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4. [5 points] Solve the equation $5 x^{4}+x^{2}+8 x-6 x^{2}|x+4|+16=0$.
The text above has been translated into English, maintaining the original text's line breaks and format. | Answer: $-\frac{4}{5} ; 1 ; \frac{1 \pm \sqrt{17}}{2}$.
Solution. The equation can be rewritten as $5 x^{4}+(x+4)^{2}-6 x^{2}|x+4|=0$ or $|x+4|^{2}-6 x^{2} \mid x+$ $4 \mid+5 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x+4|$ with a discriminant equal to $\left(6 ... | -\frac{4}{5};1;\frac{1\\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,226 |
5. [5 points] Around a hook with a worm, in the same plane as it, a carp and a minnow are swimming along two circles. In the specified plane, a rectangular coordinate system is introduced, in which the hook (the common center of the circles) is located at the point $(0 ; 0)$. At the initial moment of time, the carp and... | Answer: $(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4),(\sqrt{2}-4 ;-4-\sqrt{2}),(-\sqrt{2}-4 ; 4-\sqrt{2})$.
Solution. Let the points where the carp and the minnow are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles formed by the radius vectors of points $M$ and $N... | (4-\sqrt{2};4+\sqrt{2}),(4+\sqrt{2};\sqrt{2}-4),(\sqrt{2}-4;-4-\sqrt{2}),(-\sqrt{2}-4;4-\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,227 |
6. [5 points] a) Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omeg... | Answer: $\frac{R_{2}}{R_{1}}=\frac{5}{4}, R_{1}=\frac{2}{\sqrt{5}}, R_{2}=\frac{\sqrt{5}}{2}$.
Fig. 3: Variant 12, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\om... | \frac{R_{2}}{R_{1}}=\frac{5}{4},R_{1}=\frac{2}{\sqrt{5}},R_{2}=\frac{\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,228 |
7. [7 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
(x-a)^{2}+y^{2}=64 \\
(|x|-8)^{2}+(|y|-15)^{2}=289
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{-28\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{28\}$.
Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordinate axes. In the firs... | \in{-28}\cup(-24;-8]\cup[8;24)\cup{28} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,229 |
1. In an isosceles triangle \(ABC\), the base \(AC\) is equal to 1, and the angle \(ABC\) is equal to \(2 \operatorname{arctg} \frac{1}{2}\). Point \(D\) lies on side \(BC\) such that the area of triangle \(ABC\) is four times the area of triangle \(ADC\). Find the distance from point \(D\) to the line \(AB\) and the r... | Answer: distance $=\frac{3}{2 \sqrt{5}}, \quad$ radius $=\frac{\sqrt{265}}{32}$. | distance=\frac{3}{2\sqrt{5}},\quadradius=\frac{\sqrt{265}}{32} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,230 |
2. Solve the equation
$$
\frac{\sin 3 x \cos 5 x-\sin 2 x \cos 6 x}{\cos x}=0
$$ | Answer: $\pi n, \quad \pm \frac{\pi}{6}+\pi n, \quad n \in \mathbb{Z}$. | \pin,\quad\\frac{\pi}{6}+\pin,\quadn\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,231 |
4. Solve the system of equations
$$
\left\{\begin{aligned}
\log _{x}(y+1) & =4 \log _{x+2} \sqrt{y-1} \\
\log _{y-1}(x+2) & =\log _{x}\left(\frac{x^{3}}{y+1}\right)
\end{aligned}\right.
$$ | Answer: $\left(\frac{1+\sqrt{17}}{2}, \frac{7+\sqrt{17}}{2}\right),\left(\frac{5+\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right)$. | (\frac{1+\sqrt{17}}{2},\frac{7+\sqrt{17}}{2}),(\frac{5+\sqrt{17}}{2},\frac{3+\sqrt{17}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,232 |
6. The base of the triangular pyramid $S A B C$ is an equilateral triangle $A B C$ with a side length of 8. The lateral edge $S C$ is perpendicular to the base and has a length of 15. A sphere with center $O$ lying in the plane $S B C$ touches the edges $S A$, $A B$, and $A C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, r... | Answer: $A A_{1}=6, \quad$ distance $=\frac{18}{5}, \quad$ radius $=\frac{4 \sqrt{39}}{5}$. | AA_{1}=6,\quad | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,233 |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=2, E N=5$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, then $H M=A M=F M=2$, $C N=E N=H N=5$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\alpha... | \angleABC=60,S_{\triangleABC}=36\sqrt{3},R=2\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,234 |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 32, the sum of the numbers on the board will become 477. If the largest number is multiplied by 14, the sum of the numbers on the board will also become 477. What numbers could have been written on ... | Answer: $13,14,16,31$ or $13,30,31$.
Solution. Let there be $n$ numbers $x_{1}32 x_{1} \geqslant 32 \cdot(13 \cdot 2)>477$, which is impossible. Therefore, $k=1$, from which $x_{1}=13, x_{n}=31$.
Then $32 x_{1}+x_{n}=32 \cdot 13+31=447$, and the sum of the remaining numbers is $477-447=30$, and the remaining numbers ... | 13,14,16,31or13,30,31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,235 |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-6 a x-2 a y+2 x^{2}+2 x y+y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-8 a^{2} x-2 a^{3} y+12 a y+a^{4}+36=0$. Find all values of the parameter... | Answer: $(-\infty ;-2) \cup(-1 ; 0) \cup(3 ;+\infty)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(y^{2}+2(x-a) y\right)+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+2 \cdot y \cdot(x-a)+(x-a)^{2}\right)-(x-a)^{2}+2 x^{2}-6 a x+5 a^{2}=0 \Leftright... | (-\infty;-2)\cup(-1;0)\cup(3;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,236 |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-2 x^{2} y^{2}=3 \\
x^{4}+y^{4}+\frac{2}{3} x^{2} y^{2}=17
\end{array}\right.
$$ | Answer: $(\sqrt{2} ; \pm \sqrt{3}),(-\sqrt{2} ; \pm \sqrt{3}),(\sqrt{3} ; \pm \sqrt{2}),(-\sqrt{3} ; \pm \sqrt{2})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
3 u-2 v=3 \\
... | (\sqrt{2};\\sqrt{3}),(-\sqrt{2};\\sqrt{3}),(\sqrt{3};\\sqrt{2}),(-\sqrt{3};\\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,237 |
5. [5 points] A magician has a set of $12^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 12 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 1386.
