problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
1. If a two-digit natural number is decreased by 54, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the median of the sequence of all such numbers.
# | # Solution.
Let $\overline{x y}=10 x+y$ be the original two-digit number, then $\overline{y x}=10 y+x$ is the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$. From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,160 |
3. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 9 points, if there were 4 tens, and the results of the hits were sevens, eights, and nines. There were no misses at all. | Solution: Since the soldier scored 90 points and 40 of them were scored in 4 attempts, he scored 50 points with the remaining 6 shots. Since the soldier only hit the seven, eight, and nine, let's assume that in three shots (once each in seven, eight, and nine), he scored 24 points. Then, for the remaining three shots, ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,162 |
4. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let $x$ be the length of the roe deer's jump, then $0.78 x$ is the length of the wolf's jump; $y$ - the number of jumps the roe deer makes over the time interval specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes over the same time interval. The wolf will not be a... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,163 |
5. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61; 20\}$?... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$B=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,164 |
6. Given triangle $A B C$. Lines $O_{1} O_{2}, O_{1} O_{3}, O_{3} O_{2}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $O$ is the center of the inscribed circle of triangle $A B C$. Find the angle in degrees between the lines $O_{1} O_{2}$ and $O O_{3}$.
.
1) $\triangle I G C = \triangle G F H$ - by two legs, since $I C = G H = 2, I G = H F = 1$, therefore ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,166 |
8. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} M C_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering t... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,167 |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{25}{3}$ times. Deter... | # Solution.
1) Let the syrup content in the initial solution be $p \%$ and let $x$ liters of the solution were poured out the first time.
2) Then after pouring out the liquid, there remained $(1000-x)$ liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,168 |
1. A warehouse has coffee packed in bags of 15 kg and 8 kg. How many bags of coffee in total does the warehouseman need to prepare to weigh out 1998 kg of coffee, with the number of 8 kg bags being the smallest possible? | Solution. Let $x$ be the number of bags weighing 15 kg, and $y$ be the number of bags weighing 8 kg. We get the equation $15 x + 8 y = 1998$. $8(x + y) + 7 x = 1998$, let $x + y = k$,
$8 k + 7 x = 1998$,
$7(k + x) + k = 1998$, let $k + x = t$,
$7 t + k = 1998$, $k = 1998 - 7 t$. Substitute into (2), $x = 8 t - 1998$... | 136 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,169 |
4. Ivan Ivanovich approached a source with two empty cans, one with a capacity of 10 liters, and the other - 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched t... | Solution. Let $x$ liters of water fill the larger can while 4 liters fill the smaller can. After the switch, $(10-x)$ liters fill the larger can, and 4 liters fill the smaller can again. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equa... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,172 |
5. The clock shows 00:00, at which the hour and minute hands of the clock coincide. Counting this coincidence as number 0, determine how much time (in minutes) will pass before they coincide for the 21st time. Round your answer to the nearest hundredth. | Solution. The minute hand passes 1 circle in an hour, while the hour hand passes $1 / 12$ of a circle, so their closing speed is $11 / 12$ of a circle per hour, one closing takes $1 /(11 / 12)=12 / 11$ hours or $720 / 11$ minutes. The 21st closing will occur after $21 \cdot \frac{720}{11}=\frac{15120}{11}$ minutes.
An... | 1374.55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,173 |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} B M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,174 |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B C A=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=60^{\circ}$. On the side $\mathrm{BC}$, a point $\mathrm{M}$ is ch... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\Delta T P M = \Delta K O M$... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,175 |
8. Right triangles $M D C$ and $A D K$ have a common right angle $D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$, counting from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of the degree measures of angles $A K D$ and $M C D$, if $A D: C D=2: 5$. | Solution. Extend triangle $A D C$ to form a square $L J C D$.
Choose point $H$ on side $L J$ such that $L H: H J=2: 3$, point $N$ on side $C J$ such that $C N: N J=3: 2$, and point $B$ on si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,176 |
9. A chemistry student conducted an experiment: from a bottle filled with syrup solution, he poured out one liter of liquid, refilled the bottle with water, then poured out one liter of liquid again and refilled the bottle with water. As a result, the percentage content of syrup decreased from 36 to 1 percent. Determin... | # Solution.
1) Let $x$ be the volume of the bottle in liters.
2) Then after pouring out one liter of liquid, there are $(x-1)$ liters of solution left, and in it $(x-1) \cdot \frac{36}{100}$ liters of syrup and $(x-1) \cdot \frac{64}{100}$ liters of water.
3) After adding one liter of water to the bottle, there are $x... | 1.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,177 |
2. Vasya and Petya, participating in a school sports and entertainment game, need to cover a distance of 3 km as quickly as possible with only one pair of roller skates between them. They start simultaneously, one just running, the other running on roller skates. At any time, the one running on roller skates can leave ... | Solution: The entire distance is divided into several stages, where one of the schoolchildren runs without roller skates, and the other on roller skates. Let \( x \) be the sum of the lengths of the stages that Vasya runs on roller skates, \( t_{1} \) be the time he spends running the entire distance, and \( t_{2} \) b... | 0.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,179 |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
(16 points) | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}^{m}}$, where $l_{i} \in\left\{0,1, \l... | 25920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,180 |
4. Solve the inequality $\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0$. (20 points)
# | # Solution:
$$
\text { Domain: } 2 x^{2}+7 x+3 \geq 0, \Rightarrow x \in(-\infty ;-3] \cup[-0.5 ;+\infty) .
