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# 5. Solve the inequality
$$
\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0
$$ | # Solution:
$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0$. odZ: $2 x^{2}+7 x+3 \geq 0, \Rightarrow$
$x \in(-\infty ;-3] \cup[-0.5 ;+\infty)$. The original inequality is equivalent to the following
$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left(x^{2}-4 x+2\right)... | x\in(-\infty;-3]\cup[0;1]\cup{2}\cup[3;4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,403 |
6. Find the set of values of the function
$$
f(x)=\sqrt{g^{2}(x)-245}, \text { where } g(x)=15-2 \cos 2 x-4 \sin x
$$ | # Solution:
$$
f(x)=\sqrt{g^{2}(x)-245}, \text { where } g(x)=15-2 \cos 2 x-4 \sin x
$$
Let's find the range of the function $z=g(x)=15-2 \cos 2 x-4 \sin x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then $z=13+4 t^{2}-4 t=12+(2 t-1)^{2}$ for $t \in[-1 ; 1]$, an... | E_{f}=[0;14] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,404 |
7. The inscribed circle of a triangle divides one of its sides into segments equal to 3 and 4. Find the area of the triangle if the radius of the circumscribed circle around it is $7 / \sqrt{3}$. | # Solution:
Let $D$ be the point of tangency of the circle with side $B C$.
By the condition, $B D=a=3, D C=b=4$. Therefore, $B C=a+b=7$. Since $2 R_{\text{on}}=B C / \sin \angle A$, we have $\sin \angle A=B C / 2 R_{\text{on}}=\sqrt{3} / 2$. Thus, $\angle A=60^{\circ}$ or $\angle A=120^{\circ}$. If $F$ and $E$ are t... | 12\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,405 |
8. On the line $3 x+4 y=2$ find the point through which two perpendicular to each other tangents to the graph of the function $y=x^{2}$ pass, and write the equations of these tangents.
# | # Solution:
$y=x^{2}, 3 x+4 y=2, M\left(x_{0} ; y_{0}\right)$. The equation of the tangent line: $y=y_{0}+k\left(x-x_{0}\right)$.
The equation $x^{2}=y_{0}+k\left(x-x_{0}\right)$, or $x^{2}-k x+k x_{0}-y_{0}=0$, has a unique solution if $D=k^{2}-4 k x_{0}+4 y_{0}=0$. The two values of $k$ found from this equation mus... | M(1;-1/4),equationsofthetangents:-1/4+(2\\sqrt{5})(x-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,406 |
9. Find all values of the parameter $a$ for which the system of equations $\left\{\begin{array}{c}4 \sqrt{y}=x-a, \\ y^{2}-x^{2}+2 y-4 x-3=0 .\end{array}\right.$ has a unique solution. Specify this solution for each $a$. | # Solution:
$\left\{\begin{array}{c}4 \sqrt{y}=x-a, \\ y^{2}-x^{2}+2 y-4 x-3=0 .\end{array}\right.$ Transform the second equation of the system.
$\left(y^{2}+2 y+1\right)-\left(x^{2}+4 x+4\right)=0,(y+1)^{2}-(x+2)^{2}=0,(y+x+3)(y-x-1)=0$.
1) $y+4 \sqrt{y}+a+3=0, \sqrt{y}=-2 \mp \sqrt{4-a-3}=-2 \mp \sqrt{1-a}$. One r... | \in(-\infty;-5),4(-2+\sqrt{1-})+,(-2+\sqrt{1-})^{2};\in(-1;+\infty),4(2+\sqrt{+5})+,(2+\sqrt{+5})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,407 |
10. Rectangle $ABCD$ with sides $AB=1$ and $AD=10$ serves as the base of pyramid $SABCD$, and the edge $SA=4$ is perpendicular to the base. Find a point $M$ on edge $AD$ such that triangle $SMC$ has the smallest perimeter. Find the area of this triangle.
# | # Solution:
Let's lay off $A T=A S$ on the extension of edge $A B$. For any position of point $M$ on side $A D$, $T M=S M$, so the minimum value of the sum $S M+M C$ will be when $M=T C \cap A D$.
Let's introduce the notation
$A B=a, A D=b, A S=c, A M=x$. From
$\triangle T A M \sim \triangle C D M$ it follows that
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,408 |
1. A pedestrian left point $A$ for point $B$. When he had walked 8 km, a second pedestrian set off from point $A$ after him. When the second pedestrian had walked 15 km, the first was halfway through his journey, and both pedestrians arrived at point $B$ at the same time. What is 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 pedestrians. Then,
$\frac{s-8}{v_{1}}=\frac{s}{v_{2}}$ and $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$. From this, $\frac{s}{2(s-8)}=\frac{s-15}{s} ; s^{2}=2 s^{2}-46 s+240 ; s^{2}-46 s+240=0$;
$s_{1,2}=23 \pm 17 \cdot... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,409 |
2. Solve the inequality $\sqrt{\frac{x-4}{x+3}}-\sqrt{\frac{x+3}{x-4}}<\frac{7}{12}$. | # Solution:
$$
\begin{aligned}
& \sqrt{\frac{x-4}{x+3}}-\sqrt{\frac{x+3}{x-4}}=0 ; \frac{1}{y}-y=0 ; \\
& y_{1,2}=\frac{-7 \pm \sqrt{49+576}}{24}=\frac{-7 \pm 25}{24}, y_{1}=\frac{3}{4}, y_{2}=-\frac{4}{3} . \text { Therefore, } \\
& y>\frac{3}{4}, \frac{x+3}{x-4}>\frac{9}{16} \Leftrightarrow \frac{x+12}{x-4}>0 \Leftr... | x\in(-\infty;-12)\cup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,410 |
3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array}... | -442 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,411 |
4. Solve the equation $\sqrt{2+\cos 2 x-\sqrt{3} \tan x}=\sin x-\sqrt{3} \cos x$. Find all its roots that satisfy the condition $|x-3|<1$.
# | # Solution:
Both sides of the equation $\sqrt{2+\cos 2 x-\sqrt{3} \tan x}=\sin x-\sqrt{3} \cos x$ can be squared under the condition $\sin x-\sqrt{3} \cos x \geq 0$. Since $\sin x-\sqrt{3} \cos x=2 \sin (x-\pi / 3)$, the condition can be written as $\pi / 3+2 \pi n \leq x \leq 4 \pi / 3+2 \pi n, n \in Z$. Under the fo... | \frac{3\pi}{4},\pi,\frac{5\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,412 |
5. Solve the inequality $\frac{|x-2|-\left|x^{2}-4 x+2\right|}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq 0$. | # Solution:
$\frac{|x-2|-\left|x^{2}-4 x+2\right|}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq 0 . \quad$ Domain: $2 x^{2}+7 x+3 \geq 0, \Rightarrow x \in(-\infty ;-3] \cup[-0.5 ;+\infty)$.
The original inequality is equivalent to the following $\frac{(x-2)^{2}-\left(x^{2}-4 x+2\right)^{2}}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq ... | x\in[-0.5;0]\cup[1;2)\cup(2;3]\cup[4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,413 |
6. Find the set of values of the function
$$
f(x)=\sqrt{13-g^{2}(x)}, \text { where } g(x)=(13 / 4)-\cos ^{2} x+\sin x \text {. }
$$ | # Solution:
$f(x)=\sqrt{13-g^{2}(x)}$, where $g(x)=13 / 4-\cos ^{2} x+\sin x$.
