problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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4. Circle $\omega$ with radius 8 and center $O$ is inscribed in an acute-angled triangle $E N T$ and touches its sides $E N$ and $N T$ at points $C$ and $A$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{145}}{2}$ and center $B$ is circumscribed around triangle $A C N$.
a) Find $O N$.
b) Suppose it is additio... | Answer: a) $O N=\sqrt{145}$, b) $N L=\frac{5 \sqrt{145}}{3}, S=360$.
Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O C N$ and $O A N$ are right angles, i.e., the segment $O N$ is seen from points $A$ and $C$ at a right angle. Therefore, the circle constructed on the... | )ON=\sqrt{145},b)NL=\frac{5\sqrt{145}}{3},S=360 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,631 |
5. 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 arb... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,632 |
6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}9-x^{2}-y^{2}-2 y|+|-2 y \mid=9-x^{2}-y^{2}-4 y \\ 15 y+3 a=(4 a+15) x .\end{array}\right.\end{array}\right.$
a) Plot on the plane $(x ; y)$ the set of points satisfying the first equation of the system, and find the area of the res... | Answer: a) $S=10 \pi+3-10 \operatorname{arctg} 3$, b) $a=-5, a=-3$.
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 sy... | )S=10\pi+3-10\operatorname{arctg}3,b)=-5,=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,633 |
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A_{1} C_{1}$ intersects the edges $B_{1} C_{1}$ and $A_{1} B_{1}$ at points $N_{1}$ and $K_{1}$, respectively, which are different from the vertices of the prism. Segments $A N_{1}$ and $C K_{1}$ intersect at point $F$, and at the same... | Answer: a) $90^{\circ}$, b) $1: 3$, c) $V=56 \sqrt{3}$.
Solution. a) Points $K_{1}$ and $N_{1}$ lie on a circle with diameter $A_{1} C_{1}$; therefore, $\angle A_{1} K_{1} C_{1}=90^{\circ}, \angle A_{1} N_{1} C_{1}=90^{\circ}$ (i.e., $A_{1} N_{1}$ and $C_{1} K_{1}$ are altitudes of triangle $A_{1} B_{1} C_{1}$). The l... | 56\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,634 |
1. Solve the equation $\frac{2 \sin 3 x}{\sin x}-\frac{3 \cos 3 x}{\cos x}=7|\sin x|$. | Answer: $x= \pm \arcsin \frac{3}{4}+k \pi, k \in Z$.
Solution. The given equation, under the condition $\cos x \sin x \neq 0$, is equivalent to each of the following:
$$
\frac{2\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{3\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=7|\sin x| \Leftrightarrow 6-8 \sin ^{2} x-12... | \\arcsin\frac{3}{4}+k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,635 |
2. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}3 y-\sqrt{\frac{y}{x}}-6 \sqrt{x y}+2=0, \\ x^{2}+81 x^{2} y^{4}=2 y^{2} .\end{array}\right.$ Answer: $\left(\frac{1}{3} ; \frac{1}{3}\right),\left(\frac{\sqrt[4]{31}}{12} ; \frac{\sqrt[4]{31}}{3}\right)$ | Solution. Let $\sqrt{\frac{y}{x}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system takes the form
$$
\left\{\begin{array} ... | (\frac{1}{3};\frac{1}{3}),(\frac{\sqrt[4]{31}}{12};\frac{\sqrt[4]{31}}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,636 |
3. Solve the inequality $2^{\frac{1-x}{3+x}} \cdot 7^{3+x} \geq 56$. | Answer: $x \in(-3 ;-2] \cup\left[4 \log _{7} 2-3 ;+\infty\right)$.
Solution. By taking the logarithm of both sides of the inequality with base 7, we get:
$$
\frac{1-x}{3+x} \log _{7} 2+(3+x) \geq 1+3 \log _{7} 2 \Leftrightarrow \frac{-4 x-8}{3+x} \log _{7} 2+(2+x) \geq 0 \Leftrightarrow \frac{(x+2)\left(-4 \log _{7} ... | x\in(-3;-2]\cup[4\log_{7}2-3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,637 |
4. Circle $\omega$ with radius 6 and center $O$ is inscribed in an acute triangle $ABC$ and touches its sides $AB$ and $BC$ at points $F$ and $T$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{85}}{2}$ and center $M$ is circumscribed around triangle $BFT$.
a) Find $BO$.
b) Suppose it is additionally known tha... | Answer: a) $BO=\sqrt{85}$, b) $BQ=\frac{5 \sqrt{85}}{3}, S=210$.
Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $OFB$ and $OTB$ are right angles, i.e., from points $F$ and $T$, segment $OB$ is seen at a right angle. Therefore, the circle constructed on segment $OB$ as... | BO=\sqrt{85},BQ=\frac{5\sqrt{85}}{3},S=210 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,638 |
5. 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. We will choose three digi... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,639 |
6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}4-2 x-x^{2}-y^{2}|+|-2 x \mid=4-4 x-x^{2}-y^{2} \\ (3 a+15) x=3 y+a+4\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 result... | Answer: a) $S=5 \pi-5 \operatorname{arctg} 2+2$, b) $a=-10, a=2$.
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 syst... | )S=5\pi-5\operatorname{arctg}2+2,b)=-10,=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,640 |
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A_{1} B_{1}$ intersects the edges $A_{1} C_{1}$ and $B_{1} C_{1}$ at points $T_{1}$ and $L_{1}$, respectively, different from the vertices of the prism. Segments $B T_{1}$ and $A L_{1}$ intersect at point $S$, and $A L_{1}=7, S T_{1}=2... | Answer: a) $90^{\circ}$, b) $3: 2$, c) $V=35 \sqrt{3}$.
Solution. a) Points $T_{1}$ and $L_{1}$ lie on a circle with diameter $A_{1} B_{1}$; hence, $\angle A_{1} L_{1} B_{1}=90^{\circ}, \angle A_{1} T_{1} B_{1}=90^{\circ}$ (i.e., $A_{1} L_{1}$ and $B_{1} T_{1}$ are altitudes of triangle $A_{1} B_{1} C_{1}$). The line ... | 35\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,641 |
1. Solve the inequality $\left(x^{2}-3 x+3\right)^{4 x^{3}+5 x^{2}} \leq\left(x^{2}-3 x+3\right)^{2 x^{3}+18 x}$. | Answer: $x \in\left(-\infty ;-\frac{9}{2}\right] \cup[0 ; 1] \cup\{2\}$.
Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:
$$
\left(x^{2}-3 x+2\right)\left(\left(4 x^{3}+5 x^{2}\right)-\left(2 x^{3}+18 x\right)\right) \leq 0 \Leftrightarrow(x-1)(x-2) x... | x\in(-\infty;-\frac{9}{2}]\cup[0;1]\cup{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,642 |
2. Solve the equation $\frac{\cos 5 x-\cos 7 x}{\sin 4 x+\sin 2 x}=2|\sin 2 x|$. | Answer: $x=-\arccos \frac{\sqrt{13}-1}{4}+2 k \pi, x=-\arccos \frac{1-\sqrt{13}}{4}+2 k \pi, k \in Z$.
Solution. On the domain of definition, the given equation is equivalent to each of the following:
$\frac{2 \sin x \sin 6 x}{2 \sin 3 x \cos x}=2|\sin 2 x| \Leftrightarrow \frac{2 \sin x \sin 3 x \cos 3 x}{\sin 3 x \... | -\arccos\frac{\sqrt{13}-1}{4}+2k\pi,-\arccos\frac{1-\sqrt{13}}{4}+2k\pi,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,643 |
3. 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 | 3,644 |
4. In triangle $ABC$, medians $BD$ and $CE$ intersect at point $M$. The circle constructed on segment $BM$ as a diameter passes through vertex $C$ and is tangent to line $DE$. It is known that $CM=4$. Find the height $AH$ of triangle $ABC$, the angle $CBD$, and the area of triangle $ABC$. | Answer: $A H=12, \angle C B D=30^{\circ}, S_{A B C}=24 \sqrt{3}$.
Solution. Let the center of the circle be $O$, the point of tangency of the line $D E$ with the circle be $T$, and the radius of the circle be $R$. Since the medians intersect at a point dividing them in the ratio $2: 1$ from the vertex,
“Phystech-2016... | AH=12,\angleCBD=30,S_{ABC}=24\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,645 |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (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 (1 way).
To ensure divisibility by three, we proceed as follows. Select four digits arbitrarily (this can be done in $6 \cdot ... | 2592 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,646 |
7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 12. A sphere $\Omega$ with radius $r=\sqrt{\frac{35}{3}}$ touches all the lateral faces of the prism. Points $K$ and $L$ are chosen on segments $A A_{1}$ and $B B_{1}$, respectively, such that $K L \| A B$, and the planes $K B C$ and $L A_{1} C_{1... | Answer: $V=420 \sqrt{3}, A K=8$ or $A K=4$.
Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=2 \sqrt{35}$, the area of the base $S=\frac{\sqrt{3}}{4}(... | V=420\sqrt{3},AK=8orAK=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,647 |
1. Solve the inequality $\left(x^{2}-x+1\right)^{16 x^{3}-6 x} \leq\left(x^{2}-x+1\right)^{13 x^{2}+x^{3}}$. | Answer: $x \in\left(-\infty ;-\frac{1}{3}\right] \cup\{0\} \cup\left[1 ; \frac{6}{5}\right]$.
Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:
$$
\left.\left.\left(x^{2}-x\right)\right)\left(16 x^{3}-6 x\right)-\left(13 x^{2}+x^{3}\right)\right) \leq ... | x\in(-\infty;-\frac{1}{3}]\cup{0}\cup[1;\frac{6}{5}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,648 |
2. Solve the equation $\frac{\sin 5 x+\sin 7 x}{\sin 4 x-\sin 2 x}=-3 \cdot|\sin 2 x|$ | Answer: $x=\pi-\arcsin \frac{\sqrt{57}-3}{8}+2 k \pi, x=\pi-\arcsin \frac{3-\sqrt{57}}{8}+2 k \pi, k \in \mathrm{Z}$.
Solution. On the domain of definition, the given equation is equivalent to each of the following:
$\frac{2 \cos x \sin 6 x}{2 \sin x \cos 3 x}=-3|\sin 2 x| \Leftrightarrow \frac{2 \cos x \sin 3 x \cos... | \pi-\arcsin\frac{\sqrt{57}-3}{8}+2k\pi,\pi-\arcsin\frac{3-\sqrt{57}}{8}+2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,649 |
3. 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 | 3,650 |
4. In triangle $K L M$, medians $L D$ and $M E$ intersect at point $G$. The circle constructed on segment $L G$ as a diameter passes through vertex $M$ and is tangent to line $D E$. It is known that $G M=6$. Find the height $K T$ of triangle $K L M$, the angle $L G M$, and the area of triangle $K L M$. | Answer: $K T=18, \angle L G M=60^{\circ}, S_{K L M}=54 \sqrt{3}$.
Solution. Let the center of the circle be denoted as $O$, the point of tangency of the line $D E$ with the circle as $S$, and the radius of the circle as $R$. Since the medians intersect at a point dividing them in the ratio $2:1$ from the vertex, $D G=... | KT=18,\angleLGM=60,S_{KLM}=54\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,651 |
5. 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 | 3,652 |
6. Given the system of equations $\left\{\begin{array}{l}|3 x|+|4 y|+|48-3 x-4 y|=48, \\ (x-8)^{2}+\left(y+6 \cos \frac{a \pi}{2}\right)^{2}=(a+4)^{2} \text {. }\end{array}\right.$
a) Sketch on the $(x ; y)$ plane the set of points satisfying the first equation of the system, and find the area of the resulting figure.... | Answer: a) 96; b) $a=6, a=-14$.
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 side). Therefore, the first equation is equivalent to the system of inequalities
$$... | =6,=-14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,653 |
7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 6. A sphere $\Omega$ with radius $r=\sqrt{\frac{8}{3}}$ touches all the lateral faces of the prism. Points $M$ and $K$ are chosen on segments $A A_{1}$ and $B B_{1}$, respectively, such that $K M \| A B$, and the planes $A C K$ and $M B_{1} C_{1}$... | Answer: $V=48 \sqrt{3}, B K=5$ or $B K=1$.
Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=4 \sqrt{2}$, the area of the base $S=\frac{\sqrt{3}}{4}(4 ... | V=48\sqrt{3},BK=5orBK=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,654 |
1. Solve the inequality $\left(x^{2}+3 x+3\right)^{5 x^{3}-3 x^{2}} \leq\left(x^{2}+3 x+3\right)^{3 x^{3}+5 x}$. | Answer: $x \in(-\infty ;-2] \cup\{-1\} \cup\left[0 ; \frac{5}{2}\right]$.
Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:
$$
\left(x^{2}+3 x+2\right)\left(\left(5 x^{3}-3 x^{2}\right)-\left(3 x^{3}+5 x\right)\right) \leq 0 \Leftrightarrow(x+1)(x+2) x... | x\in(-\infty;-2]\cup{-1}\cup[0;\frac{5}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,655 |
2. Solve the equation $\frac{\sin 5 x+\sin 7 x}{\sin 4 x+\sin 2 x}=-4 \cdot|\sin 2 x|$. | Answer: $x=\pi-\arcsin \frac{\sqrt{2}-1}{2}+2 k \pi, x=\pi-\arcsin \frac{1-\sqrt{2}}{2}+2 k \pi, k \in \mathrm{Z}$.
Solution. On the domain of definition, the given equation is equivalent to each of the following:
$\frac{2 \cos x \sin 6 x}{2 \sin 3 x \cos x}=-4 \cdot|\sin 2 x| \Leftrightarrow \frac{2 \sin 3 x \cos 3 ... | \pi-\arcsin\frac{\sqrt{2}-1}{2}+2k\pi,\pi-\arcsin\frac{1-\sqrt{2}}{2}+2k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,656 |
3. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}+3 x+3 y+24=0, \\ x^{3} y-x y^{3}+3 x^{2}-3 y^{2}-48=0 .\end{array}\right.$ | Answer: $(-3, -1)$.
Solution. The given system is equivalent to the following:
$$
\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 | 3,657 |
4. In triangle $A B C$, medians $B D$ and $C E$ intersect at point $K$. The circle constructed on segment $C K$ as a diameter passes through vertex $B$ and is tangent to line $D E$. It is known that $D K=3$. Find the height $A H$ of triangle $A B C$, the radius of the circle, and the area of triangle $A B C$. | Answer: $A H=18, R=6, S_{A B C}=54 \sqrt{3}$.
Solution. Let the center of the circle be denoted as $O$, the point of tangency of the line $D E$ with the circle as $T$, and the radius of the circle as $R$. Since the medians intersect at a point dividing them in the ratio $2: 1$ from the vertex,
“Phystech-2016", mathem... | AH=18,R=6,S_{ABC}=54\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,658 |
5. 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).
To ensure divisibility by three, we proceed as follows. Choose three digits arbitrarily (this can be done in $6 \cdot ... | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,659 |
7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 12. A sphere $\Omega$ with radius $r=\frac{3 \sqrt{5}}{2}$ touches all the lateral faces of the prism. Points $T$ and $F$ are chosen on segments $C C_{1}$ and $B B_{1}$, respectively, such that $F T \| B C$, and the planes $A B T$ and $F A_{1} C_{... | Answer: $V=405 \sqrt{3}, C T=9$ or $C T=3$.
“Fiztekh-2016”, mathematics, solutions for ticket 35
Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=3 \... | V=405\sqrt{3},CT=9orCT=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,660 |
1. Solve the inequality $\left(x^{2}+x+1\right)^{4 x^{3}+x^{2}} \geq\left(x^{2}+x+1\right)^{x-2 x^{3}}$. | Answer: $x \in\left[-1 ;-\frac{1}{2}\right] \cup\{0\} \cup\left[\frac{1}{3} ;+\infty\right)$.
Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:
$$
\left(x^{2}+x\right)\left(\left(4 x^{3}+x^{2}\right)-\left(x-2 x^{3}\right)\right) \geq 0 \Leftrightarrow... | x\in[-1;-\frac{1}{2}]\cup{0}\cup[\frac{1}{3};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,661 |
2. Solve the equation $\frac{\sin 8 x+\sin 4 x}{\cos 5 x+\cos x}=6|\sin 2 x|$. | Answer: $x=-\arccos \frac{\sqrt{13}-3}{4}+2 k \pi, x=-\arccos \frac{3-\sqrt{13}}{4}+2 k \pi, x=k \pi, k \in \mathrm{Z}$.
Solution. On the domain of definition, the given equation is equivalent to each of the following:
$\frac{2 \cos 2 x \sin 6 x}{2 \cos 3 x \cos 2 x}=6 \cdot|\sin 2 x| \Leftrightarrow \frac{2 \sin 3 x... | -\arccos\frac{\sqrt{13}-3}{4}+2k\pi,-\arccos\frac{3-\sqrt{13}}{4}+2k\pi,k\pi,k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,662 |
3. Solve the system of equations $\left\{\begin{array}{l}x^{2} y-x y^{2}-5 x+5 y+3=0, \\ x^{3} y-x y^{3}-5 x^{2}+5 y^{2}+15=0 .\end{array}\right.$ | Answer: $(4 ; 1)$.
Solution. The given system is equivalent to the following:
$$
\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)=-1... | (4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,663 |
4. In triangle $K L M$, medians $L D$ and $M E$ intersect at point $T$. The circle constructed on segment $M T$ as a diameter passes through vertex $L$ and is tangent to line $D E$. It is known that $D T=2$. Find the height $K H$ of triangle $K L M$, the angle $E T D$, and the area of triangle $K L M$. | Answer: $K H=12, \angle E T D=60^{\circ}, S_{K L M}=24 \sqrt{3}$.
Solution. Let the center of the circle be denoted as $O$, the point of tangency of line $D E$ with the circle as $S$, and the radius of the circle as $R$. Since the medians intersect at a point dividing them in the ratio $2:1$ from the vertex, $E T=\fra... | KH=12,\angleETD=60,S_{KLM}=24\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,664 |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (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 (2 ways).
To ensure divisibility by 3, we proceed as follows. Choose four digits arbitrarily (this can be done in $6 \cdot ... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,665 |
7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 5. A sphere $\Omega$ with radius $r=\frac{\sqrt{33}}{4}$ touches all the lateral faces of the prism. Points $F$ and $L$ are chosen on segments $C C_{1}$ and $B B_{1}$, respectively, such that $F L \| B C$, and the planes $L A C$ and $F A_{1} B_{1}... | Answer: $V=\frac{495 \sqrt{3}}{16}, C F=2$ or $C F=3$.
Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=\frac{3 \sqrt{11}}{2}$, the area of the base $... | V=\frac{495\sqrt{3}}{16},CF=2orCF=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,666 |
21. Find the largest integer $n$ for which param 1 is a positive prime number.
| param1 | Answer |
| :---: | :---: |
| $4 n^{4}-96 n^{2}+1$ | |
| $4 n^{4}-192 n^{2}+1$ | |
| $4 n^{4}-252 n^{2}+1$ | |
| $4 n^{4}-396 n^{2}+1$ | |
| $4 n^{4}-572 n^{2}+1$ | | | 21. Find the largest integer $n$ for which param 1 is a positive prime number.
| param1 | Answer |
| :---: | :---: |
| $4 n^{4}-96 n^{2}+1$ | 5 |
| $4 n^{4}-192 n^{2}+1$ | 7 |
| $4 n^{4}-252 n^{2}+1$ | 8 |
| $4 n^{4}-396 n^{2}+1$ | 10 |
| $4 n^{4}-572 n^{2}+1$ | 12 | | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,667 | |
23. At a round table, param1 people are sitting. Some of them are knights, and the rest are liars. It is known that knights always tell the truth, while liars always lie. Each person sitting at the table said: “Among my neighbors, there is a liar.” What is the maximum number of people sitting at the table who can say: ... | 23. At a round table, param1 people are sitting. Some of them are knights, and the rest are liars. It is known that knights always tell the truth, while liars always lie. Each person sitting at the table said: "Among my neighbors, there is a liar." What is the maximum number of people sitting at the table who can say: ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 3,669 |
25. On the hypotenuse $AB$ of a right triangle $ABC$, a point $D$ is taken. Let $DE$ and $DF$ be the angle bisectors of triangles $ADC$ and $BDC$. It turns out that $CD = EF = $ param1. What is the greatest length that segment $AB$ can have?
| 7 | |
| :---: | :--- |
| 8 | |
| 9 | |
| 10 | |
| 11 | | | 25. On the hypotenuse $AB$ of a right triangle $ABC$, a point $D$ is taken. Let $DE$ and $DF$ be the angle bisectors of triangles $ADC$ and $BDC$. It turns out that $CD = EF = $ param1. What is the maximum length that segment $AB$ can have?
| 7 | 14 |
| :---: | :---: |
| 8 | 16 |
| 9 | 18 |
| 10 | 20 |
| 11 | 22 | | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,671 | |
26. Given a grid rectangle of size param1. In how many ways can it be cut into grid rectangles of size $1 \times 1$ and $1 \times 5$?
| param1 | Answer |
| :---: | :---: |
| $1 \times 29$ | |
| $1 \times 31$ | |
| $1 \times 32$ | |
| $1 \times 33$ | |
| $1 \times 34$ | | | 26. Given a rectangular grid of size param1. In how many ways can it be cut into grid rectangles of size $1 \times 1$ and $1 \times 5$?
| param1 | Answer |
| :---: | :---: |
| $1 \times 29$ | 1757 |
| $1 \times 31$ | 3084 |
| $1 \times 32$ | 4085 |
| $1 \times 33$ | 5411 |
| $1 \times 34$ | 7168 | | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,672 | |
27. How many integer solutions ( $m ; n$ ) does the equation param1 have?
| param1 | Answer |
| :---: | :---: |
| $m^{2}+9 m-27=n^{2}$ | |
| $m^{2}+5 m-104=n^{2}$ | |
| $m^{2}+3 m-279=n^{2}$ | |
| $m^{2}+11 m-26=n^{2}$ | |
| $m^{2}+7 m-139=n^{2}$ | | | 27. How many integer solutions $(m ; n)$ does the equation param1 have?
| param1 | Answer |
| :---: | :---: |
| $m^{2}+9 m-27=n^{2}$ | 16 |
| $m^{2}+5 m-104=n^{2}$ | 18 |
| $m^{2}+3 m-279=n^{2}$ | 24 |
| $m^{2}+11 m-26=n^{2}$ | 18 |
| $m^{2}+7 m-139=n^{2}$ | 12 | | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,673 |
28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| ... | 28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,674 |
29. The numbers $a, b, c, d$ satisfy the equation param1. What is the greatest value that the difference between any two of the numbers $a, b, c, d$ can take?
| param1 | Answer |
| :---: | :---: |
| $11 \cdot \sqrt{a-11^{2}}+50 \cdot \sqrt{b-50^{2}}+13 \cdot \sqrt{c-13^{2}}+35 \cdot \sqrt{d-35^{2}}=\frac{a+b+c+d}{2}$ ... | 29. The numbers $a, b, c, d$ satisfy the equation ragat1. What is the greatest value that the difference between any two of the numbers $a, b, c, d$ can take?
| param1 | Answer |
| :---: | :---: |
| $11 \cdot \sqrt{a-11^{2}}+50 \cdot \sqrt{b-50^{2}}+13 \cdot \sqrt{c-13^{2}}+35 \cdot \sqrt{d-35^{2}}=\frac{a+b+c+d}{2}$ ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,675 | |
30. On the board, there are param1 natural numbers. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all the numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | |
| 18 | 97 | |
| 19 | 107 | |
| 26 | 153... | 30. On the board, ragat1 natural numbers are written. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | 477 |
| 18 | 97 | 357 |
| 19 | 107 | 415 |
| 2... | 477 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,676 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-a x-3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=3 x^{2}+(2-2 a) x-6-b$ and $f_{4}(x)=3 x^{2}+(4-a) x-3-2 b$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C,... | Answer: 3.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this form... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,677 |
2. Solve the system of equations $\left\{\begin{array}{l}3 x-y-3 x y=-1, \\ 9 x^{2} y^{2}+9 x^{2}+y^{2}-6 x y=13 .\end{array}\right.$ | Answer: $\left(-\frac{2}{3} ; 1\right),(1 ; 1),\left(-\frac{1}{3} ;-3\right),\left(-\frac{1}{3} ; 2\right)$.
Solution. Rewrite the system as $\left\{\begin{array}{l}(3 x-y)-3 x y=-1, \\ (3 x y)^{2}+(3 x-y)^{2}=13\end{array}\right.$, then introduce new variables: $u=3 x-y, v=3 x y$. The system becomes $\left\{\begin{ar... | (-\frac{2}{3};1),(1;1),(-\frac{1}{3};-3),(-\frac{1}{3};2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,678 |
3. In a right triangle $ABC (\angle B=90^{\circ})$, a circle $\Gamma$ with center $I$ is inscribed, touching sides $AB$ and $BC$ at points $K$ and $L$ respectively. A line passing through point $I$ intersects sides $AB$ and $BC$ at points $M$ and $N$ respectively. Find the radius of the circle $\Gamma$ if $MK=144$, $NL... | Answer: $r=60, AC=390$.
