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7. The distance from the point of intersection of the diameter of a circle with a chord of length 18 cm to the center of the circle is $7 \mathrm{~cm}$. This point divides the chord in the ratio $2: 1$. Find the radius.
$$
A B=18, E O=7, A E=2 B E, R=?
$$
|
Solution: $2 B E \cdot B E=(R-7)(7+R)$
$$
A E \cdot B E=D E \cdot E C, \quad A E+B E=18, \quad B E=6
$$
$$
2 \cdot 6 \cdot 6=\left(R^{2}-7^{2}\right), R^{2}=72+49=121=11^{2}
$$
Answer: $R=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the equation $7^{-x}-3 \cdot 7^{1+x}=4$.
|
Solution: $7^{-x}-3 \cdot 7^{1+x}=4, \frac{1}{7^{x}}-21 \cdot 7^{x}=4,-21\left(7^{x}\right)^{2}-4 \cdot 7^{x}+1=0$,
$\left(7^{x}\right)=\frac{4 \pm \sqrt{16+84}}{-42}=\frac{4 \pm 10}{-42}=\left[\begin{array}{l}-\frac{14}{42}=-\frac{1}{3} \text {, not valid } \\ \frac{1}{7}\end{array}, x=-1\right.$. Answer: $x=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Calculate $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}$.
|
Solution: $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}=A, A^{2}=12-2 \sqrt{36-20}=4, A= \pm \sqrt{4}=2$. Answer: 2 .
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the equation $9 \cdot 3^{2 x-1}+3^{x}-30=0$.
|
Solution: $9 \cdot 3^{2 x-1}+3^{x}-30=0,3 \cdot 3^{2 x}+3^{x}-30=0$, $3^{x}=\frac{-1 \pm \sqrt{1+360}}{6}=\frac{-1 \pm 19}{6}=\left[\begin{array}{l}3 \\ -\frac{10}{3} \text {, not valid }\end{array}, x=1\right.$.
Answer: $x=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Solve the equation $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$.
|
Solution: $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$, Domain of Definition $\frac{x^{2}-16}{x-3} \geq 0, x>3 \Rightarrow x \geq 4$,
$\sqrt{x^{2}-16}+\sqrt{x^{2}-9}=7, x^{2}-16=t, \sqrt{t}+\sqrt{t+7}=7, t+2 \sqrt{t(t+7)}+t+7=49$,
$2 \sqrt{t(t+7)}=42-2 t, \sqrt{t^{2}+7 t}=21-t, t \leq 21, t^{2}+7 t=21^{2}-42 t+t^{2}, 49 t=21^{2}$,
$t=\frac{3^{2} \cdot 7^{2}}{49}=9, x^{2}-16=9, x^{2}=25, x= \pm 5, x=5$.
Verification: $\sqrt{\frac{9}{2}}+\sqrt{8}=\frac{7}{\sqrt{2}}, \frac{3}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{7}{\sqrt{2}}$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
the given edge. (12 points)
Solution. Let the area of each of the given faces be \( S \). If the face lies in the plane of projection, then the projection of the tetrahedron is equal to the area of this face \( \Pi = S \).
When rotated by an angle \( 0 < \varphi < 30^\circ \), the area of the projection is \( \Pi = S \cos \varphi < S \).
When rotated by an angle \( 30^\circ < \varphi < 90^\circ \), the area of the projection is
\[
\Pi = S \cos \varphi + S \cos \psi = S \cos \varphi + S \cos \left(\pi - \frac{\pi}{3} - \varphi\right) = S \cos \varphi + S \cos \left(\frac{2\pi}{3} - \varphi\right).
\]
\[
\Pi' = S \left(-\sin \varphi + \sin \left(\frac{2\pi}{3} - \varphi\right)\right), \quad \Pi' = 0 \text{ when } \varphi = \frac{\pi}{3}.
\]
The maximum of the function in the considered interval is achieved at
\[
\varphi = \frac{\pi}{3}, \quad \Pi = 2 S \cos \left(\frac{\pi}{3}\right) = 2 S \cdot \frac{1}{2} = S.
\]
When rotated by an angle \( 90^\circ < \varphi < 120^\circ \), the area of the projection is \( \Pi = S \cos \left(\frac{\pi}{2} - \varphi\right) = S \sin \varphi < S \).
When \( \varphi = \frac{2\pi}{3} \), the area \( \Pi = S \).
|
Answer. $\Pi=S=\frac{1}{2} \cdot \sqrt{2}^{2}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given 50 numbers. It is known that among their pairwise products, exactly 500 are negative. Determine the number of zeros among these numbers.
|
Solution. Let $m, n$ and $p-$ be the number of negative, zero, and positive numbers among the given 50 numbers. Then from the condition of the problem we have: $m+n+p=50$ and $m \cdot p=500$. It follows that $m$ and $p$ are divisors of 500, the sum of which does not exceed 50. Among all divisors of the number 500, only the pair 20 and 25 has this property, i.e. $m+p=45$, hence $n=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation
$$
2(x-6)=\frac{x^{2}}{(1+\sqrt{x+1})^{2}}
$$
|
Solution 1. Rewrite the equation as $2(x-6)(1+\sqrt{x+1})^{2}=x^{2}$ and make the substitution $\sqrt{x+1}=y$. Then $x=y^{2}-1$ and $y \geq 0$. After the substitution, we get $2\left(y^{2}-7\right)(y+1)^{2}=$ $\left(y^{2}-1\right)^{2}$. Factor the right side as $(y+1)^{2}(y-1)^{2}$, move it to the left, and factor out $(y+1)^{2}$. We get $(y+1)^{2}\left(2 y^{2}-14-(y-1)^{2}\right)=(y+1)^{2}\left(y^{2}+\right.$ $2 y-15)=(y+1)^{2}(y-3)(y+5)=0$. This equation has three solutions $y_{1}=3, y_{2}=$ $-1, y_{3}=-5$. The last two do not satisfy the condition $y \geq 0$, so they are extraneous. If $y=3$, then $x=8$. Clearly, this solution is valid.
Solution 2. Transform the equation to the form
$$
2(x-6)(x+2+2 \sqrt{x+1})-x^{2}=0
$$
and denote the expression on the left side by $f(x)$. We have
$$
f(x)=2 x^{2}-8 x-24+4(x-6) \sqrt{x+1}-x^{2}=x^{2}-8 x-24+4(x-6) \sqrt{x+1}
$$
We need to solve the equation $f(x)=0$. It is clear that all roots of this equation satisfy the condition $x>6$ (otherwise, the left side of the original equation is non-positive, while the right side is positive). Find the derivative $f^{\prime}(x)$ :
$$
f^{\prime}(x)=(2 x-8)+4 \sqrt{x+1}+\frac{2(x-6)}{\sqrt{x+1}}
$$
It is clear that for $x>6$ each term in the last sum is non-negative. Therefore, the function $f(x)$ is increasing on the interval $(6,+\infty)$. Hence, the equation $f(x)=$ 0 can have no more than one solution. One solution $x=8$ can be easily found by trial.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many positive numbers are there among the 2014 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}$, $\sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$
|
Solution. For $n>3$ we have
$10^{n}-1000=10^{3}\left(10^{n-3}-1\right)=25 \cdot 40 \cdot\left(10^{n-3}-1\right)$.
Since $10^{n-3}-1$ is divisible by 9, it follows that $10^{n}-1000$ is divisible by 360. Therefore, all terms of the sequence, starting from the fourth, coincide with $\sin 1000^{\circ}=\sin \left(3 \cdot 360^{\circ}-80^{\circ}\right)=\sin \left(-80^{\circ}\right)<0$. Thus, the positive terms in the sequence are $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. Mom baked a cake for Anya's birthday, which weighs an integer number of grams. Before decorating it, Mom weighed the cake on digital scales that round the weight to the nearest ten grams (if the weight ends in 5, the scales round it down). The result was 1440 g. When Mom decorated the cake with identical candles, the number of which was equal to Anya's age, the scales showed 1610 g. It is known that the weight of each candle is an integer number of grams, and if you put one candle on the scales, they will show 40 g. How old can Anya be? List all possible answers and explain why there are no others. (20 points)
|
Answer: 4 years.
Solution: From the condition, it follows that the cake weighed from 1436 to 1445 grams before decoration, and from 1606 to 1615 grams after decoration. Therefore, the total weight of the candles is between 161 and 179 grams. Since the weight of each candle is given as 40 grams, the weight of one candle is between 36 and 45 grams. If Anya is less than 4 years old, the total weight of the candles is no more than $45 \cdot 3 = 135$ grams, and if she is more than 4 years old, the total weight of the candles is no less than $5 \cdot 36 = 180$ grams. Neither of these values is possible. Therefore, Anya is exactly 4 years old.
Criteria: Correctly determined the weight limits of the cake - 3 points.
Correctly determined the weight limits of all candles - 2 points.
Correctly determined the weight limits of one candle - 3 points.
Correctly verified only one of the inequalities $n \leqslant 4$ or $n \geqslant 4-4$ points. All these criteria are cumulative.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. Timofey placed 10 grid rectangles on a grid field, with areas of $1, 2, 3, \ldots, 10$ respectively. Some of the rectangles overlapped each other (possibly completely, or only partially). After this, he noticed that there is exactly one cell covered exactly once; there are exactly two cells covered exactly twice; there are exactly three cells covered exactly three times, and exactly four cells covered exactly four times. What is the maximum number of cells that could be covered at least five times? The area of a grid rectangle is the number of cells it contains. Each rectangle lies on the field exactly along the grid cells. (20 points)
|
Answer: 5 cells.
Solution. Estimation. There are a total of $1+2+\ldots+10=55$ cell coverings. The ten cells described in the condition are covered in total exactly $1 \cdot 1+2 \cdot 2+3 \cdot 3+4 \cdot 4=30$ times. There remain $55-30=25$ cell coverings. Therefore, the number of cells covered at least five times is no more than $25 / 5=5$.
Example. Let's assume that all rectangles have the form $1 \times n$. We will combine them into rectangles of area $15=3+4+8$, $14=1+6+7$, and $12=2+10$. There are still two rectangles of area 5 and 9 left. We will overlay all five rectangles so that their leftmost cells coincide. It is easy to verify that the example works.
Criteria. Example only -10 points.
Estimation only -10 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. (20 points) How many zeros does the number $4^{5^{6}}+6^{5^{4}}$ end with in its decimal representation?
#
|
# Answer: 5
Solution. Consider the number $4^{25}+6$. Let's check that it is divisible by 5 and not divisible by 25. We have
$$
\begin{gathered}
4^{25}+6 \equiv(-1)^{25}+6 \equiv 0 \quad(\bmod 5) \\
4^{25} \equiv 1024^{5}+6 \equiv(-1)^{5}+6 \equiv 5 \quad(\bmod 25)
\end{gathered}
$$
We will write $5^{t} \| c$, if $5^{t} \mid c, 5^{t+1} \nmid c$. For arbitrary integers $a, b$, not divisible by 5, such that $5^{k} \| a-b, k>0$, we will prove that $5^{k+1} \| a^{5}-b^{5}$.
Let $a-b=s, 5^{k} \| s$. We have
$$
a^{5}-b^{5}=s\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)=\quad=s\left((b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4}\right)
$$
Expand the brackets using the binomial theorem and consider the expression $\left((b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4}\right)$ modulo 25. Clearly, from each bracket only the terms where $s$ is raised to a power no greater than 1 remain. Thus, we have
$$
\begin{aligned}
&(b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4} \equiv \\
& \equiv 5 b^{4}+(4+3+2+1) b^{3} s \equiv
\end{aligned}
$$
$$
\equiv 5 b^{4}+10 b^{3} s \equiv 5 b^{4} \quad(\bmod 25)
$$
Thus,
$$
5^{1} \|\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$
Therefore, $5^{k+1} \| a^{2}-b^{2}$.
Now, let's proceed to solving the problem. We have
$$
4^{5^{6}}+6^{5^{4}}=\left(1025^{5}\right)^{5^{4}}-(-6)^{5^{4}}
$$
Let $a=1025^{5}, b=-6$. Apply the proven statement. We get that
$$
5^{2} \|\left(1025^{5}\right)^{5}-(-6)^{5}
$$
Applying the statement three more times, we get that
$$
5^{5} \|\left(1025^{5}\right)^{5^{4}}-(-6)^{5^{4}}
$$
It remains to note that the number from the condition is divisible by $2^{5}$. Thus, the number ends with exactly 5 zeros.
Remark. If we use Hensel's Lemma, the solution becomes obvious.
