problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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3. 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 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,909 |
4. Find all values of the parameter $a$, for each of which the equation $a|2-x|+\frac{x^{2}-x-6}{3-x}=0$ has exactly one solution | Answer: $a \in(-1 ; 1] \cup\{5\}$.
Solution. Given the condition $x \neq 3$, the equation is equivalent to $a|x-2|=x+2$. The graph of the right side of the equation is the line $y=x+2$. The graph of the left side of the equation is a "V" shape with its vertex at the point $(2 ; 0)$, and the slope of its branches is de... | \in(-1;1]\cup{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,910 |
5. 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
$$
\left\{\begin{array} { l }
{ u + v = 5 , } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=5-u \\
u^{2}(5-u)+u^{2}=32
\end{array}\right.\right.
$$
"Ph... | (\frac{17}{4};\frac{5}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,911 |
6. Point $A$ lies on side $L M$ of triangle $K L M$ with an angle of $120^{\circ}$ at vertex $K$. Incircles are inscribed in triangles $A K L$ and $A K M$ with centers $F$ and $O$ respectively. Find the radius of the circumcircle of triangle $F K O$, if $A O=2, A F=7$. | Answer: $\sqrt{\frac{53}{3}}$.
Solution. The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $A F$ and $A O$ are the bisectors of angles $L A K$ and $M A K$. Since the angle between the bisectors of adjacent angles is a right angle, $\angle F A O=90^{\circ}$, and then by the Pyth... | \sqrt{\frac{53}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,912 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{4}+5 x+39$. | Answer: 12.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+5 x+39>0 \Leftrightarrow-\frac{1}{4}(x+6)(x-26)>0$, from which $-6<x<26$. On this interval, there are 25 natural values of $x: x=1, x=2, \ldots, x=25$. In this interval, $y$ takes integer values only for even $x$ - a total ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,913 |
2. Solve the inequality $\frac{\sqrt{1-x}-12}{1-\sqrt{2-x}} \geq 1+\sqrt{2-x}$. | Answer: $x \in[-8 ; 1)$.
Solution. The domain of the inequality is determined by the conditions $x \leq 1,1-\sqrt{2-x} \neq 0$, from which we get that $x<1$. Note that on the domain, the denominator of the fraction is negative, so we can multiply both sides of the inequality by it, changing the sign of the inequality.... | x\in[-8;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 2,914 |
3. 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. Choose four digits arb... | 26244 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,915 |
4. Find all values of the parameter $a$, for each of which the equation $a|2+x|+\frac{x^{2}+x-12}{x+4}=0$ has exactly one solution | Answer: $a \in(-1 ; 1] \cup\left\{\frac{7}{2}\right\}$.
Solution. Given the condition $x \neq-4$, the equation is equivalent to $a|x+2|=3-x$. The graph of the right side of the equation is the line $y=x-3$. The graph of the left side of the equation is a "V" shape with its vertex at the point $(-2 ; 0)$, and the slope... | \in(-1;1]\cup{\frac{7}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,916 |
5. 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 takes the form
“Phystech-2016”, mathematics, solution to ticket 4
$$
\left\{\begin{array}{l}
u + v = 6, \\
u^2 v + u^2 = 36
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=6-u, \\
u^2(6-u)+u^2=36
\end{array}\right.\right.
$$
... | (2,-3),(6,-18) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,917 |
6. Point $N$ lies on side $DE$ of triangle $CDE$ with an angle of $60^{\circ}$ at vertex $C$. Incircles are inscribed in triangles $CNE$ and $CDE$ with centers $K$ and $P$ respectively. Find the radius of the circumcircle of triangle $CKP$, if $KN=8, NP=7$. | Answer: $\sqrt{113}$.
Solution. The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $N K$ and $N P$ are the bisectors of angles $E N C$ and $D N C$. Since the angle between the bisectors of adjacent angles is right, then $\angle K N P=90^{\circ}$, and then by the Pythagorean theo... | \sqrt{113} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,918 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+13 x+42$. | Answer: 13.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+13 x+42>0 \Leftrightarrow-\frac{1}{3}(x+3)(x-42)>0$, from which $-3<x<42$. On this interval, there are 41 natural values of $x: x=1, x=2, \ldots, x=41$. In this case, $y$ takes integer values only when $x$ is divisible by... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,919 |
2. Find the value of the expression $\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are the largest and smallest roots of the equation $x^{3}-7 x^{2}+7 x=1$, respectively. | Answer: 34.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-7\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-7 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-6 x+1\right)=0,
$$
from which $x=1$ or $x=3 \pm \sqrt{8}$. The largest root is $a=3+\sqrt{8}$, and the smallest... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,920 |
3. 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 from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. Choose four digits arbitrarily... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,921 |
4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(x-|x|)^{2}+(y-|y|)^{2} \leq 4, \\ y+2 x \leq 0\end{array}\right.$ and find the area of the resulting figure. | Answer: $\frac{5+\pi}{4}$.
Solution. Consider the first inequality. There are four possible cases.
1) $x \geq 0, y \geq 0$ (first quadrant). Then $0 \leq 4$, the inequality is satisfied by all points in the first quadrant.
