(\frac{x-\sqrt{2}}{x+\sqrt{2}}\right)^{\prime}=1+\frac{1}{\sqrt{2}} \cdot \frac{x+\sqrt{2}}{x-\sqrt{2}} \cdot \frac{1 \cdot(x+\sqrt{2})-(x-\sqrt{2}) \cdot 1}{(x+\sqrt{2})^{2}}= \\ & =1+\frac{1}{\sqrt{2}} \cdot \frac{1}{x-\sqrt{2}} \cdot \frac{2 \sqrt{2}}{x+\sqrt{2}}=1+\frac{2}{x^{2}-2}=\frac{x^{2}-2}{x^{2}-2}+\frac{2}{x^{2}-2}=\frac{x^{2}}{x^{2}-2} \end{aligned} $$ ## Problem Kuznetsov Differentiation 8-12" 1e1c1812661d,"3. In a row, 100 knights and 100 liars are standing (in some order). The first person was asked: ""Are you a knight?"", and the rest were asked in turn: ""Is it true that the previous person answered 'Yes'?"" What is the maximum number of people who could have said ""Yes""? Knights always tell the truth, liars always lie.",See reasoning trace,medium,"Solution. 150, for example, if there are 100 knights followed by 100 liars. We will now prove that this is the maximum. Consider any liar, except possibly the first one. Either they said ""no,"" or the previous person said ""no."" Thus, to each of the 99 liars, we can associate at least one ""no"" response (either theirs or the previous person's), and each ""no"" response is associated with no more than two liars. Therefore, there are at least 50 ""no"" responses." 59a68931e9b5,"## Explanation. Denote a bag with parentheses. If there was one bag, there is only one way to form a ""bag of bags"": (). Two bags can also be formed in only one way: (()). Three bags can be formed in two different ways: (()()) and ((())), and so on. The order of the bags inside a bag does not matter. For example, the variant $((())())$ is not different from $(()(()))$.","2 \cdot 1 \cdot 1=2$ ways, while in the first case - not $\Pi_{3} \cdot \Pi_{3} \cdot \Pi_{1}=2 \cdo",medium,"Solution. If $\Pi_{n}$ denotes the number of ways for $n$ packages, then: $$ \begin{gathered} \Pi_{1}=1, \Pi_{2}=1, \Pi_{3}=2, \Pi_{4}=4, \Pi_{5}=9, \Pi_{6}=20, \Pi_{7}=48, \Pi_{8}=115, \Pi_{9}=286 \\ \Pi_{10}=719 \end{gathered} $$ The problem is solved by enumerating the cases. For example, if we take $\Pi_{5}$: $\Pi_{5}=P_{4}+P_{3}+P_{2}+P_{1}$, where $P_{k}$ is the number of ways corresponding to the case when the root package contains $k$ packages. $P_{4}=1$, there is only one way: $(()()()()$. $P_{3}=1$, there is also only one way: $((())()())$. $P_{2}=3$, there are three ways to split 4 packages into 2 groups: 3 and 1, 2 and 2. The first way corresponds to $\left(\Pi_{3}()\right)$ - two variants, the second way - one variant, $((())(()))$. $P_{4}=\Pi_{4}$ ways: $\left(\Pi_{4}\right)$. In total, it is $1+1+(2+1)+4=9$. As $n$ grows, cases will arise where one must be careful with combinatorics. For example, 8 packages can be arranged as $\left(\Pi_{3} \Pi_{3} \Pi_{1}\right)$, or as $\left(\Pi_{3} \Pi_{2} \Pi_{2}\right)$. In the second case, this is $\Pi_{3} \cdot \Pi_{2} \cdot \Pi_{2}=2 \cdot 1 \cdot 1=2$ ways, while in the first case - not $\Pi_{3} \cdot \Pi_{3} \cdot \Pi_{1}=2 \cdot 2 \cdot 1=4$, but actually only three, because the order of the triple packages does not matter." 76760186ad28,"Example 5 Let $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy for any real numbers $x$, $y$, $$ f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x) \text {. } $$ Find the analytical expression of the function $f(x)$. [2]",See reasoning trace,medium,"Let $x=y=0$. Then $f(f(0))=0$. Let $f(0)=s, x=0, y=s$. Then $2 s=s^{2} \Rightarrow s_{1}=2, s_{2}=0$. This gives us the points for case analysis. (1) When $s=2$, let $x=0, x=-y$, we get $f(f(y))=f(y)+2 y-2$, $f(x+2)+f\left(-x^{2}\right)=x+2-x f(x)$. Let $x=-1$ in equation (2), then $f(1)=1$. Let $y=1$ in the original equation, we get $$ f(x+f(x+1))=x+f(x+1) \text {. } $$ Inspired by the ""fixed point"" form in equation (3), let $$ \begin{array}{l} y=x+f(x+1) . \\ \text { Hence } f(f(y))=f(f(x)+f(x+1)) \\ =f(x+f(x+1))=f(y) \\ \Rightarrow 2 y-2=0 \\ \Leftrightarrow x+f(x+1)=1 \Leftrightarrow f(x)=2-x . \end{array} $$ (2) When $s=0$, $f(0)=0$ has a ""cancellation"" effect. Let $x=0, y=0$, we get $f(f(y))=f(y)$, $$ f(x+f(x))=x+f(x) \text {. } $$ Both have a form similar to a ""fixed point"". However, there is no obvious breakthrough when looking at each equation separately. Instead, let $x=f(y)$ in the above two equations, then $$ f(2 f(y))=2 f(y) \text {. } $$ Let $x=-1, y=1$, we get $f(-1)=-1$. By symmetry, let $x=1, y=-1$, we get $f(1)=1$. Let $x=1$, to simplify $f(x+y)$ in the original equation, replace $y$ with $y-1$, turning it into $f(y)$, then $$ f(f(y)+1)+f(y-1)=f(y)+y . $$ To create an equation around $f(f(y)+1)$, combine the ""reduction"" function of equation (4), let $x=1$, replace $y$ with $f(y)-1$, then $$ f(f(y)+1)+f(f(y)-1)=2 f(y) \text {. } $$ Let $x=-1$, replace $y$ with $y+1$, we get $$ f(f(y)-1)+f(-y-1)=f(y)-y-2 \text {. } $$ From equations (7) and (9), we get $f(f(y)+1)$ and $f(f(y)-1)$. By (9) - ( (8) - (7)), we have $$ f(y-1)=-f(-y-1)-2 \text {. } $$ To simplify one of the expressions $f(y-1)$ or $f(-y-1)$, the above equation can be simplified to $$ f(y-2)=-f(-y)-2 \text {. } $$ The best use of this equation is to simplify the $f(x+y)$ part in the original equation by letting $x=-2$ or $y=-2$. Since $x$ involves a wider range, in the original equation, let $x=-2$, and according to equation @, let $y=0$ to find $f(-2)=-2$, then $$ f(f(y-2)-2)+f(-2 y)=f(y-2)-2-2 y \text {. } $$ Using equation (10) twice to simplify $f(f(y-2)-2)$, then $$ -f(f(-y)+2)+f(-2 y)=-f(-y)-2-2 y \text {. } $$ Noticing that $y$ appears in the form of $-y$ here, replace $y$ with $-y$, then $$ f(f(y)+2)+2 y=f(2 y)+f(y)+2 \text {. (11) } $$ Replace $y$ with $f(y)$ and combine $f(f(y))=f(y)$, we get $$ f(f(y)+2)+2 f(y)=f(2 f(y))+f(y)+2 \text {. (12) } $$ In conjunction with equation (6), we get $f(f(y)+2)=f(y)+2$. Substitute into equation (11), we have $$ f(y)+2+2 y=f(2 y)+f(y)+2, $$ Thus, for any real number $y$, $f(2 y)=2 y$. Simplifying, we get for any real number $x$, $f(x)=x$. Upon verification, $f(x)=x$ or $f(x)=2-x(x \in \mathbf{R})$. Assigning values to eliminate terms focuses on handling non-iterative functions, while iterative elimination focuses on handling iterative functions. For many functional equation problems, whether they ""show their form externally"" or ""hide their momentum internally"", they all require the method of assigning values using the domain and range of the function. Appropriate assignment of values is the key to opening up ideas and solving problems." 17ca930ada81,"Example 6 As shown in Figure 1, in the isosceles right triangle $AOB$ with one leg being 1, a point $P$ is randomly taken within this triangle. Parallel lines to the three sides are drawn through $P$, forming three triangles (shaded areas in the figure) with $P$ as a vertex. Find the minimum value of the sum of the areas of these three triangles and the position of point $P$ when the minimum value is achieved.","y_{P}=\frac{1}{3}$, the area $S$ is minimized, and the minimum value is $S_{\text {min }}=\frac{1}{6",medium,"Parse: Taking $O A, O B$ as the $x$-axis and $y$-axis respectively to establish a rectangular coordinate system as shown in Figure 2, the equation of the line $A B$ is $x+y=1$. Let the coordinates of point $P$ be $P\left(x_{p}, y_{P}\right)$, then the sum of the areas of the three triangles sharing $P$ as a common vertex is $S=\frac{1}{2} x_{P}^{2}+\frac{1}{2} y_{P}^{2}+\frac{1}{2}\left(1-x_{P}-y_{P}\right)^{2}$, i.e., $2 S=x_{P}^{2}+y_{P}^{2}+\left(1-x_{P}-y_{P}\right)^{2}$. By the Cauchy-Schwarz inequality, we have $$\begin{array}{l} {\left[x_{P}^{2}+y_{P}^{2}+\left(1-x_{P}-y_{P}\right)^{2}\right] \cdot\left(1^{2}+1^{2}+1^{2}\right)} \\ \geqslant\left[x_{P}+y_{P}+\left(1-x_{P}-y_{P}\right)\right]^{2}, \end{array}$$ i.e., $2 S \times 3=6 S \geqslant 1$. Equality holds if and only if $\frac{x_{P}}{1}=\frac{y_{P}}{1}=\frac{1-x_{P}-y_{P}}{1}$, i.e., when $x_{P}=y_{P}=\frac{1}{3}$, the area $S$ is minimized, and the minimum value is $S_{\text {min }}=\frac{1}{6}$." 91b5455dc662,"15. Given that $a$, $b$, and $c$ are distinct integers. Then $$ 4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} $$ the minimum value is $\qquad$",See reasoning trace,easy,"15.8. Notice, $$ \begin{array}{l} 4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\ =(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2}, \end{array} $$ its minimum value is 8." 2efd338276ba,"34.2. Find all functions $f(x)$ that are defined for $x \neq 1$ and satisfy the relation $$ f(x) + f\left(\frac{1}{1-x}\right) = x $$",See reasoning trace,medium,"34.2. Let $\varphi_{1}(x)=\frac{1}{1-x}$. Then $\varphi_{2}(x)=1-1 / x$ and $\varphi_{3}(x)=x$. Therefore, we obtain the system of equations $$ \left\{\begin{array}{l} f(x)+f\left(\frac{1}{1-x}\right)=x \\ f\left(\frac{1}{1-x}\right)+f\left(1-\frac{1}{x}\right)=\frac{1}{1-x} \\ f\left(1-\frac{1}{x}\right)+f(x)=1-\frac{1}{x} \end{array}\right. $$ Add the first equation to the third and subtract the second equation from them. As a result, we get $$ f(x)=\frac{1}{2}\left(x+1-\frac{1}{x}-\frac{1}{1-x}\right) $$ A direct check shows that this function satisfies the required relation." 22e92ffa2bcc,"If $\frac{1}{2 n}+\frac{1}{4 n}=\frac{3}{12}$, then $n$ equals (A) 6 (B) $\frac{1}{2}$ (C) $\frac{1}{3}$ (D) 2 (E) 3",(E),easy,"Since $\frac{1}{2 n}+\frac{1}{4 n}=\frac{2}{4 n}+\frac{1}{4 n}=\frac{3}{4 n}$, then the given equation becomes $\frac{3}{4 n}=\frac{3}{12}$ or $4 n=12$. Thus, $n=3$. Answer: (E)" 1fbbe743f5eb,1. Solve the equation $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \sin ^{2} z . \quad(5$ points),". $x=\frac{\pi}{2}+\pi n, y=\pi m, z=\frac{\pi}{2}+\pi k$, where $n, k, m \in Z$",medium,"Solution. The domain of the inequality is limited by the conditions $\sin x \neq 0, \cos y \neq 0$. Note that $u+\frac{1}{u} \geq 2$ for any positive $u$, as this inequality is equivalent to $$ \frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \geq 0 $$ with equality achieved at $u=1$. Thus, the minimum value of the left side of the equation is $2^{3}+2^{3}=16$. The right side of the equation is no more than 16, so equality is possible only if both the left and right sides of the equation are equal to 16. This means that $\sin ^{2} x=\cos ^{2} y=\sin ^{2} z=1$, that is, $x=\frac{\pi}{2}+\pi n, y=\pi m, z=\frac{\pi}{2}+\pi k$, where $n, k, m \in Z$. Answer. $x=\frac{\pi}{2}+\pi n, y=\pi m, z=\frac{\pi}{2}+\pi k$, where $n, k, m \in Z$." fae8a498469e,"9. If $5 \pi$ is a period of the function $f(x)=\cos n x \cdot \sin \frac{80}{n^{2}} x$, then all possible values of the positive integer $n$ are $\qquad$","k \in \mathbf{N}^{*} \Rightarrow n=1,2,4,5,10,20$. Upon verification, $n=2,10$.",easy,"$$ f(5 \pi)=\cos 5 n \pi \sin \frac{400 \pi}{n^{2}}=f(0)=0 \Rightarrow \sin \frac{400 \pi}{n^{2}}=0 \Rightarrow \frac{400 \pi}{n^{2}}=k \pi \text {, } $$ Thus, $\frac{400}{n^{2}}=k \in \mathbf{N}^{*} \Rightarrow n=1,2,4,5,10,20$. Upon verification, $n=2,10$." 0feaf209bfce,"$$ \begin{array}{l} \log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\ =2 \log _{3}(5 x+2) \log _{5}(5 x+4) \end{array} $$ Solve the equation for \( x \): $$ \begin{array}{l} \log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\ =2 \log _{3}(5 x+2) \log _{5}(5 x+4) \end{array} $$",\frac{1}{5} \).,medium,"First, we prove the lemma: If \( n \geqslant m > 1 \), then \[ 01, \\ \therefore \log _{n}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{m}\right), \\ \therefore \log _{n}(n+1)-1 \leqslant \log _{m}(m+1)-1, \\ \log _{n}(n+1) \leqslant \log _{m}(m+1) . \\ \frac{\lg (n+1)}{\lg n} \leqslant \frac{\lg (m+1)}{\lg m}, \\ \frac{\lg m}{\lg n} \leqslant \frac{\lg (m+1)}{\lg (n+1)}, \end{array} \] Thus, \[ \log _{n} m \leqslant \log _{(n+1)}(m+1). \] Second, it is easy to verify that \( x = \frac{1}{5} \) is a root of the original equation. Finally, we prove that the original equation has no other real roots: Let \( f(x) = \frac{\log _{2}(5 x+1)}{\log _{3}(5 x+2)} + \frac{\log _{4}(5 x+3)}{\log _{5}(5 x+4)} \). Then the original equation can be transformed into \( f(x) = 2 \). Since \( 5 x + 1 > 0 \), it follows that \( 5 x + 2, 5 x + 3, 5 x + 4 \) are all greater than 1. According to the lemma: If \( 0 < 5 x + 1 < 2 \), then \[ f(x) = \frac{\log _{7}(5 x+2)}{\log _{2}(5 x+1)} + \frac{\log _{5}(5 x+4)}{\log _{4}(5 x+3)} > 1 + 1 = 2. \] If \( 5 x + 1 > 2 \), then \[ f(x) = \frac{\log _{7}(5 x+2)}{\log _{2}(5 x+1)} + \frac{\log _{5}(5 x+4)}{\log _{4}(5 x+3)} < 1 + 1 = 2. \] In summary, when \( 0 < 5 x + 1 \neq 2 \), \( f(x) \neq 2 \). Therefore, the original equation has and only has one real root \( x = \frac{1}{5} \)." 2167e81366d4,"4. From the numbers $1,2,3,4,5$, select several (at least 2) without repetition and find their sum, the number of different sums that can be obtained is ( ) . (A) 26 (B) 17 (C) 13 (D) 12","3$, and will not be greater than $1+2+3+4+5=15$. Therefore, the number of different sums will not ex",easy,"4.C. The sum formed by $1,2,3,4,5$ will not be less than $1+2$ $=3$, and will not be greater than $1+2+3+4+5=15$. Therefore, the number of different sums will not exceed $15-3+1=13$. Furthermore, since these five numbers are consecutive integers starting from 1, each of the 13 different sums can be achieved by substituting $n+1$ for $n$. Thus, there are 13 different sums." abd7713afa50,"1. $a, b$ are constants. If the parabola $C$: $$ y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+t^{2}+3 a t+b $$ passes through the fixed point $P(1,0)$ for any real number $t$, find the value of $t$ when the chord intercepted by the parabola $C$ on the $x$-axis is the longest.","-1$, the chord length $|A B|$ reaches its maximum value of 2.",medium,"$=1.2$. Substituting $P(1,0)$ into the equation of the parabola $C$ yields $$ \left(t^{2}+t+1\right)-2(a+t)^{2}+\left(t^{2}+3 a t+b\right)=0 \text {, } $$ which simplifies to $t(1-a)+\left(1-2 a^{2}+b\right)=0$, holding for all $t$. Thus, $1-a=0$, and $1-2 a^{2}+b=0$. Solving these, we get $a=1, b=1$. Substituting into the equation of the parabola $C$ gives $$ y=\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right) \text {. } $$ Setting $y=0$, we have $$ \left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right)=0 \text {. } $$ Let the roots of this equation be $x_{1}$ and $x_{2}$, then $$ \begin{array}{l} |A B|=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\ =\left|\frac{2 t}{t^{2}+t+1}\right| . \end{array} $$ Noting that the discriminant of this quadratic equation $\Delta=4 t^{2}>0$, to find the maximum value of $|A B|$, we categorize by $t$. (1) When $t>0$, $$ |A B|=\frac{2 t}{t^{2}+t+1}=\frac{2}{t+\frac{1}{t}+1} \leqslant \frac{2}{3} \text {; } $$ (2) When $t<0$, $$ \begin{array}{l} |A B|=-\frac{2 t}{t^{2}+t+1}=\frac{2}{-t-\frac{1}{t}-1} \\ \leqslant \frac{2}{-1+2 \sqrt{(-t) \frac{1}{-t}}}=2 . \end{array} $$ When $t=-1$, $|A B|_{\text {max }}=2$. From (1) and (2), we know that when $t=-1$, the chord length $|A B|$ reaches its maximum value of 2." 6dd8095d3d3b,"The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$? $\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$",\textbf{(E),easy,"The area of the larger pizza is $16\pi$, while the area of the smaller pizza is $9\pi$. Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$, which is closest to $\boxed{\textbf{(E) }78}$." 6f9d64bcf619,"Example 1 As shown in Figure 13-4, in $\triangle A B C$, $O$ is the circumcenter of $\triangle A B C$, and $I$ is its incenter. If $\angle B O C=$ $\angle B I C$, find $\angle A$. 保留了原文的格式和换行。","\frac{1}{2}(\angle B + \angle C) = 90^{\circ} - \frac{1}{2} \angle A$, hence $\angle B I C = 90^{\ci",easy,"Since $I$ is the incenter, then $\angle I B C + \angle I C B = \frac{1}{2}(\angle B + \angle C) = 90^{\circ} - \frac{1}{2} \angle A$, hence $\angle B I C = 90^{\circ} + \frac{1}{2} \angle A$. Also, $O$ is the circumcenter, so $\angle B O C = 2 \angle A$, thus $90^{\circ} + \frac{1}{2} \angle A = 2 \angle A$, which means $\angle A = 60^{\circ}$ is the solution." f49f5196fe75,"10. Right triangle $X Y Z$ has right angle at $Y$ and $X Y=228, Y Z=$ 2004. Angle $Y$ is trisected, and the angle trisectors intersect $X Z$ at $P$ and $Q$ so that $X, P, Q, Z$ lie on $X Z$ in that order. Find the value of $(P Y+Y Z)(Q Y+X Y)$.",See reasoning trace,easy,"Solution: 1370736 The triangle's area is $(228 \cdot 2004) / 2=228456$. All the angles at $Y$ are 30 degrees, so by the sine area formula, the areas of the three small triangles in the diagram are $Q Y \cdot Y Z / 4, P Y \cdot Q Y / 4$, and $X Y \cdot P Y / 4$, which sum to the area of the triangle. So expanding $(P Y+Y Z)(Q Y+X Y)$, we see that it equals $$ 4 \cdot 228456+X Y \cdot Y Z=6 \cdot 228456=1370736 . $$" 9dc99852ebd9,"$7 \cdot 2$ If $\sin x=3 \cos x$, then $\sin x \cos x$ equals (A) $\frac{1}{6}$. (B) $\frac{1}{5}$. (C) $\frac{2}{9}$. (D) $\frac{1}{4}$. (E) $\frac{3}{10}$. (39th American High School Mathematics Examination, 1988)",$(E)$,easy,"[Solution] Given $\operatorname{tg} x=3$. $$ \therefore \quad \sin x \cos x=\frac{1}{2} \sin 2 x=\frac{1}{2} \cdot \frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}=\frac{3}{10} \text {. } $$ Therefore, the answer is $(E)$." cafb0119e12b,"420. A tangent to a circle was rotated by an angle $\beta$ around the point of tangency, causing it to become a secant. Find the measure of the arc it intercepts.",248&width=263&top_left_y=670&top_left_x=385),medium,"$\triangleright$ Let $OA$ be the radius drawn to the point of tangency, and $AB$ be the secant. The tangent is perpendicular to the radius, so the angle $OAB$ between the radius and the secant is $90^{\circ}-\beta$. The triangle $OAB$ is isosceles, so the second angle in it is also $90^{\circ}-\beta$, and the third angle (with the vertex at the center of the circle) is $2 \beta$. Thus, the tangent turned by an angle $\beta$ cuts off an arc of size ![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-126.jpg?height=369&width=397&top_left_y=358&top_left_x=1321) $2 \beta$. $\triangleleft$ The statement of the previous problem can be formulated as follows: the angle between the tangent and the chord passing through the point of tangency is equal to half the arc cut off by the chord. ![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-126.jpg?height=377&width=348&top_left_y=754&top_left_x=1365) ![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-127.jpg?height=248&width=263&top_left_y=670&top_left_x=385)" b7471d83f2a6,316. $\left\{\begin{array}{l}(x+y)\left(x^{2}+y^{2}\right)=539200 \\ (x-y)\left(x^{2}-y^{2}\right)=78400 .\end{array}\right.$,See reasoning trace,easy,"316. Instruction. Dividing the 1st equation by the 2nd: $$ \frac{x^{2}+y^{2}}{(1-y)^{2}}=\frac{337}{49} $$ from which $$ 674 x y=288 x^{2}+288 y^{2} $$ or $$ 98 x y=288(x-y)^{2} $$ $$ \frac{(x-y)^{2}}{x y}=\frac{49}{144}=\frac{x}{y}+\frac{y}{x}-2 $$ $$ \frac{x}{y}+\frac{y}{x}=\frac{337}{144} $$ let $$ \frac{x}{y}=z $$ $10^{*}$ rorдa $$ 144 z^{2}-337 z+144=0 $$ from which $$ z=\frac{16}{9} $$ Now it is easy to find: $$ x=64 ; y=36 $$" 325d2f2e1e35,"Let $a$ and $b$ be positive integers such that $(2a+b)(2b+a)=4752$. Find the value of $ab$. [i]Proposed by James Lin[/i]",520,medium,"1. **Expand the given equation:** \[ (2a + b)(2b + a) = 4752 \] Expanding the left-hand side, we get: \[ 4ab + 2a^2 + 2b^2 + ab = 4752 \] Simplifying, we have: \[ 2a^2 + 5ab + 2b^2 = 4752 \] 2. **Divide the equation by 2:** \[ a^2 + \frac{5}{2}ab + b^2 = 2376 \] To simplify further, multiply the entire equation by 2: \[ 2a^2 + 5ab + 2b^2 = 4752 \] 3. **Rewrite the equation in a more manageable form:** \[ (a + b)^2 + \frac{ab}{2} = 2376 \] Let \( s = a + b \) and \( p = ab \). Then the equation becomes: \[ s^2 + \frac{p}{2} = 2376 \] 4. **Estimate \( s \):** \[ \sqrt{2376} \approx 48.7 \] Since \( s \) must be an integer, we start testing values around 48.7. 5. **Test values for \( s \):** - For \( s = 44 \): \[ 44^2 + \frac{p}{2} = 2376 \implies 1936 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 440 \implies p = 880 \] Check if \( a \) and \( b \) are integers: \[ t^2 - 44t + 880 = 0 \] The discriminant is: \[ 44^2 - 4 \cdot 880 = 1936 - 3520 = -1584 \quad (\text{not a perfect square}) \] - For \( s = 45 \): \[ 45^2 + \frac{p}{2} = 2376 \implies 2025 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 351 \implies p = 702 \] Check if \( a \) and \( b \) are integers: \[ t^2 - 45t + 702 = 0 \] The discriminant is: \[ 45^2 - 4 \cdot 702 = 2025 - 2808 = -783 \quad (\text{not a perfect square}) \] - For \( s = 46 \): \[ 46^2 + \frac{p}{2} = 2376 \implies 2116 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 260 \implies p = 520 \] Check if \( a \) and \( b \) are integers: \[ t^2 - 46t + 520 = 0 \] The discriminant is: \[ 46^2 - 4 \cdot 520 = 2116 - 2080 = 36 \quad (\text{a perfect square}) \] Solving for \( t \): \[ t = \frac{46 \pm \sqrt{36}}{2} = \frac{46 \pm 6}{2} \] Thus, \( t = 26 \) or \( t = 20 \). Therefore, \( a = 20 \) and \( b = 26 \) (or vice versa). 6. **Calculate \( ab \):** \[ ab = 20 \times 26 = 520 \] The final answer is \(\boxed{520}\)" 253defdbfe72,"6. In a cube with edge length $a$, accommodate 9 equal spheres, placing one at each of the eight corners, then the maximum radius of these equal spheres is $\qquad$ .",See reasoning trace,easy,6. $\frac{1}{2}(2 \sqrt{3}-3) a$ 2a9577eecc3e,7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers?,50,medium,"Answer: 50. Solution: Note that two consecutive numbers cannot both be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number before them (counterclockwise) is also a natural number. Since it is greater than the product of these two natural numbers, it is greater than each of them. Continuing this reasoning, we will conclude that all numbers are natural, and if we move counterclockwise, the numbers will increase. However, when the ""circle closes,"" we will arrive at a contradiction. Therefore, in any pair of adjacent numbers, there is at least one negative number. Dividing our 100 numbers into 50 pairs, we get that there are no fewer than 50 negative numbers, and thus no more than 50 positive numbers. If we alternate a positive number 2 and a negative number -2, we will get an example of the arrangement of 50 positive and 50 negative numbers that satisfy the problem's condition. Comment: Correct answer without justification - 0 points. Proving the impossibility of placing two positive numbers next to each other - 3 points. Proving that there are no more than 50 positive numbers - 1 point. Providing an example with 50 positive numbers - 3 points." 3f56827d4ba7,"Find all function $ f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that for every three real positive number $x,y,z$ : $$ x+f(y) , f(f(y)) + z , f(f(z))+f(x) $$ are length of three sides of a triangle and for every postive number $p$ , there is a triangle with these sides and perimeter $p$. [i]Proposed by Amirhossein Zolfaghari [/i]",f(x) = x,medium,"1. **Define the problem and the function:** We need to find all functions \( f: \mathbb{R}^{+} \to \mathbb{R}^{+} \) such that for every three positive real numbers \( x, y, z \): \[ x + f(y), \quad f(f(y)) + z, \quad f(f(z)) + f(x) \] are the lengths of the sides of a triangle. Additionally, for every positive number \( p \), there exists a triangle with these sides and perimeter \( p \). 2. **Consider a sequence \( p_n \) approaching zero:** Let \( p_n \) be any sequence of positive numbers such that \( p_n \to 0^+ \) as \( n \to \infty \). We need to find \( x_n, y_n, z_n \) such that: \[ x_n + f(y_n) + f(f(y_n)) + z_n + f(f(z_n)) + f(x_n) = p_n \] Since \( f \) maps positive reals to positive reals, all terms \( x_n, f(x_n), f(y_n), f(f(y_n)), z_n, f(f(z_n)) \) must approach \( 0 \) as \( n \to \infty \). 3. **Apply the triangle inequality:** Using the triangle inequality, we have: \[ x + f(y) + f(f(y)) + z > f(f(z)) + f(x) \] and \[ f(f(y)) + z + f(f(z)) + f(x) > x + f(y) \] 4. **Substitute \( (y, z) = (y_n, z_n) \) and take the limit:** Plugging \( (y, z) = (y_n, z_n) \) into these inequalities and letting \( n \to \infty \), we get: \[ x \geq f(x) \] from the first inequality, and \[ f(x) \geq x \] from the second inequality. 5. **Conclude that \( f(x) = x \):** Combining these two results, we have: \[ x \geq f(x) \quad \text{and} \quad f(x) \geq x \] which implies: \[ f(x) = x \quad \text{for all} \quad x > 0 \] The final answer is \( \boxed{ f(x) = x } \)." 1da5565fb86f,"Through the midpoints $M$ and $N$ of the edges $A A_1$ and $C_1 D_1$ of the parallelepiped $A B C D A_1 B_1 C_1 D_1$, a plane parallel to the diagonal $B D$ of the base is drawn. Construct the section of the parallelepiped by this plane. In what ratio does it divide the diagonal $A_1 C$?","3:7, measured from vertex $A 1$",medium,"Continue the median line $N K$ of triangle $B 1 C 1 D 1$ until it intersects with line $A 1 B 1$ at point $X$, and with line $A 1 D 1$ at point $Y$ (Fig.1). Let segments $M X$ and $B B 1$ intersect at point $L$, and segments $M Y$ and $D D 1$ intersect at point $P$. Then the pentagon $M L K N P$ is the desired section, since the cutting plane contains the line $K N$, which is parallel to $D 1 B 1$. Let $A 1 C 1$ and $N K$ intersect at point $Q$. Then the cutting plane and the plane of the diagonal section $A C C 1 A 1$ intersect along the line $M Q$, and the point $O$ of intersection of the lines $A 1 C$ and $M Q$, lying in the plane of the diagonal section, is the desired point of intersection of the line $A 1 C$ with the cutting plane. Consider the parallelogram $A C C 1 A 1$ (Fig.2). Let $A C=4 x$. Extend $Q M$ until it intersects with line $A C$ at point $T$. From the equality of triangles $A M T$ and $A 1 M Q$, it follows that $$ A T=A 1 Q=\frac{3}{4} A 1 C 1=\frac{3}{4} \cdot 4 x=3 x $$ Then $C T=A T+A C=3 x+4 x=7 x$, and since triangles $A 1 O Q$ and $C O T$ are similar, we have $$ \frac{O A_{1}}{O C}=\frac{A_{1} Q}{C T}=\frac{3 I}{7 I}=\frac{3}{7} $$ ## Answer 3:7, measured from vertex $A 1$. ## Problem" 821e6e820490,"## 43. Math Puzzler $12 / 68$ Mr. Dull has nothing to do. He starts counting all the numbers in natural order: $1 +$ $2 + 3 + \cdots$ and so on. The doorbell rings, and Mr. Dull just manages to note down the last intermediate result he found, which is 6328. However, he forgets the last number he added. Can you help him without being as dull as he is?",12656$. The quadratic equation has the solution $x=112$. If solving quadratic equations is not yet k,medium,"The problem relates to the task that young Gauss was supposed to solve. The teacher had assigned the task of adding all numbers from 1 to 100. Gauss was finished in 5 minutes because he noticed that there are 50 pairs of numbers that add up to 101: 100+1, 99+2, 98+3, etc. Using this trick, $$ \begin{aligned} & 1+2+3+\ldots+(x-2)+(x-1)+x=6328 \\ & x+(x-1)+(x-2)+\ldots+3+2+1=6328 \end{aligned} $$ Adding both sums, we get $x(x+1)=12656$. The quadratic equation has the solution $x=112$. If solving quadratic equations is not yet known, one can use a square table to determine that $\sqrt{12656}$ lies between 112 and 113. A check with 112 confirms the result." 6602bd7686c3,"4. Given a rhombus $ABCD$ with $\measuredangle BAD=60^{\circ}$. The angle bisectors of the angles between the diagonals intersect the sides of the rhombus at points $M, N, P$ and $Q$. a) What type of quadrilateral is quadrilateral $MNPQ$? b) If point $M$ lies on side $AB$, calculate the ratio $AM: MB$. c) Determine the ratio of those segments of the larger and smaller diagonals of the rhombus that lie outside quadrilateral $MNPQ$.","\frac{x \sqrt{3}}{2}=y \sqrt{3} \frac{\sqrt{3}}{2}=\frac{3 y}{2}$, and from triangle $B R M$ it foll",medium,"Solution. a) From the congruence of triangles $O B M$ and $O B N$ it follows that $O M=O N$, and from the congruence of triangles $O A M$ and $O A Q$ it follows that $O M=O Q$ (see diagram on the right). Similarly, $O Q=O P$. From this, we conclude that $M P=N Q$ and since $\measuredangle M O N$ is a right angle, it follows that quadrilateral $M N P Q$ is a square. b) Triangle $A M Q$ is equilateral. Let the side of this triangle be denoted by $x$. Then triangle $B R M$, which is half of an equilateral triangle, has a height $M R=\frac{x}{2}$. Therefore, $M R=M B \frac{\sqrt{3}}{2}$, and if we set $M B=y$, we get $\frac{x}{2}=y \frac{\sqrt{3}}{2}$, i.e., $x=y \sqrt{3}$. Thus, $A M: M B=x: y=y \sqrt{3}: y=\sqrt{3}: 1$. c) From triangle $A M Q$ it follows that $A S=\frac{x \sqrt{3}}{2}=y \sqrt{3} \frac{\sqrt{3}}{2}=\frac{3 y}{2}$, and from triangle $B R M$ it follows that $B R=\frac{y}{2}$. Therefore, $A S: B R=\frac{3 y}{2}: \frac{y}{2}=3: 1$." 48ec9e0ed6a1,"9. In 1978, someone introduced their family by saying: ""I have a son and a daughter, they are not twins. The cube of my son's age plus the square of my daughter's age is exactly my birth year. I was born after 1900. My children are both under 21 years old. I am 8 years older than my wife."" Calculate the age of each family member in 1978. (15 points)",", 5 points; (2) For each formula, explaining the reasoning, 10 points; (3) Even if the final answer is incorrect, for correctly reasoned formulas, appropriate points should be given",medium,"9. In 1978, the son's age was 12 years old, the daughter's age was 14 years old, the mother's age was 46 years old, and the father's age was 54 years old. Solution: Since \(13^3 = 2197\), the son's age must be less than 13 years; also, since \(11^3 = 1331\), even if we add \(21^2 = 441\), it is only \(1331 + 441 = 1772\), so \(x^2 > 1900 - 12^3\), \(x^2 > 172\), which means the daughter's age is greater than 13 years. Given that this age is less than 21 years, the possible ages for the daughter are: 14, 15, 16, 17, 18, 19, 20. If \(x = 15\), then the father's birth year would be: \(12^3 + 15^2 = 1953\), which is clearly unreasonable. Similarly, the daughter's age cannot be greater than 15, so it can only be 14 years old. At this time, the father's birth year is: \(12^3 + 14^2 = 1924\), and his age in 1978 is: \(1978 - 1924 = 54\) (years old). The mother's age in 1978 is: \(54 - 8 = 46\) (years old). Scoring reference: (1) Correctly providing the answer, 5 points; (2) For each formula, explaining the reasoning, 10 points; (3) Even if the final answer is incorrect, for correctly reasoned formulas, appropriate points should be given." 60318142ba1b,"68. Four foreign friends, each of them can speak 2 out of 4 languages: English, French, Japanese, and Chinese. One language is spoken by three of them, but there is no language that all of them can speak. It is also known that: (1) A can speak Japanese, but D cannot, yet they communicate in the same language; (2) B does not speak English, and when A and C communicate, he acts as a translator; (3) B, C, and D cannot find a common language to communicate; (4) None of the four can speak both Japanese and French. The number of people who can speak Chinese is $\qquad$.",3,easy,answer: 3 e8af026aac2a,"8. Find all continuous functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that $$ f(x+f(y))=f(x)+y $$ for every $x, y \in \mathbf{R}$. ## Geometry",See reasoning trace,medium,"8. By setting $x=0$ and then $y=0$ in the relation, we get $$ f(f(y))=y+f(0), \quad f(x+f(0))=f(x) $$ Now $$ f(x)=f(x+f(0))=f(f(f(x)))=f(x)+f(0) $$ and hence $f(0)=0$. Thus from (1) we have $f(f(y))=y$. Now $$ f(x+y)=f(x+f(f(y)))=f(x)+f(y) $$ and the continuity of $f$ implies that $f(x)=a x$ for some $a \in \mathbf{R}$. From the relation in the problem we get $$ a(x+a y)=a x+y $$ ## Figure 9 for all $x, y \in \mathbf{R}$, i.e. $a= \pm 1$. Therefore $$ f(x)=x \quad \text { or } \quad f(x)=-x $$" 8f1bc051e2ca,"[ equations in integers $]$ [GCD and LCM. Mutual simplicity ] Find all pairs of integers $(x, y)$, satisfying the equation $3 \cdot 2^{x}+1=y^{2}$.","$(0, \pm 2),(3, \pm 5),(4, \pm 7)$",easy,"The equation can be rewritten as $3 \cdot 2^{x}=(y+1)(y-1)$. The difference between the numbers $y+1$ and $y-1$ is 2, so their greatest common divisor $d$ is 1 or 2. If $d=1$, then $x=0$. If $d=2$, then either one number is $\pm 3 \cdot 2^{n}$ and the other is $\pm 2$, or one number is $\pm 3 \cdot 2$ and the other is $\pm 2^{n}$. The first case is impossible, and in the second case, $x=3$ or 4. ## Answer $(0, \pm 2),(3, \pm 5),(4, \pm 7)$" 40af4c414b08,"7. Four boys and three girls want to sit on the same bench. Is it more likely that the people at both ends of the bench will be of the same or opposite gender? Determine both probabilities. The use of a pocket calculator or any manuals is not allowed.",See reasoning trace,medium,"First method: First, calculate the total number of all possible seating arrangements: Any of the 7 people can sit in the first seat, any of the remaining 6 in the second seat, any of the remaining 5 in the third seat, and so on... The last remaining person can sit in the last seat. In total, there are $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5040$ arrangements for how the children can sit on the bench. 2 POINTS Let's assume that two boys sit at the ends. The four boys can be arranged in these two seats in $4 \cdot 3=12$ ways, and then the remaining 5 people can sit anywhere, which they can do in $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways. In total, there are $12 \cdot 120=1440$ arrangements in which two boys are at the ends. 2 POINTS The people at the ends can also be girls. The three girls can be arranged in these two seats in $3 \cdot 2=6$ ways, which together with the 120 arrangements of the remaining 5 people gives $6 \cdot 120=720$ possibilities for how the children can sit on the bench, with two girls at the ends. Together, there are $1440+720=2160$ arrangements of the children in which people of the same gender sit at the ends. The probability that this happens is $2160: 5040=\frac{3}{7}=0.428571 \ldots \approx 43 \%$ 2 POINTS The probability that people of different genders will sit at the ends is $\frac{4}{7}=0.571428 \ldots \approx 57 \%$ 1 POINT It is more likely that people of different genders will sit at the ends of the bench. 1 POINT TOTAL 10 POINTS" 45b44b58051a,"9.3. Add a digit to the left and right of the number 10 so that the resulting number is divisible by 72. $$ (5-6 \text { grade) } $$",4104,easy,"9.3. The desired number must be divisible by 8 and 9, i.e., it must be even and the sum of its digits must be divisible by 9. Only 5 numbers need to be checked: $8100, 6102, 4104, 2106, 9108$. Among these numbers, only 4104 is divisible by 72. Answer: 4104." d76c1b1b5c1a,"4. If each face of a tetrahedron is not an isosceles triangle, then the minimum number of edges with unequal lengths is (A) 3 ; (B) 4 ; (C) 5 ; (D) 6 .",See reasoning trace,easy,"4. (A) Since each face of the tetrahedron is not an isosceles triangle, at least three edges have different lengths. On the other hand, a tetrahedron with each pair of opposite edges equal can be a tetrahedron where only the lengths are all different and each face is not an isosceles triangle." 3a91872aeea8,"## Task B-2.4. The centers of two circles are $44 \mathrm{~cm}$ apart. If the radii of these circles are $15 \mathrm{~cm}$ and $37 \mathrm{~cm}$, what is the length of their common chord?","2 |A E| = 24 \, \text{cm} \).",medium,"## First Solution. Let's introduce the notation as shown in the figure. ![](https://cdn.mathpix.com/cropped/2024_05_30_ab4e7c3a83d03c059773g-09.jpg?height=663&width=854&top_left_y=565&top_left_x=521) Let's calculate the area of triangle \( A S_{1} S_{2} \) using Heron's formula. For \( a = |S S_{1}| = 44 \, \text{cm} \), \( b = |S_{2} A| = 37 \, \text{cm} \), and \( c = |S_{1} A| = 15 \, \text{cm} \), we get \( s = \frac{1}{2}(15 + 37 + 44) = 48 \, \text{cm} \), \[ P = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{48 \cdot 4 \cdot 11 \cdot 33} = 4 \cdot 3 \cdot 2 \cdot 11 = 264 \, \text{cm}^2 \] On the other hand, the area of triangle \( A S_{1} S_{2} \) is \( P = \frac{1}{2} \cdot |S_{1} S_{2}| \cdot |A E| \). From \( 264 = \frac{1}{2} \cdot 44 \cdot |A E| \), it follows that \( |A E| = 12 \, \text{cm} \). The length of the common chord of these two circles is \( |A B| = 2 |A E| = 24 \, \text{cm} \)." f654b4c31bb9,"2. Find all integer solutions of the inequality $$ x^{2} y^{2}+y^{2} z^{2}+x^{2}+z^{2}-38(x y+z)-40(y z+x)+4 x y z+761 \leq 0 $$","$(6 ; 2 ; 7),(20 ; 0 ; 19)$",medium,"# Solution: $\left(x^{2} y^{2}+2 x y z+z^{2}\right)+\left(y^{2} z^{2}+2 x y z+x^{2}\right)-38(x y+z)-40(y z+x)+761 \leq 0$. $(x y+z)^{2}+(y z+x)^{2}-38(x y+z)-40(y z+x)+761 \leq 0$, $(x y+z)^{2}-2 \cdot 19(x y+z)+361-361+(y z+x)^{2}-2 \cdot 20(y z+x)+400-400+761 \leq 0$, $(x y+z-19)^{2}+(y z+x-20)^{2} \leq 0,\left\{\begin{array}{l}x y+z=19, \\ y z+x=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ y(z-x)-(z-x)=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ (z-x)(y-1)=1 .\end{array}\right.\right.\right.$ Since we need to find integer solutions to the inequality, there are only two cases: 1) $\left\{\begin{array}{c}x y+z=19, \\ z-x=1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{c}2 x+x+1=19, \\ z=x+1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{l}x=6, \\ y=2, \\ z=7 .\end{array}\right.\right.\right.$ 2) $\left\{\begin{array}{c}x y+z=19, \\ z-x=-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x-1=19, \\ z=x-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=20, \\ y=0, \\ z=19 .\end{array}\right.\right.\right.$ Answer: $(6 ; 2 ; 7),(20 ; 0 ; 19)$." b3b1841473b2,"$4 \cdot 26$ Given that the two roots of the equation $x^{2}+p x+1=0$ are the lengths of two sides of a triangle, and the third side is 2. Then the range of the real number $p$ is (A) $p<-2$. (B) $p \leqslant-2$ or $p \geqslant 2$. (C) $-2 \sqrt{2}2, \quad r+2>s, \quad s+2>r . $$ The last two inequalities can be reduced to $|r-s|<2$, \\ { p ^ { 2 } < 8 . } \end{array} \text { which gives } \quad \left\{\begin{array}{l} p<-2, \\ -2 \sqrt{2} b$. Thus, the requirement is: $$ a b = 10(a - b) $$ and through appropriate manipulation, $$ a = \frac{-10 b}{b - 10} = -10 - \frac{100}{b - 10} = \frac{100}{10 - b} - 10 $$ From this, we get a positive integer for $a$ if and only if the fraction in the last form is an integer greater than 10, which means $10 - b$ is a positive divisor of 100 that is less than 10, and in every such case, $b$ will also be a natural number. Therefore, the solutions are: $$ \begin{array}{rrrr} 10 - b = 1, & 2, & 4, & 5 \\ b = 9, & 8, & 6, & 5 \\ a = 90, & 40, & 15, & 10 \end{array} $$ By swapping these pairs, we get 4 new solutions, as the absolute value of the difference remains unchanged after the swap. Zoltán Szabó (Budapest, I. István Gymn., II. o. t.) Note. We can also achieve the goal by collecting the terms of (1) on one side and adding 100 to both sides, so that the multi-term side can be factored: $$ 10 a - 10 b - a b + 100 = (10 - b)(a + 10) = 100 $$" 860a3da2c8ff,"4. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=1$, for $n \geqslant 1$, we have $$ \begin{array}{l} 4\left(x_{1} x_{n}+2 x_{2} x_{n-1}+3 x_{3} x_{n-2}+\cdots+n x_{n} x_{1}\right) \\ =(n+1)\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n} x_{n+1}\right) . \end{array} $$ Find the general term formula for $\left\{x_{n}\right\}$. )",n$.,medium,"For $n=1$, substituting into equation (1) yields $4 x_{1}^{2}=2 x_{1} x_{2}$, thus $x_{2}=2$. For $n=2$, substituting into equation (1) yields $4\left(x_{1} x_{2}+2 x_{2} x_{1}\right)=3\left(x_{1} x_{2}+x_{2} x_{3}\right)$, thus $x_{3}=3$. Assume $x_{k}=k$, where $1 \leqslant k \leqslant n$. From $$ 4 \sum_{k=1}^{n} k x_{k} x_{n+1-k}=(n+1) \sum_{k=1}^{n} x_{k} x_{k+1} \text {, } $$ we get $4 \sum_{k=1}^{n} k^{2}(n+1-k)$ $$ \begin{array}{l} =(n+1) \sum_{k=1}^{n-1} k(k+1)+(n+1) n x_{n+1}, \\ 4(n+1) \cdot \frac{n(n+1)(2 n+1)}{6}-4 \cdot \frac{n^{2}(n+1)^{2}}{4} \\ =(n+1) \cdot \frac{(n-1) n(2 n-1)}{6}+ \\ \quad(n+1) \cdot \frac{(n-1) n}{2}+n(n+1) x_{n+1} . \end{array} $$ Therefore, $x_{n+1}=\frac{2(n+1)(2 n+1)}{3}-n(n+1)-$ $$ \begin{aligned} & \frac{(n-1)(2 n-1)}{6}-\frac{n-1}{2} \\ = & n+1 . \end{aligned} $$ In conclusion, $x_{n}=n$." 89cf8869bdc9,"What is the maximum number of planes of symmetry a tetrahedron can have? #",6,medium,"Show that each plane of symmetry must pass through an edge of the tetrahedron. ## Solution In a regular tetrahedron, there are 6 planes of symmetry: each of them passes through one of the edges and bisects the edge that intersects it. We will show that there cannot be more than six planes of symmetry. Immediately note that a plane of symmetry passes through an even number of vertices of the tetrahedron, as the remaining vertices must be divided into pairs that are symmetric with respect to this plane of symmetry. Suppose a plane of symmetry does not pass through any of the vertices of the tetrahedron. Then the 4 vertices of the tetrahedron must be divided into two pairs of vertices that are symmetric with respect to the plane of symmetry. Consequently, the lines passing through these pairs of vertices must be perpendicular to the plane of symmetry, i.e., parallel to each other. Thus, the four vertices of the tetrahedron must lie in the same plane, which is impossible. It is clear that all four vertices of the tetrahedron cannot lie in the plane of symmetry. Therefore, the plane of symmetry must pass through exactly two of the vertices (i.e., through an edge) of the tetrahedron. Suppose a plane of symmetry passes through one of the edges of the tetrahedron. Then the remaining two vertices are symmetric with respect to this plane, so this plane must pass through the midpoint of the edge connecting the remaining two vertices. The previous arguments show that there is no more than one plane of symmetry passing through a given edge of the tetrahedron. Thus, the total number of planes of symmetry is no greater than the number of edges of the tetrahedron, i.e., no more than six. ## Answer 6.00" 996b2f2ecbca,"Example 1 In $\triangle ABC$, it is known that $x \sin A + y \sin B + z \sin C = 0$. Find the value of $(y + z \cos A)(z + x \cos B)(x + y \cos C) + (y \cos A + z)(z \cos B + x)(x \cos C + y)$.","-(y+z \cos A)(z+x \cos B)(x+y \cos C)$, which simplifies to $(y+z \cos A)(z+x \cos B)(x+y \cos C)+(y",medium,"In $\triangle A B C$, $\sin C=\sin (A+B)=\sin A \cos B+\cos A \sin B$. Substituting the known conditions, we get $x \sin A+y \sin B+z(\sin A \cos B+\cos A \sin B)=0$, which simplifies to $\sin A(x+z \cos B)=-\sin B(y+z \cos A)$. Similarly, $\sin B(y+x \cos C)=-\sin C(z+x \cos B)$, $\sin C(z+y \cos A)=-\sin A(x+y \cos C)$, Multiplying the three equations, we have: $(x+z \cos B)(y+x \cos C)(z+y \cos A)=-(y+z \cos A)(z+x \cos B)(x+y \cos C)$, which simplifies to $(y+z \cos A)(z+x \cos B)(x+y \cos C)+(y \cos A+z)(z \cos B+x)(x \cos C+y)=0$." fe2b64a8bb11,3. Find the greatest and least values of the function $y=3 x^{4}+4 x^{3}+34$ on the interval $[-2 ; 1]$.,0 ; x=-1 . y(-2)=50 ; y(1)=41 ; y(-1)=33$.,easy,"Solution: $y=3 x^{4}+4 x^{3}+34 ; y^{\prime}=12 x^{3}+12 x^{2}=12 x^{2}(x+1)$ ![](https://cdn.mathpix.com/cropped/2024_05_06_8eeeb8d6ace591217257g-14.jpg?height=120&width=571&top_left_y=1362&top_left_x=1416) $x=0 ; x=-1 . y(-2)=50 ; y(1)=41 ; y(-1)=33$." b8431edc4c23,"The bisectors $A M$ and $B N$ of triangle $A B C$ intersect at point $O$. It is known that $A O=\sqrt{3} M O, N O=(\sqrt{3}-$ 1) $B O$. Find the angles of triangle $A B C$.","$60^{\circ}, 90^{\circ}, 30^{\circ}$",medium,"$A N: N B=N O: B O$. ## Solution Let $A B=a$. By the property of the angle bisector of a triangle, $A N: A B=N O: B O$. Therefore, $A N=a(\sqrt{3}-1)$. Similarly, $B M=\frac{a}{\sqrt{3}}$. If $C M=c$ and $C N=b$, then $\frac{a}{\sqrt{3}}+c=\frac{b}{\sqrt{3}-1}, c \sqrt{3}=a(\sqrt{3}-1)+b$. Multiplying the first equation by $\sqrt{3}-1$ and subtracting the second from it, we get $c=\frac{2 a}{\sqrt{3}}$. Thus, $B C=a \sqrt{3}$, $A C=C M \cdot A M /{ }_{M B}=2 a$. Since $A C^{2}=A B^{2}+B C^{2}$, triangle $A B C$ is a right triangle, $\angle B=90^{\circ}, \angle C=30^{\circ}, \angle A=60^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-35.jpg?height=365&width=766&top_left_y=229&top_left_x=652) Answer $60^{\circ}, 90^{\circ}, 30^{\circ}$." c89a1ef6b6c2,"34. Let $S=\{1,2,3, \ldots, 20\}$ be the set of all positive integers from 1 to 20 . Suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$, and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of the digits of $N$.",2 \times(6+8+4)=36$.,medium,"34. Answer: 36 We first find out which two consecutive numbers from $S$ are not factors of $N$. Clearly 1 is a factor of $N$. Note that if an integer $k$ is not a factor of $N$, then $2 k$ is not a factor of $N$ either. Therefore for $2 \leq k \leq 10$, since $2 k$ is in $S, k$ must be a factor of $N$, for otherwise, there would be at least three numbers from $S$ (the two consecutive numbers including $k$, and $2 k$ ) that are not factors of $N$. Hence $2,3, \ldots$, 10 are factors of $N$. Then it follows that $12=3 \times 4,14=2 \times 7,15=3 \times 5,18=2$ $\times 9,20=4 \times 5$ are also factors of $N$. Consequently, since the two numbers from $S$ that are not factors of $N$ are consecutive, we deduce that 11,13 , and 19 are factors of $N$ as well. Thus we conclude that 16 and 17 are the only two consecutive numbers from $S$ that are not factors of $N$. Hence $$ N=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13 \times 19=6846840, $$ so the sum of digits of $N=2 \times(6+8+4)=36$." 256fb8118c0b,"3. The perimeter of rectangle $A B C D$ is $66 \mathrm{~cm}$. Point $E$ is the midpoint of side $\overline{A B}$. The perimeter of triangle $\triangle E B C$ is $36 \mathrm{~cm}$, and the perimeter of quadrilateral ECDA is $60 \mathrm{~cm}$. What is the area of triangle $\triangle E B C$?",54 \text{ cm}^2$.,medium,"First method: Sketch: $1 \text{POINT}$ ![](https://cdn.mathpix.com/cropped/2024_05_30_f1ff0db3c8ac61ddb2e9g-3.jpg?height=234&width=494&top_left_y=1708&top_left_x=290) With the labels as shown in the figure, the perimeter of triangle $EBC$ is $x + y + z = 36 \text{ cm}$, the perimeter of quadrilateral $ECDA$ is $z + 2x + y + x = 60 \text{ cm}$, and the perimeter of rectangle $ABCD$ is $4x + 2y = 66 \text{ cm}$. If we substitute $x + y + z = 36$ into the expression $z + 2x + y + x = 60$, we get $2x = 24 \text{ cm}, \text{i.e.,} \quad x = 12 \text{ cm}$. From the condition $4x + 2y = 66$ and the fact that $x = 12$, we get $y = 9 \text{ cm}$. The area of triangle $EBC$ is equal to half the product of the lengths of its legs, i.e., $p(EBC) = xy / 2$. Substituting the values, we get $p(EBC) = 54 \text{ cm}^2$." ce367d2f3026,"## Task Condition Find the derivative of the specified order. $$ y=(3 x-7) \cdot 3^{-x}, y^{I V}=? $$",See reasoning trace,medium,"## Solution $y^{\prime}=\left((3 x-7) \cdot 3^{-x}\right)^{\prime}=\left((3 x-7) \cdot\left(e^{\ln 3}\right)^{-x}\right)^{\prime}=$ $=\left((3 x-7) \cdot e^{-x \cdot \ln 3}\right)^{\prime}=3 e^{-x \cdot \ln 3}-\ln 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}$ $y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(3 e^{-x \cdot \ln 3}-\ln 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}\right)^{\prime}=$ $=-3 \ln 3 \cdot e^{-x \cdot \ln 3}-\ln 3 \cdot 3 \cdot e^{-x \cdot \ln 3}+\ln ^{2} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}=$ $=-6 \ln 3 \cdot e^{-x \cdot \ln 3}+\ln ^{2} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}$ $y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-6 \ln 3 \cdot e^{-x \cdot \ln 3}+\ln ^{2} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}\right)^{\prime}=$ $=6 \ln ^{2} 3 \cdot e^{-x \cdot \ln 3}+3 \ln ^{2} 3 \cdot e^{-x \cdot \ln 3}-\ln ^{3} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}=$ $=9 \ln ^{2} 3 \cdot e^{-x \cdot \ln 3}-\ln ^{3} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}$ $y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(9 \ln ^{2} 3 \cdot e^{-x \cdot \ln 3}-\ln ^{3} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}\right)^{\prime}=$ $=-9 \ln ^{3} 3 \cdot e^{-x \cdot \ln 3}-3 \ln ^{3} 3 \cdot e^{-x \cdot \ln 3}-\ln ^{4} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}=$ $=-12 \ln ^{3} 3 \cdot e^{-x \cdot \ln 3}-\ln ^{4} 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}=$ $=-12 \ln ^{3} 3 \cdot 3^{-x}-\ln ^{4} 3 \cdot(3 x-7) \cdot 3^{-x}=(7 \ln 3-12-3 \ln 3 \cdot x) \cdot \ln ^{3} 3 \cdot 3^{-x}$ ## Problem Kuznetsov Differentiation 19-20" 89d41f29c251,12.287. The edges of a rectangular parallelepiped are in the ratio $3: 4: 12$. A diagonal section is made through the larger edge. Find the sine of the angle between the plane of this section and a diagonal of the parallelepiped not lying in it.,$\frac{24}{65}$,medium,"## Solution. Let $O$ and $O_{1}$ be the points of intersection of the diagonals of the bases of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.150), $A_{1} B_{1}=3 ; B_{1} C_{1}=$ $=4 ; B B_{1}=12$. Then $A_{1} C_{1}=\sqrt{A_{1} B_{1}^{2}+B_{1} C_{1}^{2}}=5$; $$ B_{1} D=\sqrt{A_{1} B_{1}^{2}+B_{1} C_{1}^{2}+B B_{1}^{2}}=13 $$ Draw a perpendicular $B_{1} F$ to $A_{1} C_{1}$ in the plane of the rectangle $A_{1} B_{1} C_{1} D_{1}$. Since the plane of the diagonal section $A A_{1} C_{1} C$ is perpendicular to the planes of the bases, $B_{1} F$ is perpendicular to the plane of the section, and the line $F E$ ( $F$ is the intersection point of $B_{1} D$ and the section - the midpoint of $O O_{1}$ ) is the projection of $B_{1} D$ onto the plane of the section, and $\angle B_{1} E F$ is the required angle between the plane of the section $A A_{1} C_{1} C$ and $B_{1} D$. $$ S_{\triangle A_{1} B_{1} C_{1}}=\frac{1}{2} A_{1} B_{1} \cdot B_{1} C_{1}=\frac{1}{2} A_{1} C_{1} \cdot B_{1} F \Rightarrow B_{1} F=\frac{A_{1} B_{1} \cdot B_{1} C_{1}}{A_{1} C_{1}}=\frac{12}{5} $$ In $\triangle B_{1} F E\left(\angle B_{1} F E=90^{\circ}\right)$ : $\sin \angle B_{1} E F=\frac{B_{1} F}{B_{1} E}=\frac{12}{5}: \frac{13}{2}=\frac{24}{65}$. Answer: $\frac{24}{65}$." a420f9d630c1,"Let $n$ be a positive integer. For each partition of the set $\{1,2,\dots,3n\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value that $S$ can attain? (An [i]arithmetic progression[/i] is a set of the form $\{a,a+d,\dots,a+kd\}$, where $a,d,k$ are positive integers, and $k\geqslant 2$; thus an arithmetic progression has at least three elements, and successive elements have difference $d$, called the [i]common difference[/i] of the arithmetic progression.)",n^2,medium,"To find the maximal value of the sum \( S \) of the common differences of the arithmetic progressions that partition the set \(\{1, 2, \dots, 3n\}\), we need to consider the structure of these progressions and how they can be arranged to maximize \( S \). 1. **Understanding the Problem:** - We need to partition the set \(\{1, 2, \dots, 3n\}\) into arithmetic progressions. - Each arithmetic progression has the form \(\{a, a+d, a+2d, \dots, a+kd\}\) where \(a, d, k\) are positive integers and \(k \geq 2\). - We need to maximize the sum \( S \) of the common differences \( d \) of these progressions. 2. **Formulating the Strategy:** - To maximize \( S \), we need to maximize the common differences \( d \) of the arithmetic progressions. - Consider the arithmetic progression \(\{a, a+n, a+2n\}\). Here, the common difference is \( n \). 3. **Constructing the Progressions:** - We can construct \( n \) arithmetic progressions each with a common difference of \( n \). - For example, the progressions can be: \[ \{1, 1+n, 1+2n\}, \{2, 2+n, 2+2n\}, \ldots, \{n, n+n, n+2n\} \] - Each of these progressions has a common difference of \( n \). 4. **Summing the Common Differences:** - Since there are \( n \) such progressions and each has a common difference of \( n \), the sum \( S \) of the common differences is: \[ S = n + n + \cdots + n = n^2 \] 5. **Verification:** - We need to ensure that this is indeed the maximal value. - Any other partitioning strategy would result in smaller common differences or fewer progressions, leading to a smaller sum \( S \). Thus, the maximal value that \( S \) can attain is \( n^2 \). The final answer is \(\boxed{n^2}\)" 8861b618e3e8,"2. (10 points) Jia, Yi, Bing, and Ding planted a total of 60 trees. It is known that, the number of trees planted by Jia is half of the rest three, the number of trees planted by Yi is one-third of the rest three, and the number of trees planted by Bing is one-fourth of the rest three. How many trees did Ding plant $\qquad$?",D planted 13 trees,medium,"【Analysis】According to the problem, the number of trees planted by A accounts for $\frac{1}{3}$ of the total, B accounts for $\frac{1}{4}$, and C accounts for $\frac{1}{5}$. Therefore, D accounts for $1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}=\frac{47}{60}$ of the total. We can further solve the problem using the meaning of fraction multiplication. 【Solution】Solution: $60 \times\left(1-\frac{1}{1+2}-\frac{1}{1+3}-\frac{1}{1+4}\right)$ $$ \begin{array}{l} =60 \times\left(1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) \\ =60 \times \frac{13}{60} \\ =13 \text { (trees) } \end{array} $$ Answer: D planted 13 trees. Therefore, the answer is: 13." 1c6d90ba011c,"27. [13] In $\triangle A B C, D$ and $E$ are the midpoints of $B C$ and $C A$, respectively. $A D$ and $B E$ intersect at $G$. Given that $G E C D$ is cyclic, $A B=41$, and $A C=31$, compute $B C$.",See reasoning trace,easy,"Answer: 49 Solution: By Power of a Point, $$ \frac{2}{3} A D^{2}=A D \cdot A G=A E \cdot A C=\frac{1}{2} \cdot 31^{2} $$ so $A D^{2}=\frac{3}{4} \cdot 31^{2}$. The median length formula yields $$ A D^{2}=\frac{1}{4}\left(2 A B^{2}+2 A C^{2}-B C^{2}\right), $$ whence $$ B C=\sqrt{2 A B^{2}+2 A C^{2}-4 A D^{2}}=\sqrt{2 \cdot 41^{2}+2 \cdot 31^{2}-3 \cdot 31^{2}}=49 . $$" 49167e2146b8,The lateral surface area of a right circular cylinder is in the ratio of $5:3$ to its base area; the diagonal of the axial section is $39 \mathrm{~cm}$. What is the total surface area and volume of the cylinder?,See reasoning trace,medium,"Let $r$ be the radius of the base circle, and $m$ the height of the cylinder. Then $$ 2 r \pi m: r^{2} \pi=5: 3 $$ or $$ 6 m=5 r $$ and $$ (2 r)^{2}+m^{2}=39^{2} $$ or $$ 4 r^{2}+\frac{25}{36} r^{2}=39^{2} $$ from which the positive value of $r$ is: $$ r=18 \mathrm{~cm} \text { and thus } m=15 \mathrm{~cm} . $$ Therefore, the total surface area of the cylinder is $$ F=2 r \pi m+2 r^{2} \pi=1188 \pi \mathrm{cm}^{2} $$ and its volume is $$ K=r^{2} \pi m=4860 \pi \mathrm{cm}^{3} $$ (Leó Spitzer, Budapest.) The problem was also solved by: Baján A., Basch R., Breuer P., Dénes M., Erdős V., Fried E., Koffer B., Lengyel K., Rosenberg E., Sichermann F., Szántó L., Szilárd V., Szóbel I., Ungar E., Vilcsek A., Viola R., Weinberger Gy." c2c9c4da2a10,"Four. (20 points) Let $n$ be a positive integer greater than 1. If there exists a positive integer $i (i < n)$ such that $$ \sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{n} b^{2} $$ is a perfect square, then $n$ is called a ""TI number."" For example, by $\left(1^{2}+2^{2}+\cdots+9^{2}\right)-\left(10^{2}+11^{2}\right)=8^{2}$, we know that 11 is a TI number. Question: Are 46, 36, and 26 TI numbers? Why?",See reasoning trace,medium,"(1) From $\sum_{a=1}^{39} a^{2}-\sum_{b=40}^{46} b^{2}=7569=87^{2}$, we know that 46 is a TI number. (2) From $\sum_{a=1}^{30} a^{2}-\sum_{b=31}^{36} b^{2}=52^{2}$, we know that 36 is a TI number. (3) By calculation, when $1 \leqslant i \leqslant 20$, $$ \sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{26} b^{2}<0 \text {, } $$ and $\square$ $$ \begin{array}{l} \sum_{a=1}^{21} a^{2}-\sum_{b=2}^{2 \infty} b^{2}=421, \\ \sum_{a=1}^{2} a^{2}-\sum_{b=3}^{\infty} b^{2}=1389, \\ \sum_{a=1}^{2} a^{2}-\sum_{b=24}^{\infty} b^{2}=2447, \\ \sum_{a=1}^{24} a^{2}-\left(25^{2}+26^{2}\right)=3599, \\ \sum_{a=1}^{25} a^{2}-26^{2}=4849, \end{array} $$ none of them are perfect squares. Therefore, 26 is not a TI number." cce0063fe150,"In the drawing below, each square in the figure has a side length of $1 \mathrm{~m}$. Determine the sum of the angles $\angle P B C+\angle Q A P+\angle Q D C$. ![](https://cdn.mathpix.com/cropped/2024_05_01_ffb48c2069f209c1311cg-05.jpg?height=552&width=534&top_left_y=1921&top_left_x=841) #",See reasoning trace,easy,"Solution By rotating the square $90^{\circ}$ counterclockwise, we can conclude that triangle $A B Q$ is equal to triangle $B P C$, and consequently, $\angle B A Q=\angle P B C$. Furthermore, by rotating the square $90^{\circ}$ clockwise, we can also conclude that triangle $A D P$ is equal to triangle $D Q C$ and that $\angle Q D C=\angle D A P$. Therefore, $$ \begin{aligned} & \angle P B C+\angle Q A P+\angle Q D C= \\ & \angle B A Q+\angle Q A P+\angle P A D=90^{\circ} . \end{aligned} $$" 731436e6e75e,"2. The sum of all four-digit natural numbers divisible by 45, where the tens digit is three times the hundreds digit, minus the sum of all three-digit natural numbers divisible by 18, where the hundreds digit is the largest one-digit prime number.",See reasoning trace,medium,"2. The sum of all four-digit natural numbers divisible by 45, where the tens digit is three times the hundreds digit, decreased by the sum of all three-digit natural numbers divisible by 18, where the hundreds digit is the largest one-digit prime number. ## Solution. A number is divisible by 45 if it is divisible by 9 and 5. A number is divisible by 9 if the sum of its digits is divisible by 9, and by 5 if its last digit is 0 or 5. Since the tens digit is three times the hundreds digit, the possible combinations of these two digits are: hundreds digit 0, tens digit 0 hundreds digit 1, tens digit 3 hundreds digit 2, tens digit 6 hundreds digit 3, tens digit 9 $1 \text{ POINT}$ The units digit of the desired four-digit numbers can be 0 or 5. If the units digit is 0, then (due to the rule of divisibility by 9) we have the following solutions: $9000, 5130, 1260$, and 6390. If the units digit is 5, then (due to the rule of divisibility by 9) we have the following solutions: 4005, 9135, 5265, and 1395. The sum of all four-digit numbers that satisfy the desired property is: $9000 + 5130 + 1260 + 6390 + 4005 + 9135 + 5265 + 1395 = 41580$. A number is divisible by 18 if it is divisible by 9 and 2. A number is divisible by 9 if the sum of its digits is divisible by 9, and by 2 if its last digit is 0, 2, 4, 6, or 8. Since the hundreds digit is the largest one-digit prime number, the hundreds digit is 7. The sum of all three-digit numbers that satisfy the desired property is: $720 + 702 + 792 + 774 + 756 + 738 = 4482$. Note. If the student, when determining the four-digit numbers, failed to notice that the tens and hundreds digits can both be zero, i.e., omitted the numbers 9000 and 4005, the sum is 28575, the difference is 24093, and the task should be graded with a maximum of 7 POINTS." a0c6f9e0c5d1,"4. The real number $b$ satisfies $|b|<3$, and there exists a real number $a$, such that $a0 \geqslant 4[a n]$, which is a contradiction! Therefore, $a>0$. Thus, $4(a n-1)n+a(a n-1)-1$, So $4 a>1+a^{2}-\frac{a+1}{n}$. And $4 a<1+a^{2}+\frac{4}{n}$, i.e., $1+a^{2}-\frac{a+1}{n}<4 a<1+a^{2}+\frac{4}{n}$. Let $n \rightarrow+\infty$, we get $1+a^{2}=4 a$, So $a=2-\sqrt{3}$ or $a=2+\sqrt{3}$. Let $n=1$, then $4[a]=1+[a[a]]$, and $a=2-\sqrt{3}$ does not satisfy $4[a]=1+[a[a]]$. Hence $a \neq 2-\sqrt{3}$. When $a=2+\sqrt{3}$, for any $n \in \mathbf{N}_{+}$, let $\left[\frac{n}{a}\right]=b,\left\{\frac{n}{a}\right\}=c$. Then, from $1+a^{2}=4 a$, we know $a=4-\frac{1}{a}$, so $$\begin{array}{l} n+[a[a n]]=\left[n+a\left[4 n-\frac{n}{a}\right]\right] \\ =[n+a(4 n-b-1)] \\ =[a(4 n-1+c)]=\left[\left(4-\frac{1}{a}\right)(4 n-1+c)\right] \\ =\left[4(4 n-1)-4\left(\frac{n}{a}-c\right)+\frac{1-c}{a}\right] \\ =4(4 n-1-b)=4\left[4 n-\frac{n}{a}\right]=4[a n] . \end{array}$$ This means that $a=2+\sqrt{3}$ satisfies the condition. Therefore, $a=2+\sqrt{3}$ is the only solution." c3eb835c4ce9,14th ASU 1980,See reasoning trace,medium,"33 Solution There cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a * are not red, but now the bold rectangle has at most 1 red square. Contradiction. Consider a red square. One of the two diagonally adjacent squares marked * must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any 9 x 11 rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining 9 x 10 rectangle can easily be partitioned into fifteen 3 x 2 rectangles, each with 2 red squares. 14th ASU 1980 © John Scholes jscholes@kalva.demon.co.uk 4 December 2003 Last corrected/updated 4 Dec 03" b24a98d430e8,"The value of $10^{2}+10+1$ is (A) 101 (B) 1035 (C) 1011 (D) 111 (E) 31",(D),easy,"Evaluating, $10^{2}+10+1=10 \times 10+10+1=100+10+1=111$. ANSWER: (D)" 0a6eb54e49b6,"7. (Average/Difficult) A geometric sequence has a nonzero first term, distinct terms, and a positive common ratio. If the second, fourth, and fifth terms form an arithmetic sequence, find the common ratio of the geometric sequence.",See reasoning trace,medium,"Answer: $\frac{1+\sqrt{5}}{2}$ Solution: Let $a_{1}$ be the first term and $r$ the common ratio of the geometric sequence. Since the second, fourth, and fifth terms form an arithmetic sequence, $$ \begin{aligned} a_{1} r^{3}-a_{1} r & =a_{1} r^{4}-a_{1} r^{3} \\ r^{3}-r & =r^{4}-r^{3} \\ 0 & =r^{4}-2 r^{3}+r \\ 0 & =r\left(r^{3}-2 r^{2}+1\right) \\ 0 & =r(r-1)\left(r^{2}-r-1\right) \\ r & =0,1, \frac{1 \pm \sqrt{5}}{2} \end{aligned} $$ Since the geometric sequence has distinct terms and a positive common ratio, the only possible value of the common ratio is $\frac{1+\sqrt{5}}{2}$." 37c6feebd8d9,"Consider a parameterized curve $ C: x \equal{} e^{ \minus{} t}\cos t,\ y \equal{} e^{ \minus{} t}\sin t\ \left(0\leq t\leq \frac {\pi}{2}\right).$ (1) Find the length $ L$ of $ C$. (2) Find the area $ S$ of the region bounded by $ C$, the $ x$ axis and $ y$ axis. You may not use the formula $ \boxed{\int_a^b \frac {1}{2}r(\theta)^2d\theta }$ here.",\sqrt{2,medium,"### Part (1): Finding the Length \( L \) of \( C \) 1. **Compute the derivatives:** \[ \frac{dx}{dt} = -e^{-t}\cos t - e^{-t}\sin t = -e^{-t}(\cos t + \sin t) \] \[ \frac{dy}{dt} = -e^{-t}\sin t + e^{-t}\cos t = e^{-t}(\cos t - \sin t) \] 2. **Set up the arc length integral:** \[ L = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] 3. **Substitute the derivatives:** \[ L = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(-e^{-t}(\cos t + \sin t)\right)^2 + \left(e^{-t}(\cos t - \sin t)\right)^2} \, dt \] 4. **Simplify the integrand:** \[ L = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t}(\cos t + \sin t)^2 + e^{-2t}(\cos t - \sin t)^2} \, dt \] \[ = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t} \left[ (\cos t + \sin t)^2 + (\cos t - \sin t)^2 \right]} \, dt \] 5. **Expand and combine terms:** \[ (\cos t + \sin t)^2 = \cos^2 t + 2\cos t \sin t + \sin^2 t = 1 + 2\cos t \sin t \] \[ (\cos t - \sin t)^2 = \cos^2 t - 2\cos t \sin t + \sin^2 t = 1 - 2\cos t \sin t \] \[ (\cos t + \sin t)^2 + (\cos t - \sin t)^2 = (1 + 2\cos t \sin t) + (1 - 2\cos t \sin t) = 2 \] 6. **Simplify the integral:** \[ L = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t} \cdot 2} \, dt = \int_{0}^{\frac{\pi}{2}} \sqrt{2} e^{-t} \, dt \] 7. **Evaluate the integral:** \[ L = \sqrt{2} \int_{0}^{\frac{\pi}{2}} e^{-t} \, dt = \sqrt{2} \left[ -e^{-t} \right]_{0}^{\frac{\pi}{2}} \] \[ = \sqrt{2} \left( -e^{-\frac{\pi}{2}} + 1 \right) = \sqrt{2} \left( 1 - e^{-\frac{\pi}{2}} \right) \] ### Part (2): Finding the Area \( S \) of the Region Bounded by \( C \), the \( x \)-axis, and the \( y \)-axis 1. **Set up the area integral:** \[ S = \int_{0}^{\frac{\pi}{2}} y \frac{dx}{dt} \, dt \] 2. **Substitute \( y \) and \( \frac{dx}{dt} \):** \[ y = e^{-t} \sin t, \quad \frac{dx}{dt} = -e^{-t}(\cos t + \sin t) \] \[ S = \int_{0}^{\frac{\pi}{2}} e^{-t} \sin t \cdot \left( -e^{-t}(\cos t + \sin t) \right) \, dt \] \[ = -\int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t (\cos t + \sin t) \, dt \] 3. **Distribute and simplify:** \[ = -\int_{0}^{\frac{\pi}{2}} e^{-2t} (\sin t \cos t + \sin^2 t) \, dt \] 4. **Split the integral:** \[ = -\int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t \cos t \, dt - \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin^2 t \, dt \] 5. **Evaluate each integral separately:** For the first integral: \[ \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t \cos t \, dt = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin 2t \, dt \] Using integration by parts or a suitable substitution, this integral evaluates to zero because the integrand is an odd function over the symmetric interval. For the second integral: \[ \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin^2 t \, dt \] Using the identity \(\sin^2 t = \frac{1 - \cos 2t}{2}\): \[ = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-2t} (1 - \cos 2t) \, dt \] \[ = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} e^{-2t} \, dt - \int_{0}^{\frac{\pi}{2}} e^{-2t} \cos 2t \, dt \right) \] 6. **Evaluate the first part:** \[ \int_{0}^{\frac{\pi}{2}} e^{-2t} \, dt = \left[ -\frac{1}{2} e^{-2t} \right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \left( e^{-\pi} - 1 \right) = \frac{1 - e^{-\pi}}{2} \] 7. **Evaluate the second part:** \[ \int_{0}^{\frac{\pi}{2}} e^{-2t} \cos 2t \, dt \] This integral also evaluates to zero because the integrand is an odd function over the symmetric interval. 8. **Combine the results:** \[ S = \frac{1}{2} \left( \frac{1 - e^{-\pi}}{2} \right) = \frac{1 - e^{-\pi}}{4} \] The final answer is \( \boxed{ \sqrt{2} \left( 1 - e^{-\frac{\pi}{2}} \right) } \) for the length and \( \frac{1 - e^{-\pi}}{4} \) for the area." 9e158f8638ed,"Example 10 Find all functions $f: \mathbf{Q} \rightarrow \mathbf{Q}$, satisfying the condition $f[x+f(y)]=f(x) \cdot f(y)(x, y \in \mathbf{Q})$.",0$ or $f(x)=1$.,medium,"Solution: Clearly, $f=0$ satisfies the condition. Suppose $f \neq 0$, take $f(y) \neq 0$, from $f(y)=f[y-f(y)] \cdot f(y)$, we get $f[y-f(y)]=1$, i.e., there exists $a \in \mathbf{Q}$, such that $f(a)=1$. Let $y=x$ and substitute into the given condition, we have $f(x+1)=f(x), x \in \mathbf{Q}$. Therefore, $f(x)$ is a periodic function, and 1 is one of its periods. Thus, $f(x+n)=f(x), n \in \mathbf{Z}$. Also, because $f[n \cdot f(y)]=f[(n-1) f(y)+f(y)]$ $$ =f[(n-1) f(y)] \cdot f(y)=\cdots=f(0) \cdot f^{n}(y), $$ Let $f(g)=\frac{q}{p}, p>0, p, q \in \mathbf{Z}$, then $f(0)=f(q)=f(0) \cdot f(y)$. Since $f(0) \neq 0$, then $f^{2}(y)=1, f(y)=1$ or -1. If $f(y)=-1$, then $f(x-1)=f(x) \cdot f(y)=-f(x)$, which still leads to $f(x)=0$, thus for all $x \in \mathbf{Q}, f(y)=1$. Hence, the required $f(x)$ is $f(x)=0$ or $f(x)=1$." 7d860afb0a5a,"6. Let $n \in \mathbf{Z}, f: \mathbf{Z} \rightarrow \mathbf{R}$ satisfy $f(n)=\left\{\begin{array}{ll}n-10, & n>100 \\ f(f(n+11)), & n \leqslant 100\end{array}\right.$ Among the statements above, the correct one is A. When $n \leqslant 100$, the value of $f(n)$ remains unchanged, and is 91 B. When $n \leqslant 100$, the value of $f(n)$ remains unchanged, and is 90 C. When $n \leqslant 100$, the value of $f(n)$ changes with $n$ D. When $n<90$, the value of $f(n)$ changes with $n$","f^{(2)}(n+11)=f(n+11-10)=f(n+1)$, thus, $f(90)=$ $f(91)=\cdots=f(101)=91$; If $n<90$, take $m \in \m",easy,"6. A Let $90 \leqslant n \leqslant 100$, then $f(n)=f^{(2)}(n+11)=f(n+11-10)=f(n+1)$, thus, $f(90)=$ $f(91)=\cdots=f(101)=91$; If $n<90$, take $m \in \mathbf{N}^{*}$, such that $90 \leqslant n+11 m \leqslant 100$, then $f(n)=f^{(2)}(n+11)$ $=\cdots=f^{(m+1)}(n+11 m)=f^{(m)}(f(n+11 m))=f^{(m)}(91)=91$." d2462928744a,"7.5. Vasya had 101 real coins. One of them was replaced with a fake one (it differs in weight, but it is unknown whether it is lighter or heavier than a real coin). Vasya wants to find as many real coins as possible in one weighing on a balance scale without weights. What should he do and how many real coins will he be able to find?",50 coins,medium,"Answer: 50 coins. Divide the coins into piles of $25 ; 25 ; 51$ and place the piles of 25 coins on the scales. If the scales are in balance, then these piles contained genuine coins, and we have found 50 genuine coins. If the scales are not in balance, then the genuine coins are in the third pile, and we have found 51 genuine coins. Thus, in any case, it is possible to find 50 genuine coins. We will prove that more cannot be found. If the same number of coins, not more than 24, is placed on the scales, then in the case of equality, it is possible to reliably determine no more than 48 genuine coins. If the same number of coins, not less than 26, is placed on the scales, then in the case of inequality, it is possible to determine no more than 49 genuine coins - these are the coins that were not on the scales. Comment. If a correct algorithm for determining a smaller number of genuine coins than 50 is proposed, the score is 1 point. If a correct algorithm for determining 50 genuine coins (without proving the maximum) is proposed, the score is 3 points." 472c6112ca18,"4. Let $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x\sqrt{6}-2$, so the maximum value of $f(x)$ on $[2,3]$ is $f(3)=15-4 \sqrt{6}$." 4305855a7f8a,"12. As shown in Figure 2, point $A$ is on the diagonal of the parallelogram. Determine the relationship between $S_{1}$ and $S_{2}$ ( ). (A) $S_{1}=S_{2}$ (B) $S_{1}>S_{2}$ (C) $S_{1}0 $$ gives the solution $r<199 / 3$. It implies that $$ a_{0}= 1. We consider first the simpler case of minimising (A-1) 2 + (B-1) 2 subject to AB = k 2 and A, B > 1. We show that it is achieved by taking A = B. This is routine. Set f(x) = (x - 1) 2 + (k 2 /x - 1) 2 and set f'(x) = 0, leading to a quartic (x - k)(x + k)(x 2 - x + k 2 ) = 0 with real roots k, -k. Note that f'(x) 0 for x just greater than k, so x = k is a minimum. Now it follows that all four of A, B, C, D must be equal. For if any two were unequal, we could keep the others fixed and reduce the sum of squares by equalising the unequal pair whilst keeping their product fixed. Thus the minimum value is achieved with A = B = C = D = √2, which gives value 12 - 8√2. 42nd Putnam 1981 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001" 3524f8419046,24. Cut a square piece of paper with a side length of 13 and reassemble it to form three squares with different integer side lengths. The sum of the perimeters of these three squares is $\qquad$ .,76,easy,Answer: 76 a18dc40a5668,"5.3. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{5}$. In your answer, specify its numerator.",77,easy,Answer: 77. Instructions. The required fraction: $\frac{77}{96}$. 71bfd2346e61,"4. Which of the following numbers is not an integer? A $\frac{2011}{1}$ B $\frac{2012}{2}$ C $\frac{2013}{3}$ D $\frac{2014}{4}$ E $\frac{2015}{5}$","6$ so, using the divisibility rule for divisibility by $3, \frac{2013}{3}$ is also an integer. Howev",easy,"4. D The numbers in options A, B and E are clearly integers. In option C, $2+0+1+3$ $=6$ so, using the divisibility rule for divisibility by $3, \frac{2013}{3}$ is also an integer. However, while 2000 is divisible by 4,14 is not so only $\frac{2014}{4}$ is not an integer." 7a838a8b4a05,"A circle and a square have the same perimeter. Then: $\text{(A) their areas are equal}\qquad\\ \text{(B) the area of the circle is the greater} \qquad\\ \text{(C) the area of the square is the greater} \qquad\\ \text{(D) the area of the circle is } \pi \text{ times the area of the square}\qquad\\ \text{(E) none of these}$",$\fbox{B}$,easy,"Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is 4^2 =16, so the answer is $\fbox{B}$." cc98b4c97a4a,1st Iberoamerican 1985,"( (a + b + c) 2 - (a 2 + b 2 + c 2 ) )/2 = 183, so a, b, c are roots of the cubic x 3 - 24x 2 + 183x",easy,"ab + bc + ca = ( (a + b + c) 2 - (a 2 + b 2 + c 2 ) )/2 = 183, so a, b, c are roots of the cubic x 3 - 24x 2 + 183x - 440 = 0. But it easily factorises as (x - 5)(x - 8)(x - 11) = 0, so the only solutions are permutations of (5, 8, 11). 1st Ibero 1985 © John Scholes jscholes@kalva.demon.co.uk 1 July 2002" ffe195e3c65d,"7. Let the edge length of a regular tetrahedron be $2 \sqrt{6}$, and let a sphere be constructed with the center $O$ of the tetrahedron as its center. The total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron is $4 \pi$. Then the radius of the sphere $O$ is $\qquad$",See reasoning trace,medium,"7. $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$. Let the radius of sphere $O$ be $R$. If a face of the regular tetrahedron intersects the sphere as shown in Figure 3, then the circumference of the small circle is $\pi$, and the radius of the small circle is $\frac{1}{2}$. Also, the distance from the center of the sphere to the face of the tetrahedron is 1, so $$ \begin{aligned} R & =\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}} \\ & =\frac{\sqrt{5}}{2} . \end{aligned} $$ If a face of the regular tetrahedron intersects the sphere as shown in Figure 4, let $D$ be the midpoint of $A C$. According to the problem, the length of arc $\overparen{A B}$ is $\frac{\pi}{3}$. Let the radius of the small circle $\odot O_{1}$ be $r$. $$ \begin{array}{l} \text { Then } \angle A O_{1} B=\frac{\pi}{3 r} . \\ \text { Also } \angle B O_{1} C=\frac{2 \pi}{3}, O_{1} D=\sqrt{2}, \\ \angle A O_{1} D=\frac{1}{2}\left(\angle B O_{1} C-\angle A O_{1} B\right) \\ =\frac{\pi}{3}-\frac{\pi}{6 r}, \end{array} $$ Thus, $\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)=\frac{\sqrt{2}}{r}$. Let $f(r)=\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)-\frac{\sqrt{2}}{r}$. Then $f^{\prime}(r)=-\frac{\pi}{6 r^{2}} \sin \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)+\frac{\sqrt{2}}{r^{2}}>0$. Therefore, the function $f(r)$ is monotonically increasing on the interval $(0,+\infty)$ and has at most one zero. Since $f(2)=0$, the equation (1) has a unique solution 2. Thus, $R=\sqrt{r^{2}+1}=\sqrt{5}$. In summary, the radius of sphere $O$ is $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$." c222d2e43164,"- I have 8 legs. And you only have 6. - I have 8 legs, - the dark blue one was offended. - And you only have 7. - The dark blue one really has 8 legs, - supported the purple one and boasted: - And I have 9! - None of you have 8 legs, - the striped octopus joined the conversation. - Only I have 8 legs! Who among the octopuses had exactly 8 legs? [7 points] (I.V. Raskina)",The striped octopus has 8 legs,medium,"Answer: The striped octopus has 8 legs. Solution: If the purple octopus is telling the truth, then it has an even number of legs. But in this case, it cannot say that it has 9 legs. Therefore, the purple octopus is lying. So, the dark blue octopus does not have 8 legs. But the dark blue octopus says that it has 8 legs, which means it is lying. Therefore, it has an odd number of legs. By saying that the dark blue octopus has 6 legs, the green one lied. Therefore, it lied the first time as well, and it does not have 8 legs. Thus, the first statement of the striped octopus is true. So, the second statement is also true, and it indeed has 8 legs. And the other octopuses have an odd number of legs." c2b8355e7c5e,"If $x=\frac{1}{4}$, which of the following has the largest value? (A) $x$ (B) $x^{2}$ (C) $\frac{1}{2} x$ (D) $\frac{1}{x}$ (E) $\sqrt{x}$",(D),easy,"If $x=\frac{1}{4}$, which of the following has the largest value? (A) $x$ (B) $x^{2}$ (C) $\frac{1}{2} x$ (D) $\frac{1}{x}$ (E) $\sqrt{x}$ Solution If we calculate the value of the given expressions, we get (A) $\frac{1}{4}$ (B) $\left(\frac{1}{4}\right)^{2}$ (C) $\frac{1}{2}\left(\frac{1}{4}\right)$ $=\frac{1}{16}$ $=\frac{1}{8}$ (D) $\frac{1}{\frac{1}{4}}$ (E) $\sqrt{\frac{1}{4}}$ $=1 \times 4 \quad=\frac{1}{2}$ $=4$ ANSWER: (D)" 6a4f34cbb26c,"1. Find all pairs of natural numbers $a, b$ for which the set equality holds $$ \{a \cdot[a, b], b \cdot(a, b)\}=\{45,180\} $$ where $(x, y)$ and $[x, y]$ denote the greatest common divisor and the least common multiple of numbers $x$ and $y$, respectively.","The sought pairs are two, namely $a=2, b=45$ and $a=6, b=15$",medium,"1. From the given equality, it follows that the number $b$ is odd (otherwise both numbers on the left would be even), and thus the number $a$ is even (otherwise both numbers on the left would be odd). The equality of sets must therefore be satisfied as follows: $$ a \cdot[a, b]=180 \quad \text { and } \quad b \cdot(a, b)=45 . $$ Since the number $a$ divides the number $[a, b]$, the number $180=2^{2} \cdot 3^{2} \cdot 5$ must be divisible by the square of (even) number $a$, so it must be either $a=2$ or $a=6$. In the case of $a=2$ (considering that $b$ is odd), we have $$ a \cdot[a, b]=2 \cdot[2, b]=2 \cdot 2 b=4 b $$ which means that the first equality in (1) is satisfied only for $b=45$. Then $b \cdot(a, b)=45 \cdot(2,45)=45$, so the second equality in (1) is also satisfied, and thus the pair $a=2, b=45$ is a solution to the problem. In the case of $a=6$, similarly, we get $$ a \cdot[a, b]=6 \cdot[6, b]=6 \cdot 2 \cdot[3, b]=12 \cdot[3, b], $$ which means that the first equality in (1) is satisfied precisely when $[3, b]=15$. This is satisfied only by the values $b=5$ and $b=15$. However, only the value $b=15$ satisfies the second equality in (1), which is now in the form $b \cdot(6, b)=45$. The second solution to the problem is thus the pair $a=6$, $b=15$, and no other solutions exist. Answer: The sought pairs are two, namely $a=2, b=45$ and $a=6, b=15$. Another solution. Let $d=(a, b)$. Then $a=u d$ and $b=v d$, where $u, v$ are coprime natural numbers, so $[a, b]=u v d$. From the equalities $$ a \cdot[a, b]=u d \cdot u v d=u^{2} v d^{2} \quad \text { and } \quad b \cdot(a, b)=v d \cdot d=v d^{2} $$ we see that the number $a \cdot[a, b]$ is a $u^{2}$-multiple of the number $b \cdot(a, b)$, so the given set equality can only be satisfied as we wrote the relations (1) in the original solution. These can now be expressed by the equalities $$ u^{2} v d^{2}=180 \quad \text { and } \quad v d^{2}=45 \text {. } $$ Therefore, $u^{2}=\frac{180}{45}=4$ or $u=2$. From the equality $v d^{2}=45=3^{2} \cdot 5$ it follows that either $d=1$ (and $v=45$) or $d=3$ (and $v=5$). In the first case, $a=u d=2 \cdot 1=2$ and $b=v d=45 \cdot 1=45$, in the second case, $a=u d=2 \cdot 3=6$ and $b=v d=5 \cdot 3=15$. Note. Since the given equality immediately implies that both numbers $a, b$ are divisors of the number 180 (even their least common multiple $[a, b]$ is such a divisor), it is possible to provide a complete solution by various other methods based on testing a finite number of specific pairs of numbers $a$ and $b$. We can speed up this process by first determining some necessary conditions that the numbers $a, b$ must satisfy. For example, the refinement of the set equality to a pair of equalities (1) can be (even without using the argument about the parity of the numbers $a, b$) explained by the general observation: the product $a \cdot[a, b]$ is divisible by the product $b \cdot(a, b)$, because their quotient can be written in the form $$ \frac{a \cdot[a, b]}{b \cdot(a, b)}=\frac{a}{(a, b)} \cdot \frac{[a, b]}{b} $$ i.e., as a product of two integers. For a complete solution, award 6 points. If the justification of the equalities (1) is missing in an otherwise complete solution (arguments $(a, b)<[a, b]$ or $(a, b) \mid[a, b]$ alone are insufficient), deduct 1 point. For finding both valid pairs (even by guessing), award 1 point, and the remaining (up to 3) points according to the quality of the provided solution process, especially its systematic nature." 27736d2dfc12,"2. a $W e g$ in the plane leads from the point $(0,0)$ to the point $(6,6)$, whereby in each step you can either go 1 to the right or 1 upwards. How many paths are there that contain neither the point $(2,2)$ nor the point $(4,4)$? ## Solution:",See reasoning trace,medium,"For $a, b \geq 0$, the number of paths from the point $(x, y)$ to the point $(x+a, y+b)$, in which one moves 1 to the right or upwards in each step, is generally equal to $\binom{a+b}{a}=$ $\binom{a+b}{b}$. This is because such a path is defined by specifying the positions of the $a$ steps to the right (or, equivalently, the positions of the $b$ steps upwards) in the sequence of $a+b$ steps required. There are therefore a total of $\binom{12}{6}$ paths from $(0,0)$ to $(6,6)$. Of these, exactly $\binom{4}{2}\binom{8}{4}$ contain the point $(2,2)$ and just as many contain the point $(4,4)$. Finally, exactly $\binom{4}{2}^{3}$ of these contain both points. According to the switch-on/switch-off formula, the required number of paths is therefore $$ \binom{12}{6}-\left[2\binom{8}{4}\binom{4}{2}-\binom{4}{2}^{3}\right]=300 $$" c4d67f34a037,"$18 \cdot 110$ As shown, $A B C D$ is an isosceles trapezoid, $A D$ $=B C=5, A B=4, D C=10$, point $C$ is on $D F$, and $B$ is the midpoint of the hypotenuse of the right triangle $\triangle D E F$. Then $C F$ equals (A) 3.25 . (B) 3.5 . (C) 3.75 . (D) 4.0 . (E) 4.25 .",$(D)$,easy,"[Solution] Draw $B H \perp D F$, then $B H / / E F$. Since $A B C D$ is an isosceles trapezoid, $$ \therefore \quad C H=\frac{1}{2}(D C-A B)=3, $$ then $$ D H=D C-C H=10-3=7 \text {. } $$ Since $B$ is the midpoint of $D E$, and $\frac{D B}{D E}=\frac{D H}{D F}$, $$ \therefore D F=14 \text {. } $$ Then $C F=D F-C D=4$. Therefore, the answer is $(D)$." 64e834e42137,"25.24. Let among the numbers $2^{1}, 2^{2}, \ldots, 2^{n}$ exactly $a_{n}$ numbers start with the digit 1. Compute the limit $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}$.",2^{m}$ by 2 and obtain a number within the same limits. The power of two following the smallest one ,medium,"25.24. Consider all natural numbers $a$ for which $10^{k-1} \leqslant a10^{k}$. The smallest power of two within these limits starts with a one, since otherwise we could divide the number $a=2^{m}$ by 2 and obtain a number within the same limits. The power of two following the smallest one starts with the digit 2 or 3. Therefore, among the considered numbers, there is exactly one power of two starting with one. Thus, if $10^{k-1} \leqslant 2^{n}<10^{k}$, then $a_{n}=k-1$. Therefore, $a_{n} \leqslant n \lg 2 < a_{n}+1$, i.e., $\lg 2 - \frac{1}{n} < \frac{a_{n}}{n} \leqslant \lg 2$. Hence, $\lim _{n \rightarrow \infty} \frac{a_{n}}{n} = \lg 2$." 966962b9678b,"You have 2 six-sided dice. One is a normal fair die, while the other has 2 ones, 2 threes, and 2 fives. You pick a die and roll it. Because of some secret magnetic attraction of the unfair die, you have a 75% chance of picking the unfair die and a 25% chance of picking the fair die. If you roll a three, what is the probability that you chose the fair die?",\frac{1,medium,"1. **Define the events:** - Let \( F \) be the event of choosing the fair die. - Let \( U \) be the event of choosing the unfair die. - Let \( R_3 \) be the event of rolling a three. 2. **Given probabilities:** - \( P(F) = 0.25 \) - \( P(U) = 0.75 \) 3. **Calculate the probability of rolling a three with each die:** - For the fair die, the probability of rolling a three is \( P(R_3 | F) = \frac{1}{6} \). - For the unfair die, the probability of rolling a three is \( P(R_3 | U) = \frac{2}{6} = \frac{1}{3} \). 4. **Use the law of total probability to find \( P(R_3) \):** \[ P(R_3) = P(R_3 | F)P(F) + P(R_3 | U)P(U) \] Substituting the given values: \[ P(R_3) = \left(\frac{1}{6}\right)\left(\frac{1}{4}\right) + \left(\frac{1}{3}\right)\left(\frac{3}{4}\right) \] \[ P(R_3) = \frac{1}{24} + \frac{3}{12} = \frac{1}{24} + \frac{6}{24} = \frac{7}{24} \] 5. **Calculate the conditional probability \( P(F | R_3) \) using Bayes' theorem:** \[ P(F | R_3) = \frac{P(R_3 | F)P(F)}{P(R_3)} \] Substituting the known values: \[ P(F | R_3) = \frac{\left(\frac{1}{6}\right)\left(\frac{1}{4}\right)}{\frac{7}{24}} = \frac{\frac{1}{24}}{\frac{7}{24}} = \frac{1}{7} \] \(\blacksquare\) The final answer is \( \boxed{ \frac{1}{7} } \)" 1179caf441ad,"【Question 6】Using 2 unit squares (unit squares) can form a 2-connected square, which is commonly known as a domino. Obviously, dominoes that can coincide after translation, rotation, or symmetry transformation should be considered as the same one, so there is only one domino. Similarly, the different 3-connected squares formed by 3 unit squares are only 2. Using 4 unit squares to form different 4-connected squares, there are 5. Then, using 5 unit squares to form different 5-connected squares, there are $\qquad$. (Note: The blank space at the end is left as in the original text for the answer to be filled in.)",See reasoning trace,easy,"Exam Point: Shape Cutting and Pasting Analysis: There are 12 different 5-ominoes (polyominoes made of 5 unit squares) that can be intuitively remembered using the English letters F, I, L, P, T, U, V, W, X, Y, Z, which represent 11 of these shapes." 7ebd68d490c8,"Senderov V.A. Find all natural numbers $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power). #",$k=1$,medium,"Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$. Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, then we can replace $n$ with $t$ and $a$ with $a^{s}$). Note that $n>2$, since $a^{2}+1$ cannot be divisible by $3=p_{2}$. We will show that $n>p_{k}$. Indeed, otherwise $n=p_{i}$, where $i \leq k$. Then $a^{p_{i}}+1$ is divisible by $p_{i}$; on the other hand, by Fermat's Little Theorem, $a^{p_{i}}-a$ is divisible by $p_{i}$. Since $a^{p_{i}}+1=(a+1)\left(a^{p_{i}-1}-a^{p_{i}-2}+\ldots-a+1\right)$, and $a+1=\left(a^{p_{i}}+1\right)-\left(a^{p_{i}}-a\right)$ is divisible by $p_{i}$, and $a^{p_{i}-1}-a^{p_{i}-2}+\ldots-a+1 \equiv 1+1+\ldots+1=p_{i} \equiv 0\left(\bmod p_{i}\right)$, then $a^{p_{i}}+1$ is divisible by $p_{i}^{2}$, which contradicts the condition. Thus, $a>p_{k}$ and $n>p_{k}$, from which $a^{n}+1>p_{k}^{p_{k}}>p_{1} p_{2} \ldots p_{k}$, which contradicts the equation (*). ## Answer $k=1$." 2fdbdcb291f7,"2. Find all integers $m, n$ such that $m n \mid \left(3^{m}+1\right), m n \mid \left(3^{n}+1\right)$.",See reasoning trace,medium,"First prove that at least one of $m$ and $n$ is odd, then prove that at least one of them is 1, and find that the only $(m, n)$ satisfying the condition are $(1,1),(1,2),(2,1)$." cd098caf72e2,(solved by Alice Héliou). The sum of twenty consecutive integers is 1030. What is the smallest of these integers?,"1030$, which gives $n=42$.",easy,". If $n$ denotes the smallest of these integers, the sum is: $$ n+(n+1)+(n+2)+\cdots+(n+19)=20 n+190 $$ We are thus reduced to solving the equation $20 n+190=1030$, which gives $n=42$." 5307b2c63486,"10.2. Solve the equation: $1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.",is provided - 1 point,medium,"# Solution. $1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this equation are the numbers 2 and -2, so the only root of the original equation is the number 2. ## Grading Criteria - Only the correct answer is provided - 1 point. - The correct solution process, but the extraneous root is not discarded - 3 points." 279f99563234,"13. The common divisors of two numbers that are not both 0 are called their common divisors. Find all the common divisors of $26019, 826, 2065$",See reasoning trace,easy,"$1,7,59,413$" f5464d864464,"18. In the diagram, $P Q R S$ is a square of side $10 \mathrm{~cm} . T$ is a point inside the square so that $\angle S P T=75^{\circ}$ and $\angle T S P=30^{\circ}$. What is the length of $T R$ ? A $8 \mathrm{~cm}$ B $8.5 \mathrm{~cm}$ C $9 \mathrm{~cm}$ D $9.5 \mathrm{~cm}$ E $10 \mathrm{~cm}$","75^{\circ}$ and $\angle T S P=30^{\circ}$, we obtain $\angle P T S=75^{\circ}$. Therefore $\triangle",medium,"18. E Draw in lines $P T$ and $T S$ as shown. Since angles in a triangle add to $180^{\circ}$ and we are given $\angle S P T=75^{\circ}$ and $\angle T S P=30^{\circ}$, we obtain $\angle P T S=75^{\circ}$. Therefore $\triangle P T S$ is isosceles and hence $T S=P S=10 \mathrm{~cm}$. Therefore, since $R S=10 \mathrm{~cm}$ as it is a side of the square, $\triangle R S T$ is also isosceles. Since $\angle R S P=90^{\circ}$ and $\angle T S P=30^{\circ}$, we have $\angle R S T=60^{\circ}$. Therefore $\triangle R S T$ is isosceles with one angle equal to $60^{\circ}$. Hence $\triangle R S T$ is equilateral and therefore the length of $T R$ is $10 \mathrm{~cm}$." 979febf0a6cc,"62. If positive integers $x, y, z$ satisfy $2^{x}+2^{y}+2^{z}=2336$, then $x+y+z=$",24,easy,Reference answer: 24 cdcd536fc943,"Bogdanov I.I. The distance between two cells on an infinite chessboard is defined as the minimum number of moves in the path of a king between these cells. On the board, three cells are marked, the pairwise distances between which are 100. How many cells exist such that the distances from them to all three marked cells are 50?",One cell,medium,"Consider two arbitrary cells $A$ and $B$. Let the difference in the abscissas of their centers be $x \geq 0$, and the difference in the ordinates be $y \geq 0$. Then the distance $\rho(A, B)$ between these cells is $\max \{x, y\}$. Let cells $A, B, C$ be marked. Then for each pair of cells, there exists a coordinate in which they differ by exactly 100. For two pairs of cells, this will be the same coordinate; for definiteness, let these be the pairs $(A, B)$ and $(A, C)$, differing horizontally. Then the abscissas of points $B$ and $C$ either differ by 200 or coincide. The first case is impossible, since $\rho(B, C)=100$. Therefore, their abscissas coincide, and their ordinates differ by 100. We can assume that the cells have coordinates $B(0,0), C(0,100), A(100, x)(0 \leq x \leq 100)$. Consider a point $X$ that is 50 units away from points $A, B$, and $C$. Its abscissa must be 50, otherwise $\rho(X, B)>50$ or $\rho(X, A)>50$. Similarly, the ordinate of $X$ is 50. Therefore, the coordinates of $X$ are $(50,50)$, and this cell fits. Thus, the sought cell is exactly one. ## Answer One cell" 5a99647ff5d5,"For what minimum value of $t$ does the inequality $\sqrt{x y} \leq t(2 x+3 y)$ hold for all non-negative real numbers $x, y$?",See reasoning trace,medium,"Solution. In the inequality of the problem, substituting 3 for $x$ and 2 for $y$, it must hold that $t \cdot 12 \geq \sqrt{6}$, and thus the value of $t$ must be at least $\frac{\sqrt{6}}{12}=\frac{1}{2 \sqrt{6}}$. We will show that the inequality holds for all such values of $t$ if $x$ and $y$ are non-negative numbers. By the inequality between the arithmetic and geometric means, for non-negative $2x$ and $3y$, $$ 2 x+3 y \geq 2 \sqrt{2 x \cdot 3 y}=2 \sqrt{6} \sqrt{x y} $$ If $t \geq 0$, then multiplying by $t$, $$ t(2 x+3 y) \geq t \cdot 2 \sqrt{6} \sqrt{x y} $$ If $t \geq \frac{1}{2 \sqrt{6}}$, that is, $t \cdot 2 \sqrt{6} \geq 1$, then the right-hand side of (1) is greater than or equal to $\sqrt{x y}$, so the inequality in question is true. The inequality in question, therefore, holds for all non-negative $x, y$ real numbers if and only if $t$ is at least $\frac{1}{2 \sqrt{6}}$. (Aron Kalcsu, Zalaegerszeg, Zrínyi Miklós Gymnasium, 12th grade) Note. The solutions essentially followed the above path. Of course, any substitution that results in equality in the inequality between the arithmetic and geometric means is suitable in the first step, that is, $2 x=3 y$. Something is simplified by the structure of the problem if we rearrange it so that neither side depends on the variables, for example, $$ A(x, y)=\frac{\sqrt{x y}}{2 x+3 y} \leq t $$ In this way, it becomes clearer that we are actually looking for the smallest upper bound of the function $A(x, y)$. The inequality between the arithmetic and geometric means implies that the function $A(x, y)$ has a maximum value, and this maximum value is precisely $\frac{1}{2 \sqrt{6}}$, which is the solution to the problem." b9475769a525,"1. Provide an example of a natural number that is divisible by 2019, and the sum of its digits is also divisible by 2019. Don't forget to show that your example meets the condition.",". For example, $20192019 \ldots 2019$ (2019 times)",easy,"1. Answer. For example, $20192019 \ldots 2019$ (2019 times). Grading criteria. Any correct example with verification - 7 points. Just an example without verification - 2 points. Incorrect example - $\mathbf{0}$ points." f9c9cb6c2c5c,"4. If $\log _{a} \frac{2}{3}<1$, then the range of values for $a$ is $(\quad)$. A. $\left(\frac{2}{3}, 1\right)$ B. $\left(\frac{2}{3},+\infty\right)$ C. $\left(0, \frac{2}{3}\right) \cup(1,+\infty)$ D. $\left(0, \frac{2}{3}\right) \cup\left(\frac{2}{3},+\infty\right)$",See reasoning trace,easy,"4. C. $\log _{a} \frac{2}{3}1 \\ \frac{2}{3}a,$ we get $01$. The translation is as follows: 4. C. $\log _{a} \frac{2}{3}1 \\ \frac{2}{3}a,$ we get $01$." f8f4a4c87917,,"6 \text{ cm}$ and $\overline{MQ}=16 \text{ cm}$ (see the diagram). Now, from $\overline{NP}=6 \text{",medium,"Solution. Let the endpoints of the given segment be $A$ and $B$, and let the points $C, D, E$ divide it into four unequal parts. Let $M$ be the midpoint of ![](https://cdn.mathpix.com/cropped/2024_06_05_dc4d34c0a9dd2a3ba572g-04.jpg?height=113&width=639&top_left_y=1155&top_left_x=570) $AC$, $N$ the midpoint of $CD$, $P$ the midpoint of $DE$, and $Q$ the midpoint of $EB$. Then, from the condition $\overline{NP}=6 \text{ cm}$ and $\overline{MQ}=16 \text{ cm}$ (see the diagram). Now, from $\overline{NP}=6 \text{ cm}$, we have that $\overline{CD}+\overline{DE}=2 \cdot 6=12 \text{ cm}$. From this and from $\overline{MQ}=16 \text{ cm}$, we get that $\overline{MC}+\overline{EQ}=16-12=4 \text{ cm}$, and since $M$ is the midpoint of $AC$ and $Q$ is the midpoint of $EB$, it follows that $\overline{AM}+\overline{QB}=4 \text{ cm}$. Finally, the length of the given segment is $\overline{AB}=(\overline{AM}+\overline{QB})+\overline{MQ}=4+16=20 \text{ cm}$." cddf3e111a73,"9. Let $\left\{a_{n}\right\}$ be a sequence of positive terms in a geometric progression, and let $S_{n}=\lg a_{1}+\lg a_{2}+\cdots+\lg a_{n}, \forall n \in \mathbf{N}^{*}$. If there exist distinct positive integers $m 、 n$, such that $S_{n}=S_{m}$, find $S_{m+n}$.",(m+n) a_{1}+\frac{(m+n)(m+n-1)}{2} d=(m+n)\left(a_{1}+\frac{m+n-1}{2} d\right)=0$.,easy,"9. Solution: Let $b_{n}=\lg a_{n}\left(n \in \mathbf{N}^{*}\right)$, then $\left\{b_{n}\right\}$ is an arithmetic sequence, with common difference $d$. Since $S_{n}=S_{m}$, $\therefore n a_{1}+\frac{n(n-1) d}{2}=m a_{1}+\frac{m(m-1) d}{2}$, which means $(n-m)\left(a_{1}+\frac{n+m-1}{2} d\right)=0$, or $a_{1}+\frac{m+n-1}{2} d=0$. Therefore, $S_{m+n}=(m+n) a_{1}+\frac{(m+n)(m+n-1)}{2} d=(m+n)\left(a_{1}+\frac{m+n-1}{2} d\right)=0$." 497cdc259265,"\section*{ In a parent-teacher meeting, exactly 18 fathers and exactly 24 mothers were present, with at least one parent of each student in the class attending. Of exactly 10 boys and exactly 8 girls, both parents were present for each. For exactly 4 boys and exactly 3 girls, only the mother was present, while for exactly 1 boy and exactly 1 girl, only the father was present. Determine the number of all those children in this class who have siblings in the same class! (There are no children in this class who have step-parents or step-siblings.)",See reasoning trace,medium,"} According to 3., 1 or 2 single fathers were present, so 16 or 17 fathers (out of the 18 present) were with the mother (4). According to 2., at most 7 mothers were alone, so out of the total 24 mothers, at least 17 mothers were with the father (5). From these two statements (4) and (5), it follows that there are exactly 17 parent couples, thus one single father and 7 single mothers remain (6). According to 1., both parents are present for 18 students, meaning 17 parent couples (6) belong to 18 children, so one of them must be a pair of siblings (7). Now, one single father remains for 2 children according to 3. and (4), so these children are also siblings. Therefore, there are 2 pairs of siblings in the class, which means 4 children have siblings in the class. \section*{Adopted from \([5]\)}" d10f7b380815,"5. If from the numbers $1,2, \cdots, 14$, we select $a_{1}, a_{2}, a_{3}$ in ascending order, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}$ $-a_{2} \geqslant 3$ are satisfied, then the total number of different ways to select them is $\qquad$ kinds.",See reasoning trace,medium,"120 5.【Analysis and Solution】Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$; $$ \begin{array}{l} T=\left\{\left(a_{1}, a_{2}, a_{3}\right) \mid a_{1}, a_{2}, a_{3} \in S, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3\right\}, \\ T^{\prime}=\left\{\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{S}^{\prime}\right) \mid a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{S}^{\prime}, a_{1}^{\prime}1-\frac{1}{10^{6}}-\frac{1}{10^{7}}-\ldots-\frac{1}{10^{99}}>1-\frac{2}{10^{6}}$. Consequently, $$ \begin{aligned} & a b0.890011 \cdot 0.999998>0.890009 \end{aligned} $$" eceee0e6811f,"Subject 3. Let $a, b \in \mathbb{C}$ and the function $f: \mathbb{C} \rightarrow \mathbb{C}$, defined by $f(z)=z^{2}+a|z|+b$. i) Determine $a, b \in \mathbb{C}$ such that $f(1)=f(2)=0$. ii) For $a, b$ determined in the previous point, find all complex numbers $z$ such that $f(z)=0$. Mathematical Gazette 1980",136&width=1404&top_left_y=905&top_left_x=339),easy,"Subject 3. i) $1+a+b=0$ (1p); $4+2 a+b=0$ (1p); $a=-3, b=2$ (1p). ii) substitute $z=x+i y$ into $z^{2}-3|z|+2=0$ : $(x+i y)^{2}-3 \sqrt{x^{2}+y^{2}}+2=0$ (1p); ![](https://cdn.mathpix.com/cropped/2024_06_07_a049303b8b5804c00308g-2.jpg?height=136&width=1404&top_left_y=905&top_left_x=339)" 515c98dbf0ae,"A Négyszögletű Kerek Erdő fái egy szabályos háromszögrács rácspontjain állnak. El lehet-e keríteni az erdőből egy téglalap alakú részt úgy, hogy a téglalap csúcsai rácspontok legyenek, és a téglalap határán ugyanannyi rácspont legyen, mint a belsejében? The trees of the Square Round Forest stand on the points of a regular triangular grid. Can a rectangular part be cut out of the forest such that the vertices of the rectangle are grid points, and there are as many grid points on the boundary of the rectangle as there are inside it?",See reasoning trace,medium,"Solution. We will show that such a rectangle can be drawn. Let the distance between the grid points be a unit distance. Then the distance between the points on the horizontal side of the rectangle is one unit, and on the vertical side, it is $\sqrt{3}$ units. Let the length of the horizontal side be $m$, and the length of the vertical side be $n \sqrt{3}$, where $n$ and $m$ are positive integers. The number of points on the perimeter of the rectangle is then $2(m+n)$. In the interior of the rectangle, on those horizontal grid lines that do not intersect the vertical sides at grid points, there are $m$ points, and there are $n$ such lines. Thus, the number of these points is $m \cdot n$. ![](https://cdn.mathpix.com/cropped/2024_05_02_8d8deaf30e718d872653g-1.jpg?height=420&width=504&top_left_y=430&top_left_x=800) On those horizontal grid lines that intersect the vertical sides at grid points, there are $m-1$ interior points, and there are $n-1$ such lines. Thus, the number of these points is $(m-1) \cdot(n-1)$. Therefore, in the interior of the rectangle, there are $m \cdot n+(m-1) \cdot(n-1)=2 m n-m-n+1$ grid points. Our condition is: $2(m+n)=2 m n-m-n+1$, or $2 m n-3 m-3 n+1=0$. Multiplying by 2 and factoring: $$ \begin{aligned} 4 m n-6 m-6 n+2 & =0 \\ (2 m-3)(2 n-3)-9+2 & =0 \\ (2 m-3)(2 n-3) & =7 \end{aligned} $$ Here, $2 m-3=1$ and $2 n-3=7$, or vice versa. Therefore, the sides of the rectangle that meet the condition are $m=2$ and $n=5$, or $m=5$ and $n=2$. In both rectangles, there are 14 trees on the perimeter and in the interior." 39e9c92413f7,"A [i]normal magic square[/i] of order $n$ is an arrangement of the integers from $1$ to $n^2$ in a square such that the $n$ numbers in each row, each column, and each of the two diagonals sum to a constant, called the [i]magic sum[/i] of the magic square. Compute the magic sum of a normal magic square of order $8$.",260,medium,"1. First, we need to find the sum of all integers from $1$ to $n^2$ for a normal magic square of order $n$. For $n = 8$, the integers range from $1$ to $8^2 = 64$. 2. The sum of the first $k$ positive integers is given by the formula: \[ S = \frac{k(k+1)}{2} \] Here, $k = 64$, so the sum of the integers from $1$ to $64$ is: \[ S = \frac{64 \cdot 65}{2} = 32 \cdot 65 = 2080 \] 3. In a normal magic square of order $n$, the total sum of all the integers is evenly divided among the $n$ rows, $n$ columns, and the two diagonals. Therefore, the magic sum $M$ is the sum of the integers in any row, column, or diagonal. 4. Since the total sum of all integers is $2080$ and there are $8$ rows (or columns), the magic sum $M$ is: \[ M = \frac{2080}{8} = 260 \] The final answer is $\boxed{260}$" 41adf0169a1b,"I4.4 In Figure 1 , let $O D C$ be a triangle. Given that $F H, A B$, $A C$ and $B D$ are line segments such that $A B$ intersects $F H$ at $G, A C, B D$ and $F H$ intersect at $E, G E=1, E H=c$ and $F H / / O C$. If $d=E F$, find the value of $d$.",2$,medium,"Remark: there are some typing mistakes in the Chinese old version: ... AC 及 $A D$ 為綫段 ...FH// $B C \ldots$ $\triangle A G E \sim \triangle A B C$ (equiangular) Let $\frac{C E}{A E}=k, A E=x, A G=t$. $B C=k+1, E C=k x, G B=k t$ (ratio of sides, $\sim \Delta$ 's) $\triangle D E H \sim \triangle D B C$ (equiangular) $\frac{B C}{E H}=\frac{k+1}{2}=\frac{D B}{D E} \quad$ (ratio of sides, $\sim \Delta$ 's) Let $D E=2 y \Rightarrow D B=(k+1) y$ $E B=D B-D E=(k-1) y$ $\triangle A F G \sim \triangle A O B$ (equiangular) $F G=d-1, \frac{O B}{F G}=\frac{A B}{A G}$ (ratio of sides, $\sim \Delta$ 's) $O B=(d-1) \cdot \frac{(k+1) t}{t}=(d-1)(k+1)$ $\triangle D F E \sim \triangle D O B$ (equiangular) $\frac{F E}{O B}=\frac{D E}{D B}$ (ratio of sides, $\sim \Delta$ 's) $\Rightarrow d=(d-1)(k+1) \cdot \frac{2 y}{(k+1) y} \Rightarrow d=2$ Method 2 $\triangle A F G \sim \triangle A O B$ and $\triangle A G E \sim \triangle A B C$ $\frac{d-1}{1}=\frac{O B}{B C}$ (ratio of sides, $\sim \Delta$ 's) $\triangle D F E \sim \triangle D O B$ and $\triangle D E H \sim \triangle D B C$ $\frac{d}{2}=\frac{O B}{B C}$ (ratio of sides, $\sim \Delta$ 's) Equating the two equations $\frac{d-1}{1}=\frac{d}{2}$ $d=2$" 82d5cd3cb1fb,"## Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$. $A(4, -2, 0)$ $B(1, -1, -5)$ $C(-2, 1, -3)$",See reasoning trace,easy,"## Solution Let's find the vector $\overrightarrow{BC}$: $\overrightarrow{BC}=\{-2-1 ; 1-(-1) ;-3-(-5)\}=\{-3 ; 2 ; 2\}$ Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be: $$ \begin{aligned} & -3 \cdot(x-4)+2 \cdot(y+2)+2 \cdot(z-0)=0 \\ & -3 x+12+2 y+4+2 z=0 \\ & -3 x+2 y+2 z+16=0 \end{aligned} $$ ## Problem Kuznetsov Analytic Geometry 9-3" 1b9a6f4c4066,"## Among all permutations $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the set $\{1,2, \ldots, n\}$, ($n \geq 1$ integer), consider those that satisfy $2\left(a_{1}+a_{2}+\cdots+a_{m}\right)$ is divisible by $m$, for each $m=1,2, \ldots, n$. Calculate the total number of such permutations.",See reasoning trace,medium,"Solution. Let $\mathcal{P}_{n}$ be the set of permutations of $\{1,2, \ldots, n\}$ that satisfy the conditions of the problem. The problem is to calculate $\left|\mathcal{P}_{n}\right|$. Observe that, for any $n$, the conditions are always satisfied for $m=1$, for $m=2$ and for $m=n$, so $\mathcal{P}_{1}, \mathcal{P}_{2}$ and $\mathcal{P}_{3}$ are, in each case, the set of all permutations and $\left|\mathcal{P}_{1}\right|=1,\left|\mathcal{P}_{2}\right|=2$ and $\left|\mathcal{P}_{3}\right|=6$. Suppose that $\left(a_{1}, \ldots, a_{n}\right) \in \mathcal{P}_{n}$. Taking $m=n-1$, it must be that $(n-1) \mid 2\left(a_{1}+\cdots+a_{n-1}\right)=2\left(a_{1}+\cdots+a_{n}\right)-2 a_{n}=n(n+1)-2 a_{n}$. Looking at this relation in terms of congruences, we have $2-2 a_{n} \equiv 0 \bmod (n-1)$, or that $2\left(a_{n}-1\right)$ is a multiple of $n-1$, which is equivalent to $a_{n}-1$ being a multiple of $\frac{n-1}{2}$. Given the obvious bound $a_{n}-1 \leq n-1$, the only values that $a_{n}-1$ can take are $0, \frac{n-1}{2}$ or $n-1$. Therefore, $a_{n}$ can only be $1, \frac{n+1}{2}$ or $n$. If $a_{n}=\frac{n+1}{2}$, then $n$ must be odd. The property of $\mathcal{P}_{n}$ for $m=n-2$ tells us, with a similar calculation as before, that $(n-2) \mid 2\left(a_{1}+\cdots+a_{n-2}\right)=n(n+1)-2 a_{n-1}-2 a_{n}=$ $(n-1)(n+1)-2 a_{n-1}$. Looking at this relation in terms of congruences modulo $(n-2)$, we get $3-2 a_{n-1} \equiv 0$ mod $(n-2)$, so $2 a_{n-1}-3$ must be a multiple of $n-2$ and this only happens if it is $n-1$. But this leads to $a_{n-1}=\frac{n+1}{2}=a_{n}$, which is absurd. In conclusion, $a_{n}$ can only take the values 1 and $n$. Let's study these two cases. Case $a_{n}=n$. Then $\left(a_{1}, \ldots, a_{n-1}\right)$ is a permutation of $\{1,2, \ldots, n-1\}$. It is easily verified that it is from $\mathcal{P}_{n-1}$. Therefore, there will be as many permutations of $\mathcal{P}_{n}$ with $a_{n}=n$ as there are permutations in $\mathcal{P}_{n-1}$. Case $a_{n}=1$. Now $a_{1}, a_{2}, \ldots, a_{n-1}>1$ and $\left(a_{1}-1, a_{2}-1, \ldots, a_{n-1}-1\right)$ is a permutation of $\mathcal{P}_{n-1}$. The correspondence $\left(a_{1}, a_{2}, \ldots, a_{n-1}, 1\right) \rightleftarrows\left(a_{1}-1, a_{2}-1, \ldots, a_{n-1}-1\right)$ is bijective. There will be as many permutations of $\mathcal{P}_{n}$ with $a_{n}=1$ as there are permutations in $\mathcal{P}_{n-1}$. In short, $\left|\mathcal{P}_{n}\right|=2\left|\mathcal{P}_{n-1}\right|$ if $n>3$, from which, $\left|\mathcal{P}_{n}\right|=3 \cdot 2^{n-2}$. Observation: The proof above gives us the recursive algorithm to obtain all permutations that satisfy the condition. For example, we know that the permutations of $\mathcal{P}_{3}$ are all. By adding a 4 to the end of each one, we get half of those of $\mathcal{P}_{4}$. The other half comes from adding 1 to each element of each permutation and adding a 1 at the end." 531c14ede99e,"[Circles] $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their [centers] at 1, 2, and 4, respectively. Line $t_1$ is a common internal [tangent] to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive [slope], and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$",027,medium,"[2006 II AIME-9.png](https://artofproblemsolving.com/wiki/index.php/File:2006_II_AIME-9.png) Call the centers $O_1, O_2, O_3$, the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$, and $s_2$ on $\mathcal{C}_2$), and the intersection of each common internal tangent to the [X-axis](https://artofproblemsolving.com/wiki/index.php/X-axis) $r, s$. $\triangle O_1r_1r \sim \triangle O_2r_2r$ since both triangles have a [right angle](https://artofproblemsolving.com/wiki/index.php/Right_angle) and have vertical angles, and the same goes for $\triangle O_2s_2s \sim \triangle O_3s_1s$. By [proportionality](https://artofproblemsolving.com/wiki/index.php/Proportion), we find that $O_1r = 4$; solving $\triangle O_1r_1r$ by the [Pythagorean theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_theorem) yields $r_1r = \sqrt{15}$. On $\mathcal{C}_3$, we can do the same thing to get $O_3s_1 = 4$ and $s_1s = 4\sqrt{3}$. The vertical [altitude](https://artofproblemsolving.com/wiki/index.php/Altitude) of each of $\triangle O_1r_1r$ and $\triangle O_3s_1s$ can each by found by the formula $c \cdot h = a \cdot b$ (as both products equal twice of the area of the triangle). Thus, the respective heights are $\frac{\sqrt{15}}{4}$ and $2\sqrt{3}$. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: $\sqrt{15 - \frac{15}{16}} = \frac{15}{4}$, and by 30-60-90: $6$. From this information, the slope of each tangent can be uncovered. The slope of $t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{4}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}$. The slope of $t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}$. The equation of $t_1$ can be found by substituting the point $r (4,0)$ into $y = \frac{1}{\sqrt{15}}x + b$, so $y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}$. The equation of $t_2$, found by substituting point $s (16,0)$, is $y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$. Putting these two equations together results in the desired $\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$ $\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}$ $= \frac{76 - 12\sqrt{5}}{4}$ $= 19 - 3\sqrt{5}$. Thus, $p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}$." 4d79bb53dcca,"## Task A-1.7. How many five-digit natural numbers are there whose product of digits is equal to 900?",See reasoning trace,medium,"## Solution. The prime factorization of the number 900 is $900=2^{2} \cdot 3^{2} \cdot 5^{2}$. Among the non-zero decimal digits, only the digit 5 is divisible by 5, so exactly two of the digits in the five-digit number must be 5. There are $\frac{5 \cdot 4}{2}=10$ ways to place the two 5s in some of the five positions in the five-digit number. In the five-digit number, we now need to determine the remaining three digits that multiply to $2^{2} \cdot 3^{2}=36$ and arrange them in the three remaining positions. Notice that $2^{2} \cdot 3^{2}=36$ can be factored into three digits in 5 ways: $$ 1 \cdot 4 \cdot 9, \quad 2 \cdot 3 \cdot 6, \quad 1 \cdot 6 \cdot 6, \quad 2 \cdot 2 \cdot 9, \quad 3 \cdot 3 \cdot 4 . \quad 1 \text { point } $$ In the first two cases, all three digits are different, so there are $3 \cdot 2 \cdot 1=6$ different ways to arrange these three digits in the remaining positions. In the last three cases, we have three digits where two are the same. Determining one of the three positions for the unique digit uniquely determines the positions of the remaining (identical) digits. Therefore, in these cases, there are 3 different arrangements of these digits. Thus, we conclude that the total number of five-digit numbers with the desired property is $$ 10 \cdot(2+2+3+3+3)=210 $$ Note: Each of the 5 cases in the above solution is worth 1 point, corresponding to the last five points in the above scoring scheme. If students do not consider the digits 5 separately but reduce the problem to the cases $1 \cdot 4 \cdot 9 \cdot 5 \cdot 5, \quad 2 \cdot 3 \cdot 6 \cdot 5 \cdot 5, \quad 1 \cdot 6 \cdot 6 \cdot 5 \cdot 5, \quad 2 \cdot 2 \cdot 9 \cdot 5 \cdot 5, \quad 3 \cdot 3 \cdot 4 \cdot 5 \cdot 5$, each of these cases is worth 2 points. ## SCHOOL COMPETITION IN MATHEMATICS ## 2nd grade - high school - A variant ## January 26, 2024. ## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO SCORE AND EVALUATE THAT APPROACH APPROPRIATELY." 3d2bc6fcdce2,"8. Given a positive integer $n(n \geqslant 2)$. Find the minimum value of $|X|$, such that for any $n$ two-element subsets $B_{1}$, $B_{2}, \cdots, B_{n}$ of the set $X$, there exists a subset $Y$ of $X$ satisfying: (1) $|Y|=n$; (2) for $i=1,2, \cdots, n$, we have $\left|Y \cap B_{i}\right| \leqslant 1$. Here, $|A|$ denotes the number of elements in the finite set $A$.",See reasoning trace,hard,"8. $|X|_{\min }=2 n-1$. (1) When $|X|=2 n-2$, it is not necessarily true that there exists a $Y$ that satisfies the conditions. In fact, let $X=\{1,2, \cdots, 2 n-2\}$, and consider a partition of $X$: $$ \begin{array}{l} B_{1}=\{1,2\}, B_{2}=\{3,4\}, \cdots \cdots \\ B_{n-1}=\{2 n-3,2 n-2\} . \end{array} $$ Since $|Y|=n$, there must be at least two elements in $Y$ that belong to the same $B_{j}$. In this case, $\left|Y \cap B_{j}\right|>1$, which is a contradiction. (2) Next, we prove that $|X|=2 n-1$ satisfies the condition. Let $B=\bigcup_{i=1}^{n} B_{i},|B|=2 n-1-z$, then there are $z$ elements that do not appear in any $B_{i}$, denoted as $a_{1}, a_{2}, \cdots, a_{i}$. If $z \geqslant n-1$, then we can take $Y=\left\{a_{1}, a_{2}, \cdots, a_{n-1}, d\right\}(d \in B)$. Assume $z0$, thus, $a>b$. Similarly, $b>c$." efbdd16ccda2,"If $T_n=1+2+3+\cdots +n$ and \[P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}\] for $n=2,3,4,\cdots,$ then $P_{1991}$ is closest to which of the following numbers? $\text{(A) } 2.0\quad \text{(B) } 2.3\quad \text{(C) } 2.6\quad \text{(D) } 2.9\quad \text{(E) } 3.2$","\frac{n(n+1)}{2}$, so we can put this in and simplify to get that the general term of the product is",easy,"$\fbox{D}$ It is well known that $T_n = \frac{n(n+1)}{2}$, so we can put this in and simplify to get that the general term of the product is $\frac{n(n+1)}{(n+2)(n-1)}.$ Now by writing out the terms, we see that almost everything cancels (telescopes), giving $\frac{3 \times 1991}{1993}$ which is almost $3$, so closest to $2.9.$" 09e4f4674706,"Example 3 Find all triples $(x, y, n), x, y, n \in \mathbf{N}^{\cdot}$ satisfying $\frac{x!+y!}{n!}=3^{n}$. Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.","(2, 1, 1), (1, 2, 1)$.",medium,"Analysis: From the $3^{n}$ on the right, we can preliminarily estimate that $n$ cannot be too large, so we need to further determine the range of $n$. Solution (1) is $$ x! + y! = 3^{n} \cdot n! \text{.} $$ Assume without loss of generality that $x \geqslant y$. The exponent of the factor 3 in $y!$ is $$ 3(y!) = \left[\frac{y}{3}\right] + \left[\frac{y}{3^{2}}\right] + \cdots + \left[\frac{y}{3^{n}}\right] + \cdots $$ The exponent of the factor 3 in $x!$ is $$ 3(x!) = \left[\frac{x}{3}\right] + \left[\frac{x}{3^{2}}\right] + \cdots + \left[\frac{x}{3^{n}}\right] + \cdots $$ Since $x \geqslant y$, we have $3(x!) \geqslant 3(y!)$. Therefore, the highest power of 3 on the left side of (2) is $3(y!)$, so $3(y!) \geqslant n$. Also, $\left[\frac{y}{3}\right] + \left[\frac{y}{3^{2}}\right] + \cdots + \left[\frac{y}{3^{n}}\right] + \cdots \leqslant \frac{y}{3} + \frac{y}{3^{2}} + \cdots + \frac{y}{3^{n}} + \cdots n$, so $y > 2n$. Thus, the left side of (2) $>(2n)! + (2n)! = n! \cdot 2 \cdot (2n)(2n-1) \cdots (n+1)$. If $n \geqslant 2$, then the left side $> n! \cdot 2 \cdot 3^{n} >$ the right side, so there is no solution. Therefore, $n = 1$, and $x! + y! = 3$; so $x = 1, y = 2$ or $x = 2, y = 1$. In summary, the solutions are $(x, y, n) = (2, 1, 1), (1, 2, 1)$." d10eb8f1ab04,"Solve the following system of equations: $$ \sqrt[3]{a^{y}}=\sqrt{\frac{a^{13}}{a^{x}}} $$ $$ \sqrt[7]{a^{3 x}} \cdot \sqrt[9]{a^{4 y}}=a^{7} $$",See reasoning trace,easy,"From (1): $$ \frac{y}{3}=\frac{13-x}{2} $$ From (2): $$ \frac{3 x}{7}+\frac{4 y}{9}=7 $$ Solving this system of equations: $x=7, y=9$. (Alfréd Haar, VII.o. t., Budapest.) Number of solutions: 25." 3657f3f0849a,"## 261. Math Puzzle $2 / 87$ The pioneers of class 6a are going on a group trip over the weekend. If each participant pays 12 marks, there will be a surplus of 33 marks compared to the required total amount. If each pays 10 marks, however, 11 marks will be missing. How many people are participating in the trip? How much does each person have to pay?","y: x=231: 22=10.50 \mathrm{M}$, each participant has to pay 10.50 Mark.",easy,"Number of participants $=x$, Total amount $=y$, Individual price $=z$ $$ y=x \cdot 12 \mathrm{M}-33 \mathrm{M} \quad, \quad y=x \cdot 10 \mathrm{M}+11 \mathrm{M} $$ with $x=22$. There are 22 people participating in the trip. $$ y=x \cdot 12 \mathrm{M}-33 \mathrm{M}=22 \cdot 12 \mathrm{M}-33 \mathrm{M}=231 \mathrm{M} $$ The total amount is $231 \mathrm{M}$. $z=y: x=231: 22=10.50 \mathrm{M}$, each participant has to pay 10.50 Mark." b736918a4d86,"2. Given the function $y=a \cos x+b$ has a maximum value of 1 and a minimum value of $-7$, the maximum value of $a \cos x+b \sin x$ is ( ). A. 1 B. 3 C. 5 D. 7","|a|+b=1, y_{\min }=-|a|+b=-7$, so we solve to get $|a|=4, b=3$. Then the maximum value of $a \cos x+",easy,"2. C. From the conditions, we know that $y_{\max }=|a|+b=1, y_{\min }=-|a|+b=-7$, so we solve to get $|a|=4, b=3$. Then the maximum value of $a \cos x+b \sin x$ is $\sqrt{a^{2}+b^{2}}=5$." 474c31cbcf0c,"8. (15 points) A thin ray of light falls on a thin converging lens at a distance of $x=10$ cm from its optical center. The angle between the incident ray and the plane of the lens $\alpha=45^{\circ}$, and the angle between the refracted ray and the plane of the lens $\beta=30^{\circ}$. Determine its focal length.",$\approx 13,medium,"Answer: $\approx 13.7$ cm. Solution. Parallel rays intersect at the focus, so $F$ is the focus of the given lens. ![](https://cdn.mathpix.com/cropped/2024_05_06_7a4bc87492c06fc34cd6g-06.jpg?height=480&width=302&top_left_y=1502&top_left_x=957) In triangle $O A F$: angle $F A O=30^{\circ}$, angle $O F A=15^{\circ}$, angle $A O F=135^{\circ}$. Therefore, $\frac{O F}{\sin 30^{\circ}}=\frac{A O}{\sin 15^{\circ}}$. We obtain that the focal length: $F C=O F \sin 45^{\circ}=A O \frac{\sin 30^{\circ}}{\sin 15^{\circ}} \sin 45^{\circ} \approx 13.7 \mathrm{~cm}$. ## Multidisciplinary Engineering Olympiad ""Star"" in Natural Sciences 10th grade Final Round $2022-2023$ Version 2 Problems, answers, and evaluation criteria" 48bb5d02caba,"Rubanov I.S. Petya came up with 1004 reduced quadratic trinomials $f_{1}, \ldots, f_{1004}$, among the roots of which are all integers from 0 to 2007. Vasya considers all possible equations $f_{i}=f_{j}$ ( $i \neq j$ ), and for each root found, Petya pays Vasya one ruble. What is the smallest possible income for Vasya?",0,medium,"Let $f_{1}(x)=x(x-2007), f_{2}(x)=(x-1)(x-2006), f_{1004}(x)=(x-1003)(x-1004)$. All these trinomials are pairwise distinct because they have different roots, but the coefficient of $x$ in each of them is -2007. Therefore, the difference between any two of them is a constant, different from 0, which means that no equation $f_{n}(x)=f_{m}(x)$ has solutions. Answer 0. Petya thought of a natural number and for each pair of its digits, he wrote down their difference on the board. After that, he erased some differences, and the numbers $2,0,0,7$ remained on the board. What is the smallest number Petya could have thought of? ## Solution Answer: 11138. Indeed, the number 11138 could have been thought of: $2=3-1,0=1-1,7=8-1$. Suppose the number thought of is $N1$, then among the differences of the digits, only the numbers $0,1, a, a-1$ appear, which is impossible. Otherwise, the number $N$ does not contain zeros, and $N=\overline{111 b c}$, where $b=2$ or $b=3$ (since $N<11138$). Then $c \geq 1+7=8$, from which $b \neq 3$. But if $b=2$, then among the differences of the digits, only the numbers $0,1, c-1, c-2$ appear. Contradiction. ## Answer 11138" 05469d91422f,"Task 1. (5 points) Solve the equation $x^{9}-22 x^{3}-\sqrt{21}=0$. #",See reasoning trace,medium,"# Solution. Rewrite the equation as $x^{9}-21 x^{3}-x^{3}-\sqrt{21}=0$. $$ \begin{aligned} & \text { then } \\ & x^{3}\left(x^{6}-21\right)-\left(x^{3}+\sqrt{21}\right)=0 \\ & x^{3}\left(x^{3}-\sqrt{21}\right)\left(x^{3}+\sqrt{21}\right)-\left(x^{3}+\sqrt{21}\right)=0 \\ & \left(x^{3}+\sqrt{21}\right)\left(x^{6}-x^{3} \sqrt{21}-1\right)=0 \\ & x^{3}+\sqrt{21}=0, \text { or } x^{6}-x^{3} \sqrt{21}-1=0 \\ & x_{1}=-\sqrt[6]{21} ; \quad x^{3}=\frac{\sqrt{21} \pm 5}{2}, \\ & x_{2,3}=\sqrt[3]{\frac{\sqrt{21} \pm 5}{2}} . \end{aligned} $$ $$ \text { Answer. }\left\{-\sqrt[6]{21} ; \sqrt[3]{\frac{\sqrt{21} \pm 5}{2}}\right\} $$" 013c66bfc3be,"5. In a regular 100-gon $A_{1} A_{2} A_{3} \ldots A_{100}$, the diagonals $\overline{A_{2} A_{100}}$ and $\overline{A_{1} A_{3}}$ intersect at point $S$. Determine the measure of the angle $\angle A_{2} S A_{3}$. ##",See reasoning trace,medium,"First method: Sketch: ![](https://cdn.mathpix.com/cropped/2024_05_30_4e3bea37fa3879b468d2g-4.jpg?height=363&width=925&top_left_y=1897&top_left_x=591) The measure of one interior angle of a regular 100-gon is $\alpha_{100}=\frac{(100-2) \cdot 180^{\circ}}{100}=176.4^{\circ}$ 1 POINT Triangle $\Delta A_{100} A_{1} A_{2}$ is isosceles. The angle $\alpha_{100}$ is the angle opposite the base. The base angles are equal in measure $\beta$ and it holds that $2 \beta=180^{\circ}-\alpha_{100}$. It follows that $\beta=1.8^{\circ}$. 2 POINTS Similarly, due to the isosceles nature of triangle $\triangle A_{3} A_{1} A_{2}$, the angle $\left|\angle A_{3} A_{1} A_{2}\right|=1.8^{\circ}$. 1 POINT The angle $\angle A_{2} S A_{3}$ is an exterior angle of triangle $\triangle A_{1} A_{2} S$, so it is equal to the sum of the two non-adjacent interior angles, i.e., $\left|\angle A_{2} S A_{3}\right|=\left|\angle S A_{1} A_{2}\right|+\left|\angle A_{1} A_{2} S\right|=1.8^{\circ}+1.8^{\circ}=3.6^{\circ}$. TOTAL 6 POINTS ##" 173d6d984397,"Steve begins at 7 and counts forward by 3 , obtaining the list $7,10,13$, and so on. Dave begins at 2011 and counts backwards by 5, obtaining the list 2011,2006,2001, and so on. Which of the following numbers appear in each of their lists? (A) 1009 (B) 1006 (C) 1003 (D) 1001 (E) 1011",s to see if they satisfy this requirement by subtracting 7 from each of them and then determining if the resulting number is divisible by 3,medium,"Steve counts forward by 3 beginning at 7 . That is, the numbers that Steve counts are each 7 more than some multiple of 3. We can check the given answers to see if they satisfy this requirement by subtracting 7 from each of them and then determining if the resulting number is divisible by 3 . We summarize the results in the table below. | Answers | Result after subtracting 7 | Divisible by 3? | | :---: | :---: | :---: | | 1009 | 1002 | Yes | | 1006 | 999 | Yes | | 1003 | 996 | Yes | | 1001 | 994 | No | | 1011 | 1004 | No | Of the possible answers, Steve only counted 1009, 1006 and 1003. Dave counts backward by 5 beginning at 2011. That is, the numbers that Dave counts are each some multiple of 5 less than 2011. We can check the given answers to see if they satisfy this requirement by subtracting each of them from 2011 and then determining if the resulting number is divisible by 5 . We summarize the results in the table below. | Answers | Result after being subtracted from 2011 | Divisible by $5 ?$ | | :---: | :---: | :---: | | 1009 | 1002 | No | | 1006 | 1005 | Yes | | 1003 | 1008 | No | | 1001 | 1010 | Yes | | 1011 | 1000 | Yes | Of the possible answers, Dave only counted 1006, 1001 and 1011. Thus while counting, the only answer that both Steve and Dave will list is 1006. ANswer: (B)" bec7b3310fa8,"In square $A B C D$, point $M$ lies on side $B C$, and point $N$ lies on side $A B$. Lines $A M$ and $D N$ intersect at point $O$. Find the area of the square, given that $D N=4, A M=3$, and the cosine of angle $D O A$ is $q$. #",See reasoning trace,medium,"Denote $$ \angle A O D=\alpha, \angle A M B=\angle D A M=\beta, \angle A D N=\gamma $$ and form a trigonometric equation with respect to $\gamma$. ## Solution Let the side of the square be $a$. Denote, $$ \angle A O D=\alpha, \angle A M B=\angle D A M=\beta, \angle A D N=\gamma $$ From the right triangles $A B M$ and $D A N$, we find that $$ \sin \beta=\frac{A B}{A M}=\frac{a}{3}, \cos \gamma=\frac{A D}{D N}=\frac{a}{4} $$ From this, it follows that $$ 4 \cos \gamma=3 \sin \beta=3 \sin \left(180^{\circ}-\alpha-\gamma\right)=3 \sin (\alpha+\gamma)=3 \sin \alpha \cos \gamma+3 \cos \alpha \sin \gamma $$ Dividing both sides of the equation $$ 4 \cos \gamma=3 \sin \alpha \cos \gamma+3 \cos \alpha \sin \gamma $$ by $\cos \gamma$, we get $$ 3 \sin \alpha+3 \cos \alpha \operatorname{tg} \gamma=4 $$ from which $$ \operatorname{tg} \gamma=\frac{4-3 \sin \alpha}{3 \cos \alpha}=\frac{4-3 \sqrt{1-q^{2}}}{3 q} $$ Then $$ \cos ^{2} \gamma=\frac{1}{1+\operatorname{tg}^{2} \gamma}=\frac{9 q^{2}}{9 q^{2}+16-24 \sqrt{1-q^{2}}+9\left(1-q^{2}\right)}=\frac{9 q^{2}}{25-24 \sqrt{1-q^{2}}} $$ Therefore, $$ S_{\mathrm{ABCD}}=a^{2}=(4 \cos \gamma)^{2}=\frac{144 q^{2}}{25-24 \sqrt{1-q^{2}}} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-48.jpg?height=515&width=535&top_left_y=1406&top_left_x=756) ## Answer $\frac{144 q^{2}}{25-24 \sqrt{1-q^{2}}}$ Submit a comment" 3a532b65fb22,"Determine the number of pairs of integers $(m, n)$ such that $$ \sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q} $$","r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $",medium,"Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then $$ n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2} $$ and $$ (m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q} $$ Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then $$ \sqrt{n^{2}-2016}=\frac{1}{2}\left(r^{2}-2 n\right) \in \mathbb{Q} $$ Hence, there is some nonnegative integer $p$ such that $n^{2}-2016=p^{2}$ and (1) becomes $2 n+2 p=r^{2}$. It follows that $2(n+p)=r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $2016=(n-p)(n+p)$ and $n+p$ is a divisor of 2016, larger than $\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^{2}=2(n+p)$ is a divisor of $2016=2^{5} \cdot 3^{2} \cdot 7$, larger than $2 \sqrt{2016}>88$. The only possibility is $r^{2}=2^{4} \cdot 3^{2}=12^{2}$. Hence, $n+p=72$ and $n-p=28$, and we conclude that $n=m=50$. Thus, there is only one such pair." c3c98acc6121,Oleg Mushkarov,"\frac{b}{4}$. Therefore $a=\frac{3 b}{4}, c=\frac{5 b}{4}$ and then $a^{2}+b^{2}=c^{2}$, i.e. $\Vara",medium,"12.2. Let $M$ be the midpoint of $B C$, let $I$ be the center of the excircle of $\triangle A B C$ tangent to $A B$ and let $T$ be its tangent point to the line $B C$. In the standard notation for $\triangle A B C$ we have $I M=\frac{a}{2}+r_{c}, I T=r_{c}$ and $M T=p-\frac{a}{2}$ since $B T=\stackrel{p}{p}-a$. It follows from the right $\triangle M I T$ that $$ \left(\frac{a}{2}+r_{c}\right)^{2}=r_{c}^{2}+\left(p-\frac{a}{2}\right)^{2} $$ ![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-143.jpg?height=584&width=598&top_left_y=1844&top_left_x=1114) that Then $a r_{c}=p(p-a)$. Since $r_{c}=\frac{S}{p-c}$, we obtain by using Heron's formula $$ a S=p(p-a)(p-c)=\frac{S^{2}}{p-b} $$ $$ a(p-b)=S $$ Since $a, b$ and $c$ form (in this order) an arithmetic progression, we have $a=$ $b-x, c=b+x$ and $$ p=\frac{3 b}{2}, p-a=\frac{b}{2}+x, p-b=\frac{b}{2}, p-c=\frac{b}{2}-x $$ Now by (1) and the Heron formula we obtain the equation $$ (b-x)^{2}=3\left(\frac{b^{2}}{4}-x^{2}\right) $$ which has a unique solution $x=\frac{b}{4}$. Therefore $a=\frac{3 b}{4}, c=\frac{5 b}{4}$ and then $a^{2}+b^{2}=c^{2}$, i.e. $\Varangle A C B=90^{\circ}$." 317e8c4a8d1c,"## Task B-2.7. Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by, $$ f(x)=\frac{2}{\sqrt[3]{x^{2}+4 x+4}+\sqrt[3]{x^{2}-4}+\sqrt[3]{x^{2}-4 x+4}} $$ What is $f(4)+f(8)+f(12)+\cdots+f(2024)$?",See reasoning trace,medium,"## Solution. Notice that the expression on the right side can be written as $$ f(x)=\frac{2}{\sqrt[3]{(x+2)^{2}}+\sqrt[3]{(x+2)(x-2)}+\sqrt[3]{(x-2)^{2}}} $$ Also, observe that the denominator is of the form $a^{2}+a b+b^{2}$, where $a=\sqrt[3]{x+2}, b=\sqrt[3]{x-2}$. By multiplying the numerator and the denominator by $a-b=\sqrt[3]{x+2}-\sqrt[3]{x-2}$, we will rationalize the denominator of the given fraction. Then we have $$ \begin{aligned} & f(x)=\frac{2}{\sqrt[3]{(x+2)^{2}}+\sqrt[3]{(x+2)(x-2)}+\sqrt[3]{(x-2)^{2}}} \cdot \frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{\sqrt[3]{x+2}-\sqrt[3]{x-2}}= \\ & =\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{\sqrt[3]{(x+2)^{3}}-\sqrt[3]{(x-2)^{3}}}= \\ & =\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{(x+2)-(x-2)}=\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{4}=\frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{2} \end{aligned} $$ It follows that $$ \begin{aligned} & f(4)+f(8)+f(12)+\cdots+f(2024)= \\ & =\frac{1}{2}(\sqrt[3]{6}-\sqrt[3]{2}+\sqrt[3]{10}-\sqrt[3]{6}+\sqrt[3]{14}-\sqrt[3]{10}+\cdots+\sqrt[3]{2026}-\sqrt[3]{2022})= \\ & =\frac{1}{2}(\sqrt[3]{2026}-\sqrt[3]{2}) \end{aligned} $$ ## SCHOOL COMPETITION IN MATHEMATICS 3rd grade - high school - B variant January 4, 2024. ## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER." 7d211535fcaf,"Example 1 Given the equation about $x$: $\lg (x-1)+\lg (3-x)=\lg (a-x)$ has two distinct real roots, find the range of values for $a$.",See reasoning trace,easy,"Analysis We will convert the logarithmic equation into a quadratic equation problem and pay attention to the domain of the function. Solution The original equation $\Leftrightarrow\left\{\begin{array}{l}10 \\ f(1)>0 \\ f(3)>0\end{array}\right.$, i.e., $\left\{\begin{array}{l}13-4 a>0 \\ a-1>0 \\ a-3>0\end{array}\right.$, so $33$, the remainder under consideration is 4 - not square-free. So $q=3$ and $p=3^{k}+4$ for some $k \in \mathbb{N}$. Analogously, the prime divisors of $p-8$ are 5 or 7 , and these of $p-9$ are 2 or 7 . Since 7 cannot be a divisor in both cases, we get that $p=5^{m}+8$ or $p=2^{n}+9$. Hence $3^{k}=5^{m}+4$ or $3^{k}=2^{n}+5$. In the first case, $3^{k} \equiv 1(\bmod 4)$, i.e. $k=2 k_{1}$, and so $\left(3^{k_{1}}-2\right)\left(3^{k_{1}}+2\right)=5^{m}$ which gives $k_{1}=1, m=1$, i.e. $k=2$. In the second case one has that $n \geq 2$ and we conclude as above (using modulo 4 and 3) that $k=2 k_{1}$ and $n=2 n_{1}$ are even integers. Then $\left(3^{k_{1}}-2^{n_{1}}\right)\left(3^{k_{1}}+2^{n_{1}}\right)=5$ which implies $k_{1}=n_{1}=1$, i.e. $k=n=2$. Thus, in both cases, new solutions do not appear." faf1a60f67ff,"5. Inside a regular hexagon ABCDEF with an area of $30 \mathrm{~cm}^{2}$, a point $M$ is chosen. The areas of triangles $ABM$ and $BCM$ are $3 \mathrm{~cm}^{2}$ and $2 \mathrm{~cm}^{2}$, respectively. Determine the areas of triangles CDM, DEM, EFM, and FAM.",See reasoning trace,medium,"SOLUTION. The problem deals with the areas of six triangles into which a given regular hexagon is divided by connecting its vertices with point $M$ (Fig.3). The entire hexagon, with a given area denoted as $S$, can also be divided into six equilateral triangles, each with an area of $S / 6$ (Fig. 4). If we denote $r$ as the side length of these triangles, $v$ as the distance between the parallel lines $A B$ and $D E$, and $v_{1}$ as the distance from point $M$ to the line $A B$, we get $$ S_{A B M}+S_{E D M}=\frac{1}{2} r v_{1}+\frac{1}{2} r\left(v-v_{1}\right)=\frac{1}{2} r v=\frac{S}{3}, $$ since $S / 3$ is the sum of the areas of two shaded equilateral triangles. Due to symmetry, the sums $S_{B C M}+S_{E F M}$ and $S_{C D M}+S_{F A M}$ also have the same value of $S / 3$. From this, we can determine the first two unknown areas: $S_{D E M}=S / 3-S_{A B M}=7 \mathrm{~cm}^{2}$ and $S_{E F M}=S / 3- S_{B C M}=8 \mathrm{~cm}^{2}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-8.jpg?height=516&width=577&top_left_y=2152&top_left_x=385) Fig. 3 ![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-8.jpg?height=517&width=577&top_left_y=2149&top_left_x=1096) Fig. 4 How can we determine the remaining two areas $S_{C D M}$ and $S_{F A M}$, given that we only know their sum $S / 3$ so far? Notice that the sum of the given areas of triangles $A B M$ and $B C M$ has a significant value of $S / 6$, which is also the area of triangle $A B C$ (as seen in Fig.4). This equality of areas means that point $M$ lies on the diagonal $A C$. Triangles $A B M$ and $B C M$ thus have the same heights from the common vertex $B$, and the same applies to the heights of triangles $C D M$ and $F A M$ from vertices $F$ and $D$ (points that are equidistant from the line $A C$). For the ratios of the areas of these pairs of triangles, we get $$ \frac{S_{C D M}}{S_{F A M}}=\frac{|C M|}{|A M|}=\frac{S_{B C M}}{S_{A B M}}=\frac{2}{3} $$ In the sum $S_{C D M}+S_{F A M}$, which has a value of $S / 3$, the addends are in the ratio $2: 3$. Therefore, $S_{C D M}=4 \mathrm{~cm}^{2}$ and $S_{F A M}=6 \mathrm{~cm}^{2}$. GUIDING AND SUPPLEMENTARY PROBLEMS:" 7bef5a999e48,"2. For real numbers $x$ and $y$, where $x \neq 0, y \notin\{-2,0,2\}$ and $x+y \neq 0$, simplify the expression: $$ \frac{x y^{2018}+2 x y^{2017}}{y^{2016}-4 y^{2014}} \cdot\left(\left(\frac{x^{2}}{y^{3}}+x^{-1}\right):\left(x y^{-2}-\frac{1}{y}+x^{-1}\right)\right): \frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}}-\frac{y^{2}+2 y}{y+2} $$ Time for solving: 45 minutes.",See reasoning trace,medium,"2. The numerator of the first fraction $x y^{2018}+2 x y^{2017}$ is transformed into $x y^{2017}(y+2)$, and the denominator $y^{2016}-$ $4 y^{2014}$ is transformed into $y^{2014}(y-2)(y+2)$. We simplify the first fraction and get $\frac{x y^{3}}{y-2}$. The expression $\frac{x^{2}}{y^{3}}+x^{-1}$ is transformed into $\frac{x^{3}+y^{3}}{x y^{3}}$. We factorize the sum of cubes. The expression $x y^{-2}-\frac{1}{y}+x^{-1}$ is transformed into $\frac{x^{2}-x y+y^{2}}{x y^{2}}$. The expression $\frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}}$ is transformed into $\frac{x(x+y)^{2}}{x+y}=x(x+y)$. We factorize and simplify the last fraction, and get $y$. We simplify the first term and get $\frac{y^{2}}{y-2}$. We subtract $y$ from this expression and get $\frac{2 y}{y-2}$. Factoring $x y^{2018}+2 x y^{2017}=x y^{2017}(y+2)$ ..... 1 point Factoring and factorizing $y^{2016}-4 y^{2014}=y^{2014}\left(y^{2}-4\right)=y^{2014}(y-2)(y+2)$ ..... 1 point Transforming the first fraction into $\frac{x y^{3}}{y-2}$ ..... 1 point Transforming $\left(\frac{x^{2}}{y^{3}}+x^{-1}\right) \text{into} \frac{x^{3}+y^{3}}{x y^{3}}$ ..... 1 point Factorizing the sum of cubes ..... 1 point Transforming $x y^{-2}-\frac{1}{y}+x^{-1} \text{into} \frac{x^{2}-x y+y^{2}}{x y^{2}}$ ..... 1 point Transforming $\frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}} \text{into} \frac{x(x+y)^{2}}{x+y}=x(x+y)$ ..... 1 point Transforming $\frac{y^{2}+2 y}{y+2} \text{into} y$ ..... 1 point Considering division, simplifying fractions to get $\frac{y^{2}}{y-2}$ from the first term ..... 1 point Calculating $\frac{y^{2}}{y-2}-y=\frac{2 y}{y-2}$ ..... 1 point ## 18th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools Preliminary Competition, March 15, 2018 ## Solutions for the Second Year A competitor who arrives at the solution by any correct method (even if the scoring guide does not provide for it) receives all possible points. A correct method is considered to be any procedure that: - sensibly takes into account the wording of the problem, - leads to the solution of the problem, - is mathematically correct and complete. If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided for. If, however, the solution is guessed (cannot be arrived at without calculation), it is scored with 0 points. A competitor who has only partially solved the problem, but from which the correct procedures for solving the problem are not visible, cannot receive more than half of the possible points. The mark '*' next to the points means that the point or points can be awarded to the competitor for a correct procedure, even if the calculation is incorrect. " 69a808643bbc,"4. Given a spatial quadrilateral $A B C D, A B=a, B C=$ $b, C D=c, D A=d$. Then $\overrightarrow{A C} \cdot \overrightarrow{B D}=(\quad)$. (A) $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$ (B) $-\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$ (C) $\frac{1}{2}\left(a^{2}+c^{2}-b^{2}-d^{2}\right)$ (D) $\frac{1}{2}\left(b^{2}+d^{2}-a^{2}-c^{2}\right)$",See reasoning trace,medium,"4.D. $$ \begin{array}{l} \overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C}) \cdot(\overrightarrow{B C}+\overrightarrow{C D}) \\ =\overrightarrow{B C}^{2}+\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C D}+\overrightarrow{C D} \cdot \overrightarrow{A B} \\ =\overrightarrow{B C}^{2}+\frac{(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D})^{2}-\overrightarrow{A B}^{2}-\overrightarrow{B C}^{2}-\overrightarrow{C D}^{2}}{2} \\ =\overrightarrow{B C}^{2}+\frac{\left(-\overrightarrow{D A}^{2}-\overrightarrow{A B}^{2}-\overrightarrow{B C}^{2}-\overrightarrow{C D}^{2}\right)}{2} \\ =\frac{\overrightarrow{B C}^{2}+\overrightarrow{D A}^{2}-\overrightarrow{A B}^{2}-\overrightarrow{C D}^{2}}{2}=\frac{b^{2}+d^{2}-a^{2}-c^{2}}{2} . \end{array} $$" e7ec5c65ea86,"Three, find all positive integer solutions $(x, y, z)$ of the indeterminate equation $1+2^{x} \times 7^{y}=z^{2}$.",See reasoning trace,medium,"Three, the original equation can be transformed into $$ 2^{x} \times 7^{y}=z^{2}-1=(z+1)(z-1) \text {. } $$ It is easy to see that $z$ is an odd number, and $z+1$ and $z-1$ are two consecutive even numbers, with only one of them being a multiple of 4. Since $(z+1)-(z-1)=2$, they cannot both be multiples of 7. From the first equation, we get $2=2 \times 7^{y}-2^{x-1}$. Clearly, $x \neq 1,2$. When $x \geqslant 3$, $7^{y}-2^{x-2}=1$, taking both sides modulo 7 gives $$ 2^{x-2}=6(\bmod 7) \text {. } $$ However, $2^{k} \equiv 1,2,4(\bmod 7)$, so this equation does not hold. From the second equation, we get $2=2^{x-1}-2 \times 7^{y}$, it is easy to see that $x=5, y=1$ is a solution, in this case, $$ z=1+2 \times 7=15 \text {. } $$ When $x \geqslant 6$, $2^{x-2}-7^{y}=1$, taking both sides modulo 16 gives $$ 7^{y} \equiv 15(\bmod 16) \text {. } $$ However, $7^{2} \equiv 1(\bmod 16)$, so for any positive integer $y$, $$ 7^{y} \equiv 1,7(\bmod 16), $$ there is no solution. In summary, the complete solution to the indeterminate equation is $(5,1,15)$." bd635c257276,"94. Person A and Person B start from points A and B respectively, traveling towards each other at speeds of 65 meters/min and 55 meters/min. They meet after 10 minutes. The distance between A and B is $\qquad$ meters, and the meeting point is $\qquad$ meters away from the midpoint between A and B.","1200,50",easy,"Reference answer: 1200,50" 38f25b3b1fde,"V-2 If one side of the rectangle is reduced by $3 \mathrm{~cm}$, and the other side is reduced by $2 \mathrm{~cm}$, we get a square whose area is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle. Calculate the dimensions of the rectangle. ![](https://cdn.mathpix.com/cropped/2024_06_05_627a0068487f082c2c6cg-1.jpg?height=219&width=271&top_left_y=1438&top_left_x=1208)","21$ is derived, 5 points are awarded.",medium,"Solution. If we denote the length of the side of the square by $a$, then the dimensions of the rectangle are $a+3$ and $a+2$. From the diagram, it is seen that the rectangle is divided into a square with area $a^{2}$ and three rectangles with areas $2a$, $3a$, and 6. Since the area of the square is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle, it follows that $2a + 3a + 6 = 21$, or $5a = 15$. From this, we get $a = 3 \mathrm{~cm}$. Therefore, the dimensions of the rectangle are $6 \mathrm{~cm}$ and $5 \mathrm{~cm}$. Note: For a diagram from which the equation $2a + 3a + 6 = 21$ is derived, 5 points are awarded." f078d513c949,"5. Given that $n$ is a two-digit natural number. $n$ divided by 9 leaves a remainder of 1, and $n$ divided by 10 leaves a remainder of 3. Then the remainder when $n$ is divided by 11 is ( ). (A) 0 (B) 2 (C) 4 (D) 5 (E) 7",73 \equiv 7(\bmod 11)$.,easy,"5. E. From $n \equiv 3(\bmod 10)$, we know that the units digit of $n$ is 3. Let $n=10a+3$. Then $a+3 \equiv 1(\bmod 9) \Rightarrow a=7$ $\Rightarrow n=73 \equiv 7(\bmod 11)$." eebb89aa5a38,"15. (MEX 2) Determine for which positive integers $k$ the set $$ X=\{1990,1990+1,1990+2, \ldots, 1990+k\} $$ can be partitioned into two disjoint subsets $A$ and $B$ such that the sum of the elements of $A$ is equal to the sum of the elements of $B$.","4 r+3, r \geq 0$, and $k=4 r, r \geq 23$.",medium,"15. Let $S(Z)$ denote the sum of all the elements of a set $Z$. We have $S(X)=$ $(k+1)-1990+\frac{k(k+1)}{2}$. To partition the set into two parts with equal sums, $S(X)$ must be even and hence $\frac{k(k+1)}{2}$ must be even. Hence $k$ is of the form $4 r$ or $4 r+3$, where $r$ is an integer. For $k=4 r+3$ we can partition $X$ into consecutive fourtuplets $\{1990+$ $4 l, 1990+4 l+1,1990+4 l+2,1990+4 l+3\}$ for $0 \leq l \leq r$ and put $1990+4 l, 1990+4 l+3 \in A$ and $1990+4 l+1,1990+4 \bar{l}+2 \in B$ for all $l$. This would give us $S(A)=S(B)=(3980+4 r+3)(r+1)$. For $k=4 r$ the numbers of elements in $A$ and $B$ must differ. Let us assume without loss of generality $|A|<|B|$. Then $S(A) \leq(1990+2 r+1)+(1990+$ $2 r+2)+\cdots+(1990+4 r)$ and $S(B) \geq 1990+1991+\cdots+(1990+2 r)$. Plugging these inequalities into the condition $S(A)=S(B)$ gives us $r \geq 23$ and consequently $k \geq 92$. We note that $B=\{1990,1991, \ldots, 2034,2052,2082\}$ and $A=\{2035,2036, \ldots, 2051,2053, \ldots, 2081\}$ is a partition for $k=92$ that satisfies $S(A)=S(B)$. To construct a partition out of higher $k=4 r$ we use the $k=92$ partition for the first 93 elements and construct for the remaining elements as was done for $k=4 x+3$. Hence we can construct a partition exactly for the integers $k$ of the form $k=4 r+3, r \geq 0$, and $k=4 r, r \geq 23$." 988dad98888d,We consider a polyhedron with 20 triangular faces. How many edges and vertices does it have?,12$.,medium,"Let $S, A, F$ be the numbers of vertices, edges, faces respectively, so that $F=20$. We will proceed by double counting. We count the pairs $(a, f)$ where $a$ is an edge and $f$ is a face of which $a$ is a side. Each edge belongs to two faces, so this number is $2 A$. But since the faces are triangular, each face has three sides, so this number is also $3 F=60$. We deduce that $A=30$. Now that we have $A$ and $F$, we think of Euler's formula (from the previous exercise) to find $S$. We have $$ S=A+2-F=30+2-20 $$ so $S=12$." a6a2bfd74346,"In the diagram, the circle has centre $O$. The shaded sector $A O B$ has sector angle $90^{\circ}$, and $A B$ has arc length $2 \pi$ units. The area of sector $A O B$ is (A) $4 \pi$ (D) $24 \pi$ (B) $16 \pi$ (C) $6 \pi$ (E) $8 \pi$ ![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-130.jpg?height=349&width=376&top_left_y=1590&top_left_x=1357)",(A),easy,"Let $r$ be the radius of the circle. Since $\angle A O B=90^{\circ}$, then this sector is one quarter of the whole circle, so the circumference of the circle is $4 \times 2 \pi=8 \pi$. So $2 \pi r=8 \pi$, or $r=4$. Thus the area of sector $A O B$ is $\frac{1}{4} \pi r^{2}=\frac{1}{4} \pi(16)=4 \pi$. ![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-234.jpg?height=412&width=458&top_left_y=1317&top_left_x=1300) ANSWER: (A)" 5b6843c37239,"Let $N$ be the number of ways of distributing $8$ chocolates of different brands among $3$ children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of $N$.",24,medium,"To solve the problem, we need to distribute 8 different chocolates among 3 children such that each child gets at least one chocolate, and no two children get the same number of chocolates. We will consider the two possible distributions that satisfy these conditions. 1. **Distribution 1: 1, 2, and 5 chocolates** - One child gets 1 chocolate, another gets 2 chocolates, and the third gets 5 chocolates. - The number of ways to choose which child gets 1, 2, and 5 chocolates is \(3!\). - The number of ways to choose 1 chocolate out of 8 is \(\binom{8}{1}\). - The number of ways to choose 2 chocolates out of the remaining 7 is \(\binom{7}{2}\). - The number of ways to choose 5 chocolates out of the remaining 5 is \(\binom{5}{5}\). Therefore, the total number of ways for this distribution is: \[ 3! \times \binom{8}{1} \times \binom{7}{2} \times \binom{5}{5} = 6 \times 8 \times 21 \times 1 = 1008 \] 2. **Distribution 2: 1, 3, and 4 chocolates** - One child gets 1 chocolate, another gets 3 chocolates, and the third gets 4 chocolates. - The number of ways to choose which child gets 1, 3, and 4 chocolates is \(3!\). - The number of ways to choose 1 chocolate out of 8 is \(\binom{8}{1}\). - The number of ways to choose 3 chocolates out of the remaining 7 is \(\binom{7}{3}\). - The number of ways to choose 4 chocolates out of the remaining 4 is \(\binom{4}{4}\). Therefore, the total number of ways for this distribution is: \[ 3! \times \binom{8}{1} \times \binom{7}{3} \times \binom{4}{4} = 6 \times 8 \times 35 \times 1 = 1680 \] Adding the number of ways for both distributions, we get: \[ 1008 + 1680 = 2688 \] Finally, we need to find the sum of the digits of 2688: \[ 2 + 6 + 8 + 8 = 24 \] The final answer is \(\boxed{24}\)." f7705117da04,"Example 3 As shown in Figure 3, in $\triangle A B C$, $\angle B=46^{\circ}$, point $D$ is on side $B C$, and satisfies $\angle B A D=21^{\circ}$. If $A B$ $=C D$, find the degree measure of $\angle C A D$. ${ }^{[3]}$",180^{\circ}-\angle A D B=67^{\circ}$.,easy,"Solve: $\triangle D E C \cong \triangle B D A$, as shown in Figure 3. Then $\angle E D C=\angle B=46^{\circ}$, $E C=A D, D E=B D$. Thus, $\angle A D E=\angle A D C+\angle E D C$ $=113^{\circ}=\angle A D B$. Also, $A D=A D$, hence, $\triangle B D A \cong \triangle E D A$. Therefore, $\triangle D E C \cong \triangle E D A$. Then $\angle D A E=\angle D C E$. Thus, points $A, D, E, C$ are concyclic. Hence $\angle D A C=180^{\circ}-\angle D E C$ $=180^{\circ}-\angle A D B=67^{\circ}$." a50705fbeec0,"18. Let $a_{k}$ be the coefficient of $x^{k}$ in the expansion of $$ (x+1)+(x+1)^{2}+(x+1)^{3}+(x+1)^{4}+\cdots+(x+1)^{99} \text {. } $$ Determine the value of $\left\lfloor a_{4} / a_{3}\right\rfloor$.",19,easy,"18. Answer: 19 . Note that $$ a_{k}=\binom{1}{k}+\binom{2}{k}+\binom{3}{k}+\binom{4}{k}+\cdots+\binom{99}{k}=\binom{100}{k+1} $$ Thus $$ a_{4} / a_{3}=\frac{\binom{100}{5}}{\binom{100}{4}}=\frac{96}{5} $$ Thus the answer is 19 ." 1ce772f2e4f4,"1. Given the functions $$ f(x)=x^{2}-2 x, \quad g(x)=\left\{\begin{array}{ll} x-2, & x \geqslant 1, \\ -x, & x<1 . \end{array}\right. $$ Then the solution set of the inequality $f(x) \leqslant 3 g(x)$ is $\qquad$ (Supplied by An Zhenping)",See reasoning trace,medium,"1. $[-1,0] \cup[2,3]$. (Method 1) Since $g(x)=|x-1|-1$, the inequality $f(x) \leqslant 3 g(x)$ is equivalent to $$ x^{2}-2 x \leqslant 3|x-1|-3, $$ which is equivalent to $$ (x-1)^{2} \leqslant 3|x-1|-2, $$ that is, $$ |x-1|^{2}-3|x-1|+2 \leqslant 0, $$ thus $1 \leqslant|x-1| \leqslant 2$, solving this yields $-1 \leqslant x \leqslant 0$ or $2 \leqslant x \leqslant 3$. (Method 2) The original inequality is equivalent to $$ \left\{\begin{array}{l} x^{2}-2 x \leqslant 3(x-2), \\ x \geqslant 1, \end{array} \text { or } \quad \left\{\begin{array}{l} x^{2}-2 x \leqslant-3 x, \\ x<1 . \end{array}\right.\right. $$ Solving this yields $-1 \leqslant x \leqslant 0$ or $2 \leqslant x \leqslant 3$." 426699182e38,"Let $ABCD$ be a square of side length $4$. Points $E$ and $F$ are chosen on sides $BC$ and $DA$, respectively, such that $EF = 5$. Find the sum of the minimum and maximum possible areas of trapezoid $BEDF$. [i]Proposed by Andrew Wu[/i]",16,medium,"1. **Define the problem and set up coordinates:** Let the square \(ABCD\) have vertices at \(A(0, 4)\), \(B(0, 0)\), \(C(4, 0)\), and \(D(4, 4)\). Points \(E\) and \(F\) are on sides \(BC\) and \(DA\) respectively, such that \(EF = 5\). 2. **Determine the coordinates of \(E\) and \(F\):** Let \(E\) be at \((4, y_E)\) on \(BC\) and \(F\) be at \((x_F, 4)\) on \(DA\). Since \(EF = 5\), we use the distance formula: \[ \sqrt{(4 - x_F)^2 + (y_E - 4)^2} = 5 \] Simplifying, we get: \[ (4 - x_F)^2 + (y_E - 4)^2 = 25 \] 3. **Analyze the vertical and horizontal distances:** Since \(E\) is on \(BC\) and \(F\) is on \(DA\), the horizontal distance between \(E\) and \(F\) is \(4\). Therefore, the vertical distance must be: \[ \sqrt{25 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \] 4. **Determine the possible positions of \(E\) and \(F\):** - **Case 1:** \(E\) is close to \(B\) and \(F\) is close to \(D\): - \(y_E = 3\) (since \(E\) is 3 units above \(B\)) - \(x_F = 1\) (since \(F\) is 3 units to the left of \(D\)) - **Case 2:** \(E\) is far from \(B\) and \(F\) is far from \(D\): - \(y_E = 1\) (since \(E\) is 1 unit above \(C\)) - \(x_F = 3\) (since \(F\) is 1 unit to the right of \(A\)) 5. **Calculate the areas of trapezoid \(BEDF\):** - **Case 1:** - Bases: \(BE = 3\) and \(DF = 1\) - Height: \(4\) - Area: \[ \text{Area}_1 = \frac{1}{2} \times (3 + 1) \times 4 = \frac{1}{2} \times 4 \times 4 = 8 \] - **Case 2:** - Bases: \(BE = 1\) and \(DF = 3\) - Height: \(4\) - Area: \[ \text{Area}_2 = \frac{1}{2} \times (1 + 3) \times 4 = \frac{1}{2} \times 4 \times 4 = 8 \] 6. **Sum of the minimum and maximum possible areas:** Since both cases yield the same area, the sum of the minimum and maximum possible areas is: \[ 8 + 8 = 16 \] The final answer is \(\boxed{16}\)" 613028b47775,"2. In the complex plane, there are 7 points corresponding to the 7 roots of the equation $x^{7}=$ $-1+\sqrt{3} i$. Among the four quadrants where these 7 points are located, only 1 point is in ( ). (A) the I quadrant (B) the II quadrant (C) the III quadrant (D) the IV quadrant",See reasoning trace,medium,"2. (C). $$ \begin{array}{l} \because x^{7}=-1+\sqrt{3} \mathrm{i}=2\left(\cos 120^{\circ} + i \sin 120^{\circ}\right), \\ \therefore x_{n}=2^{\frac{1}{7}}\left(\cos \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}+\sin \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}\right) . \end{array} $$ where $n=0,1,2, \cdots, 6$. After calculation, we get $\arg x_{0}=\frac{2 \pi}{21}, \arg x_{1}=\frac{8 \pi}{21}, \arg x_{2}=\frac{14 \pi}{21}$, $\arg x_{3}=\frac{20 \pi}{21}, \arg x_{4}=\frac{26 \pi}{21}, \arg x_{5}=\frac{32 \pi}{21}, \arg x_{6}=\frac{38 \pi}{21}$. Clearly, there is only 1 root corresponding to a point in the third quadrant." 5231a1da9ca5,"Do either $(1)$ or $(2)$ $(1)$ $x$ and $y$ are functions of $t.$ Solve $x' = x + y - 3, y' = -2x + 3y + 1,$ given that $x(0) = y(0) = 0.$ $(2)$ A weightless rod is hinged at $O$ so that it can rotate without friction in a vertical plane. A mass $m$ is attached to the end of the rod $A,$ which is balanced vertically above $O.$ At time $t = 0,$ the rod moves away from the vertical with negligible initial angular velocity. Prove that the mass first reaches the position under $O$ at $t = \sqrt{(\frac{OA}{g})} \ln{(1 + sqrt(2))}.$",x(t) = e^{2t,medium,"We are given the system of differential equations: \[ \begin{cases} x' = x + y - 3 \\ y' = -2x + 3y + 1 \end{cases} \] with initial conditions \( x(0) = 0 \) and \( y(0) = 0 \). 1. **Transform the system into a single differential equation:** First, we express \( y \) in terms of \( x \) and its derivatives. From the first equation, we have: \[ y = x' - x + 3 \] Substitute this into the second equation: \[ y' = -2x + 3(x' - x + 3) + 1 \] Simplify: \[ y' = -2x + 3x' - 3x + 9 + 1 = 3x' - 5x + 10 \] But we also know from the second equation that: \[ y' = -2x + 3y + 1 \] Equate the two expressions for \( y' \): \[ 3x' - 5x + 10 = -2x + 3(x' - x + 3) + 1 \] Simplify: \[ 3x' - 5x + 10 = -2x + 3x' - 3x + 9 + 1 \] \[ 3x' - 5x + 10 = 3x' - 5x + 10 \] This confirms our transformation is consistent. 2. **Solve the differential equation for \( x(t) \):** The differential equation for \( x(t) \) is: \[ x'' - 4x' + 5x = 10 \] The characteristic equation is: \[ r^2 - 4r + 5 = 0 \] Solving for \( r \): \[ r = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i \] Thus, the general solution for the homogeneous equation is: \[ x_h(t) = e^{2t}(A \cos t + B \sin t) \] For the particular solution, we assume a constant solution \( x_p = C \): \[ 0 - 0 + 5C = 10 \implies C = 2 \] Therefore, the general solution is: \[ x(t) = e^{2t}(A \cos t + B \sin t) + 2 \] 3. **Determine the constants \( A \) and \( B \) using initial conditions:** Given \( x(0) = 0 \): \[ x(0) = A + 2 = 0 \implies A = -2 \] To find \( B \), we use \( x'(0) \). First, compute \( x'(t) \): \[ x'(t) = 2e^{2t}(A \cos t + B \sin t) + e^{2t}(-A \sin t + B \cos t) \] At \( t = 0 \): \[ x'(0) = 2A + B = -4 + B \] From the initial condition \( x'(0) = x(0) + y(0) - 3 = -3 \): \[ -4 + B = -3 \implies B = 1 \] Thus, the solution for \( x(t) \) is: \[ x(t) = e^{2t}(-2 \cos t + \sin t) + 2 \] 4. **Find \( y(t) \) using \( x(t) \):** From \( y = x' - x + 3 \): \[ x'(t) = 2e^{2t}(-2 \cos t + \sin t) + e^{2t}(2 \sin t + \cos t) \] Simplify: \[ x'(t) = e^{2t}(-4 \cos t + 2 \sin t + 2 \sin t + \cos t) = e^{2t}(-3 \cos t + 4 \sin t) \] Therefore: \[ y(t) = e^{2t}(-3 \cos t + 4 \sin t) - e^{2t}(-2 \cos t + \sin t) + 3 \] Simplify: \[ y(t) = e^{2t}(-3 \cos t + 4 \sin t + 2 \cos t - \sin t) + 3 = e^{2t}(- \cos t + 3 \sin t) + 3 \] The final answer is \( \boxed{ x(t) = e^{2t}(-2 \cos t + \sin t) + 2 } \) and \( y(t) = e^{2t}(- \cos t + 3 \sin t) + 3 \)." a1560eb3be4f,"## 232. Math Puzzle $9 / 84$ A motorcycle battery ( $6 \mathrm{~V}$ ) is discharged through a lamp $6 \mathrm{~V} / 0.5 \mathrm{~W}$. The lamp burns for 48 hours. How many Ah was the battery charged if we assume that the discharge current remains constant?",4 \mathrm{Ah}$ charged.,easy,"The discharge current is $I=\frac{0.5 \mathrm{~W}}{6 \mathrm{~V}}=\frac{1}{12} \mathrm{~A}$. If we assume that it remains constant during continuous discharge, the battery previously had $48 \cdot \frac{1}{12} \mathrm{Ah}=4 \mathrm{Ah}$ charged." 26d5e91ed133,"Example 7 Let the roots of the odd-degree real-coefficient equation $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}=0$ all lie on the unit circle, and $-a_{n}=a_{0} \neq 0$, find $a_{0}+a_{1}+\cdots+a_{n}$.","0$ equals $(-1)^{n} a_{n} / a_{0}=(-1)^{n} \cdot(-1)=1$ (for $n$ being odd). However, the product of",medium,"Let $f(c)=a_{0} c^{n}+a_{1} c^{n-1}+\cdots+a_{n-1} c+a_{n}=0$, i.e., the complex number $c$ is a root of $f(x)=0$, then the conjugate complex number $\bar{c}$ is also a root of $f(x)=0$. However, $f(x)=0$ is an equation of odd degree, so apart from the conjugate pairs of imaginary roots, it must have an odd number of real roots. Since all roots of $f(x)=0$ lie on the unit circle, the real roots of $f(x)=0$ can only be 1 or -1. Furthermore, by the relationship between roots and coefficients, the product of all roots of $f(x)=0$ equals $(-1)^{n} a_{n} / a_{0}=(-1)^{n} \cdot(-1)=1$ (for $n$ being odd). However, the product of conjugate imaginary numbers must be positive, so -1 is either not a root of $f(x)=0$, or it is a root of even multiplicity. Therefore, 1 must be a root of $f(x)=0$. Hence, $a_{0}+a_{1}+\cdots+a_{n}=f(1)=0$." 8bd0d070f5ef,43rd Putnam 1982,AC. 43rd Putnam 1982 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001,easy,": 2. Let N be the midpoint of AB and L the midpoint of AC. Cut AMB along MN into AMN and BMN. Then AMN is congruent to MAL and BNM is congruent ot MLC, so the two pieces can be reassembled to AMC. Clearly one piece is not sufficient unless AB = AC. 43rd Putnam 1982 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001" 9b1217591b66,"19. (2005 National High School Mathematics Competition Question) If the sum of the digits of a natural number $a$ equals 7, then $a$ is called a “lucky number”. Arrange all “lucky numbers” in ascending order as $a_{1}, a_{2}, a_{3}, \cdots$. If $a_{n}=$ 2005, then $a_{5 n}=$ $\qquad$",52000$.,medium,"19. Since the number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ is $C_{m+k-1}^{m}$, and the number of integer solutions where $x_{1} \geqslant 1, x_{i} \geqslant 0(i \geqslant 2)$ is $C_{m+k-2}^{m-1}$. Taking $m=7$, the number of $k$-digit “lucky numbers” is $P(k)=C_{k+5}^{6}$. 2005 is the smallest “lucky number” in the form $\overline{2 a b c}$, and $P(1)=C_{6}^{6}=1, P(2)=C_{7}^{6}=7, P(3)=C_{8}^{6}=28$. For four-digit “lucky numbers” $\overline{1 a b c}$, the number of such numbers is the number of non-negative integer solutions to $a+b+c=6$, which is $\mathrm{C}_{8+3-1}^{6}=28$. Since 2005 is the $1+7+28+28+1=65$-th “lucky number”, i.e., $a_{65}=2005$, thus $n=65, 5n=325$. Also, $P(4)=C_{9}^{6}=84, P(5)=C_{10}^{6}=210$, and $\sum_{k=1}^{5} P(k)=330$. Therefore, the last six five-digit “lucky numbers” in descending order are: 70000, 61000, 60100, 60010, 60001, 52000. Hence, the 325-th “lucky number” is 52000, i.e., $a_{55}=52000$." 24a76f14d697,"4. If $-\frac{2}{3}x \Rightarrow t \geqslant t^{2}$ $\Rightarrow t^{2}-t \leqslant 0 \Rightarrow t \in[0,1]$, so choose $D$." 504228bfb879,"6. Let the circumcenter of the tetrahedron $S-ABC$ be $O$, the midpoints of $SB$ and $AC$ be $N$ and $M$ respectively, and the midpoint of segment $MN$ be $P$. Given that $SA^2 + SB^2 + SC^2 = AB^2 + BC^2 + AC^2$, if $SP = 3\sqrt{7}$ and $OP = \sqrt{21}$, then the radius of the sphere $O$ is $\qquad$",See reasoning trace,medium,"6. $2 \sqrt{21}$. Let $\overrightarrow{S A}=x, \overrightarrow{S B}=y, \overrightarrow{S C}=z, \overrightarrow{B C}=a, \overrightarrow{C A}=\boldsymbol{b}$, $$ \overrightarrow{A B}=c \text {. } $$ Then $a=z-y, b=x-z, c=y-x$. Also, $S A^{2}+S B^{2}+S C^{2}=A B^{2}+B C^{2}+C A^{2}=m$ $\Rightarrow|\boldsymbol{a}|^{2}+|\boldsymbol{b}|^{2}+|\boldsymbol{c}|^{2}$ $=2\left(|x|^{2}+|y|^{2}+|z|^{2}\right)-$ $2(y \cdot z+z \cdot x+x \cdot y)$ $\Rightarrow 2(y \cdot z+z \cdot x+x \cdot y)=m$. It is easy to get $\overrightarrow{S P}=\frac{1}{4}(x+y+z)$. Let $\overrightarrow{S O}=\boldsymbol{r}$. Then $\overrightarrow{O A}=\boldsymbol{x}-\boldsymbol{r}$. $$ \begin{array}{l} \text { By }|\overrightarrow{S O}|=|\overrightarrow{O A}| \\ \Rightarrow|r|^{2}=|r|^{2}-2 r \cdot x+|x|^{2} \\ \Rightarrow|x|^{2}=2 r \cdot x . \end{array} $$ Similarly, $|y|^{2}=2 r \cdot y,|z|^{2}=2 r \cdot z$. Thus, $|x|^{2}+|y|^{2}+|z|^{2}$ $$ =2 \boldsymbol{r} \cdot(\boldsymbol{x}+\boldsymbol{y}+\boldsymbol{z})=8 \boldsymbol{r} \cdot \overrightarrow{S P} \text {. } $$ Also, $|\overrightarrow{S P}|^{2}=(3 \sqrt{7})^{2}$ $$ =\frac{1}{16}(x+y+z)^{2}=\frac{1}{8} m \text {, } $$ Then $$ \begin{array}{l} m=8 \times 63 \Rightarrow \boldsymbol{r} \cdot \overrightarrow{S P}=63 . \\ \text { By } \overrightarrow{O P}=\overrightarrow{O S}+\overrightarrow{S P} \\ \Rightarrow 21=|\boldsymbol{r}|^{2}-2 \boldsymbol{r} \cdot \overrightarrow{S P}+63=|\boldsymbol{r}|^{2}-63 \\ \Rightarrow|r|=2 \sqrt{21} . \end{array} $$" 35a8a9898235,"## 6. Light Bulbs In the room, there are two light bulbs. When the switch of the first light bulb is turned on, it lights up after 6 seconds and stays on for 5 seconds, then it is off for 6 seconds and on for 5 seconds, and this cycle repeats continuously. When the switch of the second light bulb is turned on, it lights up after 4 seconds and stays on for 3 seconds, then it is off for 4 seconds and on for 3 seconds, and this cycle repeats continuously. Linda turned on both switches at the same time and turned them off after 2021 seconds. How many seconds did both light bulbs shine simultaneously during this time? ## Result: 392",392$ seconds.,medium,"## Solution. Every 11 seconds, the first light will repeat a cycle where it will not light for 6 seconds and then light for 5 seconds. The second light will repeat such a cycle every 7 seconds. Since $V(11,7)=77$, the time interval we need to observe lasts 77 seconds. In the first two rectangles, yellow represents the seconds during which the lights are on, and gray represents the seconds during which the lights are off. The first rectangle shows the state of the first light, and the second shows the state of the second light. In the third rectangle, the seconds when both lights are on at the same time are marked. In the observed period of 77 seconds, both lights are on at the same time for $1+3+1+3+2+2+3=$ 15 seconds. Since $2021=26 \cdot 77+19$, there will be 26 such periods, and then there will be 19 seconds remaining. In these 19 seconds, the lights will be on at the same time for 2 seconds. Both lights are on at the same time for a total of $26 \cdot 15+2=392$ seconds." 41db3cbaa68c,"- the sum of the digits in each column is the same; - the sum of the digits in each row is the same; - the sum of the digits in the red cells is equal to the sum of the digits in any row. Enter the three-digit number $A B C$ as your answer. | 4 | | 3 | | :--- | :--- | :--- | | | $A$ | 1 | | 1 | | 6 | | $B$ | 2 | $C$ |",652,medium,"Answer: 652. Solution. If we add 3 to the sum of the two upper red numbers, we get the same number as if we added the sum of the two lower red numbers. Therefore, the unknown lower red number is 2. Then, in each row, the sum is $1+2+6=$ 9. This means that the unknown upper red number complements $4+3$ to 9, i.e., it is also 2. We have found that the sum of the digits in each row is 9. This means that the sum of the numbers in the entire table is 36, and in each column, it is 12. Now, it is not difficult, by sequentially considering columns and rows with one unknown number, to fill in the entire table (for example, we can first determine that $C=2$ and $A=6$, and then $B=5$): | 4 | 2 | 3 | | :--- | :--- | :--- | | 2 | 6 | 1 | | 1 | 2 | 6 | | 5 | 2 | 2 | We get that the number $A B C$ is 652." 2b398cf0ae62,"3. (20 points) Ten graduating classes have donated books to younger students, and the total number of books donated by any six of these classes is no less than 50% of the total number of donated books. What is the maximum percentage of the total number of books that the class which donated the most could have donated? $\qquad$",: $25 \%$,medium,"【Analysis】According to the problem, we can assume that Class A is the class that donates the most books, $x \%$ is the percentage of the number of books it donates out of the total, and the remaining 9 classes can be divided into three groups $A$, $B$, and $C$, each consisting of 3 classes. Using the fact that the sum of the number of books donated by any 6 classes is no less than $50 \%$, we can solve for the range of $x$, and the maximum value of $x$ is the required maximum percentage. 【Solution】Solution: According to the analysis, let Class A be the class that donates the most books, $x \%$ is the percentage of the number of books it donates out of the total, and the remaining 9 classes can be divided into three groups $A$, $B$, and $C$, with the percentages of the number of books they donate being $a \%$, $b \%$, and $c \%$. Then: $$ 2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geqslant 50+50+50, $$ Solving this, we get: $x \leqslant 25$. $\therefore$ The maximum percentage of the number of books donated by the class that donates the most books cannot exceed $25 \%$. When each of the 9 classes donates a percentage of the total number of books equal to $\frac{75}{9} \%=8 \frac{1}{3} \%$, then the sum of the number of books donated by any 6 of them is exactly $50 \%$ of the total, and Class A's donation is exactly $25 \%$. Therefore, the answer is: $25 \%$." fae4c89f578d,"6. In a three-digit number, the digit in the hundreds place is 2 more than the digit in the units place. Find the difference between this number and the number formed by the same digits but in reverse order.",\ldots = 198\).,easy,"6. Solution. Let \(x\) be the tens digit and \(y\) be the hundreds digit of a three-digit number. Then the units digit will be \(y-2\), and thus the desired number is \(100 y + 10 x + y - 2 - 100(y - 2) - 10 x - 2 = \ldots = 198\)." ef69c2508b3f,"14. For each integer $n \geqslant 2$, determine the maximum value of the sum $a_{0}+a_{1}+a_{2}+\cdots+a_{n}$ satisfying the conditions $a_{0}=1, a_{i} \geqslant a_{i+1}+a_{i+2}, i=0$. $1, \cdots, n-2$ (where $a_{0}, a_{1}, \cdots a_{n}$ are non-negative).","0, a_{n}=\frac{1}{F_{n-1}}, a_{i}=a_{i+1}+a_{i+2}(0 \leqslant i \leqslant n-2)$.",medium,"14. If there is 1 place where the inequality holds: $a_{i}>a_{i+1}+a_{i+2}$, then let $a_{i+1}^{\prime}=a_{i}-a_{i+2}, a_{j}^{\prime}=a_{1}(j \neq i)$. Then $a_{i}^{\prime}=a_{i+1}^{\prime}+a_{i+2}^{\prime}, a_{i+1}^{\prime}>a_{i+1} \geqslant a_{t+2}^{\prime}+a_{i+3}^{\prime}$, so $a_{0}^{\prime}, a_{1}^{\prime}, \cdots, a_{n}^{\prime}$ still satisfy the problem's conditions while $a_{0}^{\prime}+a_{1}^{\prime}+\cdots+a_{n}^{\prime}>a_{0}+a_{1}+\cdots+a_{n}$. Therefore, when the sum is maximized, $a_{i}=a_{i+1}+a_{i+2}$ holds for any $0 \leqslant i \leqslant n-2$. At this point, using the recursive relation, we can prove that $a_{k}=F_{n-k} \cdot a_{n-1}+F_{n-k-1} \cdot a_{n}$, where $F_{2}$ is the $i$-th Fibonacci number $\left(F_{0}=0, F_{1}=1, F_{i+2}=F_{2+1}+F_{i}\right)$, then $a_{0}+a_{1}+\cdots+a_{n}=$ $\left(F_{n+2}-1\right) a_{n-1}+F_{n-1} a_{n}$. Also, $a_{0}=1=F_{n} \cdot a_{n-1}+F_{n-1} \cdot a_{n}$, so $a_{n}=\frac{1-F_{n} \cdot a_{n-1}}{F_{n} \cdot 1}$. Thus, $a_{0}+a_{1}+\cdots+a_{n}=\frac{F_{n+2} \cdot F_{n-1}-F_{n+1} \cdot F_{n}-F_{n-1}}{F_{n-1}} \cdot a_{n-1}+\frac{F_{n+1}}{F_{n-1}}$ $$ \begin{array}{l} =\frac{(-1)^{n}-F_{n-1}}{F_{n-1}} \cdot a_{n-1}+\frac{F_{n+1}}{F_{n-1}} \\ \leqslant \frac{(-1)^{n}-F_{n-1}}{F_{n-1}} \cdot 0+\frac{F_{n+1}}{F_{n-1}}=\frac{F_{n+1}}{F_{n-1}} . \end{array} $$ Thus, the maximum value is $\frac{F_{n+1}}{F_{n-1}}$, which is achieved when $a_{n-1}=0, a_{n}=\frac{1}{F_{n-1}}, a_{i}=a_{i+1}+a_{i+2}(0 \leqslant i \leqslant n-2)$." fb5eb6cae7f4,"3 [Pythagorean Theorem in Space] Find the distance between the midpoints of the non-parallel sides of different bases of a regular triangular prism, all edges of which are equal to 2. #",$\sqrt{5}$,medium,"Let $M$ and $N$ be the midpoints of the edges $AC$ and $A1B1$ of a regular triangular prism $ABC A1B1C1$ with bases $ABC$ and $A1B1C1$ ($AA1 \parallel BB1 \parallel CC1$), all edges of which are equal to 2; $M1$ - the orthogonal projection of point $M$ onto the plane $A1B1C1$. Then $M1$ is the midpoint of $A1C1$, and $M1N$ is the midline of triangle $A1B1C1$. From the right triangle $MM1N$, we find that $$ MN = \sqrt{MM1^2 + NM1^2} = \sqrt{4 + 1} = \sqrt{5} $$ It is clear that the distances between the midpoints of any other non-parallel sides of the base are also equal to $\sqrt{5}$. ## Answer $\sqrt{5}$" 50fc4b34fdbf,"Ondra spent $\frac{2}{3}$ of his money on a trip and then gave $\frac{2}{3}$ of the remainder to a school for children from Tibet. He then spent $\frac{2}{3}$ of the new remainder on a small gift for his mother. From the remaining money, he lost $\frac{4}{5}$ due to a hole in his pocket, and when he gave half of what was left to his little sister, he was left with exactly one crown. With what amount $S$ did Ondra go on the trip? (M. Volfová)","2, 2 \cdot 5=10, 10 \cdot 3=30$, $30 \cdot 3=90$ and $90 \cdot 3=270$. Ondra had 270 crowns before t",medium,"The number of Ondra's crowns before the trip is denoted as $x$. - Ondra spent $\frac{2}{3}$ of his money on the trip, leaving him with $\frac{1}{3} x$ crowns. - He donated $\frac{2}{3}$ of the remaining money to a school in Tibet, leaving him with $\frac{1}{3} \cdot \frac{1}{3} x=\frac{1}{9} x$ crowns. - The gift for his mother cost $\frac{2}{3}$ of the remainder, leaving him with $\frac{1}{3} \cdot \frac{1}{9} x=\frac{1}{27} x$ crowns. - He lost $\frac{4}{5}$ of what was left, leaving him with $\frac{1}{5} \cdot \frac{1}{27} x=\frac{1}{135} x$ crowns. - He gave half of the remaining money to his sister, leaving him with the other half, $\mathrm{i.e.} \cdot \frac{1}{2} \cdot \frac{1}{135} x=\frac{1}{270} x$, which was 1 crown. If $\frac{1}{270} x=1$, then $x=270$. Ondra went on the trip with an amount of 270 crowns. Another solution. The problem can also be solved by working backwards according to the following scheme: ![](https://cdn.mathpix.com/cropped/2024_04_17_faa98acd9497c1119dfag-2.jpg?height=137&width=1391&top_left_y=1419&top_left_x=367) Step by step, from right to left, we get the following values: $1 \cdot 2=2, 2 \cdot 5=10, 10 \cdot 3=30$, $30 \cdot 3=90$ and $90 \cdot 3=270$. Ondra had 270 crowns before the trip." cc4c4ddfb5fe,"28. Given that $\mathbf{m}=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}$ and $\mathbf{n}=(\sqrt{2}-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are the usual unit vectors along the $x$-axis and the $y$-axis respectively, and $\theta \in(\pi 2 \pi)$. If the length or magnitude of the vector $\mathbf{m}+\mathbf{n}$ is given by $\mathbf{m}+\mathbf{n}=\frac{8 \sqrt{2}}{5}$, find the value of $5 \cos \left(\frac{\theta}{2}+\frac{\pi}{8}\right)+5$.",See reasoning trace,medium,"28. Answer: 1 $$ \begin{aligned} \frac{8 \sqrt{2}}{5}=\mathbf{m}+\mathbf{n} & =\sqrt{4+2 \sqrt{2} \cos \theta-2 \sqrt{2} \sin \theta} \\ & =2 \sqrt{1+\cos \left(\theta+\frac{\pi}{4}\right)} \\ & =2 \sqrt{2}\left|\cos \left(\frac{\theta}{2}+\frac{\pi}{8}\right)\right| \end{aligned} $$ Since $\pi<\theta<2 \pi$, we have $\frac{5}{3} \pi<\frac{\theta}{2}+\frac{\pi}{8}<\frac{9}{8} \pi$. Thus, $\cos \left(\frac{\theta}{2}+\frac{\pi}{8}\right)<0$. This implies that $\cos \left(\frac{\theta}{2}+\frac{\pi}{8}\right)=-\frac{4}{5}$ and hence $$ 5 \cos \left(\frac{\theta}{2}+\frac{\pi}{8}\right)+5=1 $$" 004fcbc4c61a,"At the ""Lukomorye"" station, they sell cards for one, five, and twenty rides. All cards cost a whole number of gold coins. Five cards for one ride are more expensive than one card for five rides, and four cards for five rides are more expensive than one card for twenty rides. It turned out that the cheapest way for 33 bogatyrs to travel is to buy cards for 35 rides, spending 33 gold coins on this. How much does a card for five rides cost?",5 coins,medium,"Since 5, 20, and 35 are all divisible by 5, the number of tickets bought for a single ride is also divisible by 5. However, it is more cost-effective to replace every five such tickets with one ticket for five rides. Therefore, there is no need to buy tickets for a single ride. For the same reason, it is most cost-effective to buy three tickets for five rides and one ticket for 20 rides, which costs exactly 33 coins, while seven tickets for five rides cost more. Therefore, a ticket for five rides costs no less than 5 coins. On the other hand, according to the condition, it is more cost-effective to pay for 35 rides than to buy two tickets for 20 rides, meaning that three tickets for five rides are cheaper than one for 20. Therefore, six tickets for five rides cost less than 33 coins, meaning that one such ticket costs no more than five coins. ## Answer 5 coins." 08c510834388,3. Find $\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^{3}+x^{2}}-\sqrt[3]{x^{3}-x^{2}}\right)$.,See reasoning trace,medium,"Solution: $2 / 3$ Observe that $$ \lim _{x \rightarrow \infty}\left[(x+1 / 3)-\sqrt[3]{x^{3}+x^{2}}\right]=\lim _{x \rightarrow \infty} \frac{x / 3+1 / 27}{\left(\sqrt[3]{x^{3}+x^{2}}\right)^{2}+\left(\sqrt[3]{x^{3}+x^{2}}\right)(x+1 / 3)+(x+1 / 3)^{2}}, $$ by factoring the numerator as a difference of cubes. The numerator is linear in $x$, while the denominator is at least $3 x^{2}$, so the limit as $x \rightarrow \infty$ is 0 . By similar arguments, $\lim _{x \rightarrow \infty}\left[(x-1 / 3)-\sqrt[3]{x^{3}-x^{2}}\right]=0$. So, the desired limit equals $$ 2 / 3+\lim _{x \rightarrow \infty}\left[(x-1 / 3)-\sqrt[3]{x^{3}-x^{2}}\right]-\lim _{x \rightarrow \infty}\left[(x+1 / 3)-\sqrt[3]{x^{3}+x^{2}}\right]=2 / 3 . $$" 2bc36a8e86a8,"In the diagram below, square $ABCD$ with side length 23 is cut into nine rectangles by two lines parallel to $\overline{AB}$ and two lines parallel to $\overline{BC}$. The areas of four of these rectangles are indicated in the diagram. Compute the largest possible value for the area of the central rectangle. [asy] size(250); defaultpen (linewidth (0.7) + fontsize (10)); draw ((0,0)--(23,0)--(23,23)--(0,23)--cycle); label(""$A$"", (0,23), NW); label(""$B$"", (23, 23), NE); label(""$C$"", (23,0), SE); label(""$D$"", (0,0), SW); draw((0,6)--(23,6)); draw((0,19)--(23,19)); draw((5,0)--(5,23)); draw((12,0)--(12,23)); label(""13"", (17/2, 21)); label(""111"",(35/2,25/2)); label(""37"",(17/2,3)); label(""123"",(2.5,12.5));[/asy] [i]Proposed by Lewis Chen[/i]",180,medium,"1. Let the side length of the square \(ABCD\) be 23. The square is divided into nine rectangles by two lines parallel to \(\overline{AB}\) and two lines parallel to \(\overline{BC}\). 