Solution. Angles $KIM$ and $LNI$ are equal as corresponding angles when lines $BC$ and $KI$ are parallel, so right triangles $KIM$ and $LNI$ are similar. Therefore, $\frac{MK}{KI}=\frac{IL}{LN}$, or (if we denote the radius of the circle by $r$) $\frac{144}{r}=\frac{r}{25}$, from which $r=60$.
... | r=60,AC=390 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,679 |
4. On the table, there are 100 different cards with numbers $3, 6, 9, \ldots 297, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 990.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 3. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the nex... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,680 |
5. Given an isosceles trapezoid \(ABCD (AD \parallel BC, AD > BC)\). A circle \(\Omega\) is inscribed in angle \(BAD\), touches segment \(BC\) at point \(C\), and intersects \(CD\) again at point \(E\) such that \(CE = 9\), \(ED = 16\). Find the radius of the circle \(\Omega\) and the area of trapezoid \(ABCD\). | Answer: $R=\frac{15}{2}, S_{A B C D}=\frac{675}{2}$.
Solution. Let the points of tangency of the circle with the sides $A B$ and $A D$ of the trapezoid be denoted as $K$ and $W$ respectively. By the tangent-secant theorem, $D W^{2}=D E \cdot D C=16 \cdot 25, D W=20$. Since $C$ and $W$ are points of tangency of the cir... | R=\frac{15}{2},S_{ABCD}=\frac{675}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,681 |
6. For what values of the parameter $a$ will there be two solutions among the solutions of the inequality $(x+2) \sqrt{a x+x-x^{2}-a} \geqslant 0$ such that the difference between them is 4? | Answer: $a \in[-6 ;-3] \cup[5 ;+\infty)$.
Solution. Note that the expression under the square root can be transformed as follows: $a x+2 x-x^{2}-2 a=a(x-1)-x(x-1)=-(x-a)(x-1)$. The domain of the inequality is determined by the condition $(x-a)(x-1) \leqslant 0$. When $a=1$, this condition becomes $(x-1)^{2} \leqslant ... | \in[-6;-3]\cup[5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,682 |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|y|+|4+y| \leqslant 4 \\
\frac{x-y^{2}-4 y-3}{2 y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $y<0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y \in \mathbb{R}$, but since $y<0$, we have $y \in (-\infty; 0)$.
2) $y=0$. Then $y+4-y \leqslant 4 \Leftrightarro... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,683 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-2 x+a, f_{2}(x)=x^{2}+b x-2, f_{3}(x)=4 x^{2}+(b-6) x+3 a-2$ and $f_{4}(x)=4 x^{2}+(3 b-2) x-6+a$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C,... | Answer: 2.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this form... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,684 |
2. Solve the system of equations $\left\{\begin{array}{l}2 y-x-2 x y=-1, \\ 4 x^{2} y^{2}+x^{2}+4 y^{2}-4 x y=61 \text {. }\end{array}\right.$ | Answer: $\left(-6 ;-\frac{1}{2}\right),(1 ; 3),\left(1 ;-\frac{5}{2}\right),\left(5 ;-\frac{1}{2}\right)$.
Solution. Rewrite the system as $\left\{\begin{array}{l}(2 y-x)-2 x y=-1, \\ (2 x y)^{2}+(2 y-x)^{2}=61\end{array}\right.$, then introduce new variables: $u=2 y-x, v=2 x y$. The system becomes $\left\{\begin{arra... | (-6;-\frac{1}{2}),(1;3),(1;-\frac{5}{2}),(5;-\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,685 |
3. In a right triangle $ABC (\angle B=90^{\circ})$, a circle $\Gamma$ with center $I$ is inscribed, touching sides $AB$ and $BC$ at points $K$ and $L$ respectively. A line passing through point $I$ intersects sides $AB$ and $BC$ at points $M$ and $N$ respectively. Find the radius of the circle $\Gamma$ if $MK=225$, $NL... | Answer: $R=120, AC=680$.
Solution. Angles $KIM$ and $LNI$ are equal as corresponding angles when lines $BC$ and $KI$ are parallel, so right triangles $KIM$ and $LNI$ are similar. Therefore, $\frac{MK}{KI}=\frac{IL}{LN}$, or (if we denote the radius of the circle by $r$) $\frac{225}{r}=\frac{r}{64}$, from which $r=120$... | 680 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,686 |
4. On the table, there are 150 different cards with numbers $2, 4, 6, \ldots 298, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 2235.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the next numb... | 2235 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,687 |
5. Given an isosceles trapezoid \(ABCD (AD \parallel BC, AD > BC)\). A circle \(\Omega\) is inscribed in angle \(BAD\), touches segment \(BC\) at point \(C\), and intersects \(CD\) again at point \(E\) such that \(CE = 16, ED = 9\). Find the radius of the circle \(\Omega\) and the area of trapezoid \(ABCD\). | Answer: $R=10, S_{A B C D}=400$.
Solution. Let the points of tangency of the circle with the sides $A B$ and $A D$ of the trapezoid be denoted as $K$ and $W$ respectively. By the tangent-secant theorem, $D W^{2}=D E \cdot D C=9 \cdot 25, D W=15$. Since $C$ and $W$ are points of tangency of the circle with parallel lin... | R=10,S_{ABCD}=400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,688 |
6. For what values of the parameter $a$ will there be two solutions among the solutions of the inequality $(x-5) \sqrt{a x+2 x-x^{2}-2 a} \leqslant 0$ such that the difference between them is 6? | Answer: $a \in(-\infty ;-4] \cup[8 ; 11]$.
Solution. Note that the expression under the square root can be transformed as follows: $a x+2 x-x^{2}-2 a=a(x-$ $2)-x(x-2)=-(x-a)(x-2)$. The domain of the inequality is determined by the condition $(x-a)(x-2) \leqslant 0$. When $a=2$, this condition becomes $(x-2)^{2} \leqsl... | \in(-\infty;-4]\cup[8;11] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,689 |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|x|+|4-x| \leqslant 4 \\
\frac{x^{2}-4 x-2 y+2}{y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 4.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $x<0$. Then $-x-x+4 \leqslant 4 \Leftrightarrow -2x \leqslant 0 \Leftrightarrow x \geqslant 0$, i.e., there are no solutions in this case.
2) $0 \leqslant x \leqslant 4$. Then $x-x+4 \leqslant 4 \Left... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,690 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)$ is $\sqrt{10}$, and the distance between the points of intersection of the graphs $y=x^{2}-1$ and $y=f(x)+1$ is $\sqrt{42}$. Find the distance between the points of intersection of th... | Answer: $\sqrt{26}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}=a x+b$, and in the second case - from the equation $x^{2}-1=a x+b+1$.
