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
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5. It is known that for all pairs of positive numbers ( $x ; y$ ), for which the equality $x+y=$ param 1 and the inequality $x^{2}+y^{2}>$ param2 hold, the inequality $x^{5}+y^{5}>m$ also holds. What is the greatest value that $m$ can take?
| param1 | param2 | |
| :---: | :---: | :---: |
| 5 | 13 | |
| 7 | 27 | |
|... | 5. It is known that for all pairs of positive numbers ( $x ; y$ ), for which the equality $x+y=$ param 1 and the inequality $x^{2}+y^{2}>$ param 2 hold, the inequality $x^{5}+y^{5}>m$ also holds. What is the greatest value that $m$ can take?
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 5 | 13 | 275 |
| 7 |... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. A sphere $\Omega$ is inscribed in a regular tetrahedron $K L M N$ with edge length $1$. A cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is positioned such that its diagonal $A_{1} C_{1}$ lies on the line $K L$, and the line $B D$ is tangent to the sphere $\Omega$ at a point lying on the segment $B D$. What is the minimum s... | 6. A sphere $\Omega$ is inscribed in a regular tetrahedron $K L M N$ with edge length $a$. A cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is positioned such that its diagonal $A_{1} C_{1}$ lies on the line $K L$, and the line $B D$ is tangent to the sphere $\Omega$ at a point lying on the segment $B D$. What is the minimum s... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :--- |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 x^{2}+15... | 7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | 3 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find the maximum value of the expression param1, if the numbers $x, y, z$ are solutions of the system param2. Round the answer to the nearest thousandth.
| param1 | param2 |
| :---: | :---: |
| $\sin (x+y+z)$ | $\left\{\begin{array}{l}\sqrt{\frac{8}{9}+\sin x}=\frac{3}{2} \cos y+\frac{1}{3} \operatorname{tg} y \\ \... | 8. Find the maximum value of the expression param 1, if the numbers $x, y, z$ are solutions of the system of equations param 2. Round the answer to the nearest thousandth.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $\sin (x+y+z)$ | $\left\{\begin{array}{l}\sqrt{\frac{8}{9}+\sin x}=\frac{3}{2} \cos y+\fra... | Other | math-word-problem | More than one problem | Yes | olympiads | false | |
9. On the table, there are param 1 externally identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee getting at least... | 9. On the table, there are param 1 visually identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee that you will get ... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the maximum possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+3 x^{2}+6... | 17. It is known that the number $a$ satisfies the equation param 1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | 2 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. In a football tournament held in a single round-robin format (each team must play every other team exactly once), $N$ teams are participating. At a certain point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by... | 19. In a football tournament held in a single round-robin format (each team must play every other team exactly once), $N$ teams are participating. At a certain point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
20. On the table, there are param 1 externally identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee getting at leas... | 20. On the table, there are param1 externally identical coins. It is known that among them, there are exactly param2 fake ones. It is allowed to point to any two coins and ask whether it is true that both these coins are fake. What is the minimum number of questions needed to guarantee getting at least one "True" answe... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. During a survey of param 1 people, each was asked to indicate one favorite movie. It turned out that among any 10 surveyed, at least 3 indicated the same movie. For what largest $M$ can we assert that there will definitely be $M$ people who indicated the same movie?
| param1 | |
| :---: | :---: |
| 64 | |
| 72 |... | 10. During a survey of param 1 people, each was asked to indicate one favorite movie. It turned out that among any 10 surveyed, at least 3 indicated the same movie. For what largest $M$ can we assert that there will definitely be $M$ people among those surveyed who indicated the same movie?
| param1 | Answer |
| :---:... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. In one notebook, Vasya wrote down 11 natural numbers. In another notebook, Petya wrote down the greatest common divisors of each pair of numbers written in Vasya's notebook. It turned out that each number written in one of the two notebooks is also in the other notebook. What is the maximum number of different nu... | Answer. 10.
Solution.
Answer. 10.
First, note that for any natural numbers $A \geq B$, the inequality $H C F(A ; B) \leq A$ holds, and equality is only achieved when $A=B$. Let $A \geq B$ be the two largest numbers in Vasya's notebook. Then in Petya's notebook, the number $A$ can only appear in the case when $A=B$; ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.6. In a chess tournament, 11th graders and 10th graders participated. There were 10 times more 11th graders, and they scored 4.5 times more points than the 10th graders. Which class did the winner of the tournament belong to, and how many points did they score? In chess, 1 point is awarded for a win, 0.5 points for ... | Answer: 10th-grader, 10 points.
Solution: Let $x$ be the number of 10th-graders and $10x$ be the number of 11th-graders in the tournament. We will estimate the maximum number of points that 10th-graders could have scored and the minimum number of points that 11th-graders could have scored. This will happen if all 10th... | 10th-grader,10points | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.8. Given real numbers $a_{1}, a_{2}$, ?, $a_{7}$ such that $a_{1}=a_{7}=0$. Is it true that one can always choose an index $k \leq 5$ such that the inequality $a_{k}+a_{k+2} \leq a_{k+1} \sqrt{3}$ holds? | Answer. Correct.
Solution. Suppose the opposite, that is, for any $k=1,2,3,4,5$ the inequality $a_{k}+a_{k+2}>a_{k+1} \sqrt{3}$ holds.
Then $\quad\left(a_{1}+a_{3}\right)+\left(a_{3}+a_{5}\right)>a_{2} \sqrt{3}+a_{4} \sqrt{3}=\left(a_{2}+a_{4}\right) \sqrt{3}>\left(a_{3} \sqrt{3}\right) \sqrt{3}=3 a_{3}$ $\left(a_{3}... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
1. When $3 x^{2}$ was added to the quadratic trinomial $f(x)$, its minimum value increased by 9, and when $x^{2}$ was subtracted from it, its minimum value decreased by 9. How will the minimum value of $f(x)$ change if $x^{2}$ is added to it? | Answer. It will increase by $\frac{9}{2}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes the 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 \cdo... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $\left|x^{3}-2 x^{2}+2\right| \geqslant 2-3 x$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Answer. $x \in\left(-\infty ; \frac{1-\sqrt{17}}{2}\right] \cup[0 ;+\infty)$.