$$
The original inequality is equivalent to the following
$$
\begin{aligned}
& \left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left(x^{2}-4 x+2\right)^{2}-(x-2)^{2}\right) \leq 0 \Leftrightarrow \\
& \left(3 x... | x\in(-\infty;-3]\cup[0;1]\cup{2}\cup[3;4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,181 |
6. Two circles touch each other and the sides of triangle $ABC$. The first circle with radius $\frac{1}{18}$ touches sides $AB$ and $AC$ at points $L$ and $K$, the second circle with radius $\frac{2}{9}$ touches sides $AC$ and $BC$ at points $N$ and $M$. Find the area of triangle $ABC$ if $AL=\frac{1}{9}, CM=\frac{1}{6... | Solution: $a=\frac{1}{9}, \quad b=\frac{1}{6}, \quad r=\frac{1}{18}, \quad R=\frac{2}{9}$
$K N=2 \sqrt{r R}=\frac{2}{9}, \quad A C=a+K N+b=\frac{1}{2}$
$\operatorname{tg} \frac{A}{2}=\frac{r}{a}, \quad \operatorname{tg} \frac{C}{2}=\frac{R}{b}$
$\cos ^{2} \frac{A}{2}=\frac{a^{2}}{a^{2}+r^{2}}, \quad \cos ^{2} \frac{... | \frac{3}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,183 |
1. Given two groups of numerical sequences, each consisting of 15 arithmetic progressions containing 10 terms each. The first terms of the progressions in the first group are $1,2,3, \ldots, 15$, and their differences are respectively $2,4,6, \ldots, 30$. The second group of progressions has the same first terms $1,2,3... | Solution. Let $a_{1}, \ldots, a_{15}$ be the first terms of the progressions, $d_{1}, \ldots, d_{15}$ be the differences of the first group, and $p_{1}, \ldots, p_{15}$ be the differences of the second group. Then the sums of the first group are $S_{1}=\frac{2 a_{1}+9 d_{1}}{2} \cdot 10=10 a_{1}+45 d_{1}, \ldots, S_{15... | \frac{160}{151} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,184 |
2. Two telephone companies agreed to release sets of all possible three-digit telephone numbers such that the first company would have all numbers consisting of odd digits, and the second company only even digits, except for 0. The first company sold its entire first set for $X$ rubles per number, and the second compan... | Solution. a) There are 5 odd digits: $1,3,5,7,9$ and any of them can stand on each of the 3 positions, the total number of combinations is $N_{\text {odd }}=5^{3}=125$,
b) Even digits are $2,4,6,8$, so 4 digits can stand in the first position, and any of the 4 even digits can stand further, the total number of combina... | (41;80),(105;205) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,185 |
3. Solve the equation $3 \sqrt{(x+1)^{2}}+\sqrt{x^{2}-4 x+4}=|1+6 x|-4|x-1| . \quad$ (16 points) | Solution: Rewrite the equation as $|3 x+3|+|x-2|+|4 x-4|=|1+6 x|$. Notice that $3 x+3-x+2+4 x-4=1+6 x$, so the solution to the equation is found from the system $\left\{\begin{array}{l}3 x+3 \geq 0, \\ -x+2 \geq 0, \\ 4 x-4 \geq 0, \\ 1+6 x \geq 0 .\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \... | x\in[1,2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,186 |
4. Solve the inequality $4+x^{2}+2 x \sqrt{2-x^{2}}<8 \sqrt{2-x^{2}}+5 x$. (20 points) | Solution. Rewrite it as $4+x^{2}-5 x+2 x \sqrt{2-x^{2}}-8 \sqrt{2-x^{2}}0\end{array}\right.$ Solve the second inequality of the system
$(x-1)+2 \sqrt{2-x^{2}}>0 \Rightarrow 2 \sqrt{2-x^{2}}>1-x \Rightarrow\left[\begin{array}{l}\left\{\begin{array}{l}1-x1-2 x+x^{2}\end{array}\right.\end{array}\right.$ $\left[\begin{arr... | (-1;\sqrt{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,187 |
6. A circle of radius 2, inscribed in triangle $A B C$, touches side $B C$ at point $D$. A circle of radius 4 touches the extensions of sides $A B$ and $A C$, as well as side $B C$ at point $E$. Find the area of triangle $A B C$, if the measure of angle $A C B$ is $120^{\circ}$. | Solution. The centers of the circles lie on the angle bisectors. Therefore, \( C D = r / \operatorname{tg} 60^{\circ} = 2 / \sqrt{3}, \quad C E = R / \operatorname{tg} 30^{\circ} = 4 \sqrt{3}, D E = 4 \sqrt{3} - 2 / \sqrt{3} = 10 / \sqrt{3} \).
 Even digits are $0,2,4,6,8$, a seven-digit number cannot start with 0, so there can be 4 digits in the first place, and any of the 5 even digits can follow, the total number of combinations is $N_{\text {even }}=4 \cdot 5^{6}$, b) there are 5 odd digits: $1,3,5,7,9$ and any of them can stand in each of the... | )62500,b)78125,)1.25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,190 |
2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N, \quad$ is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N$ (12 points) | Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$.
For the second sequence $\quad b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}, \quad$ the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.
We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,191 |
3. Solve the equation
$$
3 \sqrt{6 x^{2}+13 x+5}-6 \sqrt{2 x+1}-\sqrt{3 x+5}+2=0
$$ | Solution. Note that $6 x^{2}+13 x+5=(2 x+1)(3 x+5)$. Introduce the notations: $\left\{\begin{array}{l}m=\sqrt{2 x+1} ; n=\sqrt{3 x+5} . \\ m \geq 0 \\ n \geq 0\end{array}\right.$
Then the equation can be represented as $3 m n-6 m-n+2=0$.
Considering the domain of definition, solve this equation:
$3 m(n-2)-(n-2)=0 \R... | -1/3;-4/9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,192 |
4. Solve the inequality
$$
\frac{x(x-2)(x-3)-(x-2)^{2}+1}{(|x-1|-|x-2|) \sqrt{16-x^{2}}} \geq 0
$$ | Solution. Domain of definition: $\left\{\begin{array}{l}|x-1| \neq|x-2|, \\ 16-x^{2}>0,\end{array} \Rightarrow\left\{\begin{array}{l}x \neq 3 / 2, \\ x \in(-4,4) .\end{array}\right.\right.$
Transform the numerator
$x(x-2)(x-3)+1-(x-2)^{2}=x(x-2)(x-3)+(1-x+2)(1+x-2)=$
$=x(x-2)(x-3)+(3-x)(x-1)=(x-3)\left(x^{2}-2 x+1-x... | x\in(-4;\frac{3-\sqrt{5}}{2}]\cup(\frac{3}{2};\frac{3+\sqrt{5}}{2}]\cup[3;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,193 |
5. Determine all values of the parameter $p$ for which the system
$$
\left\{\begin{array}{l}
(x-p)^{2}=16(y-3+p) \\
y^{2}+\left(\frac{x-3}{|x|-3}\right)^{2}=1
\end{array}\right.
$$
has solutions and solve it for each of the found values of $p$.
(20 points) | Solution. Domain of definition: $|x| \neq 3$. For $x \geq 0 \quad y^{2}+1=1 \Rightarrow y=0 . \quad$ If $\quad x0, \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4)0 \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4) \geq 0, \Rightarrow \\ x_{1}+x_{2}>0\end{array}\right.$
Thus, we get the answer:
$x_{1,2}=p \pm 4 \sqrt{p-3}, p \in(3 ; ... | p\in(3;4]\cup[12;19)\cup(19;\infty),x_{1,2}=p\4\sqrt{p-3},\quad0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,194 |
6. A circle with center $O_{1}$ and radius 2, inscribed in triangle $A B C$, touches side $B C$ at point $D$. A second circle with center $O_{2}$ and radius 4 touches the extensions of sides $A B$ and $A C$, as well as side $B C$ at point $E$. Find the area of quadrilateral $O_{1} D O_{2} E$, if the measure of angle $A... | # Solution.
$S=S_{\square O_{1} E D}+S_{\square O_{2} E D}=1 / 2 \cdot E D \cdot\left(O_{1} D+O_{2} E\right)=3 E D$
$E D=E C-D C=O_{2} E / \operatorname{tg} 30-O_{1} D / \operatorname{tg} 60=4 \sqrt{3}-2 / \sqrt{3}=10 / \sqrt{3}$ $S=30 / \sqrt{3}$
Answer: $30 / \sqrt{3}$.
^{3}-x^{3}-y^{3}-z^{3}=18 .\end{array}\right.\right.$
$(x+y+z)^{3}-x^{3}-y^{3}-z^{3}=18 \quad \Leftrightarrow\left((x+y+z)^{3}-x^{3}\right)-\left(y^{3}+z^{3}\right)=18 \Leftr... | (3,3,-4),(3,-4,3),(-4,3,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,196 |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, dark blue - 87, purple - 91, black -... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text{ times}}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few value... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,198 |
4. Find all pairs of natural numbers $a$ and $b$, for which out of the four statements:
1) $a^{2}+6 a+8$ is divisible by $b$;
2) $a^{2}+a b-6 b^{2}-15 b-9=0$;
3) $a+2 b+2$ is divisible by 4;
$$
a+6 b+2 \text { is a prime number }
$$
three are true, and one is false. | Solution. Statements 3) and 4) cannot both be true. If 3) is true, then $a+6b+2=(a+2b+2)+4b$ is divisible by 4 and is not a prime number. Therefore, one of the statements 3) or 4) is false.
