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# Problem 6.
A $10 \times 10$ square was cut into rectangles, the areas of which are different and expressed as natural numbers. What is the maximum number of rectangles that can be obtained? | Solution. The area of the square is 100. If we represent 100 as the sum of natural numbers, the number of addends will be the largest if the difference between the numbers is one. Let's take rectangles with areas of $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. Their total area is 55. Therefore, the sum of the areas of the remainin... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,633 |
1. From point $A$ to point $B$, which are 8 km apart, a tourist and a cyclist set out simultaneously. The cyclist, who took no less than half an hour to travel from $A$ to $B$, without stopping, turned back and started moving towards point $A$, increasing his speed by $25 \%$. After 10 minutes from his departure from p... | Solution: Let $x$ (km/h) be the speed of the tourist, $y$ (km/h) be the initial speed of the cyclist, and $t$ (h) be the time spent by the cyclist traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{c}
x(t+1 / 6)+5 y / 24=8, \\
y t=8, \\
t \geq 0.5,
\end{array} \Rightarrow x(8 / y+1 / 6)+5 y / 24=8, \Rightarrow 5 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,634 |
2. Solve the inequality $\log _{x}\left(25-40 x+16 x^{2}\right)<0$. | Solution: $\log _{x}\left(25-40 x+16 x^{2}\right)0, \quad x \neq 1, \quad x \neq 5 / 4$.
1) $01 \Leftrightarrow 2 x^{2}-5 x+3>0 ;\left\{\begin{array}{l}x3 / 2 \\ 01, \quad x \neq 5 / 4 ; 25-40 x+16 x^{2}<1 \Leftrightarrow 2 x^{2}-5 x+3<0 \Leftrightarrow 1<x<3 / 2, x \neq 5 / 4$.
Answer: $x \in(0 ; 1) \cup(1 ; 5 / 4) ... | x\in(0;1)\cup(1;5/4)\cup(5/4;3/2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,635 |
3. Two numbers x and y satisfy the equation $280 x^{2}-61 x y+3 y^{2}-13=0$ and are the fourth and ninth terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: We factorize the expression $280 x^{2}-61 x y+3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(280\left(\frac{x}{y}\right)^{2}-61\left(\frac{x}{y}\right)+3\right)=280 y^{2}\left(\frac{x}{y}-\frac{3}{40}\right)\left(\frac{x}{y}-\frac{1}{7}\right)=(40 x-3 y)(7 x-y)$. This formula is also valid for all real numbers... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,636 |
4. Solve the equation $\frac{4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2}{\sqrt{\cos x-\sin x}}=0$. | # Solution:
$$
\frac{4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2}{\sqrt{\cos x-\sin x}}=0
$$
Given the condition $\cos x-\sin x>0$, we find the roots of the equation $4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2=0 \Leftrightarrow$
$4 \operatorname{tg}^{4} 8 x+4 \sin 2 ... | -\frac{\pi}{2}+2\pin,-\frac{\pi}{4}+2\pin,2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,637 |
5. Solve the inequality
$$
\frac{\left(2 \cdot 3^{-\log _{x} 2}-6\right) \sqrt{2-\sqrt{2 \log _{x} 2+3}}}{2+\sqrt{\log _{x} 2+2}}>\frac{\left(3^{-\log _{x} 2}-3\right) \sqrt{2-\sqrt{2 \log _{x} 2+3}}}{\sqrt{\log _{x} 2+2}-1}
$$ | Solution: Let's make the substitution $y=\log _{x} 2$.
$$
\begin{aligned}
& \frac{\left(2 \cdot 3^{-y}-6\right) \sqrt{2-\sqrt{2 y+3}}}{2+\sqrt{y+2}}>\frac{\left(3^{-y}-3\right) \sqrt{2-\sqrt{2 y+3}}}{\sqrt{y+2}-1} \Leftrightarrow \\
& \left(3^{-y}-3\right)\left(\frac{2}{2+\sqrt{y+2}}-\frac{1}{\sqrt{y+2}-1}\right) \sqr... | x\in(0;1/2)\cup(1/2;1/\sqrt[3]{4}]\cup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,638 |
6. Find the set of values of the function $f(x)=1 / g(64 g(g(\ln x)) / 1025)$, where
$$
g(x)=x^{5}+1 / x^{5} \cdot(10 \text { points })
$$ | Solution: First, consider the function $\varphi(t)=t+1 / t$. The function $\varphi(t)$ is defined for all $t \neq 0$. Let's find the extrema of the function $\varphi(t)$. To do this, we will determine the intervals of the sign constancy of the derivative of the function
$$
\varphi(t): \quad \varphi^{\prime}(t)=1-1 / t... | [-32/1025;0)\cup(0;32/1025] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,639 |
7. On side $B C$ of triangle $A B C$, a point $K$ is marked such that $A K=5, B K=16$, $K C=2$. A circle is circumscribed around triangle $A B K$. A line through point $C$ and the midpoint $D$ of side $A B$ intersects the circle at point $P$, with $C P>C D$. Find $D P$, if $\angle A P B=\angle B A C$. | # Solution:
1) $\angle A P B = \angle B A C, \angle A P B = \angle A K C, \angle A K C = \angle B A C, \angle K A C = \angle A B C$.
Segment $A C$ is a tangent to the circle.
$$
\begin{ali... | \frac{-21+12\sqrt{89}}{2\sqrt{55}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,640 |
8. On the line $x=\sqrt{3}$, find the point $M$ through which two tangents to the graph of the function $y=x^{2} / 4$ pass, such that the angle between them is $60^{\circ}$. | Solution (without using derivatives).
$y=x^{2} / 4, M\left(\sqrt{3} ; y_{0}\right)$. The equation $\frac{1}{4} x^{2}=y_{0}+k(x-\sqrt{3})$, or $x^{2}-4 k x+4 k \sqrt{3}-4 y_{0}=0$, has a unique solution if $\frac{D}{4}=4 k^{2}-4 k \sqrt{3}+4 y_{0}=0, k^{2}-k \sqrt{3}+y_{0}=0$. The two values of $k$ found from this equa... | M_{1}(\sqrt{3};0),M_{2}(\sqrt{3};-10/3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,641 |
9. Find all values of $a$ for which the system of equations $y-1=a(x-1), \quad \frac{2 x}{|y|+y}=\sqrt{x}$ has at least one solution, and solve it for each $a$.
# | # Solution:
Domain of definition: $y>0, x \geq 0$.
In the domain of definition, the second equation of the system becomes: $x=y \sqrt{x}$.
I. $x=0, y=1-a>0$, hence $a<1$.
II. $x>0, y=\sqrt{x} ; \sqrt{x}-1=a(\sqrt{x}-1)(\sqrt{x}+1)$.
1) $\sqrt{x}=1, x=1, y=1, a \in \mathbb{R}$.
$$
\text { 2) } 1=a(\sqrt{x}+1), \sqrt... | \begin{aligned}&\in(-\infty;0]\cup{\frac{1}{2}},x_{1}=0,y_{1}=1-x_{2}=1,y_{2}=1;\\&\in(0;\frac{1}{2})\cup(\frac{1}{2};1),x_{1}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,642 |
10. The lateral edges of the triangular pyramid $TABC$ form right angles with each other. What is the minimum area that the section of the pyramid by a plane passing through vertex $C$ and the midpoint of side $AB$ of the base can have, if the base edge $AC=5$ and the lateral edges $TA=4, TB=12$? Which of the lateral e... | # Solution:
If the secant plane intersects the lateral edge $T B$, the area of the section will be the smallest if $K N$ is the common perpendicular to the lines $T B$ and $C S$.