Consider the first case in more detail. The equation has the form $x^{2}-a x-b... | \sqrt{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,692 |
2. Solve the system of equations $\left\{\begin{array}{l}2 x+y+2 x y=11, \\ 2 x^{2} y+x y^{2}=15 .\end{array}\right.$ Answer: $\left(\frac{1}{2} ; 5\right),(1 ; 3),\left(\frac{3}{2} ; 2\right),\left(\frac{5}{2} ; 1\right)$. | Solution. Let's make the substitution $2 x+y=u, x y=w$. Then the system takes the form
$$
\left\{\begin{array} { l }
{ u + 2 w = 1 1 , } \\
{ u w = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ u = 1 1 - 2 w , } \\
{ w ( 1 1 - 2 w ) = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u=11-2 ... | (\frac{1}{2};5),(1;3),(\frac{3}{2};2),(\frac{5}{2};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,693 |
3. Chords $AB$ and $CD$ of a circle with center $O$ have a length of 10. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$, and $DP=3$. Line $PO$ intersects segment $AC$ at point $L$. Find the ratio $AL: LC$. | Answer: $A L: L C=3: 13$.
Solution. Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles $O P N$ and $O P H$ are equal by the leg and hypotenuse ( $O... | AL:LC=3:13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,694 |
4. There are 306 different cards with numbers $3,19,3^{2}, 19^{2}, \ldots, 3^{153}, 19^{153}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 17328.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of three are chosen. We have 76 even exponents $(2,4,6, \ldots, 152)$ and 77 odd exponents $(1,3,5, \ldots, 15... | 17328 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,695 |
5. A trapezoid $ABCD (AD \| BC)$ and a rectangle $A_{1}B_{1}C_{1}D_{1}$ are inscribed in a circle $\Omega$ with a radius of 10, such that $AC \| B_{1}D_{1}, BD \| A_{1}C_{1}$. Find the ratio of the areas of $ABCD$ and $A_{1}B_{1}C_{1}D_{1}$, given that $AD=16, BC=12$. | Answer: $\frac{49}{50}$ or $\frac{1}{2}$.
Solution. Draw a line through the center of the circle $O$ perpendicular to the bases of the trapezoid. Let it intersect $A D$ and $B C$ at points $N$ and $M$ respectively. Since a diameter perpendicular to a chord bisects the chord, $B M=M C=6, A N=N D=8$. By the Pythagorean ... | \frac{49}{50} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,696 |
6. For what values of the parameter $a$ will there be two solutions among the solutions of the inequality $\left(x^{2}-a x-x+a\right) \sqrt{x+5} \leqslant 0$ whose difference is 4? | Answer: $a \in(-\infty ;-1] \cup[5 ;+\infty)$.
Solution. The domain of the inequality is determined by the condition $x \geqslant-5$. By factoring the quadratic trinomial in the parentheses (for example, by the grouping method), we get $(x-1)(x-a) \sqrt{x+5} \leqslant 0$. We will solve this inequality using the interv... | \in(-\infty;-1]\cup[5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,697 |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x-y \geqslant|x+y| \\
\frac{x^{2}-6 x+y^{2}-8 y}{3 y-x+6} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 3.
Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant x-y, \\ x+y \geqslant y-x\end{array} \Leftrightarrow\left\{\begin{array}{l}y \leqslant 0, \\ x \geqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x-3)... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,698 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)$ is $2 \sqrt{3}$, and the distance between the points of intersection of the graphs $y=x^{2}-2$ and $y=f(x)+1$ is $\sqrt{60}$. Find the distance between the points of intersection of t... | Answer: $2 \sqrt{11}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}=a x+b$, and in the second case - from the equation $x^{2}-2=a x+b+1$.
Consider the first case in more detail. The equation has the form $x^{2}-a x... | 2\sqrt{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,699 |
2. Solve the system of equations
$$
\left\{\begin{array}{l}
x+3 y+3 x y=-1 \\
x^{2} y+3 x y^{2}=-4
\end{array}\right.
$$ | Answer: $\left(-3 ;-\frac{1}{3}\right),(-1 ;-1),\left(-1 ; \frac{4}{3}\right),\left(4 ;-\frac{1}{3}\right)$.
Solution. Let's make the substitution $x+3 y=u, x y=w$. Then the system takes the form
$$
\left\{\begin{array} { l }
{ u + 3 w = - 1 , } \\
{ u w = - 4 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }... | (-3;-\frac{1}{3}),(-1;-1),(-1;\frac{4}{3}),(4;-\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,700 |
3. Chords $AB$ and $CD$ of a circle with center $O$ have a length of 5. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$, and $DP=13$. Line $PO$ intersects segment $AC$ at point $L$. Find the ratio $AL: LC$. | Answer: $A L: L C=13: 18$.
Solution. Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles $O P N$ and $O P H$ are equal by the leg and hypotenuse ( $... | AL:LC=13:18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,701 |
4. There are 294 different cards with numbers $7, 11, 7^{2}, 11^{2}, \ldots, 7^{147}, 11^{147}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 15987.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of seven are chosen. We have 73 even exponents $(2,4,6, \ldots, 146)$ and 74 odd exponents $(1,3,5, \ldots, 14... | 15987 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,702 |
5. A trapezoid $ABCD (AD \| BC)$ and a rectangle $A_{1}B_{1}C_{1}D_{1}$ are inscribed in a circle $\Omega$ with radius 13, such that $AC \| B_{1}D_{1}, BD \| A_{1}C_{1}$. Find the ratio of the areas of $ABCD$ and $A_{1}B_{1}C_{1}D_{1}$, given that $AD=24, BC=10$. | Answer: $\frac{289}{338}$ or $\frac{1}{2}$.
Solution. Draw a line through the center of the circle $O$ perpendicular to the bases of the trapezoid. Let it intersect $A D$ and $B C$ at points $N$ and $M$ respectively. Since a diameter perpendicular to a chord bisects the chord, $B M=M C=5, A N=N D=12$. By the Pythagore... | \frac{289}{338} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,703 |
6. For what values of the parameter $a$ will there be two solutions among the solutions of the inequality $\left(x^{2}-a x+2 x-2 a\right) \sqrt{5-x} \leqslant 0$ such that the difference between them is 5? | Answer: $a \in(-\infty ;-7] \cup[0 ;+\infty)$.
