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4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 7 with integer non-negative exponents, and at the same time, their coefficients in absolute value do not exceed $343^{36}$. | Answer: 2969.
Solution. Such quadratic trinomials can be represented in the form $\left(x-7^{a}\right)\left(x-7^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(7^{a}+7^{b}\right) x+7^{a+b}$. By the condition
$$
\... | 2969 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,327 |
5. The diagonals $A C$ and $B D$ of the quadrilateral $A B C D$, inscribed in a circle, intersect at point $P$. It is known that the distances from point $P$ to the sides $A B, B C, C D, D A$ are $4, \sqrt{3}, \frac{8}{\sqrt{19}}$ and $8 \sqrt{\frac{3}{19}}$ respectively (the bases of the perpendiculars dropped from po... | Answer: $A P: P C=4, B D=\frac{35}{\sqrt{19}}$.
Solution. Since inscribed angles subtended by the same arc are equal, $\angle P B C=$ $\angle P A D, \angle P C B=\angle P D A$. Therefore, triangles $P B C$ and $P D A$ are similar. Similarly, it can be proven that $\triangle A B P \sim \triangle D C P$. Corresponding e... | AP:PC=4,BD=\frac{35}{\sqrt{19}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,328 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twenty consecutive natural numbers to some number $a$ is 4460, and the sum of the distances from these same twenty numbers to the number $a^{2}$ is 2755. Find all possible values of $a$. | Answer: $a=-\frac{37}{2}, a=\frac{39}{2}$.
Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+19$. Notice that if some number lies on the segment $[k ; k+19]$, then the sum of the distances from it to the given twenty numbers does not exceed $10 \cdot 19=190$ (the sum of the distances... | =-\frac{37}{2},=\frac{39}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,329 |
7. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 13 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ at points $E$ and $K$ respectively. The length of arc $A E$ is... | Answer: $B C=13, B K=\frac{120}{13}, P_{A K M}=\frac{340}{13}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\ang... | BC=13,BK=\frac{120}{13},P_{AKM}=\frac{340}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,330 |
1. Given 2414 cards, on which natural numbers from 1 to 2414 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: 29112.
Solution. We will take 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 24 such cards). For the sum to be divisible by 100, the second card must also end in 00. We get a total of $C_{2} 4^{2}=\frac{24 \cdot 23}{2}=... | 29112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,331 |
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=-8 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x)... | Answer: $A=8, 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=-8 g(x)$ is equivalent to the equation $(f(x))^{2}=-8 g(x)$ having exactly one solution. We get $(a x+b)^{2}=-8(a x+c), a^{2} x^{2}+2 a(b+4) x+b^{2}+8 c=0$. A quarter of the discriminant of th... | A=8,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,332 |
3. The equation $x^{2}+a x+3=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{3}-\frac{99}{2 x_{2}^{2}}=x_{2}^{3}-\frac{99}{2 x_{1}^{2}}
$$
Find all possible values of $a$. | Answer: $a=-6$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-12>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=3$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-3$.
Transform the given equation:
$$
x_{1... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,333 |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 5 with integer non-negative exponents, and at the same time, their coefficients in absolute value do not exceed $122^{20}$. | Answer: 5699.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. According to the condi... | 5699 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,334 |
5. The diagonals $A C$ and $B D$ of the quadrilateral $A B C D$, inscribed in a circle, intersect at point $P$. It is known that the distances from point $P$ to the sides $A B, B C, C D, D A$ are $5, \sqrt{3}, \frac{5}{\sqrt{7}}$ and $5 \sqrt{\frac{3}{7}}$ respectively (the feet of the perpendiculars dropped from point... | Answer: $A P: P C=5, B D=\frac{24}{\sqrt{7}}$.
Solution. Since inscribed angles subtended by the same arc are equal, $\angle P B C=$ $\angle P A D, \angle P C B=\angle P D A$. Therefore, triangles $P B C$ and $P D A$ are similar. Similarly, it can be proven that $\triangle A B P \sim \triangle D C P$. Corresponding el... | AP:PC=5,BD=\frac{24}{\sqrt{7}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,335 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twenty consecutive natural numbers to some number $a$ is 360, and the sum of the distances from these same twenty numbers to the number $a^{2}$ is 345. Find all possible values of $a$. | Answer: $a=-\frac{1}{2}, a=\frac{3}{2}$.
Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+19$. Note that if some number lies on the segment $[k ; k+19]$, then the sum of the distances from it to the given twenty numbers does not exceed $10 \cdot 19=190$ (the sum of the distances to ... | =-\frac{1}{2},=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,336 |
7. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 5 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ at points $E$ and $K$ respectively. The length of arc $A E$ is ... | Answer: $B C=5, B K=\frac{24}{5}, P_{A K M}=\frac{42}{5}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\angle A ... | BC=5,BK=\frac{24}{5},P_{AKM}=\frac{42}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,337 |
1. It is known that the roots of the quadratic polynomial $f(x)=x^{2}+a x+b$ are each 3 more than the roots of the quadratic polynomial $g(x)$. Let $M$ be the sum of the coefficients of $f(x)$, and $N$ be the sum of the coefficients of the quadratic polynomial $g(x)$. What is the smallest value that $|M-N|$ can take?
