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7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, the area of triangle $AEC$ is $\sqrt{3} / 2$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1: 2$. Fin... | Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right-angled, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{E C}{\sin 30^{\circ}}=2 E C, A E=E C \operatorname{tg} 60^{\circ}=\sqrt{3} E C ;... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,518 |
8. On the line $y=-13 / 6$ find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, the angle between which is $60^{\circ}$. | Solution (without using derivatives).
$$
y=x^{2} / 2, M\left(x_{0} ;-13 / 6\right)
$$
The equation $\frac{1}{2} x^{2}=-\frac{13}{6}+k\left(x-x_{0}\right)$, or $x^{2}-2 k x+2 k x_{0}+\frac{13}{3}=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-\frac{13}{3}=0$. The two values of $k$ found from this equation m... | M(\2;-13/6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,519 |
9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=9(y-x+a-2), \log _{(x / 2)}(y / 2)=1$ has at least one solution, and solve it for each $a$. | Solution. The second equation is equivalent to the system: $x>0, x \neq 2, y=x$. Substituting $y=x$ into the first equation, we get; $(x-a)^{2}=9(a-2)$, or $x^{2}-2 a x+a^{2}-9 a+18=0(*)$, which has $D / 4=a^{2}-a^{2}+9 a-18=9(a-2)$. The number of solutions to the given system of equations depends on the number of root... | 11,(20,20) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,520 |
10. The base of the right prism $A B C A_{1} B_{1} C_{1}$ is a triangle $A B C$ with angle $B$ equal to $90^{\circ}$ and angle $C$ equal to $30^{\circ}$. Find the area of the section of the prism by a plane passing through the center of the lateral face $A A_{1} C_{1} C$ and vertex $B$ and parallel to the diagonal of t... | Solution: Construction of the section. Through point $O$ - the center of the lateral face $A A_{1} C_{1} C$ - draw $O S \| A B_{1}, S \in(A B C), O S=A B_{1} / 2$ and $O H \perp A A_{1}, H \in A C$. Then $S H \| A B, S H=A B / 2$. Connect points $B$ and $S, D=B S \cap A C$. Extend $D O$ until it intersects the extensio... | \frac{6}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,522 |
1. A bus, starting at 6:00 AM, travels between points $A$ and $B$ at a constant speed, and upon reaching point $A$ or $B$, it immediately turns around. Petya on a moped and Vasya on a bicycle simultaneously set off from point $A$ to point $B$ at 6:00 AM, moving at constant speeds. It is known that the bus, during its s... | Solution. We solve graphically. Let's represent all movements in the coordinates $t$ (time) and $S$ (distance). The trajectory of Petya's movement from point $A$ to point $B$ is the segment $A_{1} B_{2}$. The trajectory of Vasya's movement from point $A$ to point $B$ is the segment $A_{1} B^{3}$.
The bus traveled the ... | 3:5,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,523 |
2. Find all integer solutions of the inequality
$$
x^{2} y^{2}+y^{2} z^{2}+x^{2}+z^{2}-38(x y+z)-40(y z+x)+4 x y z+761 \leq 0
$$ | # Solution:
$\left(x^{2} y^{2}+2 x y z+z^{2}\right)+\left(y^{2} z^{2}+2 x y z+x^{2}\right)-38(x y+z)-40(y z+x)+761 \leq 0$.
$(x y+z)^{2}+(y z+x)^{2}-38(x y+z)-40(y z+x)+761 \leq 0$,
$(x y+z)^{2}-2 \cdot 19(x y+z)+361-361+(y z+x)^{2}-2 \cdot 20(y z+x)+400-400+761 \leq 0$,
$(x y+z-19)^{2}+(y z+x-20)^{2} \leq 0,\left\... | (6,2,7),(20,0,19) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,524 |
3. Solve the inequality $\frac{4 x^{4}+1}{4 \sqrt{2}} \leq x \sqrt{x^{4}-\frac{1}{4}}$. | Solution. Domain of definition: $x^{4}-\frac{1}{4} \geq 0, \quad x \in(-\infty ;-1 / \sqrt{2}] \cup[1 / \sqrt{2} ;+\infty)$
1) for $x \in(-\infty ;-1 / \sqrt{2}]$ the inequality has no solutions;
2) for $x \in[1 / \sqrt{2} ;+\infty)$ we square both sides of the inequality:
$$
\frac{16 x^{8}+8 x^{4}+1}{32} \leq x^{2}\... | \sqrt{\frac{1+\sqrt{2}}{2}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,525 |
4. Prove that for any natural values of $n$ the number $5^{n}\left(2^{2 n}-3^{n}\right)+2^{n}-7^{n}$ is divisible by 65. | Solution. We will prove that the number $5^{n}\left(2^{2 n}-3^{n}\right)+2^{n}-7^{n}$ is divisible by 5 for any natural values of $n$:
$20^{n}-15^{n}+2^{n}-7^{n}=5^{n}\left(4^{n}-3^{n}\right)+2^{n}-(2+5)^{n}=5^{n}\left(4^{n}-3^{n}\right)+2^{n}-2^{n}-5 k=5^{n}\left(4^{n}-3^{n}\right)-5 k=5 N$, where $N$ is a natural nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 10,526 |
5. Find all values of the parameter $b$, for which for any value of the parameter $a \in[-1 ; 1]$ the inequality $x^{2}+6 x+2(a+b+1) \sqrt{-x^{2}-6 x-5}+8<a^{2}+b^{2}+2 a$ is not satisfied for at least one value of $x \in[-5 ;-1]$.
(20 points) | Solution: Let's make the substitution $y=\sqrt{-x^{2}-6 x-5}=\sqrt{4-(x+3)^{2}}, \quad y \in[0 ; 2]$. We get $y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3>0$. We need to determine for which values of $a$ and $b$ the inequality $y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3>0$ holds for any $y \in[0 ; 2]$. Consider the function $f(y)=y^{2}-... | [-1;4] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,527 |
6. The first circle with center at point $O$ is inscribed in triangle $A B C$. Points $A$ and $B$ lie on the second circle with the same center $O$. Line $A C$ intersects the second circle at point $D$ $(D \neq A)$, and line $B C$ intersects the second circle at point $E(B \neq E)$. It is known that angle $A B C$ is eq... | Solution. Points $D$ and $E$ lie outside segments $AC$ and $BC$ respectively (Fig. 1). Otherwise (Fig. 2), arc $DE$ is a part of arc $AE$ (and does not coincide with it), which contradicts the condition of the equality of angles $ABC$ and $CAE$.
Let $\angle ABC = \beta$. Then $\angle CAE = \beta$, $\angle CAE = \angle... | \frac{1+\sqrt{5}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,528 |
1. For chemical experiments, two identical test tubes were taken, each containing 200 ml of a liquid substance. From the first test tube, $1 / 4$ of the content was poured out and the same amount of water was added, then this procedure was repeated 3 more times, each time pouring out a quarter of the content of the tes... | Solution. The initial amount of the substance is $-V$. After pouring out $a$ part, the concentration of the substance in the test tube becomes $\frac{V-a}{V}$. After the second time, the concentration is $\left(\frac{V-a}{V}\right)^{2}$, and after the fourth time, it is $\left(\frac{V-a}{V}\right)^{4}$. Substituting th... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,529 |
3. Solve the inequality
$$
\frac{(|x+1|-|x-1|)\left(x^{3}-7 x^{2}+36\right)}{x^{8}+2 x^{6}-6 x^{4}+2 x^{2}+1} \geq 0
$$ | Solution. Let's regroup the terms in the denominator:
$$
x^{8}+2 x^{6}-6 x^{4}+2 x^{2}+1=x^{8}-2 x^{4}+1+2 x^{6}-4 x^{4}+2 x^{2}=\left(x^{4}-1\right)^{2}+2 x^{2}\left(x^{2}-1\right)^{2} \geq 0
$$
Multiply the fraction by a positive quantity, and we get $\frac{\left(|x+1|^{2}-|x-1|^{2}\right)(x+2)(x-3)(x-6)}{\left(\le... | (-\infty;-2]\cup[0;1)\cup(1;3]\cup[6;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,531 |
4. Solve the equation $\quad \sqrt{x+\sqrt{x}-\frac{71}{16}}-\sqrt{x+\sqrt{x}-\frac{87}{16}}=\frac{1}{2}$. | Solution. Taking into account that $x$ is non-negative, we make the substitution $u=\sqrt{x+\sqrt{x}-\frac{71}{16}}, v=\sqrt{x+\sqrt{x}-\frac{87}{16}}, u \geq 0, v \geq 0$.
