problem stringlengths 2 1.73k | answer stringlengths 1 313 | solution stringlengths 3 9.24k | difficulty float64 1 10 |
|---|---|---|---|
6. In the Kingdom of Mathematics, the denominations of banknotes in circulation are 1 yuan, 5 yuan, 10 yuan, 20 yuan, 50 yuan, and 100 yuan. One day, two customers each bought a chocolate bar worth 15 yuan at the same grocery store. One of them paid with two 10-yuan banknotes, while the other paid with a 20-yuan and a ... | 8 | 【Answer】D
【Analysis】Considering the four cases respectively, when the gum costs 8 yuan, one person pays 10 yuan (2 five-yuan notes), and another person pays 13 yuan (1 ten-yuan note, 3 one-yuan notes), the first person's 1 five-yuan note can be given as change to the second person, and the second person's 2 one-yuan no... | 4.67 |
4. A kilo of sausages was placed on a straight line between a dog in a kennel and a cat. The animals simultaneously rushed to the sausages. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and ate an equal amount. It is known that the cat could eat a... | 1.4 | Answer: 1.4 times closer to the dog than to the cat.
## Solution:
Let $v$ be the running speed of the dog, $u$ be the eating speed of the cat, and the volume of sausages eaten by each animal be 1.
Then, $2 v$ is the running speed of the cat, and $2 u$ is the eating speed of the dog.
Let the distance from the cat to... | 2 |
## Task 1.
If $a_{1}, a_{2}, \ldots, a_{2000}$ is a sequence of 2000 positive real numbers, for how many indices $i \in$ $\{1,2, \ldots, 2000\}$ can the equality
$$
a_{i} a_{i+3}=a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} ?
$$
hold? We consider that $a_{j+2000}=a_{j}$ for $j \in\{1,2,3\}$. | 999 | ## Solution.
For an index $i$, we say it is good if it satisfies the equality in the problem statement.
Assume there exist two consecutive good indices $i, i+1$. Then we have
$$
\begin{aligned}
a_{i} a_{i+3} & =a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} \\
a_{i+1} a_{i+4} & =a_{i+1} a_{i+2}+a_{i+2} a_{i+3}+a_{i+3... | 5.5 |
14. (12 points) Use 36 solid rectangular prisms of size $3 \times 2 \times 1$ to form a large cube of size $6 \times 6 \times 6$. Among all possible arrangements, the maximum number of small rectangular prisms that can be seen from a point outside the large cube is $\qquad$. | \documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
31
\end{document} | 【Answer】Solution: As shown in the figure,
To see the maximum number from the outside, it is necessary to make the rectangles seen from the outside as "deeply" inside the square as possible, the result is as follows: a total of $6 \times 3+3 \times 4+3 \times 1+1=31$ (pieces). Therefore, the answer is: 31. | 4 |
The elements of the sequence $\mathrm{Az}\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence. | x_n = n | The elements of the sequence $\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence.
Solution. We wi... | 5 |
18. (2 marks) Let $A_{1} A_{2} \cdots A_{2002}$ be a regular 2002-sided polygon. Each vertex $A_{i}$ is associated with a positive integer $a_{i}$ such that the following condition is satisfied: If $j_{1}, j_{2}, \cdots, j_{k}$ are positive integers such that $k<500$ and $A_{j_{1}} A_{j_{2}} \cdots A_{j_{k}}$ is a regu... | 287287 | 18. Since $2002=2 \times 7 \times 11 \times 13$, we can choose certain vertices among the 2002 vertices to form regular 7-, 11-, 13-, $\cdots$ polygons (where the number of sides runs through divisors of 2002 greater than 2 and less than 1001). Since $2002=7 \times 286$, so there are at least 286 different positive int... | 6.33 |
After a fair die with faces numbered 1 to 6 is rolled, the number on the top face is $x$. What is the most likely outcome? | x > 2 | With a fair die that has faces numbered from 1 to 6, the probability of rolling each of 1 to 6 is $\frac{1}{6}$. We calculate the probability for each of the five choices. There are 4 values of $x$ that satisfy $x>2$, so the probability is $\frac{4}{6}=\frac{2}{3}$. There are 2 values of $x$ that satisfy $x=4$ or $x=5$... | 1.2 |
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. | 2 \sqrt{2}-1 | 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}... | 9 |
6.014. $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$. | $x=0$ | ## Solution.
Domain: $x \in R$.
$\frac{2 x^{4}+9 x^{2}}{\left(x^{2}+4\right)\left(x^{2}+5\right)}=0 \Leftrightarrow 2 x^{4}+9 x^{2}=0 \Leftrightarrow x^{2}\left(2 x^{2}+9\right)=0$,
$x^{2}=0, x_{1}=0$ or $2 x^{2}+9=0, x_{2,3} \in \varnothing$.
Answer: $x=0$. | 2 |
6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers. | 4060224 | Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\... | 8 |
2. Find all positive integers $n$ $(n \geqslant 3)$ such that there exists a set $M$ with $n$ elements, where the elements are distinct non-zero vectors of equal length, and the following conditions are satisfied: $\sum_{u \in M} u=0$, and for any $v, w \in M$, $\boldsymbol{v}+\boldsymbol{w} \neq \mathbf{0}$. | \{n \in \mathbf{N} \mid n \geqslant 3, n \neq 4\} | 2. First, for any odd number $n$ not less than 3, take $M$ as the $n$ distinct complex roots of the equation $z^{n}-1=0$. Clearly, the set $M$ satisfies the requirements.
Next, consider even numbers $n$ not less than 6.
Let $\frac{n}{2}=k$, first consider the decomposition of $\frac{1}{2}$.
From $\frac{1}{n}=\frac{1}{n... | 3 |
Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$?
... | 14.6 | First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\le... | 2.75 |
497. Shuffling Cards. An elementary method of shuffling cards consists of taking a deck face down in the left hand and transferring the cards one by one to the right hand; each successive card is placed on top of the previous one: the second on top of the first, the fourth on top of the third, and so on until all the c... | 14 | 497. To shuffle 14 cards in the manner described above and return them to their original order, it takes 14 shuffles, although in the case of 16 cards, only 5 are required. We cannot delve into the nature of this phenomenon here, but the reader may find it interesting to conduct an independent investigation of this que... | 4 |
5. The number of real roots of the equation $\frac{\left(x^{2}-x+1\right)^{3}}{x^{2}(x-1)^{2}}=\frac{\left(\pi^{2}-\pi+1\right)^{3}}{\pi^{2}(\pi-1)^{2}}$ with respect to $x$ is exactly $($ )
A. 1
B. 2
C. 4
D. 6 | 6 | 5. D
From the given conditions, it is clear that $x=\pi$ is a root of the original equation. Let's denote the left side as $f(x)$. It is easy to verify:
$$
f\left(\frac{1}{x}\right)=f(x), f(1-x)=f(x), f\left(\frac{1}{1-x}\right)=f(x), f\left(1-\frac{1}{x}\right)=f
$$
$(x), f\left(\frac{x}{x-1}\right)=f(x)$. Therefore,... | 5 |
16. Given the six-digit number $\overline{9786 \square}$ is a multiple of $\mathbf{99}$, the quotient when this six-digit number is divided by $\mathbf{99}$ is ( ). | 6039 | $\begin{array}{l}\text { [Analysis] Let } 99 \mid \overline{A 9786 B} \text {, sum of pairs from right to left } \\ 99 \mid \overline{A 9}+78+\overline{6 B} \text {, i.e., } 99|78+69+\overline{A B} \Rightarrow 99| 48+\overline{A B} \\ \overline{A B}=51 \text {, i.e., } \mathrm{A}=5, \mathrm{~B}=1 \\ 597861 \div 99=6039... | 1.5 |
The base of the right prism $A B C A_{1} B_{1} C_{1}$ is an isosceles triangle $A B C$, where $A B=B C=5$, $\angle A B C=2 \arcsin 3 / 5$. A plane perpendicular to the line $A_{1} C$ intersects the edges $A C$ and $A_{1} C_{1}$ at points $D$ and $E$ respectively, such that $A D=1 / 3 A C, E C_{1}=1 / 3 A_{1} C_{1}$. Fi... | \frac{40}{3} | Let $M$ be the midpoint of the base $AC$ of the isosceles triangle $ABC$ (Fig. 1). From the right triangle $AMB$ we find that
Fig. 1
+17=42
$$
and when $x=4, y=2$, the equality can be achieved, so the maximum value of $x y+8 x+y$ is 42.