2. Denote the lengths of the segments created by the lines parallel to \(\overline{AB}\) as \(a\), \(b\), and \(c\) such that \(a + b + c = 23\). Similarly, denote the lengths of the segments created by the lines parallel to \(\overline{BC}\) as \(d\), \(e\), and \(f\) such that \(d + e + f = 23\). 3. Given the areas of four rectangles: - The rectangle with area 13 has dimensions \(a \times d\), so \(ad = 13\). - The rectangle with area 111 has dimensions \(b \times f\), so \(bf = 111\). - The rectangle with area 37 has dimensions \(c \times e\), so \(ce = 37\). - The rectangle with area 123 has dimensions \(a \times f\), so \(af = 123\). 4. We need to find the largest possible value for the area of the central rectangle, which has dimensions \(b \times e\). 5. From the given areas, we can express \(d\) and \(f\) in terms of \(a\) and \(b\): \[ d = \frac{13}{a}, \quad f = \frac{123}{a} \] 6. Using the sum of the segments: \[ d + f + e = 23 \implies \frac{13}{a} + \frac{123}{a} + e = 23 \implies \frac{136}{a} + e = 23 \implies e = 23 - \frac{136}{a} \] 7. Similarly, express \(b\) and \(c\) in terms of \(f\) and \(e\): \[ b = \frac{111}{f}, \quad e = \frac{37}{c} \] 8. Using the sum of the segments: \[ b + c + a = 23 \implies \frac{111}{f} + \frac{37}{c} + a = 23 \] 9. Substitute \(f = \frac{123}{a}\) into the equation for \(b\): \[ b = \frac{111}{\frac{123}{a}} = \frac{111a}{123} = \frac{37a}{41} \] 10. Now, substitute \(e = 23 - \frac{136}{a}\) into the equation for \(c\): \[ c = \frac{37}{e} = \frac{37}{23 - \frac{136}{a}} \] 11. To find the area of the central rectangle \(b \times e\): \[ b \times e = \left(\frac{37a}{41}\right) \times \left(23 - \frac{136}{a}\right) \] 12. Simplify the expression: \[ b \times e = \frac{37a}{41} \times \left(23 - \frac{136}{a}\right) = \frac{37a}{41} \times \frac{23a - 136}{a} = \frac{37(23a - 136)}{41} \] 13. To maximize \(b \times e\), we need to find the maximum value of the expression: \[ \frac{37(23a - 136)}{41} \] 14. Solving for \(a\) to maximize the expression, we find that the maximum value occurs when \(a = 10\) and \(b = 18\). 15. Therefore, the largest possible value for the area of the central rectangle is: \[ \boxed{180} \]" c56148abe6ba,"We call a path Valid if i. It only comprises of the following kind of steps: A. $(x, y) \rightarrow (x + 1, y + 1)$ B. $(x, y) \rightarrow (x + 1, y - 1)$ ii. It never goes below the x-axis. Let $M(n)$ = set of all valid paths from $(0,0) $, to $(2n,0)$, where $n$ is a natural number. Consider a Valid path $T \in M(n)$. Denote $\phi(T) = \prod_{i=1}^{2n} \mu_i$, where $\mu_i$= a) $1$, if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y + 1)$ b) $y$, if the $i^{th} $ step is $(x, y) \rightarrow (x + 1, y - 1)$ Now Let $f(n) =\sum _{T \in M(n)} \phi(T)$. Evaluate the number of zeroes at the end in the decimal expansion of $f(2021)$",0,medium,"1. **Define the problem and the function $f(n)$:** - We need to evaluate the number of zeroes at the end in the decimal expansion of $f(2021)$. - $f(n) = \sum_{T \in M(n)} \phi(T)$, where $M(n)$ is the set of all valid paths from $(0,0)$ to $(2n,0)$. - A valid path consists of steps $(x, y) \rightarrow (x + 1, y + 1)$ or $(x, y) \rightarrow (x + 1, y - 1)$ and never goes below the x-axis. - $\phi(T) = \prod_{i=1}^{2n} \mu_i$, where $\mu_i = 1$ if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y + 1)$ and $\mu_i = y$ if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y - 1)$. 2. **Analyze the parity of $f(n)$:** - We need to show that $f(n)$ is odd for all $n \in \mathbb{N}$. - Consider the steps as moving either in the North East (NE) direction or in the South East (SE) direction. - Reverse engineer the path starting from $(2n,0)$ and moving towards the left. - If the path touches $y=2$ at any point, $\phi(T)$ becomes even because $\mu_i = y$ for some $i$ will be 2. - To avoid touching $y=2$, the path must vary between $y=0$ and $y=1$ only. 3. **Construct the valid path:** - Since $2n$ is even, we can construct a valid path by alternating between going up and down, i.e., $(x, y) \rightarrow (x + 1, y + 1)$ followed by $(x, y) \rightarrow (x + 1, y - 1)$. - This ensures that the path never goes below the x-axis and stays between $y=0$ and $y=1$. 4. **Calculate $\phi(T)$ for the valid path:** - For the valid path that alternates between $y=0$ and $y=1$, $\phi(T)$ will be the product of 1's and 0's. - Specifically, $\phi(T) = 1 \cdot 1 \cdot 1 \cdot \ldots \cdot 1 = 1$ for all steps $(x, y) \rightarrow (x + 1, y + 1)$. - Since the path does not touch $y=2$, $\phi(T)$ remains 1 for all valid paths. 5. **Determine the parity of $f(n)$:** - Since $\phi(T) = 1$ for all valid paths $T \in M(n)$, $f(n)$ is the sum of 1's. - The number of valid paths is odd, so $f(n)$ is odd. 6. **Conclusion:** - Since $f(n)$ is odd for all $n \in \mathbb{N}$, $f(2021)$ is also odd. - An odd number has no zeroes at the end of its decimal expansion. The final answer is $\boxed{0}$" 59bbb16f4fb7,"Example 1-8: There are 5 Japanese books, 7 English books, and 10 Chinese books. If two books of different languages are taken, how many possibilities are there? If two books of the same language are taken, how many schemes are there? If two books are taken regardless of the language, how many possibilities are there?",231$ ways.,medium,"(1) Take two books of different languages, according to the multiplication rule: (1) There are $n_{1}=5 \times 7=35$ possibilities for one English and one Japanese book; (2) There are $n_{2}=10 \times 5=50$ possibilities for one Chinese and one Japanese book; (3) There are $n_{3}=10 \times 7=70$ possibilities for one Chinese and one English book. According to the addition rule, the number of ways to take two books of different languages is $$ N_{1}=35+50+70=155 \text { ways } $$ (2) The number of ways to take two books of the same language are: (1) Both Chinese: $n_{1}^{\prime}=(10 \times 9) / 2=45$, The first book has 10 possibilities, and the second book has 9 possibilities, but the case where the first book is $a$ and the second book is $b$ is the same as the case where the first book is $b$ and the second book is $a$, so it is counted twice, hence divided by 2. (2) Both English: $n_{2}^{\prime}=7 \times 6 / 2=21$, (3) Both Japanese: $n_{3}^{\prime}=5 \times 4 / 2=10$. According to the addition rule, the number of ways to take two books of the same language is $$ N_{2}=45+21+10=76 \text { ways } $$ (3) Without considering whether the languages are the same or not, according to the addition rule $$ N_{3}=155+76=231 \text { ways, } $$ or taking two books from 22 books has $22 \times 21 / 2=231$ ways." ea6a54b66707,"8,9 Given a triangle with sides 12, 15, 18. A circle is drawn that touches both smaller sides and has its center on the larger side. Find the segments into which the center of the circle divides the larger side of the triangle.",8 and 10,easy,"The center of the circle inscribed in an angle lies on the bisector of this angle. ## Solution The center of the circle lies on the bisector of the larger angle. Therefore, the ratio of the required segments is equal to the ratio of the lateral sides, i.e., $\frac{4}{5}$. Consequently, the required segments are $$ \frac{4}{9} \cdot 18=8, \frac{5}{9} \cdot 18=10 $$ ## Answer 8 and 10." 64ad3236a49c,"30. For which values of $b$, the system of inequalities $$ \left\{\begin{array}{l} 0.5(2 x+5)<2(x-2)+5 \\ 2(b x-1)<3 \end{array}\right. $$ has no solution?","$b \in\left[\frac{5}{3},+\infty\right)$ )",easy,"(Answer: $b \in\left[\frac{5}{3},+\infty\right)$ )" e8e3b004cfab,"Example 4 The center of the ellipse is the origin $O$, its minor axis length is $2 \sqrt{2}$, and the directrix $l$ corresponding to the focus $F(c, 0)(c>0)$ intersects the $x$-axis at point $A, |O F|=2|F A|$, a line passing through point $A$ intersects the ellipse at points $P$ and $Q$. (1) Find the equation of the ellipse and its eccentricity; (2) If $\overrightarrow{O P} \cdot \overrightarrow{O Q}=0$, find the equation of the line $P Q$.",0$ or $x$ $+\sqrt{5} y-3=0$.,medium,"Solution: (1) According to the problem, we can set the equation of the ellipse as $\frac{x^{2}}{a^{2}}+$ $a=\sqrt{6}, c=2$. $\therefore$ The equation of the required ellipse is $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$, and the eccentricity is $e=\frac{\sqrt{6}}{3}$. (2) From (1), we get $A(3,0)$. Let the equation of line $P Q$ be $y=k(x-3)$, From the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{6}+\frac{y^{2}}{2}=1, \\ y=k(x-3),\end{array}\right.$ we get $\left(3 k^{2}+1\right) x^{2}-$ $18 k^{2} x+27 k^{2}-6=0$, According to the problem: $\Delta=12\left(2-3 k^{2}\right)>0$, we get $-\frac{\sqrt{6}}{3}0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ). (A) -3 (B) -8 (C) 3 (D) 8",See reasoning trace,easy,"8. B. When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point, $$ x_{1}+x_{2}=-3 \text {. } $$ Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives $$ x^{2}+5 x+3=0 \text {. } $$ Therefore, $x_{3}+x_{4}=-5$. Thus, the sum of all $x$ is $$ -3+(-5)=-8 \text {. } $$" 2c400f527c92,"Example 3.15. Find the partial derivatives of the function $$ f(x, y, z)=\left\{\begin{array}{lr} x^{4}+\frac{2 x y z}{y^{2}+z^{2}}, & y^{2}+z^{2} \neq 0 \\ x^{4}, & y^{2}+z^{2} = 0 \end{array}\right. $$ at the point $A(1,0,0)$.",See reasoning trace,medium,"Solution. Consider the function $f(x, 0,0)=x^{4}$. Let's find the derivative of this function: $f_{x}^{\prime}(x, 0,0)=4 x^{3}$ and $f_{x}^{\prime}(1,0,0)=4$. Next, consider the functions $f(1, y, 0)=1$ and $f(1,0, z)=1$. We have: $f_{x}^{\prime}(1, y, 0)=0, f_{z}^{\prime}(1,0, z)=0$ and, in particular, $f_{y}^{\prime}(1,0,0)=0$ and $f_{z}^{\prime}(1,0,0)=0$. Thus, the partial derivatives of the function at point $A(1,0,0)$ exist, but as shown in Example 3.11, the function is discontinuous at point $A(1,0,0)$. Therefore, the existence of partial derivatives at a point does not guarantee the continuity of the function at that point in general. Remark. The above definition of partial derivatives usually applies to interior points of the domain of a function of several variables and, in general, is not suitable for boundary points of this domain, as it is not always possible to compute the partial increments of the function. Therefore, in the boundary points of the domain of functions of several variables, partial derivatives are defined as the limit values of these derivatives. From the definition of partial derivatives as ordinary derivatives under the condition of fixing the values of all variables except one, for which the derivative is taken, it follows, in particular, that the geometric meaning of partial derivatives of a function of two variables $z=f(x, y)$ at a point $P_{0}\left(x_{0}, y_{0}\right)$ is analogous to the geometric meaning of the derivative of a function of one variable: $f_{x}^{\prime}(x, y)$ is the tangent of the angle of inclination to the $O X$ axis (relative to this axis) of the tangent at the point $M_{0}\left(x_{0}, y_{0}, f\left(x_{0}, y_{0}\right)\right)$ to the section of the surface $z=f(x, y)$ by the plane $y=y_{0}$. Partial derivatives of higher orders at point $P_{0}$ are defined by transitioning from partial derivatives of order $r-1$ to partial derivatives of order $r$ according to the following relationship: $$ \frac{\partial}{\partial x_{i_{r}}}\left(\frac{\partial^{r-1} f}{\partial x_{i_{r-1}} \ldots \partial x_{i_{1}}}\right)=\frac{\partial^{r} f}{\partial x_{i_{r}}} \partial x_{i_{r-1}} \ldots \partial x_{i_{1}} \equiv f_{x_{1_{1}} \ldots x_{r_{r-1} x_{r_{r}}}}^{(r)} $$" a32ca81d543a,"12.94 The equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has a rational solution, the necessary and sufficient condition regarding the positive integer $n$ is what?",See reasoning trace,medium,"[Solution] The necessary and sufficient condition is $n=1$. When $n=1$, the equation $$ x^{n}+(2+x)^{n}+(2-x)^{n}=0 . $$ has a rational solution $$ x=-4 \text {. } $$ Next, we prove that $n=1$ is also necessary. First, because $x, 2+x, 2-x$ cannot all be 0, $n$ cannot be an even number. When $n$ is an odd number, suppose $x=\frac{p}{q}$ is a solution to equation (1), where $p, q$ are coprime integers. Then from (1) we get $$ \begin{array}{c} p^{n}+(2 q+p)^{n}+(2 q-p)^{n}=0, \\ p^{n}+2\left[(2 q)^{n}+C_{n}^{2}(2 q)^{n-2} p^{2}+\cdots+C_{n}^{n-1}(2 q) p^{n-1}\right]=0 . \end{array} $$ Since every term except the first can be divided by $2 q$, $p^{n}$ can also be divided by $2 q$. Because $p$ and $q$ are coprime, then $q= \pm 1$. Assume $q=1$ without loss of generality, otherwise change the signs of $p, q$ simultaneously, the value of $x$ remains unchanged. Since $p^{n}$ can be divided by $2 q=2$, $p$ is even. Let $p=2 r$, substitute into (1), we get $$ \begin{array}{c} (2 r)^{n}+(2+2 r)^{n}+(2-2 r)^{n}=0, \\ r^{n}+(1+r)^{n}+(1-r)^{n}=0 . \end{array} $$ When $n \geqslant 3$, we have $$ \begin{array}{l} r^{n}+2\left(1+C_{n}^{2} r^{2}+C_{n}^{4} r^{4}+\cdots\right)=0, \\ 2=-r^{n}-2\left(C_{n}^{2} r^{2}+C_{n}^{4} r^{4}+\cdots\right), \end{array} $$ so 2 can be divided by $r^{2}$, thus $r= \pm 1$. But when $r= \pm 1$, (2) is not satisfied, hence $n \geqslant 3$ is impossible. Therefore, $n$ can only be 1." 8a1a95e9a31f,Let's determine the maximum of the function $f(x)=\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x}$.,5$.,medium,"Solution. First, let's examine the domain of the function. The value under the square root cannot be negative, therefore $3.5 \leq x \leq 6$. We apply the inequality between the arithmetic mean and the quadratic mean: $$ \frac{a+b+c}{3} \leq \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} $$ to the three square root expressions in the function: $$ \frac{\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x}}{3} \leq \sqrt{\frac{(x-2)+(2 x-7)+(18-3 x)}{3}}=\sqrt{3} $$ Thus, $$ f(x)=\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x} \leq 3 \sqrt{3} $$ Equality holds if and only if $$ \sqrt{x-2}=\sqrt{2 x-7}=\sqrt{18-3 x} \Leftrightarrow x=5 $$ which is an element of the domain. Therefore, the maximum of the function is $3 \sqrt{3}$, which is attained (uniquely) at $x=5$." babc41859691,"2. If $a0, \therefore x_{1}<0, x_{2}>0. \\ \text { When } x \in(-\infty, -\frac{a}{3}), f^{\prime}(x)>0, \text { function } y \text { is increasing; } \\ \text { When } x \in(-\frac{a}{3}, a), f^{\prime}(x)<0, \text { function } y \text { is decreasing; } \\ \text { When } x \in(a, +\infty), f^{\prime}(x)>0, \text { function } y \text { is increasing. } \\ \therefore \text { when } x=-\frac{a}{3}, y \text { reaches a local maximum; when } x=a, y \text { reaches a local minimum. } \\ \text { To make } y_{\text {min}}<0, \text { we need } f(a)<0, \text { i.e., } \\ a^{3}-a^{3}-a^{3}+2 a^{2}+a<0, \text { which simplifies to } -a^{3}+2 a^{2}+a<0; \text { by (2) we get } \\ a^{2}-2 a-1>0, \\ \therefore[a-(1+\sqrt{2})][a+(\sqrt{2}-1)]>0. \\ \because a>0, \therefore a+(\sqrt{2}-1)>0. \end{array} $$ Therefore, the range of $a$ is $a>1+\sqrt{2}$." 6981ec3dec2c,"24) Three cyclists set off from Cesenatico along the road to Cortona. The first one leaves an hour before the other two, who start together. Each cyclist maintains a constant speed. After a certain time, the third cyclist catches up with the first, and two hours later, the second also catches up with him. The ratio of the second cyclist's speed to the third cyclist's speed is $\frac{2}{3}$. What is the ratio of the first cyclist's speed to that of the third? (A) $\frac{1}{4}$ (B) $\frac{1}{3}$ (C) $\frac{2}{5}$ (D) $\frac{1}{2}$ (E) $\frac{2}{3}$.",(D),medium,"24) The answer is (D). Indicating with $v_{1}, v_{2}$, and $v_{3}$ the speeds of the three cyclists, it is evident that $v_{1}\frac{66}{5}$, i.e., $u \geqslant 14$. Also, since $2 \nmid u$ and $3 \nmid u$, we know $u \geqslant 17$. From $v \geqslant u$ and (4), we get $\frac{66 u-1}{5 u-66} \geqslant u$, i.e., $5 u^{2}-132 u+1 \leqslant 0$. Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$. $\therefore 17 \leqslant u \leqslant 26$. Furthermore, since $2 \nmid u$ and $3 \nmid u$, $u$ can only be $17,19,23,25$. When $u=17$, from (4) we get $v=\frac{1121}{19}=59$. When $u=19$, from (4) we get $v$ is not a natural number, so it should be discarded. When $u=23$, from (4) we get $v$ is not a natural number, so it should be discarded. When $u=25$, from (4) we get $v$ is not a natural number, so it should be discarded. Therefore, only when $p=u+v=76$, the two roots of equation (*) are natural numbers, and the three roots of the original equation are natural numbers." ada968084819,"22. [12] Find the number of pairs of integers $(a, b)$ with $1 \leq a0$ and $b>0$. If in this case the equality $a^{2}=b^{2}$ holds, then $a=b$. We find $$ \begin{aligned} & a^{2}=\frac{(7+4 \sqrt{3})(19-8 \sqrt{3})}{(4-\sqrt{3})^{2}}= \\ & =\frac{(7+4 \sqrt{3})(19-8 \sqrt{3})}{19-8 \sqrt{3}}=7+4 \sqrt{3} \\ & b^{2}=(2+\sqrt{3})^{2}=7+4 \sqrt{3} \end{aligned} $$ Since $a^{2}=b^{2}$, then $a=b$, i.e., the given equality is valid. This example can be solved faster if one guesses that both expressions under the square roots in the condition are squares of positive numbers, specifically: $7+4 \sqrt{3}=(2+\sqrt{3})^{2}$ and $19-8 \sqrt{3}=(4-\sqrt{3})^{2}$. Then the left side of the given equality is $$ \frac{(2+\sqrt{3})(4-\sqrt{3})}{(4-\sqrt{3})}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2 \text { and } 2=2 $$" a2fdfab39a9f,"The 27th question: Find the smallest real number $\lambda$, such that $\left(a_{1}{ }^{n}+\prod_{i=1}^{n} a_{i}, a_{2}{ }^{n}+\prod_{i=1}^{n} a_{i}, \ldots, a_{n}{ }^{n}+\prod_{i=1}^{n} a_{i}\right) \leq$ $\lambda\left(a_{1}, a_{2}, \ldots, a_{n}\right)^{n}$ holds for any positive odd number $n$ and $n$ positive integers $a_{1}, a_{2}, \ldots, a_{n}$.",See reasoning trace,medium,"The 27th problem, solution: Without loss of generality, assume $\left(a_{1}, a_{2}, \ldots, a_{n}\right)=1$ (otherwise, replace $a_{i}$ with $\frac{a_{i}}{\left(a_{1}, a_{2}, \ldots, a_{n}\right)}$ for $1 \leq \mathrm{i} \leq \mathrm{n}$, which does not change the essence of the problem). In this case, we need to find the smallest real number $\lambda$ such that $$ \begin{array}{l} \left(a_{1}^{n}+\prod_{i=1}^{n} a_{i}, a_{2}^{n}+\prod_{i=1}^{n} a_{i}, \quad \ldots, a_{n}^{n}+\prod_{i=1}^{n} a_{i}\right) \leq \lambda \\ \text { Let }\left(a_{1}^{n}+\prod_{i=1}^{n} a_{i}, a_{2}^{n}+\prod_{i=1}^{n} a_{i}, \ldots, a_{n}^{n}+\prod_{i=1}^{n} a_{i}\right)=d, \text { then } \\ a_{1}^{n} \equiv a_{2}^{n} \equiv \ldots \equiv a_{n}^{n} \equiv-\prod_{i=1}^{n} a_{i}(\bmod d) \end{array} $$ We can prove that each $a_{i}(1 \leq \mathrm{i} \leq \mathrm{n})$ is coprime with $\mathrm{d}$. If this conclusion is not true, assume without loss of generality that $\left(\mathrm{a}_{1}, \mathrm{~d}\right)>1$, and take any prime factor $\mathrm{p}$ of $\left(\mathrm{a}_{1}, \mathrm{~d}\right)$. Then, according to (2), we have: $$ \begin{array}{l} a_{1}^{n} \equiv a_{2}^{n} \equiv \ldots \equiv a_{n}^{n} \equiv-\prod_{i=1}^{n} a_{i}(\bmod p) \\ \Rightarrow a_{1} \equiv a_{2} \equiv \ldots \equiv a_{n} \equiv 0(\bmod p) \end{array} $$ This contradicts $\left(\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}\right)=1$. According to (2), we have: $$ \prod_{i=1}^{n} a_{i}^{n} \equiv\left(-\prod_{i=1}^{n} a_{i}\right)^{n} \equiv-\prod_{i=1}^{n} a_{i}^{n}(\bmod d) \Rightarrow d \mid 2 $$ Thus, taking $\lambda=2$ in (1) satisfies the condition. On the other hand, if $a_{1}=a_{2}=\ldots=a_{n}=1$, then $\lambda \geq 2$. In conclusion, the smallest value of $\lambda$ that satisfies the condition is 2." 94997ac68f1f,"21. As shown in Figure 2, fold the square $A B C D$ with side length 1. Vertex $C$ lands on point $C^{\prime}$ on side $A D$, and $B C$ intersects $A B$ at point $E$. If $C^{\prime} D$ $=\frac{1}{3}$, then the perimeter of $\triangle A E C^{\prime}$ is ( ). (A) 2 (B) $1+\frac{2 \sqrt{3}}{3}$ (C) $\frac{7}{3}$ (D) $\sqrt{136}$ (E) $1+\frac{3 \sqrt{3}}{4}$",\frac{3+4+5}{6} = 2$.,easy,"21. A. Let $N$ be the intersection of the crease with $CD$, and let $DN=x$. Then $C'N = CN = 1 - x$. In $\triangle C' DN$, $$ \left(\frac{1}{3}\right)^{2} + x^{2} = (1 - x)^{2} \Rightarrow x = \frac{4}{9}. $$ By $\triangle AEC' \sim \triangle DC'N$, $$ \Rightarrow \frac{AC'}{DN} = \frac{AE}{C'D} = \frac{C'E}{C'N}. $$ Given $AC' = \frac{2}{3}, C'N = \frac{5}{9}$, we have $$ AE = \frac{1}{2}, C'E = \frac{5}{6}. $$ Thus, $AE + AC' + C'E = \frac{3+4+5}{6} = 2$." 95fa21315842,"Example 1. Draw a tangent line to the circle $x^{2}+y^{2}=25$ through point C $(6,-8)$, find the distance from C to the line connecting the two points of tangency. 保留了原文的格式和换行。",See reasoning trace,easy,"Solve: As shown in Figure 2, the equation of the chord of contact A.B is: $$ 6 x-8 y-25=0 \text {. } $$ The distance from C $(6,-8)$ to the chord of contact $\mathrm{AB}$ is: $$ \mathrm{d}=\frac{|36+64-25|}{\sqrt{6^{2}+(-8)^{2}}}=7.5 . $$" 1a2af24a145c,9.189. $\log _{5} \sqrt{3 x+4} \cdot \log _{x} 5>1$.,$x \in(1 ; 4)$,easy,"## Solution. Domain of definition: $01$. From this, we have $\log _{x} \sqrt{3 x+4}>1$. The last inequality is equivalent to two systems of inequalities: 1) $\left\{\begin{array}{l}00,\end{array}\right.\right.\right.$ 2); $\left\{\begin{array}{l}x>1, \\ \sqrt{3 x+4}>x\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 3 x+4>x^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ (x-4)(x+1)<0\end{array} \Leftrightarrow 1 1$. Let's find the derivative of this function and its critical points within the given interval: ![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-33.jpg?height=111&width=532&top_left_y=2077&top_left_x=763) If $h \geq 2$, then there are no critical points on this interval, and the derivative is positive on this interval, meaning the function is increasing. Therefore, $S_{\min} = S(h) = \frac{4h^2}{h - 1}$. If $1 < h < 2$, then there is one critical point $x = 2$ on the interval $[h; +\infty)$. As we pass through this point, the derivative changes sign from ""-"" to ""+"", so the point $x = 2$ is a point of minimum. In this case, $S_{\min} = S(2) = 16$. This is the minimum value of $S(x)$ on the interval $[h; +\infty)$ when $h \geq 2$. ## Answer If $h < 2$, then $S_{\min} = S(2) = 16$; if $h \geq 2$, then $S_{\min} = S(h) = \frac{4h^2}{h - 1}$." a11007df2883,"7. The function $$ f(x)=\frac{\sin x-1}{\sqrt{3-2 \cos x-2 \sin x}}(0 \leqslant x \leqslant 2 \pi) $$ has the range . $\qquad$","-\cos \theta \in[-1,0]$.",easy,"7. $[-1,0]$. Let $1-\sin x=a, 1-\cos x=b$. Then $$ y=f(x)=\frac{-a}{\sqrt{a^{2}+b^{2}}}(0 \leqslant a \leqslant 1,0 \leqslant b \leqslant 1), $$ and the point $P(a, b)$ lies on the circle $(x-1)^{2}+(y-1)^{2}=1$. Let the inclination angle of the line $O P$ be $\theta$. Then $$ 0 \leqslant \theta \leqslant \frac{\pi}{2} \text {. } $$ Therefore, $y=-\cos \theta \in[-1,0]$." a82176030c3b,"The number of positive integers $k$ for which the equation \[kx-12=3k\] has an integer solution for $x$ is $\text{(A) } 3\quad \text{(B) } 4\quad \text{(C) } 5\quad \text{(D) } 6\quad \text{(E) } 7$",3-x$,easy,"$\fbox{D}$ $kx -12 = 3k$ $-12=3k-kx$ $-12=k(3-x)$ $\frac{-12}{k}=3-x$ Positive factors of $-12$: $1,2,3,4,6,12$ 6 factors, each of which have an integer solution for $x$ in $\frac{-12}{k}=3-x$" 3c208be32083,"Example 3 Given in $\triangle A B C$, $A B=A C$, $\angle B A C=100^{\circ}, P$ is a point on the angle bisector of $\angle C$, $\angle P B C=10^{\circ}$. Find the degree measure of $\angle A P B$.","30^{\circ}$ of $\triangle PBC$, and construct an equilateral triangle, which creates conditions for ",medium,"Solution: As shown in Figure 6, take a point $D$ on the extension of $CA$ such that $CD = CB$, and connect $DP$ and $DB$. Let the intersection of $DB$ and $CP$ be $E$. It is easy to see that $B$ and $D$ are symmetric with respect to $PC$. Therefore, $\triangle PBD$ is an equilateral triangle, and $\angle PDA = \angle PBC = 10^{\circ}$. In $\triangle CBD$, we know that $\angle DBC = 70^{\circ}$. Thus, $\angle ABD = 30^{\circ} = \angle ABP$, so $AB$ is the perpendicular bisector of $PD$. $$ \therefore AP = AD. $$ Therefore, $\angle APD = \angle ADP = 10^{\circ}$. $$ \therefore \angle APB = 70^{\circ}. $$ Here, we focus on the exterior angle $\angle EPB = 30^{\circ}$ of $\triangle PBC$, and construct an equilateral triangle, which creates conditions for solving the problem." ec47af8f05e7,5. 5.1. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7.,60718293,medium,"Answer: 60718293. ## Solution. We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7. ![](https://cdn.mathpix.com/cropped/2024_05_06_8a1d5b347fbc679ac0eeg-5.jpg?height=448&width=590&top_left_y=524&top_left_x=819) We see that the vertices corresponding to the digits 4 and 5 are not connected to anything, so the maximum number that can be obtained will be 8 digits long. The vertices corresponding to the digits 3 and 6 can only be the first or last in the number. Since we need to find the maximum number, it is uniquely determined by moving along the edges of the graph and will be equal to 60718293." a0faa1ec4125,"10. (15 points) Kangaroo A and B are jumping in the area shown in the figure. A jumps in the order of $A B C D E F G H I A B C \cdots$, and B jumps in the order of $A B D E G H A B D \cdots$. If both kangaroos start from $A$ and this counts as the first time they jump together, how many times have they jumped together after 2017 jumps?","After 2017 jumps, they meet 226 times",medium,"【Analysis】First, find the least common multiple of the periods of 2 jumps, 6 and 9, which is 18. In this cycle, there are 2 meetings. Finding the number of groups and the remainder can solve the problem. 【Solution】According to the problem, we know: By enumeration, the table is as follows: \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline Kangaroo A & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $A$ & $\cdots$ \\ \hline Kangaroo B & $A$ & $B$ & $D$ & $E$ & $G$ & $H$ & $A$ & $B$ & $D$ & $E$ & $G$ & $H$ & $A$ & $B$ & $D$ & $E$ & $G$ & $H$ & $A$ & $\cdots$ \\ \hline \end{tabular} The cycle length is 18. There are two meetings in each cycle. $$ 2017 \div 18=112 \cdots 1 \text {. } $$ Therefore, after 2017 jumps, the two kangaroos meet $1+2 \times 112+1=226$ (times); Answer: After 2017 jumps, they meet 226 times." f2e9c3515b8d,"A right rectangular prism has integer side lengths $a$, $b$, and $c$. If $\text{lcm}(a,b)=72$, $\text{lcm}(a,c)=24$, and $\text{lcm}(b,c)=18$, what is the sum of the minimum and maximum possible volumes of the prism? [i]Proposed by Deyuan Li and Andrew Milas[/i]",3024,medium,"To solve this problem, we need to find the integer side lengths \(a\), \(b\), and \(c\) of a right rectangular prism such that: \[ \text{lcm}(a, b) = 72, \quad \text{lcm}(a, c) = 24, \quad \text{lcm}(b, c) = 18 \] We will then find the sum of the minimum and maximum possible volumes of the prism. 1. **Finding the prime factorizations:** \[ 72 = 2^3 \cdot 3^2, \quad 24 = 2^3 \cdot 3, \quad 18 = 2 \cdot 3^2 \] 2. **Determine the possible values of \(a\), \(b\), and \(c\):** - Since \(\text{lcm}(a, b) = 72\), \(a\) and \(b\) must have the prime factors \(2\) and \(3\) such that their least common multiple is \(72\). - Similarly, \(\text{lcm}(a, c) = 24\) and \(\text{lcm}(b, c) = 18\). 3. **Minimizing the volume:** - Let's test \(a = 8\) (since \(8 = 2^3\)). - For \(b\), we need \(\text{lcm}(8, b) = 72\). The smallest \(b\) that satisfies this is \(b = 9\) (since \(9 = 3^2\)). - For \(c\), we need \(\text{lcm}(8, c) = 24\) and \(\text{lcm}(9, c) = 18\). The smallest \(c\) that satisfies both conditions is \(c = 6\) (since \(6 = 2 \cdot 3\)). - Therefore, the minimum volume is: \[ V_{\text{min}} = 8 \cdot 9 \cdot 6 = 432 \] 4. **Maximizing the volume:** - Let's test \(a = 24\) (since \(24 = 2^3 \cdot 3\)). - For \(b\), we need \(\text{lcm}(24, b) = 72\). The largest \(b\) that satisfies this is \(b = 18\) (since \(18 = 2 \cdot 3^2\)). - For \(c\), we need \(\text{lcm}(24, c) = 24\) and \(\text{lcm}(18, c) = 18\). The largest \(c\) that satisfies both conditions is \(c = 6\) (since \(6 = 2 \cdot 3\)). - Therefore, the maximum volume is: \[ V_{\text{max}} = 24 \cdot 18 \cdot 6 = 2592 \] 5. **Sum of the minimum and maximum volumes:** \[ V_{\text{sum}} = V_{\text{min}} + V_{\text{max}} = 432 + 2592 = 3024 \] The final answer is \(\boxed{3024}\)." 0fea9351ebb0,"5. April, May and June have 90 sweets between them. May has three-quarters of the number of sweets that June has. April has two-thirds of the number of sweets that May has. How many sweets does June have? A 60 B 52 C 48 D 40 E 36",90$. So $\frac{9 x}{4}=90$. Hence $x=\frac{4}{9} \times 90=40$.,easy,"5. D Let June have $x$ sweets. Then May has $\frac{3 x}{4}$ sweets. Also, April has $\frac{2}{3} \times \frac{3 x}{4}$, that is $\frac{x}{2}$, sweets. Therefore $x+\frac{3 x}{4}+\frac{x}{2}=90$. So $\frac{9 x}{4}=90$. Hence $x=\frac{4}{9} \times 90=40$." 7fa6af114956,"Example 5 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, Q$ is a point inside the triangle, $\angle Q C A=$ $\angle Q A B=20^{\circ}$. Find the degree measure of $\angle Q B C$.","QA$, making point $D$ the symmetric point of $Q$ with respect to $AC$. By using the perpendicular bi",medium,"Solution: As shown in Figure 5, let the perpendicular bisector of $BC$ intersect $BA$ and $AC$ at $D$ and $E$, respectively, and let $F$ be the foot of the perpendicular. Connect $QE$, $BE$, and $DC$. Given $\angle ACD = 20^{\circ} = \angle ACQ$ and $\angle DAC = 80^{\circ} = \angle QAC$, we know that point $D$ is symmetric to point $Q$ with respect to $AC$. Therefore, $\angle AEQ = \angle AED = \angle FEC = 60^{\circ}$. Since $\angle BEF = \angle FEC = 60^{\circ}$, we have $\angle AEB = 60^{\circ} = \angle AEQ$. Thus, points $B$, $Q$, and $E$ are collinear. Then $\angle QBC = \angle EBC = 30^{\circ}$. Note that $AC$ is the external angle bisector of $\angle QAB$ in $\triangle AQB$ (this is not immediately obvious). On the extension of $BA$, take a point $D$ such that $DA = QA$, making point $D$ the symmetric point of $Q$ with respect to $AC$. By using the perpendicular bisector of $BC$, we appropriately ""flip"" $\angle ABC$ to the position of $\angle DCE$." dcbea7eb596f,"Example 24 Find all five-digit numbers $\overline{a b c d e}$, such that the number is divisible by 9, and $\overline{a c e}-\overline{b d a}=760$.","2$, and the corresponding five-digit number is 91269.",medium,"The number theory problem to solve the equation $\overline{a c e}-\overline{b \overline{d a}}=760$ can be rewritten as $100 a+10 c+e-100 b-10 d-a=760$, from which we can deduce that $e=a$. Dividing both sides of the equation by 10, we get $10(a-b)+(c-d)=76$. Therefore, there are only two possibilities: $c-d=6$ or $c-d=-4$. If $c-d=6$, then we have $c=d+6, a=b+7$. Since the five-digit number is divisible by 9, we have $$ \begin{aligned} a+b+c+d+e & =b+7+b+d+6+d+b+7 \\ & =3 b+2 d+20=3 b+2(d+1+9) \end{aligned} $$ which must be divisible by 9. So, $d+1$ must be divisible by 3. Since $c-d=6$, we get $d=2$, thus $c=8$. At the same time, $3(b+2)$ must be divisible by 9. Since $a=b+7$, we have $b=1$. In this case, the desired five-digit number is 81828. If $c-d=-4$, then we have $$ d=c+4, a=b+8 \text {. } $$ Therefore, $a=8, b=0$, or $a=9, b=1$. If $a=8$, by $$ 9 \mid(a+b+c+d+e)=8+2 c+4+8, $$ we can deduce that $2 c+2$ must be divisible by 9. Thus, $c=8$, and consequently $d=c+4=12$, which is not a single digit. If $a=9$, by $$ 9 \mid(a+b+c+d+e)=10+2 c+4+9, $$ we can deduce that $2 c+5$ must be divisible by 9. Thus, $c=2$, and the corresponding five-digit number is 91269." a7659f33920e,"34. Find the number of 20 -tuples of integers $x_{1}, \ldots, x_{10}, y_{1}, \ldots, y_{10}$ with the following properties: - $1 \leq x_{i} \leq 10$ and $1 \leq y_{i} \leq 10$ for each $i$; - $x_{i} \leq x_{i+1}$ for $i=1, \ldots, 9$ - if $x_{i}=x_{i+1}$, then $y_{i} \leq y_{i+1}$.","10 x_{i}+y_{i}$, we see that the problem is equivalent to choosing a nondecreasing sequence of numbe",medium,"Solution: $\left.\begin{array}{c}109 \\ 10\end{array}\right)$ By setting $z_{i}=10 x_{i}+y_{i}$, we see that the problem is equivalent to choosing a nondecreasing sequence of numbers $z_{1}, z_{2}, \ldots, z_{10}$ from the values $11,12, \ldots, 110$. Making a further substitution by setting $w_{i}=z_{i}-11+i$, we see that the problem is equivalent to choosing a strictly increasing sequence of numbers $w_{1}, \ldots, w_{10}$ from among the values $1,2, \ldots, 109$. There are $\binom{109}{10}$ ways to do this." 0aa375e4891b,"Task B-2.1. (20 points) For complex numbers $z, w$ such that $|z|=|w|=|z-w|$, calculate $\left(\frac{z}{w}\right)^{99}$.",See reasoning trace,easy,"First solution. Let $u=\frac{z}{w}=x+y i$. Then $|u|=1,|u-1|=1$. Then $x^{2}+y^{2}=1,(x-1)^{2}+y^{2}=1$. By subtracting these equations, we get $x=\frac{1}{2}, y= \pm \frac{\sqrt{3}}{2}$, so $u=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$. Now, $$ \left(\frac{z}{w}\right)^{99}=u^{99}=\left[\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right)^{3}\right]^{33}=\left(\frac{1}{8} \pm \frac{3 \sqrt{3}}{8} i-\frac{9}{8} \mp \frac{3 \sqrt{3}}{8} i\right)^{33}=(-1)^{33}=-1 $$" 2f84b72473be,"6. Let $M=\left\{a \mid a=x^{2}-y^{2}, x, y \in \mathbf{Z}\right\}$. Then for any integer $n$, among the numbers of the form $4 n, 4 n+1, 4 n+2, 4 n+3$, the number that is not an element of the set $M$ is ( ). (A) $4 n$ (B) $4 n+1$ (C) $4 n+2$ (D) $4 n+3$","(x+y)(x-y)$, and $x+y$ and $x-y$ have the same parity, $(x+y)(x-y)$ is either odd or a multiple of 4",medium,"6. C. Notice that for any integer $n$, $$ \begin{array}{l} 4 n=(n+1)^{2}-(n-1)^{2}, \\ 4 n+1=(2 n+1)^{2}-(2 n)^{2}, \\ 4 n+3=(2 n+2)^{2}-(2 n+1)^{2}, \end{array} $$ and $n-1, n+1, 2 n, 2 n+1, 2 n+2 \in \mathbf{Z}$. Thus, $4 n, 4 n+1, 4 n+3 \in M$. If $4 n+2$ is an element of set $M$, then there exist $x, y \in \mathbf{Z}$ such that $4 n+2=x^{2}-y^{2} (x, y \in \mathbf{Z})$. However, since $x^{2}-y^{2}=(x+y)(x-y)$, and $x+y$ and $x-y$ have the same parity, $(x+y)(x-y)$ is either odd or a multiple of 4, meaning $4 n+2$ is not an element of set $M$." caa5343c3c94,"2. If for all $\theta \in \mathbf{R}$, the complex number $$ z=(a+\cos \theta)+(2 a-\sin \theta) \mathrm{i} $$ has a modulus not exceeding 2, then the range of the real number $a$ is $\qquad$.",See reasoning trace,medium,"2. $\left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right]$. According to the problem, we have $$ \begin{array}{l} |z| \leqslant 2 \\ \Leftrightarrow(a+\cos \theta)^{2}+(2 a-\sin \theta)^{2} \leqslant 4 \\ \Leftrightarrow 2 a(\cos \theta-2 \sin \theta) \leqslant 3-5 a^{2} \\ \Leftrightarrow-2 \sqrt{5} a \sin (\theta-\varphi) \leqslant 3-5 a^{2} \end{array} $$ This must hold for any real number $\theta$, where $\varphi=\arcsin \frac{1}{\sqrt{5}}$. Thus, $2 \sqrt{5}|a| \leqslant 3-5 a^{2} \Rightarrow|a| \leqslant \frac{\sqrt{5}}{5}$. Therefore, $a \in\left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right]$." 3b56538e866b,"4. While working on fruit harvesting, class 5A collected 560 kg of apples, 5B - 595 kg, and 5V - 735 kg. All the collected apples were packed into boxes, with each box containing the same maximum possible number of kilograms. How many such boxes were required for each class?","2^{4} \cdot 5 \cdot 7, 595=5 \cdot 7 \cdot 17$ and $735=3 \cdot 5 \cdot 7^{2}$, then GCD $(560,595,7",easy,"4. The number of kilograms of apples in each box is the greatest common divisor of the numbers 560, 595, and 735. Since $560=2^{4} \cdot 5 \cdot 7, 595=5 \cdot 7 \cdot 17$ and $735=3 \cdot 5 \cdot 7^{2}$, then GCD $(560,595,735)=35$. This means that 35 kg of apples were placed in each box. Then the first squad needed 16 boxes $(560: 35=16)$, the second squad needed 17 boxes $(595: 35=17)$, and the third squad needed 21 boxes $(735: 35=21)$." d412fb0178c9,"8. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be 4 distinct numbers from $1,2, \cdots, 100$, satisfying $$ \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right)=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2}, $$ then the number of such ordered quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is $\qquad$ .",40,medium,"Answer: 40. Solution: By Cauchy-Schwarz inequality, we know that $\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \geq\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2}$, equality holds if and only if $\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{3}}{a_{4}}$, which means $a_{1}, a_{2}, a_{3}, a_{4}$ form a geometric sequence. Therefore, the problem is equivalent to counting the number of geometric sequences $a_{1}, a_{2}, a_{3}, a_{4}$ that satisfy $\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\} \subseteq\{1,2,3, \cdots, 100\}$. Let the common ratio of the geometric sequence be $q \neq 1$, and $q$ is a rational number. Denote $q=\frac{n}{m}$, where $m, n$ are coprime positive integers, and $m \neq n$. First, consider the case where $n>m$. In this case, $a_{4}=a_{1} \cdot\left(\frac{n}{m}\right)^{3}=\frac{a_{1} n^{3}}{m^{3}}$. Note that $m^{3}, n^{3}$ are coprime, so $l=\frac{a_{1}}{m^{3}}$ is a positive integer. Accordingly, $a_{1}, a_{2}, a_{3}, a_{4}$ are $m^{3} l, m^{2} n l, m n^{2} l, n^{3} l$, all of which are positive integers. This indicates that for any given $q=\frac{n}{m}>1$, the number of geometric sequences $a_{1}, a_{2}, a_{3}, a_{4}$ that satisfy the conditions and have $q$ as the common ratio is the number of positive integers $l$ that satisfy the inequality $n^{3} l \leq 100$, which is $\left[\frac{100}{n^{3}}\right]$. Since $5^{3}>100$, we only need to consider the cases $q=2,3, \frac{3}{2}, 4, \frac{4}{3}$. The corresponding number of geometric sequences is $\left[\frac{100}{8}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{64}\right]+\left[\frac{100}{64}\right]=12+3+3+1+1=20$. When $n1$, they are all composite numbers. Therefore, the only prime number that meets the condition is 3." a98e017ff769,3. The monotonic increasing interval of the function $f(x)$ $=\frac{(x+1)^{2}}{x^{2}+1}$ is $\qquad$,See reasoning trace,easy,"$$ f(x)=1+\frac{2 x}{x^{2}+1}, f^{\prime}(x)=\frac{2\left(x^{2}+1\right)-2 x \cdot 2 x}{\left(x^{2}+1\right)^{2}}=-\frac{2(x+1)(x-1)}{\left(x^{2}+1\right)^{2}}>0 $$ $\Rightarrow -10 \Rightarrow a_{n+1} \in\left(0, \frac{\pi}{2}\right)$. Then $a_{n+1}=\arctan \left(\sec a_{n}\right) \Rightarrow \tan a_{n+1}=\sec a_{n}=\sqrt{1+\tan ^{2} a_{n}}$ $\Rightarrow \tan ^{2} a_{n+1}-\tan ^{2} a_{n}=1$, and $\tan ^{2} a_{1}=\frac{1}{3}$, Thus $\left\{\tan ^{2} a_{n}\right\}$ is an arithmetic sequence with the first term $\frac{1}{3}$ and common difference 1 $\Rightarrow \tan ^{2} a_{n}=\frac{3 n-2}{3}$. Therefore, $\sin ^{2} a_{n}=\frac{3 n-2}{3 n+1} \Rightarrow \sin a_{n}=\sqrt{\frac{3 n-2}{3 n+1}}\left(n \in \mathbf{N}^{*}\right)$. Hence, $\sin a_{1} \cdot \sin a_{2} \cdots \sin a_{m}=\sqrt{\frac{1}{4} \cdot \frac{4}{7} \cdots \frac{3 m-2}{3 m+1}}=\sqrt{\frac{1}{3 m+1}}=\frac{1}{100}$ $\Rightarrow 3 m+1=10000 \Rightarrow m=3333$" b36608a47091,"【Question 2】 The sum of two numbers is 17, the difference between the two numbers is 7, the larger number is $\qquad$.",12$.,easy,"【Analysis and Solution】 Sum and difference problem. Larger number + Smaller number $=17$; Larger number - Smaller number $=7$; The larger number is $(17+7) \div 2=12$." 44479d7c34e2,11.5. Solve the system of equations $\left\{\begin{array}{l}3^{x}=\sqrt{y} \\ 2^{-y}=x^{3}\end{array}\right.$.,". $x=1 / 2, y=3$",medium,"Answer. $x=1 / 2, y=3$. Solution. It is directly verified that the numbers $x=1 / 2, y=3$ are a solution to the system. We will prove that the solution is unique. For this, we will show that the function $y(x)$, defined by the first equation, is strictly decreasing, and the function defined by the second equation is strictly decreasing. Indeed, the derivative of the first function is $y'=\left(3^{2 x}\right)'=\ln 3 \cdot 3^{2 x} \cdot 2>0$. The second function is defined for $x>0$ and its derivative has the form $y'=\left(-\frac{3 \ln x}{\ln 2}\right)'=-\frac{3}{\ln 2} \cdot \frac{1}{x}<0$. Thus, the system has a unique solution $x=1 / 2, y=3$. ![](https://cdn.mathpix.com/cropped/2024_05_06_ca5361dea44456c8863cg-1.jpg?height=497&width=800&top_left_y=2133&top_left_x=1162) When justifying the monotonicity of the specified functions, one can also avoid using the derivative and refer to the corresponding properties of exponential and logarithmic functions for specific bases." ab529c8a73f3,"4-123 Let $a, b, c$ be given positive real numbers, try to determine all positive real numbers $x, y, z$ that satisfy the system of equations $$\left\{\begin{array}{l} x+y+z=a+b+c \\ 4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c . \end{array}\right.$$","\frac{b+c}{2}, \quad y=\frac{c+a}{2}, \quad z=\frac{a+b}{2} .$$",medium,"［Solution］ The second equation above is equivalent to $$\frac{a^{2}}{y z}+\frac{b^{2}}{x z}+\frac{c^{2}}{x y}+\frac{a b c}{x y z}=4$$ Let $x_{1}=\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}=\frac{c}{\sqrt{x y}}$, then $$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$$ Obviously, $00$, thus $$z_{1}=-2 \sin u \sin v+2 \cos u \cos v=2 \cos (u+v)$$ Therefore, we have $$\left\{\begin{array}{l} 2 \sin u \sqrt{y z}=a \\ 2 \sin v \sqrt{z x}=b \\ 2 \cos (u+v) \sqrt{x y}=c \end{array}\right.$$ Thus, $$\begin{aligned} & (\sqrt{x} \cos v-\sqrt{y} \cos u)^{2}+(\sqrt{x} \sin v+\sqrt{y} \sin u-\sqrt{z})^{2} \\ = & x+y+z-2 \sqrt{x y} \cos u \cos v+2 \sqrt{x y} \sin u \sin v-2 \sqrt{z} x \sin v \\ & -2 \sqrt{y z} \sin u \\ = & a+b+c-c-b-a \\ = & 0 \end{aligned}$$ Therefore, $$\begin{aligned} \sqrt{z} & =\sqrt{x} \sin v+\sqrt{y} \sin u \\ & =\sqrt{x} \cdot \frac{b}{2 \sqrt{z x}}+\sqrt{y} \cdot \frac{a}{2 \sqrt{y z}} \\ & =\frac{1}{\sqrt{z}} \cdot \frac{a+b}{2} \\ z & =\frac{a+b}{2} \end{aligned}$$ By the symmetry of the original equation, we get $$x=\frac{b+c}{2}, y=\frac{c+a}{2}$$ It is easy to verify that the triplet $\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right)$ satisfies the given system of equations. Therefore, the original system of equations has only one positive real solution: $$x=\frac{b+c}{2}, \quad y=\frac{c+a}{2}, \quad z=\frac{a+b}{2} .$$" 5e99bd5a789e,$14 \cdot 20$ Find the real solutions of the equation $3 x^{3}-[x]=3$.,\sqrt[3]{\frac{4}{3}}$.,easy,"[Solution] (1) If $x \leqslant -1$, then $$ 3 x^{3}-[x] \leqslant 3 x - x + 13 x - x \geqslant 4, $$ (4) If $0 \leqslant x < 1$, then $$ 3 x^{3}-[x]=3 x^{3}<3 \text{. } $$ In all the above cases, the equation $$ 3 x^{3}-[x]=3 $$ has no real solutions. The last case. (5) If $1 \leqslant x < 2$, then $$ \begin{array}{c} 3 x^{3}-[x]=3 x^{3}-1=3 . \\ x=\sqrt[3]{\frac{4}{3}} . \end{array} $$ Therefore, the given equation has only one solution $x=\sqrt[3]{\frac{4}{3}}$." d616393d3db6,What is the sum of the squares of the roots of the equation $x^2 -7 \lfloor x\rfloor +5=0$ ?,92,medium,"1. Given the equation \(x^2 - 7 \lfloor x \rfloor + 5 = 0\), we need to find the sum of the squares of the roots of this equation. 2. By definition of the floor function, we have: \[ x - 1 < \lfloor x \rfloor \leq x \] Multiplying by 7: \[ 7x - 7 < 7 \lfloor x \rfloor \leq 7x \] 3. Substituting \(7 \lfloor x \rfloor\) in the given equation: \[ 7x - 7 < x^2 + 5 \leq 7x \] 4. Subtracting \(7x\) from all sides: \[ -7 < x^2 - 7x + 5 \leq 0 \] 5. Completing the square for the quadratic expression \(x^2 - 7x + 5\): \[ x^2 - 7x + \frac{49}{4} - \frac{49}{4} + 5 = \left(x - \frac{7}{2}\right)^2 - \frac{29}{4} \] Thus, the inequality becomes: \[ -7 < \left(x - \frac{7}{2}\right)^2 - \frac{29}{4} \leq 0 \] 6. Adding \(\frac{29}{4}\) to all sides: \[ -7 + \frac{29}{4} < \left(x - \frac{7}{2}\right)^2 \leq \frac{29}{4} \] Simplifying: \[ \frac{1}{4} < \left(x - \frac{7}{2}\right)^2 \leq \frac{29}{4} \] 7. Multiplying by 4: \[ 1 < (2x - 7)^2 \leq 29 \] 8. Since we are dealing with the floor function, we can consider the perfect square within the range: \[ 1 < (2x - 7)^2 \leq 25 \] 9. Solving the inequality \(1 < (2x - 7)^2 \leq 25\): - For the positive case: \[ 1 < 2x - 7 \leq 5 \implies 4 < 2x \leq 12 \implies 2 < x \leq 6 \] - For the negative case: \[ -5 \leq 2x - 7 < -1 \implies 2 \leq 2x < 6 \implies 1 \leq x < 3 \] 10. Considering the integer values for \(\lfloor x \rfloor\) within these ranges: - For \(2 < x \leq 6\), \(\lfloor x \rfloor\) can be 3, 4, 5, or 6. - For \(1 \leq x < 3\), \(\lfloor x \rfloor\) can be 1 or 2. 11. Checking each value of \(\lfloor x \rfloor\): - For \(\lfloor x \rfloor = 1\): \[ x^2 - 7(1) + 5 = 0 \implies x^2 - 2 = 0 \implies x^2 = 2 \] - For \(\lfloor x \rfloor = 2\): \[ x^2 - 7(2) + 5 = 0 \implies x^2 - 9 = 0 \implies x^2 = 9 \] - For \(\lfloor x \rfloor = 4\): \[ x^2 - 7(4) + 5 = 0 \implies x^2 - 23 = 0 \implies x^2 = 23 \] - For \(\lfloor x \rfloor = 5\): \[ x^2 - 7(5) + 5 = 0 \implies x^2 - 30 = 0 \implies x^2 = 30 \] - For \(\lfloor x \rfloor = 6\): \[ x^2 - 7(6) + 5 = 0 \implies x^2 - 37 = 0 \implies x^2 = 37 \] 12. Summing the valid squares: \[ 2 + 23 + 30 + 37 = 92 \] The final answer is \(\boxed{92}\)" fcba171c7dbc,"\section*{Aufgabe 2 - 171212} Man ermittle alle reellen Lösungen \((x, y)\) des Gleichungssystems \[ \begin{aligned} 2 \cdot \sqrt{x+y}+x+y & =8 \\ x^{3}+y^{3} & =40 \end{aligned} \]",See reasoning trace,medium,"} Angenommen, es sei ( \(x, y\) ) eine reelle Lösung des Gleichungssystems (1), (2). Dann gilt \(x+y \geq 0\). Setzt man \(z=\sqrt{x+y}\), so gilt \(z \geq 0\) und wegen (1) \[ z^{2}+2 z-8=0 \] Die quadratische Gleichung (3) hat genau eine nichtnegative reelle Lösung, nämlich \(z=2\). Daraus folgt \[ x+y=4 \quad ; \quad y=4-x \] also wegen \((2)\) \[ \begin{aligned} x^{3}+(4-x)^{3} & =40 \\ x^{3}+\left(64-48 x+12 x^{2}-x^{3}-40\right. & =0 \\ x^{2}-4 x+2 & =0 \end{aligned} \] Die quadratische Gleichung (5) hat genau zwei reelle Lösungen, nämlich \(x_{1}=2 \sqrt{2}\); dann ist wegen (4) \(y_{1}=2-\sqrt{2}\), und \(x_{2}=2-\sqrt{2}\), dann ist \(y_{2}=2+\sqrt{2}\). Wenn also das Gleichungssystem (1), (2) überhaupt reelle Lösungen hat, so können es nur die Paare \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\) sein. Für \((x, y)=(2+\sqrt{2}, 2-\sqrt{2})\) sowie für \((x, y)=(2-\sqrt{2}, 2+\sqrt{2})\) gilt nun \[ \begin{aligned} 2 \sqrt{x+y}+x+y & =2 \sqrt{4}+4=8 \\ x^{3}+y^{2}=(2+\sqrt{2})^{3}+(2-\sqrt{2})^{3} & =2(8+12)=40 \end{aligned} \] d.h., die Gleichungen (1) und (2) sind erfüllt. Daher hat das Gleichungssystem (1), (2) genau zwei reelle Lösungen, nämlich \((2+\sqrt{2}, 2-\sqrt{2})\) und \((x, y)=(2-\sqrt{2}, 2+\sqrt{2})\). Übernommen von \([5]\) " 06f1f177eb6d,"13. (2005 National High School Mathematics Competition) Let the set $T=\{0,1,2,3,4,5,6\}, M=$ $\left\{\left.\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}} \right\rvert\, a_{i} \in T, i=1,2,3,4\right\}$, and arrange the elements of $M$ in descending order. The 2005th number is ( ). A. $\frac{5}{7}+\frac{5}{7^{2}}+\frac{6}{7^{3}}+\frac{3}{7^{4}}$ B. $\frac{5}{7}+\frac{5}{7^{2}}+\frac{6}{7^{3}}+\frac{2}{7^{4}}$ C. $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{1}}$ D. $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{3}{7^{4}}$",C,medium,"13. Use $\left[a_{1} a_{2} \cdots a_{k}\right]_{p}$ to represent a $k$-digit number in base $p$. Multiply each number in set $M$ by $7^{4}$, we get $$ \begin{aligned} M^{\prime} & =\left\{a_{1} \cdot 7^{3}+a_{2} \cdot 7^{2}+a_{3} \cdot 7+a_{4} \mid a_{i} \in T, i=1,2,3,4\right\} \\ & =\left\{\left[a_{1} a_{2} a_{3} a_{4}\right]_{7} \mid a_{i} \in T, i=1,2,3,4\right\} . \end{aligned} $$ The maximum number in $M^{\prime}$ is $[6666]_{7}=[2400]_{10}$. In decimal numbers, the 2005th number from 2400 in descending order is $2400-2004=396$. And $[396]_{10}=[1104]_{7}$. Dividing this number by $7^{4}$, we get the number in $M$ as $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{4}}$. Therefore, the answer is C." 1d9812044c55,"$2 \cdot 88$ Using each digit from $0,1,2,3, \cdots, 9$ once, find the largest possible multiple of 12 that can be formed.",to this problem is 9876543120,easy,"[Solution] A multiple of 12 can be divided by both 3 and 4. Since $9+8+7+6+5+4$ $+3+2+1=45$, which is divisible by 3, to make the number a multiple of 12, we only need its last two digits to be divisible by 4. Since placing larger digits at the front forms a larger number, the answer to this problem is 9876543120." 477a389b235c,14. Simplify the fraction: $\frac{37373737}{81818181}$,\frac{37 \cdot 1010101}{81 \cdot 1010101}=\frac{37}{81}$.,easy,14. Solution: $\frac{37373737}{81818181}=\frac{37 \cdot 1010101}{81 \cdot 1010101}=\frac{37}{81}$. 6dda2f0002e3,"5. Borya and Vova are playing the following game on an initially white $8 \times 8$ board. Borya moves first and on each of his turns, he colors any four white cells black. After each of his moves, Vova colors an entire row (row or column) completely white. Borya aims to color as many cells black as possible, while Vova aims to prevent him. What is the maximum number of black cells that can be on the board after Borya's move, no matter how Vova plays? Answer: 25 cells.",See reasoning trace,medium,"Solution. Let Vova make white the row with the most black cells on each of his moves. Then, as soon as Borya achieves a row of no less than four black cells (we will call such a row ""rich""), Vova will remove at least four cells, meaning that Borya will not be able to increase the number of black cells compared to his previous move. And further, with the presence of ![](https://cdn.mathpix.com/cropped/2024_05_06_704e715e936cf2f990c2g-3.jpg?height=392&width=394&top_left_y=1021&top_left_x=1508) rich rows, Vova removes at least 4 black cells, and Borya adds four after that, meaning he cannot increase his maximum result. At the same time, before the first and all subsequent moments of creating a rich row when there are no rich rows on the board, there were a maximum of \(7 \cdot 3 = 21\) black cells—seven rows of 3 black cells, since Vova just made a row of this direction white. Thus, Borya always creates a structure with a maximum of 25 black cells. Now we will show that Borya can always achieve 25 black cells. Let's highlight 24 cells on the board as shown in the figure—three in each vertical and horizontal row. Suppose Borya always colors only the cells of this set, as long as there are white cells among them. Then Vova, with his next move, can remove a maximum of three black cells, meaning the number of black cells after each pair of their moves will increase by at least 1. Therefore, at some point, Borya will be able to make all these cells black (and possibly some others). Then, after Vova's move, there will be no fewer than 21 black cells left, and Borya, with his next move, will achieve at least 25 black cells. Thus, with correct play by both, the maximum number of black cells on the board that Borya can ensure is 25. Criteria. If the solution is incorrect - 0 points. If only the correct example is provided - 3 points. If only the correct estimate is provided - 3 points. If the correct solution (any correct example and justification of the estimate) - 7 points." dce8ec6ace85,"Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is: $\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \text{finite but more than 2}\qquad \textbf{(E)}\ \infty$",\textbf{(C),medium,"For two lines to be the same, their slopes must be equal and their intercepts must be equal. This is a necessary and sufficient condition. The slope of the first line is $\frac{-3}{b}$, while the slope of the second line is $\frac{c}{2}$. Thus, $\frac{-3}{b} = \frac{c}{2}$, or $bc = -6$. The intercept of the first line is $\frac{-c}{b}$, while the intercept of the second line is $6$. Thus, $6 = \frac{-c}{b}$, or $-6b = c$. Plugging $-6b = c$ into $bc = -6$ gives $b(-6b) = -6$, or $b^2 = 1$. This means $b = \pm 1$ This in turn gives $c = \mp 6$. Thus, $(b, c) = (\pm 1, \mp 6)$, for two solutions, which is answer $\boxed{\textbf{(C)}}$" 1444d74e3822,"Example 1-13 5 flags of different colors, 20 different kinds of potted flowers, arranged in a form with two flags at the ends and 3 potted flowers in the middle. How many different arrangements are there?",See reasoning trace,easy,"The number of permutations of taking 2 flags out of 5 is $$ P(5,2)=5 \times 4=20 $$ The number of permutations of taking 3 flowers out of 20 is $$ P(20,3)=20 \times 19 \times 18=6840 $$ According to the multiplication rule, the total number of schemes is $$ N=20 \times 6840=136800 $$" 194330e61313,"10. (20 points) Try to find the maximum value of the function defined on $(0, \pi)$: $$ \begin{aligned} f(x)= & \log _{\sqrt{2}-1}\left[\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)+1\right]+ \\ & \log _{\sqrt{2}+1} \frac{\sin ^{2} x}{2 \sin x-\cos x+1}+ \\ & \cos ^{3} x+\sin ^{2} x-\cos x \end{aligned} $$",\sqrt{2}$.,medium,"10. Notice that $$ \begin{aligned} f(x)= & \log _{\sqrt{2-1}} \frac{\sin x+\cos x+1}{\sin x} \cdot \frac{2 \sin x-\cos x+1}{\sin x}+ \\ & (1-\cos x) \sin ^{2} x \\ = & \log _{\sqrt{2}-1}\left(1+\cot \frac{x}{2}\right)\left(2+\tan \frac{x}{2}\right)+ \\ & 8 \sin ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2} \cdot \cos ^{2} \frac{x}{2} \\ = & -\log _{\sqrt{2}+1}\left(3+2 \cot \frac{x}{2}+\tan \frac{x}{2}\right)+ \\ & 4 \sin ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2} \cdot 2 \cos ^{2} \frac{x}{2} \\ \leqslant & -\log _{\sqrt{2}+1}(3+2 \sqrt{2})+4\left(\frac{2}{3}\right)^{3} \\ = & -2+\frac{32}{27}=-\frac{22}{27} . \end{aligned} $$ The equality holds if and only if $\tan \frac{x}{2}=\sqrt{2}$." 565528492f74,"6. Let $\xi_{1}, \ldots, \xi_{N}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=p, \mathrm{P}\left\{\xi_{i}=-1\right\}=1-p, S_{i}=\xi_{1}+\ldots+\xi_{i}, 1 \leqslant i \leqslant N$, $S_{0}=0$. Let $\mathscr{R}_{N}$ be the range, i.e., the number of distinct points visited by the random walk $S_{0}, S_{1}, \ldots, S_{N}$. Find $\mathrm{E} \mathscr{R}_{N}$. Determine for which values of $p$ the law of large numbers in the form $$ \mathrm{P}\left\{\left|\frac{\mathscr{R}_{N}}{N}-c\right|>\varepsilon\right\} \rightarrow 0, \quad N \rightarrow \infty $$ holds, where $\varepsilon>0$ and $c$ is some constant.",See reasoning trace,medium,"Solution. Let's represent $\mathscr{R}_{N}$ in the following form: $$ \begin{aligned} \mathscr{R}_{N}=1+I\left(S_{1} \neq 0\right)+I\left(S_{2} \neq\right. & \left.0, S_{2} \neq S_{1}\right)+\ldots \\ & \ldots+I\left(S_{N} \neq 0, S_{N} \neq S_{1}, \ldots, S_{N} \neq S_{N-1}\right) \end{aligned} $$ Let $V_{n}^{N}=\sum_{k=0}^{n-1} \xi_{N-k}$, then $\mathscr{R}_{N}=1+I\left(V_{1}^{1} \neq 0\right)+I\left(V_{1}^{2} \neq 0, V_{2}^{2} \neq 0\right)+\ldots+I\left(V_{1}^{N} \neq 0, \ldots, V_{N}^{N} \neq 0\right)$. Since $$ \mathrm{P}\left\{V_{1}^{i} \neq 0, \ldots, V_{i}^{i} \neq 0\right\}=\mathrm{P}\left\{S_{1} \neq 0, \ldots, S_{i} \neq 0\right\}, \quad i=1, \ldots, N $$ we obtain that $$ \mathrm{E} \mathscr{R}_{N}=1+\sum_{k=1}^{N} \mathrm{P}\left\{S_{1} \neq 0, \ldots, S_{k} \neq 0\right\} $$ Notice that on odd steps $S_{k}$ cannot be zero, so $$ \mathrm{E} \mathscr{R}_{N}=1+\sum_{k=1}^{[N / 2]} \mathrm{P}\left\{S_{2} \neq 0, \ldots, S_{2 k} \neq 0\right\} $$ For a symmetric random walk, by problems I.5.8 and I.9.3, we have $$ \mathrm{E} \mathscr{R}_{N}=1+\sum_{k=1}^{[N / 2]} 2^{-2 k} C_{2 k}^{k} \sim \sqrt{\frac{2 N}{\pi}} $$ Due to problem II. 10.3, the almost sure convergence is valid: $$ \frac{\mathscr{R}_{N}}{N} \rightarrow \mathrm{P}\left\{S_{n} \neq 0, n \geqslant 0\right\} $$" 8f4522b81da2,"2. Hudson labels each of the four vertices of a triangular pyramid with a different integer chosen from 1 to 15 . For each of the four triangular faces, he then calculates the mean of the three numbers at the vertices of the face. Given that the means calculated by Hudson are all integers, how many different sets of four numbers could he have chosen to label the vertices of the triangular pyramid?",which uses this argument,medium,"Say the numbers on the four vertices are $a, b, c$ and $d$. Hudson has chosen numbers such that $a+b+c, a+b+d, a+c+d$ and $b+c+d$ are all multiples of 3 . Say we have three numbers, $x, y$ and $z$, whose sum is a multiple of 3 , and let the remainders when $x, y$ and $z$ are divided by 3 be $p, q$ and $r$ [note that $p, q$ and $r$ can only be 0,1 or 2]. Then $p+q+r$ must also be a multiple of 3 (otherwise there would be a remainder when $x+y+z$ was divided by 3 ). So either: - $p+q+r=0$, in which case $p=q=r=0$; or - $p+q+r=3$, in which case $p=q=r=1$ or $p, q$ and $r$ take the values 0,1 and 2 in some order; or - $p+q+r=6$, in which case $p=q=r=2$. Hence either all three of $x, y$ and $z$ have the same remainder when divided by 3 , or they all have a different remainder when divided by 3 . Going back to $a, b, c$ and $d$ : if $a, b$ and $c$ all have a different remainder when divided by 3 , we can say without loss of generality that $a$ is divisible by $3, b$ has a remainder of 1 when divided by 3 , and $c$ has a remainder of 2 . Then, since $b+c+d$ is divisible by $3, d$ must be divisible by 3 (since $b+c$ is divisible by 3 ). However, then $a+b+d$ would have a remainder of 1 when divided by 3 , which isn't allowed, so $a, b$ and $c$ cannot all have a different remainder when divided by 3 . Hence $a, b$ and $c$ must have the same remainder when divided by 3 . Once this is established, it quickly becomes clear that $d$ must also have the same remainder when divided by 3 . If that remainder is 1 , then $a, b, c$ and $d$ are four of the five numbers $1,4,7,10$ and 13. There are 5 ways to choose four numbers from this set (since there are five ways to choose which number to leave out). If that remainder is 2 , then again there Hudson chose four of the five numbers $2,5,8,11$ and 14 , which gives another five ways to choose the numbers. And finally, if that remainder is 0 , then Hudson chose four of the five numbers 3, 6, 9, 12 and 15 , which gives another five ways. Hence the total number of different sets of numbers could have chosen is $5+5+5=15$. REMARK If you have seen modular arithmetic before, you will find it a very helpful way to notate an answer which uses this argument." 11c2e3c4f57d,"Let's determine all solutions of the system of equations $x^{2}+y^{2}=x, 2 x y=y$. --- Find all solutions of the system of equations $x^{2}+y^{2}=x, 2 x y=y$.",423&width=449&top_left_y=1434&top_left_x=838),medium,"Combine the two equations: $$ x^{2}+y^{2}=x, \quad (1) \quad 2xy=y, \quad (2) $$ and we get $$ (x+y)^{2}=x+y $$ If $x+y \neq 0$, we can simplify by dividing by $x+y$, leading to the equation $x+y=1$. From this, substituting $x=1-y$ we get $$ 2y(1-y)=y $$ If $y \neq 0$, simplifying by dividing by $y$ gives the equation $2(1-y)=1$. Solving this, we find $y=\frac{1}{2}$ and from (2) $x=\frac{1}{2}$. If $y=0$, then from (1) $x=1$ (since we assumed $x+y \neq 0$). Now, let's return to the case where $x+y=0$, or $x=-y$. From (2), we get $-2y^2=y$ (for $y \neq 0$), and thus $y=-\frac{1}{2}$ and $x=\frac{1}{2}$, or $y=0=x$. The system of equations has 4 pairs of solutions: $$ x_{1}=\frac{1}{2}, y_{1}=\frac{1}{2} ; \quad x_{2}=0, y_{2}=0 ; \quad x_{3}=\frac{1}{2}, y_{3}=-\frac{1}{2} ; \quad x_{4}=1, y_{4}=0 $$ By substitution, it is easy to verify that all 4 pairs are indeed solutions to the system of equations. Remark. The system of equations can be visualized geometrically, showing that we get four solutions, and the solutions themselves are almost immediately apparent: (1) can be rewritten as $\left(x-\frac{1}{2}\right)^{2}+y^{2}=\left(\frac{1}{2}\right)^{2}$, which represents a circle with center $C\left(\frac{1}{2} ; 0\right)$ and radius $\frac{1}{2}$. (2) can be factored as $2y\left(x-\frac{1}{2}\right)=0$, which is the equation of a pair of intersecting lines. (The intersection point is the center of the circle.) One pair of solutions is obtained as the intersection points of the x-axis and the circle: $(0,0)$ and $(1,0)$; the other pair is obtained as the intersection points of the line $x=\frac{1}{2}$ and the circle: $\left(\frac{1}{2},-\frac{1}{2}\right)$ and $\left(\frac{1}{2} ; \frac{1}{2}\right)$. ![](https://cdn.mathpix.com/cropped/2024_05_02_442fda23af3e66eeafbcg-1.jpg?height=423&width=449&top_left_y=1434&top_left_x=838)" c0f620e12f04,"In the diagram, the value of $x$ is (A) 40 (B) 35 (C) 150 (D) 30 (E) 25 ![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-056.jpg?height=225&width=328&top_left_y=1964&top_left_x=1321)",(D),easy,"The four angles shown, $150^{\circ}, 90^{\circ}, x^{\circ}$, and $90^{\circ}$, form a complete rotation, a $360^{\circ}$ angle. Thus, $150^{\circ}+90^{\circ}+x^{\circ}+90^{\circ}=360^{\circ}$, or $x^{\circ}=360^{\circ}-150^{\circ}-90^{\circ}-90^{\circ}=30^{\circ}$. ANSWER: (D)" 3d0130f9dcd9,"Example 4.18. Find the general solution of the equation $$ y^{\prime \prime}+9 y=0 $$",See reasoning trace,easy,"Solution. This equation corresponds to the characteristic equation $$ r^{2}+9=0 $$ which has two imaginary conjugate roots $r_{1,2}= \pm 3 i$. Using formula (5) with $\alpha=0$ and $\beta=3$, we obtain the general solution $$ y=C_{1} \cos 3 x+C_{2} \sin 3 x $$" 578bb98de247,"## 3. Math Puzzle $8 / 65$ A crane is being transported to a construction site. To measure the length of the crane, Peter walks from the tip of the crane to the end and takes 15 steps (step length 80 $\mathrm{cm}$). To walk back from the end to the tip of the crane while the train is moving at a constant speed, he needs 75 steps at the same walking speed. How long is the crane?",12 \mathrm{~m}$ and $l+v t_{2}=60 \mathrm{~m}$. Multiplying the first equation by 5 and adding both ,easy,"With $v=$ speed, $s=$ distance, $t_{1}=$ time for 15 steps, $t_{2}=$ time for 75 steps, and $l=$ crane length, we have $l-v t_{1}=12 \mathrm{~m}, l+v t_{2}=60 \mathrm{~m}$. It holds that $\frac{t_{1}}{t_{2}}=\frac{1}{5} \rightarrow t_{1}=\frac{1}{5} t_{2}$. Thus, $l-v \cdot \frac{1}{5} t_{2}=12 \mathrm{~m}$ and $l+v t_{2}=60 \mathrm{~m}$. Multiplying the first equation by 5 and adding both equations yields $6 l=120 \mathrm{~m}, l=20 \mathrm{~m}$. The length of the crane is $20 \mathrm{~m}$." d515b1d77679,"3. The maximum value of the area of an inscribed triangle in a circle with radius $R$ is $\qquad$ Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.",See reasoning trace,medium,"3. $\frac{3 \sqrt{3}}{4} R^{2}$. Let the inscribed triangle of $\odot O$ be $\triangle A B C$. Obviously, when $\triangle A B C$ is an acute or right triangle, the area can reach its maximum value (because if $\triangle A B C$ is an obtuse triangle, the side opposite to the obtuse angle (let's assume it is $A$) can be symmetrically transformed about the center of the circle to become $B^{\prime} C^{\prime}$). Therefore, $S_{\triangle A B^{\prime} C^{\prime}}>S_{\triangle A B C}$. Now, let $\angle A O B=2 \alpha, \angle B O C=2 \beta$, $\angle C O A=2 \gamma, \alpha+\beta+\gamma=\pi$. Then $S_{\triangle A B C}=\frac{1}{2} R^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma)$. From the discussion, we can assume $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $y=\sin x$ is a concave function on $(0, \pi)$. Then, by Jensen's inequality, we have $$ \frac{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma}{3} \leqslant \sin \frac{2(\alpha+\beta+\gamma)}{3}=\frac{\sqrt{3}}{2} \text {. } $$ Therefore, $S_{\triangle A B C} \leqslant \frac{1}{2} R^{2} \times 3 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} R^{2}$. The equality holds if and only if $\triangle A B C$ is an equilateral triangle." a91ac94631ce,"An urn contains $k$ balls marked with $k$, for all $k=1,2, \ldots, 2016$. What is the minimum number of balls we must withdraw, without replacement and without looking at the balls, to be sure that we have 12 balls with the same number?",See reasoning trace,medium,"Solution Let's sum the maximum number of balls that can be drawn of each type without obtaining 12 balls of each color: $$ 1+2+3+4+5+6+7+8+9+10+11+\underbrace{11+11+\ldots+11}_{2005 \text { times }}=22121 $$ Thus, it is possible that we have bad luck and draw such a number of balls without obtaining 12 balls of each color. However, if we draw 22122 balls, we will certainly have 12 balls of the same color, as the sum above counts exactly the maximum number of balls that can be drawn without this happening. Therefore, the minimum sought is 22122. #" 531c8e6b14d3,3. Given an isosceles triangle $ABC$ where $\overline{AC}=\overline{BC}$ and the height dropped from vertex $C$ is equal to half of the angle bisector of the angle at vertex $A$. Find the angle at vertex $C$.,445&width=568&top_left_y=263&top_left_x=896),medium,"Solution. Let $C C_{1}$ be the height dropped from vertex $C$ and let $A A_{1}$ be the angle bisector of angle $\alpha$ at vertex $A$ (see diagram). Since triangle $A B C$ is isosceles, point $C_{1}$ is the midpoint of side $A B$ and $C C_{1}$ is the angle bisector of angle $\gamma$ at vertex $C$. Let $C_{2}$ be a point on line $C C_{1}$ such that $\overline{C C_{1}}=\overline{C_{1} C_{2}}$ (see diagram). From the fact that the diagonals of quadrilateral $A C_{2} B C$ bisect each other, it follows that this quadrilateral is a parallelogram. Since $A C_{2} \| C A_{1}$ and $\overline{A A_{1}}=\overline{C C_{2}}$, quadrilateral $A C_{2} A_{1} C$ is an isosceles trapezoid, from which it follows that $\frac{\gamma}{2}=\measuredangle C_{2} C A_{1}=\measuredangle C A_{1} A$. From triangles $A C_{1} C$ and $A A_{1} C$, we get $\gamma=90^{\circ}-\frac{\gamma}{2}$ and $\frac{\gamma}{2}=180^{\circ}-\gamma-\frac{\gamma}{2}$ respectively. Then from the equation $$ 90^{\circ}-\frac{\gamma}{2}=2\left(180^{\circ}-\gamma-\frac{\gamma}{2}\right) $$ we obtain that $\gamma=108^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_06_05_fbbd9d9f1d0d5083b697g-2.jpg?height=445&width=568&top_left_y=263&top_left_x=896)" 2d06f28412c3,"A box has the shape of a rectangular block with dimensions $102 \mathrm{~cm}, 255 \mathrm{~cm}$, and $170 \mathrm{~cm}$. We want to store the smallest possible number of small cubes with integer edge lengths in this box, so as to fill the entire box. Hint: Note that the length of the edge of the cube must be a divisor of each of the three dimensions of the box. (a) What is the length of the edge of each block? (b) How many blocks will be needed? ##",See reasoning trace,medium,"(a) Since the number of blocks is the smallest possible, the edge of the same must be the largest possible. The measure of the edge must be a divisor of 102, 255, and 170. Since we want the largest possible edge, the measure of it must be equal to $\operatorname{gcd}(102,255,170)=17$. Therefore, the edge of the cube measures $17 \mathrm{~cm}$. (b) The number of blocks is $$ \frac{102 \cdot 255 \cdot 170}{17 \cdot 17 \cdot 17}=6 \cdot 15 \cdot 10=900 $$ Suggestion: Determine the minimum value for the sum of the digits of the number. Facts that Help: The sum of the digits of a multiple of 9 is divisible by 9. Suggestion: Determine the possible values for the product and their factorizations. Facts that Help: 101 is prime. Suggestion: Initially show that he could not have bought more than 127 items." e46c20a7eec1,"2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $$ \left\{\begin{array}{l} x^{3}+\sin x-2 a=0, \\ 4 y^{3}+\sin y \cdot \cos y+a=0 . \end{array}\right. $$ Then $\cos (x+2 y)=$ $\qquad$ .","2 a=(-2 y)^{3}+\sin (-2 y)$, let $f(t)=t^{3}+\sin t$, then $f(x)=(-2 y)$. Since the function $f(t)=t",easy,"2. From the given, $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$, let $f(t)=t^{3}+\sin t$, then $f(x)=(-2 y)$. Since the function $f(t)=t^{3}+$ $\sin t$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\therefore x=-2 y, x+2 y=0$, hence $\cos (x+2 y)=1$" 186c6eeb962f,"11. (5 points) As shown in the figure, the ""L"" shaped paper piece with a perimeter of 52 centimeters can be divided into two identical rectangles along the dotted line. If the longest side is 16 centimeters, then the area of the ""L"" shaped paper piece is $\qquad$ square centimeters.",: 120,medium,"【Analysis】According to the problem, let's assume the width of the divided rectangle is $x$ cm, then the length of the divided rectangle should be (16-x) cm. By adding the perimeters of the ""L"" shaped paper pieces, we can get an equation: $16+(16-x)+x+(16-x-x)+(16-x)+x=52$. After solving this equation, we can find the width of the divided rectangle, then substitute $16-x$ to calculate the length of the divided rectangle, and finally calculate the area of the rectangle to get the answer. 【Solution】Solution: Let the width of the divided rectangle be $x$, then the length of the divided rectangle is $16-x$, $$ \begin{array}{l} 16+(16-x)+x+(16-x-x)+(16-x)+x=52 \\ 64-2 x=52, \\ 2 x=12, \\ x=6, \end{array} $$ The length of the divided rectangle is: $16-6=10$ (cm), the area of the rectangle is: $10 \times 6=60$ (square cm), the area of the ""L"" shaped paper piece is: $60 \times 2=120$ (square cm). Therefore, the answer is: 120." ebc36f8cf311,"14. Asymmetric coin. (From 9th grade, 2 points) Billy Bones has two coins - a gold one and a silver one. One of them is symmetric, and the other is not. It is unknown which coin is asymmetric, but it is known that the asymmetric coin lands heads with a probability of \( p = 0.6 \). Billy Bones tossed the gold coin, and heads came up immediately. Then Billy Bones started tossing the silver coin, and heads came up only on the second toss. Find the probability that the asymmetric coin is the gold one. ![](https://cdn.mathpix.com/cropped/2024_05_06_a67860e8fb746bcfb6a9g-10.jpg?height=588&width=577&top_left_y=1556&top_left_x=1342)",$5 / 9$,medium,"Solution. Let's introduce notations for the events: $$ A=\{\text { the gold coin is biased }\}, $$ $B=\left\{\begin{array}{l}\text { when the gold coin is tossed, heads appear immediately, } \\ \text { and when the silver coin is tossed, heads appear on the second attempt. }\end{array}\right\}$ We need to find the conditional probability $\mathrm{P}(A \mid B)$. We will use Bayes' formula: $$ \begin{gathered} \mathrm{P}(A \mid B)=\frac{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)}{\mathrm{P}(B)}=\frac{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)}{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)+\mathrm{P}(B \mid \bar{A}) \cdot \mathrm{P}(\bar{A})}= \\ =\frac{p \cdot 0.5^{2}}{p \cdot 0.5^{2}+0.5 \cdot(1-p) p}=\frac{0.5}{0.5+(1-p)}=\frac{0.5}{0.5+0.4}=\frac{5}{9} \end{gathered} $$ Answer: $5 / 9$." 3fb991fe71fd,"In the regular hexagon $A B C D E F$, two of the diagonals, $F C$ and $B D$, intersect at $G$. The ratio of the area of quadrilateral FEDG to the area of $\triangle B C G$ is (A) $3 \sqrt{3}: 1$ (B) $4: 1$ (C) $6: 1$ (D) $2 \sqrt{3}: 1$ (E) $5: 1$ ![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-139.jpg?height=214&width=258&top_left_y=1216&top_left_x=1389)",(E),medium,"In the regular hexagon $A B C D E F$, two of the diagonals, $F C$ and $B D$, intersect at $G$. The ratio of the area of quadrilateral $F E D G$ to the area of $\triangle B C G$ is (A) $3 \sqrt{3}: 1$ (B) $4: 1$ (C) $6: 1$ (D) $2 \sqrt{3}: 1$ (E) $5: 1$ ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-278.jpg?height=350&width=393&top_left_y=1294&top_left_x=1430) ## Solution 1 Join $E$ to $B$ and $D$ to $A$ as shown. Also join $E$ to $A$ and draw a line parallel to $A E$ through the point of intersection of $B E$ and $A D$. Quadrilateral $F E D G$ is now made up of five triangles each of which has the same area as $\triangle B C G$. The required ratio is $5: 1$. ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-278.jpg?height=358&width=393&top_left_y=1691&top_left_x=1427) ## Solution 2 For convenience, assume that each side of the hexagon has a length of 2 units. Each angle in the hexagon equals $120^{\circ}$ so $\angle B C G=\frac{1}{2}\left(120^{\circ}\right)=60^{\circ}$. Now label $\triangle B C G$ as shown. Using the standard ratios for a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle we have $B G=\sqrt{3}$ and $G C=1$. ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-279.jpg?height=355&width=249&top_left_y=300&top_left_x=1491) The area of $\triangle B C G=\frac{1}{2}(1) \sqrt{3}=\frac{\sqrt{3}}{2}$. Dividing the quadrilateral $F G D E$ as illustrated, it will have an area of $2(\sqrt{3})+\frac{1}{2}(1)(\sqrt{3})=\frac{5 \sqrt{3}}{2}$. ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-279.jpg?height=263&width=371&top_left_y=698&top_left_x=1430) The required ratio is $\frac{5 \sqrt{3}}{2}: \frac{\sqrt{3}}{2}$ or $5: 1$, as in solution 1 . ANSWER: (E)" 84994a89d81a,A triangle's three side lines are tangent to four circles whose radii are four consecutive elements of a geometric sequence. What is the largest angle of the triangle?,to the question of the problem,medium,"It is known that for the area $t$ of a triangle, the radii $\varrho, \varrho_{a}, \varrho_{b}$, and $\varrho_{c}$ of the inscribed and excircles, and the semiperimeter $s$, the following relationships hold: $$ t=\varrho \cdot s=\varrho_{a}(s-a)=\varrho_{b}(s-b)=\varrho_{c}(s-c) $$ In our problem - since the inscribed circle is the smallest and any two radii have different lengths - we can assume that the sides are in the following order: $a1)$. According to these, $$ \begin{gathered} q=\frac{\varrho_{a}}{\varrho}=\frac{\varrho_{c}}{\varrho_{b}}, \quad \text { i.e., } \quad \frac{s}{s-a}=\frac{s-b}{s-c} \\ s(s-c)=(s-b)(s-a) \\ (a+b+c)(a+b-c)=(c+a-b)(c-(a-b)) \\ (a+b)^{2}-c^{2}=c^{2}-(a-b)^{2} \\ a^{2}+b^{2}=c^{2} \end{gathered} $$ Thus, by the converse of the Pythagorean theorem, the triangle can only be a right triangle - if it exists at all - and its largest angle can only be $\gamma=90^{\circ}$. However, since in this determination (1) we only used two ratios, we do not yet know if there is a right triangle for which the $$ \frac{\varrho_{b}}{\varrho_{a}}=q $$ requirement is also satisfied. Clarifying this would not have been possible within the time of the competition (see the following 3rd remark). Remarks. 1. Some may still know the following formulas, which allow for ""easier"" (i.e., using logarithm tables) calculation of angles from the sides, compared to the ""cumbersome"" cosine rule: $$ \operatorname{tg} \frac{\gamma}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}, \quad \sin \frac{\gamma}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}, \quad \cos \frac{\gamma}{2}=\sqrt{\frac{s(s-c)}{a b}} $$ (The similar formulas for the other two angles can be written by cyclic permutation of the sides, while $s$ remains unchanged.) According to the first formula, $$ \operatorname{tg} \frac{\gamma}{2}=\sqrt{\frac{\varrho \cdot \varrho_{c}}{\varrho_{a} \cdot \varrho_{c}}}=\sqrt{\frac{\varrho^{2} q^{3}}{\varrho^{2} q^{3}}}=1 $$ $\frac{\gamma}{2}=45^{\circ}, \gamma=90^{\circ}$. Here, the existence of the triangle is guaranteed because the formulas only give meaningless results for non-existent triangles. 2. The mentioned formulas - and many similar ones - are from the era when the most advanced tool for computation was the logarithm table. In the following - only in large steps - came the mechanical calculators with gears, first manual, then motorized, and today the electronic calculators. (We omitted the slide rule, which was only good for quick estimates compared to these.) - The formulas are still valid today. - However, the $\operatorname{tg} x / 2$ appears, for example, in the transformation of certain integrals; but it is impossible to list all such small but well-usable tricks in theory. 3. For a right triangle that also satisfies the $\varrho_{b}: \varrho_{a}=q$ requirement, we get a cubic equation for the tangent of half the larger acute angle. Similarly to the above, $$ \begin{aligned} & \operatorname{tg} \frac{\beta}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\frac{\varrho \cdot \varrho_{b}}{\varrho_{a} \cdot \varrho_{c}}}=\frac{1}{q} \quad \text { and } \\ & \operatorname{tg} \frac{\alpha}{2}=\frac{1}{q^{2}} \end{aligned} $$ (i.e., the tangents of the half-angles also form a geometric sequence with 3 elements and the same ratio). From these, $$ \begin{gathered} \operatorname{tg}^{2} \frac{\beta}{2}=\frac{1}{q^{2}}=\operatorname{tg} \frac{\alpha}{2}=\operatorname{tg}\left(45^{\circ}-\frac{\beta}{2}\right)= \\ =\frac{1-\operatorname{tg} \frac{\beta}{2}}{1+\operatorname{tg} \frac{\beta}{2}}=\frac{1-\frac{1}{q}}{1+\frac{1}{q}}=\frac{q-1}{q+1} \\ q^{3}-q^{2}-q-1=0 \end{gathered} $$ (This ""beautiful"" equation also pops up in the solution of many other simply formulated problems.) Its approximate solution is: $$ q=1.83929, \quad \alpha=32^{\circ} 56^{\prime} 6^{\prime \prime}, \quad \beta=57^{\circ} 3^{\prime} 54^{\prime \prime} $$ Second solution. In addition to the notation used in the previous solution, let the vertices of the triangle be $A, B, C$, the center of the inscribed circle be $O$, and the centers of the excircles be $O_{a}, O_{b}, O_{c}$ such that the circle centered at $O_{a}$ touches the extensions of the sides $A B$ and $A C$, and the side $B C$ from the outside, and so on. Thus, the line $O O_{a}$ bisects the angle $B A C$, and the line $O_{b} O_{c}$ bisects the exterior angles at $A$, making it perpendicular to $O A$. $O$ is the orthocenter of the triangle $O_{a} O_{b} O_{c}$ and an interior point of the triangle, so this triangle is acute-angled. ![](https://cdn.mathpix.com/cropped/2024_05_02_823261d193f0d3d28e5bg-2.jpg?height=1093&width=943&top_left_y=993&top_left_x=580) Let the points of tangency of the circles on the line $A B$ be $E, E_{a}, E_{b}, E_{c}$, respectively. From the similar right triangles $A O E$ and $A O_{a} E_{a}$, as well as $A O_{b} E_{b}$ and $A O_{c} E_{c}$, $$ \frac{A O_{a}}{A O}=\frac{E_{a} O_{a}}{E O}=\frac{\varrho_{a}}{\varrho}=q=\frac{\varrho_{c}}{\varrho_{b}}=\frac{E_" eeb0ca55768d,"11 On the extensions of the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$ of a regular $n$-gon $(n \geq 5) A_{1} A_{2} \ldots A_{n}$, construct points $B_{1}, B_{2}$, $\ldots, B_{n}$ such that $B_{1} B_{2}$ is perpendicular to $A_{1} A_{2}, B_{2} B_{3}$ is perpendicular to $A_{2} A_{3}, \ldots, B_{n} B_{1}$ is perpendicular to $A_{n} A_{1}$.","x_{2}=\ldots=x_{n}=$ $x$, where $x=(a+x) \cos \alpha$, that is, $x=\frac{a \cos \alpha}{1-\cos \alph",medium,"Let $x_{k}=A_{k+1} B_{k}, \alpha$ be the exterior angle of a regular $n$-gon, $a$ be the length of its side. Then $x_{1}=\left(a+x_{2}\right) \cos$ $\alpha, x_{2}=\left(a+x_{3}\right) \cos \alpha, \ldots$, $x_{n}=\left(a+x_{1}\right) \cos \alpha$. Then $x_{1}=a_{1}+b_{1} x_{2}=a_{2}+b_{2} x_{3}=\ldots=a_{n}+b_{n} x_{1}$, where $b_{n}=(\cos \alpha)^{n} \neq 1$. Therefore, $x_{1}$ (and also $x_{k}$ for any $k$) is uniquely defined. It is also clear that we will obtain a solution if we set $x_{1}=x_{2}=\ldots=x_{n}=$ $x$, where $x=(a+x) \cos \alpha$, that is, $x=\frac{a \cos \alpha}{1-\cos \alpha}$." 73cdef5740b5,"One. (20 points) A factory needs to produce two types of products, A and B. According to the process specifications, each unit of product A requires 2 hours, 3 hours, and 4 hours of processing on three different machines A, B, and C, respectively. Each unit of product B requires 4 hours, 4 hours, and 3 hours of processing on machines B, C, and D, respectively. The maximum working hours per day for machines A, B, C, and D are 12 hours, 15 hours, 16 hours, and 24 hours, respectively. If producing one unit of product A yields a profit of 700 yuan, and producing one unit of product B yields a profit of 800 yuan, how should production be arranged daily to maximize the profit?",See reasoning trace,medium,"Let the daily production of product A be $x$ units, and the production of product B be $y$ units. Then we have: $$ \left\{\begin{array}{l} 2 x \leqslant 12, \\ 3 x+4 y \leqslant 15, \\ 4 x+4 y \leqslant 16, \\ 3 y \leqslant 24 . \end{array}\right. $$ The profit obtained is $$ W=700 x+800 y=100(7 x+8 y) \text {. } $$ Thus, the problem is to find the values of $x$ and $y$ that maximize $W$. (2) + (3) gives $7 x+8 y \leqslant 31$. Therefore, $W \leqslant 3100$. Since $x$ and $y$ are non-negative integers, and from $$ \left\{\begin{array}{l} 3 x+4 y=15, \\ 4 x+4 y=16 \end{array} \text { we solve to get } x=1, y=3\right. \text {. } $$ Clearly, this solution also satisfies (1) and (4). Therefore, producing 1 unit of product A and 3 units of product B daily yields the maximum profit." 35d5cac69794,"Example 3 Let the coordinates of the three vertices of the equilateral $\triangle A B C$ be $\left(x_{A}, y_{A}\right),\left(x_{B}, y_{B}\right),\left(x_{C}, y_{C}\right)$. Please express $S_{\triangle A B C}$ using $x_{A}, x_{B}, x_{C}$.",See reasoning trace,medium,"【Analysis】Let $\triangle A B C$ be counterclockwise, and $O$ be the center of the equilateral $\triangle A B C$, with $R$ being the radius of the circumcircle. Then $$ x_{0}=\frac{x_{A}+x_{B}+x_{C}}{3} . $$ Therefore, the x-coordinate of vector $\overrightarrow{O A}$ is $$ x_{A}-x_{o}=\frac{2 x_{A}-x_{B}-x_{C}}{3}=R \cos \varphi \text {, } $$ the x-coordinate of vector $\overrightarrow{O B}$ is $$ x_{B}-x_{o}=R \cos \left(\varphi+120^{\circ}\right) \text {, } $$ and the x-coordinate of vector $\overrightarrow{O C}$ is $$ \begin{array}{l} x_{C}-x_{o}=R \cos \left(\varphi-120^{\circ}\right) . \\ \text { Hence }\left(x_{C}-x_{O}\right)-\left(x_{B}-x_{O}\right) \\ =R\left[\cos \left(\varphi-120^{\circ}\right)-\cos \left(\varphi+120^{\circ}\right)\right] \\ \Rightarrow x_{C}-x_{B}=2 R \sin \varphi \cdot \sin 120^{\circ}=\sqrt{3} R \sin \varphi . \\ \text { Then }\left[\frac{\left(2 x_{A}-x_{B}-x_{C}\right)}{3}\right]^{2}+\left(\frac{x_{C}-x_{B}}{\sqrt{3}}\right)^{2} \\ =R^{2} \cos ^{2} \varphi+R^{2} \sin ^{2} \varphi=R^{2} \\ \Rightarrow R^{2}=\frac{4}{9} \sum x_{A}^{2}-\frac{4}{9} \sum x_{A} x_{B}, \end{array} $$ where “$\sum$” denotes the cyclic sum. Then $S_{\triangle A B C}=3 S_{\triangle A O B}=3 \times \frac{1}{2} R^{2} \sin 120^{\circ}$ $$ =\frac{3 \sqrt{3}}{4} R^{2}=\frac{\sqrt{3}}{3} \sum\left(x_{A}^{2}-x_{B} x_{C}\right) . $$" 0a248ba70edb,"2. A fruit store is conducting a promotional sale, with the following combinations: Combination A: 2 kg of fruit $A$, 4 kg of fruit $B$; Combination B: 3 kg of fruit $A$, 8 kg of fruit $B$, 1 kg of fruit $C$; Combination C: 2 kg of fruit $A$, 6 kg of fruit $B$, 1 kg of fruit $C$. It is known that fruit $A$ costs 2 yuan per kg, fruit $B$ costs 1.2 yuan per kg, and fruit $C$ costs 10 yuan per kg. One day, the store earned a total of 441.2 yuan from selling these three combinations, with the sales of fruit $A$ amounting to 116 yuan. What is the sales amount for fruit $C$? $\qquad$ yuan.",150$ (yuan).,medium,"2.150. Let the number of sets of fruit A, B, and C sold on that day be $x$, $y$, and $z$ respectively. According to the problem, we have $$ \begin{array}{l} \left\{\begin{array}{l} 2(2 x+3 y+2 z)=116, \\ 8.8 x+25.6 y+21.2 z=441.2, \end{array}\right. \\ \therefore\left\{\begin{array}{l} 2 x+3 y+2 z=58, \\ 22 x+64 y+53 z=1103 . \end{array}\right. \end{array} $$ (2) - (1) $\times 11$, eliminating $x$, we get $$ 31(y+z)=465 \text {, } $$ Thus, $y+z=15$. Therefore, a total of 15 kg of fruit C was sold, and the sales revenue from fruit C is $15 \times 10=150$ (yuan)." 161937c2e80c,"87 Given the complex sequence $\left\{a_{n}\right\}$ with the general term: $$ a_{n}=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)\left(1+\frac{i}{\sqrt{3}}\right) \cdots\left(1+\frac{i}{\sqrt{n}}\right) $$ then $\left|a_{n}-a_{n+1}\right|=$ A. $\frac{\sqrt{2}}{2}$ B. $\sqrt{2}$ C. 1 D. 2",See reasoning trace,easy,"87 C. It is easy to prove by mathematical induction that $\left|a_{n}\right|=\sqrt{n+1}$. Therefore, $$ \left|a_{n+1}-a_{n}\right|=\left|a_{n}-a_{n}\left(1+\frac{i}{\sqrt{n+1}}\right)\right|=\left|a_{n}\right| \cdot\left|\frac{i}{\sqrt{n+1}}\right|=1 . $$" 186350367e1e,"12. If $(x, y, z)$ is $$ \left\{\begin{array}{l} 5 x-3 y+2 z=3, \\ 2 x+4 y-z=7, \\ x-11 y+4 z=3 \end{array}\right. $$ the value of $z$ is ( ). (A) 0 (B) -1 (C) 1 (D) $\frac{4}{3}$ (E) $*$","0$. That is, $2 x+4 y-z=0$, which contradicts (2). Therefore, the system of equations has no solutio",easy,"12. (E). (1) $-(3)$ yields $4 x+8 y-2 z=0$. That is, $2 x+4 y-z=0$, which contradicts (2). Therefore, the system of equations has no solution." 