If the secant plane intersects the lateral edge $T A$, the area of the section will be the smallest if $M E$ is the common perpendicular to... | \frac{21}{\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,643 |
1. From point $A$ to point $B$, which are 24 km apart, a pedestrian and a cyclist set out simultaneously. The cyclist, who spent no less than two hours on the journey from $A$ to $B$, without stopping, turned back and started moving towards point $A$ at a speed twice the initial speed. After 24 minutes from his departu... | Solution: Let $x$ (km/h) be the speed of the cyclist, $y$ (km/h) be the initial speed of the truck, and $t$ (h) be the time spent by the truck traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{rl}
x(t+0.4)+0.8 y=24, \\
y t & =24, \\
t & \geq 2,
\end{array} \Rightarrow x(24 / y+0.4)+0.8 y=24, \Rightarrow 2 y^{2}+... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,644 |
2. Solve the inequality $\log _{x}\left(36-60 x+25 x^{2}\right)<0$. | Solution: $\log _{x}\left(36-60 x+25 x^{2}\right)0, \quad x \neq 1, \quad x \neq 6 / 5$.
1) $01 \Leftrightarrow 5 x^{2}-12 x+7>0 ;\left\{\begin{array}{l}{\left[\begin{array}{c}x7 / 5\end{array} \Leftrightarrow 01, \quad x \neq 6 / 5 ; 36-60 x+25 x^{2}<1 \Leftrightarrow 5 x^{2}-12 x+7<0 \Leftrightarrow 1<x<7 / 5, x \ne... | x\in(0;1)\cup(1;6/5)\cup(6/5;7/5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,645 |
3. Two numbers \(x\) and \(y\) satisfy the equation \(26 x^{2} + 23 x y - 3 y^{2} - 19 = 0\) and are the sixth and eleventh terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: Factorize the expression $26 x^{2}+23 x y-3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(26\left(\frac{x}{y}\right)^{2}+23\left(\frac{x}{y}\right)-3\right)=26 y^{2}\left(\frac{x}{y}+1\right)\left(\frac{x}{y}-\frac{3}{26}\right)=(x+y)(26 x-3 y)$. This formula is valid for all real numbers $y$. According to the ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,646 |
4. Solve the equation $\frac{2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2}{\sqrt{\cos x-\sqrt{3} \sin x}}=0$. | # Solution:
$$
\frac{2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2}{\sqrt{\cos x-\sqrt{3} \sin x}}=0
$$
Given the condition $\cos x-\sqrt{3} \sin x>0$, we find the roots of the equation $2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2=0 \Leftrightarrow$
$2 \operatorname{tg}... | 2\pin,-\frac{\pi}{6}+2\pin,-\frac{\pi}{3}+2\pin,-\frac{\pi}{2}+2\pin,-\frac{2\pi}{3}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,647 |
5. Solve the inequality
$$
\frac{\left(5 \cdot 2^{-\log _{x} 3}-2.5\right) \sqrt{2-\sqrt{\log _{x} 3+1}}}{1+\sqrt{\log _{x} 3+8}}>\frac{\left(2^{-\log _{x} 3}-0.5\right) \sqrt{2-\sqrt{\log _{x} 3+1}}}{\sqrt{\log _{x} 3+8}-3}
$$ | Solution: Let's make the substitution $y=\log _{x} 3$.
$$
\frac{\left(5 \cdot 2^{-y}-2.5\right) \sqrt{2-\sqrt{y+1}}}{1+\sqrt{y+8}}>\frac{\left(2^{-y}-0.5\right) \sqrt{2-\sqrt{y+1}}}{\sqrt{y+8}-3}
$$
$\left(2^{-y}-0.5\right)\left(\frac{5}{1+\sqrt{y+8}}-\frac{1}{\sqrt{y+8}-3}\right) \sqrt{2-\sqrt{y+1}}>0$ Find the doma... | x\in(0;1/3]\cup(\sqrt[3]{3};3)\cup(3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,648 |
6. Find the set of values of the function $f(x)=1 / g\left(64 g\left(16 g\left(\log _{2} x\right)\right) / 5\right)$, where
$g(x)=\sqrt[5]{x}+1 / \sqrt[5]{x}$. | Solution: First, consider the function $\varphi(t)=t+1 / t$. The function $\varphi(t)$ is defined for all $t \neq 0$. Let's find the extrema of the function $\varphi(t)$. To do this, we will determine the intervals of constancy of the derivative of the function $\varphi(t): \quad \varphi^{\prime}(t)=1-1 / t^{2}=(t-1)(t... | [-2/5;0)\cup(0;2/5] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,649 |
7. On side $BC$ of triangle $ABC$, a point $K$ is marked such that $AK=8$, $BK=24$, $KC=3$. A circle is circumscribed around triangle $ABK$. A line through point $C$ and the midpoint $D$ of side $AB$ intersects the circle at point $P$, with $CP > CD$. Find $DP$ if $\angle APB = \angle BAC$. | # Solution:
$\angle A P B = \angle B A C, \angle A P B = \angle A K C, \angle A K C = \angle B A C, \angle K A C = \angle A B C$. Segment $A C$ is a tangent to the circle.
$\triangle A B C... | \frac{-6+18\sqrt{13}}{\sqrt{29}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,650 |
8. On the line $x=1$, find the point $M$ through which two tangents to the graph of the function $y=x^{2} / 4$ pass, such that the angle between them is $45^{\circ}$. | # Solution:
$y=x^{2} / 4, \quad M\left(1 ; y_{0}\right)$. The equation $\frac{1}{4} x^{2}=y_{0}+k(x-1)$, or $x^{2}-4 k x+4 k-4 y_{0}=0$, has a unique solution if $\frac{D}{4}=4 k^{2}-4 k+4 y_{0}=0$. The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=1(1), k_{1} \cdot k_{2}=y_{0}$ (... | M_{1}(1;0),M_{2}(1;-6) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,651 |
9. Find all values of $a$ for which the system of equations $2 y-2=a(x-1), \quad \frac{2 x}{|y|+y}=\sqrt{x}$ has at least one solution, and solve it for each $a$.
# | # Solution.
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. $x=0, y=1-\frac{a}{2}>0$, hence $a<2$, and $a \leq 0$.
II. $x>0, y=\sqrt{x} ; 2(\sqrt{x}-1)=a(\sqrt{x}-1)(\sqrt{x}+1)$.
1) $\sqrt{x}=1, x=1, y=1, a \in \mathbb{R}$.
2) ... | \begin{aligned}&\in(-\infty;0]\cup{1},x_{1}=0,y_{1}=1-\frac{}{2};x_{2}=1,y_{2}=1;\\&\in(0;1)\cup(1;2),x_{1}=0,y_{1}=1-\frac{}{2};x_{2}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,652 |
Problem 1. Which number is larger, $2^{5^{4^{4}}}$ or $3^{4^{2^{5}}}$? | Solution:
$2^{5^{4^{3}}} \vee 3^{4^{2^{5}}} \Leftrightarrow 2^{5^{64}} \vee 3^{4^{32}} \Leftrightarrow 2^{5^{64}}>2^{6^{64}}=2^{4^{63 \cdot 4}}=\left(2^{4}\right)^{4^{63}}=16^{4^{63}}>3^{4^{63}}>3^{4^{32}}$
Answer: $2^{5^{4^{3}}}>3^{4^{4^{5}}}$ | 2^{5^{4^{3}}}>3^{4^{4^{5}}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,653 |
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 between 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 | 10,654 |
Problem 3. Given segments $a, b, c$. Construct a segment of length $\frac{a b}{c \sqrt{5}}$ using a compass and straightedge.