Then we obtain the system $\left\{\begin{array}{l}u-v=1 / 2, \\ u^{2}-v^{2}=1,\end{array} \Rightarrow\left\{\begin{array}{l}u-v=1 / 2, \\ u+v=2,\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,532 |
6. For what values of the parameter a does the equation $\left(x^{2}-a\right)^{2}+2\left(x^{2}-a\right)+(x-a)+2=0$ have exactly one solution? Specify the solution for the found values of the parameter a. (20 points) | Solution.
Transform the equation $\quad\left(x^{2}-a\right)^{2}+2\left(x^{2}-a\right)+(x-a)+2=0$ to
the form $a^{2}-a\left(2 x^{2}+3\right)+x^{4}+2 x^{2}+x+2=0$.
This is a quadratic equation in terms of the parameter a. Let's find its discriminant $D=\left(2 x^{2}+3\right)^{2}-4\left(x^{4}+2 x^{2}+x+2\right)=(2 x-1)... | =0.75,\quadx_{1}=-0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,534 |
1. In the "Clumsy Hands" club, unbalanced lever scales with arms of different lengths and pans of different weights were made in a slapdash manner. As a result of four weighings on these scales, the following "equilibria" were obtained:
$$
\begin{array}{ll}
{[\text { left } 3 \text { kg }=\text { right melon ]; }} & \... | Solution. The mass ratio [left $x=$ right $y$] is linear: $y=k x+b$, where $k>0$ and $b$ are constants. If [left $x=$ right $y$] and [left $y=$ right $z$], then $z=K x+B$, where $K=k^{2}, B=(k+1) b$. We have
$$
\left\{\begin{array} { l }
{ K \cdot 3 + B = 5 . 5 } \\
{ K \cdot 5 + B = 1 0 }
\end{array} \Longrightarrow... | melon=4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,536 |
2. What is the maximum possible number of consecutive terms of an increasing geometric progression that can be three-digit natural numbers? Provide an example of such a sequence. (16 points) | Solution. Let the required members of the progression be $a_{0}, a_{1}, \ldots, a_{n}, a_{k}=a_{0} q^{k}$, the common ratio - an irreducible fraction $q=r / s, r>s$. Then $a_{0}=b s^{n}, a_{n}=b r^{n}, b \in \mathbb{N}$, since $r^{n}$ and $s^{n}$ are coprime. We obtain the restriction
$$
r^{n}<1000 / b, \quad s^{n} \g... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,537 |
3. In triangle $A B C$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn, $L$ is the intersection point of segments $B_{1} C_{1}$ and $A A_{1}$, $K$ is the intersection point of segments $B_{1} A_{1}$ and $C C_{1}$. Find the ratio $L M: M K$, if $M$ is the intersection point of the angle bisector $B B_{1}$ wit... | # Solution.
$A B: B C: A C=2: 3: 4$,
$a=B C=3 x, b=A C=4 x$,
$$
c=A B=2 x
$$
Draw the lines
$L L_{1}\left\|B B_{1}, K K_{1}\right\| B B_{1}$,
$L_{1} \in C_{1} B, K_{1} \in A_{1} B$.
$\angle C_{1} L_{1} L=\angle A_{1} K_{1} K=\angle B / 2$.
# | # Solution:
Let $u=a-3|x+0.5|+x+2.5, v=a-x^{2}$. Then the original equation will have the form $|u|+|v|=u-v$. The solutions to the latter equation are all $u$ and $v$ such that $u \geq 0, v \leq 0$, or $a-3|x+0.5|+x+2.5 \geq 0$ and $a-x^{2} \leq 0$. From this, we have $3|x+0.5|-x-2.5 \leq a \leq x^{2}$.
In the Oha sy... | \in{0,1,4}\cup[7,8)\cup(9,11) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,539 |
5. The cross-section of a regular hexagonal pyramid SABCDEF is formed by a plane passing through the center of the base $A B C D E F$ and parallel to the median $C M$ of the lateral face $S C D$ and the apothem $S N$ of the lateral face $S A F$. The side of the base of the pyramid is 8, and the distance from the vertex... | Solution. In the plane $S N U$ ($S U$ - the apothem of the face $S C D$) through the point $O$ draw a line $O Y$, parallel to $S N, Y \in S U, O Y$ - the midline of the triangle $2 N U$.
$S ... | \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,540 |
1. A die is rolled twice, and the sum of the points that come up is calculated and recorded. This procedure is repeated three times (a total of six rolls). Find the probability that only one of the three recorded sums is divisible by three. (12 points) | Solution. The sum of points with two throws equals 3 in two variants: $(1 ; 2)$ and (2; 1). The sum of points with two throws equals 6 in five variants: $(1 ; 5),(5 ; 1),(2 ; 4),(4 ; 2)$ and $(3 ; 3)$. The sum of points with two throws equals 9 in four variants: $(4 ; 5),(5 ; 4)$, $(3 ; 6)$ and ( $6 ; 3$ ). The sum of ... | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,542 |
2. Find the smallest natural number $m$, for which the expression $148^{n}+m \cdot 141^{n}$ is divisible by 2023 for any odd natural $n$. (16 points) | Solution. $2023=7 \cdot 289$, GCD $(7 ; 289)=1$. Since $n-$ is an odd number, then $148^{n}+m \cdot 141^{n}=(289-141)^{n}+m \cdot 141^{n}=289 l+(m-1) 141^{n}, l \in \square$. Then $(m-1) 141^{n}$ must be divisible by 289. Since 289 and 141 are coprime, then $m-1=289 k, k \in\{0\} \cup \square$. On the other hand $148^{... | 1735 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,543 |
3. In triangle $A B C$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn, $L$ is the intersection point of segments $B_{1} C_{1}$ and $A A_{1}$, $K$ is the intersection point of segments $B_{1} A_{1}$ and $C C_{1}$, $M$ is the intersection of $B K$ and $A A_{1}$, $N$ is the intersection of $B L$ and $C C_{1}$.... | # Solution.
$$
\begin{gathered}
A B: B C: A C=2: 3: 4 \\
a=B C=3 x, b=A C=4 x \\
c=A B=2 x
\end{gathered}
$$
Draw the lines
$$
\begin{gathered}
L L_{1} \| B B_{1}, K K_{1} \| B B_{1} \\
L_{1} \in C_{1} B, K_{1} \in A_{1} B
\end{gathered}
$$
 | Solution:
Let $\quad u=2+|x|-a$, $v=a-|x+1|-|x-1| . \quad$ Then the original equation will have the form $|u|-|v|=u-v$. The solutions to the latter equation are all $u$ and $v$ such that $u \geq 0, v \geq 0$, or $u=v$, $\left[\begin{array}{c}\left\{\begin{array}{c}2+|x|-a \geq 0, \\ a-|x+1|-|x-1| \geq 0,\end{array}\ri... | \in(2;3]withsolutionsx_1=-1,x_2=1;1+\frac{3n}{2},n\in\mathbb{Z},n\geq2,withsolutionsx_1=-n,x_2=n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,545 |
5. The cross-section of a regular hexagonal pyramid SABCDEF is formed by a plane passing through the vertex $C$ of the base $A B C D E F$ and parallel to the median $B M$ of the lateral face $S A B$ and the apothem $S N$ of the lateral face $S A F$. The side length of the base of the pyramid is 2, and the distance from... | Solution. Construct the section of the pyramid. In the plane $S A F$, through the point $M$, draw a line $M Q$ parallel to $S N$, where $Q$ lies on the line $A F$, and $M Q$ is the midline of the triangle $S A N$. Given $A F = a$, $A Q = Q N = \frac{a}{4}$, where $a$ is the side length of the base of the pyramid.
The ... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,546 |
6. Astronomers have discovered a new celestial body beyond the planet Saturn, moving in a circular orbit, for the study of which a scientific research probe - an autonomous robot equipped with rocket engines, its own power plant, communication and navigation systems, scientific instruments, photo and video equipment, w... | Solution. Let the angle POK be $\alpha$. Triangle PKO is isosceles, so the two angles at the base are equal to $\beta=90-\alpha / 2, \quad$ angle POT is equal to $\frac{3 \cdot \alpha}{2}$, then $\alpha+\left(\alpha+\frac{3 \alpha}{2}\right)+90-\frac{\alpha}{2}=180 \Rightarrow 3 \alpha=90 \Rightarrow \alpha=30$.