Alternatively,... | 5 |
Let $x_{1}=y_{1}=x_{2}=y_{2}=1$, then for $n \geq 3$ let $x_{n}=x_{n-1} y_{n-2}+x_{n-2} y_{n-1}$ and $y_{n}=y_{n-1} y_{n-2}- x_{n-1} x_{n-2}$. What are the last two digits of $\left|x_{2012}\right|$ ? | 84 | Let $z_{n}=y_{n}+x_{n} i$. Then the recursion implies that: $$\begin{aligned} & z_{1}=z_{2}=1+i \\ & z_{n}=z_{n-1} z_{n-2} \end{aligned}$$ This implies that $$z_{n}=\left(z_{1}\right)^{F_{n}}$$ where $F_{n}$ is the $n^{\text {th }}$ Fibonacci number $\left(F_{1}=F_{2}=1\right)$. So, $z_{2012}=(1+i)^{F_{2012}}$. Notice ... | 7.4 |
10. (5 points) According to the pattern shown in the figure, deduce that $M=$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 1692 | 【Solution】Solution: According to the problem, we have:
First, observe the pattern: $12+3=15$;
$15+5=20 ; \cdots$ Each square plus the circle equals the next number.
At the same time, we notice $20=12+3+5$
$$
27=12+3+5+7
$$
Summarizing the pattern, the numbers in the circles form an arithmetic sequence with the first t... | 6.25 |
A3. A square with a side length of 2 is divided into 4 triangles (see the figure). All 3 shaded triangles have the same area. What is the area of the white triangle?
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{8}{5}$
(C) 2
(D) $3 \sqrt{5}-5$
(E) $6-2 \sqrt{5}$
.
Let $x = |A E|$. Since the right triangles $A E D$ and $... | 4 |
Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 504 | The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \(1\), and assume the side length of the octagon is \(2\).
Let \(r\) denote the radius of the circle, \(O\) be the center of the circle. Then:
\[r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.\]
No... | 6.8 |
22. Let $x>1, y>1$ and $z>1$ be positive integers for which the following equation
$$
1!+2!+3!+\ldots+x!=y^{2}
$$
is satisfied. Find the largest possible value of $x+y+z$. | 8 | 22. Answer. 8
Solution. We first prove that if $x \geq 8$, then $z=2$. To this end, we observe that the left hand side of the equation $1!+2!+3!+\ldots+x!$ is divisible by 3 , and hence $3 \mid y^{2}$. Since 3 is a prime, $3 \mid y$. So, $3^{z} \mid y^{z}$ by elementary properties of divisibility.
On the other hand, wh... | 8 |
6. To reduce heating costs in a residential building, the tenants decided to change the facade, and they were granted non-repayable funds from the Environmental Protection and Energy Efficiency Fund, which cover 60% of all total costs. The total cost for the new facade is 1,200,000 kn. The new facade guarantees that th... | 9 | 6. The cost for the facade is 1200000 kn. The non-refundable portion of funds is $60\%$, which means that the residents have to cover $40\%$ of the costs, amounting to $480000 \mathrm{kn}$.
1 POINT
The average annual heating cost was 168000 kn, and with a savings of 35%, it will amount to 109200 kn.
1 POINT
Let $n$... | 1 |
8. There are two teams competing, and the probability of each team winning a match is $\frac{1}{2}$. It is stipulated that a team must win four consecutive matches to end the series. The number of matches played is a random variable $\xi$, then the expected value E $\xi=$ $\qquad$ . | \frac{27}{2} | 8. $\frac{27}{2}$.
Let the probability that the match ends exactly after the $k(k \geqslant 1)$-th game be $p_{k}$. Then
$$
p_{1}=0, p_{2}=0, p_{3}=0, p_{4}=\frac{1}{2^{3}} \text {. }
$$
When $n \geqslant 6$,
$$
\begin{array}{l}
p_{n-1}=\frac{1}{2} p_{n-2}+\frac{1}{2^{2}} p_{n-3}+\frac{1}{2^{3}} p_{n-4} \\
p_{n}=\fra... | 5.5 |
Task 2. What is $\frac{31}{71}$ of the number
$$
\frac{1-\frac{1}{3}:\left(2+\frac{1}{6}\right)}{3 \frac{2}{5}+\frac{10-\frac{1}{4}}{3}: \frac{5}{8}} \cdot 8 \frac{3}{5}-\frac{1.5 \cdot \frac{15}{4} \cdot 2.5+\frac{3}{5-\frac{2}{3}}}{1+\frac{1}{7}+\frac{6}{\frac{12}{11} \cdot\left(\frac{8}{3}-\frac{7}{4}\right) \cdot ... | 0 | Solution. Let's first calculate the value of the given expression. We have:
$$
\begin{aligned}
& \frac{1-\frac{1}{3}:\left(2+\frac{1}{6}\right)}{3 \frac{2}{5}+\frac{10-\frac{1}{4}}{3}: \frac{5}{8}} \cdot 8 \frac{3}{5}-\frac{1.5: \frac{15}{4} \cdot 2.5+\frac{3}{5-\frac{2}{3}}}{1+\frac{1}{7}+\frac{6}{\frac{12}{11} \cdot... | 2 |
Example 14 Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be a family of distinct three-element subsets of $I=\{1,2,3, \cdots, 36\}$, satisfying:
(1) For any $1 \leqslant i<j \leqslant n, A_{i} \cap A_{j} \neq \varnothing$,
(2) $A_{1} \cap A_{2} \cap \cdots \cap A_{n}=\varnothing$.
