problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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4. Answer: $P(A)=\frac{1285}{9000}=\frac{257}{1800}$ | Solution.
Any integer $A$, divisible by 14 and ending in 2, has the form $A=14(5k+3), k \geq 0, k \in Z$. If it ends in 32, it can be represented as $A=100m+32, m \geq 0, m \in Z$. Combining these two conditions, we arrive at the equation $100m+32=70k+42 \rightarrow 10m-7k=1$.
Its general solution is $\left\{\begin{a... | \frac{257}{1800} | Logic and Puzzles | math-word-problem | Not a problem | Yes | olympiads | false |
5. Answer: $a=13 t+4, t \in Z$ | Solution
The first equation of the system is a linear equation with integer solutions, so $\left\{\begin{array}{c}{[x]=3 t+2} \\ {[y]=2 t-1, t \in Z}\end{array}\right.$. Then for each $t \in Z$, the solutions to the first equation are pairs $(x ; y)$ for which $x \in[3 t+2 ; 3 t+3), y \in[2 t-1 ; 2 t)$ (squares with s... | 13+4,\inZ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
6. $\lambda=5: 9, \alpha=\operatorname{arctg} 5$
$x=\frac{2}{3} ; \operatorname{tg} \beta=1 ; \cos \alpha=\frac{1}{\sqrt{26}} ; \cos \beta=\frac{1}{\sqrt{2}} ; \frac{\cos \alpha+\cos \beta}{1+\cos \alpha}=\frac{\sqrt{13}+1}{\sqrt{26}+1} ; \frac{(x+1)^{2}}{4}=\frac{25}{36}$ | Answer: $S_{\text {EABMN }}: S_{\text {EABCD }}=\frac{(x+1)^{2}(\cos \alpha+\cos \beta)}{4(1+\cos \alpha)}=\frac{25(\sqrt{13}+1) \sqrt{2}}{36(\sqrt{26}+1)}$
Solution
Notation $E M: E C=E N: E D=x, V=V_{\text {EABCD }}, S=S_{E A B C D}, \sigma=S_{E A B}, \omega=S_{A B C D}, H=E O, a=A B$
\sqrt{2}}{36(\sqrt{26}+1)} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
1. Answer: $n=1,2,3,4,6,12$ | Solution
Transforming the equation:
$1+3+5+\ldots+(2 n-1)=n^{2} \rightarrow n^{2} \log _{2} x=5040=2^{4} \cdot 3^{2} \cdot 5 \cdot 7$
The equation has a rational solution if 5040 is divisible by $n^{2}$. This occurs when $n=2,4,3,6,12$ | 1,2,3,4,6,12 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: 1$\left.)(|x|+|y|+|z|)_{\max }=\pi ; 2\right)$ six triples $(x ; y ; z)= \pm\left(\frac{\pi}{3} ; \frac{\pi}{3} ; \frac{\pi}{3}\right), \pm\left(\frac{\pi}{3} ;-\frac{\pi}{3} ; \pm \frac{\pi}{3}\right)$ | Solution.
The expression $|x|+|y|+|z| \leq \pi$, for all triples $(x ; y ; z)$, satisfying the second equation of the system (on the sphere), since any of the planes $\pm x \pm y \pm z=a$ for $a>\pi$ is at a distance greater than $\frac{\pi}{\sqrt{3}}$ from the origin and has no common points with the sphere. If $x, y... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
4. Answer: $P(A)=\frac{3}{7}$ | Solution
The area of the section by a plane containing at least three vertices of a cube can take three possible values. The first $s_{1}=\frac{a^{2} \sqrt{3}}{2}$, if no two vertices belong to the same edge of the cube (an equilateral triangle with side $a \sqrt{2}$). In the first variant, $a=2$, so $s_{1}=2 \sqrt{7}... | \frac{3}{7} | Logic and Puzzles | math-word-problem | Not a problem | Yes | olympiads | false |
5. Answer: $n=4 m+1, m \in Z$ | Solution.
From the first condition $\cos \left(n^{2} \pi+\frac{(n+1) \pi}{2}\right)=1 \rightarrow(-1)^{n^{2}} \cos \left(\frac{(n+1) \pi}{2}\right)=1$. If $\cos \left(\frac{(n+1) \pi}{2}\right)=1$, then $\frac{(n+1) \pi}{2}=2 \pi m \rightarrow n=4 m-1, m \in Z$. Then $(-1)^{n^{2}}=-1, \forall m$ and the equation has n... | 4+1,\inZ | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
6. Answer: $L_{\max }=\frac{55}{6}$
Option 0
Two parallel lines, the distance between which is $h$, intersect a rectangle of dimensions $a \times b$ at an angle $\alpha$ to its side. Find the maximum possible value of the sum of the lengths of the segments of these lines that belong to the rectangle. | Solution. For definiteness, on Fig. $a \leq b$.
Fig 1
In Fig 1, segments $A M$ and $C N$ are drawn at an angle $\alpha_{1}$ to side $A D$. The value of $\alpha_{1}$ is determined by the co... | \frac{55}{6} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. Answer: $x=2^{17+3 t}, y=3^{13-2 t}, t \in Z$ | Solution
If $\log _{\sqrt{2}} x=\frac{p}{q}$ is an irreducible fraction, $q \neq 1$, then $x=2^{\frac{p}{2 q}}$ can only be rational if $\frac{p}{2 q}=k$ is an integer. Then $p=2 k q$ and $p$ is divisible by $q$, making the fraction $\frac{p}{q}$ reducible. Thus, $q=1$ and $p=2 k=\log _{\sqrt{2}} x, k \in Z$. Then $x=... | 2^{17+3t},\quad3^{13-2t},\quad\in\mathbb{Z} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: 1) $n=1+6 t, t \geq 0, t \in Z \quad$ 2) $n=3+6 t, t \geq 0, t \in Z$ | Solution
$\left\{\begin{array}{c}\cos \pi x \geq 0 \\ 10 \cos ^{2}(\pi x)=4 \cos (\pi x)-\cos (2 \pi x)\end{array} \rightarrow\left\{\begin{array}{c}\cos \pi x \geq 0 \\ 12 \cos ^{2}(\pi x)-4 \cos (\pi x)-1=0\end{array} \rightarrow \cos (\pi x)=\frac{1}{2}\right.\right.$
The numbers $n$ of the terms of the progressio... | n=1+6t,\geq0,\inZ\quador\quadn=3+6t,\geq0,\inZ | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: $\hat{a}_{\max }=9^{2} \cdot 11^{2} \cdot 10^{8}$ | Solution
Let $m_{1}$ and $m_{2}$ be the sum of the digits in the representation of numbers $a_{1}$ and $a_{2}$ respectively. By the condition, $m_{1}+m_{2}=36$.
