problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 5
values | source stringclasses 11
values | synthetic bool 2
classes |
|---|---|---|---|---|---|---|---|---|
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
y=x^{2}+a \\
x^{2}+y^{2}+2 b^{2}=2 b(x-y)+1
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer: $a \leq \sqrt{2}+\frac{1}{4}$.
Solution: The second equation of the system can be transformed into $(x-b)^{2}+(y+b)^{2}=1^{2}$, which means it represents a circle with radius 1 and center $(b, -b)$. For all possible $b \in \mathbf{R}$, the graphs of these functions sweep out the strip $-x-\sqrt{2} \leq y \leq ... | \leq\sqrt{2}+\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In the corners $C$ and $A$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $AC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $C K_{1}=6, A K_{2}=8$, and $A C=21$.
a) Fin... | Answer. a) $C K=9$, b) $\angle A C B=60^{\circ}$.
Solution. a) The lines $\mathrm{CO}_{1}$ and $A O_{2}$ are the bisectors of angles $C$ and $A$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1}$ and $O_{2}$ be $r$, and the radius of ... | )CK=9,b)\angleACB=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 700. The answer should be presented as an integer. | Answer: 2520.
Solution. Since $700=7 \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, one seven, and three ones, or (b) one four, two fives, one seven, and four ones. We will calculate the number of variants in each case.
(a) First, we choose two places out of... | 2520 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. [4 points] Given a geometric progression $b_{1}, b_{2}, \ldots, b_{3000}$, all of whose terms are positive, and their sum is $S$. It is known that if all its terms with indices divisible by 3 (i.e., $b_{3}, b_{6}, \ldots, b_{3000}$) are increased by 50 times, the sum $S$ will increase by 10 times. How will $S$ chang... | Answer: It will increase by $\frac{11}{8}$ times.
Solution. Let the common ratio of the geometric progression be $q$. Since all its terms are positive, $q>0$. If $q=1$, then $S=3000 b_{1}$, and when the terms with numbers divisible by 3 are increased by 50 times, we get the sum $S+49\left(b_{3}+b_{6}+\ldots+b_{3000}\r... | \frac{11}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. [4 points] Solve the equation $\left(\frac{x}{\sqrt{2}}+3 \sqrt{2}\right) \sqrt{x^{3}-4 x+80}=x^{2}+10 x+24$. | Answer: $4, \sqrt{13}-1$.
Solution. Factoring the right-hand side and multiplying both sides by $\sqrt{2}$, we get $(x+6) \sqrt{x^{3}-4 x+80}=\sqrt{2}(x+4)(x+6)$. From here, there are two possibilities: either $x+6=0$ (then $x=-6$, which does not fit the domain of definition, as the expression under the square root is... | 4,\sqrt{13}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [6 points] Solve the inequality $2 x^{4}+x^{2}-4 x-3 x^{2}|x-2|+4 \geqslant 0$. | Answer: $(-\infty ;-2] \cup\left[\frac{-1-\sqrt{17}}{4} ; \frac{-1+\sqrt{17}}{4}\right] \cup[1 ;+\infty)$.
Solution. The inequality can be rewritten as $2 x^{4}+(x-2)^{2}-3 x^{2}|x-2| \geqslant 0$ or $\mid x-$ $\left.2\right|^{2}-3 x^{2}|x+2|+2 x^{4} \geqslant 0$. To factorize the left side, note that it represents a ... | (-\infty;-2]\cup[\frac{-1-\sqrt{17}}{4};\frac{-1+\sqrt{17}}{4}]\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A water strider and a water beetle slide counterclockwise around a float on two circles. A rectangular coordinate system is introduced on the water surface, in which the float (the common center of the circles) is at the point ( $0 ; 0$ ). The speed of the water strider is twice the speed of the beetle. A... | Answer: $\left(\frac{5}{\sqrt{2}}(1-\sqrt{7}) ; \frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right)$, $\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)... | (\frac{5}{\sqrt{2}}(1-\sqrt{7});\frac{5}{\sqrt{2}}(1+\sqrt{7})),(-\frac{5}{\sqrt{2}}(1+\sqrt{7});\frac{5}{\sqrt{2}}(1-\sqrt{7})),(\frac{5}{\sqrt{2}}(\sqrt{7}- | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. [6 points] a) Two circles of the same radius 5 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a... | Answer: $C F=10, S_{\triangle A C F}=7$.
Fig. 1: Variant 3, Problem 6
Solution. a) Let $R=5$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. ... | CF=10,S_{\triangleACF}=7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. [6 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y-6-x|+|y-6+x|=12 \\
(|x|-8)^{2}+(|y|-6)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{4 ; 100\}$.
Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To remove the absolute values, we find the sets of points where the expressions under the absolute values are zero. These are the lines $y-6-x=0$ and $y-6+x=0$. They divide th... | \in{4;100} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 4900. The answer should be presented as an integer. | Answer: 4200.
Solution. Since $4900=7^{2} \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, two sevens, and two ones, or (b) a four, two fives, two sevens, and three ones. We will calculate the number of variants in each case.
(a) First, we choose two places ou... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. [4 points] Given a geometric progression $b_{1}, b_{2}, \ldots, b_{3000}$, all of whose terms are positive, and their sum is $S$. It is known that if all its terms with indices divisible by 3 (i.e., $b_{3}, b_{6}, \ldots, b_{3000}$) are increased by 40 times, the sum $S$ will increase by 5 times. How will $S$ change... | Answer: It will increase by $\frac{11}{7}$ times.