# | # Solution.
1) The segment of length $c \sqrt{5}$ is constructed as the hypotenuse of a right triangle with legs $c$ and $2c$.
2) The segment of length $x=\frac{a b}{c \sqrt{5}}$ is constructed according to Thales' theorem as a proportional segment, cut off on the sides of an angle
, hence $A O: O D=A B: C D=3: 4$.
Let $A O=3 y, \quad O D=4 y$, then, by the Pythagorean theorem for the right triangle $A O D$ we have $9 y^{2}+16 y^{2}=A D^{2}=64, \quad$ from which $\q... | AB=10.2,CD=13.6,BC=15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,656 |
9. Rhombus $ABCD$ is the base of a pyramid with vertex $S$. All its lateral faces are inclined to the base plane at the same angle of $60^{\circ}$. Points $M, N, K$, and $L$ are the midpoints of the sides of rhombus $ABCD$. A rectangular parallelepiped is constructed on rectangle $MNKL$ as its base. The edges of the up... | # Solution Variant № 1 (11th grade, qualifying stage) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,658 |
1. The production of ceramic items consists of 3 sequential stages: forming a ceramic item on a potter's wheel for 15 minutes, drying for 10 minutes, and firing for 30 minutes. It is required to produce 75 items. How should 13 masters be distributed between molders and firers to work on stages 1 and 3 respectively for ... | Solution. Molders - 4, decorators - 8. The thirteenth master can work at any stage, or not participate in the work at all. In this case, the working time is 325 minutes $\left(55+\left(\left[\frac{75}{4}\right]+1\right) \cdot 15=325\right)$. We will show that with other arrangements, the working time is longer. Suppose... | 325 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,659 |
2. Solve the equation $\frac{15}{x\left(\sqrt[3]{35-8 x^{3}}\right)}=2 x+\sqrt[3]{35-8 x^{3}}$. Write the sum of all obtained solutions in the answer.
(5 points)
# | # Solution.
$$
u=\sqrt[3]{35-8 x^{3}}, 8 x^{3}=35-u^{3}
$$
$\left\{\begin{array}{c}\frac{15}{x u}=2 x+u, \\ 8 x^{3}+u^{3}=35,\end{array} t=2 x u, v=2 x+u \Rightarrow\left\{\begin{array}{c}\frac{30}{t}=v, \\ v\left(v^{2}-3 t\right)=35,\end{array} \Rightarrow\left\{\begin{array}{l}t=6, \\ v=5,\end{array}\right.\right.\... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,660 |
3. Find all natural values of $n$ for which
$$
\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9}, \text { and } \log _{2}^{2} n+45<\log _{2} 8 n^{13}
$$
In your answer, write the sum of the obtained values of $n$.
(6 points) | # Solution.
$$
\begin{aligned}
& \cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9} \quad \Leftrightarrow 2 \sin \frac{\pi}{9} \cos \frac{2 \pi}{9}+2 \sin \frac{\pi}{9} \cos \frac{4 \pi}{9}+\cdots+ \\
& 2 \sin \frac{\pi}{9} \cos \frac{2 \pi n}{9}=2 \sin \frac{\pi}{9} \cos \frac{... | 644 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,661 |
4. In how many ways can a bamboo trunk (a non-uniform natural material) 4 m long be sawn into three parts, the lengths of which are multiples of 1 dm, and from which a triangle can be formed?
(12 points) | Solution.
Let $A_{n}$ be the point on the trunk at a distance of $n$ dm from the base. We will saw at points $A_{p}$ and $A_{q}, p<q$. To satisfy the triangle inequality, it is necessary and sufficient that each part is no longer than 19 dm:
$$
p \leq 19, \quad 21 \leq q \leq p+19
$$
Thus, the number of ways to choo... | 171 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,662 |
5. How many solutions in natural numbers $x, y$ does the inequality $x / 76 + y / 71 < 1$ have?
(12 points) | # Solution.
All solutions lie in the rectangle
$$
\Pi=\{0<x<76 ; \quad 0<y<41\}
$$
We are interested in the number of integer points lying inside Π below its diagonal $x / 76 + y / 41=1$. There are no integer points on the diagonal itself, since 76 and 41 are coprime. So we get half the number of integer points insi... | 1500 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,663 |
6. What is the maximum area that a rectangle can have, the coordinates of whose vertices satisfy the equation
$$
|y+1|\left(y^{2}+2 y+28\right)+|x-2|=9\left(y^{2}+2 y+4\right)
$$
and whose sides are parallel to the coordinate axes?
(12 points)
# | # Solution.
$$
|y+1|\left((y+1)^{2}+27\right)+|x-2|=9(y+1)^{2}+27
$$
Substitution: $x_{1}=y+1, y_{1}=x-2$.
Then $\quad\left|y_{1}\right|=-\left(\left|x_{1}\right|-3\right)^{3}$
The area of the rectangle does not change with this substitution. The area is calculated using the formula $S(x)=-4 x(x-3)^{3}, x \in[0 ; 3... | 34.171875 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,664 |
7. The inscribed circle of triangle $A B C$ with radius 3 touches side $A C$ at point $D$. On the extensions of sides $A C$ and $B C$ beyond point $C$, points $K$ and $L$ are taken respectively, such that angle $C K L$ is $30^{\circ}$. The lengths of segments $A K$ and $B L$ are equal to the semiperimeter of triangle $... | Solution. We will prove that $\angle OMC = 90^{\circ}$. Construct a circle $S$ with diameter $OC$. Denote the intersection point (different from $C$) of this circle with the circumcircle of triangle $ABC$ as $M_1$. Denote the intersection point (different from $C$) of circle $S$ with the line parallel to $KL$ as $M_2$.... | 0.875 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,665 |
8. Indicate the smallest value of the parameter $a$ for which the equation has at least one solution
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)+1=a+2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\frac{\pi x^{2}}{12}\right)$. | Solution.
Rewrite the equation as
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)-2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\pi-\frac{\pi x^{2}}{12}\right)=a-1$, or
$\sin \left(\pi-\frac{\pi x^{2}}{12}-\frac{\pi}{6} \sqrt{9-x^{2}}\right)=\frac{a-1}{2}$.
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,666 |
9. Rhombus $ABCD$ is the base of a pyramid with vertex $S$. All its lateral faces are inclined to the base plane at the same angle of $60^{\circ}$. Points $M, N, K$, and $L$ are the midpoints of the sides of rhombus $ABCD$. A rectangular parallelepiped is constructed on rectangle $MNKL$ as its base. The edges of the up... | Solution.
The height of the pyramid $SO = h$. The diagonals of the rhombus $AC = d_1, BD = d_2$.
The height of the parallelepiped is $h / 2$, and the sides of the base of the parallelepiped are $d_1 / 2$ and $d_2 / 2$.