However, proving the lemma is more complicated than solving this problem.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\cos \alpha = \frac{2}{5}$?
|
Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{2}{5} \cdot 10=8$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $8$ and $\cos \alpha = \frac{3}{4}$?
|
Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{3}{4} \cdot 8=12$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $6$ and $\cos \alpha = \frac{2}{3}$?
|
Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{2}{3} \cdot 6=8$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\cos \alpha = \frac{1}{4}$?
|
Answer: 6.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{1}{4} \cdot 12=6$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $9$ and $\cos \alpha = \frac{1}{3}$?
|
Answer: 6.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{1}{3} \cdot 9=6$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) Find all pairs of natural numbers $x$ and $y$ such that
$$
\log _{2} a x+\log _{2} b y=\log _{2}\left(b x+a y+p_{1} p_{2}-1\right), \text { where } p_{1}, p_{2} \in \mathbb{P} \quad a, b>2
$$
In your answer, write the smallest possible value of $x+y$.
|
Solution: Let's get rid of the logarithms in the equation
$$
a x b y=b x+a y+p_{1} p_{2}-1
$$
Transform and factorize
$$
(a y-1)(b x-1)=p_{1} p_{2}
$$
Since $p_{1}, p_{2} \in \mathbb{P}$, we get four possible solutions
$$
\left\{\begin{array} { l }
{ x = \frac { p _ { 1 } p _ { 2 } + 1 } { b } } \\
{ y = \frac { 2 } { a } }
\end{array} \quad \left\{\begin{array} { l }
{ x = \frac { 2 } { b } } \\
{ y = \frac { p _ { 1 } p _ { 2 } + 1 } { a } }
\end{array} \left\{\begin{array} { l }
{ x = \frac { p _ { 1 } + 1 } { b } } \\
{ y = \frac { p _ { 2 } + 1 } { a } }
\end{array} \quad \left\{\begin{array}{l}
x=\frac{p_{2}+1}{b} \\
y=\frac{p_{1}+1}{a}
\end{array}\right.\right.\right.\right.
$$
Since $a, b>2$ and $x, y \in \mathbb{N}$, the first two solutions are definitely not suitable. From the remaining two, we choose the one where $x, y$ are natural numbers and the sum $x+y$ is minimal.
Answers:
| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 5 | 4 | 5 | 6 | 7 | 6 | 4 | 6 | 10 | 4 | 8 |
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) Find the sum of the real roots of the equation
$$
2 \cdot 3^{3 x}-a \cdot 3^{2 x}-3(a+4) \cdot 3^{x}+18=0
$$
|
Solution: Let's make the substitution $t=3^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-3(a+4) t+18=0
$$
It is obvious that the number $t_{1}=-3$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+3)\left(2 t^{2}-(a+6) t+6\right)=0
$$
The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get
$$
3^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{6}{2}=3
$$
Therefore, $x_{2}+x_{3}=1$.
(10 points) Find the sum of the real roots of the equation
$$
2 \cdot 4^{3 x}-a \cdot 4^{2 x}-4(a+6) \cdot 4^{x}+32=0
$$
Solution: Let's make the substitution $t=4^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-4(a+6) t+32=0
$$
It is obvious that the number $t_{1}=-4$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+4)\left(2 t^{2}-(a+8) t+8\right)=0
$$
The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get
$$
4^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{8}{2}=4
$$
Therefore, $x_{2}+x_{3}=1$.
Answers:
| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0.5 |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$?
|
Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 15=12$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{24}}{5}$?
|
Answer: 4.
## Solution:
Consider the point $B_{1}$ symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{1}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{1}{5} *$ $10=4$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{21}}{5}$?
|
Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 10=8$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\sin \alpha = \frac{\sqrt{35}}{6}$?
|
Answer: 4.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{1}{6}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{1}{6} *$ $12=4$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $16$, and $\sin \alpha = \frac{\sqrt{55}}{8}$?
|
Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{3}{8}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{3}{8} * 16=12$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. (20 points) Find all values of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2|x+7|+|y-4|=c \\
|x+4|+2|y-7|=c
\end{array}\right.
$$
|
Answer: $c=3$.
Solution. Let $\left(x_{0} ; y_{0}\right)$ be the unique solution of the system. Then $\left(-y_{0} ;-x_{0}\right)$ also satisfies the conditions of the system. This solution coincides with the first, so $y_{0}=-x_{0}$. The equation $2|x+7|+|x+4|=$ $c$ has a unique solution $x_{0}=-7$ when $c=|7-4|$, since the function $f(x)=2|x+7|+|x+4|$ decreases on the interval $\left(-\infty ; x_{0}\right]$ and increases on $\left[x_{0} ;+\infty\right)$. Similarly, $2|y-7|+|y-4| \geqslant|7-4|$. Therefore, when $c=|7-4|$, the equality $f(x)+f(-y)=2 c$ holds only when $x=-y=-7$, i.e., the system has a unique solution.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) Find the remainder when $20^{16}+201^{6}$ is divided by 9.
|
Answer: 7
Solution. $201^{16}$ is divisible by 9. 20 gives a remainder of $2.2^{6}$ gives a remainder of $1,2^{16}=2^{6} \cdot 2^{6} \cdot 2^{4}$ gives the same remainder as 16.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 70 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=70+62$, which means $x=12$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(12-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(11-y)$ times. Then $y(y-1)+(12-y)(11-y)=62 \Rightarrow$ $y^{2}-12 y+35=0 \Rightarrow y=5$ or $y=7$. Both answers satisfy the condition (in both cases, 70 negative numbers are recorded).
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 68 positive and 64 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=64+68$, which means $x=12$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(12-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(11-y)$ times. Then $y(y-1)+(12-y)(11-y)=68 \Rightarrow$ $y^{2}-12 y+32=0 \Rightarrow y=4$ or $y=8$. Both answers satisfy the condition (in both cases, 64 negative numbers are recorded).
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 78 positive and 54 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=54+78$, which means $x=12$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(12-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(11-y)$ times. Then $y(y-1)+(12-y)(11-y)=78 \Rightarrow$ $y^{2}-12 y+27=0 \Rightarrow y=3$ or $y=9$. Both obtained answers satisfy the condition (in both cases, 54 negative numbers are recorded).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=21
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$.
|
Answer: 12
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=12$, $B C^{2}=9, A C^{2}=21$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 2 \sqrt{3} \cdot 3$. From this, we obtain the answer.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 92 positive and 40 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 2
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=40+92$, which means $x=12$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(12-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(11-y)$ times. Then $y(y-1)+(12-y)(11-y)=92 \Rightarrow$ $y^{2}-12 y+20=0 \Rightarrow y=2$ or $y=10$. Both obtained answers satisfy the condition (in both cases, 40 negative numbers are recorded).
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 50 positive and 60 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=60+50$, which means $x=11$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(11-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(10-y)$ times. Then $y(y-1)+(11-y)(10-y)=50 \Rightarrow$ $y^{2}-11 y+30=0 \Rightarrow y=5$ or $y=6$. Both answers satisfy the condition (in both cases, 60 negative numbers are recorded).
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 54 positive and 56 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=56+54$, which means $x=11$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(11-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(10-y)$ times. Then $y(y-1)+(11-y)(10-y)=54 \Rightarrow$ $y^{2}-11 y+28=0 \Rightarrow y=4$ or $y=7$. Both obtained answers satisfy the condition (in both cases, 56 negative numbers are recorded).
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 48 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+62$, which means $x=11$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(11-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(10-y)$ times. Then $y(y-1)+(11-y)(10-y)=62 \Rightarrow$ $y^{2}-11 y+24=0 \Rightarrow y=3$ or $y=8$. Both obtained answers satisfy the condition (in both cases, 48 negative numbers are recorded).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 42 positive and 48 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+42$, which means $x=10$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(10-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(9-y)$ times. Then $y(y-1)+(10-y)(9-y)=42 \Rightarrow$ $y^{2}-10 y+24=0 \Rightarrow y=4$ or $y=6$. Both answers satisfy the condition (in both cases, 48 negative numbers are recorded).
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 48 positive and 42 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each of them gave $x-1$ answers, so $x(x-1)=42+48$, which means $x=10$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(10-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(9-y)$ times. Then $y(y-1)+(10-y)(9-y)=48 \Rightarrow$ $y^{2}-10 y+21=0 \Rightarrow y=3$ or $y=7$. Both obtained answers satisfy the condition (in both cases, 42 negative numbers are recorded).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 36 positive and 36 negative numbers were recorded. What is the smallest number of times a positive temperature could have been announced?
|
Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=36+36$, which means $x=9$.
2) Let there be $y$ people with "positive temperature" at the conference, then there were $(9-y)$ people with "negative temperature". Each "positive" person recorded a positive number $(y-1)$ times, and each "negative" person recorded a negative number $-(8-y)$ times. Then $y(y-1)+(9-y)(8-y)=36 \Rightarrow$ $y^{2}-9 y+18=0 \Rightarrow y=3$ or $y=6$. Both obtained answers satisfy the condition (in both cases, 36 negative numbers are recorded).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. (20 points) Given various natural numbers $a, b, c, d$, for which the following conditions are satisfied: $a>d, a b=c d$ and $a+b+c+d=a c$. Find the sum of all four numbers.
|
Answer: 12
Solution. From the relations $a>d$ and $ab=cd$, we have the inequality $ba+d$ and $\frac{ac}{2} \geqslant 2c > b+c$. Adding them, we get a contradiction with the condition.
Therefore, without loss of generality, assume that $a \geqslant 3$. Since $a$ is a natural number, then $a \in\{1,2,3\}$. Moreover, $d$ is a natural number less than $a$, so $a \neq 1$. Let's consider two cases.
Case 1. $a=3$. The relations can be rewritten as
$$
3b=cd, 3+b+c+d=3c, 3>d
$$
From the last inequality, in particular, it follows that $(3, d)=1$. Therefore, $3 \mid c$. Since all numbers are distinct, $c \geqslant 6$. Then we have a chain of inequalities
$$
3+b+c+d < 3c \Rightarrow b+c+d < 2c \Rightarrow b+d < c
$$
Since $3>d$ implies $d=1$. The relations can be rewritten as $2b=c, 3+b=c$. The only solution to this system of equations is the pair $(b, c)=(3,6)$. We find the solution $(2,3,6,1)$, the sum of which is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (10 points) The area of a sector of a circle is 100. For what value of the radius of the circle will the perimeter of this sector be minimal? If the answer is not an integer, round to the tenths.
Answer: 10
|
Solution. The area of the sector corresponding to $\alpha$ radians $S_{\alpha}=\frac{1}{2} R^{2} \alpha=100$. The length of the perimeter is $L(R)=2 R+R \alpha=2 R+\frac{200}{R}$, which reaches a minimum at the point $R=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2(x+7)^{2}+(y-4)^{2}=c \\
(x+4)^{2}+2(y-7)^{2}=c
\end{array}\right.
$$
|
Answer: $c=6.0$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$
\begin{aligned}
& \left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right) \geqslant(|x+\alpha|+|y-\beta|)^{2} \\
& \left(1+\frac{1}{2}\right)\left((x+\beta)^{2}+2(y-\alpha)^{2}\right) \geqslant(|x+\beta|+|y-\alpha|)^{2}
\end{aligned}
$$
Therefore, for any solution $(x, y)$ of the system, we have
(3) $3 c \geqslant(|x+\alpha|+|y-\beta|)^{2}+(|x+\beta|+|y-\alpha|)^{2} \geqslant$
$$
\begin{aligned}
\geqslant(\alpha-\beta+x+y)^{2}+(\alpha-\beta-(x+y))^{2} & = \\
& =2(\alpha-\beta)^{2}+2(x+y)^{2} \geqslant 2(\alpha-\beta)^{2}
\end{aligned}
$$
Thus, if $c<\frac{2}{3}(\alpha-\beta)^{2}$, the system has no solutions.
If $c=\frac{2}{3}(\alpha-\beta)^{2}$, then equality must be achieved in all inequalities (1)-(3). Therefore, $x+y=0$, the numbers $x+\alpha$ and $x+\beta$ must have different signs, and
$$
\frac{2(x+\alpha)^{2}}{(x+\beta)^{2}}=\frac{1}{2}
$$
From these conditions, it follows that when $c=\frac{2}{3}(\alpha-\beta)^{2}$, there is only one solution $\left(-\frac{2 \alpha+\beta}{3}, \frac{2 \alpha+\beta}{3}\right)$. Remark. One can immediately notice that the solution of the system is symmetric with respect to the line $y=-x$ (two ellipses with centers at points $(-\alpha ; \beta)$ and $(-\beta ; \alpha)$). The solution will be unique if one of the ellipses touches this line. This is satisfied if the equation $2(x+\alpha)^{2}+(x+\beta)^{2}=c$ has a unique solution. Setting the discriminant to zero, we find the solution. https://ggbm.at/FVmUGS4Y
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Several people played a round-robin table tennis tournament. At the end of the tournament, it turned out that for any four participants, there would be two who scored the same number of points in the games between these four participants. What is the maximum number of tennis players that could have participated in this tournament? In table tennis, there are no ties; one point is awarded for a win, and zero points for a loss.
(from materials of international olympiads)
|
Answer: 7 tennis players.
Solution. Let $n \geqslant 8$ be the number of tennis players in the tournament. Then the total number of matches played is $\frac{1}{2} n(n-1)$, and thus the total number of victories is also $\frac{1}{2} n(n-1)$. Therefore, there must be a participant who has won at least $\frac{n-1}{2}$ matches. Hence, some participant must have won at least 4 matches.
Let's call this participant Andrei. Consider those he has defeated. Let these be Borya, Vasya, Grisha, and Dima. In the matches among them, 6 matches were played, so at least one of them must have won at least two matches.
Let's assume, for definiteness, that this is Borya, and he won against Vasya and Grisha. Then consider the quartet of players Andrei, Borya, Vasya, and Grisha. Andrei won all his matches among these four players, so he scored three points, Borya won exactly two matches, so he scored two points, the winner of the match between Vasya and Grisha scored one point, and the loser scored no points.