Solution. The domain of the inequality is defined by the condition $x \leqslant 5$. Factoring the quadratic trinomial in the parentheses (for example, by the grouping method), we get $(x+2)(x-a) \sqrt{5-x} \leqslant 0$. We will solve this inequality using the interval met... | \in(-\infty;-7]\cup[0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,704 |
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y+x \geqslant|x-y| \\
\frac{x^{2}-8 x+y^{2}+6 y}{x+2 y-8} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. The first inequality is equivalent to the system $\left\{\begin{array}{l}x-y \leqslant x+y, \\ x-y \geqslant-x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$.
Consider the second inequality. It can be written as $\frac{(x-4)^{2}+(y+3... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,705 |
1. Find all values of $p$, for each of which the numbers $4 p+5, 2 p$ and $|p-3|$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=-1, p=\frac{15}{8}$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 p)^{2}=(4 p+5)|p-3|$. For $p \geq 3$, we get $4 p^{2}=4 p^{2}-7 p-15$, from which $p=-\frac{15}{7}$ (does not fit). If $p<3$, then $4 p^{2}=-... | p=-1,p=\frac{15}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,707 |
2. Find the value of the expression $\cos ^{4} \frac{5 \pi}{24}+\cos ^{4} \frac{11 \pi}{24}+\sin ^{4} \frac{19 \pi}{24}+\sin ^{4} \frac{13 \pi}{24}$. | Answer: $\frac{3}{2}$.
Solution. Using reduction formulas ( $\sin (\pi-\alpha)=\sin \alpha$ ), the given expression can be rewritten as $\cos ^{4} \frac{5 \pi}{24}+\cos ^{4} \frac{11 \pi}{24}+\sin ^{4} \frac{5 \pi}{24}+\sin ^{4} \frac{11 \pi}{24}$. Next, we notice that
$$
\begin{gathered}
\cos ^{4} \gamma+\sin ^{4} \... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,708 |
4. Solve the system of equations $\left\{\begin{array}{l}x+\sqrt{x+2 y}-2 y=\frac{7}{2}, \\ x^{2}+x+2 y-4 y^{2}=\frac{27}{2}\end{array}\right.$. | Answer: $\left(\frac{19}{4} ; \frac{17}{8}\right)$.
Solution. Let $\sqrt{x+2 y}=u, x-2 y=v$. Then the system takes the form
$$
\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}\lef... | (\frac{19}{4};\frac{17}{8}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,709 |
5. 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 | 3,710 |
6. Around an acute isosceles triangle $N P Q$ with base $N Q$, a circle $\Omega$ is circumscribed. Point $F$ is the midpoint of the arc $P N$ not containing point $Q$. It is known that the distances from point $F$ to the lines $P N$ and $Q N$ are 5 and $\frac{20}{3}$, respectively. Find the radius of the circle $\Omega... | Answer: $R=6, S=\frac{35 \sqrt{35}}{9}$.
Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $O F$ and $N P$ (then $G$ is the midpoint of $N P$ and $O G \perp N P$); $P H$ be the height of the triangle ($O \in P H$), $F D$ be the perpendicular dropped from point $F$ to line $N Q$, ... | R=6,S=\frac{35\sqrt{35}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,711 |
1. Find all values of $p$, for each of which the numbers $|p-8|, 2p-1$, and $4p+5$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=-1, p=\frac{39}{8}$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 p-1)^{2}=(4 p+5)|p-8|$. For $p \geq 8$, we get $4 p^{2}-4 p+1=4 p^{2}-27 p-40$, from which $p=-\frac{41}{23}$ (does not fit). If $p<8$, then ... | p=-1,p=\frac{39}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,712 |
2. Find the value of the expression $\sin ^{4} \frac{\pi}{24}+\cos ^{4} \frac{5 \pi}{24}+\sin ^{4} \frac{19 \pi}{24}+\cos ^{4} \frac{23 \pi}{24}$. | Answer: $\frac{3}{2}$.
Solution. Using reduction formulas ( $\sin (\pi-\alpha)=\sin \alpha$ ), the given expression can be rewritten as $\sin ^{4} \frac{\pi}{24}+\cos ^{4} \frac{5 \pi}{24}+\sin ^{4} \frac{5 \pi}{24}+\sin ^{4} \frac{\pi}{24} \cdot$ Next, we notice that
$$
\begin{gathered}
\cos ^{4} \gamma+\sin ^{4} \g... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,713 |
4. Solve the system of equations $\left\{\begin{array}{l}y+\sqrt{y-3 x}+3 x=12, \\ y^{2}+y-3 x-9 x^{2}=144 .\end{array}\right.$ | Answer: $(-24 ; 72),\left(-\frac{4}{3} ; 12\right)$.
Solution. Let $\sqrt{y-3 x}=u, y+3 x=v$. Then the system takes the form
$$
\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 | 3,714 |
5. 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 $(-12 ; 4)$, the diagonals of which are 6 and parallel to the coordinate axes. The second defi... | =16,=53+4\sqrt{123} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,715 |
6. Around an acute isosceles triangle $F K T$ with base $K T$, a circle $\Omega$ is described. Point $M$ is the midpoint of the arc $F T$ that does not contain point $K$. It is known that the distances from point $M$ to the lines $K T$ and $F T$ are $\frac{9}{5}$ and 1, respectively. Find the radius of the circle $\Ome... | Answer: $R=\frac{5}{3}, S=\frac{56 \sqrt{7}}{25 \sqrt{3}}$.
Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $O M$ and $F T$ (then $G$ is the midpoint of $F T$ and $O G \perp F T$); $F H$ be the height of the triangle ($O \in F H$), $M Q$ be the perpendicular dropped from point ... | R=\frac{5}{3},S=\frac{56\sqrt{7}}{25\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,716 |
1. Find all values of $p$, for each of which the numbers $9 p+10, 3 p$ and $|p-8|$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=-1, p=\frac{40}{9}$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 p)^{2}=(9 p+10) p-8 \mid$. For $p \geq 8$, we get $9 p^{2}=9 p^{2}-62 p-80$, from which $p=-\frac{40}{31} \quad$ (does not fit). If $p<8$, th... | p=-1,p=\frac{40}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,717 |
2. Find the value of the expression $\sin ^{4} \frac{5 \pi}{24}+\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{17 \pi}{24}+\cos ^{4} \frac{19 \pi}{24}$. | Answer: $\frac{3}{2}-\frac{\sqrt{3}}{4}$.