Denote by $r_{1}$ and $r_{2}$ the remainders of the division of $m_{1}$ and $m_{2}$ by 9, i.e., $m_{1}=9 k_{1}+r_{1}, m_{2}=9 k_{2}+r_{2}$.
Then $9\left(k_{... | 9^{2}\cdot11^{2}\cdot10^{8} | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
4. Answer: $P(A)=\frac{1}{12}$ | Solution
Let $p_{k}, k=1,2,3,4,5$ be the probability that the die covers the digit $k$ in one throw.
Let $\sigma$ be the area of the side face, $\omega$ be the area of the base, $4 \sigma \cos \alpha=\omega$, $p_{1}=p_{2}=p_{3}=p_{4}=\lambda \sigma, p_{5}=\lambda \omega$
By the condition, $\quad \frac{p_{4}}{p_{5}}=... | \frac{1}{12} | Combinatorics | math-word-problem | Not a problem | Yes | olympiads | false |
Variant 1. $p=\frac{1}{2}, q=\frac{2}{3}, r=\frac{1}{3}, z=\frac{2}{7} V_{E M N K T}=\frac{1}{2} p r(q+z) \cdot V_{E A B C D}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3}\left(\frac{2}{3}+\frac{2}{7}\right) V_{\text {EABCD }}=\frac{5}{63} V_{\text {EABCD }}$ | Answer: 5:58
## Answers and solutions | \frac{5}{63} | Algebra | math-word-problem | Incomplete | Problem not solved | olympiads | false |
1. Answer: $f_{\max }=6$ | Solution
$z=27-x-y \rightarrow f=\log _{3} x+\log _{3} y+\log _{3}(27-x-y)$
Fig 1
The figure shows the location of points $a$ and $b$, giving the maximum value of $f$ at $x=x^{*}, y=y^{*},... | 6 | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: $x= \pm \frac{\pi}{6}+\pi k, x=\frac{\pi}{2}+\pi k, k \in Z$ | Solution
Transformation:
$\cos (\arcsin (\sin x))=\cos \left(\frac{\pi}{2}-\arccos (\cos 2 x)\right)=\cos (\arcsin (\cos 2 x))$ | \\frac{\pi}{6}+\pik,\frac{\pi}{2}+\pik,k\inZ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer 1) $x=4, y=11$ 2) $x=7, y=16$ | Solution
Factorization $1232=2^{4} \cdot 7 \cdot 11$.
Solution of the equation $5 x-3 y+13=0 \rightarrow\left\{\begin{array}{c}x=3 t-2 \\ y=5 t+1,1 \leq t \leq 246, t \in Z\end{array}\right.$
Numbers $x \neq 1$ and $y \neq 1$ are coprime and each of them is divisible by one of the numbers 2, 7, or 11.
Case 1. $x=2 ... | 4,117,16 | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
4. Answer: $s=10$ and $s=11$ | Solution
Let $x, y, z$ be the numbers rolled on the first, second, and third dice. The random variable $s=x+y+z$ takes values from 3 to 18. The total number of outcomes of the experiment is $n=6^{3}$. Let $m(s)$ denote the number of trials (the number of different triples $(x; y; z)$) that favor the event $s=x+y+z$. W... | =10=11 | Logic and Puzzles | math-word-problem | Not a problem | Yes | olympiads | false |
5. Answer: 1) $S=\frac{86}{3}$;
$\left(b \in(-\infty ; k-5] \cup[1-2 k ;+\infty), \forall k \in\left[-2 ;-\frac{1}{3}\right]\right.$
2) $b \in(-\infty ; k-5] \cup[k+2 ;+\infty), \forall k \in\left(-\frac{1}{3} ; 2\right]$ $b \in(-\infty ; 1-2 k] \cup[k+2 ;+\infty), \forall k \in(2 ; 5]$ | Solution
Notation: $f(x, y)=y-k x-b$ A line has common points with the triangle if it has a common point with at least one of its sides.
Case 1. The line intersects side $A C$
$$
f\left(x_{A}, y_{A}\right) \cdot f\left(x_{C}, y_{C}\right) \leq 0 \rightarrow(2+k-b)(-5+k-b) \leq 0
$$
In Figure 1, the strip between th... | Algebra | math-word-problem | Incomplete | Yes | olympiads | false | |
6. Answer: $p=2 R\left(1+4 \sin \frac{\pi}{8}\right)=2 R(1+2 \sqrt{2-\sqrt{2}})$ | Solution
Figure 1
Let $A B C D E$ be the desired pentagon, and $M N$ be the diameter parallel to $A E$. We will prove that $A B = B C = C D = D E$. In triangle $A B C$, the sum of the squar... | 2R(1+2\sqrt{2-\sqrt{2}}) | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
3. Solution. Since $\left(1-\frac{1}{k^{2}}\right)=\left(1-\frac{1}{k}\right)\left(1+\frac{1}{k}\right)=\frac{k-1}{k} \cdot \frac{k+1}{k}$, then
$$
\begin{aligned}
& \left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{2021^{2}}\right)\left(1-\frac{1}{2022^{2}}\right)=... | Answer: $\frac{2023}{4044}$. | \frac{2023}{4044} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solution. First, let's determine the values that $x$ and $y$ can take:
$$
1-\sqrt{x-1}=\sqrt{y-1}
$$
from which $\left\{\begin{array}{c}x \geq 1, \\ 1-\sqrt{x-1} \geq 0 .\end{array} \rightarrow x \in[1,2]\right.$. Similarly, $y \in[1,2]$.