01538a859942,"## Task Condition Find the derivative of the specified order. $y=(2 x+3) \ln ^{2} x, y^{\prime \prime \prime}=?$",See reasoning trace,medium,"## Solution $y^{\prime}=2 \cdot \ln ^{2} x+(2 x+3) \cdot 2 \cdot \ln x \cdot \frac{1}{x}=2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}$ $y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}\right)^{\prime}=4 \cdot \ln x \cdot \frac{1}{x}+\left(4 \ln x+\frac{6 \cdot \ln x}{x}\right)^{\prime}=$ $=4 \cdot \ln x \cdot \frac{1}{x}+4 \cdot \frac{1}{x}+6 \cdot \frac{\frac{1}{x} \cdot x-\ln x \cdot 1}{x^{2}}=\frac{4 \cdot \ln x}{x}+\frac{4}{x}+\frac{6}{x^{2}}-\frac{6 \ln x}{x^{2}}$ $y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{4 \cdot \ln x}{x}+\frac{4}{x}+\frac{6}{x^{2}}-\frac{6 \ln x}{x^{2}}\right)^{\prime}=$ $=4 \frac{\frac{1}{x} \cdot x-\ln x \cdot 1}{x^{2}}-\frac{4}{x^{2}}-2 \cdot \frac{6}{x^{3}}-6 \cdot \frac{\frac{1}{x} \cdot x^{2}-\ln x \cdot 2 x}{x^{4}}=\frac{4-4 \ln x}{x^{2}}-\frac{4}{x^{2}}-\frac{12}{x^{3}}-\frac{6-12 \ln x}{x^{3}}=$ $=\frac{-4 \ln x \cdot x-12-6+12 \ln x}{x^{3}}=\frac{4 \ln x \cdot(3-x)-18}{x^{3}}$ ## Problem Kuznetsov Differentiation 19-9" 0f31301f2d60,"$29 \cdot 61$ The polynomial $x^{2 n}+1+(x+1)^{2 n}$ is not divisible by $x^{2}+x+1$ if $n$ is equal to (A) 17 . (B) 20 . (C) 21 . (D) 64 . (E) 65 . (31st American High School Mathematics Examination, 1980)",$(C)$,medium,"［Solution］Let $f(x)=x^{2}+x+1$, and let $g_{n}(x)=x^{2 n}+1+(x+1)^{2 n}$. Also, $f(x)=(x-r)\left(x-r^{\prime}\right)$, where $r=\frac{-1+\sqrt{3} i}{2}$ and $r^{\prime}=\frac{-1-\sqrt{3} i}{2}$ are the two roots of $f(x)=0$. Thus, $f(x)$ can divide $g_{n}(x)$ if and only if both $x-r$ and $x-r^{\prime}$ can divide $g_{n}(x)$, i.e., $g_{n}(r)=g_{n}\left(r^{\prime}\right)=0$. Since $r$ and $r^{\prime}$ are conjugate complex numbers, it is sufficient to determine the $n$ for which $g_{n}(r)=0$. Note that $$ g_{n}(x)=\left[x^{2}\right]^{n}+\left[(x+1)^{2}\right]^{n}+1, $$ and note that $r$ and $r+1=\frac{1+\sqrt{3} i}{2}$ are both 6th roots of 1 (this can be easily seen from the unit circle in the complex plane). Thus, $r^{2}$ and $(r+1)^{2}$ are both cube roots of 1. Therefore, the value of $g_{n}(r)$ depends only on the remainder when $n$ is divided by 3. Direct calculation (or from the unit circle in the complex plane) yields $$ g_{0}(r)=3, \quad g_{1}(r)=0, \quad g_{2}(r)=0 . $$ That is, when $3 \mid n$, $f(x) \nmid g_{n}(x)$. Hence, the answer is $(C)$." 404ae60599b8,"On the plane, nine parallel lines intersect with $n$ other parallel lines, all of which are perpendicular to the first nine. The lines together form a total of 756 rectangles. What is the value of $n$?","756$ for $n$, we get two roots: -6 and 7. The former is clearly not a solution, so the value of $n$ ",medium,"Solution. The problem actually involves determining the number of rectangles formed by 9, and respectively $n$ lines. We will present two possibilities for this below: a) A rectangle is uniquely determined by its two pairs of parallel side lines. From the 9 ""horizontal"" lines, we can choose 2 in $\binom{9}{2}$ ways, and from the $n$ ""vertical"" lines, we can choose 2 in $\binom{n}{2}$ ways. Thus, the number of rectangles is $$ \binom{9}{2} \cdot\binom{n}{2}=18 n(n-1) $$ b) In the drawn grid, each rectangle is uniquely determined by its diagonals. Since the number of grid points is $9 n$, we can choose one end of the diagonal in this many ways. The other end cannot be in the same row or column as the chosen point, so we can choose it in $(9-1)(n-1)=8(n-1)$ ways. Thus, on the one hand, we counted each diagonal twice, and on the other hand, we counted each rectangle according to both of its diagonals, so the number of resulting rectangles is $$ \frac{9 n \cdot 8(n-1)}{4}=18 n(n-1) $$ Solving the equation $18 n(n-1)=756$ for $n$, we get two roots: -6 and 7. The former is clearly not a solution, so the value of $n$ is 7." 1ae43435b525,"1. Given $\sin \alpha \cdot \cos \beta=-\frac{1}{2}$. Then the range of $\cos \alpha \cdot \sin \beta$ is ( ). (A) $\left[-1, \frac{1}{2}\right]$ (B) $\left[-\frac{1}{2}, 1\right]$ (C) $\left[-\frac{3}{4}, \frac{3}{4}\right]$ (D) $\left[-\frac{1}{2}, \frac{1}{2}\right]$","-\frac{\pi}{4}, \beta=\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta$ can take the maximum value $\fr",medium,"$-1 .(D)$ $$ \text { Method i: } \begin{aligned} & \cos ^{2} \alpha \cdot \sin ^{2} \beta=\left(1-\sin ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right) \\ & =1-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right)+\sin ^{2} \alpha \cdot \cos ^{2} \beta \\ & =\frac{5}{4}-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right) \\ & =\frac{5}{4}-(\sin \alpha+\cos \beta)^{2}+2 \sin \alpha \cdot \cos \beta \\ & =\frac{1}{4}-(\sin \alpha+\cos \beta)^{2} \\ & \leqslant \frac{1}{4} . \\ \therefore-\frac{1}{2} \leqslant & \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2} . \end{aligned} $$ When $\alpha=-\frac{\pi}{4} ， \beta=\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta=\frac{1}{2}$; when $\alpha=-\frac{\pi}{4}, \beta=-\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta=-\frac{1}{2}$. Therefore, the range of $\cos \alpha \cdot \sin \beta$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$. Method 2: From the given information, $$ \begin{array}{l} \sin (\alpha+\beta)=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\ =-\frac{1}{2}+\cos \alpha \cdot \sin \beta \in[-1,1] . \end{array} $$ Similarly, $$ \begin{array}{l} \sin (\alpha-\beta)=-\frac{1}{2}-\cos \alpha \cdot \sin \beta \in[-1,1] \\ \therefore\left\{\begin{array}{l} -\frac{1}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{3}{2} \\ -\frac{3}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2} \end{array}\right. \\ -\frac{1}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2} . \end{array} $$ When $\alpha=-\frac{\pi}{4}, \beta=\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta$ can take the maximum value $\frac{1}{2}$; when $\alpha=-\frac{\pi}{4}, \beta=-\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta$ can take the minimum value $-\frac{1}{2}$." c7ed71a2382c,"11.6. Given a tetrahedron \(ABCD\), all of whose faces are similar right triangles with acute angles at vertices \(A\) and \(B\). The edge \(AB\) is equal to 1. Find the length of the shortest edge of the tetrahedron.",$\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{3}{2}}$,medium,"Answer: $\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{3}{2}}$. Solution. Let $\angle C B D=\alpha, \angle C D B=\beta, \alpha+\beta=90^{\circ}$ (see Fig. 11.6a). Right triangles $D B C$ and $A B C$ share the side $B C$. If $\angle A B C=\alpha$, then triangles $A B C$ and $D B C$ are congruent, which implies $C D=A C$, which is impossible because $A C$ is the hypotenuse of the right triangle $A C D$. Therefore, $\angle A B C=\beta, \angle B A C=\alpha$. Since $B A C$ is the angle between the line $A B$ and the plane $A C D$, then $\alphaC D$, hence $C DA D$. Therefore, $\angle B A D=\beta, \quad \angle A B D=\alpha, \quad$ then $A D=\cos \beta$; $C D=B D \cos \beta=A B \cos \alpha \cos \beta=\cos \alpha \cos \beta$. Thus, $C Dr$, and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $AB/BC=$ $\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$",\frac{1,medium,"First note that arc length equals $r\theta$, where $\theta$ is the central angle in radians. Call the center of the circle $O$. Then $\angle{BOC} = 1$ radian because the minor arc $BC$ has length $r$. Since $ABC$ is isosceles, $\angle{AOB} = \pi - \tfrac{1}{2}$. We use the Law of Cosines to find that \[\frac{AB}{BC} = \frac{\sqrt{2r^2 - 2r^2\cos{(\pi - \frac{1}{2})}}}{\sqrt{2r^2 - 2r^2\cos1}} = \frac{\sqrt{1 + \cos{(\frac{1}{2})}}}{\sqrt{1 - \cos1}}.\] Using half-angle formulas, we have that this ratio simplifies to \[\frac{\cos\frac{1}{4}}{\sin{\frac{1}{2}}} = \frac{\cos\frac{1}{4}}{\sqrt{1 - \cos^2{\frac{1}{2}}}} = \frac{\cos\frac{1}{4}}{\sqrt{(1 + \cos{\frac{1}{2}})(1 - \cos{\frac{1}{2}})}} = \frac{\cos{\frac{1}{4}}}{2\cos{\frac{1}{4}}\sin{\frac{1}{4}}}\] \[= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}\]" 15639589bb29,"6. Let the function $f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}$, where $k$ is a positive integer. If for any real number $a$, it holds that $\{f(x) \mid a5 \pi$, any open interval $(a, a+1)$ contains a complete period of $f(x)$, at which point $\{f(x) \mid a y\). Then: \[ x > y = \frac{x^3 + 1}{2} \] Since \(x > y\), it follows that: \[ x > \frac{x^3 + 1}{2} \] Now, consider \(y\) and \(z\): \[ y = \frac{x^3 + 1}{2} > \frac{y^3 + 1}{2} = z \] Thus: \[ x > \frac{x^3 + 1}{2} > \frac{y^3 + 1}{2} > \frac{z^3 + 1}{2} = x \] This is a contradiction because it implies \(x > x\). 5. Similarly, if \(x < y\), we would get a similar contradiction. Therefore, the only possibility is \(x = y = z\). 6. Substituting \(x = y = z\) into the original equations, we get: \[ x^3 = 2x - 1 \] Solving for \(x\), we rearrange the equation: \[ x^3 - 2x + 1 = 0 \] Factoring, we find: \[ (x - 1)(x^2 + x - 1) = 0 \] The solutions to this equation are \(x = 1\) or the roots of \(x^2 + x - 1 = 0\). The quadratic equation \(x^2 + x - 1 = 0\) has roots: \[ x = \frac{-1 \pm \sqrt{5}}{2} \] However, substituting these roots back into the original system does not satisfy all three equations simultaneously. Therefore, the only real solution is \(x = 1\). The final answer is \( \boxed{ (1, 1, 1) } \)." 7255c74a4332,"5. Given the hyperbola $C_{1}: 2 x^{2}-y^{2}=1$, and the ellipse $C_{2}$ : $4 x^{2}+y^{2}=1$. If $M$ and $N$ are moving points on the hyperbola $C_{1}$ and the ellipse $C_{2}$ respectively, $O$ is the origin, and $O M \perp O N$, then the distance from point $O$ to the line $M N$ is $\qquad$",See reasoning trace,medium,"5. $\frac{\sqrt{3}}{3}$. Let the inclination angles of the lines $O M$ and $O N$ be $\alpha$ and $\beta$, respectively. Substitute the points $M(|O M| \cos \alpha,|O M| \sin \alpha)$, $$ N(|O N| \cos \beta,|O N| \sin \beta) $$ into the equations of the hyperbola $C_{1}$ and the ellipse $C_{2}$, respectively, to get $$ \begin{array}{l} 2|O M|^{2} \cos ^{2} \alpha-|O M|^{2} \sin ^{2} \alpha=1, \\ 4|O N|^{2} \cos ^{2} \beta+|O N|^{2} \sin ^{2} \beta=1 . \end{array} $$ Then $\frac{1}{|O M|^{2}}+\frac{1}{|O N|^{2}}$ $$ =2 \cos ^{2} \alpha-\sin ^{2} \alpha+4 \cos ^{2} \beta+\sin ^{2} \beta \text {. } $$ Since $O M \perp O N$, we have $$ \begin{array}{l} |\alpha-\beta|=\frac{\pi}{2}, \\ \cos ^{2} \beta=\sin ^{2} \alpha, \sin ^{2} \beta=\cos ^{2} \alpha . \\ \text { Therefore, } \frac{1}{|O M|^{2}}+\frac{1}{|O N|^{2}} \\ =2 \cos ^{2} \alpha-\sin ^{2} \alpha+4 \sin ^{2} \alpha+\cos ^{2} \alpha=3 \text {. } \end{array} $$ Let the distance from point $O$ to $M N$ be $h$. $$ \begin{array}{l} \text { Therefore, } \frac{1}{h^{2}}=\frac{1}{|O M|^{2}}+\frac{1}{|O N|^{2}}=3 \\ \Rightarrow h=\frac{\sqrt{3}}{3} . \end{array} $$" 474f36432d4a,"18. (6 points) Pour 100 grams of 40% sugar water into $a$ grams of 20% sugar water, resulting in 25% sugar water, then $a=$ $\qquad$",: 300,easy,"【Answer】Solution: According to the problem, we have: Based on the cross method of concentration: The ratio of the difference in concentration is equal to the ratio of the solution's mass That is, 1: $3=100: a$, so $a=300$ grams Therefore, the answer is: 300" fcca4bfbd228,"Example 4. The necessary and sufficient condition for $\arccos (-x)$ to be greater than $\arccos x$ is: (A) $x \in(0,1)$; (B) $x \in(-1,0)$; (C) $x \in[0,1]$; (D) $x \in\left[0, \frac{\pi}{2}\right]$.","s, the domain of $\arccos x$, $|x| \leqslant 1$, directly eliminates (D); then by $\arccos (-x) > \arccos x$, using the monotonicity of the arccosine function, we can deduce that $-x < 0$, thus eliminating (B)",easy,"Among the four answers, the domain of $\arccos x$, $|x| \leqslant 1$, directly eliminates (D); then by $\arccos (-x) > \arccos x$, using the monotonicity of the arccosine function, we can deduce that $-x < 0$, thus eliminating (B). When $x=0$, we have $\arccos (-x) = \arccos x$, which eliminates (C); therefore, (A) is the correct answer." f0c98083be63,"10. If $x, y \in R$, and $x^{2}+2 \sqrt{3} x y-y^{2}=$ 3 , then the minimum value of $x^{2}+y^{2}$ is $\qquad$",See reasoning trace,easy,10. $\frac{3}{2}$. b783c187ef4f,"3. Find such an integer $\mathrm{n}$ that for any integer $\mathrm{k} \neq 2013$, the number $n-k^{2014}$ is divisible by $2013-k$. Answer: $2013^{2014}$.","2013^{2014}$ satisfies the condition of the problem, but it is not proven that there are no other nu",medium,"# Solution Let $t=2013-k$, then $\frac{n-k^{2014}}{2013-k}=\frac{n-(2013-t)^{2014}}{t}$. If we raise the number $2013-t$ to the power of 2014, we get a sum of terms, each of which contains a factor $t^{m}$ (for some $m=1, \ldots, 2014$) except for $2013^{2014}$, then we get $\frac{n-(2013-t)^{2014}}{t}=\frac{n-2013^{2014}-t S}{t}=\frac{n-2013^{2014}}{t}-S$, where S is an integer, meaning $n-2013^{2014}: t$ for any integer t. This is only possible when $n-2013^{2014}=0 \Rightarrow n=2013^{2014}$. ## Recommendations for checking 1) It is proven that the number $\mathrm{n}=2013^{2014}$ satisfies the condition of the problem, but it is not proven that there are no other numbers that satisfy the condition - 2 points." 08e08ebd1b0b,,See reasoning trace,medium,"Solution. We use the notations as in the diagram on the right. Let $$ \overline{A C}=3, \overline{B C}=4 \text { and } \overline{A B}=5 . $$ Triangles $A B C$ and $A P N$ are right-angled and share a common acute angle, so $\triangle A B C \sim \triangle A P N$, which means $\frac{3-x}{x}=\frac{3}{4}$. ![](https://cdn.mathpix.com/cropped/2024_06_05_b353751bdd87cc544500g-12.jpg?height=286&width=503&top_left_y=400&top_left_x=979) From the last equation, we get $x=\frac{12}{7}$. Therefore, $$ \overline{P N}=\overline{P Q}=x=\frac{12}{7}, \overline{A N}=3-x=\frac{9}{7} \text { and } \overline{B Q}=4-x=\frac{16}{7} . $$ Now, using the Pythagorean theorem, we get $$ \overline{A P}=\sqrt{\overline{A N}^{2}+\overline{P N}^{2}}=\frac{15}{7} \text { and } \overline{P B}=\sqrt{\overline{P Q}^{2}+\overline{B Q}^{2}}=\frac{20}{7} $$ Finally, $$ \begin{aligned} & L_{\triangle A P N}=\overline{A P}+\overline{P N}+\overline{A N}=\frac{15}{7}+\frac{12}{7}+\frac{9}{7}=\frac{36}{7} \text{ cm} \\ & L_{\triangle P B Q}=\overline{P B}+\overline{P Q}+\overline{B Q}=\frac{20}{7}+\frac{12}{7}+\frac{16}{7}=\frac{48}{7} \text{ cm} \\ & P_{\triangle A P N}=\frac{\overline{P N} \cdot \overline{A N}}{2}=\frac{1}{2} \cdot \frac{12}{7} \cdot \frac{9}{7}=\frac{54}{49} \text{ cm}^{2} \\ & P_{\triangle P B Q}=\frac{\overline{P Q} \cdot \overline{B Q}}{2}=\frac{1}{2} \cdot \frac{12}{7} \cdot \frac{16}{7}=\frac{96}{49} \text{ cm}^{2} \end{aligned} $$" afb419759ae9,"Task B-3.4. A group of children found a wooden board in the shape of a quadrilateral and decided to use it for the game ""pick-a-point"". Since the target was of an unusual shape, they had to adapt the game rules. They asked for advice from Mark's older brother, a good mathematician. He measured, calculated, and wrote something... and came to an important conclusion. He established that there is a point $O$ equally distant from all sides of the quadrilateral. Then he divided the board into areas by drawing lines from point $O$ to all vertices $A, B, C, D$. He assigned points to the areas inversely proportional to their area. Which area yields the most points and why? Mark's brother added a condition that points are only awarded if the distance from the hit point $T$ to point $O$ is not greater than the distance from point $O$ to any side of the board. Determine the ratio of the area of the part of the board that does not yield points to the area of the part that does yield points. Mark's brother measured these values $|A B|=5 \mathrm{dm},|B C|=3 \mathrm{dm},|A D|=4 \mathrm{dm}$, $|\angle B A D|=60^{\circ}$.",See reasoning trace,medium,"## Solution. ![](https://cdn.mathpix.com/cropped/2024_05_30_2595713ee0b414c2653fg-13.jpg?height=745&width=1022&top_left_y=1015&top_left_x=519) If the point $O$ is equidistant from the sides of the quadrilateral, then a circle can be inscribed in that quadrilateral. In a tangential quadrilateral, $a+c=b+d$ holds. We conclude that $c=2 \mathrm{dm}$. The tile is divided into 4 regions, triangles $A B O, B C O, C D O, A D O$. The triangle with the smallest area is triangle $C D O$ because it has the shortest side on which $r$ is the height, so this region scores the most points. Points are earned by hitting a point inside the inscribed circle of the quadrilateral. Let's calculate the radius of the inscribed circle, i.e., the distance from point $O$ to the sides of the quadrilateral. From $\triangle A B D$ we get $$ \begin{aligned} |B D|^{2} & =16+25-2 \cdot 4 \cdot 5 \cdot \cos 60^{\circ}=21 \\ |B D| & =\sqrt{21} \mathrm{dm} \end{aligned} $$ From $\triangle B C D$ we get $$ \cos \gamma=\frac{9+4-21}{2 \cdot 2 \cdot 3}=\frac{-2}{3}, \text { and } \sin \gamma=\frac{\sqrt{5}}{3} $$ Consider the area of the quadrilateral. $$ P(A B C D)=r \cdot \frac{a+b+c+d}{2}=r \cdot s=7 r $$ Similarly, $$ P(A B C D)=P(A B D)+P(B C D)=\frac{3 \cdot 2 \sin \gamma}{2}+\frac{4 \cdot 5 \sin 60^{\circ}}{2} $$ so $$ \begin{aligned} 7 r & =\sqrt{5}+5 \sqrt{3} \\ r & =\frac{\sqrt{5}+5 \sqrt{3}}{7} \mathrm{dm} \end{aligned} $$ The ratio of the desired areas is $$ \frac{7 r-r^{2} \pi}{r^{2} \pi}=\frac{7}{r \pi}-1=\frac{7}{\pi} \cdot \frac{7}{\sqrt{5}+5 \sqrt{3}}-1=\frac{7}{10 \pi}(5 \sqrt{3}-\sqrt{5})-1 $$ Note. The radius can also be calculated without using areas. From $\triangle A E O$ we have $\operatorname{tg} 30^{\circ}=\frac{r}{x}, x=r \sqrt{3}$, $$ \begin{aligned} y & =5-r \sqrt{3} \\ \operatorname{tg} \frac{\gamma}{2} & =\frac{\sin \frac{\gamma}{2}}{\cos \frac{\gamma}{2}}=\sqrt{\frac{1-\cos \gamma}{1+\cos \gamma}}=\sqrt{5} \end{aligned} $$ From $\triangle O G C$ we have $\operatorname{tg} \frac{\gamma}{2}=\frac{r}{z}, z=\frac{r \sqrt{5}}{5}$. $$ \begin{aligned} z & =3-y=3-(5-r \sqrt{3})=r \sqrt{3}-2 \\ \frac{r \sqrt{5}}{5} & =r \sqrt{3}-2 \\ r & =\frac{10}{5 \sqrt{3}-\sqrt{5}}=\frac{5 \sqrt{3}+\sqrt{5}}{7} \mathrm{dm} \end{aligned} $$" b4bb96cff0b7,"97. a) It is clear that any plane broken line of length 1 can be enclosed in a circle of radius 1: for this, it is sufficient for the center $O$ of the circle to coincide with one of the ends of the broken line. What is the radius of the smallest circle in which any broken line of length $l$ can be enclosed? b) It is clear that any plane closed broken line of perimeter 1 can be enclosed in a circle of radius $\leqslant \frac{1}{2}$: for this, it is sufficient for the center $O$ of the circle to coincide with any point of the broken line. (In this case, for any point $M$ of the broken line, the length of one of the two segments $O M$ of the broken line will be $\leqslant \frac{1}{2}$, and therefore the distance $O M$ will be $\leqslant \frac{1}{2}$.) What is the radius of the smallest circle inside which any closed broken line of length 1 can be enclosed?",See reasoning trace,medium,"97. a) Let $S$ be the midpoint of the broken line $AB$, i.e., such a point that the broken lines $AS$ and $SB$ have the same length $1/2$. In this case, each point $M$ of the broken line is such that the length of the broken line $SM$ does not exceed $1/2$, and even more so, the distance $SM \leqslant 1/2$, from which it follows that a circle of radius $1/2$ with center $S$ completely covers the broken line. Thus, every broken line of length 1 can be covered by a circle of radius $1/2$, and the number $1/2$ cannot be reduced, for it is obvious that a segment of length 1 cannot be covered by a circle of radius $<1/2$. ![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-314.jpg?height=566&width=578&top_left_y=883&top_left_x=310) Fig. 176. b) We will prove that any closed plane broken line of perimeter 1 can be enclosed within a circle of radius 1/4. Let $A$ be an arbitrary point of our broken line, and $B$ be such a point on the broken line that both parts of the broken line connecting points $A$ and $B$ have the same length $1/2$, and let $Q$ be the midpoint of the segment $AB$ (Fig. 176). Draw a circle of radius $1/4$ with center at point $Q$. We claim that the entire broken line will be enclosed within this circle, i.e., that the distance from any point $M$ of the broken line to point $Q$ is no more than 1/4. Indeed, let $M'$ be the point symmetric to $M$ with respect to point $Q$. Connect $M$ and $M'$ with points $A$ and $B$. Since $AM$ is no greater than the part of the broken line enclosed between points $A$ and $M$, and $BM$ is no greater than the part of the broken line enclosed between points $B$ and $M$, then $AM + BM$ is no greater than the part of the broken line enclosed between points $A$ and $B$, i.e., no greater than $1/2$. But from Fig. 176, it is not difficult to see that $AM = BM', \quad BM = AM', \quad QM = QM'$. From the consideration of $\triangle AMM'$ it follows that $$ MM' = 2QM \leqslant AM + AM' = AM + BM \leqslant 1/2 $$ from which $$ QM \leqslant 1/4 $$ which is what we needed to prove. On the other hand, the diameter of a broken line of perimeter 1 can be arbitrarily close to $1/2$ (this will be the case, for example, if the broken line represents a rhombus with side $1/4$ and a very small acute angle). Since the diameter of a circle containing the broken line is obviously not less than the diameter of the broken line, no circle of radius less than $1/4$ can cover any closed plane broken line of perimeter 1. Therefore, the required value of the radius of the circle is $1/4$." 7c2c0f1ca4d8,"Knowing that $a b c=8$, with $a, b, c$ positive, what is the minimum value of $a+b+c$?",6$. This bound is indeed achieved with $a=b=c=2$.,easy,"By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality): $\frac{a+b+c}{3} \geqslant \sqrt[3]{a b c}$. Therefore, $a+b+c \geqslant 3 \cdot \sqrt[3]{8}=6$. This bound is indeed achieved with $a=b=c=2$." cf8887ed4da3,"1. Let $p$ be a prime number greater than 3, and $p+8$ is also a prime number. Let $S_{1}=p(p+8)$, and $S_{i+1} (i=1,2, \cdots, k-1)$ denote the sum of the digits of $S_{i}$. If $S_{k}$ is a single-digit number, for all possible prime numbers $p$, $S_{k}$ has ( ) different values. (A) 1 (B) 2 (C) 3 (D) 4","2$, which means $S_{k}$ has only one value.",easy,"$-1 . A$. Since $p$ is a prime number greater than 3, and $p+8$ is also a prime number, therefore, $p$ is a prime of the form $3M+2$. So, $$ p(p+8)=(p+4)^{2}-16 \equiv 2(\bmod 9) \text {. } $$ Also, $S_{i+1} \equiv S_{i}(\bmod 9)$, then $S_{k}=2$, which means $S_{k}$ has only one value." dfe2f8b44650,"5. On a long stick, there are three types of graduation lines. The first type divides the stick into 10 equal parts; The second type divides the stick into 12 equal parts; The third type divides the stick into 15 equal parts. If each graduation line cuts the stick, how many pieces will the stick be cut into?",See reasoning trace,medium,"5. 28 5. 30 【Solution】The least common multiple of 10, 12, and 15 is 60. Taking $\frac{1}{60}$ of the length of the stick as a unit, each 10th part of the stick is 6 units long; each 12th part is 5 units long; and each 15th part is 4 units long. Excluding the two endpoints of the stick, the internal division points of the stick are 9, 11, and 14 (corresponding to 10, 12, and 15 divisions), totaling 34. Since the least common multiple of 5 and 6 is 30, the division points of the 10 and 12 divisions coincide at 30 units, so 1 must be subtracted from 34. Since the least common multiple of 4 and 5 is 20, the division points of the 12 and 15 divisions coincide at 20 and 40 units, so 2 must be subtracted. Similarly, since the least common multiple of 6 and 4 is 12, the division points of the 15 and 10 divisions coincide at 12, 24, 36, and 48 units, so 4 must be subtracted. Since these coinciding points are all different, 1 is subtracted from 34, then 2, and then 4, resulting in 27 division points. Cutting the stick along these division points results in 28 segments." 86c6ceb45e60,"[b]p1.[/b] Meghan Trainor is all about those base-systems. If she weighs $451$ pounds in some base, and $127$ pounds in a base that is twice as big, how much does she weigh in base $10$? [b]p2.[/b] Taylor Swift made the song $15$ in $2008$, and the song $22$ in $2012$. If she had continued at the same rate, in what year would her song $1989$ have come out? [b]p3.[/b] Sir Mix-A-Lot likes big butts and cannot lie. He is with his three friends, who all always lie. If each of the following statements was said by a different person, who is Sir Mix-A-Lot? A: B and D both like big butts. B: C and D either both like big butts, or neither of them do. C: B does not like big butts. D: A or C (or both) like big butts. [b]p4.[/b] Mark Ronson is going to give some uptown funk-tions to you. Bruno Mars is at $1$ degree Fahrenheit right now (that is, time, in seconds, $t = 0$), and his heat h(t) is given by $h(t) = 2 \cdot h(t - 1) + t$. If he will explode when he is over $1,000,000$ degrees Fahrenheit, at what time t will Bruno Mars explode? [b]p5.[/b] Jay-Z, who is a emcee, took $4$ AMC contests, for a total of $100$ PS. You had better use hide for answers.","(2916, 54)",medium,"### Problem 1: 1. Let the base in which Meghan weighs 451 pounds be \( b \). Then, in base \( b \), the number 451 represents: \[ 4b^2 + 5b + 1 \] 2. In a base that is twice as big, \( 2b \), she weighs 127 pounds. In base \( 2b \), the number 127 represents: \[ 1(2b)^2 + 2(2b) + 7 = 4b^2 + 4b + 7 \] 3. Since both expressions represent the same weight, we set them equal to each other: \[ 4b^2 + 5b + 1 = 4b^2 + 4b + 7 \] 4. Simplify the equation: \[ 5b + 1 = 4b + 7 \implies b = 6 \] 5. Substitute \( b = 6 \) back into the original base \( b \) expression to find the weight in base 10: \[ 4(6)^2 + 5(6) + 1 = 144 + 30 + 1 = 175 \] The final answer is \( \boxed{(2916, 54)} \)" 98954133b3b2,"7. If $p$ and $q$ are both prime numbers, the number of divisors $d(a)$ of the natural number $a=p^{\alpha} q^{\beta}$ is given by the formula $$ d(a)=(\alpha+1)(\beta+1) $$ For example, $12=2^{2} \times 3^{1}$, so the number of divisors of 12 is $$ d(12)=(2+1)(1+1)=6, $$ with the divisors being $1, 2, 3, 4, 6$, and 12. According to the given formula, please answer: Among the divisors of $20^{30}$ that are less than $20^{15}$, how many are not divisors of $20^{15}$? $\qquad$",450$ in number.,medium,"7.450 . Notice that, $20^{30}=2^{60} \times 5^{30}$. Then $d\left(20^{30}\right)=(60+1)(30+1)=1891$. Also, $20^{30}=20^{15} \times 20^{15}$ is a perfect square, and except for $20^{15}$, for any divisor greater than $20^{15}$, there is a corresponding divisor less than $20^{15}$, whose product is exactly $20^{30}$. Thus, among the divisors of $20^{30}$, there are $\frac{1891-1}{2}=945$ divisors less than $20^{15}$. Furthermore, the divisors of $20^{15}$ are all divisors of $20^{30}$, and $20^{15}=2^{30} \times 5^{15}$, so $d\left(20^{15}\right)=(30+1)(15+1)=496$. Therefore, among the divisors of $20^{30}$ that are less than $20^{15}$, those that are not divisors of $20^{15}$ total $945-496+1=450$ in number." 7ecce5c4c70d,"## T-3 Find the smallest integer $b$ with the following property: For each way of colouring exactly $b$ squares of an $8 \times 8$ chessboard green, one can place 7 bishops on 7 green squares so that no two bishops attack each other. Remark. Two bishops attack each other if they are on the same diagonal. Answer. 41",See reasoning trace,medium," Solution. Let us place 40 bishops on 6 diagonals as shown in Figure 2. If we select any 7 of the placed bishops, by Pigeon hole principle, at least two of the selected bishops are on the same diagonal, so they attack each other. Thus, the number $b$ of selected bishops is at least 41. ![](https://cdn.mathpix.com/cropped/2024_06_05_775ea5fd909fa7632c76g-14.jpg?height=437&width=443&top_left_y=1095&top_left_x=812) Figure 1 Now, suppose for a contrary, that there is a placement of 41 bishops such there are not 7 non-attacking bishops. Divide all tiles to 8 groups as shown in Figure ??. It is easy to see that any two bishops belonging to the same group do not attack each other. Therefore, each group contains at most 6 bishops. Moreover, groups 7 and 8 contain at most 2 bishops due to their size. Therefore, we have at most $6 \cdot 6+2 \cdot 2=40$ bishops, which is a contradiction. Therefore, from any placement of 41 bishops, it is possible to select some 7 of them such that no two attack each other. This, together with the lower bound of $b \geq 41$ finishes this solution. ![](https://cdn.mathpix.com/cropped/2024_06_05_775ea5fd909fa7632c76g-14.jpg?height=429&width=422&top_left_y=2127&top_left_x=817) Figure 2 Comment (1). A weaker upper bound of 49 can be shown as follows: Consider a placement of 49. bishops. We have 8 rows and $49 / 8>6$, so by Pigeon hole principle, there is a row with at least 7 bishops. Clearly, bishops in the same row do not attack each other. Comment (2). This problem can be generalised for larger dimensions of the chessboard and also larger number of sought non-attacking bishops. " fb53d6ccf0cc,"8.28 Let \(a, b\) and \(x\) be positive real numbers not equal to 1. Then, the equation \[ 4\left(\log _{a} x\right)^{2}+3\left(\log _{b} x\right)^{2}=8\left(\log _{a} x\right)\left(\log _{b} x\right) \] holds if and only if (A) for all values of \(a, b\) and \(x\). (B) if and only if \(a=b^{2}\). (C) if and only if \(b=a^{2}\). (D) if and only if \(x=a b\). (E) none of these. (27th American High School Mathematics Examination, 1976)",$(E)$,medium,"［Solution］The given equation can be written in the following form $$ 4\left(\log _{a} x\right)^{2}-8\left(\log _{a} x\right)\left(\log _{b} x\right)+3\left(\log _{b} x\right)^{2}=0 \text {, } $$ which is $$ \left(2 \log _{a} x-\log _{b} x\right)\left(2 \log _{a} x-3 \log _{b} x\right)=0 . $$ Therefore, $\log _{a} x^{2}=\log _{b} x$, or $\log _{a} x^{2}=\log _{b} x^{3}$. If we set $r=\log _{a} x^{2}$, then from the above equations we have $a^{r}=x^{2}$ and $b^{r}=x$, or $a^{r}=x^{2}$ and $b^{r}=x^{3}$, which means $a^{r}=b^{2 r}$, or $a^{3 r}=b^{2 r}$, thus $a=b^{2}$ or $a^{3}=b^{2}$. Therefore, the answer is $(E)$." 5d7c203f5b5e,"8. For all non-negative integers $x, y$, find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, satisfying $f(3 x+2 y)=f(x) f(y)$, where $\mathbf{N}$ is the set of non-negative integers. )",See reasoning trace,medium,"Solution: Let $x=y=0$, we get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$. If $f(0)=0$, for $x=0$ or $y=0$, we get $f(2 y)=$ $f(3 x)=0$ for all $x, y \in \mathrm{N}$. Let $f(1)=a$, then $$ \begin{array}{l} f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\ f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} . \end{array} $$ Since $f(25)=f(2 \times 2+3 \times 7)=f(2) f(7)=0$, we have $a=0$. For any odd number $k>4$, there exists $y \in \mathbb{N}$, such that $k=3+2 y$. Then $f(k)=0$, hence for all $x \in \mathbb{N}, f(x)=0$ satisfies the condition. If $f(0)=1$, let $x=0$ or $y=0$, we get $f(2 y)=f(y)$ or $f(3 x)=f(x)$. Let $f(1)=a$, then $$ \begin{array}{l} f(2)=a, \\ f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\ f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} . \end{array} $$ Since $f(25)=f(3 \times 3+2 \times 8)=f(3) f(8)=$ $f(1) f(4)=f(1) f(2)=a^{2}$, we have $a=0$ or $a=1$. Similarly, we have $$ f(x)=\left\{\begin{array}{ll} 1, & x=0, \\ 0, & x>0 \end{array} \text { or } f(x)=1, x \in \mathbb{N} .\right. $$ In summary, there are 3 functions that satisfy the conditions." a5518e8ac1bf,"9. Let $f(x)=x^{2}+p x+q, p 、 q \in \mathbf{R}$. If the maximum value of $|f(x)|$ on $[-1,1]$ is $M$, find the minimum value of $M$. Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.","\max _{-1 \leq x \leq 1}|f(x)|$, then we have $M \geqslant|f(1)|=|1+p+q|, M \geqslant|f(-1)|=|1-p+q|",easy,"9. Let $M=\max _{-1 \leq x \leq 1}|f(x)|$, then we have $M \geqslant|f(1)|=|1+p+q|, M \geqslant|f(-1)|=|1-p+q|$, $M \geqslant|f(0)|=|q|$, thus $4 M \geqslant|1+p+q|+2|-q|+|1-p+q| \geqslant |(1+p+q)+2(-q)+(1-p+q)|=2$, hence $M \geqslant \frac{1}{2}$. Example: when $f(x)=x^{2}-\frac{1}{2}$, $\max _{-1 \leq x \leq 1}|f(x)|=\frac{1}{2}$, so the minimum value of $M$ is $\frac{1}{2}$." 2aaa92f90055,"5. $A, B$ are distinct points on the parabola $y=3-x^{2}$ that are symmetric with respect to the line $x+y=0$, then $|A B|$ equals A. 3 B. 4 C. $3 \sqrt{2}$ D. $4 \sqrt{2}$",$C$,easy,"Let $A(a, b), B(-b,-a)$, then we have $\left\{\begin{array}{l}b=3-a^{2}, \\ -a=3-b^{2}\end{array} \Rightarrow a^{4}-6 a^{2}-a+6=0\right.$ $\Rightarrow(a+2)(a-1)\left(a^{2}-a-3\right)=0$. If $a=-2 \Rightarrow b=-1 \Rightarrow|A B|=3 \sqrt{2}$; If $a=1 \Rightarrow b=2 \Rightarrow|A B|=3 \sqrt{2}$; If $a^{2}-a-3=0 \Rightarrow a=a^{2}-3=-b \Rightarrow A, B$ lie on the line $x+y=0$, which is a contradiction. Therefore, the answer is $C$." 7d762f059f78,"Example 5 Let $x, y, z, w$ be four real numbers, not all zero. Find: $S=\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$'s maximum value.","\frac{1}{2 \lambda}+1$, solving this gives $\lambda=1+\sqrt{2}$. Substituting $\lambda=1+\sqrt{2}$, ",medium,"Notice that, in the given function expression, swapping $x$ with $w$ and $y$ with $z$ leaves the expression unchanged. Thus, by the mean value inequality with parameter $\lambda$, we have $$ 2 x y \leqslant \lambda x^{2}+\frac{1}{\lambda} y^{2}, \quad 2 z w \leqslant \lambda w^{2}+\frac{1}{\lambda} z^{2}, \quad 2 y z \leqslant y^{2}+z^{2}, $$ where equality holds if and only if $y=\lambda x, z=\lambda w, y=z$. Therefore, $x y+2 y z+z w \leqslant \frac{\lambda}{2} x^{2}+\left(\frac{1}{2 \lambda}+1\right) y^{2}+\left(\frac{1}{2 \lambda}+1\right) z^{2}+\frac{\lambda}{2} w^{2}$. To make the right-hand side symmetric so that it can cancel out with the denominator of $S$, choose $\lambda$ such that $\frac{\lambda}{2}=\frac{1}{2 \lambda}+1$, solving this gives $\lambda=1+\sqrt{2}$. Substituting $\lambda=1+\sqrt{2}$, we get $S \leqslant \frac{1+\sqrt{2}}{2}$, where equality holds if and only if $x=w=1, y=z=1+\sqrt{2}$. Thus, the maximum value sought is $\frac{1}{2}(1+\sqrt{2})$." 0afd8d93f028,"1. Let the polynomial $f(x)$ satisfy for any $x \in \mathbf{R}$, $f(x+1)+f(x-1)=2 x^{2}-4 x$, then the minimum value of $f(x)$ is $\qquad$ .",See reasoning trace,easy,"1. -2 . Since $f(x)$ is a polynomial, then $f(x+1), f(x-1)$ and $f(x)$ have the same degree, so $f(x)$ is a quadratic polynomial. Let $f(x)=a x^{2}+b x+c$, then $$ f(x+1)+f(x-1)=2 a x^{2}+2 b x+2(a+c), $$ Therefore, $$ 2 a x^{2}+2 b x+2(a+c)=2 x^{2}-4 x, $$ Thus $a=1, b=-2, c=-1$. Therefore, $$ f(x)=x^{2}-2 x-1=(x-1)^{2}-2 \geqslant-2 . $$" c83d2365ccba,"$31 \cdot 26$ satisfies the equation $m^{3}+6 m^{2}+5 m=27 n^{3}+9 n^{2}+9 n+1$ for the number of integer pairs $(m, n)$ is (A) 0. (B) 1. (C) 3. (D) 9. (E) infinitely many. (30th American High School Mathematics Examination, 1979)",$(A)$,easy,"[Solution] Since $m^{3}+6 m^{2}+5 m=m(m+1)(m+5)$, for all integers $m$, it can be divisible by 3, which only requires noting that when the remainder of $m$ divided by 3 is 0, 1, or 2, $m$, $m+5$, or $m+1$ can be divisible by 3, respectively. However, $27 n^{3}+9 n^{2}+9 n+1$ always leaves a remainder of 1 when divided by 3. Thus, there are no integer pairs $(m, n)$ that satisfy the given equation. Therefore, the answer is $(A)$." 858fc7488d8c,"[Coordinate method on the plane $]$ [ Orthocenter and orthotriangle ] Given points $A(5, -1), B(4, -8), C(-4, -4)$. Find the coordinates of the intersection point of the altitudes of triangle $ABC$.","$(3, -5)$",medium,"Applying the condition of perpendicularity of two lines ($k_{1} \cdot k_{2}=-1$), find the equations of the lines on which two altitudes of the triangle lie. Then find the coordinates of the intersection point of these lines by solving the corresponding system of equations. ## Solution Let's find the equation of the line BC through two points: $$ \frac{y-(-8)}{-4-(-8)}=\frac{x-4}{-4-4}, \text { or } y=-\frac{1}{2} x-6 $$ Then its slope $k_{1}=-\frac{1}{2}$. If $k_{2}$ is the slope of the line containing the altitude $AP$, then $k_{1} \cdot k_{2}=-1$. Therefore, $$ k_{2}=-\frac{1}{k_{1}}=2 $$ The equation of the line containing the altitude $AP$ of triangle $ABC$ will be found using point $A(5, -1)$ and the slope $k_{2}=2$: $$ y+1=2(x-5) \text {, or } y=2 x-11. $$ Let's find the equation of the line $AC$ through two points: $$ \frac{y-(-1)}{-4-(-1)}=\frac{x-5}{-4-5}, \text { or } y=\frac{1}{3} x-\frac{8}{3} $$ Then its slope $k_{3}=\frac{1}{3}$. If $k_{4}$ is the slope of the line containing the altitude $BQ$, then $k_{4} \cdot k_{3}=-1$. Therefore, $$ k_{4}=-\frac{1}{k_{3}}=-3 $$ The equation of the line containing the altitude $BQ$ of triangle $ABC$ will be found using point $B(4, -8)$ and the slope $k_{4}=-3$: $$ y+8=-3(x-4), \text { or } y=-3 x+4 $$ The coordinates of the point $H$ of intersection of the altitudes of triangle $ABC$ will be found by solving the system of equations defining the lines $AP$ and $BQ$: $$ \left\{\begin{array}{l} y=2 x-11 \\ y=-3 x+4 \end{array}\right. $$ We get: $x=3, y=-5$. ![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-02.jpg?height=431&width=560&top_left_y=1694&top_left_x=752) ## Answer $(3, -5)$." 028446508fd9,"2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .",4\).,easy,"2.4 . From the problem, we have $$ \left\{\begin{array}{l} 3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\ 3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 . \end{array}\right. $$ Solving this, we get \(3 \frac{8}{21}0$). Finally, the height of the prism is $h=\sqrt{L_{1} N^{2}-L N^{2}}=\sqrt{4(14+2 \sqrt{13})-32}=2 \sqrt{6+2 \sqrt{13}}$; then the volume of the prism is $V=K L^{2} \cdot h=32 \sqrt{6+2 \sqrt{13}}$. ""Phystech-2016"", mathematics, instructions for checking, tickets 9-12 The problem is considered fully solved (and the maximum possible number of points is awarded for it) only if all necessary transformations are provided in the solution text and all logical steps are fully explained, and the answers obtained are reduced to a simplified form. The presence of a correct answer does not guarantee a positive score for the problem. 1.(6) For each correctly considered case (""base > 1"", ""base <1"") 3 points, - non-equivalent transformation of the inequality 0 points for all subsequent actions, - the answer includes a value of $x$ for which the base of the logarithm is 1 ... deduct 1 point, - the answer includes a value of $x$ for which the base of the logarithm is 0 ... deduct 1 point, - the answer includes values of $x$ for which the base of the logarithm is negative no more than 3 points for the problem. 2.(6) It is proved that the minimum of one part of the equation is equal to the maximum of the other ... 1 point, - only one of the cases of the sign of the expression in parentheses is correctly considered ... 1 point, 3.(6) The substitution of variables (as in the solution) is made ... 1 point, - a cubic equation is obtained with respect to one of the new variables ... 1 point, - the cubic equation is solved ... 2 points, - extraneous solutions are obtained ... deduct 1 point. 4.(8) The ratio of the distances from the centers of the circles to the line $A B$ is found ... 1 point, - the ratio of the radii of the circles is found ... 4 points. 5.(6) All possible options for the last digit of the number are indicated ... 1 point, - points are not added for formulating divisibility criteria by 12 (by 75), - if the answer is obtained by enumeration and is incorrect ... no more than 1 point for the problem, - the answer is written in the form $9^{k} \cdot 2$ and so on, points are not deducted. 6.(8) The set of points satisfying the first equation is constructed ... 2 points, From this $^{1}$ : - for each found value of the parameter ... 3 points, - 1 extra value of the parameter is obtained ... deduct 1 point, - 2 extra values of the parameter are obtained ... deduct 3 points. 7.(8) - The ratio of the segments of part a) is found ... 2 points, - it is indicated that the distance between the planes is equal to the diameter of the sphere ... 1 point, - the segment of part b) is found ... 3 points, - the volume of the prism is found ... 2 points." 5a5172e5cfe9,Example 2 Find all such prime numbers that are both the sum of two prime numbers and the difference of two prime numbers.,"3, p=5, p+2=7$, does it satisfy the conditions, hence $p=5$.",medium,"Analysis Consider the prime number 2, which is the only even number among primes. Solution Let the prime number sought be $p$. Since it is the sum of two primes, $p>2$. Thus, $p$ is odd, so one of the two primes that sum to $p$ must be 2. At the same time, in the two primes with a difference of $p$, the subtrahend is also 2, i.e., $p=q+2, p=r-2$, where $q, r$ are primes. Thus, $p-2, p, p+2$ are all primes. And $p-2, p, p+2$ are three consecutive odd numbers, one of which must be a multiple of 3. Since this number is a prime, it can only be 3. Upon inspection, only when $p-2=3, p=5, p+2=7$, does it satisfy the conditions, hence $p=5$." b75e555de169,"For example, $13 x_{1}, x_{2}, \cdots, x_{1993}$ satisfy $\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{1992}-x_{1993}\right|=1993$, $y_{k}=\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}(k=1,2, \cdots, 1993)$, then what is the maximum possible value of $\left|y_{1}-y_{2}\right|+\left|y_{2}-y_{3}\right|+\cdots+\left|y_{1922}-y_{1993}\right|$?",1}^{1992} \mid y_{k}-y_{k+1} \mathrm{I}$ is 1992.,medium,"Solve $\left|y_{k}-y_{k+1}\right|=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right|$ $$ \begin{array}{l} =\frac{1}{k(k+1)}\left|\left(x_{1}-x_{k+1}\right)+\left(x_{2}-x_{k+1}\right)+\cdots+\left(x_{k}-x_{k+1}\right)\right| \\ \leqslant \frac{1}{k(k+1)}\left(\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|\right) \\ =\frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\right|, \end{array} $$ Thus, $\left|y_{1}-y_{2}\right|+\left|y_{2}-y_{3}\right|+\cdots+\left|y_{1922}-y_{1993}\right|=\sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right|$ $$ \begin{array}{l} =\sum_{k=1}^{1992} \sum_{i=1}^{k} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right| \\ =\sum_{i=1}^{192} \sum_{k=1}^{1992} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right| \\ =\sum_{i=1}^{1922} i \cdot\left[\frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\cdots+\frac{1}{1992 \cdot 1993}\right] \cdot\left|x_{i}-x_{i+1}\right| \\ =\sum_{i=1}^{192} i \cdot\left(\frac{1}{i}-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \leqslant \sum_{i=1}^{1992}\left(1-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \\ =\left(1-\frac{1}{1993}\right) \cdot \sum_{i=1}^{1992}\left|x_{i}-x_{i+1}\right|=\left(1-\frac{1}{1993}\right) \cdot 1993=1992 . \end{array} $$ On the other hand, if $x_{1}=t+1993, x_{2}=x_{3}=\cdots=x_{1993}=t$, then $$ \begin{array}{l} \left|y_{k}-y_{k+1}\right|=\frac{i}{k(k+1)} \cdot\left|x_{1}-x_{2}\right|=\frac{1993}{k(k+1)}, \\ \sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right|=1993 \cdot \sum_{k=1}^{1922} \frac{1}{k(k+1)}=1992 . \end{array} $$ Therefore, the maximum value of $\sum_{k=1}^{1992} \mid y_{k}-y_{k+1} \mathrm{I}$ is 1992." 63cbcf560c30,"$[\underline{\text { Properties of Sections }}]$ The edge of the cube $A B C D A 1 B 1 C 1 D 1$ is 12. Point $K$ lies on the extension of edge $B C$ at a distance of 9 from vertex $C$. Point $L$ on edge $A B$ is 5 units away from $A$. Point $M$ divides the segment $A 1 C 1$ in the ratio $1: 3$, counting from $A 1$. Find the area of the section of the cube by the plane passing through points $K, L, M$.",156,medium,"Let the line $L K$ intersect the edge $C D$ at point $F$ (Fig.1). According to the theorem on the intersection of two parallel planes by a third intersecting plane, the plane intersects the base $A 1 B 1 C 1 D 1$ along a line $a$, passing through point $M$ parallel to $L F$. Let the line $a$ intersect the lines $A 1 B 1$ and $C 1 D 1$ at points $N$ and $E$ respectively, and let points $L 1$, $F 1$, and $K 1$ be the orthogonal projections of points $L$, $F$, and $K$ onto the plane $A 1 B 1 C 1 D 1$. Denote $\angle B K L = \angle B 1 K 1 L 1 = \alpha$. Then $$ \operatorname{tg} \alpha = \frac{B L}{B K} = \frac{7}{21} = \frac{1}{3}, \quad C 1 F 1 = C F = C K \operatorname{tg} \alpha = 9 \cdot \frac{1}{3} = 3, \quad \cos \alpha = \frac{3}{\sqrt{10}} $$ Let $F 1 E = L 1 N = x$ (Fig.2). Then $$ A 1 N = A 1 L 1 - L 1 N = A L - L 1 N = 5 - x, \quad C 1 E = C 1 F + F 1 E = 3 + x, $$ $$ \frac{A_{1} N}{C_{1} E} = \frac{A_{1} M}{M C_{1}} = \frac{1}{3}, \quad \frac{5 - x}{3 + x} = \frac{1}{3} $$ From this, we find that $L 1 N = x = 3$. Then $C 1 E = 6$, $A 1 N = 2$. Points $E$ and $N$ lie on the sides $C 1 D 1$ and $A 1 B 1$. Therefore, the section of the given cube by the plane passing through points $K, L$, and $M$ is a parallelogram $L F E N$. Drop a perpendicular $L H$ from point $L$ to the line $NE$ (Fig.1). Then $L H$ is the height of the parallelogram $L F E N$. By the theorem of three perpendiculars, $L 1 H \perp N E$. From the right triangle $L 1 H N$ we find (Fig.2) that $$ L 1 H = L 1 N \cos \angle N L 1 H = L 1 N \cos \alpha = 3 \cdot \frac{3}{\sqrt{10}} = \frac{9}{\sqrt{10}} $$ Thus, $$ L H = \sqrt{L_{1} L^{2} + L_{1} H^{2}} = \sqrt{144 + \frac{81}{10}} = \frac{39}{\sqrt{10}} $$ Drop a perpendicular $F P$ to $A B$. Then $$ \begin{gathered} L P = B L - B P = 7 - 3 = 4 \\ N E = L F = \sqrt{F P^{2} + L P^{2}} = \sqrt{144 + 16} = 4 \sqrt{10} \end{gathered} $$ Therefore, ![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-14.jpg?height=120&width=602&top_left_y=2101&top_left_x=728) ## Answer 156.00" a3dd545ad45e,"$ A$ takes $ m$ times as long to do a piece of work as $ B$ and $ C$ together; $ B$ takes $ n$ times as long as $ C$ and $ A$ together; and $ C$ takes $ x$ times as long as $ A$ and $ B$ together. Then $ x$, in terms of $ m$ and $ n$, is: $ \textbf{(A)}\ \frac {2mn}{m \plus{} n} \qquad \textbf{(B)}\ \frac {1}{2(m \plus{} n)} \qquad \textbf{(C)}\ \frac {1}{m \plus{} n \minus{} mn} \qquad \textbf{(D)}\ \frac {1 \minus{} mn}{m \plus{} n \plus{} 2mn} \qquad \textbf{(E)}\ \frac {m \plus{} n \plus{} 2}{mn \minus{} 1}$",\frac{2 + m + n,medium,"1. Let \( A, B, \) and \( C \) respectively work at rates of \( a, b, \) and \( c \). Furthermore, let the total amount of work be \( 1 \). Then, we have the following relationships: \[ \frac{1}{a} = m \cdot \frac{1}{b+c} \] \[ \frac{1}{b} = n \cdot \frac{1}{a+c} \] \[ \frac{1}{c} = x \cdot \frac{1}{a+b} \] 2. Take the reciprocals of each equation: \[ a = \frac{b+c}{m} \] \[ b = \frac{a+c}{n} \] \[ c = \frac{a+b}{x} \] 3. Multiply all three equations: \[ a \cdot b \cdot c = \left(\frac{b+c}{m}\right) \cdot \left(\frac{a+c}{n}\right) \cdot \left(\frac{a+b}{x}\right) \] 4. Simplify the right-hand side: \[ abc = \frac{(b+c)(a+c)(a+b)}{mnx} \] 5. Expand the numerator: \[ (b+c)(a+c)(a+b) = (b+c)(a^2 + ab + ac + bc) \] \[ = a^2b + a^2c + ab^2 + abc + abc + b^2c + ac^2 + bc^2 \] \[ = a^2b + a^2c + ab^2 + 2abc + b^2c + ac^2 + bc^2 \] 6. Substitute back into the equation: \[ abc = \frac{a^2b + a^2c + ab^2 + 2abc + b^2c + ac^2 + bc^2}{mnx} \] 7. Divide both sides by \( abc \): \[ 1 = \frac{a^2b + a^2c + ab^2 + 2abc + b^2c + ac^2 + bc^2}{abc \cdot mnx} \] 8. Group the terms in the numerator: \[ 1 = \frac{2abc + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2}{abc \cdot mnx} \] 9. Simplify the numerator: \[ 1 = \frac{2abc + abc \left(\frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b}\right)}{abc \cdot mnx} \] 10. Divide both sides by \( abc \): \[ 1 = \frac{2 + \frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b}}{mnx} \] 11. Group the RHS numerator by similar denominators: \[ 1 = \frac{2 + \frac{a+c}{b} + \frac{b+c}{a} + \frac{a+b}{c}}{mnx} \] 12. We already know that: \[ \frac{b+c}{a} = m \] \[ \frac{a+c}{b} = n \] \[ \frac{a+b}{c} = x \] 13. Substituting these back into the equation: \[ 1 = \frac{2 + m + n + x}{mnx} \] 14. Rearrange the equation: \[ mnx = 2 + m + n + x \] \[ mnx - x = 2 + m + n \] \[ x(mn - 1) = 2 + m + n \] \[ x = \frac{2 + m + n}{mn - 1} \] The final answer is \(\boxed{\frac{2 + m + n}{mn - 1}}\)" 87bcd6ab143f,"$31 \cdot 17$ The number of integer pairs $(m, n)$ that satisfy the equation $m+n=mn$ is (A) 1 . (B) 2 . (C) 3 . (D) 4 . (E) greater than 4. (28th American High School Mathematics Examination, 1977)",$(B)$,easy,"[Solution] If $m+n=mn$, then $m+n-mn=0$, i.e., $\square$ we get $$ \begin{array}{l} m+n(1-m)=0, \\ n=\frac{m}{m-1} \quad(m \neq 1) . \end{array} $$ When $m=1$, there is no solution; only when $m=0$ or 2, the solution $\left[m, \frac{m}{m-1}\right]$ is an integer pair, i.e., there are two solutions $(0,0)$ and $(2,2)$. Therefore, the answer is $(B)$." eb7a3598f2f1,"""Pre-assembled"" unit cube has a base and a top, as well as its sides. The top is rotated by an acute angle $\alpha$ around the line connecting the centers of the two faces, relative to the base. By how much does the top get closer to the base if the side edges remain the same length and the connection between the vertices also remains unchanged?",1008&width=574&top_left_y=1083&top_left_x=754),medium,"The height of the cube before unscrewing is $m=1$, and the diagonal length of the square plates is $\sqrt{2}$. After the unscrewing (only the top plate rotates), the projections of the rotated points $A^{\prime}$ and $D^{\prime}$ onto the base plane are denoted as $B^{\prime}$ and $C^{\prime}$, respectively, and the center of the base square is $O$ (Figure 1). In the top view of the original and rotated squares (Figure 2), $O B=O B^{\prime}=\frac{\sqrt{2}}{2}, B B^{\prime}=x$. In the isosceles triangle $O B B^{\prime}$, $\angle B O B^{\prime}=\alpha$ and $\sin \frac{\alpha}{2}=\frac{\frac{x}{2}}{\frac{\sqrt{2}}{2}}$, from which $x=\sqrt{2} \sin \frac{\alpha}{2}$. Since the side edges do not change in length during the unscrewing, according to the notation in Figure 1, $A^{\prime} B=1, A^{\prime} B^{\prime}=m^{\prime}$ is the new height, and from the right triangle $A^{\prime} B B^{\prime}$: $m^{\prime 2}+x^{2}=1$, from which $$ m^{\prime}=\sqrt{1-2 \sin ^{2} \frac{\alpha}{2}}=\sqrt{1-2 \frac{1-\cos \alpha}{2}}=\sqrt{\cos \alpha} $$ The difference between the original and the new height gives how much closer the top plate has come to the base plate: $$ m-m^{\prime}=1-\sqrt{\cos \alpha} $$ Remark. The result also reveals something else. If $\alpha=0$, the value of $\cos \alpha$ is 1, so if nothing is unscrewed, the distance does not change. The unscrewing has a maximum, and this is $90^{\circ}$. In this case, if we imagine the unscrewing, the previously ""vertical"" (perpendicular to the base plane) edges will lie in the base plane, and the base plate and the top plate will now lie on top of each other. Várallyay György (Budapest, Kodály Z. Hungarian Chorus, 12th grade) ![](https://cdn.mathpix.com/cropped/2024_05_02_5d1e04bd74043e79a8efg-1.jpg?height=1008&width=574&top_left_y=1083&top_left_x=754)" 0a8f1345dbea,"The unit-radius circles $k_{1}$ and $k_{2}$ touch each other at point $P$. One of the common tangents of the two circles, which does not pass through $P$, is the line $e$. For $i>2$, let $k_{i}$ be the circle different from $k_{i-2}$ that touches $k_{1}$, $k_{i-1}$, and $e$. Determine the radius of $k_{1999}$.",460&width=727&top_left_y=1624&top_left_x=707),medium,"If two circles and a line touch each other as shown in the first figure, the radii of the circles are $R$ and $r$, and the distance between the points of tangency on the line is $d$, then the centers of the circles and the points of tangency form a right trapezoid, for which - using the Pythagorean theorem - the relationship $d^{2}=4 R r$ holds. Let the point of tangency of the circle $k_{i}$ and the line $e$ be $E_{i}$. From the definition of the circle $k_{i}$, it follows that for $i>2$, $E_{i}$ is an interior point of the segment $E_{1} E_{i-1}$. Therefore, $$ E_{1} E_{i}=E_{1} E_{i+1}+E_{i+1} E_{i} $$ Thus, the distance between the points of tangency, expressed using the radii of the circles, and using the result proven in the previous paragraph, $$ \sqrt{4 r_{1} r_{i}}=\sqrt{4 r_{1} r_{i+1}}+\sqrt{4 r_{i+1} r_{i}} $$ and therefore $$ \sqrt{r_{i+1}}=\frac{\sqrt{r_{1} r_{i}}}{\sqrt{r_{1}}+\sqrt{r_{i}}} $$ We will prove by complete induction that this relationship implies $\sqrt{r_{i+1}}=\frac{1}{i}$. The statement is true for $i=1$, because $r_{2}=1$. Suppose the statement is true for $2 \leq i \leq k$. Then $$ \sqrt{r_{k+1}}=\frac{\sqrt{r_{1} \cdot r_{k}}}{\sqrt{r_{1}}+\sqrt{r_{k}}}=\frac{1 \cdot \frac{1}{k-1}}{1+\frac{1}{k-1}}=\frac{1}{1+(k-1)}=\frac{1}{k} $$ Therefore, $\sqrt{r_{i+1}}=\frac{1}{i}$ is true for all $i$, so $r_{1999}=\frac{1}{1998^{2}}$. Gergely Kiss (Fazekas M. Főv. Gyak. Gimn., 12th grade) ![](https://cdn.mathpix.com/cropped/2024_05_02_25026db3e10a517736d3g-1.jpg?height=442&width=716&top_left_y=1134&top_left_x=726) $d$ ![](https://cdn.mathpix.com/cropped/2024_05_02_25026db3e10a517736d3g-1.jpg?height=460&width=727&top_left_y=1624&top_left_x=707)" 83e472ce86fc,37. (10-11 grades) How many planes are equidistant from four points that do not lie in the same plane?,See reasoning trace,easy,"37. Taking these points as the vertices of a tetrahedron, it is easy to establish that only seven planes can be drawn equidistant from its vertices." c1da768ad7d1,"Example 5 As shown in Figure 5, given that $\odot O$ is inside rectangle $ABCD$, and tangents to $\odot O$ are drawn through vertices $A, B, C,$ and $D$, touching points are $A_{1}, B_{1}, C_{1}, D_{1}$ respectively. If $AA_{1} = 3, BB_{1} = 4, CC_{1} = 5$, find the length of $DD_{1}$.",See reasoning trace,medium,"Connect $A O$, $B O$, $C O$, $D O$, $A_{1} O$, $B_{1} O$, $C_{1} O$, $D_{1} O$. Then $O A_{1} \perp A A_{1}$, $O B_{1} \perp B B_{1}$, $O C_{1} \perp C C_{1}$, $O D_{1} \perp D D_{1}$. Let the radius of $\odot O$ be $r$. Then by the Pythagorean theorem, we have $A O^{2}=A A_{1}^{2}+r^{2}$, $B O^{2}=B B_{1}^{2}+r^{2}$, $C O^{2}=C C_{1}^{2}+r^{2}$, $D O^{2}=D D_{1}^{2}+r^{2}$. Draw $O E \perp A B$, $O F \perp D C$, then $B E=C F$. Apply the property to Rt $\triangle A O E$ and point $B$, and to Rt $\triangle O D F$ and point $C$ respectively, we get $$ \begin{array}{l} A O^{2}=O B^{2}+A B^{2}-2 B A \cdot B E, \\ D O^{2}=O C^{2}+C D^{2}-2 C D \cdot C F . \end{array} $$ Subtracting the two equations, we get $$ A O^{2}-D O^{2}=O B^{2}-O C^{2} \text {, } $$ which means $A O^{2}+O C^{2}=D B^{2}+O D^{2}$. Thus, $A A_{1}^{2}+C C_{1}^{2}=B B_{1}^{2}+D D_{1}^{2}$. Therefore, $D D_{1}=\sqrt{A A_{1}^{2}+C C_{1}^{2}-B B_{1}^{2}}=3 \sqrt{2}$. [Note] Equation (1) indicates: the sum of the squares of the distances from a point inside a rectangle to two pairs of opposite vertices is equal." e5be5aa49768,"Subject 2. Determine the integer values of $x$ and $y$ such that $$ x-3 y+4=0 \quad \text { and } \quad \sqrt{x^{2}+7 y^{2}+8 x+8 y+4} \in \mathbb{Q} $$",See reasoning trace,medium,"## Solution: From $x-3 y+4=0 \Rightarrow x+4=3 y$ From $x^{2}+7 y^{2}+8 x+8 y+4=x^{2}+7 y^{2}+8 x+8 y+4+12-12=x^{2}+8 x+16+7 y^{2}+8 y-12=$ $$ \begin{aligned} & =(x+4)^{2}+7 y^{2}+8 y-12=(3 y)^{2}+7 y^{2}+8 y-12=9 y^{2}+7 y^{2}+8 y-12= \\ & =16 y^{2}+8 y+1-13=(4 y+1)^{2}-13 \end{aligned} $$ From $\sqrt{x^{2}+7 y^{2}+8 x+8 y+4} \in \mathbb{Q}$ we have that $(4 y+1)^{2}-13=k^{2}$, for $k \in \mathbb{Q}$. | Details of solution | Associated grading | | :---: | :---: | | Makes the substitution $x+4=3 y \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \ldots . \ldots . \ldots . \ldots$ | $1 \mathrm{p}$ | | ![](https://cdn.mathpix.com/cropped/2024_06_07_5dba17adea0b2d4d33dcg-1.jpg?height=69&width=1576&top_left_y=2581&top_left_x=185) | $2 \mathrm{p}$ | Since $(4 y+1)^{2}-k^{2}=13 \Leftrightarrow(4 y+1-k)(4 y+1+k)=13$ with $x$ and $y$ from $\mathbb{Z}$. By analyzing the possible cases, we obtain the solution $x=-10$ and $y=-2$. | ![](https://cdn.mathpix.com/cropped/2024_06_07_5dba17adea0b2d4d33dcg-2.jpg?height=129&width=1585&top_left_y=226&top_left_x=189) | | | :---: | :---: | | Analyzes the cases | |" fb646ea7b088,"Find the $(x, y)$ natural integers such that $x^{2}=y^{2}+7 y+6$",y+3$. It remains to solve $y^{2}+6 y+9=y^{2}+7 y+6$. Thus $y=3$ and then $x=6$. The only solution is,easy,"We find the order of magnitude of x relative to y: $(y+2)^{2}=y^{2}+4 y+4$ Therefore $x>y+2$ $(y+4)^{2}=y^{2}+8 y+16$ Therefore $x0$. What are the limits within which the value of $a$ can vary to ensure that the",See reasoning trace,medium,"One part is denoted by $x$, the other by $17-x$, and the conditional equation is $$ \frac{3}{4} x=\frac{5}{6}(17-x)+a $$ Rearranging the equation, we get $$ x=\frac{170+12 a}{19} $$ From the condition, it follows that $x$ is greater than 0 and less than 17, that is, $$ 0<\frac{170+12 a}{19}<17 $$ The first condition is satisfied for all positive $a$, and the second condition holds if $$ 170+12 a<17 \cdot 19 $$ or $$ 4 a<51 ; \quad a<12 \frac{3}{4} $$ Solved by: Dömölki B., Főző Éva, Kántor S., Kovács L., Szabó Magda, Zobor E." bc2793a12ed7,"$2 \cdot 90$ Given the equation $|x-2 n|=k \sqrt{x} \ (n \in N)$ has two distinct real roots in the interval $(2 n-1,2 n+1]$. Then the range of $k$ is (A) $k>0$. (B) $00$. Also, from $(x-2 n)^{2}=k^{2} x$, we know that the parabola $y=(x-2 n)^{2}$ intersects the line $y=k^{2} x$ at two distinct points in the interval $(2 n-1,2 n+1]$, Therefore, when $x=2 n-1$, we have $(x-2 n)^{2}>k^{2} x$, and when $x=2 n+1$, we have $(x-2 n)^{2} \geqslant k^{2} x$. Thus, $k^{2}(2 n+1) \leqslant 1$, which means $k \leqslant \frac{1}{\sqrt{2 n+1}}$. Hence, the answer is $(B)$." e2c675ccda0f,"Murashkin M.V. On each cell of a $10 \times 10$ board, there is a chip. It is allowed to choose a diagonal with an even number of chips and remove any chip from it. What is the maximum number of chips that can be removed from the board using such operations?",See reasoning trace,medium,"Let's call a (non-)even diagonal the one on which there is a (non-)even number of chips (at the moment). After removing a chip, an even diagonal becomes odd, and an odd one becomes even. Therefore, the number of odd diagonals does not decrease. At the beginning, there are 20 odd diagonals on the board, so there will be no fewer than 20 at the end. Among them, there are no fewer than 10 parallel ones, and on them, there will be no fewer than 10 chips left. It is possible to remove 90 chips, for example, in the following order. Remove all the chips from the left column. Now we can remove all the chips from the 2nd column, except for the top and bottom ones, then all the chips from the 3rd column... and so on. ## Otvet ## Task" 7648308a6d53,"Jack is jumping on the number line. He first jumps one unit and every jump after that he jumps one unit more than the previous jump. What is the least amount of jumps it takes to reach exactly $19999$ from his starting place? [i]2017 CCA Math Bonanza Lightning Round #2.3[/i]",201,medium,"1. **Determine the total distance Jack jumps after \( n \) jumps:** On his \( n \)-th jump, Jack jumps \( n \) units. The total distance after \( n \) jumps is the sum of the first \( n \) natural numbers: \[ S = \frac{n(n+1)}{2} \] We need to find the smallest \( n \) such that: \[ \frac{n(n+1)}{2} \geq 19999 \] 2. **Solve the inequality for \( n \):** \[ n(n+1) \geq 2 \times 19999 = 39998 \] This is a quadratic inequality. To solve it, we first solve the corresponding quadratic equation: \[ n^2 + n - 39998 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 1 \), and \( c = -39998 \): \[ n = \frac{-1 \pm \sqrt{1 + 4 \times 39998}}{2} = \frac{-1 \pm \sqrt{159993}}{2} \] Approximating \( \sqrt{159993} \approx 400 \): \[ n \approx \frac{-1 + 400}{2} = 199.5 \] Since \( n \) must be an integer, we take \( n = 200 \). 3. **Verify the total distance for \( n = 200 \):** \[ S = \frac{200 \times 201}{2} = 20100 \] This is 101 units more than 19999. 4. **Adjust the jumps to reach exactly 19999:** To reduce the total distance by 101 units, we need to change one of the jumps. Since changing any jump by \( k \) units will change the total by \( 2k \) units (because we can either add or subtract \( k \)), we need to adjust by an odd number. We can achieve this by adding one more jump (making it 201 jumps) and then adjusting one of the jumps to reduce the total by 101 units. Specifically, we can add a jump of 201 units and then change a jump of 151 units to -151 units: \[ 20100 + 201 - 151 = 20150 - 151 = 19999 \] Thus, the least number of jumps required is \( \boxed{201} \)." 5a22981deb1a,"Three, (10 points) Let $a+b+c=0, a^{2}+b^{2}+c^{2}=1$. (1) Find the value of $ab+bc+ca$; (2) Find the value of $a^{4}+b^{4}+c^{4}$.",See reasoning trace,medium,"Three, (1) Since $a+b+c=0$, then, $$ \begin{array}{l} (a+b+c)^{2} \\ =a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=0 . \end{array} $$ Given $a^{2}+b^{2}+c^{2}=1$, then $a b+b c+c a=-\frac{1}{2}$. $$ \begin{array}{l} \text { (2) } a b+b c+c a=-\frac{1}{2} \\ \Rightarrow(a b+b c+c a)^{2}=\frac{1}{4} \\ \Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=\frac{1}{4} \\ \Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=\frac{1}{4} . \end{array} $$ Given $a^{2}+b^{2}+c^{2}=1$, squaring it we get $$ \begin{array}{l} a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}=1 \\ \Rightarrow a^{4}+b^{4}+c^{4}=1-2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \\ \quad=1-2 \times \frac{1}{4}=\frac{1}{2} . \end{array} $$" 76888ea7acda,"14. A square $A B C D$ has side 40 units. Point $F$ is the midpoint of side $A D$. Point $G$ lies on $C F$ such that $3 C G=2 G F$. What is the area of triangle $B C G$ ?",320$.,easy,"Solution 320 Start by drawing a diagram. By Pythagoras' Theorem, $F C=20 \sqrt{5}$ and hence $C G=8 \sqrt{5}$. Now, $\sin \angle B C G=\sin \angle C F D=$ $\frac{40}{20 \sqrt{5}}=\frac{2}{\sqrt{5}}$. Hence the area of triangle $B C G$ is $\frac{1}{2} \times 40 \times 8 \sqrt{5} \times \frac{2}{\sqrt{5}}=320$." 5389bc3c2313,"A rectangular plot is almost square: its width and length are integers in meters that differ by exactly 1 meter. The area of the plot, in square meters, is a four-digit number, with the thousands and hundreds digits being the same, and the same is true for the tens and units digits. What are the possible dimensions of the plot?",See reasoning trace,medium,"The area is a number of the form $a a b b$, where $a$ and $b$ represent digits; now recall that $$ a a b b=1100 a+11 b=11(100 a+b) $$ Let $x$ be the width of the plot, then $$ x(x+1)=11(100 a+b) \quad(\mathrm{I}) $$ and we deduce that $x$ or $x+1$ is a multiple of 11. Searching for multiples of 11 that satisfy condition (I) is quite laborious, so to simplify, let's establish which values $x$ can have. We will look for the minimum and maximum values for $x$: - Minimum: the smallest possible area is $1111$, so $x(x+1)=1111 \Rightarrow x>32$ (II). - Maximum: the largest possible area is 9999, so $x(x+1)=9999 \Rightarrow x<100$ (III). Now we look for $x$ and $x+1$ satisfying (I), (II), and (III). $$ \begin{aligned} & 33 \times 34=1122 ; 43 \times 44=1892 ; 44 \times 45=1980 ; 54 \times 55=2970 ; 55 \times 56=2970 \\ & 65 \times 66=4290 ; 66 \times 67=4422 ; 76 \times 77=5852 ; 77 \times 78=6006 \\ & 87 \times 88=7656 ; 88 \times 89=7832 ; 99 \times 100=9900 \end{aligned} $$ We find 3 possibilities for $x: 33, 66$, and 99." e968f57393ad,"Example 16 Divide a circle into $n(\geqslant 2)$ sectors $S_{1}, S_{2}, \cdots, S_{n}$. Now, color these sectors using $m(\geqslant 2)$ colors, with each sector being colored one color and requiring that adjacent sectors have different colors. How many different coloring methods are there?","\frac{a_{n}}{(m-1)^{n}}$, then $b_{n}+\frac{1}{m-1} b_{n-1}=\frac{m}{m-1}$, i.e., $b_{n}-1=-\frac{1}",medium,"Let the number of coloring methods be $a_{n}$. (1) Find the initial value. When $n=2$, there are $m$ ways to color $S_{1}$, and then there are $m-1$ ways to color $S_{2}$ (since $S_{1}$ and $S_{2}$ must be different colors), so $a_{2}=m(m-1)$. (2) Find the recurrence relation. Since $S_{1}$ has $m$ coloring methods, $S_{2}$ has $m-1$ coloring methods, $S_{3}$ has $m-1$ coloring methods, $\cdots, S_{n}$ has $m-1$ coloring methods (only ensuring that $S_{i+1}$ and $S_{i}$ are different colors, $i=1,2, \cdots, n-1$, without ensuring that $S_{n}$ and $S_{1}$ are different colors), there are a total of $m(m-1)^{n}$ coloring methods. These coloring methods can be divided into two categories: One category is where $S_{n}$ and $S_{1}$ are different colors, and there are $a_{n}$ such methods; the other category is where $S_{n}$ and $S_{1}$ are the same color, in which case $S_{n}$ and $S_{1}$ can be merged into one sector, and note that $S_{n-1}$ and $S_{1}$ must be different colors, so there are $a_{n-1}$ such coloring methods. By the principle of addition, we get $a_{n}+a_{n-1}=m(m-1)^{n-1}$. (3) Find $a_{n}$. Let $b_{n}=\frac{a_{n}}{(m-1)^{n}}$, then $b_{n}+\frac{1}{m-1} b_{n-1}=\frac{m}{m-1}$, i.e., $b_{n}-1=-\frac{1}{m-1} \cdot \left(b_{n-1}-1\right)$, which shows that $b_{n}-1$ is the $(n-1)$-th term of a geometric sequence with the first term $b_{2}-1=\frac{a_{2}}{(m-1)^{2}}-1=\frac{1}{m-1}$ and the common ratio $-\frac{1}{m-1}$. Therefore, $b_{n}-1=\frac{1}{m-1}\left(-\frac{1}{m-1}\right)^{n-2}=(-1)^{n} \frac{1}{(m-1)^{n-1}}$. From this, we can get $a_{n}=(m-1)^{n} b_{n}=(m-1)^{n}+(-1)^{n}(m-1)$." b35971f1a1e4,"## Task 2 - 260832 Determine all pairs $(p ; q)$ of prime numbers that satisfy the following conditions (1), (2), (3)! (1) The difference $q-p$ is greater than 0 and less than 10. (2) The sum $p+q$ is the square of a natural number $n$. (3) If you add the number $n$ to the sum of $p$ and $q$, you get 42.","0$, only $n_{1}=6$ is a natural number.",medium,"I. If a pair $(p ; q)$ of prime numbers together with a natural number $n$ satisfies conditions (1), (2), (3), then: From (2) we have $p+q=n^{2}$, and from (3) we have $n+p+q=42$. Therefore, due to the previous equation, $n+n^{2}=42$, i.e., $n(n+1)=42$ (4). The only ways to express 42 as a product of two natural numbers are (disregarding the order of the factors) $42=1 \cdot 42=2 \cdot 21=3 \cdot 14=6 \cdot 7$. Among these representations, only $6 \cdot 7$ contains two factors of the form $n$ and $n+1$, so from (4) (for natural numbers $n$) it follows that $n=6$. From (2) it follows that $p+q=36$. Therefore, $p$ is one of the prime numbers less than 36, i.e., one of the numbers 2, $3,5,7,11,13,17,19,23,29,31$, and $q$ is respectively one of the numbers 34, 33, 31, 29, 25, $23,19,17,13,7,5$. Of these possibilities, all except $p=17, q=19$ are eliminated, as for them either $q-p \geq 10$ or $q-p6$ is $n(n+1)>6 \cdot 7$. Therefore, among the natural numbers $n$, only $n=6$ can satisfy the equation $n(n+1)=42$. 2. With knowledge of the solution formula for quadratic equations (or by derivation, for example, using $\left.\left(n+\frac{1}{2}\right)^{2}-\frac{1}{4}=42\right)$, we obtain: Of the two solutions $$ n_{1 ; 2}=-\frac{1}{2} \pm \sqrt{\frac{1}{4}+42}=\frac{1}{2}(-1 \pm 13) $$ of the quadratic equation $n^{2}+n-42=0$, only $n_{1}=6$ is a natural number." a2dfd8fcf076,"The sum of the (decimal) digits of a natural number $n$ equals $100$, and the sum of digits of $44n$ equals $800$. Determine the sum of digits of $3n$.",300,medium,"1. Let \( n \) be a natural number such that the sum of its digits is 100. We can represent \( n \) in its decimal form as: \[ n = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0 \] where \( a_i \) are the digits of \( n \). Therefore, the sum of the digits of \( n \) is: \[ a_k + a_{k-1} + \ldots + a_1 + a_0 = 100 \] 2. Given that the sum of the digits of \( 44n \) equals 800, we need to analyze the representation of \( 44n \). We can write: \[ 44n = 44 \cdot (10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0) \] Expanding this, we get: \[ 44n = 10^{k+1} (4a_k) + 10^k (4a_k + 4a_{k-1}) + 10^{k-1} (4a_{k-1} + 4a_{k-2}) + \ldots + 10 (4a_1 + 4a_0) + 4a_0 \] 3. The sum of the digits in this representation is: \[ 4a_k + 4a_k + 4a_{k-1} + 4a_{k-1} + \ldots + 4a_0 + 4a_0 = 800 \] Simplifying, we get: \[ 8(a_k + a_{k-1} + \ldots + a_1 + a_0) = 800 \] Since \( a_k + a_{k-1} + \ldots + a_1 + a_0 = 100 \), we have: \[ 8 \cdot 100 = 800 \] 4. To ensure there are no carries in the sum of the digits of \( 44n \), we must have \( 4a_i + 4a_{i-1} < 10 \) for all \( i \). This implies: \[ a_i + a_{i+1} \leq 2 \] for all \( i \). 5. Now, we need to determine the sum of the digits of \( 3n \). Since \( a_i + a_{i+1} \leq 2 \), multiplying by 3 will not cause any carries. Therefore, the sum of the digits of \( 3n \) is: \[ 3 \cdot (a_k + a_{k-1} + \ldots + a_1 + a_0) = 3 \cdot 100 = 300 \] The final answer is \(\boxed{300}\)." 6031f0ea044c,![](https://cdn.mathpix.com/cropped/2024_05_06_b46fbd582cc3a82460aeg-39.jpg?height=321&width=517&top_left_y=195&top_left_x=468),-6,medium,"Answer: -6. ![](https://cdn.mathpix.com/cropped/2024_05_06_b46fbd582cc3a82460aeg-39.jpg?height=359&width=614&top_left_y=600&top_left_x=420) Fig. 11: to the solution of problem 10.7 Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$. Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$. Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$. Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$. Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$. ![](https://cdn.mathpix.com/cropped/2024_05_06_b46fbd582cc3a82460aeg-40.jpg?height=521&width=571&top_left_y=83&top_left_x=437) Fig. 12: to the solution of problem 10.7 It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$." c39895b73afa,"$1 \cdot 14$ Let $S$ be the unit circle in the complex plane (i.e., the set of all complex numbers with modulus 1), and $f$ be a mapping from $S$ to $S$. For any $z \in S$, define $$ \begin{array}{l} f^{(1)}=f(z), f^{(2)}(z)=f\left(f^{(1)}(z)\right), \cdots, \\ f^{(k+1)}(z)=f\left(f^{(k)}(z)\right), k=2,3, \cdots . \end{array} $$ If $c \in S$, such that $$ f^{(1)}(c) \neq c, \cdots, f^{(n-1)}(c) \neq c, f^{(n)}(c)=c, $$ then $c$ is called an $n_{-}$periodic point of $f$. Let $m$ be a natural number greater than 1, and $f$ is defined as follows: $$ f(z)=z^{m}, z \in S \text {. } $$ Try to find the total number of 1989-periodic points of $f$.",See reasoning trace,medium,"[Solution] Let the set of all 1989-periodic points of $f$ be denoted by $T$ and let $B_{j}=\{z \in S \mid f^{(j)}(z)=z\}, j=1,2, \cdots$. (1) If $z_{0} \in S$ is a fixed point of $f^{(n)}$, i.e., $f^{(n)}\left(z_{0}\right)=z_{0}$, and $z_{0}$ is a Suppose $n=p h+q, 0 \leqslant qk$, let $n=p k+q, 0 \leqslant q0$ is achieved at $x=\frac{-b}{2 a}$. Thus, the minimum area will be achieved if $$ x=-\frac{\left(-\frac{5}{4}\right)}{2 \frac{1}{8}}=5 $$ In other words, if the rope is cut exactly in the middle! c) From the previous part, we know that to minimize the sum of the areas, it is necessary to cut the rope exactly in the middle. Well, we claim that to minimize the area with nine cuts (i.e., creating ten squares), it is necessary that all pieces of rope be equal. To show this, consider the following argument: if two of the ten pieces of rope were different, it would be possible to reduce the area by cutting the pieces of rope so that these two were equal (we are using the previous part). Therefore, any two pieces of rope must be equal. Hence, all must be equal!" 5ea9fc120483,"8. (10 points) Among the three given phrases “尽心尽力”, “力可拔山”, and “山穷水尽”, each Chinese character represents a number between 1 and 8, with the same character representing the same number and different characters representing different numbers. If the sum of the numbers represented by the characters in each phrase is 19, and “尽” > “山” > “力”, then the maximum value of “水” is $\qquad$",: 7,medium,"【Analysis】Through analysis, we can know: From “尽心尽力” (exerting oneself to the utmost), “力可拔山” (strength can move mountains), and “山穷水尽” (reaching the end of one's resources), the sum of the numbers represented by the Chinese characters in each word is 19, which gives us the equations: $$ \left\{\begin{array}{l} \text { 尽+心+尽+力=19(1) } \\ \text { 力+可+拔+山=19(2) } \\ \text { 山+穷+水+尽 }=19(3) \end{array}\right. $$ We can get 3 尽 + 心 + 2 力 + 可 + 拔 + 2 山 + 穷 + 水 = 19 × 3 = 57 The sum of 1 to 8 is 36, so 2 尽 + 1 力 + 1 山 = 57 - 36 = 21, comparing with (1) we get 山 - 心 = 2. “尽” > “山” > “力”, “力” should be as large as possible, and “尽” should be as small as possible, Assuming “力”, “山”, “尽” are consecutive natural numbers, we have 2 (力 + 2) + 力 + 1 + 力 = 21 “力” is 4, at this time 山 = 5, 心 = 3, 尽 = 6; (1) is satisfied: 6 + 3 + 6 + 4 = 19; (3): 5 + 穷 + 水 + 6 = 19, at this time the maximum value of 水 is 7, 穷 is 1, to deduce (2): (2): 4 + 可 + 拔 + 5 = 19, and now only 2 and 8 are left, which satisfies the condition. At this time, the maximum value of 水 is 7. But at this time, 6 (尽), 4 (山), 5 (力) do not satisfy “尽” > “山” > “力”, so it does not meet the requirements. Therefore, the maximum value of 水 is 7. According to this, we can solve the problem. 【Solution】From the words “尽心尽力” (exerting oneself to the utmost), “力可拔山” (strength can move mountains), and “山穷水尽” (reaching the end of one's resources), the sum of the numbers represented by the Chinese characters in each word is 19, which gives us the equations: $$ \left\{\begin{array}{l} \text { 尽+心+尽+力=19(1) } \\ \text { 力+可+拔+山=19(2) } \\ \text { 山+穷+水+尽 }=19(3) \end{array}\right. $$ (1) + (2) + (3) gives: 3 尽 + 心 + 2 力 + 可 + 拔 + 2 山 + 穷 + 水 = 19 × 3 = 57 The sum of 1 to 8 is 36, So 2 尽 + 1 力 + 1 山 = 57 - 36 = 21, comparing with (1) we get 山 - 心 = 2. “尽” > “山” > “力”, “力” should be as large as possible, and “尽” should be as small as possible, Assuming “力”, “山”, “尽” are consecutive natural numbers, we have 2 (力 + 2) + 力 + 1 + 力 = 21 “力” is 4, at this time 山 = 5, 心 = 3, 尽 = 6; (1) is satisfied: 6 + 3 + 6 + 4 = 19; (3): 5 + 穷 + 水 + 6 = 19, at this time the maximum value of 水 is 7, 穷 is 1, to deduce (2): (2): 4 + 可 + 拔 + 5 = 19, and now only 2 and 8 are left, which satisfies the condition. At this time, the maximum value of 水 is 7. But at this time, 6 (尽), 4 (山), 5 (力) do not satisfy “尽” > “山” > “力”, so it does not meet the requirements. Therefore, the maximum value of 水 is 7. The answer is: 7." 35f2b16db83c,"$\left[\begin{array}{l}\text { Properties of medians. The centroid of a triangle. } \\ {[\text { Pythagorean Theorem (direct and inverse). }]} \end{array}\right]$ Find the area of triangle $ABC$, if $AC=3$, $BC=4$, and the medians $AK$ and $BL$ are perpendicular to each other.",$\sqrt{11}$,medium,"The medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex. ## Solution Let $M$ be the point of intersection of the medians of triangle $ABC$. Denote $M L=x, M K=y$. Then $M B=2 x, M A=2 y$. By the Pythagorean theorem from right triangles $B M K$ and $A M L$, we find: $$ \left\{\begin{array}{l} 4 x^{2}+y^{2}=4 \\ x^{2}+4 y^{2}=\frac{9}{4} \end{array}\right. $$ From the obtained system, we find that $$ x=\frac{\sqrt{11}}{2 \sqrt{3}}, y=\frac{1}{\sqrt{3}} $$ $$ S_{\triangle \mathrm{AMB}}=\frac{1}{2} A M \cdot M B=\frac{1}{2} \cdot 2 x \cdot 2 y=2 x y=\frac{\sqrt{11}}{3} $$ Therefore, $$ S_{\triangle \mathrm{ABC}}=3 S_{\triangle \mathrm{AMB}}=\sqrt{11} $$ ![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-25.jpg?height=443&width=828&top_left_y=356&top_left_x=615) ## Answer $\sqrt{11}$." b6ba7f8c59c7,"Example 1 Given real numbers $a, b, c$ satisfy $$ \begin{array}{l} a b c=-1, a+b+c=4, \\ \frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 . \end{array} $$ Find the value of $a^{2}+b^{2}+c^{2}$.",(a+b+c)^{2}-2(a b+b c+c a)=14$.,medium,"From the conditions given in the problem, we have $$ \frac{1}{a}=-b c, \quad a=4-b-c \text {. } $$ Notice that, $$ \begin{array}{l} \frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\ =\frac{1}{(b-1)(c-1)} . \end{array} $$ Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$, $\frac{c}{c^{2}-3 c-1}=\frac{1}{(a-1)(b-1)}$. Therefore, $\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}$ $=\frac{1}{(b-1)(c-1)}+\frac{1}{(c-1)(a-1)}+\frac{1}{(b-1)(c-1)}$ $=\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$ $=\frac{1}{(a-1)(b-1)(c-1)}=1$ $\Rightarrow (a-1)(b-1)(c-1)$ $=a b c-(a b+b c+c a)+(a+b+c)-1$ $=1$ $\Rightarrow a b+b c+c a=1$ $\Rightarrow a^{2}+b^{2}+c^{2}$ $=(a+b+c)^{2}-2(a b+b c+c a)=14$." 003e443888bc,"[ Pigeonhole Principle (continued).] What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other #",16 kings,easy,"Answer: 16 kings. Divide the board into 16 squares, in each of which there can be no more than one king." 0941b013eb9b,"4. 247 Find all positive integers $n$ such that the following equation has an integer solution. $$x^{n}+(2+x)^{n}+(2-x)^{n}=0$$",0$ has integer solutions if and only if $n=1$.,medium,"[Solution] When $n=1$, we get $x+(2+x)+(2-x)=0, x=-4$ as its integer solution. When $n$ is even, the original equation has integer solutions if and only if: $$\left\{\begin{array}{l} x=0 \\ 2+x=0 \\ 2-x=0 \end{array}\right.$$ But (1) is a contradictory equation, hence we know that when $n$ is even, the original equation has no integer solutions. When $n$ is an odd number greater than 1, using the binomial theorem to expand the left side and then combining and organizing, we get a polynomial $p(x)$, whose highest degree coefficient is 1, all coefficients are non-negative integers, and the constant term is $2^{n+1}$. By Vieta's theorem, the possible integer solutions of $p(x)=0$ can be set as $-2^{t}, t$ being a non-negative integer: If $t=0$, then $x=-1$, the left side of the original equation is the sum of three odd numbers, so it cannot be zero. If $t=1$, then $x=-2$, the left side of the original equation is $(-2)^{n}+0+4^{n}$, which also cannot be 0. If $t \geqslant 2$, let $t=p+1, p \geqslant 1$, then the left side of the original equation is $$\begin{aligned} & 2^{n}\left[-2^{p n}+\left(1-2^{p}\right)^{n}+\left(1+2^{p}\right)^{n}\right] \\ = & 2^{n}\left\{-2^{p n}+2\left[1+\binom{n}{2} 2^{2 p}+\binom{n}{4} 2^{4 p}+\cdots\right]\right\} \end{aligned}$$ But $n \geqslant 3$, the value inside the large braces is 2 modulo 4, so it cannot be 0, thus the left side cannot be 0. Therefore, when $n$ is an odd number greater than or equal to 3, the original equation has no integer solutions. In summary, the equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has integer solutions if and only if $n=1$." beb88207a452,"## Task 2 - 060612 At the beginning of the school year, Heinz bought two different types of notebooks, one costing 8 Pf, the other 15 Pf per piece. He paid 1.31 MDN for 12 notebooks in total. How many notebooks did he buy of each type?",See reasoning trace,easy,"Let $a$ be the number of notebooks costing 8 Pf and $b$ the number of notebooks costing 15 Pf. Then we have $$ a \cdot 8 + b \cdot 15 = 131 \quad ; \quad a + b = 12 \quad \text{or} \quad a = 12 - b $$ Substituting the second equation into the first, we get $8 \cdot (12 - b) + 15b = 131$. From this, we find $b = 5$ and further $a = 7$. Heinz buys 7 notebooks costing 8 Pf and 5 notebooks costing 15 Pf." 1d0d0d7aa49a,"6. (8 points) On the board, 33 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 33 minutes?",528,medium,"Answer: 528. Solution: Let's represent 33 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups "" $x$ "" and "" $y$ "" will be connected by $xy$ line segments. This is the same number of candies that Karlson eats. After 33 minutes, all points will be connected. In total, $\frac{33 \cdot (33-1)}{2}=528$ line segments will be drawn. Therefore, Karlson will eat 528 candies." c0e366af0884,"4. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, the sum of the first $n$ terms is $S_{n}$, for any integer $n \geqslant 2, S_{n-1}-8, a_{n}, 8 S_{n}-4$ are always in arithmetic progression. Then $\lim _{n \rightarrow \infty}\left(a_{n}+S_{n}\right)=$ $\qquad$ .",See reasoning trace,medium,"4. $\frac{4}{3}$. From the problem, we get $$ 2 a_{n}=S_{n-1}-8+8 S_{n}-4=9 S_{n}-a_{n}-12, $$ Thus, $a_{n}=3 S_{n}-4(n \geqslant 2)$. Given $a_{1}=1$, we have $a_{2}=\frac{1}{2}$. When $n \geqslant 2$, $$ 3 S_{n}=a_{n}+4, \quad 3 S_{n+1}=a_{n+1}+4 . $$ Subtracting the two equations, we get $$ 3 a_{n+1}=a_{n+1}-a_{n} . $$ Therefore, $\frac{a_{n+1}}{a_{n}}=-\frac{1}{2}(n \geqslant 2)$. Hence, $$ a_{n}=\left\{\begin{array}{ll} 1, & n=1, \\ -\left(-\frac{1}{2}\right)^{n-1}, & n \geqslant 2 . \end{array}\right. $$ Thus, when $n \geqslant 2$, we have $a_{n}+S_{n}=\frac{4}{3} a_{n}+\frac{4}{3}$. Therefore, $$ \lim _{n \rightarrow \infty}\left(a_{n}+S_{n}\right)=\frac{4}{3} \lim _{n \rightarrow \infty} a_{n}+\frac{4}{3}=\frac{4}{3} . $$" dd4b027e390a,"6. For the equation in terms of $x$ $$ \frac{1}{x^{2}-x}+\frac{k-5}{x^{2}+x}=\frac{k-1}{x^{2}-1} $$ if it has an extraneous root $x=1$. Then the value of $k$ is ( ). (A) 2 (B) 3 (C) 6 (D) 10",1$ yields $k=3$.,easy,"6. B. Eliminating the denominator, we get $$ x+1+(k-5)(x-1)=(k-1) x \text {. } $$ Substituting $x=1$ yields $k=3$." 863f05d045b0,"12. There are two buckets, a large one and a small one, both containing some water. It is known that the weight of the water in the large bucket is half that in the small bucket. If 30 kilograms of water are added to the large bucket, the weight of the water in the large bucket becomes three times that in the small bucket. Originally, the large bucket contained $\qquad$ water.",5 parts. So the original amount of water in the large bucket = 30 ÷ (6-1) = 6 kilograms.,easy,"【Answer】6 【Analysis】According to the line segment diagram, it is found that 30 kilograms corresponds to (6-1) = 5 parts. So the original amount of water in the large bucket = 30 ÷ (6-1) = 6 kilograms." d8af52021deb,"(Poland 1995) Let $n$ be a strictly positive integer. Find the minimum value that the quantity $$ x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{3}}{3}+\ldots+\frac{x_{n}^{n}}{n} $$ can take, where $x_{1}, \ldots, x_{n}$ are strictly positive real numbers whose sum of reciprocals is exactly $n$.",See reasoning trace,medium,"The constraint given in the statement tells us that the harmonic mean of $x_{i}$ is equal to 1. Thus, by the HM-GM inequality, $$ x_{1} \cdots x_{n} \geq 1 $$ Once this is obtained, it suffices to apply the weighted AM-GM inequality with weights $1, \frac{1}{2}, \ldots, \frac{1}{n}$ to the $x_{i}^{i}$ to get ![](https://cdn.mathpix.com/cropped/2024_05_10_a954a7760d797f08770eg-512.jpg?height=112&width=1073&top_left_y=1646&top_left_x=500) $$ \begin{aligned} & \geq 1+\frac{1}{2}+\ldots+\frac{1}{n} \end{aligned} $$ We then see that this minimum is achieved when all $x_{i}$ are equal to 1." d7ee6c4dbf03,"Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, $$ f(f(x+1)+y-1)=f(x)+y $$ ## Solutions to the exercises","B$, where $A$ and $B$ are expressions that depend on a variable $x$. If $B$ covers the entire codoma",medium,"Let's show that $f$ is injective: let $a, b$ be two real numbers such that $f(a)=f(b)$. When $x=0$ and $y=a+1-f(1)$, the equation becomes $f(a)=f(0)+a-1+f(1)$. Similarly, if $x=0$ and $y=b+1-f(1)$, we have $f(b)=f(0)+b-1+f(1)$, so $f(0)+a-1+f(1)=f(0)+b-1+f(1)$, hence $a=b$. We have shown that the function $f$ is injective. To use this property, we set $y=0$, then $f(f(x+1)-1)=f(x)$, and since the function is injective, $f(x+1)=x+1 \Longrightarrow f(z)=z$ for all $z$ where we have made the change of variable $z=x-1$. We verify that the function $f: x \mapsto x$ is indeed a solution to the equation. ## Remark 4. In functional equations, showing that a function is injective goes hand in hand with finding an equality of the form $f(""bidule"")=f(""machin"")$ to deduce ""bidule"" = ""machin"". It is necessary that such equalities be non-trivial for them to constitute progress in the exercise, which implies that ""bidule"" or ""machin"" must contain $f$. Thus, functional equations containing nested $f$s are the most conducive to using the notion of injectivity. A simple idea that works well to obtain equalities of the form $f(""bidule"")=f(""machin"")$ is to eliminate all the output variables. In the previous exercise, we did this by setting $y=0$. This kind of reasoning that motivates variable changes is valuable, as it saves time by avoiding trying 10 useless substitutions. However, there will always be functional equations that will require you to try many substitutions blindly until you find the right ones. ## Surjective and Bijective Functions Definition 5. We say that a function $f: E \rightarrow F$ is surjective if for every $t \in F$, there exists $a \in E$ such that $f(a)=t$. In other words, a function is surjective if every element of the codomain has at least one preimage. Definition 6. A function is said to be bijective if it is both injective and surjective. ## Remark 7. A function is therefore bijective if every element of the codomain has a unique preimage. ## Remark 8. To show that a function is surjective, we generally try to reduce it to an equation of the form $f(A)=B$, where $A$ and $B$ are expressions that depend on a variable $x$. If $B$ covers the entire codomain of $f$ as $x$ covers the same set, then $f$ is surjective." 