... | 1. It is known that the roots of the quadratic polynomial $f(x)=x^{2}+a x+b$ are each 3 more than the roots of the quadratic polynomial $g(x)$. Let $M$ be the sum of the coefficients of $f(x)$, and $N$ be the sum of the coefficients of the quadratic polynomial $g(x)$. What is the smallest value that $|M-N|$ can take?
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,338 | |
3. Points $K, L$ and $M$ are the midpoints of sides $B C, C A$ and $A B$ of triangle $A B C$. Denote $R(X Y Z)$ - the radius of the circumcircle of triangle $X Y Z$. It turns out that $R(A K C): R(A K B)=$ param $1, R(C L B): R(B L A)=$ param2. Find the ratio $R(A M C): R(B M C)$.
| param1 | param2 | |
| :---: | :---... | 3. Points $K, L$ and $M$ are the midpoints of sides $B C, C A$ and $A B$ of triangle $A B C$. Denote $R(X Y Z)$ - the radius of the circumcircle of triangle $X Y Z$. It turns out that $R(A K C): R(A K B)=$ param1, $R(C L B): R(B L A)=$ param2. Find the ratio $R(A M C): R(B M C)$.
| param1 | param2 | Answer |
| :---: |... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,340 | |
5. The numbers $x$ and $y$ are solutions to the system of equations param 1, where $a$ is a parameter. What param 2 value does the expression param 3 take?
| param1 | param2 | param3 | |
| :---: | :---: | :---: | :---: |
| $\left\{\begin{array}{l}a x+y=a+1 \\ x+4 a y=3\end{array}\right.$ | maximum | $x^{2}-6 y^{2}$ |... | 5. The numbers $x$ and $y$ are solutions to the system of equations param 1, where $a$ is a parameter. What param 2 value does the expression param 3 take?
| param1 | param2 | param3 | Answer |
| :---: | :---: | :---: | :---: |
| $\left\{\begin{array}{l}a x+y=a+1 \\ x+4 a y=3\end{array}\right.$ | maximum | $x^{2}-6 y^... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,342 |
6. In how many ways can all asterisks be replaced with param1 even and param2 odd digits (not necessarily distinct) in the number param3 so that the resulting number is divisible by 12?
| param1 | param2 | param3 | |
| :---: | :---: | :---: | :---: |
| 3 | 3 | $2017^{*} 7^{* *} 13^{* *} 112^{*}$ | |
| 5 | 2 | $2017^... | 6. In how many ways can all asterisks be replaced with param1 even and param2 odd digits (not necessarily distinct) in the number param3 so that the resulting number is divisible by 12?
| param1 | param2 | param3 | Answer |
| :---: | :---: | :---: | :---: |
| 3 | 3 | $2017^{*} 7^{* *} 13^{* *} 112^{*}$ | 31250 |
| 5 |... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,343 |
7. An acute triangle $A B C$ is inscribed in a circle $\Omega$. The extensions of the altitudes of triangle $A B C$, drawn from vertices $A$ and $C$, intersect $\Omega$ at points $M$ and $N$. Find the radius of the circle $\Omega$, if the ratio is 1.
| param1 | |
| :---: | :---: |
| $A C=7, M N=4 \sqrt{6}$ | |
| $A ... | 7. An acute triangle $ABC$ is inscribed in a circle $\Omega$. The extensions of the altitudes of triangle $ABC$, drawn from vertices $A$ and $C$, intersect $\Omega$ at points $M$ and $N$. Find the radius of the circle $\Omega$ if param1.
| param1 | Answer |
| :---: | :---: |
| $AC=7, MN=4 \sqrt{6}$ | 4.9 |
| $AC=11, M... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,344 | |
8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among those they named. Find the smallest po... | 8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among the ones they named. Find the smallest... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,345 |
9. Given a regular param1. Find the number of its vertex quadruples that are vertices of a convex quadrilateral with exactly two angles equal to $90^{\circ}$. (Two quadruples of vertices that differ in the order of vertices are considered the same.)
| param1 | |
| :---: | :---: |
| 16-gon | |
| 18-gon | |
| 20-gon ... | 9. Given a regular param1. Find the number of its vertex quadruples that are vertices of a convex quadrilateral with exactly two angles equal to $90^{\circ}$. (Two quadruples of vertices that differ in the order of vertices are considered the same.)
| param1 | Answer |
| :---: | :---: |
| 16-gon | 336 |
| 18-gon | 504... | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,346 |
10. Find the sum of the roots of the equation param1 that lie in the interval param2. Write the answer in degrees.
| param 1 | param2 | |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 540^{\cir... | 10. Find the sum of the roots of the equation param1 lying in the interval param2. Write the answer in degrees.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | 1800 |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 54... | -360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,347 |
1. [5 points] $S$ is the sum of the first 10 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{6} a_{12}>S+1, a_{7} a_{11}<S+17$. Determine all possible values of $a_{1}$. | Answer: $-6; -5; -4; -2; -1; 0$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 5 d ) ( a _ { 1 } + 1 1 d ) > S + 1 , } \\
{ ( a _ { 1 } + 6 d ) ( a _ { 1 } + 1 0 d ) S+1 \\
a_{1}^{2}+... | -6;-5;-4;-2;-1;0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,348 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=2$, $AC=CB=5$, and $AD=DB=6$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is t... | Answer: $\sqrt{34} \pm \sqrt{23}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence... | \sqrt{34}\\sqrt{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,349 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x ; y)$ such that there exists a pair of real numbers $a, b$, for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 2 \\
a^{2}+b^{2} \leqslant \min (2 a+2 b ; 2)
\end{array}\right.
