problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{1}{3}\right)$, $\log _{x-\frac{1}{3}}(x-3)$, and $\log _{x-3} x$ is equal to the product of the other two. | Answer: $x=\frac{10}{3}, x=\frac{3+\sqrt{13}}{2}$.
Solution. Note that on the domain of definition, the product of all three logarithms is
$$
\log _{x}\left(x-\frac{1}{3}\right) \cdot \log _{x-\frac{1}{3}}(x-3) \cdot \log _{x-3} x=1
$$
Let the number that is the product of the other two be denoted by $c$, and the tw... | \frac{10}{3},\frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,529 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=-6 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x)... | Answer: $A=6, A=0$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=-6 g(x)$ is equivalent to the equation $(f(x))^{2}=-6 g(x)$ having exactly one solution. We get $(a x+b)^{2}=-6(a x+c), a^{2} x^{2}+2 a(b+3) x+b^{2}-7 c=0$. The quarter of the discriminant of ... | A=6,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,530 |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root,
all their roots are powers of the number 11 with integer non-negative exponents, and their coefficients in absolute value do not exceed $1331^{38}$... | Answer: 3363.
Solution. Such quadratic trinomials can be represented in the form $\left(x-11^{a}\right)\left(x-11^{b}\right)$, where $a \geqslant 0, b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(11^{a}+11^{b}\right) x+11^{a+b}$. According... | 3363 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,531 |
4. The numbers $x$ and $y$ are such that the equations $\sin y + \sin x + \cos 3x = 0$ and $\sin 2y - \sin 2x = \cos 4x + \cos 2x$ are satisfied. What is the greatest value that the sum $\cos y + \cos x$ can take? | Answer: $1+\sin \frac{3 \pi}{8}=1+\frac{\sqrt{2+\sqrt{2}}}{2}$.
Solution. Transform the second equality:
$$
\sin 2 y=2 \sin x \cos x+2 \cos 3 x \cos x \Leftrightarrow \sin 2 y=2 \cos x(\sin x+\cos 3 x) \text {. }
$$
Substituting $-\sin y$ for $\sin x+\cos 3 x$, we get $\sin 2 y=-2 \cos x \sin y, 2 \sin y \cos y+2 \c... | 1+\frac{\sqrt{2+\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,532 |
5. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 13 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ again at points $E$ and $K$ respectively. The length of arc $A... | Answer: $A D=13, B K=\frac{120}{13}, P_{E B M}=\frac{340}{13}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\ang... | AD=13,BK=\frac{120}{13},P_{EBM}=\frac{340}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,533 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twenty-five consecutive natural numbers to some number $a$ is 1270, and the sum of the distances from these same twenty-five numbers to the number $a^2$ is 1234. Find all possible values of ... | Answer: $a=-\frac{4}{5}$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+24$. Note that if some number lies on the segment $[k ; k+24]$, then the sum of the distances from it to the given twenty-five numbers does not exceed $\frac{25}{2} \cdot 24=300$ (the sum of the distances to ... | -\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,534 |
7. On the edge $B C$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a point $M$ is chosen. A sphere constructed on the segment $C_{1} M$ as its diameter touches the planes of four faces of the parallelepiped, and one of them at a point lying on the edge $B_{1} B$. It is known that $B M=1, C M=3$. Find the len... | Answer: $A A_{1}=5, R=2, V=32$.
Solution. Since the center of the sphere is the midpoint of the segment $C_{1} M$ and lies in the face $B C C_{1} B_{1}$, the sphere does not touch this face. Note that if the segment $C_{1} M$ is not perpendicular to the planes $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, then the sphere d... | AA_{1}=5,R=2,V=32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,535 |
9.3. Non-zero numbers $a, b$, and $c$ are such that the equalities $a^{2}(b+c-a)=b^{2}(a+c-b)=c^{2}(b+a-c)$ hold. What is the greatest value that the expression $\frac{2 b+3 c}{a}$ can take? | Answer: 5.
Solution: By equating the first and second expressions, after transformation, we get: $(a-b)\left(a^{2}+b^{2}-a c-b c\right)=0$. Similarly, we obtain the equalities $(b-c)\left(b^{2}+c^{2}-a b-a c\right)=0$ and $(a-c)\left(a^{2}+c^{2}-a b-c b\right)=0$.
We will prove that $a=b=c$.
Assume that $a=b \neq c$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,536 |
9.5. Does there exist thirteen consecutive natural numbers such that their sum is a 2021st power of a natural number? | Answer. They exist.
Solution. Let's denote 13 consecutive numbers as $N-6, N-5$, ..., $N+5, N+6$. Then their sum is $13N$. If $N=13^{2020}$, then the sum will be $13N = 13 \cdot 13^{2020} = 13^{2021}$.
Comment. A correct answer without justification - 0 points.
9.6-1. On a given circle $\omega$, points $A, B$, and $... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,537 |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 3375. The answer should be presented as an integer. | Answer: 1680.
Solution. Since $3375=3^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three threes, three fives, and two ones, or (b) one three, one nine, three fives, and three ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight fo... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,538 |
2. [5 points] Solve the equation $\cos 11 x - \cos 3 x - \sin 11 x + \sin 3 x = \sqrt{2} \cos 14 x$. | Answer: $x=-\frac{\pi}{28}+\frac{\pi k}{7}, x=\frac{\pi}{12}+\frac{2 \pi k}{3}, x=\frac{5 \pi}{44}+\frac{2 \pi k}{11}, k \in \mathbb{Z}$.
Solution. The given equation is equivalent to the following:
$-2 \sin 4 x \sin 7 x-2 \sin 4 x \cos 7 x=\sqrt{2} \cos 14 x \Leftrightarrow-2 \sin 4 x(\sin 7 x+\cos 7 x)=\sqrt{2}\lef... | -\frac{\pi}{28}+\frac{\pik}{7},\frac{\pi}{12}+\frac{2\pik}{3},\frac{5\pi}{44}+\frac{2\pik}{11},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,539 |
3. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\left(\frac{y^{5}}{x}\right)^{\lg x}=y^{2 \lg x y} \\
x^{2}-2 x y-4 x-3 y^{2}+12 y=0
\end{array}\right.
$$ | Answer: $(2 ; 2),(9 ; 3),\left(\frac{9-\sqrt{17}}{2} ; \frac{\sqrt{17}-1}{2}\right)$.
Solution. We take the logarithm of the first equation of the system to base 10:
$$
\lg \left(\frac{y^{5}}{x}\right) \cdot \lg x=\lg y^{2} \cdot \lg (x y)
$$
This equation on the domain of valid values is equivalent to the following... | (2;2),(9;3),(\frac{9-\sqrt{17}}{2};\frac{\sqrt{17}-1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,540 |
4. [5 points] A sphere with center $O$ is inscribed in a trihedral angle with vertex $S$ and touches its faces at points $K, L, M$ (all plane angles of the trihedral angle are different). Find the angle $K S O$ and the area of the section of the given trihedral angle by the plane $K L M$, if it is known that the areas ... | Answer: $\angle K S O=\arcsin \frac{1}{3}, S=\frac{16}{9}$.
Solution. Let the points of intersection of the line $S O$ with the sphere be $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). Let the radius of the sphere be $r$. Triangles $O K S, O L S$, and $O M S$ are right triangles (angles at... | \angleKSO=\arcsin\frac{1}{3},S=\frac{16}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,541 |
5. [5 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
|y-3-x|+|y-3+x|=6 \\
(|x|-4)^{2}+(|y|-3)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{1 ; 25\}$.