Using... | 0.6\cdot10^{6}(\sqrt{2}+\sqrt{6}-2)\kappa\mathrm{},\quad6\sqrt{2-\sqrt{3}}\approx3.11\kappa\mathrm{}/\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,547 |
2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$
Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | Answer: 0.
Problem 8 (2nd version).
Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.
## Solution.
Consider the function $f(x)=g^{3}(x)-g\left(\frac{1}{x}\right)-5 x^{3}-\frac{1}{x}$, then the roots of the original equation are the roots of the... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,548 |
2. $g(-x)=(-x)^{3}+(-x)=-x^{3}-x=-\left(x^{3}+x\right)=-g(x)$
$$
f(-x)=g^{3}(-x)-g\left(\frac{1}{-x}\right)-5(-x)^{3}-\frac{1}{-x}=-g^{3}(x)+g\left(\frac{1}{x}\right)+5 x^{3}+\frac{1}{x}=-f(x)
$$
Therefore, $f$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | # Answer: 0.
Task 9 (1st variant). In each vertex of an equilateral triangle with side $\sqrt{10}$, circles of radius $\sqrt{5}$ were constructed, the inner areas of which were painted gray, brown, and raspberry. Find the area of the gray-brown-raspberry region.
Solution. By symmetry, the areas of the figures formed ... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 10,549 |
Problem 1. Student Vasya, who lives in the countryside, arrives at the station by train every evening after classes at 6 PM. By this time, his father picks him up by car and takes him home. One day, Vasya's last class at the institute was canceled, and he arrived at the station an hour earlier. Unfortunately, he forgot... | # Solution:
Vasya arrived home 20 minutes earlier than usual, during which time his father would have driven the distance Vasya walked twice. Therefore, on the way to the station, his father saved 10 minutes and met Vasya 10 minutes earlier than usual, that is, at 17:50.
Answer: $17: 50$. | 17:50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,550 |
Problem 2. Given that $1580!=a, \quad$ calculate: $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+1580 \cdot 1580!$ $(n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n)$ | Solution:
$1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+1580 \cdot 1580!=$
$=(2-1) \cdot 1!+(3-1) \cdot 2!+(4-1) \cdot 3!+\ldots+(1581-1) \cdot 1580!=$
$=2 \cdot 1!-1!+3 \cdot 2!-2!+4 \cdot 3!-3!+\ldots+1581 \cdot 1580!-1580!=$
$=2!-1!+3!-2!+4!-3!+\ldots+1581!-1580!=$
$=-1!+1581!=1581 \cdot 1580!-1=1581 a-1$
Answer: $... | 1581a-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,551 |
Problem 3. Given segments a and $b(a>b)$. Construct a segment of length $\frac{a^{2}+b^{2}}{a-b}$ using a compass and a straightedge. | Solution:
Construct an equilateral triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=a$. On the line $\mathrm{AB}$, mark segments $\mathrm{BD}=\mathrm{DE}=b$. Construc... | -b+\frac{2ab}{-b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,552 |
Problem 4. Find integer solutions of the equation: $2 x^{4}-4 y^{4}-7 x^{2} y^{2}-27 x^{2}+63 y^{2}+85=0$. Solution: Make the substitution: $x^{2}=a, \quad y^{2}=b$, then the equation becomes: $2 a^{2}-4 b^{2}-7 a b-27 a+63 b+85=0$
$2 a^{2}-(7 b+27) a-4 b^{2}+63 b+85=0$
$D=(9 b-7)^{2}$, the roots of the equation are ... | Answer: $(3 ; \pm 1),(-3 ; \pm 1),(2 ; \pm 3),(-2 ; \pm 3)$ | (3;\1),(-3;\1),(2;\3),(-2;\3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,553 |
Problem 5. In triangle $\mathrm{KLM}$ with angle $\mathrm{L}=120^{\circ}$, the angle bisectors LA and $\mathrm{KB}$ of angles KLM and LKM are drawn respectively. Find the measure of angle KBA. | # Solution.
1). Let KS be the extension of KL beyond point L. Then LM is the bisector of angle MLS, since $\angle M L S = \angle M L A = \angle A L K = 60^{\circ}$. Point B is the intersecti... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,554 |
Problem 6. Plot the figure on the coordinate plane defined by the system of inequalities and find its area $\left\{\begin{array}{c}|x+5|+\sqrt{3}|y-1| \leq 3 \\ y \leq \sqrt{4-4 x-x^{2}}+1 \\ |2 y-1| \leq 5\end{array}\right.$. | # Solution:
The first inequality defines the area inside a rhombus centered at the point $(-5 ; 1)$ with diagonals of 6 and $2 \sqrt{3}$.
The second inequality defines the area below the upper semicircle centered at the point $(-2 ; 1)$ with a radius of $2 \sqrt{2}$.
The third inequality: the strip $-2 \leq y \leq 3... | \frac{2}{3}\pi+\frac{4}{3}\sqrt{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,555 |
4. $\frac{2 \operatorname{tg}^{4} 8 x+4 \sin 3 x \sin 5 x-\cos 6 x-\cos 10 x+2}{\sqrt{\sin x-\cos x}}=0$.
Given the condition $\sin x-\cos x>0$, we find the roots of the equation
$2 \operatorname{tg}^{4} 8 x+4 \sin 3 x \sin 5 x-\cos 6 x-\cos 10 x+2=0 \Leftrightarrow$
$2 \operatorname{tg}^{4} 8 x+4 \sin 3 x \sin 5 x-... | Answer: $x=\frac{\pi}{2}+2 \pi n, x=\frac{3 \pi}{4}+2 \pi n, x=\pi+2 \pi n, n \in Z$. | \frac{\pi}{2}+2\pin,\frac{3\pi}{4}+2\pin,\pi+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,558 |
5. $\frac{\left(2 \cdot 2^{-\log _{x} 3}-4\right) \sqrt{2-\sqrt{\log _{x} 3+2}}}{1+\sqrt{\log _{x} 3+5}}>\frac{\left(2^{-\log _{x} 3}-2\right) \sqrt{2-\sqrt{\log _{x} 3+2}}}{\sqrt{\log _{x} 3+5}-2}$
Let's make the substitution $y=\log _{x} 3$.
$$
\begin{aligned}
& \frac{\left(2 \cdot 2^{-y}-4\right) \sqrt{2-\sqrt{y+2... | Answer: $x \in (0 ; 1 / 3) \cup (1 / 3 ; 1 / \sqrt{3}] \cup (\sqrt{3} ; +\infty)$. | x\in(0;1/3)\cup(1/3;1/\sqrt{3}]\cup(\sqrt{3};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,559 |
# 7.
1) $\angle A P B=\angle B A C, \angle A P B=\angle A K C, \angle A K C=\angle B A C, \angle K A C=\angle A B C$.
Segment $A C$ is a tangent segment to the circle.
$\triangle A B C \approx \triangle A K C \Rightarrow$
$\frac{A B}{A K}=\frac{A C}{K C}=\frac{B C}{A C} \Rightarrow \frac{A B}{4}=\frac{A C}{3}=\frac... | Answer: $\frac{-11+3 \sqrt{145}}{\sqrt{74}}$.
 | \frac{-11+3\sqrt{145}}{\sqrt{74}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,561 |
8. On the line $x=\sqrt{3} / 2$, find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, such that the angle between them is $60^{\circ}$ (without using derivatives).
$$
y=x^{2} / 2, \quad M\left(\sqrt{3} / 2 ; y_{0}\right) . \text { The equation } \frac{1}{2} x^{2}=y_{0}+k\left... | Answer: $M_{1}(\sqrt{3} / 2 ; 0), M_{2}(\sqrt{3} / 2 ;-5 / 3)$. | M_{1}(\sqrt{3}/2;0),M_{2}(\sqrt{3}/2;-5/3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,562 |
9. Find all values of $a$ for which the system of equations
$$
y-2=a(x-4), \quad \frac{2 x}{|y|+y}=\sqrt{x}
$$
has at least one solution, and solve it for each $a$.
Domain of definition: $y>0, x \geq 0$.