$... | Answer: $6 \pi-\sqrt{3}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant 2 a+2 b \\
a^{2}+b^{2} \leqslant 2
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x - a ) ^ { 2 }... | 6\pi-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,350 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=6 \\
\operatorname{LCM}(a ; b ; c)=2^{15} \cdot 3^{16}
\end{array}\right.
$$ | Answer: 7560.
Solution. Let $a=2^{\alpha_{1}} \cdot 3^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 3^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 3^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{... | 7560 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,351 |
5. [5 points] Given the numbers $\log _{\sqrt{5 x-1}}(4 x+1), \log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, \log _{\frac{x}{2}+2}(5 x-1)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=2$.
Solution. From the condition, it follows that the functions $4 x+1, \frac{x}{2}+2,5 x-1$ are positive and do not take the value 1 for all $x$ from the domain of admissible values. Let $a=\log _{\sqrt{5 x-1}}(4 x+1), b=\log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, c=$ $\log _{\frac{x}{2}+2}(5 x-1)$. Then
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,352 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=25 ;$ b) $A C=\frac{50}{\sqrt{42}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. Therefore, points $A$ and $C$ lie on a circle with diameter $O T$ (let's call this ... | S_{ABC}=25;AC=\frac{50}{\sqrt{42}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,353 |
1. [5 points] $S$ - the sum of the first 7 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{7} a_{12}>S+20, a_{9} a_{10}<S+44$. Indicate all possible values of $a_{1}$. | Answer: $-9; -8; -7; -6; -4; -3; -2; -1$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 6 d ) ( a _ { 1 } + 11 d ) > S + 20 , } \\
{ ( a _ { 1 } + 8 d ) ( a _ { 1 } + 9 d ) S+20 \\
a_... | -9;-8;-7;-6;-4;-3;-2;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,354 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=2$, $AC=CB=5$, and $AD=DB=7$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is t... | Answer: $\sqrt{47} \pm \sqrt{23}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence... | \sqrt{47}\\sqrt{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,355 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 5 \\
a^{2}+b^{2} \leqslant \min (4 a-2 b ; 5)
\end{array}\right.
$$
... | Answer: $15 \pi-\frac{5 \sqrt{3}}{2}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant 4 a-2 b \\
a^{2}+b^{2} \leqslant 5
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x ... | 15\pi-\frac{5\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,356 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=15 \\
\operatorname{LCM}(a ; b ; c)=3^{15} \cdot 5^{18}
\end{array}\right.
$$ | Answer: 8568.
Solution. Let $a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 5^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{... | 8568 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,357 |
5. [5 points] Given the numbers $\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), \log _{6 x-14}(x-1)^{2}, \log _{x-1}\left(\frac{x}{3}+3\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=3$.
Solution. From the condition, it follows that the functions $6 x-14, x-1, \frac{x}{3}+3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), b=\log _{6 x-14}(x-1)^{2}, c=$ $\log _{x-1}\left(\frac{x}{3}+3\right)$. Then
$$
... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,358 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=\frac{121}{5}$; b) $A C=\frac{5 \sqrt{33}}{6}$.
Solution. Since the lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. Therefore, points $A$ and $C$ lie on a circle with diameter $O T$ (l... | S_{ABC}=\frac{121}{5};AC=\frac{5\sqrt{33}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,359 |
1. [5 points] $S$ is the sum of the first 14 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{9} a_{17}>S+12, a_{11} a_{15}<S+47$. Indicate all possible values of $a_{1}$. | Answer: $-9; -8; -7; -6; -4; -3; -2; -1$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the conditions can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 8 d ) ( a _ { 1 } + 1 6 d ) > S + 1 2 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 4 d ) S+1... | -9;-8;-7;-6;-4;-3;-2;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,360 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=2$, $AC=CB=6$, and $AD=DB=7$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is t... | Answer: $\sqrt{47} \pm \sqrt{34}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence... | \sqrt{47}\\sqrt{34} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,361 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 25 \\
a^{2}+b^{2} \leqslant \min (-8 a-6 b ; 25)
\end{array}\right.
... | Answer: $75 \pi-\frac{25 \sqrt{3}}{2}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant-8 a-6 b \\
a^{2}+b^{2} \leqslant 25
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( ... | 75\pi-\frac{25\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,362 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=21 \\
\operatorname{LCM}(a ; b ; c)=3^{17} \cdot 7^{15}
\end{array}\right.
$$ | Answer: 8064.
Solution. Let $a=3^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 7^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{\m... | 8064 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,363 |
5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right), \log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2}$. For which $x$ are two of these numbers equal, and the third one greater than them by 1? | Answer: $x=5$.