Solution. The given inequality is equivalent to the following system:
$$
\begin{aligned}
& {\left[\begin{array} { l }
{ x ^ { 3 } - 2 x ^ { 2 } + 2 \geqslant 2 - 3 x , } \\
{ x ^ { 3 } - 2 x ^ { 2 } + 2 \leqslant - 2 + 3 x }
\end{array} \L... | x\in(-\infty;\frac{1-\sqrt{17}}{2}]\cup[0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. The continuation of the height $B H$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$ (points $B$ and $D$ lie on opposite sides of line $A C$). The degree measures of arcs $A D$ and $C D$, not containing point $B$, are $60^{\circ}$ and $90^{\circ}$, respectively. Determine in what ratio... | Answer: $\sqrt{3}: 1$.
Solution. By the inscribed angle theorem, angle $D C A$ is half of arc $A D$, and angle $D B C$ is half of arc $C D$. Therefore, $\angle D C H=30^{\circ}, \angle H B C=45^{\circ}$. Then triangle $B H C$ is a right and isosceles triangle, $B H=H C$. But $H D=C H \operatorname{tg} 30^{\circ}=\frac... | \sqrt{3}:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. 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 "lower" vertex of the square can be located in 45 ways: $... | 2070 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that one of the roots of the equation $x^{2}-4 a^{2} b^{2} x=4$ is $x_{1}=\left(a^{2}+b^{2}\right)^{2}$. Find $a^{4}-b^{4}$. Answer. $\pm 2$. | Solution. Substituting $x=x_{1}$ into the given equation, we get $\left(a^{2}+b^{2}\right)^{4}-4 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}=4$, from which $\left(a^{2}+b^{2}\right)^{2}\left(\left(a^{2}+b^{2}\right)^{2}-4 a^{2} b^{2}\right)=4,\left(a^{2}+b^{2}\right)^{2}\left(a^{2}-b^{2}\right)^{2}=4,\left(a^{4}-b^{4}\righ... | \2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}4|x|+3|y|=12, \\ x^{2}+y^{2}-2 x+1-a^{2}=0\end{array} \quad\right.$ a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a|=2$; b) $|a| \in\left\{\frac{8}{5}\right\} \bigcup\left(2 ; \frac{16}{5}\right) \bigcup\{\sqrt{17}\}$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced by $-y$. Therefore, the set of points defined by the first equation is symmetric with resp... | )||=2;b)||\in{\frac{8}{5}}\bigcup(2;\frac{16}{5})\bigcup{\sqrt{17}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $ABC$, the median $BM$ is drawn; $MD$ and $ME$ are the angle bisectors of triangles $AMB$ and $CMB$ respectively. Segments $BM$ and $DE$ intersect at point $P$, and $BP=2$, $MP=4$.
a) Find the segment $DE$.
b) Suppose it is additionally known that a circle can be circumscribed around quadrilateral $ADE... | Answer. a) $D E=8 ;$ b) $R=2 \sqrt{85}$.
Solution. a) By the property of the angle bisector of a triangle, we get $A D: D B=A M: M B, C E: E B=C M: M B$, and since $A M=C M$, it follows that $A D: D B=C E: E B$, therefore $A C \| D E$. Then $\angle P D M=\angle A M D=\angle B M D$, which means that triangle $P D M$ is... | DE=8;R=2\sqrt{85} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. When $2 x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by 10, and when $5 x^{2}$ was subtracted from it, its maximum value decreased by $\frac{15}{2}$. How will the maximum value of $f(x)$ change if $3 x^{2}$ is added to it? | Answer. It will increase by $\frac{45}{2}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its maximum value, its leading coefficient is negative, and the maximum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \cd... | \frac{45}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $\left|x^{3}+2 x^{2}-2\right| \geqslant-2-3 x$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Answer. $x \in\left(-\infty ; \frac{-1-\sqrt{17}}{2}\right] \cup[-1 ;+\infty)$.
Solution. The given inequality is equivalent to the following system:
$$
\begin{aligned}
& {\left[\begin{array} { l }
{ x ^ { 3 } + 2 x ^ { 2 } - 2 \geqslant - 2 - 3 x , } \\
{ x ^ { 3 } + 2 x ^ { 2 } - 2 \leqslant 2 + 3 x }
\end{array} ... | x\in(-\infty;\frac{-1-\sqrt{17}}{2}]\cup[-1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. The continuation of the height $B H$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$ (points $B$ and $D$ lie on opposite sides of line $A C$). The degree measures of arcs $A D$ and $C D$, not containing point $B$, are $120^{\circ}$ and $90^{\circ}$, respectively. Determine in what rati... | Answer: $1: \sqrt{3}$.
Solution. By the inscribed angle theorem, angle $DCA$ is half of arc $AD$, and angle $DBC$ is half of arc $CD$. Therefore, $\angle DCH=60^{\circ}, \angle HBC=45^{\circ}$. Then triangle $BHC$ is a right and isosceles triangle, $BH=HC$. But $HD=CH \operatorname{tg} 60^{\circ}=CH \sqrt{3}$. Therefo... | 1:\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On the coordinate plane, consider squares all of whose vertices 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 |
5. It is known that one of the roots of the equation $x^{2}+4 a^{2} b^{2} x=4$ is $x_{1}=\left(a^{2}-b^{2}\right)^{2}$. Find $b^{4}-a^{4}$. Answer. $\pm 2$. | Solution. Substituting $x=x_{1}$ into the given equation, we get $\left(a^{2}-b^{2}\right)^{4}+4 a^{2} b^{2}\left(a^{2}-b^{2}\right)^{2}=4$, from which $\left(a^{2}-b^{2}\right)^{2}\left(\left(a^{2}-b^{2}\right)^{2}+4 a^{2} b^{2}\right)=4,\left(a^{2}-b^{2}\right)^{2}\left(a^{2}+b^{2}\right)^{2}=4,\left(a^{4}-b^{4}\righ... | \2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}5|x|+12|y|=60, \\ x^{2}+y^{2}-2 y+1-a^{2}=0\end{array} \quad\right.$ a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a|=4$; b) $|a| \in\left\{\frac{48}{13}\right\} \cup\left(4 ; \frac{72}{13}\right) \cup\{\sqrt{145}\}$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced by $-y$. Therefore, the set of points defined by the first equation is symmetric with respec... | )||=4;b)||\in{\frac{48}{13}}\cup(4;\frac{72}{13})\cup{\sqrt{145}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $ABC$, the median $BM$ is drawn; $MD$ and $ME$ are the angle bisectors of triangles $AMB$ and $CMB$ respectively. Segments $BM$ and $DE$ intersect at point $P$, and $BP=1$, $MP=3$.