Find the maximum value of $n$, a... | (100, 34 C_{36}^{3}) | Let $X$ satisfy the conditions of the problem, and without loss of generality, let $A_{1}=\{1,2,3\}$. By condition (2), there exists an element of $X$ that does not contain 1, and we can assume $A_{2}=\left\{a_{1}, a_{2}, a_{3}\right\}$, and $1 \notin A_{2}$. Similarly, we can set $A_{3}=\left\{b_{1}, b_{2}, b_{3}\righ... | 9 |
1.3. A game of Jai Alai has eight players and starts with players $P_{1}$ and $P_{2}$ on court and the other players $P_{3}, P_{4}, P_{5}, P_{6}, P_{7}, P_{8}$ waiting in a queue. After each point is played, the loser goes to the end of the queue; the winner adds 1 point to his score and stays on the court; and the pla... | P_{4} | 1.3 Each time a player loses a match, he has to wait six games before his turn comes again. If $x$ is the number of games before his first turn, then the player will win if $x+7 r+7=37$, where $r \geq 0$ is an integer and $0 \leq x \leq 6$. Here $r$ counts the number of times he lost. From this, we obtain $x=2$ and $r=... | 2.67 |
4. Find the product of all roots of the equation $z^{3}+|z|^{2}=10 i$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | -5+10i | 4. From $z^{3}=-|z|^{2}+10 i$ we get $z^{3}=-|z|^{2}-10 i$, then $|z|^{6}=|z|^{4}+100$, which means $|z|^{2}=$ 5 , so the original equation becomes $z^{3}+5-10 i=0$, thus the product of all roots is $-5+10 i$. | 4 |
3.2. A quadrilateral $ABCD$ is inscribed in a circle $\omega$ such that $AB = AD$ and $BC = CD$ (a deltoid). It turns out that the radius of the inscribed circle of triangle $ABC$ is equal to the radius of the circle that touches the smaller arc $BC$ of circle $\omega$ and the side $BC$ at its midpoint (see Fig. 3). Fi... | 6 | 3.2. Let $I$ be the center of the inscribed circle of triangle $ABC$, $P$ the point of tangency of this inscribed circle with side $AB$, $M$ the midpoint of side $AB$, and $Q$ the midpoint of arc $AB$ (see Fig. 5).
Note that triangle $IPB$ is isosceles and right-angled. Moreover, by the trident theorem, triangle $IQB$... | 4.24 |
6. Let $k$ be a real number, in the Cartesian coordinate system $x O y$ there are two point sets $A=\left\{(x, y) \mid x^{2}+y^{2}=\right.$ $2(x+y)\}$ and $B=\{(x, y) \mid k x-y+k+3 \geqslant 0\}$. If $A \cap B$ is a singleton set, then the value of $k$ is $\qquad$ | -2 - \sqrt{3} | The point set $A$ represents the circle $\odot C:(x-1)^{2}+(y-1)^{2}=2$, and the point set $B$ represents the region below the line $y=k(x+1)+3$. Since the line is tangent to $\odot C$, the distance from the center $C(1,1)$ to the line is
$$
d=\frac{|2 k+2|}{\sqrt{k^{2}+1}}=\sqrt{2} \Rightarrow k^{2}+4 k+1=0 \Rightarro... | 6 |
2. (24th Canadian Mathematical Olympiad) Solve the equation $x^{2}+\left(\frac{x^{2}}{x+1}\right)^{2}=3$ in the set of complex numbers. | \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2}, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2} | 2. Since $x^{2}-2 \cdot \frac{x^{2}}{x+1}+\left(\frac{x}{x+1}\right)^{2}=\left(x-\frac{x}{x+1}\right)^{2}=\left(\frac{x^{2}}{x+1}\right)^{2}$, the original equation can be transformed into $x^{2}+\left(\frac{x}{x+1}\right)^{2}-2 \cdot$ $\frac{x^{2}}{x+1}+2 \cdot \frac{x^{2}}{x+1}=3$
which simplifies to $\left(\frac{x^... | 2.6 |
The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$?
$\text... | 8 | The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$:
If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Th... | 3.75 |
5 Find all real numbers $a$ such that any positive integer solution of the inequality $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)$ are the lengths of the sides of some triangle. | [1, \frac{6}{5}) | 5. Taking $x=2, y=z=1$, we have $a \geqslant \frac{6}{5}$. Therefore, when $a \geqslant \frac{6}{5}$, the original inequality has integer roots $(2,1,1)$, but $(x, y, z)$ cannot form a triangle, so $a<\frac{6}{5}$.
When $a<1$, $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)<x y+y z+z x$, which is a contradiction! Hence, $... | 5 |
$2.351 A=\frac{x^{8}+x^{4}-2 x^{2}+6}{x^{4}+2 x^{2}+3}+2 x^{2}-2$. | A=x^{4} | Solution. Let's divide the polynomial $x^{8}+x^{4}-2 x^{6}+6$ by the polynomial $x^{4}+2 x^{2}+3$.
$$
\begin{aligned}
& \begin{array}{r}
-\frac{-2 x^{6}-4 x^{4}-6 x^{2}}{-2 x^{4}+4 x^{2}+6} ... | 2.33 |
80. One day, Xiao Ben told a joke. Except for Xiao Ben himself, four-fifths of the classmates in the classroom heard it, but only three-quarters of the classmates laughed. It is known that one-sixth of the classmates who heard the joke did not laugh. Then, what fraction of the classmates who did not hear the joke laugh... | \frac{5}{12} | Reference answer: $5 / 12$ | 2.33 |
18. Collinear points $A, B$, and $C$ are given in the Cartesian plane such that $A=(a, 0)$ lies along the $x$-axis, $B$ lies along the line $y=x, C$ lies along the line $y=2 x$, and $A B / B C=2$. If $D=(a, a)$, the circumcircle of triangle $A D C$ intersects $y=x$ again at $E$, and ray $A E$ intersects $y=2 x$ at $F$,... | 7 | Answer: $\square$
Let points $O, P$, and $Q$ be located at $(0,0),(a, 2 a)$, and $(0,2 a)$, respectively. Note that $B C / A B=1 / 2$ implies $[O C D] /[O A D]=1 / 2$, so since $[O P D]=[O A D],[O C D] /[O P D]=1 / 2$. It follows that $[O C D]=[O P D]$. Hence $O C=C P$. We may conclude that triangles $O C Q$ and $P C A... | 5 |
14. Let the curve $C:\left|x^{2}-16 y\right|=256-16|y|$ enclose a closed region $D$.
(1) Find the area of region $D$; (2) Let a line passing through the point $M(0,-16)$ intersect the curve $C$ at two points $P, Q$, find the maximum value of $|P Q|$. | $16 \sqrt{20-10 \sqrt{3}}$ | Solution (1) From the given, we have $256-16|y| \geq 0$, thus $-16 \leq y \leq 16$.
If $\left|x^{2}-16 y\right|=x^{2}-16 y$, then when $0 \leq y \leq 16$, $\left|x^{2}-16 y\right|=x^{2}-16 y=256-16 y, x^{2}=256$, at this time $x= \pm 16(0 \leq y \leq 16)$, the image is two line segments;
When $-16 \leq y \leq 0$, $\lef... | 8.33 |
I3.2 In figure $1, P Q R$ is a right-angled triangle and $R S T U$ is a rectangle. Let $A, B$ and $C$ be the areas of the corresponding regions. If $A: B=m: 2$ and $A: C=n: 1$, find the value of $n$. | 9 | $$
A: B=3: 2, A: C=n: 1 \Rightarrow A: B: C=3 n: 2 n: 3
$$
Let $T S=U R=x, Q U=y$
$\triangle P T S \sim \triangle T Q U \sim \triangle P Q R$ (equiangular)
$$
\begin{array}{l}
S_{\triangle P P S}: S_{\triangle T Q U}: S_{\triangle P Q R}=A: C:(A+B+C)=3 n: 3:(5 n+3) \\
x^{2}: y^{2}:(x+y)^{2}=3 n: 3:(5 n+3) \\
\frac{y}{... | 3.2 |
4. Given that the radius of the circumcircle of a triangle is $4 \mathrm{~cm}$, one of its internal angles is $60^{\circ}$, and the difference between the two sides enclosing this angle is $4 \mathrm{~cm}$, then, the area of this triangle is $\mathrm{cm}$.