2) $x<0, y \geq 0$ (second quadrant). Then $4 x^{2} \leq 4,|x| \leq 1$; in this case $x$ is neg... | \frac{5+\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,922 |
5. 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: $\left(\frac{1}{3} ; 3\right),\left(\sqrt[4]{\frac{21}{76}} ; 2 \cdot \sqrt[4]{\frac{84}{19}}\right)$. | Solution. Let $\sqrt{\frac{y}{x}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system takes the form
$$
\left\{\begin{array}{... | (\frac{1}{3};3),(\sqrt[4]{\frac{21}{76}};2\cdot\sqrt[4]{\frac{84}{19}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,923 |
6. A circle passes through vertices $A$ and $C$ of triangle $ABC$ and intersects its sides $AB$ and $BC$ at points $K$ and $T$ respectively, such that $AK: KB = 3: 2$ and $BT: TC = 1: 2$. Find $AC$, if $KT = \sqrt{6}$. | Answer: $3 \sqrt{5}$.
Solution. Let $B K=2 x, B T=y$; then $A K=3 x, C T=2 y$. By the theorem of two secants $B K \cdot B A=B T \cdot B C$, from which $2 x \cdot 5 x=y \cdot 3 y, y=x \sqrt{\frac{10}{3}}$. Triangles $A B C$ and $T B K$ are similar by two sides and the angle between them ( $B A: B T=B C: B K, \angle B-$... | 3\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,924 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+7 x+54$. | Answer: 8.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+7 x+54>0 \Leftrightarrow-\frac{1}{3}(x+6)(x-27)>0$, from which $-6<x<27$. On this interval, there are 26 natural values of $x: x=1, x=2, \ldots, x=26$. In this interval, $y$ takes integer values only when $x$ is divisible ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,925 |
2. Find the value of the expression $\frac{p}{q}+\frac{q}{p}$, where $p$ and $q$ are the largest and smallest roots of the equation $x^{3}+6 x^{2}+6 x=-1$, respectively. | Answer: 23.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}+1\right)+6\left(x^{2}+x\right)=0 \Leftrightarrow(x+1)\left(x^{2}-x+1\right)+6 x(x+1)=0 \Leftrightarrow(x+1)\left(x^{2}+5 x+1\right)=0 \text {, }
$$
from which $x=-1$ or $x=\frac{-5 \pm \sqrt{21}}{2}$. The largest root is $p=\frac{... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,926 |
3. 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 | 2,927 |
4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(|x|-x)^{2}+(|y|-y)^{2} \leq 16, \\ 2 y+x \leq 0\end{array}\right.$, and find the area of the resulting figure. | Answer: $5+\pi$.
Solution. Consider the first inequality. There are four possible cases.
1) $x \geq 0, y \geq 0$ (first quadrant). Then $0 \leq 16$, the inequality is satisfied by all points in the first quadrant. In this case, $x$ is positive, so $x \leq 2$.
2) $x<0, y \geq 0$ (second quadrant). Then $4 x^{2} \leq 1... | 5+\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,928 |
5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}x-3 \sqrt{x y}-2 \sqrt{\frac{x}{y}}+6=0, \\ x^{2} y^{2}+x^{4}=82 .\end{array}\right.$ Answer: $\left(3 ; \frac{1}{3}\right),\left(\sqrt[4]{66} ; \frac{4}{\sqrt[4]{66}}\right)$ | Solution. Let $\sqrt{\frac{x}{y}}=u, \quad \sqrt{x y}=v \quad$ (with $\quad 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 $x$ and $y$ are positive by condition. The system takes the form
$$
\left\{\begi... | (3;\frac{1}{3}),(\sqrt[4]{66};\frac{4}{\sqrt[4]{66}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,929 |
6. A circle passes through vertices $A$ and $B$ of triangle $ABC$ and intersects its sides $AC$ and $BC$ at points $Q$ and $N$ respectively, such that $AQ: QC = 5: 2$ and $CN: NB = 5: 2$. Find $AB$, if $QN = 5 \sqrt{2}$. | Answer: $7 \sqrt{5}$.
Solution. Let $C Q=2 x, C N=5 y$; then $A Q=5 x, C T=2 y$. By the theorem of two secants $C Q \cdot C A=C N \cdot C B$, from which $2 x \cdot 7 x=5 y \cdot 7 y, y=x \sqrt{\frac{2}{5}}$. Triangles $A B C$ and $N Q C$ are similar by two sides and the angle between them ( $C A: C N=C B: C Q, \quad \... | 7\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,930 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+5 x+72$. | Answer: 7.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+5 x+72>0 \Leftrightarrow-\frac{1}{3}(x+9)(x-24)>0$, from which $-9<x<24$. On this interval, there are 23 natural values of $x: x=1, x=2, \ldots, x=23$. During this time, $y$ takes integer values only when $x$ is divisible ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,931 |
2. Find the value of the expression $\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are the largest and smallest roots of the equation $x^{3}-9 x^{2}+9 x=1$, respectively. | Answer: 62.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-9\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-9 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-8 x+1\right)=0
$$
from which $x=1$ or $x=4 \pm \sqrt{15}$. The largest root is $a=4+\sqrt{15}$, the smallest is... | 62 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,932 |
3. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | # Answer: 864.
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 from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits a... | 864 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,933 |
4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}|| x \mid+x)^{2}+(|y|+y)^{2} \leq 4, \\ 3 y+x \leq 0\end{array}\right.$ | # Answer: $\infty$.
Solution. Consider the first inequality. There are four possible cases.
1) $x \geq 0, y \geq 0$ (first quadrant). Then $4 x^{2}+4 y^{2} \leq 4, x^{2}+y^{2} \leq 1^{2}$. We obtain points lying on the circle centered at $O(0 ; 0)$ with radius 1 or inside it.