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 find the sets of points where the expressions under the absolute values are zero. These are the lines $y-3-x=0$ and $y-3+x=0$. They divide the p... | \in{1;25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,542 |
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: a) $C F=10 ;$ b) $S_{A C F}=49$.
Fig. 3: variant 5, problem 6
Solution. a) Let $R=5$ - the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Th... | CF=10;S_{ACF}=49 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,543 |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 16875. The answer should be presented as an integer. | Answer: 1120.
Solution. Since $16875=3^{3} \cdot 5^{4}$, the sought numbers can consist of the following digits: (a) three threes, four fives, and one one, or (b) one three, one nine, four fives, and two ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight for th... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,544 |
2. [5 points] Solve the equation $\cos 7 x+\cos 3 x-\sqrt{2} \cos 10 x=\sin 7 x+\sin 3 x$. | Answer: $x=\frac{\pi}{20}+\frac{\pi k}{5}, x=\frac{\pi}{12}+\frac{2 \pi k}{3}, x=\frac{\pi}{28}+\frac{2 \pi k}{7}, k \in \mathbb{Z}$.
Solution. The given equation is equivalent to the following:
$2 \cos 5 x \cos 2 x-2 \sin 5 x \cos 2 x=\sqrt{2} \cos 10 x \Leftrightarrow 2 \cos 2 x(\cos 5 x-\sin 5 x)=\sqrt{2}\left(\co... | \frac{\pi}{20}+\frac{\pik}{5},\frac{\pi}{12}+\frac{2\pik}{3},\frac{\pi}{28}+\frac{2\pik}{7},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,545 |
3. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\left(\frac{x^{4}}{y^{2}}\right)^{\lg y}=(-x)^{\lg (-x y)} \\
2 y^{2}-x y-x^{2}-4 x-8 y=0
\end{array}\right.
$$ | Answer: $(-4 ; 2),(-2 ; 2),\left(\frac{\sqrt{17}-9}{2} ; \frac{\sqrt{17}-1}{2}\right)$.
Solution. We take the logarithm of the first equation of the system to base 10:
$$
\lg \left(\frac{x^{4}}{y^{2}}\right) \cdot \lg y=\lg (-x) \cdot \lg (-x y)
$$
This equation on the domain of admissible values is equivalent to th... | (-4;2),(-2;2),(\frac{\sqrt{17}-9}{2};\frac{\sqrt{17}-1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,546 |
4. [5 points] A sphere with center $O$ is inscribed in a trihedral angle with vertex $S$ and touches its faces at points $K, L, M$ (all plane angles of the trihedral angle are different). Find the angle $K S O$ and the area of the section of the given trihedral angle by the plane $K L M$, if it is known that the areas ... | Answer: $\angle K S O=\arcsin \frac{1}{5} ; S=\frac{144}{25}$.
Solution. Let the points of intersection of the line $S O$ with the sphere be $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). Let the radius of the sphere be $r$. Triangles $O K S, O L S$, and $O M S$ are right triangles (angles... | \angleKSO=\arcsin\frac{1}{5};S=\frac{144}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,547 |
5. [5 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
|x-6-y|+|x-6+y|=12 \\
(|x|-6)^{2}+(|y|-8)^{2}=a
\end{array}\right.
$$
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, find the sets of points where the expressions under the absolute values are zero. These are the lines $x-6-y=0$ and $x-6+y=0$. They divide the p... | \in{4;100} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,548 |
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: a) $C F=26 ;$ b) $S_{A C F}=289$.
Fig. 6: Variant 6, Problem 6
Solution. a) Let $R=13$ - the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. ... | CF=26;S_{ACF}=289 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,549 |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 9261. The answer should be presented as an integer. | Answer: 1680.
Solution. Since $9261=3^{3} \cdot 7^{3}$, the sought numbers can consist of the following digits: (a) three threes, three sevens, and two ones, or (b) one three, one nine, three sevens, and three ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight ... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,550 |
2. [5 points] Solve the equation $\cos 9 x-\cos 5 x-\sqrt{2} \cos 4 x+\sin 9 x+\sin 5 x=0$. | Answer: $x=\frac{\pi}{8}+\frac{\pi k}{2}, x=\frac{\pi}{20}+\frac{2 \pi k}{5}, x=\frac{\pi}{12}+\frac{2 \pi k}{9}, k \in \mathbb{Z}$.
Solution. The given equation is equivalent to the following:
$-2 \sin 2 x \sin 7 x+2 \sin 7 x \cos 2 x=\sqrt{2} \cos 4 x \Leftrightarrow 2 \sin 7 x(\cos 2 x-\sin 2 x)=\sqrt{2}\left(\cos... | \frac{\pi}{8}+\frac{\pik}{2},\frac{\pi}{20}+\frac{2\pik}{5},\frac{\pi}{12}+\frac{2\pik}{9},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,551 |
3. $[5$ points] Solve the system of equations
$$
\left\{\begin{array}{l}
\left(x^{2} y^{4}\right)^{-\ln x}=y^{\ln \left(y / x^{7}\right)} \\
y^{2}-x y-2 x^{2}+8 x-4 y=0
\end{array}\right.
$$ | Answer: $(2 ; 2),(2 ; 4),\left(\frac{\sqrt{17}-1}{2} ; \frac{9-\sqrt{17}}{2}\right)$.
Solution. We take the logarithm of the first equation of the system with base $e$:
$$
\ln \left(x^{2} y^{4}\right) \cdot(-\ln x)=\ln y \cdot \ln \left(y / x^{7}\right)
$$
This equation on the domain of valid values is equivalent to... | (2;2),(2;4),(\frac{\sqrt{17}-1}{2};\frac{9-\sqrt{17}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,552 |
4. [5 points] A sphere with center $O$ is inscribed in a trihedral angle with vertex $S$ and touches its faces at points $K, L, M$ (all plane angles of the trihedral angle are different). Find the angle $K S O$ and the area of the section of the trihedral angle by the plane $K L M$, given that the areas of the sections... | Answer: $\angle K S O=\arcsin \frac{3}{5}, S=\frac{64}{25}$.
Solution. Let the points of intersection of the line $S O$ with the sphere be $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). Let the radius of the sphere be $r$. Triangles $O K S, O L S$, and $O M S$ are right triangles (angles a... | \angleKSO=\arcsin\frac{3}{5},S=\frac{64}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,553 |
5. [5 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
|x+y+5|+|y-x+5|=10 \\
(|x|-12)^{2}+(|y|-5)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{49 ; 169\}$.
Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To expand the absolute values, we find the sets of points where the expressions under the absolute values are zero. These are the lines $x+y+5=0$ and $y-x+5=0$. They divide t... | \in{49;169} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,554 |
6. [6 points] a) Two circles of the same radius 10 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: a) $C F=20 ;$ b) $S_{A C F}=196$.
Fig. 9: Option 7, Problem 6
Solution. a) Let $R=10$ - the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. T... | CF=20;S_{ACF}=196 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,555 |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 64827. The answer should be presented as an integer. | Answer: 1120.