In the domain of definition, the second equation of the system takes the form: $x=y \sqrt{x}$.
I. $x=0, y=2-4 a>... | # Answer:
$$
\begin{aligned}
& a \in(-\infty ; 0] \cup\left\{\frac{1}{4}\right\}, x_{1}=0, y_{1}=2-4 a ; x_{2}=4, y_{2}=2 \\
& a \in\left(0 ; \frac{1}{4}\right) \cup\left(\frac{1}{4} ; \frac{1}{2}\right), x_{1}=0, y_{1}=2-4 a ; x_{2}=4, y_{2}=2 ; x_{3}=\left(\frac{1-2 a}{a}\right)^{2}, y_{3}=\frac{1-2 a}{a} ; \\
& a \... | \begin{aligned}&\in(-\infty;0]\cup{\frac{1}{4}},x_{1}=0,y_{1}=2-4x_{2}=4,y_{2}=2\\&\in(0;\frac{1}{4})\cup(\frac{1}{4};\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,563 |
1. A pedestrian left point $A$ for point $B$, and after some delay, a second pedestrian followed. When the first pedestrian had walked half the distance, the second had walked 15 km, and when the second pedestrian had walked half the distance, the first had walked 24 km. Both pedestrians arrived at point $B$ simultaneo... | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the pedestrians. Then $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$ and $\frac{s-24}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24 ;$ $s^{2}-52 s+480... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,564 |
2. Solve the inequality $\sqrt{\frac{x-12}{x+4}}-\sqrt{\frac{x+4}{x-12}}<\frac{16}{15}$. | # Solution:
$$
\sqrt{\frac{x+4}{x-12}}=y>0 ; \frac{1}{y}-y>0 ; y_{1,2}=\frac{-8 \pm \sqrt{64+225}}{15}=\frac{-8 \pm 17}{15}
$$
$y_{1}=\frac{3}{5}, y_{2}=-\frac{5}{3}$. Therefore, $y>\frac{3}{5}, \frac{x+4}{x-12}>\frac{9}{25} \Leftrightarrow \frac{x+13}{x-12}>0 \Leftrightarrow\left[\begin{array}{c}x<12 \\ x>-13 .\end{... | x<12orx>-13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,565 |
3. Three numbers form a geometric progression. If the second term is increased by 8, the progression turns into an arithmetic one, but if then the third term of the obtained progression is increased by 64, it turns back into a geometric progression. Find these numbers. | Solution: Let $a, b, c$ be the required numbers. Then
$$
\begin{gathered}
\left\{\begin{array} { c }
{ a c = b ^ { 2 } , } \\
{ a + c = 2 ( b + 8 ) , } \\
{ a ( c + 6 4 ) = ( b + 8 ) ^ { 2 } , }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ a c = b ^ { 2 } , } \\
{ a + c = 2 ( b + 8 ) , } \\
{ a c + 6 4 a... | a_{1}=\frac{4}{9},\quadb_{1}=-\frac{20}{9},\quadc_{1}=\frac{100}{9},\quada_{2}=4,\quadb_{2}=12,\quadc_{2}=36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,566 |
4. Solve the equation $\frac{\cos x}{\sqrt{3}}-\sqrt{\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2}}=0$.
# | # Solution:
$$
\frac{\cos x}{\sqrt{3}}=\sqrt{\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2}} . \text { Given } \cos x \geq 0 \text {, we square both sides of the equation }
$$
$\frac{\cos ^{2} x}{3}=\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2} \Leftrightarrow \frac{\cos ^{2} x}{3}=\frac{2 \sin ^{2} x-2 \sin ^{3} x}{6 \si... | \frac{\pi}{2}+2\pin,\frac{\pi}{6}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,567 |
5. Solve the inequality $\sqrt{x+2-|x+1|} \leq x+5-|2 x+3|$.
# | # Solution:
$$
\left\{\begin{array} { c }
{ x + 5 - | 2 x + 3 | \geq 0 , } \\
{ x + 2 - | x + 1 | \geq 0 , } \\
{ x + 2 - | x + 1 | \leq ( x + 5 - | 2 x + 3 | ) ^ { 2 } , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
|2 x+3| \leq x+5 \\
|x+1| \leq x+2, \\
x+2-|x+1| \leq(x+5-|2 x+3|)^{2}
\end{array} \Leftright... | x\in[-1.5;1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,568 |
6. Find the set of values of the function $f(x)=g\left(\sqrt{25-g^{2}(x)}\right)$, where $g(x)=|| x|-2|-1$. | Solution: The function $g(x)=|| x|-2|-1 \quad$ is defined on the entire number line and takes all values from the interval $[-1 ;+\infty)$. The graph of the function $g(x)$ is shown in the figure. The function $\phi(t)=\sqrt{25-t^{2}}$ is defined for $t \in[-5 ; 5]$. When $\quad t=g(x)$, the function $\quad \phi(t)=\sq... | [-1;2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,569 |
7. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A B$ at point $M$, and $A M=18$. Find the sides of the trapezoid if its perimeter is 112, and the area is 672. | Solution: $M, E, K, F$ are the points of tangency of the circle with sides $A B$, $B C, C D, A D$ respectively. Then
$$
A M=A F=a=18, B M=B E=x, C E=C K=y, D K=D F=z
$$
$P_{A B C D}=2(18+x+... | 1)AB=26,BC=14,CD=30,AD=42,2)AB=26,BC=32,CD=30,AD=24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,570 |
9. Specify all values of $a$ for which the equation $(x-a)^{2}=\frac{x}{|x|}+a+1$ has at least one solution, and solve it for each $a$.
# | # Solution:
I. For $x>0 \quad x^{2}-2 a x+a^{2}-a-2=0 \quad(*)$. Equation (*) has two distinct positive roots $x_{1,2}=a \pm \sqrt{a+2}$, if: $\left\{\begin{array}{c}D / 4=a+2>0, \\ a>0, \\ a^{2}-a-2>0\end{array} \Leftrightarrow\left\{\begin{array}{c}a>-2, \\ a>0, \\ {\left[\begin{array}{c}a2\end{array}\right.}\end{ar... | \in(-1;0]\cup[1;2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,572 |
10. The base of the pyramid is a rectangle with sides $AB=24$ and $BC=30$, and the lateral edge of the pyramid $TA=16$ is perpendicular to the plane of the base. What is the minimum area that the section of the pyramid by a plane passing through the center of symmetry of the base $O$, the vertex of the pyramid, and a p... | # Solution:
Regardless of the position of point $M$ on side $B C$, the face $T A B$ is the orthogonal projection of the section $T M N$. The area of the section will be the smallest if the angle between the cutting plane and the face $T A B$ is the smallest. Since the cutting plane passes through the center of symmetr... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,573 |
1. Two cyclists set off simultaneously from point A to point B. When the first cyclist had covered half the distance, the second cyclist still had 24 km to go, and when the second cyclist had covered half the distance, the first cyclist still had 15 km to go. Find the distance between points A and B. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}} \quad$ and $\quad \frac{s-15}{v_{1}}=\frac{s}{2 v_{2}} . \quad$ From this, $\quad \frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; \quad s^{2}=4 s^{2}-4 \cdot 39 s+60... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,574 |
3. Three numbers, the sum of which is 114, are, on the one hand, three consecutive terms of a geometric progression, and on the other - the first, fourth, and twenty-fifth terms of an arithmetic progression, respectively. Find these numbers. | Solution: Let $a, b, c$ be the required numbers, and $d$ be the common difference of the arithmetic progression. Then
$$
\begin{gathered}
\left\{\begin{array} { c }
{ a + b + c = 114 , } \\
{ a c = b ^ { 2 } , } \\
{ b = a + 3 d , } \\
{ c = a + 24 d , }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ 3 a +... | a_{1}=38,b_{1}=38,c_{1}=38ora_{2}=2,b_{2}=14,c_{2}=98 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,575 |
4. Solve the equation $\operatorname{ctg} 2 x \sqrt{\sin x \cos x}-\sqrt{1-\sin x \cos x}=0$. | # Solution:
$\operatorname{ctg} 2 x \sqrt{\sin x \cos x}=\sqrt{1-\sin x \cos x}$. Given $\operatorname{ctg} 2 x \geq 0$, we square both sides of the equation ($1-\sin x \cos x>0$ for all $x$): $\operatorname{ctg}^{2} 2 x \sin x \cos x=1-\sin x \cos x \Leftrightarrow$ $\operatorname{ctg}^{2} 2 x \sin 2 x=2-\sin 2 x \Le... | \frac{\pi}{12}+\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,576 |
6. Find the set of values of the function $f(x)=g(2 \sqrt{2.5-g(x)})$, where $g(x)=\frac{3}{|x-2|+1}$. | # Solution:
The function $g(x)=\frac{3}{|x-2|+1}$ is defined on the entire number line and takes all values from
the interval $(0 ; 3]$. The function $g(x)$ reaches its maximum value at th... | [1;3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,578 |
7. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A B$ at point $M$, and the lateral side $C D$ at point $K$, with $A M=9, C K=3$. Find the diagonals of the trapezoid, if its perimeter is 56. | Solution: $M, E, K, F$ are the points of tangency of the circle with sides $A B, B C, C D, A D$ respectively. Then
$$
A M=A F=a=9
$$
$$
B M=B E=x, \quad C E=C K=y=3,
$$
$D K=D F=z$.
$$
\begin{aligned}
& P_{A B C D}=2(9+x+3+z)=56 \\
& x+z=16
\end{aligned}
$$
Let $B P \perp A D, C N \perp A D$.