The volume of the parallelepiped is
$$
V_{\Pi} = \frac{d_1}{2} \cdot \frac{d_2}{2} \cdot \frac{h}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,667 |
9. The lateral face of a regular triangular pyramid $S A B C$ is inclined to the base plane $A B C$ at an angle $\alpha=\operatorname{arctg} \frac{3}{4}$. Points $M, N, K$ are the midpoints of the sides of the base $A B C$. Triangle $M N K$ serves as the lower base of a right prism. The edges of the upper base of the p... | # Solution for Variant № 2 (11th grade, qualifying stage) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,668 |
1. A workshop produces items of types $A$ and $B$. For one item of type $A$, 10 kg of steel and 23 kg of non-ferrous metals are used, and for one item of type $B-70$ kg of steel and 40 kg of non-ferrous metals are used. The profit from selling one item of type $A$ is 80 thousand rubles, and for type $B-100$ thousand ru... | Solution. Let $x$ be the number of items of type $A$, and $y$ be the number of items of type $B$. Then the profit per shift is calculated by the formula $D=80 x+100 y$, with the constraints $10 x+70 y \leq 700$, $23 x+40 y \leq 642$, and $x$ and $y$ are non-negative integers.
 | Solution. $\quad x=\frac{3}{\sin t}, t \in\left(0 ; \frac{\pi}{2}\right)$
$$
\begin{gathered}
\frac{1}{\sin t}+\frac{1}{\cos t}=\frac{35}{12}, \quad 24(\sin t+\cos t)=35\left((\sin t+\cos t)^{2}-1\right) \\
z=\sin t+\cos t, 35 z^{2}-24 z-35=0 \\
z_{1}=-\frac{5}{7}, z_{2}=\frac{7}{5}, \sin t+\cos t=\frac{7}{5}
\end{gat... | 8.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,670 |
3. Solve the inequality $x^{2} \leq 2([\sqrt[3]{x}+0.5]+[\sqrt[3]{x}])$, where $[x]$ is the integer part of $x$, i.e., $[x]$ is the integer for which the inequality $[x] \leq x<[x]+1$ holds. In the answer, write the difference between the largest and smallest solutions of the inequality. | Solution. Let's make the substitution $t=\sqrt[3]{x}$. We have $t^{6} \leq 2([t+0.5]+[t])$. | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,671 | |
4. In how many ways can the number 210 be factored into the product of four natural numbers? The order of the factors does not matter.
$(12$ points) | # Solution.
Factorize 210 into a product of prime numbers: $2 \cdot 3 \cdot 5 \cdot 7$. Let's see how 4 prime
#### Abstract
divisors can be distributed among the desired factors
1) $4+0+0+0-1$ way.
2) $3+1+0+0-C_{4}^{3}=4$ ways.
3) $2+2+0+0-C_{4}^{2} / 2=3$ ways.
4) $2+1+1+0-C_{4}^{2}=6$ ways.
5) $1+1+1+1-1$ way... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,672 |
5. The sequence is defined recursively:
$x_{0}=0, x_{n+1}=\frac{\left(n^{2}+n+1\right) x_{n}+1}{n^{2}+n+1-x_{n}} . \quad$ Find $x_{8453}$. (12 points) | Solution.
$$
\text { We calculate } x_{1}=\frac{1}{1}=1, x_{2}=\frac{4}{2}=2, x_{3}=\frac{15}{5}=3 \text {, a hypothesis emerges: } x_{n}=n \text {. }
$$
Let's verify by induction:
$$
x_{n+1}=\frac{\left(n^{2}+n+1\right) n+1}{n^{2}+n+1-n}=\frac{n^{3}+n^{2}+n+1}{n^{2}+1}=n+1
$$
## Answer: 8453 | 8453 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,673 |
6. What is the greatest area that a rectangle can have, the coordinates of whose vertices satisfy the equation $|y-x|=(y+x+1)(5-x-y)$, and whose sides are parallel to the lines $y=x$ and $y=-x$? Write the square of the found area in your answer. $\quad(12$ points $)$
# | # Solution.
Substitution: $x_{1}=x+y, y_{1}=y-x$. This substitution increases all dimensions by a factor of $\sqrt{2}$. We have $\left|y_{1}\right|=\left(x_{1}+1\right)\left(5-x_{1}\right), \quad S\left(x_{1}\right)=4\left(x_{1}-2\right)\left(x_{1}+1\right)\left(5-x_{1}\right), x_{1} \in(2 ; 5)$. $S^{\prime}\left(x_{1... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,674 |
7. The inscribed circle of triangle $ABC$ with radius 2 touches side $AC$ at point $D$, and angle $C$ of this triangle is $\arcsin \frac{\sqrt{15}}{8}$. On the extensions of sides $AC$ and $BC$ beyond point $C$, points $K$ and $L$ are taken, respectively. The lengths of segments $AK$ and $BL$ are equal to the semiperim... | Solution. We will prove that $\angle OMC = 90^{\circ}$. Construct a circle $S$ with diameter $OC$. Denote the intersection point (different from $C$) of this circle with the circumcircle of triangle $ABC$ as $M_1$. Denote the intersection point (different from $C$) of circle $S$ with the line parallel to $KL$ as $M_2$.... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,675 |
8. Specify the greatest value of the parameter $p$ for which the equation has at least one solution
$2 \cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)-3=p-2 \sin \left(-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)$. (16 points) | # Solution.
Let's rewrite the equation as
$$
\cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)+\sin \left(2 \pi-\frac{\pi x^{2}}{6}\right) \sin \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)=\frac{p+3}{2} \text {, or }
$$
$\cos \left(2 \pi-\frac{\pi x^{2}}{6}-\frac{\pi}{3} \sqrt{9... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,676 |
9. The lateral face of a regular triangular pyramid $S A B C$ is inclined to the base plane $A B C$ at an angle $\alpha=\operatorname{arctg} \frac{3}{4}$. Points $M, N, K$ are the midpoints of the sides of the base $A B C$. Triangle $M N K$ is the lower base of a right prism. The edges of the upper base of the prism in... | # Solution.
The height of the pyramid $SO = h$. The side of the base of the pyramid $AC = a$.
The height of the prism is $3h/4$, and the sides of the base of the prism are $a/2$.
The area of triangle $MNK$:
$$
S_{MNK} = \frac{a^2 \sqrt{3}}{16}
$$
The area of triangle $FPR$:
$$
S_{FPR} = \frac{a^2 \sqrt{3}}{64}
$$... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,677 |
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events
$$
A=\{\text { There is a cat }\} \text { and } B=\{\text { There is a dog }\} .
$$
| | There is a dog | There is no dog |
| :--- | :---: | :---: |
| There is a cat | $a$ | $b$ |
| There is no cat | $c$ | $d$ |
These events ... | Answer: approximately 4272 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a... | 4272 | Other | math-word-problem | Yes | Yes | olympiads | false | 10,678 |
9. Solution. Suppose Olga Pavlovna has \( x \) liters of jam left, and Maria Petrovna has \( y \) liters of jam left. The numbers \( x \) and \( y \) are randomly and independently chosen from the interval from 0 to 1. We will consider that a random point with coordinates \((x; y)\) is selected from the unit square \( ... | Answer: 0.375.
Note. Other solution methods are possible.
## Grading Criteria
| Solution is complete and correct | 3 points |
| :--- | :--- |
| The solution contains correct reasoning for individual cases and an enumeration of these cases, and the formula for total probability is applied to them. However, the answer... | 0.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,679 |
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events $A=\{$ Has a card $\}$ and $B=\{$ Makes online purchases $\}$. These events are independent if the proportion of cardholders among online shoppers is the same as among those who do not make online purchases, that is,
| | Has a... | Answer: approximately 1796 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a... | 1796 | Other | math-word-problem | Yes | Yes | olympiads | false | 10,680 |
# 4. Calculator (6-9).
a) (1 pt.) On the calculator keyboard, there are digits from 0 to 9 and symbols for two operations (see figure). Initially, the display shows the number 0. You can press any keys. The calculator performs operations in the sequence of key presses. If the operation symbol is pressed several times ... | Solution. a) Note that at least one addition operation is performed, even if the Scholar entered only one number - thereby adding this number to zero.
Let $A$ be the event "the result is odd". The parity of the result is determined by the last addend. Let's explain this in more detail.