Let's find the range of the function $z=g(x)=13 / 4-\cos ^{2} x+\sin x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then $z=9 / 4+t^{2}+t=2+(t+1 / 2)^{2}$ for $t \in[-1 ; 1]$, and $E_{... | E_{f}=[0;3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,414 |
7. The circle inscribed in triangle $A B C$ touches side $B C$ at point $D$, and $B D=5$. Find the radius of this circle and the area of triangle $A B C$, if angle $A$ is $60^{\circ}$ and the radius of the circumscribed circle around triangle $A B C$ is $7 / \sqrt{3}$. | # Solution:
Since $2 R_{\text {on }}=B C / \sin \angle A$, then
$B C=2 R_{\text {on }} \sin \angle A=7$. Then $B D=a=5$, $D C=b=2$. If $F$ and $E$ are the points of tangency of the circle with sides $B A$ and $A C$ respectively, then $B F=a=5, E C=b=2$, and $F A=A E=x$. By the cosine rule we have:
$B C^{2}=A B^{2}+A... | S_{ABC}=10\sqrt{3},r_{}=\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,415 |
8. On the Oy axis, find the point through which two perpendicular to each other tangents to the graph of the function $y=1-2 x-x^{2}$ pass, and write the equations of these tangents.
# | # Solution:
$y=1-2 x-x^{2}, M\left(0 ; y_{0}\right)$. Equation of the tangent: $y=y_{0}+k x$.
The equation $1-2 x-x^{2}=y_{0}+k x$, or $x^{2}+(k+2) x+y_{0}-1=0$, has a unique solution if $D=k^{2}+4 k+8-4 y_{0}=0$. The two values of $k$ found from this equation must satisfy the condition of perpendicularity $k_{1} \cd... | M(0;9/4),\,9/4+(-2\\sqrt{5})x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,416 |
9. Find all values of the parameter $a$ for which the system of equations $x=4 \sqrt{y}+a, y^{2}-x^{2}+3 y-5 x-4=0$ has a unique solution. Specify this solution for each $a$.
# | # Solution:
$\left\{\begin{array}{c}x=4 \sqrt{y}+a, \\ y^{2}-x^{2}+3 y-5 x-4=0 .\end{array}\right.$
Transform the second equation of the system.
$$
\left(y^{2}+3 y+\frac{9}{4}\right)-\left(x^{2}+5 x+\frac{25}{4}\right)=0,\left(y+\frac{3}{2}\right)^{2}-\left(x+\frac{5}{2}\right)^{2}=0,(y+x+4)(y-x-1)=0
$$
1) $y+4 \sq... | \in(-\infty;-5),4(-2+\sqrt{-})+,(-2+\sqrt{-})^{2};\in(-1;+\infty),4(2+\sqrt{+5})+,(2+\sqrt{+5})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,417 |
10. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ with edges $A B=4, A D=2$, $A A_{1}=3 \sqrt{2}$, a plane is drawn through the diagonal $B D_{1}$, intersecting the edge $A A_{1}$ such that the section of the parallelepiped by this plane has the smallest perimeter. Find the area of this section. | # Solution:
Extend the edge $AD$ and mark $AL = AB$; similarly, extend the edge $CD$ and mark $CK = BC$. Clearly, $B \in KL$. For any position of point $M$ on the edge $AA_1$, $BM = LM$, so... | 8\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,418 |
Problem 1. (Option 1)
Calculate $x^{3}+3 x$, where $x=\sqrt[3]{2+\sqrt{5}}-\frac{1}{\sqrt[3]{2+\sqrt{5}}}$. | Solution: Let $\sqrt[3]{2+\sqrt{5}}=a$, then $x=a-\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}+3 x=\left(a-\frac{1}{a}\right)^{3}+3\left(a-\frac{1}{a}\right)=a^{3}-\frac{1}{a^{3}}=2+\sqrt{5}-\frac{1}{2+\sqrt{5}}=\frac{(2+\sqrt{5})^{2}-1}{2+\sqrt{5}}= \\
& =\frac{8+4 \sqrt{5}}{2+\sqrt{5}}=4
\end{aligned}
$$
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,419 |
Problem 1. (Option 2)
Calculate $x^{3}-3 x$, where $x=\sqrt[3]{7+4 \sqrt{3}}+\frac{1}{\sqrt[3]{7+4 \sqrt{3}}}$. | Solution: Let $\sqrt[3]{7+4 \sqrt{3}}=a$, then $x=a+\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}-3 x=\left(a+\frac{1}{a}\right)^{3}-3\left(a+\frac{1}{a}\right)=a^{3}+\frac{1}{a^{3}}=7+4 \sqrt{3}+\frac{1}{7+4 \sqrt{3}}=\frac{(7+4 \sqrt{3})^{2}+1}{7+4 \sqrt{3}}= \\
& =\frac{98+56 \sqrt{3}}{7+4 \sqrt{3}}=14
\end{aligned}
$$
... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,420 |
Task 2. (Option 1)
At the bus depot, there are 5 red, 6 blue, and 5 yellow buses. A convoy of 7 buses is randomly formed. What is the probability that the first bus in the convoy is red, and among the rest there are no red buses, but exactly 4 blue ones? | Solution: There are a total of $5+6+5=16$ buses. It is necessary for the first bus to be red - the probability is $\frac{5}{16}$. From the remaining 15, 4 blue, 2 yellow, and 0 red buses are chosen, so the probability is $\frac{C_{6}^{4} \cdot C_{5}^{2} \cdot C_{4}^{0}}{C_{15}^{6}}$. The desired probability is obtained... | \frac{75}{808} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,421 |
Problem 2. (Option 2)
In a box, there are 3 red, 4 gold, and 5 silver stars. Randomly, one star is taken from the box and hung on the Christmas tree. What is the probability that a red star will end up on the top of the tree, there will be no more red stars on the tree, and there will be exactly 3 gold stars, if a tot... | Solution: There are a total of $3+4+5=12$ toys. It is necessary for a red toy to be on top - the probability is $\frac{3}{12}$. From the remaining 11, 3 golden, 2 silver, and 0 red toys are chosen, so the probability is $\frac{C_{4}^{3} \cdot C_{5}^{2} \cdot C_{2}^{0}}{C_{11}^{5}}$.
The desired probability is obtained... | \frac{5}{231} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,422 |
Problem 3. (Option 1)
Find the sum of the squares of the roots of the equation $\left(x^{2}+4 x\right)^{2}-2016\left(x^{2}+4 x\right)+2017=0$. | Solution: Let's make the substitution: $x^{2}+4 x+4=t$, then $x^{2}+4 x=t-4$ and the equation will take the form:
$(t-4)^{2}-2016(t-4)+2017=0$
$t^{2}-2024 t+10097=0$
The discriminant of the equation is greater than zero, therefore, the equation has two roots. By Vieta's theorem: $t_{1}+t_{2}=2024, t_{1} \cdot t_{2}=... | 4064 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,423 |
Problem 3. (Option 2)
Find the sum of the squares of the roots of the equation $\left(x^{2}+6 x\right)^{2}-1580\left(x^{2}+6 x\right)+1581=0$. | Solution: Let's make the substitution: $x^{2}+6 x+9=t$, then $x^{2}+6 x=t-9$ and the equation will take the form:
$$
\begin{aligned}
& (t-9)^{2}-1580(t-9)+1581=0 \\
& t^{2}-1598 t+15882=0
\end{aligned}
$$
The discriminant of the equation is greater than zero, so the equation has two roots. By Vieta's theorem, $t_{1}+... | 3232 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,424 |
# Problem 4. (Variant 1)
On a circle, 5 points $A, B, C, D$ and $E$ are located at equal intervals. Given two vectors $\overrightarrow{D A}=\vec{a}, \overrightarrow{D B}=\vec{b}$. Express the vector $\overrightarrow{E C}$ in terms of $\vec{a}$ and $\vec{b}$. | # Solution:
Connect B and E.