2) $x<0, y \geq 0$ (second quadrant). The... | \infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 2,934 |
5. 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 $u v=\sqrt{\frac{x}{y}} \cdot \sqrt{x y}=\sqrt{x^{2}}=|x|=x$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{x}{y}}=\sqrt{y^{2}}=|y|=y$, since $x$ and $y$ are positive b... | (2;2),(\frac{\sqrt[4]{286}}{4};\sqrt[4]{286}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,935 |
6. A circle passes through the vertices $A$ and $K$ of triangle $A K T$ and intersects its sides $A T$ and $K T$ at points $C$ and $N$ respectively, such that $A C: C T=4: 1, T N: N K=1: 2$. Find $A K$, if $C N=\sqrt{10}$. | Answer: $5 \sqrt{6}$.
Solution. Let $C T=x, T N=y$; then $A C=4 x, K N=2 y$. By the theorem of two secants $T C \cdot T A=T N \cdot T K$, from which $x \cdot 5 x=y \cdot 3 y, y=x \sqrt{\frac{5}{3}}$. Triangles $A K T$ and $C N T$ are similar by two sides and the angle between them ( $A T: N T=K T: C T, \angle T-$ is c... | 5\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,936 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+20 x+63$. | Answer: 20.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+20 x+63>0 \Leftrightarrow-\frac{1}{3}(x+3)(x-63)>0$, from which $-3<x<63$. On this interval, there are 62 natural values of $x: x=1, x=2, \ldots, x=62$. In this case, $y$ takes integer values only when $x$ is divisible by 3... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,937 |
2. Find the value of the expression $\frac{p}{q}+\frac{q}{p}$, where $p$ and $q$ are the largest and smallest roots of the equation $x^{3}-8 x^{2}+8 x=1$, respectively. | # Answer: 47.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-8\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-8 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-7 x+1\right)=0 \text {, }
$$
from which $x=1$ or $x=\frac{7 \pm \sqrt{45}}{2}$. The largest root is $p=\frac{... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,938 |
3. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
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 $0, 2, 4, 8$ (4 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits arbit... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,939 |
4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(|x|+x)^{2}+(|y|-y)^{2} \leq 16, \\ y-3 x \leq 0\end{array}\right.$, and find the area of the resulting figure. | Answer: $\frac{20}{3}+\pi$.
Solution. Consider the first inequality. There are four possible cases.
1) $x \geq 0, y \geq 0$ (first quadrant). Then $4 x^{2} \leq 16,|x| \leq 2$; in this case, $x$ is positive, so $x \leq 2$.
2) $x<0, y \geq 0$ (second quadrant). Then $0 \leq 16$, the inequality is satisfied by all poin... | \frac{20}{3}+\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,940 |
5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}3 y-\sqrt{\frac{y}{x}}-6 \sqrt{x y}+2=0, \\ x^{2}+81 x^{2} y^{4}=2 y^{2}\end{array}\right.$. Answer: $\left(\frac{1}{3} ; \frac{1}{3}\right),\left(\frac{\sqrt[4]{31}}{12} ; \frac{\sqrt[4]{31}}{3}\right)$ | Solution. Let $\quad \sqrt{\frac{y}{x}}=u, \quad \sqrt{x y}=v \quad$ (with $\left.\quad u>0, \quad v>0\right) . \quad$ Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system tak... | (\frac{1}{3};\frac{1}{3}),(\frac{\sqrt[4]{31}}{12};\frac{\sqrt[4]{31}}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,941 |
6. A circle passes through the vertices $K$ and $P$ of triangle $K P M$ and intersects its sides $K M$ and $P M$ at points $F$ and $B$ respectively, such that $K F: F M=3: 1, P B: B M=6: 5$. Find $K P$, if $B F=\sqrt{15}$. | Answer: $2 \sqrt{33}$.
Solution. Let $F M=x, B M=5 y$; then $K F=3 x, B P=6 y$. By the theorem of two secants $M F \cdot M K=M B \cdot M P$, from which $x \cdot 4 x=5 y \cdot 11 y, y=\frac{2 x}{\sqrt{55}}$. Triangles $K P M$ and $B F M$ are similar by two sides and the angle between them ( $K M: B M=P M: F M, \angle M... | 2\sqrt{33} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,942 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+50$. | Answer: 7.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+50>0 \Leftrightarrow x^{2}<450$, from which $-\sqrt{450}<x<\sqrt{450}$. On this interval, there are 21 natural values of $x: x=1, x=2, \ldots, x=21$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,943 |
2. Solve the inequality $8|x-\sqrt{x}+2|+2 x \sqrt{x}<x^{2}+x+28$. | Answer: $x \in[0 ; 4) \cup(9 ;+\infty)$.
Solution. Note that the expression under the modulus is non-negative on the domain of definition (it is a quadratic trinomial in terms of $\sqrt{x}$ and $D0$, from which $t \in(-\infty ; 2) \cup(6 ;+\infty)$.
If $t6$, then $x-\sqrt{x}-6>0 \Leftrightarrow(\sqrt{x}-3)(\sqrt{x}+2... | x\in[0;4)\cup(9;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 2,944 |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, 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 75. In how many ways can this be done? | Answer: 2592.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit from the available options (1 way).
To ensure divisibility by three, we proceed as follows. Select four digits arbitrarily (th... | 2592 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,945 |
4. Plot on the plane $(x ; y)$ the set of points satisfying the equation $|15 x|+|8 y|+|120-15 x-8 y|=120$, and find the area of the resulting figure. | Answer: 60.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ 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 inequalities
$$
\left\{\begin{array} { l... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,946 |
5. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}-2 x-2 y+10=0, \\ x^{3} y-x y^{3}-2 x^{2}+2 y^{2}-30=0 .\end{array}\right.$ | Answer: $(-4, -1)$.