Therefore, they scored different numbers of points in their matches against each other, which means that such a quartet does not satisfy the condition of the problem. Hence, there are no more than seven players.
Let's provide an example of a tournament with seven participants. A victory is denoted by a plus, and a loss by a minus.
| | + | + | + | - | - | - |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| - | | + | - | + | + | - |
| - | - | | + | + | - | + |
| - | + | - | | - | + | + |
| + | - | - | + | | + | - |
| + | - | + | - | - | | + |
| + | + | - | - | + | - | |
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. On the side $AB$ of triangle $ABC$, a point $M$ is taken. It starts moving parallel to $BC$ until it intersects with $AC$, then it moves parallel to $AB$ until it intersects with $BC$, and so on. Is it true that after a certain number of such steps, point $M$ will return to its original position? If this is true, what is the minimum number of steps sufficient for the return?
|
Solution. Let the length of side $AB$ be 1, and let point $M$ be at a distance $a$ from point $B$. From the properties of the parallelogram, the small triangles are equal, so after 3 steps, point $M$ will be at a distance $a$ from point $A$, which is at a distance $1-a$ from point $B$. After another 3 steps, the point will be at a distance $1-(1-a)=a$ from point $B$, meaning it will return to its initial position.
A special case is when $a=1/2$. Then $1-a=a$, and the return will occur after 3 steps.
Answer: Correct.
It takes 3 steps if point $M$ divides side $AB$ in half, and 6 steps in other cases.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$.
|
Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, \quad k=1 \ldots n \text {. }
$$
Thus, all elements of the set are equal to each other. Since the sum does not change when one addend is replaced, this addend must be equal to what it is replaced with, i.e.,
$$
x_{1}=x_{2}+x_{3}+\cdots+x_{n}
$$
Considering the equality of the elements, we get $x_{1}=(n-1) x_{1}$, hence, $x_{1}=0$. Therefore, the product of all numbers in the set $M$ is 0.
## Answer: 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x+3}{2}\right]^{2}-x=1$.
|
# Solution.
From the equation, it follows that $x=\left[\frac{x+3}{2}\right]^{2}-1$ is an integer. Therefore, $x=n \in \mathbb{Z}$, but we need to consider the cases of even and odd $n$ separately.
First, let's move all terms to the left side of the equation.
1) If $x=2 k$.
$$
\left[\frac{2 k+3}{2}\right]^{2}-2 k-1=\left[k+1+\frac{1}{2}\right]^{2}-2 k-1=\quad=(k+1)^{2}-2 k-1=k^{2}=0
$$
The obtained equation has a solution $k=0$, which gives $x=0$.
2) If $x=2 k+1$.
$$
\begin{aligned}
& {\left[\frac{2 k+4}{2}\right]^{2}-(2 k+1)-1=(k+2)^{2}-2 k-2=k^{2}+2 k+2=} \\
& \quad=\left(k^{2}+k\right)+(k+1)+1=k(k+1)+(k+1)+1=(k+1)(k+1)+1 \geqslant 1
\end{aligned}
$$
Since the obtained expression is strictly positive, solutions of the form $x=2 k+1$ are not possible.
Answer. $x=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 2.
What is the last digit of the value of the sum $2019^{2020}+2020^{2019} ?$
|
# Solution.
The number $2019^{n}$ for $n \in \mathbb{N}$ ends in 9 if $n$ is odd, and in 1 if $n$ is even. Therefore, $2019^{2020}$ ends in 1. The number $2020^{n}$ ends in 0 for any $n \in \mathbb{N}$, so $2020^{2019}$ ends in 0. Thus, the sum $2019^{2020} +$ $2020^{2019}$ ends in the digit 1.
Answer. The digit 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3.
On the coordinate plane, a square $K$ is marked with vertices at points $(0,0)$ and $(10,10)$. Inside this square, draw the set $M$ of points $(x, y)$ whose coordinates satisfy the equation
$$
[x]=[y],
$$
where $[a]$ denotes the integer part of the number $a$ (i.e., the greatest integer not exceeding $a$; for example, $[10]=10,[9.93]=9,[1/9]=0,[-1.7]=-2$). What fraction of the area of square $K$ is the area of set $M$?
|
Solution. Let $n \leq x < n+1$, where $n$ is an integer from 0 to 9. Then $[x]=n$ and $[y]=n$. The solution to the latter equation is all $y \in [n, n+1)$. Thus, the solution will be the union of unit squares
$$
\{x \in [n, n+1), y \in [n, n+1), n \in \mathbb{Z}\}
$$
Inside the square $K$ specified in the problem, ten such squares will fit. Since $K$ consists of 100 unit squares, the ratio of the areas is $\frac{S_{M}}{S_{K}}=\frac{10}{100}=0.1=10\%$.
The answer is represented by dotted hatching in the figure below (the left and lower boundaries are included in $M$, the right and upper boundaries are not). The area of the set $M$ is $10\%$ of the area of the square $K$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that word. Is it possible to encode the letters Е, О, С, Т, Ш, Ь with elementary codes, each consisting of a single digit from 0 to 9, so that the weight of the word "СТО" is not less than the weight of the word "ШЕСТЬСОТ"? If such encoding is possible, in how many ways can it be implemented? If such encoding is possible, does it allow for the unambiguous restoration of a word from its code?
|
# Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have
$k(C)+k(T)+k(O) \geq k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)$, which is equivalent to
$$
k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})=0 .
$$
Thus,
$$
k(\amalg)=k(\mathrm{E})=k(C)=k(T)=k(\mathrm{~b})=0 .
$$
The elementary code $k(O)$ can be any digit from 0 to 9. Therefore, there are exactly 10 ways to encode.
However, for any choice of $k(O)$, the words "TOCT" and "TOT" (for example) will have the same code $k(O)$, which cannot be decoded uniquely.
Answer. The inequality for the weights holds if and only if $k(\amalg)=k(\mathrm{E})=k(C)=k(T)=k(\mathrm{~b})=0$.
There are exactly 10 ways to encode.
In any of them, unique decoding is impossible.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. A triangle was cut into two triangles. Find the greatest value of $N$ such that among the 6 angles of these two triangles, exactly $N$ are the same.
|
Solution. For $N=4$, an example is an isosceles right triangle divided into two isosceles right triangles: four angles of $45^{\circ}$. Suppose there are five equal angles. Then in one of the triangles, all three angles are equal, meaning all of them, and two angles of the other triangle are $60^{\circ}$. But then both these triangles are equilateral, and it is impossible to form a triangle from two equilateral triangles.
Answer: $\mathrm{N}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements
from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$.
|
# Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, \quad k=1 \ldots n
$$
Thus, all elements of the set are equal to each other. Since the sum does not change when one addend is replaced, this addend must be equal to what it is replaced with, i.e.,
$$
x_{1}=x_{2}+x_{3}+\cdots+x_{n}
$$
Considering the equality of the elements, we get $x_{1}=(n-1) x_{1}$, hence, $x_{1}=0$. Therefore, the product of all numbers in the set $M$ is 0.
Answer: 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A table of numbers with 20 rows and 15 columns, $A_{1}, \ldots, A_{20}$ are the sums of the numbers in the rows, $B_{1}, \ldots, B_{15}$ are the sums of the numbers in the columns.
a) Is it possible that $A_{1}=\cdots=A_{20}=B_{1}=\cdots=B_{15}$?
b) If the equalities in part a) are satisfied, what is the sum $A_{1}+\cdots+A_{20}+$ $B_{1}+\cdots+B_{15}?$
|
Let $A_{i}=B_{j}=X$ for $i=1, \ldots 20$ and $j=1, \ldots, 15$. Consider the sum $S$ of all elements in the table. We have $S=20 X=15 X, X=0$ and $A_{1}+\cdots+A_{20}+B_{1}+\cdots+B_{15}=0$. An example of such a table is, for instance, a table consisting entirely of zeros. There is no need to consider other cases.
Answer: a) - yes, b) - 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Early in the morning, the pump was turned on and they started filling the pool. At 10 am, another pump was connected and by 12 pm the pool was half full. By 5 pm, the pool was full. What could be the latest time the first pump was turned on?
|
Solution. Let the volume of the pool be $V$. Denote by $x$ and $y$ the capacities of the pumps, and by $t$ the time the first pump operates before the second pump is turned on.
Then $t x + 2 x + 2 y = V / 2.5 x + 5 y = V / 2$.
From this, $t x + 2 x + 2 y = 5 x + 5 y$ or $t x = 3 x + 3 y$.
In the end, $t = 3 + 3 y / x$. Since $x > 0$ and $y > 0$, then $t \geqslant 3$.
Answer. The first pump was turned on no later than 7 a.m.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x-1}{2}\right]^{2}+2 x+2=0$.
|
Solution. From the equation, it follows that $2 x=2-\left[\frac{x-1}{2}\right]^{2}-$ is an integer. Therefore, either $2 x=n \in \mathbb{Z}$, or $x=n+\frac{1}{2} \quad(n \in \mathbb{Z})$. In this case, we need to separately consider the cases of even and odd $n$.
1) If $x=2 k$, then
$$
\begin{aligned}
& {\left[\frac{2 k-1}{2}\right]^{2}+2(2 k)+2=\left[k-\frac{1}{2}\right]^{2}+4 k+2=(k-1)^{2}+4 k+2=} \\
& =k^{2}-2 k+1+4 k+2=\left(k^{2}+2 k+1\right)+2=(k+1)^{2}+2 \geqslant 2
\end{aligned}
$$
Since the obtained expression is strictly positive, solutions of the form $x=2 k$ are impossible.
2) If $x=2 k+1$, then
$$
\left[\frac{2 k}{2}\right]^{2}+2(2 k+1)+2=k^{2}+4 k+4=(k+2)^{2}=0
$$
The obtained equation has a solution $k=-2$, which gives $x=-3$.
3) If $x=2 k+\frac{1}{2}$, then
$$
\begin{aligned}
& {\left[\frac{2 k-\frac{1}{2}}{2}\right]^{2}+2\left(2 k+\frac{1}{2}\right)+2=\left[k-\frac{1}{4}\right]^{2}+4 k+3=(k-1)^{2}+4 k+3=} \\
& =k^{2}-2 k+1+4 k+3=\left(k^{2}+2 k+1\right)+3=(k+1)^{2}+3 \geqslant 3
\end{aligned}
$$
Since the obtained expression is strictly positive, solutions of the form $x=2 k+\frac{1}{2}$ are impossible.
4) If $x=2 k+1+\frac{1}{2}$, then
$$
\begin{aligned}
{\left[\frac{2 k+\frac{1}{2}}{2}\right]^{2}+2\left(2 k+\frac{3}{2}\right)+2=\left[k+\frac{1}{4}\right]^{2} } & +4 k+3+2=k^{2}+4 k+5= \\
& =k^{2}+4 k+4+1=(k+2)^{2}+1 \geqslant 1
\end{aligned}
$$
Since the obtained expression is strictly positive, solutions of the form $x=2 k+1+\frac{1}{2}$ are impossible.
Answer. $x=-3$.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The Atelier "Heavy Burden" purchased a large batch of cast iron buttons. If they sew two buttons on each coat or if they sew three buttons on each coat, in each case 1 piece will remain from the entire batch. If, however, they sew four buttons on each coat or if they sew five buttons on each coat, in each case 3 pieces will remain from the entire batch. How many buttons will remain if they sew twelve buttons on each coat?
|
# Solution
Let $a$ be the desired number. From the condition, it follows that the number $a-1$ is divisible by 2 and 3. Therefore, $a=6k+1$. Also, the number $a-3$ is divisible by 4 and 5. Therefore, $a=20n+3$. We solve the equation
$$
6k+1=20n+3
$$
Or, equivalently,
$$
3k=10n+1
$$
Its general solution is
$$
k=7+10s, \quad n=2+3s, \quad \Rightarrow \quad a=60s+43
$$
We find the remainder of the division of $a$ by 12. Since 60 is divisible by 12, the desired remainder is the remainder of the division of 43 by 12, which is 7.
Answer: 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4.
Four brigades were developing an open coal deposit for three years, working with a constant productivity for each brigade. In the second year, due to weather conditions, work was not carried out for four months, and for the rest of the time, the brigades worked in rotation (one at a time). The ratio of the working times of the first, second, third, and fourth brigades and the amount of coal mined respectively are: in the first year 4:1:2:5 and 10 million tons; in the second year $2: 3: 2: 1$ and 7 million tons; in the third year 5:2:1:4 and 14 million tons. How much coal would these four brigades have mined in 4 months if they worked together?
|
Solution. Let the $i$-th brigade mine $x_{i}$ coal per month. Then we have the system
$$
\left\{\begin{array}{l}
4 x_{1}+x_{2}+2 x_{3}+5 x_{4}=10 \\
2 x_{1}+3 x_{2}+2 x_{3}+x_{4}=7 \\
5 x_{1}+2 x_{2}+x_{3}+4 x_{4}=14
\end{array}\right.
$$
By adding twice the first equation to thrice the second and subtracting the third equation, we get
$$
9\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=27 \text {. }
$$
Then
$$
4\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=12 \text {. }
$$
Answer: 12 million tons.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 2.
What is the last digit of the value of the sum $5^{2020}+6^{2019} ?$
|
# Solution.
The number 5 to any power ends in 5.