Consider the closed broken line BKDMEFB.
By Menelaus' Theorem: $\frac{B K}{K D} \cdot \frac{D M}{M F} \cdot \frac{F E}{E B}=1$.
Let $K D=x, K M=y$,
then fro... | \overrightarrow{EC}=\frac{1+\sqrt{5}}{2}\vec{b}-\frac{1+\sqrt{5}}{2}\vec{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,425 |
Problem 4. (Option 2)
On a circle at equal intervals, there are 5 points $A, B, C, D$ and $E$. Given two vectors $\overrightarrow{D A}=\vec{a}, \overrightarrow{D B}=\vec{b}$. Express the vector $\overrightarrow{A C}$ in terms of $\vec{a}$ and $\vec{b}$. | Solution.
The line $\mathrm{BE}$ intersects $\mathrm{AC}$ and $\mathrm{AD}$ at points F and $\mathrm{M}$, respectively. Consider the closed broken line BKDMAFB.
By Menelaus' Theorem: $\frac{B K}{K D} \cdot \frac{F M}{B F} \cdot \frac{A D}{M A}=1$.
Let $K B=x, F M=y$,
then from the condition of symmetry of points
A... | \overrightarrow{AC}=\vec{b}-\frac{(1+\sqrt{5})}{2}\vec{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,426 |
Problem 5. (Option 1). Find all values of the parameters $a, b$, and $c$, for which the set of real roots of the equation $x^{5}+4 x^{4}+a x^{2}=b x+4 c$ consists exactly of the two numbers 2 and -2. | Solution: Since the numbers 2 and -2 are roots of the equation, then
$\left\{\begin{array}{c}32+64+4 a=2 b+4 c \\ -32+64+4 a=-2 b+4 c\end{array} \Leftrightarrow\left\{\begin{array}{c}a=c-16 \\ b=16\end{array}\right.\right.$, substitute into the original equation:
$x^{5}+4 x^{4}+(c-16) x^{2}-16 x-4 c=0$
$x\left(x^{4}... | =-48,b=16,=-32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,427 |
Problem 5. (Option 2). Find all values of the parameters $a, b$, and $c$, for which the set of real roots of the equation $x^{5}+4 x^{4}+a x=b x^{2}+4 c$ consists exactly of the two numbers 2 and -2. | Solution: Since the numbers 2 and -2 are roots of the equation, then
\[
\begin{aligned}
& \left\{\begin{array} { c }
{ 3 \cdot 2 + 6 \cdot 4 + 2a = 4b + 4c } \\
{ -3 \cdot 2 + 6 \cdot 4 - 2a = 4b + 4c }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
b=16-c \\
a=-16
\end{array},\right.\right. \text { substitute i... | =-16,b=48,=-32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,428 |
Problem 6. (Option 1).
In the right triangle $ABC: \angle ACB=90^{\circ}, AC=6, BC=4$. On the line $BC$, a point $D (CD>BD)$ is marked such that $\angle ADC=45^{\circ}$. On the line $AD$, a point $E$ is marked such that the perimeter of triangle $CBE$ is the smallest possible. Then, on the line $DC$, a point $F$ is ma... | # Solution:
$C D=A C=6 \Rightarrow B D=B_{1} D=6-4=2$. Since triangles $\mathrm{ACE}$ and $\mathrm{B}_{1} \mathrm{ED}$
are similar ( $\angle A E C=\angle B_{1} E D$ - vertical angles and $\angle E A C=\angle B_{1} D E=45^{\circ}$ ), then
$E C: B_{1} E=A C: D B_{1}=A E: E D=6: 2=3: 1$
Let $E H \perp D C \Rightarrow$... | 3.6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,429 |
Problem 6. (Variant 2).
$C A K D$ is a square with a side length of 6. On side $C D$, a point $B(B D=2)$ is chosen, and on line $A D$, a point $E$ is chosen such that the perimeter of triangle $B E C$ is the smallest possible. Then, on line $D C$, a point $F$ is marked such that the perimeter of triangle $F E A$ is th... | # Solution:
$C D=A C=6, \quad B D=B_{1} D=2$. Since triangles $A C E$ and $\mathrm{B}_{1} \mathrm{ED}$ are similar ( $\angle A E C=\angle B_{1} E D$ - vertical, $\angle E A C=\angle B_{1} D E=45^{0}$ ),
then $E C: B_{1} E=A C: D B_{1}=A E: E D=6: 2=3: 1$.
Let $E H \perp D C \Rightarrow$ right triangles ACD and EHD a... | 0.3\sqrt{34} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,430 |
1. Seeing a fox several meters away, the dog chased after it along a straight dirt road. The dog's jump is $23 \%$ longer than the fox's jump. There is a time interval during which both the fox and the dog make a whole number of jumps. Each time it turns out that the dog manages to make $t \%$ fewer jumps than the fox,... | Solution: Let $\mathrm{x}$ be the length of the fox's jump, and $y$ be the number of jumps it makes in some unit of time. Then $xy$ is the distance the fox covers in this time. The distance covered by the dog in the same time is $1.23 x\left(1-\frac{t}{100}\right) y$. The fox will escape from the dog if $1.23 x\left(1-... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,431 |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=150, d=0.5$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,433 |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
## Find $\angle C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $\mathrm{B}$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and is therefore equidistant from its sides, we get that $A_{... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,434 |
5. Buyers pay pearl divers with gold and corn. After consulting, the buyers decided to reduce the price "in gold" by 5% on the first island. Then the price "in corn" will drop by $7 \%$, as the prices of gold and corn are linked in the market.
For the pearl divers of the second island, due to the arrival of a cold cur... | Solution. Let x be the original price of the pearl; for the first island: $0.95 x$ in "gold"; $0.93 x$ - in "corn". Their ratio $\frac{95}{93}$ is constant in the market. For the second island: 0.99 x in "gold", $y$ - in "corn"; $\quad \frac{0.99 x}{y}=\frac{95}{93} ; y=\frac{0.99 x \cdot 93}{95}$. The pearl became che... | 3.08 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,435 |
6. What is the minimum value that the function $F(x ; y)=x^{2}+8 y+y^{2}+14 x-6$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.
# | # Solution.
$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=$ M, then $(x+7)^{2}+(y+4)^{2}=(M+71)$ - this is a circle with center $(-7 ;-4)$ and radius $(M+71)^{0.5}$.
Since the center of the second circle lies outside the first, ... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,436 |
7. What is the minimum number of cells that need to be painted in a square with a side of 35 cells (a $35 \times 35$ square, with a total of 1225 cells), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
# | # Solution.
You should color every 3rd cell diagonally (see the figure). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be colored. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be colored. $\left[\frac{35^{2}}{3}\right]=408$.
. Find the minimum possible length of the text.
# | # Solution:
Let the length of the text be L. Let a character appear in the text $x$ times. The problem can be reformulated as: find the smallest natural number $\mathrm{L}$, for which there exists a natural number $x$ such that $\frac{10.5}{100}19$ when $x \geq 3$.