Suppose the penultimate number ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,682 |
5. Garland (1 b. 6-11). A string of 100 lights was hung on a Christmas tree in a row. Then the lights started switching according to the following algorithm: all lights turned on, after a second, every second light turned off, after another second, every third light switched: if it was on, it turned off and vice versa.... | Solution. Obviously, the bulb with number $n$ will remain on only if it has been switched an odd number of times, that is, if the number $n$ has an odd number of natural divisors. It is clear that only squares satisfy this condition: $n=1,4,9, \ldots, 100$. Thus, 10 bulbs out of 100 will remain on. Therefore, the proba... | 0.1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,683 |
6. Probabilistic Voting (6-9). In the final of a play competition for March 8, two plays made it to the finals. In the first play, $n$ students from 5th grade A participated, and in the second play, $n$ students from 5th grade B participated. At the play, $2n$ mothers of all $2n$ students were present. The best play is... | # Solution.
a) Let's call a mother confident if her child is playing in the best play. A confident mother will vote for the best play with a probability of 1. There are exactly $n$ confident mothers, so the best play will receive at least half of the votes - it will get, at the very least, the votes of all confident m... | 1-(\frac{1}{2})^{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,684 |
7. Mothers' Voting (9-11). In the final of the play competition for March 8, two plays made it to the final. The first play featured $n$ students from 5th grade A, and the second play featured $n$ students from 5th grade B. At the play, there were $2 n$ mothers of all $2 n$ students. The best play is chosen by the moth... | # Solution
a) Let's call a mother honest if she votes for the best play in any case. If a mother votes for her child, we will consider her dishonest. It is known that $n$ honest mothers will vote for the best play, so it will not lose for sure (half of the votes are definitely for it). The only case when the best play... | \frac{(2n)!-(n!)^{2}}{(2n)!} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,685 |
8. Selection of sailors for a submarine. ( $\mathbf{1}$ p. 6-9) A sailor can serve on a submarine if their height does not exceed 168 cm. There are four teams A, B, C, and D. All sailors in these teams want to serve on a submarine and have passed a strict selection process. The last selection is based on height. In tea... | # Solution.
Example of Team A: three sailors with heights of 160, 169, and 169 cm. The average height is 166 cm, but two out of three are not fit for service on a submarine. Therefore, in Team A, not necessarily half of the sailors are fit.
Example of Team B: two sailors with heights of 169 cm. None can serve on the ... | B | Other | math-word-problem | Yes | Yes | olympiads | false | 10,686 |
9. Right Triangle (7-11). The length of the hypotenuse of a right triangle is 3.
a) (1 pt.) The Absent-Minded Scholar calculated the variance of the side lengths of this triangle and found it to be 2. Could he have made a mistake in his calculations?
b) (2 pts.) What is the smallest standard deviation of the side len... | Solution. a) Let $a$ and $b$ be the legs of the triangle. Then $a^{2}+b^{2}=9$. We will find the variance of the set of side lengths. For this, we will use the formula $S^{2}=\overline{x^{2}}-\bar{x}^{2}$ - the mean of the squares minus the square of the mean:
$$
S^{2}=\frac{a^{2}+b^{2}+9}{3}-\left(\frac{a+b+3}{2}\rig... | ) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,687 |
10. The jury of the play competition (3 p. 10-11). The math teacher suggested changing the voting scheme for the play competition (see problem 7 a). In her opinion, out of all $2 n$ mothers, a jury of $2 m$ people should be randomly selected $(2 m \leq n)$. Find the probability that the best play will win under these v... | Solution. As before, we will consider a mother honest if she votes for the best play and dishonest if she votes for the play in which her child is acting, regardless of its quality.
Since there are $n$ dishonest mothers and the same number of honest ones, the probability that $q$ dishonest mothers and $2 m - q$ honest... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,688 |
11. Scattering function (2 p. 7-11). Given a numerical set $x_{1}, \ldots, x_{n}$. Consider the function $d(t)=\frac{\min _{i=1, \ldots, n}\left|x_{i}-t\right|+\max _{i=1, \ldots, n}\left|x_{i}-t\right|}{2}$.
a) Is it true that the function $d(t)$ takes its minimum value at a unique point, regardless of the set of num... | # Solution
a) Consider the set $\{2,4,7,11\}$. We will construct the function $d(t)$ for this set. First, we will plot the graphs of the functions
$$
y=|x-2|, y=|x-4|, y=|x-7|, y=|x-11| \text {. }
$$
Now we will plot the graphs of the functions $y=\min _{i=1, \ldots, n}\left|x_{i}-t\right|$ (red) and $y=\max _{i=1, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,689 |
12. A problem inspired by A.D. Sakharov's problem (7-11). Vasya, in a fit of rage, is cutting a rectangular sheet of paper with scissors. Every second, he cuts the first piece he comes across with a random straight cut into two parts.
a) (1 pt.) Find the expected number of sides of the polygon that will randomly fall ... | Solution. Let's solve problem b). Suppose that as a result of Vasya's activity, there are $p_{3}$ triangles, $p_{4}$ quadrilaterals, and so on, with the number of $m$-gons being $p_{m}$. Here, $m$ is the largest possible number of sides of the resulting polygon, and it doesn't even matter what $m$ is. Then the expected... | \frac{n+4k}{k+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,690 |
14. Postman (2 p. 7-11). On the street, there are $n$ houses. Every day, the postman goes to the post office, takes letters for the residents of one house, and delivers them. Then he returns to the post office, takes letters for the residents of another house, and delivers them again. He continues this process until he... | # Solution.
a) Let's introduce a coordinate line, on which we will place the houses on Pushkin Street with coordinates $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ ordered in ascending order.
Let $t$ be the coordinate of the post office on the number line.
. A fair die is rolled many times. Find the expected number of rolls made until the sum of all the points rolled reaches 2010. | Solution. Let $X_{n}$ be the number of rolls required to reach a sum of $n$. Introduce the random variable
$$
I_{k}=\left\{\begin{array}{l}
1, \text { if the first roll gives } k \text { points, } \\
0 \text { otherwise. }
\end{array}\right.
$$
Obviously, $\mathrm{E} I_{k}=0 \cdot \frac{5}{5}+1 \cdot \frac{1}{6}=\fra... | 574.761904 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,694 |
18. The Absent-Minded Scientist on the Bus (4 p. 6-11). In the bus, there are $n$ seats, and all tickets have been sold to $n$ passengers. The first to enter the bus is the Absent-Minded Scientist, who, without looking at the ticket, takes the first available seat. Subsequently, passengers enter one by one. If an enter... | Solution. First method. Let's number all passengers, starting with the Scientist, in the order in which they entered the bus. The last passenger has the number $n$. For simplicity, let's number the seats in the same way. Suppose all passengers except the last one have already entered and taken their seats. There is one... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,695 |
19. Strict Backgammon (5 p. 10-11). A fair die is rolled many times. It is known that at some point the sum of the points became exactly 2010. Find the expected number of rolls made by this point. | Solution. First method. Let event $A_{n}$ be "the sum of points is $n$", and $X_{n}$ be the number of rolls made to achieve this. We consider event $A_{n}$ as already having occurred, and we are interested in the expectation $\mathrm{E} X_{n}$. Let, as before,
$$
I_{k}=\left\{\begin{array}{l}
1, \text { if the first r... | 574.5238095 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,696 |
7. Solution. To increase the average score in both groups, it is necessary to transfer students from group A to group B whose average score is higher than the average in group B but lower than the average in group A. Two students have a score higher than the average in B but lower than the average in A: Kalinina and Si... | Answer: Kalinina and Sidorov need to be transferred from Group A to Group B.