Solution. Let the common ratio of the geometric progression be $q$. Since all its terms are positive, $q>0$. If $q=1$, then $S=3000 b_{1}$, and when the terms with numbers divisible by 3 are increased by 40 times, we get the sum $S+39\left(b_{3}+b_{6}+\ldots+b_{3000}\r... | \frac{11}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. [4 points] Solve the equation $\left(\frac{x}{2 \sqrt{2}}+\frac{5 \sqrt{2}}{2}\right) \sqrt{x^{3}-64 x+200}=x^{2}+6 x-40$. | Answer: $6, \sqrt{13}+1$.
Solution. Factoring the right-hand side and multiplying both sides by $2 \sqrt{2}$, we get $(x+10) \sqrt{x^{3}-64 x+200}=2 \sqrt{2}(x-4)(x+10)$. From here, there are two possibilities: either $x+10=0$ (then $x=-10$, which does not fit the domain of definition, as the expression under the squa... | 6,\sqrt{13}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [6 points] Solve the inequality $4 x^{4}+x^{2}+4 x-5 x^{2}|x+2|+4 \geqslant 0$. | Answer: $(-\infty ;-1] \cup\left[\frac{1-\sqrt{33}}{8} ; \frac{1+\sqrt{33}}{8}\right] \cup[2 ;+\infty)$.
Solution. The inequality can be rewritten as $4 x^{4}+(x+2)^{2}-5 x^{2}|x+2| \geqslant 0$ or $\mid x+$ $\left.2\right|^{2}-5 x^{2}|x+2|+4 x^{4} \geqslant 0$. To factorize the left side, note that it represents a qu... | (-\infty;-1]\cup[\frac{1-\sqrt{33}}{8};\frac{1+\sqrt{33}}{8}]\cup[2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] Around a hook with a worm, in the same plane as it, a carp and a minnow are swimming along two circles. In the specified plane, a rectangular coordinate system is introduced, in which the hook (the common center of the circles) is located at the point $(0 ; 0)$. At the initial moment of time, the carp and... | Answer: $(\sqrt{2}-4 ;-4-\sqrt{2}),(-4-\sqrt{2} ; 4-\sqrt{2}),(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4)$.
Solution. Let the points where the carp and the minnow are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles formed by the radius vectors of points $M$ and $N... | (\sqrt{2}-4;-4-\sqrt{2}),(-4-\sqrt{2};4-\sqrt{2}),(4-\sqrt{2};4+\sqrt{2}),(4+\sqrt{2};\sqrt{2}-4) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. [6 points] a) Two circles of the same radius 13 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, ... | Answer: $C F=26, S_{\triangle A C F}=119$.
Fig. 3: variant 4, problem 6
Solution. a) Let $R=13$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$... | CF=26,S_{\triangleACF}=119 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. [6 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y+x+8|+|y-x+8|=16 \\
(|x|-15)^{2}+(|y|-8)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{49 ; 289\}$.
Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To open the absolute values, we need to find the sets of points where the expressions under the absolute values are zero. These are the lines $y+x+8=0$ and $y-x+8=0$. They di... | \in{49;289} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. [5 points] We rolled 80 fair dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) and counted the sum of the numbers that came up. Which of the following probabilities is smaller: the probability that this sum is greater than 400, or the probability that this sum is no... | Answer: It is more likely that the sum is no more than 160.
Solution. The result of rolling the dice can be described by a set of 80 numbers from 1 to 6. Consider any such set. If each number in the set is replaced from $x$ to 7 - $x$, we get a new set consisting of numbers from 1 to 6. In this case, if the sum of the... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. [4 points] Given a finite arithmetic progression $a_{1}, a_{2} \ldots, a_{n}$ with a positive common difference, such that the sum of all its terms is $S$, and $a_{1}>0$. It is known that if the common difference of the progression is tripled, while the first term remains unchanged, then the sum $S$ doubles. By what... | Answer: It will increase by $\frac{5}{2}$ times.
Solution. The sum of the terms of the original progression is determined by the formula $S_{0}=\frac{2 a_{1}+(n-1) d}{2} \cdot n$, where $d-$ is the common difference of the progression. If $d$ is tripled, the sum of the terms will become $S_{1}=\frac{2 a_{1}+(n-1) 3 d}... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+2 x-58}+5\right)\left|x^{3}-7 x^{2}+13 x-3\right| \leqslant 0$. | Answer: $2+\sqrt{3}$.
Solution. The first factor on the left side is positive for any valid value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}+2 x-58 \geqslant 0$) is equivalent to the equation $x^{3}-7 x^{2}+13 x-3=0$. One of the ... | 2+\sqrt{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$.
翻译结果：
4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$. | Answer: $-\frac{4}{3} ; 1 ; \frac{-1 \pm \sqrt{17}}{2}$.
Solution. The equation can be rewritten as $3 x^{4}+(x-4)^{2}-4 x^{2}|x-4|=0$ or $|x-4|^{2}-4 x^{2} \mid x-$ $4 \mid+3 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x-4|$ with a discriminant equal to $\left(4... | -\frac{4}{3};1;\frac{-1\\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A water strider and a water beetle slide counterclockwise around a float along two circles. A rectangular coordinate system is introduced on the water surface, in which the float (the common center of the circles) is at the point ( $0 ; 0$ ). The speed of the water strider is twice the speed of the beetle... | Answer: $\left(\frac{5}{\sqrt{2}}(1-\sqrt{7}) ; \frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right)$, $\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)... | (\frac{5}{\sqrt{2}}(1-\sqrt{7});\frac{5}{\sqrt{2}}(1+\sqrt{7})),(-\frac{5}{\sqrt{2}}(1+\sqrt{7});\frac{5}{\sqrt{2}}(1-\sqrt{7})),(\frac{5}{\sqrt{2}}(\sqrt{7}- | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] a) Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omeg... | Answer: $\frac{R_{2}}{R_{1}}=\frac{4}{3}, R_{1}=\frac{\sqrt{3}}{4}, R_{2}=\frac{1}{\sqrt{3}}$.