Solution. Since the magician has a set of $12^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 12,1 \leqslant j \leqslant 12$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 1386 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,238 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=2, AD=3$. Find... | Answer: $\frac{19}{25}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{19}{25} | Geometry | proof | Yes | Yes | olympiads | false | 3,239 |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=2, E N=11$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circl... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=2$, $C N=E N=H N=11$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\a... | \angleABC=60,S_{\triangleABC}=120\sqrt{3},R=14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,240 |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 30, the sum of the numbers on the board will become 450. If the largest number is multiplied by 14, the sum of the numbers on the board will also become 450. What numbers could have been written on ... | Answer: $13,14,17,29$ or $13,15,16,29$.
Solution. Let there be $n$ numbers $x_{1}30 x_{1} \geqslant$ $30 \cdot(13 \cdot 2)>450$, which is impossible. Therefore, $k=1$, from which $x_{1}=13, x_{n}=29$.
Then $30 x_{1}+x_{n}=30 \cdot 13+29=419$, and the sum of the remaining numbers is $450-419=31$, and the remaining num... | 13,14,17,29or13,15,16,29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,241 |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $2 a^{2}+2 a x+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+a^{2} y^{2}-$ $2 a^{3} x-6 a x-2 a^{2} y+a^{4}+9=0$. Find all values of the parameter $a$ for... | Answer: $(-\infty ;-2) \cup(0 ; 1) \cup(3 ; \infty)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
&\left(x^{2}+2(a-y) x\right)+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}+2 \cdot x \cdot(a-y)+(a-y)^{2}\right)-(a-y)^{2}+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Le... | (-\infty;-2)\cup(0;1)\cup(3;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,242 |
5. [5 points] A magician has a set of $15^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 15 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 2835.
Solution. Since the magician has a set of $15^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 15,1 \leqslant j \leqslant 15$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 2835 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,244 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=3, AD=4$. Find... | Answer: $\frac{37}{49}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{37}{49} | Geometry | proof | Yes | Yes | olympiads | false | 3,245 |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=1, E N=7$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=1$, $C N=E N=H N=7$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\al... | \angleABC=60,S_{\triangleABC}=45\sqrt{3},R=2\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,246 |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 32, the sum of the numbers on the board will become 581. If the largest number is multiplied by 17, the sum of the numbers on the board will also become 581. What numbers could have been written on ... | Answer: $16,17,21,31$ or $16,18,20,31$.
Solution. Let there be $n$ numbers $x_{1}32 x_{1} \geqslant$ $32 \cdot(16 \cdot 2)>581$, which is impossible. Therefore, $k=1$, from which $x_{1}=16, x_{n}=31$.
Then $32 x_{1}+x_{n}=32 \cdot 16+31=543$, and the sum of the remaining numbers is $581-543=38$, with the remaining nu... | 16,17,21,31or16,18,20,31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,247 |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-6 a x-4 a y+2 x^{2}+2 x y+y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-6 a^{2} x-2 a^{3} y+4 a y+a^{4}+4=0$. Find all values of the parameter $... | Answer: $(-1 ; 0) \cup(1 ; 2)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(y^{2}+2(x-2 a) y\right)+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+2 \cdot y \cdot(x-2 a)+(x-2 a)^{2}\right)-(x-2 a)^{2}+2 x^{2}-6 a x+5 a^{2}=0 \Leftrightarrow \\
& \qua... | (-1;0)\cup(1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,248 |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-x^{2} y^{2}=3 \\
x^{4}+y^{4}-x^{2} y^{2}=31
\end{array}\right.
$$ | Answer: $(\sqrt{5} ; \pm \sqrt{6}),(-\sqrt{5} ; \pm \sqrt{6}),(\sqrt{6} ; \pm \sqrt{5}),(-\sqrt{6} ; \pm \sqrt{5})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
3 u-v=3 \\
u^... | (\sqrt{5};\\sqrt{6}),(-\sqrt{5};\\sqrt{6}),(\sqrt{6};\\sqrt{5}),(-\sqrt{6};\\sqrt{5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,249 |
5. [5 points] A magician has a set of $20^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 20 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 7030.
Solution. Since the set contains $20^{2}$ cards, the magician has all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two t... | 7030 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,250 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=4, AD=5$. Find... | Answer: $\frac{61}{81}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\... | \frac{61}{81} | Geometry | proof | Yes | Yes | olympiads | false | 3,251 |
1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=1, E N=4$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle... | Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, we have $H M=A M=F M=1$, $C N=E N=H N=4$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\al... | \angleABC=60,S_{\triangleABC}=18\sqrt{3},R=2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,252 |
2. [5 points] Several pairwise distinct natural numbers are written on the board. If the smallest number is multiplied by 35, the sum of the numbers on the board will become 592. If the largest number is multiplied by 16, the sum of the numbers on the board will also become 592. What numbers could have been written on ... | # Answer: $15,16,17,34$ or $15,33,34$.
Solution. Let there be $n$ numbers $x_{1}35 x_{1} \geqslant$ $35 \cdot(15 \cdot 2)>592$, which is impossible. Therefore, $k=1$, from which $x_{1}=15, x_{n}=34$.