a) Find the segment $DE$.
b) Suppose it is additionally known that a circle can be circumscribed around quadrilateral $ADE... | Answer. a) $D E=6$; b) $R=3 \sqrt{65}$.
Solution. a) By the property of the angle bisector of a triangle, we get $A D: D B=A M: M B, C E: E B=C M: M B$, and since $A M=C M$, it follows that $A D: D B=C E: E B$, so $A C \| D E$. Then $\angle P D M=\angle A M D=\angle B M D$, which means that triangle $P D M$ is isoscel... | 3\sqrt{65} | 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 13, 13, and 35, respectively. Find the smallest possible value of $f(x)$. | Answer: $\frac{41}{4}$.
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 $x_{\mathrm{B}}$, then $x_{\mathrm{B}}=n+0.5$, and thus $f(x)$ can be represente... | \frac{41}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $x^{2}-2 x+1-\left|x^{3}-1\right|-2\left(x^{2}+x+1\right)^{2} \geqslant 0$. | Answer. $x \in\left[-1 ;-\frac{1}{2}\right]$.
Solution. The given inequality is equivalent to the inequality
$$
|x-1|^{2}-|x-1|\left(x^{2}+x+1\right)-2\left(x^{2}+x+1\right)^{2} \geqslant 0
$$
Let $|x-1|=u, x^{2}+x+1=v$ (note that $u \geqslant 0, v>0$, since $v$ is a quadratic trinomial with a negative discriminant)... | x\in[-1;-\frac{1}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=12, P_{1} S_{1}=3$. Find the area of triangle $... | Answer: $\frac{225}{2}$.
Solution: Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=9$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of he... | \frac{225}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $x^{2}+x y=30000000$. | Answer: 256.
Solution. By factoring the left and right sides of the equation, we get $x(x+y)=$ $3 \cdot 2^{7} \cdot 5^{7}$. Then, if $x>0$, $x$ is one of the divisors of the right side. The right side has a total of $2 \cdot 8 \cdot 8=128$ divisors (since any divisor can be represented as $3^{a} \cdot 2^{b} \cdot 5^{c... | 256 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and at the same time,
the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b \vdots 5$, so one of the numbers $a$ or $b$ mus... | 1260 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, it is known that $A B=3, A C=4, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$ ( $O$ - the center of the circumcircle of triangle $\left.A B C\ri... | Answer. $S_{\triangle O A_{2} C}=\frac{13 \sqrt{3}}{12} ; S_{\triangle A_{1} A_{2} C}=\frac{13 \sqrt{3}}{21}$.
Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+9-2 \cdot \frac{1}{2} \cdot 3 \cdot 4=13$, $B C=\sqrt{13}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C... | S_{\triangleOA_{2}C}=\frac{13\sqrt{3}}{12};S_{\triangleA_{1}A_{2}C}=\frac{13\sqrt{3}}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
3|y|-4|x|=6 \\
x^{2}+y^{2}-14 y+49-a^{2}=0
\end{array}\right.
$$
a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a| \in\{5 ; 9\} ;$ b) $|a| \in\{3\} \cup(5 ; 9)$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ by $-y$. Therefore, the set of points defined by the first equation is symmetric with respect to both coordinate axes. In the first quadrant, we get a part o... | )||\in{5,9};b)||\in{3}\cup(5,9) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 14, and 14, respectively. Find the greatest possible value of $f(x)$. | Answer: 15.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text{v}}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the inequality $x^{2}+2 x+1-\left|x^{3}+1\right|-2\left(x^{2}-x+1\right)^{2} \leqslant 0$. | Answer. $x \in\left(-\infty ; \frac{1}{2}\right] \cup[1 ;+\infty)$.
Solution. The given inequality is equivalent to the inequality
$$
|x+1|^{2}-|x+1|\left(x^{2}-x+1\right)-2\left(x^{2}-x+1\right)^{2} \leqslant 0
$$
Let us denote here $|x+1|=u, x^{2}-x+1=v$ (note that $u \geqslant 0, v>0$, since $v$ is a quadratic tr... | x\in(-\infty;\frac{1}{2}]\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=3, P_{1} S_{1}=9$. Find the area of triangle $A... | Answer: 72.
Solution. Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=6$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of the height to t... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $y^{2}-x y=700000000$. | Answer: 324.
Solution. Factoring the left and right sides of the equation, we get $y(y-x)=$ $7 \cdot 2^{8} \cdot 5^{8}$. Then if $y>0$, $y$ is one of the divisors of the right side. The right side has a total of $2 \cdot 9 \cdot 9=162$ divisors (since any divisor can be represented as $7^{a} \cdot 2^{b} \cdot 5^{c}$, ... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, it is known that $A B=4, A C=6, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$. ( $O$ - the center of the circumcircle of triangle $\left.A B C\r... | Answer: $S_{O A_{2} C}=\frac{7}{\sqrt{3}} ; S_{A_{1} A_{2} C}=\frac{7 \sqrt{3}}{5}$.
Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+36-2 \cdot \frac{1}{2} \cdot 6 \cdot 4=28$, $B C=\sqrt{28}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C}{2 \sin 60^{\circ}}=\fra... | S_{OA_{2}C}=\frac{7}{\sqrt{3}};S_{A_{1}A_{2}C}=\frac{7\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
5|x|-12|y|=5 \\
x^{2}+y^{2}-28 x+196-a^{2}=0
\end{array}\right.
$$
a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a| \in\{13 ; 15\} ;$ b) $|a| \in\{5\} \cup(13 ; 15)$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced by $-y$. Therefore, the set of points defined by the first equation is symmetric with respect to both coordinate axes. In the first quadrant,... | )||\in{13,15};b)||\in{5}\cup(13,15) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{3 x}{2}\right)^{\log _{3}(8 x)}=\frac{x^{7}}{8}$. | Answer: $x=\frac{729}{8}, x=2$.
Solution. By taking the logarithm to base 3, we obtain an equation equivalent to the original:
$$
\log _{3}(8 x) \cdot \log _{3}\left(\frac{3 x}{2}\right)=\log _{3}\left(\frac{x^{7}}{8}\right) \Leftrightarrow\left(\log _{3} x+3 \log _{3} 2\right)\left(1+\log _{3} x-\log _{3} 2\right)=7... | \frac{729}{8},2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $A A_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $A T: A_{1} T=4: 1$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{\frac{5}{3}}$; b) $V=\frac{1280 \pi}{29 \sqrt{29}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each... | \frac{1280\pi}{29\sqrt{29}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x=|y-b|+\frac{3}{b} \\
x^{2}+y^{2}+32=a(2 y-a)+12 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in(-\infty, 0) \cup\left[\frac{3}{8} ;+\infty\right)$.
Solution. The second equation of the system can be transformed into the form $(x-6)^{2}+(y-a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center $(6 ; a)$. For all possible $a \in \mathbb{R}$, these circles sweep out the strip $4 \leq x \... | b\in(-\infty,0)\cup[\frac{3}{8};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $BAD$ and $BCD$ respectively, with the first touching side $AD$ at point $K$, and the second touching side $BC$ at point $T$.
a) Find the r... | Answer. a) $r=4$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{5}-1}{2}$ or $\angle B D C=\pi-\operatorname{arctg} \frac{\sqrt{5}+1}{2}$.
Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (the center of the circle inscribed in an angle lies on the bisector of this angle... | )r=4,b)\angleBDC=\operatorname{arctg}\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{x}{243}\right)^{\log _{2}\left(\frac{9 x}{4}\right)}=\frac{729}{x^{4}}$. | Answer: $x=\frac{243}{4}, x=\frac{1}{9}$.
Solution. By taking the logarithm with base 2, we obtain an equation equivalent to the original:
$$
\log _{2}\left(\frac{x}{243}\right) \cdot \log _{2}\left(\frac{9 x}{4}\right)=\log _{2}\left(\frac{729}{x^{4}}\right) \Leftrightarrow\left(\log _{2} x-5 \log _{2} 3\right)\left... | \frac{243}{4},\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 5$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{\frac{7}{3}} ;$ b) $V=\frac{3500 \pi}{37 \sqrt{37}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to eac... | \frac{3500\pi}{37\sqrt{37}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x=|y+a|+\frac{4}{a} \\
x^{2}+y^{2}+24+b(2 y+b)=10 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in(-\infty, 0) \cup\left[\frac{2}{3} ;+\infty\right)$.
“Phystech-2015”, 11th grade, solutions for ticket 2
Solution. The second equation of the system can be transformed into the form $(x-5)^{2}+(y+b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center $(5 ;-b)$. For all possible $b \in \math... | \in(-\infty,0)\cup[\frac{2}{3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $ABC$ and $ADC$ respectively, with the first touching side $BC$ at point $K$, and the second touching side $AD$ at point $T$.
a) Find the r... | Answer. a) $r=3$, b) $\angle B D C=30^{\circ}$.
Solution. a) Segments $B O_{1}$ and $D O_{2}$ are the angle bisectors of angles $A B C$ and $A D C$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite an... | )r=3,b)\angleBDC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{x}{400}\right)^{\log _{5}\left(\frac{x}{8}\right)}=\frac{1024}{x^{3}}$. | Answer: $x=\frac{8}{5}, x=16$.
Solution. By logarithmizing with base 3, we obtain an equation equivalent to the original:
$$
\log _{5}\left(\frac{x}{400}\right) \cdot \log _{5}\left(\frac{x}{8}\right)=\log _{5}\left(\frac{1024}{x^{3}}\right) \Leftrightarrow\left(\log _{5} x-2-4 \log _{5} 2\right)\left(\log _{5} x-3 \... | \frac{8}{5},16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $C C_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $C T: C_{1} T=1: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{2}$; b) $V=\frac{576 \pi \sqrt{3}}{11 \sqrt{11}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each o... | \frac{576\pi\sqrt{3}}{11\sqrt{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x=\frac{7}{b}-|y+b| \\
x^{2}+y^{2}+96=-a(2 y+a)-20 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in\left(-\infty ;-\frac{7}{12}\right] \cup(0 ;+\infty)$.
Solution. The second equation of the system can be transformed into the form $(x+10)^{2}+(y+a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center $(-10 ;-a)$. For all possible $a \in \mathrm{R}$, these circles sweep out the strip $-12 \... | b\in(-\infty;-\frac{7}{12}]\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $BAD$ and $BCD$ respectively, with the first touching side $AB$ at point $L$, and the second touching side $BC$ at point $F$.
a) Find the r... | Answer. a) $r=2$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{3}-1}{\sqrt{2}}$.
Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a ci... | )r=2,b)\angleBDC=\operatorname{arctg}\frac{\sqrt{3}-1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{x}{4}\right)^{\log _{5}(50 x)}=x^{6}$. | Answer: $x=\frac{1}{2}, x=2500$.