Fig. 1: Variant 11, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\om... | \frac{R_{2}}{R_{1}}=\frac{4}{3},R_{1}=\frac{\sqrt{3}}{4},R_{2}=\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. [7 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
x^{2}+(y-a)^{2}=64 \\
(|x|-6)^{2}+(|y|-8)^{2}=100
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{-8-12 \sqrt{2}\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{12 \sqrt{2}+8\}$.
Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordi... | \in{-8-12\sqrt{2}}\cup(-24;-8]\cup[8;24)\cup{12\sqrt{2}+8} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. [5 points] We rolled 90 fair dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) and counted the sum of the numbers that came up. Which probability is greater: that this sum is not less than 500, or that this sum is less than 130? | Answer: It is more likely that the sum is not less than 500.
Solution. The result of rolling the dice can be described by a set of 90 numbers ranging from 1 to 6. Consider any such set. If each number in the set is replaced from $x$ to 7 - $x$, we get a new set consisting of numbers from 1 to 6. In this case, if the s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. [4 points] Given a finite arithmetic progression $a_{1}, a_{2} \ldots, a_{n}$ with a positive common difference, such that the sum of all its terms is $S$, and $a_{1}>0$. It is known that if the common difference of the progression is increased by 4 times, while keeping the first term unchanged, then the sum $S$ inc... | Answer: It will increase by $\frac{5}{3}$ times.
Solution. The sum of the terms of the original progression is determined by the formula $S_{0}=\frac{2 a_{1}+(n-1) d}{2} \cdot n$, where $d$ is the common difference of the progression. If $d$ is increased by four times, the sum of the terms will become $S_{1}=$ $\frac{... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+x-90}+7\right)\left|x^{3}-10 x^{2}+31 x-28\right| \leqslant 0$. | Answer: $3+\sqrt{2}$.
Solution. The first factor on the left side is positive for any valid value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}+x-90 \geqslant 0$) is equivalent to the equation $x^{3}-10 x^{2}+31 x-28=0$. One of the ... | 3+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Solve the equation $5 x^{4}+x^{2}+8 x-6 x^{2}|x+4|+16=0$.
The text above has been translated into English, maintaining the original text's line breaks and format. | Answer: $-\frac{4}{5} ; 1 ; \frac{1 \pm \sqrt{17}}{2}$.
Solution. The equation can be rewritten as $5 x^{4}+(x+4)^{2}-6 x^{2}|x+4|=0$ or $|x+4|^{2}-6 x^{2} \mid x+$ $4 \mid+5 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x+4|$ with a discriminant equal to $\left(6 ... | -\frac{4}{5};1;\frac{1\\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] Around a hook with a worm, in the same plane as it, a carp and a minnow are swimming along two circles. In the specified plane, a rectangular coordinate system is introduced, in which the hook (the common center of the circles) is located at the point $(0 ; 0)$. At the initial moment of time, the carp and... | Answer: $(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4),(\sqrt{2}-4 ;-4-\sqrt{2}),(-\sqrt{2}-4 ; 4-\sqrt{2})$.
Solution. Let the points where the carp and the minnow are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles formed by the radius vectors of points $M$ and $N... | (4-\sqrt{2};4+\sqrt{2}),(4+\sqrt{2};\sqrt{2}-4),(\sqrt{2}-4;-4-\sqrt{2}),(-\sqrt{2}-4;4-\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] a) Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omeg... | Answer: $\frac{R_{2}}{R_{1}}=\frac{5}{4}, R_{1}=\frac{2}{\sqrt{5}}, R_{2}=\frac{\sqrt{5}}{2}$.
Fig. 3: Variant 12, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\om... | \frac{R_{2}}{R_{1}}=\frac{5}{4},R_{1}=\frac{2}{\sqrt{5}},R_{2}=\frac{\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. [7 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
(x-a)^{2}+y^{2}=64 \\
(|x|-8)^{2}+(|y|-15)^{2}=289
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{-28\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{28\}$.
Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordinate axes. In the firs... | \in{-28}\cup(-24;-8]\cup[8;24)\cup{28} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In an isosceles triangle \(ABC\), the base \(AC\) is equal to 1, and the angle \(ABC\) is equal to \(2 \operatorname{arctg} \frac{1}{2}\). Point \(D\) lies on side \(BC\) such that the area of triangle \(ABC\) is four times the area of triangle \(ADC\). Find the distance from point \(D\) to the line \(AB\) and the r... | Answer: distance $=\frac{3}{2 \sqrt{5}}, \quad$ radius $=\frac{\sqrt{265}}{32}$. | distance=\frac{3}{2\sqrt{5}},\quadradius=\frac{\sqrt{265}}{32} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\frac{\sin 3 x \cos 5 x-\sin 2 x \cos 6 x}{\cos x}=0
$$ | Answer: $\pi n, \quad \pm \frac{\pi}{6}+\pi n, \quad n \in \mathbb{Z}$. | \pin,\quad\\frac{\pi}{6}+\pin,\quadn\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system of equations
$$
\left\{\begin{aligned}
\log _{x}(y+1) & =4 \log _{x+2} \sqrt{y-1} \\
\log _{y-1}(x+2) & =\log _{x}\left(\frac{x^{3}}{y+1}\right)
\end{aligned}\right.