Solution. Since $64827=3^{3} \cdot 7^{4}$, the sought numbers can consist of the following digits: (a) three threes, four sevens, and one one, or (b) one three, one nine, four sevens, and two ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight for ... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,556 |
2. [5 points] Solve the equation $\cos 7 x+\cos 3 x+\sin 7 x-\sin 3 x+\sqrt{2} \cos 4 x=0$. | Answer: $x=-\frac{\pi}{8}+\frac{\pi k}{2}, x=-\frac{\pi}{4}+\frac{2 \pi k}{3}, x=\frac{3 \pi}{28}+\frac{2 \pi k}{7}, k \in \mathbb{Z}$.
Solution. The given equation is equivalent to the following:
$2 \cos 5 x \cos 2 x+2 \sin 2 x \cos 5 x=-\sqrt{2} \cos 4 x \Leftrightarrow 2 \cos 5 x(\cos 2 x+\sin 2 x)=-\sqrt{2}\left(... | -\frac{\pi}{8}+\frac{\pik}{2},-\frac{\pi}{4}+\frac{2\pik}{3},\frac{3\pi}{28}+\frac{2\pik}{7},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,557 |
3. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\left(-\frac{x^{7}}{y}\right)^{\ln (-y)}=x^{2 \ln \left(x y^{2}\right)} \\
y^{2}+2 x y-3 x^{2}+12 x+4 y=0
\end{array}\right.
$$ | Answer: $(2 ;-2),(3 ;-9),\left(\frac{\sqrt{17}-1}{2} ; \frac{\sqrt{17}-9}{2}\right)$.
Solution. We take the natural logarithm of the first equation in the system:
$$
\ln \left(-\frac{x^{7}}{y}\right) \cdot \ln (-y)=\ln x^{2} \cdot \ln \left(x y^{2}\right)
$$
This equation, within the domain of valid values, is equiv... | (2,-2),(3,-9),(\frac{\sqrt{17}-1}{2},\frac{\sqrt{17}-9}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,558 |
4. [5 points] A sphere with center $O$ is inscribed in a trihedral angle with vertex $S$ and touches its faces at points $K, L, M$ (all plane angles of the trihedral angle are different). Find the angle $K S O$ and the area of the section of the given trihedral angle by the plane $K L M$, if it is known that the areas ... | Answer: $\angle K S O=\arcsin \frac{1}{7} ; S=\frac{576}{49}$.
Solution. Let the points of intersection of the line $S O$ with the sphere be $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). Let the radius of the sphere be $r$. Triangles $O K S, O L S$, and $O M S$ are right triangles (angles... | \angleKSO=\arcsin\frac{1}{7};S=\frac{576}{49} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,559 |
5. [5 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
|x+y+8|+|x-y+8|=16 \\
(|x|-8)^{2}+(|y|-15)^{2}=a
\end{array}\right.
$$
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 expand the absolute values, find the sets of points where the expressions under the absolute values are zero. These are the lines $x+y+8=0$ and $x-y+8=0$. They divide the ... | \in{49;289} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,560 |
6. [6 points] a) Two circles of the same radius 17 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: $C F=34, S_{A C F}=529$.
Fig. 12: variant 8, problem 6
Solution. a) Let $R=17$ - the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\a... | CF=34,S_{ACF}=529 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,561 |
1. [3 points] A coin is flipped 90 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear no fewer than 55 times, and $q$ be the probability that heads will appear fewer than 35 times. Find $p-q$. | Answer: $\frac{1}{2^{90}} \cdot C_{90}^{35}$.
Solution. Since the probability of getting heads or tails is equal, the probability of getting 90 heads is equal to the probability of getting 90 tails (i.e., 0 heads); the probability of getting 89 heads is equal to the probability of getting 89 tails (i.e., one head), an... | \frac{1}{2^{90}}\cdotC_{90}^{35} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,562 |
2. [5 points] Solve the equation $\frac{\cos 8 x}{\cos 3 x+\sin 3 x}+\frac{\sin 8 x}{\cos 3 x-\sin 3 x}=\sqrt{2}$. | Answer: $x=\frac{\pi}{44}+\frac{2 \pi k}{11}, k \neq 11 p+4, k \in \mathbb{Z}, p \in \mathbb{Z}$.
Solution. By bringing the fractions on the left side of the equation to a common denominator, we obtain
$$
\frac{\cos 8 x \cos 3 x-\cos 8 x \sin 3 x+\cos 3 x \sin 8 x+\sin 3 x \sin 8 x}{\cos ^{2} 3 x-\sin ^{2} 3 x}=\sqrt... | \frac{\pi}{44}+\frac{2\pik}{11},k\neq11p+4,k\in\mathbb{Z},p\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,563 |
3. [5 points] Solve the inequality $27 \sqrt{\log _{3} x}-11 \cdot 3 \sqrt{4 \log _{3} x}+40 \cdot x \sqrt{\log _{x} 3} \leqslant 48$. | Answer: $x \in(1 ; 3] \cup\left\{4^{\log _{3} 4}\right\}$.
Solution. Note that $x \sqrt{\sqrt{\log _{x} 3}}=\left(3^{\log _{3} x}\right)^{\sqrt{\log _{x} 3}}=3^{\sqrt{\log _{3}^{2} x \cdot \log _{x} 3}}=3 \sqrt{\log _{3} x}$. Therefore, the inequality can be rewritten as $3^{3 \sqrt{\log _{3} x}}-11 \cdot 3^{2 \sqrt{\... | x\in(1;3]\cup{4^{\log_{3}4}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,564 |
4. [5 points] a) A sphere with center $O$ touches the lateral edges $S A, S B, S C$ of the pyramid $S A B C$ at points $K, L, M$ respectively, and also touches its base $A B C$. A plane is drawn through the point on the sphere closest to point $S$, tangent to the sphere. The area of the section of the pyramid $S A B C$... | Answer: a) $S_{K L M}=9.8 ;$ b) $V_{S A B C}=\frac{1372}{3}$.
Fig. 1: Variant 13, Problem 4
Solution. a) Let the radius of the sphere be $R$. Denote the points of intersection of the line ... | S_{KLM}=9.8;V_{SABC}=\frac{1372}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,565 |
5. [6 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
y=|x-\sqrt{a}|+\sqrt{a}-2 \\
(|x|-4)^{2}+(|y|-3)^{2}=25
\end{array}\right.
$$
has exactly three solutions.
^{2}\right\}$.
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 t... | \in{1;16;(\frac{5\sqrt{2}+1}{2})^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,566 |
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: a) $\frac{R_{2}}{R_{1}}=\frac{3}{2} ;$ b) $R_{1}=\sqrt{\frac{2}{3}}, R_{2}=\sqrt{\frac{3}{2}}$.
Solution. a) Let $R_{1}, R_{2}$ be the radii of circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B O_{2}=\alpha$, and the lines $D O_{2}$ and $\ell_{2}$ intersect at point $P$. Then, from the condition... | \frac{R_{2}}{R_{1}}=\frac{3}{2};R_{1}=\sqrt{\frac{2}{3}},R_{2}=\sqrt{\frac{3}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,567 |
1. [3 points] A coin is flipped 70 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear more than 42 times, and $q$ be the probability that heads will appear no more than 28 times. Find $p-q$. | Answer: $-\frac{1}{2^{70}} \cdot C_{70}^{28}$.
Solution. Since the probability of getting heads or tails is equal, the probability of getting 70 heads is equal to the probability of getting 70 tails (i.e., 0 heads); the probability of getting 69 heads is equal to the probability of getting 69 tails (i.e., one head), a... | -\frac{1}{2^{70}}\cdotC_{70}^{28} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,568 |
2. [5 points] Solve the equation $\frac{\cos 4 x}{\cos 3 x - \sin 3 x} + \frac{\sin 4 x}{\cos 3 x + \sin 3 x} = \sqrt{2}$. | Answer: $x=\frac{\pi}{52}+\frac{2 \pi k}{13}, k \neq 13 p-5, k \in \mathbb{Z}, p \in \mathbb{Z}$.