Then $35 x_{1}+x_{n}=32 \cdot 15+34=559$, and the sum of the remaining numbers is $592-559=33$, and the remaining numb... | 15,16,17,34or15,33,34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,253 |
3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-4 a x+6 a y+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-4 a^{3} x-2 a x+2 a^{2} y+4 a^{4}+1=0$. Find all values of the parameter... | Answer: $\left(0 ; \frac{1}{2}\right) \cup(1 ; 3)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(x^{2}-2(2 a+y) x\right)+2 y^{2}+6 a y+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}-2 \cdot x \cdot(2 a+y)+(2 a+y)^{2}\right)-(2 a+y)^{2}+2 y^{2}+6 a y+5 a^{2}=0 \Left... | (0;\frac{1}{2})\cup(1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,254 |
4. [4 points] Solve the system of equations
$$
\left\{\begin{array}{l}
2 x^{2}+2 y^{2}-x^{2} y^{2}=2 \\
x^{4}+y^{4}-\frac{1}{2} x^{2} y^{2}=19
\end{array}\right.
$$ | Answer: $(2 ; \pm \sqrt{3}),(-2 ; \pm \sqrt{3}),(\sqrt{3} ; \pm 2),(-\sqrt{3} ; \pm 2)$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system becomes
$$
\left\{\begin{array}{l}
2 u-v=2 \\
u^{2}-2 v-\frac{1}{2} v=19
\en... | (2;\\sqrt{3}),(-2;\\sqrt{3}),(\sqrt{3};\2),(-\sqrt{3};\2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,255 |
5. [5 points] A magician has a set of $16^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 16 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 3480.
Solution. Since the magician has a set of $16^{2}$ cards, there are all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 16,1 \leqslant j \leqslant 16$ there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two ... | 3480 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,256 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=3, AD=5$. Find... | Answer: $\frac{49}{64}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid, so a circle (let's call it $\Omega$) can be circumscribed around it. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^{... | \frac{49}{64} | Geometry | proof | Yes | Yes | olympiads | false | 3,257 |
1. When $x^{2}$ was added to the quadratic trinomial $f(x)$, its minimum value increased by 1, and when $x^{2}$ was subtracted from it, its minimum value decreased by 3. How will the minimum value of $f(x)$ change if $2 x^{2}$ is added to it? | # Answer. It will increase by $\frac{3}{2}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its minimum value, its leading coefficient is positive, and the minimum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \c... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,258 |
2. Solve the inequality $x^{\log _{3} x}-2 \leqslant(\sqrt[3]{3})^{\log _{\sqrt{3}}^{2} x}-2 \cdot x^{\log _{3}} \sqrt[3]{x}$. | Answer. $x \in\left(0 ; 3^{-\sqrt{\log _{3} 2}}\right] \cup\{1\} \cup\left[3 \sqrt{\log _{3} 2} ;+\infty\right)$.
Matching the third with the second, and the second with the fourth, we facto... | x\in(0;3^{-\sqrt{\log_{3}2}}]\cup{1}\cup[3^{\sqrt{\log_{3}2}};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,259 |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{2}{5}\right)$, and the numbers $3+\sin x, 3+\sin y, 3+\sin z$ form a non-constant geometric progression in the given order. Find $\sin y$. | Answer: $-\frac{1}{10}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
$... | -\frac{1}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,260 |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are 4 and 9, respectively.
a) Find the distance from point $A$ to the line $BC$.
b... | Answer. a) $5 ;$ b) $R=\frac{32}{7} ; A B=\frac{16}{\sqrt{7}}$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater ... | )5;b)R=\frac{32}{7};AB=\frac{16}{\sqrt{7}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,261 |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $60 ; 45$ ). Find the number of such squares. | Answer: 2070.
Solution. Draw through the given point $(60 ; 45)$ vertical and horizontal lines $(x=60$ and $y=45)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "bottom" vertex of the square can be located in 45 ways: ... | 2070 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,262 |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a - 2 \leqslant 0 \\
x^{2} + y^{2} + 6x - 2y - b^{2} + 4b + 6 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ; 4-\sqrt{10}] \cup[\sqrt{10} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a-2=0$ is 2, and the point $(0 ; 0)$ satisfies this inequality. Therefore, the inequality defines a half-... | b\in(-\infty;4-\sqrt{10}]\cup[\sqrt{10};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,263 |
7. The base of the triangular pyramid $A B C D$ is an equilateral triangle $A B C$. The volume of the pyramid is $\frac{25}{\sqrt{3}}$, and its height, drawn from vertex $D$, is 3. Point $M$ is the midpoint of edge $C D$. It is known that the radii of the spheres inscribed in pyramids $A B C M$ and $A B D M$ are equal.... | Answer. a) $\angle=\arccos \left( \pm \frac{4}{5}\right)$; b) $C D=\frac{\sqrt{31}}{\sqrt{3}}$ or $C D=3 \sqrt{13}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $A B C M$ and $A B D... | \arccos(\\frac{4}{5}),\frac{\sqrt{31}}{\sqrt{3}},3\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,264 |
1. When $x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by $\frac{27}{2}$, and when $4 x^{2}$ was subtracted from it, its maximum value decreased by 9. How will the maximum value of $f(x)$ change if $2 x^{2}$ is subtracted from it? | Answer. It will decrease by $\frac{27}{4}$.
Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its maximum value, its leading coefficient is negative, and the maximum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \cd... | \frac{27}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,265 |
2. Solve the inequality $(\sqrt[10]{125})^{\log _{\sqrt{5}}^{2} x}+3 \geqslant x^{\log _{5} x}+3(\sqrt[5]{x})^{\log _{5} x}$. | Answer. $x \in\left(0 ; 5^{-\sqrt{\log _{5} 3}}\right] \cup\{1\} \cup\left[5 \sqrt{\log _{5} 3} ;+\infty\right)$.