$$ | Answer: $\left(\frac{1+\sqrt{17}}{2}, \frac{7+\sqrt{17}}{2}\right),\left(\frac{5+\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right)$. | (\frac{1+\sqrt{17}}{2},\frac{7+\sqrt{17}}{2}),(\frac{5+\sqrt{17}}{2},\frac{3+\sqrt{17}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The base of the triangular pyramid $S A B C$ is an equilateral triangle $A B C$ with a side length of 8. The lateral edge $S C$ is perpendicular to the base and has a length of 15. A sphere with center $O$ lying in the plane $S B C$ touches the edges $S A$, $A B$, and $A C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, r... | Answer: $A A_{1}=6, \quad$ distance $=\frac{18}{5}, \quad$ radius $=\frac{4 \sqrt{39}}{5}$. | AA_{1}=6,\quad | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=2, E N=5$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, then $H M=A M=F M=2$, $C N=E N=H N=5$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\alpha... | \angleABC=60,S_{\triangleABC}=36\sqrt{3},R=2\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 32, the sum of the numbers on the board will become 477. If the largest number is multiplied by 14, the sum of the numbers on the board will also become 477. What numbers could have been written on ... | Answer: $13,14,16,31$ or $13,30,31$.
Solution. Let there be $n$ numbers $x_{1}32 x_{1} \geqslant 32 \cdot(13 \cdot 2)>477$, which is impossible. Therefore, $k=1$, from which $x_{1}=13, x_{n}=31$.
Then $32 x_{1}+x_{n}=32 \cdot 13+31=447$, and the sum of the remaining numbers is $477-447=30$, and the remaining numbers ... | 13,14,16,31or13,30,31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-6 a x-2 a y+2 x^{2}+2 x y+y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-8 a^{2} x-2 a^{3} y+12 a y+a^{4}+36=0$. Find all values of the parameter... | Answer: $(-\infty ;-2) \cup(-1 ; 0) \cup(3 ;+\infty)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(y^{2}+2(x-a) y\right)+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+2 \cdot y \cdot(x-a)+(x-a)^{2}\right)-(x-a)^{2}+2 x^{2}-6 a x+5 a^{2}=0 \Leftright... | (-\infty;-2)\cup(-1;0)\cup(3;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-2 x^{2} y^{2}=3 \\
x^{4}+y^{4}+\frac{2}{3} x^{2} y^{2}=17
\end{array}\right.
$$ | Answer: $(\sqrt{2} ; \pm \sqrt{3}),(-\sqrt{2} ; \pm \sqrt{3}),(\sqrt{3} ; \pm \sqrt{2}),(-\sqrt{3} ; \pm \sqrt{2})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
3 u-2 v=3 \\
... | (\sqrt{2};\\sqrt{3}),(-\sqrt{2};\\sqrt{3}),(\sqrt{3};\\sqrt{2}),(-\sqrt{3};\\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $12^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 12 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 1386.
Solution. Since the magician has a set of $12^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 12,1 \leqslant j \leqslant 12$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 1386 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=2, AD=3$. Find... | Answer: $\frac{19}{25}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{19}{25} | Geometry | proof | Yes | Yes | olympiads | false |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=2, E N=11$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circl... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=2$, $C N=E N=H N=11$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\a... | \angleABC=60,S_{\triangleABC}=120\sqrt{3},R=14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 30, the sum of the numbers on the board will become 450. If the largest number is multiplied by 14, the sum of the numbers on the board will also become 450. What numbers could have been written on ... | Answer: $13,14,17,29$ or $13,15,16,29$.
Solution. Let there be $n$ numbers $x_{1}30 x_{1} \geqslant$ $30 \cdot(13 \cdot 2)>450$, which is impossible. Therefore, $k=1$, from which $x_{1}=13, x_{n}=29$.
Then $30 x_{1}+x_{n}=30 \cdot 13+29=419$, and the sum of the remaining numbers is $450-419=31$, and the remaining num... | 13,14,17,29or13,15,16,29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $2 a^{2}+2 a x+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+a^{2} y^{2}-$ $2 a^{3} x-6 a x-2 a^{2} y+a^{4}+9=0$. Find all values of the parameter $a$ for... | Answer: $(-\infty ;-2) \cup(0 ; 1) \cup(3 ; \infty)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
&\left(x^{2}+2(a-y) x\right)+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}+2 \cdot x \cdot(a-y)+(a-y)^{2}\right)-(a-y)^{2}+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Le... | (-\infty;-2)\cup(0;1)\cup(3;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
7 x^{2}+7 y^{2}-3 x^{2} y^{2}=7 \\
x^{4}+y^{4}-x^{2} y^{2}=37
\end{array}\right.
$$ | Answer: $(\sqrt{7} ; \pm \sqrt{3}),(-\sqrt{7} ; \pm \sqrt{3}),(\sqrt{3} ; \pm \sqrt{7}),(-\sqrt{3} ; \pm \sqrt{7})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
7 u-3 v=7 \\
... | (\sqrt{7};\\sqrt{3}),(-\sqrt{7};\\sqrt{3}),(\sqrt{3};\\sqrt{7}),(-\sqrt{3};\\sqrt{7}) | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. [5 points] A magician has a set of $15^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 15 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 2835.
Solution. Since the magician has a set of $15^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 15,1 \leqslant j \leqslant 15$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 2835 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=3, AD=4$. Find... | Answer: $\frac{37}{49}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{37}{49} | Geometry | proof | Yes | Yes | olympiads | false |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=1, E N=7$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=1$, $C N=E N=H N=7$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\al... | \angleABC=60,S_{\triangleABC}=45\sqrt{3},R=2\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 32, the sum of the numbers on the board will become 581. If the largest number is multiplied by 17, the sum of the numbers on the board will also become 581. What numbers could have been written on ... | Answer: $16,17,21,31$ or $16,18,20,31$.
Solution. Let there be $n$ numbers $x_{1}32 x_{1} \geqslant$ $32 \cdot(16 \cdot 2)>581$, which is impossible. Therefore, $k=1$, from which $x_{1}=16, x_{n}=31$.