366334214b29,"1. If $x=\frac{23+\sqrt{469}}{5}$, then $25 x^{4}-1996 x^{2}+144$ $=(\quad)$. (A) 0 (B) 1 (C) 469 (D) 1996",0$.,easy,",- 1. A. From the difference in the number of times the letters appear in the condition and conclusion, we know that we should either increase the degree of the condition or decrease the degree of the conclusion (factorize). Given $5 x-23=\sqrt{469}$, squaring gives $5 x^{2}-46 x+12=0$. Rearranging, $46 x=5 x^{2}+12$. Squaring again gives $25 x^{4}-1996 x^{2}+144=0$." 7fb111d074ea,"6. Evaluate $$ \prod_{\theta=1}^{89}\left(\tan \theta^{\circ} \cos 1^{\circ}+\sin 1^{\circ}\right) $$",See reasoning trace,easy,6. $\csc 1^{\circ}$ or $\sec 89^{\circ}$ or equivalent a853dca59540,"1. Determine the number of pairs $(a, b)$ of natural numbers $(1 \leqq a1$; then we can set $a^{\prime}=2 a-3 b, b^{\prime}=2 b-a$. Then $(a+b \sqrt{3})=(2+\sqrt{3})\left(a^{\prime}+b^{\prime} \sqrt{3}\right)$ and $a^{\prime 2}-3 b^{2}=\left(4 a^{2}-12 a b+9 b^{2}\right)-3\left(4 b^{2}-4 a b+a^{2}\right)=a^{2}-3 b^{2}=1$. Moreover, we claim $a^{\prime}$ and $b^{\prime}$ are also nonnegative, with $a^{\prime}0 ; 3 b^{2}=a^{2}-1>0 \Rightarrow b>0 \Rightarrow a^{2}=3 b^{2}+1 \leq 4 b^{2} \Rightarrow b^{\prime}=2 b-a \geq 0$; and $b>0 \Rightarrow 9 b^{2}>3 b^{2}+1=a^{2} \Rightarrow a-a^{\prime}=3 b-a>0$. Thus, a solution $(a, b)$ in nonnegative integers with $a>1$ yields a new solution $\left(a^{\prime}, b^{\prime}\right)$ such that $a^{\prime}1$, we can repeat this process to obtain nonnegative integers $a^{\prime \prime}, b^{\prime \prime}$ such that $a^{\prime \prime}a^{\prime}>a^{\prime \prime}>\cdots$. But this sequence of nonnegative integers cannot decrease forever, so eventually we get to some $n$ for which $a^{(n)}=1$. The equation then implies $b^{(n)}=0$, so $a+b \sqrt{3}=(2+\sqrt{3})^{n}\left(a^{(n)}+b^{(n)} \sqrt{3}\right)=(2+\sqrt{3})^{n}$. Then, it follows that $(2-\sqrt{3})^{n}=a-b \sqrt{3}$. (This can be proven by comparing the binomial expansions of $(2+\sqrt{3})^{n}$ and $(2-\sqrt{3})^{n}$, or by a straightforward induction on $n$; note that the equation $a+b \sqrt{3}=(2+\sqrt{3})^{n}$ uniquely determines the integers $a$ and b.) Thus, we conclude that $$ a=\frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad b=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}} . $$ If we had wanted all integer solutions to the equation $a^{2}-3 b^{2}=1$, rather than all nonnegative integer solutions, we simply note that the equation is invariant under switching signs of $a$ and/or $b$, so the set of all solutions is simply the set of solutions given above with possible sign changes thrown in. Now we consider the given equation, $2 x^{2}-6 x y+3 y^{2}=-1$. This can be rewritten as $x^{2}-3(y-x)^{2}=1$. Thus, by letting $a=x, b=y-x$, we have an equivalence between our given equation and the Pell equation, so we can express our solutions to the former in terms of those of the latter: $x=a$ and $y=a+b$ gives $$ x= \pm \frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad y=x \pm \frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}} $$ where $n$ is any nonnegative integer. It can be checked by computation that any such pair is a valid, integral solution, and the foregoing argument shows that all solutions are of this form." 2c79abd5f6f0,"1 Convex quadrilateral $E F G H$ has vertices $E, F, G, H$ on the sides $A B, B C, C D, D A$ of convex quadrilateral $A B C D$, respectively, satisfying $$ \frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1 $$ and points $A, B, C, D$ are on the sides $H_{1} E_{1}, E_{1} F_{1}, F_{1} G_{1}, G_{1} H_{1}$ of convex quadrilateral $E_{1} F_{1} G_{1} H_{1}$, respectively, such that $E_{1} F_{1} \parallel E F, F_{1} G_{1} \parallel F G, G_{1} H_{1} \parallel G H, H_{1} E_{1} \parallel H E$. Given $\frac{E_{1} A}{A H_{1}}=\lambda$, find the value of $\frac{F_{1} C}{C G_{1}}$.",See reasoning trace,medium,"(1) If $E F / / A C$, as shown in Figure 1, then $$ \frac{B E}{E A}=\frac{B F}{F C} $$ Substituting the known conditions, we can get $\frac{D H}{H A}=\frac{D G}{G C}$, so $H G / / A C$, thus $E_{1} F_{1} / / A C / / H_{1} G_{1}$, hence $$ \frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda $$ (2) If $E F$ is not parallel to $A C$, as shown in Figure 2, let the extension of $F E$ intersect the extension of $C A$ at point $T$, then by Menelaus' theorem we get $$ \begin{array}{l} \frac{C F}{F B} \cdot \frac{B E}{E A} \cdot \frac{A T}{T C}=1 \\ \frac{C G}{G D} \cdot \frac{D H}{H A} \cdot \frac{A T}{T C}=1 \end{array} $$ Combining the given conditions, by the converse of Menelaus' theorem, points $T, H, G$ are collinear. Let $T F, T G$ intersect $E_{1} H_{1}$ at points $M, N$ respectively. By $E_{1} B / / E F$, we get $E_{1} A=\frac{B A}{E A} \cdot A M$, similarly $H_{1} A=\frac{A D}{A H} \cdot A N$, so Also $$ \frac{E_{1} A}{H_{1} A}=\frac{A M}{A N} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D} $$ From (1) and (2), we get $\frac{E_{1} A}{A H_{1}}=\frac{E Q}{Q H} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}=\frac{\triangle A B C}{\triangle A D C}$ Similarly, we can get $$ \frac{F_{1} C}{C G_{1}}=\frac{\triangle A B C}{\triangle A D C} $$ Therefore $$ \frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda $$" ac5fd558eb4d,"$$ x^{2} y+y^{2}=x^{3} \text {. } $$ (Daniel Paleka)",See reasoning trace,medium,"First Solution. Firstly, let us consider the case $x=0$. In that case, it is trivial to see that $y=0$, and thus we get the solution $(0,0)$. From now on, we can assume $x \neq 0$. From the trivial $x^{2} \mid x^{3}$, the equation gives $x^{2}\left|x^{2} y+y^{2} \Rightarrow x^{2}\right| y^{2}$, which means $x \mid y$. 1 point. We use the substitution $y=k x$, where $k \in \mathbb{Z}$. 1 point. The substitution gives us $$ \begin{gathered} k x^{3}+k^{2} x^{2}=x^{3} \\ k x+k^{2}=x \\ k^{2}=x(1-k) \end{gathered} $$ 2 points. Considering the greatest common divisor of $k^{2}$ and $1-k$, we get $$ G C D\left(k^{2}, 1-k\right)=G C D\left(k^{2}+k(1-k), 1-k\right)=G C D(k, 1-k)=G C D(k, 1-k+k)=G C D(k, 1)=1 $$ 3 points. That leaves us with two possibilities. a) $1-k=1 \Rightarrow k=0 \Rightarrow x=0$ which is not possible since $x \neq 0$. 1 point. b) $1-k=-1 \Rightarrow k=2 \Rightarrow x=-4, y=-8$, which gives a solution to the original equation. 1 point. Second Solution. We rearrange the equation into: $$ y^{2}=x^{2}(x-y) $$ It can easily be shown that if $y \neq 0, x-y$ must be square. 1 point. If $y=0$, from the starting equation we infer $x=0$, and we have a solution $(x, y)=(0,0)$. In the other case, we set $x=y+a^{2}$, where $a$ is a positive integer. Taking the square root of the equation gives: $$ |y|=|x| a $$ 1 point. Because $x=y+a^{2}>y$, it is impossible for $y$ to be a positive integer, because then the equation would say $y=x a>x$, which is false. That means $y<0$, and also: $$ -y=|x| a $$ 2 points. If $x$ is positive, we can write: $$ -y=x a=\left(y+a^{2}\right) a=a y+a^{3} $$ which rearranges into $$ -y(a+1)=a^{3} $$ so $a^{3}$ is divisible by $a+1$, which is not possible for positive $a$ due to $a^{3}=(a+1)\left(a^{2}-a+1\right)-1$. 2 points. We see that $x$ cannot be zero due to $y$ being negative, so the only remaining option is that $x<0$ also. We write: $$ -y=x a=-\left(y+a^{2}\right) a=-a y+a^{3} $$ which can similarly be rearranged into $$ -y(a-1)=a^{3} $$ and this time $a^{3}$ is divisible by $a-1$. 1 point. Analogously, we decompose $a^{3}=(a-1)\left(a^{2}+a+1\right)+1$, so $a-1$ divides 1 and the unique possibility is $a=2$. 2 points. The choice $a=2$ gives $y=-8$ and $x=-4$, which is a solution to the original equation. 1 point. ## Notes on marking: - Points awarded for different solutions are not additive, a student should be awarded the maximal number of points he is awarded following only one of the schemes. - Saying that $(0,0)$ is a solution is worth $\mathbf{0}$ points. The point is awarded only if the student argues that, disregarding the solution $(0,0)$, we must only consider $x \neq 0$, or a similar proposition. - Failing to check that $(0,0)$ is a solution shall not be punished. Failing to check that $(-4,-8)$ is a solution shall result in the deduction of 1 point only if a student did not use a chain of equivalences to discover the solution." 1ca0b95e441c,"The base of the triangle is 20; the medians drawn to the lateral sides are 18 and 24. Find the area of the triangle. #",## Problem,medium,"If $M$ is the point of intersection of the medians of triangle $ABC$, then its area is three times the area of triangle $BMC$. ## Solution Let $BB_1$ and $CC_1$ be the medians of triangle $ABC$, $M$ be their point of intersection, $BB_1 = 18$, $CC_1 = 24$, and $BC = 20$. Then the sides of triangle $BMC$ are: $$ BC = 20, BM = \frac{2}{3} BB_1 = 12, CM = \frac{2}{3} CC_1 = 16 $$ This triangle is a right triangle ( $BC^2 = BM^2 + CM^2$ ). Its area is 96, and the area of triangle $ABC$ is three times larger, i.e., 288. ![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-51.jpg?height=391&width=483&top_left_y=222&top_left_x=793) ## Answer ## Problem" ce0f4703ffce,"11. Given the quadratic function $$ y=a x^{2}+b x+c \geqslant 0(a0, b^{2}-4 a c \leqslant 0$. Notice that $$ \begin{array}{l} M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\ \geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} . \end{array} $$ Let $t=\frac{b}{a}$. Then $t>1$. Thus, $$ \begin{array}{l} M \geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)}=\frac{t^{2}+2 t+1}{t-1} \\ =(t-1)+\frac{4}{t-1}+4 \\ \geqslant 2 \sqrt{4}+4=8 . \end{array} $$ The equality holds if and only if $t=3, b^{2}=$ $4 a c$, i.e., $b=3 a, c=\frac{9}{4} a$. Therefore, the minimum value of $M$ is 8." 5f1268a63e87,"1. In a magic square, the sum of the numbers in each row, column, and on both diagonals is the same. The figure below shows a $3 \times 3$ square in which four cells have been filled with rational numbers. Determine whether it is possible to fill the remaining empty cells with rational numbers so that the square becomes a magic square. | 3.6 | | | | :--- | :--- | :--- | | | | $5 \frac{2}{5}$ | | $4 \frac{4}{5}$ | | 2.4 |",See reasoning trace,medium,"1. In a magic square, the sum of the numbers in each row, column, and on both diagonals is the same. The image shows a $3 \times 3$ square with four fields filled with rational numbers. Investigate whether it is possible to fill the remaining empty fields with rational numbers so that the square becomes magic. ## Solution. Let the unknown numbers in the magic square be denoted by $a, b, c, d$, and $e$. The sum of the numbers in each row/column/diagonal is the same, so from the first column and the second row we get: $3.6 + c + 4 \frac{4}{5} = c + d + 5 \frac{2}{5}$ $8.4 = d + 5.4$ 2 POINTS $d = 8.4 - 5.4 = 3$ 1 POINT | 3.6 | $a$ | $b$ | | :---: | :---: | :---: | | $c$ | $d$ | $5 \frac{2}{5}$ | | $4 \frac{4}{5}$ | $e$ | 2.4 | The sum of the numbers on one diagonal is $3.6 + 3 + 2.4 = 9$. The square will be magic if the sum of the elements in all columns, rows, and both diagonals is 9. From the second row, we get: $c = 9 - 3 - 5 \frac{2}{5} = 0.6$ 1 POINT From the third column, we get: $b = 9 - 3.6 - 2.4 = 1.2$ 1 POINT From the third row, we get: $e = 9 - 4 \frac{4}{5} - 2.4 = 1.8$ From the first row, we get: $a = 9 - 3.6 - 1.2 = 4.2$ We need to check if the sum is magic in the first and second columns and on the other diagonal: $3.6 + 0.6 + 4.8 = 4.2 + 3 + 1.8 = 4.8 + 3 + 1.2 = 9$ Since the sum of the numbers in each row, column, and on both diagonals is 9, the square is magic. Note 1. If the student arrives at the correct results in a different order, evaluate as follows: determining the first unknown number 3 POINTS, sum by columns/rows/diagonals 1 POINT, each subsequent number 1 POINT, and checking the sum in the remaining rows/columns/diagonals not used for calculating the unknown numbers 2 POINTS. Note 2. For a complete solution, it is not necessary to show the calculation process with equations; it is sufficient to determine the numbers that are missing (8 POINTS) and verify that the sum in all columns/rows and diagonals is 9 (2 POINTS). | 3.6 | 4.2 | 1.2 | | :---: | :---: | :---: | | 0.6 | 3 | $5 \frac{2}{5}$ | | $4 \frac{4}{5}$ | 1.8 | 2.4 |" 86faed49ff9a,"(a) Show that for every $n\in\mathbb{N}$ there is exactly one $x\in\mathbb{R}^+$ so that $x^n+x^{n+1}=1$. Call this $x_n$. (b) Find $\lim\limits_{n\rightarrow+\infty}x_n$.",\lim_{n \to +\infty,medium,"### Part (a) 1. Consider the function \( f(x) = x^n + x^{n+1} \) for \( x \in \mathbb{R}^+ \). 2. The function \( f(x) \) is continuous and strictly increasing on \( \mathbb{R}^+ \). This is because both \( x^n \) and \( x^{n+1} \) are continuous and strictly increasing functions for \( x > 0 \). 3. As \( x \) approaches 0, \( f(x) \) approaches 0. As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( +\infty \). 4. By the Intermediate Value Theorem, since \( f(x) \) is continuous and strictly increasing, for any \( y \) in the interval \( (0, +\infty) \), there exists a unique \( x \in \mathbb{R}^+ \) such that \( f(x) = y \). 5. Specifically, for \( y = 1 \), there exists a unique \( x_n \in \mathbb{R}^+ \) such that \( x_n^n + x_n^{n+1} = 1 \). Thus, we have shown that for every \( n \in \mathbb{N} \), there is exactly one \( x \in \mathbb{R}^+ \) such that \( x^n + x^{n+1} = 1 \). We denote this unique \( x \) as \( x_n \). \(\blacksquare\) ### Part (b) 1. We need to find \( \lim_{n \to +\infty} x_n \). 2. First, observe that \( 0 < x_n < 1 \). This is because if \( x_n \geq 1 \), then \( x_n^n + x_n^{n+1} \geq 1 + x_n > 1 \), which contradicts \( x_n^n + x_n^{n+1} = 1 \). 3. Assume \( x_{n+1} < x_n \). Then \( 1 = x_n^n + x_n^{n+1} > x_{n+1}^{n+1} + x_{n+1}^{n+2} \). This is a contradiction because \( x_{n+1}^{n+1} + x_{n+1}^{n+2} \) should also equal 1. 4. Therefore, \( x_{n+1} \geq x_n \), meaning \( \{x_n\} \) is an increasing sequence. 5. Since \( \{x_n\} \) is increasing and bounded above by 1, it converges to some limit \( \alpha \) where \( 0 < \alpha \leq 1 \). 6. Suppose \( 0 < \alpha < 1 \). Then \( \lim_{n \to +\infty} \ln x_n^n = \lim_{n \to +\infty} n \ln x_n = -\infty \). Hence, \( \lim_{n \to +\infty} x_n^n = 0 \). 7. Since \( 0 < x_n^{n+1} < x_n^n \), by the Squeeze Theorem, \( \lim_{n \to +\infty} x_n^{n+1} = 0 \). 8. Thus, \( \lim_{n \to +\infty} (x_n^n + x_n^{n+1}) = 0 \), which contradicts \( x_n^n + x_n^{n+1} = 1 \). 9. Therefore, \( \alpha = 1 \). \(\blacksquare\) The final answer is \( \boxed{ \lim_{n \to +\infty} x_n = 1 } \)." f483246bd7f8,"60. Solve the matrix equation $$ \left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right) x=\binom{7}{17} $$","\binom{3}{2}$, by the definition of equal matrices, we get $x_{1}=3, x_{2}=2$.",medium,"Solution. $1^{0}$. We will find the inverse matrix $A^{-1}$. Let's find the determinant of matrix $A$: $$ D=\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|=1 \cdot 4-2 \cdot 3=4-6=-2 \neq 0 $$ Calculate the algebraic complements of each element of matrix $A$: $A_{11}=(-1)^{1+1} \cdot 4=4, A_{12}=(-1)^{1+2} \cdot 3=-3 ; A_{21}=(-1)^{2+1} \cdot 2=$ $=-2 ; A_{22}=(-1)^{2+2} \cdot 1=1$. Write the matrix $\left(\begin{array}{rr}4 & -3 \\ -2 & 1\end{array}\right)$ and transpose it: $\left(\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right)$. Considering that $1 / D=-1 / 2$, write the inverse matrix: $$ A^{-1}=-\frac{1}{2}\left(\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right)=\left(\begin{array}{rr} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right) $$ $2^{0}$. Multiply matrix $A^{-1}$ by matrix $B$: $$ X=A^{-1} B=\left(\begin{array}{rr} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right) \cdot\binom{7}{17}=\binom{(-2) \cdot 7+1 \cdot 17}{\frac{3}{2} \cdot 7+\left(-\frac{1}{2}\right) \cdot 17}=\binom{3}{2} $$ $3^{0}$. Since $\binom{x_{1}}{x_{2}}=\binom{3}{2}$, by the definition of equal matrices, we get $x_{1}=3, x_{2}=2$." c94c4939b8b7,"4. Given the sequence $\left\{a_{n}\right\}$ satisfies: $$ a_{1}=1 \text {, } $$ $S_{n+1}=4 a_{n}+2$ ( $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$). Then $a_{n}=$ . $\qquad$",(3 n-1) \cdot 2^{n-2}$.,easy,"4. $(3 n-1) \cdot 2^{n-2}$. From the problem, we know $$ \begin{array}{l} 4 a_{n+1}+2=S_{n+2}=a_{n+2}+S_{n+1} \\ =a_{n+2}+4 a_{n}+2 \\ \Rightarrow a_{n+2}=4 a_{n+1}-4 a_{n} . \end{array} $$ Therefore, $a_{n}=(3 n-1) \cdot 2^{n-2}$." 02b4e76ff113,"Andrea has finished the third day of a six-day canoe trip. If she has completed $\frac{3}{7}$ of the trip's total distance of $168 \mathrm{~km}$, how many km per day must she average for the remainder of her trip? (A) 29 (B) 24 (C) 27 (D) 32 (E) 26",(D),easy,"Since Andrea has completed $\frac{3}{7}$ of the total $168 \mathrm{~km}$, then she has completed $\frac{3}{7} \times 168 \mathrm{~km}$ or $3 \times 24=72 \mathrm{~km}$. This means that she has $168-72=96 \mathrm{~km}$ remaining. To complete the $96 \mathrm{~km}$ in her 3 remaining days, she must average $\frac{96}{3}=32 \mathrm{~km}$ per day. ANSWER: (D)" d97f641fc900,"There are a certain number of red balls, green balls and blue balls in a bag. Of the balls in the bag, $\frac{1}{3}$ are red and $\frac{2}{7}$ are blue. The number of green balls in the bag is 8 less than twice the number of blue balls. The number of green balls in the bag is (A) 12 (B) 16 (C) 20 (D) 24 (E) 28",(B),medium,"Solution 1 Suppose there are $x$ balls in total in the bag. Then there are $\frac{1}{3} x$ red balls and $\frac{2}{7} x$ blue balls. This tells us that the number of green balls is $x-\frac{1}{3} x-\frac{2}{7} x=\frac{21}{21} x-\frac{7}{21} x-\frac{6}{21} x=\frac{8}{21} x$. But we know that the number of green balls is $2 \times \frac{2}{7} x-8$. Thus, $\frac{8}{21} x=2 \times\left(\frac{2}{7} x\right)-8$ or $\frac{8}{21} x=\frac{12}{21} x-8$ or $\frac{4}{21} x=8$ or $x=42$. Since $x=42$, the number of green balls is $\frac{8}{21} x=\frac{8}{21}(42)=16$. ## Solution 2 Suppose that there were 21 balls in the bag. (We choose 21 since there are fractions with denominator 3 and fractions with denominator 7 in the problem.) Since $\frac{1}{3}$ of the balls are red, then 7 balls are red. Since $\frac{2}{7}$ of the balls are blue, then 6 balls are red. Thus, there are $21-7-6=8$ green balls in the bag. However, this is only 4 less than twice the number of blue balls, so there cannot be 21 balls in the bag. To get from "" 4 less"" to "" 8 less"", we try doubling the number of balls in the bag. If there are 42 balls in the bag, then 14 are red and 12 are blue, so 16 are green, which is 8 less than twice the number of blue balls. Therefore, the number of green balls is 16 . ANSwER: (B)" 8d77e1a21b06,"## Task Condition Calculate the area of the figure bounded by the graphs of the functions: $$ x=(y-2)^{3}, x=4 y-8 $$",See reasoning trace,medium,"## Solution $$ \begin{aligned} & (y-2)^{3}=4(y-2) \\ & (y-2)\left[4-(y-2)^{2}\right]=0 \\ & \begin{array}{l} 4-y^{2}+4 y-4=0 \\ y(y-4)=0 \end{array} \\ & \frac{S}{2}=\int_{2}^{4}(4 y-8) d y-\int_{2}^{4}(y-3)^{3} d y= \\ & =\left.2 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}-\int_{2}^{4}\left(y^{3}-6 y^{2}+12 y-8\right) d y= \\ & =32-8-32+16-\left(\left.\frac{y^{4}}{4}\right|_{4} ^{2}-\left.6 \frac{y^{3}}{3}\right|_{4} ^{2}+\left.6 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}\right)= \\ & =8-\left(\frac{256}{4}-4-128+16+96-24-32+16\right)=4 \\ & S=2 \cdot \frac{S}{2}=8 \end{aligned} $$ ![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-40.jpg?height=446&width=1371&top_left_y=1065&top_left_x=571) Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-21$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals Ukrainian Banner Network - Last edited on this page: 08:22, 11 June 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 14-22 ## Material from Plusi" e0cb06609251,"3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied $$ \left\{\begin{array}{l} (x-a)^{2}+(y-b)^{2} \leqslant 8 \\ a^{2}+b^{2} \leqslant \min (-4 a+4 b ; 8) \end{array}\right. $$ Find the area of the figure $M$.",$24 \pi-4 \sqrt{3}$,medium,"Answer: $24 \pi-4 \sqrt{3}$. Solution. The second inequality is equivalent to the system of inequalities $$ \left\{\begin{array}{l} a^{2}+b^{2} \leqslant-4 a+4 b \\ a^{2}+b^{2} \leqslant 8 \end{array}\right. $$ Thus, the original system is equivalent to the following: $$ \left\{\begin{array} { l } { ( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } \leqslant 8 , } \\ { a ^ { 2 } + b ^ { 2 } \leqslant - 4 a + 4 b , } \\ { a ^ { 2 } + b ^ { 2 } \leqslant 8 } \end{array} \Leftrightarrow \left\{\begin{array}{l} (a-x)^{2}+(b-y)^{2} \leqslant 8 \\ (a+2)^{2}+(b-2)^{2} \leqslant 8 \\ a^{2}+b^{2} \leqslant 8 \end{array}\right.\right. $$ The sets of points defined by these inequalities on the plane $(a ; b)$ (where $x$ and $y$ act as parameters) are circles $\omega_{1}, \omega_{2}, \omega_{3}$ with radius $\sqrt{8}$ and centers at $P(x ; y), B(-2 ; 2), A(0 ; 0)$, respectively. The condition of the problem means that the system (31) must have a solution with respect to $(a ; b)$, that is, all three circles must have at least one common point. Let the circles bounding $\omega_{2}$ and $\omega_{3}$ intersect at points $C$ and $D$ (then triangles $A B C$ and $A B D$ are equilateral). The intersection of circles $\omega_{2}$ and $\omega_{3}$ is a figure $F$, representing the union of two smaller segments of these circles, bounded by the chord $C D$. Then the figure $M$ consists of all possible points $(x ; y)$ located at a distance of no more than $\sqrt{8}$ from the figure $F$. (This is the union of all circles of radius $\sqrt{8}$, the centers of which belong to the figure $F$.) Let points $P$ and $Q$ be symmetric to points $A$ and $B$ (respectively) relative to point $C$; points $T$ and $R$ are symmetric to points $A$ and $B$ (respectively) relative to point $D$. It is not difficult to understand that $M$ is the union of four sectors (the central angle of all sectors is less than $180^{\circ}$): - sector $P A T$ of the circle with center at point $A$ and radius $A P$, - sector $Q B R$ of the circle with center at point $B$ and radius $B Q$, - sector $P C Q$ of the circle with center at point $C$ and radius $C P$, - sector $R D T$ of the circle with center at point $D$ and radius $D T$. Note that the first two sectors intersect along the rhombus $A C B D$, and there are no other intersections between the sectors. At the same time, the first two sectors are equal to each other, and the last two sectors are also equal to each other. Thus, the area of the figure $M$ is $$ S_{M}=S_{P A T}+S_{Q B R}+S_{P C Q}+S_{R D T}-S_{A C B D}=2 \cdot \frac{\pi(2 \sqrt{8})^{2}}{3}+2 \cdot \frac{\pi(\sqrt{8})^{2}}{6}-\frac{\sqrt{3}}{2} \cdot(\sqrt{8})^{2}=24 \pi-4 \sqrt{3} $$" 2e07894a5306,B3. Find all real solutions of the equation $\left(x^{2}+x+3\right)^{2}+3\left(x^{2}+x-1\right)=28$.,0$ and introduce a new variable $u$ for the expression in parentheses. The equation $u^{2}+3 u-40=0$,easy,"B3. We write the equation in the form $\left(x^{2}+x+3\right)^{2}+3\left(x^{2}+x+3\right)-40=0$ and introduce a new variable $u$ for the expression in parentheses. The equation $u^{2}+3 u-40=0$ has two solutions $u_{1}=-8$ and $u_{2}=5$. From this, we get two equations for $x$: $x^{2}+x+3=-8$, which has no solution, and $x^{2}+x+3=5$, which has two real solutions $x_{1}=-2$ and $x_{2}=1$." 50db5e9bd494,22. Find the square root of $25 \cdot 26 \cdot 27 \cdot 28+1$.,"25$, the square root is $25^{2}+3 \cdot 25+1=701$.)",easy,"(ans 701 . $$ \begin{array}{l} a(a+1)(a+2)(a+3)+1=a(a+3)(a+1)(a+2)+1=\left[a^{2}+3 a\right]\left[a^{2}+\right. \\ 3 a+2]+1=\left[\left(a^{2}+3 a+1\right)-1\right]\left[\left(a^{2}+3 a+1\right)+1\right]+1=\left(a^{2}+3 a+1\right)^{2} . \end{array} $$ If $a=25$, the square root is $25^{2}+3 \cdot 25+1=701$.)" 64dbb7263562,"8.59 For each pair of real numbers $x, y$, the function $f$ satisfies the functional equation $$ f(x)+f(y)=f(x+y)-x y-1 . $$ If $f(1)=1$, then the number of integers $n$ (where $n \neq 1$) that satisfy $f(n)=n$ is (A) 0. (B) 1. (C) 2. (D) 3. (E) infinitely many. (30th American High School Mathematics Examination, 1979)",$(B)$,medium,"[Solution] Substituting $x=1$ into the functional equation, we get $$ f(y+1)=f(y)+y+2 \text {. } $$ Since $f(1)=1$, substituting $y=2,3,4, \cdots$ consecutively, we can see that for $y$ being a positive integer, $f(y)>0$. Therefore, for $y$ being a positive integer, $$ f(y+1)>y+2>y+1 ; $$ Thus, for integers $n>1$, $f(n)=n$ has no solution. Solving the equation for $f(y)$: From $\circledast$ we have $$ f(y)=f(y+1)-(y+2), $$ Substituting $y=0,-1,-2, \cdots$ consecutively into this equation, we get $$ \begin{array}{l} f(0)=-1, \quad f(-1)=-2, \quad f(-2)=-2, \quad f(-3)=-1, \\ f(-4)=1 . \end{array} $$ Since $f(-4)>0$ and for $y0$. Thus, for $y0$. Therefore, for $n<-4, f(n) \neq n$. Hence, $f(n)=n$ has only the solutions $n=1,-2$. Therefore, the answer is $(B)$." 0c3c8248f196,"20) At a birthday party, four toys are drawn by lot among the three boys present. The draws are independent, i.e., all the boys participate in all the draws. What is the probability $p$ that at least one boy ends up without a toy? (A) $p=\frac{4}{9}$ (B) $\frac{4}{9}0$, for any $k$-element subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ among the $f_{k}$ $k$-element subsets, in the permutations of $1,2, \cdots, n$, the permutations that exactly form the subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ are $k!(n-k)!$ in number. Since any two elements in $A$ are distinct, the permutations obtained by the above method for all elements in $A$ are necessarily distinct. Therefore, we have $\sum_{k=1}^{n} f_{k} \cdot k!(n- \sum_{k=1}^{n} \frac{f_{k}}{C_{n}^{k}}=C_{n}^{\left\lfloor\frac{n}{2}\right\rfloor} \cdot \frac{1}{n!} \sum_{k=1}^{n} f_{k} \cdot k!(n-k)!\leqslant C_{n}^{\left\{\frac{n}{2}\right]}$. When $A$ is the set of all $\left[\frac{n}{2}\right]$-element subsets of $\{1,2, \cdots, n\}$, it exactly reaches $|A|=C_{n}^{\left\{\frac{n}{2}\right]}$. In summary, the maximum number of elements in $A$ is $C_{n}^{\left[\frac{n}{2}\right)}$ (where $\left[\frac{n}{2}\right]$ denotes the greatest integer not exceeding $\frac{n}{2}$)." 9c813eaf784b,Example 10 Arrange all positive integers that are coprime with 105 in ascending order. Try to find the 1000th term of this sequence.,See reasoning trace,medium,"Let $S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S$ and $3 \mid x\}$, $A_{2}=\{x \mid x \in S$, and $5 \mid x\}, A_{3}=\left\{x \mid x \in S_{1}\right.$ and $\left.7 \mid x\right\}$, then $$ |S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15,\left|A_{1} \cap A_{2}\right|=7 \text {, } $$ $\left|A_{1} \cap A_{3}\right|=5,\left|A_{2} \cap A_{3}\right|=3,\left|A_{1} \cap A_{2} \cap A_{3}\right|=1$, so, the number of natural numbers not greater than 105 and coprime with 105 is $\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|=|S|-\left(\left|A_{1}\right|+\right.$ $\left.\left.\left|A_{2}\right|+\left|A_{3}\right|\right)+\left(\left|A_{1} \cap A_{2}\right|+\right) A_{2} \cap A_{3}|+| A_{3} \cap A_{1} \mid\right)-\left(A_{1} \cap A_{2} \cap\right.$ $\left.A_{3}\right)=48$. Let the sequence of positive integers that are coprime with 105, arranged in ascending order, be $a_{1}$, $a_{2}, a_{3}, \cdots, a_{n}$, then $a_{1}=1, a_{2}=2, a_{3}=4, \cdots, a_{48}=104$. Let $P=\left\{a_{1}, a_{2}, \cdots, a_{48}\right\}$, on one hand, for $n \geqslant 1$, let $a_{n}=105 k+r$ $(k \geqslant 0 ; 0 \leqslant r \leqslant 104, k, r \in Z)$, since $\left(a_{n}, 105\right)=1$, we have $(r, 105)=1$, hence $r \in$ $P$; on the other hand, for any non-negative integer $k$ and $r \in P$, since $(r, 105)=1$, we have $(105 k+r, 105)=1$. Thus, there must be an $n$ such that $a_{n}=105 k+r$. This indicates that the sequence $\left\{a_{n}\right\}$ consists of and only of numbers of the form $105 k+r(k \geqslant 0, k \in Z, r \in P)$ arranged in ascending order. Since the sequence is increasing, and for each fixed non-negative integer $k$, when $r$ takes all values in $P$, a total of 48 numbers are obtained, and $1000=48 \times 20+40$, hence $a_{1000}=105 \times 20+$ $a_{40}$. $$ \begin{array}{c} \text { Also, } a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94, a_{43}=92, a_{42} \\ =89, a_{41}=88, a_{40}=86, \text { so, } a_{1000}=105 \times 20+86=2186 . \end{array} $$" dff66980c4c1,"3. (8 points) In another 12 days, it will be 2016, Hao Hao sighs: I have only experienced 2 leap years so far, and the year I was born is a multiple of 9, so in 2016, Hao Hao is $\qquad$ years old.",】Solution: Huhu has only experienced 2 leap years so far,easy,"【Answer】Solution: Huhu has only experienced 2 leap years so far. Tracing back two leap years from 2015, they are 2012 and 2008, so Huhu's birth year is between 2015 and 2004, among which the year that is a multiple of 9 is 2007, so Huhu was born in 2007, $$ 2016-2007=9 \text { (years old) } $$ Answer: In 2016, Huhu is 9 years old. Therefore, the answer is: 9." 45f71b834854,"10. If $f(x)=f\left(\frac{1}{x}\right) \lg x+10$, then the value of $f(10)$ is $\qquad$ .","f\left(\frac{1}{10}\right) \lg 10+10, f\left(\frac{1}{10}\right)=f(10) \lg \frac{1}{10}+10$, elimina",easy,"10. 10 . $f(10)=f\left(\frac{1}{10}\right) \lg 10+10, f\left(\frac{1}{10}\right)=f(10) \lg \frac{1}{10}+10$, eliminating $f\left(\frac{1}{10}\right)$ we get $f(10)=10$." e9ef0b500d57,"19.2.15 ** Let $s$ be the set of all rational numbers $r$ that satisfy the following conditions: $00(j \geqslant 5)$, and $\alpha_{j} \leqslant$ 2. To make $n$ as small as possible, it is advisable to take $0 \leqslant \alpha_{5} \leqslant 2$. By $$ \begin{array}{l} \left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right) \\ =144\left(\alpha_{5} \neq 0 \text { when }\right) \end{array} $$ or $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right)$ $$ =144 \text { (when } \alpha_{5}=0 \text {). } $$ Considering the possible factorizations of $144=2^{4} \times 3^{2}$, and comparing the corresponding sizes of $n$, we find that the smallest $n$ is $$ n=2^{5} \times 3^{2} \times 5 \times 7 \times 11=110880 \text {. } $$" 83dbdaf76c32,"A circle is inscribed in a sector that is one sixth of a circle of radius 6 . (That is, the circle is tangent to both segments and the arc forming the sector.) Find, with proof, the radius of the small circle. ![](https://cdn.mathpix.com/cropped/2024_04_17_45b306ab1b407a5798d6g-2.jpg?height=352&width=396&top_left_y=244&top_left_x=859)","r$ and $O A=2 r$, but $O T=r$ so $A T=3 r$. But $A T$ is the radius of the large circle, so $3 r=6$ ",medium,"Because the circles are tangent, we can draw the line $A T$, which passes through the center $O$ of the other circle. Note that triangles $A D O$ and $A E O$ are symmetric (this is HL congruence: $A O$ is shared, radii $O D$ and $O E$ are equal, and angles $A D O$ and $A E O$ are right). Therefore, since $\angle B A C$ is $60^{\circ}$, angles $B A T$ and $T A C$ are each $30^{\circ}$. Now $A D O$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle. If we let $r$ be the radius of the small circle, then $O D=r$ and $O A=2 r$, but $O T=r$ so $A T=3 r$. But $A T$ is the radius of the large circle, so $3 r=6$ and $r=2$." 5140f070b0a4,,11,medium,"Answer: 11. Solution. First, let's show that $n^{3} \equiv n(\bmod 3)$ and $n^{5} \equiv n(\bmod 5)$ for any natural $n$. This can be done in several ways. We will present only three of them. First method. We will divide all natural numbers into groups based on their remainders when divided by 3 and 5: | $n$ | $n^{3}$ | $n^{3} \bmod 3$ | | :---: | :---: | :---: | | 0 | 0 | 0 | | 1 | 1 | 1 | | 2 | 8 | 2 | | $n$ | $n^{5}$ | $n^{5} \bmod 5$ | | :---: | :---: | :---: | | 0 | 0 | 0 | | 1 | 1 | 1 | | 2 | 32 | 2 | | 3 | 243 | 3 | | 4 | 1024 | 4 | From the table, it is clear that the statement to be proved is true. Second method. Note that $n^{3}-n=(n-1) n(n+1)-$ is the product of three consecutive integers, and $$ n^{5}-n=(n-1) n(n+1)\left(n^{2}+1\right) \equiv(n-1) n(n+1)\left(n^{2}-4\right)=(n-2)(n-1) n(n+1)(n+2) \quad(\bmod 5) $$ - is the product of five consecutive integers. In the first case, the expression is always divisible by 3, and in the second case, by 5, which is what we needed to prove. The third method is to say that these statements are special cases of Fermat's Little Theorem for $3$ and $5$. Now let's look at the original expression modulo 3: $$ 10 n^{3}+3 n^{5}+7 a n \equiv 10 n+7 a n \equiv(a+1) \cdot n \quad(\bmod 3) $$ This expression is divisible by 3 for any integer $n$ if and only if $a+1$ is divisible by 3, i.e., when $a$ gives a remainder of 2 when divided by 3. Similarly, $10 n^{3}+3 n^{5}+7 a n \equiv(2 a+3) \cdot n(\bmod 5)$, i.e., $2 a+3$ is divisible by 5, so $a$ gives a remainder of 1 when divided by 5. Thus, $a \equiv 2(\bmod 3)$ and $a \equiv 1(\bmod 5)$, i.e., $a \equiv 11(\bmod 15)$. Another solution. Substitute $n=1$ and we get that if such an $a$ exists, then $13+7 a$ must be divisible by 15, i.e., $a$ must give a remainder of 11 when divided by 15. It remains to check that if $a \equiv 11(\bmod 15)$, then the given expression is divisible by 15 for any natural $n$. This can be done in many ways (see, for example, the first solution). Here we will present another one. We will prove this statement by induction on $n$ (for $n=0$ the divisibility is obvious, for negative $n$ it is proved similarly or reduced to the case of positive $n$ by replacing $n \rightarrow-n$). If $n=1$, the statement is already verified. Suppose now that we have already proved that $10 n^{3}+3 n^{5}+7 a n$ is divisible by 15 and we will prove that $10(n+1)^{3}+3(n+1)^{5}+7 a(n+1)$ is also divisible by 15. Let's look at the difference between these two expressions: $10\left((n+1)^{3}-n^{3}\right)+3\left((n+1)^{5}-n^{5}\right)+7 a((n+1)-n)=10\left(3 n^{2}+3 n+1\right)+3\left(5 n^{4}+10 n^{3}+10 n^{2}+5 n+1\right)+7 a$. After expanding the brackets, all terms on the right side, except for $10+3+7 a$, are divisible by 15, but $13+7 a$ is divisible by 15, since $a \equiv 11(\bmod 15)$." 3c78856e2320,"Five students ran a race. Ryan was faster than Henry and Faiz. Henry was slower than Faiz. Toma was faster than Ryan but slower than Omar. Which student finished fourth? (A) Faiz (B) Henry (C) Omar (D) Ryan (E) Toma",(A),easy,"Henry was slower than Faiz and thus finished the race behind Faiz. Ryan was faster than Henry and Faiz and thus finished the race ahead of both of them. From fastest to slowest, these three runners finished in the order Ryan, Faiz and then Henry. Toma was faster than Ryan but slower than Omar. Therefore, from fastest to slowest, the runners finished in the order Omar, Toma, Ryan, Faiz, and Henry. The student who finished fourth was Faiz. ANSwer: (A)" 227eda855ebc,"6.2. On the plate, there were 15 doughnuts. Karlson took three times more doughnuts than Little Man, and Little Man's dog Bibbo took three times fewer than Little Man. How many doughnuts are left on the plate? Explain your answer.",2 doughnuts are left,medium,"Answer: 2 doughnuts are left. From the condition of the problem, it follows that Little One took three times as many doughnuts as Bimbo, and Karlson took three times as many doughnuts as Little One. We can reason in different ways from here. First method. If Bimbo took one doughnut, then Little One took three doughnuts, and Karlson took nine doughnuts, so together they took $1+3+9=13$ doughnuts. Thus, two doughnuts are left on the plate. If Bimbo had taken two doughnuts or more, then Little One would have taken six doughnuts or more, and Karlson would have taken 18 doughnuts or more, which is impossible. Second method. Let Bimbo take $x$ doughnuts, then Little One took $3x$ doughnuts, and Karlson took $3 \cdot 3x = 9x$ doughnuts. Then together they took $x + 3x + 9x = 13x$ doughnuts. Since the number $13x$ must not exceed 15, the only possible natural value for $x$ is 1. Therefore, together they ate 13 doughnuts, and two doughnuts are left on the plate. + a complete and justified solution $\pm$ the correct answer is provided, it is shown that it satisfies the condition, and there is a statement that it is the only one $\pm$ the number of doughnuts each person ate is justified, but the explicit answer to the question is not provided 干 the correct answer is provided, it is shown that it satisfies the condition, but there is no mention of its uniqueness 干 only the correct answer is provided" eb4b3468128f,"18. As shown in the figure, the lengths of the base edges of the rectangular prism are $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$, and the height is $6 \mathrm{~cm}$. If a thin string is used to start from point $A$, pass through the four sides and wrap around once to reach point $B$, then the shortest length of the string used is $\qquad$ $\mathrm{cm}$.",10,easy,answer: 10 7c7c4f5c2523,"9. (20 points) Find all values of $x$ and $y$ for which the following equality holds: $$ (x-13)^{2}+(y-14)^{2}+(x-y)^{2}=\frac{1}{3} $$","14-y=$ $y-x=\frac{1}{3}$. Then $x=13 \frac{1}{3}, y=13 \frac{2}{3}$.",easy,"Answer: $x=13 \frac{1}{3}, y=13 \frac{2}{3}$ Solution: We use the inequality between the quadratic mean and the arithmetic mean: $$ \sqrt{\frac{(x-13)^{2}+(14-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-13|+|14-y|+|x-y|}{3} \geqslant \frac{x-13+14-y+y-x}{3}=\frac{1}{3} $$ That is, $(x-13)^{2}+(14-y)^{2}+(x-y)^{2} \geqslant \frac{1}{3}$. Equality is achieved only when $x-13=14-y=$ $y-x=\frac{1}{3}$. Then $x=13 \frac{1}{3}, y=13 \frac{2}{3}$." ba52660a5e44,3. How many positive perfect cubes are divisors of the product $1!\cdot 2!\cdot 3!\cdots 10$ !?,"1!\cdot 2!\cdot 3!\cdots 10!=2^{38} 3^{17} 5^{7} 7^{4}$. Thus, a positive divisor of $N$ that is a p",medium,"Answer: 468 Solution: We have $N:=1!\cdot 2!\cdot 3!\cdots 10!=2^{38} 3^{17} 5^{7} 7^{4}$. Thus, a positive divisor of $N$ that is a perfect cube must be of the form $2^{3 a} 3^{3 b} 5^{3 c} 7^{3 d}$ for some nonnegative integers $a, b, c, d$. We see that $3 a \leq 38,3 b \leq 17,3 c \leq 7$ and $3 d \leq 4$. Thus, there are $\left\lceil\frac{38}{3}\right\rceil=13$ choices for $a,\left\lceil\frac{17}{3}\right\rceil=6$ choices for $b$, $\left\lceil\frac{7}{3}\right\rceil=3$ choices for $c$, and $\left\lceil\frac{4}{3}\right\rceil=2$ choices for $d$. Hence, there are $13 \cdot 6 \cdot 3 \times 2=468$ positive perfect cube divisors of $N$." eb0d4893ffb0,"How many positive integer solutions does the following equation have? $$ \left[\frac{x}{10}\right]=\left[\frac{x}{11}\right]+1 $$",110$ solutions.,medium,"Every sought $x$ value can be uniquely written in the form $x=11 \cdot k+m$, where $k$ is a non-negative integer, and $m$ takes one of the values 0, $1, \ldots, 10$. Then the equation is: $$ \left[\frac{11 \cdot k+m}{10}\right]=\left[\frac{11 \cdot k+m}{11}\right]+1 $$ Based on the interpretation of the integer part function, $$ \left[\frac{11 \cdot k+m}{10}\right]=\left[k+\frac{k+m}{10}\right]=k+\left[\frac{k+m}{10}\right] $$ and $$ \left[\frac{11 \cdot k+m}{11}\right]=k $$ Thus, $$ k+\left[\frac{k+m}{10}\right]=k+1 $$ which implies $$ \left[\frac{k+m}{10}\right]=1 $$ The solutions to this are given by $10 \leq k+m \leq 19$. That is, $10-m \leq k \leq 19-m$, where $m=0,1, \ldots, 10$ can be. Thus, for each value of $m$, there are ten possible values of $k$, and different $(k, m)$ pairs determine different $x$ values. This means that the equation has $10 \cdot 11=110$ solutions." 44dc90ca5cdd,7.255. $25^{\log _{4} x}-5^{\log _{16} x^{2}+1}=\log _{\sqrt{3}} 9 \sqrt{3}-25^{\log _{16} x}$.,4,easy,"## Solution. Domain of definition: $x>0$. Rewrite the equation as $5^{\log _{2} x}-5 \cdot 5^{\frac{1}{2} \log _{2} x}=5-5^{\frac{1}{2} \log _{2} x} \Leftrightarrow 5^{\log _{2} x}-4 \cdot 5^{\frac{1}{2} \log _{2} x}-5=0$. Solving this equation as a quadratic equation in terms of $5^{\frac{1}{\log _{2} x}}$, we get $5^{\frac{1}{2} \log _{2} x}=-1$ (no solutions), or $5^{\frac{1}{2} \log _{2} x}=5 \Rightarrow \frac{1}{2} \log _{2} x=1, \log _{2} x=2$, $x=2^{2}=4$. Answer: 4." b2c2c0403c51,"$:$ Folkiore It is known that among 63 coins there are 7 counterfeit ones. All counterfeit coins weigh the same, all genuine coins also weigh the same, and a counterfeit coin is lighter than a genuine one. How can you determine 7 genuine coins in three weighings using a balance scale without weights?",See reasoning trace,medium,"1) Let's set aside one coin and put 31 coins on each pan of the scales. If the pans balance, then we have set aside the counterfeit coin, and there are 3 counterfeit coins on each pan. If one of the pans is heavier, then there are no more than three counterfeit coins on it. Thus, after the first weighing, we will be able to select 31 coins, among which there are no more than three counterfeit ones. 2) Take this group of coins and perform a similar operation: set aside one coin again and put 15 coins on each pan of the scales. After this weighing, we will be able to select 15 coins, among which there is no more than one counterfeit. 3) Repeating a similar operation for the third time, we will end up with 7 genuine coins. At an international congress, 578 delegates from different countries arrived. Any three delegates can communicate with each other without the help of others (although one of them may have to translate the conversation between the other two). Prove that all delegates can be accommodated in double rooms of a hotel in such a way that any two living in the same room can communicate without external help. ## Solution Take any three delegates; some two of them can definitely communicate with each other without a translator. We will house them in one room. We will repeat this operation until only 4 delegates remain. As it is not difficult to verify by enumeration, they can always be divided into 2 rooms in such a way that those living in each room can communicate with each other without external help." 1e3ab5855299,"315*. Find all roots of the equation $$ 8 x\left(2 x^{2}-1\right)\left(8 x^{4}-8 x^{2}+1\right)=1 $$ satisfying the condition $0-1+\sqrt{2}$. In summary, the inequality is satisfied precisely when $$ x \leq-1-\sqrt{2}, \quad \text { or } \quad x=\frac{-1-\sqrt{5}}{2}, \quad \text { or } \quad x \geq-1+\sqrt{2} $$ Remark. Many solvers reasoned after the factorization that the square of a number is non-negative, so the inequality is equivalent to the inequality $x^{2}+2 x-1 \geq 0$. This line of thought is flawed, and such solutions were awarded 3 points." bc57b40bcda0,"5. If $00$ and $a_{2} x^{2}+b_{2} x+c_{2}>0$ with respect to $x$ are the same; Proposition $Q: \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$. Then proposition $Q$ ( ). (A) is a sufficient and necessary condition for proposition $P$ (B) is a sufficient condition for proposition $P$ but not a necessary condition (C) is a necessary condition for proposition $P$ but not a sufficient condition (D) is neither a sufficient condition nor a necessary condition for proposition $P$",See reasoning trace,easy,"4.D. For example, the solution sets of $x^{2}-3 x+2>0$ and $-x^{2}+3 x-2>0$ are different, thus eliminating (A) and (B); also, the solution sets of $x^{2}+x+1>0$ and $x^{2}+x+3>0$ are the same, thus eliminating (C)." 4489947a90fb,"# Find the minimum value of the expression $$ \sqrt{x^{2}-2 \sqrt{3} \cdot|x|+4}+\sqrt{x^{2}+2 \sqrt{3} \cdot|x|+12} $$ as well as the values of $x$ at which it is achieved. #",$2 \sqrt{7}$ when $x= \pm \frac{\sqrt{3}}{2}$,medium,"# Solution: Let $t=\frac{|x|}{2} \geq 0$. Then we need to find the minimum value of the function $$ f(t)=2\left(\sqrt{t^{2}-\sqrt{3} \cdot t+4}+\sqrt{t^{2}+\sqrt{3} \cdot t+12}\right) $$ for $t \geq 0$. Transform the function $f(t)$ to the form: $$ f(t)=2\left(\sqrt{\left(t-\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{4}}+\sqrt{\left(t+\frac{\sqrt{3}}{2}\right)^{2}+\frac{9}{4}}\right) $$ Let points $M(t, 0), A\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right), B\left(-\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ be given. It is not difficult to understand that the expression we are considering is nothing but twice the sum of the lengths of segments $A M$ and $B M$, that is, $f(t)=2(A M+B M)$. If point $M$ does not lie on segment $A B$, then $A M+B M>A B$. On the other hand, if $M \in A B$, then $A M+B M=A B$, so the sum $A M+B M$ attains its minimum value if and only if point $M$ lies on segment $A B$. Therefore, point $M$ is the intersection of segment $A B$ with the $O x$ axis. ![](https://cdn.mathpix.com/cropped/2024_05_06_e2fedf2b675f713cdba7g-11.jpg?height=746&width=739&top_left_y=181&top_left_x=270) To find the coordinates of point $M$, consider similar triangles $M D A$ and $M C B$: $$ \frac{M D}{C M}=\frac{A D}{B C}=\frac{1}{3} $$ Therefore, if $M D=a$, then $M C=3 a$ and from this $$ \begin{gathered} C D=\sqrt{3}=M D+M C=a+3 a=4 a \\ a=\frac{\sqrt{3}}{4} . \end{gathered} $$ From this, we find the coordinates of point $M\left(\frac{\sqrt{3}}{4}, 0\right)$. Thus, the minimum value of the function $f(t)$ is achieved when $t=\frac{|x|}{2}=\frac{\sqrt{3}}{4}$ and is equal to: $$ f\left(\frac{\sqrt{3}}{4}\right)=2\left(\sqrt{\frac{3}{16}+\frac{1}{4}}+\sqrt{\frac{27}{16}+\frac{9}{4}}\right)= $$ $$ =2\left(\frac{\sqrt{7}}{4}+\frac{3 \sqrt{7}}{4}\right)=2 \sqrt{7} $$ Answer: $2 \sqrt{7}$ when $x= \pm \frac{\sqrt{3}}{2}$. ## Criteria:" 6c70fef52b12,"Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} \plus{} r_{j} \plus{} r_{k}$.",n = 4,medium,"To determine all integers \( n > 3 \) for which there exist \( n \) points \( A_1, A_2, \ldots, A_n \) in the plane, no three collinear, and real numbers \( r_1, r_2, \ldots, r_n \) such that for \( 1 \leq i < j < k \leq n \), the area of \( \triangle A_i A_j A_k \) is \( r_i + r_j + r_k \), we will analyze the cases for \( n = 4 \) and \( n = 5 \). 1. **Case \( n = 4 \)**: - Consider four points \( A_1, A_2, A_3, A_4 \) forming a convex quadrilateral. - We can take \( A_1, A_2, A_3, A_4 \) to be the vertices of a square. - Let the area of the square be \( S \). The area of any triangle formed by three vertices of the square is \( \frac{S}{2} \). - Set \( r_i = \frac{S}{6} \) for \( i = 1, 2, 3, 4 \). Then, for any three points \( A_i, A_j, A_k \), the area of \( \triangle A_i A_j A_k \) is \( r_i + r_j + r_k = \frac{S}{6} + \frac{S}{6} + \frac{S}{6} = \frac{S}{2} \), which matches the area of the triangle. - Therefore, \( n = 4 \) works. 2. **Case \( n = 5 \)**: - Suppose \( A_1, A_2, A_3, A_4, A_5 \) form a convex pentagon. - For a convex pentagon, the sum of the areas of the triangles formed by any three points must be consistent with the given condition. - If \( A_1, A_2, A_3, A_4, A_5 \) form a convex pentagon, then \( r_i + r_{i+2} \) must be the same for all \( i \) (indices modulo 5). - This implies that all \( r_i \) are equal, say \( r_i = r \) for all \( i \). - However, this leads to a contradiction because the area of \( \triangle A_1 A_2 A_3 \) cannot be equal to the area of \( \triangle A_1 A_2 A_5 \), as the latter includes a different set of points. - Therefore, \( n = 5 \) is impossible. 3. **Case \( n > 5 \)**: - If \( n > 5 \), then we can always find a subset of 5 points among the \( n \) points. - Since \( n = 5 \) is impossible, any \( n > 5 \) is also impossible. Conclusion: - The only integer \( n > 3 \) for which there exist \( n \) points \( A_1, A_2, \ldots, A_n \) in the plane, no three collinear, and real numbers \( r_1, r_2, \ldots, r_n \) such that for \( 1 \leq i < j < k \leq n \), the area of \( \triangle A_i A_j A_k \) is \( r_i + r_j + r_k \), is \( n = 4 \). The final answer is \( \boxed{ n = 4 } \)." 7a595f823bdf,"Suppose that $c\in\left(\frac{1}{2},1\right)$. Find the least $M$ such that for every integer $n\ge 2$ and real numbers $0 m \)**: Similarly, since \( 0 < a_{m+1} \le a_{m+2} \le \ldots \le a_n \) and \( 0 < m+1 - cn \le \ldots \le n - cn \), we have: \[ \sum_{k=m+1}^{n}(k-cn)a_{k} \ge \frac{1}{n-m}\sum_{k=m+1}^{n}(k-cn)\sum_{k=m+1}^{n}a_{k} \ge (\frac{m+n+1}{2} - cn)\sum_{k=m+1}^{n}a_{k} \] 6. **Combining the inequalities**: \[ (\frac{m+n+1}{2} - cn)\sum_{k=m+1}^{n}a_{k} \le (cn - \frac{m+1}{2})\sum_{k=1}^{m}a_{k} \] 7. **Simplifying**: \[ \sum_{k=1}^{n}a_k \le \frac{n}{m+n+1-2cn}\sum_{k=1}^{m}a_k = \frac{1}{1 + \frac{m+1}{n} - 2c}\sum_{k=1}^{m}a_k \] Since \( m + 1 > cn \), we have: \[ \frac{1}{1 + \frac{m+1}{n} - 2c} < \frac{1}{1 - c} \] 8. **Constructing a counterexample**: If \( n \ge \frac{1}{2c-1} \), we can take \( a_1 = a_2 = \ldots = a_m = 1 \) and \( a_{m+1} = a_{m+2} = \ldots = a_n = \frac{2cnm - m(m+1)}{(n-m)(n+m+1) - 2cn(n-m)} \). This ensures \( a_{m+1} \ge 1 \) and satisfies: \[ \frac{1}{n}\sum_{k=1}^{n}ka_{k} = c\sum_{k=1}^{n}a_{k} \] Thus, \( M \ge \frac{1}{1-c} - \frac{1}{n(1-c)^2} \). When \( n \) is sufficiently large, \( M \ge \frac{1}{1-c} \). 9. **Conclusion**: Therefore, the least \( M \) is \( \frac{1}{1-c} \). The final answer is \( \boxed{\frac{1}{1-c}} \)." a5796993bb7d,"How many $2 \times 2 \times 2$ cubes must be added to a $8 \times 8 \times 8$ cube to form a $12 \times 12 \times 12$ cube? [i]Proposed by Evan Chen[/i]",152,medium,"To solve the problem, we need to determine how many $2 \times 2 \times 2$ cubes must be added to an $8 \times 8 \times 8$ cube to form a $12 \times 12 \times 12$ cube. 1. **Calculate the volume of the $8 \times 8 \times 8$ cube:** \[ V_{\text{small}} = 8^3 = 512 \] 2. **Calculate the volume of the $12 \times 12 \times 12$ cube:** \[ V_{\text{large}} = 12^3 = 1728 \] 3. **Determine the additional volume needed:** \[ V_{\text{additional}} = V_{\text{large}} - V_{\text{small}} = 1728 - 512 = 1216 \] 4. **Calculate the volume of a single $2 \times 2 \times 2$ cube:** \[ V_{\text{small\_cube}} = 2^3 = 8 \] 5. **Determine the number of $2 \times 2 \times 2$ cubes needed to fill the additional volume:** \[ \text{Number of small cubes} = \frac{V_{\text{additional}}}{V_{\text{small\_cube}}} = \frac{1216}{8} = 152 \] Thus, the number of $2 \times 2 \times 2$ cubes required is $\boxed{152}$." 9750dee48d12,"For the real pairs $(x,y)$ satisfying the equation $x^2 + y^2 + 2x - 6y = 6$, which of the following cannot be equal to $(x-1)^2 + (y-2)^2$? $ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 23 \qquad\textbf{(E)}\ 30 $",2,medium,"1. Start with the given equation: \[ x^2 + y^2 + 2x - 6y = 6 \] We can complete the square for both \(x\) and \(y\). 2. For \(x\): \[ x^2 + 2x = (x+1)^2 - 1 \] 3. For \(y\): \[ y^2 - 6y = (y-3)^2 - 9 \] 4. Substitute these into the original equation: \[ (x+1)^2 - 1 + (y-3)^2 - 9 = 6 \] 5. Simplify the equation: \[ (x+1)^2 + (y-3)^2 - 10 = 6 \] \[ (x+1)^2 + (y-3)^2 = 16 \] This represents a circle centered at \((-1, 3)\) with radius 4. 6. We need to determine the possible values of \((x-1)^2 + (y-2)^2\). This represents the square of the distance from \((x, y)\) to \((1, 2)\). 7. The distance from the center \((-1, 3)\) to the point \((1, 2)\) is: \[ \sqrt{(1 - (-1))^2 + (2 - 3)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \] 8. Using the triangle inequality, the distance from any point \((x, y)\) on the circle to \((1, 2)\) must be between: \[ 4 - \sqrt{5} \quad \text{and} \quad 4 + \sqrt{5} \] 9. Squaring these distances: \[ (4 - \sqrt{5})^2 = 16 - 8\sqrt{5} + 5 = 21 - 8\sqrt{5} \] \[ (4 + \sqrt{5})^2 = 16 + 8\sqrt{5} + 5 = 21 + 8\sqrt{5} \] 10. Therefore, the possible values of \((x-1)^2 + (y-2)^2\) lie in the interval: \[ 21 - 8\sqrt{5} \quad \text{to} \quad 21 + 8\sqrt{5} \] 11. Calculate the approximate values: \[ 21 - 8\sqrt{5} \approx 21 - 8 \times 2.236 \approx 21 - 17.888 \approx 3.112 \] \[ 21 + 8\sqrt{5} \approx 21 + 8 \times 2.236 \approx 21 + 17.888 \approx 38.888 \] 12. From the given options, the value 2 is not within the interval \([3.112, 38.888]\). The final answer is \(\boxed{2}\)" fd6569b198ed,"2. Let real numbers $x, y$ satisfy $$ x^{2}-8 x+y^{2}-6 y+24=0 \text {. } $$ Then the maximum value of $x-2 y$ is $\qquad$ $\therefore$",-2 \pm \sqrt{5}$ (negative value discarded).,easy,"2. $\sqrt{5}-2$. From $x^{2}-8 x+y^{2}-6 y+24=0$, we get $$ (x-4)^{2}+(y-3)^{2}=1 \text {. } $$ When the line $x-2 y=k$ is tangent to the circle, it is the desired maximum value. $$ \text { By } \frac{|4-6-k|}{\sqrt{1+2^{2}}}=1 $$ $\Rightarrow k=-2 \pm \sqrt{5}$ (negative value discarded)." cd9018c0106d,"1. Simplify $$ \sqrt{1+2 \sin \alpha \cdot \cos \alpha}+\sqrt{1-2 \sin \alpha \cdot \cos \alpha} $$ $\left(0^{\circ}<\alpha \leqslant 90^{\circ}\right)$ The result is $\qquad$ .",See reasoning trace,medium,"II. 1.2 $2 \cos \alpha$ or $2 \sin \alpha$. It is easy to know that $1+2 \sin \alpha \cdot \cos \alpha$ $$ \begin{array}{l} =\sin ^{2} \alpha+2 \sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha \\ =(\sin \alpha+\cos \alpha)^{2} . \end{array} $$ Similarly, $1-2 \sin \alpha \cdot \cos \alpha=(\sin \alpha-\cos \alpha)^{2}$. Therefore, the original expression is $$ \begin{array}{l} =\sqrt{(\sin \alpha+\cos \alpha)^{2}}+\sqrt{(\sin \alpha-\cos \alpha)^{2}} \\ =|\sin \alpha+\cos \alpha|+|\sin \alpha-\cos \alpha| . \end{array} $$ Since $0^{\circ}<\alpha<90^{\circ}$, we have $\sin \alpha>0, \cos \alpha>0$, and when $0^{\circ}<\alpha<45^{\circ}$, $\sin \alpha<\cos \alpha$; when $45^{\circ} \leqslant \alpha \leqslant 90^{\circ}$, $\sin \alpha \geqslant \cos \alpha$. Thus, (1) when $0^{\circ}<\alpha<45^{\circ}$, the original expression is $(\sin \alpha+\cos \alpha)+(\cos \alpha-\sin \alpha)$ $$ =2 \cos \alpha \text {; } $$ (2) when $45^{\circ} \leqslant \alpha \leqslant 90^{\circ}$, $$ \begin{array}{l} \text { the original expression }=(\sin \alpha+\cos \alpha)+(\sin \alpha-\cos \alpha) \\ =2 \sin \alpha . \end{array} $$" eceb6809ad3b,"13.2.5* Consider all line segments of length 4, one end of which lies on the line $y=x$, and the other end on the line $y=2x$. Find: the equation of the locus of the midpoints of these line segments.","2(y-x), b=2(2x-y)$. Substituting into (2) yields $[2(y-x)-2(2x-y)]^{2}+[4(y-x)-2(2x-y)]^{2}=16$, whi",easy,"Let $A(a, 2a)$ be any point on the line $y=2x$, and $B(b, b)$ be any point on the line $y=x$. The midpoint $M(x, y)$ of the line segment $AB$ is such that $x=\frac{a+b}{2}, y=\frac{2a+b}{2}$. (1) From the length of line segment $AB$ being 4, we get $(a-b)^{2}+(2a-b)^{2}=4^{2}$. (2) From (1), we have $a=2(y-x), b=2(2x-y)$. Substituting into (2) yields $[2(y-x)-2(2x-y)]^{2}+[4(y-x)-2(2x-y)]^{2}=16$, which simplifies to $25x^{2}-36xy+13y^{2}=4$." ca33e267e5c6,"[ Coordinate method on the plane [ Ratio of areas of triangles with a common base or common height] On the coordinate plane, points $A(1 ; 9)$, $C(5 ; 8)$, $D(8 ; 2)$, and $E(2 ; 2)$ are given. Find the area of the pentagon $A B C D E$, where $B$ is the intersection point of the lines $E C$ and $A D$.",See reasoning trace,easy,"If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ is $$ \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} $$ ## Otвет 27.00" 1bc1bba23701,"In $\triangle A B C$, $\angle B=50^{\circ}, \angle C=30^{\circ}, D$ is a point inside $\triangle A B C$, satisfying $\angle D B C=\angle D C B=20^{\circ}$. Find the degree measure of $\angle D A C$.",\angle DFC = 20^{\circ}$.,medium,"Solution: As shown in the figure, let $E$ be the circumcenter of $\triangle ABC$, and connect $AE$, $BE$, and $CE$. Since $\angle AEB = 2 \angle ACB = 60^{\circ}$, $\triangle ABE$ is an equilateral triangle, and $\angle ECB = \angle EBC = 10^{\circ}$, $\angle BEC = 160^{\circ}$. On the extension of $BD$, take a point $F$ such that $CF = CB$, and connect $AF$ and $EF$. $$ \because \angle BEC + \angle BFC = 160^{\circ} + 20^{\circ} = 180^{\circ}, $$ $\therefore B$, $E$, $C$, and $F$ are concyclic. Thus, $\angle BFE = \angle BCE = 10^{\circ}$. By the symmetry of the equilateral $\triangle ABE$, $$ \angle AFB = \angle BFD = 10^{\circ}. $$ $\vec{x} \angle DCA = 10^{\circ}$, $$ \therefore \angle AFD = \angle DCA. $$ $\therefore A$, $D$, $C$, and $F$ are concyclic. Thus, $\angle DAC = \angle DFC = 20^{\circ}$." a304156d47c1,"Example 1 What is the maximum number of rational points (points with both coordinates being rational numbers) that can lie on a circle in the plane, given that the center of the circle is not a rational point.",See reasoning trace,medium,"【Analysis】If $A, B, C$ are three rational points on a circle, then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisector of $BC$ is also a linear equation with rational coefficients. Therefore, the intersection of the perpendicular bisector of $AB$ and the perpendicular bisector of $BC$ is a rational point, i.e., the center of the circle is a rational point, which does not meet the condition. Two examples of rational points: $(0,0)$ and $(1,0)$, the center of the circle is $\left(\frac{1}{2}, \sqrt{2}\right)$, which satisfies the condition. Therefore, at most two rational points can exist on a circle such that the center of the circle is not a rational point." f7ab28ee5f62,"4. Given that $S$ is a set composed of $n(n \geqslant 3)$ positive numbers. If there exist three distinct elements in $S$ that can form the three sides of a triangle, then $S$ is called a ""triangle number set"". Consider the set of consecutive positive integers $\{4,5, \cdots, m\}$, all of whose 10-element subsets are triangle number sets. Then the maximum possible value of $m$ is","253$, all ten-element sets $S$ satisfy: there exist three elements in $S$ that can form the three si",medium,"4. 253 Analysis: When $m=254$, the ten-element set $\{4,5,9,14,23,37,60,97,157,254\}$ does not contain three numbers that can form the three sides of a triangle, so $m \leq 253$. We now prove that $m=253$ is feasible. The following proof uses proof by contradiction: Assume that when $m=253$, there exists a ten-element set $S$ such that no three elements in $S$ can form the three sides of a triangle. Without loss of generality, let $S=\left\{a_{1}, a_{2}, \ldots, a_{10}\right\}, 4 \leq a_{1} < a_{2} < \ldots < a_{10} \leq 253$. Then, $a_{10}+a_{9}>2 a_{9}+a_{8}>2 a_{8}+a_{7}>3 a_{7}+2 a_{6}>\ldots>34 a_{2}+21 a_{1} \geq 34 \times 5+21 \times 4=254$, which is a contradiction. Therefore, when $m=253$, all ten-element sets $S$ satisfy: there exist three elements in $S$ that can form the three sides of a triangle." 9e5685c90788,"$13 \cdot 58$ In $\triangle A B C$, if $(\sin B+\sin C):(\sin C+\sin A):(\sin A+\sin B)=4: 5: 6$. Then the degree of the largest angle of the triangle is (A) $72^{\circ}$. (B) $84^{\circ}$. (C) $1.5^{\circ}$. (D) $120^{\circ}$. (E) $150^{\circ}$. (Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1984)",$(D)$,medium,"［Solution］By the Law of Sines and the given conditions, $$ \begin{array}{r} (b+c):(c+a):(a+b)=4: 5: 6 . \\ \text { Let } \quad b+c=4 k, \quad c+a=5 k, \quad a+b=6 k, \end{array} $$ Adding the three equations and simplifying, we get $\quad a+b+c=7.5 k$. Thus, $a=3.5 k, b=2.5 k, c=1.5 k$. Let $k=2 t$, then $a=7 t, b=5 t, c=3 t$, and we know the largest angle is $A$. Also, $$ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\left(3^{2}+5^{2}-7^{2}\right) t^{2}}{2 \cdot 3 \cdot 5 \cdot t^{2}}=-\frac{1}{2}, $$ so $A=120^{\circ}$. Therefore, the answer is $(D)$." 487c3567a4ab,"3. $$ \begin{aligned} x^{2}+y^{2} & =2, \\ \frac{x^{2}}{2-y}+\frac{y^{2}}{2-x} & =2 . \end{aligned} $$",\left(x^{2}+y^{2}\right)-2(x+y)+2=2-4+2=0$. Therefore $x=y=1$.,medium,"Solution: (Viet Hoang) From the second equation, after a few steps of algebraic manipulation, one has $$ \begin{aligned} x^{2}(2-x)+y^{2}(2-y) & =2(2-x)(2-y) \\ 2\left(x^{2}+y^{2}\right)-(x+y)\left(x^{2}+y^{2}-x y\right) & =8-4(x+y)+2 x y \\ 4-(x+y)(2-x y) & =8-4(x+y)+2 x y \\ -2(x+y)+x y(x+y) & =4-4(x+y)+2 x y \\ 2(x+y)+x y(x+y) & =4+2 x y \\ (x+y-2)(2+x y) & =0 \end{aligned} $$ In the case $x y=-2$, substituting this into the first equation, we have $x^{2}+x y+y^{2}=0$. However, this means $$ \left(x+\frac{y}{2}\right)^{2}+\frac{3 y^{2}}{4}=0 \Longrightarrow x=y=0 $$ Substituting these values into the second equation, it yields a contradiction. For the second case, using the substitution $y=2-x$, we get $$ \begin{array}{r} x^{2}+y^{2}=2, \\ x+y=2 . \end{array} $$ From here we get: $(x-1)^{2}+(y-1)^{2}=\left(x^{2}+y^{2}\right)-2(x+y)+2=2-4+2=0$. Therefore $x=y=1$." 0aaccd08c917,"What is the largest amount of postage in cents that cannot be made using only 3 cent and 5 cent stamps? (A) 7 (B) 13 (C) 4 (D) 8 (E) 9",must be 7,medium,"Solution 1 We look at each of the choices and try to make them using only 3 cent and 5 cent stamps: (A): 7 cannot be made, since no more than one 5 cent and two 3 cent stamps could be used (try playing with the possibilities!) (B): $13=5+5+3$ (C): 4 cannot be the answer since a larger number (7) already cannot be made (D): $8=5+3$ (E): $9=3+3+3$ Therefore, the answer must be 7 . (We have not really justified that 7 is the largest number that cannot be made using only $3 \mathrm{~s}$ and 5 s; we have, though, determined that 7 must be the answer to this question, since it is the only possible answer from the given possibilities! See Solution 2 for a justification that 7 is indeed the answer.) ## Solution 2 We make a table to determine which small positive integers can be made using $3 \mathrm{~s}$ and $5 \mathrm{~s}$ : | Integer | Combination | | :---: | :---: | | 1 | Cannot be made | | 2 | Cannot be made | | 3 | 3 | | 4 | Cannot be made | | 5 | 5 | | 6 | $3+3$ | | 7 | Cannot be made | | 8 | $5+3$ | | 9 | $3+3+3$ | | 10 | $5+5$ | | 11 | $5+3+3$ | Every integer larger than 11 can also be made because the last three integers in our table can be made and we can add a 3 to our combinations for 9,10 and 11 to get combinations for 12, 13 and 14 , and so on. From the table, the largest amount of postage that cannot be made is 7 . ANSWER: (A)" c8f52107f76d,"Group Event 7 $O A B C$ is a tetrahedron with $O A, O B$ and $O C$ being mutually perpendicular. Given that $O A=O B=O C=6 x$. G7.1 If the volume of $O A B C$ is $a x^{3}$, find $a$. G7.2 If the area of $\triangle A B C$ is $b \sqrt{3} x^{2}$, find $b$. G7.3 If the distance from $O$ to $\triangle A B C$ is $c \sqrt{3} x$, find $c$. By finding the volume of $O A B C$ in two different ways. G7.4 If $\theta$ is the angle of depression from $C$ to the midpoint of $A B$ and $\sin \theta=\frac{\sqrt{d}}{3}$, find $d$.",See reasoning trace,medium,"$$ \begin{array}{l} a x^{3}=\frac{1}{3} \cdot \frac{1}{2}(6 x)^{2} \cdot(6 x)=36 x^{3} \\ \Rightarrow a=36 \\ A B=B C=A C=\sqrt{(6 x)^{2}+(6 x)^{2}}=6 x \sqrt{2} \end{array} $$ $\triangle A B C$ is equilateral $$ \angle B A C=60^{\circ} $$ Area of $\triangle A B C=b \sqrt{3} x^{2}=\frac{1}{2}(6 x \sqrt{2})^{2} \sin 60^{\circ}=18 \sqrt{3} x^{2}$ $$ \begin{array}{l} b=18 \\ \frac{1}{3} \cdot 18 \sqrt{3} x^{2} \times(c \sqrt{3} x)=36 x^{3} \\ c=2 \\ \frac{1}{3} \cdot 18 \sqrt{3} x^{2} \times(c \sqrt{3} x)=36 x^{3} \end{array} $$ Let the midpoint of $A B$ be $M$. $$ \begin{array}{l} O C=6 x, \frac{O M \times A B}{2}=\frac{O A \times O B}{2} \\ \Rightarrow 6 x \sqrt{2} \cdot O M=(6 x)^{2} \\ \Rightarrow O M=3 \sqrt{2} x \\ C M=\sqrt{O M^{2}+O C^{2}} \\ =\sqrt{(3 \sqrt{2} x)^{2}+(6 x)^{2}} \\ =3 \sqrt{6} x \\ \sin \theta=\frac{\sqrt{d}}{3}=\frac{O C}{C M} \\ =\frac{6 x}{3 \sqrt{6} x}=\frac{\sqrt{6}}{3} \\ d=6 \\ \end{array} $$" db81fd4681b7,"4. Roll a die twice, and let $X$ denote the maximum of the two rolls. Then, among the following numbers, the one closest to the expectation $\mathrm{E}(X)$ is ( ). (A) 4 (B) $4 \frac{1}{2}$ (C) 5 (D) $5 \frac{1}{2}$",See reasoning trace,medium,"4. B. $$ \begin{array}{l} \text { It is easy to see that } P(X=1)=\frac{1}{36}, \\ P(X=2)=\frac{4}{36}-\frac{1}{36}=\frac{3}{36}, \\ P(X=3)=\frac{9}{36}-\frac{4}{36}=\frac{5}{36}, \\ P(X=4)=\frac{16}{36}-\frac{9}{36}=\frac{7}{36}, \\ P(X=5)=\frac{25}{36}-\frac{16}{36}=\frac{9}{36}, \\ P(X=6)=1-\frac{25}{36}=\frac{11}{36} . \end{array} $$ Therefore, $\mathrm{E}(X)=1 \times \frac{1}{36}+2 \times \frac{3}{36}+3 \times \frac{5}{36}+$ $$ \begin{array}{l} 4 \times \frac{7}{36}+5 \times \frac{9}{36}+6 \times \frac{11}{36} \\ =\frac{161}{36}=4 \frac{17}{36}, \end{array} $$ which is closest to $4 \frac{1}{2}$." d1e8c1cf4704,"Example 3. Solve the equation $$ 2 f(x+y)+f(x-y)=f(x)\left(2 e^{y}+e^{-y}\right) $$ 28",See reasoning trace,medium,"Solution. Performing the following substitutions: $$ x=0, y=t ; x=t, y=2 t ; x=t, y=-2 t $$ we obtain the equations: $$ \begin{aligned} & 2 f(t)+f(-t)=a\left(2 e^{t}+e^{-t}\right) \\ & 2 f(3 t)+f(-t)=f(t)\left(2 e^{2 t}+e^{-2 t}\right) \\ & 2 f(-t)+f(3 t)=f(t)\left(2 e^{-2 t}+e^{2 t}\right) \end{aligned} $$ where $a=f(0)$. We eliminate $f(-t)$ and $f(3 t)$ from these equations. For this, we subtract the third equation, multiplied by 2, from the sum of the first equation, multiplied by 3, and the second equation. We obtain the general solution in implicit form: $$ 6 f(t)=3 a\left(2 e^{t}+e^{-t}\right)-3 f(t) \cdot e^{-2 t} $$ from which $$ f(t)=a \cdot \frac{2 e^{t}+e^{-t}}{2+e^{-2 t}}=a \cdot e^{t} $$ This function indeed satisfies the original equation, as can be easily verified by checking: $$ 2 a e^{x+y}+a e^{x-y}=a \cdot e^{x}\left(2 e^{y}+e^{-y}\right) $$ ## EXERCISES 53. Show that the following functional equations a) $f(x+y)=f(x)+f(y)$ b) $f(x y)=f(x) \cdot f(y)$ c) $f(x+y)=f(x) \cdot f(y)$; d) $f(x y)=f(x)+f(y)$ e) $f(x+y)+f(x-y)=2 f(x) \cdot f(y)$; f) $f(x+y)=\frac{f(x)+f(y)}{1-f(x) f(y)}$ are satisfied respectively by the functions $$ a x, x^{a}, a^{x}, \log _{a} x, \cos x, \operatorname{tg} x $$ Consider some properties of these functions based on the corresponding functional equation (e.g., evenness, oddness, etc.). 54. Solve the following functional equations: a) $f\left(\frac{x}{x-1}\right)=2 f\left(\frac{x-1}{x}\right)$; b) $f(x)+f\left(\frac{4}{2-x}\right)=x$; c) $a f(x)+f\left(\frac{1}{x}\right)=a x$; d) $f(n)=f(n-1)+a^{n}, \quad f(1)=1, \quad n \in \mathbf{N}, a \in \mathbf{R}$; e) $n^{f(n)-1}=(n-1)^{\{(n-1)}, n \in \mathbf{N}$; f) $f(x+y)+f(x-y)=2 f(x) \cdot \cos y$. 55. Prove that if for $x \in \mathbf{R}$ and a constant $m \in \mathbf{R}$ the equality $f(x+m)=\frac{1+f(m)}{1-f(m)}$ holds, then $f(x)$ is a periodic function. Find an example of such a function. 56*. Solve the following functional equations: a) $f^{2}(x)=f(2 x)$ b) $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ c) $f(x+y)+2 f(x-y)=3 f(x)-y$; d) $f(x+y)=f(x)+y$ e) $f(x+y)-2 f(x-y)+f(x)-2 f(y)=y-2$. ## § 5. ""Non-standard"" equations and systems of equations There is no general theory for solving equations and systems of equations of the considered type. In the process of solving them, such properties of the functions involved in the equation are used as boundedness (from above or below), monotonicity, evenness or oddness, etc. Consider examples." f2afc04f92cc,"3、Clock at 5 o'clock strikes 5 times using 8 seconds, then at 10 o'clock strikes 10 times using how many seconds?",18$ (seconds),easy,$3.8 \div(5-1) \times(10-1)=18$ (seconds) d84502625f9e,"Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$ For what values of $x$ does the equality hold?",x = 1,medium,"To prove the inequality \[ \sqrt{\frac{1}{3x+1}} + \sqrt{\frac{x}{x+3}} \ge 1 \] for all \( x > 0 \), we will use the method of substitution and algebraic manipulation. 1. **Substitution and Simplification:** Let \( y = \sqrt{\frac{1}{3x+1}} \). Then, we have: \[ y^2 = \frac{1}{3x+1} \] which implies: \[ 3x + 1 = \frac{1}{y^2} \] Solving for \( x \), we get: \[ x = \frac{1}{3y^2} - \frac{1}{3} \] 2. **Substitute \( x \) into the second term:** We need to express \( \sqrt{\frac{x}{x+3}} \) in terms of \( y \). Using the expression for \( x \): \[ \sqrt{\frac{x}{x+3}} = \sqrt{\frac{\frac{1}{3y^2} - \frac{1}{3}}{\frac{1}{3y^2} - \frac{1}{3} + 3}} \] Simplify the denominator: \[ x + 3 = \frac{1}{3y^2} - \frac{1}{3} + 3 = \frac{1}{3y^2} - \frac{1}{3} + \frac{9}{3} = \frac{1}{3y^2} + \frac{8}{3} \] Thus: \[ \sqrt{\frac{x}{x+3}} = \sqrt{\frac{\frac{1}{3y^2} - \frac{1}{3}}{\frac{1}{3y^2} + \frac{8}{3}}} = \sqrt{\frac{1 - y^2}{8y^2 + 1}} \] 3. **Combine the terms:** We need to show: \[ y + \sqrt{\frac{1 - y^2}{8y^2 + 1}} \ge 1 \] Let \( z = \sqrt{\frac{1 - y^2}{8y^2 + 1}} \). Then, we need to prove: \[ y + z \ge 1 \] 4. **Analyze the function:** Consider the function \( f(y) = y + \sqrt{\frac{1 - y^2}{8y^2 + 1}} \). We need to show that \( f(y) \ge 1 \) for \( y \in (0, 1) \). 5. **Check boundary conditions:** - When \( y = 0 \): \[ f(0) = 0 + \sqrt{\frac{1 - 0}{8 \cdot 0 + 1}} = 1 \] - When \( y \to 1 \): \[ f(1) = 1 + \sqrt{\frac{1 - 1}{8 \cdot 1 + 1}} = 1 \] 6. **Intermediate values:** For \( 0 < y < 1 \), we need to show that \( f(y) \ge 1 \). This can be done by analyzing the derivative of \( f(y) \) and showing that it is non-negative in the interval \( (0, 1) \). 7. **Equality condition:** The equality holds when \( y = 1 \), which corresponds to \( x = 1 \). Therefore, the inequality is proven for all \( x > 0 \), and the equality holds when \( x = 1 \). The final answer is \( \boxed{ x = 1 } \)" af7e3d9845f5,![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-10.jpg?height=261&width=506&top_left_y=606&top_left_x=468),$38^{\circ}$,medium,"Answer: $38^{\circ}$. Solution. Let $a$ be the length of the lateral side of the trapezoid. Note that point $T$ lies on the perpendicular bisector of the bases of the trapezoid, that is, on its axis of symmetry. From symmetry, we get that $B T=T C=a$ (Fig. 5). ![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-10.jpg?height=394&width=515&top_left_y=1109&top_left_x=469) Fig. 5: to the solution of problem 10.4 Next, note that $\angle B A D=\angle C D A=2 \angle C A D$, so $A C$ is the bisector of angle $B A D$. Since $\angle C A D=\angle A C B$ due to parallelism, triangle $A B C$ is isosceles. Therefore, $B C$ is also equal to $a$. We have obtained that triangle $B T C$ is equilateral, and its angles are each $60^{\circ}$. Now it is not difficult to calculate the required angle: $$ \angle T C D=\angle B C D-\angle B C T=\left(180^{\circ}-\angle A D C\right)-60^{\circ}=120^{\circ}-\angle A D C=38^{\circ} . $$" 3e423b68b5df,"3-ча 1. Construct two progressions: an arithmetic and a geometric one, each consisting of four terms; in such a way that if the corresponding terms of both progressions are added, the resulting numbers should be: $27, 27, 39, 87$.",See reasoning trace,medium,"Let $a, a+d, a+2 d, a+3 d$ be the desired arithmetic progression, and $b, b q, b q^{2}, b q^{3}$ be the desired geometric progression. According to the problem, \[ \begin{aligned} a+b & =27 \\ a+d+b q & =27 \\ a+2 d+b q^{2} & =39 \\ a+3 d+b q^{3} & =87 \end{aligned} \] Subtract the first equation from the second, the second from the third, and the third from the fourth: \[ \begin{aligned} d+b(q-1) & =0 \\ d+b q(q-1) & =12 \\ d+b q^{2}(q-1) & =48 \end{aligned} \] From the first equation, we get $b(q-1)=-d$; substitute this expression into the second and third equations: \[ \begin{aligned} d-d q & =12 \\ d-d q^{2} & =48 \end{aligned} \] Dividing the last equation by the second-to-last equation, we get $q=3$. Therefore, $d=-6, b=3$ and $a=24$. Thus, the desired progressions are \[ \begin{array}{cccc} 24, & 18, & 12, & 6 \\ 3, & 9, & 27, & 81 \end{array} \]" ff11cc0190bc,Let $N = 2^{\left(2^2\right)}$ and $x$ be a real number such that $N^{\left(N^N\right)} = 2^{(2^x)}$. Find $x$.,66,medium,"1. First, we need to determine the value of \( N \). Given \( N = 2^{(2^2)} \), we can simplify this as follows: \[ N = 2^{(2^2)} = 2^4 = 16 \] 2. Next, we are given the equation \( N^{(N^N)} = 2^{(2^x)} \). Substituting \( N = 16 \) into the equation, we get: \[ 16^{(16^{16})} = 2^{(2^x)} \] 3. We need to simplify the left-hand side of the equation. First, we calculate \( 16^{16} \): \[ 16 = 2^4 \implies 16^{16} = (2^4)^{16} = 2^{(4 \cdot 16)} = 2^{64} \] 4. Now, substituting \( 16^{16} = 2^{64} \) back into the original equation, we get: \[ 16^{(16^{16})} = 16^{(2^{64})} \] 5. We know that \( 16 = 2^4 \), so we can rewrite \( 16^{(2^{64})} \) as: \[ 16^{(2^{64})} = (2^4)^{(2^{64})} = 2^{(4 \cdot 2^{64})} = 2^{(2^{64 + 2})} = 2^{(2^{66})} \] 6. Therefore, the equation \( 16^{(16^{16})} = 2^{(2^x)} \) simplifies to: \[ 2^{(2^{66})} = 2^{(2^x)} \] 7. Since the bases are the same, we can equate the exponents: \[ 2^{66} = 2^x \] 8. This implies: \[ x = 66 \] The final answer is \( \boxed{66} \)" b6336efee26d,"Simplify the following expression as much as possible: $$ \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} $$",See reasoning trace,easy,"After rationalizing the denominators under the square roots: $$ \sqrt{\frac{(2-\sqrt{2})^{2}}{4-2}}+\sqrt{\frac{(2+\sqrt{2})^{2}}{4-2}}=\frac{2-\sqrt{2}}{\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2 \sqrt{2} $$ Barnabás Takács (Kecskemét, J. Katona II. Grammar School)" 09b637ac4fff,"123. How many right-angled triangles are there with sides of integer lengths, if one of the legs of these triangles is equal to 15?",4 triangles,medium,"Solution. If $x$ is the hypotenuse and $y$ is the unknown leg of a right triangle, then $x^{2}-y^{2}=15^{2} ;(x-y)(x+y)=$ $=3 \cdot 3 \cdot 5 \cdot 5$ Since $(x-y)$ and $(x+y)$ are natural numbers, and $x+y > x-y$, there are only 4 cases: $$ \left\{\begin{array} { l } { x - y = 1 } \\ { x + y = 225 } \end{array} \left\{\begin{array} { l } { x - y = 3 } \\ { x + y = 75 } \end{array} \quad \left\{\begin{array}{l} x-y=5 \\ x+y=45 \end{array} ;\left\{\begin{array}{l} x-y=9 \\ x+y=25 \end{array}\right.\right.\right.\right. $$ By solving these 4 systems of equations, we find that there are 4 right triangles that satisfy the given property. Their sides are: 1) $15 ; 112 ; 113 ; 2) 15 ; 36 ; 39 ; 3) 15 ; 20$; $25 ; 4) 15 ; 8 ; 17$. Answer: 4 triangles." 77d9712f38c8,"An $8 \mathrm{~cm}$ cube has a $4 \mathrm{~cm}$ square hole cut through its centre, as shown. What is the remaining volume, in $\mathrm{cm}^{3}$ ? (A) 64 (D) 384 (B) 128 (E) 448 (C) 256 ![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-144.jpg?height=382&width=441&top_left_y=1652&top_left_x=1292)",(D),medium,"An $8 \mathrm{~cm}$ cube has a $4 \mathrm{~cm}$ square hole cut through its centre, as shown. What is the remaining volume, in $\mathrm{cm}^{3}$ ? (A) 64 (B) 128 (C) 256 (D) 384 (E) 448 ![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-243.jpg?height=453&width=528&top_left_y=1107&top_left_x=1357) ## Solution Remaining volume $=8 \times 8 \times 8-8 \times 4 \times 4\left(\right.$ in $\left.\mathrm{cm}^{3}\right)$ $$ \begin{aligned} & =8(64-16) \\ & =8 \times 48 \\ & =384 \end{aligned} $$ ANSWER: (D)" 32649058cf3c,"$29 \cdot 26\left(2^{48}-1\right)$ can be divided by two numbers between 60 and 70, these two numbers are (A) 61,63 . (B) 61,65 . (C) 63,65 . (D) 63,67 . (E) 67,69 . (22nd American High School Mathematics Examination, 1971)",$(C)$,easy,"[Solution] Notice the transformation of the following algebraic expression: $$ \begin{aligned} 2^{48}-1 & =\left(2^{24}-1\right)\left(2^{24}+1\right) \\ & =\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right) \\ & =\left(2^{6}-1\right)\left(2^{6}+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right) \\ & =63 \cdot 65 \cdot\left(2^{12}+1\right)\left(2^{24}+1\right) . \end{aligned} $$ Therefore, the answer is $(C)$." 9f21a178ed11,"Initial 293 Given the equation $$ x^{3}-(2 a+11) x^{2}+\left(a^{2}+11 a+28\right) x-28 a=0 $$ all of whose roots are positive integers. Find the value of $a$ and the roots of the equation.","5$, the roots of the original equation are $5,2,14$.",medium,"It is known that $x=a$ is a root of the original equation. Thus, $a$ is a positive integer. Factoring the left side of the equation, we get $$ (x-a)\left[x^{2}-(a+11) x+28\right]=0 \text {. } $$ Since all roots of the original equation are positive integers, then $$ x^{2}-(a+11) x+28=0 $$ has a discriminant $$ \Delta=(a+11)^{2}-112 $$ which should be a perfect square. Let $(a+11)^{2}-112=k^{2}\left(k \in \mathbf{Z}_{+}\right)$. Then $$ (a+11)^{2}-k^{2}=112 \text {, } $$ which means $(a+11+k)(a+11-k)=112$. Obviously, $a+11+k$ and $a+11-k$ have the same parity, and $$ a+11+k \geqslant 11 \text {. } $$ Since $112=56 \times 2=28 \times 4=14 \times 8$, then $$ \left\{\begin{array}{l} a+11+k=56,28,14, \\ a+11-k=2,4,8 . \end{array}\right. $$ Solving these, we get $(a, k)=(18,27),(5,12),(0,3)$. Since $a$ is a positive integer, $a=18,5$. When $a=18$, the roots of the original equation are $18,1,28$. When $a=5$, the roots of the original equation are $5,2,14$." dbf1e23b8326,"【Question 7】 A boat is sailing in a river, traveling downstream at a speed of 15 kilometers per hour. It is known that the boat travels the same distance in 4 hours downstream as it does in 5 hours upstream. What is the speed of the current in kilometers per hour?",1.5$ km/h.,easy,"【Analysis and Solution】 Travel problem, boat in running water. Downstream speed is 15 km/h; Upstream speed is $15 \times 4 \div 5=12$ km/h; Water speed is $(15-12) \div 2=1.5$ km/h." f4ffc6d463bc,(given to Ambroise Marigot). Solve $5 x^{3}+11 y^{3}+13 z^{3}=0$ in relative integers.,"13 x^{\prime}$ and $y=13 y^{\prime}$ and substituting into the original equation, we realize that $z",medium,". We consider this equation modulo 13. An exhaustive study shows that a cube modulo 13 can only take the following values: $0,1,5,8$ and 12. Thus the quantity $5 x^{3}$ can only take the values $0,1,5,8$ and 12, while $11 y^{3}$ can only take the values $0,2,3,10$ and 11. The term $13 z^{2}$, on the other hand, obviously vanishes modulo 13 regardless of the value of $z$. From the above, we deduce that the only way to have a zero sum is $5 x^{3} \equiv 11 y^{3}(\bmod 1) 3$. It follows, since 13 is prime, that $x$ and $y$ are both multiples of 13. By writing $x=13 x^{\prime}$ and $y=13 y^{\prime}$ and substituting into the original equation, we realize that $z$ is also necessarily a multiple of 13, i.e., can be written as $z=13 z^{\prime}$. However, the triplet $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ is again a solution to the same equation, and an application of the principle of infinite descent ensures that the only solution is the triplet $(0,0,0)$." f573f7cacb8a,"10. Given the set $A=\{x|| x \mid \leqslant 2, x \in \mathbf{R}\}, B=\{x \mid x \geqslant a\}$, and $A \subseteq B$; then the range of the real number $a$ is $\qquad$ .",See reasoning trace,easy,10. $a \leqslant -2$ c7f6775cd94b,13. Which decimal digits occur as the final digit of a fourth power of an integer?,See reasoning trace,easy,"13. $0,1,5,6$" 924b2df2154c,"## Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$. $A(7, -5, 1)$ $B(5, -1, -3)$ $C(3, 0, -4)$",See reasoning trace,easy,"## Solution Let's find the vector $\overrightarrow{B C}:$ $\overrightarrow{B C}=\{3-5 ; 0-(-1) ;-4-(-3)\}=\{-2 ; 1 ;-1\}$ Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be: $$ \begin{aligned} & -2 \cdot(x-7)+(y-(-5))-(z-1)=0 \\ & -2 x+14+y+5-z+1=0 \\ & -2 x+y-z+20=0 \end{aligned} $$ ## Problem Kuznetsov Analytic Geometry 9-5" 15dfdc8e4a62,"In how many ways can $10001$ be written as the sum of two primes? $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$",\textbf{(A),medium,"For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers. The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$, so the number of ways $10001$ can be written as the sum of two primes is $\boxed{\textbf{(A)}\ 0}$" 55594f41b879,"Example 2 Given points $A_{1}(-1,0), A_{2}(1, -1), A_{3}(2,0), A_{4}(3,3), A_{5}(0,5)$ on a plane, find the area of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$.",See reasoning trace,easy,"Solve: Since points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are arranged counterclockwise, by Theorem 2, the area of pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ is $$\begin{array}{l} S=\frac{1}{2}\left|\begin{array}{rrrrrr} 1 & 1 & 2 & 3 & 0 & -1 \\ 0 & -1 & 0 & 3 & 5 & 0 \end{array}\right| \\ \quad=\frac{1}{2}[(-1) \times(-1)+1 \times 0+2 \times 3+3 \times 5+ \\ 0 \times 0-1 \times 0-2 \times(-1)-3 \times 0-0 \times 3-(-1) \times 5] \\ \quad=14.5 \end{array}$$" a51b2ebedadb,"19. (15 points) As shown in Figure 4, there is a pointer on a disk, initially pointing to the top of the disk. The pointer rotates clockwise around the center of the disk by an angle $\alpha$ each time, and $3.6^{\circ}<\alpha<180^{\circ}$. After 2,004 rotations, it returns to its initial position for the first time, pointing to the top of the disk again. How many different possible values can $\alpha$ have?",See reasoning trace,medium,"19. Clearly, if $3.6^{\circ}1$, then let $n_{1}=\frac{n}{d}, n_{2}=\frac{2004}{d}$. At this point, we have $\alpha=\frac{n_{1} \times 360^{\circ}}{n_{2}}$. This means that after the pointer rotates $n_{2}$ times, each time rotating by $\alpha$, the pointer will rotate $n_{1}$ full circles and return to its initial position, which contradicts the problem statement. From the above discussion, we know that for any $n$ satisfying $21 \leqslant n \leqslant 1001$ and $(n, 2004)=1$, there corresponds a possible $\alpha$. Conversely, this is also true. Therefore, the problem becomes finding the number of all $n$ that satisfy the above two conditions. Since $2004=2^{2} \times 3 \times 167$, we have $(n, 2004)=1 \Leftrightarrow 2 \chi_{n}, 3 \psi_{n}, 167 \uparrow_{n}$. Among the positive integers not greater than 1001, the number of positive integers that are not divisible by 2 or 3 is $$ \begin{array}{l} 1001-\left(\left[\frac{1001}{2}\right]+\left[\frac{1001}{3}\right]-\left[\frac{1001}{2 \times 3}\right]\right) \\ =1001-(500+333-166)=334 \text { (numbers). } \end{array} $$ (The symbol $[a]$ represents the greatest integer not exceeding $a$.) Among them, only $1 \times 167$ and $5 \times 167$ are divisible by 167, so the number of $n$ not greater than 1001 and satisfying the conditions is $334-2=332$. Removing the 7 numbers $1, 5, 7, 11, 13, 17, 19$ that are not greater than 20, we find that the number of $n$ satisfying both conditions is $332-7=325$. Therefore, $\alpha$ has 325 possible distinct values." 0e0e2c79e519,"A vertical-axis, rotational paraboloid-shaped tall container has its focus 0.05 meters above the axis point. We pour a little water into the container. At what angular velocity $\omega$ should we rotate the container around its axis so that the water flows out at the top of the container?",See reasoning trace,medium,"The surface of the fluid is a surface of rotation, so it is sufficient to examine a single cross-section passing through the axis. It is useful to visualize the motion in a coordinate system fixed to the container, rotating with angular velocity $\omega$, where the fluid is at rest. The surface of the fluid is perpendicular to the resultant force acting on a small patch of the surface. $1987-03-129-1$. eps Two forces act on the selected fluid element: the gravitational force and the centrifugal force. According to the diagram, the slope of the surface at a distance $x_{0}$ from the axis is: $\operatorname{tg} \alpha=\frac{\omega^{2} \cdot x_{0}}{g}$. We know that the functions whose derivative with respect to $x$ is $\frac{\omega^{2} \cdot x}{g}$ can be given by the formula $y=\frac{\omega^{2}}{2 g} x^{2}+C$. These functions define a parabola, whose equation is $y=\frac{\omega^{2}}{2 g} \cdot x^{2}+C$. (The origin is fixed to the bottom of the container, $C=0$ can be taken if the amount of fluid is sufficiently small.) The fluid will flow out of the container, regardless of its height, when its surface is parallel to the wall of the container's paraboloid, where the equation of the cross-section is $y=\frac{1}{4 \cdot 0.05 \mathrm{~m}} x^{2}$. The condition for parallelism is $\frac{\omega^{2}}{2 g} \cdot x^{2}=\frac{1}{0.2 \mathrm{~m}} \cdot x^{2}$, from which the required minimum angular velocity is $$ \omega=\sqrt{\frac{2 g}{0.2 \mathrm{~m}}}=9.9 \mathrm{~s}^{-1} $$ Remarks. 1. Some solvers cleverly used the small amount of water in the solution. Some referred to the entire amount of water as a ""point mass,"" thus turning the problem into one of equilibrium for a ball placed in the container. Another solver had the small amount of water condensing into a ring in a horizontal plane. 2. Some did not see that the water can flow out regardless of the height of the container, and thus believed that the outflow of water could be prevented with a sufficiently tall container." 66413c8c27a0,"One. (20 points) Given the equations about $x$: $4 x^{2}-8 n x -3 n=2$ and $x^{2}-(n+3) x-2 n^{2}+2=0$. Does there exist a value of $n$ such that the square of the difference of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain the reason.","0$, the square of the difference of the two real roots of the first equation equals the integer root",medium,"$$ \begin{aligned} -\Delta_{1} & =(-8 n)^{2}-4 \times 4 \times(-3 n-2) \\ & =(8 n+3)^{2}+23>0 . \end{aligned} $$ Then $n$ is any real number, the first equation always has real roots. Let the two roots of the first equation be $\alpha, \beta$, then $$ \begin{array}{l} \alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} . \\ \therefore(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=4 n^{2}+3 n+2 . \end{array} $$ From the second equation, we get $$ [x-(2 n+2)][x+(n-1)]=0 \text {. } $$ Solving, we get $x_{1}=2 n+2, x_{2}=1-n$. If $x_{1}$ is an integer, then $$ 4 n^{2}+3 n+2=2 n+2 \text {. } $$ Solving, we get $n_{1}=0, n_{2}=-\frac{1}{4}$. When $n=0$, $x_{1}=2$; When $n=-\frac{1}{4}$, $x_{1}=\frac{3}{2}$ is not an integer. Discard. If $x_{2}$ is an integer, then $4 n^{2}+3 n+2=1-n$. Solving, we get $n_{3}=n_{4}=-\frac{1}{2}$. When $n=-\frac{1}{2}$, $x_{2}=\frac{3}{2}$ is not an integer. Discard. In summary, when $n=0$, the square of the difference of the two real roots of the first equation equals the integer root of the second equation." 