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 sentence i... | 8 \sqrt{3} | 4. $8 \sqrt{3}$
Solution Without loss of generality, let $\triangle A B C$ have $\angle A=60^{\circ}, b-c=4, R=4$. By the cosine and sine theorems, we get
$$
a^{2}=b^{2}+c^{2}-2 b c \cos 60^{\circ}=(b-c)^{2}+b c=16+b c,
$$
and $a=2 R \sin \angle A=4 \sqrt{3}$.
Then $b c=a^{2}-16=32$.
$$
S=\frac{1}{2} b c \sin A=8 \sq... | 4.25 |
(16) As shown in the figure, the polyhedron is obtained by making a section $A_{1} B C_{1} D_{1}$ through the vertex $B$ of the base of a regular quadrilateral prism, given that $A A_{1}=O C_{1}$, the dihedral angle between the section $A_{1} B C_{1} D_{1}$ and the base $A B C D$ is $45^{\circ}$, $A B=1$, then the volu... | \frac{\sqrt{2}}{2} | 16 Extend $D_{1} A_{1} 、 D A 、 D_{1} C_{1} 、 D C$ to complete the polyhedron into a triangular pyramid. The volume of the polyhedron can be found by the difference between the volume of the large triangular pyramid and the volumes of the two smaller triangular pyramids, which is $\frac{\sqrt{2}}{2}$. | 5 |
7.58 $2 \lg x^{2}-\lg ^{2}(-x)=4$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.58 $2 \log x^{2}-\log ^{2}(-x)=4$. | x=-100 | 7.58 Here it should be $-x>0$, hence $x<0$. Then $\lg x^{2}=2 \lg (-x)$ and the given equation will take the form $4 \lg (-x)-\lg ^{2}(-x)=4$.
Letting $\lg (-x)=y$, we find $y=2$, hence $\lg (-x)=2$, i.e. $-x=100, x=-100$.
Answer: $x=-100$. | 6 |
[ Systems of points and segments. Examples and counterexamples ] [ Classical combinatorics (miscellaneous). ]
On a plane, 10 equal segments were drawn, and all their points of intersection were marked. It turned out that each point of intersection divides any segment passing through it in the ratio $3: 4$. What is the... | 10 | On each segment, there are no more than two points. On the other hand, each intersection point belongs to at least two segments. Therefore, there are no more than $10 \cdot 2: 10=10$ points. An example with 10 points is shown in the figure.
 | 11 | Answer: $n=11$. Solution. Obviously, $\mathrm{n} \leq 12$, since there is only one pair of numbers with a difference of 13. Suppose the required is possible for $n=12$. The number 12 can be represented as the difference of numbers from 1 to 14 in exactly two ways: 13-1 and 14-2. Let the number 1 be red for definiteness... | 7 |
Example 2. In the stamping of plastic plates, the defect rate is $3 \%$. Find the probability that when checking a batch of 1000 plates, the deviation from the established defect rate will be less than $1 \%$. | 0.709 | Solution. From the condition of the problem, it follows that $n=1000, \varepsilon=0.01$, $p=0.03, q=1-p=0.97$.
In accordance with formula (4.2.5.), we obtain
$$
P\left(\left|\frac{m}{n}-p\right| \leq 0.01\right) \geq 1-\frac{p q}{n \varepsilon^{2}}=1-\frac{0.03 \cdot 0.97}{10000 \cdot(0.01)^{2}}=1-\frac{0.0291}{0.1}=... | 2.33 |
Two random points are chosen on a segment and the segment is divided at each of these two points. Of the three segments obtained, find the probability that the largest segment is more than three times longer than the smallest segment. | \frac{27}{35} | We interpret the problem with geometric probability. Let the three segments have lengths $x, y, 1-x-y$ and assume WLOG that $x \geq y \geq 1-x-y$. The every possible $(x, y)$ can be found in the triangle determined by the points $\left(\frac{1}{3}, \frac{1}{3}\right),\left(\frac{1}{2}, \frac{1}{2}\right),(1,0)$ in $\ma... | 3.8 |
$2 \cdot 77$ In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is $x_{i j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). However, the numbers in each column of Table 1 are rearranged in descending order from top ... | 97 | [Solution] The minimum value of $k$ is 97.
In fact, if we take
$$x_{i j}=\left\{\begin{array}{l}
0,4,(j-1)+1 \leqslant i \leqslant 4 j \\
\frac{1}{24} . \text { for the rest of } i
\end{array} \quad(j=1,2, \cdots, 25)\right.$$
Then, we have
$$\sum_{j=1}^{25} x_{i j}=0+24 \times \frac{1}{24}=1 \quad(i=1,2, \cdots, 100)... | 7.25 |
21 Find all positive integer solutions to the indeterminate equation $7^{x}-3 \cdot 2^{y}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | (1,1), (2,4) | 21. When $x=1$, it is known that $y=1$. Now consider the case $x \geqslant 2$, at this time $y \geqslant 4$, taking both sides modulo 8, we should have
$$(-1)^{x} \equiv 1(\bmod 8)$$
Thus $x$ is even, let $x=2 m$, then
$$\left(7^{m}-1\right)\left(7^{m}+1\right)=3 \cdot 2^{y}$$
Since $7^{m}-1$ and $7^{m}+1$ differ by ... | 8 |
5.1. Insert parentheses and operation signs in the record 22222 so that the result is 24.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | (2+2+2) \times(2+2) | Solution. One of the possible options: $(2+2+2) \times(2+2)$.
Comment. Any correct option is scored 7 points. | 4.22 |
4. As shown in Figure $2, A B, C D$ are two chords of $\odot O$, $C D=2$, $A B=m+n \sqrt{5}(m, n$ are rational numbers), the degree measures of arcs $\overparen{A B}, \overparen{C D}$ are $108^{\circ}, 36^{\circ}$ respectively. Then $108 m-36 n=$ $\qquad$ | 72 | 4. 72.
As shown in Figure 8, on $AB$, take $AE=OA$, and connect $OE$. Let $OA=r$. It is easy to find that
$$
\begin{array}{l}
\angle AOB=108^{\circ} \\
\angle OAB=\angle ABO \\
=\angle COD=36^{\circ} \\
\angle AEO=72^{\circ} \\
\angle BOE=36^{\circ} \\
=\angle OBE .
\end{array}
$$
Therefore, $\triangle OAE \cong \tri... | 4 |
Jóska gave his friends the following task.
I multiplied an integer by 23 and added 1, then multiplied the sum by 11. I multiplied the product by 23 again and added 1, then multiplied by 33, and finally by 77. I moved the last digit of the latest product to the beginning of the number, divided the resulting number by i... | 68611344063777 | a) Since Jóska's number ended in a 7, the last digit of its 23-fold increased by 1 was 2, because the last digit of the product (and sum) is determined solely by the last digits of the factors (or terms), and the last digit of $7 \cdot 3 + 1$ is 2. Similarly, the last digit of the 11-fold sum is also 2. Moving forward,... | 10 |
6.58 $n(\geqslant a)$ football teams participate in a round-robin tournament, and the team ranked third from the bottom scores fewer points than the teams ahead of it and more points than the two teams behind it. Additionally, this team has more wins than the teams ahead of it but fewer wins than the two teams behind i... | 13 | [Solution] Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}, C_{2}$, where team B is the third from the bottom, and $C_{1}$ and $C_{2}$ are the last two.