Solution. Rewrite the inequality as $5^{\frac{6}{5} \log _{5}^{2} x}+3-5^{\log _{5}^{2} x}-3 \cdot 5^{\frac{1}{5} \log _{5}^{2} x} \geqslant 0$. By grouping the first term with the third and the second wit... | x\in(0;5^{-\sqrt{\log_{5}3}}]\cup{1}\cup[5\sqrt{\log_{5}3};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,266 |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{1}{9}$, and the numbers $5+\cos x, 5+\cos y, 5+\cos z$ form a non-constant geometric progression in the given order. Find $\cos y$. | Answer: $-\frac{1}{9}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
\[... | -\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,267 |
4. In triangle $ABC$, side $AB$ is equal to $\sqrt{11}$, and the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are in the ratio $9: 25$.
a) Find the ratio... | Answer. a) $9: 16$, b) $d_{C}=\frac{275}{54}, R=3$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\... | )9:16,b)d_{C}=\frac{275}{54},R=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,268 |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 40)$. Find the number of such squares. | Answer: 1560.
Solution. Draw through the given point $(55 ; 40)$ vertical and horizontal lines $(x=55$ and $y=40)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 39 ways: $... | 1560 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,269 |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a + 4 \leqslant 0 \\
x^{2} + y^{2} + 10 x + 2 y - b^{2} - 8 b + 10 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-8-\sqrt{26}] \cup[\sqrt{26} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a+4=0$ is 3, and the point $(0 ; 0)$ does not satisfy this inequality. Therefore, the inequality defines ... | b\in(-\infty;-8-\sqrt{26}]\cup[\sqrt{26};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,270 |
7. The base of the triangular pyramid $A B C D$ is an equilateral triangle $A B C$. The volume of the pyramid is $\frac{100}{3 \sqrt{3}}$, and its height, drawn from vertex $D$, is 4. Point $M$ is the midpoint of edge $C D$. It is known that the radii of the spheres inscribed in pyramids $A B C M$ and $A B D M$ are equ... | Answer. a) $\angle=\arccos \left( \pm \frac{3}{5}\right)$; b) $C D=\frac{8}{\sqrt{3}}$ or $C D=\frac{4 \sqrt{19}}{\sqrt{3}}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $A B C M$ a... | \arccos(\\frac{3}{5}),\frac{8}{\sqrt{3}},\frac{4\sqrt{19}}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,271 |
2. Solve the inequality $x^{\log _{13} x}+7(\sqrt[3]{x})^{\log _{13} x} \leqslant 7+(\sqrt[3]{13})^{\log _{\sqrt{13}}^{2} x}$ | Answer. $x \in\left(0 ; 13^{-\sqrt{\log _{13} 7}}\right] \cup\{1\} \cup\left[13 \sqrt{\log _{13} 7} ;+\infty\right)$.
Solution. Rewrite the inequality as $13^{\log _{13}^{2} x}+7 \cdot 13^{\frac{1}{3} \log _{13}^{2} x}-7-13^{\frac{4}{3} \log _{13}^{2} x} \leqslant 0$. By grouping the first term with the fourth and the... | x\in(0;13^{-\sqrt{\log_{13}7}}]\cup{1}\cup[13\sqrt{\log_{13}7};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,272 |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{5}\right)$, and the numbers $2+\sin x, 2+\sin y, 2+\sin z$ form a non-constant geometric progression in the given order. Find $\sin y$. | Answer: $-\frac{1}{5}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of the two adjacent terms), we obtain the equatio... | -\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,273 |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are 5 and 12, respectively.
a) Find the distance from point $A$ to the line $BC$.
... | Answer. a) 7; b) $R=\frac{25}{4}; BC=5\sqrt{6}$.
Solution. Let $\angle ACB = \gamma$, then $\angle BAC = 2\gamma$. By the generalized sine theorem, we find that $AB = 2R \sin \gamma, BC = 2R \sin 2\gamma$. Note that $\gamma < 90^\circ$ (otherwise, the sum of the angles in the triangle would be greater than $180^\circ$... | )7;b)R=\frac{25}{4};BC=5\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,274 |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $25 ; 60$ ). Find the number of such squares. | Answer: 650.
Solution. Draw through the given point $(25 ; 60)$ vertical and horizontal lines $(x=25$ and $y=60)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 25 ways: $(0... | 650 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,275 |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a - y \sin a - 3 \leqslant 0 \\
x^{2} + y^{2} - 8 x + 2 y - b^{2} - 6 b + 8 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-\sqrt{17}] \cup[\sqrt{17}-6 ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a-y \sin a-3=0$ is 3, and the point $(0 ; 0)$ satisfies this inequality. Therefore, the inequality defines a half-... | b\in(-\infty;-\sqrt{17}]\cup[\sqrt{17}-6;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,276 |
7. The base of the triangular pyramid $K L M N$ with a volume of 75 is an equilateral triangle $K L M$ with a side length of 10. Point $T$ is the midpoint of edge $M N$. It is known that the radii of the spheres inscribed in pyramids $K L M T$ and $K L N T$ are equal to each other.
a) Find all possible values of the a... | Answer. a) $\angle=\arccos \left( \pm \frac{4}{5}\right)$; b) $M N=\sqrt{31}$ or $M N=3 \sqrt{39}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $K L M T$ and $K L N T$ are equal (th... | \arccos(\\frac{4}{5}),\sqrt{31},3\sqrt{39} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,277 |
2. Solve the inequality $(\sqrt[7]{4})^{\log _{\sqrt{2}}^{2} x}+6 \geqslant x^{\log _{2} x}+6(\sqrt[7]{x})^{\log _{2} x}$. | Answer. $x \in\left(0 ; 2^{-\sqrt{\log _{2} 6}}\right] \cup\{1\} \cup\left[2 \sqrt{\log _{2} 6} ;+\infty\right)$.
Solution. Rewrite the inequality as $2^{\frac{8}{7} \log _{2}^{2} x}+6-2^{\log _{2}^{2} x}-6 \cdot 2^{\frac{1}{7} \log _{2}^{2} x} \geqslant 0$. Grouping the first term with the third and the second with t... | x\in(0;2^{-\sqrt{\log_{2}6}}]\cup{1}\cup[2\sqrt{\log_{2}6};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,278 |
3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{5}{9}$, and the numbers $1+\cos x, 1+\cos y, 1+\cos z$ form a non-constant geometric progression in the given order. Find $\cos y$. | Answer: $-\frac{7}{9}$.
Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation
$$... | -\frac{7}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,279 |
4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The radius of the circle $\Omega$ is $5 \sqrt{2}$, and the distances from points $A$ and $C$ to the tangent line $\ell$ are in the ratio $2:5$... | Answer. a) $2: 3$; b) $A B=5 \sqrt{3}, d_{C}=\frac{75}{4 \sqrt{2}}$.
Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be gre... | )2:3;b)AB=5\sqrt{3},d_{C}=\frac{75}{4\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,280 |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(35 ; 65)$. Find the number of such squares. | Answer: 1190.
Solution. Draw through the given point $(35 ; 65)$ vertical and horizontal lines $(x=35$ and $y=65)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 34 ways: $(... | 1190 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,281 |
6. Find all values of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
x \cos a + y \sin a + 3 \leqslant 0 \\
x^{2} + y^{2} + 8x - 4y - b^{2} + 6b + 11 = 0
\end{array}\right.
$$
has at least one solution for any value of the parameter $a$. | Answer. $b \in(-\infty ;-2 \sqrt{5}] \cup[6+2 \sqrt{5} ;+\infty)$.
Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a+3=0$ is 3, and the point $(0 ; 0)$ does not satisfy this inequality. Therefore, the inequality define... | b\in(-\infty;-2\sqrt{5}]\cup[6+2\sqrt{5};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,282 |
7. The base of the triangular pyramid $K L M N$ with a volume of 100 is an equilateral triangle $K L M$ with a side length of 10. Point $T$ is the midpoint of edge $M N$. It is known that the radii of the spheres inscribed in pyramids $K L M T$ and $K L N T$ are equal.
a) Find all possible values of the angle between ... | Answer. a) $\angle=\arccos \left( \pm \frac{3}{5}\right)$; b) $C D=8$ or $C D=4 \sqrt{19}$.
Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $K L M T$ and $K L N T$ are equal (the face $... | \arccos(\\frac{3}{5}),8,4\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,283 |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 5, and 5, respectively. Find the smallest possible value of $f(x)$. | Answer: $\frac{39}{8}$.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text {v }}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented ... | \frac{39}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,284 |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{3}{7}\right)$, and the numbers $\frac{1}{\cos x}, \frac{7}{\cos y}, \frac{1}{\cos z}$ also form an arithmetic progression in the given order. Find $\cos ^{2} y$. | Answer: $\frac{10}{13}$.
Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{14}{\cos y}=\frac{1}{\cos x}+\frac{1}{\cos z}$. Transform the second equation:
$$
\frac{14}{\cos y}=\frac{\cos x+\cos z}{\cos x \cos z} \Leftrightarrow \frac{14}{\cos y}=\frac{2 \cos \fr... | \frac{10}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,285 |
3. Solve the inequality $\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3 \leqslant 64^{\log _{4}^{2} x}-15 \cdot x^{\log _{4} x}$. | Answer. $x \in\left(0 ; \frac{1}{4}\right] \cup[4 ;+\infty)$.
Solution. Transform the left side of the inequality:
$$
\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\log _{3} 2\right)}{\log _{3} 162}=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\lo... | x\in(0;\frac{1}{4}]\cup[4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,286 |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 7. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=14 ;$ b) $\angle D A C=45^{\circ}, S_{A B C D}=97$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\... | AC=14,\angleDAC=45,S_{ABCD}=97 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,287 |
5. Given the number $5300 \ldots 0035$ (100 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 495. In how many ways can this be done? | Answer: 22100.
Solution. $495=5 \cdot 9 \cdot 11$. Divisibility by 5 is always satisfied since the number ends in five. For investigating divisibility by 11, it is significant which positions the replaceable digits occupy.
First case. We replace two zeros in positions of the same parity (both even or both odd). For d... | 22100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,288 |
6. Find all values of the parameter $a$ for which there exists a value of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
\arcsin \left(\frac{a-y}{3}\right)=\arcsin \left(\frac{4-x}{4}\right) \\
x^{2}+y^{2}-8 x-8 y=b
\end{array}\right.
$$
has exactly two solutions. | Answer. $a \in\left(-\frac{13}{3} ; \frac{37}{3}\right)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{a-y}{3}=\frac{4-x}{4}, y=\frac{3 x}{4}-3+a$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4-x}{4} \leqslant 1,0 \leqslant x \leqslant 8... | \in(-\frac{13}{3};\frac{37}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,289 |
7. Consider quadrilateral pyramids $MABCD$ with the following properties: the base of the pyramid is a convex quadrilateral $ABCD$, where $AB = BC = 1$, $CD = DA = 2$, and each of the planes of the lateral faces $MAB$, $MBC$, $MCD$, $MDA$ forms an angle of $45^{\circ}$ with the plane of the base.
a) Find the volume of... | Answer. a) $\frac{27}{25} ;$ b) $\frac{4}{3}$.
Let $M H$ be the height of the pyramid, $(M H=h), P$ be the projection of $M$ onto the line $A B$. Then $M H P$ is a right triangle with angle $\angle M P H=45^{\circ}$, from which $H P=h \cdot \operatorname{ctg} 45^{\circ}=h$. Similarly, it can be proved that the point $... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,290 |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values $-9, -9$, and $-15$ respectively. Find the greatest possible value of $f(x)$. | Answer: $-\frac{33}{4}$.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text {v }}$, then $x_{\mathrm{B}}=n+0.5$, and thus $f(x)$ can be represent... | -\frac{33}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,291 |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arcsin \frac{\sqrt{7}}{4}$, and the numbers $\frac{1}{\sin x}, \frac{4}{\sin y}, \frac{1}{\sin z}$ also form an arithmetic progression in the given order. Find $\sin ^{2} y$. | Answer: $\frac{7}{13}$.