Solution. Using reduction formulas ( $\sin (\pi-\alpha)=\sin \alpha$ ), the given expression can be rewritten as $\sin ^{4} \frac{5 \pi}{24}+\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{7 \pi}{24}+\cos ^{4} \frac{5 \pi}{24}$. Next, we notice that
$$
\begin{gathered}
\cos ^{4} \... | \frac{3}{2}-\frac{\sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,718 |
4. Solve the system of equations $\left\{\begin{array}{l}2 x+\sqrt{2 x+3 y}-3 y=5, \\ 4 x^{2}+2 x+3 y-9 y^{2}=32 .\end{array}\right.$ | Answer: $\left(\frac{17}{4} ; \frac{5}{2}\right)$.
Solution. Let $\sqrt{2 x+3 y}=u, 2 x-3 y=v$. Then the system becomes
$$
\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.
$$
Fro... | (\frac{17}{4};\frac{5}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,719 |
5. Find all values of the parameter $a$, for each of which the system of equations
$$
\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 | 3,720 |
6. Around an acute isosceles triangle $A B C$ with base $B C$, a circle $\Omega$ is circumscribed. Point $T$ is the midpoint of the arc $A C$ that does not contain point $B$. It is known that the distances from point $T$ to the lines $A C$ and $B C$ are 3 and 7, respectively. Find the radius of the circle $\Omega$ and ... | Answer: $R=9, S=40 \sqrt{5}$.
Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $OT$ and $AC$ (then $G$ is the midpoint of $AC$ and $OG \perp AC$); $AH$ be the height of the triangle ($O \in AH$), $TQ$ be the perpendicular dropped from point $T$ to line $BC$, and $OJ$ be the perp... | R=9,S=40\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,721 |
1. Find all values of $p$, for each of which the numbers $|p-3|, 3p+1$, and $9p+10$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=-1, p=\frac{29}{18}$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 p+1)^{2}=(9 p+10)|p-3|$. For $p \geq 3$, we get $9 p^{2}+6 p+1=9 p^{2}-17 p-30$, from which $p=-\frac{31}{23}$ (does not fit). If $p<3$, the... | p=-1,p=\frac{29}{18} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,722 |
2. Find the value of the expression $\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{11 \pi}{24}+\sin ^{4} \frac{17 \pi}{24}+\cos ^{4} \frac{13 \pi}{24}$. | Answer: $\frac{3}{2}$.
Solution. Using reduction formulas ( $\sin (\pi-\alpha)=\sin \alpha$ ), the given expression can be rewritten as $\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{13 \pi}{24}+\sin ^{4} \frac{7 \pi}{24}+\cos ^{4} \frac{13 \pi}{24}$. Next, we notice that
$$
\begin{gathered}
\cos ^{4} \gamma+\sin ^{4} \... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,723 |
4. 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 o... | (2,-3),(6,-18) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,724 |
5. 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 | 3,725 |
6. Around an acute isosceles triangle $C L E$ with base $L E$, a circle $\Omega$ is described. Point $N$ is the midpoint of the arc $C E$ not containing point $L$. It is known that the distances from point $N$ to the lines $C E$ and $E L$ are 6 and 9, respectively. Find the radius of the circle $\Omega$ and the area of... | Answer: $R=8, S=15 \sqrt{15}$.
Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $O N$ and $C E$ (then $G$ is the midpoint of $C E$ and $O G \perp C E$); $C H$ be the height of the triangle ($O \in C H$), $N Q$ be the perpendicular dropped from point $N$ to line $E L$, and $O J$ ... | R=8,S=15\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,726 |
1. Find all values of $p$, for each of which the numbers $p-2$, $2 \cdot \sqrt{p}$, and $-3-p$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(p-2)(-p-3)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+5 p-6=0
\end{array} \Leftrightarrow p=1\right.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,727 |
2. Plot on the plane $(x ; y)$ the set of points satisfying the equation $\left|16+6 x-x^{2}-y^{2}\right|+|6 x|=16+12 x-x^{2}-y^{2}$, and find the area of the resulting figure. | Answer: $25 \pi-25$ arcsin $0.8+12$.
Solution. 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 of inequalities
$$
\lef... | 25\pi-25\arcsin0.8+12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,728 |
3. Find the value of the expression $\operatorname{tg} 20^{\circ}+4 \sin 20^{\circ}$. | Answer: $\sqrt{3}$.
Solution. $\operatorname{tg} 20^{\circ}+4 \sin 20^{\circ}=\frac{\sin 20^{\circ}+4 \sin 20^{\circ} \cos 20^{\circ}}{\cos 20^{\circ}}=\frac{\sin 20^{\circ}+2 \sin 40^{\circ}}{\cos 20^{\circ}}=\frac{\left(\sin 20^{\circ}+\sin 40^{\circ}\right)+\sin 40^{\circ}}{\cos 20^{\circ}}=$ $=\frac{2 \sin 30^{\ci... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,729 |
4. 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. We will choose four digits arb... | 5184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,730 |
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 4 \sqrt{\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^{... | (\frac{1}{3};3),(\sqrt[4]{\frac{21}{76}};2\cdot4\sqrt[4]{\frac{84}{19}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,731 |
6. Around an acute isosceles triangle $N P Q$ with base $N Q$, a circle $\Omega$ is circumscribed. The distance from the midpoint of the arc $P N$, not containing point $Q$, to the side $P N$ is 4, and the distance from the midpoint of the arc $Q N$, not containing point $P$, to the side $Q N$ is 0.4. Find the radius o... | Answer: $R=5, S=\frac{192 \sqrt{6}}{25}$.
Solution. Let $C$ and $D$ be the midpoints of arcs $N P$ and $N Q$; $A$ and $H$ be the midpoints of segments $P N$ and $N Q$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O C \perp P N, O D \perp Q N ; O A=O C-A C=R-4, O H=O D-D H=R-\frac{2}{5}$.
From th... | R=5,S=\frac{192\sqrt{6}}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,732 |
1. Find all values of $p$, for each of which the numbers $-p-12, 2 \cdot \sqrt{p}$, and $p-5$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(-p-12)(p-5)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+11 p-60=0
\end{array} \Leftrightarrow p=4\right.
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,733 |
2. Plot on the plane $(x ; y)$ the set of points satisfying the equation $\left|9+8 y-x^{2}-y^{2}\right|+|8 y|=16 y+9-x^{2}-y^{2}$, and find the area of the resulting figure. | Answer: $25 \pi-25 \arcsin 0.6+12$.
Solution. Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system of inequalities
$$
\left... | 25\pi-25\arcsin0.6+12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,734 |
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