Firstly, note that each team in group A scores more than team B, and team B scores more than each team in group C; each team in group A has fewer wins t... | 4 |
10000 korona increased to 17800 korona over 10 years with compound interest. What is the interest rate if the rate for the first 5 years was $\frac{1}{2}$ higher than for the last 5 years.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
10000 korona increased to 17800 korona over 10 years with compound interest. What is the inter... | 5.685 \% | Suppose that in the last 5 years the interest rate was $x$, then in the first 5 years it was $x+\frac{1}{2}$. At the end of the fifth year, the compounded capital
$$
C=10000\left(1+\frac{x+\frac{1}{2}}{100}\right)^{5}
$$
became, by the end of the 10th year:
$$
C\left(1+\frac{x}{100}\right)^{5}=10000\left(1+\frac{x+\... | 3.8 |
8. Given $a_{k}$ as the number of integer terms in $\log _{2} k, \log _{3} k, \cdots, \log _{2018} k$. Then $\sum_{k=1}^{2018} a_{k}=$ $\qquad$ . | 4102 | 8.4102.
Let $b_{m}$ be the number of integer terms in $\log _{m} 1, \log _{m} 2, \cdots, \log _{m} 2018$.
Then, $\sum_{k=1}^{2018} a_{k}=\sum_{m=2}^{2018} b_{m}$.
Notice that, $\log _{m} t$ is an integer if and only if $t$ is a power of $m$.
\[
\begin{array}{l}
\text { Then } b_{2}=11, b_{3}=7, b_{4}=6, b_{5}=b_{6}=5,... | 7.5 |
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the an... | 50000 | Answer: 50000.
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 = 150000$ rubles. Profit $=$ revenue costs $=150000-100000=50000$ rubles. | 3 |
5. A mouse lives in a circular cage with completely reflective walls. At the edge of this cage, a small flashlight with vertex on the circle whose beam forms an angle of $15^{\circ}$ is centered at an angle of $37.5^{\circ}$ away from the center. The mouse will die in the dark. What fraction of the total area of the ca... | \frac{3}{4} | Answer: $\frac{3}{4}$ We claim that the lit region is the entire cage except for a circle of half the radius of the cage in the center, along with some isolated points on the boundary of the circle and possibly minus a set of area 0 . Note that the region is the same except for a set of area 0 if we disallow the light ... | 4.2 |
Find the sum of the ages of everyone who wrote a problem for this year's HMMT November contest. If your answer is $X$ and the actual value is $Y$, your score will be $\max (0,20-|X-Y|)$ | 258 | There was one problem for which I could not determine author information, so I set the author as one of the problem czars at random. Then, I ran the following command on a folder containing TeX solutions files to all four contests: ``` evan@ArchMega ~/Downloads/November $ grep --no-filename "Proposed by: " *.tex | sort... | 1.25 |
$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$. | x \in(-1 ; \sqrt[3]{4}) | Solution.
Domain of definition: $4-x^{3}>0, x^{3}<0$. Expanding the absolute values, we have the following three cases:
$$
\text { 1) }\left\{\begin{array} { l }
{ x 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x0
\end{array} \varnothing\right.\right.
$$
2) $\left\{\begin{array}{l}-2 \leq x0\end{array} ... | 5 |
For dessert, Melinda eats a spherical scoop of ice cream with diameter 2 inches. She prefers to eat her ice cream in cube-like shapes, however. She has a special machine which, given a sphere placed in space, cuts it through the planes $x=n, y=n$, and $z=n$ for every integer $n$ (not necessarily positive). Melinda cent... | $7+\frac{13 \pi}{3}$ | Note that if we consider the division of \mathbb{R}^{3}$ into unit cubes by the given planes, we only need to compute the sum of the probabilities that the ice cream scoop intersects each cube. There are three types of cubes that can be intersected: - The cube $0 \leq x, y, z \leq 1$ in which the center lies, as well a... | 6.5 |
Students were surveyed about their favourite season. The results are shown in the bar graph. What percentage of the 10 students surveyed chose Spring?
(A) 50
(B) 10
(C) 25
(D) 250
(E) 5
 | 10\% | Reading from the bar graph, only 1 student chose spring.
Since 10 students were surveyed, then the percentage of students that chose spring was $\frac{1}{10} \times 100 \%$ or $10 \%$.
ANsWER: (B) | 1.8 |
Find the value of $\frac{\sin^{2}B+\sin^{2}C-\sin^{2}A}{\sin B \sin C}$ given that $\frac{\sin B}{\sin C}=\frac{AC}{AB}$, $\frac{\sin C}{\sin B}=\frac{AB}{AC}$, and $\frac{\sin A}{\sin B \sin C}=\frac{BC}{AC \cdot AB}$. | \frac{83}{80} | Using the Law of Sines, we have $$\frac{\sin^{2}B+\sin^{2}C-\sin^{2}A}{\sin B \sin C}=\frac{\sin B}{\sin C}+\frac{\sin C}{\sin B}-\frac{\sin A}{\sin B} \frac{\sin A}{\sin C}=\frac{AC}{AB}+\frac{AB}{AC}-\frac{BC}{AC} \frac{BC}{AB}=\frac{83}{80}$$ | 3.8 |
3.3. Write the number 100 using four fives and arithmetic signs.
$(4-5$ grade.)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (5+5)(5+5)=100 | 3.3. The two solutions to this problem are given below:
1) $(5+5)(5+5)=100$
2) $(5 \cdot 5-5) \cdot 5=100$ | 1.33 |
10. Suppose point $P$ is inside quadrilateral $A B C D$ such that
$$
\begin{array}{l}
\angle P A B=\angle P D A, \\
\angle P A D=\angle P D C, \\
\angle P B A=\angle P C B, \text { and } \\
\angle P B C=\angle P C D .
\end{array}
$$
If $P A=4, P B=5$, and $P C=10$, compute the perimeter of $A B C D$. | \frac{9 \sqrt{410}}{5} | Answer:
$\frac{9 \sqrt{410}}{5}$
Solution:
First of all, note that the angle conditions imply that $\angle B A D+\angle A B C=180^{\circ}$, so the quadrilateral is a trapezoid with $A D \| B C$. Moreover, they imply $A B$ and $C D$ are both tangent to $(P A D)$ and $(P B C)$; in particular $A B=C D$ or $A B C D$ is iso... | 9.2 |
403. Determine the force of water pressure on a vertical parabolic segment, the base of which is $4 \mathrm{m}$ and is located on the water surface, while the vertex is at a depth of $4 \mathrm{m}$ (Fig. 204). | 167424 \text{ N} | Solution. We have $|B A|=2 x=4$ (m). Point $A$ in the chosen coordinate system has coordinates $(2 ; 4)$. The equation of the parabola relative to this system is $y = a x^{2}$ or $4=a \cdot 2^{2}$, from which $a=1$, i.e., $y=x^{2}$.
Consider an elementary area $d S$ at a distance $y$ from the origin. The length of thi... | 5.5 |
11. A $3 \mathrm{~cm}$ wide strip is grey on one side and white on the other. Maria folds the strip, so that it fits inside a rectangle of length $27 \mathrm{~cm}$, as shown. The grey trapeziums are identical. What is the length of the original strip?