Solution: Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{8}{\sin y}=\frac{1}{\sin x}+\frac{1}{\sin z}$. Transform the second equation:
$$
\frac{8}{\sin y}=\frac{\sin x+\sin z}{\sin x \sin z} \Leftrightarrow \frac{8}{\sin y}=\frac{2 \sin \frac{x... | \frac{7}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,292 |
3. Solve the inequality $\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5 \leqslant 125^{\log _{5}^{2} x}-24 \cdot x^{\log _{5} x}$. Answer. $x \in\left(0 ; \frac{1}{5}\right] \cup[5 ;+\infty)$. | Solution. Transform the left side of the inequality:
$\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{\log _{5} 50}=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{2+\log _{5} 2}=5$
Let $5^{\log _{5}^{2} x}=v... | x\in(0;\frac{1}{5}]\cup[5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,293 |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 4. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=8$; b) $\angle D A C=45^{\circ}, S_{A B C D}=31$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\an... | AC=8,\angleDAC=45,S_{ABCD}=31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,294 |
5. Given the number $800 \ldots 008$ (80 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 198. In how many ways can this be done? | Answer: 14080.
Solution. $198=2 \cdot 9 \cdot 11$. Divisibility by 2 is always satisfied since the number ends in eight. For investigating divisibility by 11, it is significant on which positions the replaceable digits stand.
First case. We replace two zeros in positions of the same parity (both even or both odd). Fo... | 14080 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,295 |
6. Find all values of the parameter $a$ such that the system
$$
\left\{\begin{array}{l}
\arccos \left(\frac{4+y}{4}\right)=\arccos (x-a) \\
x^{2}+y^{2}-4 x+8 y=b
\end{array}\right.
$$
has no more than one solution for any value of the parameter $b$. | Answer. $a \in(-\infty ;-15] \cup[19 ;+\infty)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{4+y}{4}=x-a, y=4 x-4 a-4$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4+y}{4} \leqslant 1,-8 \leqslant y \leqslant 0$. Thus, the first equatio... | \in(-\infty;-15]\cup[19;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,296 |
7. Consider the pyramids $T A B C D$ with the following properties: the base of the pyramid is a convex quadrilateral $A B C D$, where $A B=B C=2, C D=D A=3$, and each of the planes of the lateral faces $T A B, T B C, T C D, T D A$ forms an angle of $30^{\circ}$ with the plane of the base.
a) Find the volume of such a... | Answer. a) $\frac{4}{\sqrt{3}}$; b) $4 \sqrt{3}$.
Let $TH$ be the height of the pyramid, $(TH=h)$, and $P$ be the projection of $T$ onto the line $AB$. Then $THP$ is a right triangle with angle $\angle TPH=30^{\circ}$, from which $HP=h \cdot \operatorname{ctg} 30^{\circ}=h \sqrt{3}$. Similarly, it can be proven that t... | \frac{4}{\sqrt{3}};4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,297 |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{3}\right)$, and the numbers $\frac{1}{\cos x}, \frac{3}{\cos y}, \frac{1}{\cos z}$ also form an arithmetic progression in the given order. Find $\cos ^{2} y$. | Answer: $\frac{4}{5}$.
Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{14}{\cos y}=\frac{1}{\cos x}+\frac{1}{\cos z}$. Transform the second equation:
$$
\frac{6}{\cos y}=\frac{\cos x+\cos z}{\cos x \cos z} \Leftrightarrow \frac{6}{\cos y}=\frac{2 \cos \frac{x... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,298 |
3. Solve the inequality $27^{\log _{3}^{2} x}-8 \cdot x^{\log _{3} x} \geqslant \log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5$. | Answer. $x \in\left(0 ; \frac{1}{3}\right] \cup[3 ;+\infty)$.
Solution. Transform the right side of the inequality:
$$
\log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\log _{5} 2\right)}{\log _{5} 250}=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\l... | x\in(0;\frac{1}{3}]\cup[3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,299 |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 7. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=14 ;$ b) $\angle D A C=45^{\circ}, S_{A B C D}=94$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\... | AC=14,\angleDAC=45,S_{ABCD}=94 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,300 |
5. Given the number $500 \ldots 005$ (80 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 165. In how many ways can this be done? | Answer: 17280.
Solution. $165=3 \cdot 5 \cdot 11$. Divisibility by 5 is always satisfied since the number ends in a five. For divisibility by 11, it is significant on which positions the replaceable digits stand.
First case. We replace two zeros in positions of the same parity (both even or both odd). For divisibilit... | 17280 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,301 |
6. Find all values of the parameter $a$ for which there exists a value of the parameter $b$ such that the system
$$
\left\{\begin{array}{l}
\arcsin \left(\frac{a+y}{2}\right)=\arcsin \left(\frac{x+3}{3}\right) \\
x^{2}+y^{2}+6 x+6 y=b
\end{array}\right.
$$
has exactly two solutions. | Answer. $a \in\left(-\frac{7}{2} ; \frac{19}{2}\right)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{a+y}{2}=\frac{x+3}{3}, y=\frac{2 x}{3}+2-a$. The domain of definition is determined by the inequality $-1 \leqslant \frac{x+3}{3} \leqslant 1, -6 \leqslant x \leqslant ... | \in(-\frac{7}{2};\frac{19}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,302 |
7. Consider quadrilateral pyramids $KABCD$ with the following properties: the base of the pyramid is a convex quadrilateral $ABCD$, where $AB = BC = 3$, $CD = DA = 4$, and each of the planes of the lateral faces $KAB$, $KBC$, $KCD$, $KDA$ forms an angle of $45^{\circ}$ with the plane of the base.
a) Find the volume of... | Answer: a) $\frac{4}{3}$; b) 48.