Solution. The given system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ x y ( x + y ) - 2 ( x + y ) + 10 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 2 ( x ^ { 2 } - y ^ { 2 } ) - 30 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y-2)(x+y)=-10 \\
(x y-2)(x-y)(x+y)... | (-4,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,947 |
6. A circle passes through the vertices $A$ and $N$ of triangle $A C N$ and intersects its sides $A C$ and $C N$ at points $B$ and $K$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $B C K$ to the area of triangle $A C N$ is $\frac{1}{4}$.
a) Find the ratio $A N: B K$.
b... | Answer: a) $A N: B K=2$, b) $N K: A B=2: 5$.
Solution. a) By the theorem of two secants $C K \cdot C N=C B \cdot C A$. Therefore, triangles $A C N$ and $K C B$ are similar by two sides and the angle between them ($A C: K C=C N: C B, \angle C$ - common). The areas of similar figures are in the ratio of the square of th... | )AN:BK=2,b)NK:AB=2:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,948 |
1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+70$. | Answer: 4.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+70>0 \Leftrightarrow x^{2}<210$, from which $-\sqrt{210}<x<\sqrt{210}$. On this interval, there are 14 natural values of $x: x=1, x=2, \ldots, x=14$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,949 |
3. In the number $2 * 0 * 1 * 6 * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can repeat) so that the resulting 11-digit number is divisible by 12. In how many ways can this be done? | Answer: 1296.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0, 4, or 8 as the last digit (3 ways).
To ensure divisibility by 3, we proceed as follows. We will choose three digits arbitrarily (this can be done... | 1296 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,951 |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|3 x|+|4 y|+|48-3 x-4 y|=48$, and find the area of the resulting figure. | Answer: 96.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ 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 inequalities
$$
\left\{\begin{array} { l... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,952 |
5. Solve the system of equations $\left\{\begin{array}{l}x^{2} y-x y^{2}-3 x+3 y+1=0, \\ x^{3} y-x y^{3}-3 x^{2}+3 y^{2}+3=0 .\end{array}\right.$ | Answer: $(2 ; 1)$.
Solution. The given system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ x y ( x - y ) - 3 ( x - y ) + 1 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 3 ( x ^ { 2 } - y ^ { 2 } ) + 3 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y-3)(x-y)=-1, \\
(x y-3)(x-y)(x+y)=-3... | (2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,953 |
6. A circle passes through the vertices $L$ and $M$ of triangle $F L M$ and intersects its sides $F L$ and $F M$ at points $A$ and $H$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $F L M$ to the area of triangle $A F H$ is $\frac{49}{9}$.
a) Find the ratio $L M: A H$.
... | Answer: a) $L M: A H=7: 3$, b) $A L: M H=11$.
Solution. a) By the theorem of two secants $F L \cdot F A=F M \cdot F H$. Therefore, triangles $F L M$ and $A F H$ are similar by two sides and the angle between them ($F L: F H=F M: F A, \angle F$ - common). The areas of similar figures are in the ratio of the square of t... | )7:3,b)11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,954 |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+33$. | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+33>0 \Leftrightarrow x^{2}<297$, from which $-\sqrt{297}<x<\sqrt{297}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,955 |
3. In the number $2 * 0 * 1 * 6 * 07 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,6,7$ (digits can repeat) so that the resulting 11-digit number is divisible by 75. In how many ways can this be done? | # Answer: 432.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit (1 way) from the available options.
To ensure divisibility by three, we proceed as follows. We will choose three digits arbit... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,957 |
4. Plot on the plane $(x ; y)$ the set of points satisfying the equation $|5 x|+|12 y|+|60-5 x-12 y|=60$, and find the area of the resulting figure. | Answer: 30.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ 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 inequalities
$$
\left\{\begin{array} { l... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,958 |
5. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}+3 x+3 y+24=0, \\ x^{3} y-x y^{3}+3 x^{2}-3 y^{2}-48=0 .\end{array}\right.$ | Answer: $(-3, -1)$.
Solution. The given system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ x y ( x + y ) + 3 ( x + y ) + 24 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) + 3 ( x ^ { 2 } - y ^ { 2 } ) - 48 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y+3)(x+y)=-24 \\
(x y+3)(x-y)(x+y)... | (-3,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,959 |
6. A circle passes through the vertices $Q$ and $E$ of triangle $M Q E$ and intersects its sides $M Q$ and $M E$ at points $B$ and $D$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $B D M$ to the area of triangle $M Q E$ is $\frac{9}{121}$.
a) Find the ratio $Q E: B D$.
... | Answer: a) $Q E: B D=11: 3$, b) $B Q: D E=5: 19$.
Solution. a) By the theorem of two secants $M Q \cdot M B=M E \cdot M D$. Therefore, triangles $M Q E$ and $M D B$ are similar by two sides and the angle between them ( $M Q: M D=M E: M B, \angle M-$ is common). The areas of similar figures are in the ratio of the squa... | )QE:BD=11:3,b)BQ:DE=5:19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,960 |
1. Find the number of points in the plane $x O y$ that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+98$ | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+98>0 \Leftrightarrow x^{2}<294$, from which $-\sqrt{294}<x<\sqrt{294}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,961 |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, 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 12. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0 or 4 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose four digits arb... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,963 |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|4 x|+|3 y|+|24-4 x-3 y|=24$, and find the area of the resulting figure.
# | # Answer: 24.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ 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 inequalities
$$
\left\{\begin{array} {... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,964 |
5. Solve the system of equations $\left\{\begin{array}{l}x^{2} y-x y^{2}-5 x+5 y+3=0, \\ x^{3} y-x y^{3}-5 x^{2}+5 y^{2}+15=0 .\end{array}\right.$ | Answer: $(4 ; 1)$.