Answer: 19.
## Solution variant № 2 | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,439 |
1. Two sisters were picking currants: the elder one in a four-liter bucket, and the younger one in a bucket with a capacity of 3.5 liters. The elder sister always worked faster than the younger one. When the elder sister had collected three quarters of her bucket, and the younger one had collected more than half of her... | Solution: $0.75 \cdot 4=3$ liters of berries were collected by the older sister by the time they exchanged buckets. Let x liters be the amount collected by the younger sister by this time. Since the work rate of the sisters is the same before and after exchanging buckets, we can set up the equation: $\frac{3}{x}=\frac{... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,440 |
3. In the arithmetic progression $\left(a_{n}\right)$, $a_{1}=1$, $d=3$.
Calculate $A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}$.
In the answer, write the smallest integer greater than $A$. | Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,442 |
4. In rectangle $A B C D$, $A B: A D=1: 2$. Point $M$ is the midpoint of $A B$, and point $K$ lies on $A D$ and divides it in the ratio $3: 1$ starting from point $A$. Find the sum of $\angle C A D$ and $\angle A K M$.
# | # Solution:
Complete the rectangle $A B C D$ to a square $A E F D$ with side $A D$.
$$
\begin{aligned}
& \text { Let } L \in E F, E L: L F=1: 3, \\
& \triangle M E L=\triangle A K M \Rightarrow \angle E M L=\angle A K M \\
& N \in F D, F N=N C, M R\|A D \Rightarrow M N\| A C \Rightarrow \\
& \Rightarrow \angle N M R=... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,443 |
5. There are two lead-tin alloys. In the first alloy, the mass of lead is to the mass of tin as $1: 2$; in the second - as $2: 3$. How many grams of the first alloy are needed to obtain 22 g of a new alloy with the mass ratio of lead to tin being 4:7? | Solution. Let the first alloy contain x g of lead and 2x g of tin. In the second alloy, there are 2y g of lead and 3y g of tin. Then $k \cdot 3x + n \cdot 5y = 22 ; \frac{kx + n \cdot 2y}{k \cdot 2x + n \cdot 3y} = \frac{4}{7} ; \quad$ we need to find $k \cdot 3x$ and $5ny$. Let $ny = b ; kx = a \cdot \frac{a + 2b}{2a ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,444 |
6. For all non-negative values of the real variable $x$, the function $f(x)$ satisfies the condition $f(x+1)+1=f(x)+\frac{43}{(x+1)(x+2)}$. Calculate $\frac{101}{f(2020)}$, if $\quad f(0)=2020$. | # Solution.
Notice that
$f(x+2020)-f(x)=(f(x+2020)-f(x+2019))+(f(x+2019)-f(x+2018))+\cdots+(f(x+$ 1) $-f(x))=\frac{43}{(x+2020)(x+2021)}-1+\frac{43}{(x+2019)(x+2020)}-1+\cdots+\frac{43}{(x+1)(x+2)}-1$. Thus, $f(2020)-f(0)=43\left(\frac{1}{2020}-\frac{1}{2021}+\cdots+1-\frac{1}{2}\right)-2020=43\left(1-\frac{1}{2021}\... | 2.35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,445 |
7. What is the minimum number of cells that need to be painted in a square with a side of 35 cells (a total of $35 \times 35$ cells, which is 1225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess? | # Solution.
Cells should be shaded in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be shaded. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table (larger than $4 \times 4$) such that the... | 612 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,446 |
9. A table consisting of 2019 rows and 2019 columns is filled with natural numbers from 1 to 2019 such that each row contains all numbers from 1 to 2019. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the filling of the table is symmetric with respect to... | # Solution:
We will show that all numbers from 1 to 2019 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 2019\}$ is not on the diagonal. Then, due to the symmetry of the table, the number $a$ appears an even number of times. On the other hand, since the number $a$ appears once in each row, the to... | 2039190 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,448 |
1. Prove that the polynomial $P(t)=t^{3}-2 t^{2}-10 t-3$ has three distinct real roots. Find the polynomial $R(t)$ of the third degree with roots $u=x^{2} y^{2} z, v=x^{2} z^{2} y, w=y^{2} z^{2} x$, where $x, y, z$ are the distinct roots of the polynomial $P(t)$. (12 points) | Solution. The polynomial has 3 different real roots, since $P(-100)0, P(0)0$. By Vieta's theorem, $x+y+z=2, xy+xz+yz=-10, xyz=3$.
$$
\begin{gathered}
u+v+w=xyz(xy+xz+yz)=3 \cdot(-10)=-30 \\
uv+uw+vw=x^3 y^3 z^3(x+y+z)=3^3 \cdot 2=54 \\
uvw=x^5 y^5 z^5=3^5=243
\end{gathered}
$$
Answer: $R(t)=t^{3}+30 t^{2}+54 t-243$. | R()=^{3}+30^{2}+54-243 | Algebra | proof | Yes | Yes | olympiads | false | 10,449 |
2. There is one power outlet connected to the network, two extension cords with three outlets each, and one table lamp included. Nosy Nick randomly plugged all three plugs into 3 out of 7 outlets. What is the probability that the lamp will light up? (16 points) | Solution. The number of equally probable ways to plug in $A_{7}^{3}=7 \cdot 6 \cdot 5=210$. The lamp can be powered through 0, 1, or 2 extension cords.
0) The lamp is plugged into the socket, the other 2 plugs are plugged in randomly. The number of such possibilities $A_{6}^{2}=30$.
1) The lamp is plugged into one of ... | \frac{13}{35} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,450 |
3. The bisectors of the external angles of triangle $ABC$, $AA_{1}$, $BB_{1}$, and $CC_{1}$, intersect the extensions of the opposite sides of the triangle at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Find the angle $A_{1} C_{1} B_{1}$ and the length of the segment $A_{1} B_{1}$, if $AC=5$, $BC=2$, and the an... | Solution.
$b=AC=5, a=BC=2, c=AB$,
$$
\gamma=\angle ACB=\arccos \frac{13}{20}
$$
1) We will prove that $\frac{BA_{1}}{CA_{1}}=\frac{c}{b}$. For the areas of triangles $AA_{1}B$ and $AA_{1}C$, we have the following relationships:
$$
\begin{gathered}
\frac{S_{AA_{1}B}}{S_{AA_{1}C}}=\frac{BA_{1}}{CA_{1}}, \\
S_{AA_{1}B... | 180,\sqrt{190} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,451 |
4. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}y=|a-3| x+1|+x+3|+3|x+1|, \\ 2^{2-y} \log _{\sqrt{3}}\left((x+|a+2 x|)^{2}-6(x+1+|a+2 x|)+16\right)+2^{x+|a+2 x|} \log _{1 / 3}\left(y^{2}+1\right)=0, \text { has a unique } \\ x+|a+2 x| \leq 3,\end{array}\right.$
solution $(x ; y)$, ... | # Solution:
Transform the second equation of the system as follows:
$$
\begin{aligned}
& 2^{3-y} \log _{3}\left((x+|a+2 x|)^{2}-6(x+|a+2 x|)+9+1\right)-2^{x+|a+2 x|} \log _{3}\left(y^{2}+1\right)=0 \\
& 2^{3-y} \log _{3}\left((x+|a+2 x|-3)^{2}+1\right)=2^{x+|a+2 x|} \log _{3}\left(y^{2}+1\right) \\
& 2^{3-x-|a+2 x|} ... | =-2,-1,0;=-1,-1,1;=1,-1,3;=3,-2,4;=4,-2,5;=6,-3,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,452 |
5. Find the area of the cross-section of a regular hexagonal pyramid $S A B C D E F$ by a plane passing through the vertex $F$ 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 $S A F$, if the side of the base of the pyramid is $4 \sqrt{7}$,... | Solution. We will construct the section of the pyramid. In the plane $SCD$, through point $S$, draw a line $SQ$ parallel to $CM$, where $Q$ lies on line $CD$, and $CM$ is the midline of triangle $SQD$, $QC=a$, where $a$ is the side of the base of the pyramid.