Let $KH$ be the height of the pyramid $(KH=h)$, and $P$ be the projection of $K$ onto the line $AB$. Then $KHP$ is a right triangle with an angle $\angle KPH=45^{\circ}$, from which $HP=h \cdot \operatorname{ctg} 45^{\circ}=h$. Similarly, it can be proven that the point $H$ is at a dis... | \frac{4}{3};48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,303 |
2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{2}{3}$, and the numbers $\frac{1}{\sin x}, \frac{6}{\sin y}, \frac{1}{\sin z}$ also form an arithmetic progression in the given order. Find $\sin ^{2} y$. | Answer: $\frac{5}{8}$.
Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{12}{\sin y}=\frac{1}{\sin x}+\frac{1}{\sin z}$. Transform the second equation:
$$
\frac{12}{\sin y}=\frac{\sin x+\sin z}{\sin x \sin z} \Leftrightarrow \frac{12}{\sin y}=\frac{2 \sin \frac... | \frac{5}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,304 |
3. Solve the inequality $8^{\log _{2}^{2} x}-2 \cdot x^{\log _{2} x} \geqslant \log _{6} 108+\left(4-\log _{6}^{2} 3\right) \log _{108} 6$. | Answer. $x \in\left(0 ; \frac{1}{2}\right] \cup[2 ;+\infty)$.
Solution. Transform the right side of the inequality:
$\log _{6} 108+\left(4-\log _{6}^{2} 3\right) \log _{108} 6=2+\log _{6} 3+\frac{\left(2-\log _{6} 3\right)\left(2+\log _{6} 3\right)}{\log _{6} 108}=2+\log _{6} 3+\frac{\left(2-\log _{6} 3\right)\left(2... | x\in(0;\frac{1}{2}]\cup[2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,305 |
4. Quadrilateral $ABCD$ is inscribed in a circle of radius 6. The rays $AB$ and $DC$ intersect at point $P$, and the rays $BC$ and $AD$ intersect at point $Q$. It is known that triangles $ADP$ and $QAB$ are similar (vertices are not necessarily in the corresponding order).
a) Find $AC$.
b) Suppose additionally that t... | Answer. a) $A C=12 ;$ b) $\angle D A C=45^{\circ}, S_{A B C D}=68$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\... | AC=12,\angleDAC=45,S_{ABCD}=68 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,306 |
5. Given the number $200 \ldots 002$ (100 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 66. In how many ways can this be done? | Answer: 27100.
Solution. $66=3 \cdot 2 \cdot 11$. Divisibility by 2 is always satisfied since the number ends in two. To investigate divisibility by 11, it is significant which positions the replaceable digits occupy.
First case. We replace two zeros in positions of the same parity (both even or both odd). For divisi... | 27100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,307 |
6. Find all values of the parameter $a$ such that the system
$$
\left\{\begin{array}{l}
\arccos \left(\frac{4-y}{4}\right)=\arccos \left(\frac{a+x}{2}\right) \\
x^{2}+y^{2}+2 x-8 y=b
\end{array}\right.
$$
has no more than one solution for any value of the parameter $b$. | Answer. $a \in(-\infty ;-9] \cup[11 ;+\infty)$.
Solution. The first equation on the domain of definition is equivalent to the equation $\frac{4-y}{4}=\frac{a+x}{2}, y=4-2 a-2 x$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4-y}{4} \leqslant 1,0 \leqslant y \leqslant 8$. Thus, the first... | \in(-\infty;-9]\cup[11;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,308 |
7. Consider the pyramids $T A B C D$ with the following properties: the base of the pyramid is a convex quadrilateral $A B C D$, where $A B=B C=1, C D=D A=2$, and each of the planes of the lateral faces $T A B, T B C, T C D, T D A$ forms an angle of $60^{\circ}$ with the plane of the base.
a) Find the volume of such a... | Answer: a) $\frac{4}{3 \sqrt{3}}$; b) $\frac{4}{\sqrt{3}}$.
Let $TH$ be the height of the pyramid, $(TH=h)$, and $P$ be the projection of $T$ onto the line $AB$. Then $THP$ is a right triangle with angle $\angle TPH=60^{\circ}$, from which $HP=h \cdot \operatorname{ctg} 60^{\circ}=\frac{h}{\sqrt{3}}$. Similarly, it ca... | \frac{4}{3\sqrt{3}};\frac{4}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,309 |
1. Which whole numbers from 1 to $8 \cdot 10^{20}$ (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: the number of numbers containing only odd digits is greater by $\frac{5^{21}-5}{4}$.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 20$ ). The number of numbers consisting only of odd digits is $5^{k}$ (for each of the $k$ positions, any of the digits $1,3,5,7$, 9 can be chosen); the number of... | \frac{5^{21}-5}{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,310 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}+$ $2 f(x)$, if the minimum value of the function $(f(x))^{2}+2 g(x)$ is 5. | Answer: -7.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+2 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+2(a x+c)=a^{2} x^{2}+2 a(b+1) x+b^{2}+2 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,311 |
3. The equation $x^{2}+a x+5=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{2}+\frac{250}{19 x_{2}^{3}}=x_{2}^{2}+\frac{250}{19 x_{1}^{3}}
$$
Find all possible values of $a$. | Answer: $a=10$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-20>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=5$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-5$.
Transform the given equality:
$$
x_{1... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,312 |
4. On each of the lines $x=0$ and $x=2$, there are 62 points with ordinates $1, 2, 3, \ldots, 62$. In how many ways can three points be chosen from the marked 124 points so that they form the vertices of a right triangle? | Answer: 7908.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the ... | 7908 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,313 |
5. In a convex quadrilateral $ABCD$, diagonal $BD$ is drawn, and a circle is inscribed in each of the resulting triangles $ABD$ and $BCD$. A line passing through vertex $B$ and the center of one of the circles intersects side $DA$ at point $M$. Here, $AM=\frac{8}{5}$ and $MD=\frac{12}{5}$. Similarly, a line passing thr... | Answer: a) $A B: C D=4: 5, A B=4, C D=5$.