Solution. By bringing the fractions on the left side of the equation to a common denominator, we obtain
$$
\frac{\cos 4 x \cos 3 x+\cos 4 x \sin 3 x+\cos 3 x \sin 4 x-\sin 3 x \sin 4 x}{\cos ^{2} 3 x-\sin ^{2} 3 x}=\sqrt... | \frac{\pi}{52}+\frac{2\pik}{13},k\neq13p-5,k\in\mathbb{Z},p\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,569 |
3. [5 points] Solve the inequality $8^{\sqrt{\log _{2} x}}-7 \cdot 2^{1+\sqrt{4 \log _{2} x}}+60 \cdot x \sqrt{\log _{x} 2} \leqslant 72$. | Answer: $x \in(1 ; 2] \cup\left\{6^{\log _{2} 6}\right\}$.
Solution. Note that $x^{\sqrt{\log _{x} 2}}=\left(2^{\log _{2} x}\right)^{\sqrt{\log _{x} 2}}=2^{\sqrt{\log _{2}^{2} x \cdot \log _{x} 2}}=2^{\sqrt{\log _{2} x}}$. Therefore, the inequality can be rewritten as $2^{3 \sqrt{\log _{2} x}}-14 \cdot 2^{2 \sqrt{\log... | x\in(1;2]\cup{6^{\log_{2}6}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,570 |
4. [5 points] a) A sphere with center $O$ touches the lateral edges $S A, S B, S C$ of the pyramid $S A B C$ at points $K, L, M$ respectively, and also touches its base $A B C$. A plane is drawn through the point on the sphere closest to point $S$, tangent to the sphere. The area of the section of the pyramid $S A B C$... | Answer: a) $S_{K L M}=\frac{49}{8} ;$ b) $V_{S A B C}=\frac{1029}{2}$.
Fig. 4: Variant 14, Problem 4
Solution. a) Let the radius of the sphere be $R$. Denote the points of intersection of ... | S_{KLM}=\frac{49}{8};V_{SABC}=\frac{1029}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,571 |
5. [6 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
x=|y-\sqrt{a}|+\sqrt{a}-4 \\
(|x|-6)^{2}+(|y|-8)^{2}=100
\end{array}\right.
$$
has exactly three solutions.
 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 first quadrant (including its bou... | \in{4;64;51+10\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,572 |
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: a) $\frac{R_{2}}{R_{1}}=\frac{5}{3}$; b) $R_{1}=\frac{3 \sqrt{3}}{2 \sqrt{5}}, R_{2}=\frac{\sqrt{15}}{2}$.
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\omega_{1}, \omega_{2}$ respectively, $\angle O_{1} B O_{2}=\alpha$, and the lines $D O_{2}$ and $\ell_{2}$ intersect at point $P$. Then, from ... | \frac{R_{2}}{R_{1}}=\frac{5}{3};R_{1}=\frac{3\sqrt{3}}{2\sqrt{5}},R_{2}=\frac{\sqrt{15}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,573 |
1. [3 points] A coin is flipped 110 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear no fewer than 61 times, and $q$ be the probability that heads will appear fewer than 49 times. Find $p-q$. | Answer: $\frac{1}{2^{110}} \cdot C_{100}^{61}$.
Solution. Since the probability of getting heads or tails is equal, the probability of getting 110 heads is equal to the probability of getting 110 tails (i.e., 0 heads); the probability of getting 109 heads is equal to the probability of getting 109 tails (i.e., one hea... | \frac{1}{2^{110}}\cdotC_{110}^{61} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,574 |
2. [5 points] Solve the equation $\frac{\cos 14 x}{\cos 5 x-\sin 5 x}-\frac{\sin 14 x}{\cos 5 x+\sin 5 x}=-\sqrt{2}$. | Answer: $x=\frac{3 \pi}{76}+\frac{2 \pi k}{19}, k \neq 19 p+2, k \in \mathbb{Z}, p \in \mathbb{Z}$.
Solution. By bringing the fractions on the left side of the equation to a common denominator, we get $\frac{\cos 14 x \cos 5 x+\cos 14 x \sin 5 x-\cos 5 x \sin 14 x+\sin 5 x \sin 14 x}{\cos ^{2} 5 x-\sin ^{2} 5 x}=-\sqr... | \frac{3\pi}{76}+\frac{2\pik}{19},k\neq19p+2,k\in\mathbb{Z},p\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,575 |
3. [5 points] Solve the inequality $27^{\sqrt{\log _{3} x}}-13 \cdot 3 \sqrt{\sqrt{4 \log _{3} x}}+55 \cdot x^{\sqrt{\log _{x} 3}} \leqslant 75$. | Answer: $x \in(1 ; 3] \cup\left\{5^{\log _{3} 5}\right\}$.
Solution. Note that $x \sqrt{\sqrt{\log _{x} 3}}=\left(3^{\log _{3} x}\right)^{\sqrt{\log _{x} 3}}=3^{\sqrt{\log _{3}^{2} x \cdot \log _{x} 3}}=3 \sqrt{\log _{3} x}$. Therefore, the inequality can be rewritten as $3^{3 \sqrt{\log _{3} x}}-13 \cdot 3^{2 \sqrt{\... | x\in(1;3]\cup{5^{\log_{3}5}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,576 |
4. [5 points] a) A sphere with center $O$ touches the lateral edges $S A, S B, S C$ of the pyramid $S A B C$ at points $K, L, M$ respectively, and also touches its base $A B C$. A plane is drawn through the point on the sphere closest to point $S$, tangent to the sphere. The area of the section of the pyramid $S A B C$... | Answer: a) $S_{K L M}=12.25 ;$ b) $V_{S A B C}=\frac{343}{2}$.
Fig. 7: Variant 15, Problem 4
Solution. a) Let the radius of the sphere be $R$. Denote the points of intersection of the line... | 12.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,577 |
5. [6 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
y=-|x-\sqrt{a}|+2-\sqrt{a} \\
(|x|-12)^{2}+(|y|-5)^{2}=169
\end{array}\right.
$$
has exactly three solutions. | Answer: $a \in\left\{1 ; 36 ;\left(\frac{13 \sqrt{2}-5}{2}\right)^{2}\right\}$.
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 ... | \in{1;36;(\frac{13\sqrt{2}-5}{2})^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,578 |
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: a) $\frac{R_{2}}{R_{1}}=\frac{7}{4}$; b) $R_{1}=\frac{4}{\sqrt{7}}, R_{2}=\sqrt{7}$.
Solution. a) Let $R_{1}, R_{2}$ be the radii of circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B O_{2}=\alpha$, and the lines $D O_{2}$ and $\ell_{2}$ intersect at point $P$. Then, from the condition of tangenc... | \frac{R_{2}}{R_{1}}=\frac{7}{4};R_{1}=\frac{4}{\sqrt{7}},R_{2}=\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,579 |
1. [3 points] A coin is flipped 80 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear more than 51 times, and $q$ be the probability that heads will appear no more than 29 times. Find $p-q$ | Answer: $-\frac{1}{2^{80}} \cdot C_{80}^{29}$.