Solution. The given system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ x y ( x - y ) - 5 ( x - y ) + 3 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 5 ( x ^ { 2 } - y ^ { 2 } ) + 1 5 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y-5)(x-y)=-3, \\
(x y-5)(x-y)(x+y)=... | (4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,965 |
6. A circle passes through the vertices $P$ and $T$ of triangle $MPT$ and intersects its sides $MP$ and $MT$ at points $D$ and $E$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $MDE$ to the area of triangle $MPT$ is $\frac{1}{4}$.
a) Find the ratio $DE: TP$.
b) Suppose ... | Answer: a) $D E: T P=1: 2$, b) $T E: P D=1: 4$.
Solution. a) By the theorem of two secants $M T \cdot M E=M P \cdot M D$. Therefore, triangles $M P T$ and $M E D$ are similar by two sides and the angle between them ($M P: M E=M T: M D, \angle M$ - common). The areas of similar figures are proportional to the square of... | DE:TP=1:2,\quadTE:PD=1:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,966 |
1. Solve the equation $\frac{|\cos x|-\cos 3 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}}$. | Answer. $x=\frac{\pi}{6}+2 k \pi, x=\frac{5 \pi}{6}+2 k \pi, x=\frac{4 \pi}{3}+2 k \pi, k \in Z$.
Solution. There are two possible cases.
a) $\cos x \geq 0$. Then $\frac{\cos x-\cos 3 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{2 \sin x \sin 2 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \o... | \frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi,\frac{4\pi}{3}+2k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,967 |
2. Given a regular 20-gon $M$. Find the number of quadruples of vertices of this 20-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 765.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{20}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{20}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 765 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,968 |
3. Find the number of natural numbers $k$, not exceeding 291000, such that $k^{2}-1$ is divisible by 291. | Answer: 4000.
Solution. By factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 97)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 97. Let's consider two cases.
a) $(k+1): 97$, i.e., $k=97 p+96, p \in \mathrm{Z}$. Then we get $(97 p+95)(97 p+97):(3 \cdot 97) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,969 |
4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 2, \\ 81 x^{4}-18 x^{2} y^{2}+y^{4}-360 x^{2}-40 y^{2}+400=0 .\end{array}\right.$ | Answer. $\left(-\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.
Solution. Transform the equation of the system:
$$
\begin{gathered}
81 x^{4}-18 x^{2} y^{2}+y... | (-\frac{3}{\sqrt{5}};\frac{1}{\sqrt{5}}),(-\frac{3}{\sqrt{5}};-\frac{1}{\sqrt{5}}),(\frac{3}{\sqrt{5}};-\frac{1}{\sqrt{5}}),(\frac{3}{\sqrt{5}};\frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,970 |
5. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\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 at $(6 ; a)$. For all possible $a \in \mathbb{R}$, these circles sweep out the strip $4 \leq ... | b\in(-\infty,0)\cup[\frac{3}{8};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,971 |
6. 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 | 2,972 |
1. Solve the equation $\frac{|\sin x|-\sin 3 x}{\cos x \cos 2 x}=2 \sqrt{3}$. | Answer. $x= \pm \frac{2 \pi}{3}+2 k \pi, x=-\frac{\pi}{6}+2 k \pi, k \in Z$.
Solution. There are two possible cases.
a) $\sin x \geq 0$. Then $\frac{\sin x-\sin 3 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \frac{-2 \sin x \cos 2 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \operatorname{tg} x=-\sqrt{3} \Leftrig... | \\frac{2\pi}{3}+2k\pi,-\frac{\pi}{6}+2k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,973 |
2. Given a regular 16-gon $M$. Find the number of quadruples of vertices of this 16-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 364.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{16}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{16}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 364 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,974 |
3. Find the number of natural numbers $k$, not exceeding 445000, such that $k^{2}-1$ is divisible by 445. Answer: 4000. | Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(5 \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):(5 \cdot 89) \Leftrightarrow(89 p+87)(p+... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,975 |
4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 1, \\ 16 x^{4}-8 x^{2} y^{2}+y^{4}-40 x^{2}-10 y^{2}+25=0 .\end{array}\right.$ | Answer. $\left(-\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.
Solution. Transform the equation of the system:
$$
\begin{gathered}
16 x^{4}-8 x^{2} y^{2}+y^... | (-\frac{2}{\sqrt{5}};\frac{1}{\sqrt{5}}),(-\frac{2}{\sqrt{5}};-\frac{1}{\sqrt{5}}),(\frac{2}{\sqrt{5}};-\frac{1}{\sqrt{5}}),(\frac{2}{\sqrt{5}};\frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,976 |
5. 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. $a \in(-\infty, 0) \cup\left[\frac{2}{3} ;+\infty\right)$.
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 $b \in \mathbf{R}$, these circles sweep the strip $4 \leq x \leq ... | \in(-\infty,0)\cup[\frac{2}{3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,977 |
6. 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 | 2,978 |
1. Solve the equation $\frac{|\cos x|+\cos 3 x}{\sin x \cos 2 x}=-2 \sqrt{3}$. | Answer. $x=\frac{2 \pi}{3}+2 k \pi, x=\frac{7 \pi}{6}+2 k \pi, x=-\frac{\pi}{6}+2 k \pi, k \in \mathrm{Z}$.