Solution. By logarithmizing with base 5, we obtain an equation equivalent to the original:
$$
\log _{5}(50 x) \cdot \log _{5}\left(\frac{x}{4}\right)=\log _{5}\left(x^{6}\right) \Leftrightarrow\left(\log _{5} x+2+\log _{5} 2\right)\left(\log _{5} x-2 \log _{5} 2\right)=6 \log _{5} x
$... | \frac{1}{2},2500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. Answer: 4000. | Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.
a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathrm{Z}$. Then we get $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrightarrow(89 p+87)(p+... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{5}$; b) $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that the three vertices of the prism are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equa... | \frac{180\pi\sqrt{3}}{23\sqrt{23}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x=\frac{6}{a}-|y-a| \\
x^{2}+y^{2}+b^{2}+63=2(b y-8 x)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in\left(-\infty ;-\frac{2}{3}\right] \cup(0+\infty)$.
“Phystech-2015”, 11th grade, solution to ticket 4
Solution. The second equation of the system can be transformed into the form $(x+8)^{2}+(y-b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center at $(-8 ; b)$. For all possible $b \in \mat... | \in(-\infty;-\frac{2}{3}]\cup(0+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $ABC$ and $ADC$ respectively, with the first touching side $BC$ at point $F$, and the second touching side $AD$ at point $P$.
a) Find the r... | Answer. a) $r=\sqrt{6}$, b) $\angle B D C=30^{\circ}$.
Solution. a) Segments $B O_{1}$ and $D O_{2}$ are the bisectors of angles $A B C$ and $A D C$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite a... | )r=\sqrt{6},b)\angleBDC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{\log _{3}\left(x^{4}\right) \cdot \log _{\frac{1}{3}}\left(x^{2}\right)+\log _{3}\left(x^{2}\right)-\log _{\frac{1}{3}}\left(x^{4}\right)+2}{\left(\log _{\frac{1}{3}}\left(x^{2}\right)\right)^{3}+64} \leq 0$.
“Phystech-2015”, 11th grade, solutions for ticket 5 | Answer. $x \in(-9 ;-3] \cup\left[-\frac{1}{\sqrt[4]{3}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[4]{3}}\right] \cup[3 ; 9)$.
Solution. The given inequality is equivalent to the following:
$$
\frac{2 \log _{3} x^{2} \cdot\left(-\log _{3} x^{2}\right)+\log _{3} x^{2}+2 \log _{3} x^{2}+2}{64-\log _{3}^{3} x^{2}} \leq 0
$... | x\in(-9;-3]\cup[-\frac{1}{\sqrt[4]{3}};0)\cup(0;\frac{1}{\sqrt[4]{3}}]\cup[3;9) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{7}{2} \cos 2 x+2\right) \cdot|2 \cos 2 x-1|=\cos x(\cos x+\cos 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\cos x(\cos x+\cos 5 x)=\cos x \cdot 2 \cos 3 x \cos 2 x=\cos 2 x \cdot(2 \cos x \cos 3 x)=\cos 2 x \cdot(\cos 2 x+\cos 4 x)=
$$
$$
=\cos 2 x\left(2 \cos ^{2} 2 x+\cos 2 x-1\right)=\cos 2 x(2 \cos... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=-\frac{2}{15} \\ \frac{1}{y}+\frac{1}{x+z}=-\frac{2}{3}, \\ \frac{1}{z}+\frac{1}{x+y}=-\frac{1}{4}\end{array}\right.$, | Answer. $(5; -1; -2)$.
Solution. Multiplying both sides of the first equation by $-\frac{15}{2} x(y+z)$, both sides of the second by $-\frac{3}{2} y(x+z)$, and the third by $-4 z(x+y)$, we obtain the system
“MIPT-2015”, 11th grade, solutions for ticket 5
$$
\left\{\begin{array}{l}
-7.5(x+y+z)=xy+xz \\
-1.5(x+y+z)=xy... | (5;-1;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a+y-x)=49 \\
y=15 \cos (x-b)-8 \sin (x-b)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in[-24 ; 24]$.
Solution. The first equation of the system can be transformed into the form $(x-a)^{2}+(y+a)^{2}=7^{2}$, hence it represents a circle of radius 7 with center $(a ;-a)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=17 \cos (x-b-\theta)$. ... | \in[-24;24] | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $A C=4$ and $\angle D B C=30^{\circ}$. A sphere passes through the vertices $D, A, B, B_{1}, C_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $16 \pi$, b) $90^{\circ}$, c) $48 \sqrt{3}$.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, we get that the acute angle of the rhombus is $60^{\circ}$. In the cross-section of the sphere by the plane $B C D$, we obtain a circle circumscribed around the triangle $A B D$. The c... | )16\pi,b)90,)48\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{125+\left(\log _{\frac{1}{2}}\left(x^{4}\right)\right)^{3}}{\log _{2}\left(x^{4}\right) \cdot \log _{2}\left(x^{2}\right)+6 \log _{\frac{1}{2}}\left(x^{4}\right)+17 \log _{2}\left(x^{2}\right)-3} \geq 0$. | Answer. $x \in[-2 \sqrt[4]{2} ;-\sqrt[4]{2}) \cup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup(\sqrt[4]{2} ; 2 \sqrt[4]{2}]$.
Solution. The given inequality is equivalent to the following:
$$
\frac{125-\left(2 \log _{2}\left(x^{2}\right)\right)^{3}}{2 \log _{2}\left(x^{2}\right... | x\in[-2\sqrt[4]{2};-\sqrt[4]{2})\cup(-\frac{1}{2\sqrt{2}};0)\cup(0;\frac{1}{2\sqrt{2}})\cup(\sqrt[4]{2};2\sqrt[4]{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{7}{4}-2 \cos 2 x\right) \cdot|2 \cos 2 x+1|=\cos x(\cos x-\cos 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in Z$.