Solution. Since the probability of getting heads or tails is equal, the probability of getting 80 heads is the same as the probability of getting 80 tails (i.e., 0 heads); the probability of getting 79 heads is the same as the probability of getting 79 tails (i.e., one he... | -\frac{1}{2^{80}}\cdotC_{80}^{29} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,580 |
2. [5 points] Solve the equation $\frac{\cos 4 x}{\cos 5 x - \sin 5 x} + \frac{\sin 4 x}{\cos 5 x + \sin 5 x} = -\sqrt{2}$. | Answer: $x=\frac{5 \pi}{76}+\frac{2 \pi k}{19}, k \neq 19 p-3, k \in \mathbb{Z}, p \in \mathbb{Z}$.
Solution. By bringing the fractions in the left side of the equation to a common denominator, we get
$$
\frac{\cos 4 x \cos 5 x+\cos 4 x \sin 5 x+\cos 5 x \sin 4 x-\sin 5 x \sin 4 x}{\cos ^{2} 5 x-\sin ^{2} 5 x}=-\sqrt... | \frac{5\pi}{76}+\frac{2\pik}{19},k\neq19p-3,k\in\mathbb{Z},p\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,581 |
3. [5 points] Solve the inequality $8^{\sqrt{\log _{2} x}}-2^{\sqrt{4 \log _{2} x}+3}+21 \cdot x^{\sqrt{\log _{x} 2}} \leqslant 18$. | Answer: $x \in(1 ; 2] \cup\left\{3^{\log _{2} 3}\right\}$.
Solution. Note that $x^{\sqrt{\log _{x} 2}}=\left(2^{\log _{2} x}\right)^{\sqrt{\log _{x} 2}}=2^{\sqrt{\log _{2}^{2} x \cdot \log _{x} 2}}=2^{\sqrt{\log _{2} x}}$. Therefore, the inequality can be rewritten as $2^{3 \sqrt{\log _{2} x}}-8 \cdot 2^{2 \sqrt{\log ... | x\in(1;2]\cup{3^{\log_{2}3}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,582 |
4. [5 points] a) A sphere with center $O$ touches the lateral edges $S A, S B, S C$ of the pyramid $S A B C$ at points $K, L, M$ respectively, and also touches its base $A B C$. A plane is drawn through the point on the sphere closest to point $S$, tangent to the sphere. The area of the section of the pyramid $S A B C$... | Answer: a) $S_{K L M}=6.25 ;$ b) $V_{S A B C}=\frac{250}{3}$.
Fig. 10: Variant 16, Problem 4
Solution. a) Let the radius of the sphere be $R$. Denote the points of intersection of the line... | S_{KLM}=6.25;V_{SABC}=\frac{250}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,583 |
5. [6 points] Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
x=-|y-\sqrt{a}|+6-\sqrt{a} \\
(|x|-8)^{2}+(|y|-15)^{2}=289
\end{array}\right.
$$
has exactly three solutions. | Answer: $a \in\left\{9 ; 121 ;\left(\frac{17 \sqrt{2}-1}{2}\right)^{2}\right\}$.
Fig. 11: Variant 16, Problem 5
Consider the second equation of the system. It is invariant under the subs... | \in{9;121;(\frac{17\sqrt{2}-1}{2})^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,584 |
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: a) $\frac{R_{2}}{R_{1}}=\frac{9}{5}$; b) $R_{1}=\frac{5 \sqrt{5}}{6}, R_{2}=\frac{3 \sqrt{5}}{2}$.
Solution. a) Let $R_{1}, R_{2}$ be the radii of circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B O_{2}=\alpha$, and the lines $D O_{2}$ and $\ell_{2}$ intersect at point $P$. Then, from the condit... | \frac{R_{2}}{R_{1}}=\frac{9}{5};R_{1}=\frac{5\sqrt{5}}{6},R_{2}=\frac{3\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,585 |
1. Solve the inequality $\log _{\frac{x^{2}-2}{2 x-3}}\left(\frac{\left(x^{2}-2\right)(2 x-3)}{4}\right) \geq 1$. | Answer: $x \in\left[\frac{1}{2} ; 1\right) \cup(1 ; \sqrt{2}) \cup\left[\frac{5}{2} ;+\infty\right)$.
Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-2}{2 x-3}>0, \frac{x^{2}-2}{2 x-3} \neq 1$ (then the expression under the logarithm is also positive), from which $x \in(-\sqrt{2} ; \sq... | x\in[\frac{1}{2};1)\cup(1;\sqrt{2})\cup[\frac{5}{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,586 |
2. Solve the equation $(\cos x-3 \cos 4 x)^{2}=16+\sin ^{2} 3 x$. | Answer: $x=\pi+2 k \pi, k \in \mathrm{Z}$.
Solution. Note that for any $x$, the inequality $-4 \leq \cos x-3 \cos 4 x \leq 4$ holds, which implies that the left side of the equation does not exceed 16. At the same time, the right side of the equation is not less than 16. Therefore, equality can only be achieved when b... | \pi+2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,587 |
3. Solve the system of equations $\left\{\begin{array}{l}x+\sqrt{x+2 y}-2 y=\frac{7}{2}, \\ x^{2}+x+2 y-4 y^{2}=\frac{27}{2}\end{array}\right.$. | Answer: $\left(\frac{19}{4} ; \frac{17}{8}\right)$.
Solution. Let $\sqrt{x+2 y}=u, x-2 y=v$. Then the system becomes
$$
\left\{\begin{array} { l }
{ u + v = \frac { 7 } { 2 } } \\
{ u ^ { 2 } v + u ^ { 2 } = \frac { 2 7 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=\frac{7}{2}-u \\
u^{2}\left(\frac... | (\frac{19}{4};\frac{17}{8}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,588 |
4. Points $A, B, C, D, E$ are sequentially located on a line, such that $A B=B C=2, C D=1, D E=3$. Circles $\Omega$ and $\omega$, touching each other, are such that $\Omega$ passes through points $A$ and $E$, and $\omega$ passes through points $B$ and $C$. Find the radii of circles $\Omega$ and $\omega$, given that the... | Answer: $R=8 \sqrt{\frac{3}{11}}, r=5 \sqrt{\frac{3}{11}}$.
Solution. Let the centers of circles $\Omega$ and $\omega$ be $O$ and $Q$ respectively. Since $C$ is the midpoint of chord $A E$ of circle $\Omega$, segment $O C$ is perpendicular to $O E$. Drop a perpendicular $Q H$ from point $Q$ to line $A E$. Then $B H=H ... | R=8\sqrt{\frac{3}{11}},r=5\sqrt{\frac{3}{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,589 |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can be repeated) so that the resulting 12-digit number is divisible by 45. In how many ways can this be done | Answer: 13122.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by nine, we proceed as follows. We will choose four digits arb... | 13122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,590 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
(|y+9|+|x+2|-2)\left(x^{2}+y^{2}-3\right)=0 \\
(x+2)^{2}+(y+4)^{2}=a
\end{array}\right.
$$
has exactly three solutions. | Answer: $a=9, a=23+4 \sqrt{15}$.