Solve the equation:
$$
1-\sqrt{x-1}=\sqrt{y-1}, \rightarrow y-1=1-2 \sqrt{... | Answer: $(x+y)_{\max }=3$ when $x=2, y=1$ or $x=1, y=2, (x+y)_{\min }=\frac{5}{2}$ when $x=y=\frac{5}{4}$. | (x+y)_{\max}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Solution. Let $M A=1=a, \quad M C=2=b . \quad$ Angle $\angle$ AMC is $120^{\circ}$, so by the cosine rule
$$
A C^{2}=a^{2}+b^{2}+a b
$$
In the diagram, equal angles are marked with the same numbers. Triangles $A N M$ and $B N C$ are similar (by two equal angles), so $\frac{B C}{A M}=\frac{N C}{M N}$. Triangles $C ... | Answer: $A B=\sqrt{7} ; M N=\frac{2}{3}$. | AB=\sqrt{7};MN=\frac{2}{3} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
3. Calculate the value of the product $\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdot\left(1-\frac{1}{16}\right) \cdot \ldots \cdot\left(1-\frac{1}{2023^{2}}\right)$. | Answer: $\frac{1012}{2023}$. | \frac{1012}{2023} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the greatest and least values of the expression $2 x+y$, if $(x ; y)$ are related by the relation $\sqrt{x-1}+\sqrt{y-4}=2$. For which $(x ; y)$ are they achieved? | Answer: $(2 x+y)_{\max }=14$ when $x=5, y=4,(2 x+y)_{\min }=\frac{26}{3}$ when $x=\frac{13}{9} y=\frac{52}{9}$. | (2x+y)_{\max}=14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the arc $A C$ of the circumcircle of an equilateral triangle $A B C$, a point $M$ is chosen such that the lengths of the segments $M A$ and $M C$ are 2 and 3, respectively. The line $B M$ intersects the side $A C$ at point $N$. Find the length of the segment $M N$ and the side of the triangle $A B C$. | Answer: $A B=\sqrt{19} ; M N=\frac{6}{5}$. | AB=\sqrt{19};MN=\frac{6}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the product $\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdot\left(1-\frac{1}{16}\right) \cdot \ldots \cdot\left(1-\frac{1}{2021^{2}}\right)$. | Answer: $\frac{1011}{2021}$. | \frac{1011}{2021} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the greatest and least values of the expression $x-2 y$, if $(x ; y)$ are related by the relation $\sqrt{x-2}+\sqrt{y-3}=3$. For which $(x ; y)$ are they achieved? | Answer: $(x-2 y)_{\max }=5$ when $x=11, y=3,(x-2 y)_{\min }=-22$ when $x=2, y=12$. | (x-2y)_{\max}=5when11,3,(x-2y)_{\}=-22when2,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the arc $A C$ of the circumcircle of an equilateral triangle $A B C$, a point $M$ is chosen such that the lengths of the segments $M A$ and $M C$ are 3 and 4, respectively. The line $B M$ intersects the side $A C$ at point $N$. Find the length of the segment $M N$ and the side of the triangle $A B C$. | Answer: $A B=\sqrt{37} ; M N=\frac{12}{7}$. | AB=\sqrt{37};MN=\frac{12}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the product $\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdot\left(1-\frac{1}{16}\right) \cdot \ldots \cdot\left(1-\frac{1}{2020^{2}}\right)$. | Answer: $\frac{2021}{4040}$. | \frac{2021}{4040} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the greatest and least values of the expression $2 x+3 y$, if $(x ; y)$ are related by the relation $\sqrt{x-3}+\sqrt{y-4}=4$. For which $(x ; y)$ are they achieved? | Answer: $(2 x+3 y)_{\min }=37.2$ when $x=\frac{219}{25}, y=\frac{264}{25}, \quad(2 x+3 y)_{\max }=66$ when $x=3, y=$ 20. | (2x+3y)_{\}=37.2when\frac{219}{25},\frac{264}{25},\quad(2x+3y)_{\max}=66when3,20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the arc $A C$ of the circumcircle of an equilateral triangle $A B C$, a point $M$ is chosen such that the lengths of the segments $M A$ and $M C$ are 4 and 5, respectively. The line $B M$ intersects the side $A C$ at point $N$. Find the length of the segment $M N$ and the side of the triangle $A B C$. | Answer: $AB=\sqrt{61}; MN=\frac{20}{9}$.
Criteria for checking works, 9th grade
Preliminary round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics
# | AB=\sqrt{61};MN=\frac{20}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The number $A$ in decimal form is represented as $A=\overline{7 a 631 b}$, where $a, b$ are non-zero digits. The number $B$ is obtained by summing all distinct six-digit numbers, including $A$, that are formed by cyclic permutations of the digits of $A$ (the first digit moves to the second position, the second to th... | 2. Solution. The sum of the digits of number $A$ and the numbers obtained from $A$ by cyclic permutations of its digits is $a+b+17$. After summing these numbers (there are 6 of them), in each digit place of number $B$ we get
$a+b+17$, so $B=(a+b+17)\left(10^{5}+10^{4}+10^{3}+10^{2}+10+1\right)=(a+b+17) \cdot 111111$. ... | 796317 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For the polynomial $p(x)=2x+1$, find a polynomial $q(x)$ of the first degree such that $p^{2}(q(x))=q\left(p^{2}(x)\right)$ for any $x$. | 3. Solution. Let $q(x)=a x+b$. Then
$$
p^{2}(q(x))=(2 q(x)+1)^{2}=4(a x+b)^{2}+4(a x+b)+1=4 q^{2}(x)+4 q(x)+1=4 a^{2} x^{2}+(8 a b+4 a) x+4 b^{2}+4 b+1
$$
and
$q\left(p^{2}(x)\right)=a\left(4 x^{2}+4 x+1\right)+b=4 a x^{2}+4 a x+a+b$.