009440c4d939,A car and a minivan drive from Alphaville to Betatown. The car travels at a constant speed of $40 \mathrm{~km} / \mathrm{h}$ and the minivan travels at a constant speed of $50 \mathrm{~km} / \mathrm{h}$. The minivan passes the car 10 minutes before the car arrives at Betatown. How many minutes pass between the time at which the minivan arrives in Betatown and the time at which the car arrives in Betatown?,2 minutes,medium,"The car takes 10 minutes to travel from the point at which the minivan passes it until it arrives in Betatown. Since the car drives at $40 \mathrm{~km} / \mathrm{h}$ and since 10 minutes equals $\frac{1}{6}$ hour, then the car travels $40 \mathrm{~km} / \mathrm{h} \cdot \frac{1}{6} \mathrm{~h}=\frac{20}{3} \mathrm{~km}$ in these 10 minutes. Thus, the distance between the point where the vehicles pass and Betatown is $\frac{20}{3} \mathrm{~km}$. Since the minivan travels at $50 \mathrm{~km} / \mathrm{h}$, it covers this distance in $\frac{203 \mathrm{~km}}{50 \mathrm{~km} / \mathrm{h}}=\frac{2}{15} \mathrm{~h}$. Now $\frac{2}{15} \mathrm{~h}=\frac{8}{60} \mathrm{~h}$ which equals 8 minutes, and so the minivan arrives in Betatown $10-8=2$ minutes before the car. ANSWER: 2 minutes" 41478c956699,"$7 \cdot 48$ In an acute triangle $\triangle ABC$, the three sides are $a, b, c$, and the distances from its circumcenter to the three sides are $m, n, p$. Then, $m: n: p$ equals (A) $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$. (B) $a: b: c$. (C) $\cos A: \cos B: \cos C$. (D) $\sin A: \sin B: \sin C$. (China Junior High School Mathematics League, 1993)",$(C)$,medium,"[Solution 1] Connect $O A, O B, O C$, then $O A=O B=O C=R$ (circumradius). Clearly, by the Pythagorean theorem we have $$ \begin{array}{l} m^{2}=R^{2}-\left(\frac{a}{2}\right)^{2}=R^{2}-\frac{a^{2}}{4}, \\ n^{2}=R^{2}-\frac{b^{2}}{4} \\ p^{2}=R^{2}-\frac{c^{2}}{4} \end{array} $$ Thus, $m: n: p=\sqrt{R^{2}-\frac{a^{2}}{4}}: \sqrt{R^{2}-\frac{b^{2}}{4}}: \sqrt{R^{2}-\frac{c^{2}}{4}}$ (divide by $R$ ) $$ \begin{array}{l} =\sqrt{1-\left(\frac{a}{2 R}\right)^{2}}: \sqrt{1-\left(\frac{b}{2 R}\right)^{2}}: \sqrt{1-\left(\frac{c}{2 R}\right)^{2}} \\ =\sqrt{1-\sin ^{2} A}: \sqrt{1-\sin ^{2} B}: \sqrt{1-\sin ^{2} C} \\ =\cos A: \cos B: \cos C . \end{array} $$ Here, note that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$, and $\triangle A B C$ is an acute triangle. Therefore, the answer is $(C)$. [Solution 2] Connect $O A, O B, O C$, clearly $$ O A=O B=O C=R \text { (circumradius). } $$ Also, $\angle C O B=2 \angle A$, and $O M$ is perpendicular and bisects $B C$, so $O M$ bisects $\angle C O B$. In the right triangle $\triangle C O M$, $\angle C O M=\frac{1}{2} \angle C O B=$ $\angle A$, then $m=R \cos A$. Similarly, $n=R \cos B, p=R \cos C$. $$ \begin{aligned} \therefore m: n: p & =R \cos A: R \cos B: R \cos C \\ & =\cos A: \cos B: \cos C . \end{aligned} $$ Therefore, the answer is $(C)$." a98ea35925f8,How many positive integers $n$ exist such that $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer?,See reasoning trace,easy,"As $$ \frac{2 n^{2}+4 n+18}{3 n+3}=\frac{2}{3} \frac{\left(n^{2}+2 n+1\right)+8}{n+1}=\frac{1}{3}\left(2 n+2+\frac{16}{n+1}\right) $$ it follows that $n+1$ must divide 16. Thus, $n$ must belong to the set $\{1,3,7,15\}$. In each of these cases, we have | $n$ | $\frac{2 n^{2}+4 n+18}{3 n+3}$ | | :---: | :---: | | 1 | 4 | | 3 | 4 | | 7 | 6 | | 15 | 11 | Therefore, for the four values of $n$, 1, 3, 7, and 11, $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer." 566a6e928468,"Task solved by cyrix # Task 2 - 261222 Determine all triples $(p, q, r)$ of prime numbers that satisfy the following conditions (1), (2): (1) In the sequence of all prime numbers, $p, q, r$ are consecutive prime numbers in this order. (2) The number $s=p^{2}+q^{2}+r^{2}$ is a prime number.",See reasoning trace,medium,"I. If $(p, q, r)$ is a triple that satisfies conditions (1) and (2), then $(p, q, r)$ is not the triple $(2,3,5)$; because this does not satisfy condition (2) due to $2^{2}+3^{2}+5^{2}=38$. Furthermore, it follows that: $(p, q, r)$ is not a triple with $p>3$ (and thus also $q>3, r>3$); because every prime number greater than 3, when divided by 3, leaves a remainder of either 1 or 2, and thus its square leaves a remainder of 1 in both cases. For every triple $(p, q, r)$ of prime numbers $p, q, r>3$, $p^{2}+q^{2}+r^{2}$ is therefore divisible by 3 (and greater than 3), and thus not a prime number. By (1), the only remaining possibility is that $p, q, r$ is the triple $(3,5,7)$. II. This triple satisfies condition (1) as a triple of three consecutive prime numbers, and since $3^{2}+5^{2}+7^{2}=83$, it also satisfies condition (2). With I. and II., it is shown that exactly the triple $(3,5,7)$ satisfies conditions (1) and (2). ## Adopted from $[5]$ #### Abstract" f7d457bdd895,"For [tip=see hungarian]Az $X$ ́es$ Y$ valo ́s ́ert ́eku ̋ v ́eletlen v ́altoz ́ok maxim ́alkorrel ́acio ́ja az $f(X)$ ́es $g(Y )$ v ́altoz ́ok korrela ́cio ́j ́anak szupr ́emuma az olyan $f$ ́es $g$ Borel m ́erheto ̋, $\mathbb{R} \to \mathbb{R}$ fu ̈ggv ́enyeken, amelyekre $f(X)$ ́es $g(Y)$ v ́eges sz ́ora ́su ́. Legyen U a $[0,2\pi]$ interval- lumon egyenletes eloszl ́asu ́ val ́osz ́ınu ̋s ́egi v ́altozo ́, valamint n ́es m pozit ́ıv eg ́eszek. Sz ́am ́ıtsuk ki $\sin(nU)$ ́es $\sin(mU)$ maxim ́alkorrela ́ci ́oja ́t. [/tip] Edit: [hide=Translation thanks to @tintarn] The maximal correlation of two random variables $X$ and $Y$ is defined to be the supremum of the correlations of $f(X)$ and $g(Y)$ where $f,g:\mathbb{R} \to \mathbb{R}$ are measurable functions such that $f(X)$ and $g(Y)$ is (almost surely?) finite. Let $U$ be the uniformly distributed random variable on $[0,2\pi]$ and let $m,n$ be positive integers. Compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$. (Remark: It seems that to make sense we should require that $E[f(X)]$ and $E[g(Y)]$ as well as $E[f(X)^2]$ and $E[g(Y)^2]$ are finite. In fact, we may then w.l.o.g. assume that $E[f(X)]=E[g(Y)]=0$ and $E[f(Y)^2]=E[g(Y)^2]=1$.)[/hide]",0,medium,"1. **Understanding the Problem:** The problem asks us to compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$, where $U$ is a uniformly distributed random variable on $[0, 2\pi]$, and $m$ and $n$ are positive integers. The maximal correlation of two random variables $X$ and $Y$ is defined as the supremum of the correlations of $f(X)$ and $g(Y)$ over all measurable functions $f$ and $g$ such that $f(X)$ and $g(Y)$ are finite. 2. **Correlation Definition:** The correlation $\rho$ between two random variables $A$ and $B$ is given by: \[ \rho(A, B) = \frac{\text{Cov}(A, B)}{\sqrt{\text{Var}(A) \text{Var}(B)}} \] where $\text{Cov}(A, B)$ is the covariance of $A$ and $B$, and $\text{Var}(A)$ and $\text{Var}(B)$ are the variances of $A$ and $B$, respectively. 3. **Uniform Distribution Properties:** Since $U$ is uniformly distributed on $[0, 2\pi]$, its probability density function is: \[ f_U(u) = \frac{1}{2\pi}, \quad 0 \leq u \leq 2\pi \] 4. **Expectation and Variance Calculations:** We need to calculate the expectations and variances of $\sin(nU)$ and $\sin(mU)$: \[ E[\sin(nU)] = \int_0^{2\pi} \sin(nu) \cdot \frac{1}{2\pi} \, du = 0 \] because the integral of $\sin(nu)$ over a full period is zero. Similarly, \[ E[\sin(mU)] = 0 \] The variances are: \[ \text{Var}(\sin(nU)) = E[\sin^2(nU)] - (E[\sin(nU)])^2 = E[\sin^2(nU)] \] \[ E[\sin^2(nU)] = \int_0^{2\pi} \sin^2(nu) \cdot \frac{1}{2\pi} \, du = \frac{1}{2} \] because $\sin^2(x)$ has an average value of $\frac{1}{2}$ over a full period. Thus, \[ \text{Var}(\sin(nU)) = \frac{1}{2} \] and similarly, \[ \text{Var}(\sin(mU)) = \frac{1}{2} \] 5. **Covariance Calculation:** The covariance $\text{Cov}(\sin(nU), \sin(mU))$ is: \[ \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)] - E[\sin(nU)]E[\sin(mU)] \] Since $E[\sin(nU)] = 0$ and $E[\sin(mU)] = 0$, we have: \[ \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)] \] Using the product-to-sum identities: \[ \sin(nU) \sin(mU) = \frac{1}{2} [\cos((n-m)U) - \cos((n+m)U)] \] Therefore, \[ E[\sin(nU) \sin(mU)] = \frac{1}{2} \left( E[\cos((n-m)U)] - E[\cos((n+m)U)] \right) \] Since $U$ is uniformly distributed over $[0, 2\pi]$, the expectations of $\cos((n-m)U)$ and $\cos((n+m)U)$ are zero: \[ E[\cos((n-m)U)] = 0 \quad \text{and} \quad E[\cos((n+m)U)] = 0 \] Thus, \[ \text{Cov}(\sin(nU), \sin(mU)) = 0 \] 6. **Correlation Calculation:** Finally, the correlation $\rho(\sin(nU), \sin(mU))$ is: \[ \rho(\sin(nU), \sin(mU)) = \frac{\text{Cov}(\sin(nU), \sin(mU))}{\sqrt{\text{Var}(\sin(nU)) \text{Var}(\sin(mU))}} = \frac{0}{\sqrt{\frac{1}{2} \cdot \frac{1}{2}}} = 0 \] 7. **Maximal Correlation:** Since the correlation is zero, the maximal correlation is also zero. \[ \boxed{0} \]" ceecde58f829,"2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-b\right|<\frac{1}{125}$ is what?","-\left(a_{n}-1\right), a_{1}-1=8$, it can be known that the sequence $\left\{a_{n}-1\right\}$ is a g",easy,"2. $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, it can be known that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$, from which we can deduce that $n-1>5, n>6$, hence $n \geqslant 7$, so the smallest integer $n=7$." c73f1ada14b5,"8. Teacher Xiao has a total of 50 blue and red chalks, the number of blue chalks is 5 more than twice the number of red chalks. Teacher Xiao has $\qquad$ blue chalks.",See reasoning trace,easy,$35$ f10116de92f1,"Example 1 Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$.","r_{2}=4, r_{3}=2$, $n$ takes the minimum value, which is $n=2^{4} \times 3^{4} \times 5^{2}=32400$.",medium,"Solution: Let the prime factorization of $n$ be $$ n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}}, $$ where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers. Thus, the number of distinct positive divisors of $n$ is $$ \left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right). $$ From the problem, we have $$ \left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right)=75=3 \times 5^{2}. $$ Therefore, $n$ can have at most 3 distinct prime factors. To make $n$ the smallest and a multiple of 75, the prime factors of $n$ should be $2, 3, 5$, and 3 should appear at least once, and 5 should appear at least twice, i.e., $$ \begin{array}{l} n=2^{r_{1}} \times 3^{r_{2}} \times 5^{r_{3}}, \\ \left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)=75, \\ r_{1} \geqslant 0, r_{2} \geqslant 1, r_{3} \geqslant 2 . \end{array} $$ The tuples $\left(r_{1}, r_{2}, r_{3}\right)$ that satisfy the above conditions are $$ \begin{array}{l} (4,4,2),(4,2,4),(2,4,4),(0,4,14), \\ (0,14,4),(0,2,24),(0,24,2) . \end{array} $$ After calculation, it is found that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ takes the minimum value, which is $n=2^{4} \times 3^{4} \times 5^{2}=32400$." 97804c80c7ea,"5. Solve the system of equations: $$ \left\{\begin{array}{l} x y-2 y=x+106 \\ y z+3 y=z+39 \\ z x+3 x=2 z+438 \end{array}\right. $$",See reasoning trace,easy,"Solution. Note that $y \neq 1$ (since substituting into the first equation would then yield $x-2=x+106$). Then from the first two equations, we can express $x$ and $z$: $$ \begin{aligned} & x=\frac{106+2 y}{y-1} \\ & z=\frac{39-3 y}{y-1} \end{aligned} $$ Substitute into the last equation, multiply by $(y-1)^{2}$, and expand the brackets: $$ \begin{aligned} -432 y^{2}+864 y+3456 & =0 \\ -432(y-4)(y-2) & =0 \end{aligned} $$ Thus, the solutions are $(38,4,9)$ and $(-34,-2,-15)$." 40a9fdfbbd08,"Find 100 times the area of a regular dodecagon inscribed in a unit circle. Round your answer to the nearest integer if necessary. [asy] defaultpen(linewidth(0.7)); real theta = 17; pen dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1); draw(unitcircle,dg); for(int i = 0; i < 12; ++i) { draw(dir(30*i+theta)--dir(30*(i+1)+theta), db); dot(dir(30*i+theta),Fill(rgb(0.8,0,0))); } dot(dir(theta),Fill(dr)); dot((0,0),Fill(dr)); [/asy]",300,medium,"1. **Partition the Dodecagon into Triangles:** A regular dodecagon can be divided into 12 congruent isosceles triangles, each with a central angle of \( \frac{360^\circ}{12} = 30^\circ \). 2. **Calculate the Area of One Triangle:** Each triangle has a central angle of \(30^\circ\) and two sides of length 1 (the radius of the unit circle). The area \(A\) of one such triangle can be calculated using the formula for the area of a triangle with two sides \(a\) and \(b\) and included angle \(\theta\): \[ A = \frac{1}{2}ab \sin(\theta) \] Here, \(a = 1\), \(b = 1\), and \(\theta = 30^\circ\). Thus, \[ A = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(30^\circ) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] 3. **Calculate the Total Area of the Dodecagon:** Since there are 12 such triangles, the total area \(A_{\text{total}}\) of the dodecagon is: \[ A_{\text{total}} = 12 \cdot \frac{1}{4} = 3 \] 4. **Calculate 100 Times the Area:** To find 100 times the area of the dodecagon, we multiply the total area by 100: \[ 100 \cdot 3 = 300 \] The final answer is \(\boxed{300}\)." e9c1eb168df3,"3. Substituting the natural number $n$ into $n^{3}-n$, the following result can be obtained ( ). A. 678927 B. 695644 C. 728910 D. 874231 E. 934508",See reasoning trace,easy,"C Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly." f5e7ac2c6e02,"14. Gauss and his classmates were punished at school for being mischievous, and the teacher assigned a math From the natural numbers 1 to 100, remove two consecutive odd numbers, the average of the remaining numbers is 51, what is the product of the two removed odd numbers? Whoever answers correctly can go home. Gauss once again quickly gave the correct answer. Gauss's answer is $\qquad$.",See reasoning trace,easy,"\begin{tabular}{l} 14 \\ \hline 675 \end{tabular}" bd47582d29a4,"Baranov d..V. The hare bought seven drums of different sizes and seven pairs of sticks of different lengths for her seven baby hares. If a baby hare sees that both its drum is larger and its sticks are longer than those of one of its brothers, it starts to drum loudly. What is the maximum number of baby hares that can start drumming?",6 bunnies,easy,"Not all the bunnies can play the drum, as the baby bunny that gets the smallest drum will not play it. On the other hand, if the same baby bunny is also given the shortest drumsticks, all the other bunnies will play the drum. ## Answer 6 bunnies." e7aaf0771736,"![](https://cdn.mathpix.com/cropped/2024_05_06_d55ea3e8288d58a3f771g-07.jpg?height=420&width=485&top_left_y=1241&top_left_x=484) Construct the correspondence - In circle $A$ - In circle $B$ - In circle $C$ - In circle $D$ - the number 6 is placed. - the number 7 is placed. - the number 8 is placed. - the number 9 is placed.","$A=6, B=8, C=7, D=9$",medium,"Answer: $A=6, B=8, C=7, D=9$. Solution. From the condition, it follows that $A+C+3+4=5+D+2+4$, from which $D+4=A+C$. Note that $13 \geqslant D+4=A+C \geqslant 6+7$. Therefore, this is only possible when $D=9$, and $A$ and $C$ are 6 and 7 in some order. Hence, $B=8$. The sum of the numbers along each side is $5+9+3+4=20$. Since $5+1+8+A=20$, then $A=6$. Since $6+C+3+4=20$, then $C=7$. ![](https://cdn.mathpix.com/cropped/2024_05_06_d55ea3e8288d58a3f771g-08.jpg?height=418&width=474&top_left_y=433&top_left_x=494)" fce6ee331177,"\section*{ All ordered pairs of real numbers \((x, y)\) are to be determined for which the system of equations \[ \begin{aligned} x \cdot\left(a x^{2}+b y^{2}-a\right) & =0 \\ y \cdot\left(a x^{2}+b y^{2}-b\right) & =0 \end{aligned} \] is satisfied. Here, \(a\) and \(b\) are real numbers with \(a \neq 0, b \neq 0\) and \(a \neq b\).",See reasoning trace,medium,"} We perform a case distinction: 1. Case: \(x=0\). Then the second equation becomes \(y \cdot b \cdot\left(y^{2}-1\right)=0\), which, due to \(b \neq 0\), leads to \(y=0\) or \(y= \pm 1\). For all three elements \((x, y) \in\{(0,-1),(0,0),(0,1)\}\), verification confirms that they are indeed solutions to the system of equations. 2. Case: \(x \neq 0\). Then it follows from the first equation that \(a x^{2}+b y^{2}-a=0\), and thus, due to \(a \neq b\), \(a x^{2}+b y^{2}-b \neq 0\). Therefore, from the second equation, we directly get \(y=0\). Substituting this into the equation just obtained, we get \(a x^{2}-a=0\) or \(x= \pm 1\). Here, too, all elements of the set \(\{(-1,0),(1,0)\}\) are solutions to the system of equations, as the verification confirms. Thus, the given system of equations has a total of five solutions, which are noted in the two cases." ddadbc920a9c,"## Condition of the To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$. $y=\frac{x}{x^{2}+1}, x_{0}=-2$",See reasoning trace,medium,"## Solution Let's find $y^{\prime}:$ $$ \begin{aligned} & y^{\prime}=\left(\frac{x}{x^{2}+1}\right)^{\prime}=\frac{x^{\prime}\left(x^{2}+1\right)-x\left(x^{2}+1\right)^{\prime}}{\left(x^{2}+1\right)^{2}}= \\ & =\frac{x^{2}+1-x \cdot 2 x}{\left(x^{2}+1\right)^{2}}=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}} \end{aligned} $$ Then: $y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1-(-2)^{2}}{\left((-2)^{2}+1\right)^{2}}=\frac{-3}{5^{2}}=-\frac{3}{25}$ Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is: $y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$, where $y_{0}^{\prime}=-\frac{3}{25}$ $y_{0}=y\left(x_{0}\right)=\frac{-2}{(-2)^{2}+1}=-\frac{2}{5}$ We get: $y-\left(-\frac{2}{5}\right)=-\frac{3}{25} \cdot(x-(-2))$ $y=-\frac{3}{25} \cdot(x+2)-\frac{2}{5}$ $y=-\frac{3}{25} \cdot x-\frac{6}{25}-\frac{2}{5}$ $y=-\frac{3}{25} \cdot x-\frac{16}{25}$ Thus, the equation of the tangent line is: $$ y=-\frac{3}{25} \cdot x-\frac{16}{25} $$ ## Problem Kuznetsov Differentiation 3-21" e9b2f137dd9e,"13. Given that $b$ and $c$ are positive integers, and the three quadratic equations $x^{2}+b x+c=k$ (for $k=0,1,2$) all have real roots but no integer roots, find the minimum value of $b+c$.","k(k=0,1,2)$ all have no integer roots, $\therefore$ the minimum value of $b+c$ is 8.",medium,"13. The discriminant of $x^{2}+b x+c=k$ is $D_{k}=b^{2}-4 c+4 k(k=0,1,2)$. According to the problem, $D_{0} \geqslant 0$ and $\sqrt{D_{0}} \cdot \sqrt{D_{1}}=$ $\sqrt{D_{0}+4}, \sqrt{D_{2}}=\sqrt{D_{0}+8}$ are not integers. $$ \therefore D_{0} \neq 1,4,5,8,9,12 \text {. } $$ (1) If $D_{0}=b^{2}-4 c=2$ or 6 or 10, then $b$ is even. Let $b=2 m$, then $4 m^{2}-4 c=2$ or 6 or $4, \therefore 4 \mid 2$ or 6 or 4, a contradiction. (2) If $D_{0}=b^{2}-4 c=3$ or 7 or 11, then $b$ is odd. Let $b=2 m+1$, then $4 m(m+1)-4 c=2$ or 6 or 10, thus $4 \mid 2$ or 6 or 4, a contradiction. $\therefore D_{0} \geqslant 13$, i.e., $b^{2}-4 c \geqslant 13$. Also, the equation $x^{2}+b x+c=k(k=0,1,2)$, when $c=k$, has integer solutions $x=0$ and $x=-6$. $$ \begin{array}{l} \therefore c \geqslant 3, b^{2} \geqslant 13+4 c \geqslant 25, \therefore b \geqslant 5 \\ \therefore b+c \geqslant 8 . \end{array} $$ Also, $x^{2}+5 x+3=k(k=0,1,2)$ all have no integer roots, $\therefore$ the minimum value of $b+c$ is 8." a03df43a1e0b,"1(!). The perimeter of a triangle is \(12 \mathrm{~cm}\), and the length of one of its sides is 5 cm. Find the relationship between the lengths of the other two sides and plot the graph of this relationship.","7\), and therefore \(y = 7 - x\). The graph of the dependence is shown in Figure 29. The dependence ",medium,"1. Solution. Let the length of the second side be \(x\) cm, and the third side be \(y\) cm. The length of each side is a positive number that is less than half the perimeter of the triangle. Therefore, ![](https://cdn.mathpix.com/cropped/2024_05_21_ffd37dfe6eab44561342g-75.jpg?height=680&width=685&top_left_y=251&top_left_x=340) Fig. 29 ![](https://cdn.mathpix.com/cropped/2024_05_21_ffd37dfe6eab44561342g-75.jpg?height=665&width=668&top_left_y=267&top_left_x=1048) Fig. 30 \(0 < x < 6\), \(0 < y < 6\). Additionally, from the conditions of the problem, it follows that \(x + y = 7\), and therefore \(y = 7 - x\). The graph of the dependence is shown in Figure 29. The dependence between \(x\) and \(y\) can be written as: \(y = 7 - x\) and \(1 < x < 6\)." 2a55b72ccac4,"5. On the website of the football club ""Rostov,"" a poll is being conducted to determine which of the $m$ football players the website visitors consider the best at the end of the season. Each visitor votes once for one player. The website displays the rating of each player, which is the percentage of votes cast for them, rounded to the nearest whole number. After several visitors have voted, the total rating of the nominees was $95 \%$. What is the smallest $m$ for which this is possible?",11,medium,"Solution. Let $a$ be the greatest loss of a percentage point due to rounding when determining a footballer's rating. Then, according to rounding rules, $aa m \geqslant 5$, which means $0.5 m>5$ or $m>10$. We will show that a solution exists when $m=11$. For example, let $m=11$ and 73 visitors voted, with 33 of them voting for one footballer, while the remaining 40 visitors evenly distributed their votes among 10 footballers, giving each 4 votes. In this case, the total rating is $\frac{33}{73} \cdot 100 \%+10 \cdot \frac{4}{73} \cdot 100 \approx 45 \%+10 \cdot 5 \%=95 \%$. Thus, the minimum $m=11$. Answer: 11." 5de76bcc61e6,Let's consider one vertex of a cube and look at the rays leading from it to the other vertices. How many different magnitudes of angles do we get if we pair these rays in every possible way?,See reasoning trace,medium,"Let one face of the cube be $ABCD$, the third edges starting from its vertices be $AE, BF, CG, DH$, and the selected vertex be $A$. Among the 7 rays leading to the other vertices, 3 are edges of the cube: $AB, AD$, and $AE$, 3 are face diagonals: $AC, AF$, $AH$, and the seventh, $AG$, is a space diagonal. The mentioned 3 edges can be transformed into each other by $120^{\circ}$ rotations around $AG$, and the same applies to the mentioned face diagonals, so it is sufficient to determine the magnitudes of the angles whose one side is $AB$ or $AC$. $AB$ forms a right angle with $AD$ and $AE$, and similarly, the diagonal $AH$ does as well. The diagonals $AC$ and $AF$ form a $45^{\circ}$ angle, because, for example, $ABF$ is an isosceles right triangle. Finally, in the right triangle $ABG$, if $AB$ is taken as a unit, then $BG = \sqrt{2}$, so the tangent of the angle $BAG$ is $\sqrt{2}$. (This angle is clearly not $45^{\circ}$.) The angles formed by $AC$ with the edges are already known from the above. The angle $CAF$ is $60^{\circ}$, because the triangle $CAF$ is equilateral (face diagonals). The angle $CAG$ is such that its tangent is $1 / \sqrt{2}$, since the triangle $CAG$ is congruent to the previously mentioned triangle $FAG$. According to this, this angle differs from the previous ones, because $\tan 60^{\circ} = \sqrt{3}$. According to these, 5 different angles appear among the pairs formed by the 7 rays drawn from $A$ to the other vertices: $90^{\circ}, 60^{\circ}, 45^{\circ}$, and those whose tangents are $\sqrt{2}$ and $1 / \sqrt{2}$. (Their values in degrees are $54^{\circ} 44'$ and $35^{\circ} 16'$.) György Ponácz (Székesfehérvár, József A. Gymnasium II. o. t.)" 62cdedfafdf5,"Three balls marked $1,2$ and $3$ are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is $6$, what is the probability that the ball numbered $2$ was drawn all three times? $\textbf{(A)} \ \frac{1}{27} \qquad \textbf{(B)} \ \frac{1}{8} \qquad \textbf{(C)} \ \frac{1}{7} \qquad \textbf{(D)} \ \frac{1}{6} \qquad \textbf{(E)}\ \frac{1}{3}$",\textbf{(C),easy,"Since a ball is drawn three times and the sum is $6$, the only ways this can happen are the permutations of $1, 2, 3$ and $2, 2, 2$. Therefore, with there being $3! + 1 = 7$ equally-likely possibilities in total, the probability that the ball numbered $2$ was drawn all three times is $\boxed{\textbf{(C)}\ \frac{1}{7}}$." 41dfebc1983d,"5.3. In three piles, there are 22, 14, and 12 nuts. It is required to equalize the number of nuts in all piles by making three moves, while adhering to the following condition: from one pile to another, only as many nuts can be moved as there are in the pile to which the nuts are being moved. $$ \text { (4-6 grades) } $$",See reasoning trace,easy,"5.3. Move 14 nuts from the first pile to the second. After this, the piles will contain 8, 28, and 12 nuts, respectively. Next, move 12 nuts from the second pile to the third. Now the piles will contain 8, 16, and 24 nuts, respectively. Finally, move 8 nuts from the third pile to the first, after which each of the three piles will contain 16 nuts." 3ed1709c4960,3. Given the line $y=\frac{1}{2} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Then the equation of the perpendicular bisector of line segment $A B$ is . $\qquad$,-2 x-\frac{3}{2}$.,medium,"3. $y=-2 x-\frac{3}{2}$. As shown in Figure 6, let the perpendicular bisector of line segment $AB$ intersect $AB$ at $C$ and the negative half of the $x$-axis at $D$. Connect $BD$. Then $AD=BD$. It is easy to see that $A(-2,0)$ and $B(0,1)$. By the Midline Theorem of a triangle, we know $C\left(-1, \frac{1}{2}\right)$. Let $D(a, 0)$. Then $AD=a+2$. In the right triangle $\triangle BOD$, by the Pythagorean theorem, we have $BD^2=OD^2+OB^2$. Therefore, $(a+2)^2=a^2+1^2$. Solving for $a$ gives $a=-\frac{3}{4}$. Thus, $D\left(-\frac{3}{4}, 0\right)$. Let the equation of line $CD$ be $y=kx+b$. Then $\left\{\begin{array}{l}\frac{1}{2}=-k+b, \\ 0=-\frac{3}{4} k+b .\end{array}\right.$ Solving these equations gives $k=-2, b=-\frac{3}{2}$. Therefore, the equation of the perpendicular bisector of line segment $AB$ is $y=-2 x-\frac{3}{2}$." f36db22cb889,"24. Points $A_{1}, \cdots, A_{5}$ are on a sphere with radius 1, what is the maximum value of $\min _{1 \leqslant \mathrm{i}, \mathrm{j} \leqslant 5} A_{1} A_{\mathrm{j}}$? Determine all cases where the maximum value is achieved.",See reasoning trace,medium,"24. We can use the size of $\angle A_{1} O A_{1}$ to replace the distance $A_{1} A_{\mathrm{j}}$, where $O$ is the center of the sphere. There exists a set of points $A_{1}, A_{2}, \cdots, A_{5}$, such that $$ \mathrm{mn}_{1 \leqslant \mathrm{i}, \mathrm{j}} \leqslant A_{1} O A_{\mathrm{j}} \leqslant \frac{\pi}{2} . $$ For example, take 5 vertices of a regular octahedron. Now let $A_{1}, A_{2}, \cdots, A_{5}$ be any set of points that satisfy (24.1) - we assert that there must be two antipodal points. First, set $A_{5}$ as the South Pole. In this case, $A_{1}, A_{2}, A_{\mathrm{j}}, A_{4}$ are all in the Northern Hemisphere (including the equator). If there are no antipodal points, then the North Pole has no points. If a quadrant in the Northern Hemisphere, bounded by a quarter of the equator and two mutually perpendicular meridians, contains two $A_{\mathrm{t}}$, then due to (24.1), one must be at the ""corner"" (i.e., the intersection of the equator and a meridian), and the other must be on the opposite meridian. Thus, the difference in longitude between any two of $A_{1}, \cdots, A_{4}$ is less than $\frac{\pi}{2}$. Therefore, they lie on the four meridians that divide the Northern Hemisphere into four quadrants. Finally, if one point is not on the equator, then its two ""neighbors"" must be on the equator (each in a quadrant's ""corner""), and these two points are antipodal. Since there are always two antipodal points $A_{1}, A_{2}$, and the third point $A_{3}$ cannot simultaneously satisfy $\angle A_{1} O A_{3} > \frac{\pi}{2}$ and $\angle A_{2} O A_{3} > \frac{\pi}{2}$, the strict inequality in (24.1) cannot hold. Thus, the maximum value of $\min \angle A_{1} O A_{j}$ is $\frac{\pi}{2}$. When the equality holds, let $A_{1}, A_{2}$ be the two poles, then $A_{3}, A_{4}, A_{5}$ are all on the equator, and the angles between $O A_{3}, O A_{4}$, and $O A_{5}$ are not less than $\frac{\pi}{2}$. The maximum value of $\min A_{1} A_{\mathrm{j}}$ is $\sqrt{2}$, which is achieved in the above situation." fe24c64d5f6b,"1. Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$. Find the maximum value of $\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$.",See reasoning trace,medium,"Let $P=\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$, Since $a_{i}>\frac{11}{9} a_{i+1}^{2} \Rightarrow a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2}=\frac{2}{9} a_{i+1}^{2} \geqslant 0$, If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$; If $a_{2016}-a_{1}^{2}>0$, let $a_{2017}=a_{1}$, then $P^{\frac{1}{2016}} \leqslant \frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right)=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i+1}^{2}\right)$ $=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016}\left[-\left(a_{i}-\frac{1}{2}\right)^{2}+\frac{1}{4}\right]$ $\leqslant \frac{1}{2016} \sum_{i=1}^{2016} \frac{1}{4}=\frac{1}{4}$, equality holds when $a_{i}=\frac{1}{2}, i=1,2, \cdots, 2016$. At this point, $9 a_{i}=\frac{9}{2}>\frac{11}{4}=11 a_{i+1}^{2}$ is clearly satisfied, so the maximum value of $P^{\frac{1}{2016}}$ is $\frac{1}{4}$. In summary, the maximum value of $\left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)$ is $\frac{1}{4^{2016}}$." 7e2aaa6b1218,"2. The three sides of a triangle are all integers, and its perimeter equals 10. Then this triangle is ( ). (A) a right triangle (B) an obtuse triangle (C) a triangle with exactly two equal sides (D) a triangle with exactly one interior angle of $60^{\circ}$",See reasoning trace,medium,"2. C. Decompose 10 into the sum of three positive integers, we have $$ \begin{array}{l} 10=1+1+8=1+2+7=1+3+6=1+4+5 \\ =2+2+6=2+3+5=2+4+4=3+3+4 \end{array} $$ There are a total of eight cases. According to the ""sum of the lengths of any two sides of a triangle is greater than the length of the third side,"" only the groups $(2,4,4)$ and $(3,3,4)$ can form triangles. Since the two base angles of an isosceles triangle are acute, in the isosceles triangle with sides $2,4,4$, the angle opposite the smallest side 2 is also acute. In the isosceles triangle with sides $3,3,4$, since $3^{2}+3^{2}>4^{2}$, the angle opposite the base is also acute. Therefore, the isosceles triangles with sides $2,4,4$ and $3,3,4$ are both acute triangles, eliminating options (A) and (B). Furthermore, since an isosceles triangle has exactly one angle of $60^{\circ}$ when it becomes an equilateral triangle, which contradicts the conditions of sides $(2,4,4)$ and $(3,3,4)$, option (D) is also eliminated. The triangles with sides $(2,4,4)$ and $(3,3,4)$ are isosceles triangles with exactly two equal sides." ba3a62a83094,"1. In the four-digit number 4753 , three two-digit numbers are formed by successive pairs of digits $(47,75,53)$. Exactly two of these two-digit numbers are prime. Find all four-digit numbers in which all four digits are prime, and all three two-digit numbers formed by successive digits are prime.",See reasoning trace,medium,"SolUTION Let $P$ represent such a four-digit number. As the digits of $P$ must each be prime, the only choices of digits we can make are $2,3,5$ or 7 . It will be helpful to list the two digit prime numbers which use only these digits: $23,37,53$, and 73 . The first two digits of $P$ must form a prime number when taken as a two-digit number (from left to right), so we should consider each of the four options in turn. If the first two digits are 2 and 3 respectively, this means that the third digit must be 7 as the second and third digits must form a two-digit prime number and our only choice is 37 . The final digit must be 3 , as the third and fourth digits must also form a two-digit prime number and our only choice is 73. This gives 2373 as a possible value for $P$. If we consider the other three options for the first two digits in turn, similar reasoning yields 3737, 5373 and 7373 as possible values for $P$. In conclusion, we have four possible values for $P$ which are: $2373,3737,5373$, and 7373 ." 2f13794e5e21,"1. Which of these fractions has the largest value? A $\frac{8+5}{3}$ B $\frac{8}{3+5}$ C $\frac{3+5}{8}$ D $\frac{8+3}{5}$ E $\frac{3}{8+5}$","4 \frac{1}{3}, \frac{8}{8}=1, \frac{8}{8}=1, \frac{11}{5}=2 \frac{1}{5}$ and $\frac{3}{13}$. Hence t",easy,"Solution A The values of the fractions shown are $\frac{13}{3}=4 \frac{1}{3}, \frac{8}{8}=1, \frac{8}{8}=1, \frac{11}{5}=2 \frac{1}{5}$ and $\frac{3}{13}$. Hence the fraction which has the largest value is $\frac{8+5}{3}$." 73f30174e384,"8.58 If $m, n, p$ and $q$ are real numbers and $f(x)=m x+n, \quad g(x)=p x+$ $q$, then the condition for the equation $f(g(x))=g(f(x))$ to have a solution is (A) $m, n, p$ and $q$ can be any real numbers. (B) If and only if $m=p$ and $n=q$. (C) If and only if $m q-n p=0$. (D) If and only if $n(1-p)-q(1-m)=0$. (E) If and only if $(1-n)(1-p)-(1-q)(1-m)=0$. (27th American High School Mathematics Examination, 1976)",See reasoning trace,easy,"[Solution] For any real number $x$, the following equations are equivalent: $$ \begin{aligned} f(g(x)) & =g(f(x)), \\ m(p x+q)+n & =p(m x+n)+q, \end{aligned} $$ or thus $$ m p x+m q+n=m p x+n p+q . $$ $$ n(1-p)=q(1-m) \text {, } $$ Therefore, the correct choice is $(D)$." d2bf629a6c5e,"[Square] $ABCD$ has side length $2$. A [semicircle] with [diameter] $\overline{AB}$ is constructed inside the square, and the [tangent]) to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$? $\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$",\mathrm{(D),medium,"Solution 1 Let the point of tangency be $F$. By the [Two Tangent Theorem](https://artofproblemsolving.com/wiki/index.php/Two_Tangent_Theorem) $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on $\triangle CDE$ yields \begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*} Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$. Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF$. Thus, $\angle EGF=\angle FCG$ and $\tan EGF=\tan FCG=\frac{1}{2}$. Solving $EF=\frac{1}{2}$. Adding, the answer is $\frac{5}{2}$. Solution 3 [2004 AMC12A-18.png](https://artofproblemsolving.com/wiki/index.php/File:2004_AMC12A-18.png) Clearly, $EA = EF = BG$. Thus, the sides of [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $CDE$ are in arithmetic progression. Thus it is [similar](https://artofproblemsolving.com/wiki/index.php/Similar_triangles) to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$. Solution 4 Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\frac{1}{2}$, so the answer is $\frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$. Solution 5 Using the diagram as drawn in Solution 4, let the total area of square $ABCD$ be divided into the triangles $DCE$, $EAG$, $CGB$, and $EGC$. Let x be the length of AE. Thus, the area of each triangle can be determined as follows: \[DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 2-x\] \[EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}\] \[CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1\] \[EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}\] (the length of CE is calculated with the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), lines GE and CE are perpendicular by definition of tangent) Adding up the areas and equating to the area of the total square \[(2 \cdot 2=4)\], we get \[x = \frac{1}{2}\] So, \[CE = 2 + \frac{1}{2} = 5/2\]. ~Typo Fix by doulai1 Video Solution [https://youtu.be/pM0zICtH6Lg](https://artofproblemsolving.comhttps://youtu.be/pM0zICtH6Lg) Education, the Study of Everything" 0a668b911689,"9. (16 points) For a given positive integer $M$, define $f_{1}(M)$ as the square of the sum of the digits of $M$. When $n>1$ and $n \in \mathbf{N}$, $f_{n}\left(f_{n-1}(M)\right)$ represents the $r_{n}$-th power of the sum of the digits of $f_{n-1}(M)$, where, when $n$ is odd, $r_{n}=2$; when $n$ is even, $r_{n}=3$. Find the value of $f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)$.",See reasoning trace,medium,"Let the positive integer $M=a_{1} a_{2} \cdots a_{m}$, where $a_{i} \in \mathbf{N}, a_{1}>0 (i=1,2, \cdots, m)$. Then $$ f_{1}(M)=\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2} \equiv M^{2}(\bmod 9) . $$ When $M=3^{2010}$, $f_{1}(M) \equiv M^{2} \equiv 0(\bmod 9)$. By mathematical induction, it is easy to see that for $n \in \mathbf{N}$, and $n \geqslant 2$, we always have $$ f_{n}\left(f_{n-1}(M)\right) \equiv 0(\bmod 9) . $$ Since $3^{2010}=9^{1005}<10^{1005}$, the sum of its digits is less than $9 \times 1005=9045$. Thus, $f_{1}\left(3^{2010}\right)<9045^{2}<9 \times 10^{7}$. Therefore, the sum of the digits of $f_{1}\left(3^{2010}\right)$ is less than $9 \times 8=72$. Hence, $f_{2}\left(f_{1}\left(3^{2010}\right)\right)<72^{3}<400000$, and $f_{3}\left(f_{2}\left(3^{2010}\right)\right)<(3+9 \times 5)^{2}=48^{2}<50^{2}$. Let the sum of the digits of $f_{3}\left(f_{2}\left(3^{2010}\right)\right)$ be $a$. Then $1 \leqslant a<1+9 \times 3=28$, and $a \equiv f_{3}\left(f_{2}\left(3^{2010}\right)\right) \equiv 0(\bmod 9)$. Therefore, $a \in\{9,18,27\}$. Thus, $f_{4}\left(f_{3}\left(3^{2010}\right)\right)=a^{3} \in\left\{9^{3}, 18^{3}, 27^{3}\right\}$ $$ =\{729,5832,19683\} \text {. } $$ The sums of the digits of $729$, $5832$, and $19683$ are $18$, $18$, and $27$, respectively. Hence, $$ f_{5}\left(f_{4}\left(3^{2010}\right)\right) \in\left\{18^{2}, 27^{2}\right\}=\{324,729\} \text {. } $$ Furthermore, $$ \begin{array}{l} f_{6}\left(f_{5}\left(3^{2010}\right)\right) \in\left\{9^{3}, 18^{3}\right\}=\{729,5832\} \\ \Rightarrow f_{7}\left(f_{6}\left(3^{2010}\right)\right)=18^{2}=324 \\ \Rightarrow f_{8}\left(f_{7}\left(3^{2010}\right)\right)=9^{3}=729 \\ \Rightarrow f_{9}\left(f_{8}\left(3^{2010}\right)\right)=18^{2}=324 \\ \Rightarrow \cdots \cdots \\ \Rightarrow f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)=729 \end{array} $$" 78a498a1fac1,"11. Consider a $2n$-digit number, where the first $n$ digits and the last $n$ digits are each treated as an $n$-digit number. If the square of the sum of these two $n$-digit numbers is exactly equal to the $2n$-digit number, then this $2n$-digit number is called a Kabulek (Kabulek) strange number. For example, $(30+25)^{2}=3025$, so 3025 is a Kabulek strange number. Please find which four-digit numbers are Kabulek strange numbers?","The Kaprekar numbers in four-digit numbers are: $2025, 3025, 9801$",medium,"【Analysis】Let the first two digits of the number be $x$, and the last two digits be $y$. Thus, we have $(x+y)^{2}=100 x+y=x+y+99 x$, which means: $(x+y)$ $$ (x+y-1)=99 x \text {, } $$ From this, we can see that one of $x+y$ and $x+y-1$ is a multiple of 9, and the other is a multiple of 11 (of course, depending on the number of digits, it could also be other factors), thus finding the three numbers that satisfy the condition: 45, 55, and 99, and then calculating their squares. 【Solution】Solution: Let the first two digits of the number be $x$, and the last two digits be $y$. Thus, we have $(x+y)^{2}=100 x+y=x+y+99 x$, which means: $(x+y)(x+y-1)=99 x$, From this, we can see that one of $x+y$ and $x+y-1$ is a multiple of 9, and the other is a multiple of 11 (of course, depending on the number of digits, it could also be other factors), We can find the three numbers that satisfy the condition: 45, 55, and 99, and their squares are $2025, 3025, 9801$; That is: $(20+25)^{2}=2025$; $(30+25)^{2}=3025$; $(98+1)^{2}=9801$; Answer: The Kaprekar numbers in four-digit numbers are: $2025, 3025, 9801$." f691cd9a1015,Let $n$ be a positive integer such that $12n^2+12n+11$ is a $4$-digit number with all $4$ digits equal. Determine the value of $n$.,21,medium,"1. We start with the given expression \(12n^2 + 12n + 11\) and note that it must be a 4-digit number with all digits equal. The possible 4-digit numbers with all digits equal are \(1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999\). 2. We rewrite the expression \(12n^2 + 12n + 11\) in a more convenient form: \[ 12n^2 + 12n + 11 = 12(n^2 + n + 1) - 1 \] This implies that \(12(n^2 + n + 1) - 1\) must be one of the 4-digit numbers listed above. 3. Adding 1 to each of these numbers, we get: \[ 1112, 2223, 3334, 4445, 5556, 6667, 7778, 8889, 10000 \] We need to find which of these numbers is a multiple of 12. 4. Checking each number: - \(1112 \div 12 = 92.6667\) (not an integer) - \(2223 \div 12 = 185.25\) (not an integer) - \(3334 \div 12 = 277.8333\) (not an integer) - \(4445 \div 12 = 370.4167\) (not an integer) - \(5556 \div 12 = 463\) (an integer) - \(6667 \div 12 = 555.5833\) (not an integer) - \(7778 \div 12 = 648.1667\) (not an integer) - \(8889 \div 12 = 740.75\) (not an integer) - \(10000 \div 12 = 833.3333\) (not an integer) The only number that is a multiple of 12 is \(5556\). 5. Setting \(12(n^2 + n + 1) = 5556\), we solve for \(n\): \[ n^2 + n + 1 = \frac{5556}{12} = 463 \] This gives us the quadratic equation: \[ n^2 + n + 1 = 463 \] \[ n^2 + n - 462 = 0 \] 6. Solving the quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-462)}}{2 \cdot 1} \] \[ n = \frac{-1 \pm \sqrt{1 + 1848}}{2} \] \[ n = \frac{-1 \pm \sqrt{1849}}{2} \] \[ n = \frac{-1 \pm 43}{2} \] This gives us two solutions: \[ n = \frac{42}{2} = 21 \quad \text{and} \quad n = \frac{-44}{2} = -22 \] Since \(n\) must be a positive integer, we have \(n = 21\). The final answer is \( \boxed{21} \)." 4d600001f1a7,8.384. $\log _{\sin x} 4 \cdot \log _{\sin ^{2} x} 2=4$.,"$x=(-1)^{k} \frac{\pi}{4}+\pi k, k \in Z$",medium,"Solution. Domain of definition: $0<\sin x<1$. Switch to the base $\sin x$. We have $$ \log _{\sin x} 2^{2} \cdot \frac{1}{2} \log _{\sin x} 2=4, \log _{\sin x}^{2} 2=4 $$ ## Hence 1) $\log _{\sin x} 2=2, \sin ^{2} x=2, \varnothing$; 2) $\log _{\sin x} 2=-2, \frac{1}{\sin ^{2} x}=2, \sin ^{2} x=\frac{1}{2}, \sin x= \pm \frac{\sqrt{2}}{2}$. Considering the domain of definition $\sin x=\frac{\sqrt{2}}{2}, x=(-1)^{k} \frac{\pi}{4}+\pi k, k \in Z$. Answer: $x=(-1)^{k} \frac{\pi}{4}+\pi k, k \in Z$." c9e3fb2515f7,"In 1998, the population of Canada was 30.3 million. Which of the options below represents the population of Canada in 1998? (a) 30300000 (b) 303000000 (c) 30300 (d) 303300 (e) 30300000000","1000000$, we have 30.3 million $=30.3 \times 1000000=30300000$.",easy,"The correct option is (a). Since 1 million $=1000000$, we have 30.3 million $=30.3 \times 1000000=30300000$." c1b291e24494,"1. For a given positive integer $k$, we call an integer $n$ a $k$-number if both of the following conditions are satisfied: (i) The integer $n$ is the product of two positive integers which differ by $k$. (ii) The integer $n$ is $k$ less than a square number. Find all $k$ such that there are infinitely many $k$-numbers.",4$ then setting $m=k+2$ for $k \geq 0$ shows that there are infinitely many 4-numbers.,medium,"Solution Note that $n$ is a $k$-number if and only if the equation $$ n=m^{2}-k=r(r+k) $$ has solutions in integers $m, r$ with $k \geq 0$. The right-hand equality can be rewritten as $$ k^{2}-4 k=(2 r+k)^{2}-(2 m)^{2}, $$ so $k$-numbers correspond to ways of writing $k^{2}-4 k$ as a difference of two squares, $N^{2}-M^{2}$ with $N>r$ and $M$ even (which forces $N$ to have the same parity as $k$ ). Any non-zero integer can only be written as a difference of two squares in finitely many ways (e.g. since the gaps between adjacent squares grow arbitrarily large), so if $k \neq 4$ then there are only finitely many $k$-numbers. Conversely, if $k=4$ then setting $m=k+2$ for $k \geq 0$ shows that there are infinitely many 4-numbers." 671ef077f249,"5. The absolute value of a number $x$ is equal to the distance from 0 to $x$ along a number line and is written as $|x|$. For example, $|8|=8,|-3|=3$, and $|0|=0$. For how many pairs $(a, b)$ of integers is $|a|+|b| \leq 10$ ?",221,medium,"5. Since $|a|$ is at least 0 and $|b|$ is at least 0 and $|a|+|b| \leq 10$, then $|a|$ is at most 10 and $|b|$ is at most 10 . We count the number of possible pairs $(a, b)$ by working through the possible values of $|a|$ from 0 to 10 . Suppose that $|a|=0$. This means that $a=0$. There is 1 possible value of $a$ in this case. Since $|a|=0$ and $|a|+|b| \leq 10$, then $|b| \leq 10$ which means that the possible values for $b$ are $-10,-9,-8, \ldots,-1,0,1, \ldots, 8,9$, or 10 . There are 21 possible values of $b$ in this case. Since there is 1 possible value for $a$ and there are 21 possible values for $b$, then overall there are $1 \times 21=21$ pairs $(a, b)$ when $|a|=0$. Suppose that $|a|=1$. This means that $a=1$ or $a=-1$. There are 2 possible values of $a$ in this case. Since $|a|=1$ and $|a|+|b| \leq 10$, then $|b| \leq 9$ which means that the possible values of $b$ are $-9,-8,-7, \ldots,-1,0,1, \ldots, 7,8$, or 9 . There are 19 possible values of $b$ in this case. Since there are 2 possible values for $a$ and 19 possible values for $b$, then overall there are $2 \times 19=38$ pairs $(a, b)$ when $|a|=1$. Suppose that $|a|=2$. This means that $a=2$ or $a=-2$. There are 2 possible values of $a$ in this case. Here, $|b| \leq 8$ which means that $b$ could equal $-8,-7,-6, \ldots,-1,0,1, \ldots, 6,7$, or 8 . There are 17 possible values of $b$ in this case. Overall, there are $2 \times 17=34$ pairs $(a, b)$ when $|a|=2$. As $|a|$ increases from 2 to 9 , at each step, the largest possible value of $|b|$ will decrease by 1 , which means that there will be 2 fewer possible values of $b$ from each step to the next. Since there are 2 possible values for $a$ at each step, this means that there will be $2 \times 2=4$ fewer pairs $(a, b)$ at each step. We check the final case $|a|=10$ to verify that nothing different happens in the last case. Suppose that $|a|=10$. This means that $a=10$ or $a=-10$. There are 2 possible values of $a$ in this case. Here, $|b| \leq 0$ which means that $b$ can only equal 0 . There is 1 possible value of $b$ in this case. Overall, there are $2 \times 1=2$ pairs $(a, b)$ when $|a|=10$. In total, this means that there are $$ 21+38+34+30+26+22+18+14+10+6+2 $$ pairs $(a, b)$ with $|a|+|b| \leq 10$. Grouping the last 10 numbers in pairs from the outside towards the middle we obtain $$ 21+(38+2)+(34+6)+(30+10)+(26+14)+(22+18) $$ which equals $21+5 \times 40$ or 221 . Thus, there are 221 pairs. (This problem can also be solved using a neat result called Pick's Theorem. We encourage you to look this up and think about how you might apply it here.) ANSWER: 221" f00dcd6932ab,"5. Observe the array: (1),(3,5),(7,9,11),(13,15,17,19), ..... In which group is 2003?","1035$ numbers. The last number of the 45th group is 2069, and the first number is 1981, so, 2003 is ",easy,"(Tip: Use the trial method. The first 45 groups have a total of $1+2+\cdots$ $+45=1035$ numbers. The last number of the 45th group is 2069, and the first number is 1981, so, 2003 is in the 45th group.)" 753092576e27,"7.1. The sum of the minuend, subtrahend, and difference is 25. Find the minuend.",12,easy,"Answer: 12.5. Since the sum of the subtrahend and the difference equals the minuend, the condition means that twice the minuend equals 25, hence the minuend is 12.5." 440157024431,"1. Set $A=\{2,0,1,4\}$, $$ B=\left\{k \mid k \in \mathbf{R}, k^{2}-2 \in A, k-2 \notin A\right\} \text {. } $$ Then the product of all elements in set $B$ is $\qquad$ .",See reasoning trace,easy,"$-、 1.72$ From $k^{2}-2 \in A$, we get $k^{2}=4,2,3,6$, that is, $k= \pm 2, \pm \sqrt{2}, \pm \sqrt{3}, \pm \sqrt{6}$. Since $k-2 \notin A$, thus, $k \neq 2$. Therefore, $k=-2, \pm \sqrt{2}, \pm \sqrt{3}, \pm \sqrt{6}$. Hence, the set $B=\{-2, \pm \sqrt{2}, \pm \sqrt{3}, \pm \sqrt{6}\}$. Thus, the product of all elements in set $B$ is 72." 30fc712484d9,"## Task Condition Write the decomposition of vector $x$ in terms of vectors $p, q, r$: $x=\{-19 ;-1 ; 7\}$ $p=\{0 ; 1 ; 1\}$ $q=\{-2 ; 0 ; 1\}$ $r=\{3 ; 1 ; 0\}$",See reasoning trace,medium,"## Solution The desired decomposition of vector $x$ is: $x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$ Or in the form of a system: $$ \left\{\begin{array}{l} \alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\ \alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\ \alpha \cdot p_{3}+\beta \cdot q_{3}+\gamma \cdot r_{3}=x_{3} \end{array}\right. $$ We obtain: $$ \left\{\begin{array}{l} -2 \beta+3 \gamma=-19 \\ \alpha+\gamma=-1 \\ \alpha+\beta=7 \end{array}\right. $$ Subtract the second row from the third: $$ \left\{\begin{array}{l} -2 \beta+3 \gamma=-19 \\ \alpha+\gamma=-1 \\ \beta-\gamma=8 \end{array}\right. $$ Add the third row multiplied by 2 to the first row: $$ \left\{\begin{array}{l} \gamma=-3 \\ \alpha+\gamma=-1 \\ \beta-\gamma=8 \end{array}\right. $$ $$ \begin{aligned} & \left\{\begin{array}{l} \gamma=-3 \\ \alpha-3=-1 \\ \beta-(-3)=8 \end{array}\right. \\ & \left\{\begin{array}{l} \gamma=-3 \\ \alpha=2 \\ \beta=5 \end{array}\right. \end{aligned} $$ The desired decomposition: $x=2 p+5 q-3 r$ ## Problem Kuznetsov Analytic Geometry 2-7" f62bb5391d48,"8. Given the vertices of $\triangle O A B$ are $O(0,0), A(4,4 \sqrt{3}), B(8,0)$, and its incenter is $I$. Let circle $C$ pass through points $A$ and $B$, and intersect circle $I$ at points $P$ and $Q$. If the tangents to the two circles at points $P$ and $Q$ are perpendicular, then the radius of circle $C$ is $\qquad$ .","\frac{8}{4-t}, \\ y=\frac{\sqrt{3}}{3} x\end{array} \Rightarrow x_{D}=\frac{6}{4-t}+6\right.$, by th",medium,"Given that $C$ is on $O I: y=\frac{\sqrt{3}}{3} x$, let $C\left(t, \frac{\sqrt{3} t}{3}\right)$ $\bigodot I:(x-4)^{2}+\left(y-\frac{4 \sqrt{3}}{3}\right)^{2}=\frac{16}{3}, \bigodot C:$ $$ (x-t)^{2}+\left(y-\frac{\sqrt{3} t}{3}\right)^{2}=(t-8)^{2}+\left(\frac{\sqrt{3} t}{3}\right)^{2} \text {, } $$ Subtracting the two circle equations, we get $P Q: x+\frac{\sqrt{3}}{3} y-8=\frac{8}{4-t}$. Solving the system $\left\{\begin{array}{l}x+\frac{\sqrt{3}}{3} y-8=\frac{8}{4-t}, \\ y=\frac{\sqrt{3}}{3} x\end{array} \Rightarrow x_{D}=\frac{6}{4-t}+6\right.$, by the projection theorem we have $I P^{2}=I D \cdot I C$ $\Rightarrow \frac{16}{3}=\frac{4}{3}\left(\frac{6}{4-t}+6-4\right)(t-4) \Rightarrow t=9$, thus $C A^{2}=1+27=28$. Therefore, the radius of circle $C$ is $2 \sqrt{7}$." 454e29bfa5f1,"A totó requires tipping 12 matches, in each case with the symbols $\gg 1 \ll, » 2 \ll$ or $\gg \mathrm{x} \ll$. What is the expected number of correct tips if we fill out the ticket randomly (e.g., by drawing lots) so that each of the three possibilities has an equal chance for every match?",See reasoning trace,medium,"I. solution: The probability of exactly $k$ (and only $k$) hits $$ \left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k} \quad \text { where } \quad k=0,1,2, \ldots, 12 $$ Since $k$ hits can occur in $\binom{12}{k}$ ways out of 12, the probability of exactly $k$ hits (in any order) is $$ v_{k}=\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k}\binom{12}{k}=\frac{1}{3^{12}} \cdot\binom{12}{k} 2^{12-k} \quad(k=0,1,2, \ldots, 12) $$ Therefore, the expected number of hits is $$ \begin{gathered} M=0 \cdot v_{0}+1 \cdot v_{1}+2 \cdot v_{2}+\ldots+11 \cdot v_{11}+12 \cdot v_{12}= \\ =\frac{1}{3^{12}}\left[1 \cdot\binom{12}{1} 2^{11}+2\binom{12}{2} 2^{10}+\ldots+10\binom{12}{10} 2^{2}+11 \cdot\binom{12}{11} 2+12\binom{12}{12} 2^{0}\right]= \\ =\frac{1}{3^{12}}[24576+135168+337920+506880+506880+354816+177408+ \\ +63360+15840+2640+254+12]=\frac{2125764}{531441}=4 \end{gathered} $$ Antal Rockenbauer (Bp., X., I. László g. II. o. t.) II. solution: We can also reach the goal without any numerical calculations by using the binomial theorem. $$ \begin{aligned} & v_{k}=\binom{12}{k}\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k}=\frac{12!}{k!(12-k)!}\left(\frac{2}{3}\right)^{12-k}\left(\frac{1}{3}\right)^{k}= \\ & =\frac{12 \cdot 11!}{k(k-1)![11-(k-1)]!}\left(\frac{2}{3}\right)^{11-(k-1)}\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{k-1} \end{aligned} $$ From this $$ k v_{k}=12 \cdot \frac{1}{3}\left(\frac{11}{k-1}\right)\left(\frac{2}{3}\right)^{11-(k-1)}\left(\frac{1}{3}\right)^{k-1} \quad(k=1,2, \ldots, 12) $$ Therefore $$ \begin{aligned} & M=4\left[\binom{11}{0}\left(\frac{2}{3}\right)^{11}+\binom{11}{1}\left(\frac{2}{3}\right)^{10} \frac{1}{3}+\binom{11}{2}\left(\frac{2}{3}\right)^{9}\left(\frac{1}{3}\right)^{2}+\ldots+\binom{10}{10} \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{10}+\binom{11}{11}\left(\frac{1}{3}\right)^{11}\right]= \\ & =4\left(\frac{2}{3}+\frac{1}{3}\right)^{11}=4 \end{aligned} $$ Remark: This solution can be easily generalized to $n$ instead of 12, and $v$ instead of $\frac{1}{3}$, in which case for $M$ we get $$ M=n v[(1-v)+v]^{n-1}=n v $$ Zoltán Zsombok (Bp., IV., Kálmán Könyves g. IV. o. t.)" 4f0d7acfb1dc,"11. Satisfy the equation $$ \begin{array}{l} \sqrt{x-2009-2 \sqrt{x-2010}}+ \\ \sqrt{x-2009+2 \sqrt{x-2010}}=2 . \end{array} $$ All real solutions are",See reasoning trace,easy,"$$ \text { II, 11.2 } 010 \leqslant x \leqslant 2011 \text {. } $$ Transform the original equation to get $$ \begin{array}{l} \sqrt{(\sqrt{x-2010}-1)^{2}}+\sqrt{(\sqrt{x-2010}+1)^{2}}=2 \\ \Rightarrow 0 \leqslant \sqrt{x-2010} \leqslant 1 \\ \Rightarrow 2010 \leqslant x \leqslant 2011 . \end{array} $$" f33be2d2d7e8,"p1. Juan and Carlos draw the tadpoles in their design class (they are the ones with contour continuous). Which of the two has a greater area, if all the circles have the same radius and the equilateral triangle has a side equal to one diameter? [img]https://cdn.artof p2. Find some positive $N$ of $10$ digits, all different from $0$ and such that, by adding $N$ with the product of its digits (which we will denote by $P$), we obtain a number $M$ that has a sum of digits equal to $P$. p3. In each square of a $ 3\times 3$ square one of the numbers $1,2,3,...,9$ is written . It is known that the sum of the numbers of each is equal to $15$, and that the same happens with the numbers of each column or diagonal. $\bullet$ What number is in the center box? $\bullet$ Give examples of configurations that verify the properties. p4. A mathematician tells another: ''I think you can guess how many grandchildren I have and what age are, if I tell you that the product of their ages is $36$ and the sum of their ages coincides with the number of floors of the front building.'' His friend thinks for a while and replies: ''Your information is not enough, Caramba!.'' Then, the first adds smiling: '' My older grandson is called Sandro .'' ''Now it is another thing! '' says the second one, and gives the correct answer. What are the ages then? p5. Calculate the area of the triangle $ABC$, knowing that $BD = 4$, $DE = 2$, $EC = 6$, $BF = FC = 3$. [img]https://cdn.artof p6. Maca's family has $4$ people: grandfather, mother, father, and herself. If we double the size of Maca's monthly scholarship, the family income will increase by $5\%$. If instead of the Maca scholarship, her mother's salary would be doubled, the family income will grow by $15\%$. The same procedure gives $25\%$ in the case of Dad now. In what percentage does the family income grow, if only the grandfather's salary is doubled? PS. Juniors P6 was also [url=https://artof",18\sqrt{3,medium,"To solve the problem, we need to calculate the area of triangle \( \triangle ABC \) given the lengths \( BD = 4 \), \( DE = 2 \), \( EC = 6 \), and \( BF = FC = 3 \). 1. **Identify the Equilateral Triangle:** - Given \( BF = FC = 3 \), we can see that \( BFC \) is an equilateral triangle with side length 3. - The area of \( \triangle BFC \) can be calculated using the formula for the area of an equilateral triangle: \[ \text{Area of } \triangle BFC = \frac{\sqrt{3}}{4} \times s^2 = \frac{\sqrt{3}}{4} \times 3^2 = \frac{9\sqrt{3}}{4} \] 2. **Calculate the Area of \( \triangle BCE \):** - Since \( BC = CE = EB = 6 \), \( \triangle BCE \) is also an equilateral triangle with side length 6. - The area of \( \triangle BCE \) is: \[ \text{Area of } \triangle BCE = \frac{\sqrt{3}}{4} \times s^2 = \frac{\sqrt{3}}{4} \times 6^2 = 9\sqrt{3} \] 3. **Calculate the Area of \( \triangle BFD \):** - \( BD = 4 \) and \( BF = 3 \), and the angle between them is \( 60^\circ \) (since \( \triangle BFC \) is equilateral). - The area of \( \triangle BFD \) is: \[ \text{Area of } \triangle BFD = \frac{1}{2} \times BF \times BD \times \sin(60^\circ) = \frac{1}{2} \times 3 \times 4 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \] 4. **Calculate the Area of Quadrilateral \( CFDE \):** - The area of \( \triangle CDE \) can be calculated by subtracting the area of \( \triangle BFD \) from the area of \( \triangle BCE \): \[ \text{Area of } \triangle CDE = \text{Area of } \triangle BCE - \text{Area of } \triangle BFD = 9\sqrt{3} - 3\sqrt{3} = 6\sqrt{3} \] 5. **Calculate the Total Area of \( \triangle ABC \):** - Since \( D \) is the centroid of \( \triangle ABC \), it divides the triangle into three smaller triangles of equal area. - Therefore, the total area of \( \triangle ABC \) is: \[ \text{Area of } \triangle ABC = 2 \times \text{Area of } \triangle BCE = 2 \times 9\sqrt{3} = 18\sqrt{3} \] The final answer is \( \boxed{18\sqrt{3}} \)." 14698c9052d6,"What is the minimum length of the edge of a cube that can fit 2 unit-radius spheres, which do not have any common interior points?",See reasoning trace,medium,"Solution. Let's consider the minimal-edged cube sought in the problem. Its edge length is greater than 2 units, since the edge length of a cube circumscribing a sphere with a radius of 1 unit is exactly this much. Therefore, there exists a smaller cube inside the minimal-edged cube, which is bounded by planes that are 1 unit away from the faces of the larger cube, and these planes separate the center of the cube from the face planes. ![](https://cdn.mathpix.com/cropped/2024_05_02_7fc5086b22abd123f1dcg-1.jpg?height=805&width=808&top_left_y=346&top_left_x=648) A sphere with a radius of 1 unit will not protrude from the larger cube if and only if its center is inside or on the boundary of the smaller cube. Furthermore, the two spheres do not have any common interior points, so the distance between their centers is at least 2 units. Therefore, the smaller cube has two points whose distance is at least 2 units. Among the points of a cube, the farthest apart are the endpoints of a space diagonal, so the space diagonal of the smaller cube is at least 2 units, meaning its edge is at least $2 / \sqrt{3}$ units. The edge of the larger cube is exactly 2 units longer, so it is at least $2 + 2 / \sqrt{3}$ units. In a cube with an edge of $2 + 2 / \sqrt{3}$, two spheres can already be placed such that their centers lie on a space diagonal, 1 unit away from the center of the cube. Based on the work of Gergely Faragó (Bp., Fazekas M. Gyak. Gimn., I. o. t.)" 0d0fd861885d,"4. The apothem of a regular hexagonal pyramid is $m$, and the angle between the base and the lateral face is $\alpha$. Find the area and volume of the pyramid.",553&width=459&top_left_y=257&top_left_x=991),medium,"Solution. Let $M$ and $N$ be the midpoints of sides $A F$ and $C D$ respectively (see diagram). Triangle $M N S$ is isosceles with leg $m=\overline{M S}=\overline{N S}$ and angle at the base $\alpha$. Therefore, $M N=2 m \cos \alpha$. If $a$ is the side of the base, then $$ \overline{M N}=2 \overline{O M}=a \sqrt{3} $$ Thus, $$ a=\frac{2 \sqrt{3}}{3} m \cos \alpha $$ For the area $P$ of the pyramid, we have $$ P=6\left(\frac{a^{2} \sqrt{3}}{4}+\frac{a m}{2}\right)=2 m^{2} \sqrt{3}(1+\cos \alpha) \cos \alpha $$ The height of the pyramid is $H=m \sin \alpha$, so for the volume $V$ we get: $$ V=\frac{2 \sqrt{3}}{3} m^{2} \sin \alpha \cos ^{2} \alpha $$ ## Fourth Year ![](https://cdn.mathpix.com/cropped/2024_06_05_e3f82bd39eecc817d651g-5.jpg?height=553&width=459&top_left_y=257&top_left_x=991)" 9b4891ef65a9,"10.256. The lengths of two sides of an acute triangle are $\sqrt{13}$ and $\sqrt{10} \mathrm{~cm}$. Find the length of the third side, knowing that this side is equal to the height drawn to it.",3 cm,medium,"Solution. Let in $\triangle A B C$ (Fig. 10.61) $A B=\sqrt{10}$ cm, $B C=\sqrt{13}$ cm, height $B D=A C$. Since $\triangle A B C$ is an acute-angled triangle, point $D$ lies on the segment $A C$. Let $B D=A C=x$ cm. From $\triangle A D B\left(\angle A D B=90^{\circ}\right)$: $A D=\sqrt{A B^{2}-B D^{2}}=\sqrt{10-x^{2}} \cdot$ And from $\triangle C D B\left(\angle C D B=90^{\circ}\right): C D=\sqrt{B C^{2}-B D^{2}}=$ $=\sqrt{13-x^{2}} \cdot$ Therefore, $A D+D C=A C ; \sqrt{10-x^{2}}+\sqrt{13-x^{2}}=x$; $\sqrt{13-x^{2}}=x-\sqrt{10-x^{2}} ; 13-x^{2}=x^{2}-2 x \cdot \sqrt{10-x^{2}}+10-x^{2} ; 2 x \sqrt{10-x^{2}}=x^{2}-3$; $\left\{\begin{array}{l}4 x^{2}\left(10-x^{2}\right)=x^{2}-6 x^{2}+9 \\ x^{2} \geq 3 ;\end{array}\left\{\begin{array}{l}5 x^{4}-46 x^{2}+9=0, \\ x^{2} \geq 3 ;\end{array}\left\{\begin{array}{l}{\left[\begin{array}{l}x^{2}=9, \\ x^{2}=\frac{1}{5},\end{array} x^{2}=9\right.} \\ x^{2} \geq 3 ;\end{array}\right.\right.\right.$ From this, $A C=3 \text{ cm}$. Answer: 3 cm." 6231fb966313,"5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$. (points)",2018,medium,"Solution. The number $N$ can be represented as $$ \begin{aligned} & N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\ & =\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2 \cdot 2 \cdot \ldots \cdot 2 \cdot(1 \cdot 2 \cdot \ldots \cdot 2017 \cdot 2018)}{2018!}=(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2^{2018} \end{aligned} $$ We obtained the product of odd numbers and a power of two. Answer: 2018." e3e5cf2469b7,"Bogdanov I.I. On the hypotenuse $AB$ of a right triangle $ABC$, a point $K$ is chosen such that $CK = BC$. The segment $CK$ intersects the bisector $AL$ at its midpoint. Find the angles of triangle $ABC$. #","$36^{\circ}, 54^{\circ}, 90^{\circ}$",medium,"Let $\angle A=2 \alpha$, and $O$ be the intersection point of segments $C K$ and $A L$ (see the figure). Then $C O$ is the median to the hypotenuse of the right triangle $A C L$. Therefore, $A O=O C=O L$, and $\angle O C A=\angle O A C=\angle O A K=\alpha$. ![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-19.jpg?height=372&width=512&top_left_y=543&top_left_x=790) Since triangle $C B K$ is isosceles, $\angle B=\angle B K C=\angle A C K+\angle K A C=3 \alpha$. Thus, $2 \alpha+3 \alpha=\angle A+\angle B=90^{\circ}$. Therefore, $\alpha=18^{\circ}$. Consequently, $\angle B=3 \cdot 18^{\circ}=54^{\circ}$, and $\angle A=2 \cdot 18^{\circ}=36^{\circ}$. ## Answer $36^{\circ}, 54^{\circ}, 90^{\circ}$." 9542d7f1fa7c,"12. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first 15 terms $S_{15}=30$. Then $a_{1}+a_{8}+a_{15}=$ $\qquad$",3\left(a_{1}+7 d\right)=6$.,easy,"12.6. From $S_{15}=30 \Rightarrow a_{1}+7 d=2$. Therefore, $a_{1}+a_{8}+a_{15}=3\left(a_{1}+7 d\right)=6$." fe0c9ba5976a,21st Putnam 1960,"1, 2, 3, 4, 5, let p i (n) = prob that the final score is n+i if the player stops when his total sco",medium,": 2/7. For i = 1, 2, 3, 4, 5, let p i (n) = prob that the final score is n+i if the player stops when his total score is at least n. We note that p(n) is also the probability that the player's total equals n on some throw if he throws repeatedly. Now we can see that p 5 (n) = 1/6 p(n-1), because the only way to achieve a final score of n+5 without passing through n, n+1, n+2, n+3, n+4 is to reach n-1 and then throw a 6. Similarly, p 4 (n) = 1/6 p(n-1) + 1/6 p(n-2), because to reach n+4 without passing through n, n+1, n+2, n+3 you must either to through n-1, which requires reaching n-1 and then throwing a 5, or not, in which case you must reach n-2 and then throw a 6. Similarly, p 3 (n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3), p 2 = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4), and p 1 (n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4) + 1/6 p(n-5). Adding, we get: 1 = p(n) + p 1 (n) + p 2 (n) + p 3 (n) + p 4 (n) + p 5 (n) = p(n) + 5/6 p(n-1) + 4/6 p(n-2) + 3/6 p(n-3) + 2/6 p(n-4) + 1/6 p(n-5). In the limit, p(n-5) = p(n-4) = ... = p(n). Hence we get: p 1 (n) = 5/6 p(n), p 2 (n) = 4/6 p(n), p 3 (n) = 3/6 p(n), p 4 (n) = 2/6 p(n), p 5 (n) = 1/6 p(n). But they sum to 1, so p(n) = 2/7. The objection to the above is that it fails to establish that lim p(n) exists. So one should arguably add the following. We have (as above) p(n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4) + 1/6 p(n-5) + 1/6 p(n-6) (*). Let m(n) = min{ p(n-1), p(n-2), p(n-3), p(n-4), p(n-5), p(n-6) }. Then (*) establishes that p(n) ≥ m(n), and so m(n+1) ≥ m(n). Similarly, let M(n) = max{ p(n-1), p(n-2), p(n-3), p(n-4), p(n-5), p(n-6) }. Then (*) shows that p(n) ≤ M(n), so M(n+1) ≤ M(n). Thus m(n) is a monotonic increasing sequence and M(n) is a monotonic decreasing sequence. But m(n) is obviously bounded above by any M(m), and M(n) is bounded below by any m(m). So both sequences converge. Suppose they converged to different limits. So m(n) converges to m and M(n) converges to M with M - m > 36k > 0. Take n sufficiently large that m(n) > m - 6k. At least one of the terms on the rhs of (*) must equal M(n) and the others are at least m(n), so p(n) ≥ 5/6 m(n) + 1/6 M(n) > 5/6 (m - 6k) + 1/6 M > 5/6 m - 5k + 1/6 (m + 36k) = m+ k. But that means that m(n) > m+k for all sufficiently large n. Contradiction. Hence M and m are the same and p(n) must have the same limit. 21st Putnam 1960 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002" 300ae8ec51e0,"17. (10 points) There are small sticks of lengths 1 cm, 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, and 9 cm, with one stick of each length. Choose several sticks to form a square (no breaking allowed). How many different ways can this be done? Please draw four of these ways and label the lengths of the sticks.",There are 9 ways to form the squares,medium,"【Solution】Solution: $1+2+\cdots+9=45$, The multiples of 4 less than 45 are $4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44$, Therefore, the corresponding side lengths of the squares should be 1 cm, 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm, 10 cm, 11 cm. According to the analysis of the problem, the only squares that can be formed using the sticks mentioned in the problem are those with side lengths of 7 cm, 8 cm, 9 cm, 10 cm, and 11 cm, Side length 7 cm: 7, 1+6, 2+5, 3+4 can form one square Side length 8 cm: 8, 1+7, 2+6, 3+5 can form one square Side length 9 cm: 9, 1+8, 2+7, 3+6, 4+5 can form five squares Side length 10 cm: 1+9, 2+8, 3+7, 4+6 can form one square Side length 11 cm: 2+9, 3+8, 4+7, 5+6 can form one square In total, 9 can be formed. Answer: There are 9 ways to form the squares. The construction methods are shown in the figure below:" b074a339912a,"1. The smallest positive period of the function $f(x)=|\sin 2 x+\cos 2 x|$ is ( ). (A) $2 \pi$ (B) $\pi$ (C) $\frac{\pi}{2}$ (D) $\frac{\pi}{4}$",See reasoning trace,easy,"- 1. C. Notice, $$ \begin{array}{l} f(x)=|\sin 2 x+\cos 2 x| \\ \Leftrightarrow f(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)\right| . \end{array} $$ By the relationship between function graphs, it is known that $f(x)$ and $g(x)=|\sin 2 x|$ have the same smallest positive period. Since the smallest positive period of $|\sin x|$ is $\pi$, therefore, the smallest positive period of $g(x)$ is $\frac{\pi}{2}$." a04d002ebc6e,"12. In the Cartesian coordinate system $x O y$, the area of the plane region corresponding to the point set $S=\left\{(x, y) \mid(|x|+|y|-1)\left(x^{2}+y^{2}-1\right) \leqslant 0\right\}$ is $\qquad$ .",See reasoning trace,easy,"According to the problem, we have \[ \left\{\begin{array}{l} |x|+|y|-1 \geqslant 0, \\ x^{2}+y^{2}-1 \leqslant 0, \end{array}\right. \] The plane region corresponding to $S$ is shown in the shaded area in the figure, and its area is $\pi-2$." f5bfae111fc3,"A trapezoid's base is $a$, the sum of the other three sides is $d$. What are the lengths of the sides if the area of the trapezoid is maximized?",480&width=488&top_left_y=123&top_left_x=816),medium,"A cyclic trapezoid is a trapezoid whose sides are chords of a circle. Since the opposite angles of a cyclic quadrilateral sum to $180^{\circ}$, it follows that the trapezoid is isosceles. Let the larger base of the trapezoid be $AB = a$, the smaller base $DC = c$, and the legs $AD = BC = b$ (Figure 1). According to the condition: $$ d = 2b + c $$ The area of the trapezoid is: $T = \frac{a + c}{2} \cdot m$. By the Pythagorean theorem: $$ m = \sqrt{b^2 - \left(\frac{a - c}{2}\right)^2} $$ From (1), $b = \frac{d - c}{2}$; substituting these values, we get: $$ T = \frac{a + c}{2} \sqrt{\left(\frac{d - c}{2}\right)^2 - \left(\frac{a - c}{2}\right)^2} $$ Transform the expression, using the known identity $x^2 - y^2 = (x + y)(x - y)$: $$ T = \frac{a + c}{4} \sqrt{(d + a - 2c)(d - a)} $$ The value of $(d - a)$ is constant and does not affect the maximum value; $d + a = k$ is the perimeter of the trapezoid. Further transforming the expression, we get: $$ T = \frac{1}{4} \sqrt{d - a} \sqrt{(a + c)(a + c)(k - 2c)} $$ Each factor under the second square root is positive, and we can apply the arithmetic mean-geometric mean inequality to the product of three factors: $$ \begin{gathered} \sqrt[3]{(a + c)(a + c)(k - 2c)} \leq \\ \leq \frac{(a + c) + (a + c) + (k - 2c)}{3} = \frac{k + 2a}{3} \end{gathered} $$ The product is maximized when the factors are equal, i.e., $a + c = k - 2c = d + a - 2c$, from which $3c = d$, and thus $c = \frac{d}{3}$ and $b = \frac{d - c}{2} = \frac{d}{3}$, meaning the legs and the shorter base have the same length. Such a trapezoid exists if $d > a$ (Figure 2). The opposite angles are equal because they correspond to equal-length chords, so $AB \parallel CD$, meaning the quadrilateral is indeed a trapezoid. Katalin Ásványi (Ajka, Bródy I. Gymnasium, 11th grade) Remarks. 1. The problem can also be solved as an extremum problem using derivatives (see the online solution outline). 2. The unusually high number of zero-point solutions is due to the following incorrect conclusions: a) ""Since the perimeter of the quadrilateral is given, and among quadrilaterals with a given perimeter, the square has the largest area, the quadrilateral must be a square."" This reasoning is flawed because not only the perimeter but also one side is given, so (except for the case $d = 3a$) the trapezoid cannot be a square. b) ""Reflecting the cyclic trapezoid over its longer parallel side, we get a hexagon. For a given perimeter, the regular hexagon has the largest area, so the trapezoid has three equal sides, each of length $\frac{d}{3}$."" This error is similar to the previous one: the resulting trapezoid has not only a given perimeter but also a predetermined diagonal, so we cannot refer to the general extremum problem. c) Some students decomposed the area into two independent terms and tried to find the maximum of each term separately, and from this, they attempted to determine the common maximum. ![](https://cdn.mathpix.com/cropped/2024_05_02_fead467f9254c86dacf1g-1.jpg?height=436&width=480&top_left_y=2254&top_left_x=817) ![](https://cdn.mathpix.com/cropped/2024_05_02_fead467f9254c86dacf1g-2.jpg?height=480&width=488&top_left_y=123&top_left_x=816)" fcbc79a2b5b6,"3. Given the complex number $z$ satisfies $3 z^{6}+2 \mathrm{i} z^{5}-2 z-3 \mathrm{i}=$ 0 . Then the modulus of $z$ ( ). (A) greater than 1 (B) equal to 1 (C) less than 1 (D) cannot be determined","1$, i.e., $|z|=1$.",medium,"3. (B). The original equation can be transformed into $z^{5}=\frac{2 z+3 \mathrm{i}}{3 z+2 \mathrm{i}}$. Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$, then $$ \begin{aligned} \left|z^{5}\right| & =\left|\frac{2 a+(2 b+3) \mathrm{i}}{3 a+(3 b+2) \mathrm{i}}\right|=\sqrt{\frac{4 a^{2}+(2 b+3)^{2}}{9 a^{2}+(3 b+2)^{2}}} \\ & =\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} . \end{aligned} $$ If $a^{2}+b^{2}>1$, then the left side of equation (1) $=|z|^{5}>1$, while the right side $1$, so equation (1) does not hold. Therefore, only $a^{2}+b^{2}=1$, i.e., $|z|=1$." 69f60c1b1b81,"6. As shown in Figure 1, the radius of $\odot O$ is $6, M$ is a point outside $\odot O$, and $O M=12$. A line through $M$ intersects $\odot O$ at points $A, B$. The points $A, B$ are symmetric to points $C, D$ with respect to $O M$. $A D$ and $B C$ intersect at point $P$. Then the length of $O P$ is ( ) (A) 4 (B) 3.5 (C) 3 (D) 2.5",O M-P M=3$.,easy,"6. C. As shown in Figure 4, extend $M O$ to intersect $\odot O$ at point $N$, and connect $O B$ and $O D$. By symmetry, point $P$ lies on line $M N$. Since $\angle B A D=\frac{1}{2} \angle B O D=\angle B O N$, points $A$, $B$, $O$, and $P$ are concyclic. Thus, $O M \cdot P M=M A \cdot M B=O M^{2}-O N^{2}$ $\Rightarrow P M=O M-\frac{O N^{2}}{O M}=12-3=9$. Therefore, $O P=O M-P M=3$." 0d0987d2e5d8,"Question 11: Given the mapping f: $\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ is both injective and surjective, and satisfies $f(x)+f(f(x))=6$, then $f(1)=$ $\qquad$ _.",5$.,easy,"Question 11, Solution: Let $S=\{1,2,3,4,5\}$, for any $\mathrm{a} \in S$, according to the conditions: $$ \left\{\begin{array}{c} f(a)+f(f(a))=6 \\ f(f(a))+f(f(f(a)))=6 \end{array}\right. $$ Thus, $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{a})))$. Since $f$ is a bijection, $\{f(a) \mid a \in S\}=S$, hence for any $a \in S$, we have $a=f(f(a))$. Therefore, $6=f(a)+f(f(a))=f(a)+a$, which means $f(a)=6-a$. Based on the above analysis, we know $f(1)=5$." 851ab3ca1467,"11.5. Find all numbers $a$ such that for any natural $n$ the number $a n(n+2)(n+3)(n+4)$ is an integer. (O. Podlipsky)",". $a=\frac{k}{6}$, where $k$ is any integer",medium,"Answer. $a=\frac{k}{6}$, where $k$ is any integer. Solution. Substituting $n=1, n=3$ and $n=4$, we get that the numbers $60a, 630a$, and $24 \cdot 56a$ are integers. Therefore, $a$ is a rational number, and the denominator $q$ in its irreducible form is a divisor of the numbers 60, 630, and $24 \cdot 56$. Consequently, their greatest common divisor (GCD) is also divisible by $q$. Since GCD $(60,630)=$ $=30$, and $24 \cdot 56$ is not divisible by 5, then $q$ is a divisor of the number 6, and $a=k / 6$ for some integer $k$. It remains to show that all numbers of this form are suitable. Indeed, one of the three consecutive numbers $n+2, n+3$, $n+4$ is divisible by 3, and one of the consecutive numbers $n+2$, $n+3$ is divisible by 2; therefore, $n(n+2)(n+3)(n+4)$ is divisible by 2 and by 3, and hence by 6. Therefore, $\operatorname{an}(n+2)(n+3)(n+4)=$ $=k \frac{n(n+2)(n+3)(n+4)}{6}-$ is an integer. Comment. It is proven that any number of the specified form is suitable - 2 points. It is proven that the number $a$ must have the specified form - 4 points. Only the correct answer - 1 point (this point is not combined with others)." 90e66a1cdfcd,"4. The bases $AB$ and $CD$ of the trapezoid $ABCD$ are extended on both sides. The bisectors of the exterior angles of the trapezoid at vertices $A$ and $D$ intersect at point $M$, and the bisectors of the exterior angles at vertices $B$ and $C$ intersect at point $N$. Determine the perimeter of trapezoid $ABCD$ if $MN=2k$.",See reasoning trace,medium,"Solution. Since point $M$ lies on the angle bisectors at vertices $B$ and $C$, it is equidistant from line $A B$ and line $C D$, which means that it lies on the midline of trapezoid $A B C D$. Similarly, point $N$ lies on the midline of trapezoid $A B C D$. Furthermore, triangles $N A P$ and $N D P$ are isosceles (Why?), so $N P=A P=P D$. Similarly, $M Q=B Q=C Q$. Finally, for the perimeter of the trapezoid, we get: $$ \begin{aligned} A B+C D+A D+B C & =2 \cdot\left(\frac{A B}{2}+\frac{C D}{2}+A P+B Q\right) \\ & =2 \cdot\left(\frac{A B+C D}{2}+N P+Q M\right) \\ & =2(P Q+N P+Q M) \\ & =2 M N=4 k \end{aligned} $$" 5b95ab605f3a,"# Task 4. In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the letter code. Then, the weight of a word can be defined as the sum of the codes of all the letters in that word. Is it possible to encode the letters O, P, S, T, Ь, Я with elementary codes, each consisting of a single digit from 0 to 9, so that the weight of the word ""СТО"" is not less than the weight of the word ""ПЯТЬСОТ""? If such encoding is possible, does it allow for the unambiguous restoration of a word from its code?",See reasoning trace,medium,"# Solution. Let $k(x)$ denote the elementary code of the letter $x$. We have: $$ k(C)+k(T)+k(O) \geq k(\Pi)+k(\text { ( })+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T) $$ which is equivalent to $$ k(\Pi)+k(T)+k(\mathrm{~b})+k(\text { Я) }=0 $$ from which it follows that $$ k(\Pi)=k(T)=k(\mathrm{~b})=k(\text { Я })=0 $$ The elementary codes of the two remaining letters $k(C)$ and $k(O)$ can be any digits from 0 to 9. Therefore, there are exactly $10^{2}$ ways to encode them. However, for any choice of $k(O)$ and $k(C)$, the words ""ПОП"" and ""ТОТ"" will have the same codes $k(O)$, which cannot be decoded uniquely. Answer. The inequality for the weights holds if and only if $k(\Pi)=k(T)=k(\mathrm{~b})=k($ Я $)=0$ There are exactly 100 ways to encode them. In any of these ways, unique decoding is impossible." 07c7e5edf90c,Each term of a sequence of positive integers is obtained from the previous term by adding to it its largest digit. What is the maximal number of successive odd terms in such a sequence?,5,medium,"1. **Understanding the Problem:** Each term of a sequence of positive integers is obtained from the previous term by adding to it its largest digit. We need to determine the maximal number of successive odd terms in such a sequence. 2. **Initial Observations:** - If a number is odd, its largest digit must be added to it to form the next term. - For the next term to remain odd, the largest digit added must be even. - The largest digit of an odd number cannot be 0 (since the number is positive and odd). 3. **Case Analysis:** - **Max digit is 2:** - If the last digit is 1, the next number ends with \(1 + 2 = 3\). The new max digit is 3, and the next number will be even. Thus, we get two successive odd numbers. - **Max digit is 4:** - If the last digit is 1, the next number ends with \(1 + 4 = 5\). The new max digit is 5, and the next number will be even. Thus, we get two successive odd numbers. - If the last digit is 3, the next number ends with \(3 + 4 = 7\). The new max digit is 7, and the next number will be even. Thus, we get two successive odd numbers. - **Max digit is 6:** - If the last digit is 1, the next number ends with \(1 + 6 = 7\). The new max digit is 7, and the next number will be even. Thus, we get two successive odd numbers. - If the last digit is 3, the next number ends with \(3 + 6 = 9\). The new max digit is 9, and the next number will be even. Thus, we get two successive odd numbers. - If the last digit is 5, the next number ends with \(5 + 6 = 11\). The new max digit is 1, and the next number will be odd. Thus, we get three successive odd numbers. - **Max digit is 8:** - If the last digit is 1, the next number ends with \(1 + 8 = 9\). The new max digit is 9, and the next number will be even. Thus, we get two successive odd numbers. - If the last digit is 3, the next number ends with \(3 + 8 = 11\). The new max digit is 1, and the next number will be odd. Thus, we get three successive odd numbers. - If the last digit is 5, the next number ends with \(5 + 8 = 13\). The new max digit is 3, and the next number will be odd. Thus, we get four successive odd numbers. - If the last digit is 7, the next number ends with \(7 + 8 = 15\). The new max digit is 5, and the next number will be odd. Thus, we get five successive odd numbers. 4. **Conclusion:** - The maximum number of successive odd terms in such a sequence is 5. This can be achieved with the sequence \(807, 815, 823, 831, 839\). The final answer is \(\boxed{5}\)." 170818bdd438,"6. (15 points) A pedestrian is moving towards a crosswalk along a straight path at a constant speed of 3.6 km/h. At the initial moment, the distance from the pedestrian to the crosswalk is 20 m. The length of the crosswalk is $5 \mathrm{~m}$. At what distance from the crosswalk will the pedestrian be after half a minute?",5 m,easy,"Answer: 5 m. Solution. The pedestrian's speed is 3.6 km/h = 1 m/s. In half a minute, he walked $s = v t = 1 \cdot 30 s = 30$ m. From the crossing, he is at a distance of $l = s - 20 - 5 = 5$ m. ## Tasks, answers, and evaluation criteria" ee8813b2cc15,"## Calculate the area of the figure bounded by the graphs of the functions: $$ x=\frac{1}{y \sqrt{1+\ln y}}, x=0, y=1, y=e^{3} $$",See reasoning trace,medium,"## Solution The desired area $S$ is: $S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y$ We make a substitution of variables: $t=\ln y$, hence $d t=\frac{d y}{y}$ When $t=\ln 1 \Rightarrow t=0$ and when $t=\ln e^{3} \Rightarrow t=3$ Then we get ![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-47.jpg?height=1357&width=1168&top_left_y=989&top_left_x=752) $$ \begin{aligned} & S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y= \\ & =\int_{0}^{3} \frac{1}{\sqrt{t+1}} d t=\left.2 \sqrt{1+t}\right|_{0} ^{3}=2 \sqrt{4}-2 \sqrt{1}=2 \end{aligned} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-25$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals Ukrainian Banner Network - Last edited on this page: 10:36, 11 June 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 14-26 ## Material from PlusPi" 6ffe335ec789,"3. In a circle of radius 3, triangles $A B C$ and $A M N$ are inscribed. The line $A M$ passes through the midpoint $E$ of segment $B C$, and the line $B N$ passes through the midpoint $F$ of segment $A C$. Find the perimeter of triangle $A B C$, if $A M: A E=2: 1$ and $B N: B F=17: 13$.",$\frac{72}{5}$,medium,"# Problem 3. Answer: $\frac{72}{5}$. Solution. According to the intersecting chords theorem: $A E \cdot E M = B E \cdot E C$. But since by the condition $A E = E M$ and $B E = E C$, then $A E = B E = C E$, i.e., $E$ is the center of the circumscribed circle around $\triangle A B C$. Therefore, angle $A$ is a right angle. Similarly, $A F \cdot F C = B F \cdot F N$ and by the condition: $B F = 13 x, F N = 4 x$, then $A F = F C = 2 \sqrt{13} x$. Consequently, from the right triangle $A B F: A B^{2} = B F^{2} - A F^{2} = 9 \cdot 13 x^{2}$, and from triangle $A B C: 9 \cdot 13 x^{2} + 16 \cdot 13 x^{2} = 36 \Rightarrow x = \frac{6}{5 \sqrt{13}}$. Therefore, the perimeter is $7 \sqrt{13} x + 6 = \frac{72}{5}$." 172ef42367e1,"5. Given a real number $x$ between 0 and 1, consider its decimal representation $0, c_{1} c_{2} c_{3} \ldots$ We call $B(x)$ the set of different subsequences of six consecutive digits that appear in the sequence $c_{1} c_{2} c_{3} \ldots$ For example, $B(1 / 22)=\{045454,454545,545454\}$. Determine the minimum number of elements of $B(x)$ as $x$ varies among the irrational numbers between 0 and 1 (i.e., those whose decimal expansion is neither finite nor periodic from a certain point onward.)",1$ is immediate. Suppose we know the statement for $k$ and prove that it holds for $k+1$. We will re,medium,"SOLUTION: The number sought is seven. To prove this, we will show that (a) there exists an irrational number $x$ such that $B(x)$ has no more than seven elements, and that (b) every real number $y$ such that $B(y)$ has six or fewer elements is rational. Part (a). Consider a number $x$ constructed as follows. The first digit in the decimal expansion of $x$ is a 1, followed by five zeros, then another 1, followed by six zeros, then another 1 followed by seven zeros, and so on, with the decimal expansion of $x$ consisting of 1s interspersed with sequences of zeros of increasing length. Clearly, $$ B(x)=\{000000,000001,000010,000100,001000,010000,100000\} $$ Moreover, $x$ is irrational. Indeed, if $x$ were rational, then from a certain digit onward, say the $k$-th, its decimal expansion would be periodic. Let $l$ be the length of the period. Since the expansion of $x$ contains infinitely many sequences of zeros of length greater than $2l$, one of these would occur after the $k$-th digit and would contain a complete period. This implies that the period would consist of $l$ zeros, which contradicts the fact that the expansion of $x$ contains infinitely many 1s. Part (a), alias. Let $[x]_{i}$ denote the $i$-th digit after the decimal point of a number $x$ between 0 and 1. Consider any irrational number $\alpha$ in this interval, for example, $\alpha=\sqrt{2} / 2$. For $i=1 \ldots 6$ and $j=0 \ldots 9$, define the numbers $\alpha_{i, j}$ as follows: $$ \left[\alpha_{i, j}\right]_{k}= \begin{cases}1 & \text { if } k=6 n+i \text { for some } n \text { and }[\alpha]_{k}0$, the graph of $y=A x^{2}$ is a parabola opening upwards with its vertex at the origin. Therefore, the parabola intersects the lower branch of the hyperbola at two distinct points. The upper branch of the hyperbola is given by $y=2+\sqrt{x^{2}+1}$. For any given positive number $A$, if $|x|$ is sufficiently large, then $A x^{2}>2+\sqrt{x^{2}+1}$ $\approx 2+x$. On the other hand, when $x=0$, $A x^{2}<2+\sqrt{x^{2}+1}$. Thus, for any given positive number $A$, the parabola must intersect the upper branch of the hyperbola at two distinct points. (A) True." 61f76aab47d3,"Let $b_m$ be numbers of factors $2$ of the number $m!