A $36 \mathrm{~cm}$
B $48 \mathrm{~cm}$
C $54 \mathrm{~cm}$
D $57 \m... | 57 \mathrm{~cm} | 11. D Let the length of the shorter of the two parallel sides of the grey trapeziums be $x$ $\mathrm{cm}$. Since the folded shape is $27 \mathrm{~cm}$ long and the strip is $3 \mathrm{~cm}$ wide, we have $3+x+3+x+3+x+3+x+3=27$ which has solution $x=3$. Hence the length of the longer of the two parallel sides of the gre... | 3.8 |
3.501 January ** Proof: For each integer $s>1$, there exists a positive integer $m_{s}$, such that when $n \geqslant m_{s}$, there is at least one number between $n$ and $2n$ that is the $s$-th power of some integer. Find the smallest $m_{x}$ for $s=2$ and $s=3$. | $m_2=5, m_3=33$ | Let $m_{s}=\left[\frac{1}{(\sqrt[3]{2}-1)^{s}}\right]+1$, for integers $n \geqslant m_{s}$, we have $n>\frac{1}{(\sqrt[3]{2}-1)^{s}}$, which means $\sqrt[3]{2 n}-$ $\sqrt[3]{n}=\sqrt[3]{n}(\sqrt[3]{2}-1)>1$. Therefore, there exists a positive integer $k$, such that $\sqrt[n]{n}<k<\sqrt[0]{2 n}$. That is, $n<k^{s}<2 n$.... | 5 |
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$. | x=1 | ## Solution.
We have:
$$
\begin{aligned}
& \left(x^{2}+2 x+1\right)(x+2)+\left(x^{2}-2 x+1\right)(x-2)-12=0 \Leftrightarrow \\
& \Leftrightarrow x^{3}+2 x^{2}+2 x^{2}+4 x+x+2+x^{3}-2 x^{2}-2 x^{2}+4 x+x-2-12=0 \\
& 2 x^{3}+10 x-12=0, x^{3}+5 x-6=0
\end{aligned}
$$
The last equation can be rewritten as:
$x^{3}+5 x-5... | 4 |
Problem 3. Senya cannot write some letters and always makes mistakes in them. In the word TETRAHEDRON he would make five mistakes, in the word DODECAHEDRON - six, and in the word ICOSAHEDRON - seven. How many mistakes would he make in the word OCTAHEDRON?
 \mid x>0, y>0, x+y<200$, and $x, y \in \mathbb{Z}\}$. Find the number of parabolas $\mathcal{P}$ with vertex $V$ that satisfy the following conditions: - $\mathcal{P}$ goes through both $(100,100)$ and at least one point in $S$, - $V$ has integer coordinates, and - $\mathcal{P}$ is tangent to the line $... | 264 | We perform the linear transformation $(x, y) \rightarrow(x-y, x+y)$, which has the reverse transformation $(a, b) \rightarrow\left(\frac{a+b}{2}, \frac{b-a}{2}\right)$. Then the equivalent problem has a parabola has a vertical axis of symmetry, goes through $A=(0,200)$, a point $B=(u, v)$ in $S^{\prime}=\{(x, y) \mid x... | 5.8 |
1. The number $N$ has the smallest positive divisor 1, the second largest positive divisor $k$, and the third largest positive divisor $m$. Given that $k^{k}+m^{m}=N$. What is $N$? | 260 | Solution. Answer: $N=260$. If $N$ were odd, then both $k$ and $m$ would be odd, which means $k^{k}+m^{m}$ would be an even number. Therefore, the equation $N=k^{k}+m^{m}$ would be impossible in this case.
Now let $N$ be even. Then $k=2$ and $m$ is also even. Since $4=N-m^{m}$ is divisible by $m$, we have $m=4$ (since ... | 4 |
3. Find all values of $c$ for which the inequality $a+\sqrt{b+c}>b+\sqrt{a+c}$ holds for any positive $a, b$ and $a>b$.
(25 points.) | $c = \frac{1}{4}$ | Answer: $c=\frac{1}{4}$.
Write the inequality in the form $a-b>\sqrt{a+c}-\sqrt{b+c}$. By multiplying and dividing the right part by its conjugate, we can represent the inequality as
$$
a-b>\frac{a-b}{\sqrt{a+c}+\sqrt{b+c}} \quad \Longleftrightarrow \quad \sqrt{a+c}+\sqrt{b+c}>1
$$
The last inequality must hold for ... | 3 |
2. Let $a$ be a real number, and the sequence $a_{1}, a_{2}, \cdots$ satisfies:
$$
\begin{array}{l}
a_{1}=a, \\
a_{n+1}=\left\{\begin{array}{ll}
a_{n}-\frac{1}{a_{n}}, & a_{n} \neq 0 ; \\
0, & a_{n}=0
\end{array}(n=1,2, \cdots) .\right.
\end{array}
$$
Find all real numbers $a$ such that for any positive integer $n$, w... | $a=0, \pm \frac{\sqrt{2}}{2}$ | 2. Discuss in two cases.
(1) The sequence $\left\{a_{n}\right\}$ contains a term that is zero. Let $m$ be the smallest positive integer such that $a_{m}=0$.
If $m \geqslant 2$, then $a_{m-1} \neq 0$. Therefore,
$$
a_{m-1}-\frac{1}{a_{m-1}}=a_{m}=0 \Rightarrow a_{m-1}= \pm 1 \text {, }
$$
which contradicts the problem ... | 4.33 |
4. 20 balls of the same mass are moving along a chute towards a metal wall with the same speed. Coming towards them at the same speed are 16 balls of the same mass. When two balls collide, they fly apart with the same speed. After colliding with the wall, a ball bounces off it with the same speed. (The balls move only ... | 510 | # Answer: 510.
Solution. We will assume that initially, each ball moving towards the wall has a red flag, and the rest of the balls have blue flags. Imagine that when the balls collide, they exchange flags. Then each blue flag moves at a constant speed in one direction (away from the wall), and each red flag reaches t... | 4 |
7. Given real numbers $x, y$ satisfy:
$$
x>\max \{-3, y\},(x+3)\left(x^{2}-y^{2}\right)=8 \text {. }
$$
Then the minimum value of $3 x+y$ is $\qquad$ . | 4 \sqrt{6} - 6 | 7. $4 \sqrt{6}-6$.
From the problem, we know $x+3>0, x-y>0$.
Then $x+y=\frac{8}{(x+3)(x-y)}>0$.
Thus, $8=(x+3)(x+y)(x-y)$
$$
\begin{array}{l}
=\lambda(x+3) \mu(x+y)(x-y) \\
\leqslant\left(\frac{\lambda(x+3)+\mu(x+y)+(x-y)}{3}\right)^{3},
\end{array}
$$
where, $\lambda \mu=1, \lambda-2 \mu+4=0$.
Therefore, $\lambda=\s... | 7 |
4. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
y=b-x^{2}, \\
x^{2}+y^{2}+2 a^{2}=4-2 a(x+y)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | b \geq -2 \sqrt{2} - \frac{1}{4} | Answer. $b \geq-2 \sqrt{2}-\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x+a)^{2}+(y+a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $( -a ;-a )$. For all possible $a \in \mathrm{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt{2} \... | 4.67 |
7. Modify one digit of 31743, you can get a multiple of 823, what is the modified number?
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | 33743 | 7. 33743
7.【Solution】 $31743 \div 823=38 \cdots \cdots \cdot 469$
No matter which of the last three digits 7, 4, 3 is changed, it cannot make the remainder increase or decrease to become a multiple of 823. If the thousand's digit 1 is changed to 3, then since $2469=823 \times 3$, we get that 33743 is divisible by 823. ... | 3.4 |
# 3. Option 1.
The Ivanov family consists of three people: dad, mom, and daughter. Today, on the daughter's birthday, the mother calculated the sum of the ages of all family members and got 74 years. It is known that 10 years ago, the total age of the Ivanov family members was 47 years. How old is the mother now, if s... | 33 | Answer: 33.