The plane $SNQ$ is parallel to the plane of the section. Th... | \frac{202\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,453 |
6. The game of hockey is popular all over the world. Much in the game depends on the goalkeeper. To hone the skills of goalkeepers and ensure a training process that does not depend on other players, a puck launcher was created. The machine can be set to eject pucks at a specified frequency, speed, and angle. The goal ... | Solution. Introduce a coordinate system with the center at point O. The x-axis is directed towards the goal line.
Express time from the first equation of the system and substitute into the second:
$y(x)=\frac{V_{0}(x-d)}{V_{0} \cos \alpha} \sin \alpha-\frac{g\left(\frac{x-d}{V_{0} \cos \alpha}\right)^{2}}{2}=(x-d) \op... | \frac{5+2\sqrt{121.2}}{121} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,454 |
2. It is known that the graphs of the functions $f(x)=x^{p}$ and $g(x)=\ln x$ touch (have a common point where the tangents to both graphs coincide). Find the constant $p$ and the point of tangency. (16 points) | Solution. Since the graphs have a common tangent, the following relationships hold for the abscissa of the point of tangency:
$$
\left\{\begin{array} { c }
{ f ( x ) = g ( x ) } \\
{ f ^ { \prime } ( x ) = g ^ { \prime } ( x ) }
\end{array} \Rightarrow \left\{\begin{array} { c }
{ x ^ { p } = \operatorname { l n } x... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,455 |
3. The bisectors $A A_{1}$ and $B B_{1}$ of triangle $A B C$ intersect at point $O$, and $A O: O A_{1} = 2: 1$. The bisector of the external angle at vertex $C$ of triangle $A B C$ intersects line $A B$ at point $C_{1}$. Find the angle $B_{1} A_{1} C_{1}$ and the length of segment $A_{1} C_{1}$, if $A B=2, A C=4$. (16 ... | # Solution.
$$
\begin{gathered}
a=BC, b=AC=4, c=AB=2 \\
AO: OA_{1}=2: 1
\end{gathered}
$$
1) By the properties of angle bisectors, we have
$$
\begin{gathered}
AB: BA_{1}=AO: OA_{1}=2: 1, BA_{1}=\frac{c}{2} \\
AC: CA_{1}=AO: OA_{1}=2: 1, CA_{1}=\frac{b}{2} \\
BC=\frac{b+c}{2}=3
\end{gathered}
$$
2) We will prove tha... | 180,\sqrt{34} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,456 |
4. Find all values of the parameter $a$ for which the inequality $\log _{\sqrt{a / 10}}\left(\frac{a+8 x-x^{2}}{20}\right) \geq 2$
has at least one solution $x \geq 2$, and every solution of the inequality is also a solution
of the equation $|a+2 x-16|+|a-2 x+9|=|2 a-7|$.
(16 points) | # Solution:
$\log _{\sqrt{a / 10}}\left(\frac{a+8 x-x^{2}}{20}\right) \geq 2 \Leftrightarrow$
$\left\{\begin{array}{c}a+8 x-x^{2}>0, \\ a>0, \\ a \neq 10, \\ (\sqrt{a / 10}-1)\left(\frac{a+8 x-x^{2}}{20}-\frac{a}{10}\right) \geq 0\end{array} \Leftrightarrow\right.$
$\left\{\begin{array}{cl}a>x^{2}-8 x, & \\ a>0, & \... | \in[9;10)\cup[12;16] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,457 |
5. Find the area of the cross-section of a regular hexagonal pyramid $S A B C D E F$ 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$, if the side of the base of the pyramid is 2, and the di... | Solution. We will 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 pyrami... | \frac{34\sqrt{3}}{35} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,458 |
6. The game of hockey is popular all over the world. Much in the game depends on the goalkeeper. To hone the skills of goalkeepers and ensure a training process that does not depend on other players, a puck launcher was created. The machine can be set to eject pucks at a specified frequency, speed, and angle.
Let the ... | Solution. Express time from the first equation of the system and substitute into the second
$y(x)=\frac{V_{0} x}{V_{0} \cos \alpha} \sin \alpha-\frac{g\left(\frac{x}{V_{0} \cos \alpha}\right)^{2}}{2}=x \operatorname{tg} \alpha-\frac{g}{2}\left(\frac{x}{V_{0} \cos \alpha}\right)^{2} \Rightarrow$ $y(x)=x \operatorname{t... | 120\leqV_{0}^{2}\leq10+10\cdot\sqrt{145} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,459 |
1. We will call a decimal number interesting if it is divisible by the number 1111 and all its digits are distinct. How many interesting numbers exist?
(15 points) | # Problem 1.
Justified and correct solution - 20 points
One of the proof points is insufficiently justified - 15 points
Proved that the sum $\overline{\mathrm{a}_{9} \ldots \mathrm{a}_{5}}+\overline{\mathrm{a}_{4} \ldots \mathrm{a}_{0}}=99999$ - 10 points
Proved that the number is divisible by 99999 - 5 points
# | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,460 |
3. The numbers from 1 to 2015 are written on a board. Two players take turns erasing one number at a time. The game ends when two numbers remain on the board. If their sum is divisible by 3, the player who makes the first move wins. If not, the other player wins. Who will win with the correct play? Justify your answer.... | # Task 3.
Correct answer provided with justification - 15 points
Correct answer provided - 10 points
# | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,462 |
6. Among 11 visually identical coins, 10 are genuine, weighing 20 grams each, and one is counterfeit, weighing 21 grams. There are balance scales that balance if the load on the right pan is exactly twice as heavy as the load on the left (if the load on the right pan is less than twice the load on the left, the left pa... | # Task 6.
Correct solution is indicated - 15 points
Indicated how to find the fake coin from five in two weighings -10 points | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,465 |
6. A symphony orchestra has hired three musicians: Boris, Semyon, and Vasily, who can play the violin, flute, viola, clarinet, oboe, and trumpet. It is known that:
1) Semyon is the tallest;
2) the violinist is shorter than the flutist;
3) the violinist, flutist, and Boris love borscht;
4) when there is a quarrel betwee... | # Solutions for the correspondence round, 8th grade. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,466 |
1. Calculate: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}$. | Solution: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}=$
$$
\begin{aligned}
& =\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\ldots+\left(2015^{2}-2016^{2}\right)+2017^{2}= \\
& =(1-2)(1+2)+(3-4)(3+4)+\ldots+(2015-2016)(2015+2016)+2017^{2}= \\
& =-(1+2)-(3+4)-\ldots-(2015+2016)+2017^{2}= \\
& =-(1+2+3+4+\ldots+2015+20... | 2035153 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,467 |
2. Solve the equation: $4(x+5)(x+6)(x+10)(x+12)-3 x^{2}=0$ | Solution: $4(x+5)(x+6)(x+10)(x+12)-3 x^{2}=0$
$4\left(x^{2}+17 x+60\right)\left(x^{2}+16 x+60\right)-3 x^{2}=0$
Let's make a substitution: $t=x^{2}+16 x+60$, then the equation will take the form
$4(t+x) t-3 x^{2}=0$
$4 t^{2}+4 t x-3 x^{2}=0$
$(2 t-x)(2 t+3 x)=0$
$\left[\begin{array}{c}2 t-x=0 \\ 2 t+3 x=0\end{arr... | -7.5;8;\frac{-35\\sqrt{265}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,468 |
4. The train is traveling at a speed of 60 kilometers per hour, making stops every 48 kilometers. The duration of each stop, except the fifth, is 10 minutes, and the fifth stop is half an hour. How far has the train traveled if it departed at noon on September 29 and arrived at its destination on October 1 at 10:00 PM? | Solution: The train was on the way for 58 hours.