The number 6 to any power ends in 6.
Therefore, the sum $5^{2020}+6^{2019}$ ends in the digit 1.
Answer. The digit 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5.
The carriages of the express train "Moscow - Yalta" must be numbered consecutively, starting from one. But in a hurry, two adjacent carriages received the same number. As a result, it turned out that the sum of the numbers of all carriages is 111. How many carriages are in the train and which number was used twice?
#
|
# Solution.
Let's start calculating the sums sequentially
$$
\begin{aligned}
& 1+2=3 \\
& 1+2+3=6 \\
& \cdots \\
& 1+2+\ldots+14=105 \\
& 1+2+\ldots+14+15=120
\end{aligned}
$$
From this, it is clear that the first 14 wagons had the first 14 sequential numbers, which sum up to 105, and one more wagon had a number equal to \(111-105=6\).
Answer. There were 15 wagons in the train, two of which had the number 6.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1.
Three electric generators have powers $x_{1}, x_{2}, x_{3}$, the total power of all three does not exceed 2 MW. In the power system with such generators, a certain process is described by the function
$$
f\left(x_{1}, x_{2}, x_{3}\right)=\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} .
$$
Find the maximum and minimum values of this function.
|
# Solution.
It is clear that the minimum value of the function is zero (achieved when $\left.x_{1}=x_{2}=x_{3}=0\right)$.
Let's find the maximum. We can assume that $x_{1} \geq x_{2} \geq x_{3} \geq 0$. We will prove two inequalities:
$$
\begin{gathered}
\sqrt{x_{1}^{2}+x_{2} x_{3}} \leq x_{1}+\frac{x_{3}}{2} \\
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \frac{2 x_{1}+3 x_{2}+x_{3}}{2}
\end{gathered}
$$
The first inequality is equivalent to the inequality $4 x_{3}\left(x_{1}-x_{2}\right)+x_{3}^{2} \geq 0$.
To prove the second inequality, we use the fact that for $u \geq 0$ and $v \geq 0$, $\sqrt{u}+\sqrt{v} \leq \sqrt{2(u+v)}$ holds.
Thus,
$$
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \sqrt{2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right)}
$$
It is not difficult to verify that
$2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right) \leq\left(\frac{x_{1}+3 x_{2}+2 x_{3}}{2}\right)^{2} \Leftrightarrow\left(x_{1}-x_{2}-2 x_{3}\right)^{2}+8 x_{3}\left(x_{2}-x_{3}\right) \geq 0$.
We have:
$$
\begin{aligned}
\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq & x_{1}+\frac{x_{3}}{2}+\frac{x_{1}+3 x_{2}+2 x_{3}}{2}= \\
& =\frac{3}{2}\left(x_{1}+x_{2}+x_{3}\right) \leq \frac{3}{2} \cdot 2=3 .
\end{aligned}
$$
Answer: $\quad \max f\left(x_{1}, x_{2}, x_{3}\right)=3, \min f\left(x_{1}, x_{2}, x_{3}\right)=0$
for $x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3.
Two swimmers are training in a rectangular quarry. The first swimmer finds it more convenient to exit at a corner of the quarry, so he swims along the diagonal to the opposite corner and back. The second swimmer finds it more convenient to start from a point that divides one of the quarry's shores in the ratio $2018: 2019$. He swims along a quadrilateral, visiting one point on each shore, and returns to the starting point. Can the second swimmer choose points on the other three shores such that his path is shorter than the first swimmer's? What is the minimum value that the ratio of the length of the longer path to the shorter can have?
|
# Solution.
Let's draw a rectangular quarry $A D C D$ and the inscribed quadrilateral route of the second swimmer $N L M K$.
Reflect the drawing symmetrically first with respect to side $C D$, then with respect to side $C B^{\prime}$, and finally with respect to side $A^{\prime} B^{\prime}$.
By construction, segments $N D$ and $N^{\prime} D^{\prime}$ are equal and parallel to each other. Therefore, $A A^{\prime} N^{\prime} N$ is a parallelogram, and its side $N N^{\prime}$ is equal to twice the diagonal of the original rectangle $A C$ (i.e., the length of the first swimmer's path).
Olympiad for schoolchildren "Nadezhda Energetiki". Final stage. Solutions
It is clear that the sum of the lengths of segments $N K, K M^{\prime}, M^{\prime} L^{\prime}$, and $L^{\prime} N^{\prime}$ is equal to the perimeter of the quadrilateral $N L M K$ (i.e., the length of the second swimmer's path).
But the length of the broken line $N K M^{\prime} L^{\prime} N^{\prime}$ cannot be less than the length of the segment $N N^{\prime}$, which is the shortest distance between points $N$ and $N^{\prime}$. From this, it follows that the second swimmer's path cannot be shorter than the first swimmer's path.
The paths can be made equal if we transfer (using the symmetry of the construction) the points of intersection of the segment $N N^{\prime}$ with the sides $D C, C B^{\prime}$, and $B^{\prime} A^{\prime}$ to the original rectangle. (It can be shown that the resulting figure will be a parallelogram, but this is not part of the problem statement.)
With such a construction, the swimmers' paths will be equal, and their ratio will be 1.
Note that nowhere in the reasoning is it used that the initial point $N$ divides the corresponding side of the rectangle $A D$ in a certain proportion.
## Answer:
the second swimmer's path cannot be shorter than the first swimmer's path; the minimum ratio of the length of the longer path to the shorter one is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be equal to the same value k. Find the value of k. Provide at least one set of four numbers that satisfy the condition. Describe all possible sets of such numbers and determine how many there are.
|
Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2}+x_{3}}{x_{1}+x_{4}}=k
\end{aligned}
$$
Let $A=x_{1}+x_{2}, B=x_{3}+x_{4}$. Then from (2) we get $A=k B, B= k A, A B \neq 0$, from which $(k^{2}-1) A=0, k= \pm 1$. The same values will be obtained by analyzing relations (3) and (4).
If $k=1$, then equations (2)-(4) take the form $x_{1}+x_{2}=x_{3}+x_{4}, x_{1}+x_{3}= x_{2}+x_{4}, x_{1}+x_{4}=x_{2}+x_{3}$, from which we find the general solution $x_{1}=x_{2}=x_{3}= x_{4}=C \neq 0$. This case does not meet the condition (all numbers turned out to be equal).
If $k=-1$, then each of equations (2)-(4) takes the form $x_{1}+x_{2}+x_{3}+ x_{4}=0$, and the general solution is
$x_{1}=A, x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$.
Answer: $k=-1$.
$x_{1}=A, \quad x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$.
The set of tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ is infinite.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5.
An electric cable 21 meters long is cut into 21 pieces. For any two pieces, their lengths differ by no more than a factor of three. What is the smallest $m$ such that there will definitely be two pieces whose lengths differ by no more than a factor of $m$?
|
# Solution.
If among the pieces of cable there are at least two that differ in length, then by taking the ratio of the smaller to the larger, we get that $m \leq 1$.
However, the cable can be cut in such a way that all pieces are equal. This implies that if $m<1$, then a way of cutting has been found where no two pieces differ by more than a factor of $m$.
The constructed example proves that the smallest value of $m$ is 1.
Answer: $m=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. Boys and girls formed a circle in such a way that the number of children whose right neighbor is of the same gender is equal to the number of children whose right neighbor is of a different gender. What could be the total number of children in the circle?
|
Solution. Let $n$ be the number of children next to whom stands a child of the opposite gender, and $m$ be the number of children next to whom stands a child of the same gender. Initially, $n=m$, i.e., the total number of children ($n+m$) is even. We will swap the positions of two adjacent children so that all boys gather in a row on one side of the circle, and all girls on the other. Then $n$ will become 2. With each such swap of children, the numbers $n$ and $m$ either do not change, or one increases by 2, and the other decreases by 2. This means that the remainder of the division of the difference ($m \sim n$) by 4 does not change.
Initially, this remainder was 0.
If the total number of children is $2k$, $n=2$, then $m=2k \curlyvee 2$. Then $m^{\llcorner} n=2k^{\llcorner} 4$.
For the remainder of the division of $m^{\llcorner} n$ by 4 to be zero, $k$ must be divisible by 2. That is, the total number of children must be a multiple of 4.
Answer. Any natural number that is a multiple of four.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be the same value $\mathrm{k}$. Find the value of $\mathrm{k}$. Provide at least one set of four numbers that satisfy the condition. Describe all possible sets of such numbers and determine how many there are.
|
Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2}+x_{3}}{x_{1}+x_{4}}=k
\end{aligned}
$$
Let $A=x_{1}+x_{2}, B=x_{3}+x_{4}$. Then from (2) we get $A=k B, B= k A, A B \neq 0$, from which $\left(k^{2}-1\right) A=0, k= \pm 1$. The same values will be obtained by analyzing relations (3) and (4).
If $k=1$, then equations (2)-(4) take the form $x_{1}+x_{2}=x_{3}+x_{4}, x_{1}+x_{3}=$ $x_{2}+x_{4}, x_{1}+x_{4}=x_{2}+x_{3}$, from which we find the general solution $x_{1}=x_{2}=x_{3}=$ $x_{4}=C \neq 0$. This case does not meet the condition (all numbers turned out to be equal).
If $k=-1$, then each of equations (2)-(4) takes the form $x_{1}+x_{2}+x_{3}+$ $x_{4}=0$, and the general solution is
$x_{1}=A, \quad x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$.
Answer: $k=-1$.
$x_{1}=A, \quad x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$. The set of tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ is infinite.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1.
The Triassic Discoglossus tadpoles have five legs, while the saber-toothed frog tadpoles grow several tails (all have the same number of tails). An employee of the Jurassic Park scooped up several tadpoles with water. It turned out that the captured tadpoles had a total of 100 legs and 64 tails. How many tails does each saber-toothed frog tadpole have, if all five-legged tadpoles have one tail, and all multi-tailed tadpoles have four legs?
|
# Solution.
Let $x$ be the number of tails of the saber-toothed frog's pollywog. Suppose $n$ five-legged and $k$ many-tailed pollywogs were caught. Counting the total number of legs and tails gives the equations
$$
\left\{\begin{aligned}
5 n+4 k & =100 \\
n+x k & =64
\end{aligned}\right.
$$
From the first equation, we have $n \leq 100 / 5=20, \quad k \leq 100 / 4=25$.
Moreover, $4 k=5(20-n)$, so $k$ must be a multiple of 5.
Now, multiply the second equation of the system by 5 and subtract the first from the result. We get
$$
(5 x-4) k=64 \cdot 5-100=220=2 \cdot 2 \cdot 5 \cdot 11 .
$$
Since the left side is the product of two natural numbers and considering the obtained limits, we find that $k$ can only take the values
$$
k=5 \quad \text { or } \quad k=10 \quad \text { or } \quad k=20 .
$$
Substitute them one by one into the system of equations. For $k=5$, we have
$$
\left\{\begin{aligned}
5 n+20 & =100 \\
n+5 x & =64
\end{aligned}\right.
$$
from which $n=16$. But $64-16=48$ is not a multiple of 5. There are no solutions.
For $k=10$, we get
$$
\left\{\begin{array}{r}
5 n+40=100 \\
n+10 x=64
\end{array}\right.
$$
from which $n=12$. But $64-12=52$ is not a multiple of 10. There are no solutions again.
Olympiad for Schoolchildren "Hope of Energy". Final Stage. Solutions
Finally, for $k=20$, we find
$$
\left\{\begin{array}{l}
5 n+80=100 \\
n+20 x=64
\end{array}\right.
$$
from which $n=4$ and $n=3$.
Answer: 3 tails.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5.
Once upon a time, Baba Yaga and Koschei the Deathless tried to divide a magical powder that turns everything into gold equally. Baba Yaga took out a scale and weighed all the powder. The scales showed 6 zolotniks. Then she started removing the powder until the scales showed 3 zolotniks. However, Koschei suspected that the scales were lying and weighed the portion that had been removed separately on the same scales (there were no others). The scales showed 2 zolotniks. Determine the exact weight of the two parts into which Baba Yaga divided the powder. Assume that if the scales are lying, they always lie by the same amount.
|
# Solution.
Let $A$ be the weight of the first part (the one that remained on the scales), $B$ be the weight of the second part (the one that was poured off), and $d$ be the error of the scales.
Then the result of the first weighing (of the entire powder) gives
$$
A+B+d=6,
$$
the result of the second weighing (after pouring off) gives
$$
A+d=3
$$
and the result of the last weighing (of the poured-off part) gives
$$
B+d=2
$$
Adding the last two equations, we get
$$
A+B+2 d=5
$$
Subtracting the first equation from the obtained one, we get
$$
d=-1
$$
Thus, the scales reduce all readings by 1 zolotnik. From this, the answer immediately follows.
Answer: 4 and 3 zolotniks.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4.