Let's consider statement 2): $a^{2}+a b-6 b^{2}-15 b-9=0$; Consider this equation as a quadratic in terms of $a$. Calculate the d... | =5,b=1;=17,b=7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,199 |
5. Find all values of the parameter $a$ for which the equation $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}-\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}=5-2 a-(3 a+2) x^{2}-2 a^{2}-\left(1-x^{2}\right)^{2}$ has two distinct solutions on the interval $\left[-\frac{\sqrt{6}}{2} ; \sqrt{2}\right]$. Specify the... | Answer: $a \in[0.25 ; 1), x_{1}=\sqrt{2-2 a}, x_{2}=-\sqrt{2-2 a}$.
$$
a \in[-3.5 ;-2), x_{1}=\sqrt{-a-2}, x_{2}=-\sqrt{-a-2}
$$ | \in[0.25;1),x_{1}=\sqrt{2-2},x_{2}=-\sqrt{2-2}\\\in[-3.5;-2),x_{1}=\sqrt{--2},x_{2}=-\sqrt{--2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,200 |
6. In triangle $A B C$ with angle $A$ equal to $60^{\circ}$, the angle bisector $A D$ is drawn. The radius of the circumcircle of triangle $A D C$ with center at point $O$ is $\sqrt{3}$. Find the length of the segment $O M$, where $M$ is the intersection point of segments $A D$ and $B O$, if $A B=1.5$. | Solution: 1) $D C=\sqrt{3}$
$\triangle O D C \quad-\quad$ is equilateral,
since
$2 \angle D A C=\angle D O C=60^{\circ}$.
2) Let $A D=l, B D=x, A C=z$. By the properties of the angle bisector, we have $\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow \frac{3}{2 z}=\frac{x}{\sqrt{3}}$. Since $S_{\triangle A B D}+S_{\tria... | \frac{\sqrt{21}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,201 |
Task 2. (Variant 1)
## Proof:
First method. Using the Law of Sines.
Let $\quad M \in A C, \quad \angle A B M=\beta_{1}, \quad \angle C B M=\beta_{2}$, $A B=a, R$ - the radius of the circumscribed circle, $A M=m$, $C M=n, B M=b$. We need to prove that $m+n=b$.
We have: $\left\{\begin{array}{l}m=2 R \sin \beta_{1} \\... | Answer: the statement is proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 10,202 |
Task 2. (Option 2)
The bisector $D E$ of triangle $A D C$ is extended to intersect the circumscribed circle at point $B$. It is known that the length $B D=1$ and the value $\sin \angle A D C=a$. Find the area of quadrilateral $A B C D$. | # Solution:
For the sought area, we have (see fig.):
$S=\frac{1}{2} B D \cdot A C \cdot \sin \angle D E C=\frac{1}{2} B D \cdot 2 R \cdot \sin \angle D E C$
From $\triangle A D E$ and $\triangle A B E$, it is clear that $\angle D E C=\alpha+\frac{\beta}{2}=\angle D A B$,
Find the smallest length of the segment $\mathrm{AB}$, if point A belongs to the set defined by the equation $y^{2}-3 x^{2}-2 x y-9-12 x=0$, and point B belongs to the set defined by the equation $x^{2}-8 y+23+6 x+y^{2}=0$. | Solution: We will transform these equations into canonical forms and plot the graphs of these equations:
1) $y^{2}-3 x^{2}-2 x y-9-12 x=0 \Leftrightarrow y^{2}-2 x y+x^{2}-4 x^{2}-12 x-9=0 \Leftrightarrow$
$$
(y-x)^{2}-(2 x+3)^{2}=0 \Leftrightarrow(y-x-2 x-3) \cdot(y-x+2 x+3)=0 \Leftrightarrow
$$
$$
(y-3 x-3) \cdot(... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,204 |
Problem 3. (Option 2)
Find the smallest length of the segment $AB$, if point $A$ belongs to the set defined by the equation $y^{2}-9+2 y x-12 x-3 x^{2}=0$, and point $B$ belongs to the set defined by the equation $y^{2}+3-4 x-2 y+x^{2}=0$. | Solution: We will transform these equations into canonical forms and plot the graphs of these equations:
1) $y^{2}-9+2 x y-12 x-3 x^{2}=0 \Leftrightarrow y^{2}+2 x y+x^{2}-4 x^{2}-12 x-9=0 \Leftrightarrow$
$(y+x)^{2}-(2 x+3)^{2}=0 \Leftrightarrow(y+x-2 x-3) \cdot(y+x+2 x+3)=0 \Leftrightarrow$
$(y-x-3) \cdot(y+3 x+3)... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,205 |
Task 4. (Option 1)
For which values of a does the inequality $x^{2}+|x+a|<2$ have at least one positive solution? | Solution: On the xOa plane, we will plot the set of points whose coordinates satisfy the inequality $x^{2}+|x+a|x^{2}-x-2$. In the shaded area, points with positive abscissas exist when $a \in\left[-\frac{9}{4} ; 2\right)$.
 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,206 |
Task 4. (Option 2)
For what values of $a$ does the inequality $x^{2}<4-|x-a|$ have at least one negative solution? | Solution: On the plane $x O a$, plot the set of points whose coordinates satisfy the inequality $x^{2}x^{2}+x-4$; when $x-a<0$, the form is $a<-x^{2}+x+4$. In the shaded area, points with positive abscissas exist when $a \in\left[-\frac{17}{4} ; 4\right)$.
 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,207 |
Task 5. (Option 1).
Away from a straight section of the railway, 20 km from it, lies village V. Where should a halfway station C be set up so that the total travel time from A to V by railway AC and by road CB is minimized. The speed of movement by railway is 0.8, and by road is 0.2 km per minute. | Solution: Let the distance $AD$ (from $A$ to the base of the perpendicular $BD$ to $AD$) be denoted by $a$, and $CD$ by $x$. Then $AC = AD - CD = a - x$, and $CB = \sqrt{CD^2 + BD^2} = \sqrt{x^2 + 20^2}$. The time it takes for the train to travel the path $AC$ is $\frac{AC}{0.8} = \frac{a - x}{0.8}$.
The time it takes... | 5.16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,208 |
Problem 5. (Option 2).
How to build a highway? (an old problem) From a riverside city A, goods need to be transported to point B, located $a$ kilometers downstream and $d$ kilometers from the riverbank. How should the highway be built from B to the river so that the transportation of goods from A to B is as cost-effec... | Solution: Let the distance $AD$ be denoted by $x$ and the length of the highway $DB$ by $y$: by assumption, the length of $AC$ is $a$ and the length of $BC$ is $d$.
Since transportation along the highway is twice as expensive as along the river, the sum $x + 2y$ should be the smallest according to the problem's requir... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,209 |
Task 6. (Option 2).
Translate the text above into English, preserving the original text's line breaks and formatting, and output the translation result directly. | Solution: A total of $68+59+46=173$ "yes" answers were given to the three questions. Note that since each resident of Kashino lives in exactly one district, if every resident were a knight (i.e., told the truth only), the number of "yes" answers would be equal to the number of residents in the city (a knight says "yes"... | 12 | Number Theory | proof | Yes | Yes | olympiads | false | 10,210 |
1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$
considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)... | Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.