The train covers a section of 48 kilometers in $\frac{4}{5}$ of an hour.
Let the train make $N$ stops during its journey, then the time of its movement will be $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$, where $t$ is the time ... | 2870 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,470 |
5. Calculate $f(2)$, if $25 f\left(\frac{x}{1580}\right)+(3-\sqrt{34}) f\left(\frac{1580}{x}\right)=2017 x$. Round the answer to the nearest integer. | Solution: Let's make the substitution: $\mathrm{t}=\frac{x}{1580}$, then the equation will take the form:
$25 f(t)+(3-\sqrt{34}) f\left(\frac{1}{t}\right)=2017 \cdot 1580 \cdot t \quad(1)$,
substitute $\frac{1}{t}$ for $t$ in the equation, we get
$$
25 f\left(\frac{1}{t}\right)+(3-\sqrt{34}) f(t)=2017 \cdot 1580 \cd... | 265572 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,471 |
6. On a plane, a segment of length 1 cm is given. Using a compass and a straightedge, construct a segment of length $\sqrt{\sqrt{3}+\sqrt{2}+1}$ cm on this plane. | Solution: On the line $\mathrm{AC}$, we mark off unit segments $\mathrm{AB}$ and $\mathrm{BC}$. We construct the perpendicular bisector $\mathrm{PB}$ of segment $\mathrm{AC}$ using a compass and straightedge. We mark off a unit segment $\mathrm{BK}$ on it and connect points $\mathrm{A}$ and $\mathrm{K}$. The length of ... | \sqrt{\sqrt{3}+\sqrt{2}+1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,472 |
7. Dad was driving his son to a math club lesson. On the way, they were overtaken by 4 cars. The son admitted to his father that he could only remember the numbers of the two smallest cars - 119 and 179, and that the others definitely had the digit 3. The father, however, remembered all of them and gave his son the fol... | Solution: Let x and y be the numbers of the two remaining cars. Then there are two possible cases.
1st case. $\quad x y+119 \cdot 179=105080 \Rightarrow x y=83779=199 \cdot 421-\quad$ this $\quad$ is $\quad$ two $\quad$ prime factors, so there are no other three-digit factors. But in this case, there is no digit 3.
2... | 337,363 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,473 |
1. The number $\overline{6 x 62 y 4}$ is divisible by 11, and when divided by 9, it gives a remainder of 6. Find the remainder when this number is divided by $13 .(15$ points) | # Solution
By the divisibility rule for 11, we get
$((x+2+4)-(6+6+y)) \vdots 11$ or $(x-6-y) \vdots 11$
Let's find suitable options: $(0 ; 5)(1 ; 6)(2 ; 7)(3 ; 8)(4 ; 9)(6 ; 0)(7 ; 1)(8 ; 2)$ $(9 ; 3)$.
If the number $\overline{6 x 62 y 4}$ gives a remainder of 6 when divided by 9, then by the divisibility rule for... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,476 |
2. Five people are performing a certain task. The first, second, and third, working together, complete the work in the same time as the second, fourth, and fifth. The first, working with the fifth, completes the work in the same time as the third, working with the fourth, which is 2 hours. How long will it take for all... | Solution:
Let V be the volume of work, $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the speed of completing the work respectively.
$$
\left\{\begin{array} { c }
{ ( x _ { 1 } + x _ { 2 } + x _ { 3 } ) * 1 = V } \\
{ ( x _ { 2 } + x _ { 4 } + x _ { 5 } ) * 1 = V } \\
{ ( x _ { 1 } + x _ { 5 } ) * 2 = V } \\
{ ( x _ { 3 } +... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,477 |
4. The numbers x and y are solutions to the system $\left\{\begin{array}{l}a x-y=2 a+1 \\ -x+a y=a\end{array}\right.$, where a is a parameter. What is the smallest value that the expression $2 y^{2}$ $x^{2}$ can take if $a \in[-0.5 ; 2] .(20$ points) | Solution: $\left\{\begin{array}{l}a x-y=2 a+1 \\ x=a(y-1)\end{array} \Leftrightarrow\left\{\begin{array}{l}a^{2}(y-1)-y=2 a+1 \\ x=a(y-1)\end{array} \Leftrightarrow\left\{\begin{array}{l}y\left(a^{2}-1\right)=a^{2}+2 a+1 \\ a y-a=x\end{array}\right.\right.\right.$
For $a \neq \pm 1$
$\left\{\begin{array}{l}y=\frac{a+... | -\frac{2}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,479 |
5. Nif-Nif, Naf-Naf, Nuf-Nuf, and the Wolf decided to weigh themselves. It turned out that the Wolf weighs more than Nif-Nif; Nuf-Nuf and Nif-Nif weigh more than the Wolf and Naf-Naf; Nuf-Nuf and Naf-Naf weigh as much as the Wolf and Nif-Nif. Arrange the piglets and the Wolf in descending order of weight. (15 points) | Solution.
Let the masses of the Wolf be $a$, Nif-Nif be $b$, Naf-Naf be $c$, and Nuf-Nuf be $d$. From the condition, we get $a>b(1)$,
$$
d+b>a+c(2), d+c=a+b(3)
$$
From conditions (1) and (2), it follows that $a+c>b+c$, so $d>c$. From conditions (2) and (3), we get: $2 d+b+c>2 a+c+b$, then $d>a$. If $d>a$, then from ... | >>b> | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,480 |
6. Three friends - Masha, Olya, and Svetlana - enrolled in the economic, information, and mathematics classes at the lyceum. If Masha is an economist, then Svetlana is not an informatician. If Olya is not an informatician, then Masha is an economist. If Svetlana is not an economist, then Olya is a mathematician. Determ... | Solution.
Let Olya not be an informatics student, then according to condition 2, Masha is an economist. If Masha is an economist, then Svetlana, according to condition 1, is not an informatics student. We have a contradiction (Masha is both an informatics student and an economist). Therefore, Olya is an informatics st... | Masha:Mathematics,Olya:Informatics,Svetlana:Economics | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,481 |
6. A circle is circumscribed around an isosceles triangle with a $45^{0}$ angle at the vertex. A second circle is internally tangent to the first circle and to the two lateral sides of the given triangle. The distance from the center of the second circle to the given vertex of the triangle is 4 cm. Find the distance fr... | # Solutions for the Correspondence Round for 10th Grade. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,482 |
Problem 1. Buratino, Karabas-Barabas, and Duremar are running along a path around a circular pond. They start simultaneously from the same point, with Buratino running in one direction and Karabas-Barabas and Duremar running in the opposite direction. Buratino runs three times faster than Duremar and four times faster ... | # Solution:
Let the length of the path be S.