Solution. From the condition, it follows that the functions $\frac{x}{2}-1, \frac{x}{2}-\frac{1}{4}, x-\frac{11}{4}$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), b=\log _{\frac{x}{2}-\frac{1}{4}}\... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,364 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=\frac{128}{3}$; b) $A C=\frac{4 \sqrt{26}}{\sqrt{3}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter ... | S_{ABC}=\frac{128}{3};AC=\frac{4\sqrt{26}}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,365 |
1. [5 points] $S$ - the sum of the first 5 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{6} a_{11}>S+15, a_{9} a_{8}<S+39$. Indicate all possible values of $a_{1}$. | Answer: $-9; -8; -7; -6; -4; -3; -2; -1$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 5 d ) ( a _ { 1 } + 10 d ) > S + 15 , } \\
{ ( a _ { 1 } + 8 d ) ( a _ { 1 } + 7 d ) S+15 \\
a_... | -9;-8;-7;-6;-4;-3;-2;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,366 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=2$, $AC=CB=7$, and $AD=DB=8$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is t... | Answer: $\sqrt{62} \pm \sqrt{47}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence... | \sqrt{62}\\sqrt{47} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,367 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 13 \\
a^{2}+b^{2} \leqslant \min (-4 a-6 b ; 13)
\end{array}\right.
... | Answer: $39 \pi-\frac{13 \sqrt{3}}{2}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant-4 a-6 b \\
a^{2}+b^{2} \leqslant 13
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( ... | 39\pi-\frac{13\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,368 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=10 \\
\operatorname{LCM}(a ; b ; c)=2^{17} \cdot 5^{16}
\end{array}\right.
$$ | Answer: 8640.
Solution. Let $a=2^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 5^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{... | 8640 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,369 |
5. [5 points] Given the numbers $\log _{\sqrt{2 x-8}}(x-4), \log _{(x-4)^{2}}(5 x-26), \log _{\sqrt{5 x-26}}(2 x-8)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1? | Answer: $x=6$.
Solution. From the condition, it follows that the functions $x-4, 5x-26$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-8}}(x-4), b=\log _{\sqrt{5 x-26}}(2 x-8), c=\log _{(x-4)^{2}}(5 x-26)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,370 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=\frac{81}{2}$; b) $A C=\frac{3 \sqrt{17}}{2}$.
Solution. Since the lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter $O T... | S_{ABC}=\frac{81}{2};AC=\frac{3\sqrt{17}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,371 |
1. [5 points] $S$ - the sum of the first 7 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{8} a_{17}>S+27, a_{11} a_{14}<S+60$. Indicate all possible values of $a_{1}$. | Answer: $-11; -10; -9; -7; -6; -5$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the problem can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 7 d ) ( a _ { 1 } + 1 6 d ) > S + 2 7 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 3 d ) S+27 \\
a_{1... | -11;-10;-9;-7;-6;-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,372 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=4, AC=CB=5, AD=DB=6$. Each such tetrahedron is inscribed in a cylinder so that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. We select the tetrahedron for which the radius of the cylinder is the smal... | Answer: $2 \sqrt{7} \pm \sqrt{17}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, henc... | 2\sqrt{7}\\sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,373 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 20 \\
a^{2}+b^{2} \leqslant \min (8 a-4 b ; 20)
\end{array}\right.
$... | Answer: $60 \pi-10 \sqrt{3}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant 8 a-4 b \\
a^{2}+b^{2} \leqslant 20
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x - a ) ^ ... | 60\pi-10\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,374 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=35 \\
\operatorname{LCM}(a ; b ; c)=5^{18} \cdot 7^{16}
\end{array}\right.
$$ | Answer: 9180.
Solution. Let $a=5^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=5^{\beta_{1}} \cdot 7^{\beta_{2}}, c=5^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=5^{\m... | 9180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,375 |
5. [5 points] Given the numbers $\log _{\sqrt{2 x-3}}(x+1), \log _{2 x^{2}-3 x+5}(2 x-3)^{2}, \log _{x+1}\left(2 x^{2}-3 x+5\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=4$.
Solution. From the condition, it follows that the functions $x+1, 2x-3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-3}}(x+1), b=\log _{2 x^{2}-3 x+5}(2 x-3)^{2}, c=\log _{x+1}\left(2 x^{2}-3 x+5\right)$. Then
$$
\begin{aligned}
a b c=... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,376 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=49$; b) $A C=\frac{7 \sqrt{5}}{\sqrt{6}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. Therefore, points $A$ and $C$ lie on a circle with diameter $O T$ (let's call... | S_{ABC}=49;AC=\frac{7\sqrt{5}}{\sqrt{6}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,377 |
1. [5 points] $S$ is the sum of the first 15 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{7} a_{16}>S-24, a_{11} a_{12}<S+4$. Determine all possible values of $a_{1}$. | Answer: $-5; -4; -2; -1$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 6 d ) ( a _ { 1 } + 1 5 d ) > S - 2 4 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 1 d ) S-24 \\
a_{1}^{2}+21... | -5;-4;-2;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,378 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=4, AC=CB=5, AD=$ $DB=7$. Each such tetrahedron is inscribed in a cylinder so that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. We select the tetrahedron for which the radius of the cylinder is the s... | Answer: $\sqrt{41} \pm \sqrt{17}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence... | \sqrt{41}\\sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,379 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 50 \\
a^{2}+b^{2} \leqslant \min (14 a+2 b ; 50)
\end{array}\right.
... | Answer: $150 \pi-25 \sqrt{3}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant 14 a+2 b \\
a^{2}+b^{2} \leqslant 50
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x - a ) ... | 150\pi-25\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,380 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$, satisfying the system of equations
$$
\left\{\begin{array}{l}
\text { GCD }(a ; b ; c)=14 \\
\text { LCM }(a ; b ; c)=2^{17} \cdot 7^{18}
\end{array}\right.
$$ | Answer: 9792.