At 9:00 AM, ships "Anin" and "Vinin" departed from port O to port E. At the same moment, ship "Sanin" departed from port E to port O. All three vessels are traveling on the same course ("Sanin" heading towards "Anin" and "Vinin") at constant but different speeds. At 10:00 AM, "Vinin" was at the same distance from "Anin" and "Sanin". At 10:30 AM, "Sanin" was at the same distance from "Anin" and "Vinin". At what moment will "Anin" be exactly in the middle between "Vinin" and "Sanin"?
|
# Solution.
For simplicity, let's assume the distance between the ports is 1. Let $x$ be the speed of "Anina", $y$ be the speed of "Vanina", and $z$ be the speed of "Sanina". Then at 10 o'clock (i.e., one hour after departure),
$$
2 y = x + 1 - z,
$$
and one and a half hours after departure,
$$
2(1 - 1.5 z) = 1.5 x + 1.5 y.
$$
After time $t$ from departure, the condition should be met:
$$
2 x t = 1 - t z + t y,
$$
which simplifies to
$$
t(2 x + z - y) = 1.
$$
From (1) and (2),
$$
3 z + 1.5 x + 1.5 y = 4 y + 2 z - 2 x
$$
or $z = 2.5 y - 3.5 x$. Substituting this expression into (1), we get $y - x = 2 / 9$. Considering this, from (1) we have $y + z + y - x = y + z + 2 / 9$. Therefore, $t + z = 7 / 9$. Thus, from (3) we get
$$
\begin{gathered}
t(2 x + z - y) = t(z + y - 2 x + 2 y) = 1 \\
t(7 / 9 - 4 / 9) = 1
\end{gathered}
$$
which gives $t = 3$.
Answer: at 12 o'clock noon.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3. (15 points) In one move, you can choose a natural number $x$ and cross out all natural numbers $y$ such that $|x-y|$ is a natural composite number. At the same time, $x$ can be chosen from already crossed out numbers.
What is the minimum number of moves needed to cross out all numbers from the natural number sequence?
|
Answer: 2.
Solution. First, note that one move is not enough, because if we choose some natural number $x$, then the number $y=x$ will remain unchecked. Let's prove that two moves are enough.
We will look for two suitable numbers $x_{1}$ and $x_{2}$ of different parity. Then one of the differences $\left|y-x_{1}\right|$ or $\left|y-x_{2}\right|$ will be even. This means it will be composite - except for the cases when it equals 2 or 0. These cases need to be considered separately. For $y=x_{1}-2, y=x_{1}$, and $y=x_{1}+2$, the difference $\left|y-x_{2}\right|$ must be a composite number; and for $y=x_{2}-2, y=x_{2}$, and $y=x_{2}+2$, the difference $\left|y-x_{1}\right|$ must be a composite number.
In any case, it is sufficient for the differences $\left|x_{1}-x_{2}-2\right|, \left|x_{1}-x_{2}\right|$, and $\left|x_{1}-x_{2}+2\right|$ to be odd composite numbers. Since the expressions under the modulus cannot have different signs (otherwise one of them would be equal to 1), they must be three consecutive odd composite numbers.
Now it is enough to find three consecutive odd composite numbers $a-2, a, a+2$ (or even just prove that such numbers exist). Then we can choose, for example, $x_{1}=a+1$ and $x_{2}=1$ and get the required result.
There are different ways to prove that three consecutive odd composite numbers exist. Here are some of them.
Method 1. Consider the consecutive odd numbers $7!+3, 7!+5$, and $7!+7$. They are divisible by the prime numbers 3, 5, and 7, respectively, but are not equal to them, so they are composite.
Method 2. If we assume that $a-2$ is divisible by 3, $a$ is divisible by 5, and $a+2$ is divisible by 7, then the number $a$ must give remainders 2, 0, and 5 when divided by 3, 5, and 7, respectively, and a remainder of 1 when divided by 2. By the Chinese Remainder Theorem, there are infinitely many such $a$, and they form an arithmetic progression with a common difference of $2 \cdot 3 \cdot 5 \cdot 7 = 210$. The first such number is obviously $a=5$, but it does not work; the next one is $a=215$, which gives the triplet of composites $213, 215, 217$.
Method 3. We can also try odd numbers and find a suitable triplet. The smallest such triplet is $91, 93, 95$, which corresponds to $x_{1}=1$ and $x_{2}=94$.
## Criteria
The highest applicable criterion is used:
15 p. Any complete solution to the problem.
0 p. Proved that one move is not enough.
0 p. Only the answer is provided.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7. Problem 7.1*
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha, in any order, tells the digits that stand in these places. The order in which to tell the digits is chosen by Misha. For example, if the number thought of is 67890, and Igor asked about the digits in the 1st and 5th places, then Misha can answer either "6 and 0" or "0 and 6". In what minimum number of moves can Igor guarantee to find out the number? Points for the problem: 13.
#
|
# 7. Problem $7.1^{*}$
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha reports the digits in these positions in any order. The order in which to report the digits is chosen by Misha. For example, if the number thought of is 67890, and Igor asks about the digits in the 1st and 5th positions, Misha can answer either "6 and 0" or "0 and 6". What is the minimum number of moves Igor needs to guarantee he can determine the number? Points for the problem: 13.
## Answer: 3
#
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.1. (15 points) Do there exist polynomials $P(x), Q(x)$, and $R(x)$ with real coefficients such that the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have the same degree, while the polynomials $P(x)+Q(x), P(x)+R(x)$, and $Q(x)+R(x)$ have pairwise distinct degrees? (We consider that the zero polynomial has no degree, so the specified polynomials cannot be equal to it.)
|
Answer: $: 2$.
Solution. It is sufficient to take, for example, $P(x)=x^{2}, Q(x)=-x^{2}+1, R(x)=x^{2}+x$. Then the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have degree 4 (these are products of two polynomials of degree 2); the polynomial $P(x)+Q(x)$ equals 1 and has degree 0; the polynomial $P(x)+R(x)$ equals $2 x^{2}+x$ and has degree 2; the polynomial $Q(x)+R(x)$ equals $x+1$ and has degree 1.
## Criteria
The highest applicable criterion is used:
15 p. A correct example of polynomials $P(x), Q(x), R(x)$ is provided (or it is otherwise proven that such polynomials exist).
0 p. Only the answer is provided.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.4. (15 points) Once, 45 friends living in different parts of the world decided to exchange news with each other. To do this, they plan to arrange $k$ video meetings, at each of which every person will share their news, as well as all the news from other people they have learned previously.
For the video meetings, 10 days were proposed, but it turned out that each friend can only be present on 8 of them. What is the smallest natural number $k$ that guarantees we can select $k$ days for the video meetings from the proposed 10 so that everyone learns the news from everyone else?
(Between the proposed days, no new news arises among the people, and they do not communicate with each other in any other way. On each of the proposed days, one video meeting is held, which is attended by everyone who can be present on that day.)
|
Answer: 5 days.
Solution. Let's provide an example of a situation where 4 days would not be enough. Suppose each of the 45 people has their own unique pair of days when they cannot participate in the meeting. Since the number of ways to choose a pair of days from the 10 offered is $\mathrm{C}_{10}^{2}=45$, for any pair of days, there will be a person who cannot be present on exactly those two days. Suppose we managed to choose some 4 days such that everyone learns all the news. Then there exists a person $A$ who cannot be present on the first two of these four days, and a person $B$ who cannot be present on the last two of these four days. Notice that then $B$ will not be able to learn the news from $A$. Contradiction.
Now let's understand that 5 days will always be enough. Choose 5 days arbitrarily. We will prove that any two people will be present together at some meeting. Indeed, among these 5 days, there are no more than 2 days when the first person cannot be present, and no more than 2 days when the second person cannot be present. Therefore, there will be a day when both can be present. Thus, each pair of people will be able to exchange news, i.e., everyone will learn the news from everyone else.
## Criteria
## Points for evaluation and example are summed up:
+9 p. Evaluation - it is proven that 4 days may not be enough.
In the absence of this proof, the following progress is evaluated:
+1 p. Mentioned that different pairs of missed days can correspond to all people.
+6 p. Example - it is proven that 5 days are guaranteed to be enough.
The following progress is not evaluated:
0 p. Only the correct answer is provided.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 10.5. (20 points) Find all composite natural numbers $n$ that have the following property: each natural divisor of the number $n$ (including $n$ itself), decreased by 1, is a square of an integer.
|
Answer: 10.
Solution. Suppose that $n$ is divisible by the square of some prime number $p$. Then it has a divisor $p^{2}=b^{2}+1$; but two squares of integers can only differ by 1 if they are 0 and 1.
Let $n$ be divisible by some two prime numbers $p$ and $q$. Without loss of generality, we can assume that $p>q$. From the condition that any divisor, decreased by 1, is a square, we can write
\[
\begin{gathered}
p=a^{2}+1 \\
q=b^{2}+1 \\
p q=c^{2}+1
\end{gathered}
\]
Subtract the first equation from the third, we get $p q-p=c^{2}-a^{2}$. This can be rewritten as
\[
p(q-1)=(c-a)(c+a)
\]
Since $p$ is a prime number, one of the factors on the right side must be divisible by $p$. Note that from the condition $p>q$ it follows that $p>q-1$, so $p$ cannot divide $q-1$. Therefore, $p$ must divide $c-a$ or $c+a$. Since $p>q$, we get that $q=2$.
Substituting $q=2$, we get $2=c-a+1=p-2a+1$, from which, first, $c=a+1$, and second, $p=2a+1$. Then $p q=2 p=4a+2$ and $p q=c^{2}+1=(a+1)^{2}+1$. Equating, we get the quadratic equation $a^{2}+2a+2=4a+2$, the roots of which are 2 and 0, from which $p$ equals 5 or 1. But since $p$ must be prime, the only option left is $p=5$.
Thus, the only possible case is $p=5, q=2$. It is clear that there cannot be any other prime numbers in the factorization of $n$.
## Criteria
The highest applicable criterion is used:
20 points. Any complete solution to the problem is provided.
16 points. It is proven that $n$ is even.
4 points. The case of even $n$ is correctly analyzed.
In the absence of the above advancements, the following criteria are summed up:
+1 point. The correct answer is provided.
+1 point. It is proven that $n$ is square-free.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In an acute-angled triangle $A B C$, the altitude $A A_{1}$ is drawn. $H$ is the orthocenter of triangle $A B C$. It is known that $A H=3, A_{1} H=2$, and the radius of the circumcircle of triangle $A B C$ is 4. Find the distance from the center of this circle to $H$.
|
Solution. Draw the altitudes $B B_{1}$ and $C C_{1}$ in the triangle. Then the quadrilateral $A C_{1} H B_{1}$ is cyclic, since its opposite angles $C_{1}$ and $B_{1}$ are right angles. Therefore: $\angle B H C=$ $\angle C_{1} H B_{1}=180^{\circ}-\angle C_{1} A B_{1}$. Reflect point $H$ symmetrically with respect to the line $B C$ and denote the resulting point by $H_{1}$. Then $\angle B H_{1} C=\angle B H C=180^{\circ}-\angle C_{1} A B_{1}$,
so the quadrilateral $B A C H_{1}$ is cyclic and point $H_{1}$ lies on the circumcircle of triangle $A B C$. Denote the center of this circle by $O$.
By construction, we have $H_{1} A_{1}=H A_{1}=2$, therefore, in the isosceles triangle $O A H_{1}$, the base $A H_{1}$ is $3+2+2=7$, and the lateral side is equal to the radius of the circle: that is, 4. The height dropped to the base is
$$
\sqrt{4^{2}-(7 / 2)^{2}}=\sqrt{15} / 2
$$
and the distance from the base $O_{1}$ of this height to $H$ is $A H_{1} / 2-A H=7 / 2-3=1 / 2$. Therefore,
$$
O H=\sqrt{O O_{1}^{2}+O_{1} H^{2}}=\sqrt{(\sqrt{15} / 2)^{2}+(1 / 2)^{2}}=2
$$
Answer: 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9-4. From $n$ regular hexagons with side 1, a polygon on the plane was made by gluing hexagons along their sides. Any two hexagons either have exactly one common side or have no common points at all. There are no holes inside the polygon. Moreover, each hexagon has at least one side lying on the boundary of the polygon. What is the smallest perimeter that the polygon can have under these conditions?
|
Answer: $2 n+6$ for $n \geq 2$, $6$ for $n=1$.
Solution. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.
Let's draw a bit. Imagine that in each point that is an endpoint for three sides of a hexagon, we place a red dot. And each path along the sides of the hexagons from a red dot to a red dot we outline with a blue marker. Thus, the blue lines go along the sides of the hexagons and through vertices from which only two sides emerge (in other words, which are vertices of exactly one hexagon). Note that the blue lines do not intersect except at their endpoints, which are red dots. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.
Let's draw a bit. Imagine that in each point that is an endpoint for three sides of a hexagon, we place a red dot. And each path along the sides of the hexagons from a red dot to a red dot we outline with a blue marker. Thus, the blue lines go along the sides of the hexagons and through vertices from which only two sides emerge (in other words, which are vertices of exactly one hexagon). Note that the blue lines do not intersect except at their endpoints, which are red dots.