Since Buratino runs three times faster than Duremar, by the time they meet, Buratino has run three-quarters of the circle ($3 / 4$ S), while Duremar has run one-quarter. Since Buratino runs four times faster than Karabas-Barabas, by the time they meet, Buratino has run fou... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,483 |
Problem 2. It is known that $a, b$ and $c$ are natural numbers, $LCM(a, b)=945$, $LCM(b, c)=525$. What can $\operatorname{LCM}(a, c)$ be equal to? | # Solution:
Let's factorize the numbers into prime factors, since $3^{3}$ only appears in $H O K(a, b)$, it follows that $a: 3^{3}$, similarly we get that $b: 5^{2}$
$$
\left\{\begin{array} { l }
{ \operatorname { H O K } ( a , b ) = 9 4 5 = 3 ^ { 3 } \cdot 5 \cdot 7 } \\
{ \operatorname { H O K } ( b , c ) = 5 2 5 ... | 675or4725 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,484 |
Problem 3. On a plane, there are 3 different points M, G, and T such that the figure formed by points M, G, and T has no axis of symmetry. Construct a point U on this plane so that the figure formed by points M, G, T, and U has at least one axis of symmetry. How many different such points exist in this plane? | # Solution:
If points M, G, and T are the vertices of a right triangle, then there are 5 such points; otherwise, there are 6. Indeed, 3 such points are the points symmetric to the given ones with respect to the line containing the other two, since all three points do not lie on the same line (by the condition). For ex... | 5or6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,485 |
Problem 4. For what values of $a$ and $b$ do the equations $2 x^{3}+a x-12=0$ and $x^{2}+b x+2=0$ have two common roots? | # Solution:
Let $x_{0}$ be the common root of the two equations, that is, the following equalities hold $\left\{\begin{array}{c}2 x_{0}^{3}+a x_{0}-12=0 \\ x_{0}^{2}+b x_{0}+2=0\end{array}\right.$
Notice that $x_{0}=0$ is not a solution to the quadratic equation, and multiply both sides of this equality by $2 x_{0}$.... | =-14,b=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,486 |
Problem 5. Solve the system of equations $\left\{\begin{array}{l}\sqrt{2016 \frac{1}{2}+x}+\sqrt{2016 \frac{1}{2}+y}=114 \\ \sqrt{2016 \frac{1}{2}-x}+\sqrt{2016 \frac{1}{2}-y}=56\end{array}\right.$. | # Solution:
We will use the inequality $\sqrt{\frac{a^{2}+b^{2}}{2}} \geq \frac{a+b}{2}$ (the quadratic mean of two numbers is greater than or equal to their arithmetic mean). Note that the right-hand side of the inequality equals the left-hand side when $a$ and $b$ are equal.
By taking one of the roots of the equati... | (1232\frac{1}{2};1232\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,487 |
1. At the same time, a cyclist leaves point $A$ for point $B$ at a speed of $15 \mathrm{km} / \mathrm{u}$, and a tourist leaves point $B$ for point $C$ at a speed of 5 km/h. After 1 hour and 24 minutes from the start of their journey, they were at the minimum distance from each other. Find the distance between the poin... | Solution: Let $AB = BC = AC = S$. Denote the distance between the cyclist and the tourist as $r = r(t)$, where $t$ is the time from the start of the movement. Then, by the cosine theorem, we have: $r^{2} = (S - 15t)^{2} + (5t)^{2} - 5t(S - 15t)$. To find the time when the distance between the cyclist and the tourist wa... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,490 |
2. Solve the inequality $\log _{x}(6 x-5)>2$.
# | # Solution:
$\log _{x}(6 x-5)>2$. Domain: $x \in(5 / 6 ; 1) \cup(1 ;+\infty) . \log _{x} \frac{x^{2}}{6 x-5}1\end{array} \Leftrightarrow\left\{\begin{array}{c}5 / 60\end{array} \Leftrightarrow\left\{\begin{array}{c}5 / 65\end{array} \Leftrightarrow 5 / 6<x<1\right.\right.\right.$.
2) $\left\{\begin{array}{c}1<x<+\inft... | x\in(5/6;1)\cup(1;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,491 |
3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 463. | Solution: If we consider that a single number forms an increasing geometric progression, then the single number 463 is a solution to the problem. Let the progression have no fewer than two terms. According to the condition, for some natural number $n \geq 2$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}... | {463},{1;462},{1;21;441} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,492 |
4. Solve the equation $\sin 7 x+\sqrt[4]{1-\cos ^{11} 3 x \cos ^{2} 7 x}=0 . \quad(8$ points) | Solution: Given the condition $\sin 7 x \leq 0$, both sides of this equation can be squared. Under the found constraints, the equation is equivalent to the following: $\sin ^{4} 7 x+\cos ^{11} 3 x \cos ^{2} 7 x-1=0$, $\left(1-\cos ^{2} 7 x\right)^{2}+\cos ^{11} 3 x \cos ^{2} 7 x-1=0,-2 \cos ^{2} 7 x+\cos ^{4} 7 x+\cos ... | -\frac{\pi}{14}+\frac{2\pin}{7},n\inZ\quad2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,493 |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(40 \cos 2 x+41)$ for $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]$ (10 points) | Solution: Since $\cos (\pi-\arcsin (-0.8))=\cos (\pi+\arcsin 0.8)=-\cos (\arcsin 0.8)=-0.6$, then $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]=[-\operatorname{arctg}(1 / 5) ; \operatorname{arctg} 3]$. $2 x \in[-2 \operatorname{arctg}(1 / 5) ; 2 \operatorname{arctg} 3]$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,495 |
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}$, $AE = \sqrt{3}$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, ... | # 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 triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{A E}{\cos 30^{\circ}}=2, E C=A C \sin 30^{\circ}=1$
$y=x^{2} / 2, M\left(x_{0} ;-15 / 2\right)$. The equation
$\frac{1}{2} x^{2}=-\frac{15}{2}+k\left(x-x_{0}\right)$, or
$x^{2}-2 k x+2 k x_{0}+15=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-15=0$.
The solutions of this equation for $k$ satisfy the conditions $k_{... | M(\2/\sqrt{3};-15/2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,497 |
9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=18(y-x+a-4), \log _{(x / 4)}(y / 4)=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 4, y=x$. Substituting $y=x$ into the first equation, we get; $(x-a)^{2}=18(a-4)$, or $x^{2}-2 a x+a^{2}-18 a+72=0(*)$, which has $D / 4=a^{2}-a^{2}+18 a-72=18(a-4)$. The number of solutions to the given system of equations depends on the number of ... | 22,(40;40) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,498 |
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:
In $O S \| A B_{1}, S \in(A B C), O S=A B_{1} / 2$ and $O H \| A A_{1}, H \in A C$. Then $S H \| A B, \quad A \quad A B / 2$. We connect points $B$ and $S, D=B S \cap A C$. Ext... | \frac{5\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,500 |
1. At the same time, a car departs from point $A$ to point $B$ at a speed of 80 km/h, and from point $B$ to point $C$ - a train at a speed of $50 \kappa м / ч$. After 7 hours of travel, they were at the shortest distance from each other. Find the distance between the points, if all three points are equidistant from eac... | Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 80t)^{2} + (50t)^{2} - 50t(S - 80t)$. To find the time at which the distance between the car and the train was the sm... | 860 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,501 |
2. Solve the inequality $\log _{x}(4 x-3)>2$. | Solution: $\quad \log _{x}(4 x-3)>2$. Domain: $x \in(3 / 4 ; 1) \cup(1 ;+\infty) . \log _{x} \frac{x^{2}}{4 x-3}1\end{array} \Leftrightarrow\left\{\begin{array}{c}3 / 40\end{array} \Leftrightarrow\left\{\begin{array}{c}3 / 43\end{array} \Leftrightarrow 3 / 4<x<1\right.\right.\right.$.