Note: This solution may not be the only one.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 2 |
| Solution is generally correct, but it is not shown that both students need to be transferred, not just ... | Kalinina | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,697 |
9. Solution. Let's introduce the notations for the events:
$A=\{$ the first basket has an idea $\}, B=\{$ the second basket has an idea $\}$.
We need to find the probability of the event $A \cap B: \mathrm{P}(A \cap B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cup \dot{\boldsymbol{\phi}}$. From the condition, we find... | Answer: $1-2 p^{5}+(2 p-1)^{5}$.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Solution is generally correct, but an error has been made in transformations or in the transition to the probability of the opposite event. Possibly, an incorrect answer | 2 |
| The formula for... | 1-2p^{5}+(2p-1)^{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,698 |
7. Solution. To increase the average score in both groups, it is necessary to transfer students from group A to group B whose average score is higher than the average in group B but lower than the average in group A. Two students have a score higher than the average in B but lower than the average in A: Lopatin and Fil... | Answer: It is necessary to transfer Lapatin and Fylin from Group A to Group B.
Note: The solution may not be the only one.
| Evaluation Criterion | Score |
| :--- | :---: |
| The solution is complete and correct | 2 |
| The solution is basically correct, but it is not shown that both students need to be transferred, ... | ItisnecessarytotransferLapatinFylinfromGroupAtoGroupB | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,699 |
9. Solution. Let's introduce the notations for the events:
$A=\{$ an idea is in the first basket $\}, B=\{$ an idea is in the second basket $\}$.
We need to find the probability of the event $A \cap B: \quad \mathrm{P}(A \cap B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cup B)$. The probability that the Scientist wil... | Answer: $1-2(1-p)^{6}+(1-2 p)^{6}$.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Solution is basically correct, but there is an error in transformations or in the transition to the probability of the opposite event. Possibly, the wrong answer | 2 |
| The formula for the ... | 1-2(1-p)^{6}+(1-2p)^{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,700 |
9. Solution. To get a complete collection, it is necessary that among Vanya's 10 stickers, there are 6 that Varya is missing. The probability of this is
$$
\frac{C_{6}^{6} \cdot C_{12}^{4}}{C_{18}^{10}}=\frac{5}{442} \approx 0.011
$$ | Answer: approximately $0.011$.
## Grading Criteria
| Complete Solution | 3 points |
| :--- | :--- |
| Reasoning is correct, the necessary ratio or product of fractions is set up, but there is a computational error | 2 points |
| No solution or incorrect. | 0 points |
Variant 2 | 0.011 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,703 |
2. Solution. A one-way trip takes 2 hours and 40 minutes, so a round trip takes 5 hours and 20 minutes. Therefore, after a round trip, the driver must rest for at least one hour.
Let's call the drivers of the first, second, and third trips A, B, and C, respectively. Driver A returns to the station at 12:40. At this ti... | Answer: 4 drivers; at $21:30$.
## Grading Criteria
| Full Solution | 2 points |
| :--- | :--- |
| It is shown that there should be four drivers, but there is no explanation that the driver of the first trip was on duty at 16:10 | 1 point |
| No solution, or incorrect (even if the correct answer is given) | 0 points | | 4drivers;at21:30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,704 |
9. Solution. To get a complete collection, it is necessary that among the 9 coupons from Semy, there are 6 that Temy is missing. The probability of this is
$$
\frac{C_{6}^{6} \cdot C_{11}^{3}}{C_{17}^{9}}=\frac{3}{442} \approx 0.007
$$ | Answer: approximately $0.007$.
## Grading Criteria
| Complete Solution | 3 points |
| :--- | :--- |
| Reasoning is correct, the necessary ratio or product of fractions is set up, but a computational error is made | 2 points |
| No solution or incorrect solution. | 0 points | | 0.007 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,706 |
1. Peers (from 6th grade, 1 point). Vitya and Masha were born in the same year in June. Find the probability that Vitya is at least one day older than Masha. | Answer: $\frac{29}{60}$.
Solution. $\frac{1}{2} \cdot \frac{900-30}{900}=\frac{29}{60} \approx 0.483$. | \frac{29}{60} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,707 |
2. King Arthur's Round Table. Three knights are randomly seated on chairs arranged around the table at King Arthur's court. What is the probability that there will be an empty chair on both sides of each knight if:
a) there are a total of 7 chairs? (From 6th grade, 1 point).
b) there are a total of $n$ chairs? (From ... | Answer: a) 0.2; b) $\frac{(n-4)(n-5)}{(n-1)(n-2)}$.
Solution. The problem with seven chairs can be solved by enumeration. Let's solve the problem in the general case (part b). For $n<6$, the probability is 0. Consider the case $n \geq 6$. Suppose, for definiteness, that one of these knights is King Arthur himself, and... | \frac{(n-4)(n-5)}{(n-1)(n-2)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,708 |
3. Overcrowded carriages. The histogram (Fig. 1) shows the distribution of passenger carriages by the number of passengers (in $4 \%$ of carriages, passengers are from 10 to 19, in $6 \%$ of carriages - from 20 to 29, etc.). If a carriage has 60 passengers or more, it is called overcrowded.
a) (for 6th grade, 1 point)... | Answer: a) $40 \%$; b) approx. $49 \%$; c) cannot.
Solution. b) To make the share of passengers in overcrowded cars as small as possible, there should be as few passengers as possible in them, and as many as possible in non-overcrowded cars. Let the total number of cars be $N$. We find the maximum possible number of p... | )40;b)\approx49;) | Other | math-word-problem | Yes | Yes | olympiads | false | 10,709 |
5. Cards (from 7th grade, 2 points). When Vitya was in the first grade, he had a set of digits with 12 cards: two cards with the digit 1, two cards with the digit 2, and so on up to the digit 6. Vitya laid them out in a row on the table in a random order from left to right, and then removed the first one, the first two... | Answer: $\frac{1}{720}$.
Solution. The total number of sequences that Vitya could lay out from 12 cards is
$$
C_{12}^{2} C_{10}^{2} C_{8}^{2} C_{6}^{2} C_{4}^{2} C_{2}^{2}=66 \cdot 45 \cdot 28 \cdot 15 \cdot 6 \cdot 1
$$
Let's find how many sequences, after removing six cards according to Vitya's rule, result in 123... | \frac{1}{720} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,711 |
6. Vitya's Torments (from 7th grade, 2 points). Vitya has five math lessons in a week, one each day from Monday to Friday. Vitya knows that with a probability of $1 / 2$ the teacher will not check his homework during the school week, and with a probability of $1 / 2$ the teacher will check, but only during one of the m... | Answer: $\frac{1}{6}$.
Solution. Let's construct a tree for this random experiment (Fig.3).
Fig. 3
Event $A$ "homework is not checked until Thursday" means that either there will be no c... | \frac{1}{6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,712 |
7. Highway (from 7th grade, 3 points). A highway running from west to east intersects with $n$ equal roads, numbered from 1 to $n$ in order. Cars travel on these roads from south to north and from north to south. The probability that a car will approach the highway from each of these roads is $\frac{1}{n}$. Similarly, ... | Answer: $\frac{2 k n-2 k^{2}+2 k-1}{n^{2}}$.
Solution. The specified event will occur in one of three cases. | \frac{2kn-2k^{2}+2k-1}{n^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,713 |
8. The Lonely Ball (from 8th grade, 2 points). In 100 boxes, exactly 100 balls are placed randomly. What is the probability that the last box will contain only one ball? | Answer: approximately $0.370$.