Thus, we have obtained a drawing of a multigraph on the plane. For it, Euler's formula $F-E+V=2$ holds, where $F, E, V$ are the number of faces, edges, and vertices, respectively. Since all vertices have degree $3, 3V = 2E$. Additionally, $F = n + 1$, since these are all the hexagons and the external face. From these three equations, we derive $E = 3n - 3$. Traversing the external cycle, we walked along all $n$ hexagons, so for $n > 1$, we changed the hexagon we were walking along at least $n$ times (note: this statement is not true if $n = 1$: we walked along one hexagon and never changed it). Thus, in the external cycle, there are at least $n$ edges, so in the rest of the graph, there are no more than $2n - 3$ edges. Each of these consists of exactly one segment that was a side for two hexagons, since inside the polygon, there cannot be points that are endpoints for exactly two side segments. Thus, inside there are no more than $4n - 6$ sides of hexagons, glued in pairs, so on the boundary, there are at least $2n + 6$ sides.
| Evaluation | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Minor errors in the proof |
| $+/-$ | 14 | Significant gaps in the proof (e.g., existence of a hexagon touching others by no more than 2 sides is not proven) |
| $+/2$ | 10 | There is a statement that the minimum perimeter for $n$ and $n+1$ differs by 2, but the proof is missing. Example is provided |
| $-/+$ | 6 | There is an example for $2n + 6$. There is a statement that the minimum perimeter for $n$ and $n+1$ differs by 2, but the proof is missing |
| - | 0 | Solution is completely incorrect/only the answer |
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number.
|
Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to the choice of $q$, we get $q=3$ and $n=12$.
B) Suppose the smallest divisors are of the form $p$ and $p^{2}$, where $p$ is a prime number. This case is analyzed similarly.
Remark. It is possible that a number has only three proper divisors. Then the difference mentioned in the condition is the difference between the largest and the smallest of the proper divisors. But any number with three proper divisors is a power of a prime $p^{4}$, and the difference $p^{3}-p$ cannot be a prime number.
| Evaluation | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 16 | Minor errors |
| $+/ 2$ | 10 | Significant gaps in the solution |
| .- | 6 | The idea that $p n q=(p+q)(n-p q)$ is present |
| - | 0 | The solution is completely incorrect/only the answer |
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. Along the shore of a circular lake with a perimeter of 1 km, two salmons are swimming - one at a constant speed of $500 \mathrm{m} /$ min clockwise, the other at a constant speed of 750 m/min counterclockwise. Along the edge of the shore, a bear is running, always moving along the shore at a speed of 200 m/min in the direction of the nearest salmon to it. How many full laps around the lake will the bear make in one hour?
ANSWER: 7.
CRITERIA
-/. For a specific initial position of the bear and the fish, direct calculations of their movements over the course of two or more laps of the bear around the lake are provided, followed by an explicit but unproven (though intuitively correct) statement that the average speed of the bear's rotation around the lake is approximately equal to that obtained in this numerical experiment. Based on this, the answer is obtained with an accuracy of $10 \%$.
|
Solution. Note that the bear runs to the nearest salmon if and only if it runs from a point on the shore that is equidistant from the salmons and not separated from the bear by the salmons. Since both points equidistant from the salmons move counterclockwise at a speed of $|750-500| / 2=125<200$ m/min, they will never catch up to the bear, and the bear will never be at such a point (except possibly at the initial moment). Since each of these points will run $125 \cdot 60$ meters in an hour, they will make exactly $125 \cdot 60 / 1000=15 / 2=7.5$ laps around the lake. Therefore, the bear, which never crosses either of these points, will make strictly more than 7 but strictly less than 8 laps, that is, 7 full laps.
|
7
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively, such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.
|
# 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.
## Answer: 7
#
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1.
|
Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow \quad(t-1)\left(t^{2}-(\lambda-1) t+(10-\lambda)\right)=0
$$
From this, it is clear that for any $\lambda$ there is a root $t=1$, so the value $x=1$ works. It remains to prove that there are no other values that are roots for any $\lambda$ (although this is obvious). Indeed, $t^{2}-(\lambda-1) t+(10-\lambda)=0$ implies $t=\frac{\lambda-1 \pm \sqrt{\lambda^{2}+2 \lambda-39}}{2}$. Take any pair of values of $\lambda$ for which the discriminant takes the same positive value, for example, for $\lambda=10$ and $\lambda=-12$ we have $t \in\{0,9\}$ and $t \in\{-11,-2\}$ - there are no intersections. Therefore, the answer is $x=1$.
Second Solution. Subtract the second equation from the first, transforming it, we get $(a-1)(b-1)(c-1)=0$. From this, it follows that one of $a, b, c$ is equal to one. Other $x$ do not work, as the triples $(a, b, c)=(4,1,1)$ and $(a, b, c)=(0,9,1)$ satisfy the condition.
| Grade | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Not proven why there are no others besides 1, or there is a minor error in the proof. |
| $+/-$ | 16 | Significant errors in the proof (several transitions with division by possibly zero, misunderstanding of the condition (with necessary calculations for the proof)). |
| $-/+$ | 10 | Two specific cases are considered, which show that it can only be equal to 1. But there is no proof that $= 1$. |
| .- | 6 | Only one case is found (it is claimed that $\in\{1,4\}$ or $\in\{0,1,9\}$. |
| - | 0 | The solution is completely incorrect/only the answer. |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not intersect pairwise?
Answer: 2.
|
Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+q$ or $x^{2}+x+q$.
The ranges of any two polynomials of different forms intersect: indeed, the values of the polynomials $x^{2}+q$ and $x^{2}+x+q^{\prime}$ coincide at $x=q-q^{\prime}$. Therefore, polynomials of different forms cannot be chosen.
Polynomials of the first form can be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+q$ and $f_{2}(x)=x^{2}+q^{\prime}$ do not intersect, then $q-q^{\prime}=4 k+2$ for some $k \in \mathbb{Z}$. Indeed, for an odd difference of the constant terms $q-q^{\prime}=2 k+1$, we have $f_{1}(k)=f_{2}(k+1)$. For a difference of the constant terms divisible by 4, $q-q^{\prime}=4 k$, we have $f_{1}(k-1)=f_{2}(k+1)$. But if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one does not have the form $4 k+2$.
Polynomials of the second form can also be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+x+q$ and $f_{2}(x)=x^{2}+x+q^{\prime}$ do not intersect, then $q-q^{\prime}=2 k+1$ for some $k \in \mathbb{Z}$. Indeed, for an even difference of the constant terms $q-q^{\prime}=2 k$, we have $f_{1}(k-1)=f_{2}(k)$. Again, if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one is even.
Thus, more than two polynomials cannot be chosen. Example for two: $f_{1}(x)=x^{2}$ and $f_{2}(x)=x^{2}+2$.
| Grade | Score | Content criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Solution is correct modulo minor inaccuracies |
| $+/-$ | 14 | There is a proof that for each type of values $\left(x^{2}+q\right.$ and $\left.x^{2}+x+q\right)$ no more than two trinomials can be taken |
| $-/+$ | 8 | Complete solution, but only for the case $x^{2}+q$ |
| .- | 4 | Example of two trinomials whose ranges do not intersect |
| - | 0 | Solution is completely incorrect/only answer |
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7-5. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3.
|
Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question for the 4th turtle. If both answers are 1, then turtles 1 and 3 are of the same gender. Then we ask about $1, 3, 4$. We will hear an answer of at least 2 if turtles 1 and 3 were males, and an answer of no more than 1 otherwise. In this case, the gender of turtle 4 is also uniquely determined, and one question remains to determine the gender of turtle 2.
We will prove that two questions are insufficient. First, note that: a) in one question, without knowing the total number of males, we cannot figure out even two turtles; b) knowing the total number of males, we can figure out two, but not three turtles. Therefore, in the situation with four turtles and two questions, asking the first question about all turtles at once or about one turtle is meaningless. Since when asking about a group of three or two turtles, we can hear an answer of 1, such a question will also not lead to success.
| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| $+/-$ | 14 | Correct answer, but not proven that it cannot be less |
| $+/ 2$ | 10 | Correct answer with minor flaws in the solution and not proven that it cannot be less |
| .- | 4 | Attempt to analyze cases with 2 and 3 questions, but made an incorrect conclusion that the minimum number is 4 |
| - | 0 | Solution is completely incorrect/only answer |
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.6. (20 points) Quadratic trinomials $P(x)$ and $Q(x)$ with real coefficients are such that together they have 4 distinct real roots, and each of the polynomials $P(Q(x))$ and $Q(P(x))$ has 4 distinct real roots. What is the smallest number of distinct real numbers that can be among the roots of the polynomials $P(x), Q(x), P(Q(x))$, and $Q(P(x))$?
|
# Answer: 6.
Solution. Note that if among the roots of the polynomial $P(Q(x))$ there is a root of $Q(x)$, say, the number $x_{0}$, then $P\left(Q\left(x_{0}\right)\right)=P(0)=0$, from which 0 is a root of $P(x)$. Similarly, if among the roots of $Q(P(x))$ there is a root of $P(x)$, then 0 is a root of $Q(x)$. However, $P(x)$ and $Q(x)$ cannot both have the root 0, otherwise, in total, they would have fewer than 4 roots.
From this, we can obtain an estimate of the total number of distinct roots. If there are no more than 5, then $P(Q(x))$ and $Q(x)$ have a common root, and $Q(P(x))$ and $P(x)$ have a common root, which cannot be according to the above.
Now let's construct an example where there are exactly 6 distinct roots. Let $P(x)=\frac{1}{2} x(x-3), Q(x)=$ $-\frac{3}{2}(x+1)(x-2)$. Then the roots of $P(x)$ will be the numbers 0 and 3; the roots of $Q(x)$ will be the numbers -1 and 2; the roots of $P(Q(x))$ will be the numbers -1, 0, 1, 2; the roots of $Q(P(x))$ will be the numbers -1, 1, 2, 4. In total, the roots of all polynomials together are the integers from -1 to 4.
Remark. There are other examples where there are exactly 6 distinct roots; in all examples, one of the quadratic polynomials $P(x)$ and $Q(x)$ has the root 0. For example, the pair of quadratic polynomials $P(x)=$ $x(x-3)$ and $Q(x)=-\frac{3}{4}(x+1)(x-4)$ also works.
## Criteria
The points for the estimate and the example are summed.
+6 points. Estimate - it is proven that there cannot be fewer than 6 distinct numbers among the roots of the considered polynomials.
In the absence of this proof, the following progress is evaluated:
+3 points. It is noted that if the polynomials $P(Q(x))$ and $Q(x)$ have a common root, then the polynomial $P(x)$ has the root 0.
+14 points. Example - suitable quadratic polynomials $P(x)$ and $Q(x)$ are provided, and it is shown that among the roots of all four considered polynomials, there are exactly 6 distinct numbers (or it is otherwise proven that such polynomials exist).
In the absence of such an example with justification, the highest of the following criteria is used:
+10 points. Correct example of quadratic polynomials $P(x)$ and $Q(x)$.
+12 points. Correct example of quadratic polynomials $P(x)$ and $Q(x)$, and the work also includes a correct estimate. (Thus, for a solution with a correct estimate and a correct pair of $P(x)$ and $Q(x)$ without proof that they fit, a total of 18 points is given.)
The following progress is not evaluated:
0 points. Only the answer is provided.
0 points. A correct example with 7 or more roots is provided.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8-4. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3.
|
Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question for the 4th turtle. If both answers are 1, then turtles 1 and 3 are of the same gender. Then we ask about $1, 3, 4$. We will hear an answer of at least 2 if turtles 1 and 3 were males, and an answer of no more than 1 otherwise. In this case, the gender of turtle 4 is also uniquely determined, and one question remains to determine the gender of turtle 2.
We will prove that two questions are insufficient. First, note that: a) in one question, without knowing the total number of males, we cannot figure out even two turtles; b) knowing the total number of males, we can figure out two, but not three turtles. Therefore, in the situation with four turtles and two questions, asking the first question about all turtles at once or about one turtle is meaningless. Since when asking about a group of three or two turtles, we can hear an answer of 1, such a question will also not lead to success.
| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| $+/-$ | 14 | Correct answer, but not proven that it cannot be less |
| $+/ 2$ | 10 | Correct answer with minor flaws in the solution and not proven that it cannot be less |
| .- | 4 | Attempt to analyze cases with 2 and 3 questions, but made an incorrect conclusion that the minimum number is 4 |
| - | 0 | Solution is completely incorrect/only answer |
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8-5. 8 8-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number.
|
Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to the choice of $q$, we get $q=3$ and $n=12$.
B) Suppose the smallest divisors are of the form $p$ and $p^{2}$, where $p$ is a prime number. This case is analyzed similarly.
Remark. It is possible that a number has only three proper divisors. Then the difference mentioned in the condition is the difference between the largest and the smallest of the proper divisors. But any number with three proper divisors is a power of a prime $p^{4}$, and the difference $p^{3}-p$ cannot be a prime number.
| Evaluation | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 16 | Minor errors |
| $+/ 2$ | 10 | Significant gaps in the solution |
| .- | 6 | The idea that $p n q=(p+q)(n-p q)$ is present |
| - | 0 | The solution is completely incorrect/only the answer |
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Natural numbers $x$ and $y$ are such that the following equality holds:
$$
x^{2}-3 x=25 y^{2}-15 y
$$
How many times greater is the number $x$ than the number $y$?
|
Answer. 5
Solution. The equality $x^{2}-3 x=25 y^{2}-15 y$ is equivalent to the equality $(5 y-x)(5 y+x-3)=$ 0. Since $x$ and $y$ are natural numbers, $x, y \geq 1$. Therefore, the second bracket $5 y+x-3 \geq 3$, in particular, is non-zero. Therefore, the first bracket $5 y-x$ is zero, meaning $x$ is 5 times $y$.