Then $32 x_{1}+x_{n}=32 \cdot 16+31=543$, and the sum of the remaining numbers is $581-543=38$, with the remaining nu... | 16,17,21,31or16,18,20,31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-6 a x-4 a y+2 x^{2}+2 x y+y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-6 a^{2} x-2 a^{3} y+4 a y+a^{4}+4=0$. Find all values of the parameter $... | Answer: $(-1 ; 0) \cup(1 ; 2)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(y^{2}+2(x-2 a) y\right)+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+2 \cdot y \cdot(x-2 a)+(x-2 a)^{2}\right)-(x-2 a)^{2}+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \qua... | (-1;0)\cup(1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-x^{2} y^{2}=3 \\
x^{4}+y^{4}-x^{2} y^{2}=31
\end{array}\right.
$$ | Answer: $(\sqrt{5} ; \pm \sqrt{6}),(-\sqrt{5} ; \pm \sqrt{6}),(\sqrt{6} ; \pm \sqrt{5}),(-\sqrt{6} ; \pm \sqrt{5})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
3 u-v=3 \\
u^... | (\sqrt{5};\\sqrt{6}),(-\sqrt{5};\\sqrt{6}),(\sqrt{6};\\sqrt{5}),(-\sqrt{6};\\sqrt{5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $20^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 20 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 7030.
Solution. Since the set contains $20^{2}$ cards, the magician has all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two t... | 7030 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=4, AD=5$. Find... | Answer: $\frac{61}{81}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{61}{81} | Geometry | proof | Yes | Yes | olympiads | false |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=1, E N=4$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=1$, $C N=E N=H N=4$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\al... | \angleABC=60,S_{\triangleABC}=18\sqrt{3},R=2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 35, the sum of the numbers on the board will become 592. If the largest number is multiplied by 16, the sum of the numbers on the board will also become 592. What numbers could have been written on ... | # Answer: $15,16,17,34$ or $15,33,34$.
Solution. Let there be $n$ numbers $x_{1}35 x_{1} \geqslant$ $35 \cdot(15 \cdot 2)>592$, which is impossible. Therefore, $k=1$, from which $x_{1}=15, x_{n}=34$.
Then $35 x_{1}+x_{n}=32 \cdot 15+34=559$, and the sum of the remaining numbers is $592-559=33$, and the remaining numb... | 15,16,17,34or15,33,34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-4 a x+6 a y+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-4 a^{3} x-2 a x+2 a^{2} y+4 a^{4}+1=0$. Find all values of the parameter... | Answer: $\left(0 ; \frac{1}{2}\right) \cup(1 ; 3)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(x^{2}-2(2 a+y) x\right)+2 y^{2}+6 a y+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}-2 \cdot x \cdot(2 a+y)+(2 a+y)^{2}\right)-(2 a+y)^{2}+2 y^{2}+6 a y+5 a^{2}=0 \Left... | (0;\frac{1}{2})\cup(1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
2 x^{2}+2 y^{2}-x^{2} y^{2}=2 \\
x^{4}+y^{4}-\frac{1}{2} x^{2} y^{2}=19
\end{array}\right.
$$ | Answer: $(2 ; \pm \sqrt{3}),(-2 ; \pm \sqrt{3}),(\sqrt{3} ; \pm 2),(-\sqrt{3} ; \pm 2)$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
2 u-v=2 \\
u^{2}-2 v-\frac{1}{2} v=19
\en... | (2;\\sqrt{3}),(-2;\\sqrt{3}),(\sqrt{3};\2),(-\sqrt{3};\2) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $16^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 16 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 3480.
Solution. Since the magician has a set of $16^{2}$ cards, there are all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 16,1 \leqslant j \leqslant 16$ there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two ... | 3480 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=3, AD=5$. Find... | Answer: $\frac{49}{64}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^{... | \frac{49}{64} | Geometry | proof | Yes | Yes | olympiads | false |
1. When $x^{2}$ was added to the quadratic trinomial $f(x)$, its minimum value increased by 1, and when $x^{2}$ was subtracted from it, its minimum value decreased by 3. How will the minimum value of $f(x)$ change if $2 x^{2}$ is added to it? | # Answer. It will increase by $\frac{3}{2}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its minimum value, its leading coefficient is positive, and the minimum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \c... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $x^{\log _{3} x}-2 \leqslant(\sqrt[3]{3})^{\log _{\sqrt{3}}^{2} x}-2 \cdot x^{\log _{3}} \sqrt[3]{x}$. | Answer. $x \in\left(0 ; 3^{-\sqrt{\log _{3} 2}}\right] \cup\{1\} \cup\left[3 \sqrt{\log _{3} 2} ;+\infty\right)$.
Matching the third with the second, and the second with the fourth, we facto... | x\in(0;3^{-\sqrt{\log_{3}2}}]\cup{1}\cup[3^{\sqrt{\log_{3}2}};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{2}{5}\right)$, and the numbers $3+\sin x, 3+\sin y, 3+\sin z$ form a non-constant geometric progression in the given order. Find $\sin y$. | Answer: $-\frac{1}{10}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
$... | -\frac{1}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are 4 and 9, respectively.
a) Find the distance from point $A$ to the line $BC$.
b... | Answer. a) $5 ;$ b) $R=\frac{32}{7} ; A B=\frac{16}{\sqrt{7}}$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater ... | )5;b)R=\frac{32}{7};AB=\frac{16}{\sqrt{7}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $60 ; 45$ ). Find the number of such squares. | Answer: 2070.