Solution. Transform the right side of the equation:
$\cos x(\cos x-\cos 5 x)=\cos x \cdot(2 \sin 2 x \sin 3 x)=\sin 2 x \cdot(2 \sin 3 x \cos x)=\sin 2 x \cdot(\sin 4 x+\sin 2 x)=$
$$
=\sin 2 x(2 \sin 2 x \cos 2 x+\sin 2 x)=\sin ^{2} 2 x(1+2 \cos 2 x)=\left(1-... | \\frac{\pi}{6}+\frac{k\pi}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 b(b-x+y)=4 \\
y=5 \cos (x-a)-12 \sin (x-a)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in[-15 ; 15]$.
Solution. The first equation of the system can be transformed into the form $(x-b)^{2}+(y+b)^{2}=2^{2}$, hence it represents a circle of radius 9 with center $(-b ;-b)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=13 \cos (x-a-\theta)$.... | b\in[-15;15] | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $C D=3$ and $\angle A B D=30^{\circ}$. A sphere passes through the vertices $D, C, B, B_{1}, A_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $9 \pi$, b) $90^{\circ}$, c) 81.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, we get that the acute angle of the rhombus is $60^{\circ}$. In the cross-section of the sphere by the plane $A C D$, we obtain a circle circumscribed around the triangle $B C D$. The center of thi... | )9\pi,b)90,)81 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{\log _{2}\left(x^{6}\right) \cdot \log _{\frac{1}{2}}\left(x^{2}\right)-\log _{\frac{1}{2}}\left(x^{6}\right)-8 \log _{2}\left(x^{2}\right)+2}{8+\left(\log _{\frac{1}{2}}\left(x^{2}\right)\right)^{3}} \leq 0$. | Answer. $x \in(-2 ;-\sqrt[6]{2}] \cup\left[-\frac{1}{2} ; 0\right) \cup\left(0 ; \frac{1}{2}\right] \cup[\sqrt[6]{2} ; 2)$.
Solution. The given inequality is equivalent to the following:
$$
\frac{3 \log _{2} x^{2} \cdot\left(-\log _{2} x^{2}\right)+3 \log _{2} x^{2}-8 \log _{2} x^{2}+2}{8-\log _{2}^{3} x^{2}} \leq 0
... | x\in(-2;-\sqrt[6]{2}]\cup[-\frac{1}{2};0)\cup(0;\frac{1}{2}]\cup[\sqrt[6]{2};2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(3 \cos 2 x+\frac{9}{4}\right) \cdot|1-2 \cos 2 x|=\sin x(\sin x-\sin 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$\sin x(\sin x-\sin 5 x)=\sin x \cdot(-2 \sin 2 x \cos 3 x)=-\sin 2 x \cdot(2 \sin x \cos 3 x)=-\sin 2 x \cdot(\sin 4 x-\sin 2 x)=$
$$
=-\sin 2 x(2 \sin 2 x \cos 2 x-\sin 2 x)=\sin ^{2} 2 x(1-2 \cos ... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a-x-y)=64 \\
y=8 \sin (x-2 b)-6 \cos (x-2 b)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in[-18 ; 18]$.
Solution. The first equation of the system can be transformed into the form $(x-a)^{2}+(y-a)^{2}=8^{2}$, hence it represents a circle of radius 8 with center $(a ; a)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=10 \cos (x-2 b-\theta)$... | \in[-18;18] | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $B D=12$ and $\angle B A C=60^{\circ}$. A sphere passes through the vertices $D, A, B, B_{1}, C_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing th... | Answer. a) $48 \pi$, b) $90^{\circ}$, c) $192 \sqrt{3}$.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, the acute angle of the rhombus is $60^{\circ}$, and $A C=4 \sqrt{3}$. In the cross-section of the sphere by the plane $A_{1} C_{1} B_{1}$, we obtain a circle circumscribed around the ... | )48\pi,b)90,)192\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{64+\left(\log _{\frac{1}{5}}\left(x^{2}\right)\right)^{3}}{\log _{\frac{1}{5}}\left(x^{6}\right) \cdot \log _{5}\left(x^{2}\right)+5 \log _{5}\left(x^{6}\right)+14 \log _{\frac{1}{5}}\left(x^{2}\right)+2} \leq 0$. | Answer. $x \in[-25 ;-\sqrt{5}) \cup\left(-\frac{1}{\sqrt[3]{5}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[3]{5}}\right) \cup(\sqrt{5} ; 25]$.
Solution. The given inequality is equivalent to the following:
$$
\frac{64-\left(\log _{5}\left(x^{2}\right)\right)^{3}}{-3 \log _{5}\left(x^{2}\right) \cdot \log _{5}\left(x^{2}... | x\in[-25;-\sqrt{5})\cup(-\frac{1}{\sqrt[3]{5}};0)\cup(0;\frac{1}{\sqrt[3]{5}})\cup(\sqrt{5};25] | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\frac{7}{4}-3 \cos 2 x\right) \cdot|1+2 \cos 2 x|=\sin x(\sin x+\sin 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in Z$.
Solution. Transform the right side of the equation:
$\sin x(\sin x+\sin 5 x)=\sin x \cdot 2 \sin 3 x \cos 2 x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)=$
$$
=\cos 2 x\left(-2 \cos ^{2} 2 x+\cos 2 x+1\right)=\cos 2 x(1-\cos 2 x)(2 \cos... | \\frac{\pi}{6}+\frac{k\pi}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=3: 7$. The bisector $B L$ of the given triangle and the segment $A M$ intersect at point $P$ at an angle of $90^{\circ}$.
a) Find the ratio of the area of triangle $A B P$ to the area of quadrilateral $L P M C$.
b) On the segment $M C$... | Answer. a) $39: 161$, b) $\arccos \sqrt{\frac{13}{15}}$.