By the condition of the problem $p^{2}(q(x))=q\left(p^{2}(x)\right)$. Equating th... | q(x)=x | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For which natural numbers $n$ does the equality $\operatorname{GCD}(6, n)+\operatorname{GCD}(8,2 n)=10$ hold? | 4. Solution. Note that the possible values of $GCD(6, n)$ are $1,2,3,6$, and the possible values of $GCD(8, n)$ are $1,2,4,8$. Their sum, equal to 10, is possible in two cases.
Case 1. $\left\{\begin{array}{l}G C D(8,2 n)=8 \\ G C D(6, n)=2\end{array}\right.$. From the first equation, we get $n=4 k$, and from the seco... | n=12t+4,n=12t+6,n=12t+8,\geq0,\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a rectangle, a segment $A B$ of length $l=4$, shorter than the lengths of its sides, is positioned such that its ends lie on the sides of the rectangle. As point $A$ makes a complete revolution around the rectangle, it travels a path equal to its perimeter. During this, point $C$ - the midpoint of segment $A B$, ... | 5. Solution. From the conditions of the problem (the length of segment $AB$ is less than the lengths of the sides of the rectangle), it follows that points $A$ and $B$ are either on the same side of the rectangle or on adjacent sides of the rectangle. If points $A$ and $B$ move along the same side of the rectangle, poi... | L(4-\pi) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Answer: For $a= \pm 2$ the solutions are $x_{1}= \pm 1, x_{2}= \pm 4$, for $a= \pm 8$ the solutions are $x_{1}= \pm 1, x_{2}= \pm 4$ (the signs of $x$ match the sign of $a$) | Solution.
When $a=0$, the equation has one integer solution $x=0$.
When $a \neq 0$, the equation has solutions $x_{1}=\frac{8}{a}$ and $x_{2}=\frac{a}{2}$, which coincide when $a^{2}=16 \rightarrow a= \pm 4$.
The solution $x_{1}$ is an integer if $a= \pm 1, \pm 2, \pm 4, \pm 8$. The solution $x_{2}$ is an integer if... | For\=\2,\x_{1}=\4,\x_{2}=\1;\for\=\8,\x_{1}=\1,\x_{2}=\4 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: $\frac{22}{3}$ when $p=-1$.
переведено как:
3. Answer: $\frac{22}{3}$ when $p=-1$. | Solution.
Let's discard the condition of the integer nature of $p$ and consider the deviation $\left|\frac{19 p-3}{p-2}-5\right|$ as a function of $p$: Transformation: $f(p)=\frac{19 p-3}{p-2}-5=7 \cdot \frac{2 p+1}{p-2}$, a fractional-linear function. Its graph is shown in Fig. 1.
+1=0 \Leftrightarrow \\
& \Leftrightarrow 4 x^{2}-4 x+4 x-4 x \sin (x \pi+y)+1=0 \Leftrightarrow \\
& \Leftrightarrow(2 x-1)^{2}+4 ... | Answer: $S_{\min }=2 \pi$. | 2\pi | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Let the real numbers $x_{1}, x_{2}$, and $x_{3}$ be the roots of the equation $3 x^{3}-p x^{2}+28 x-p=0$, taking into account their possible multiplicities. Then $3 x^{3}-p x^{2}+28 x-p \equiv 3\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=$
$$
=3 x^{3}-3 x^{2}\left(x_{1}+x_{2}+x_{3}\right)+3 x\left(... | The smallest value $\frac{4}{9}$ is achieved when $p=16$. | \frac{4}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Consider some intermediate step in Kuzia's movement. If she is at point $A$ at this step, the probability of being in $A$ on the next step is zero. If, however, she is at any of the remaining points, $B, C$ or $D$, the probability of being in $A$ on the next step is $1 / 3$, since from each such point there are thre... | Answer: $\quad P(A)=\frac{3^{2019}+1}{4 \cdot 3^{2019}}$. | \frac{3^{2019}+1}{4\cdot3^{2019}} | Algebra | proof | Yes | Yes | olympiads | false |
6. Let point $O$ be the center of the circle. Then $O A \perp P, O B \perp Q$, $O A=O B=1$.
By the condition, the chord $A B$ subtends an arc of the circle at $60^{\circ}$, so $\measuredangle A O B=60^{\circ}$.
Furthermore, $A F=F B$, since these are segments of tangents drawn to the given circle from one point.