^{2}=y+a$ has at least one solution, and solve it for each $a$.
# | # Solution:
I. When $x>0 \quad y=2, \quad x^{2}-2 a x+a^{2}-a-2=0(*)$. Equation $(*)$ has two distinct positive roots $x_{1,2}=a \pm \sqrt{a+2}$, if: $\left\{\begin{array}{c}D / 4=a+2>0, \\ a>0, \\ a^{2}-a-2>0\end{array} \Leftrightarrow\left\{\begin{array}{c}a>-2, \\ a>0, \\ {\left[\begin{array}{c}a2\end{array}\right.... | \in(-1;0]\cup[1;2]\quad+\sqrt{+2},2;\quad(0;1)x_{1}=+\sqrt{+2},y_{1}=2;x_{2}=-\sqrt{};y_{2}=0;\quad\in(2;+\infty)x_{1,2}=\\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,581 |
10. The base of the pyramid $T A B C D$ is a rectangle with sides $A B=12$ and $A D=4$, and the lateral edges are respectively equal to $T A=3, T D=5, T C=13$. What is the minimum area that the cross-section of the pyramid can have when cut by a plane passing through vertex $T$, the center of symmetry of the base, and ... | # Solution:
When constructing the drawing, the following should be taken into account.
$$
\begin{aligned}
& T A^{2}+A D^{2}=T D^{2},\left(3^{2}+4^{2}=5^{2}\right) \Rightarrow \angle T A D=\pi / 2 \\
& A B=D C=12 \\
& D C^{2}+T D^{2}=T C^{2},\left(12^{2}+5^{2}=13^{2}\right) \Rightarrow \angle T D C=\pi / 2 \\
& D C \p... | \frac{42}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,582 |
1. Two trucks were transporting fertilizers, making the same number of trips. It turned out that 4 tons less could be loaded onto the first truck and 3 tons less onto the second truck than planned, so each truck had to make 10 extra trips. As a result, the first truck transported 60 tons more than the second, as planne... | # Solution:
Let $x, y$ - capacity, $t-$ number of trips as planned.
$$
\left\{\begin{array}{l}
x t=(x-4)(t+10), \\
y t=(y-3)(t+10), \Leftrightarrow\left\{\begin{array} { l }
{ 1 0 x - 4 t = 4 0 } \\
{ \quad x t - y t = 6 0 }
\end{array} \quad \left\{\begin{array}{l}
10 y-3 t=30, \\
(x-y) t=60
\end{array} \Rightarrow... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,583 |
2. Solve the inequality $\frac{\sqrt{x^{6}-5}}{x^{3}}-\frac{x^{3}}{\sqrt{x^{6}-5}}<\frac{5}{6}$. | Solution: Domain of definition: $|x|>\sqrt[6]{5} . \frac{\sqrt{x^{6}-5}}{x^{3}}-\frac{x^{3}}{\sqrt{x^{6}-5}}\sqrt[6]{5}$ inequality is true.
2) For $x<-\sqrt[6]{5}$, or $\left(-x^{3} \sqrt{x^{6}-5}\right)^{2}>36$, $x^{12}-5 x^{6}-36>0,\left(x^{6}+4\right)\left(x^{6}-9\right)>0,\left(x^{3}+3\right)\left(x^{3}-3\right)>0... | x\in(-\infty;-\sqrt[3]{3})\cup(\sqrt[6]{5};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,584 |
5. Solve the inequality $\quad \frac{|x|+21-7 \sqrt{|x|+9}}{x^{2}-8|x|}>0$ | # Solution:
Domain of definition: $x \neq 0, \quad x \neq \pm 8$.
Factorize the numerator:
$$
|x|+21-7 \sqrt{|x|+9}=|x|+9-7 \sqrt{|x|+9}+12=(\sqrt{|x|+9})^{2}-7 \sqrt{|x|+9}+12=(\sqrt{|x|+9}-3)(\sqrt{|x|+9}-4)
$$
We have:
$$
\frac{(\sqrt{|x|+9}-3)(\sqrt{|x|+9}-4)}{|x|(|x|-8)}>0 \Leftrightarrow \frac{(|x|+9-9)(|x|+... | x\in(-\infty;-8)\cup(-7;0)\cup(0;7)\cup(8;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,586 |
6. Find the set of values of the function $f(x)=\sqrt{1-g^{2}(x)}$, where $g(x)=\frac{\cos 6 x+2 \sin ^{2} 3 x}{2-2 \cos 3 x}$. | Solution: $\quad g(x)=\frac{\cos 6 x+2 \sin ^{2} 3 x}{2-2 \cos 3 x}=\frac{1}{2-2 \cos 3 x}$.
The function $t=\cos 3 x$ takes values $t \in[-1 ; 1]$. Consider the function $y=\frac{1}{2-2 t}$, defined on the half-interval $[-1 ; 1)$. The graph of this function is a hyperbola with
asymptotes $t=1$ and $y=0$. The functio... | E_{f}=[0;\frac{\sqrt{15}}{4}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,587 |
7. In trapezoid $ABCD$, the bases $AD=9, BC=2$, angles $A$ and $D$ at the base are $\operatorname{arctg} 4$ and $\operatorname{arctg}(2 / 3)$, respectively. Find the radius of the circle circumscribed around triangle $CBE$, where $E$ is the point of intersection of the diagonals of the trapezoid. | # Solution:
$$
\begin{aligned}
& B F \perp A D ; C G \perp A D ; B F=C G=h, F G=B C . \\
& \text { Let } A F=x \Rightarrow \\
& x \cdot \operatorname{tg} A=((A D-B C)-x) \operatorname{tg} D ... | \frac{5\sqrt{5}}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,588 |
9. Find all values of $a$ for which the system of equations $2 y-2=a(x-2), \frac{4 y}{|x|+x}=\sqrt{y}$ has at least one solution, and solve it for each $a$.
# | # Solution:
Domain of definition: $x>0, y \geq 0$.
In the domain of definition, the second equation of the system takes the form: $2 y=x \sqrt{y}$.