Grading criteria.
| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but there ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,211 |
5. Solution: Let Masha have $x$ rubles, and Petya have $y$ rubles, then
$n(x-3)=y+3$
$x+n=3(y-n)$
Express $x$ from the second equation and substitute into the first:
$n(3 y-4 n-3)=y+3$,
$3 n y-y=4 n^{2}+3 n+3$,
$y=\left(4 n^{2}+3 n+3\right):(3 n-1)$
For $y$ to be an integer, $\left(4 n^{2}+3 n+3\right)$ must be ... | Answer: $1 ; 2 ; 3 ; 7$.
| 20 points | Complete justified solution |
| :--- | :--- |
| 15 points | Values of n are found, it is proven that there are no other values, but it is not shown why these values meet the condition of the problem. |
| 12 points | Correct approach to the solution, one value of n is lost. |
| 10... | 1;2;3;7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,215 |
1. According to the inverse theorem of Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,221 |
4. $\quad$ Transform the expression $\left|\frac{x^{2}-6 x+9}{3-x}+\frac{4 x^{2}-5 x}{x}\right|=\left|\frac{(3-x)^{2}}{3-x}+\frac{x(4 x-5)}{x}\right|=|3 x-2|$, we get $f(x)=|3 x-2|, D(f): x \neq 3, x \neq 0$.
Solve the equation $|3 x-2|=|x+a|$ graphically in the system $x O a$ $(3 x-2)^{2}=(x+a)^{2}$
$$
\begin{gather... | Answer: $a=-\frac{2}{3} ; a=2$.
| 15 points | Correct justified solution |
| :--- | :--- |
| 12 points | Functions are correctly transformed, the equation is set up, but a computational err... | =-\frac{2}{3};=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,224 |
6. The expression contains the following products:
$(2 \cdot 5) \cdot(10) \cdot(12 \cdot 15) \cdot(25 \cdot 4) \cdot(20 \cdot 30) \cdot(35 \cdot 14) \ldots \Rightarrow$ since each of these grouped products ends in 0, the number will end in 8 zeros $\Rightarrow B=0$
Let's break down the expression into the following p... | Answer: $A=\mathbf{2}, \mathrm{B}=\mathbf{0}, \mathrm{C}=\mathbf{1}, \mathrm{D}=\mathbf{4}$.
| 20 points | Correct answer obtained with justification |
| :---: | :--- |
| 15 points | Three out of four digits are correctly found. |
| 10 points | Two out of four digits are correctly found. |
| 5 points | One out of four... | A=2,B=0,C=1,D=4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,226 |
1. Solve the equation
$$
\left(x^{2}-x+1\right) \cdot\left(3 x^{2}-10 x+3\right)=20 x^{2}
$$
Solution:
$$
\left(x^{2}-x+1\right) \cdot\left(3 x^{2}-10 x+3\right)=20 x^{2} \Leftrightarrow\left(x-1+\frac{1}{x}\right) \cdot\left(3 x-10+3 \frac{1}{x}\right)=20
$$
Substitution $\boldsymbol{x}+\frac{1}{x}=\boldsymbol{t}$... | Answer: $x=\frac{5 \pm \sqrt{21}}{2}$ | \frac{5\\sqrt{21}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,227 |
2. (5 points) A company of 10 friends sits down at a round table in any order. Among them, there is one Vanya and one Dima. What is the probability that they will end up sitting next to each other?
## Solution:
Let's count the number of seating arrangements where Vanya and Dima are sitting next to each other. If Vany... | # Answer: $\frac{2}{9}$
# | \frac{2}{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,228 |
# 3. (10 points) Determine the sign of the expression
$\sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38} \quad$ or $\quad(\sqrt{25 \sqrt{7}-28 \sqrt{6}}-\sqrt{17 \sqrt{5}-38})^{* * *}$
## Solution:
1st method: Each radicand is the cube of the difference of two numbers:
$$
\begin{gathered}
25 \sqrt{7}-27 \sqrt{6}... | Answer: $\sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38}<0$
2nd method: Compare the expressions under the square roots $25 \sqrt{7}-27 \sqrt{6} \vee 17 \sqrt{5}-38$
Since both numbers are positive, we square them and gradually eliminate the roots. As a result, we find that the minuend is greater than the subtrah... | \sqrt{25\sqrt{7}-27\sqrt{6}}-\sqrt{17\sqrt{5}-38}<0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,229 |
4. (10 points) Solve the equation
$$
\sqrt{7-x^{2}+6 x}+\sqrt{6 x-x^{2}}=7+\sqrt{x(3-x)}
$$ | # Solution:
The domain of the variable x in our problem is the interval [0;3]. By completing the square in the expressions under the square roots on the left side of the equation or by plotting the graphs, we notice that the values of the first root do not exceed 4, and the second does not exceed 3, with the minimum v... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,230 |
5. (10 points) Find all values of parameters $k$ and $n$ for which the system of equations has infinitely many solutions.
$$
\left\{\begin{array}{c}
k y + x + n = 0 \\
|y-2| + |y+1| + |1-y| + |y+2| + x = 0
\end{array}\right.
$$
Solution:
It is convenient to introduce a coordinate system, directing the Ox axis upward... | Answer: $(k ; n)=(4 ; 0) ;(-4 ; 0) ;(2 ; 4) ;(-2 ; 4) ;(0 ; 6)$ | (k;n)=(4;0);(-4;0);(2;4);(-2;4);(0;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,231 |
7. (20 points) Given $\boldsymbol{m}$ natural numbers, not exceeding $\boldsymbol{n}$, arranged in non-decreasing order: $a_{1} \leq a_{2} \leq \cdots \leq a_{m}$. Similarly, $n$ natural numbers, not exceeding $\boldsymbol{m}$, are arranged in non-decreasing order: $\boldsymbol{b}_{1} \leq \boldsymbol{b}_{2} \leq \cdot... | Solution:
Since $1 \leq \boldsymbol{a}_{1} \leq \boldsymbol{a}_{2} \leq \cdots \leq \boldsymbol{a}_{\boldsymbol{m}} \leq \boldsymbol{n}$, then $2 \leq \mathrm{a}_{1}+1<\mathrm{a}_{2}+2<\cdots<\mathrm{a}_{\mathrm{m}}+\mathrm{m} \leq \mathrm{n}+\mathrm{m}$. Similarly, $2 \leq \mathrm{b}_{1}+1<\mathrm{b}_{2}+2<\cdots<\ma... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 10,232 |
8. (25 points) Let $h$ and $l$ be the height and the angle bisector drawn from one vertex of a triangle, and $r$ and $R$ be the radii of its inscribed and circumscribed circles. Prove that $\frac{h}{l} \geq \sqrt{\frac{2 r}{R}}$. | # Solution:
Let CL be the bisector of $\angle \mathrm{ACB}$, $\mathrm{P}$ be the intersection point of the bisector with the circumcircle, $\mathrm{I}$ be the incenter, and $\mathrm{O}$ be th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 10,233 |
1. Friends Vasya, Petya, and Kolya live in the same house. One day, Vasya and Petya went fishing at the lake on foot. Kolya stayed home, promising to meet his friends on a bicycle on their way back. Vasya was the first to head home, and at the same time, Kolya set out on a bicycle to meet him. Petya, moving at the same... | Solution: Let $x$ be the speed of Vasya and Petya, $v$ be the speed of cyclist Kolya, and $S$ be the distance from home to the lake. Then Vasya and Kolya met after time $t=\frac{S}{x+v}$ from the start of the journey. After Kolya's second departure from home, the meeting between Petya and Kolya occurred after time $t_{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,234 |
2. Solve the inequality $\frac{\sqrt{x^{2}-16}}{x}-\frac{x}{\sqrt{x^{2}-16}}<\frac{16}{15}$. (8 points) | Solution: Domain of definition: $|x|>4 . \frac{\sqrt{x^{2}-16}}{x}-\frac{x}{\sqrt{x^{2}-16}}4$ inequality is true.