![]... | Answer: $m_{B}=\sqrt{3}-\frac{1}{2}$. | \sqrt{3}-\frac{1}{2} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. Each athlete participating in the marathon can be categorized into at least one of three categories: cheerful, trained, or beginners. It is known that every sixteenth cheerful athlete is a beginner, and one-fifth of the beginners are cheerful. One-fifth of the trained athletes are cheerful, and only one-tenth of the... | Answer: 130 athletes. | 130 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
2. On the plane, there are quadrilaterals, the coordinates $(x ; y)$ of the vertices of which satisfy the equation $16 x^{2}-8 x \sin (2 x \pi+3 y)+1=0$. Find the minimum possible value of the areas of such quadrilaterals. | Answer: $\quad S_{\min }=\frac{\pi}{3}$. | \frac{\pi}{3} | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
3. The real numbers $x_{1}, x_{2}$, and $x_{3}$ are the three roots of the equation $x^{3}-3 x^{2}+2(1-p) x+4=0$, considering their possible multiplicities. Find the smallest value of the expression $\left(x_{1}-1\right)^{2}+\left(x_{2}-1\right)^{2}+\left(x_{3}-1\right)^{2}$ under these conditions. For which $p$ is it ... | The smallest value 6 is achieved when $p=1$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Flea Kuzya can make jumps from each vertex of a regular tetrahedron $A B C D$ to the three adjacent vertices, and the choice of these vertices is random and equally probable. Kuzya started jumping from vertex $A$ and, after 2024 jumps, ended up not in vertex $A$. What is the probability that this could happen? | Answer: $\quad P(A)=\frac{3^{2024}-1}{4 \cdot 3^{2023}}$. | \frac{3^{2024}-1}{4\cdot3^{2023}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the plane, there are quadrilaterals, the coordinates $(x ; y)$ of the vertices of which satisfy the equation $9 x^{2}-6 x \cos \left(\frac{3 \pi x}{2}+2 y\right)+1=0$. Find the minimum possible value of the areas of such quadrilaterals. | Answer: $\quad S_{\min }=\frac{2 \pi}{3}$. | \frac{2\pi}{3} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. The real numbers $x_{1}, x_{2}$, and $x_{3}$ are the three roots of the equation $2 x^{3}-(p-4) x^{2}-(2 p-1) x-p+8=0$, considering their possible multiplicities. Find the smallest value of the expression $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ under these conditions. For which $p$ is it realized? | Answer: the minimum value $\frac{417}{64}$ is achieved at $p=\frac{15}{4}$. | \frac{417}{64} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The flea Kuzya can make jumps from each vertex of a regular tetrahedron $A B C D$ to the three adjacent vertices, and the choice of these vertices is random and equally probable. Kuzya started jumping from vertex $A$ and, after 2019 jumps, ended up at vertex $B$. With what probability could this have happened? | Answer: $\quad P(A)=\frac{3^{201}+1}{4 \cdot 3^{201}}$. | \frac{3^{201}+1}{4\cdot3^{201}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+1944=0$ has positive integer roots? | Answer: $\quad 108+24=132$ pairs. | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A circle of radius 3 touches line $P$ at point $A$ and line $Q$ at point $B$ such that the chord $A B$ subtends an arc of $60^{\circ}$. Lines $P$ and $Q$ intersect at point $F$. Point $C$ is located on ray $A F$, and point $D$ is on ray $F B$ such that $A C=B D=4$. Find the length of the median of triangle $C A D$, ... | Answer: $\quad m_{A}=2 \sqrt{3}+\frac{3}{2}$. | 2\sqrt{3}+\frac{3}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Quadrilaterals are located on a plane, the coordinates $(x ; y)$ of the vertices of which satisfy the equation $36 x^{2}-12 x \cos (3 \pi x+4 y)+1=0$. Find the minimum possible value of the areas of such quadrilaterals. | Answer: $S_{\min }=\frac{\pi}{6}$. | \frac{\pi}{6} | Geometry | math-word-problem | Incomplete | Incomplete | olympiads | false |
3. The real numbers $x_{1}, x_{2}$, and $x_{3}$ are the three roots of the equation $x^{3}-(p+4) x^{2}+(4 p+5) x-4 p-5=0$, considering their possible multiplicities. Find the smallest value of the expression $\left(x_{1}+1\right)^{2}+\left(x_{2}+1\right)^{2}+\left(x_{3}+1\right)^{2}$ under these conditions. For which $... | Answer: the minimum value $\frac{257}{16}$ is achieved at $p=-\frac{5}{4}$. | \frac{257}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Kuzya the flea can make jumps from each vertex of a regular tetrahedron \(ABCD\) to the three adjacent vertices, with the choice of these vertices being random and equally probable. Kuzya started jumping from vertex \(A\) and, after 2018 jumps, ended up at vertex \(C\). What is the probability that this could happen... | Answer: $\quad P(A)=\frac{3^{2018}-1}{4 \cdot 3^{2018}}$. | \frac{3^{2018}-1}{4\cdot3^{2018}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+432=0$ has positive integer roots | Answer: $\quad 78+20=98$ pairs. | 98 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A circle of radius 4 touches line $P$ at point $A$ and line $Q$ at point $B$ such that the chord $A B$ subtends an arc of $60^{\circ}$. Lines $P$ and $Q$ intersect at point $F$. Point $C$ is located on ray $F A$, and point $D$ is on ray $B F$ such that $A C = B D = 5$. Find the length of the median of triangle $C A ... | Answer: $\quad m_{A}=\frac{5 \sqrt{3}}{2}-2$.
## On-site Selection Round of the "Rosatom" Olympiad in Regions, Autumn
2019
11th grade, set 1
# | \frac{5\sqrt{3}}{2}-2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. By the condition
$$
T(t)=\frac{270-s(t)}{s(t) / t}=\frac{t(270-s(t))}{s(t)}=C>1, t \in[0.5 ; 1]
$$
Then $s(t)=\frac{270 t}{t+C}$ on this interval. The speed of movement
$$
\begin{aligned}
& v(t)=s^{\prime}(t)=\frac{270 C}{(t+C)^{2}}=60 \text { when } t=1 \text {, i.e. } \\
& \qquad 2 c^{2}-5 c+2=0 \rightarrow C_{... | Answer: 1) 90 km; 2) 86.4 km/hour | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The following cases are possible.
Case 1. $\cos x=1, x=2 \pi k, k \in Z, \cos x y=1$
$$
\cos (x / 2)=\cos (\pi k)=(-1)^{k}
$$
Case 1.a $k=2 m, x=4 \pi m$
$$
15 y^{2}-y-6=0 \rightarrow\left\{\begin{array}{c}
y_{1}=2 / 3 \\
y_{2}=-3 / 5
\end{array}\right.
$$
For $y=2 / 3$
$$
\begin{aligned}
& \cos \left(\frac{2 ... | # Answer:
$$
\left\{\begin{array}{l}
x=12 \pi t \\
y=2 / 3
\end{array},\left\{\begin{array}{l}
x=20 \pi t \\
y=-3 / 5
\end{array},\left\{\begin{array}{c}
x=6 \pi(2 t-1) \\
y=-2 / 3
\end{array},\left\{\begin{array}{c}
x=10 \pi(2 t-1) \\
y=3 / 5
\end{array}, t \in \mathbb{Z}\right.\right.\right.\right.