Solution. a) Since the bisector of a triangle divides its side proportionally to the other two sides, $A B: B D=A M: M D=2: 3$ and $B D: D C=B N: N C=6: 5$. Therefore, $A B: C D=\frac{2}{3} \cdot \frac{6}{5}=4: 5$.
b) Let the points of tangency of the incircle of triangle $A ... | AB=4,CD=5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,314 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from nine consecutive natural numbers to some number $a$ is 294, and the sum of the distances from these same nine numbers to some number $b$ is 1932. Find all possible values of $a$, given that ... | Answer: $a=\frac{13}{3}, a=\frac{755}{3}$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+8$. Notice that if some number lies on the segment $[k ; k+8]$, then the sum of the distances from it to the given nine numbers does not exceed $\frac{9}{2} \cdot 8=36$ (the sum of the distan... | =\frac{13}{3},=\frac{755}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,315 |
7. In triangle $A B C$, side $A C$ is equal to 6, and angle $A C B$ is $120^{\circ}$. Circle $\Omega$ with radius $\sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ($M$ lies between $A$ and $N$) such that segment $M K$ is par... | Answer: $C L=1, M K=3, A B=2 \sqrt{13}, S_{\triangle A L N}=\frac{125 \sqrt{3}}{52}$.
Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ of the triangle as $H$, and the angle $B A C$ as $\alpha$.
The center of the circle inscribed in an angle lies on the bisector of tha... | CL=1,MK=3,AB=2\sqrt{13},S_{\triangleALN}=\frac{125\sqrt{3}}{52} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,316 |
1. Which whole numbers from 1 to $4 \cdot 10^{25}$ (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: the number of numbers containing only odd digits is greater by $\frac{5^{26}-5}{4}$.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 25$ ). The number of numbers consisting only of odd digits is $5^{k}$ (for each of the $k$ positions, any of the digits $1,3,5,7$, 9 can be chosen); the number of... | \frac{5^{26}-5}{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,317 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}+$ $8 f(x)$, if the minimum value of the function $(f(x))^{2}+8 g(x)$ is -29. | Answer: -3.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+8 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+8(a x+c)=a^{2} x^{2}+2 a(b+4) x+b^{2}+8 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,318 |
3. The equation $x^{2}+a x+4=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{2}-\frac{20}{3 x_{2}^{3}}=x_{2}^{2}-\frac{20}{3 x_{1}^{3}}
$$
Find all possible values of $a$. | Answer: $a=-10$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-16>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=4$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-4$.
Transform the given equation:
$$
x_{... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,319 |
4. On each of the lines $y=0$ and $y=2$, there are 64 points with abscissas $1,2,3, \ldots, 64$. In how many ways can three points be chosen from the marked 128 points so that they form the vertices of a right triangle? | Answer: 8420.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the ... | 8420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,320 |
5. In a convex quadrilateral $ABCD$, diagonal $BD$ is drawn, and a circle is inscribed in each of the resulting triangles $ABD$ and $BCD$. A line passing through vertex $B$ and the center of one of the circles intersects side $DA$ at point $M$. Here, $AM=\frac{25}{7}$ and $MD=\frac{10}{7}$. Similarly, a line passing th... | Answer: a) $A B: C D=5: 6 ;$ b) $A B=5, C D=6$.
Solution. a) Since the bisector of a triangle divides its side proportionally to the other two sides, $A B: B D=A M: M D=5: 2$ and $B D: D C=B N: N C=1: 3$. Therefore, $A B: C D=\frac{5}{2} \cdot \frac{1}{3}=5: 6$.
b) Let the points of tangency of the circle inscribed i... | AB=5,CD=6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,321 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from eight consecutive natural numbers to some number $a$ is 612, and the sum of the distances from these same eight numbers to some number $b$ is 240. Find all possible values of $a$, given that... | Answer: $a=27, a=-3$.
Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+7$. Notice that if some number lies in the interval $[k ; k+8]$, then the sum of the distances from it to the given eight numbers does not exceed $4 \cdot 7=28$ (the sum of the distances to the two extreme number... | =27,=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,322 |
7. In triangle $A B C$, side $B C$ is equal to 4, and angle $A C B$ is equal to $\frac{2 \pi}{3}$. Circle $\Gamma$ with radius $2 \sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that seg... | Answer: $C K=2, M K=6, A B=4 \sqrt{13}, S_{\triangle C M N}=\frac{72 \sqrt{3}}{13}$.
Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ as $H$, and the angle $A B C$ as $\beta$.
The center of the circle inscribed in an angle lies on the angle bisector, so $\angle O C K ... | CK=2,MK=6,AB=4\sqrt{13},S_{\triangleCMN}=\frac{72\sqrt{3}}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,323 |
1. Given 2117 cards, on which natural numbers from 1 to 2117 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done? | Answer: 22386.
Solution. We will take the cards in turn. There are several cases depending on the number written on the first card.
1) The number on the card ends in 00 (there are 21 such cards). For the sum to be divisible by 100, the second card must also end in 00. We get $C_{2} 1^{2}=\frac{21 \cdot 20}{2}=210$ op... | 22386 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,324 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=11 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x)... | Answer: $A=-11, A=0$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=11 g(x)$ is equivalent to the equation $(f(x))^{2}=11 g(x)$ having exactly one solution. We get $(a x+b)^{2}=11(a x+c), a^{2} x^{2}+a(2 b-11) x+b^{2}-11 c=0$. The discriminant of this equati... | A=-11,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,325 |
3. The equation $x^{2}+a x+2=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{3}+\frac{14}{x_{2}^{2}}=x_{2}^{3}+\frac{14}{x_{1}^{2}}
$$
Find all possible values of $a$. | Answer: $a=4$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-8>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=2$. Then $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-2$.
Transform the given equation:
$$
x_{1}^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,326 |
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