Solution: If the daughter had been born no less than 10 years ago, then 10 years ago the total age would have been $74-30=44$ years. But the total age is 3 years less, which means the daughter was born 7 years ago. The mother is now $26+7=33$ years old. | 3 |
Points $X$ and $Y$ are inside a unit square. The score of a vertex of the square is the minimum distance from that vertex to $X$ or $Y$. What is the minimum possible sum of the scores of the vertices of the square? | \frac{\sqrt{2}+\sqrt{6}}{2} | Let the square be $A B C D$. First, suppose that all four vertices are closer to $X$ than $Y$. Then, by the triangle inequality, the sum of the scores is $A X+B X+C X+D X \geq A B+C D=2$. Similarly, suppose exactly two vertices are closer to $X$ than $Y$. Here, we have two distinct cases: the vertices closer to $X$ are... | 5.75 |
How many integers $n$ are there for which the value of expression (1) is an integer? How many positive integers $n$ are there for which the value of $K$ is a positive integer?
$$
K=\frac{2610+201 n+110 n^{2}-n^{3}}{10+n}
$$ | \text{144 and 27} | a) The given expression can be transformed as follows:
$$
\begin{gathered}
K=\frac{\left(n^{3}+10 n^{2}\right)-\left(120 n^{2}+1200 n\right)+(999 n+9990)-12600}{-(n+10)}= \\
=-n^{2}+120 n-999+\frac{12600}{n+10} .
\end{gathered}
$$
Substituting an integer for \( n \), the first three terms on the right side are intege... | 5 |
37. A mechanical beetle crawls along a square spiral on the ground. It starts from the lower left corner and crawls in the direction of the arrows, sequentially for $1 \mathrm{~cm}, 1 \mathrm{~cm}, 1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 2 \mathrm{~cm}, 1 \mathrm{~cm}, 3 \mathrm{~cm}, 3 \mathrm{~cm}$, $\qquad$.... | 1156 | \begin{tabular}{c}
$\mathbf{3 7}$ \\
\hline 1156
\end{tabular} | 4 |
$$
\begin{array}{l}
16\left(\frac{1}{5}-\frac{1}{3} \times \frac{1}{5^{3}}+\frac{1}{5} \times \frac{1}{5^{5}}-\frac{1}{7} \times \frac{1}{5^{7}}+ \\
\frac{1}{9} \times \frac{1}{5^{9}}-\frac{1}{11} \times \frac{1}{5^{11}}\right)-4\left(\frac{1}{239}-\frac{1}{3} \times \frac{1}{239^{3}}\right) \\
=\quad(\text { (to } 8 \... | 3.14159265 | -、1.3.14159265
(No translation needed as the text is a number and a separator, which are universal and do not require translation.) | 2.4 |
6. Let $f$ be a function from $\{1,2, \ldots, 22\}$ to the positive integers such that $m n \mid f(m)+f(n)$ for all $m, n \in\{1,2, \ldots, 22\}$. If $d$ is the number of positive divisors of $f(20)$, compute the minimum possible value of $d$. | 2016 | Answer: 2016
Solution: Let $L=\operatorname{lcm}(1,2, \ldots, 22)$. We claim that the possible values of $f(20)$ are the multiples of $20 L$. If we can prove this, we will be done, since the minimum value of $d$ will be the number of divisors of $20 L=2^{6} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cd... | 6 |
10,11 On the edge $AC$ of a regular triangular prism $ABC A1B1C1$, a point $K$ is taken such that $AK=\frac{1}{4}, CK=\frac{3}{4}$. A plane is drawn through point $K$, forming an angle $\operatorname{arctg}^{\frac{7}{6}}$ with the plane $ABC$ and dividing the prism into two polyhedra with equal surface areas. Find the ... | \frac{3}{8} | Let $\phi 1$ and $\phi 2$ be polyhedra into which a plane $\alpha$ cuts a prism, such that a sphere can be circumscribed around $\phi 1$, but not around $\phi 2$. Let $S 1$ and $S 2$ be the areas of their surfaces, respectively. Each face of a polyhedron inscribed in a sphere is an inscribed polygon, since the intersec... | 8 |
8. Given a natural number $k$. In the city, there are several children who attend several clubs. It is known that no more than $3 k$ children attend each club, each child attends exactly three clubs, and for any two children, there is a club that both of them attend. What is the maximum number of children that can be i... | $7k$ | Answer. $7 k$. Solution. Example. Let's divide $7 k$ children into 7 groups (numbered from 1 to 7) with $k$ children in each, and form 7 clubs from children of groups $(1,2,3) ;(1,4,5) ;(1,6,7) ;(2,4,6) ;(2,5,7) ;(3,4,7) ;(3,5,6)$. It is not hard to verify that all conditions are met.
Estimate. If all clubs consist of... | 8 |
9. If the equation $9^{-x^{x}}=4 \cdot 3^{-x^{x}}+m$ has real solutions for $x$, then the range of real values for $m$ is | [-3, 0) | 9. $m \in[-3,0) \quad m=\left(3^{-x^{2}}\right)^{2}-4 \cdot 3^{-x^{2}}$, let $t=3^{-x^{2}}, m=t^{2}-4 t, t \in(0,1]$. It is easy to find that $m \in[-3,0)$ | 4.2 |
3. A frog starts climbing from the bottom of a 12-meter deep well at 8 o'clock. It climbs up 3 meters and then slides down 1 meter due to the slippery well wall. The time it takes to slide down 1 meter is one-third of the time it takes to climb up 3 meters. At 8:17, the frog reaches 3 meters below the well's mouth for ... | 22 | Answer: A.