$$ | \begin{pmatrix}12\pi,\quad2/3\\20\pi,\quad-3/5\\6\pi(2-1),\quad-2/3\\10\pi(2-1),\quad3/5\end{pmatrix},\quad\in\mathbb{Z} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. If $m+2019 n$ and $n+2019 m$ are divisible by $d$, then the number
$$
2019(m+2019 n)-(n+2019 m)=(2019^2-1) n
$$
is also divisible by $d$. If $n$ is divisible by $d$, and $m+2019 n$ is divisible by $d$, then $m$ is divisible by $d$ and the numbers $m$ and $n$ are not coprime. Therefore, $d$ divides the number
$$
2... | Answer: $d_{\text {min }}=101$. | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The discriminant of a quadratic trinomial
$$
D=a^{2}-4 a+12=(a-2)^{2}+8>0
$$
and therefore the equation always has two distinct roots. We will use the conditions for the placement of the roots of a quadratic trinomial. In the case of a positive discriminant, the roots of
$$
f(x)=x^{2}+p x+q
$$
lie to the right o... | Answer: $P(A)=\frac{2}{3}$. | \frac{2}{3} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. Solving the equation with respect to the parameter, we get
$$
a=\frac{x^{3}+6 x-6}{3(x-1)}
$$
The derivative
$$
a^{\prime}(x)=\frac{2 x^{3}-3 x^{2}}{3(x-1)^{2}}=\frac{x^{2}(2 x-3)}{3(x-1)^{2}}
$$
has only one point of sign change (from negative to positive) at $x=3 / 2$. Therefore, this is the only extremum (min... | Answer: $a=17 / 4 ; x=-3$ and $x=3 / 2$. | \frac{17}{4};-3 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
1. Kostya is making a car trip from point A to point B, which are 320 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Kostya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (h... | Answer: 1) 128 km; 2) 38.4 km/h. | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+2024 n$ and $n+2024 m$ have a common prime divisor $d>7$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=17$.
For example,
$$
\begin{aligned}
& m=16, n=1 \rightarrow 2024 m+n=2024 \cdot 16+1=32385=17 \cdot 1905 \\
& m+2024 n=16+2024=17 \cdot 120
\end{aligned}
$$ | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A random variable $a$ is uniformly distributed on the interval $[-4 ; 4]$. Find the probability that at least one of the roots of the quadratic equation $x^{2}-2 a x-a-4=0$ does not exceed 1 in absolute value. | Answer: $P(A)=\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For what values of $a$ does the equation
$$
x^{3}-12(a-1) x+4(a+2)=0
$$
have exactly two roots? Find these roots. | Answer: 1) $a=2$; 2) $x_{1}=-4, x_{2}=2$. | =2;x_{1}=-4,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the side $A C$ of triangle $A B C$, there is a point $D$ such that $A D: A C=1: 3$, and $B D+2 B C=4 A B$. The inscribed circle of the triangle with center at point $O$ intersects $B D$ at points $M$ and $N$. Find the angle $M O N$. | Answer: $2 \arccos \frac{1}{6}$. | 2\arccos\frac{1}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Vasya is making a car trip from point A to point B, which are 360 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Vasya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (hou... | Answer: 1) 120 km; 2) 115.2 km/h. | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the system $\left\{\begin{array}{c}\sin (x y) \cdot \cos x=1 \\ 8 y^{2}=2 y \sin (x / 2)+15\end{array}\right.$. | Answer: $\left\{\begin{array}{c}x=\pi(4 t+1) \\ y=\frac{3}{2}\end{array},\left\{\begin{array}{c}x=\pi(4 t+3) \\ y=-\frac{3}{2}\end{array}, t \in Z\right.\right.$ | {\begin{pmatrix}\pi(4+1)\\\frac{3}{2}\end{pmatrix},{\begin{pmatrix}\pi(4+3)\\-\frac{3}{2}\end{pmatrix},\inZ..}} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+1941 n$ and $n+1941 m$ have a common prime divisor $d>8$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=97$.
For example,
$$
\begin{aligned}
& m=96, n=1 \rightarrow 1941 m+n=1941 \cdot 96+1=97 \cdot 1921 \\
& m+1941 n=96+1941=97 \cdot 21
\end{aligned}
$$ | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The random variable $a$ is uniformly distributed on the interval $[-2 ; 1]$. Find the probability that all roots of the quadratic equation $a x^{2}-x-4 a+1=0$ in absolute value exceed 1. | Answer: $P(A)=\frac{7}{9}$. | \frac{7}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For what values of $a$ does the equation
$$
x^{3}+27(a+1) x+18(a-1)=0
$$
have exactly two roots? Find these roots. | Answer: 1) $a=-2$; 2) $x_{1}=-3, x_{2}=6$. | =-2;x_{1}=-3,x_{2}=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Danya is making a car trip from point A to point B, which are 300 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Danya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (hou... | Answer: 1) 180 km; 2) 48 km/h. | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the system $\left\{\begin{array}{c}\sin x \cdot \sin (x y)=-1 \\ 6 y^{2}=y \operatorname{tg}(x / 2)+1\end{array}\right.$. | Answer: $\left\{\begin{array}{c}x=\frac{3 \pi}{2}(4 t+1), \\ y=\frac{1}{3},\end{array} t \in Z\right.$. | {\begin{pmatrix}\frac{3\pi}{2}(4+1),\\\frac{1}{3},\end{pmatrix}\inZ.} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+1947 n$ and $n+1947 m$ have a common prime divisor $d>9$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=139$.