Solution. Consider the placement of balls into boxes as a series of $n$ independent Bernoulli trials. A success will be considered as a ball landing in the last box. The probability of exactly one success is
$$
C_{n}^{1}\left(\frac{n-1}{n}\right)^{n-1} \cdot \frac{1}{n}=\left(\frac{n-1}... | 0.370 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,714 |
9. Two friendly series (from 8th grade, 2 points). Two football teams A and B play football equally well. The coaches agreed on two friendly matches. A team is awarded 2 points for a win, 1 point for a draw, and 0 points for a loss. The probability of a draw in each match is the same and equal to $p$.
The following ye... | Answer: No.
Solution. In each match, the probabilities of winning and losing for each team are $(1-p) / 2$. Both teams will accumulate equal points if both matches end in a draw or if one team wins one match and the other team wins the second match. The probability of this event can be represented as a function $f(p)$... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,715 |
10. Pills for Forgetfulness (from 8th grade, 3 points). In January, the doctor gave the Forgetful Scientist a pack of 10 pills for forgetfulness. The Scientist keeps the pills in a cabinet. When the Scientist is overwhelmed by a bout of forgetfulness (which happens several times a week at random moments), he opens the ... | Answer: approximately $0.1998$.
Solution. The Scholar cannot have more than two packages at any moment. Indeed, he orders a new package only when there is only one pill left, which means he has one package at that moment. Therefore, at any moment, he either has one package or two packages, one of which contains one pi... | \frac{1023}{5120}\approx0.1998 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,716 |
11. Three Triangles (8th grade, 3 points). Inside triangle $ABC$, a random point $M$ is chosen. What is the probability that the area of one of the triangles $ABM$, $BCM$, and $CAM$ will be greater than the sum of the areas of the other two? | Answer: $0.75$.
Solution. Draw the midlines $A_{1} B_{1}, B_{1} C_{1}$, and $C_{1} A_{1}$ in the triangle. For the area of triangle $A C M$ to be greater than half the area of triangle $A B C$, point $M$ must lie inside triangle $A_{1} B C_{1}$. Similarly, two other cases are considered. The desired probability is
$$... | 0.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,717 |
12. Stefan Banach ${ }^{1}$ and matches (8th grade, 4 points). An incredible legend says that one day in a small shop in the center of Lviv, Stefan Banach bought two boxes of matches and put them in his jacket pocket. Each box contained 60 matches.
When Banach needed a match, he would randomly pick a box and take a ma... | Answer: approximately 7.795.
Solution proposed by Olympiad participant Alexandra Nesterenko. Let there be $n$ matches in the box. Banach took a box (let's call it red for definiteness) and found it to be empty. Banach knows that he took the red box $n$ times, and the second (blue) box he could have taken $k$ times, wh... | 7.795 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,718 |
13. Clean Plane (from 9th grade, 3 points). The vacuum robot "Orthogonal" decided to vacuum the coordinate plane. It starts from point $O$, located at the origin, and travels a straight distance $X_{1}$. At the point it arrives at, the vacuum turns and describes a circle centered at the origin. After completing the cir... | Answer: $\pi n\left(d+a^{2}\right)$.
Solution. Exclude the circles from the vacuum cleaner's trajectory. It turns out that at each point it reaches after each step, the vacuum cleaner turns so that the direction of its movement forms a right angle with the direction to the origin (Fig. 7). Therefore, after the $n$-th ... | \pin(+^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,719 |
15. Minimum Sum (9th grade, 4 points). There are a lot of symmetric dice. They are thrown simultaneously. With some probability $p>0$, the sum of the points can be 2022. What is the smallest sum of points that can fall when these dice are thrown with the same probability $p$? | Answer: 337.
Solution. The distribution of the sum rolled on the dice is symmetric. To make the sum $S$ the smallest possible, the sum of 2022 should be the largest possible. This means that the sum of 2022 is achieved when sixes are rolled on all dice. Therefore, the total number of dice is $2022: 6=337$. The smalles... | 337 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,720 |
16. Extra Crocodiles (from 9th grade, 6 points). A chocolate egg manufacturer with a toy inside announced that a new collection is being released, featuring ten different crocodiles. The crocodiles are uniformly and randomly distributed among the chocolate eggs, meaning that in a randomly chosen egg, each crocodile can... | Answer: approximately 3.59.
Solution. Let's assume there are a total of $n$ crocodiles in the collection. We will number the exhibits in the order in which they appeared in the collection. For example, the crocodile with glasses was the first, so it is a crocodile of type 1. After some number of attempts, a crocodile ... | 3.59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,721 |
1. $(1 \mathbf{1}, 7-9)$ A and B are shooting at a shooting range, but they only have one six-chamber revolver with one bullet. Therefore, they agreed to take turns randomly spinning the cylinder and shooting. A starts. Find the probability that the shot will occur when the revolver is with A. | # Solution.
The shot occurs on an odd attempt (when the revolver is with A), and no shot occurs on the other attempts. Therefore, the desired probability is the sum
$$
\frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\ldots... | \frac{6}{11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,722 |
2. (9-11) Kolya and Zhenya agreed to meet at the metro in the first hour of the day. Kolya arrives at the meeting place between noon and one o'clock, waits for 10 minutes, and leaves. Zhenya does the same.
a) (1b) What is the probability that they will meet?
b) (1b) How will the probability of meeting change if Zheny... | # Solution.
a) Let's graphically represent the arrival times. We will plot Zhenya's arrival time on the X-axis and Kolya's arrival time on the Y-axis.
The figure outlined in black is the s... | \frac{11}{36},\frac{11}{36},\frac{19}{60} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,723 |
3. (1b, 8-11) The probability that a purchased light bulb will work is 0.95. How many light bulbs need to be bought to ensure that with a probability of 0.99, there are at least 5 working ones among them? | # Solution.
Let's take 6 light bulbs. The probability that at least 5 of them will work is the sum of the probabilities that 5 will work and 1 will not, and that all 6 will work, i.e.,
$$
C_{6}^{5} \cdot 0.95^{5} \cdot 0.05 + C_{6}^{6} \cdot 0.95^{6} = 6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672
$$
Let's take 7 ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,724 |
4. (1b, 7-9) A hunter has two dogs. One day, getting lost in the forest, he came to a fork in the road. The hunter knows that each dog will choose the way home with probability $p$. He decided to release the dogs one after the other. If both choose the same road, he will follow them; if they split, the hunter will choo... | # Solution.
The probability of choosing the correct path with one dog is $p$. The probability of choosing the correct path, if acting in the specified way, with two dogs is:
$$
\tilde{p}=p \cdot p+p(1-p) \cdot \frac{1}{2}+(1-p) p \cdot \frac{1}{2}=p \cdot p+2 \cdot \frac{1}{2} p(1-p)=p
$$
where $p \cdot p$ is the pr... | willnot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,725 |
5. (1b, 7-9) When studying a foreign language, the class is divided into two groups. The lists of the groups and the students' semester grades are given below. Can the English teacher transfer one student from the first group to the second group so that the average grade of students in both groups increases?
| | 1st ... | # Solution.
The person with a rating of 4 should be transferred, as 4 is below the average in the first group and above the average in the second.
## Answer: possible. | possible | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,726 |
6. (2b, 7-11) A marketing company decided to conduct a sociological study to find out what portion of the urban population primarily gets their news from radio broadcasts, what portion from television programs, what portion from the press, and what portion from the internet. For the study, it was decided to use a sampl... | # Solution.
The sample is not representative, as the two events:
$$
\begin{gathered}
A=\{y \text { of the person has an email }\} \\
\text { and } \\
B=\{\text { the person uses the Internet }\}
\end{gathered}
$$
are dependent. On the other hand, people who use the Internet mostly use it, among other things, for rea... | cannot | Other | math-word-problem | Yes | Yes | olympiads | false | 10,727 |
7. (2b, 8-11) In a box, there are 2009 socks - blue and red. Can there be such a number of blue socks that the probability of randomly pulling out two socks of the same color is equal to $0.5$? | # Solution.