Whether up or down, the time used for each meter is the same. According to the problem, for every 3 meters climbed up, there is a 1-meter slide down. We can create a table.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline Frog's actual height & 3 & 2 & 5 & 4 & 7 & 6 & 9 & 8 & 9 & 11 & 10 & 12 \\
\hline Cl... | 4.67 |
$12.24 y=x^{3} e^{-x}$. | (3, \frac{27}{e^3}) | 12.24 The function is defined for all $x$. We find
$$
y^{\prime}=3 x^{2} e^{-x}-x^{3} e^{-x}=x^{2} e^{-x}(3-x)
$$
The equation $y^{\prime}=0$ has only one root $x=3$. Since $y^{\prime}>0$ for $x<3$ and $y^{\prime}<0$ for $x>3$, the function reaches a maximum at the point $x=3$, which is equal to $3^{3} e^{-3}=\frac{2... | 2.33 |
Mr. and Mrs. Seventh have 7 children, all born on April 1st, actually over six consecutive April 10ths. This year, for their birthdays, Mrs. Seventh made a cake with candles for each one - the number of candles equal to the number of years of each one. João Seventh, the son who loves Math the most, noticed that this ye... | 26 | The births occurred on six 1st of April, so there are twin siblings. Since this year we have 2 more cakes than 2 years ago, it means that 2 years ago the youngest had not been born yet, the second youngest had just been born, and the twins had already been born. Currently, the youngest is 1 year old and the twins are $... | 2.5 |
5、In Rt $\triangle A B C$, $C=\frac{\pi}{2}, B C=a, C A=b$, point $P$ is on side $A B$, along line $P C$ fold
$R t \triangle A B C$ into tetrahedron $P A B C$, the maximum volume of this tetrahedron is $\qquad$ (expressed in terms of $\boldsymbol{a}, \boldsymbol{b}$) | \frac{a^{2} b^{2}}{6\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}} | 5. $\frac{a^{2} b^{2}}{6\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}}$
Analysis: As shown in the right figure, $P D \perp A C, B E \perp P C$, let $|P C|=x, \angle P C A=\theta$, then we know $|B E|=a \cdot \cos \theta$, and we have
$\frac{|P D|}{a}=\frac{b-|C D|}{b}$, i.e., $\frac{x \cdot \sin \theta}{a... | 5 |
Find all prime numbers $p, q, r$ (not necessarily positive) such that
$$
\frac{1}{p-q-r}=\frac{1}{q}+\frac{1}{r}
$$ | \{(5, 2, 2), (-5, -2, -2), (-5, 3, -2), (-5, -2, 3), (5, 2, -3), (5, -3, 2)\} | If the equality holds for $p, q$, and $r$ (not necessarily positive) prime numbers, then $p-q-r \neq 0 (q \neq 0, r \neq 0)$; multiplying both sides by $q r(p-q-r)$:
$$
r q=(p-q-r) \cdot r+(p-q-r) q, \quad \text { i.e., } r q=(p-q-r)(r+q)
$$
Since $r$ and $q$ are primes, the product $r q$ can have 8 divisors: $1, -1,... | 5 |
Problem 11.8. In each cell of a strip $1 \times N$ there is either a plus or a minus. Vanya can perform the following operation: choose any three cells (not necessarily consecutive), one of which is exactly in the middle between the other two cells, and change the three signs in these cells to their opposites. A number... | 1396 | Answer: 1396.
Solution. We will prove that all the numbers under consideration are positive, except for 4 and 5. Then the answer to the problem will be $1398-2=1396$.
For each $N$, we will number the cells of the strip $1 \times N$ from left to right with numbers from 1 to $N$.
- Let $N=3$. By applying the operation... | 5.2 |
13. Five square tiles are put together side by side. A quarter circle is drawn on each tile to make a continuous curve as shown. Each of the smallest squares has side-length 1 .
What is the total length of the curve?
A $6 \pi$
B $6.5 \pi$
С $7 \pi$
D $7.5 \pi$
E $8 \pi$ | 6 \pi | Solution
A
The side lengths of the 5 squares are $1,1,2,3$ and 5 . So the curve is made up of five quarter circles with these radii. The circumference of a circle with radius $r$ is $2 \pi r$. Therefore the length of a quarter circle of radius $r$ is $\frac{1}{4}(2 \pi r)$, that is, $\frac{1}{2} \pi r$.
Therefore the ... | 2.67 |
Example 3 Find all values of $a$ such that the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy
$$
\left(x_{1}-3\right)^{2}+
\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {. }
$$
(1983 Austrian Olympiad Problem) | $a=-9$ | Let $y=x-3$, then $y_{1}=x_{1}-3, y_{2}=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
y_{1}+y_{2}+y_{3}=-3, \\
y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}=a-9, \\
y_{1} y_{2} y_{3}=27-4 a .
\... | 5 |
8 Let the function $f(x)=x^{3}+3 x^{2}+6 x+14$, and $f(a)=1, f(b)=19$, then $a$ $b=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -2 | -2 | 8 From
$$
\begin{aligned}
f(x) & =x^{3}+3 x^{2}+6 x+14 \\
& =(x+1)^{3}+3(x+1)+10,
\end{aligned}
$$
let $g(y)=y^{3}+3 y$, then $g(y)$ is an odd function and monotonically increasing. And
$$
\begin{array}{l}
f(a)=(a+1)^{3}+3(a+1)+10=1, \\
f(b)=(b+1)^{3}+3(b+1)+10=19,
\end{array}
$$
so $g(a+1)=-9, g(b+1)=9, g(-b-1)=-9$,... | 8 |
7. $\arcsin \left(\sin 2000^{\circ}\right)=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7. $\arcsin \left(\sin 2000^{\circ}\right)=$ | -20^{\circ} | $$
-20^{\circ}
$$
7. [Analysis and Solution] According to the problem, we have $2000^{\circ}=180^{\circ} \times 11+20^{\circ}$,
$$
\begin{array}{l}
\therefore \sin 2000^{\circ}=\sin \left(180^{\circ} \times 11+20^{\circ}\right)=\sin \left(-20^{\circ}\right), \\
\therefore \arcsin \left(2000^{\circ}\right)=-20^{\circ}
\... | 1 |
G6.4 A straight line with slope $=-2$ passes through the origin cutting the curve at two points $E$ and
$F$. If $\frac{7}{s}$ is the $y$-coordinate of the midpoint of $E F$, find the value of $s$. | -2 | $$
\begin{array}{l}
\text { Sub. } y=-2 x \text { into } y=-2 x^{2}+5 x-2 \\
-2 x=-2 x^{2}+5 x-2 \\
2 x^{2}-7 x+2=0
\end{array}
$$
Let $E=\left(x_{1}, y_{1}\right), F=\left(x_{2}, y_{2}\right)$.
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{7}{2} \\
\frac{7}{s}=\frac{y_{1}+y_{2}}{2}=\frac{-2 x_{1}-2 x_{2}}{2}=-\left(x_{1}+x_{... | 4.5 |
1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of ch... | 1347 | Answer: 1347.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 8 |
9.4. $N$ different natural numbers, not exceeding 1000, are written in a circle such that the sum of any two of them, standing one apart, is divisible by 3. Find the maximum possible value of $N$.
# | 664 | # Answer: 664.
Solution. Consider the remainders of the numbers when divided by 3. Divisibility by 3 means that in each pair of numbers standing one apart, either both numbers are divisible by 3, or one has a remainder of 1 and the other has a remainder of 2 when divided by 3. Among the numbers from 1 to 1000, 333 are... | 6 |
Pebbles are placed on a $2021 \times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way th... | 3031 | Let $N \geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2 N+1) \times(2 N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3 N+1$. The problem has $N=1010$, so at least 3031 pebbles are needed.
We begin by placing $(2 N+1)... | 6.4 |
2. Points $M$ and $N$ divide side $A C$ of triangle $A B C$ into three parts, each of which is equal to 5, such that $A B \perp B M, B C \perp B N$. Find the area of triangle $A B C$. | \frac{75\sqrt{3}}{4} | # Problem 2.
Answer: $3 \sqrt{3}$ (Option IV-1), $27 \sqrt{3} / 4$ (Option IV-2), $12 \sqrt{3}$ (Option IV-3), $75 \sqrt{3} / 4$ (Option IV-4).
Solution. Points $M$ and $N$ are located on side $A C$ in the following order: $A, N$, $M, C$ (otherwise, angle $B$ would be greater than $\left.180^{\circ}\right)$.
Let $A ... | 4 |
$6.284 \sqrt[4]{18+5 x}+\sqrt[4]{64-5 x}=4$ | x_{1}=\frac{63}{5}, x_{2}=-\frac{17}{5} | Solution. Let $u=\sqrt[4]{18+5 x}, v=\sqrt[4]{64-5 x}$, then $\left\{\begin{array}{l}u+v=4, \\ u^{4}+v^{4}=82,\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ \left(u^{2}+v^{2}\right)^{2}-2 u^{2} v^{2}=82\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ \left((u+v)^{2}-2 u ... | 4.2 |
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