2) For $x15$, or $\left(-x \sqrt{x^{2}-16}\right)^{2}>225$,
$$
\begin{aligned}
& x^{4}-16 x^{2}-225>0,\left(x^{2}+9\right)\left(x^{2}-25\right)>0,(x+5)(x-5)>0 \text{. Since } x<-4 \text{, then } \\
& x<-5... | x\in(-\infty;-5)\cup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,235 |
3. All members of an infinite geometric progression are natural numbers. The sum of the third, fifth, and seventh terms of this progression is equal to $7371 \cdot 2^{2016}$. Find the common ratio of the progression. (8 points) | Solution: We have a geometric progression $b_{1}, b_{1} q, b_{1} q^{2}, \cdots, b_{1} q^{n-1}, \cdots$, and $b_{1} q^{n-1} \in N$ for any number $n \in N$. Thus, $b_{1}$ and $q$ are natural numbers. According to the condition $b_{3}+b_{5}+b_{7}=7371 \cdot 2^{2016}$, or $b_{1} q^{2}+b_{1} q^{4}+b_{1} q^{6}=2^{2016} \cdo... | q=1,q=2,q=3,q=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,236 |
4. Solve the inequality $\left(\log _{x}^{2}(3 x-2)-4\right)(\sin \pi x-1) \leq 0$. (8 points)
# | # Solution:
$\left(\log _{x}^{2}(3 x-2)-4\right)(\sin \pi x-1) \leq 0 \Leftrightarrow\left(\log _{x}(3 x-2)-2\right)\left(\log _{x}(3 x-2)+2\right)(\sin \pi x-1) \leq 0$ Domain: $x \in(2 / 3,1) \cup(1,+\infty)$
\cup(1,2]\cup{1/2+2n,n\in\mathbb{N}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,237 |
5. Solve the system of equations $\left\{\begin{array}{l}4 \cos ^{2} x \cdot \sin ^{2} \frac{x}{6}+4 \sin \frac{x}{6}-4 \sin ^{2} x \cdot \sin \frac{x}{6}+1=0, \\ \sin \frac{x}{4}=\sqrt{\cos y} .\end{array}\right.$ (10 points)
# | # Solution:
Solve the 1st equation of the system:
$$
\begin{aligned}
& 4 \cos ^{2} x \cdot \sin ^{2} \frac{x}{6}+4 \sin \frac{x}{6}-4 \sin ^{2} x \cdot \sin \frac{x}{6}+1=0 \quad \Leftrightarrow \quad 4 \cos ^{2} x \cdot \sin ^{2} \frac{x}{6}+4 \cos ^{2} x \cdot \sin \frac{x}{6}+1=0 \\
& \Leftrightarrow\left(2 \cos x... | (11\pi+24\pi,\\frac{\pi}{3}+2\pin),(-5\pi+24\pi,\\frac{\pi}{3}+2\pin),,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,238 |
Problem 6. Find the set of values of the function $f(x)=\operatorname{arctg} \sqrt{\log _{0.5}^{-1}\left(\frac{\sin x}{\sin x+7}\right)}$. (10 points)
# | # Solution:
Let $\quad t=\sin x, t \in[-1 ; 1] \Rightarrow u=\log _{0.5}(z), \quad z=\frac{t}{t+7}=1-\frac{7}{t+7} \quad$ (the function is increasing). For $\mathrm{t} \in[-1 ; 1]$ we have $\mathrm{z} \in\left[-\frac{1}{6} ; \frac{1}{8}\right]$, but by the properties of logarithms $\mathrm{z}>0$, hence, $\mathrm{z} \i... | (0;\pi/6] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,239 |
7. On the diagonal $AC$ of the rhombus $ABCD$, a point $K$ is chosen, which is at a distance of 12 and 2 from the lines $AB$ and $BC$ respectively. The radius of the circle inscribed in triangle $ABC$ is 5. Find the side of the rhombus $ABCD$ and the radius of the circle inscribed in this rhombus.
(12 points)
# | # Solution:
According to the problem: $K N=a=2$ - the distance from point $K$ to line $B C$, $K M=b=12$ - the distance from point $K$ to line $A B$, $r=5$ - the radius of the circle inscribed in triangle $A B C$. Let $O$ be the intersection point of the diagonals of the rhombus, and $E$ be the center of the circle ins... | \frac{25\sqrt{21}}{6},7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,240 |
8. On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5(x-(\sqrt{3} / 2))^{2}$ pass, the angle between which is $60^{\circ} . \quad$ (12 points) | # Solution:
On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5 \cdot(x-\sqrt{3} / 2)^{2}$ pass, with the angle between them being $60^{\circ}$.
Solution (without using derivatives).
$y=0.5 \cdot(x-\sqrt{3} / 2)^{2}, M\left(0 ; y_{0}\right) \cdot U$ The equation $0.5 ... | M(0;0)orM(0;-\frac{5}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,241 |
9. Determine all values of $a$ for which the equation
$$
4 x^{2}-8|x|+(2 a+|x|+x)^{2}=4
$$
has exactly two distinct roots. Specify these roots for each of the found values of $a$.
(12 points) | # Solution:
I. $x \geq 0, x^{2}-2 x+(a+x)^{2}=1 ; 2 x^{2}+2(a-1) x+a^{2}-1=0$ ；
$D / 4=1-2 a+a^{2}-2 a^{2}+2=3-2 a-a^{2}$.
The equation has two distinct non-negative roots $x_{1,2}=\left(1-a \pm \sqrt{3-2 a-a^{2}}\right) / 2$, if $\left\{\begin{array}{c}3-2 a-a^{2}>0, \\ a-10\end{array}\right.\end{array} \Leftrighta... | \in(-3;-\sqrt{2}),\quadx_{1,2}=(1-\\sqrt{3-2-^{2}})/2;\quad\in(-1;1],x_{1}=(1-+\sqrt{3-2-^{2}})/2,x_{2}=-1-\sqrt{2-^{2}}; | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,242 |
10. A regular triangular pyramid is inscribed in a sphere of radius $R$, with the height of the pyramid equal to $4 R / 3$. What is the minimum area that the cross-section of the pyramid by a plane passing through the median of the base can have? Find the ratio of the volumes of the parts into which the cutting plane d... | # Solution:
Since $T K=4 R / 3, O K=R / 3, B K=\sqrt{1-(1 / 3)^{2}} R=\sqrt{8} R / 3 ; A B=A K \sqrt{3}=\sqrt{8 / 3} R$, $A T=\sqrt{(4 R / 3)^{2}+(\sqrt{8} R / 3)^{2}}=\sqrt{8 / 3} R$. Therefore, $A B=A T$, all edges of the pyramid are equal.