For example,
$$
\begin{aligned}
& m=138, n=1 \rightarrow 1947 m+n=1947 \cdot 138+1=139 \cdot 1933 \\
& m+1947 n=138+1947=139 \cdot 15
\end{aligned}
$$ | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The random variable $a$ is uniformly distributed on the interval $[-1 ; 4]$. Find the probability that both roots of the quadratic equation $a x^{2}+4 x+a-3=0$ are negative. | Answer: $P(A)=\frac{1}{5}$. | \frac{1}{5} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. For what values of $a$ does the equation
$$
x^{3}+48(a+2) x+32(a-1)=0
$$
have exactly two roots? Find these roots. | Answer: 1) $a=-3$; 2) $x_{1}=-4, x_{2}=8$. | =-3;x_{1}=-4,x_{2}=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the side $A C$ of triangle $A B C$, there is a point $D$ such that $A D: A C=2: 5$, and $B D+2 B C=3 A B$. The inscribed circle of the triangle with center at point $O$ intersects $B D$ at points $M$ and $N$. Find the angle $M O N$. | Answer: $2 \arccos \frac{1}{5}$.
Examples of tasks from the database of the remote qualifying round of the "Rosatom" Olympiad, 11th grade
The database of the remote qualifying round of the "Rosatom" Olympiad (which is only for 11th-grade students) contains more than 300 problems with numerical answers (which are the ... | 2\arccos\frac{1}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The acute angle of triangle $ABC$ is $30^{\circ}$. Find the maximum possible value of the ratio $r: R$, where $r$ and $R$ are the radii of the inscribed and circumscribed circles.
## Answers and Solutions
Problem 1 Answer: For $x=3 k, k>0, k \in Z \rightarrow$ GCD $\left(P_{1} ; P_{2}\right)=3$, for $x \neq 3 k, k... | Solution
Note that for admissible $x$, the value $\operatorname{tg} x \neq 0$. Divide both sides of the equation by $\operatorname{tg}^{2} x$:
$$
\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=4 \rightarrow \frac{\sin ^{2} x}{\cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}=4 \rightarrow \frac{\left(\cos ^{2} x-\sin ^{... | notfound | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
2. Let's consider 6 different cases
Case 1. $b_{5}=x_{2}, b_{7}=x_{1}, b_{9}=x_{3}$ or $b_{5}=x_{3}, b_{7}=x_{1}, b_{9}=x_{2}$
Then from the condition of the progression $x_{2} \cdot x_{3}=x_{1}^{2} \rightarrow \frac{a(a+2)}{4}=1 \rightarrow a^{2}+2 a-4=0 \rightarrow\left[\begin{array}{l}a_{1}=-1-\sqrt{5} \\ a_{2}=-1... | Solution
$a=2 R \sin \alpha, b=\frac{a \sin \beta}{\sin \alpha}, c=\frac{a \sin (\alpha+\beta)}{\sin \alpha}, r=\frac{b c \sin \alpha}{a+b+c}=\frac{a \sin \beta \sin (\alpha+\beta)}{\sin \al... | 2\sqrt{2+\sqrt{3}}+\sqrt{3}-2 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. The lengths of the sides of a parallelogram are 3 and 2. The bisectors of all its internal angles limit a polygon on the plane. Find the ratio of its area to the area of the parallelogram.
## Answers and solutions
Problem 1 Answer: in 16 min. | Solution $v_{A}, v_{k}, v_{c}$ - the speeds of the bus, Kolya, and Sasha respectively, $T$ - the required time.
The distance between Kolya and Sasha at the moment the bus stops: $2\left(v_{A}+v_{k}\right)$. The speed of their approach: $v_{C}-v_{k}, v_{C}=3 v_{k}, v_{A}=5 v_{C}=15 v_{k}-$ conditions of the problem. $T... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. $\frac{\sqrt{13+\sqrt{48}}-1}{\sqrt{7-\sqrt{24}}+1}=\frac{\sqrt{12}}{\sqrt{6}}=\sqrt{2}$.
Problem 3 . Answer: $4031: 4029$ | Solution.
$S_{2}=2 a_{1}+d, S_{1}=a_{1} \rightarrow S_{2}: S_{1}=\frac{2 a_{1}+d}{a_{1}}=\frac{8}{3} \rightarrow \frac{d}{a_{1}}=\frac{2}{3}$
$\frac{a_{2015}}{a_{2014}}=\frac{a_{1}+d \cdot 2014}{a_{1}+d \cdot 2013}=\frac{3+2 \cdot 2014}{3+2 \cdot 2013}=\frac{4031}{4029}$ | \frac{4031}{4029} | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
Problem 4. Answer: $\left\{\begin{array}{l}x= \pm 3 \\ y=2\end{array}\right.$ | Solution
Factorize the left side: $\left(x^{2}-y\right)\left(x^{2}-2 y\right)=35$. In integers, this is possible in the following eight cases | {\begin{pmatrix}\3\\2\end{pmatrix}.} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $\frac{(a-b)^{2}}{2 a b}=1: 12$ | Solution:
In the figure, a parallelogram $ABCD$ with an acute angle $2\alpha$ and a quadrilateral $MNPQ$ formed by the bisectors of the internal angles are shown. Since $\alpha + \beta = 90^... | \frac{(-b)^2}{2ab}=1:12 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$ with parallel sides $A D$ and $B C$, a line $L$ parallel to $A D$ is drawn, intersecting sides $A B$ and $C D$ at points $M$ and $N$ respectively. It is known that quadrilaterals $A M N D$ and $M B S N$ are similar, and the sum of the lengths of sides $A D$ and $B C$ is no more th... | Solution: $v$ - the speed of the escalator's movement, $W \downarrow=2 v, W \uparrow=1.5 v$ - the speeds of Petya's movement relative to the ladder going down and up respectively, $S$ - the length of the escalator. | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
3. Comparison: $T_{1}: T_{2}=5: 3$
## Task 2 Answer: 10 candies | Solution
If Petya pulls out 11 or more candies, the condition of the problem cannot be met. Indeed, with the set: 3 yellow + 8 red = 11 candies, it can happen that 2 candies remain in yellow and red wrappers, and the condition of the problem is not met. If Petya pulls out 10 candies, then no less than 3 yellow or red ... | notfound | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
Task 1. Answer: 6 hours. | Solution.