Solution. There are two possible cases.
a) $\cos x \geq 0$. Then $\frac{\cos x+\cos 3 x}{\sin x \cos 2 x}=-2 \sqrt{3} \Leftrightarrow \frac{2 \cos x \cos 2 x}{\sin x \cos 2 x}=-2 \sqrt{3} \Leftrightarrow \opera... | \frac{2\pi}{3}+2k\pi,\frac{7\pi}{6}+2k\pi,-\frac{\pi}{6}+2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,979 |
2. Given a regular 22-gon $M$. Find the number of quadruples of vertices of this 22-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 1045.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{22}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{22}$.
Consider a chord connecting two adjacent vertices of the polygon, for example,... | 1045 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,980 |
3. Find the number of natural numbers $k$, not exceeding 485000, such that $k^{2}-1$ is divisible by 485. | Answer: 4000.
Solution. By factoring the dividend and divisor, we get the condition $(k-1)(k+1):(5 \cdot 97)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 97. Let's consider two cases.
a) $(k+1): 97$, i.e., $k=97 p+96, p \in \mathrm{Z}$. Then we get $(97 p+95)(97 p+97):(5 \cdot 97) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,981 |
4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 2, \\ x^{4}-8 x^{2} y^{2}+16 y^{4}-20 x^{2}-80 y^{2}+100=0 .\end{array}\right.$ | Answer. $\left(-\frac{\sqrt{2}}{\sqrt{5}} ; \frac{2 \sqrt{2}}{\sqrt{5}}\right),\left(\frac{\sqrt{2}}{\sqrt{5}} ; \frac{2 \sqrt{2}}{\sqrt{5}}\right),\left(\frac{\sqrt{2}}{\sqrt{5}} ;-\frac{2 \sqrt{2}}{\sqrt{5}}\right),\left(-\frac{\sqrt{2}}{\sqrt{5}} ;-\frac{2 \sqrt{2}}{\sqrt{5}}\right)$.
Solution. Transform the equati... | (-\frac{\sqrt{2}}{\sqrt{5}};\frac{2\sqrt{2}}{\sqrt{5}}),(\frac{\sqrt{2}}{\sqrt{5}};\frac{2\sqrt{2}}{\sqrt{5}}),(\frac{\sqrt{2}}{\sqrt{5}};-\frac{2\sqrt{2}}{\sqrt{5}}),\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,982 |
5. 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 | 2,983 |
6. 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 | 2,984 |
1. Solve the equation $\frac{|\sin x|+\sin 3 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}}$. | Answer. $x=\frac{\pi}{12}+2 k \pi, x=\frac{7 \pi}{12}+2 k \pi, x=-\frac{5 \pi}{6}+2 k \pi, k \in \mathrm{Z}$.
Solution. There are two possible cases.
a) $\sin x \geq 0$. Then $\frac{\sin x+\sin 3 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{2 \cos x \sin 2 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftr... | \frac{\pi}{12}+2k\pi,\frac{7\pi}{12}+2k\pi,-\frac{5\pi}{6}+2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,985 |
2. Given a regular 18-gon $M$. Find the number of quadruples of vertices of this 18-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 540.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{18}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{18}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,986 |
3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. | Answer: 4000.
Solution. By 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., $k=89 p+88, p \in \mathbb{Z}$. Then we get $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,987 |
4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 1, \\ x^{4}-18 x^{2} y^{2}+81 y^{4}-20 x^{2}-180 y^{2}+100=0 .\end{array}\right.$ | Answer. $\left(-\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right),\left(-\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right)$.
Solution. Transform the equation of the system:
$$
\begin{gathered}
x^{4}-18 x^{2} y^... | (-\frac{1}{\sqrt{10}};\frac{3}{\sqrt{10}}),(-\frac{1}{\sqrt{10}};-\frac{3}{\sqrt{10}}),(\frac{1}{\sqrt{10}};\frac{3}{\sqrt{10}}),(\frac{1}{\sqrt{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,988 |
5. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\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)$.
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 \mathbf{R}$, these circles sweep the strip $-9 \leq x \... | \in(-\infty;-\frac{2}{3}]\cup(0+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,989 |
6. 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 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 oppo... | )r=\sqrt{6},b)\angleBDC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,990 |
1. Solve the inequality $\sqrt{x^{2}-25} \cdot \sqrt{-2 x-1} \leq x^{2}-25$. | Answer. $x \in(-\infty ;-6] \cup\{-5\}$.
Solution. The domain of definition of the given inequality is the set $x \in(-\infty ;-5]$. Consider two cases.
a) When $x=-5$, the inequality is satisfied (we get $0=0$).
b) When $x<-5$, we divide both sides of the inequality by the positive number $\sqrt{x^{2}-25}$ and obta... | x\in(-\infty;-6]\cup{-5} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 2,991 |
2. Given the function $g(x)=\frac{4 \sin ^{4} x+5 \cos ^{2} x}{4 \cos ^{4} x+3 \sin ^{2} x}$. Find:
a) the roots of the equation $g(x)=\frac{7}{5}$;
b) the maximum and minimum values of the function $g(x)$. | Answer. a) $x=\frac{\pi}{4}+\frac{k \pi}{2}, k \in \mathrm{Z}, x= \pm \frac{\pi}{3}+k \pi, k \in \mathrm{Z}$; b) $g_{\min }=\frac{5}{4}, g_{\max }=\frac{55}{39}$.
Solution. Transform the given function:
$$
g(x)=\frac{4\left(1-\cos ^{2} x\right)^{2}+5 \cos ^{2} x}{4 \cos ^{4} x+3-3 \cos ^{2} x}=\frac{4 \cos ^{4} x-3 \... | )\frac{\pi}{4}+\frac{k\pi}{2},k\in\mathrm{Z},\\frac{\pi}{3}+k\pi,k\in\mathrm{Z};b)g_{\}=\frac{5}{4},g_{\max}=\frac{55}{39} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,992 |
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
$$
\left\{\begin{array}{l}
-7.5(x+y+z)=x y+x z \\
-1.5(x+y+z)=x y+y z \\
-4(x+y+z)=x z+y z
\end{array}\right.