We will assume that the cutting plane passes through the median $T G$ and i... | \frac{2\sqrt{2}}{\sqrt{33}}R^{2};3:19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,243 |
1. Friends Vasya, Petya, and Kolya live in the same house. One day, Vasya and Petya went fishing at the lake on foot. Kolya stayed home, promising to meet his friends on a bicycle on their way back. Vasya was the first to head home, and at the same time, Kolya set out on a bicycle to meet him. Petya, moving at the same... | Solution: Let $x$ be the speed of Vasya and Petya, $v$ be the speed of cyclist Kolya, and $S$ be the distance from home to the lake. Then Vasya and Kolya met after time $t=\frac{S}{x+v}$ from the start of the journey. After Kolya's second departure from home, the meeting between Petya and Kolya occurred after time $t_{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,244 |
2. Solve the inequality $\frac{\sqrt{x^{2}-5}}{x}-\frac{x}{\sqrt{x^{2}-5}}<\frac{5}{6}$. (8 points) | Solution: Domain of definition: $|x|>\sqrt{5} . \frac{\sqrt{x^{2}-5}}{x}-\frac{x}{\sqrt{x^{2}-5}}\sqrt{5}$ inequality is true.
2) For $x<6$, or $\left(-x \sqrt{x^{2}-5}\right)^{2}>36$,
$$
\begin{aligned}
& x^{4}-5 x^{2}-36>0,\left(x^{2}+4\right)\left(x^{2}-9\right)>0,(x+3)(x-3)>0 \text{. Since } x<-\sqrt{5} \text{, th... | x\in(-\infty;-3)\cup(\sqrt{5};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,245 |
3. All members of an infinite geometric progression are natural numbers. The sum of the third, fifth, and seventh terms of this progression is equal to $819 \cdot 6^{2016}$. Find the common ratio of the progression. (8 points) | Solution: We have a geometric progression $b_{1}, b_{1} q, b_{1} q^{2}, \cdots, b_{1} q^{n-1}, \cdots$, and $b_{1} q^{n-1} \in N$ for any number $n \in N$. Thus, $b_{1}$ and $q$ are natural numbers. According to the condition $b_{3}+b_{5}+b_{7}=819 \cdot 6^{2016}$, or $b_{1} q^{2}+b_{1} q^{4}+b_{1} q^{6}=2^{2016} \cdot... | q=1,q=2,q=3,q=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,246 |
5. Solve the system of equations $\left\{\begin{array}{l}4 \cos ^{2} 2 x \cdot \sin ^{2} \frac{x}{3}+4 \sin \frac{x}{3}-4 \sin ^{2} 2 x \cdot \sin \frac{x}{3}+1=0, \\ \sin \frac{x}{2}+\sqrt{\cos y}=0 .\end{array}\right.$
(10 points) | Solution: Let's solve the first equation of the system:
$4 \cos ^{2} 2 x \cdot \sin ^{2} \frac{x}{3}+4 \sin \frac{x}{3}-4 \sin ^{2} 2 x \cdot \sin \frac{x}{3}+1=0$
$4 \cos ^{2} 2 x \cdot \sin ^{2} \frac{x}{3}+4 \cos ^{2} 2 x \cdot \sin \frac{x}{3}+1=0 \quad \Leftrightarrow\left(2 \cos 2 x \sin \frac{x}{3}+\cos 2 x\ri... | (-\frac{\pi}{2}+12\pi,\\frac{\pi}{3}+2\pin),(\frac{7\pi}{2}+12\pi,\\frac{\pi}{3}+2\pin),\quad,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,248 |
6. Find the set of values of the function $\quad f(x)=\operatorname{arctg} \sqrt{6 \log _{0,25}^{-1}\left(\frac{\sin x}{\sin x+15}\right)}$.
points)
# | # Solution:
Let
$t=\sin x, t \in[-1 ; 1] \Rightarrow u=\log _{0.25}(z), \quad z=\frac{t}{t+15}=1--$ (function $z(t)$ is increasing). For $t \in[-1 ; 1]$, we have $z \in\left[-\frac{1}{14} ; \frac{1}{16}\right]$, but by the properties of logarithms, $z>0$, hence, $z \in\left(0 ; \frac{1}{16}\right]$. The function $u=\... | (0;\pi/3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,249 |
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 $45^{\circ}$. | # 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... | M(0;0)orM(0;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,251 |
9. Determine all values of $a$ for which the equation $4 x^{2}-16|x|+(2 a+|x|-x)^{2}=16$ has exactly two distinct roots. Specify these roots for each of the found values of $a$. | # Solution:
I. $x \leq 0, x^{2}+4 x+(a-x)^{2}=4 ; 2 x^{2}-2(a-2) x+a^{2}-4=0$;
$D / 4=a^{2}-4 a+4-2 a^{2}+8=12-4 a-a^{2}$.
The equation has two distinct non-positive roots $x_{1,2}=\left(a-2 \pm \sqrt{12-4 a-a^{2}}\right) / 2$,
if $\left\{\begin{array}{c}12-4 a-a^{2}>0, \\ a-2 \leq 0\end{array} \Leftrightarrow\left... | \in(-6;-\sqrt{8}),x_{1,2}=(-2\\sqrt{12-4-^{2}})/2;\in(-2;2],x_{1}=(-2-\sqrt{12-4-^{2}})/2,x_{2}=2+\sqrt{8-^{2}};\in | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,252 |
10. A regular triangular pyramid is inscribed in a sphere of radius $R$, with the height of the pyramid equal to $4 R / 3$. What is the smallest area that the cross-section of the pyramid by a plane passing through the median of the base can have? Find the ratio of the volumes of the parts into which the cutting plane ... | # Solution:
Since $T K=4 R / 3, O K=R / 3, B K=\sqrt{1-(1 / 3)^{2}} R=\sqrt{8} R / 3 ; A B=A K \sqrt{3}=\sqrt{8 / 3} R$, $A T=\sqrt{(4 R / 3)^{2}+(\sqrt{8} R / 3)^{2}}=\sqrt{8 / 3} R$. Therefore, $A B=A T$, all edges of the pyramid are equal.
We will assume that the cutting plane passes through the median $T G$ and i... | \frac{2\sqrt{2}}{\sqrt{33}}R^{2};3:19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,253 |
1. Find the value of the expression $2 a-\left(\frac{2 a-3}{a+1}-\frac{a+1}{2-2 a}-\frac{a^{2}+3}{2 a^{2-2}}\right) \cdot \frac{a^{3}+1}{a^{2}-a}+\frac{2}{a}$ when $a=1580$. | Solution:
1) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)(a+1)} \cdot \frac{(a+1)\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
2) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)} \cdot \frac{\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
3) $2 a-\frac{\left(-4 a+2+2 a^{2}\right)}{(a... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,254 |
2. For a rectangle, the sum of two sides is 11, and the sum of three sides is 19.5. Find the product of all possible different values of the perimeter of such a rectangle. | Solution:
Let the sides of the rectangle be $a$ and $b$.
If the sum of adjacent sides is 11, then the system describing the condition of the problem is $\left\{\begin{array}{l}a+b=11 \\ 2 a+b=19.5\end{array}\right.$, its solution is $a=8.5, b=2.5$, the perimeter of the rectangle is $P_{1}=22$.