2) $\left\{\begin{array}{c}1<x<+\i... | x\in(3/4;1)\cup(1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,502 |
3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 157. | Solution: If we consider that one number forms an increasing geometric progression, then the number 157 is a solution to the problem. According to the condition, for some natural number $n$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}=157$ holds, or $b_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right)=157$. Sin... | {157},{1;156},{1;12;144} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,503 |
4. Solve the equation $\sqrt[4]{1-\cos ^{15} 3 x \cos ^{2} 5 x}=\sin 5 x$.
# | # Solution:
Given the condition $\sin 5 x \geq 0$, both sides of this equation can be squared. Under the found constraints, the equation is equivalent to the following: $\sin ^{4} 5 x+\cos ^{15} 3 x \cos ^{2} 5 x-1=0$, $\left(1-\cos ^{2} 5 x\right)^{2}+\cos ^{15} 3 x \cos ^{2} 5 x-1=0,-2 \cos ^{2} 5 x+\cos ^{4} 5 x+\c... | \frac{\pi}{10}+\frac{2\pin}{5},n\inZ,2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,504 |
5. Solve the inequality $\left(3^{x^{2}-1}-9 \cdot 3^{5 x+3}\right) \log _{\cos \pi}\left(x^{2}-6 x+9\right) \geq 0$.
# | # Solution:
Domain: $(x-3)^{2}>0, \quad \cos \pi x>0, \cos \pi x \neq 1, \Rightarrow x \in(-0.5+2k ; 2k) \cup(2k ; 0.5+2k), k \in Z$.
The original inequality on the domain is equivalent to the following
$$
\left(x^{2}-5 x-6\right)(\cos \pi x-1)\left(x^{2}-6 x+8\right) \geq 0 \Leftrightarrow(x-6)(x+1)(x-2)(x-4) \leq ... | x\in(-0.5;0)\cup(0;0.5)\cup(1.5;2)\cup(4;4.5)\cup(5.5;6) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,505 |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(10 \cos 2 x+17)$ for $x \in[1,25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3] . \quad(10$ points $)$ | # Solution:
Since $\quad$ $\cos (\pi-\arcsin (-0.6))=\cos (\pi+\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3]=[-\operatorname{arctg} 0.25 ; \operatorname{arctg} 3] \quad$ Therefore, $2 x \in[-2 \operatorname{arctg} 0.25 ; 2 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,506 |
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 $30^{\circ}$, $EC=1$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, where $O$... | # 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 triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$,
$$
A C=\frac{E C}{\sin 30^{\circ}}=2, \quad A E=E C \operatorname{tg} 60^{\circ}=\sqrt... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,507 |
8. On the line $y=-5 / 3$ find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, the angle between which is $60^{\circ}$. | Solution (without using derivatives).
$$
y=x^{2} / 2, \quad M\left(x_{0} ;-5 / 3\right)
$$
The equation $\frac{1}{2} x^{2}=-\frac{5}{3}+k\left(x-x_{0}\right)$, or $x^{2}-2 k x+2 k x_{0}+\frac{10}{3}=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-\frac{10}{3}=0$. The two values of $k$ found from this equati... | M(\\sqrt{3}/2;-5/3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,508 |
9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=16(y-x+a-3), \log _{(x / 3)}(y / 3)=1$ has at least one solution, and solve it for each $a$. | Solution. The second equation is equivalent to the system: $x>0, x \neq 3, y=x$. Substituting $y=x$ into the first equation, we get; $(x-a)^{2}=16(a-3)$, or $x^{2}-2 a x+a^{2}-16 a+48=0(*)$, which has $D / 4=a^{2}-a^{2}+16 a-48=16(a-3)$. The number of solutions to the given system of equations depends on the number of ... | 19,(35,35) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,509 |
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{21}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,511 |
1. At the same time, a car departs from point $A$ to point $B$ at a speed of $90 \mathrm{~km} / \mathrm{h}$, and from point $B$ to point $C$ - a train at a speed of $60 \mathrm{~km} /$ h. After 2 hours of travel, they found themselves at the minimum distance from each other. Find the distance between the points, if all... | Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 90t)^{2} + (60t)^{2} - 60t(S - 90t)$. To find the time at which the distance between the car and the train
>2$. | Solution: $\log _{x}(5 x-4)>2$. Domain: $x \in(4 / 5 ; 1) \cup(1 ;+\infty) . \log _{x} \frac{x^{2}}{5 x-4}1\end{array} \Leftrightarrow\left\{\begin{array}{c}4 / 50\end{array} \Leftrightarrow\left\{\begin{array}{c}4 / 54\end{array} \Leftrightarrow 4 / 5<x<1\right.\right.\right.$.
2) $\left\{\begin{array}{c}1<x<+\infty, ... | x\in(4/5;1)\cup(1;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,513 |
3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 211. | Solution: If we consider that one number forms an increasing geometric progression, then the number 211 is a solution to the problem. According to the condition, for some natural number $n$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}=211$ holds, or $b_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right)=211$. Sin... | {211},{1;210},{1;14;196} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,514 |
4. Solve the equation $\sqrt[4]{1-\cos ^{7} 15 x \cos ^{2} 9 x}=\sin 9 x . \quad$ (8 points) | Solution: Given the condition $\sin 9 x \geq 0$, both sides of this equation can be squared. Under the found constraints, the equation is equivalent to the following: $\sin ^{4} 9 x+\cos ^{7} 15 x \cos ^{2} 9 x-1=0,\left(1-\cos ^{2} 9 x\right)^{2}+\cos ^{7} 15 x \cos ^{2} 9 x-1=0$, $-2 \cos ^{2} 9 x+\cos ^{4} 9 x+\cos ... | \frac{\pi}{18}+\frac{2\pin}{9},n\inZ,\frac{2\pi}{3},\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,515 |
5. Solve the inequality $\left(2^{x^{2}-6}-4 \cdot 2^{x+4}\right) \log _{\cos \pi x}\left(x^{2}-2 x+1\right) \geq 0$. (10 points) | # Solution:
Domain: $(x-1)^{2}>0, \quad \cos \pi x>0, \cos \pi x \neq 1, \Rightarrow x \in(-0.5+2k ; 2k) \cup(2k ; 0.5+2k), k \in Z$. The original inequality on the domain is equivalent to the following $\left(x^{2}-x-12\right)(\cos \pi x-1)\left(x^{2}-2 x\right) \geq 0 \Leftrightarrow(x-4)(x+3)(x-2) x \leq 0 \Rightar... | x\in(-2.5;-2)\cup(-2;-1.5)\cup(-0.5;0)\cup(2;2.5)\cup(3.5;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,516 |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{2}(5 \cos 2 x+11)$ for $x \in[1,25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2] \quad$ (10 points) | # Solution:
Since $\quad$ $\cos (\pi+\arcsin (-0.6))=\cos (\pi-\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]=[-\operatorname{arctg}(1 / 3) ; \operatorname{arctg} 2]$ Therefore, $2 x \in[-2 \operatorname{arctg}(1 / 3) ; 2 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,517 |
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