Let there be $n+m$ socks in the box, of which $n$ are of one color and $m$ are of another. Then the probability of pulling out two socks of the same color is the sum of the probabilities $\mathrm{P}_{11}+\mathrm{P}_{22}$ of pulling out two socks of the first and second colors.
$$
\begin{gathered}
\mathrm{... | cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,728 |
8. (20,8-11) Aya and Vasya have 3 coins. On different sides of one coin are scissors and paper, on the sides of another coin are rock and scissors, and on the sides of the third coin are paper and rock. Scissors beat paper, paper beats rock, and rock beats scissors. First, Aya chooses a coin for herself, then Vasya, th... | # Solution.
Let's number the coins: the coin "Scissors-Paper" has number 1, the coin "Rock-Scissors" - 2, and the coin "Paper-Rock" - 3. The probabilities of winning for each coin when any pair is chosen are as follows:
| | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: |
| 1,2 | $1 / 4$ | $1 / 2$ | - |
| 1,3 | $1 / 2$... | yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,729 |
9. (20,9-11) Aся and Vasya are cutting rectangles out of graph paper. Vasya is lazy; he rolls a die once and cuts out a square with a side equal to the number of points that come up. Aся rolls the die twice and cuts out a rectangle with length and width equal to the numbers that come up. Whose expected area of the rect... | # Solution.
Let $A$ and $B$ be independent random variables taking values 1, 2, ..., 6 with equal probabilities. Then the expected value of the area of A's rectangle is $\mathrm{M} A B=\mathrm{M} A \cdot \mathrm{M} B$ (since $A$ and $B$ are independent). The expected value of the area of Vasya's rectangle is $\mathrm{... | Vasya | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,730 |
10. (9-11) On the exam, there are three trigonometry problems, two algebra problems, and five geometry problems. Vanya solves trigonometry problems with a probability of $p_{1}=0.2$, geometry problems with a probability of $p_{2}=0.4$, and algebra problems with a probability of $p_{3}=0.5$. To get a grade of three, Van... | # Solution.
a) Let the random variable $X$ be the number of trigonometry problems solved, $Y$ be the number of geometry problems solved, and $Z$ be the number of algebra problems solved. Then the sum $U=X+Y+Z$ is the total number of problems solved. We want to find the probability $\mathrm{P}(U \geq 5)$. We will use t... | geometry | Other | math-word-problem | Yes | Yes | olympiads | false | 10,731 |
11. (3b,9-11) In the conditions of a chess match, the winner is declared as the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation. | # Solution.
Let $\mathrm{X}$ be the number of successful games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two successful games, denoting a win by the first participant as 1 and a win by the second participant as 2: $11, 12... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,732 |
12. (3b, 8-11) In belly-pushing competitions, the chances of opponents winning are proportional to their body masses. Jumbo weighs more than Dumbo, and Pink weighs less than Bonk. There can be no draws in a match. Jumbo and Dumbo must take turns pushing Pink and Bonk. Which of the following events is more likely:
$$
\... | # Solution.
Let's denote the masses of the opponents as follows: $u, j, p, b$ corresponding to the first letters of their names. Then $u>j, b>p$. The probability that Yumbo defeats only Pink is the product of the probabilities that he defeats Pink and loses to Bonk, i.e., $\frac{u}{u+p} \cdot \frac{b}{u+b}$. Similarly... | the\probabilities\\equal | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,733 |
13. (3b, 8-11) A coin is tossed 10 times. Find the probability that two heads never appear consecutively.
# | # Solution.
## First Method:
The total number of outcomes when tossing a coin ten times is $2^{10}$.
Let's find the number of combinations where no two heads appear consecutively. If there are no heads at all, then the sequence consists of ten tails, and there is only one such sequence. If there is one head, then th... | \frac{9}{64} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,734 |
14. (3b, 9-11) n people are lined up back-to-back. Taller people block the view of shorter ones, making them invisible. What is the expected number of people who are visible? | # Solution.
Let $X_{n}$ be the random variable "The number of visible people among $n$ people." Suppose we have calculated $\mathrm{M} X_{n-1}$. What happens when we add the $n$-th person to the end of the queue? With probability $\frac{1}{n}$, this person is taller than all the others. And with probability $\frac{n-1... | 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,735 |
15. (4b, 8-11) In Anchuria, a checkers championship is being held in several rounds. The days and cities for the rounds are determined by a draw. According to the championship rules, no two rounds can take place in the same city, and no two rounds can take place on the same day. Among the fans, a lottery is organized: ... | # Solution.
In an $8 \times 8$ tour table, you need to select $k$ cells such that no more than one cell is chosen in any row or column. The value of $k$ should be chosen to maximize the number of combinations.
The number of combinations is given by $C_{8}^{k} A_{8}^{k}=\frac{8!\cdot 8!}{(8-k)!\cdot(8-k)!\cdot k!}$, w... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,736 |
16. (4b, 8-11) In a building with $n$ floors and two staircases leading from the first to the last floor, there is a door on each intermediate landing between any two floors (you can go from the staircase to the floor even if the door is locked). The caretaker decided that too many open doors is bad and locked exactly ... | # Solution.
Doors leading from floor $k$ to the next floor will be called the doors of floor $k$ $(k=1,2, \ldots, n-1)$.
If both doors of floor $k$ are closed (on both staircases), it is impossible to get from floor $k$ to floor $k+1$, and consequently, it is impossible to get from the first to the last floor. If at ... | \frac{2^{n-1}}{C_{2n-2}^{n-1}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,737 |
17. (5b, 10-11) In the center of a rectangular billiard table that is 3 meters long and 1 meter wide, a billiard ball is placed. It is struck with a cue in a random direction. After the strike, the ball stops after traveling exactly 2 meters. Find the expected number of reflections from the table edges. | # Solution.
Let's draw a grid with a cell of 3 m $\times 1$ m. The central cell of the grid is the table, highlighted in light green. We need to extend the grid in all directions so that it can accommodate a circle with a radius of 2 m centered at the center of the table $O$.
Allow the ball to "pass through the edges... | \frac{2}{\pi}(3\arccos\frac{1}{4}-\arcsin\frac{3}{4}+\arccos\frac{3}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,738 |
19. (6b, 9-11) A die is rolled six times. Find the expected value of the number of different faces that appear.
# | # Solution.
Let $\zeta_{i}$ be a random variable equal to 1 if the face with $i$ points has appeared at least once and 0 if it has never appeared. Then the number of different faces that have appeared is $\zeta=\zeta_{1}+\ldots+\zeta_{6}$. Transitioning to expectations, we get: $\mathrm{M} \zeta=6 \mathrm{M} \zeta_{1}... | \frac{6^{6}-5^{6}}{6^{5}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,739 |
4. Cities of Anchuria (from 6th grade, 1 point). In Anchuria, there is only one river, Rio-Blanco, which originates somewhere in the mountains and flows into the ocean, and there are only five cities: San-Mateo, Alasan, Coralio, Alforan, and Solitas.
In the map of Anchuria, the city names are missing, and the cities a... | Solution. Alasan is the only city not located on the banks of the Rio Blanco, but gets its water from wells. Therefore, Alasan has the number 2.
The lower a city is along the river, the lower its elevation above sea level. Only the "Elevation above sea level" value is required from all available statistics.
Answer: 1... | 1-Alforan,2-Alasan,3-Solitas,4-San-Mateo,5-Coralio | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,740 |
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