If, however, the numbe... | 15400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,255 |
3. A cyclist traveled the path from A to B and back at a constant speed. A pedestrian walked the path from A to B at a speed three times slower than the cyclist's speed, but on the way back, he took a bus, which traveled at a speed five times greater than the cyclist's speed. How much time did the pedestrian spend on t... | Solution:
$$
v_{\mathrm{B}}=x, t_{\mathrm{B}}=\frac{S}{x}+\frac{S}{x}=\frac{2 S}{x}
$$
$v_{n} \cdot A B=\frac{x}{3}, v_{n} \cdot \mathrm{BA}=5 x$, then
$$
t_{n}=\frac{S}{x / 3}+\frac{S}{5 x}=\frac{3 S}{x}+\frac{S}{5 x}=\frac{15 S+S}{5 x}=\frac{16 S}{5 x}
$$
We get $\frac{16 S}{5 x}-\frac{2 S}{x}=\frac{36}{60} ; \fr... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,256 |
4. In a convex quadrilateral $A B C D$, the angles at vertices $B, C$, and $D$ are $30^{\circ}, 90^{\circ}$, and $120^{\circ}$ respectively. Find the length of segment $A B$, if $A D=C D=2$. | Solution:
Extend lines $A B$ and $C D$ to intersect at point $E$, the resulting triangle $A D E$ is equilateral, so $E D=E A=2$, the leg $E C=E D+D C=2+2=4$, since it lies in the right triangle $B C E$ opposite the angle $30^{\circ}$, then the hypotenuse $B E=8$, and the segment $A B=B E-E A=8-2=6$.
Answer: 6. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,257 |
5. Find the greatest value of the parameter a for which the equation:
$$
(|x-2|+2 a)^{2}-3(|x-2|+2 a)+4 a(3-4 a)=0 \text { has three solutions. }
$$
In the answer, specify the greatest of them. | Selection (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
Let $|\mathrm{x}-2|+2 \mathrm{a}=\mathrm{t}$, then
$$
t^{2}-3 t+4 a(3-4 a)=0
$$
$t^{2}-2 \cdot \frac{3}{2} \mathrm{t}+\frac{4}{9}-\frac{4}{9}+12 \mathrm{a}-16 a^{2}=0$
$\left(\mathrm{t}-\frac{3}{2}\right)^... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,258 |
8. In an acute-angled triangle $ABC$ with sides $AB=4, AC=3$, a point $N$ is marked on the median $AM$ such that $\angle BNM = \angle MAC$. Find the length of the segment $BN$. | Solution:
Let's make an additional construction, doubling the median $A M$ beyond point $M$, thereby obtaining point $K$ such that $K \in A M, K M=A M$. Triangles $K M B$
Preliminary (correspondence) online stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
and $A M C$ are eq... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,259 |
9. Solve the equation $a^{2}+2=b$ !, given that a, b belong to N. In the answer, indicate the sum of the product of all possible a and the product of all possible b (if the equation has no solutions, indicate 0; if there are infinitely many solutions, indicate 1000). | Solution:
$b!-2=a^{2} ; x, y \in N$
$a \geq 1$, i.e. $a^{2} \geq 1 \Rightarrow b!\geq 3$, i.e. $b \geq 3$
If $x \geq 5$, then $x!$ ends in 0, then $y^{2}$ ends in 8, but there is no number whose square ends in 8, i.e. $x<5$.
This gives us:
$\left[\begin{array}{l}b=3 \\ b=4\end{array} \Rightarrow\left[\begin{array}... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,260 |
2. (15 points) Find the area of a convex quadrilateral with equal diagonals, if the lengths of the segments connecting the midpoints of its opposite sides are 13 and 7. | Solution. Let $MK$ and $PH$ be segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$, with $MK = PH$, $AC = 18$, and $BD = 7$.
We have: $MP \| AC$, $MP = \frac{1}{2} AC$ (as the midline of $\triangle ABC$); $HK \| AC$, $HK = \frac{1}{2} AC$ (as the midline of $\triangle ADC$). $\Rightarr... | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,262 |
3. (15 points) In a box, there are 22 red and 25 blue balls. They are distributed into two boxes: the first box should have 24 balls, and the second box should have 23. After the distribution, the percentage of blue balls in each box was calculated and the obtained numbers were added together. What should be the distri... | # Solution.
Solution 1. Instead of the total percentage, we will calculate the total share of blue balls, which are obviously 100 times different and reach their maximum simultaneously. Each blue ball in a box of 24 balls constitutes $1 / 24$ of the total number of balls in that box, and in a box of 23 balls, it const... | In\the\\box\-\23\blue\balls,\in\the\first\box\-\2\blue\balls\\22\red\balls | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,263 |
4. (20 points) For what values of the parameter $a$ does the equation
$$
(a+1)(|x-2.3|-1)^{2}-2(a-3)(|x-2.3|-1)+a-1=0
$$
have exactly two distinct solutions? | Solution. Let $|x-2.3|-1=t$ (1), then the original equation will take the form: $(a+1) t^{2}-2(a-3) t+a-1=0$ (2). Let's analyze equation (1): when $t>-1$, it corresponds to two different values of x. Thus, the original equation can have from zero to four solutions. It has two distinct roots in the following three cases... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,264 |
5. (20 points) Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $E$, and the rays $D A$ and $C B$ intersect at point $F$. The ray $B A$ intersects the circumcircle of triangle $D E F$ at point $L$, and the ray $B C$ intersects the same circle at point $K$. The length of segment $L K$ ... | # Solution.
$\angle F L E = \angle F D E = \angle F K E = \alpha$, since these angles subtend the arc $F E$.
$\angle E B K = \angle F D E = \alpha$, since quadrilateral $A B C D$ is cyclic... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,265 |
6. (20 points) Do there exist five pairwise distinct integers such that the sum of any four of them is a square of a natural number? | Solution. We will show a way to construct the desired sequence. Consider the first five squares: $1, 4, 9, 16, 25$ (any five different ones can be taken).
Then the first number is $1 + 4 + 9 + 16 + 25 - 4 \cdot 25 = -45$, the second is $1 + 4 + 9 + 16 + 25 - 4 \cdot 16 = -9$, the third is $1 + 4 + 9 + 16 + 25 - 4 \cdo... | -45,-9,19,39,51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,266 |
1. (10 points) Solve the inequality: $\frac{\left|3 x^{2}+8 x-3\right|+\left|3 x^{4}+2 x^{3}-10 x^{2}+30 x-9\right|}{|x-2|-2 x-1} \leq 0$. | Solution. Domain of definition: $|x-2|-2 x-1 \neq 0 \Leftrightarrow\left[\begin{array}{c}x<2 \\ x \neq \frac{1}{3} \\ \left\{\begin{array}{l}x \geqslant 2 \\ x \neq-3\end{array}\right.\end{array} \Leftrightarrow x \in\left(-\infty ; \frac{1}{3}\right) \cup\left(\frac{1}{3} ;+\infty\right)\right.$.