Solution. Let $a=2^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 7^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{\m... | 9792 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,381 |
5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right), \log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2}, \log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)$. For which $x$ are two of these numbers equal, and the third one less than ... | Answer: $x=7$.
Solution. From the condition, it follows that the functions $\left(\frac{x}{2}+1\right),\left(\frac{7 x}{2}-\frac{17}{4}\right),\left(\frac{3 x}{2}-6\right)$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,382 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=\frac{144}{5} ;$ b) $A C=\frac{4 \sqrt{17}}{\sqrt{7}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter... | S_{ABC}=\frac{144}{5};AC=\frac{4\sqrt{17}}{\sqrt{7}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,383 |
1. [5 points] $S$ - the sum of the first 6 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{10} a_{16}>S+39, a_{11} a_{15}<S+55$. Indicate all possible values of $a_{1}$. | Answer: $-12; -11; -10; -8; -7; -6$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the problem can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 9 d ) ( a _ { 1 } + 1 5 d ) > S + 3 9 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 4 d ) S+39 \\
a_{... | -12;-11;-10;-8;-7;-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,384 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=4, AC=CB=6, AD=$ $DB=7$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is the sm... | Answer: $\sqrt{41} \pm 2 \sqrt{7}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that segment $AB$ is perpendicular to the plane $CDE$, hence $A... | \sqrt{41}\2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,385 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 8 \\
a^{2}+b^{2} \leqslant \min (-4 a+4 b ; 8)
\end{array}\right.
$$... | Answer: $24 \pi-4 \sqrt{3}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant-4 a+4 b \\
a^{2}+b^{2} \leqslant 8
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x - a ) ^ { ... | 24\pi-4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,386 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=22 \\
\operatorname{LCM}(a ; b ; c)=2^{16} \cdot 11^{19}
\end{array}\right.
$$ | Answer: 9720.
Solution. Let $a=2^{\alpha_{1}} \cdot 11^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 11^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 11^{\gamma_{2}}$ (the numbers $a, b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{\... | 9720 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,387 |
5. [5 points] Given the numbers $\log _{\sqrt{x+34}}(2 x+23), \log _{(x+4)^{2}}(x+34), \log _{\sqrt{2 x+23}}(-x-4)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1? | Answer: $x=-9$.
Solution. From the condition, it follows that the functions $2 x+23, x+34,-x-4$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{x+34}}(2 x+23), b=\log _{(x+4)^{2}}(x+34), c=\log _{\sqrt{2 x+23}}(-x-4)$. Then
$$
\begin{aligned}
& a b c=\log _... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,388 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer: a) $S_{A B C}=\frac{784}{13}$; b) $A C=\frac{14}{\sqrt{3}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter $O T$ (l... | S_{ABC}=\frac{784}{13};AC=\frac{14}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,389 |
1. [5 points] $S$ - the sum of the first 9 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{5} a_{18}>S-4, a_{13} a_{10}<S+60$. Indicate all possible values of $a_{1}$. | Answer: $-10; -9; -8; -7; -5; -4; -3; -2$.
Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:
$$
\left\{\begin{array} { l }
{ ( a _ { 1 } + 4 d ) ( a _ { 1 } + 1 7 d ) > S - 4 , } \\
{ ( a _ { 1 } + 1 2 d ) ( a _ { 1 } + 9 d ) S-4 \\
... | -10;-9;-8;-7;-5;-4;-3;-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,390 |
2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=4$, $AC=CB=7$, and $AD=DB=8$. Each such tetrahedron is inscribed in a cylinder such that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. Choose the tetrahedron for which the radius of the cylinder is t... | Answer: $2 \sqrt{14} \pm \sqrt{41}$.
Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hen... | 2\sqrt{14}\\sqrt{41} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,391 |
3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 10 \\
a^{2}+b^{2} \leqslant \min (-6 a-2 b ; 10)
\end{array}\right.
... | Answer: $30 \pi-5 \sqrt{3}$.
Solution. The second inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant-6 a-2 b \\
a^{2}+b^{2} \leqslant 10
\end{array}\right.
$$
Thus, the original system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ ( x - a ) ^ {... | 30\pi-5\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,392 |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\text { GCD }(a ; b ; c)=33, \\
\text { LCM }(a ; b ; c)=3^{19} \cdot 11^{15} .
\end{array}\right.
$$ | Answer: 9072.
Solution. Let $a=3^{\alpha_{1}} \cdot 11^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 11^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 11^{\gamma_{2}}$ (the numbers $a, b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{\... | 9072 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,393 |
5. [5 points] Given the numbers $\log _{\sqrt{29-x}}\left(\frac{x}{7}+7\right), \log _{(x+1)^{2}}(29-x), \log _{\sqrt{\frac{x}{7}+7}}(-x-1)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1? | Answer: $x=-7$.
Solution. From the condition, it follows that the functions $\frac{x}{7}+7, 29-x, -x-1$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{29-x}}\left(\frac{x}{7}+7\right), b=\log _{(x+1)^{2}}(29-x), c=$ $\log _{\sqrt{\frac{x}{7}+7}}(-x-1)$. The... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,394 |
6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is kno... | Answer. a) $S_{A B C}=\frac{450}{7}$; b) $A C=\frac{5 \sqrt{41}}{\sqrt{14}}$.
Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter... | S_{ABC}=\frac{450}{7};AC=\frac{5\sqrt{41}}{\sqrt{14}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,395 |
1. Which whole numbers from 1 to 60000 (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: There are 780 more numbers containing only odd digits.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 4$ ). 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 numbers consisting only of eve... | 780 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,396 |
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 $(g(x))^{2}+$ $f(x)$, if the minimum value of the function $(f(x))^{2}+g(x)$ is -6. | # Answer: $\frac{11}{2}$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+(a x+c)=a^{2} x^{2}+a(2 b+1) x+b^{2}+c$. The graph of $y=h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coor... | \frac{11}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,397 |
3. The equation $x^{2}+a x+6=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}-\frac{72}{25 x_{2}^{3}}=x_{2}-\frac{72}{25 x_{1}^{3}}
$$
Find all possible values of $a$. | Answer: $a= \pm 9$.
Solution. For the equation to have roots, its discriminant must be positive, from which $a^{2}-24>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=6$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-6$.
Transform the given equality:... | \9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,398 |
4. On each of the lines $y=3$ and $y=4$, there are 73 points with abscissas $1,2,3, \ldots, 73$. In how many ways can three points be chosen from the marked 146 so that they form the vertices of a right triangle? | Answer: 10654.
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... | 10654 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,399 |
5. On the extension of side $A C$ of triangle $A B C$ beyond point $A$, point $T$ is marked such that $\angle B A C=2 \angle B T C$. Find the area of triangle $A B C$, given that $A B=A C, B T=70$, $A T=37$. | Answer: 420.
Solution. Let $\angle B T A=\alpha$, then by the condition $\angle B A C=2 \alpha$. Triangle $A B C$ is isosceles with base $B C$, so $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. By the sum of angles in triangle $T B C$, we get that $\angle T B C=180^{\circ}-... | 420 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,400 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from sixteen consecutive natural numbers to some number $a$ is 276, and the sum of the distances from these same sixteen numbers to some number $b$ is 748. Find all possible values of $a$, given ... | Answer: $a=-0.75, a=46.25$.
Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+15$. Notice that if some number lies on the segment $[k ; k+15]$, then the sum of the distances from it to the given sixteen numbers does not exceed $15 \cdot 8=120$ (the sum of the distances to the two ext... | =-0.75,=46.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,401 |
7. Given a convex quadrilateral $A B C D$. Let $P$ be the center of the circle inscribed in triangle $A B D$, and $Q$ be the center of the circle inscribed in triangle $C B D$. Ray $B P$ intersects side $D A$ at point $M$, and ray $D Q$ intersects side $B C$ at point $N$. It turns out that $A M=\frac{9}{7}, D M=\frac{1... | Answer: a) $A B: C D=3: 5$; b) $A B=3, C D=5$.
Solution. a) Since the bisector of a triangle divides its side proportionally to the other two sides, $A B: B D=A M: M D=3: 4$ and $B D: D C=B N: N C=4: 5$. Therefore, $A B: C D=\frac{3}{4} \cdot \frac{4}{5}=3: 5$.
b) Let the points of tangency of the incircle of triangl... | AB=3,CD=5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,402 |
1. Which whole numbers from 1 to 80000 (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: There are 780 more numbers containing only odd digits.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 4$ ). 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 numbers consisting only of eve... | 780 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,403 |
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 $(g(x))^{2}+$ $f(x)$, if the minimum value of the function $(f(x))^{2}+g(x)$ is 4. | # Answer: $-\frac{9}{2}$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+(a x+c)=a^{2} x^{2}+a(2 b+1) x+b^{2}+c$. The graph of $y=h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coor... | -\frac{9}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,404 |
3. The equation $x^{2}+a x+8=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}-\frac{64}{17 x_{2}^{3}}=x_{2}-\frac{64}{17 x_{1}^{3}}
$$
Find all possible values of $a$. | Answer: $a= \pm 12$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-32>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=8$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-8$.
Transform the given equality:
$$... | \12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,405 |
4. On each of the lines $x=5$ and $x=6$, there are 58 points with ordinates $1, 2, 3, \ldots, 58$. In how many ways can three points be chosen from the marked 116 so that they form the vertices of a right triangle? | Answer: 6724.
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 ... | 6724 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,406 |
5. On the extension of side $A C$ of triangle $A B C$ beyond point $A$, point $T$ is marked such that $\angle B A C=2 \angle B T C$. Find the area of triangle $A B C$, given that $A B=A C, B T=42$, $A T=29$. | Answer: 420.
Solution. Let $\angle B T A=\alpha$, then by the condition $\angle B A C=2 \alpha$. Triangle $A B C$ is isosceles with base $B C$, so $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. By the sum of angles in triangle $T B C$, we get that $\angle T B C=180^{\circ}-... | 420 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,407 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twelve consecutive natural numbers to some number $a$ is 358, and the sum of the distances from these same twelve numbers to some number $b$ is 212. Find all possible values of $a$, given th... | Answer: $a=\frac{190}{3}$.
Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+11$. Notice that if some number lies on the segment $[k ; k+11]$, then the sum of distances from it to the given twelve numbers does not exceed $11 \cdot 6=66$ (the sum of distances to the two extreme number... | \frac{190}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,408 |
7. Given a convex quadrilateral $A B C D$. Let $P$ be the center of the circle inscribed in triangle $A B D$, and $Q$ be the center of the circle inscribed in triangle $C B D$. Ray $B P$ intersects side $D A$ at point $M$, and ray $D Q$ intersects side $B C$ at point $N$. It turns out that $A M=\frac{8}{3}, D M=\frac{4... | Answer: a) $A B: C D=4: 3$; b) $A B=4, C D=3$.