I. $y=0, x=2-\frac{2}{a}=\frac{2(a-1)}{a}>0$, hence $\left[\begin{array}{l}a1 \text {. }\end{array}\right.$
II. $y>0, y=x^{2} / 4, x>0 ;(x-2)(x+2)=2 a(x-2)$.
1) $x=2, y... | \in(-\infty;0)\bigcup{2},x_{1}=2-2/,y_{1}=0;x_{2}=2,y_{2}=1;\in[0;1],2,1;\in(1;2)\cup(2;+\infty),x_{1}=2-2/,y_{1}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,589 |
10. The base of the pyramid $T A B C D$ is a rectangle $A B C D$. The height of the pyramid, equal to $h$, coincides with the lateral edge $T A$, and the lateral edge $T C$ is inclined to the base plane at an angle of $30^{\circ}$. The plane passing through the edge $T C$ and parallel to the diagonal of the base $B D$,... | # Solution:
Let $\alpha = \angle TCA$, then $AC = BD = TA \operatorname{ctg} \angle TCA = h \operatorname{ctg} \alpha$. Draw $EC \parallel BD$, $AK \perp EC$, $K \in EC$, and let $\beta = ... | \frac{^2\sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,590 |
1. A batch of shoes, purchased for 180 thousand rubles, was sold in the first week at a price higher than the purchase price by $25 \%$, then the markup was reduced to $16 \%$ of the purchase price; and the entire batch of shoes was sold for $20 \%$ more than it was purchased for. For what amount was the shoes sold in ... | # Solution:
$x$ thousand rubles - the purchase cost of shoes sold in the first week, $y$ - the remainder.
$$
\left\{\begin{array} { c }
{ x + y = 180 } \\
{ 25 x + 16 y = 20 ( x + y ) ; }
\end{array} \left\{\begin{array} { c }
{ 5 x = 4 y , } \\
{ x + 5 / 4 x = 180 ; }
\end{array} \left\{\begin{array}{l}
x=80 \\
y=... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,591 |
2. Solve the inequality $\frac{\sqrt{x^{6}-21}}{x^{3}}-\frac{x^{3}}{\sqrt{x^{6}-21}}<\frac{21}{10}$. | # Solution:
Domain of definition: $|x|>\sqrt[6]{21} . \frac{\sqrt{x^{6}-21}}{x^{3}}-\frac{x^{3}}{\sqrt{x^{6}-21}}\sqrt[6]{21}$ inequality is true.
2) For $x10$, or $\left(-x^{3} \sqrt{x^{6}-21}\right)^{2}>100$,
$$
\begin{aligned}
& x^{12}-21 x^{6}-100>0,\left(x^{6}+4\right)\left(x^{6}-25\right)>0,\left(x^{3}+5\right)... | x\in(-\infty;-\sqrt[3]{5})\cup(\sqrt[6]{21};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,592 |
3. What is the greatest value that the sum of the first $n$ terms of the arithmetic progression $113,109,105, \ldots$ can take?
# | # Solution:
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ takes its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. Since $a_{n}=a_{1}+d(n-1)$, from the inequality $113-4(n-1)>0$ we find $n=[117 / 4]=29$.
Then $\max S_{n}=S_{29}=0.5 \cdot(113+113-4 \cdot 28) \cdot 29=1653$.
Answer: 1653. | 1653 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,593 |
4. Solve the equation $\frac{\sin 2 x-\cos 2 x+\sqrt{3} \cos x+\sqrt{3} \sin x+1}{\sqrt{\sqrt{3} \cos x-\sin x}}=0 . \quad(8$ points) | # Solution:
$\frac{\sin 2 x-\cos 2 x+\sqrt{3} \cos x+\sqrt{3} \sin x+1}{\sqrt{\sqrt{3} \cos x-\sin x}}=0 \Leftrightarrow \frac{\sin 2 x+2 \sin ^{2} x+\sqrt{3} \cos x+\sqrt{3} \sin x}{\sqrt{\sqrt{3} \cos x-\sin x}}=0$.
Given the condition $\sqrt{3} \cos x-\sin x>0$, we find the roots of the equation
$\sin 2 x+2 \sin ... | -\frac{\pi}{3}+2\pin,-\frac{\pi}{4}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,594 |
5. Solve the inequality $\quad \frac{19-|x-3|}{\sqrt{|x-3|-1}-2} \leq 1$. (10 points) | # Solution:
Substitution: $\sqrt{|x-3|-1}=t \geq 0, x=t^{2}+1$.
$\frac{19-t^{2}-1}{t-2} \leq 1 \Leftrightarrow \frac{18-t^{2}-t+2}{t-2} \leq 0 \Leftrightarrow \frac{(t-4)(t+5)}{t-2} \geq 0 \stackrel{t \geq 0}{\Leftrightarrow} t \in[0 ; 2) \cup[4 ; \infty)$.
$|x-3|-1 \in[0 ; 4) \cup[16 ; \infty) \Leftrightarrow|x-3| ... | x\in(-\infty;-14]\cup(-2;2]\cup[4;8)\cup[20;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,595 |
6. Find the set of values of the function $f(x)=\sqrt{36-g^{2}(x)}$, where $g(x)=-8-2 \cos 8 x-4 \cos 4 x$.
# | # Solution:
$g(x)=-8-2 \cos 8 x-4 \cos 4 x=-8-4 \cos ^{2} 4 x+2-4 \cos 4 x=$
$-4 \cos ^{2} 4 x-4 \cos 4 x-6=-(2 \cos 4 x+1)^{2}-5$.
The function $t=2 \cos 4 x$ takes values $t \in[-2 ; 2]$. Consider the function $y=-5-(t+1)^{2}$, defined on the interval [-2; 2]. The graph of this function is a parabola with its vert... | E_{f}=\lfloor0;\sqrt{11}\rfloor | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,596 |
7. The area of a right triangle is 1, and its hypotenuse is $\sqrt{5}$. Find the cosine of the acute angle between the medians of the triangle drawn to its legs.
 | # Solution:
Let the legs of the triangle be $a$ and $b$. Then $\left\{\begin{array}{c}\frac{a b}{2}=1, \\ a^{2}+b^{2}=5,\end{array} \Leftrightarrow\left\{\begin{array}{c}a=\frac{2}{b}, \\ \frac{4}{b^{2}}+b^{2}=5\end{array}\right.\right.$. Solving the equation $b^{4}-5 b^{2}+4=0$, we get: $b^{2}=4$ or $b^{2}=1$. Then $... | \frac{5\sqrt{34}}{34} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,597 |
8. On the line $x-y=5$ find the point through which two perpendicular to each other tangents to the graph of the function $y=x^{2} / 8$ pass. Write the equations of these tangents. | # Solution:
Let $y=a x^{2}, A\left(x_{1} ; y_{1}\right), B\left(x_{2} ; y_{2}\right)$ be the points of tangency, $C\left(x_{0} ; y_{0}\right)$ be the point of intersection of the tangents.
Equations of the tangents
$$
\begin{aligned}
& y=a x_{1}^{2}+2 a\left(x-x_{1}\right), \text { or } y=2 a x_{1} x-a x_{1}^{2} \\
... | x_{1}=8,x_{2}=-2,x_{0}=3,y_{0}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,598 |
10. A plane is drawn through the diagonal of a rectangular parallelepiped and a point lying on a lateral edge that does not intersect this diagonal, such that the area of the section of the parallelepiped by this plane is the smallest. Find the lengths of the sides of the base of the parallelepiped, given that the diag... | # Solution:
Draw $D L \perp A C, L K\left\|C C_{1}\left(K \in A C_{1}\right), P K\right\| D L$. By laying off the segment $B Q=P D_{1}$ on the lateral edge $B B_{1}$, we obtain the parallelogram $P A Q C_{1}$, which will be the section of the smallest area; in this case, $A C_{1}$ is its larger diagonal, and $P Q$ is ... | 2\sqrt{5};\sqrt{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,601 |
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 2$) in a total of 100 pieces, with at least three flasks of each size. The lab assistant randomly selects three flasks in sequence, and fills the first one with an 80% salt solution, the second one with a 50% salt solution, and the third on... | Solution. If $N$ is the number of large flasks in the laboratory, $N=3,4, \ldots, 97$, then $n=100-N$ is the number of small flasks in the laboratory, $n=3,4, \ldots, 97$. For the event $A=\{$ the salt content in the dish is between $45 \%$ and $55 \%$ inclusive $\}$, it is necessary to find the smallest $N$ such that ... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,602 |
4. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{c}
(a y-a x+2)(4 y-3|x-a|-x+5 a)=0 \\
\left(\log _{a} x^{2}+\log _{a} y^{2}-2\right) \log _{2} a^{2}=8
\end{array}\right.
$$
has six distinct solutions. (16 points) | Solution. Simplify the second equation of the system: $\left(\log _{a} x^{2}+\log _{a} y^{2}-2\right) \log _{2} a^{2}=8 \Leftrightarrow a>$ $0, a \neq 1, \log _{a} x^{2} y^{2}=2+4 \log _{a} 2,|x y|=4 a$. We have
\cup(4;32) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,603 |
5. A sphere with radius $\frac{4}{9}$ lies inside a regular quadrilateral pyramid $SABCD$ with a base side of 8 and height 3. This sphere touches the base plane $ABCD$ of the pyramid and the lateral faces $SBC$ and $SCD$. The plane $\gamma$ touches the sphere, passes through point $B$, the midpoint $K$ of edge $CD$, an... | Solution. Since the pyramid $S A B C D$ is regular, the center $O$ of the given sphere lies in the plane $S H C$, where $S H$ is the height of the pyramid. Let $R P \| S H, R \in S C, P \in H C, O \in R P$.