The numerator of the... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,267 | |
2. (15 points) The bisectors of the obtuse angles at the base of a trapezoid intersect on the other base of the trapezoid. Find the area of the trapezoid if its height is 12 cm, and the lengths of the bisectors are 15 cm and $13 \mathrm{~cm}$. | Solution. Let segment $B K$ be the height of the given trapezoid $A B C D$ ( $\mathrm{BK}=\mathrm{CH}=12$ ), $\mathrm{BM}$ and $\mathrm{CM}$ be the bisectors of angles $\angle ABC$ and $\angle BCD$, respectively, with $\mathrm{BM}=15, \mathrm{CM}=13$. In the right triangles $B K M$ and $C H M$, by the Pythagorean theor... | 260.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,269 |
4. (20 points) Determine the number of solutions of the equation
$$
a(x+|x|-2)=x^{2}+4 x-5
$$
depending on the values of the parameter $a$. | Solution. Let's consider the graphical solution of the problem. Express $a$ in terms of $x$ and plot the corresponding graph in the $(x; a)$ axes (this correspondence is not functional). Expand the modulus:
1) $x \geq 0, 2a(x-1)=(x-1)(x+5) \Leftrightarrow\left[\begin{array}{l}x=1 \\ a=\frac{x}{2}+\frac{5}{2}\end{array... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,272 |
5. (20 points) A circle passes through the vertices $A$ and $C$ of an isosceles triangle $ABC (AB = BC)$ and intersects the sides $AB$ and $BC$ at points $M$ and $N$, respectively. $MK$, a chord of this circle, equal in length to $2 \sqrt{5}$, contains point $H$, which lies on $AC$ and is the foot of the altitude of tr... | # Solution.
Quadrilateral $A M N C$ is an isosceles trapezoid. $\triangle A M H = \Delta H N C$ - by two sides and the angle between them.
$$
\begin{gathered}
\angle A H M = \angle H M N =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,273 |
6. (20 points) Does there exist a natural number whose square is equal to the sum of five pairwise distinct squares of integers, such that among them there is the number $7^{2}$? | Solution. Let's show a possible way to construct the desired number. We will use the formula $n^{2}+\left(\frac{n^{2}-1}{2}\right)^{2}=\left(\frac{n^{2}+1}{2}\right)^{2}$. That is,
$$
7^{2}+24^{2}=25^{2}, 25^{2}+312^{2}=313^{2}, 313^{2}+48984^{2}=48985^{2}
$$
Therefore, $48985^{2}=0^{2}+7^{2}+24^{2}+312^{2}+48984^{2}... | 48985^{2}=0^{2}+7^{2}+24^{2}+312^{2}+48984^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,274 |
1. Four elevators of a skyscraper, differing in color (red, blue, green, and yellow), are moving in different directions and at different but constant speeds. Observing the elevators, someone started a stopwatch and, looking at its readings, began to record: 36th second - the red elevator caught up with the blue one (m... | Solution. Let's number the elevators: red - first, blue - second, green - third, yellow - fourth. The elevators move at constant speeds, so the distance traveled $S_{i}, i=1,2,3,4$, in some coordinate system depends on time according to the law $S_{i}=k_{i} t+b_{i}$. According to the problem, the red and blue elevators... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,275 |
2. Find the minimum value of the expression $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}$, if $f(x)=a x^{2}+b x+c$ is an arbitrary quadratic function satisfying the condition $b>2 a$ and taking non-negative values for all real $x$.
(12 points) | Solution. We have $f(1)=a+b+c, \quad f(0)=c, \quad f(-1)=a-b+c, f(-2)=4 a-2 b+c$, $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}=\frac{3(a+b+c)+6 c-a+b-c}{c-4 a+2 b-c}=\frac{2 a+4 b+8 c}{2 b-4 a}=\frac{a+2 b+4 c}{b-2 a}$.
Since $f(x)=a x^{2}+b x+c \quad-$ is an arbitrary quadratic function that takes non-negative values for ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,276 |
3. Find all natural numbers $n$ for which the number $2^{10}+2^{13}+2^{14}+3 \cdot 2^{n}$ is a square of a natural number.
$(16$ points) | # Solution.
8) Let $n>12$. Then $N=2^{10}\left(1+2^{3}+2^{4}+3 \cdot 2^{n-10}\right)=2^{10} \cdot(2 k+1)^{2}$, and $25+3 \cdot 2^{n-10}=4 k^{2}+4 k+1$, $3 \cdot 2^{n-10}=4 k^{2}+4 k-24, \quad 3 \cdot 2^{n-12}=k^{2}+k-3 \cdot 2^{n-12}=(k+3)(k-2)$, The numbers $k+3$ and $k-2$ have different parities, so one of them is a... | 13,15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,277 |
4. In triangle $A B C$, the bisector $A D$ is drawn. It is known that the centers of the inscribed circle of triangle $A B D$ and the circumscribed circle of triangle $A B C$ coincide. Find $C D$, if $A C=\sqrt{5}+1$. The answer should not include trigonometric function notations or their inverses.
(20 points) | Solution: Let $\angle A=\alpha, \quad B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \angle ABO=\beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO=\angle ABO, \beta=\alpha / 2$. Triangl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,278 |
5. Find all values of the parameter $b$, for which for any value of the parameter $a \in[-2 ; 1]$ the inequality $a^{2}+b^{2}-\sin ^{2} 2 x-2(a+b) \cos 2 x-2>0$ is not satisfied for at least one value of $x$.
(20 points) | Solution: Let's make the substitution $y=\cos 2 x, \quad y \in[-1 ; 1]$. We get $a^{2}+b^{2}-1+y^{2}-2(a+b) y-2>0$, or $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$. We need to determine for which values of $a$ and $b$ the inequality $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$ holds for any $y \in[-1 ; 1]$. Consider the function $f(y)=y^{2}-2... | [-1.5;\sqrt{3}-1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,279 |
6. The base of the pyramid $T A B C D$ is a rhombus $A B C D$. The height of the pyramid $T K$ is 1, point $K$ lies on the line containing the diagonal of the base $AC$, and $KC = KA + AC$. The lateral edge $TC$ is $2 \sqrt{2}$, and the lateral faces are inclined to the plane of the base at angles of $30^{\circ}$ and $... | # Solution:
$K N \perp A B, N=K N \cap A B$,
1)
$K M \perp C D, M=K M \cap C D . \quad K N=\frac{T K}{\operatorname{tg} 60^{\circ}}=\frac{T K \sqrt{3}}{3}$,
$K M=\frac{T K}{\operatorname{tg} 30^{\circ}}=T K \sqrt{3}, \quad N M=\frac{2 \sqrt{3}}{3} T K$.
$K C=\sqrt{T C^{2}-T K^{2}}$,
$M C=\sqrt{K C^{2}-K M^{2}}=\... | \frac{7}{6},\arcsin\frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,280 |
1. A workshop produces transformers of types $A$ and $B$. For one transformer of type $A$, 5 kg of transformer iron and 3 kg of wire are used, and for a transformer of type $B$, 3 kg of iron and 2 kg of wire are used. The profit from selling a transformer of type $A$ is 12 thousand rubles, and for type $B$ it is 10 tho... | Solution. Let $x$ be the number of transformers of type $A$, and $y$ be the number of transformers of type $B$. Then the profit per shift is calculated by the formula $D=12 x+10 y$, with the conditions
=7+9+... | 2,5,7,14,15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,282 |
4. In triangle $A B C$, the bisector $A D$ is drawn. It is known that the centers of the inscribed circle of triangle $A B D$ and the circumscribed circle of triangle $A B C$ coincide. Find the area of triangle $A B C$, if $C D=4$. The answer should not include trigonometric function notations and their inverses. | Solution. Let $\angle A=\alpha, \angle B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \quad \angle ABO=\beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO=\angle ABO, \beta=\alpha / 2$. ... | 2\sqrt{130+58\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,283 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.