Solution. a) In triangle $A B M$, segment $B P$ is both a bisector and an altitude, so triangle $A B M$ is isosceles, and $B P$ is also its median. Let $B M=3 x$, then $A B=3 x, M C=7 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=3 x: 10 ... | )\frac{39}{161},b)\arccos\sqrt{\frac{13}{15}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 b(b+x+y)=81 \\
y=4 \cos (x+3 a)-3 \sin (x+3 a)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in[-14 ; 14]$.
Solution. The first equation of the system can be transformed into the form $(x+b)^{2}+(y+b)^{2}=9^{2}$, hence it represents a circle of radius 9 with center $(-b ;-b)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=5 \cos (x+3 a-\theta)$... | b\in[-14;14] | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $B D=3$ and $\angle A D C=60^{\circ}$. A sphere passes through the vertices $D, C, B, B_{1}, A_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $3 \pi$, b) $90^{\circ}$, c) $3 \sqrt{3}$.
Solution. a) Since $B D=3$ and the acute angle of the rhombus is $60^{\circ}$, then $A C=\sqrt{3}$. In the section of the sphere by the plane $A_{1} C_{1} D_{1}$, we obtain a circle circumscribed around the triangle $A_{1} B_{1} D_{1}$. The center of this circle is... | 3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $x^{\log _{2}(8 x)}=\frac{x^{7}}{8}$. | Answer: $x=2, x=8$.
Solution. By logarithmizing with base 2, we get $\log _{2} x \cdot \log _{2}(8 x)=\log _{2} x^{7}-\log _{2} 8$, which is equivalent to the following: $\log _{2}^{2} x+3 \log _{2} x=7 \log _{2} x-3 \Leftrightarrow \log _{2}^{2} x-4 \log _{2} x+3=0$, from which $\log _{2} x=1$ or $\log _{2} x=3 ; x=2... | 2,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x+\frac{1}{2}\right|=\sin ^{2} 3 x-\sin x \sin 3 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\begin{gathered}
\sin ^{2} 3 x-\sin x \sin 3 x=\sin 3 x(\sin 3 x-\sin x)=\sin 3 x \cdot 2 \cos 2 x \sin x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)= \\
=\cos 2 x\left(-2 \... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 242400, such that $k^{2}+2 k$ is divisible by 303. | Answer: 3200.
Solution. By factoring the dividend and divisor, we get the condition $k(k+2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k+2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+2):(3 \cdot 101) \Leftrightarrow p(101 p... | 3200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{l}3 x \geq 2 y+16 \\ x^{4}+2 x^{2} y^{2}+y^{4}+25-26 x^{2}-26 y^{2}=72 x y .\end{array}\right.$ | Answer. $(6 ; 1)$.
Solution. Transform the equation of the system (add $36 x^{2}+36 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+25+10 x^{2}+10 y^{2}=36 x^{2}+36 y^{2}+72 x y \Leftrightarrow\left(x^{2}+y^{2}+5\right)^{2}=(6 x+6 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { ... | (6;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $S A$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $K$ is marked such that $A K: K S=2: 3$. The point $K$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $S A B C D$.
a) Find the ratio $C S: C D$.
b) Suppose ... | Answer. a) $\sqrt{3}$, b) $V=\frac{9 \pi}{\sqrt{5}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H K, B H K, C H K, D H K$. They have the side $H K$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H K$ and $D H K$ are right angles, ang... | )\sqrt{3},b)V=\frac{9\pi}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
y=-b-x^{2}, \\
x^{2}+y^{2}+8 a^{2}=4+4 a(x+y)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \leq 2 \sqrt{2}+\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x-2 a)^{2}+(y-2 a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $(2 a ; 2 a)$. For all possible $a \in \mathrm{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt... | b\leq2\sqrt{2}+\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Circles with centers \(O_{1}\) and \(O_{2}\) of equal radius are inscribed in the corners \(A\) and \(B\) of triangle \(ABC\), respectively, and point \(O\) is the center of the circle inscribed in triangle \(ABC\). These circles touch side \(AB\) at points \(K_{1}\), \(K_{2}\), and \(K\) respectively, with \(AK_{1}... | Answer. a) $A K=\frac{32}{5}$, b) $\angle C A B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.
Solution. a) The lines $A O_{1}$ and $B O_{2}$ are the angle bisectors of angles $A$ and $B$ 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... | )AK=\frac{32}{5},b)\angleCAB=2\arcsin\frac{3}{5}=\arccos\frac{7}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $x^{\log _{3}\left(27 x^{2}\right)}=\frac{x^{9}}{81}$. | Answer: $x=3, x=9$.
Solution. By taking the logarithm with base 3, we get $\log _{3} x \cdot \log _{3}\left(27 x^{2}\right)=\log _{3} x^{9}-\log _{3} 81$, which is equivalent to the following: $2 \log _{3}^{2} x+3 \log _{3} x=9 \log _{3} x-4 \Leftrightarrow \log _{3}^{2} x-3 \log _{3} x+2=0$, from which $\log _{3} x=1... | 3,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x-\frac{1}{2}\right|=\cos ^{2} 3 x+\cos x \cos 3 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\begin{gathered}
\cos ^{2} 3 x+\cos x \cos 3 x=\cos 3 x(\cos x+\cos 3 x)=\cos 3 x \cdot 2 \cos 2 x \cos x=\cos 2 x \cdot(2 \cos x \cos 3 x)=\cos 2 x \cdot(\cos 2 x+\cos 4 x)= \\
=\cos 2 x\left(2 \c... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 353500, such that $k^{2}+k$ is divisible by 505. | Answer: 2800.