$ (that is, $2^{b_m}|m!$ and $2^{b_m+1}\nmid m!$). Find the least $m$ such that $m-b_m = 1990$.",2^{1990,medium,"1. We start by defining \( b_m \) as the number of factors of 2 in \( m! \). This means \( 2^{b_m} \mid m! \) and \( 2^{b_m+1} \nmid m! \). 2. The number of factors of 2 in \( m! \) can be calculated using the formula: \[ b_m = v_2(m!) = \left\lfloor \frac{m}{2} \right\rfloor + \left\lfloor \frac{m}{4} \right\rfloor + \left\lfloor \frac{m}{8} \right\rfloor + \cdots \] 3. There is a well-known relation that \( v_2(m!) = m - s_2(m) \), where \( s_2(m) \) is the sum of the digits of \( m \) in binary representation. This is because each power of 2 contributes to the sum of the digits in binary. 4. We are given that \( m - b_m = 1990 \). Substituting \( b_m = m - s_2(m) \) into this equation, we get: \[ m - (m - s_2(m)) = 1990 \] Simplifying, we find: \[ s_2(m) = 1990 \] 5. The sum of the digits in the binary representation of \( m \) is 1990. Since the binary representation consists only of 1's and 0's, the smallest \( m \) with 1990 ones is the number with 1990 ones in binary, which is: \[ m = 2^{1990} - 1 \] 6. To verify, the binary representation of \( 2^{1990} - 1 \) is a string of 1990 ones. The sum of these digits is indeed 1990, satisfying \( s_2(m) = 1990 \). Conclusion: \[ \boxed{2^{1990} - 1} \]" 71ea83f59d98,"4. How many orderings $\left(a_{1}, \ldots, a_{8}\right)$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$ ?",4608 We can divide the numbers up based on whether they have a + or - before them,medium,"Answer: 4608 We can divide the numbers up based on whether they have a + or - before them. Both the numbers following +'s and -'s must add up to 18. Without loss of generality, we can assume the +'s contain the number 1 (and add a factor of 2 at the end to account for this). The possible 4 -element sets containing a 1 which add to 18 are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\}$. Additionally, there are 4 ! ways to order the numbers following a + and 4 ! ways to order the numbers following $\mathrm{a}-$. Thus the total number of possibilities is $4 \times 2 \times 4!\times 4!=4608$." e2fd480f0e98,"8. Given the right triangle $A B C$ with a right angle at $A$, we construct a square $B C D E$ on the hypotenuse (with $D, E$ on the opposite side of $A$ relative to $B C$). Knowing that the areas of triangles $A B E$ and $A C D$ are $6 \mathrm{~m}^{2}$ and $27 \mathrm{~m}^{2}$ respectively, what is the area of triangle $A B C$? (A) $3 \sqrt{2} m^{2}$ (B) $6 \mathrm{~m}^{2}$ (C) $12 \mathrm{~m}^{2}$ (D) $9 \sqrt{2} \mathrm{~m}^{2}$ (E) $18 \mathrm{~m}^{2}$",$(\mathbf{D})$,medium,"8. The answer is $(\mathbf{D})$. FIRST SOLUTION We show that the product of the two given areas is equal to the square of the area of the triangle $A B C$ we are looking for. Let $P$ be the foot of the altitude from vertex $A$ of triangle $A C D$, $Q$ the foot of the altitude from vertex $A$ of triangle $A B E$, and $H$ the foot of the altitude from vertex $A$ of triangle $A B C$. The quadrilateral $A P C H$ is, by construction, a rectangle (the angles at $P$ and $H$ are right angles because they are formed by altitudes, the angle at $C$ is complementary to a right angle) so $A P$ is congruent to $C H$; similarly, $A Q$ is congruent to $B H$. We therefore know that $S(A C D)=\frac{1}{2} C D \cdot C H$ and $S(A B E)=\frac{1}{2} B E \cdot B H$; but, since $C D=B E=C B$ and since $C H \cdot B H=A H$ by the theorem of Euclid, the product of the two areas is $$ \frac{C D \cdot C H}{2} \frac{B E \cdot B H}{2}=\frac{C B^{2} \cdot C H \cdot B H}{4}=\left(\frac{C B \cdot A H}{2}\right)^{2}=S(A B C)^{2} . $$ Therefore, the area of $A B C$ is $\sqrt{27 m^{2} \cdot 6 m^{2}}=9 \sqrt{2} m^{2}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_8e3f264af5d5452f36c1g-10.jpg?height=620&width=1406&top_left_y=427&top_left_x=354) ## SECOND SOLUTION Draw the square $A X Y Z$ such that $B$ is on side $Z A$, $C$ is on side $A X$, $D$ is on side $X Y$, and $E$ is on side $Y Z$. Note that the triangles $A B C$, $X C D$, $D Y E$, and $E Z B$ are congruent: they have congruent angles and the same hypotenuse. On the other hand, $D X$ (which is congruent to $A C$) is the altitude relative to $A C$ in triangle $A C D$, so the area of triangle $A C D$ is $A C^{2} / 2$. Similarly, since $C Z$ and $A B$ are congruent, the area of triangle $A B E$ is $A B^{2} / 2$. Therefore, for the area $S$ of triangle $A B C$ we have $$ S=\frac{A B \cdot A C}{2}=\sqrt{\frac{A B^{2}}{2} \cdot \frac{A C^{2}}{2}}=\sqrt{27 m^{2} \cdot 6 m^{2}}=9 \sqrt{2} \mathrm{~m}^{2} . $$" fe02f75dd8e0,"9. [7] Let $n$ be an integer and $$ m=(n-1001)(n-2001)(n-2002)(n-3001)(n-3002)(n-3003) \text {. } $$ Given that $m$ is positive, find the minimum number of digits of $m$.",See reasoning trace,easy,"Answer: 11 Solution: One can show that if $m>0$, then we must either have $n>3003$ or $n1000^{5}$, meaning that $m$ has at least 16 digits. However, if $n>3003$, it is clear that the minimal $m$ is achieved at $n=3004$, which makes $$ m=2002 \cdot 1002 \cdot 1001 \cdot 3 \cdot 2 \cdot 1=12 \cdot 1001 \cdot 1001 \cdot 1002, $$ which is about $12 \cdot 10^{9}$ and thus has 11 digits." 616a50106bc9,"What are the signs of $\triangle{H}$ and $\triangle{S}$ for a reaction that is spontaneous only at low temperatures? $ \textbf{(A)}\ \triangle{H} \text{ is positive}, \triangle{S} \text{ is positive} \qquad\textbf{(B)}\ \triangle{H}\text{ is positive}, \triangle{S} \text{ is negative} \qquad$ $\textbf{(C)}\ \triangle{H} \text{ is negative}, \triangle{S} \text{ is negative} \qquad\textbf{(D)}\ \triangle{H} \text{ is negative}, \triangle{S} \text{ is positive} \qquad $",\textbf{(C),medium,"To determine the signs of $\Delta H$ and $\Delta S$ for a reaction that is spontaneous only at low temperatures, we need to analyze the Gibbs free energy equation: \[ \Delta G = \Delta H - T \Delta S \] A reaction is spontaneous if $\Delta G$ is negative. Let's consider the conditions under which $\Delta G$ can be negative: 1. **Low Temperature Condition**: - For the reaction to be spontaneous only at low temperatures, $\Delta G$ must be negative when $T$ is small. - This implies that the term $-T \Delta S$ should not dominate the $\Delta H$ term at low temperatures. 2. **Sign Analysis**: - If $\Delta H$ is negative and $\Delta S$ is positive, the reaction would be spontaneous at all temperatures because both terms would contribute to making $\Delta G$ negative. - If $\Delta H$ is positive and $\Delta S$ is negative, the reaction would not be spontaneous at any temperature because both terms would contribute to making $\Delta G$ positive. - If $\Delta H$ is positive and $\Delta S$ is positive, the reaction would be spontaneous at high temperatures because the $-T \Delta S$ term would eventually dominate and make $\Delta G$ negative. - If $\Delta H$ is negative and $\Delta S$ is negative, the reaction would be spontaneous at low temperatures because the $-T \Delta S$ term would be small and $\Delta H$ would dominate, making $\Delta G$ negative. Given that the reaction is spontaneous only at low temperatures, the correct signs for $\Delta H$ and $\Delta S$ are: \[ \Delta H < 0 \quad \text{and} \quad \Delta S < 0 \] Thus, the correct answer is: \[ \boxed{\textbf{(C)}\ \Delta H \text{ is negative}, \Delta S \text{ is negative}} \]" 4f2b54652e3d,"Suppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2-44x+12=p^n$. Find the largest value of $p$.",47,medium,"1. We start with the given equation: \[ 7x^2 - 44x + 12 = p^n \] We can factorize the left-hand side: \[ 7x^2 - 44x + 12 = (x-6)(7x-2) \] Therefore, we have: \[ (x-6)(7x-2) = p^n \] 2. We need to find the largest prime number \( p \) such that the equation holds for some integer \( x \) and natural number \( n \). 3. Consider the case when \( x = 7 \): \[ 7(7)^2 - 44(7) + 12 = 7 \cdot 49 - 308 + 12 = 343 - 308 + 12 = 47 \] Here, \( 47 \) is a prime number, and we can write: \[ 47 = 47^1 \] Thus, \( p = 47 \) and \( n = 1 \). 4. Next, we need to check if there could be any larger prime \( p \) that satisfies the equation. For this, we analyze the factors of \( (x-6) \) and \( (7x-2) \). 5. Since \( (x-6)(7x-2) = p^n \), both \( x-6 \) and \( 7x-2 \) must be powers of \( p \). Let: \[ x - 6 = p^a \quad \text{and} \quad 7x - 2 = p^b \] for some non-negative integers \( a \) and \( b \). 6. From \( x - 6 = p^a \), we get: \[ x = p^a + 6 \] Substituting \( x \) in \( 7x - 2 = p^b \): \[ 7(p^a + 6) - 2 = p^b \] Simplifying: \[ 7p^a + 42 - 2 = p^b \] \[ 7p^a + 40 = p^b \] 7. We need to check if there are any values of \( p \) larger than 47 that satisfy this equation. For \( p > 47 \), the left-hand side \( 7p^a + 40 \) grows much faster than the right-hand side \( p^b \), making it unlikely for the equation to hold for any \( a \) and \( b \). 8. Therefore, the largest prime \( p \) that satisfies the equation is \( p = 47 \). The final answer is \( \boxed{47} \)" b16f508dab8b,"4-15 Given that $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ are the roots of $x^{2}+p x+q=0$, try to find $$ \sin ^{2}(\alpha+\beta)+p \sin (\alpha+\beta) \cos (\alpha+\beta)+q \cos ^{2}(\alpha+\beta) \text { } $$ value.",See reasoning trace,medium,"[Solution] By the relationship between roots and coefficients, we have $$ \operatorname{tg} \alpha + \operatorname{tg} \beta = -p \text{ and } \operatorname{tg} \alpha \operatorname{tg} \beta = q, $$ Therefore, $\operatorname{tg}(\alpha+\beta) = \frac{-p}{1-q}$. Thus, the value we seek is $$ \begin{aligned} & \sin ^{2}(\alpha+\beta) + p \sin (\alpha+\beta) \cos (\alpha+\beta) + q \cos ^{2}(\alpha+\beta) \\ = & \cos ^{2}(\alpha+\beta)\left[\operatorname{tg}^{2}(\alpha+\beta) + p \operatorname{tg}(\alpha+\beta) + q\right] \\ = & \frac{1}{1+\operatorname{tg}^{2}(\alpha+\beta)}\left[\operatorname{tg}^{2}(\alpha+\beta) + p \operatorname{tg}(\alpha+\beta) + q\right] \\ = & \frac{1}{1+\frac{p^{2}}{(1-q)^{2}}}\left[\frac{p^{2}}{(1-q)^{2}} - \frac{p^{2}}{1-q} + q\right] \\ = & q . \end{aligned} $$" 6cf7fe69329a,3.32. At the base of a right prism lies a rhombus with an acute angle $\alpha$. The ratio of the height of the prism to the side of the base is $k$. A plane is drawn through the side of the base and the midpoint of the opposite lateral edge. Find the angle between this plane and the plane of the base.,See reasoning trace,medium,"3.32. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right prism, $A B C D$ is a rhombus, $\angle B A D=\alpha\left(\alpha<90^{\circ}\right), A A_{1}: A B=k, C_{1} E=E C, A D E F$ is a section of the prism (Fig. 3.34); it is required to find the angle between the section and the base $(A B C)$; (FAD). Let $A B=a$; then $A A_{1}=k a$. Draw $B K \perp A D$; then $F K \perp A D$ (by the theorem of three perpendiculars) and $\angle F K B$ is the linear angle between the section and the base $A B C D$. From $\triangle A K B$ we find $B K=a \sin \alpha ;$ since $E F \| A D$, then $F B=\frac{1}{2} B B_{1}=\frac{1}{2} k a$. Finally, from $\triangle F B K$ we get $\operatorname{tg} \angle F K B=\frac{F B}{B K}=$ $=\frac{k a}{2 a \sin \alpha}=\frac{k}{2 \sin \alpha}$, hence $\angle F K B=\operatorname{arctg} \frac{k}{2 \sin \alpha}$. ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-519.jpg?height=529&width=654&top_left_y=107&top_left_x=256) Fig. 3.34 ![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-519.jpg?height=506&width=515&top_left_y=1135&top_left_x=58) Fig. 3.35" d73bd4864c4d,"1.7 density, 400 gr weight liquid is mixed with 1.2 density $600 \mathrm{gr}$ solution. Measuring the density $D$ of the newly formed solution, we find that $D=1.4$. What is the contraction that occurs when mixing the two initial liquids?",See reasoning trace,medium,"If there were no contraction, then the density $\delta$ of the 1000 g mixture could be calculated from the equality of volumes: $$ \frac{1000}{\delta}=\frac{400}{1.7}+\frac{600}{1.2}, \text { hence } \delta=1.36 $$ Due to the contraction, the density is greater. Therefore, the volume of the mixture is: $\frac{1000}{D}=\frac{1000}{1.4}$, and thus the contraction is: $$ \Delta V=\frac{400}{1.7}+\frac{600}{1.2}-\frac{1000}{1.4}=735.3-714.3=21 \mathrm{~cm}^{3} $$ Branyitskai Nagy Iván (Ág. ev. g. VIII. o. Bp.)" 8a745bb011ab,"5. Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy $$ \begin{array}{l} a_{1}=\sqrt{2}-1, \\ a_{n+1}=\frac{2 n+1}{S_{n}+S_{n+1}+2}(n=1,2, \cdots), \end{array} $$ where $S_{n}$ is the sum of the first $n$ terms of $a_{n}$. Then the general term formula of the sequence is $\qquad$ .",S_{n}-S_{n-1}=\sqrt{n^{2}+1}-\sqrt{(n-1)^{2}+1}$.,medium,"5. $a_{n}=\sqrt{n^{2}+1}-\sqrt{(n-1)^{2}+1}$. By $a_{n+1}=S_{n+1}-S_{n}=\left(S_{n+1}+1\right)-\left(S_{n}+1\right)$ $\Rightarrow\left(S_{n+1}+1\right)-\left(S_{n}+1\right)=\frac{2 n+1}{\left(S_{n+1}+1\right)+\left(S_{n}+1\right)}$ $\Rightarrow\left(S_{n+1}+1\right)^{2}-\left(S_{n}+1\right)^{2}=(n+1)^{2}-n^{2}$. Let $T_{n}=\left(S_{n}+1\right)^{2}-n^{2}$. Then $T_{n+1}=T_{n}$, i.e., $\left\{T_{n}\right\}$ is a constant sequence. It is easy to see that $T_{1}=1$. Therefore, $\left(S_{n}+1\right)^{2}-n^{2}=1$ $\Rightarrow S_{n}=\sqrt{n^{2}+1}-1(n=1,2, \cdots)$. Notice that, when $n=0$, $S_{0}=0$ also satisfies. Hence $a_{n}=S_{n}-S_{n-1}=\sqrt{n^{2}+1}-\sqrt{(n-1)^{2}+1}$." 9d0bff771b5d,"There are $1000$ students in a school. Every student has exactly $4$ friends. A group of three students $ \left \{A,B,C \right \}$ is said to be a [i]friendly triplet[/i] if any two students in the group are friends. Determine the maximal possible number of friendly triplets. [i]Proposed by Nikola Velov[/i]",2000,medium,"1. **Graph Interpretation**: - Consider the problem in terms of graph theory. Let \( G \) be a graph where each vertex represents a student and each edge represents a friendship. - Given that there are \( n = 1000 \) students and each student has exactly \( d = 4 \) friends, \( G \) is a 4-regular graph with 1000 vertices. 2. **Counting Triangles**: - A *friendly triplet* corresponds to a triangle in the graph \( G \). - For a vertex \( v \) in \( G \), the number of triangles that include \( v \) is determined by the number of edges between the neighbors of \( v \). - Since \( v \) has 4 neighbors, the number of edges between these neighbors is at most \( \binom{4}{2} = 6 \). 3. **Total Number of Triangles**: - Each triangle in \( G \) is counted three times (once at each vertex of the triangle). - Therefore, the total number of triangles in \( G \) is at most: \[ \frac{1}{3} \sum_{v \in V(G)} \binom{d}{2} \] - Since \( G \) is 4-regular, each vertex \( v \) has exactly 4 neighbors, and thus: \[ \binom{4}{2} = 6 \] - The sum over all vertices is: \[ \sum_{v \in V(G)} \binom{4}{2} = 1000 \times 6 = 6000 \] - Dividing by 3 to account for the overcounting of each triangle: \[ \frac{6000}{3} = 2000 \] 4. **Achievability**: - The equality case occurs when \( G \) can be decomposed into disjoint copies of \( K_5 \) (the complete graph on 5 vertices), as each \( K_5 \) contains exactly \( \binom{5}{3} = 10 \) triangles. - Since \( K_5 \) is 4-regular and has 5 vertices, we can have \( \frac{1000}{5} = 200 \) copies of \( K_5 \). - Each \( K_5 \) contributes 10 triangles, so the total number of triangles is: \[ 200 \times 10 = 2000 \] Conclusion: \[ \boxed{2000} \]" cfe1800b645a,"Elmo bakes cookies at a rate of one per 5 minutes. Big Bird bakes cookies at a rate of one per 6 minutes. Cookie Monster [i]consumes[/i] cookies at a rate of one per 4 minutes. Together Elmo, Big Bird, Cookie Monster, and Oscar the Grouch produce cookies at a net rate of one per 8 minutes. How many minutes does it take Oscar the Grouch to bake one cookie?",120,medium,"1. Let's denote the rates of baking and consuming cookies as follows: - Elmo's rate: $\frac{1}{5}$ cookies per minute - Big Bird's rate: $\frac{1}{6}$ cookies per minute - Cookie Monster's rate: $-\frac{1}{4}$ cookies per minute (since he consumes cookies) - Oscar the Grouch's rate: $\frac{1}{o}$ cookies per minute 2. According to the problem, the net rate of producing cookies by all four together is $\frac{1}{8}$ cookies per minute. Therefore, we can set up the equation: \[ \frac{1}{5} + \frac{1}{6} + \frac{1}{o} - \frac{1}{4} = \frac{1}{8} \] 3. To solve this equation, we first find a common denominator for the fractions on the left-hand side. The least common multiple of 5, 6, 4, and 8 is 120. We rewrite each fraction with this common denominator: \[ \frac{1}{5} = \frac{24}{120}, \quad \frac{1}{6} = \frac{20}{120}, \quad \frac{1}{4} = \frac{30}{120}, \quad \frac{1}{8} = \frac{15}{120} \] 4. Substituting these values into the equation, we get: \[ \frac{24}{120} + \frac{20}{120} + \frac{1}{o} - \frac{30}{120} = \frac{15}{120} \] 5. Combine the fractions on the left-hand side: \[ \frac{24 + 20 - 30}{120} + \frac{1}{o} = \frac{15}{120} \] \[ \frac{14}{120} + \frac{1}{o} = \frac{15}{120} \] 6. Isolate $\frac{1}{o}$ by subtracting $\frac{14}{120}$ from both sides: \[ \frac{1}{o} = \frac{15}{120} - \frac{14}{120} \] \[ \frac{1}{o} = \frac{1}{120} \] 7. Taking the reciprocal of both sides, we find: \[ o = 120 \] The final answer is $\boxed{120}$" c6c3a8c05cfa,"G2.1 If 191 is the difference of two consecutive perfect squares, find the value of the smallest square number, $a$.",See reasoning trace,easy,"Let $a=t^{2}$, the larger perfect square is $(t+1)^{2}$ $$ \begin{array}{l} (t+1)^{2}-t^{2}=191 \\ 2 t+1=191 \\ t=95 \\ a=95^{2}=9025 \end{array} $$" fadc5b97c7e3,"80. Given that $\mathrm{n}$ is a positive integer, and the tens digit of $n^{2}$ is 7, find $n-100\left[\frac{n}{100}\right]$. (Note: $[x]$ denotes the integer part of $\mathrm{x}$)",See reasoning trace,medium,"Solution: $n-100\left[\frac{n}{100}\right]$ is to find the tens and units digit of $\mathrm{n}$. The tens digit of $n^{2}$ depends on the tens and units digit of $\mathrm{n}$. Let the tens digit of $\mathrm{n}$ be $\mathrm{a}$, and the units digit be $\mathrm{b}, (10 a+b)^{2}=100 a^{2}+20 a b+b^{2}$. If the tens digit of $b^{2}$ is 0 or even, then the tens digit of $n^{2}$ is even, so the units digit $\mathrm{b}$ of $\mathrm{n}$ can only be 4 or 6. When $b=4$, $(10 a+4)^{2}=100 a^{2}+80 a+16=100 a^{2}+10(8 a+1)+6$, only when the units digit of $8 a+1$ is 7, the tens digit of $n^{2}$ is 7, so $\mathrm{a}=2$ or 7; When $b=6$, $(10 a+6)^{2}=100 a^{2}+120 a+36=100\left(a^{2}+a\right)+10(2 a+3)+6$, only when the units digit of $2 a+3$ is 7, the tens digit of $n^{2}$ is 7, so $\mathrm{a}=2$ or 7. In summary, $n-100\left[\frac{n}{100}\right]$ has four possible results, which are $24, 26, 74, 76$." 3f3e48ec1f65,"## Task 4 Peter has three sisters. His father gives each child in the family two books. How many books does the father give his children?",See reasoning trace,easy,"$2+2+2+2=8$. The father gives 8 books to his children. ### 10.13.2 2nd Round 1977, Class 1" 573f1603414d,"- 1st head: “To my left is a green dragon.” - 2nd head: “To my right is a blue dragon.” - 3rd head: “There is no red dragon next to me.” What is the maximum number of red dragons that could have been at the table?",176,medium,"Answer: 176. Solution. Consider an arbitrary red dragon. To the right of this dragon, at least one head must tell the truth. Note that the 1st and 3rd heads cannot tell the truth (since there is a red dragon to the left), so the 2nd head must tell the truth, and to the right of this dragon, there must be a blue dragon. Now consider the dragon to the left of the chosen red dragon. By similar reasoning, we understand that only the 1st head can tell the truth, and to the left of this dragon, there must be a green one. From the previous reasoning, we get that for any red dragon, one position to the right there is definitely a blue dragon, and one position to the left - a green one. For each red dragon, we combine these three dragons into one group. Note that any dragon can be in at most one such group. Then we get that there are no more than $530 / 3$ < 177, i.e., no more than 176 such groups. Therefore, there are no more than 176 red dragons. Now let's provide an example. Arrange 176 red dragons in a circle, and in 175 gaps between them, place a green and a blue dragon: ...RGRB..., and in one gap, place two blue and two green dragons: ...RBGBGR.... It is easy to check that all conditions are met: each dragon has at least one head that will tell the truth, and there are exactly 530 dragons in total. ![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-17.jpg?height=492&width=525&top_left_y=852&top_left_x=464) ## 7th grade" 4e1aca5eaaaf,"1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying $$ \sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, $$ where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$. Then the number of 1's in this sequence is $\qquad$",See reasoning trace,medium,"$=, 1.510$. Let the number of -1's be $x$, and the number of 0's be $y$. Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with $$ \begin{array}{c} \text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\ \left\{\begin{array}{l} x+(x+5)+y=2015, \\ 4(x+5)+y=3040 . \end{array}\right. \end{array} $$ Solving these equations, we get $x=505$. Therefore, the number of 1's is 510." 28b91b028cfb,"Determine all polynomials $P$ satisfying the following equation for all real $x$: $$ \left(x^{2}-6 x+8\right) P(x)=\left(x^{2}+2 x\right) P(x-2) $$","x^{2}\left(x^{2}-4\right) c$ with $c$ a real number, and these are the only solutions, noting that t",medium,"First, note that the zero polynomial is a solution to the equation in the statement. Now suppose that $P$ is a solution to the equation and that $P$ is non-zero. Let's first rewrite the equation from the hypothesis. It is indeed equivalent to $$ (x-2)(x-4) P(x)=x(x+2) P(x-2) $$ By substituting $x$ with 0, we get $8 P(0)=0$ which means $P(0)=0$. By substituting $x$ with -2, we get $24 P(-2)=0$ which means $P(-2)=0$. By substituting $x$ with 4, we have $0=24 P(2)$ which means $P(2)=0$. The polynomial $P$ thus has at least 3 roots: $-2, 0, 2$. Therefore, $x, x-2$, and $x+2$ all divide the polynomial $P$, so there exists a polynomial $Q$ such that $P(x)=x(x-2)(x+2) Q(x)$ for all real $x$. By replacing $P$ with $x(x-2)(x+2) Q(x)$ in the equation, we get $$ (x-2)(x-4) x(x+2)(x-2) Q(x)=x(x+2)(x-4) x(x-2) Q(x-2) $$ which is equivalent to $$ (x-4)(x+2)(x-2) x[(x-2) Q(x)-x Q(x-2)]=0 $$ Thus, for all $x$ different from $-2, 0, 2, 4$, we have $(x-2) Q(x)-x Q(x-2)=0$. The polynomial $R(x)=(x-2) Q(x)-x Q(x-2)$ has an infinite number of roots (it suffices to take all integers different from $-2, 0, 2, 4$), so it is the zero polynomial. Therefore, for all real $x$, we have $(x-2) Q(x)-x Q(x-2)=0$. In particular, for $x=0$, we have $-2 Q(0)=0$ so $Q(0)=0$ and 0 is a root of the polynomial $Q$ so $x$ divides $Q$. Therefore, there exists a polynomial $S$ such that $Q(x)=x S(x)$ and the equation becomes $$ (x-2) x S(x)=x(x-2) S(x-2) $$ or equivalently $$ x(x-2)[S(x-2)-S(x)]=0 $$ so for $x$ different from 0 and 2, we have $S(x-2)=S(x)$. In particular, by immediate induction, we find that $S(x)=S(4)$ for all even integers strictly greater than 2. By setting $c=S(4)$, we see that the polynomial $T$ defined by $T(x)=S(x)-c$ has an infinite number of roots, so it is the zero polynomial. Therefore, $S(x)=c$ for all real $x$, which implies that $P(x)=x(x-2)(x+2) x S(x)=x^{2}\left(x^{2}-4\right) c$ with $c$ a non-zero real number. Conversely, we verify that this polynomial is indeed a solution to the polynomial equation since we have $$ (x-2)(x-4) P(x)=(x-2)(x-4) x^{2}\left(x^{2}-4\right) c=x(x+2) P(x-2) $$ The solutions are therefore all polynomials $P$ defined by $P(x)=x^{2}\left(x^{2}-4\right) c$ with $c$ a real number, and these are the only solutions, noting that the zero polynomial corresponds to the case where $c=0$." 436a7c69505b,"$$ x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}, $$ where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$.","0$ gives $a \geqslant 19$, and $t=1$ gives $a \geqslant 128$, etc.",medium,"By using the AM-GM inequality, we have $$ x(t)=\underbrace{(t+1)^{2}+\cdots+(t+1)^{2}}_{5 \uparrow}+\frac{a}{2(t+1)^{5}}+\frac{a}{2(t+1)^{5}} \geqslant 7 \sqrt[7]{\frac{a^{2}}{4}}, $$ and the equality holds when $(t+1)^{2}=\frac{a}{2(t+1)^{5}}$. Therefore, we have $7 \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24$, which implies $a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$. On the other hand, when $a=2 \sqrt{\left(\frac{24}{7}\right)^{7}}, t=\sqrt{\frac{24}{7}}-1$, we have $x(t)=24$. Thus, the minimum value of $a$ is $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$. Note: If we take special values of $t$, for example, $t=0$ gives $a \geqslant 19$, and $t=1$ gives $a \geqslant 128$, etc." 4e8a3c58120c,"What is the least possible number of cells that can be marked on an $n \times n$ board such that for each $m>\frac{n}{2}$ both diagonals of any $m \times m$ sub-board contain a marked cell? ## Answer: $n$.",See reasoning trace,medium,"For any $n$ it is possible to set $n$ marks on the board and get the desired property, if they are simply put on every cell in row number $\left\lceil\frac{n}{2}\right\rceil$. We now show that $n$ is also the minimum amount of marks needed. If $n$ is odd, then there are $2 n$ series of diagonal cells of length $>\frac{n}{2}$ and both end cells on the edge of the board, and, since every mark on the board can at most lie on two of these diagonals, it is necessary to set at least $n$ marks to have a mark on every one of them. If $n$ is even, then there are $2 n-2$ series of diagonal cells of length $>\frac{n}{2}$ and both end cells on the edge of the board. We call one of these diagonals even if every coordinate $(x, y)$ on it satisfies $2 \mid x-y$ and odd else. It can be easily seen that this is well defined. Now, by symmetry, the number of odd and even diagonals is the same, so there are exactly $n-1$ of each of them. Any mark set on the board can at most sit on two diagonals and these two have to be of the same kind. Thus, we will need at least $\frac{n}{2}$ marks for the even diagonals, since there are $n-1$ of them and $2 \nmid n-1$, and, similarly, we need at least $\frac{n}{2}$ marks for the the odd diagonals. So we need at least $\frac{n}{2}+\frac{n}{2}=n$ marks to get the desired property. #" 0ef13cb43aad,Let $a$ and $b$ be positive integers such that $(a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020$. Find $10a + b$.,53,medium,"1. We start with the given equation: \[ (a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020 \] 2. Factorize 2020: \[ 2020 = 2^2 \times 5 \times 101 \] 3. We need to find values of \(a\) and \(b\) such that: \[ (a^3 - a^2 + 1)(b^3 - b^2 + 2) = 4 \times 5 \times 101 \] 4. Let's consider the factors of 2020. We observe that: \[ b^3 - b^2 + 2 = 4 \quad \text{or} \quad b^3 - b^2 + 2 = 5 \] 5. If \(b^3 - b^2 + 2 = 4\): \[ b^3 - b^2 + 2 = 4 \implies b^3 - b^2 = 2 \implies b(b^2 - b) = 2 \] This equation does not have integer solutions for \(b\). 6. If \(b^3 - b^2 + 2 = 5\): \[ b^3 - b^2 + 2 = 5 \implies b^3 - b^2 = 3 \implies b(b^2 - b) = 3 \] This equation does not have integer solutions for \(b\). 7. If \(b^3 - b^2 + 2 = 101\): \[ b^3 - b^2 + 2 = 101 \implies b^3 - b^2 = 99 \implies b(b^2 - b) = 99 \] This equation does not have integer solutions for \(b\). 8. If \(b^3 - b^2 + 2 = 2020\): \[ b^3 - b^2 + 2 = 2020 \implies b^3 - b^2 = 2018 \implies b(b^2 - b) = 2018 \] This equation does not have integer solutions for \(b\). 9. If \(b^3 - b^2 + 2 = 2\): \[ b^3 - b^2 + 2 = 2 \implies b^3 - b^2 = 0 \implies b(b^2 - b) = 0 \] This equation has the solution \(b = 1\). 10. If \(b = 1\): \[ b^3 - b^2 + 2 = 1 - 1 + 2 = 2 \] Then: \[ (a^3 - a^2 + 1) \times 2 = 2020 \implies a^3 - a^2 + 1 = 1010 \] 11. We need to solve: \[ a^3 - a^2 + 1 = 1010 \] Testing small values of \(a\): \[ a = 10 \implies 10^3 - 10^2 + 1 = 1000 - 100 + 1 = 901 \quad (\text{not } 1010) \] \[ a = 11 \implies 11^3 - 11^2 + 1 = 1331 - 121 + 1 = 1211 \quad (\text{not } 1010) \] \[ a = 12 \implies 12^3 - 12^2 + 1 = 1728 - 144 + 1 = 1585 \quad (\text{not } 1010) \] 12. We observe that if \(b = 3\): \[ b^3 - b^2 + 2 = 3^3 - 3^2 + 2 = 27 - 9 + 2 = 20 \quad (\text{not a factor of } 2020) \] 13. We need to re-evaluate the factorization approach. Let's try: \[ b = 3 \implies b^3 - b^2 + 2 = 27 - 9 + 2 = 20 \] Then: \[ (a^3 - a^2 + 1) \times 20 = 2020 \implies a^3 - a^2 + 1 = 101 \] 14. Solving: \[ a^3 - a^2 + 1 = 101 \] Testing small values of \(a\): \[ a = 5 \implies 5^3 - 5^2 + 1 = 125 - 25 + 1 = 101 \] 15. Therefore, \(a = 5\) and \(b = 3\). 16. Finally, calculate \(10a + b\): \[ 10a + b = 10 \times 5 + 3 = 50 + 3 = 53 \] The final answer is \(\boxed{53}\)." 2403af73c74a,"G3.4 若 $a 、 b$ 及 $y$ 為實數, 並满足 $\left\{\begin{array}{l}a+b+y=5 \\ a b+b y+a y=3\end{array}\right.$, 求 $y$ 的最大值。 If $a, b$ and $y$ are real numbers and satisfy $\left\{\begin{array}{l}a+b+y=5 \\ a b+b y+a y=3\end{array}\right.$, determine the greatest possible value of $y$.",\frac{13}{3}$.,easy,"From (1): $a=5-(b+y)$ Sub. into (2): $[5-(b+y)] b+b y+[5-(b+y)] y=3$ $b^{2}+(y-5) b+\left(y^{2}-5 y+3\right)=0$, this is a quadratic equation in $b$. For real values of $b, \Delta=(y-5)^{2}-4\left(y^{2}-5 y+3\right) \geq 0$ $$ \begin{array}{l} -3 y^{2}+10 y+13 \geq 0 \\ 3 y^{2}-10 y-13 \leq 0 \\ (3 y-13)(y+1) \leq 0 \\ -1 \leq y \leq \frac{13}{3} \end{array} $$ The maximum value of $y=\frac{13}{3}$." 48ac0e49027f,"5. Find all positive integers $m, n$ such that $m^{2}+1$ is a prime number, and $$ 10\left(m^{2}+1\right)=n^{2}+1 \text {. } $$","(2,7),(4,13)$ satisfy the conditions.",medium,"5. From the given condition, we have $$ 9\left(m^{2}+1\right)=(n-m)(n+m) \text {. } $$ Notice that $m^{2}+1$ is a prime number, and $m^{2}+1 \equiv 1$ or $2(\bmod 3)$, so $m^{2}+1$ is not a multiple of 3. Therefore, $$ \left\{\begin{array}{l} n-m=1,3,9, m^{2}+1, \\ n+m=9\left(m^{2}+1\right), 3\left(m^{2}+1\right), m^{2}+1,9 . \end{array}\right. $$ Subtracting the two equations gives $9 m^{2}+8=2 m$, which is impossible. Subtracting the two equations gives $3 m^{2}=2 m$, which is impossible. (3) If $\left\{\begin{array}{l}n-m=9, \\ n+m=m^{2}+1\end{array}\right.$. Subtracting the two equations gives $m^{2}-8=2 m$, so $m=4$. (4) If $\left\{\begin{array}{l}n-m=m^{2}+1, \\ n+m=9 .\end{array}\right.$ Subtracting the two equations gives $m^{2}-8=-2 m$, so $m=2$. When $m=2$ or 4, $m^{2}+1=5$ or 17 are both prime numbers, and the corresponding $n$ is 7 or 13. Therefore, the pairs $(m, n)=(2,7),(4,13)$ satisfy the conditions." 631eddd01be3,"## Task 1 - 060831 The edge of one cube has a length of $a_{1}=2 \mathrm{~cm}$, and that of another cube has a length of $a_{2}=6 \mathrm{~cm}$. Calculate the ratio of the edge lengths of these two cubes, the ratio of their surface areas, and the ratio of their volumes!",See reasoning trace,easy,"Let the surface areas of the two cubes be $O_{1}$ and $O_{2}$, and the volumes be $V_{1}$ and $V_{2}$. Then we have: $$ \begin{aligned} a_{1}: a_{2} & =2: 6=1: 3 \\ O_{1}: O_{2} & =6 a_{1}^{2}: 6 a_{2}^{2}=24: 216=1: 9 \\ V_{1}: V_{2} & =a_{1}^{3}: a_{2}^{3}=8: 216=1: 27 \end{aligned} $$" 6d589e518e5a,"A piece of paper is flipped over, the numbers $0,1,8$ remain unchanged after a $180^{\circ}$ rotation, 6 becomes 9, 9 becomes 6, and other numbers have no meaning after a $180^{\circ}$ rotation. How many 7-digit numbers remain unchanged after a $180^{\circ}$ rotation? Among these, how many are divisible by 4? The total sum of these 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $\qquad$","75$ times, the positions of $B, C, E, F$ each have all digits appearing $4 \times 5 \times 3=60$ tim",medium,"If $\overline{\mathrm{ABCDEFG}}$ remains unchanged after a $180^{\circ}$ rotation, then $D$ must be one of $0, 1, 8$. The pairs $C$ and $E$, $B$ and $F$, $A$ and $G$ can only be $0, 1, 8$ simultaneously, or one $6$ and one $9$. Therefore, we only need to determine the four digits $A, B, C, D$, so the total number of 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $4 \times 5 \times 5 \times 3=300$. Among these, the last two digits of the numbers that are divisible by 4 can only be $00, 08, 16, 60, 68, 80, 88, 96$. Among these, the numbers ending in $00, 80, 60$ will have a leading zero after a $180^{\circ}$ rotation, which is a contradiction, so they are excluded. Therefore, the numbers that are divisible by 4 are $5 \times 3 \times 5=75$. The positions of $A, G$ each have $1, 8, 6, 9$ appearing $5 \times 5 \times 3=75$ times, the positions of $B, C, E, F$ each have all digits appearing $4 \times 5 \times 3=60$ times, and the position of $D$ has $0, 1, 8$ each appearing $4 \times 5 \times 5=100$ times. Therefore, the total sum of all 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $24 \times 75 \times 1000001 + 24 \times 60 \times 110110 + 9 \times 100 \times 1000 = 1959460200$." 41aecbb18008,"A sand pit is constructed in the shape of a rectangular prism $10 \mathrm{~m}$ long, $50 \mathrm{~cm}$ deep and $2 \mathrm{~m}$ wide. If the pit is already half-full, how much more sand, in $\mathrm{m}^{3}$, is needed to completely fill the pit? (A) 6 (B) 5 (C) 20 (D) 7.5 (E) 10",(B),easy,"The volume of the entire pit is $$ (10 \mathrm{~m}) \times(50 \mathrm{~cm}) \times(2 \mathrm{~m})=(10 \mathrm{~m}) \times(0.5 \mathrm{~m}) \times(2 \mathrm{~m})=10 \mathrm{~m}^{3} $$ Since the pit starts with $5 \mathrm{~m}^{3}$ in it, an additional $5 \mathrm{~m}^{3}$ of sand is required to fill it. ANswer: (B)" fcf7cc9d4f56,"4. Find all pairs of prime numbers $(p, q)$ for which $$ 7 p q^{2}+p=q^{3}+43 p^{3}+1 . $$","(2,7)$.",medium,"4. First, let's show that in the equation $7 p q^{2}+p=q^{3}+43 p^{3}+1$, the numbers $p$ and $q$ cannot both be odd. In that case, $7 p q^{2}+p$ would be even, while $q^{3}+43 p^{3}+1$ would be odd. Therefore, $p=2$ or $q=2$, because 2 is the only even prime number. In the case that $p=2$, we get the equation $14 q^{2}+2=q^{3}+344+1$, or $q^{3}-14 q^{2}=-343$. We see that $q$ must be a divisor of $343=7^{3}$, and thus $q=7$. Substituting gives that $(p, q)=(2,7)$ indeed satisfies the equation, since $14 q^{2}+2=2 \cdot 7 \cdot 7^{2}+2$ and $q^{3}+344+1=7^{3}+\left(7^{3}+1\right)+1=2 \cdot 7^{3}+2$ are equal. Now let's consider the case $q=2$ and $p$ odd. The equation for $p$ then becomes $28 p+p=8+43 p^{3}+1$. Since $28 p+p$ is odd and $8+43 p^{3}+1$ is even, there are no solutions in this case. The only solution is thus $(p, q)=(2,7)$." f4c7f36c02bb,"4. Given the vector $\boldsymbol{a}=(1,1)$. Then the vector $\boldsymbol{b}=$ $\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$ is obtained by rotating vector $a$ ( ). (A) clockwise by $60^{\circ}$ (B) clockwise by $120^{\circ}$ (C) counterclockwise by $60^{\circ}$ (D) counterclockwise by $120^{\circ}$","\frac{\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) \cdot(1,1)}{2}=\frac{1}{2}$. Since $\t",easy,"4. C. Let the angle between the two vectors be $\theta$. Then $\cos \theta=\frac{\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) \cdot(1,1)}{2}=\frac{1}{2}$. Since $\theta \in\left[0^{\circ}, 180^{\circ}\right]$, we have $\theta=60^{\circ}$. Note that, $\frac{1-\sqrt{3}}{2}<0$. Thus, vector $\boldsymbol{b}$ can be obtained by rotating $\boldsymbol{a}$ counterclockwise by $60^{\circ}$." 1155ce8353c7,"4. As shown in the figure, $P, Q$ are points on the side $AD$ and the diagonal $AC$ of the square $ABCD$, respectively, and $PD: AP=4: 1, QC$: $\mathrm{AQ}=2: 3$. If the area of the square $\mathrm{ABCD}$ is 25, then the area of triangle $\mathrm{PBQ}$ is $\qquad$.",See reasoning trace,medium,"Parse: Connect $\mathrm{QD}$, draw $\mathrm{QE} \perp \mathrm{BC}$ at $\mathrm{E}$, $\mathrm{QF} \perp \mathrm{AD}$ at $\mathrm{F}$, $\mathrm{QG} \perp \mathrm{CD}$ at $\mathrm{G}$, the area of square $A B C D$ is 25, so $A D=E F=5, Q C: A Q=2: 3$, according to the symmetry of the square, so $Q E=Q G=2, Q F=3, P D: A P=4: 1, A P=1, P D=4$. $$ \begin{aligned} \mathrm{S} \triangle \mathrm{PQB} & =\mathrm{S}_{\text {square }}-\mathrm{S} \triangle \mathrm{CQB}-\mathrm{S} \triangle \mathrm{DQC}-\mathrm{S} \triangle \mathrm{PQD}-\mathrm{S} \triangle \mathrm{PAB} \\ & =25-2 \times 5 \div 2 \times 2-4 \times 3 \div 2-1 \times 5 \div 2 \\ & =25-10-6-2.5 \\ & =6.5 \end{aligned} $$" fdb219215652,"Let $\alpha, \beta, \gamma$ be three given quantities and $x_{1}, x_{2}$ be the two roots of the following equation: $$ x^{2}+p x+q=0 $$ What relationship must hold between $p$ and $q$ so that $$ \alpha x_{1}^{2}+\beta x_{1}+\gamma=\alpha x_{2}^{2}+\beta x_{2}+\gamma $$ is true?",See reasoning trace,medium,"The proposition can also be written as: $$ \alpha\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)+\beta\left(x_{1}-x_{2}\right)=0 $$ or $$ \left(x_{1}-x_{2}\right)\left[\alpha\left(x_{1}+x_{2}\right)+\beta\right]=0 $$ The left side is $=0$ if $$ x_{1}=x_{2}, \quad \text { or } \quad \alpha\left(x_{1}+x_{2}\right)+\beta=0 \text {. } $$ In the first case, the discriminant of the given equation is 0, so $$ p^{2}=4 q $$ In the second case, $x_{1}+x_{2}=-p$, thus $$ p=\frac{\beta}{\alpha} $$ (István Strasser, Budapest.) Number of solutions: 46." c52d3f22f209,"3. When $x=\frac{1}{\sqrt{3}-2}$, the value of the function $f(x)=x^{3}+4 x^{2}-$ $2 x-6$ is ( ). (A) $-3 \sqrt{3}$ (B) $3 \sqrt{3}$ (C) $-6 \sqrt{3}$ (D) $6 \sqrt{3}$",See reasoning trace,easy,"3. B. Notice that $x=\frac{1}{\sqrt{3}-2}=-(2+\sqrt{3})$, i.e., $x+2=-\sqrt{3}$. Therefore, when $x=\frac{1}{\sqrt{3}-2}$, $$ \begin{array}{l} f(x)=x(x+2)^{2}-6(x+1) \\ \quad=3 x-6(x+1)=-3(x+2)=3 \sqrt{3} . \end{array} $$" e2fe5f7f43cd,"4. Let three different lines be: $$ \begin{array}{l} l_{1}: a x+2 b y+3(a+b+1)=0, \\ l_{2}: b x+2(a+b+1) y+3 a=0, \\ l_{3}:(a+b+1) x+2 a y+3 b=0, \end{array} $$ Then the necessary and sufficient condition for them to intersect at one point is $\qquad$",See reasoning trace,easy,"The necessary and sufficient condition for $l_{1}, l_{2}, l_{3}$ to intersect at one point is \[ \left|\begin{array}{ccc} a & 2 b & 3(a+b+1) \\ b & 2(a+b+1) & 3 a \\ a+b+1 & 2 a & 3 b \end{array}\right|=0 \] \[ \begin{array}{l} \Rightarrow\left|\begin{array}{ccc} a & b & a+b+1 \\ b & a+b+1 & a \\ a+b+1 & a & b \end{array}\right|=3 a b(a+b+c)-a^{3}-b^{3}-(a+b+1)^{3}=0 \\ \Rightarrow 2 a+2 b+1=0 \text { or } a+b=-\frac{1}{2} . \end{array} \]" ae14a779d8e0,"5. If $0\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} \text {, } $$ (2) $\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}>\frac{a+b}{2}$, (3) $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\sqrt{a b}$, (4) $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, the true propositions are ( ). (A) (1), (3) (B) (1), (4) (C) (2), (3) (D) (2), (4)",See reasoning trace,medium,"3. (B). Solution 1: From the arithmetic sequence, we have $x_{1}+x_{n}=x_{2}+x_{n-1}=\cdots=a+b$, thus $$ \begin{array}{l} \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \\ =\frac{\left(x_{1}+x_{n}\right)+\left(x_{2}+x_{n-1}\right)+\cdots+\left(x_{n}+x_{1}\right)}{2 n} \\ =\frac{n(a+b)}{2 n}=\frac{a+b}{2}=\sqrt{a b}+2\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} . \\ >\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} . \end{array} $$ It is clear that (1) is true, (2) is false. From the geometric sequence, we know $y_{1} y_{n}=y_{2} y_{n-1}=\cdots=a b$, thus $$ \begin{array}{l} \sqrt[n]{y_{1} y_{2} \cdots y_{n}}=\sqrt[2 n]{\left(y_{1} y_{n}\right)\left(y_{2} y_{n-1}\right) \cdots\left(y_{n} y_{1}\right)} \\ =\sqrt[2 n]{(a b)^{n}}=\sqrt{a b}=\frac{a+b}{2}-2\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} \\ <\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} . \end{array} $$ It is clear that (3) is false, (4) is true. In summary, (1) and (4) are true. Solution 2: Take $a=1, b=64, n=2$, we can verify that (2) and (3) do not hold, thus negating (A), (C), (D), and hence (B) is true." 5196005ae9e8,"## Find the coordinates of point $A$, which is equidistant from points $B$ and $C$. $A(0 ; 0 ; z)$ $B(7 ; 0 ;-15)$ $C(2 ; 10 ;-12)$",See reasoning trace,medium,"## Solution Let's find the distances $A B$ and $A C$: \[ \begin{aligned} & A B=\sqrt{(7-0)^{2}+(0-0)^{2}+(-15-z)^{2}}=\sqrt{49+0+225+30 z+z^{2}}=\sqrt{z^{2}+30 z+274} \\ & A C=\sqrt{(2-0)^{2}+(10-0)^{2}+(-12-z)^{2}}=\sqrt{4+100+144+24 z+z^{2}}=\sqrt{z^{2}+24 z+248} \end{aligned} \] Since by the condition of the problem $A B=A C$, then \[ \sqrt{z^{2}+30 z+274}=\sqrt{z^{2}+24 z+248} \] \[ z^{2}+30 z+274=z^{2}+24 z+248 \] \[ 6 z=-26 \] \[ z=-\frac{13}{3}=-4 \frac{1}{3} \] Thus, $A\left(0 ; 0 ;-4 \frac{1}{3}\right)$. Problem Kuznetsov Analytical Geometry 11-11" 396aebe2bf82,"2. Given non-zero real numbers $x, 3, z$ form an arithmetic sequence, $x+1, y, z$ and $x, y, z+2$ each form a geometric sequence. Then the value of $y$ is ( ). (A) 8 (B) 16 (C) 12 (D) 20",12$.,easy,"2. (C). From the given information, we can obtain the system of equations $\left\{\begin{array}{l}2 y=x+z, \\ y^{2}=z(x+1), \\ y^{2}=x(z+2) .\end{array}\right.$ From (2) - (3) we get $z=2 x$. Substituting into (1) gives $x=\frac{2}{3} y$. Therefore, $z=\frac{4}{3} y$. Substituting back into (2) yields the value of $y$ that satisfies the conditions is $y=12$." 84d08443782f,"Solve the following system of equations: $$ \begin{aligned} & |x-3|=5-y \\ & |y-5|+|x-3|=x+y . \end{aligned} $$",See reasoning trace,medium,"I. solution. (1) the left side is non-negative, so the right side is also: $5-y \geq 0$ and thus (2) the first term on the left side is equal to the left side of (1). Adding the two equations, $y$ cancels out: $$ 3 \cdot|x-3|=x+5 $$ From here, depending on whether $$ \begin{gathered} x<3, \quad \text { or } \quad x \geq 3, \\ 3(3-x)=x+5, \quad 3(x-3)=x+5 \\ x_{1}=1, \quad x_{2}=7 . \end{gathered} $$ Thus, according to (1), $$ y_{1}=3, \quad y_{2}=1, $$ so the system of equations is satisfied by two pairs of $x, y$ values. Attila Herczegh (Budapest, Konyves Kalman Gymnasium, 3rd grade) II. solution. Let's plot our equations in the $x, y$ Cartesian coordinate system. (1) in the case of $x-3<0$ takes the form $y=x+2$, and in the case of $x-3 \geq 0$ it takes the form $y=8-x$. The common starting point of the two half-lines is the point $x=3, y=5$ (the thick broken line in the diagram). ![](https://cdn.mathpix.com/cropped/2024_05_02_df20e869379887cf2ecdg-1.jpg?height=445&width=637&top_left_y=1008&top_left_x=728) (2) the left side cannot have non-negative numbers in both terms, otherwise we would end up with the contradiction $-8=0$. The form of the equation is $$ \begin{array}{lclll} x-3 \geq 0 & \text { and } & y-5<0 & \text { in which case } & y=1 \\ x-3<0 & \text { and } & y-5<0 & \text { in which case } & y=4-x \\ x-3<0 & \text { and } & y-5 \geq 0 & \text { in which case } & x=-1 \end{array} $$ in the plot, two half-lines connect to the endpoints of a segment, to $(3,1)$ and $(-1,4)$ (drawn with a thick dashed line in the diagram; the thin dashed lines delimit the regions of the plane corresponding to the 4 possible sign variations of the expressions $x-3$ and $y-5$). The two broken lines have two common points, the coordinate pairs of these points give the solutions to our system of equations: $$ M_{1} \text { gives } \quad x_{1}=1, \quad y_{1}=3, \quad M_{2} \text { gives } \quad x_{2}=7, \quad y_{2}=1 \text {. } $$ Oszkar Biro (Budapest, 11th district, Vali Uti Secondary School, 8th grade)" 71a3afb9a531,"7. (10 points) Five football teams compete, with each pair of teams playing one match: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. After all the matches are completed, it is found that the five teams' points are five consecutive natural numbers. Let the 1st, 2nd, 3rd, 4th, and 5th place teams have drawn $A$, $B$, $C$, $D$, and $E$ matches respectively. Then the five-digit number $\overline{\mathrm{ABCDE}}=$ $\qquad$","2$, The fourth place gets 4 points, with 4 matches, if it is 4 draws, then $B \geqslant 3, D=4, E=3$",medium,"【Solution】Solution: Five football teams compete, for a total of $C_{5}^{3}=10$ matches. The sum of points for each match is 2 or 3, so the total points are between 20 and 30. The points of the five teams are exactly five consecutive natural numbers, and the sum of the points of the five teams is 20 to 30, with the following three scenarios: 2 to 6, 3 to 7, 4 to 8. If the points of the five teams are $2 \sim 6$, then the total points are 20, which means all matches are draws, and each team gets 4 points, a contradiction. If the points of the five teams are $4 \sim 8$, then the total points are 30, which means there are no draws, and each team's points are multiples of 3, a contradiction. Therefore, the points of the five teams are $3 \sim 7$, with a total of 25 points, 5 draws, and $A+B+C+D+E=2 \times 5=10$. The first place gets 7 points, with 4 matches, which can only be 2 wins, 1 draw, and 1 loss, so $A=1$, The third place gets 5 points, with 4 matches, which can only be 1 win, 2 draws, and 1 loss, so $C=2$, The fourth place gets 4 points, with 4 matches, if it is 4 draws, then $B \geqslant 3, D=4, E=3$, so $A+B+C+D+E \geqslant 1+3+2+4+1 > 10$, a contradiction, so the fourth place is 1 win, 1 draw, and 2 losses, thus $D=1$, so $B+E=10-A-C-D=6$, and $B \leqslant 3, E \leqslant 3$, which can only be $B=E=3$, In summary, the five-digit number $\overline{\mathrm{ABCDE}}=13213$." 3983fbbb8d9b,"8.5. Eleven of the best football teams each played one match against each other. It turned out that each team scored 1 goal in their first match, 2 goals in their second match, ..., 10 goals in their tenth match. What is the maximum number of matches that could have ended in a draw?",- 0 points,medium,"Solution: According to the condition, each team scored 1 goal in the first game. In the case of a draw, it also conceded 1 goal. Then for the other team, this match was also the first. Since the number of teams is odd, they cannot be paired. Therefore, at least one of the teams played a non-draw in their first match. The same is true for the second, third, ..., tenth matches. Since each match is counted twice, there were at least 5 non-draw matches, and then the number of draws was no more than $11 \cdot 10 / 2 - 5 = 50$. Let's provide an example of a tournament with 50 draws. Number the teams from 1 to 11. Form 5 pairs of adjacent teams (leaving the 11th team without a pair). First, the teams in the pairs play against each other, and then the 11th team plays against the 1st (with a score of $1:2$). Again, we pair the teams, this time excluding the 1st, and let the teams in the pairs play again (resulting in 5 draws). Now arrange the teams in a circle with a step of 2, specifically in the order 1357911246810 (each team has two new neighbors). Applying the same match scheme, we get another 11 games, one of which is non-draw. Then apply this scheme for steps of 3, 4, and 5. We can even write out a specific table: | All play the first game, all draws | $12,34,56,78,910$ | | :--- | :--- | | 11 - first game, 1 - second game | 111 | | All play the second game, all draws | $23,45,67,89,1011$ | | All play the third game, all draws | $13,57,911,24,68$ | | 10 - third game, 1 - fourth game | 101 | | All play the fourth game, all draws | $35,79,112,46,810$ | | All play the fifth game, all draws | $14,710,25,811,36$ | | 9 - fifth game, 1 - sixth game | 91 | | All play the sixth game, all draws | $47,102,58,113,69$ | | All play the seventh game, all draws | $15,92,610,37,114$ | | 8 - seventh game, 1 - eighth game | 81 | | All play the eighth game, all draws | $59,26,103,711,48$ | | All play the ninth game, all draws | $16,115,104,93,82$ | | 7 - ninth game, 1 - tenth game | 71 | | All play the tenth game, all draws | $611,510,49,38,27$ | ## Criteria: Only the answer - 0 points. Only the estimate that there are no more than $50-2$ points. Only an example with 50 draws - 3 points. There is an estimate and a correct idea for the example, but there are inaccuracies - 5 or 6 points depending on the size of the inaccuracies." 10b523787bc7,"11. (15 points) Let real numbers $a, b$ be such that the equation $a x^{3}-x^{2} + b x-1=0$ has three positive real roots. For all real numbers $a, b$ that satisfy the condition, find the minimum value of $P=\frac{5 a_{2}^{2}-3 a b+2}{a^{2}(b-a)}$.",See reasoning trace,medium,"11. Solution: On the one hand, when $a=\frac{\sqrt{3}}{9}, b=\sqrt{3}$, the equation $\frac{\sqrt{3}}{9} x^{3} - x^{2} + \sqrt{3} x - 1 = 0 \Leftrightarrow (x - \sqrt{3})^{3} = 0$, so this equation has three equal positive real roots $\sqrt{3}$. At this time, $P = 12 \sqrt{3}$. Below, we prove that $12 \sqrt{3}$ is the minimum value sought. Let the three positive real roots of the equation $a x^{3} - x^{2} + b x - 1 = 0$ be $\alpha, \beta, \gamma$, respectively. Then, by the relationship between roots and coefficients, we have: $\alpha + \beta + \gamma = \frac{1}{a}, \alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{a}, \alpha \beta \gamma = \frac{1}{a}$. Therefore, $a > 0, b > 0$. From $(\alpha + \beta + \gamma)^{2} \geqslant 3(\alpha \beta + \beta \gamma + \gamma \alpha)$, we know that $\frac{1}{a^{2}} \geqslant \frac{3 b}{a}$, which gives: $b \leqslant \frac{1}{3 a}$. Also, from $\alpha \beta + \beta \gamma + \gamma \alpha \geqslant 3 \sqrt[3]{\alpha^{2} \beta^{2} \gamma^{2}}$, we know that $\frac{b}{a} \geqslant 3 \sqrt[3]{\frac{1}{a^{2}}}$, which gives: $b \geqslant 3 \sqrt[3]{a}$. Therefore, we have $3 \sqrt[3]{a} \leqslant b \leqslant \frac{1}{3 a}$, so $3 \sqrt[3]{a} \leqslant \frac{1}{3 a}$, solving this gives $a \leqslant \frac{\sqrt{3}}{9}$, hence $a = \sqrt[3]{a^{2}} \sqrt[3]{a}0$, so $00$, so we get $P \geqslant \frac{15 a^{2} + 3}{a - 3 a^{3}} \geqslant 12 \sqrt{3}$. In summary, the minimum value of $P$ is $12 \sqrt{3}$." 228fbb81198e,"745. Find the derivatives of the following functions: 1) $\left.y^{3}-3 y+2 a x=0 ; 2\right) x^{2}+3 x y+y^{2}+1=0$. Calculate $y^{\prime}$ at the point $(2 ;-1) ;$; 3) $\sin \varphi+r \varphi-5 r=0$. Calculate $\frac{d r}{d \psi}$; 4) $e^{y}+x y=e$. Calculate $y^{\prime}$ at the point $(0 ; 1)$.",See reasoning trace,medium,"Solution. 1) Differentiating both sides of the equality with respect to $x$, taking into account that $y$ is a function of $x$, we get $$ 3 y^{2} y^{\prime}-3 y^{\prime}+2 a=0 $$ from which $$ y^{\prime}=\frac{2 a}{3\left(1+y^{2}\right)} $$ 2) Differentiating with respect to $x$, we get $$ 2 x+3 y+3 x y^{\prime}+2 y y^{\prime}=0 $$ from which $$ y^{\prime}=-\frac{2 x+3 y}{3 x+2 y} $$ Now substituting $x=2, y=-1$, we find $$ y^{\prime}(2)=-\frac{2 \cdot 2+3(-1)}{3 \cdot 2+2(-1)}=-\frac{1}{4} $$ 3) Differentiating with respect to $\varphi$, we get $$ \cos \varphi+\frac{d r}{d \varphi} \cdot \varphi+r-5 \frac{d r}{d \varphi}=0 $$ from which $$ \frac{d r}{d \varphi}=\frac{r+\cos \varphi}{5-\varphi} $$ 4) Differentiating with respect to $x$, we find $$ e^{y} y^{\prime}+y+x y^{\prime}=0 $$ from which $y^{\prime}=-\frac{y}{e^{y}+x}, a y^{\prime}(0)=-\frac{1}{e}$. Find the derivatives of the following implicit functions:" 8e852f74dc88,"15. (3 points) There are five numbers: $2$, $3$, $5$, $6$, $7$. Use them to form one-digit or two-digit numbers to create an equation, resulting in $2010=$ $\qquad$",: $2 \times 3 \times 5 \times 67$,easy,"15. (3 points) There are five numbers: $2$, $3$, $5$, $6$, and $7$. Use them to form one-digit or two-digit numbers to create an equation, resulting in $2010=$ $\qquad$ $2 \times 3 \times 5 \times 67$. 【Solution】Solution: Since the unit digit of 2010 is 0, 2 and 5 should be used as one-digit numbers, and since 2010 is divisible by 3, 3 should also be used as a one-digit number, so we get $2 \times 3 \times 5 \times 67=2010$. Therefore, the answer is: $2 \times 3 \times 5 \times 67$." c51a75afb16c,"## Task B-1.4. Among all natural numbers divisible by 8, determine those for which the sum of the digits is 7, and the product of the digits is 6.",See reasoning trace,medium,"## Solution. Notice that the numbers do not contain the digit 0, as this would make the product of the digits equal to 0. For the product of the digits to be 6, the number must have: - one digit 2 and one digit 3, with the rest being 1s, or - one digit 6, with the rest being 1s. Since the sum of the digits is also 7, the desired number can be: - a two-digit number with digits 1 and 6, or - a four-digit number with digits 1, 1, 2, 3. Multiples of 8 are divisible by 4, which means the last two digits of the desired number must be divisible by 4. Therefore, the desired number ends with 12, 16, or 32. If the number is a two-digit number with digits 1 and 6, the only possible number is 16. If the number is a four-digit number (with digits 1, 1, 2, 3), the possible numbers are 1132, 1312, or 3112, of which only the first is not divisible by 8, and the remaining two are. Therefore, all solutions are the numbers 16, 1312, and 3112. Note: A student earns 4 points based on the recorded conclusions expressed in words or symbols that show: - that 0 cannot be a digit (1 point) - that they use the fact that the product of the digits is 6, meaning the only possible digits are 1, 6, or 1, 2, 3 (1 point) - that they use the fact that the sum of the digits is 7, so the number must be a two-digit number with digits 1, 6 or a four-digit number with digits 1, 1, 2, 3 (2 points) Listing the possibilities and checking the divisibility by 8 to find the desired numbers, 16, 1312, and 3112, is worth 2 points. If a student has the correct solution but lacks the reasoning or process showing how they concluded these are the only possibilities, they receive 2 points." 3e25fc40ee0b,1. Rok used 500 equally large cubes to build a cuboid. What is the minimum number of faces of the cubes that made up the faces of the cuboid?,52&width=1690&top_left_y=1887&top_left_x=186),medium,"II/1. Denimo, da so bile Rokove kocke enotske. Kvader je imel tedaj prostornino 500 . Če je bil dolg $a$, širok $b$ in visok $c$ enot, je veljalo $a b c=500$. Njegova površina je bila $P=2(a b+b c+a c)$, vrednost izraza $P$ pa pove tudi, koliko mejnih ploskev kock je na površju kvadra. Poiskati moramo torej najmanjšo vrednost površine kvadra. Predpostavimo lahko, da je $a \geq b \geq c$. Ker je $500=2^{2} \cdot 5^{3}$, so delitelji števila 500 števila $1,2,4,5,10,20,25,50,100,125,250$ in 500 (glej preglednico). Razberemo lahko, da je na površju kvadra najmanj 400 mejnih ploskev kock, iz katerih je sestavljen. | $a$ | | $c$ | $2(a b+b c+a c)$ | | ---: | ---: | :---: | :---: | | 500 | 1 | 1 | 2002 | | 250 | 2 | 1 | 1504 | | 125 | 4 | 1 | 1258 | | 125 | 2 | 2 | 1008 | | 100 | 5 | 1 | 1210 | | 50 | 10 | 1 | 1120 | | 50 | 5 | 2 | 720 | | 25 | 20 | 1 | 1090 | | 25 | 10 | 2 | 640 | | 25 | 5 | 4 | 490 | | 20 | 5 | 5 | 450 | | 10 | 10 | 5 | 400 | Ugotovitev, da ima kvader z dolžinami robov $a, b$ in $c$ prostornino 500 (enot) ........ 1 točka Idea o deliteljih števila 500 in njegovi delitelji ............................................................................ Povezava števila mejnih ploskev (enotskih) kock s površino kvadra ........................ 1 točka Iskanje najmanjše vrednosti površine pri različnih vrednostih dolžin robov ............. 2 točki ![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-07.jpg?height=52&width=1690&top_left_y=1887&top_left_x=186) --- Let's assume that Roko's cubes were unit cubes. The volume of the cuboid was then 500. If its length was $a$, width $b$, and height $c$ units, then $a b c=500$. Its surface area was $P=2(a b+b c+a c)$, and the value of the expression $P$ also indicates how many boundary faces of the unit cubes are on the surface of the cuboid. We need to find the smallest value of the surface area of the cuboid. We can assume that $a \geq b \geq c$. Since $500=2^{2} \cdot 5^{3}$, the divisors of 500 are the numbers $1,2,4,5,10,20,25,50,100,125,250$ and 500 (see the table). We can see that the cuboid has at least 400 boundary faces of the unit cubes it is composed of. | $a$ | | $c$ | $2(a b+b c+a c)$ | | ---: | ---: | :---: | :---: | | 500 | 1 | 1 | 2002 | | 250 | 2 | 1 | 1504 | | 125 | 4 | 1 | 1258 | | 125 | 2 | 2 | 1008 | | 100 | 5 | 1 | 1210 | | 50 | 10 | 1 | 1120 | | 50 | 5 | 2 | 720 | | 25 | 20 | 1 | 1090 | | 25 | 10 | 2 | 640 | | 25 | 5 | 4 | 490 | | 20 | 5 | 5 | 450 | | 10 | 10 | 5 | 400 | Determining that a cuboid with edge lengths $a, b$ and $c$ has a volume of 500 (units) ........ 1 point The idea of the divisors of the number 500 and its divisors ............................................................................ The connection between the number of boundary faces (unit) of the cubes and the surface area of the cuboid ........................ 1 point Finding the smallest value of the surface area for different edge lengths ............. 2 points ![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-07.jpg?height=52&width=1690&top_left_y=1887&top_left_x=186)" ab73c9244ced,"7. The right figure is composed of 4 regular hexagons, each with an area of 6. Using the vertices of these 4 hexagons as vertices, the number of equilateral triangles that can be formed with an area of 4 is $\qquad$ .",See reasoning trace,easy,"8," 686cc6f75126,"$25 \cdot 26$ A line passes through the point $(-a, 0)$ and divides the second quadrant into a triangular region with area $T$. The equation of this line is (A) $2 T x+a^{2} y+2 a T=0$. (B) $2 T x-a^{2} y+2 a T=0$. (C) $2 T x+a^{2} y-2 a T=0$. (D) $2 T x-a^{2} y-2 a T=0$. (E) None of the above. (15th American High School Mathematics Examination, 1964)",$(B)$,easy,"[Solution] Clearly, the intersection point of the line with the $x$-axis is $(-a, 0)$, and let its intersection point with the $y$-axis be $(0, b)$. Given that $S_{\triangle O A B}=\frac{1}{2} a b=T$, then $b=\frac{2 T}{a}$. Thus, the slope of the line is $\frac{b}{a}=\frac{2 T}{a^{2}}$, and the equation of the line is $$ y=\frac{2 T}{a^{2}} x+\frac{2 T}{a}, $$ or $2 T x-a^{2} y+2 a T=0$. Therefore, the answer is $(B)$." 1227d5942614,"## Calculate the lengths of the arcs of the curves given by the parametric equations. $$ \begin{aligned} & \left\{\begin{array}{l} x=8(\cos t+t \sin t) \\ y=8(\sin t-t \cos t) \end{array}\right. \\ & 0 \leq t \leq \frac{\pi}{4} \end{aligned} $$",See reasoning trace,medium,"## Solution The length of the arc of a curve defined by parametric equations is determined by the formula $$ L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t $$ Let's find the derivatives with respect to $t$ for the given curve: $$ \begin{aligned} & x=8(\cos t+t \sin t) ; x^{\prime}(t)=8(-\sin t+\sin t+t \cos t)=8 t \cos t \\ & y=8(\sin t-t \cos t) ; y^{\prime}(t)=8(\cos t-\cos t+t \sin t)=8 t \sin t \end{aligned} $$ We get: $$ \begin{aligned} L & =\int_{0}^{\pi / 4} \sqrt{(8 t \cos t)^{2}+(8 t \sin t)^{2}} d t= \\ & =\int_{0}^{\pi / 4} 8 t \sqrt{\cos ^{2} t+\sin ^{2} t} d t= \\ & =8 \int_{0}^{\pi / 4} t \sqrt{1} d t=\left.4 \cdot t^{2}\right|_{0} ^{\pi / 4}=4 \cdot\left(\left(\frac{\pi}{4}\right)^{2}-0\right)=\frac{\pi^{2}}{4} \end{aligned} $$ Source — ""http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+18-21 »$ Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals - Last edited on this page: 12:02, 21 May 2010. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 18-22 ## Material from PlusPi" 7d1afc50d9bf,"8. (10 points) In $\triangle A B C$, $B D=D E=E C$, $C F: A C=1: 3$. If the area of $\triangle A D H$ is 24 square centimeters more than the area of $\triangle H E F$, find the area of triangle $A B C$ in square centimeters?","】Solution: If the area of $\triangle A D H$ is 24 square centimeters more than the area of $\triangle H E F$, then the area of triangle $A D E$ is 24 square centimeters more than the area of triangle $F D E$",medium,"【Answer】Solution: If the area of $\triangle A D H$ is 24 square centimeters more than the area of $\triangle H E F$, then the area of triangle $A D E$ is 24 square centimeters more than the area of triangle $F D E$. Since the area of triangle $F D E$ is equal to the area of triangle $F E C$, this means the area of triangle $A E C$ is 24 square centimeters more than the area of triangle $F E C$. Given that the additional 24 square centimeters is $\frac{2}{3}$ of the area of triangle $A E C$, the area of triangle $A E C$ is $24 \div \frac{2}{3}=36$ square centimeters. Therefore, the area of triangle $A B C$ is $36 \div \frac{1}{3}=108$ (square centimeters). Answer: The area of triangle $A B C$ is 108 square centimeters." cd616cf53be2,12.027. A trapezoid with angles $\alpha$ and $\beta$ at the larger base is circumscribed around a circle of radius $R$. Find the area of this trapezoid.,See reasoning trace,medium,"Solution. Let $\angle BAD = \alpha, \angle CDA = \beta, OE = R$ (Fig. 12.30). Then $S_{ABCD} = \frac{BC + AD}{2} \cdot 2R \cdot S_{ABCD} = (BC + AD) \cdot R \cdot$ Since $BC + AD = AB + CD$, we have $BC + AD = 2R\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right)\left(AB = \frac{2R}{\sin \alpha}, CD = \frac{2R}{\sin \beta}\right)$. Therefore, ![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-735.jpg?height=322&width=502&top_left_y=95&top_left_x=137) Fig. 12.30 ![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-735.jpg?height=316&width=530&top_left_y=102&top_left_x=685) Fig. 12.31 \[ \begin{aligned} & S_{ABCD} = 2R^2\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right) = \frac{2R^2(\sin \alpha + \sin \beta)}{\sin \alpha \sin \beta} = \frac{4R^2 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\sin \alpha \sin \beta} \\ & \text{Answer: } \frac{4R^2 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\sin \alpha \sin \beta} \end{aligned} \]" 5c60a9a8ebf7,"3. Find all two-digit numbers of the form $\overline{X Y}$, if the number with a six-digit decimal representation $\overline{64 X 72 Y}$ is divisible by 72.","$80,98$",easy,"Answer: $80,98$. Solution. Since $\overline{64 X 72 Y}$ is divisible by 8, then $Y \in\{0,8\}$, and since $\overline{64 X 72 Y}$ is divisible by 9, then $6+4+X+7+2+Y=19+X+Y$ is a multiple of 9. If $Y=0$, then $X=8$, and if $Y=8$, then either $X=0$ (cannot be the first digit of the number), or $X=9$. Therefore, the desired two-digit numbers are $\overline{X Y}=80$ and $\overline{X Y}=98$." 4585c4ac2be2,"[ Isosceles, Inscribed, and Circumscribed Trapezoids ] [Properties and characteristics of isosceles triangles. ] Let $M$ be the point of intersection of the diagonals of a convex quadrilateral $ABCD$, in which sides $AB$, $AD$, and $BC$ are equal to each other. Find the angle $CMD$, given that $DM = MC$, and $\angle CAB \neq \angle DBA$.",$120^{\circ}$,medium,"Let $\angle A B D=\angle A D B=\alpha, \angle B A C=\angle A C B=\beta$. By the external angle theorem of a triangle, $\angle B M C=\alpha+\beta$. Draw a line through point $A$ parallel to side $C D$. Let this line intersect line $D B$ at point $K$. Triangle $AMK$ is isosceles because it is similar to the isosceles triangle $CMD$. Therefore, $\angle D K=D M+M K=C M+M A=C A$, which means trapezoid $A K C D$ is isosceles. Thus, $C K=A D=B C$, so triangle $B C K$ is also isosceles (by the condition that point $K$ does not coincide with point $B$). Additionally, $\angle K C M=\angle A D M=\alpha$. Consider two cases. 1) Point $K$ lies on diagonal $D B$. Then $\angle K B C=\angle B K C=\angle K M C+\angle K C M=2 \alpha+\beta$. Therefore, $180^{\circ}=\angle B M C+\angle M B C+\angle M C B=(\alpha+\beta)+(2 \alpha+\beta)+\beta=3 \alpha+3 \beta$. 2) Point $K$ lies on the extension of $D B$ beyond point $B$. Then $\angle B K C=\angle K B C=\angle B M C+\angle B C M=\alpha+2 \beta$. Therefore, $180^{\circ}=\angle K M C+\angle M K C+\angle K C M=(\alpha+\beta)+(\alpha+2 \beta)+\alpha=3 \alpha+3 \beta$. Thus, in either case, $\alpha+\beta=60^{\circ}$. Consequently, $\angle C M D=180^{\circ}-\angle K M C=180^{\circ}-(\alpha+\beta)=120^{\circ}$. ## Answer $120^{\circ}$." 9f102e0917bc,"623. Find $f^{\prime}(0), f^{\prime}(1), f^{\prime}(2)$, if $f(x)=2 x^{3}+4 x-5$.",See reasoning trace,easy,"Solution. First, we find the derivative in general form. By rule II, and then I, we will have $$ f^{\prime}(x)=\left(2 x^{3}\right)^{\prime}+(4 x)^{\prime}-(5)^{\prime}=2\left(x^{3}\right)^{\prime}+4(x)^{\prime}-(5)^{\prime} $$ Using the derivative formulas (1), (2), (3), we finally get $$ f^{\prime}(x)=6 x^{2}+4 $$ Now we find the values of the derivative at the specified points: $$ f^{\prime}(0)=4 ; \quad f^{\prime}(1)=10 ; \quad f^{\prime}(2)=28 $$" 728015212e06,"2.1. Find the sum of all integer values of $a$ from the interval $[-2012 ; 2013]$, for which the equation $(a-3) x^{2}+2(3-a) x+\frac{a-7}{a+2}=0$ has at least one solution.",2011,easy,"Answer: 2011. Solution. When $a=3$ there are no solutions. For other $a$, it should be $\frac{D}{4}=(a-3)^{2}-\frac{(a-3)(a-7)}{a+2} \geq 0$. The last inequality (plus the condition $a \neq 3$) has the solution $a \in(-\infty ;-2) \bigcup\{1\} \cup(3 ;+\infty)$. The desired sum is: $-2012-2011-\ldots-5-4-3+1+4+5+\ldots 2013=-3+1+2013=2011$." b55d9541a5ac,"Example 4 As shown in Figure $17-5$, in $\triangle A B C$, $\angle A C B=30^{\circ}, \angle A B C=$ $50^{\circ}, M$ is a point inside the triangle, $\angle M A C=40^{\circ}, \angle M C B=20^{\circ}$. Find the degree measure of $\angle M B C$.","80^{\circ}$, we know that $D M$ bisects $\angle B D C$. Since $C M$ bisects $\angle B C D$, $M$ is t",medium,"Take a point $E$ on the extension of $C M$ such that $\angle E B C=60^{\circ}$, and take a point $D$ on the extension of $B E$ such that $\angle D C B=40^{\circ}$. Connect $D A, D M, D C$, and $E A$. It is easy to see that $C A \perp D B, B A \perp D C$, which means $A$ is the orthocenter of $\triangle B C D$, so $\angle A D B=30^{\circ}$. In $\triangle C D E$, it is easy to see that $\angle E C D=20^{\circ}, \angle B D C=80^{\circ}$, so $\angle D E C=80^{\circ}=\angle E D C$. Since $E C = D C$, and $A C$ is the angle bisector of $\angle D C E$, $A C$ is the perpendicular bisector of $D E$, thus $A D=A E, \angle A E D=30^{\circ}$, $\angle A E M=50^{\circ}$. Also, $\angle A M E=\angle M A C+\angle M C A=50^{\circ}$, so $A M=A E=A D$, which means $A$ is the circumcenter of $\triangle D E M$, thus $\angle M D E=\frac{1}{2} \angle M A E=40^{\circ}$. From $\angle B D C=80^{\circ}$, we know that $D M$ bisects $\angle B D C$. Since $C M$ bisects $\angle B C D$, $M$ is the incenter of $\triangle B C D$, so $\angle M B C=\frac{1}{2} \angle D B C=30^{\circ}$." d4dce806e9b6,"Example 5 Let $f(n)$ be a function defined on $\mathbf{N}_{+}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}_{+}$, we have $$ \begin{array}{l} f(m+n)-f(m)-f(n)=0 \text{ or } 1, \\ f(2)=0, f(3)>0, f(6000)=2000 . \\ \text{Find } f(5961) . \end{array} $$",See reasoning trace,medium,"Solution: From the analysis, we know that $f(1)=0$. From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$. But $f(3)>0$, so $f(3)=1$. By the problem statement, we have $$ f(3 n+3)=f(3 n)+f(3)+0 \text { or } 1 \text {, } $$ which means $f(3 n+3)=f(3 n)+1$ or 2. Therefore, $f(3(n+1)) \geqslant f(3 n)+1$. Let $n=1,2, \cdots, k$, we get $$ \begin{array}{l} f(3 \times 2) \geqslant f(3 \times 1)+1, \\ f(3 \times 3) \geqslant f(3 \times 2)+1, \\ \cdots . . \end{array} $$ $$ f(3 \times k) \geqslant f(3(k-1))+1 \text {. } $$ Adding all the above inequalities, we get $$ f(3 k) \geqslant f(3)+(k-1)=k \text {. } $$ Next, we prove that when $k \leqslant 2000$, $f(3 k)=k$. Conversely, if $f(3 k) \geqslant k+1$, then $$ \begin{array}{l} f(6000) \geqslant f(6000-3 k)+f(3 k) \\ \geqslant 2000-k+k+1>2000 . \end{array} $$ This is a contradiction. Thus, $f(3 k)=k$. Therefore, $$ f(5961)=f(3 \times 1987)=1987 . $$" c66181809d1b,9. Solve the congruence equation $x^{2} \equiv 59(\bmod 125)$,See reasoning trace,medium,"9. Solution: First, it is easy to see that $59 \equiv 9\left(\bmod 5^{2}\right)$, hence $$x^{2} \equiv 59\left(\bmod 5^{2}\right)$$ has solutions $x \equiv \pm 3(\bmod 25)$. Next, we solve $$x^{2} \equiv 59\left(\bmod 5^{3}\right)$$ (1) Let $x=25 t+3$, substituting we get $$(25 t+3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$ Thus, $$(6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$ Dividing both sides by 5, we get $$6 t \equiv 2(\bmod 5)$$ This simplifies to $$t \equiv 2(\bmod 5)$$ Substituting $x=25 t+3$ gives one solution as $x=25(5 k+2)+3=53\left(\bmod 5^{3}\right)$. (2) Let $x=25 t-3$, substituting we get $$(25 t-3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$ Expanding, we get $$(-6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$ Dividing by 25, we get $$-6 t \equiv 2(\bmod 5)$$ This simplifies to $$t \equiv -2(\bmod 5)$$ Thus, the second solution is $x_{2}=25(5 k-2)-3 \equiv -53\left(\bmod 5^{3}\right)$. In summary, the solutions are $x \equiv \pm 53\left(\bmod 5^{3}\right)$. Note: Regarding the solution of higher-order congruence equations of the form $$f(x) \equiv 0\left(\bmod p^{\alpha}\right)$$ where $f(x)$ is an $n$-degree polynomial with integer coefficients, $p$ is a prime, and $x \geqslant 1$, it is closely related to the corresponding congruence equation $$f(x) \equiv 0(\bmod p)$$ Due to the depth of the involved knowledge, it will not be elaborated here. Readers interested in this topic can refer to Professor Hua Luogeng's ""Introduction to Number Theory"" and other specialized works." e7264c70958f,"$1 \cdot 20$ If positive numbers $a, b, c$ satisfy the equation $a^{3}+b^{3}+c^{3}-3 a b c=0$, then (A) $a=b=c$. (B) $a=b \neq c$. (C) $b=c \neq a$. (D) $a, b, c$ are all unequal. (1st ""Five Sheep Cup"" Junior High School Mathematics Competition, 1989)",$(A)$,easy,"[Solution] $a^{3}+b^{3}+c^{3}-3 a b c$ $$ =\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0 $$ Given $a, b, c>0$, then $a+b+c \neq 0$, $$ \begin{array}{c} \therefore \quad(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0, \\ a-b=0, \quad b-c=0, \quad c-a=0, \\ a=b=c . \end{array} $$ Therefore, the answer is $(A)$." 023db80c19a2,(8) The solution to the equation $3 \cdot 16^{x}+2 \cdot 81^{x}=5 \cdot 36^{x}$ is,"t$, further transforming it into $3 t^{2}-5 t+2=0$. Solving this, we get $t_{1}=1$, $t_{2}=\frac{2}{",medium,"$80, \frac{1}{2}$ Hint: The original equation can be transformed into $3 \cdot\left(4^{x}\right)^{2}-5 \cdot 4^{x} \cdot 9^{x}+2 \cdot\left(9^{x}\right)^{2}=$ 0 . Dividing both sides by $\left(9^{x}\right)^{2}$, we get $$ 3\left[\left(\frac{4}{9}\right)^{x}\right]^{2}-5 \cdot\left(\frac{4}{9}\right)^{x}+2=0 . $$ Let $\left(\frac{4}{9}\right)^{x}=t$, further transforming it into $3 t^{2}-5 t+2=0$. Solving this, we get $t_{1}=1$, $t_{2}=\frac{2}{3}$. That is, $\left(\frac{4}{9}\right)^{x}=1$ or $\left(\frac{4}{9}\right)^{x}=\frac{2}{3}, x=0$ or $\frac{1}{2}$." 4697e50dff68,"2. Andrey, Boris, Vasily, Gennady, and Dmitry played table tennis in pairs such that every two of them played with every other pair exactly once. There were no draws in the tennis matches. It is known that Andrey lost exactly 12 times, and Boris lost exactly 6 times. How many times did Gennady win? Om vem: Gennady won 8 times.",See reasoning trace,medium,"Solution. The first pair can be formed in $5 \times 4: 2=10$ ways, the second pair can be formed in $3 \times 2: 2=3$ ways. In total, we get $10 \times 3: 2=15$ games. Andrei played in 4 pairs, and they played with 3 pairs. Therefore, Andrei played $4 \times 3=12$ times. According to the problem, he lost 12 times, which means he lost all his games. Together with him, Boris lost 3 times in a pair. Since Boris won 6 times against Andrei when playing with Vasily (2 times), with Gennady, and with Dmitry, the rest of the games he lost, that is, 3 times with Andrei and once each with Vasily (against Gennady and Dmitry), with Gennady, and with Dmitry. Therefore, Gennady lost 3 times with Andrei and once with Boris, a total of 4 times. Thus, he won 8 times. Criteria. If the solution is incorrect - 0 points. If the reasoning is correct but there is a computational error - 3 points. If the solution is correct - 7 points." 793fa244589c,"The formula $N=8 \times 10^{8} \times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. The lowest income, in dollars, of the wealthiest $800$ individuals is at least: $\textbf{(A)}\ 10^4\qquad \textbf{(B)}\ 10^6\qquad \textbf{(C)}\ 10^8\qquad \textbf{(D)}\ 10^{12} \qquad \textbf{(E)}\ 10^{16}$",\textbf{(A),easy,"Plug $800$ for $N$ because $800$ is the number of people who has at least $x$ dollars. \[800 = 8 \times 10^{8} \times x^{-3/2}\] \[10^{-6} = x^{-3/2}\] In order to undo raising to the $-\frac{3}{2}$ power, raise both sides to the $-\frac{2}{3}$ power. \[(10^{-6})^{-2/3} = (x^{-3/2})^{-2/3}\] \[x = 10^4 \text{ dollars}\] The answer is $\boxed{\textbf{(A)}}$." 348930770037,"2. Given $a, b, c > 0$, find the maximum value of the expression $$ A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}} $$",3,medium,"Answer: 3. Solution. Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where \[ \begin{aligned} & B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\ & \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b c)^{4}}+36 \sqrt[3]{(a b c)^{4}}=78(a b c)^{4 / 3} \end{aligned} \] (we used the Cauchy inequality). Then \[ a^{4}+b^{4}+c^{4} \leqslant(a+b+c)^{4}-78(a b c)^{4 / 3} \] Let $t=\frac{(a+b+c)^{4}}{(a b c)^{4 / 3}}$. By the Cauchy inequality, $t \geqslant 81$, hence \[ A \leqslant \frac{(a+b+c)^{4}-78(a b c)^{4 / 3}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}=\frac{t-78}{t-80}=1+\frac{2}{t-80} \leqslant 3 \] Equality is achieved when $a=b=c$." 227aa334093b,8.1. Find the maximum value of the expression $\frac{\sin \left(x+\frac{\pi}{4}\right)}{2 \sqrt{2}(\sin x+\cos x) \cos 4 x-\cos 8 x-5}$.,0,medium,"Answer: 0.5. Solution. Let's estimate the modulus of the given expression in the condition: $$ \begin{aligned} & \left|\frac{\sin \left(x+\frac{\pi}{4}\right)}{2 \sqrt{2}(\sin x+\cos x) \cos 4 x-\cos 8 x-5}\right| \leq \frac{1}{5-2 \sqrt{2}(\sin x+\cos x) \cos 4 x+\cos 8 x} \\ & =\frac{1}{4-4 \sin \left(x+\frac{\pi}{4}\right) \cos 4 x+2 \cos ^{2} 4 x}=\frac{1}{4+2\left(\cos 4 x-\sin \left(x+\frac{\pi}{4}\right)\right)^{2}-2 \sin ^{2}\left(x+\frac{\pi}{4}\right)} \leq \frac{1}{2} \\ & \text { The upper bound can be achieved if }\left\{\begin{array}{l} \cos 4 x=\sin \left(x+\frac{\pi}{4}\right), \\ \sin \left(x+\frac{\pi}{4}\right)=-1, \end{array}\right. \text { and indeed } \end{aligned} $$ it is achieved at $x=\frac{5 \pi}{4}+2 \pi k$." 5631a0321efd,"For each prime $p$, let $\mathbb S_p = \{1, 2, \dots, p-1\}$. Find all primes $p$ for which there exists a function $f\colon \mathbb S_p \to \mathbb S_p$ such that \[ n \cdot f(n) \cdot f(f(n)) - 1 \; \text{is a multiple of} \; p \] for all $n \in \mathbb S_p$. [i]Andrew Wen[/i]",2,medium,"1. **Primitive Root and Function Definition:** - For any prime \( p \), there exists a primitive root \( g \) modulo \( p \). - Let \( n = g^k \) for some integer \( k \). - Define a function \( c \) such that \( g^{c(k)} = f(n) \). This implies \( g^{c(c(k))} = f(f(n)) \). 2. **Rewriting the Problem:** - Given the condition \( n \cdot f(n) \cdot f(f(n)) - 1 \) is a multiple of \( p \), we can rewrite it using the primitive root: \[ g^k \cdot g^{c(k)} \cdot g^{c(c(k))} \equiv 1 \pmod{p} \] - Simplifying, we get: \[ g^{k + c(k) + c(c(k))} \equiv 1 \pmod{p} \] - Since \( g \) is a primitive root, \( g^r \equiv 1 \pmod{p} \) if and only if \( r \equiv 0 \pmod{p-1} \). Therefore: \[ k + c(k) + c(c(k)) \equiv 0 \pmod{p-1} \] 3. **Claim: \( c \) is Bijective:** - **Injectivity:** - Assume \( c(a) = c(b) \). This implies \( c(c(a)) = c(c(b)) \). - Let \( Q = c(a) + c(c(a)) = c(b) + c(c(b)) \). Then: \[ Q + a \equiv Q + b \equiv 0 \pmod{p-1} \] - Since the domain has each residue modulo \( p-1 \) exactly once, \( a = b \). Thus, \( c \) is injective. - **Surjectivity:** - Since \( c \) has a finite domain and codomain, each of which have equal cardinality, injectivity implies surjectivity. 4. **Summing Over All \( k \):** - Since \( c(k) \) is bijective, \( c(c(k)) \) is also bijective. Summing \( k + c(k) + c(c(k)) \equiv 0 \pmod{p-1} \) for each \( k \) in the domain: \[ \sum_{i=0}^{p-1} (i + c(i) + c(c(i))) \equiv 0 \pmod{p-1} \] - Since \( c(i) \) and \( c(c(i)) \) are bijective, the sum evaluates to: \[ \sum_{i=0}^{p-1} i + \sum_{i=0}^{p-1} c(i) + \sum_{i=0}^{p-1} c(c(i)) \equiv 0 \pmod{p-1} \] - Each sum is an arithmetic series sum: \[ \sum_{i=0}^{p-1} i = \frac{(p-1)p}{2} \] - Therefore: \[ 3 \cdot \frac{(p-1)p}{2} \equiv 0 \pmod{p-1} \] - Simplifying: \[ \frac{3(p-1)p}{2} \equiv 0 \pmod{p-1} \] - This is only possible if \( p = 2 \). Conclusion: \[ \boxed{2} \]" 206f6e1a9ba4,2.068. $\frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}} ; \quad a=7.4 ; b=\frac{5}{37}$.,1,medium,"Solution. $$ \begin{aligned} & \frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}}=\frac{\frac{a+b-2 c}{a b} \cdot(a+b+2 c)}{\frac{a^{2}+2 a b+b^{2}-4 c^{2}}{a^{2} b^{2}}}= \\ & =\frac{\frac{(a+b-2 c)(a+b+2 c)}{a b}}{\frac{(a+b)^{2}-(2 c)^{2}}{a^{2} b^{2}}}=\frac{(a+b-2 c)(a+b+2 c) a^{2} b^{2}}{\left((a+b)^{2}-(2 c)^{2}\right) a b}= \\ & =\frac{(a+b-2 c)(a+b+2 c) a b}{(a+b-2 c)(a+b+2 c)}=a b=7.4 \cdot \frac{5}{37}=\frac{37}{5} \cdot \frac{5}{37}=1 \end{aligned} $$ Answer: 1." 4a7d041a738d,"## Task Condition Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line. $M(2 ; 1 ; 0)$ $\frac{x-2}{0}=\frac{y+1.5}{-1}=\frac{z+0.5}{1}$",See reasoning trace,medium,"## Solution We find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector: $\vec{n}=\vec{s}=\{0 ;-1 ; 1\}$ Then the equation of the desired plane is: $0 \cdot(x-2)-1 \cdot(y-1)+1 \cdot(z-0)=0$ $-y+1+z=0$ $-y+z+1=0$ We find the point $M_{0}$ of intersection of the line and the plane. We write the parametric equations of the line. $$ \begin{aligned} & \frac{x-2}{0}=\frac{y+1.5}{-1}=\frac{z+0.5}{1}=t \Rightarrow \\ & \left\{\begin{array}{l} x=2 \\ y=-1.5-t \\ z=-0.5+t \end{array}\right. \end{aligned} $$ Substitute into the equation of the plane: $-(-1.5-t)+(-0.5+t)+1=0$ $1.5+t-0.5+t+1=0$ $2 t+2=0$ $t=-1$ We find the coordinates of the point of intersection of the line and the plane: $\left\{\begin{array}{l}x=2 \\ y=-1.5-(-1)=-0.5 \\ z=-0.5+(-1)=-1.5\end{array}\right.$ We get: $M_{0}(2 ;-0.5 ;-1.5)$ Since $M_{0}$ is the midpoint of the segment $M M^{\prime}$, then $x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2-2=2$ $y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-0.5)-1=-2$ $z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot(-1.5)-0=-3$ We get: $M^{\prime}(2 ;-2 ;-3)$" 0a42215af0d4,"7. Let $[x]$ denote the greatest integer not greater than $x$, and the sets $$ \begin{array}{l} A=\left\{x \mid x^{2}-2[x]=3\right\}, \\ B=\left\{x \left\lvert\, \frac{1}{8}<2^{x}<8\right.\right\} . \end{array} $$ Then $A \cap B=$ $\qquad$","\{-1, \sqrt{7}\}$.",medium,"$=7 . \mid-1, \sqrt{7}\}$. Since the solution to the inequality $\frac{1}{8}<2^{x}<8$ is $-33$, then both $p-1$ and $p+1$ are even and one of them is also divisible by 3, so their product is divisible by 12. Moreover, $p^{2}+11>12$, so it has at least 7 different divisors, which are $1,2,3,4$, 612 and $p^{2}+11$. Therefore, $p$ cannot be greater than 3. If $p=2$, then $p^{2}+11=15$, which has only 4 different positive divisors. If $p=3$, then $p^{2}+11=20$, and as we have already seen, it has 6 different positive divisors, which is the only solution." c1c018bfaeaa,Determine all polynomials $p(x)$ with non-negative integer coefficients such that $p (1) = 7$ and $p (10) = 2014$.,2x^3 + x + 4,easy,"** - Check the sum of coefficients: \[ a_3 + a_2 + a_1 + a_0 = 2 + 0 + 1 + 4 = 7 \] - Check the polynomial value at \( x = 10 \): \[ p(10) = 2 \cdot 10^3 + 0 \cdot 10^2 + 1 \cdot 10 + 4 = 2000 + 0 + 10 + 4 = 2014 \] Thus, the polynomial \( p(x) = 2x^3 + x + 4 \) satisfies all the given conditions. The final answer is \( \boxed{ 2x^3 + x + 4 } \)" 869e52df94a4,"Example $\mathbf{3}$ Given the quadratic equation in $x$, $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$, has real roots, find the value of the real number $a$. The equation is $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$.",-1$.,medium,"Analysis: Let the real root of the equation be $x_{0}$, substitute it into the original equation, and according to the necessary and sufficient conditions for the equality of complex numbers, transform it into a real number problem to solve. Solution: Let the real root of the equation be $x_{0}$, then $a(1+\mathrm{i}) x_{0}^{2}+\left(1+a^{2} \mathrm{i}\right) x_{0}+a^{2}+\mathrm{i}=0$, that is, $$ \left(a x_{0}^{2}+x_{0}+a^{2}\right)+\left(a x_{0}^{2}+a^{2} x_{0}+1\right) \mathrm{i}=0 . $$ According to the necessary and sufficient conditions for the equality of complex numbers, we get $\left\{\begin{array}{l}a x_{0}^{2}+x_{0}+a^{2}=0 \\ a x_{0}^{2}+a^{2} x_{0}+1=0\end{array}\right.$ (2) - (1), and rearrange, we get $\left(x_{0}-1\right)\left(a^{2}-1\right)=0$. Therefore, $x_{0}=1$ or $a^{2}=1$. When $x_{0}=1$, substituting it into (1), we have $a^{2}+a+1=0$, Clearly, this equation has no real roots, which contradicts $a \in \mathbf{R}$. Hence, $x_{0} \neq 1$. When $a^{2}=1$, we have $a= \pm 1$. Similarly, $a=1$ is also not suitable. Therefore, $a=-1$, at this time $x_{0}=\frac{1 \pm \sqrt{5}}{2}$. In summary, $a=-1$." 9fbfa1eb965d,We drew all the diagonals in three polygons with different numbers of sides. The second polygon has 3 more sides and 3 times as many diagonals as the first one. The third one has 7 times as many diagonals as the second one. How many more sides does the third polygon have compared to the second one?,See reasoning trace,medium,"Let the number of sides in the second polygon be $m$, in the third $m+x$, then in the first polygon it is $m-3$ (> $>$, since there are no diagonals in a triangle) and we need to determine $x$. It is known that the number of diagonals in an $n$-sided polygon is $n(n-3) / 2$, so according to our data, $$ \begin{gathered} \frac{m(m-3)}{2}=3 \cdot \frac{(m-3)(m-6)}{2} \\ \frac{(m+x)(m+x-3)}{2}=7 \cdot \frac{m(m-3)}{2} \end{gathered} $$ From the first equation, $m=3(m-6), m=9$ (since $m-3 \neq 0$), and thus from the second equation, $x=12$ (only the positive root can be used). Haeffner Erika (Veszprém, Lovassy L. Gymn., I. o. t.) Mátrai István (Szombathely, Nagy Lajos Gymn., I. o. t.)" b7839b05995a,"Parallelogram $A B C D$ is made up of four equilateral triangles of side length 1. The length of diagonal $A C$ is (A) $\sqrt{5}$ (B) $\sqrt{7}$ (C) 3 (D) $\sqrt{3}$ (E) $\sqrt{10}$ ![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-144.jpg?height=228&width=444&top_left_y=1003&top_left_x=1296)",(B),medium,"Parallelogram $A B C D$ is made up of four equilateral triangles of side length 1 . The length of diagonal $A C$ is (A) $\sqrt{5}$ (D) $\sqrt{3}$ (B) $\sqrt{7}$ (C) 3 ## Solution From $C$, we draw a line perpendicular to $A D$ extended so that they meet at point $E$ as shown in the diagram. This construction makes $\triangle C D E$ a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $\angle C D E=60^{\circ}$ and $C D=1$. Thus $C E=\frac{\sqrt{3}}{2}$ and $D E=\frac{1}{2}$. Using pythagoras in $\triangle A C E$, we have $A E=\frac{5}{2}$ ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-293.jpg?height=266&width=529&top_left_y=211&top_left_x=1432) and $C E=\frac{\sqrt{3}}{2}, A C=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}}=\sqrt{7}$. ![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-293.jpg?height=266&width=556&top_left_y=580&top_left_x=1424) ANSWER: (B)" feb879f6701b,"Use the four numbers $8, 8, 8, 10$, and the four basic arithmetic operations to form an expression $\qquad$ that equals 24 (parentheses can be added).",See reasoning trace,easy,"【Analysis and Solution】Calculate, make 24 points. $$ (10-8) \times 8+8=24 $$" c1152938ae19,"Three merchants' wives - Sosipatra Titovna, Olympiada Karpovna, and Polikseina Uvarovna - sat down to drink tea. Olympiada Karpovna and Sosipatra Titovna drank 11 cups together, Polikseina Uvarovna and Olympiada Karpovna - 15, and Sosipatra Titovna and Polikseina Uvarovna - 14. How many cups of tea did all three merchants' wives drink together? #",20 cups,easy,"Notice, the cup drunk by each merchant's wife was mentioned twice in the problem's condition. ## Solution |If we add up all the cups accounted for, we get twice the sum of the cups drunk. Therefore, we need to divide this sum by two. ## Answer 20 cups." 7b18d0ff0039,"9. $[7]$ Let $a_{0}=\frac{6}{7}$, and $$ a_{n+1}=\left\{\begin{array}{ll} 2 a_{n} & \text { if } a_{n}<\frac{1}{2} \\ 2 a_{n}-1 & \text { if } a_{n} \geq \frac{1}{2} \end{array}\right. $$ Find $a_{2008}$.",a_{0}=\frac{6}{7}$. Then $a_{2008}=\frac{5}{7}$.,easy,"$$ a_{1}=\frac{5}{7}, a_{2}=\frac{3}{7}, a_{3}=\frac{6}{7}=a_{0} $$ So this sequence repeats every three terms, so $a_{2007}=a_{0}=\frac{6}{7}$. Then $a_{2008}=\frac{5}{7}$." 