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: 1$ and $B D: D C=B N: N C=2: 3$. Therefore, $A B: C D=2 \cdot \frac{2}{3}=4: 3$.
b) Let the points of tangency of the incircle of triangle $A B D$ ... | AB=4,CD=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,409 |
1. Given 6000 cards, on which natural numbers from 1 to 6000 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: 179940.
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 60 such cards). For the sum to be divisible by 100, the second card must also have a number ending in 00. We get a total of $C_{60}^{2}=\... | 179940 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,410 |
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=20 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x)... | Answer: $-0.05$.
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=20 g(x)$ is equivalent to the equation $(f(x))^{2}=20 g(x)$ having exactly one solution. We get $(a x+b)^{2}=20(a x+c), a^{2} x^{2}+2 a(b-10) x+b^{2}-20 c=0$. One quarter of the discriminant of t... | -0.05 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,411 |
3. The equation $x^{2}+a x+3=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{3}-\frac{39}{x_{2}}=x_{2}^{3}-\frac{39}{x_{1}}
$$
Find all possible values of $a$. | Answer: $a= \pm 4$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-12>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=3$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-3$.
Transform the given equality:
$$
... | \4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,412 |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 3 with natural exponents, and at the same time, their coefficients in absolute value do not exceed \(27^{45}\). | Answer: 4489.
Solution. Such quadratic trinomials can be represented as $\left(x-3^{a}\right)\left(x-3^{b}\right)$, where $a, b$ are natural numbers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(3^{a}+3^{b}\right) x+3^{a+b}$. According to the condition
$$
\left\{\begin{array... | 4489 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,413 |
5. A circle with center $O$, inscribed in triangle $P Q R$, touches its sides $P Q, Q R$ and $R P$ at points $C, A$ and $B$ respectively. Lines $B O$ and $C O$ intersect sides $P Q$ and $P R$ at points $K$ and $L$ respectively. Find the ratio $Q A: A R$, if $K Q=3, Q R=16, L R=1$. | Answer: $9: 7$.
Solution. Triangles $P B K$ and $P C L$ are equal ($P B=P C$ as tangents drawn from the same point to a circle; $\angle P$ is common; angles at vertices $B$ and $C$ are right angles). Therefore, $P K=P L$, and since $P C=P B$, then $C K=B L$. Let $C K=B L=x$. Then $Q A=Q C=$ $Q K+K C=3+x, R A=R B=R L+L... | 9:7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,414 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from thirteen consecutive natural numbers to some number $a$ is 260, and the sum of the distances from these same thirteen numbers to the number $a^2$ is 1768. Find all possible values of $a$. | Answer: $a=-12, a=13$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+12$. Notice that if some number lies on the segment $[k ; k+12]$, then the sum of distances from it to the given thirteen numbers does not exceed $\frac{13}{2} \cdot 12=78$ (the sum of distances to the two extre... | =-12,=13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,415 |
7. A quadrilateral $K L M N$ is inscribed in a circle with diagonals $K M$ and $L N$, which intersect at point $T$. The bases of the perpendiculars dropped from point $T$ to the sides of the quadrilateral lie on these sides. The distances from point $T$ to the sides $K L, L M, M N, N K$ are $4 \sqrt{2}$, $\sqrt{2}$, $\... | Answer: $K T: T M=4, B D=\frac{50}{\sqrt{34}}$.
Solution. Since inscribed angles subtended by the same arc are equal, $\angle T L K=$ $\angle T M N, \angle T K L=\angle T N M$. Therefore, triangles $T K L$ and $T N M$ are similar. Similarly, it can be proven that $\triangle T K N \sim \triangle T L M$. Corresponding e... | KT:TM=4,LN=\frac{50}{\sqrt{34}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,416 |
1. Given 5000 cards, on which natural numbers from 1 to 5000 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: 124950.
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 50 such cards). For the sum to be divisible by 100, the second card must be chosen so that the number on it also ends in 00. We get a tot... | 124950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,417 |
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=-50 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x... | Answer: 0.02.
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=-50 g(x)$ is equivalent to the equation $(f(x))^{2}=-50 g(x)$ having exactly one solution. We get $(a x+b)^{2}=-50(a x+c), a^{2} x^{2}+2 a(b+25) x+b^{2}+50 c=0$. One quarter of the discriminant of t... | 0.02 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,418 |
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{22}{x_{2}}=x_{2}^{3}+\frac{22}{x_{1}}
$$
Find all possible values of $a$. | Answer: $a= \pm 3$.
Solution. For the equation to have roots, its discriminant must be positive, from which $a^{2}+8>0$ (this inequality is true for any $a$). By Vieta's theorem, $x_{1}+x_{2}=-a$, $x_{1} x_{2}=-2$. Then $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}+2$.
Transform the ... | \3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,419 |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 5 with natural exponents, and at the same time, their coefficients in absolute value do not exceed $125^{48}$. | Answer: 5112.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a, b$ are natural numbers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. By the condition
$$
\left\{\begin{array}... | 5112 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,420 |
5. A circle with center $O$, inscribed in triangle $P Q R$, touches its sides $P Q, Q R$ and $R P$ at points $C, A$ and $B$ respectively. Lines $B O$ and $C O$ intersect sides $P Q$ and $P R$ at points $K$ and $L$ respectively. Find the ratio $Q A: A R$, if $K Q=1, Q R=11, L R=2$. | Answer: $5: 6$.