Solution: By factoring the dividend and divisor, we get the condition $k(k+1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k+1)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+1):(5 \cdot 101) \Leftrightarrow p(101 p... | 2800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $S B$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $B L: L S=2: 5$. The point $L$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $S A B C D$.
a) Find the ratio $A S: C D$.
b) Suppose ... | Answer. a) $\sqrt{\frac{5}{3}}$, b) $V=\frac{125 \pi}{\sqrt{21}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H L, B H L, C H L, D H L$. They have the side $H L$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $A H L$ and $C H L$ are righ... | \frac{125\pi}{\sqrt{21}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
y=x^{2}-a \\
x^{2}+y^{2}+8 b^{2}=4 b(y-x)+1
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \geq-\sqrt{2}-\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x+2 b)^{2}+(y-2 b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center $(-2 b ; 2 b)$. For all possible $b \in \mathbf{R}$, the graphs of these functions sweep out the strip $-x-\sqrt{2} \... | \geq-\sqrt{2}-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In the corners $B$ and $C$ 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 $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $B K_{1}=4, C K_{2}=8$, and $B C=18$.
a) Fin... | Answer. a) $C K=12$, b) $\angle A B C=60^{\circ}$.
Solution. a) The lines $\mathrm{CO}_{2}$ and $\mathrm{BO}_{1}$ are the bisectors of angles $C$ and $B$ 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 r... | )CK=12,b)\angleABC=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $x^{\log _{2}\left(0.25 x^{3}\right)}=512 x^{4}$. | Answer: $x=\frac{1}{2}, x=8$.
Solution. By logarithmizing with base 2, we get $\log _{2} x \cdot \log _{2}\left(\frac{1}{4} x^{3}\right)=\log _{2} x^{4}+\log _{2} 512$, which is equivalent to the following: $3 \log _{2}^{2} x-2 \log _{2} x=4 \log _{2} x+9 \Leftrightarrow \log _{2}^{2} x-2 \log _{2} x-3=0$, from which ... | \frac{1}{2},8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x+\frac{1}{2}\right|=\sin ^{2} x+\sin x \sin 5 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$\sin ^{2} x+\sin x \sin 5 x=\sin x(\sin x+\sin 5 x)=\sin x \cdot 2 \cos 2 x \sin 3 x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)=$
$$
=\cos 2 x\left(-2 \cos ^{2} 2 x+\cos 2 x... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400. | Solution. Factoring the dividend and divisor, we get the condition $k(k-2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k-2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-2):(3 \cdot 101) \Leftrightarrow p(101 p-2) \vdots 3$. The... | 4400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{l}2 x+y+8 \leq 0, \\ x^{4}+2 x^{2} y^{2}+y^{4}+9-10 x^{2}-10 y^{2}=8 x y .\end{array}\right.$ | Answer. $(-3 ;-2)$.
Solution. Transform the equation of the system (add $4 x^{2}+4 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+9-6 x^{2}-6 y^{2}=4 x^{2}+4 y^{2}+8 x y \Leftrightarrow\left(x^{2}+y^{2}-3\right)^{2}=(2 x+2 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l }
{ ... | (-3,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $S A$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $K$ is marked such that $A K: K S=1: 4$. The point $K$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $2 A B C D$.
a) Find the ratio $D S: B C$.
b) Suppose ... | Answer. a) $\frac{2}{\sqrt{3}}$, b) $V=\frac{64 \pi}{\sqrt{15}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H K, B H K, C H K, D H K$. They have the side $H K$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H K$ and $D H K$ are right... | \frac{64\pi}{\sqrt{15}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
y=b-x^{2} \\
x^{2}+y^{2}+2 a^{2}=4-2 a(x+y)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \geq-2 \sqrt{2}-\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x+a)^{2}+(y+a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $( -a ;-a )$. For all possible $a \in \mathbb{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt{2} \... | b\geq-2\sqrt{2}-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In the corners $C$ and $B$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $C K_{1}=3, B K_{2}=7$, and $B C=16$.
a) Fin... | Answer. a) $C K=\frac{24}{5}$, b) $\angle A C B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.
Solution. a) The lines $\mathrm{CO}_{1}$ and $\mathrm{BO}_{2}$ are the angle bisectors of angles $C$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles wi... | )CK=\frac{24}{5},b)\angleACB=2\arcsin\frac{3}{5}=\arccos\frac{7}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $x^{\log _{5}(0.008 x)}=\frac{125}{x^{5}}$. | Answer: $x=5, x=\frac{1}{125}$.
Solution. By logarithmizing with base 5, we get $\log _{5} x \cdot \log _{5}\left(\frac{x}{125}\right)=\log _{5} 125-\log _{5} x^{3}$, which is equivalent to the following: $\log _{5}^{2} x-3 \log _{5} x=3-5 \log _{5} x \Leftrightarrow \log _{5}^{2} x+2 \log _{5} x-3=0$, from which $\lo... | 5,\frac{1}{125} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x-\frac{1}{2}\right|=\cos ^{2} x+\cos x \cos 5 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\begin{gathered}
\cos ^{2} x+\cos x \cos 5 x=\cos x(\cos x+\cos 5 x)=\cos x \cdot 2 \cos 2 x \cos 3 x=\cos 2 x \cdot(2 \cos x \cos 3 x)=\cos 2 x \cdot(\cos 2 x+\cos 4 x)= \\
=\cos 2 x\left(2 \cos ^... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505. | Answer: 3600.
Solution. By factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.
a) $k \vdots: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow ... | 3600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{l}x+3 y+14 \leq 0, \\ x^{4}+2 x^{2} y^{2}+y^{4}+64-20 x^{2}-20 y^{2}=8 x y .\end{array}\right.$ | Answer: $(-2, -4)$.
Solution. Transform the equation of the system (add $4 x^{2}+4 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+64-16 x^{2}-16 y^{2}=4 x^{2}+4 y^{2}+8 x y \Leftrightarrow\left(x^{2}+y^{2}-8\right)^{2}=(2 x+2 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } ... | (-2,-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the edge $S C$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $C L: L S=1: 5$. The point $L$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $2 A B C D$.
a) Find the ratio $A S: A B$.
b) Suppose ... | Answer. a) $\frac{\sqrt{5}}{2}$, b) $V=\frac{125 \pi \sqrt{2}}{3 \sqrt{3}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H L, B H L, C H L, D H L$. They have the side $H L$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H L$ and $D H L... | \frac{\sqrt{5}}{2},\frac{125\pi\sqrt{2}}{3\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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