Solution. The first equation of the given system is equivalent to the combination of two equations $|y+9|+|x+2|=2$ and $x^{2}+y^{2}=3$. The first of these defines a square $G$ with center $(-2, -9)$, the diagonals of which are 4 and parallel to the coordinate axes. The second defines a... | =9,=23+4\sqrt{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,591 |
7. Given a regular prism $A B C D A_{1} B_{1} C_{1} D_{1}$ with base $A B C D$. Planes $\alpha$ and $\beta$ are perpendicular to $B_{1} D$ and pass through vertices $A$ and $D_{1}$, respectively. Let $F$ and $H$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $B_{1} D$, such that $D F < D... | Answer: a) $2: 1$, b) $B_{1} D=3(1+\sqrt{13}), V=108 \sqrt{6+2 \sqrt{13}}$.
Solution. a) By symmetry (with respect to the plane $B D D_{1} B_{1}$), the plane $\alpha$ passes through the point $C$ - and, therefore, through the center $O$ of the face $A B C D$. Segments $B_{1} H$ and $D F$ are projections of parallel se... | )2:1,b)B_{1}D=3(1+\sqrt{13}),V=108\sqrt{6+2\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,592 |
1. Solve the inequality $\log _{\frac{x^{2}-3}{6 x-12}}\left(\frac{\left(x^{2}-3\right)(6 x-12)}{25}\right) \geq 1$. | Answer: $x \in\left[\frac{7}{6} ; \sqrt{3}\right) \cup\left[\frac{17}{6} ; 3\right) \cup(3 ;+\infty)$.
Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-3}{6 x-12}>0, \frac{x^{2}-3}{6 x-12} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{3} ; \sqrt{3}... | x\in[\frac{7}{6};\sqrt{3})\cup[\frac{17}{6};3)\cup(3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,593 |
2. Solve the equation $(\cos 2 x-2 \cos 4 x)^{2}=9+\cos ^{2} 5 x$. | Answer: $x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$.
Solution. Note that for any $x$, the inequality $-3 \leq \cos x-3 \cos 4 x \leq 3$ holds, which implies that the left side of the equation does not exceed 9. At the same time, the right side of the equation is not less than 9. Therefore, equality can only be achieved ... | \frac{\pi}{2}+k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,594 |
3. Solve the system of equations $\left\{\begin{array}{l}y+\sqrt{y-3 x}+3 x=12, \\ y^{2}+y-3 x-9 x^{2}=144 .\end{array}\right.$ | Answer: $(-24 ; 72),\left(-\frac{4}{3} ; 12\right)$.
Solution. Let $\sqrt{y-3 x}=u, y+3 x=v$. Then the system takes the form
$$
\left\{\begin{array} { l }
{ u + v = 1 2 } \\
{ u ^ { 2 } v + u ^ { 2 } = 1 4 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=12-u \\
u^{2}(12-u)+u^{2}=144
\end{array}\right.\righ... | (-24,72),(-\frac{4}{3},12) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,595 |
5. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 18. In how many ways can this be done? | Answer: 3645.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6$ or 8 (5 ways).
To ensure divisibility by nine, we proceed as follows. Choose three digits ... | 3645 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,597 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
(|y-4|+|x+12|-3)\left(x^{2}+y^{2}-12\right)=0 \\
(x+5)^{2}+(y-4)^{2}=a
\end{array}\right.
$$
has exactly three solutions. | Answer: $a=16, a=53+4 \sqrt{123}$.
Solution. The first equation of the given system is equivalent to the combination of two equations $|y-4|+|x+12|=3$ and $x^{2}+y^{2}=12$. The first of these defines a square $G$ with center at $(-12; 4)$, the diagonals of which are 6 and parallel to the coordinate axes. The second de... | =16,=53+4\sqrt{123} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,598 |
7. Given a regular prism $K L M N K_{1} L_{1} M_{1} N_{1}$ with base $K L M N$. Planes $\alpha$ and $\beta$ are perpendicular to $L_{1} N$ and pass through vertices $K$ and $N_{1}$, respectively. Let $A$ and $B$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $L_{1} N$, such that $A N < B... | Answer: a) $2: 1$, b) $L_{1} N=\frac{1}{2}(1+\sqrt{13}), V=\frac{1}{2} \sqrt{6+2 \sqrt{13}}$.
Solution. a) By symmetry (with respect to the plane $L N N_{1} L_{1}$), the plane $\alpha$ passes through the point $M$ - and, therefore, through the center $O$ of the face $K L M N$. The segments $L_{1} B$ and $A N$ are proj... | )2:1,b)L_{1}N=\frac{1}{2}(1+\sqrt{13}),V=\frac{1}{2}\sqrt{6+2\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,599 |
1. Solve the inequality $\log _{\frac{x^{2}-5}{2 x-6}}\left(\frac{\left(x^{2}-5\right)(2 x-6)}{25}\right) \geq 1$. | Answer: $x \in\left[\frac{1}{2} ; 1\right) \cup(1 ; \sqrt{5}) \cup\left[\frac{11}{2} ;+\infty\right)$.
Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-5}{2 x-6}>0, \frac{x^{2}-5}{2 x-6} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{5} ; \sqrt{5}) ... | x\in[\frac{1}{2};1)\cup(1;\sqrt{5})\cup[\frac{11}{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,600 |
2. Solve the equation $(\cos x+2 \cos 6 x)^{2}=9+\sin ^{2} 3 x$. | Answer: $x=2 k \pi, k \in \mathrm{Z}$.
Solution. Note that for any $x$, the inequality $-3 \leq \cos x+2 \cos 6 x \leq 3$ holds, which implies that the left side of the equation does not exceed 9. At the same time, the right side of the equation is not less than 9. Therefore, equality can only be achieved when both co... | 2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,601 |
3. Solve the system of equations $\left\{\begin{array}{l}2 x+\sqrt{2 x+3 y}-3 y=5, \\ 4 x^{2}+2 x+3 y-9 y^{2}=32 .\end{array}\right.$ | Answer: $\left(\frac{17}{4} ; \frac{5}{2}\right)$.
Solution. Let $\sqrt{2 x+3 y}=u, 2 x-3 y=v$. Then the system becomes
$$
\left\{\begin{array} { l }
{ u + v = 5 } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=5-u \\
u^{2}(5-u)+u^{2}=32
\end{array}\right.\right.
$$
From ... | (\frac{17}{4};\frac{5}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,602 |
4. Points $A, B, C, D, E$ are sequentially located on a line, such that $A B=C D=1, B C=D E=2$. Circles $\Omega$ and $\omega$, touching each other, are such that $\Omega$ passes through points $A$ and $E$, and $\omega$ passes through points $B$ and $C$. Find the radii of circles $\Omega$ and $\omega$, given that their ... | Answer: $R=\frac{27}{2 \sqrt{19}}, r=\frac{8}{\sqrt{19}}$.[^2]
“Fiztekh-2016”, mathematics, solutions to ticket 11
Solution. Let the centers of the circles $\Omega$ and $\omega$ be denoted by $O$ and $Q$ respectively. Since $C$ is the midpoint of the chord $A E$ of the circle $\Omega$, the segment $O C$ is perpendicu... | R=\frac{27}{2\sqrt{19}},r=\frac{8}{\sqrt{19}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,603 |
5. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 45. In how many ways can this be done? | Answer: 1458.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by nine, we proceed as follows. We select three digits arbitrarily (this can be done in $9... | 1458 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,604 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
(|y-10|+|x+3|-2)\left(x^{2}+y^{2}-6\right)=0 \\
(x+3)^{2}+(y-5)^{2}=a
\end{array}\right.
$$
has exactly three solutions. | Answer: $a=49, a=40-4 \sqrt{51}$.