Denote $S H=h, A B=a, R P=k h$. Draw $P N \| A B, N \in B C . E$ - the point of tangency of the sphere with the ... | \frac{192}{37} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,604 |
6. In 2022, it will be 65 years since the launch of the first artificial satellite of the Earth (ASZ). Currently, various types of satellites located at different orbits and at different altitudes are used to ensure the uninterrupted operation of cellular communication, television, and radio broadcasting systems.
The ... | Solution. a) The coverage area is the part of the sphere lying inside the cone.
$S=2 \pi R \cdot h$, where $h=$ AZ - the height of the segment. $h=R-R \cos \alpha$, here the angle $\alpha$ is the angle between the radius OG and the line OA, connecting the center of the sphere with the center of the circle, which is th... | 2,3,4,5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,605 |
1. The numerical sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is defined such that $a_{1}=\log _{2}\left(\log _{2} f(2)\right), \quad a_{2}=$ $\log _{2}\left(\log _{2} f(f(2))\right), \ldots, a_{n}=\log _{2}(\log _{2} \underbrace{f(f(\ldots f}_{n}(2)))), \ldots$, where $f(x)=x^{x}$. Determine the index $n$ for which ... | # Solution.
If $\log _{2}\left(\log _{2} u\right)=t$, then $u=2^{2^{t}}, f(u)=\left(2^{2^{t}}\right)^{2^{2^{t}}}=2^{2^{t+2^{t}}}, \log _{2}\left(\log _{2} f(u)\right)=t+2^{t} . \quad$ If $u=2,2=2^{2^{t}}, t=0, a_{1}=\log _{2}\left(\log _{2} f(2)\right)=0+2^{0}=1$. If $u=f(2)$, then $t=\log _{2}\left(\log _{2} u\right)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,606 |
2. Clowns Plukha and Shmyaka have six pairs of valenki (traditional Russian felt boots) between them. Each pair of valenki is painted in a unique color, and the valenki in a pair are identical (they are not distinguished as left or right). In how many ways can both clowns be simultaneously wearing mismatched valenki? (... | Solution. We can use valenki (traditional Russian felt boots) from two, three, or four pairs.
1) We choose two pairs of valenki $C_{6}^{2}=15$ ways. Each clown puts on one valenok from different pairs, choosing which one for which foot. This gives us $15 \cdot 2 \cdot 2=60$ ways.
2) We use three pairs of valenki. We c... | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,607 |
3. Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, and $A M=2, M C=6$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Find the l... | Solution. 1. A circle can be circumscribed around quadrilateral $A B C D$ with diameter $A B$ (angles $A D B$ and $A C B$ are right angles). Then $\angle A B D=\angle A C D, \angle H A D=90^{\circ}-\angle A B D, \angle A D H=\angle A B D=\angle A C D$.
Triangles $A C D$ and $A D M$ are similar, and $\frac{A D}{A C}=\f... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,608 | |
4. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{c}
\left(a\left|\log _{2} y\right|+a\left|\log _{2} x\right|-2\right)\left(\left(\log _{2} x\right)^{2}+\left(\log _{2} y\right)^{2}-48 a^{2}\right)=0 \\
\left(\log _{a} x\right)^{2}\left(\log _{a} y\right)^{2}\left(\log _{2} a^{2}\ri... | Solution. Let's make a change of variables: $\log _{2} x=u, \log _{2} y=v$. Variables $u$ and $v$ can take any values, $x$ and $y$ are uniquely determined. Simplify the second equation of the system: $\left(\log _{a} x\right)^{2}\left(\log _{a} y\right)^{2}\left(\log _{2} a^{2}\right)^{4}=256 a^{2} \Leftrightarrow a>0,... | \in(0;1/6)\cup{1/2}\cup(1/\sqrt[3]{4};1)\cup(1;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,610 |
5. Inside a regular quadrilateral pyramid $S A B C D$ with base $A B C D$, there is a regular quadrilateral prism $K L M N K_{1} L_{1} M_{1} N_{1}$, the base $K L M N$ of which lies in the plane $A B C$. The center of the base $K L M N$ of the prism is located on the segment $A C$, $K L\|A C$, $K N\|$ $B D$ (points $K$... | Solution. Plane $\gamma$ contains $BD$.
Given $AB = a = 8\sqrt{2}$, $SO = h = 4$, $KL = b = 2$, $KK_1 = h_0 = 1$.
$PR \parallel AC \parallel KL$, $RF \parallel SO$, $RF = h_0$.
$\triangle RCF \sim \triangle SCO$, $\frac{CF}{CO} = \frac{RF}{SO} = \frac{h_0}{h} = \frac{1}{4}$, $CF = \frac{CO}{4}$.
Plane $PQR$ is para... | \frac{512}{15},\frac{2048}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,611 |
6. More and more countries are exploring space. The number of states that launch their satellites using their own launch vehicles is already 12. There are also countries that use the services of the main space powers to launch their satellites for economic purposes. Due to the increasing number of participants in space... | Solution. According to the condition, at the current moment, all satellites should be located on a sphere with radius $R+H$. Let $O$ be the center of the sphere, and its radius $R_{H}=R+H$. Denote the satellites by points $C_{i}, i=1, \ldots, n$. We need to determine the maximum value of $n$.
Let's find the distance $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,612 |
# Problem 1. Option I.
The decimal representation of a natural number $N$ contains 1580 digits. Among these digits are threes, fives, and sevens, and no other digits. It is known that the number of sevens is 20 less than the number of threes. Find the remainder when the number $N$ is divided by 3. | Solution. Let $x$ be the number of threes in the number $N$. The sum of the digits of the number $N$ is $S=3 x+7(x-20)+5(1580-(2 x-20))=7860$.
The remainder of $S$ divided by 3 is equal to the remainder of $N$ divided by 3 and is 0.
Answer: 0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,613 |
Task 1. II variant.
The decimal representation of a 2015-digit natural number $N$ contains the digits 5, 6, 7 and no other digits. Find the remainder of the division of the number $N$ by 9, given that the number of fives in the representation of the number is 15 more than the number of sevens. | Solution. Let the number $N$ contain $x$ sevens. Then the sum of the digits of the number $N$ is $S=7x+5(x+15)+6(2015-(2x+15))=12075$.
$N \equiv S \equiv 6(\bmod 9)$.
Answer: 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,614 |
Problem 2. Option I.
Solve the system of equations $\left\{\begin{array}{l}p^{2}+q^{2}+r^{2}=6, \\ p q-s^{2}-t^{2}=3 .\end{array}\right.$ | Solution. $\left\{\begin{array}{l}p^{2}+q^{2}+r^{2}=6, \\ 2 p q-2 s^{2}-2 t^{2}=6\end{array} \Rightarrow(p-q)^{2}+2 s^{2}+2 t^{2}+r^{2}=0\right.$.
From this $\left\{\begin{array}{l}p=q, \\ s=t=p=0 .\end{array}\right.$
From the equation $p^{2}+q^{2}=6$ we find $p=q= \pm \sqrt{3}$.
Answer: $(p ; q ; r ; s ; t) \in\{( ... | (\\sqrt{3};\\sqrt{3};0;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,615 |
Problem 2. II variant.
Solve the system of equations $\left\{\begin{array}{l}x+y+\sqrt{z}=4, \\ \sqrt{x} \sqrt{y}-\sqrt{w}=2 .\end{array}\right.$
Solution. $\left\{\begin{array}{l}(\sqrt{x})^{2}+(\sqrt{y})^{2}+\sqrt{z}=4, \\ 2 \sqrt{x} \sqrt{y}-2 \sqrt{w}=4\end{array} \Rightarrow(\sqrt{x}-\sqrt{y})^{2}+\sqrt{z}+2 \sq... | Answer: $(x ; y ; z ; w) \in\{(2 ; 2 ; 0 ; 0)\}$. | (2;2;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,616 |
Task 3. Option I.