Solution. Draw through the given point $(60 ; 45)$ vertical and horizontal lines $(x=60$ and $y=45)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "bottom" vertex of the square can be located in 45 ways: ... | 2070 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a - 2 \leqslant 0 \\
x^{2} + y^{2} + 6x - 2y - b^{2} + 4b + 6 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ; 4-\sqrt{10}] \cup[\sqrt{10} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a-2=0$ is 2, and the point $(0 ; 0)$ satisfies this inequality. Therefore, the inequality defines a half-... | b\in(-\infty;4-\sqrt{10}]\cup[\sqrt{10};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The base of the triangular pyramid $A B C D$ is an equilateral triangle $A B C$. The volume of the pyramid is $\frac{25}{\sqrt{3}}$, and its height, drawn from vertex $D$, is 3. Point $M$ is the midpoint of edge $C D$. It is known that the radii of the spheres inscribed in pyramids $A B C M$ and $A B D M$ are equal.... | Answer. a) $\angle=\arccos \left( \pm \frac{4}{5}\right)$; b) $C D=\frac{\sqrt{31}}{\sqrt{3}}$ or $C D=3 \sqrt{13}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $A B C M$ and $A B D... | \arccos(\\frac{4}{5}),\frac{\sqrt{31}}{\sqrt{3}},3\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. When $x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by $\frac{27}{2}$, and when $4 x^{2}$ was subtracted from it, its maximum value decreased by 9. How will the maximum value of $f(x)$ change if $2 x^{2}$ is subtracted from it? | Answer. It will decrease by $\frac{27}{4}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its maximum value, its leading coefficient is negative, and the maximum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \cd... | \frac{27}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $(\sqrt[10]{125})^{\log _{\sqrt{5}}^{2} x}+3 \geqslant x^{\log _{5} x}+3(\sqrt[5]{x})^{\log _{5} x}$. | Answer. $x \in\left(0 ; 5^{-\sqrt{\log _{5} 3}}\right] \cup\{1\} \cup\left[5 \sqrt{\log _{5} 3} ;+\infty\right)$.
Solution. Rewrite the inequality as $5^{\frac{6}{5} \log _{5}^{2} x}+3-5^{\log _{5}^{2} x}-3 \cdot 5^{\frac{1}{5} \log _{5}^{2} x} \geqslant 0$. By grouping the first term with the third and the second wit... | x\in(0;5^{-\sqrt{\log_{5}3}}]\cup{1}\cup[5\sqrt{\log_{5}3};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{1}{9}$, and the numbers $5+\cos x, 5+\cos y, 5+\cos z$ form a non-constant geometric progression in the given order. Find $\cos y$. | Answer: $-\frac{1}{9}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
\[... | -\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, side $AB$ is equal to $\sqrt{11}$, and the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are in the ratio $9: 25$.
a) Find the ratio... | Answer. a) $9: 16$, b) $d_{C}=\frac{275}{54}, R=3$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\... | )9:16,b)d_{C}=\frac{275}{54},R=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 40)$. Find the number of such squares. | Answer: 1560.
Solution. Draw through the given point $(55 ; 40)$ vertical and horizontal lines $(x=55$ and $y=40)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 39 ways: $... | 1560 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a + 4 \leqslant 0 \\
x^{2} + y^{2} + 10 x + 2 y - b^{2} - 8 b + 10 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-8-\sqrt{26}] \cup[\sqrt{26} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a+4=0$ is 3, and the point $(0 ; 0)$ does not satisfy this inequality. Therefore, the inequality defines ... | b\in(-\infty;-8-\sqrt{26}]\cup[\sqrt{26};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The base of the triangular pyramid $A B C D$ is an equilateral triangle $A B C$. The volume of the pyramid is $\frac{100}{3 \sqrt{3}}$, and its height, drawn from vertex $D$, is 4. Point $M$ is the midpoint of edge $C D$. It is known that the radii of the spheres inscribed in pyramids $A B C M$ and $A B D M$ are equ... | Answer. a) $\angle=\arccos \left( \pm \frac{3}{5}\right)$; b) $C D=\frac{8}{\sqrt{3}}$ or $C D=\frac{4 \sqrt{19}}{\sqrt{3}}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $A B C M$ a... | \arccos(\\frac{3}{5}),\frac{8}{\sqrt{3}},\frac{4\sqrt{19}}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $x^{\log _{13} x}+7(\sqrt[3]{x})^{\log _{13} x} \leqslant 7+(\sqrt[3]{13})^{\log _{\sqrt{13}}^{2} x}$ | Answer. $x \in\left(0 ; 13^{-\sqrt{\log _{13} 7}}\right] \cup\{1\} \cup\left[13 \sqrt{\log _{13} 7} ;+\infty\right)$.
Solution. Rewrite the inequality as $13^{\log _{13}^{2} x}+7 \cdot 13^{\frac{1}{3} \log _{13}^{2} x}-7-13^{\frac{4}{3} \log _{13}^{2} x} \leqslant 0$. By grouping the first term with the fourth and the... | x\in(0;13^{-\sqrt{\log_{13}7}}]\cup{1}\cup[13\sqrt{\log_{13}7};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{5}\right)$, and the numbers $2+\sin x, 2+\sin y, 2+\sin z$ form a non-constant geometric progression in the given order. Find $\sin y$. | Answer: $-\frac{1}{5}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of the two adjacent terms), we obtain the equatio... | -\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are 5 and 12, respectively.
a) Find the distance from point $A$ to the line $BC$.
... | Answer. a) 7; b) $R=\frac{25}{4}; BC=5\sqrt{6}$.
Solution. Let $\angle ACB = \gamma$, then $\angle BAC = 2\gamma$. By the generalized sine theorem, we find that $AB = 2R \sin \gamma, BC = 2R \sin 2\gamma$. Note that $\gamma < 90^\circ$ (otherwise, the sum of the angles in the triangle would be greater than $180^\circ$... | )7;b)R=\frac{25}{4};BC=5\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $25 ; 60$ ). Find the number of such squares. | Answer: 650.