$... | (5,-1,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,993 |
4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=2: 5$. The bisector $B L$ of the given triangle and the segment $A M$ intersect at point $P$ at an angle of $90^{\circ}$.
a) Find the ratio of the area of triangle $A B P$ to the area of quadrilateral $L P M C$.
b) On the segment $M C$... | Answer. a) $9: 40$, b) $\arccos \frac{3 \sqrt{3}}{2 \sqrt{7}}$.
Solution. a) In triangle $A B M$, segment $B P$ is both a bisector and an altitude, so triangle $A B M$ is isosceles, and $B P$ is also its median. Let $B M=2 x$, then $A B=2 x, M C=5 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=2... | )\frac{9}{40},b)\arccos\frac{3\sqrt{3}}{2\sqrt{7}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,994 |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $5 x^{2}-6 x y+y^{2}=6^{100}$. | Answer: 19594.
Solution: By factoring the left and right sides of the equation, we get $(5 x-y)(x-y)=2^{100} \cdot 3^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
5 x - y = 2 ^ { k } \cdot 3 ^ { l }, \\
x - y = 2 ^ { 1 0 0 - k } \cdot 3 ^ { 1 0 0 - l }
\end{array... | 19594 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,995 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a+y-x)=49 \\
y=\frac{8}{(x-b)^{2}+1}
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in[-15 ; 7)$.
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 $(a ;-a)$.
Consider the function defined by the second equation when $b=0$. At the point $x=0$, it takes the maximum value of 8. As $x... | \in[-15;7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,996 |
1. Solve the inequality $\sqrt{x^{2}-16} \cdot \sqrt{2 x-1} \leq x^{2}-16$. | Answer. $x \in\{4\} \cup[5 ;+\infty)$.
Solution. The domain of definition of the given inequality is the set $x \in[4 ;+\infty)$. Consider two cases.
a) When $x=4$, the inequality is satisfied (we get $0=0$).
b) When $x>4$, divide both sides of the inequality by the positive number $\sqrt{x^{2}-16}$ and we get $\sqr... | x\in{4}\cup[5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 2,997 |
2. Given the function $g(x)=\frac{2 \cos ^{4} x+\sin ^{2} x}{2 \sin ^{4} x+3 \cos ^{2} x}$. Find:
a) the roots of the equation $g(x)=\frac{1}{2}$;
b) the maximum and minimum values of the function $g(x)$. | Answer. a) $x=\frac{\pi}{4}+\frac{k \pi}{2}, k \in \mathrm{Z}, x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$; b) $g_{\min }=\frac{7}{15}, g_{\max }=\frac{2}{3}$.
Solution. Transform the given function:
$$
g(x)=\frac{2 \cos ^{4} x+1-\cos ^{2} x}{2\left(1-\cos ^{2} x\right)^{2}+3 \cos ^{2} x}=\frac{2 \cos ^{4} x-\cos ^{2} x... | )\frac{\pi}{4}+\frac{k\pi}{2},k\in\mathrm{Z},\frac{\pi}{2}+k\pi,k\in\mathrm{Z};b)g_{\}=\frac{7}{15},g_{\max}=\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,998 |
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{1}{12} \\ \frac{1}{y}+\frac{1}{x+z}=\frac{1}{6} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{1}{2}\end{array}\right.$, | Answer: $(-4 ; 2 ; 1)$.
Solution. Multiplying both sides of the first equation by $12 x(y+z)$, both sides of the second by $6 y(x+z)$, and the third by $2 z(x+y)$, we obtain the system
$$
\left\{\begin{array}{l}
12(x+y+z)=x y+x z \\
6(x+y+z)=x y+y z \\
2(x+y+z)=x z+y z
\end{array}\right.
$$
Adding all three equation... | (-4,2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,999 |
4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=2: 7$. The bisector $B L$ of the given triangle and the segment $A M$ intersect at point $P$ at an angle of $90^{\circ}$.
a) Find the ratio of the area of triangle $A B P$ to the area of quadrilateral $L P M C$.
b) On the segment $M C$... | Answer. a) $11: 70$, b) $\arccos \frac{\sqrt{11}}{2 \sqrt{3}}$.
Solution. a) In triangle $A B M$, segment $B P$ is both a bisector and an altitude, so triangle $A B M$ is isosceles, and $B P$ is also its median. Let $B M=2 x$, then $A B=2 x, M C=7 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=2... | )\frac{11}{70},b)\arccos\frac{\sqrt{11}}{2\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,000 |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $6 x^{2}-7 x y+y^{2}=10^{100}$. | Answer: 19998.
Solution: By factoring the left and right sides of the equation, we get $(6 x-y)(x-y)=2^{100} \cdot 5^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
6 x - y = 2 ^ { k } \cdot 5 ^ { l } , \\
x - y = 2 ^ { 1 0 0 - k } \cdot 5 ^ { 1 0 0 - l }
\end{arra... | 19998 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,001 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 b(b-x+y)=4 \\
y=\frac{9}{(x+a)^{2}+1}
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in[-11 ; 2)$.
“Phystech-2015”, 10th grade, solutions for ticket 6
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 2 with center $(b ;-b)$.