For what values of the parameter $a$ does the function $y=\frac{5}{x^{2}-2 x+20}$ decrease on the interval $[2 a ; 2-a]?$ | Solution. The given function is decreasing on the interval $[1 ;+\infty)$. The function will be decreasing on the interval $[2 a ; 2-a]$ if the conditions of the system are satisfied
$\left\{\begin{array}{l}2 a \geq 1, \\ 2-a>2 a\end{array} \Leftrightarrow \frac{1}{2} \leq a<\frac{2}{3}\right.$.
Answer: $a \in\left[\... | [\frac{1}{2};\frac{2}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,617 |
Task 3. II variant.
For what values of the parameter $a$ does the function $y=\frac{8}{x^{2}+4 x+44}$ increase on the interval $[a-3 ; 3 a]?$ | Solution. The function is increasing on the interval $(-\infty ;-2]$. The function will be decreasing on the segment $[a-3 ; 3 a]$ under the conditions
$\left\{\begin{array}{l}3 a \leq-2, \\ a-3<3 a\end{array} \Leftrightarrow \quad-\frac{3}{2}<a \leq-\frac{2}{3}\right.$.
Answer: $a \in\left(-\frac{3}{2} ;-\frac{2}{3}... | \in(-\frac{3}{2};-\frac{2}{3}] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,618 |
# Problem 4. Option I.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}-a x+a+2=0$ is equal to -8.
# | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=a\left(a^{2}-3(a+2)\right)=a^{3}-3 a(a+2)$.
2) $a^{3}-3 a(a+2)=-8 \Leftrightarrow\left[\begin{array}{l}a=-2, \\ a=1, \\ a=4 .\end{array}\right.$
3) $D=a^{2}-4 a-8$.
For $a=-2 \quad D=4+8-8=4>0$, therefo... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,619 |
Problem 4. II variant.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}+a x+a+1=0$ is equal to 1. | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=-a\left(a^{2}-3(a+1)\right)=-a^{3}+3 a(a+1)$.
2) $-a^{3}+3 a(a+1)=1 \Leftrightarrow\left[\begin{array}{l}a=-1, \\ a=2 \pm \sqrt{3} \text {. }\end{array}\right.$
3) $D=a^{2}-4 a-4$.
For $a=-1 \quad D=1+4... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,620 |
Problem 5. Option I.
Two different natural numbers are written on the board, the larger of which is 2015. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board. | Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $2015-1024=991$.
Answer: 991. | 991 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,621 |
# Problem 5. II variant.
Two different natural numbers are written on the board, the larger of which is 1580. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board. | Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $1580-1024=556$.
Answer: 556. | 556 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,622 |
# Problem 6. Option I.
On the side $A B$ of an equilateral triangle $A B C$, isosceles triangles $A B D$ with angle $D$ equal to $90^{\circ}$ and $A B E$ with angle $E$ equal to $150^{\circ}$ are constructed such that points $D$ and $E$ lie inside triangle $A B C$. Prove that $C D=D E$.
# | # Solution.
Method 1. Draw $C H \perp A B, H \in A B$. Let $A H=a$. Then
$C H=a \operatorname{tg} 60^{\circ}=a \sqrt{3}$,
$D H=a \operatorname{tg} 45^{\circ}=a$,
$E H=a \operatorname{tg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 10,623 |
# Problem 7. Option 1.
Given a triangle with sides 6, 8, and 10. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas. | Solution. 1) Note that the given triangle is a right triangle: $6^{2}+8^{2}=10^{2}$.
2) Suppose first (and we will justify this later) that the ends of the desired segment $D E=t$ lie on the larger leg $A C=8$ and the hypotenuse $A B=10$ (see figure). Let $A D=x$, $A E=y, \angle B A C=\alpha$. Then
$S_{A D E}=\frac{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,624 |
Task 7. II variant.
Given a triangle with sides 5, 12, and 13. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas. | Solution. 1) By the converse of the Pythagorean theorem, the given triangle is a right triangle.
2) Consider triangle $ABC$, where $BC=5$, $AC=12$, $AB=13$ (see figure). Take points $D \in AC$, $E \in AB$. Let $AD=x$, $AE=y$, $DE=t$, $\angle BAC=\alpha$. Then
$S_{ADE}=\frac{1}{2} x y \sin \alpha=\frac{5}{26} x y$.
3)... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,625 |
Task 8. I variant.
To plot on the coordinate plane $O a b$ the set of points for which the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=a^{2} \\
|x|+|y|=|b|
\end{array}\right.
$$
has at least one solution | Solution. The graph of the first equation is a circle of radius $|a|$ centered at the origin; the graph of the second equation is a square with vertices on the coordinate axes, half the diagonal of which is $|b|$ (see figure). The system has a solution if the condition $|a| \leq|b| \leq \sqrt{2}|a|$ is satisfied.
![](... | ||\leq|b|\leq\sqrt{2}|| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,626 |
# Problem 8. II variant.
On the coordinate plane $O a b$, plot the set of points for which the system $\left\{\begin{array}{l}x^{2}+y^{2}=a^{2}, \\ x+|y|=b\end{array}\right.$
has at least one solution. | Solution. The graph of the first equation is a circle of radius $|a|$ centered at the origin; the graph of the second equation is a "corner" with its vertex at coordinates $(b; 0)$ (see figure). The system has a solution if the condition $-|a| \leq b \leq \sqrt{2}|a|$ is satisfied.
The set of points in the plane $O a ... | -||\leqb\leq\sqrt{2}|| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,627 |
# Problem 1.
A librarian at a physics and mathematics school noticed that if the number of geometry textbooks in the school library is increased by several (integer) times and the number of algebra textbooks is added to the resulting number, the total is 2015. If the number of algebra textbooks is increased by the sam... | Solution. Let the number of geometry textbooks be $x$, and the number of algebra textbooks be $y$. We can set up the system
$$
\left\{\begin{array}{l}
x n+y=2015 \\
y n+x=1580
\end{array}\right.
$$
We can write an equivalent system, where the equations represent the sum and difference of the equations in the original... | 287 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,628 |
# Task 2.
Find the set of values of the parameter $a$ for which the discriminant of the equation $a x^{2}+2 x+1=0$ is 9 times the square of the difference of its two distinct roots. | Solution. $D=4-4 a$.
$\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\left(\frac{2}{a}\right)^{2}-4 \cdot \frac{1}{a}=\frac{4-4 a}{a^{2}}=\frac{D}{a^{2}}$.
We obtain the equation: $\frac{D}{a^{2}} \cdot 9=D$. The condition $D>0$ is satisfied only by the root $a=-3$.
Answer: $a \in\{-3\}$. | \in{-3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,629 |
# Problem 3.
One side of the parallelogram is $\sqrt{3}$ times larger than the other side. One diagonal of the parallelogram is $\sqrt{7}$ times larger than the other diagonal. How many times larger is one angle of the parallelogram than the other angle? | Solution. Let $x$ be the smaller side, then $\sqrt{3} x$ is the larger side. Let $y$ be the smaller diagonal, then $\sqrt{7} y$ is the larger diagonal. We have:
$2 x^{2}+2(\sqrt{3} x)^{2}=y^{2}+(\sqrt{7} y)^{2}$,
from which $x=y$.
We get: the acute angle of the parallelogram is $30^{\circ}$, the obtuse angle is $150^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,630 |
# Problem 4.
In the Oxy plane, find the smallest and largest distances between two points $(x ; y)$, the coordinates of which are integers and satisfy the equation $y^{2}=4 x^{2}-15$.
| Solution. Rewrite the given equation as $(2 x-y)(2 x+y)=1 \cdot 3 \cdot 5$. We see that the integer points satisfying this equation and lying in the first quadrant are solutions to the following systems:
$$
\left\{\begin{array} { l }
{ 2 x - y = 1 , } \\
{ 2 x + y = 1 5 }
\end{array} \text { or } \left\{\begin{array}... | 2;2\sqrt{65} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,631 |
# Problem 5.
Find the smallest and largest value of the expression $|x+2|+|y+3|$ under the condition $(|x|-3)^{2}+(|y|-2)^{2}=1$. | Solution. The graph of the equation $(|x|-3)^{2}+(|y|-2)^{2}=1$ is the union of four unit-radius circles centered at points $(\pm 3, 2)$, $(\pm 3, -2)$. The graph of the equation $|x+2|+|y+3|=a$ (for $a>0$) is a family of squares centered at the point $(-2, -3)$, with diagonals parallel to the coordinate axes and equal... | 2-\sqrt{2};10+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,632 |
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