Solution. Draw through the given point $(25 ; 60)$ vertical and horizontal lines $(x=25$ and $y=60)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 25 ways: $(0... | 650 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a - y \sin a - 3 \leqslant 0 \\
x^{2} + y^{2} - 8 x + 2 y - b^{2} - 6 b + 8 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-\sqrt{17}] \cup[\sqrt{17}-6 ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a-y \sin a-3=0$ is 3, and the point $(0 ; 0)$ satisfies this inequality. Therefore, the inequality defines a half-... | b\in(-\infty;-\sqrt{17}]\cup[\sqrt{17}-6;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The base of the triangular pyramid $K L M N$ with a volume of 75 is an equilateral triangle $K L M$ with a side length of 10. Point $T$ is the midpoint of edge $M N$. It is known that the radii of the spheres inscribed in pyramids $K L M T$ and $K L N T$ are equal to each other.
a) Find all possible values of the a... | Answer. a) $\angle=\arccos \left( \pm \frac{4}{5}\right)$; b) $M N=\sqrt{31}$ or $M N=3 \sqrt{39}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $K L M T$ and $K L N T$ are equal (th... | \arccos(\\frac{4}{5}),\sqrt{31},3\sqrt{39} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $(\sqrt[7]{4})^{\log _{\sqrt{2}}^{2} x}+6 \geqslant x^{\log _{2} x}+6(\sqrt[7]{x})^{\log _{2} x}$. | Answer. $x \in\left(0 ; 2^{-\sqrt{\log _{2} 6}}\right] \cup\{1\} \cup\left[2 \sqrt{\log _{2} 6} ;+\infty\right)$.
Solution. Rewrite the inequality as $2^{\frac{8}{7} \log _{2}^{2} x}+6-2^{\log _{2}^{2} x}-6 \cdot 2^{\frac{1}{7} \log _{2}^{2} x} \geqslant 0$. Grouping the first term with the third and the second with t... | x\in(0;2^{-\sqrt{\log_{2}6}}]\cup{1}\cup[2\sqrt{\log_{2}6};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{5}{9}$, and the numbers $1+\cos x, 1+\cos y, 1+\cos z$ form a non-constant geometric progression in the given order. Find $\cos y$. | Answer: $-\frac{7}{9}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
$$... | -\frac{7}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The radius of the circle $\Omega$ is $5 \sqrt{2}$, and the distances from points $A$ and $C$ to the tangent line $\ell$ are in the ratio $2:5$... | Answer. a) $2: 3$; b) $A B=5 \sqrt{3}, d_{C}=\frac{75}{4 \sqrt{2}}$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be gre... | )2:3;b)AB=5\sqrt{3},d_{C}=\frac{75}{4\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(35 ; 65)$. Find the number of such squares. | Answer: 1190.
Solution. Draw through the given point $(35 ; 65)$ vertical and horizontal lines $(x=35$ and $y=65)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 34 ways: $(... | 1190 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a + 3 \leqslant 0 \\
x^{2} + y^{2} + 8x - 4y - b^{2} + 6b + 11 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-2 \sqrt{5}] \cup[6+2 \sqrt{5} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a+3=0$ is 3, and the point $(0 ; 0)$ does not satisfy this inequality. Therefore, the inequality define... | b\in(-\infty;-2\sqrt{5}]\cup[6+2\sqrt{5};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The base of the triangular pyramid $K L M N$ with a volume of 100 is an equilateral triangle $K L M$ with a side length of 10. Point $T$ is the midpoint of edge $M N$. It is known that the radii of the spheres inscribed in pyramids $K L M T$ and $K L N T$ are equal.
a) Find all possible values of the angle between ... | Answer. a) $\angle=\arccos \left( \pm \frac{3}{5}\right)$; b) $C D=8$ or $C D=4 \sqrt{19}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $K L M T$ and $K L N T$ are equal (the face $... | \arccos(\\frac{3}{5}),8,4\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 5, and 5, respectively. Find the smallest possible value of $f(x)$. | Answer: $\frac{39}{8}$.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text {v }}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented ... | \frac{39}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{3}{7}\right)$, and the numbers $\frac{1}{\cos x}, \frac{7}{\cos y}, \frac{1}{\cos z}$ also form an arithmetic progression in the given order. Find $\cos ^{2} y$. | Answer: $\frac{10}{13}$.
Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{14}{\cos y}=\frac{1}{\cos x}+\frac{1}{\cos z}$. Transform the second equation:
$$
\frac{14}{\cos y}=\frac{\cos x+\cos z}{\cos x \cos z} \Leftrightarrow \frac{14}{\cos y}=\frac{2 \cos \fr... | \frac{10}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the inequality $\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3 \leqslant 64^{\log _{4}^{2} x}-15 \cdot x^{\log _{4} x}$. | Answer. $x \in\left(0 ; \frac{1}{4}\right] \cup[4 ;+\infty)$.
Solution. Transform the left side of the inequality:
$$
\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\log _{3} 2\right)}{\log _{3} 162}=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\lo... | x\in(0;\frac{1}{4}]\cup[4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 7. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=14 ;$ b) $\angle D A C=45^{\circ}, S_{A B C D}=97$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\... | AC=14,\angleDAC=45,S_{ABCD}=97 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $5300 \ldots 0035$ (100 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 495. In how many ways can this be done? | Answer: 22100.
Solution. $495=5 \cdot 9 \cdot 11$. Divisibility by 5 is always satisfied since the number ends in five. For investigating divisibility by 11, it is significant which positions the replaceable digits occupy.