Consider the function defined by the second equation when $a=0$. At th... | b\in[-11;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,002 |
1. Solve the inequality $\sqrt{x^{2}-9} \cdot \sqrt{-2 x-1} \leq x^{2}-9$. | Answer. $x \in(-\infty ;-4] \cup\{-3\}$.
Solution. The domain of definition of the given inequality is the set $x \in(-\infty ;-5]$. Let's consider two cases.
a) When $x=-3$, the inequality is satisfied (we get $0=0$).
b) When $x<-3$, we divide both sides of the inequality by the positive number $\sqrt{x^{2}-9}$ and... | x\in(-\infty;-4]\cup{-3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,003 |
2. Given the function $g(x)=\frac{4 \cos ^{4} x+5 \sin ^{2} x}{4 \sin ^{4} x+3 \cos ^{2} x}$. Find:
a) the roots of the equation $g(x)=\frac{4}{3}$;
b) the maximum and minimum values of the function $g(x)$. | Answer. a) $x=\frac{k \pi}{3}, k \in \mathrm{Z}$; b) $g_{\min }=\frac{5}{4}, g_{\max }=\frac{55}{39}$.
Solution. Transform the given function:
$$
g(x)=\frac{4 \cos ^{4} x+5-5 \cos ^{2} x}{4\left(1-\cos ^{2} x\right)^{2}+3 \cos ^{2} x}=\frac{4 \cos ^{4} x-5 \cos ^{2} x+5}{4 \cos ^{4} x-5 \cos ^{2} x+4}=1+\frac{1}{4 \c... | )\frac{k\pi}{3},k\in\mathrm{Z};b)g_{\}=\frac{5}{4},g_{\max}=\frac{55}{39} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,004 |
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=1, \\ \frac{1}{y}+\frac{1}{x+z}=\frac{4}{3}, \\ \frac{1}{z}+\frac{1}{x+y}=-\frac{4}{5}\end{array}\right.$. | Answer: $(2 ; 3 ;-1)$.
Solution. Multiplying both sides of the first equation by $x(y+z)$, both sides of the second by $\frac{3}{4} y(x+z)$, and both sides of the third by $-\frac{5}{4} z(x+y)$, we obtain the system
$$
\left\{\begin{array}{l}
(x+y+z)=x y+x z \\
\frac{3}{4}(x+y+z)=x y+y z \\
-\frac{5}{4}(x+y+z)=x z+y ... | (2;3;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,005 |
4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=3: 8$. 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) $21: 100$, b) $\arccos \frac{2 \sqrt{7}}{\sqrt{33}}$.
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 $A B=3 x, M C=8 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=... | )\frac{21}{100},b)\arccos\frac{2\sqrt{7}}{\sqrt{33}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,006 |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $x^{2}+6 x y+5 y^{2}=10^{100}$. | Answer: 19594.
Solution: By factoring the left and right sides of the equation, we get $(x+5 y)(x+y)=2^{100} \cdot 5^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
x+5 y=2^{k} \cdot 5^{l}, \\
x+y=2^{100-k} \cdot 5^{100-l}
\end{array} \text { or } \left\{\begin{arr... | 19594 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,007 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a-x-y)=64 \\
y=\frac{7}{(x+b)^{2}+1}
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in(-8 ; 15]$.
Solution. The first equation of the system can be transformed into the form $(x-a)^{2}+(y-a)^{2}=8^{2}$, hence it represents a circle with radius 8 and center at $(a ; a)$.
Consider the function defined by the second equation when $b=0$. At the point $x=0$, it attains its maximum value of 7.... | \in(-8;15] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,008 |
1. Solve the inequality $\sqrt{x^{2}-4} \cdot \sqrt{2 x-1} \leq x^{2}-4$. | Answer. $x \in\{2\} \cup[3 ;+\infty)$.
Solution. The domain of definition of the given inequality is the set $x \in[2 ;+\infty)$. Consider two cases.
a) For $x=2$, the inequality is satisfied (we get $0=0$).
b) For $x>2$, divide both sides of the inequality by the positive number $\sqrt{x^{2}-4}$ and we get $\sqrt{2... | x\in{2}\cup[3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,009 |
2. Given the function $g(x)=\frac{4 \sin ^{4} x+7 \cos ^{2} x}{4 \cos ^{4} x+\sin ^{2} x}$. Find:
a) the roots of the equation $g(x)=4$;
b) the maximum and minimum values of the function $g(x)$. | Answer. a) $x= \pm \frac{\pi}{3}+k \pi, k \in \mathrm{Z}, x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$; b) $g_{\min }=\frac{7}{4}, g_{\max }=\frac{63}{15}$.
Solution. Transform the given function:
$$
g(x)=\frac{4\left(1-\cos ^{2} x\right)^{2}+7 \cos ^{2} x}{4 \cos ^{4} x+1-\cos ^{2} x}=\frac{4 \cos ^{4} x-\cos ^{2} x+4}{... | )\\frac{\pi}{3}+k\pi,k\in\mathrm{Z},\frac{\pi}{2}+k\pi,k\in\mathrm{Z};b)g_{\}=\frac{7}{4},g_{\max}=\frac{63}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,010 |
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{6}{5} \text {, } \\ \frac{1}{y}+\frac{1}{x+z}=\frac{3}{4} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{2}{3}\end{array}\right.$, | Answer. $(1 ; 2 ; 3)$.
Solution. Multiplying both sides of the first equation by $\frac{5}{6} x(y+z)$, both sides of the second by $\frac{4}{3} y(x+z)$, and both sides of the third by $\frac{3}{2} z(x+y)$, we obtain the system
$$
\left\{\begin{array}{l}
\frac{5}{6}(x+y+z)=x y+x z \\
\frac{4}{3}(x+y+z)=x y+y z \\
\fra... | (1;2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,011 |
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