Criteria.
## $(-)$
$(-$.$) An example of specific numbers is considered.$
$(-/+)$ From the equality $x(x-3)=5 y(5 y-3)$, the conclusion $x=5 y$ is drawn without satisfactory explanations.
(+/2) A complete square is factored out on both sides, but an incorrect conclusion is made $(x-1.5)^{2}=$ $(5 y-1.5)^{2} \Rightarrow x-1.5=5 y-1.5$. The second case $(x-1.5)=-(5 y-1.5)$ is not considered.
(+/-) The equality $(x-5 y)(x+5 y-3) \neq 0$ is obtained, from which it is concluded that $x-5 y=0$, but it is not explained why $(x+5 y-3) \neq 0$.
(+.) Minor logical steps are omitted
$(+)$
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha played 7 matches, and Misha played 11 matches. How many times did Pete win against Sasha?
|
Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and Misha in $15-11=4$ games.
Notice now that in a round-robin tournament, one player cannot skip two consecutive games. Since Pasha did not participate in 8 out of 15 games, it means he did not participate in the very first game and then skipped every second game. This means Pasha lost all his games.
Therefore, the number of Petya's wins over Pasha is equal to the number of games in which Petya and Pasha met. This number is equal to the number of games in which Misha did not participate, which is 4, as found earlier.
Criteria.
$(-$.$) Correct answer for a specific case, or it is proven that 15 games were played.
(-/+) Correct answer for a specific case + it is proven that 15 games were played
(+/2) It is proven that Sasha lost all his games, and the correct answer for a specific example, without sufficient justification.
$(+/-)$
(+.) The problem is solved correctly with one flaw: there is no strict proof that 15 games were played.
$(+)$ The answer is correct and strictly justified.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. How many different (i.e., not equal to each other) acute triangles with integer side lengths and a perimeter of 24 exist? List the lengths of the three sides of all these triangles and prove that there are no others.
ANSWER: 6.
CRITERIA
b - correct criterion for acuteness and either a correct list of acute triangles but without full justification (it is stated without justification that only these sets of lengths satisfy the criterion, or the enumeration is incomplete), or a complete enumeration with an arithmetic error leading to the erroneous inclusion or exclusion of a triangle.
$d$ - correct criterion for acuteness and either an incorrect list without detailed proof or with significant errors in the proof, or no list.
* - a complete list of all triangles, not just acute ones.
|
Solution. Note that a triangle with sides $a \geqslant b \geqslant c$ is obtuse if and only if $a^{2}>b^{2}+c^{2}$. Indeed, if $A B C$ is a triangle with sides $a \geqslant b \geqslant c$ opposite vertices $A, B$, and $C$ respectively, and $A B^{\prime} C$ is a triangle with a right angle $\angle A$ and $A B^{\prime}=A B$, then $B$ and $C$ lie on opposite sides of the line $A B^{\prime}$, otherwise in the triangle $C B B^{\prime}$ it would be $\angle C B B^{\prime}>\angle A B B^{\prime}=\angle A B^{\prime} B>\angle B B^{\prime} C$, and $B C=ab^{2}+c^{2}$ follows that $\angle C A B>\angle C A B^{\prime}=90^{\circ}$, the converse is proved similarly.
Let's count the triples of natural numbers satisfying the conditions $a>b \geqslant c, a^{2}a^{2}$.
For $a=8$ there is one pair of numbers $(b, c)$ such that $8 \geqslant b \geqslant c$ and $b+c=16$: this is $(8,8)$, and the corresponding triangle is equilateral.
For $a=9$ there are two pairs of numbers $(b, c)$ such that $9 \geqslant b \geqslant c$ and $b+c=15$: these are $(8,7)$ and $(9,6)$, both satisfy the inequality $b^{2}+c^{2}>a^{2}$.
For $a=10$ there are four pairs of numbers $(b, c)$ such that $10 \geqslant b \geqslant c$ and $b+c=14$: these are $(7,7),(8,6),(9,5)$ and $(10,4)$. The last two satisfy the inequality $b^{2}+c^{2}>a^{2}$, while the first two do not.
For $a=11$ there are five pairs of numbers $(b, c)$ such that $11 \geqslant b \geqslant c$ and $b+c=13$: these are $(7,6),(8,5),(9,4),(10,3)$ and $(11,2)$. Only the last one satisfies the inequality $b^{2}+c^{2}>a^{2}$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. (20 points) There were $n$ identical-looking coins weighing $x_{1}, x_{2}, \ldots, x_{n}$ grams (the weights of the coins are pairwise distinct positive real numbers), and also weightless stickers with numbers $x_{1}, x_{2}, \ldots, x_{n}$. At night, a lab assistant weighed the coins and labeled them with stickers. It is required to check using a balance scale that he did not mix anything up. For example, if $n=6, x_{1}=1, \ldots, x_{6}=6$, then this can be done in 2 weighings, verifying that
$$
\begin{aligned}
& 1+2+3=6 \\
& 1+6<3+5
\end{aligned}
$$
Does there exist for $n=8$ a set of weights $x_{1}, x_{2}, \ldots, x_{8}$, the correct labeling of which can be verified in 2 weighings?
|
Answer: 2 yes
Solution. Consider the weights 10, 20, 25, 30, 35, 201, 203, 207 (weights will be measured in grams hereafter). We will perform two checks:
$$
\begin{aligned}
10+20+30 & =25+35 \\
10+25+201 & <30+207
\end{aligned}
$$
First, consider the first weighing. We will prove that if some three coins balance some two coins, then these must be 10, 20, 30 on one side and 25, 35 on the other.
We will call the coins 10, 20, 25, 30, 35 small, and the coins 201, 203, 207 - large.
If among the five coins participating in the first weighing there are large ones, then on each side there must be exactly one such coin (otherwise, the side with more such coins would outweigh). At the same time, the weights of all small coins are divisible by 5, while the weights of large coins give different remainders when divided by 5. Then the total weights on the sides give different remainders when divided by 5, which is impossible in the case of equality.
Therefore, all five coins are small. Then each side weighs $\frac{1}{2}(10+20+25+30+35)=$ $=60$. It is easy to understand that the sum of the weights of two coins can be 60 only if these two coins weigh 25 and 35. Then the other three coins weigh 10, $20,30$.
Thus, if the first weighing shows equality, then 10, 20, 30 are on the side with 3 coins (this group will be called $A$), and 25, 35 - on the side with two coins (this group will be called $B$). At the same time, the coins 201, 203, 207 (this group will be called $C$) do not participate in the first weighing.
Now consider the second weighing. On one side, one coin from each of the three groups $A, B, C$ is taken, the minimum sum of the weights of such coins is $10+25+201=236$. On the other side, one coin from groups $A$ and $C$ is taken, the maximum sum of the weights of such coins is $30+$ $207=237$. Given that 237 is only 1 more than 236, the inequality in the second weighing will not hold for any other coins. Therefore, on one side lie the coins $10,25,201$, and on the other - $30,207$.
Thus, in each weighing, we uniquely determined the set of coins on each side. At the same time, for all 8 coins, the pairs of groups they belong to in the first and second weighings (either on the side with 3 coins, or on the side with 2 coins, or not weighed) are different. Therefore, the weight of each coin is uniquely determined.
Remark. In any correct algorithm, one weighing must establish the equality of the weights of two coins on one side with three coins on the other, and the other weighing must establish that two coins on one side are heavier than three coins on the other. This can be understood from the following considerations.
- Coins that are in one of the three groups (on the first side, on the second, or not weighed) in one weighing must end up in different groups in the second weighing. (Indeed, if two coins ended up in the same groups in both weighings, the lab assistant could mix them up and we would not notice.)
- From the previous point, it follows that no weighing can create a group of 4 or more coins. This means that in each weighing, there are either 3 coins on each side, or 3 and 2.
- Also from the first point, it follows that the coins can be arranged in a $3 \times 3$ table, with the groups of the first weighing along the rows and the groups of the second weighing along the columns; exactly one cell will remain without a coin.
| $a_{1}$ | $a_{2}$ | $a_{3}$ |
| :--- | :--- | :--- |
| $b_{1}$ | $b_{2}$ | $b_{3}$ |
| $c_{1}$ | | $c_{3}$ |
- Both weighings cannot have the same format and outcome. (For example, it cannot be that in both weighings 2 coins on one side outweigh 3 coins on the other.) Indeed, in this case, if the rows and columns in the table above are ordered in the same way (for example, in the order light/heavy/unweighed), the empty cell would be on the main diagonal; then the lab assistant could mix up the labels in such a way that the coins would be reflected relative to this diagonal, and we would not notice this.
- There cannot be a weighing in which there are 3 coins on each side, and they balance each other. For example, let the first weighing be set up this way; its sides correspond to the first and second rows of our table. Then the lab assistant could mix up the labels in such a way that the first and second rows would swap places. This would not affect the results of our weighings, that is, we would not notice such an error.
- There cannot be a weighing in which there are 3 coins on each side, and one side is heavier than the other. To prove this, assume the contrary - let this be the first weighing. The lighter side of this weighing corresponds to the first row, and the heavier side to the third row; the columns are ordered so that the group of 2 coins from the second weighing is in the last column (regardless of the structure of the second weighing).
| $a_{1}$ | $a_{2}$ | $a_{3}$ |
| :--- | :--- | :--- |
| $b_{1}$ | $b_{2}$ | |
| $c_{1}$ | $c_{2}$ | $c_{3}$ |
If we now reorder the numbers in the columns from top to bottom in ascending order (leaving the empty cell empty), the weight readings will not change: the composition of the columns will not change, and the bottom row will be heavier than the top row. But if we then swap, for example, the two lower coins in the central column, the bottom row will still be heavier than the top row! It turns out that we cannot uniquely determine the placement of the coins, that is, the lab assistant has a chance to deceive us again.
- Similarly, it can be shown that there cannot be a weighing in which the side with three coins outweighs the side with two.
It is also not difficult to understand from similar considerations that for 9 coins, it is impossible to get by with two weighings in any situation.
## Criteria
The highest suitable criterion is used:
20 points. A correct example of a set of masses and a description of two weighings that allow checking the correctness of the labeling is provided.
3 points. The work contains an indication that coins that end up in the same group during the first weighing must end up in different groups during the second weighing.
3 points. The work provides an example of a set of masses and a description of two weighings that do not allow checking the correctness of the labeling, but coins that end up in the same group during each weighing end up in different groups during the other weighing.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha - 7 matches, Misha - 11 matches. How many times did Pete win against Sasha?
|
Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and Misha in $15-11=4$ games.
Notice now that in a round-robin tournament, one player cannot skip two consecutive games. Since Pasha did not participate in 8 out of 15 games, it means he did not participate in the very first game and then skipped every second game. This means Pasha lost all his games.
Therefore, the number of Petya's wins over Pasha is equal to the number of games in which Petya and Pasha met. This number is equal to the number of games in which Misha did not participate, which is 4, as found earlier.
Criteria.
$(-$.$) Correct answer for a specific case, or it is proven that 15 games were played.
$(-/+)$ Correct answer for a specific case + it is proven that 15 games were played
(+/2) It is proven that Pasha lost all his games, and the correct answer for a specific example, without sufficient justification.
$(+/-)$
(+.) The problem is solved correctly with one flaw: there is no strict proof that 15 games were played.
$(+)$ The answer is correct and strictly justified.
## Mathematics - solutions
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1.
|
Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow \quad(t-1)\left(t^{2}-(\lambda-1) t+(10-\lambda)\right)=0
$$
From this, it is clear that for any $\lambda$ there is a root $t=1$, so the value $x=1$ works. It remains to prove that there are no other values that are roots for any $\lambda$ (although this is obvious). Indeed, $t^{2}-(\lambda-1) t+(10-\lambda)=0$ implies $t=\frac{\lambda-1 \pm \sqrt{\lambda^{2}+2 \lambda-39}}{2}$. Take any pair of values of $\lambda$ for which the discriminant takes the same positive value, for example, for $\lambda=10$ and $\lambda=-12$ we have $t \in\{0,9\}$ and $t \in\{-11,-2\}$ - there are no intersections. Therefore, the answer is $x=1$.
Second Solution. Subtract the second equation from the first, transforming it, we get $(a-1)(b-1)(c-1)=0$. From this, it follows that one of $a, b, c$ is equal to one. Other $x$ do not work, as the triples $(a, b, c)=(4,1,1)$ and $(a, b, c)=(0,9,1)$ satisfy the condition.
| Grade | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Not proven why there are no others besides 1, or there is a minor error in the proof. |
| $+/-$ | 16 | Significant errors in the proof (several transitions with division by possibly zero, misunderstanding of the condition (with necessary calculations for the proof)). |
| $-/+$ | 10 | Two specific cases are considered, which show that it can only be equal to 1. But there is no proof that $= 1$. |
| .- | 6 | Only one case is found (it is claimed that $\in\{1,4\}$ or $\in\{0,1,9\}$. |
| - | 0 | The solution is completely incorrect/only the answer. |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not intersect pairwise?
Answer: 2.
|
Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+q$ or $x^{2}+x+q$.
The ranges of any two polynomials of different forms intersect: indeed, the values of the polynomials $x^{2}+q$ and $x^{2}+x+q^{\prime}$ coincide at $x=q-q^{\prime}$. Therefore, polynomials of different forms cannot be chosen.