Solution. Triangles $P B K$ and $P C L$ are equal ($P B=P C$ as tangents drawn from the same point to a circle; $\angle P$ is common; angles at vertices $B$ and $C$ are right angles). Therefore, $P K=P L$, and since $P C=P B$, then $C K=B L$. Let $C K=B L=x$. Then $Q A=Q C=$ $Q K+K C=1+x, R A=R B=R L+L... | 5:6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,421 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twelve consecutive natural numbers to a certain number $a$ is 3306, and the sum of the distances from these same twelve numbers to the number $a^{2}$ is 1734. Find all possible values of $a$... | Answer: $a=21, a=-20$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+11$. Note that if some number lies on the segment $[k ; k+11]$, then the sum of the distances from it to the given twelve numbers does not exceed $6 \cdot 11=66$ (the sum of the distances to the two extreme numb... | =21,=-20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,422 |
7. A quadrilateral $K L M N$ is inscribed in a circle with diagonals $K M$ and $L N$, which intersect at point $T$. The bases of the perpendiculars dropped from point $T$ to the sides of the quadrilateral lie on these sides. The distances from point $T$ to the sides $K L, L M, M N, N K$ are $5 \sqrt{2}$, $\sqrt{2}$, $5... | Answer: $K T: T M=5, L N=\frac{36}{\sqrt{13}}$.
Solution. Since inscribed angles subtended by the same arc are equal, $\angle T L K=$ $\angle T M N, \angle T K L=\angle T N M$. Therefore, triangles $T K L$ and $T N M$ are similar. Similarly, it can be proven that $\triangle T K N \sim \triangle T L M$. Corresponding e... | KT:TM=5,LN=\frac{36}{\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,423 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-a x+2, f_{2}(x)=x^{2}+3 x+b, f_{3}(x)=3 x^{2}+(3-2 a) x+4+b$ and $f_{4}(x)=3 x^{2}+(6-a) x+2+2 b$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, ... | Answer: $\frac{1}{3}$.
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 $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|}$. Ap... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,424 |
2. It is known that $\frac{\cos x-\sin x}{\sin y}=\frac{1}{3 \sqrt{2}} \operatorname{ctg} \frac{x+y}{2}$ and $\frac{\sin x+\cos x}{\cos y}=-6 \sqrt{2} \operatorname{tg} \frac{x+y}{2}$. Find all possible values of the expression $\operatorname{tg}(x+y)$, given that there are at least three. | Answer: $-1, \frac{12}{35}$ or $-\frac{12}{35}$.
Solution. Multiplying the two given equations, we get $\frac{\cos ^{2} x-\sin ^{2} x}{\cos y \sin y}=-2$, which is equivalent to the following on the domain of definition:
$$
\cos 2 x=-2 \sin y \cos y \Leftrightarrow \cos 2 x+\cos \left(2 y-\frac{\pi}{2}\right)=0 \Left... | -1,\frac{12}{35},-\frac{12}{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,425 |
3. On the table, there are 130 different cards with numbers $502, 504, 506, \ldots, 758, 760$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 119282
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next num... | 119282 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,426 |
4. Circles $\Omega$ and $\omega$ touch externally at point $F$, and their common external tangent touches circles $\Omega$ and $\omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\omega$ again at point $C$, and intersects $\Omega$ at points $D$ and $E$ (point $D$ is be... | Answer: $H P=7 \sqrt{46}, r=5 \sqrt{\frac{138}{7}}, R=5 \sqrt{\frac{322}{3}}$.
Solution. We apply the theorem of the tangent and secant three times:
$$
\begin{gathered}
H F^{2}=H C \cdot H B=4 \cdot 46 \Rightarrow H F=2 \sqrt{46} \\
H F^{2}=H D \cdot H E \Rightarrow H E=\frac{H F^{2}}{H D}=46 \\
B A^{2}=B D \cdot B E... | HP=7\sqrt{46},r=5\sqrt{\frac{138}{7}},R=5\sqrt{\frac{322}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,427 |
5. Solve the inequality
$$
\left(\log _{\frac{3}{2} x^{2}-\frac{4}{3} x+\frac{5}{6}}\left(1+4 x^{2}\right) \cdot \log _{\frac{3}{2} x^{2}-\frac{4}{3} x+\frac{5}{6}}\left(1-4 x^{2}\right)+1\right) \log _{1-16 x^{4}}\left(\frac{3 x^{2}}{2}-\frac{4 x}{3}+\frac{5}{6}\right) \geqslant 1
$$ | Answer: $x \in\left(-\frac{1}{2} ;-\frac{1}{3}\right] \cup\left[-\frac{1}{5} ;-\frac{1}{9}\right) \cup\left(-\frac{1}{11} ; 0\right) \cup\left(0 ; \frac{1}{3}\right]$.
Solution. The domain of definition of the logarithms in the inequality is determined by the conditions
$$
\begin{gathered}
1-4 x^{2}>0 \Leftrightarrow... | x\in(-\frac{1}{2};-\frac{1}{3}]\cup[-\frac{1}{5};-\frac{1}{9})\cup(-\frac{1}{11};0)\cup(0;\frac{1}{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,428 |
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