Solution. The first equation of the given system is equivalent to the combination of two equations $|y-10|+|x+3|=2$ and $x^{2}+y^{2}=6$. The first of these defines a square $G$ with center at $(-3, 10)$, the diagonals of which are 4 and parallel to the coordinate axes. The second defi... | =49,=40-4\sqrt{51} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,605 |
7. Given a regular prism $A B C D A_{1} B_{1} C_{1} D_{1}$ with base $A B C D$. Planes $\alpha$ and $\beta$ are perpendicular to $B_{1} D$ and pass through vertices $A$ and $D_{1}$, respectively. Let $F$ and $H$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $B_{1} D$, such that $D F < D... | Answer: a) $2: 1$, b) $B_{1} D=\frac{3}{2}(1+\sqrt{13}), V=\frac{27}{2} \sqrt{6+2 \sqrt{13}}$.
Solution. a) By symmetry (with respect to the plane $B D D_{1} B_{1}$), the plane $\alpha$ passes through the point $C$ - and, therefore, through the center $O$ of the face $A B C D$. Segments $B_{1} H$ and $D F$ are project... | )2:1,b)B_{1}D=\frac{3}{2}(1+\sqrt{13}),V=\frac{27}{2}\sqrt{6+2\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,606 |
1. Solve the inequality $\log _{\frac{x^{2}-6}{4 x-10}}\left(\frac{\left(x^{2}-6\right)(4 x-10)}{9}\right) \geq 1$. | Answer: $x \in\left[\frac{7}{4} ; 2\right) \cup(2 ; \sqrt{6}) \cup\left[\frac{13}{4} ;+\infty\right)$.
Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-6}{4 x-10}>0, \frac{x^{2}-6}{4 x-10} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{6} ; \sqrt{6}... | x\in[\frac{7}{4};2)\cup(2;\sqrt{6})\cup[\frac{13}{4};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,607 |
2. Solve the equation $(\cos 2 x-3 \cos 4 x)^{2}=16+\cos ^{2} 5 x$. | Answer: $x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$.
Solution. Note that for any $x$, the inequality $-4 \leq \cos 2 x-3 \cos 4 x \leq 4$ holds, which implies that the left side of the equation does not exceed 16. At the same time, the right side of the equation is no less than 16. Therefore, equality can only be achiev... | \frac{\pi}{2}+k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,608 |
3. Solve the system of equations $\left\{\begin{array}{l}3 x+\sqrt{3 x-y}+y=6 \\ 9 x^{2}+3 x-y-y^{2}=36 .\end{array}\right.$ | Answer: $(2, -3), (6, -18)$.
Solution. Let $\sqrt{3 x-y}=u, 3 x+y=v$. Then the system becomes
$$
\left\{\begin{array} { l }
{ u + v = 6 } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=6-u \\
u^{2}(6-u)+u^{2}=36
\end{array}\right.\right.
$$
From the second equation of the... | (2,-3),(6,-18) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,609 |
4. Points $A, B, C, D, E$ are sequentially located on a line, such that $A B=4, B C=D E=2, C D=3$. Circles $\Omega$ and $\omega$, touching each other, are such that $\Omega$ passes through points $D$ and $E$, and $\omega$ passes through points $B$ and $C$. Find the radii of circles $\Omega$ and $\omega$, given that the... | Answer: $r=\frac{9 \sqrt{3}}{2 \sqrt{17}}, R=\frac{8 \sqrt{3}}{\sqrt{17}}$.
Solution. Let the centers of circles $\Omega$ and $\omega$ be $O$ and $Q$ respectively. Drop perpendiculars $OH$ and $QT$ from points $O$ and $Q$ to line $AB$, then points $H$ and $T$ are the midpoints of chords $DE$ and $BC$ respectively (a d... | r=\frac{9\sqrt{3}}{2\sqrt{17}},R=\frac{8\sqrt{3}}{\sqrt{17}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,610 |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $1,2,3,4,5,6,7,8,9$ (digits can repeat) so that the resulting 12-digit number is divisible by 18. In how many ways can this be done | Answer: 26244.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $2, 4, 6$ or 8 (4 ways).
To ensure divisibility by nine, we proceed as follows. We select four digits ... | 26244 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,611 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
(|y+2|+|x-11|-3)\left(x^{2}+y^{2}-13\right)=0 \\
(x-5)^{2}+(y+2)^{2}=a
\end{array}\right.
$$
has exactly three solutions. | Answer: $a=9, a=42+2 \sqrt{377}$.
Solution. The first equation of the given system is equivalent to the combination of two equations $|y+2|+|x-11|=3$ and $x^{2}+y^{2}=13$. The first of these equations defines a square $G$ with center at $(11, -2)$, the diagonals of which are 6 and parallel to the coordinate axes. The ... | =9,=42+2\sqrt{377} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,612 |
7. Given a regular prism $K L M N K_{1} L_{1} M_{1} N_{1}$ with base $K L M N$. Planes $\alpha$ and $\beta$ are perpendicular to $L_{1} N$ and pass through vertices $K$ and $N_{1}$, respectively. Let $A$ and $B$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $L_{1} N$, such that $A N < B... | Answer: a) $2: 1$, b) $L_{1} N=2(1+\sqrt{13}), V=32 \sqrt{6+2 \sqrt{13}}$.
Solution. a) By symmetry (with respect to the plane $L N N_{1} L_{1}$), the plane $\alpha$ passes through the point $M$ - and, therefore, through the center $O$ of the face $K L M N$. The segments $L_{1} B$ and $A N$ are projections of the para... | L_{1}N=2(1+\sqrt{13}),V=32\sqrt{6+2\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,613 |
1. Solve the equation $\frac{2 \sin 3 x}{\sin x}-\frac{\cos 3 x}{\cos x}=5|\cos x|$. | Answer: $x= \pm \arccos \frac{1}{4}+k \pi, k \in Z$.
Solution. The given equation, under the condition $\cos x \sin x \neq 0$, is equivalent to each of the following:
$$
\frac{2\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=5|\cos x| \Leftrightarrow 6-8 \sin ^{2} x-4 \... | \\arccos\frac{1}{4}+k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,614 |
2. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}y-2 \sqrt{x y}-\sqrt{\frac{y}{x}}+2=0, \\ 3 x^{2} y^{2}+y^{4}=84 .\end{array}\right.$ | Answer: $\left(\frac{1}{3} ; 3\right),\left(\sqrt[4]{\frac{21}{76}} ; 2 \cdot 4 \sqrt{\frac{84}{19}}\right)$.
Solution. Let $\sqrt{\frac{y}{x}}=u, \quad \sqrt{x y}=v \quad$ (with $\quad u>0, \quad v>0$ ). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x... | (\frac{1}{3};3),(\sqrt[4]{\frac{21}{76}};2\cdot4\sqrt{\frac{84}{19}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,615 |
3. Solve the inequality $17^{\frac{5 x-3}{3-x}} \cdot 2^{3-x} \leq 68$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Answer: $x \in\left[3-6 \log _{2} 17 ; 1\right] \cup(3 ;+\infty)$.