Polynomials of the first form can be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+q$ and $f_{2}(x)=x^{2}+q^{\prime}$ do not intersect, then $q-q^{\prime}=4 k+2$ for some $k \in \mathbb{Z}$. Indeed, for an odd difference of the constant terms $q-q^{\prime}=2 k+1$, we have $f_{1}(k)=f_{2}(k+1)$. For a difference of the constant terms divisible by 4, $q-q^{\prime}=4 k$, we have $f_{1}(k-1)=f_{2}(k+1)$. But if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one does not have the form $4 k+2$.
Polynomials of the second form can also be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+x+q$ and $f_{2}(x)=x^{2}+x+q^{\prime}$ do not intersect, then $q-q^{\prime}=2 k+1$ for some $k \in \mathbb{Z}$. Indeed, for an even difference of the constant terms $q-q^{\prime}=2 k$, we have $f_{1}(k-1)=f_{2}(k)$. Again, if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one is even.
Thus, more than two polynomials cannot be chosen. Example for two: $f_{1}(x)=x^{2}$ and $f_{2}(x)=x^{2}+2$.
| Grade | Score | Content criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Solution is correct modulo minor inaccuracies |
| $+/-$ | 14 | There is a proof that for each type of values $\left(x^{2}+q\right.$ and $\left.x^{2}+x+q\right)$ no more than two trinomials can be taken |
| $-/+$ | 8 | Complete solution, but only for the case $x^{2}+q$ |
| .- | 4 | Example of two trinomials whose ranges do not intersect |
| - | 0 | Solution is completely incorrect/only answer |
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. On side $AB$ of triangle $ABC$, a point $D$ is taken. In triangle $ADC$, the angle bisectors $AP$ and $CQ$ are drawn. On side $AC$ of triangle $ADC$, a point $R$ is taken such that $PR \perp CQ$. It is known that the angle bisector of angle $D$ of triangle $BCD$ is perpendicular to segment $PB$, $AB=18$, $AP=12$. Find $AR$. 8.
|
Solution. Let $A B=n, A P=m, A D=c, A C=b, D P=p, C P=q, B D=D P=p, C P=C R=q$.
Using the property of the angle bisector and the formula for the length of the angle bisector, we get $\left\{\begin{array}{l}\frac{c}{b}=\frac{p}{q}, \\ c+p=A B=n, \\ m^{2}=b c-p q,\end{array}\right.$
$\left\{\begin{array}{l}c=\frac{b p}{q}, \\ \frac{b p}{q}+p=n, \\ m^{2}=b \frac{b p}{q}-p q:\end{array}\left\{\begin{array}{l}c=\frac{b p}{q}, \\ b p+p q=q n, \\ m^{2}=b \frac{b p}{q}-p q,\end{array}\left\{\begin{array}{l}c=\frac{b p}{q}, \\ p=\frac{q n}{b q}, \\ m^{2}=b \frac{b}{q} \frac{q n}{b+q}-q \frac{q n}{b+q}:\end{array} m^{2}=\frac{b^{2} n}{b+q}-\frac{q^{2} n}{b+q}, m^{2}=\frac{\left(b^{2}-q^{2}\right) n}{b+q}:\right.\right.\right.$ $m^{2}=\frac{(b-q)(b+q) n}{b+q}, m^{2}=(b-q) n, b-q=\frac{m^{2}}{n}, b-q=\frac{12^{2}}{18}=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. With the resulting set of 4 points, the same procedure is carried out - two points are chosen, the perpendicular bisector is drawn, and all points are reflected across it. What is the maximum number of distinct points that can be obtained by repeatedly performing this procedure?
|
Answer: 12 points.
## Solution. FIRST WAY.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
#
|
# 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
#
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Each move of a chess knight is a move of one square horizontally and two squares vertically, or vice versa - one square vertically and two squares horizontally. (In the diagram on the right, the knight marked with the letter $\mathrm{K}$ can move to any of the shaded squares in one move.)
A chess knight is placed in an arbitrary square of a rectangular board measuring $2 \times$ 2016 squares. Moving according to the described rules (and not going beyond the edges of the board), it can move from this square to some other squares on the board, but not to all. What is the minimum number of squares that need to be added to the board so that the knight can move from any square of the board to all other squares? (Adding a square is done so that it shares a common side with one of the existing squares. Any number of squares can be added, and the resulting board does not necessarily have to be rectangular).
|
Answer: 2
Solution. Let's number the cells as shown in Figure 1. The knight can move from any cell to any cell with the same number, and cannot move to other cells.
| $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\ldots$ |
Fig. 1
Thus, all cells are divided into 4 sets, such that the knight cannot jump from one set to another, but can freely move within one set.
Let's add two cells as shown in Figure 2. We will prove that now the knight can move from any cell to any other cell. Cell A "connects" the cells marked in pink, i.e., through it, one can jump from set 1 to set 4. (Being in any cell with number 1, one can reach the pink cell with number 1, then through A reach the pink cell with number 4, and from there - to any cell with number 4. In the end, from any cell with number 1, one can reach any cell with number 4 using A.)
Cell B connects the cells marked in green, i.e., through it, one can jump from any of the sets 2, 3, 4 to any of these sets.
In the end, by combining cells A and B, one can move from any set to any other set. For example, to move from set 1 to set 2, first use A to move from 1 to 4, then use B to move from 4 to 2.
| | $\mathrm{B}$ | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 3 | 1 | 4 | 2 | 3 | 1 | 4 | 2 | 3 | $\ldots$ | | | | | | | | | | | | | | | | | | | |
| 1 | 4 | 2 | 3 | 1 | 4 | 2 | 3 | 1 | 4 | $\ldots$ | | | | | | | | | | | | | | | | | | | |
Fig. 2
It remains to be convinced that one cell is not enough. From Figure 3, it is clear that all added cells are divided into two types: cells A can be reached from no more than 3 cells on the board, cells B can be reached from 4 cells, but among them, there are always two from the same set (marked with the same number). That is, a cell that can be reached from all 4 sets does not exist.
| | $A$ | $A$ | $B$ | $B$ | $\cdots$ | | | | | $\cdots$ | B | B | A | A | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| A | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\cdots$ | $\mathbf{O}$ | $\mathbf{1}$ | $\mathbf{4}$ | A |
| A | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\cdots$ | | $\mathbf{2}$ | $\mathbf{3}$ | A |
| | A | A | B | B | $\cdots$ | | | | | $\cdots$ | B | B | A | A | |
Fig. 3
Criteria for evaluating solutions. The solution is evaluated based on the presence of three components:
a) The correct method of adding two cells is shown,
b) It is proven that with this method of adding two cells, the knight can indeed traverse the entire board,
c) It is proven that one cell is not enough.
(-) The solution does not meet any of the criteria listed below.
$(-/+)$ The solution only contains a).
(+/2) a) and b) are present, c) is missing or incorrect.
(+/-) a) and c) are present, b) is missing or incorrect.
$(+)$ a), b), and c) are present.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27. On a perfectly competitive market for a good, the market demand and market supply functions are given respectively as: $Q^{d}=60-14 P$ and $Q^{s}=20+6 P$. It is known that if the amount of labor used by a typical firm is $\mathbf{L}$ units, then the marginal product of labor is $\frac{160}{L^{2}}$. Determine how many units of labor a typical firm will hire in a competitive labor market if the equilibrium price of a unit of labor is 5 rubles.
|
Solution: find the equilibrium price of the good $60-14 P=20+6 P, P=2$. Optimal choice of the firm: MPL $x P=w ; \frac{160}{L^{2}} \times 2=5 ; \boldsymbol{L}=\boldsymbol{8}$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. With the resulting set of 4 points, the same procedure is repeated - two points are chosen, the perpendicular bisector is drawn, and all points are reflected across it. What is the maximum number of distinct points that can be obtained by repeatedly performing this procedure?
|
Answer: 12 points.
## Solution. FIRST WAY.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (3 points) In triangle $A B C$, a square $K L M N$ with side length 1 is inscribed: points $K$ and $L$ lie on side $A C$, points $M$ and $N$ lie on sides $A B$ and $B C$ respectively. The area of the square is half the area of the triangle. Find the length of the height $B H$ of triangle $A B C$.
|
Answer: 2.
Solution 1:
$A C \cdot B H=2 S_{A B C}=4 S_{K L M N}=4$.
Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute. From the condition, it follows that
$$
\begin{gathered}
S_{A M L}+S_{C K N}+S_{B M N}=S_{K L M N} \\
\frac{A L \cdot 1}{2}+\frac{C K \cdot 1}{2}+\frac{(B H-1) \cdot 1}{2}=1 \\
A L+C K+B H-1=2 \\
A C-1+B H-1=2 \\
A C+B H=4=A C \cdot B H=2+\frac{A C \cdot B H}{2} . \\
2 A C+2 B H=A C+B H+4 \\
(A C-2)(B H-2)=0 \\
{\left[\begin{array}{l}
A C=2 \\
B H=2
\end{array}\right.}
\end{gathered}
$$
If only one of the equalities in this system is satisfied, then $A C \cdot B H$ will be either greater or less than 4, which is impossible. Therefore, $A C=B H=2$.
## Solution 2:
Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute.
On ray $L K$, take a point $X$ such that $A L=L X$. Then triangles $A M L$ and $X M L$ are equal. Similarly, on ray $K L$, take a point $Y$ such that $C K=K X$. Then triangles $C K N$ and $Y N L$ are equal. Let $Z$ be the intersection of $M X$ and $N Y$. Triangle $M N Z$ is equal to triangle $M N B$ by side $M N$ and the angles adjacent to it.
We obtain the equality:
$$
S_{K L M N}=S_{A M L}+S_{C K N}+S_{B M N}=S_{X M L}+S_{Y K N}+S_{M N Z}=S_{K L M N} \pm S_{X Y Z}
$$
where the sign $\pm$ depends on whether point $Z$ lies inside or outside triangle $A B C$. This area must be equal to 0, i.e., point $Z$ lies on $A C$. From this, the height of triangle $Z M N$, dropped from $Z$ to $M N$, is equal to the side of the square. But this height is equal to the height of triangle $B M N$, dropped from $B$ to $M N$. If we add the side of the square to this height, we get the height $B H$, from which $B H=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given a rectangle $\mathrm{ABCD}$. The length of side $\mathrm{BC}$ is one and a half times less than the length of side $\mathrm{AB}$. Point $\mathrm{K}$ is the midpoint of side $\mathrm{AD}$. Point $\mathrm{L}$ on side $\mathrm{CD}$ is such that $\mathrm{CL}=\mathrm{AK}$. Point $\mathrm{M}$ is the intersection of line $\mathrm{BL}$ and the perpendicular bisector of segment $\mathrm{DL}$. It is known that $\mathrm{KL}=4$. Find the length of segment $\mathrm{BM}$.
|
Answer: 8
## Examples of answer recording: 14 0.5
#
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a watch repair shop, there is a certain number of electronic watches (more than one), displaying time in a 12-hour format (the number of hours on the watch face changes from 1 to 12). All of them run at the same speed, but show completely different times: the number of hours on the face of any two different watches is different, and the number of minutes is also different.
One day, the master added up the number of hours on the faces of all the watches, then added up the number of minutes on the faces of all the watches, and remembered the two resulting numbers. After some time, he did the same again and found that both the total number of hours and the total number of minutes had decreased by 1. What is the maximum number of electronic watches that could have been in the shop?
|
Answer: 11
## Examples of answer notation:
14
#
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. (3 points)
In triangle $A B C$, the median $B M$ is drawn, in triangle $M C B$ - the median $B N$, in triangle $B N A$ - the median $N K$. It turned out that $N K \perp B M$. Find $A C: B C$.
Answer: 2
#
|
# Solution:
Let $\vec{b}=\overrightarrow{C B}$ and $\vec{a}=\overrightarrow{C A}$.
$\overrightarrow{N K}=\overrightarrow{C K}-\overrightarrow{C N}=\frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}}{4}=\frac{\vec{a}+2 \vec{b}}{4}$.
$\overrightarrow{B M}=\overrightarrow{C M}-\overrightarrow{C B}=\frac{\vec{a}}{2}-\vec{b}$
$0=\overrightarrow{B M} \cdot \overrightarrow{N K}=\frac{\vec{a}^{2}-4 \vec{b}^{2}}{8}$, from which $|\vec{a}|=2|\vec{b}|$, that is, $A C: B C=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 7. (4 points)
Natural numbers $a, b, c$ are such that $\operatorname{GCD}(\operatorname{LCM}(a, b), c) \cdot \operatorname{LCM}(\operatorname{GCD}(a, b), c)=200$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(a, b), c)$ can take?
|
Answer: 10
## Solution:
Note that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$200=2^{3} \cdot 5^{2}$ and in the second factor, each prime number enters with an exponent no less than in the first. Therefore, the maximum possible value of $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ is $2 \cdot 5=10$. This value is achieved when $x=y=10, z=20$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The three heights of a tetrahedron are three, four, and four times greater than the radius of its inscribed sphere, respectively. How many times greater is the fourth height than the radius of the inscribed sphere?
|
Answer: 6
## Examples of answer notations:
14
$1 / 4$
#
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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