Using a bicycle reduces the operation time. For example, if Kolya starts by getting on the bicycle, he should ride it until passing it to Maksim at point $D$, which lies on segment ... | 6 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 4. Answer: $a \in[-1 ; 0)$ | Solution. Transformation:
$$
\left\{\begin{array} { c }
{ \operatorname { tg } x + \operatorname { tg } y = 1 } \\
{ \operatorname { ctg } x + \operatorname { ctg } y = a }
\end{array} \rightarrow \left\{\begin{array} { c }
{ \operatorname { tg } x + \operatorname { tg } y = 1 } \\
{ \frac { \operatorname { tg } x +... | \in[-1;0) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $P(A)=\frac{207}{512}$. | Solution. The equation $a x^{2}+3 x+2=0$ has two negative roots if
$$
\left\{\begin{array}{c}
D=9-8 a>0 \\
a>0
\end{array} \rightarrow a \in\left(0 ; \frac{9}{8}\right)\right.
$$
$$
P(A)=\in... | \frac{207}{512} | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 6. Answer: $|L|_{\text {min }}=\frac{31}{2}=15.5$.
## Variant 0
The base $A B C$ of the pyramid $A B C D$ lies on the plane $S$. Point $M$ is located on the extension of the line $A B$ beyond point $B$ such that $M B: A B=\lambda$. Point $N$ is the midpoint of the edge $D C$. The curve $L$, connecting points ... | Answer: $|L|_{\min }=\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}$.
Solution. Let the curve $L$ be the desired one. It cannot fail to intersect the line $A B$ at some point $E$. Then the part of the curve from point $M$ to point $E$ is a straight line segment. Point $F$ is the midpoint of edge $A C$. The geometric loc... | \frac{}{2}\sqrt{4\lambda^{2}+6\lambda+3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Answer: $a \in(-1 ; 1)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided translation is already in English and retains the original format. The note is for your information and should not be inc... | Solution. $\left\{\begin{array}{c}\operatorname{tg} x+\operatorname{tg} y=a-1 \\ \operatorname{ctg} x+\operatorname{ctg} y=a+1\end{array} \rightarrow\left\{\begin{array}{l}\operatorname{tg} x+\operatorname{tg} y=a-1 \\ \operatorname{tg} x \cdot \operatorname{tg} y=\frac{a-1}{a+1}\end{array}\right.\right.$
Existence of... | \in(-1;1) | Algebra | math-word-problem | Not a problem | Yes | olympiads | false |
Problem 5. Answer: $P(A)=\frac{5}{18}$. | Solution. The equation has roots and they are less than 1 if
$$
\left\{\begin{array}{c}
D=a^{2}-16 a \geq 0 \\
x_{b}=\frac{a}{2}0
\end{array} \rightarrow a \in\left(-\frac{1}{3} ; 0\right]\right.
$$
 \cup[2 ; 3)$ | Solution. $\left\{\begin{array}{c}\operatorname{tg} x+\operatorname{tg} y=a \\ \operatorname{ctg} x+\operatorname{ctg} y=2\end{array} \rightarrow\left\{\begin{array}{l}\operatorname{tg} x+\operatorname{tg} y=a \\ \operatorname{tg} x \cdot \operatorname{tg} y=\frac{a}{2}\end{array}\left(^{*}\right)\right.\right.$
Exist... | \in(-2;0)\cup[2;3) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $P(A)=\frac{1}{50}$. | Solution. The equation $x^{2}+3 x+a=0$ has a root in the interval $[1 ; 2]$, if $a \in[-10 ;-4]$.
$$
P(A)=\int_{-5}^{-4} p(x) d x=\frac{1}{50}
$$ | \frac{1}{50} | Other | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 6. Answer: $|L|_{\min }=\frac{43}{2}=21.5$ | Solution. $\lambda=\frac{5}{2}, a=\sqrt{43} \rightarrow|L|_{\min }=\left.\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}\right|_{\lambda=2.5, a=\sqrt{43}}=\frac{43}{2}=21.5$
## Option 4 | 21.5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 3. Answer: 13.
Problem 4. Answer: $a \in(-\infty ;-\sqrt{2}) \cup[\sqrt{2} ; 2)$ | Solution. $\left\{\begin{array}{c}\operatorname{tg} x+\operatorname{tg} y=2 a \\ \operatorname{ctg} x+\operatorname{ctg} y=a\end{array} \rightarrow\left\{\begin{array}{c}\operatorname{tg} x+\operatorname{tg} y=2 a \\ \operatorname{tg} x \cdot \operatorname{tg} y=2\end{array}\right.\right.$
Existence of solutions ( ( )... | \in(-\infty;-\sqrt{2})\cup[\sqrt{2};2) | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
Problem 6. Answer: $|L|_{\min }=\frac{21}{2}=10.5$. | Solution. $\lambda=\frac{3}{2}, a=\sqrt{21} \rightarrow|L|_{\min }=\left.\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}\right|_{\lambda=1.5, a=\sqrt{21}}=\frac{21}{2}=10.5$. | 10.5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
1. The figure shows a cake in the form of a rectangle with sides $3 \times 3$. A straight line (cut) is also depicted,
intersecting the maximum number of squares, which is 5. The maximum numb... | Answer: 14 pieces and 2 cuts. | 14 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. For Vasya's birthday, a cake with a square base of $4 \times 4$ dm was bought, divided into 16 equal pieces with square bases of $1 \times 1$ dm. Vasya decided to cut the cake with a knife along a straight line so that the maximum number of separate pieces, not necessarily equal, would be formed. What is the maximum... | Answer: 23 pieces and 3 cuts. | 23 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
2. Find the integer values of $a$ for which the equation $(x-2 a+1)^{2}-2 x+4 a-10=0$ has no solutions on the interval $[-1 ; 7]$. | Answer: $(-\infty ;-3] \cup[6 ;+\infty)$. | (-\infty;-3]\cup[6;+\infty) | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
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