First case. We replace two zeros in positions of the same parity (both even or both odd). For d... | 22100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$ for which there exists a value of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
\arcsin \left(\frac{a-y}{3}\right)=\arcsin \left(\frac{4-x}{4}\right) \\
x^{2}+y^{2}-8 x-8 y=b
\end{array}\right.
$$
has exactly two solutions. | Answer. $a \in\left(-\frac{13}{3} ; \frac{37}{3}\right)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{a-y}{3}=\frac{4-x}{4}, y=\frac{3 x}{4}-3+a$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4-x}{4} \leqslant 1,0 \leqslant x \leqslant 8... | \in(-\frac{13}{3};\frac{37}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Consider quadrilateral pyramids $MABCD$ with the following properties: the base of the pyramid is a convex quadrilateral $ABCD$, where $AB = BC = 1$, $CD = DA = 2$, and each of the planes of the lateral faces $MAB$, $MBC$, $MCD$, $MDA$ forms an angle of $45^{\circ}$ with the plane of the base.
a) Find the volume of... | Answer. a) $\frac{27}{25} ;$ b) $\frac{4}{3}$.
Let $M H$ be the height of the pyramid, $(M H=h), P$ be the projection of $M$ onto the line $A B$. Then $M H P$ is a right triangle with angle $\angle M P H=45^{\circ}$, from which $H P=h \cdot \operatorname{ctg} 45^{\circ}=h$. Similarly, it can be proved that the point $... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values $-9, -9$, and $-15$ respectively. Find the greatest possible value of $f(x)$. | Answer: $-\frac{33}{4}$.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text {v }}$, then $x_{\mathrm{B}}=n+0.5$, and thus $f(x)$ can be represent... | -\frac{33}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arcsin \frac{\sqrt{7}}{4}$, and the numbers $\frac{1}{\sin x}, \frac{4}{\sin y}, \frac{1}{\sin z}$ also form an arithmetic progression in the given order. Find $\sin ^{2} y$. | Answer: $\frac{7}{13}$.
Solution: Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{8}{\sin y}=\frac{1}{\sin x}+\frac{1}{\sin z}$. Transform the second equation:
$$
\frac{8}{\sin y}=\frac{\sin x+\sin z}{\sin x \sin z} \Leftrightarrow \frac{8}{\sin y}=\frac{2 \sin \frac{x... | \frac{7}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the inequality $\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5 \leqslant 125^{\log _{5}^{2} x}-24 \cdot x^{\log _{5} x}$. Answer. $x \in\left(0 ; \frac{1}{5}\right] \cup[5 ;+\infty)$. | Solution. Transform the left side of the inequality:
$\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{\log _{5} 50}=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{2+\log _{5} 2}=5$
Let $5^{\log _{5}^{2} x}=v... | x\in(0;\frac{1}{5}]\cup[5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 4. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=8$; b) $\angle D A C=45^{\circ}, S_{A B C D}=31$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\an... | AC=8,\angleDAC=45,S_{ABCD}=31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $800 \ldots 008$ (80 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 198. In how many ways can this be done? | Answer: 14080.
Solution. $198=2 \cdot 9 \cdot 11$. Divisibility by 2 is always satisfied since the number ends in eight. For investigating divisibility by 11, it is significant on which positions the replaceable digits stand.
First case. We replace two zeros in positions of the same parity (both even or both odd). Fo... | 14080 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$ such that the system
$$
\left\{\begin{array}{l}
\arccos \left(\frac{4+y}{4}\right)=\arccos (x-a) \\
x^{2}+y^{2}-4 x+8 y=b
\end{array}\right.
$$
has no more than one solution for any value of the parameter $b$. | Answer. $a \in(-\infty ;-15] \cup[19 ;+\infty)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{4+y}{4}=x-a, y=4 x-4 a-4$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4+y}{4} \leqslant 1,-8 \leqslant y \leqslant 0$. Thus, the first equatio... | \in(-\infty;-15]\cup[19;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Consider the pyramids $T A B C D$ with the following properties: the base of the pyramid is a convex quadrilateral $A B C D$, where $A B=B C=2, C D=D A=3$, and each of the planes of the lateral faces $T A B, T B C, T C D, T D A$ forms an angle of $30^{\circ}$ with the plane of the base.
a) Find the volume of such a... | Answer. a) $\frac{4}{\sqrt{3}}$; b) $4 \sqrt{3}$.
Let $TH$ be the height of the pyramid, $(TH=h)$, and $P$ be the projection of $T$ onto the line $AB$. Then $THP$ is a right triangle with angle $\angle TPH=30^{\circ}$, from which $HP=h \cdot \operatorname{ctg} 30^{\circ}=h \sqrt{3}$. Similarly, it can be proven that t... | \frac{4}{\sqrt{3}};4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{3}\right)$, and the numbers $\frac{1}{\cos x}, \frac{3}{\cos y}, \frac{1}{\cos z}$ also form an arithmetic progression in the given order. Find $\cos ^{2} y$. | Answer: $\frac{4}{5}$.
Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{14}{\cos y}=\frac{1}{\cos x}+\frac{1}{\cos z}$. Transform the second equation:
$$
\frac{6}{\cos y}=\frac{\cos x+\cos z}{\cos x \cos z} \Leftrightarrow \frac{6}{\cos y}=\frac{2 \cos \frac{x... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the inequality $27^{\log _{3}^{2} x}-8 \cdot x^{\log _{3} x} \geqslant \log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5$. | Answer. $x \in\left(0 ; \frac{1}{3}\right] \cup[3 ;+\infty)$.
Solution. Transform the right side of the inequality:
$$
\log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\log _{5} 2\right)}{\log _{5} 250}=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\l... | x\in(0;\frac{1}{3}]\cup[3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.