Solution. By taking the logarithm of both sides of the inequality with base 2, we get:
$$
\frac{5 x-3}{3-x} \log _{2} 17+(3-x) \leq 2+\log _{2} 17 \Leftrightarrow \frac{6 x-6}{3-x} \log _{2} 17+(1-x) \leq 0 \Leftrightarrow \frac{(x-1)\left(6 \log _{2}... | x\in[3-6\log_{2}17;1]\cup(3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,616 |
4. Circle $\omega$ with radius 6 and center $O$ is inscribed in an acute-angled triangle $C F M$ and touches its sides $C M$ and $F M$ at points $P$ and $K$ respectively. Circle $\Omega$ with radius $\frac{5 \sqrt{13}}{2}$ and center $T$ is circumscribed around triangle $P K M$.
a) Find $O M$.
b) Suppose it is additi... | Answer: a) $O M=5 \sqrt{13}$, b) $M A=\frac{20 \sqrt{13}}{3}, S=204$.
Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O K M$ and $O P M$ are right angles, i.e., from points $K$ and $P$, the segment $O M$ is seen at a right angle. Therefore, the circle constructed on t... | )OM=5\sqrt{13},b)MA=\frac{20\sqrt{13}}{3},S=204 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,617 |
5. In the number $2016^{* * * *} 02 * *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 15. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by three, we proceed as follows. Choose four digits arbitrarily (this can be done in $6 \c... | 5184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,618 |
6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}16+6 x-x^{2}-y^{2}|+| 6 x \mid=16+12 x-x^{2}-y^{2} \\ (a+15) y+15 x-a=0 .\end{array}\right.\end{array}\right.$
a) Plot on the plane $(x ; y)$ the set of points satisfying the first equation of the system, and find the area of the re... | Answer: a) $25 \pi-25 \arcsin 0.8+12$, b) $a=-20, a=-12$.
Solution. a) Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system ... | )25\pi-25\arcsin0.8+12,b)=-20,=-12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,619 |
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $B C$ intersects the edges $A C$ and $A B$ at points $P$ and $Q$, respectively, different from the vertices of the prism. Segments $B_{1} P$ and $C_{1} Q$ intersect at point $T$, and $B_{1} P=5, T Q=2$.
a) Find the angle $T P A$.
b) F... | Answer: a) $90^{\circ}$, b) $2: 1$, c) $V=15$.
Solution. a) Points $P$ and $Q$ lie on the circle with diameter $BC$; therefore, $\angle BPC=90^{\circ}, \angle BQC=90^{\circ}$ (i.e., $BP$ and $CQ$ are altitudes of triangle $ABC$). The line $BP$ is the projection of the line $TP$ onto the base plane, and since $BP \perp... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,620 |
1. Solve the equation $\frac{\sin 3 x}{\sin x}-\frac{2 \cos 3 x}{\cos x}=5|\sin x|$. | Answer: $x= \pm \arcsin \frac{1}{4}+k \pi, k \in Z$.
Solution. The given equation, under the condition $\cos x \sin x \neq 0$, is equivalent to each of the following:
$$
\frac{\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{2\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=5|\sin x| \Leftrightarrow 3-4 \sin ^{2} x-8 \... | \\arcsin\frac{1}{4}+k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,621 |
3. Solve the inequality $5^{\frac{x+5}{x+4}} \cdot 3^{x+4} \geq 75$. | Answer: $x \in(-4 ;-3] \cup\left[\log _{3} 5-4 ;+\infty\right)$.
Solution. Taking the logarithm of both sides of the inequality with base 3, we get:
$$
\frac{x+5}{x+4} \log _{3} 5+(x+4) \geq 1+2 \log _{3} 5 \Leftrightarrow \frac{-x-3}{x+4} \log _{3} 5+(x+3) \geq 0 \Leftrightarrow \frac{(x+3)\left(-\log _{3} 5+x+4\rig... | x\in(-4;-3]\cup[\log_{3}5-4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,623 |
4. Circle $\omega$ with radius 4 and center $O$ is inscribed in an acute-angled triangle $E F Q$ and touches its sides $F Q$ and $E Q$ at points $M$ and $P$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{65}}{2}$ and center $T$ is circumscribed around triangle $P Q M$.
a) Find $O Q$.
b) Suppose it is addition... | Answer: a) $O Q=\sqrt{65}$, b) $Q A=\frac{3 \sqrt{65}}{2}, S=84$.
Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O M Q$ and $O P Q$ are right angles, i.e., from points $M$ and $P$, the segment $O Q$ is seen at a right angle. Therefore, the circle constructed on the s... | OQ=\sqrt{65},QA=\frac{3\sqrt{65}}{2},S=84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,624 |
5. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,625 |
6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}9+8 y-x^{2}-y^{2}|+| 8 y \mid=16 y+9-x^{2}-y^{2} \\ (a+4) x-13 y+a=0 .\end{array}\right.\end{array}\right.$
a) Plot on the plane $(x ; y)$ the set of points satisfying the first equation of the system, and find the area of the resul... | Answer: a) $25 \pi-25 \arcsin 0.6+12$, b) $a=-6, a=-3$.
Solution. a) Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of ineq... | )25\pi-25\arcsin0.6+12,b)=-6,=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,626 |
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A C$ intersects the edges $A B$ and $B C$ at points $F$ and $N$, respectively, different from the vertices of the prism. Segments $C_{1} F$ and $A_{1} N$ intersect at point $P$, and at the same time $A_{1} N=7, C_{1} P=6$.
a) Find the... | Answer: a) $90^{\circ}$, b) $5: 1$, c) $V=21 \sqrt{10}$.
Solution. a) Points $F$ and $N$ lie on the circle with diameter $A C$; therefore, $\angle A F C=90^{\circ}, \angle A N C=90^{\circ}$ (i.e., $A N$ and $C F$ are altitudes of triangle $A B C$). The line $C F$ is the projection of the line $C_{1} F$ onto the base p... | 21\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,627 |
1. Solve the equation $\frac{3 \sin 3 x}{\sin x}-\frac{2 \cos 3 x}{\cos x}=7|\cos x|$. | Answer: $x= \pm \arccos \frac{3}{4}+k \pi, k \in Z$.
Solution. The given equation, under the condition $\cos x \sin x \neq 0$, is equivalent to each of the following:
$$
\frac{3\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{2\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=7|\cos x| \Leftrightarrow 9-12 \sin ^{2} x-8... | \\arccos\frac{3}{4}+k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,628 |
2. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}2 x-\sqrt{x y}-4 \sqrt{\frac{x}{y}}+2=0, \\ 2 x^{2}+x^{2} y^{4}=18 y^{2} .\end{array}\right.$ Answer: $(2 ; 2),\left(\frac{\sqrt[4]{286}}{4} ; \sqrt[4]{286}\right)$. | Solution. Let $\sqrt{\frac{x}{y}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{x}{y}} \cdot \sqrt{x y}=\sqrt{x^{2}}=|x|=x$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{x}{y}}=\sqrt{y^{2}}=|y|=y$, since by condition $x$ and $y$ are positive. The system takes the form
$$
\left\{\begin{array} { l ... | (2;2),(\frac{\sqrt[4]{286}}{4};\sqrt[4]{286}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,629 |
3. Solve the inequality $11^{\frac{3 x-4}{4-x}} \cdot 3^{4-x} \leq 99$. | Answer: $x \in\left[4-4 \log _{3} 11 ; 2\right] \cup(4 ;+\infty)$.
Solution. By taking the logarithm of both sides of the inequality with base 3, we get:
$$
\frac{3 x-4}{4-x} \log _{3} 11+(4-x) \leq 2+\log _{3} 11 \Leftrightarrow \frac{4 x-8}{4-x} \log _{3} 11+(2-x) \leq 0 \Leftrightarrow \frac{(x-2)\left(4 \log _{3}... | x\in[4-4\log_{3}11;2]\cup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,630 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.