41c3678d0271,"3. Given $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are positive integers. The set of sums obtained by taking any four of these numbers is $\{44, 45, 46, 47\}$. Then these five numbers are $\qquad$",See reasoning trace,medium,"3. $10,11,11,12,13$. From these five positive integers, if we take any four numbers and find their sum, we can get five sum values. Since there are only four different sum values in this problem, it indicates that two of the sum values must be equal. The sum of these five sum values is $4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)$. Thus, $226=44+44+45+46+47$ $\leqslant 4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)$ $\leqslant 44+45+46+47+47=229$. Solving this, we get $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=57$. Therefore, four of the five numbers are $$ \begin{array}{l} 57-44=13,57-45=12, \\ 57-46=11,57-47=10 . \end{array} $$ Thus, the remaining number is $$ 57-(10+11+12+13)=11 \text {. } $$ Hence, the five numbers are $10, 11, 11, 12, 13$." ff79773b863d,"Example 6. Let the three sides of $\triangle A B C$ be $a, b, c$ with opposite angles $A, B, C$, respectively, and satisfy the condition $a \cos A+b \cos B=c \cos C$. Then $\triangle A B C$ must be ( ). (A) an equilateral triangle (B) a right triangle with $a$ as the hypotenuse (C) a right triangle with $b$ as the hypotenuse (D) none of the above (A), (B), (C) is correct. This is a","the multiple-choice question, just get it right, no need to argue",medium,"Answer the multiple-choice question, just get it right, no need to argue. Therefore, it is valuable to be quick and accurate. From the condition $$ a \cos A + r \cos B = c \cos C $$ it is easy to see that $A(a), [B(b)]$ and $C(c)$ cannot be interchanged, so we know that conclusion (A) is incorrect. Since the above equation is symmetric with respect to $A(a)$ and $B(b)$, we also know that conclusion (B) $\Leftrightarrow$ conclusion (C). Neither of them is correct. Therefore, the answer should be (D). Of course, using the cosine rule, we can derive a more explicit conclusion $a^2 = b^2 + c^2$ or $b^2 = a^2 + c^2$, which means $\triangle ABC$ must be a right triangle, with either $a$ or $b$ as the hypotenuse." c21d116c2c53,"Let's answer the following questions: a) How many ways can 9 people be seated I) on a bench II) around a circular table? b) How many ways can 5 men and 4 women be seated on a bench so that I.) no two people of the same gender sit next to each other? II.) the men and women sit in separate groups (only 1 man and 1 woman sit next to each other)?",See reasoning trace,medium,"a) I. The number of permutations of 9 elements is $9!=362880$. II. Here, the following 9 groups are clearly one to consider: abcdefghi, bcdefghia, cdefghiab, ..., iabcdefgh. Since these cyclic permutations can be done with every permutation group, the number of all possible different arrangements is $\frac{9!}{9}=8!=40320$. (We can also reason as follows: we fix one element and by permuting the remaining 8 elements, we get all possible arrangements around the table.) b) I. They can only sit according to the scheme $f n f n f n f n f$. The 5 men can occupy the $f$-marked places in 5! ways, and the 4 women can sit in the $n$-marked places in 4! ways in each case. Therefore, they can arrange themselves in a total of $5!4!=120 \cdot 24=2880$ ways. II. Here, they can occupy places according to 2 schemes: fffffnnnn or nnnnfffff. Therefore, the number of all possible different arrangements is $2 \cdot 5!4!=5760$. Kovács László (Debrecen, Ref. g. II. o. t.)" e7b8cda20807,"3. (physics) A wire is made into a right-angled triangle with legs of length $l=1$ m. From vertices A and B, two bugs start crawling along the legs simultaneously with speeds $v_{A}=5$ cm/s and $v_{B}=2 v_{A}=10$ cm/s (see figure). What is the minimum time after the start of their movement when the distance between the bugs reaches a minimum? ![](https://cdn.mathpix.com/cropped/2024_05_06_25c5d6100f66704b4a85g-2.jpg?height=354&width=354&top_left_y=1770&top_left_x=1545)",See reasoning trace,medium,"Solution. Let the time that has passed since the beetles started moving be $t$. Then beetle A will travel a distance of $v_{A} t$, beetle B - a distance of $v_{B} t=2 v_{A} t$. Therefore, the distance $S$ between the beetles at this moment will be $$ S=\sqrt{\left(l-v_{A} t\right)^{2}+4 v_{A}^{2} t^{2}} $$ We will find the minimum value of $S$ as a function of time. For this, we will differentiate the function $S(t)$, set the derivative to zero, and solve the resulting equation. Moreover, it is technically more convenient to differentiate $S^{2}$ rather than $S$, as the expressions will be much simpler, and both $S$ and $S^{2}$ reach their maximum at the same value of the argument. Thus, $$ \left(S^{2}\right)^{\prime}=-2\left(l-v_{A} t\right) v_{A}+8 v_{A}^{2} t $$ Setting the derivative (**) to zero, we get an equation for the time that has passed from the start of the movement until the moment when the distance between the beetles is the smallest $$ l-v_{A} t=4 v_{A} t $$ From which $$ t=\frac{l}{5 v_{A}}=4 \mathrm{~s} $$" fd801246af24,"Example 3 Let $a \in \mathbf{R}$, and the function $f(x)=a x^{2}-2 x-2 a$. If the solution set of $f(x)>0$ is $A, B=\{x \mid 10$. (i) When $a>0$, $A=\left\{x \mid xx_{2}\right\}$. The necessary and sufficient condition for $A \cap B \neq \varnothing$ is $x_{2}<3$, which simplifies to $\frac{1}{a}+\sqrt{2+\frac{1}{a^{2}}}<3$. Solving this, we get $a>\frac{6}{7}$. (ii) When $a<0$, $A=\left\{x \mid x_{1}0$, $f(x)$ opens upwards, and as long as $f(1)>0$ or $f(3)>0$, i.e., $a-2-2a>0$ or $9a-6-2a>0$, and given $a>0$, we solve to get $a>\frac{6}{7}$. When $a<0$, $f(x)$ opens downwards, and we need $f(1)<0$ and $f(3)<0$, i.e., $a-2-2a>0$, solving this gives $a<-2$. In summary, the range of $a$ for which $A \cap B=\varnothing$ is $(-\infty,-2) \cup\left(\frac{6}{7},+\infty\right)$." 170401b92b8c,For how many one-digit positive integers $k$ is the product $k \cdot 234$ divisible by 12 ?,4,easy,"Note that $234=9 \cdot 26=2 \cdot 3^{2} \cdot 13$. Thus, $k \cdot 234=k \cdot 2 \cdot 3^{2} \cdot 13$. This product is divisible by $12=2^{2} \cdot 3$ if and only if $k$ contributes another factor of 2 (that is, if and only if $k$ is even). Since $k$ is a one-digit positive integer, then $k=2,4,6,8$. Therefore, there are 4 one-digit positive integers $k$ for which $k \cdot 234$ is divisible by 12 . ANSWER: 4" 933ff2d09183,4. (5 points) The last three digits of the expression $1 \times 1+11 \times 11+111 \times 111+\cdots+111 \cdots 111$ (2010 1’s) $\times 111 \cdots 111$ (2010 1’s) are $\qquad$ .,: 690,medium,"【Analysis】This problem seems difficult, but we can start by calculating the value of each multiplication expression from the first one in the formula to find a pattern: $1 \times 1=1, 11 \times 11=121, 111 \times 111=12321, 1111 \times 1111=1234321, 11111 \times$ $11111=123454321 \cdots$, their products are: $1, 121, 12321, 1234321, 123454321, 12345654321, \cdots$, From this, we can observe that, apart from the first two multiplication expressions whose products are 1 and 121, the last three digits of the products of the subsequent multiplication expressions are all 321. Based on this pattern, we can determine the last three digits of the product of this expression. 【Solution】Solution: By calculation, we get the products of each multiplication expression as follows: 1, 121, 12321, 1234321, 123454321, $12345654321, \cdots$, From this, we can observe that, apart from the first two multiplication expressions whose products are 1 and 121, the last three digits of the products of the subsequent multiplication expressions are all 321; Thus, the sum of the last three digits of each expression in the formula is: $$ \begin{array}{l} 1 + 121 + 321 \times (2010 - 2) \\ = 122 + 64568 \\ = 644690 . \end{array} $$ Therefore, the last three digits of the result of the expression $1 \times 1 + 11 \times 11 + 111 \times 111 + \cdots + 111 \cdots 111$ (2010 ones) $\times 111 \cdots 111$ (2010 ones) are 690. The answer is: 690." 9384d175cbb5,"4. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and $$ a_{1}=3, S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right) \text {. } $$ If $\left\{a_{n}\right\}$ contains three terms $a_{1} 、 a_{p} 、 a_{q}(p 、 q \in$ $\left.\mathbf{Z}_{+}, 1\Delta_{2}$, then $a^{2}-b-8=0$. Thus, the repeated root of equation (2) is $a$, and the two roots of equation (1) are $a+4$ and $a-4$. Therefore, $a^{2}+(a-4)^{2}=(a+4)^{2}$ $$ \begin{array}{l} \Rightarrow a=16 \Rightarrow b=a^{2}-8=248 \\ \Rightarrow a+b=264 . \end{array} $$" 7348a9194627,"The numbers $1,2, \ldots, 2002$ are written in order on a blackboard. Then the 1st, 4th, 7th, $\ldots$, $3k+1$th, $\ldots$ numbers in the list are erased. Then the 1st, 4th, 7th, $\ldots$, $3k+1$th numbers in the remaining list are erased (leaving $3,5,8,9,12, \ldots$ ). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased?",1598,medium,"Answer: 1598. Let $\mathrm{a}_{\mathrm{n}}$ be the first number remaining after $\mathrm{n}$ iterations, so $\mathrm{a}_{0}=1, \mathrm{a}_{1}=2, \mathrm{a}_{3}=3, \mathrm{a}_{4}=5$ etc. We claim that: ``` \(a_{n+1}=3 / 2 a_{n} \quad\) if \(a_{n}\) is even, and \(3 / 2\left(a_{n}+1\right)-1\) if \(a_{n}\) is odd. ``` We use induction on $n$. Suppose $a_{n}=2 N$. Consider the number $3 N$. There are initially $\mathrm{N}$ smaller numbers $=1 \mathrm{mod} 3$. So after the first iteration, it will lie in 2Nth place. Hence, it will lie in first place after $n+1$ iterations. Similarly, suppose $\mathrm{a}_{\mathrm{n}}=2 \mathrm{~N}+1$. Consider $3 N+2$. There are initially $\mathrm{N}+1$ smaller numbers $=1 \bmod 3$. So after the first iteration, it will lie in $2 \mathrm{~N}+1$ st place. Hence, it will lie in first place after $\mathrm{n}+1$ iterations. That completes the induction. We may now calculate successively the members of the sequence: $1,2,3,5,8,12,18,27,41$, $62,93,140,210,315,473,710,1065,1598,2397$. Hence 1598 is the last surviving number from $1,2, \ldots, 2002$." 7f2fa3c7c27c,"21. The formula for the volume of a sphere is $V=\frac{4}{3} \pi r^{3}$, where $r$ is the radius of the sphere. In a cylindrical container, several solid iron spheres with the same radius as the cylinder's base can be placed. When water is poured into the container, the volume of water is six times the volume of one iron sphere when the water level just reaches the top of the container. The number of iron spheres in the container is $\qquad$.",See reasoning trace,easy,$12$ 52b53b83d069,"15 . In a $3 \times 3$ grid, fill in the numbers $1,2,3,4,5,6,7,8,9$ so that the sum of the numbers in any three squares in a row, column, or diagonal is equal.","45,45 \div 3=15$, the sum of numbers in each row and each column is 15.",easy,"【Key Point】Magic Square【Difficulty】Comprehension 【Answer】 【Analysis】 $1+2+3+\cdots \cdots+9=45,45 \div 3=15$, the sum of numbers in each row and each column is 15." 233996b14c03,"In a circle with a radius of 1.4, determine the distance from the center to the chord if it intercepts an arc of $120^{\circ}$. #",0,easy,"The leg opposite the $30^{\circ}$ angle is half the hypotenuse. ## Solution Let $M$ be the foot of the perpendicular dropped from the center $O$ to the chord $A B$. Then $O M$ is the leg of the right triangle $O M A$, lying opposite the angle of $30^{\circ}$. Therefore, $O M=1 / 2 O A=0.7$. ## Answer 0.7." c8a1e0474fdc,"1.29 Calculate the sum of the cubes of two numbers if their sum and product are 11 and 21, respectively.",638,easy,"1.29 Let $a$ and $b$ be the required numbers; then $a+b=11$ and $ab=21$. According to formula (1.11), we have $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$. Therefore, $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$, i.e., $a^{3}+b^{3}=$ $=11^{3}-3 \cdot 21 \cdot 11=638$. Answer: 638." a7ae9cfa9e40,"4. Fold a triangle with side lengths of $10$, $12$, and $14$ along its three midlines to form a tetrahedron. Then the diameter of the circumscribed sphere of the tetrahedron is $\qquad$ .",See reasoning trace,medium,"4. $\sqrt{55}$. The tetrahedron has equal opposite edges, and the lengths of the three pairs of opposite edges are 5, 6, and 7. As shown in Figure 4, let $A B=C D=5, A C=B D=7$, $$ A D=B C=6 \text {. } $$ Take the midpoints $M$ and $N$ of $B C$ and $A D$ respectively. It is easy to see that $M N$ is the common perpendicular segment of $A D$ and $B C$, and noting that $A D=B C$, the midpoint $O$ of $M N$ is equidistant from points $A, B, C, D$, i.e., $O$ is the center of the circumscribed sphere of the tetrahedron $A B C D$, and the radius of the sphere is $$ R=O A=\sqrt{\left(\frac{d}{2}\right)^{2}+\left(\frac{A D}{2}\right)^{2}}(d=M N) . $$ Draw $N E \perp M C$. It is easy to see that $C E \perp$ plane $A E D$. Since $N A=N D=N E=3$, we know $\angle A E D=90^{\circ}$. Then $A E^{2}=A C^{2}-C E^{2}=7^{2}-d^{2}$, $D E^{2}=D C^{2}-C E^{2}=5^{2}-d^{2}$, $A E^{2}+D E^{2}=A D^{2}=6^{2}$. Thus $7^{2}-d^{2}+5^{2}-d^{2}=6^{2} \Rightarrow d^{2}=19$ $$ \Rightarrow 2 R=\sqrt{d^{2}+A D^{2}}=\sqrt{55} \text {. } $$" f1f03e6ca615,"Question 1 In $\triangle A B C$, $A B=A C, \angle C A B$ and $\angle A B C$'s internal angle bisectors intersect the sides $B C$ and $C A$ at points $D$ and $E$ respectively. Let $K$ be the incenter of $\triangle A D C$. If $\angle B E K=45^{\circ}$, find all possible values of $\angle C A B$ ${ }^{[1,2]}$.",60^{\circ}$ or $90^{\circ}$.,medium,"As shown in Figure 1, if $\angle C E K=45^{\circ}$, then because $\angle B E K=45^{\circ}$, we have $\angle B E C=90^{\circ}$. Also, $\angle A B E=\angle C B E$, so $A B=B C$. Therefore, $A B=A C=B C$. Thus, $\angle B A C=60^{\circ}$. As shown in Figure 2, if $\angle C E K \neq 45^{\circ}$, then $\angle B E C \neq 90^{\circ}$. Let $A D$ intersect $B E$ at point $I$. Draw $I F \perp A C$ at point $F$. Connect $D K$ and $F K$. Since $I$ and $K$ are the incenters of $\triangle A B C$ and $\triangle A C D$ respectively, $I$, $K$, and $C$ are collinear. Also, since $A B=A C$ and $A D$ bisects $\angle B A C$, we have $A D \perp B C$. Since $C I$ is the angle bisector of $\angle A C D$, it is easy to see that $\triangle I D K \cong \triangle I F K$. Thus, $\angle I F K=\angle I D K$, and $\angle F K I=\angle D K I$. Since $K$ is the incenter of $\triangle A C D$, we have $\angle I D K=\frac{1}{2} \angle A D C=45^{\circ}$. Therefore, $\angle I F K=45^{\circ}=\angle I E K$. Hence, $I$, $K$, $E$, and $F$ are concyclic. Thus, $\angle F E I=\angle F K I=\angle D K I$. Let $\angle I C D=\alpha$. Then $\angle I B C=\alpha$, and $\angle E C B=2 \alpha$. Therefore, $\angle F E I=\angle I B C+\angle E C B=3 \alpha$, $\angle D K I=\angle I C D+\angle K D C=45^{\circ}+\alpha$. Thus, $3 \alpha=\alpha+45^{\circ} \Rightarrow 2 \alpha=45^{\circ}$ $\Rightarrow \angle A C B=45^{\circ} \Rightarrow \angle B A C=90^{\circ}$. In summary, $\angle B A C=60^{\circ}$ or $90^{\circ}$." 2eb8020e6e36,"Let $B = (20, 14)$ and $C = (18, 0)$ be two points in the plane. For every line $\ell$ passing through $B$, we color red the foot of the perpendicular from $C$ to $\ell$. The set of red points enclose a bounded region of area $\mathcal{A}$. Find $\lfloor \mathcal{A} \rfloor$ (that is, find the greatest integer not exceeding $\mathcal A$). [i]Proposed by Yang Liu[/i]",157,medium,"1. **Determine the distance \( BC \):** - The coordinates of points \( B \) and \( C \) are \( B = (20, 14) \) and \( C = (18, 0) \). - The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] - Applying this formula to points \( B \) and \( C \): \[ BC = \sqrt{(18 - 20)^2 + (0 - 14)^2} = \sqrt{(-2)^2 + (-14)^2} = \sqrt{4 + 196} = \sqrt{200} = 10\sqrt{2} \] 2. **Identify the set of red points:** - For every line \(\ell\) passing through \( B \), the foot of the perpendicular from \( C \) to \(\ell\) will trace out a circle centered at the midpoint of \( BC \) with radius equal to half the distance \( BC \). - The midpoint \( M \) of \( BC \) is: \[ M = \left( \frac{20 + 18}{2}, \frac{14 + 0}{2} \right) = (19, 7) \] - The radius of the circle is: \[ r = \frac{BC}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \] 3. **Calculate the area of the circle:** - The area \( \mathcal{A} \) of a circle with radius \( r \) is given by: \[ \mathcal{A} = \pi r^2 \] - Substituting \( r = 5\sqrt{2} \): \[ \mathcal{A} = \pi (5\sqrt{2})^2 = \pi \cdot 25 \cdot 2 = 50\pi \] 4. **Find the greatest integer not exceeding \( \mathcal{A} \):** - Using the approximation \( \pi \approx 3.14159 \): \[ 50\pi \approx 50 \times 3.14159 = 157.0795 \] - The greatest integer not exceeding \( 157.0795 \) is \( 157 \). The final answer is \( \boxed{157} \)" 57d2dc0f2776,"6.53 rounded to the thousandth place, $\lg 2$ is $0.301, \lg 3$ is $0.477 . \log _{5} 10$ the best approximation is (A) $\frac{8}{7}$. (B) $\frac{9}{7}$. (C) $\frac{10}{7}$. (D) $\frac{11}{7}$. (E) $\frac{12}{7}$. (30th American High School Mathematics Examination, 1979)",$(C)$,easy,"[Solution] Since $\log _{b} a=\frac{1}{\log _{a} b}$, then $$ \log _{5} 10=\frac{1}{\lg 5}=\frac{1}{\lg 10-\lg 2} \approx \frac{1}{0.699} \approx \frac{10}{7} . $$ Therefore, the answer is $(C)$." feca2b16737b,Russian,"1. The new points added at step n+1 have twice the sum of the points after step n, because each old ",easy,": 2.3 n-1 . True for n=1. The new points added at step n+1 have twice the sum of the points after step n, because each old point contributes to two new points. hence the total after step n+1 is three times the total after step n. Russian 31-40 (C) John Scholes jscholes@kalva.demon.co.uk 18 Sep 1998" a6a6fe4248d1,"6. 166 Let $f$ be a function defined on the set of non-negative real numbers and taking values in the same set. Find all functions $f$ that satisfy the following conditions: (i) $f(x f(y)) \cdot f(y)=f(x+y)$; (ii) $f(2)=0$; (iii) $f(x) \neq 0$, when $0 \leqslant x<2$.",See reasoning trace,medium,"[Solution] Let $f$ satisfy the given conditions. When $x \geqslant 2$, we have $$ \begin{aligned} f(x) & =f((x-2)+2) \\ & =f((x-2) f(2)) \cdot f(2) \\ & =0 . \end{aligned} $$ Therefore, $f(x)=0$; If $0 \leqslant x<2$. We set $y \geqslant 2-x$. Then $$ \begin{array}{l} x+y \geqslant 2, \\ f(x+y)=0 . \\ f(y f(x)) \cdot f(x)=f(y+x)=f(x+y)=0 . \end{array} $$ By the given condition, $f(x) \neq 0$, when $0 \leqslant x<2$. Thus $$ \begin{array}{l} f(y f(x))=0, \\ y f(x) \geqslant 2, \\ y \geqslant \frac{2}{f(x)} . \end{array} $$ This means, from $y \geqslant 2-x$, we get $y \geqslant \frac{2}{f(x)}$. But by the reversibility of the above reasoning, from $y \geqslant \frac{2}{f(x)}$, we can also get $y \geqslant 2-x$. Therefore, we have $$ \begin{array}{l} 2-x=\frac{2}{f(x)}, \\ f(x)=\frac{2}{2-x}, \quad 0 \leqslant x<2 \text { when. } \end{array} $$ In summary, the function that satisfies the conditions is $$ f(x)=\left\{\begin{array}{lc} \frac{2}{2-x}, & 0 \leqslant x<2 \text { when; } \\ 0, & x \geqslant 2 \text { when. } \end{array}\right. $$" 2f51e77ca1ea,"3. If $a>b>0$, then the minimum value of $a^{3}+\frac{1}{b(a-b)}$ is $\qquad$ .",See reasoning trace,medium,"$$ \begin{array}{l} \text { 3. } \frac{5}{3} \sqrt[5]{144} . \\ a^{3}+\frac{1}{b(a-b)} \geqslant a^{3}+\frac{4}{a^{2}}=2 \times \frac{a^{3}}{2}+3 \times \frac{4}{3 a^{2}} \\ \geqslant 5 \sqrt[5]{\left(\frac{1}{2}\right)^{2} \times\left(\frac{4}{3}\right)^{3}}=\frac{5}{3} \sqrt[5]{144} . \end{array} $$ The equality holds if and only if $\left\{\begin{array}{l}b=a-b, \\ \frac{a}{2}=\frac{4}{3 a^{2}},\end{array}\right.$ which is $\left\{\begin{array}{l}a=\sqrt[3]{\frac{8}{3}}, \\ b=\frac{1}{2} \sqrt[3]{\frac{8}{3}}\end{array}\right.$. " e63e89b60325,"I2.4 If $d=1-2+3-4+\ldots-c$, find the value of $d$.",(1-2)+(3-4)+\ldots+(99-100) \\ & =-1-1-\ldots-1(50 \text { times }) \\ & =-50\end{aligned}$,easy,$\begin{aligned} d & =(1-2)+(3-4)+\ldots+(99-100) \\ & =-1-1-\ldots-1(50 \text { times }) \\ & =-50\end{aligned}$ 5acbdb2ef8d4,"In a triangle $ABC$ with $ \angle A = 36^o$ and $AB = AC$, the bisector of the angle at $C$ meets the oposite side at $D$. Compute the angles of $\triangle BCD$. Express the length of side $BC$ in terms of the length $b$ of side $AC$ without using trigonometric functions.",BC = \frac{b,medium,"1. **Identify the given information and draw the triangle:** - Given: \( \angle A = 36^\circ \) - \( AB = AC \) (Isosceles triangle) - The angle bisector of \( \angle C \) meets \( AB \) at \( D \). 2. **Determine the angles of the triangle:** - Since \( \triangle ABC \) is isosceles with \( AB = AC \), the base angles are equal. - Let \( \angle B = \angle C = x \). - Using the angle sum property of a triangle, we have: \[ \angle A + \angle B + \angle C = 180^\circ \] \[ 36^\circ + x + x = 180^\circ \] \[ 2x = 144^\circ \] \[ x = 72^\circ \] - Therefore, \( \angle B = \angle C = 72^\circ \). 3. **Determine the angles of \( \triangle BCD \):** - The angle bisector of \( \angle C \) divides \( \angle C \) into two equal parts. - Therefore, \( \angle DCB = \frac{72^\circ}{2} = 36^\circ \). - Since \( \angle BCD \) is an external angle for \( \triangle BCD \), we have: \[ \angle BCD = \angle B + \angle DCB = 72^\circ + 36^\circ = 108^\circ \] - The remaining angle \( \angle CBD \) can be found using the angle sum property of a triangle: \[ \angle BCD + \angle CBD + \angle DCB = 180^\circ \] \[ 108^\circ + \angle CBD + 36^\circ = 180^\circ \] \[ \angle CBD = 180^\circ - 144^\circ = 36^\circ \] 4. **Express the length of side \( BC \) in terms of \( b \):** - Since \( \angle A = 36^\circ \) and \( AB = AC \), \( \triangle ABC \) can be inscribed in a circle with radius \( b \). - The side \( BC \) is the side length of a regular decagon inscribed in a circle of radius \( b \). - The side length \( s \) of a regular decagon inscribed in a circle of radius \( R \) is given by: \[ s = R \left( \frac{\sqrt{5} - 1}{2} \right) \] - Therefore, for \( R = b \): \[ BC = b \left( \frac{\sqrt{5} - 1}{2} \right) \] The final answer is \( \boxed{ BC = \frac{b}{2} (\sqrt{5} - 1) } \)" d3640a74a4d3,"13. 6 different points are given on the plane, no three of which are collinear. Each pair of points is to be joined by a red line or a blue line subject to the following restriction: if the lines joining $A B$ and $A C$ (where $A, B, C$ denote the given points) are both red, then the line joining $B C$ is also red. How many different colourings of the lines are possible? (2 marks) 在平面上給定 6 個不同的點, 當中沒有三點共線。現要把任意兩點均以一條紅線或監線連起, 並須符合以下規定: 若 $A B$ 和 $A C$ (這裡 $A 、 B 、 C$ 代表給定的點）均以紅線連起, 則 $B C$ 亦必須以紅線連起。那麼, 連線的顏色有多少個不同的組合？",$1+6+15+15+10+60+20+15+45+15+1=203$,medium,"13. 203 13. We only have to consider which points are to be connected by red lines, and the rest of the lines must be blue. Note that the restriction means that the 6 points must be divided into some number of groups, such that two points are joined by a red line if and only if they are in the same group. So the question becomes counting the number of such groupings, and we count using the pattern of grouping, as follows: \begin{tabular}{|c|l|} \hline Pattern & \\ \hline 6 & 1 \\ \hline $5-1$ & $C_{1}^{6}=6$ \\ \hline $4-2$ & $C_{4}^{6}=15$ \\ \hline $4-1-1$ & $C_{4}^{6}=15$ \\ \hline $3-3$ & $C_{3}^{6} \div 2=10$ \\ \hline $3-2-1$ & $C_{3}^{6} \times C_{2}^{3}=60$ \\ \hline $3-1-1-1$ & $C_{3}^{6}=20$ \\ \hline $2-2-2$ & $C_{2}^{6} \times C_{2}^{4} \div 3!=15$ \\ \hline $2-2-1-1$ & $C_{2}^{6} \times C_{2}^{4} \div 2=45$ \\ \hline $2-1-1-1-1$ & $C_{2}^{6}=15$ \\ \hline $1-1-1-1-1-1$ & 1 \\ \hline \end{tabular} Hence the answer is $1+6+15+15+10+60+20+15+45+15+1=203$. Remark. This question essentially asks for the number of partitions of the set $\{1,2,3,4,5$, 6\}. The answer is in fact the 6th Bell mumber." 88ba30bcdf2e,"2. Several numbers form an arithmetic progression, where their sum is 63, and the first term is one and a half times the common difference of the progression. If all terms of the progression are decreased by the same amount so that the first term of the progression is equal to the common difference of the progression, then the sum of all numbers will decrease by no more than 8, but no less than 7. Determine what the common difference of this progression can be.",the difference is $21 / 8$ or 2,medium,"# Problem 2. Answer: the difference is $21 / 8$ or 2. Solution. Let the numbers $a_{1}, a_{2}, \ldots, a_{n}$ form an arithmetic progression with a common difference $d$. Then $a_{1}+a_{2}+\cdots+a_{n}=63, a_{1}=3 d / 2 \Rightarrow 0.5 d(n+2) n=63$. After the described change in the condition, these numbers will again form an arithmetic progression with the same difference $d$, but with a different first term, now equal to $d$. Therefore, their sum will be equal to $0.5 d(n+1) n$, and according to the condition, the inequality $55 \leqslant 0.5 d(n+1) n \leqslant 56$ is satisfied. Considering the first relation, we get $$ \frac{55}{63} \leqslant \frac{n+1}{n+2} \leqslant \frac{56}{63} \Longleftrightarrow \frac{47}{8} \leqslant n \leqslant 7 $$ Therefore, $n=6$ or $n=7$, and the common difference $d$ is respectively equal to $21 / 8$ or 2." 4d5e6c842210,"Determine the number of pairs of integers $(a, b)$ such that $1 \leqslant a \leqslant 30, 3 \leqslant b \leqslant 30$ and such that $a$ is divisible by $b$ and by $b-2$.",22,easy,"Answer: 22. For $b=3$, the number of multiples of $b(b-2)=3$ which are $\leqslant 30$ is equal to 10. For $b=4$, the condition is that 4 divides $a$, there are 7 solutions. We find similarly 2 solutions for $b=5$, 2 solutions for $b=6$ and 1 solution for $b=8$. It is easy to show that there are no other solutions." 4fad12048002,"2. Simplify the expression $$ A=\left(1+\frac{1+i}{2}\right)\left(1+\left(\frac{1+i}{2}\right)^{2}\right)\left(1+\left(\frac{1+i}{2}\right)^{4}\right)\left(1+\left(\frac{1+i}{2}\right)^{8}\right) $$ ( $i$ is the imaginary unit)",See reasoning trace,easy,"Solution. Since $$ \left(\frac{1+i}{2}\right)^{2}=\frac{i}{2}, \quad\left(\frac{1+i}{2}\right)^{4}=\frac{-1}{4}, \quad\left(\frac{1+i}{2}\right)^{8}=\frac{1}{16} $$ we have $$ A=\left(1+\frac{1+i}{2}\right)\left(1+\left(\frac{1+i}{2}\right)^{2}\right)\left(1+\left(\frac{1+i}{2}\right)^{4}\right)\left(1+\left(\frac{1+i}{2}\right)^{8}\right)=\frac{1}{2}(3+i) \frac{1}{2}(2+i) \cdot \frac{3}{4} \cdot \frac{17}{16}=\frac{255}{256}(1+i) $$" 03fa2594e35e,"In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals: $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$",\textbf{(D),easy,"Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$. Because $\triangle CAE$ and $\triangle CEB$ share an altitude, \[c + 3b = \tfrac{3}{2} (3a+a+2b)\] \[c + 3b = 6a + 3b\] \[c = 6a\] Because $\triangle ACD$ and $\triangle ABD$ share an altitude, \[6a+3a = 3(a+2b+3b)\] \[9a = 3a+15b\] \[6a = 15b\] \[a = \tfrac{5}{2}b\] Thus, $[CAP] = 15b$, and since $[APE] = 3b$, $r = \tfrac{CP}{PE} = 5$, which is answer choice $\boxed{\textbf{(D)}}$." 89194d8e4e91,"At the base of the pyramid $A B C D$ lies an isosceles right triangle $A B C$ with hypotenuse $A B=4$. The height of the pyramid is 2, and its base coincides with the midpoint of $A C$. Find the dihedral angle between the faces $A B D$ and $A D C$.",$\arcsin \sqrt{\frac{3}{5}}$,medium,"Triangle $ABC$ is isosceles and right-angled, with its hypotenuse $AB$ equal to 4, so $AC = BC = 2\sqrt{2}$ and $$ S_{\triangle ABC} = \frac{1}{2} \cdot (2\sqrt{2})^2 = 4 $$ Let $H$ be the midpoint of $AC$, and $K$ be the projection of point $H$ onto the hypotenuse $AB$. Then $DH = 2$, and $HK = 1$ as half the height $CL$ of triangle $ABC$. From the right triangles $DKH$ and $ADH$, we find that $$ DK = \sqrt{HK^2 + DH^2} = \sqrt{1 + 4} = \sqrt{5}, \quad AD = \sqrt{AH^2 + DH^2} = \sqrt{2 + 4} = \sqrt{6} $$ Then $$ S_{\triangle ACD} = \frac{1}{2} AC \cdot DH = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = 2\sqrt{2} $$ From the theorem of three perpendiculars, it follows that $DK \perp AB$, so $$ S_{\triangle ABD} = \frac{1}{2} AB \cdot DK = \frac{1}{2} \cdot 4 \cdot \sqrt{5} = 2\sqrt{5} $$ We will calculate the volume of the pyramid in two ways. On one hand, $$ V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DH = \frac{1}{3} \cdot 4 \cdot 2 = \frac{8}{3} $$ On the other hand, if $\alpha$ is the angle between the faces $ABD$ and $ACD$, then $$ V_{ABCD} = \frac{2}{3} \cdot \frac{S_{\triangle ABD} \cdot S_{\triangle ACD} \sin \alpha}{AD} = \frac{2}{3} \cdot \frac{2\sqrt{5} \cdot 2\sqrt{2}}{\sqrt{6}} = \frac{8\sqrt{10}}{3\sqrt{6}} \sin \alpha $$ From the equation $$ \frac{8\sqrt{10}}{3\sqrt{6}} \sin \alpha = \frac{8}{3} $$ we find that $\sin \alpha = \sqrt{\frac{3}{5}}$. ## Answer $\arcsin \sqrt{\frac{3}{5}}$" b01e0bfd1ff7,"15. Given $f(x)=x^{2}+c$, and $f(f(x))=f\left(x^{2}+1\right)$. (1) Let $g(x)=f(f(x))$, find the analytical expression of the function $g(x)$; (2) Let $\varphi(x)=g(x)-\lambda f(x)$, try to find the value of the real number $\lambda$ such that $\varphi(x)$ is a decreasing function on $(-\infty,-1]$ and an increasing function on $[-1,0)$.","g(x)-\lambda f(x)=x^{4}+(2-\lambda) x^{2}+(2-\lambda)$, set $x_{1}1+1+2-\lambda=4-\lambda$, to make ",medium,"15. (1) $f(f(x))=f\left(x^{2}+c\right)=\left(x^{2}+c\right)^{2}+c, f\left(x^{2}+1\right)=\left(x^{2}+1\right)^{2}+c$, we get $c=1$, $g(x)=x^{4}+2 x^{2}+2$. (2) $\varphi(x)=g(x)-\lambda f(x)=x^{4}+(2-\lambda) x^{2}+(2-\lambda)$, set $x_{1}1+1+2-\lambda=4-\lambda$, to make $\varphi(x)$ decreasing on $(-\infty, -1]$, it is only necessary that $4-\lambda \geqslant 0$, i.e., $\lambda \leqslant 4$. Similarly, it can be proved that when $\lambda \geqslant 4$, $\varphi(x)$ is increasing on $(-1,0)$. Therefore, $\lambda=4$." fb59d385cd69,"## Task B-4.3. The lines $p_{1} \ldots y=\frac{1}{4} x, p_{2} \ldots y=9 x$ and the point $T(6,-1)$ are given. Determine the equation of the line that passes through the point $T$, and intersects the x-axis, lines $p_{1}, p_{2}$, and the y-axis at points $A, B, C, D$ respectively, such that $|A B|=|C D|$.",See reasoning trace,medium,"## Solution. ![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-17.jpg?height=922&width=736&top_left_y=767&top_left_x=660) The line passing through point $T$ has the equation: $$ \begin{aligned} & y+1=k(x-6) \\ & y=k x-(6 k+1) \end{aligned} $$ To intersect the $x$-axis, $p_{1}, p_{2}$, and the $y$-axis, the line must not be parallel to the given lines and coordinate axes, nor pass through the origin, i.e., it must satisfy: $$ k \neq \frac{1}{4}, k \neq 9, k \neq 0, k \neq-\frac{1}{6} $$ The coordinates of points $A, B, C, D$ are determined by solving the system of equations of the respective lines: $$ \begin{aligned} & A\left(\frac{6 k+1}{k}, 0\right) \\ & B\left(\frac{4(6 k+1)}{4 k-1}, \frac{6 k+1}{4 k-1}\right) \\ & C\left(\frac{6 k+1}{k-9}, \frac{9(6 k+1)}{k-9}\right) \\ & D(0,-6 k-1) \end{aligned} $$ Since $|A B|=|C D|$, we have: $$ \begin{aligned} & \left(\frac{4(6 k+1)}{4 k-1}-\frac{6 k+1}{k}\right)^{2}+\left(\frac{6 k+1}{4 k-1}\right)^{2}=\left(\frac{6 k+1}{k-9}\right)^{2}+\left(\frac{9(6 k+1)}{k-9}+(6 k+1)\right)^{2} /:(6 k+1)^{2} \\ & \left(\frac{4}{4 k-1}-\frac{1}{k}\right)^{2}+\left(\frac{1}{4 k-1}\right)^{2}=\left(\frac{1}{k-9}\right)^{2}+\left(\frac{9}{k-9}+1\right)^{2} \\ & \frac{1}{(4 k-1)^{2} k^{2}}+\frac{1}{(4 k-1)^{2}}=\frac{1}{(k-9)^{2}}+\frac{k^{2}}{(k-9)^{2}} \\ & \frac{1+k^{2}}{(4 k-1)^{2} k^{2}}=\frac{1+k^{2}}{(k-9)^{2}} \\ & (4 k-1)^{2} k^{2}=(k-9)^{2} \\ & |(4 k-1) k|=|k-9| \end{aligned} $$ We have the following possibilities: 1. $$ 4 k^{2}-k=k-9 \Rightarrow 4 k^{2}-2 k+9=0 \Rightarrow \text { no real solutions. } $$ 2. $$ 4 k^{2}-k=9-k \Rightarrow 4 k^{2}=9 \Rightarrow k_{1,2}= \pm \frac{3}{2} $$ There are two lines with the desired property: $$ p^{\prime} \ldots y=\frac{3}{2} x-10, p^{\prime \prime} \ldots y=-\frac{3}{2} x+8 $$" 7c68af7c41e3,"4.38. Find the highest power of two that divides the number $(n+1)(n+2) \cdot \ldots \cdot 2 n$. $$ * * * $$",1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n-1)$ and $(2 n)!!=2 \cdot 4 \cdot 6 \cdot \ldots \cd,easy,"4.38. Let $(2 n-1)!!=1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n-1)$ and $(2 n)!!=2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n$. It is clear that $(2 n)!=(2 n-1)!!(2 n)!!$ and $(2 n)!!=2^{n} n!$. Therefore, $(n+1)(n+2) \times \ldots$ $\cdots \times 2 n=\frac{(2 n)!}{n!}=\frac{(2 n-1)!!(2 n)!!}{n!}=2^{n}(2 n-1)!!$, where the number $(2 n-1)!!$ is odd." c17f891830b9,"1. Petar and Toso came to visit Misha in city $A$. After some time, they decided to return to their city $B$ and organized the transport in the following way: Toso cycled from city $A$ to city $B$. Misha drove Petar towards city $B$ for some time and then left him to continue on foot, while he immediately turned back towards his city $A$. All of them left city $A$ at the same time and arrived in their respective cities at the same time. If Petar walks at $6 \mathrm{~km} / \mathrm{h}$ and Toso cycles at $22 \mathrm{~km} / \mathrm{h}$, at what speed did Misha drive the motor?",38$.,medium,"Solution. Let's denote by $s$ the distance from $A$ to $B$ and by $t$ the time they needed to reach their destinations. The speeds of the motor, bicycle, and walking, i.e., of Misha, Tosh, and Petar, let's denote them by $v_{M}, v_{T}$, and $v_{\Pi}$ respectively. Tosh was riding the bicycle from $A$ to $B$ the whole time, so, $$ v_{T} \cdot t=s $$ Misha, for the time $t$, drove a certain amount of time towards $B$ and then immediately returned to his city $A$. This means that he spent half of the time driving towards $B$, and in the second half, he returned to $A$. This means that Petar spent half of the time riding on the motor with Misha, and in the second half of the time, he continued walking to $B$. Since Petar traveled the distance $s$, it can be written as $$ v_{M} \frac{t}{2}+v_{\Pi} \frac{t}{2}=s $$ The right sides of (1) and (2) are equal, so, $$ v_{T} \cdot t=v_{M} \frac{t}{2}+v_{\Pi} \frac{t}{2} $$ from which we get $v_{M}=38$." d923636c7508,"6. If the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=2$, and $$ \begin{aligned} a_{n+2}= & 2(2 n+3)^{2} a_{n+1}- \\ & 4(n+1)^{2}(2 n+1)(2 n+3) a_{n} \\ & (n \geqslant 0), \end{aligned} $$ then the general term formula of the sequence $\left\{a_{n}\right\}$ is $\qquad$ .",See reasoning trace,medium,"$$ \begin{array}{l} \text { 6. } a_{n}=\left\{\begin{array}{ll} 0, & n=0 ; \\ (2 n)!\sum_{i=1}^{n} \frac{1}{i}, & n \geqslant 1 . \end{array}\right. \\ \text { Let } v_{n}=\frac{a_{n}}{(2 n)!} . \end{array} $$ Then $v_{0}=0, v_{1}=1$, and $$ \begin{array}{l} (n+2) v_{n+2}=(2 n+3) v_{n+1}-(n+1) v_{n} \\ \Rightarrow(n+2)\left(v_{n+2}-v_{n+1}\right) \\ \quad=(n+1)\left(v_{n+1}-v_{n}\right)=\cdots=v_{1}-v_{0}=1 \\ \Rightarrow v_{n}=\sum_{i=1}^{n} \frac{1}{i} \Rightarrow a_{n}=(2 n)!\sum_{i=1}^{n} \frac{1}{i} . \end{array} $$" 54cda2d17cfc,"If $a, b$ and $c$ are positive integers with $a \times b=13, b \times c=52$, and $c \times a=4$, the value of $a \times b \times c$ is (A) 2704 (B) 104 (C) 676 (D) 208 (E) 52",See reasoning trace,medium,"If $a$ and $b$ are positive integers with $a \times b=13$, then since 13 is a prime number, we must have $a$ and $b$ equal to 1 and 13 , or 13 and 1 , respectively. If $b=1$, then since $b \times c=52$, we must have $c=52$. But then we could not have $c \times a=4$. So $b$ cannot be 1 . Thus, $b=13$, so $a=1$, and $c=4$ (which we can get either from knowing $b=13$ and $b \times c=52$, or from knowing $a=1$ and $c \times a=4$. Therefore, $a \times b \times c=1 \times 13 \times 4=52$. ANSWER: (E) #" 869ac9821776,"3. The integer solutions $(x, y)$ of the equation $2 x^{2}+x y+y^{2}-x+2 y+1=0$ are $\qquad$ _.",See reasoning trace,easy,"3. $(0,-1),(1,-1),(1,-2)$. The original equation can be transformed into $$ 2 x^{2}+(y-1) x+y^{2}+2 y+1=0 \text {. } $$ Considering $\Delta=(y-1)^{2}-8(y+1)^{2} \geqslant 0$. Solving yields $\frac{-9-4 \sqrt{2}}{7} \leqslant y \leqslant \frac{-9+4 \sqrt{2}}{7}$. Thus, $y=-1,-2$. Therefore, when $y=-1$, $x=0,1$; when $y=-2$, $x=1$. Hence, the integer solutions are $(0,-1),(1,-1)$, $(1,-2)$." fdfb463fe0a1,"Folklore Given a regular nonagon. In how many ways can three of its vertices be chosen so that they form the vertices of an isosceles triangle?",30 ways,easy,"For every two vertices of the nonagon, there exists exactly one vertex that is equidistant from them, so each of the resulting isosceles but not equilateral triangles is uniquely determined by its base. The number of ways to choose two points out of nine is $C_{9}^{2}=36$. However, with this method of counting, each of the three equilateral triangles is counted three times. Therefore, the desired number is $36-6=30$. ## Answer 30 ways" 1dd46a9ad0a9,"On a circular table are sitting $ 2n$ people, equally spaced in between. $ m$ cookies are given to these people, and they give cookies to their neighbors according to the following rule. (i) One may give cookies only to people adjacent to himself. (ii) In order to give a cookie to one's neighbor, one must eat a cookie. Select arbitrarily a person $ A$ sitting on the table. Find the minimum value $ m$ such that there is a strategy in which $ A$ can eventually receive a cookie, independent of the distribution of cookies at the beginning.",2^n,medium,"1. **Numbering and Weight Assignment**: - Number the people \(0, 1, \ldots, 2n-1\) such that person \(A\) is numbered \(0\). - Assign weights to each person: the weight of person \(i\) is \(2^{-|i \mod 2n|}\). - Specifically, the weight of person \(0\) is \(1\), the weights of persons \(1\) and \(2n-1\) are \(1/2\), and the weight of person \(n\) is \(1/2^n\). 2. **Value of a Scenario**: - Define the value of a scenario as the sum of the product of the weight and the number of cookies held by each person. - To prove that \(m \geq 2^n\), consider giving all \(m\) cookies to person \(n\). The initial value of this scenario is \(\frac{m}{2^n}\). 3. **Operations and Value**: - Operations (giving cookies to neighbors) do not increase the value of the scenario. - If person \(A\) (person \(0\)) receives a cookie, the value of the scenario must be at least \(1\). Therefore, \(m \geq 2^n\) is necessary. 4. **Sufficiency of \(m = 2^n\)**: - Consider an arbitrary distribution of cookies such that there are \(x\) cookies not held by person \(n\). - Either persons \(1, \ldots, n-1\) have at least \(x/2\) cookies, or persons \(n+1, \ldots, 2n-1\) do. Without loss of generality, assume the former. - Delete persons \(n+1, \ldots, 2n-1\) and their cookies. The resulting value of the scenario is at least: \[ \frac{2^n - x}{2^n} + \frac{x/2}{2^{n-1}} = 1 \] 5. **Restated Problem**: - Suppose we have \(n+1\) people in a line numbered \(0\) through \(n\). - Define weights and value as before. If the value is at least \(1\), we need to show that person \(0\) can receive a cookie. 6. **Strategy**: - Make moves ""towards \(0\)"" (i.e., person \(i\) gives cookies to person \(i-1\) but never to \(i+1\)) until no more moves can be made. - These moves do not change the value of the scenario, so it remains at least \(1\). 7. **Contradiction**: - Suppose person \(0\) does not have a cookie. Since no more moves can be made, person \(0\) has no cookies, and persons \(1, \ldots, n\) have at most \(1\) cookie each. - The value of the scenario is at most \(1 - 2^{-n} < 1\), which is a contradiction. Thus, the minimum value \(m\) such that there is a strategy in which person \(A\) can eventually receive a cookie, independent of the initial distribution, is \(2^n\). The final answer is \(\boxed{2^n}\)." 2c5e970349b2,"$28 \cdot 30$ Divide the 8 numbers $12,30,42,44,57,91,95,143$ into two groups so that the product of the numbers in each group is equal. The correct grouping is (A) $12,42,57,143$ and $30,44,91,95$. (B) $12,30,95,143$ and $42,44,57,91$. (C) $12,42,95,143$ and $30,44,57,91$. (D) $12,44,95,143$ and $30,42,57,91$. (3rd ""Five Sheep Cup"" Junior High School Mathematics Competition, 1991)",$(C)$,medium,"[Solution] Given $12=2^{2} \cdot 3, 30=2 \cdot 3 \cdot 5, 42=2 \cdot 3 \cdot 7$, $\begin{array}{ll}44=2^{2} \cdot 11, & 57=3 \cdot 19, \\ 95=5 \cdot 19, & 143=11 \cdot 13,\end{array}$ $91=7 \cdot 13$, we can create the following factor table: \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline$y$ & 2 & 3 & 5 & 7 & 11 & 13 & 19 \\ \hline 12 & $\vee v$ & $\checkmark$ & & & & & \\ \hline 30 & $\checkmark$ & $\checkmark$ & $v$ & & & & \\ \hline 42 & $\vee$ & $v$ & & $\checkmark$ & & & \\ \hline 44 & $\vee v$ & & & & $v$ & & \\ \hline 57 & & $\checkmark$ & & & & & $v$ \\ \hline 91 & & & & $\checkmark$ & & $\vee$ & \\ \hline 95 & & & $v$ & & & & $v$ \\ \hline 143 & & & & & $v$ & $v$ & \\ \hline \end{tabular} From the table, we can see: For groups with the same product, 95 and 57 should be in different groups. Similarly, 91 and 143; 44 and 143; 42 and 91; 30 and 95, etc., should all be in different groups. This way, we can divide them into two groups: 12, 42, 95, 143 and 30, 44, 57, 91. Therefore, the answer is $(C)$." 93612912e315,"1. Before leaving for the store, Mom asked Anya, Varya, and Svetlana to do some work. The girls work with the same productivity. At first, they worked together, but after 20 minutes, Anya had to leave, and Varya and Svetlana continued working alone. After another 15 minutes, Varya also had to leave, and Svetlana finished the work alone for 10 more minutes. When Mom returned home, she brought the girls 10 identical apples. How should the girls divide these apples among themselves in accordance with the work they did?",- 0 points,medium,"# Solution If the entire work is divided into equal parts, each of which one girl completes in 5 minutes, then Anya completed 4 parts (worked for 20 minutes), Varya - 7 parts (worked for 35 minutes), and Svetlana - 9 parts (worked for 45 minutes). In total, there are 20 such parts. Therefore, the apples should be distributed as follows: Anya gets $-10 \times \frac{4}{20}=2$ apples, Varya gets $-10 \times \frac{7}{20}=3.5$ apples, and Svetlana gets $-10 \times \frac{9}{20}=4.5$ apples. ## Grading Criteria Only the correct answer - 0 points. The solution process is correct, but the wrong answer is obtained due to an arithmetic error - 3 points. Correct solution - 7 points." 702280a6409b,"Example 36 (1992 Shanghai High School Competition Question) Let $n$ be a given natural number, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$. Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$. Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$. Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$. Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$. ![](https://cdn.mathpix.com/cropped/2024_05_06_86512448fd32765ac040g-40.jpg?height=521&width=571&top_left_y=83&top_left_x=437) Fig. 12: to the solution of problem 10.7 It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$." 7ab19b297ef0,"1) Two nonnegative real numbers $x, y$ have constant sum $a$. Find the minimum value of $x^m + y^m$, where m is a given positive integer. 2) Let $m, n$ be positive integers and $k$ a positive real number. Consider nonnegative real numbers $x_1, x_2, . . . , x_n$ having constant sum $k$. Prove that the minimum value of the quantity $x^m_1+ ... + x^m_n$ occurs when $x_1 = x_2 = ... = x_n$.",\frac{k^m,medium,"### Part 1: 1. Given two nonnegative real numbers \( x \) and \( y \) such that \( x + y = a \), we need to find the minimum value of \( x^m + y^m \) where \( m \) is a given positive integer. 2. By the Power Mean Inequality, for nonnegative real numbers \( x \) and \( y \) and for \( m \geq 1 \): \[ \sqrt[m]{\frac{x^m + y^m}{2}} \geq \frac{x + y}{2} \] 3. Since \( x + y = a \), we have: \[ \sqrt[m]{\frac{x^m + y^m}{2}} \geq \frac{a}{2} \] 4. Raising both sides to the power of \( m \): \[ \frac{x^m + y^m}{2} \geq \left(\frac{a}{2}\right)^m \] 5. Multiplying both sides by 2: \[ x^m + y^m \geq 2 \left(\frac{a}{2}\right)^m = 2 \cdot \frac{a^m}{2^m} = \frac{a^m}{2^{m-1}} \] 6. Equality holds if and only if \( x = y \). Given \( x + y = a \), this occurs when \( x = y = \frac{a}{2} \). Therefore, the minimum value of \( x^m + y^m \) is: \[ \boxed{\frac{a^m}{2^{m-1}}} \] ### Part 2: 1. Let \( x_1, x_2, \ldots, x_n \) be nonnegative real numbers with a constant sum \( k \). We need to prove that the minimum value of \( x_1^m + x_2^m + \cdots + x_n^m \) occurs when \( x_1 = x_2 = \cdots = x_n \). 2. By the Power Mean Inequality, for nonnegative real numbers \( x_1, x_2, \ldots, x_n \) and for \( m \geq 1 \): \[ \sqrt[m]{\frac{x_1^m + x_2^m + \cdots + x_n^m}{n}} \geq \frac{x_1 + x_2 + \cdots + x_n}{n} \] 3. Since \( x_1 + x_2 + \cdots + x_n = k \), we have: \[ \sqrt[m]{\frac{x_1^m + x_2^m + \cdots + x_n^m}{n}} \geq \frac{k}{n} \] 4. Raising both sides to the power of \( m \): \[ \frac{x_1^m + x_2^m + \cdots + x_n^m}{n} \geq \left(\frac{k}{n}\right)^m \] 5. Multiplying both sides by \( n \): \[ x_1^m + x_2^m + \cdots + x_n^m \geq n \left(\frac{k}{n}\right)^m = n \cdot \frac{k^m}{n^m} = \frac{k^m}{n^{m-1}} \] 6. Equality holds if and only if \( x_1 = x_2 = \cdots = x_n \). Given \( x_1 + x_2 + \cdots + x_n = k \), this occurs when \( x_1 = x_2 = \cdots = x_n = \frac{k}{n} \). Therefore, the minimum value of \( x_1^m + x_2^m + \cdots + x_n^m \) is: \[ \boxed{\frac{k^m}{n^{m-1}}} \]" 9548bc2b7c09,"There are lily pads in a row numbered $0$ to $11$, in that order. There are predators on lily pads $3$ and $6$, and a morsel of food on lily pad $10$. Fiona the frog starts on pad $0$, and from any given lily pad, has a $\frac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$? $\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14$",\textbf{(A),medium,"Firstly, notice that if Fiona jumps over the predator on pad $3$, she must land on pad $4$. Similarly, she must land on $7$ if she makes it past $6$. Thus, we can split the problem into $3$ smaller sub-problems, separately finding the probability. Fiona skips $3$, the probability she skips $6$ (starting at $4$) and the probability she doesn't skip $10$ (starting at $7$). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be $1 - \text{the probability obtained in the first sub-problem}$. In the analysis below, we call the larger jump a $2$-jump, and the smaller a $1$-jump. For the first sub-problem, consider Fiona's options. She can either go $1$-jump, $1$-jump, $2$-jump, with probability $\frac{1}{8}$, or she can go $2$-jump, $2$-jump, with probability $\frac{1}{4}$. These are the only two options, so they together make the answer $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$. We now also know the answer to the last sub-problem is $1-\frac{3}{8}=\frac{5}{8}$. For the second sub-problem, Fiona must go $1$-jump, $2$-jump, with probability $\frac{1}{4}$, since any other option would result in her death to a predator. Thus, since the three sub-problems are independent, the final answer is $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \boxed{\textbf{(A) }\frac{15}{256}}$." b77c066b6e77,"Rice grains were placed on the cells of a chessboard. The number of grains on any two adjacent cells differed by exactly 1. At the same time, there were three grains on one of the cells of the board, and 17 grains on another. A rooster pecked all the grains from one of the main diagonals of the board, and a hen - from the other. How many grains did the rooster get and how many did the hen get? #",80 grains each,medium,"Let, for example, cell $A$, which contains three grains, is $k$ cells to the left and $n$ cells below cell $B$, which contains 17 grains. Consider the shortest paths leading from cell $A$ to cell $B$. Each such path consists of $k$ steps to the neighboring cell to the right and $n$ steps to the neighboring cell upwards, made in any order, that is, in any case, such a path consists of $k+n$ steps. At each step, the number of grains changes by exactly 1, and over the entire path, the number of grains changes by $17-3=14$. However, given the size of the chessboard, the number of possible steps both upwards and to the right cannot exceed 7, meaning the shortest path cannot contain more than 14 steps. Therefore, there must be exactly 7 steps to the right and exactly 7 steps upwards, and the number of grains must increase at each step. Thus, if cell $A$ is below and to the left of cell $B$, then both lie on the main diagonal leading to the upper right ($A$ in the lower left corner, and $B$ in the upper right). In this case, the number of grains in each cell is uniquely determined (see figure). | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Then, on the specified main diagonal, there are $3+5+7+9+11+13+15+17=80$ grains. On the other main diagonal, there are $8 \cdot 10=80$ grains. Other possible cases of the placement of these cells differ from the one considered only by symmetry (axial or central). ## Answer 80 grains each." 2ed6d8a4c0ca,"On a board, the numbers from 1 to 2009 are written. A couple of them are erased and instead of them, on the board is written the remainder of the sum of the erased numbers divided by 13. After a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number.",8,medium,"1. **Define the invariant:** Let \( I \) be the remainder when the sum of all the integers on the board is divided by 13. Initially, the numbers from 1 to 2009 are written on the board. The sum of these numbers is: \[ S = \sum_{k=1}^{2009} k = \frac{2009 \times 2010}{2} = 2009 \times 1005 \] We need to find \( S \mod 13 \). 2. **Calculate \( 2009 \mod 13 \) and \( 1005 \mod 13 \):** \[ 2009 \div 13 = 154 \quad \text{remainder} \quad 7 \quad \Rightarrow \quad 2009 \equiv 7 \pmod{13} \] \[ 1005 \div 13 = 77 \quad \text{remainder} \quad 4 \quad \Rightarrow \quad 1005 \equiv 4 \pmod{13} \] 3. **Calculate \( S \mod 13 \):** \[ S = 2009 \times 1005 \equiv 7 \times 4 = 28 \equiv 2 \pmod{13} \] Therefore, \( I = 2 \). 4. **Determine the third number \( a \):** After several operations, only three numbers are left on the board: 9, 999, and \( a \). The sum of these three numbers modulo 13 must be equal to the invariant \( I \): \[ a + 9 + 999 \equiv 2 \pmod{13} \] 5. **Calculate \( 999 \mod 13 \):** \[ 999 \div 13 = 76 \quad \text{remainder} \quad 11 \quad \Rightarrow \quad 999 \equiv 11 \pmod{13} \] 6. **Set up the equation:** \[ a + 9 + 11 \equiv 2 \pmod{13} \] Simplify the equation: \[ a + 20 \equiv 2 \pmod{13} \] Subtract 20 from both sides: \[ a \equiv 2 - 20 \pmod{13} \] \[ a \equiv -18 \pmod{13} \] Since \(-18 \equiv -18 + 26 \equiv 8 \pmod{13}\), we have: \[ a \equiv 8 \pmod{13} \] 7. **Find the possible values of \( a \) between 1 and 2009:** The possible values of \( a \) are of the form: \[ a = 8 + 13k \quad \text{for integer} \quad k \] We need \( 1 \leq a \leq 2009 \): \[ 1 \leq 8 + 13k \leq 2009 \] \[ -7 \leq 13k \leq 2001 \] \[ 0 \leq k \leq 154 \] Therefore, the possible values of \( a \) are: \[ a = 8, 21, 34, \ldots, 2002 \] The final answer is \( \boxed{8} \)." 3f196f6e6ffb,12.289. All lateral faces of the pyramid form the same angle with the base plane. Find this angle if the ratio of the total surface area of the pyramid to the area of the base is $k$. For what values of $k$ does the,"$\arccos \frac{1}{k-1}, k>2$",medium,"Solution. Let $S$ be the total surface area, $S_{1}$ be the lateral surface area of the pyramid, and $S_{2}$ be the area of its base. Since all lateral faces form the same angle $\alpha$ with the plane of the base, then $S_{1}=\frac{S_{2}}{\cos \alpha}$. Further, $S=S_{1}+S_{2}=\frac{S_{2}}{\cos \alpha}+S_{2}=\frac{S_{2}(1+\cos \alpha)}{\cos \alpha}$. By the condition $\frac{S_{1}}{S_{2}}=k \Rightarrow \frac{1+\cos \alpha}{\cos \alpha}=k \Leftrightarrow(k-1) \cos \alpha=1 \Rightarrow \alpha=\arccos \frac{1}{k-1}$. Since $k>0$ and $0<\alpha<\frac{\pi}{2}$, it follows that $k>2$. Answer: $\arccos \frac{1}{k-1}, k>2$." e09d07b1e791,"## Calculate the indefinite integral: $$ \int \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x $$",See reasoning trace,medium,"## Solution $$ \int \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x= $$ Substitution: $$ \begin{aligned} & y=x \cdot \sin x \\ & d y=(\sin x+x \cdot \cos x) d x \end{aligned} $$ We get: $$ =\int \frac{d y}{y^{2}}=-\frac{1}{y}+C= $$ Reverse substitution: $$ =-\frac{1}{x \cdot \sin x}+C $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+3-12 » Categories: Kuznetsov's Problem Book Integrals Problem $3 \mid$ Integrals Ukrainian Banner Network - Last modified on this page: 17:12, 1 March 2009. - Content is available under CC-BY-SA 3.0. Created by GeeTeatoo ## Problem Kuznetsov Integrals 3-13 ## Material from Plusi" cc9535f38331,Answer. 9.,"$28<30$, a contradiction. An example is shown in the drawing.",easy,"Solution. Evaluation. Besides the ends, the broken line has 30 vertices, each of which is the intersection point of two lines. If the lines are no more than 8, then the intersection points are no more ![](https://cdn.mathpix.com/cropped/2024_06_03_376d2fa4074cfcd15764g-2.jpg?height=700&width=722&top_left_y=1569&top_left_x=1027) than 7.8/2 = $28<30$, a contradiction. An example is shown in the drawing." 29240b1fdcd1,"7. If $\alpha, \beta, \gamma$ are acute angles, and $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$, then the maximum value of $\frac{\sin \alpha+\sin \beta+\sin \gamma}{\cos \alpha+\cos \beta+\cos \gamma}$ is $\qquad$ .",See reasoning trace,medium,"7. $\frac{\sqrt{2}}{2}$ Detailed Explanation: From $\frac{a^{2}+b^{2}}{2} \geqslant\left(\frac{a+b}{2}\right)^{2} \Rightarrow a+b \leqslant \sqrt{2\left(a^{2}+b^{2}\right)}$, thus $\sin \alpha+\sin \beta \leqslant \sqrt{2\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)}=\sqrt{2} \cos \gamma$. Similarly: $\sin \beta+\sin \gamma \leqslant \sqrt{2} \cos \alpha, \sin \gamma+\sin \alpha \leqslant \sqrt{2} \cos \beta$. Therefore, $2(\sin \alpha+\sin \beta+\sin \gamma) \leqslant \sqrt{2}(\cos \alpha+\cos \beta+\cos \gamma)$, hence $\frac{\sin \alpha+\sin \beta+\sin \gamma}{\cos \alpha+\cos \beta+\cos \gamma} \leqslant \frac{\sqrt{2}}{2}$." 7fac70a8e408,"In five bags, there are a total of 100 walnuts. The first two bags together have 52, the second and third bags together have 43, the third and fourth bags together have 34, while the last two bags together contain 30 walnuts. How many walnuts are in each of the bags?",See reasoning trace,medium,"The second and third, fourth and fifth bags together contain $43+30=73$ sunflower seeds. Since the five bags together contain 100 sunflower seeds, there are 27 seeds in the first bag. Thus, the second bag has $52-27=25$, the third bag has $43-25=18$, the fourth bag has $34-18=16$, and the last bag has $30-16=14$ seeds. With this, we have solved the problem. Note: Many solvers unnecessarily set up long equations. This problem can be solved with simple logical reasoning, and using a system of five equations is overly complicated." 194845d2a49f,"Example 4 There are $n$ squares arranged in a row, to be painted with red, yellow, and blue. Each square is painted one color, with the requirement that no two adjacent squares are the same color, and the first and last squares are also different colors. How many ways are there to paint them? (1991, Jiangsu Mathematics Competition)",See reasoning trace,medium,"Solution: Let there be $a_{n}$ ways of coloring, it is easy to see that $a_{1}=3, a_{2}=$ $6, a_{3}=6$, and when $n \geqslant 4$, after numbering the $n$ cells in sequence, cell 1 and cell $(n-1)$ are not adjacent. Case 1: The color of cell $(n-1)$ is different from that of cell 1. In this case, cell $n$ has only one color to choose from, and the first $(n-1)$ cells satisfy the condition that the first and last cells are of different colors, so there are $a_{n-1}$ ways of coloring. Case 2: The color of cell $(n-1)$ is the same as that of cell 1. In this case, the color of cell $(n-2)$ must be different from that of cell 1, otherwise, cell $(n-1)$ and cell $(n-2)$ would be the same color. Thus, the first $(n-2)$ cells have $a_{n-2}$ ways of coloring. Since cell $n$ must be a different color from cell 1, there are two ways to color it, so there are $2 a_{n-2}$ ways of coloring in total. In summary, we get the recursive equation $$ a_{n}=a_{n-1}+2 a_{n-2} \cdot(n \geqslant 4) $$ Solving this, we get $a_{n}=2^{n}+2(-1)^{n} .(n \geqslant 2, n \in \mathbf{N})$ For $n$ positions in a circular arrangement, the solution can be derived similarly. This idea is also applicable to solving probability problems." 4af87abb0879,"1. Let $x, y$ be positive real numbers. Find the minimum value of $$ x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x} $$ (Gu Bin, Jin Aiguo)","2$, hence the minimum value is 2.",medium,"1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$. If $x \geqslant 1, y \geqslant 1$, then $$ f(x, y) \geqslant x+y \geqslant 2 \text {; } $$ If $00, y>0$, we have $f(x, y) \geqslant 2$. Also, $f(1,1)=2$, hence the minimum value is 2." 5d80deabcf13,"B4. A sheet of paper has the shape of a rectangle with sides measuring $12.6 \mathrm{~cm}$ and $10 \mathrm{~cm}$. a) This sheet is rolled into the mantle of a cylinder such that the shorter side of the rectangle is the height of the cylinder. Calculate the volume of the cylinder formed by this rolled paper, accurate to the nearest cubic centimeter. b) If we cut out squares with a side length of $3 \mathrm{~cm}$ from the corners of the rectangle, we get a net of a box without a lid. Draw a sketch of this net, determine the lengths of the edges of the box, and calculate the volume of this box. c) Calculate the area of the smallest lid that will cover the box. ## Solutions to ## A contestant who arrives at the correct solution by any valid method (even if the scoring guide does not anticipate it) receives all possible points. A valid method is one that - sensibly takes into account the wording of the - leads to the solution of the - is mathematically correct and complete. A contestant who has only partially solved the ## PART I The correct answers are recorded in the table. A correct answer by the contestant is scored with 2 points, an incorrect answer with -1 point, and a blank field in the table with 0 points. | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | Answer | C | A | B | C | D | E |",See reasoning trace,medium,"B4. The height of the cylinder is $10 \mathrm{~cm}$, and the circumference of its base is $12.6 \mathrm{~cm}$. From the circumference, we express the radius $r=\frac{12.6}{2 \pi}=\frac{6.3}{\pi} \mathrm{cm}$. The volume of the cylinder is $V=\pi r^{2} \cdot h=\pi \frac{(6.3)^{2}}{\pi^{2}} \cdot 10=\frac{(6.3)^{2}}{\pi} \cdot 10 \doteq 126 \mathrm{~cm}^{3}$. The lengths of the edges of the box are $6.6 \mathrm{~cm}, 4 \mathrm{~cm}$, and $3 \mathrm{~cm}$ (see the figure). The volume of the box is $6.6 \cdot 4 \cdot 3=79.2 \mathrm{~cm}^{3}$. The smallest cover that can cover the box is a rectangle with side lengths of $6.6 \mathrm{~cm}$ and $4 \mathrm{~cm}$. Its area is $26.4 \mathrm{~cm}^{2}$. ![](https://cdn.mathpix.com/cropped/2024_06_07_e4711ece0d44ad9bc5c3g-8.jpg?height=372&width=491&top_left_y=565&top_left_x=1405) Scoring: Total: 7 points $\longmapsto 12.6 \longrightarrow$ a) The height and circumference of the base of the cylinder ($h=10 \mathrm{~cm}, o=12.6 \mathrm{~cm}$) and the radius $r=\frac{12.6}{2 \pi} \mathrm{cm}$ ..... $1 \mathrm{p}$ Calculated volume of the cylinder: $V \doteq 126 \mathrm{~cm}^{3}$ ..... 1 p b) Drawn sketch: ..... 1 p The lengths of the edges of the box are: $6.6 \mathrm{~cm}, 4 \mathrm{~cm}$, and $3 \mathrm{~cm}$ ..... $2 \mathrm{p}$ Volume of the box: $V=79.2 \mathrm{~cm}^{3}$ ..... $1 \mathrm{p}$ c) The area of the cover is $26.4 \mathrm{~cm}^{2}$. ..... $1 \mathrm{p}$" e2c0f42c3fc1,"Zhenodarov R.G. Eight cells of one diagonal of a chessboard are called a fence. The rook moves around the board, not stepping on the same cell twice and not stepping on the cells of the fence (intermediate cells are not considered visited). What is the maximum number of jumps over the fence that the rook can make?",See reasoning trace,medium,"Evaluation. Note that if the rook jumps over a fence, then either the initial or the final cell of the jump is marked gray on the diagram. Since there are 24 gray cells and through each one a maximum of two jumps can be made, there can be no more than 48 jumps in total. At the same time, if there are exactly 48 jumps, then from each gray cell, two jumps to white cells must be made (in the previous count, a jump from a gray cell to a gray cell would be counted twice!). Then all moves from dark-gray cells will lead to a white square below the diagonal, and from there - only to dark-gray cells (or to other cells in the same square). This means that in this way, the rook will never reach the light-gray cells. Contradiction. Thus, the number of jumps does not exceed 47. An example with 47 jumps is shown in the diagram (the numbers in the cells indicate the order in which the rook passes through them)." 028f88dab5dd,"## 9. Diagonal Squares Vlado covered the diagonal of a large square with a side length of $2020 \mathrm{~cm}$ using a row of smaller squares with a side length of $4 \mathrm{~cm}$ cut from green collage paper. The diagonals of the green squares align with the diagonal of the large square, and the intersection of any two consecutive green squares is a square with a side length of 1 cm. Calculate the area of the shape formed by the green squares. Express the result in square decimeters and round to the nearest whole number. ## Result: $\quad 101$",See reasoning trace,medium,"## Solution, First Method. Let's calculate how many squares of side length $4 \mathrm{~cm}$ there are in total. Start by drawing squares starting from the bottom left corner of the large square, which we will label as $A_{0}$. The top right corner of the first drawn square will be labeled as $A_{1}$, the top right corner of the second drawn square as $A_{2}$, and so on. From vertex $\mathrm{A}_{0}$ to vertex $\mathrm{A}_{1}$, one can move $4 \mathrm{~cm}$ to the right and $4 \mathrm{~cm}$ up. To reach vertex $A_{2}$, one needs to move another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up, and to reach vertex $A_{3}$, another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up. Thus, after the first square, each subsequent square moves to a point that is $3 \mathrm{~cm}$ to the right and $3 \mathrm{~cm}$ up relative to the last point on the diagonal of the large square. Since adding squares moves equally upwards and to the right, it is sufficient to look at the rightward movements. From the starting point, the sum of all movements must equal the length of the side of the large square, which is $2020 \mathrm{~cm}$. For the first square, the movement is $4 \mathrm{~cm}$, and for each subsequent square, it is $3 \mathrm{~cm}$. Let $n$ be the total number of squares of side length $4 \mathrm{~cm}$ in this sequence. Then we have $4 \mathrm{~cm} + (n-1) \cdot 3 \mathrm{~cm} = 2020 \mathrm{~cm}$. This simplifies to $3n + 1 = 2020$, from which we get $n = 673$. When calculating the area, we observe that the first square is complete, and all subsequent squares are squares of side length $4 \mathrm{~cm}$ with a square of side length $1 \mathrm{~cm}$ cut out. The area is $16 + 672 \cdot 15 = 10096 \mathrm{~cm}^{2} = 100.96 \mathrm{~dm}^{2}$. Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$. ## Solution, Second Method. From the area of the initial square with side length $2020 \mathrm{~cm}$, we will subtract the parts that are not green. We have two congruent parts, each consisting of rectangles of height $3 \mathrm{~cm}$ and lengths $3 \mathrm{~cm}, 6 \mathrm{~cm}, \ldots, 2016 \mathrm{~cm}$. The sum of their areas in $\mathrm{cm}^{2}$ is $$ \begin{gathered} 2 \cdot (3 \cdot 3 + 3 \cdot 6 + \cdots + 3 \cdot 2016) = 2 \cdot 3 \cdot 3 \cdot (1 + 2 + \cdots + 672) = 2 \cdot 3 \cdot 3 \cdot \frac{672 \cdot 673}{2} \\ = 2016 \cdot 2019 \end{gathered} $$ Thus, the final result in $\mathrm{cm}^{2}$ is: $$ \begin{aligned} 2020 \cdot 2020 - & 2016 \cdot 2019 = 2020 \cdot 2020 - (2020 - 4) \cdot (2020 - 1) \\ & = 2020 \cdot 2020 - 2020 \cdot 2020 + 4 \cdot 2020 + 1 \cdot 2020 - 4 \\ & = 5 \cdot 2020 - 4 \\ & = 10096 \end{aligned} $$ The result is $100.96 \mathrm{~dm}^{2}$. Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$." f322c8eb5191,"5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number.",$\mathbf{k}=49$,medium,"Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$. We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that (for $k=46$) $2013 / 462013$ ); (for $k=47$) $2013 / 472013$ ); (for $k=48$) $2013 / 482013$ ); but (for $k=49$) $2013 / 49 > 41$ (since $\quad 41 \cdot 49 = 45^2 - 1^2 = 2025 - 16 < 2013$). Thus, the interval $[2013 / 49, 2013 / 48)$ does not contain any integers, and for $k=49$, there are no suitable $n$. ## Answer: $\mathbf{k}=49$. ## Criteria. The problem is complex, so don't be stingy with points for reasonable thoughts. Any answer in the vicinity of $\sqrt{2014}$, which is preceded by something reasonable, is worth points. A point is deducted for the absence of what is in small print." e2d9dbb68749,"## Task B-4.6. Determine the area of the set of all points associated with complex numbers $z$ for which $$ |z| \leqslant\left|\frac{1}{z}\right| \quad \text{and} \quad \frac{28277 \pi}{12} \leqslant \arg \left(\frac{z^{2020}}{1+i}\right) \leqslant \frac{40397 \pi}{12} $$",\frac{1}{4} r^{2} \pi=\frac{\pi}{4}$.,medium,"## Solution. Let $z=r(\cos \varphi+i \sin \varphi)$ be the trigonometric form of the complex number $z$. From $|z| \leqslant\left|\frac{1}{z}\right|$ it follows that $|z|^{2} \leqslant 1$, so $r \leqslant 1$. We have $$ \begin{aligned} \arg \left(\frac{z^{2020}}{1+i}\right) & =\arg \frac{r^{2020}(\cos 2020 \varphi+i \sin 2020 \varphi)}{\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)} \\ & =\arg \left[\frac{r^{2020}}{\sqrt{2}}\left(\cos \left(2020 \varphi-\frac{\pi}{4}\right)+i \sin \left(2020 \varphi-\frac{\pi}{4}\right)\right)\right] \\ & =2020 \varphi-\frac{\pi}{4} \end{aligned} $$ Therefore, from the given inequality it follows that $$ \frac{28277 \pi}{12} \leqslant 2020 \varphi-\frac{\pi}{4} \leqslant \frac{40397 \pi}{12} $$ or $\frac{7 \pi}{6} \leqslant \varphi \leqslant \frac{5 \pi}{3}$. ![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-25.jpg?height=605&width=628&top_left_y=1562&top_left_x=634) The given set of points is a circular sector with radius 1 and central angle $\frac{5 \pi}{3}-\frac{7 \pi}{6}=\frac{\pi}{2}$. The area is $P=\frac{1}{4} r^{2} \pi=\frac{\pi}{4}$." 174082b2b948,"10. (20 points) Let $x_{1}, x_{2}, x_{3}$ be non-negative real numbers, satisfying $x_{1}+x_{2}+x_{3}=1$. Find the minimum and maximum values of $$ \left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) $$","\frac{1}{2}, x_{2}=0, x_{3}=\frac{1}{2}$, the equality holds, so the desired maximum value is $\frac",medium,"10. By Cauchy's inequality, we have $$ \begin{array}{l} \left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\ \geqslant\left(\sqrt{x_{1}} \sqrt{x_{1}}+\sqrt{3 x_{2}} \sqrt{\frac{x_{2}}{3}}+\sqrt{5 x_{3}} \sqrt{\frac{x_{3}}{5}}\right)^{2} \\ =\left(x_{1}+x_{2}+x_{3}\right)^{2}=1, \end{array} $$ When $x_{1}=1, x_{2}=0, x_{3}=0$, the equality holds, so the desired minimum value is 1. $$ \begin{array}{l} \text { Also, }\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\ =\frac{1}{5}\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(5 x_{1}+\frac{5 x_{2}}{3}+x_{3}\right) \\ \leqslant \frac{1}{5} \times \frac{1}{4}\left(\left(x_{1}+3 x_{2}+5 x_{3}\right)+\left(5 x_{1}+\frac{5 x_{2}}{3}+x_{3}\right)\right)^{2} \\ =\frac{1}{20}\left(6 x_{1}+\frac{14}{3} x_{2}+6 x_{3}\right)^{2} \\ \leqslant \frac{1}{20}\left(6 x_{1}+6 x_{2}+6 x_{3}\right)^{2}=\frac{9}{5}, \end{array} $$ When $x_{1}=\frac{1}{2}, x_{2}=0, x_{3}=\frac{1}{2}$, the equality holds, so the desired maximum value is $\frac{9}{5}$." 2d869c117990,"## Task 5 $50-24+36$; $$ 24+36+32 $$ $$ 45+26-11 ; \quad 100-27-42 $$",See reasoning trace,easy,$62 ; 60 ; 92 ; 31$ d19c5dc68874,"B2. The table shows the hair color of the girls in a certain class. | Hair Color | Number of Girls | | :--- | :---: | | blonde | 8 | | brown | 7 | | red | 3 | | black | 2 | a) Write the percentage of girls who have red or black hair. b) Write the percentage of girls who would need to change their hair color to black so that $20 \%$ of the girls in the class would have black hair. c) How many redheads would need to join the class so that $32 \%$ of the class would be redheads?",48&width=1687&top_left_y=1575&top_left_x=230),medium,"B2. a) 5 students have red or black hair, which is $\frac{5}{20}$ or $25 \%$ of all students. b) $20 \%$ of black-haired students in the class corresponds to 4 students, which means that two more students need to dye their hair black, i.e., $10 \%$. c) If $x$ students with red hair join the class, there will be $x+3$ students with red hair. The total number of students will be $x+20$. Since $32 \%=\frac{8}{25}$, we can write the equation $\frac{3+x}{20+x}=\frac{8}{25}$. The solution to the equation is 5. ![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=69&width=1688&top_left_y=1322&top_left_x=227) ![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=62&width=1685&top_left_y=1371&top_left_x=228) ![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=63&width=1685&top_left_y=1422&top_left_x=228) Calculating the percentage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point ![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=68&width=1685&top_left_y=1525&top_left_x=228) ![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=48&width=1687&top_left_y=1575&top_left_x=230)" 214af4ebf757,"A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals. Suppose the 2006-gon has been dissected into triangles by 2003 nonintersecting diagonals. Find the maximum possible number of isosceles triangles with two odd sides. (Serbia)",See reasoning trace,medium,"Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity. Lemma. Let \( AB \) be one of the dissecting diagonals and let \( \mathcal{L} \) be the shorter part of the boundary of the 2006-gon with endpoints \( A, B \). Suppose that \( \mathcal{L} \) consists of \( n \) segments. Then the number of iso-odd triangles with vertices on \( \mathcal{L} \) does not exceed \( n / 2 \). Proof. This is obvious for \( n=2 \). Take \( n \) with \( 2 < n \leq 1003 \) and assume the claim to be true for every \( \mathcal{L} \) of length less than \( n \). Let now \( \mathcal{L} \) (endpoints \( A, B \)) consist of \( n \) segments. Let \( PQ \) be the longest diagonal which is a side of an iso-odd triangle \( PQS \) with all vertices on \( \mathcal{L} \) (if there is no such triangle, there is nothing to prove). Every triangle whose vertices lie on \( \mathcal{L} \) is obtuse or right-angled; thus \( S \) is the summit of \( PQS \). We may assume that the five points \( A, P, S, Q, B \) lie on \( \mathcal{L} \) in this order and partition \( \mathcal{L} \) into four pieces \( \mathcal{L}_{AP}, \mathcal{L}_{PS}, \mathcal{L}_{SQ}, \mathcal{L}_{QB} \) (the outer ones possibly reducing to a point). By the definition of \( PQ \), an iso-odd triangle cannot have vertices on both \( \mathcal{L}_{AP} \) and \( \mathcal{L}_{QB} \). Therefore every iso-odd triangle within \( \mathcal{L} \) has all its vertices on just one of the four pieces. Applying to each of these pieces the induction hypothesis and adding the four inequalities we get that the number of iso-odd triangles within \( \mathcal{L} \) other than \( PQS \) does not exceed \( n / 2 \). And since each of \( \mathcal{L}_{PS}, \mathcal{L}_{SQ} \) consists of an odd number of sides, the inequalities for these two pieces are actually strict, leaving a \( 1 / 2 + 1 / 2 \) in excess. Hence the triangle \( PQS \) is also covered by the estimate \( n / 2 \). This concludes the induction step and proves the lemma. The remaining part of the solution in fact repeats the argument from the above proof. Consider the longest dissecting diagonal \( XY \). Let \( \mathcal{L}_{XY} \) be the shorter of the two parts of the boundary with endpoints \( X, Y \) and let \( XYZ \) be the triangle in the dissection with vertex \( Z \) not on \( \mathcal{L}_{XY} \). Notice that \( XYZ \) is acute or right-angled, otherwise one of the segments \( XZ, YZ \) would be longer than \( XY \). Denoting by \( \mathcal{L}_{XZ}, \mathcal{L}_{YZ} \) the two pieces defined by \( Z \) and applying the lemma to each of \( \mathcal{L}_{XY}, \mathcal{L}_{XZ}, \mathcal{L}_{YZ} \) we infer that there are no more than 2006/2 iso-odd triangles in all, unless \( XYZ \) is one of them. But in that case \( XZ \) and \( YZ \) are odd diagonals and the corresponding inequalities are strict. This shows that also in this case the total number of iso-odd triangles in the dissection, including \( XYZ \), is not greater than 1003. This bound can be achieved. For this to happen, it just suffices to select a vertex of the 2006-gon and draw a broken line joining every second vertex, starting from the selected one. Since 2006 is even, the line closes. This already gives us the required 1003 iso-odd triangles. Then we can complete the triangulation in an arbitrary fashion." 6dce25a81f40,"3. The three points $A, B, C$ form a triangle. $A B=4, B C=5 A C=6$. Let the angle bisector of $\angle A$ intersect side $B C$ at $D$. Let the foot of the perpendicular from $B$ to the angle bisector of $\angle A$ be $E$. Let the line through $E$ parallel to $A C$ meet $B C$ at $F$. Compute $D F$.","5$, and since $E F \| A C, \frac{D F}{D C}=\frac{D E}{D A}$, so $D F=\frac{D C}{6}=\frac{1}{2}$.",medium,"Answer: $1 / 2$ Since $A D$ bisects $\angle A$, by the angle bisector theorem $\frac{A B}{B D}=\frac{A C}{C D}$, so $B D=2$ and $C D=3$. Extend $B E$ to hit $A C$ at $X$. Since $A E$ is the perpendicular bisector of $B X, A X=4$. Since $B, E, X$ are collinear, applying Menelaus' Theorem to the triangle $A D C$, we have $$ \frac{A E}{E D} \cdot \frac{D B}{B C} \cdot \frac{C X}{X A}=1 $$ This implies that $\frac{A E}{E D}=5$, and since $E F \| A C, \frac{D F}{D C}=\frac{D E}{D A}$, so $D F=\frac{D C}{6}=\frac{1}{2}$." 59e69696db45,"Solve for real $a$ and $b$: $$ a+b < ab+1 $$",See reasoning trace,easy,"The inequality is equivalent to $a b-a-b+1>0$, or $(a-1)(b-1)>0$. Thus, $(a-1)$ and $(b-1)$ must be non-zero and of the same sign. The solutions to this inequality are then the pairs $(a, b)$ such that either $a>1$ and $b>1$, or $a<1$ and $b<1$." afce00fb71e6,"2. (8 points) A shepherd was herding a flock of sheep to graze. After one ram ran out, he counted the number of sheep and found that the ratio of rams to ewes among the remaining sheep was 7:5. After a while, the ram that ran out returned to the flock, but a ewe ran out instead. The shepherd counted the sheep again and found that the ratio of rams to ewes was 5:3. How many sheep were there in the flock originally?",: 25,easy,"【Answer】Solution: According to the analysis, at the beginning, one ram was missing, the ratio was $7: 5=14: 10$, later, the ram returned to the flock, then the number of rams should be increased by 1, and the number of ewes should be decreased by 1, at this point the ratio is $15: 10=(14+1):(10-1)$, therefore, the original number of rams was 15, and the number of ewes was: 10, the total number of sheep was: $15+10=25$. So the answer is: 25." f0f0a753ee50,"If a right triangle is drawn in a semicircle of radius $1 / 2$ with one leg (not the hypotenuse) along the diameter, what is the triangle's maximum possible area?","3 x^{2}-4 x^{3}=0$, or $x=3 / 4$. Therefore the maximum area is $\frac{3 \sqrt{3}}{32}$.",medium,"Solution: It is easy to see that we will want one vertex of the triangle to be where the diameter meets the semicircle, so the diameter is divided into segments of length $x$ and $1-x$, where $x$ is the length of the leg on the diameter. The other leg of the triangle will be the geometric mean of these two numbers, $\sqrt{x(1-x)}$. Therefore the area of the triangle is $\frac{x \sqrt{x(1-x)}}{2}$, so it will be maximized when $\frac{d}{d x}\left(x^{3}-x^{4}\right)=3 x^{2}-4 x^{3}=0$, or $x=3 / 4$. Therefore the maximum area is $\frac{3 \sqrt{3}}{32}$." 6e46da264d8d,32. How many five-digit numbers are there that read the same from left to right and right to left?,900$.,easy,"32. The first digit can be chosen in nine ways, and the second and third digits can be chosen in ten ways each. Since the fourth digit must match the second, and the fifth must match the first, each of the last two digits can be chosen in only one way. Thus, the number of such numbers will be $9 \cdot 10 \cdot 10 \cdot 1 \cdot 1=900$." 017c4ce16c00,(10) The solution set of the inequality $\frac{8}{(x+1)^{3}}+\frac{10}{x+1}-x^{3}-5 x>0$ is,"t^{3}+5 t$, it is easy to know that $f(t)$ is monotonically increasing on $\mathbf{R}$, and when $x ",medium,"$10(-\infty,-2) \cup(-1,1)$ Hint: Transform the original inequality into $$ \left(\frac{2}{x+1}\right)^{3}+5 \cdot \frac{2}{x+1}>x^{3}+5 x \text {, } $$ Let $f(t)=t^{3}+5 t$, it is easy to know that $f(t)$ is monotonically increasing on $\mathbf{R}$, and when $x \neq-1$, the original inequality can be transformed into $f\left(\frac{2}{x+1}\right)>f(x)$, thus $\frac{2}{x+1}>x$, which is equivalent to $\left\{\begin{array}{l}x+1>0, \\ 2>x^{2}+x\end{array}\right.$ or $\left\{\begin{array}{l}x+1<0, \\ 21$; b) convergent geometric series $\sum_{n=1}^{\infty} c q^{n}$ for $0 AC$, then $AF$ must pass through the ( ) of $\triangle ABC$. (A) Incenter (B) Circumcenter (C) Centroid (D) Orthocenter",$\mathrm{A}$,easy,"(3) Let the three sides and the semi-perimeter of $\triangle A B C$ be $a, b, c, p$, respectively, then $$ B Q=A E=p-b, B P=C D=p-b, $$ Therefore, $B Q=B P$. Since $L M$ is parallel to $A B$, we have $$ L F=L P=B P-B L=p-b-\frac{a}{2}=\frac{c-b}{2}, $$ Thus, $$ \begin{array}{l} F M=L M-L F=\frac{b}{2}=A M, \\ \angle M A F=\angle A F M=\angle F A B, \end{array} $$ Therefore, $A F$ is the angle bisector of $\angle A$. Hence, the answer is $\mathrm{A}$." cd5b22b31fc0,"How many triangles exist whose sides are integers and the perimeter measures 12 units? (a) 1 (b) 3 (c) 5 (d) 7 (e) 9","5$, we have $a + b = 7$ and the possible values of $a$ and $b$ are $a = 2$ and $b = 5$, or $a = 3$ a",medium,"The correct option is (b). For three numbers $a, b$, and $c$ to be the lengths of the sides of a triangle, each of them must be greater than the difference and less than the sum of the other two. Let $a \leq b \leq c$ be the lengths of the sides of the triangle, so that $c < a + b$. Now, adding $c$ to both sides, we have $2c < a + b + c = 12$, that is, $2c < 12$, so $c < 6$. Moreover, since $3c \geq a + b + c = 12$, we have that $c \geq 4$, so $4 \leq c < 6$. In the case $c = 5$, we have $a + b = 7$ and the possible values of $a$ and $b$ are $a = 2$ and $b = 5$, or $a = 3$ and $b = 4$. In the case $c = 4$, we have $a + b = 8$, and therefore, the only solution is $a = b = 4$. Conclusion: we have 3 possible triangles." 00b292412d4e,1. Find the locus of the intersection point of two perpendicular intersecting tangents to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.,"0\) is the condition that any tangent slope \( k \) must satisfy, and there are only two roots for \",medium,"This problem can be solved in several ways. When explaining this problem, I combine various methods provided by students to give the following solution: Given the equation of the tangent line with a known slope: $$ \begin{aligned} & y=k x \pm \sqrt{a^{2} k^{2}-b^{2}} \\ \Longrightarrow & y-k x= \pm \sqrt{a^{2} k^{2}}-b^{2} \\ \Rightarrow & \left(a^{2}-x^{2}\right) k^{2}+2 x k+b^{2}-y^{2}=0 . \end{aligned} $$ Since the tangents are perpendicular to each other, and according to Vieta's formulas, we have $$ \begin{aligned} \frac{b^{2}-y^{2}}{a^{2}-x^{2}}=-1 \\ \Rightarrow x^{2}+y^{2}=a^{2}+b^{2} . \end{aligned} $$ This is the locus of the intersection points of the two tangents. Some students are half-convinced and half-doubtful, finding it hard to understand. The fundamental reason is the lack of deep understanding of the equation and the meaning of the parameter \( k \). In fact, the equation \(\left(a^{2}-x^{2}\right) k^{2}+2 x k+b^{2}-y^{2}=0\) is the condition that any tangent slope \( k \) must satisfy, and there are only two roots for \( k \) in the equation. These two roots are the slopes of the two tangents drawn through the point \((x, y)\). Since the two tangents are perpendicular, according to Vieta's formulas, the equation of the circle can be derived. If the idea of combining numbers and shapes is not used to understand \( k \), it would naturally be difficult to grasp the essence of this solution." f5d4aca6b348,"## Calculate the definite integral: $$ \int_{0}^{3} \frac{e^{\sqrt{(3-x) /(3+x)}} d x}{(3+x) \sqrt{9-x^{2}}} $$",See reasoning trace,medium,"## Solution $$ \int_{0}^{3} \frac{e^{\sqrt{(3-x) /(3+x)}} d x}{(3+x) \sqrt{9-x^{2}}}= $$ Substitution: $$ \begin{aligned} & t=\sqrt{\frac{3-x}{3+x}} \\ & d t=\frac{1}{2} \cdot \sqrt{\frac{3+x}{3-x}} \cdot\left(\frac{3-x}{3+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{3+x}{3-x}} \cdot \frac{-6}{(3+x)^{2}} \cdot d x=-\frac{3 d x}{\sqrt{9-x^{2}}(3+x)} \\ & x=0 \Rightarrow t=\sqrt{\frac{3-0}{3+0}}=1 \\ & x=3 \Rightarrow t=\sqrt{\frac{3-3}{3+3}}=0 \end{aligned} $$ We get: $$ =\int_{1}^{0} e^{t} \cdot\left(\frac{-d t}{3}\right)=-\frac{1}{3} \cdot \int_{1}^{0} e^{t} d t=-\left.\frac{e^{t}}{3}\right|_{1} ^{0}=-\frac{e^{0}}{3}+\frac{e^{1}}{3}=\frac{e-1}{3} $$ Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-30$ » Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals - Last edited: 13:58, July 3, 2009. - Content is available under CC-BY-SA 3.0. ## Problem Kuznetsov Integrals 11-31 ## Material from PlusPi" 450f3130f487,"For real numbers $x, y$ and $z$ it is known that $x + y = 2$ and $xy = z^2 + 1$. Find the value of the expression $x^2 + y^2+ z^2$.",2,medium,"1. Given the equations: \[ x + y = 2 \] \[ xy = z^2 + 1 \] 2. Consider the quadratic equation \( t^2 - (x+y)t + xy = 0 \) with roots \( x \) and \( y \). By Vieta's formulas, we know: \[ x + y = 2 \quad \text{(sum of the roots)} \] \[ xy = z^2 + 1 \quad \text{(product of the roots)} \] 3. Substitute \( x + y = 2 \) and \( xy = z^2 + 1 \) into the quadratic equation: \[ t^2 - 2t + (z^2 + 1) = 0 \] 4. Solve the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{2 \pm \sqrt{4 - 4(z^2 + 1)}}{2} \] \[ t = \frac{2 \pm \sqrt{4 - 4z^2 - 4}}{2} \] \[ t = \frac{2 \pm \sqrt{-4z^2}}{2} \] \[ t = \frac{2 \pm 2i|z|}{2} \] \[ t = 1 \pm i|z| \] 5. For \( x \) and \( y \) to be real numbers, the imaginary part must be zero. Therefore: \[ i|z| = 0 \implies |z| = 0 \implies z = 0 \] 6. With \( z = 0 \), the quadratic equation simplifies to: \[ t^2 - 2t + 1 = 0 \] \[ (t - 1)^2 = 0 \] \[ t = 1 \] 7. Thus, \( x = 1 \) and \( y = 1 \). 8. Finally, calculate \( x^2 + y^2 + z^2 \): \[ x^2 + y^2 + z^2 = 1^2 + 1^2 + 0^2 = 1 + 1 + 0 = 2 \] The final answer is \(\boxed{2}\)" 44ccfbaaea3e,"4. In Rt $\triangle A B C$, $C D$ is the altitude to the hypotenuse $A B$, then the sum of the inradii of the three right triangles ( $\triangle A B C, \triangle A C D$, $\triangle B C D$ ) is equal to ( ). (A) $C D$ (B) $A C$ (C) $B C$ (D) $A B$",See reasoning trace,easy,4. A f5dd51288650,Example 2.34. $I=$ V.p. $\int_{-\infty}^{\infty} \frac{(1+x) d x}{1+x^{2}}$.,See reasoning trace,medium,"Solution. $$ \begin{gathered} I=\lim _{a \rightarrow \infty}\left(\int_{-a}^{a} \frac{d x}{1+x^{2}}-\int_{-a}^{a} \frac{x d x}{1+x^{2}}\right)= \\ =\left.\lim _{a \rightarrow \infty}\left(\operatorname{arctg} x+0.5 \ln \left(1+x^{2}\right)\right)\right|_{-a} ^{a}= \\ =\lim _{a \rightarrow \infty}\left(\operatorname{arctg} a-\operatorname{arctg}(-a)+0.5\left(\ln \left(1+a^{2}\right)-\ln \left(1+(-a)^{2}\right)\right)\right)= \\ =\lim _{a \rightarrow \infty}\left(\operatorname{arctg} a+\operatorname{arctg} a+0.5\left(\ln \left(1+a^{2}\right)-\ln \left(1+a^{2}\right)\right)\right)= \\ =\frac{\pi}{2}+\frac{\pi}{2}=\pi . \end{gathered} $$" a3c016079905,"Example 7 Find all positive integers $n$ such that there exists a sequence $\left\{x_{n}\right\}$ where $1,2, \cdots, n$ each appear once and for $k=1,2, \cdots, n$ we have $k \mid x_{1}+x_{2}+\cdots+x_{n}$.","1, n=3$ are the positive integers that satisfy the conditions.",medium,"Analysis: This problem should make full use of special elements. Solution: When $k=n$, $n \left\lvert\, \frac{n(n+1)}{2}\right.$, so $n+1$ is even, which means $n$ is odd. Let $n=2 p+1, p \in \mathbf{N}^{+}$. When $k=n-1=2 p$, $2 p \left\lvert\,\left(\frac{n(n+1)}{2}-x_{n}\right)\right.$, i.e., $2 p \mid\left((2 p+1)(p+1)-x_{n}\right)$ Thus, $2 p \mid(p+1)-x_{n}$ Also, $2 p \geqslant p+1>p+1-x_{n} \geqslant p+1-(2 p+1)=-p>-2 p$. Therefore, $p+1-x_{n}=0$. Thus, $x_{n}=p+1$. Similarly, when $k=2 p-1$, $(2 p-1) \mid\left(2 p^{2}+2 p-x_{n-1}\right)$, Thus, $(2 p-1) \mid(p+1)-x_{n-1}$ Also, $p+1>p+1-x_{n-1}>p+1-(2 p+1)=-p$. Since $x_{n-1} \neq x_{n}=p+1$, we have $2 p-1