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2. On the first day, Masha collected $25 \%$ fewer berries than Vanya, and on the second day, $20 \%$ more than Vanya. Over the two days, Masha collected $10 \%$ more berries than Vanya. What is the smallest number of berries they could have collected together? | # Solution
Masha collected $3 / 4$ on the first day and $-6 / 5$ of the number of berries collected by Vanya over these days. Let Vanya collect $4 x$ berries on the first day and $5 y$ on the second day, then Masha collected $3 x$ and $6 y$ berries respectively. According to the condition, $3 x + 6 y = 11 / 10 (4 x + ... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,091 |
3. Find the angle at the vertex of the axial section of a right circular cone, given that there exist three generatrices of the lateral surface of the cone that are pairwise perpendicular to each other.
# | # Solution
From the problem statement, it follows that a triangular pyramid $P A B C$ can be inscribed in the given cone, where the lateral edges $P A, P B$, and $P C$ are equal, and all planar angles at vertex $P$ are right angles. Therefore, this pyramid is regular, and its height is the segment $P O$, where $O$ is ... | 2\arcsin\frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,092 |
4. Prove that if $\alpha, \beta$ and $\gamma$ are the angles of an acute triangle, then $\sin \alpha + \sin \beta + \sin \gamma > 2$.
# | # Solution
The first method ("trigonometric"). First, we prove an auxiliary statement: If $\alpha, \beta$, and $\gamma$ are the angles of an arbitrary triangle, then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma=1$.
Indeed, since $\gamma=180^{\circ}-(\alpha+\beta)$, we have $... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,093 |
5. In triangle $A B C$, point $D$ is the midpoint of side $A B$. Can points $E$ and $F$ be placed on sides $A C$ and $B C$ respectively, such that the area of triangle $D E F$ is greater than the sum of the areas of triangles $A E D$ and $B F D$? | # Solution
## First Method.
Consider an arbitrary triangle $ABC$ with points $E$ and $F$ on sides $AC$ and $BC$. Let $C'$ be the image of point $C$, and $F'$ be the image of point $F$ under the symmetry with center at point $D$ (see Fig. 1). Then the quadrilateral $ACBC'$ is a parallelogram, and point $F'$ lies on it... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,094 |
Task No. 1.1
# Condition:
The figure shows 4 circles.
Find the sum of the numbers that are in exactly two circles.
# | # Answer: 22
## Exact match of the answer -1 point
## Solution.
The gray areas on the diagram represent the regions that are included in exactly two circles. The number 10 is in the orange and blue circles, the number 2 is in the blue and brown circles, the number 1 is in the green and brown circles, and the number ... | 59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,095 |
# Task No. 3.1
## Condition:
The letters A, B, V, G, D, E, and Z represent the digits $2, 3, 4, 5, 6, 7$, and 8 in some order. The same letters correspond to the same digits, different letters to different digits.
| A | $:$ | B | $=$ | V | |
| :---: | :---: | :---: | :---: | :---: | :---: |
| | | | + | | |
| B... | # Answer:
| 8 | : | 2 | $=$ | 4 |
| :---: | :---: | :---: | :---: | :---: |
| | | + | | |
| 2 | $\cdot$ | 3 | $=$ | 6 |
| | | $=$ | | |
| 7 | - | 5 | $=$ | 2 |
## Exact match of the answer -1 point
Solution.
From the given set of digits, only 8 and 6 can be represented as the product of two different digits... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,096 |
9.1. Several numbers are written on the board. It is known that the square of any written number is greater than the product of any two other written numbers. What is the maximum number of numbers that can be on the board
# | # Answer. 3 numbers.
Solution. Suppose there are at least four numbers, and $a-$ is the number with the smallest absolute value. Among the remaining numbers, at least two have the same sign (both non-negative or both non-positive). Let these numbers be $b$ and $c$; then $bc = |bc| \geqslant |a|^2 = a^2$, which contrad... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 13,101 |
9.2. Circles $\omega_{1}$ and $\omega_{2}$ touch each other externally at point $P$. Through the center of $\omega_{1}$, a line $\ell_{1}$ is drawn, tangent to $\omega_{2}$. Similarly, a line $\ell_{2}$ is tangent to $\omega_{1}$ and passes through the center of $\omega_{2}$. It turns out that the lines $\ell_{1}$ and ... | Solution. Let $O_{1}, r_{1}$ and $O_{2}, r_{2}$ be the centers
Fig. 1
 and asks the Spectator to mentally choose a card and remember the column it is in. After this, the Magician collects the cards in a certain way, lays them out again in a $6 \times 6$ square, and asks the Spectator to name the numbers... | Solution. Let the Magician, after the first action, not shuffle the cards but collect them, preserving the order in the columns, and stack them into a deck one column at a time. The second time, he lays out the cards row by row, i.e., the former columns become rows (this is easy to do by laying out the same deck horizo... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,104 |
9.6. The numbers $a$ and $b$ are such that $a^{3}-b^{3}=2, a^{5}-b^{5} \geqslant 4$. Prove that $a^{2}+b^{2} \geqslant 2$
(I. Bogdanov) | Solution. Note that $2\left(a^{2}+b^{2}\right)=\left(a^{2}+b^{2}\right)\left(a^{3}-b^{3}\right)=$ $=\left(a^{5}-b^{5}\right)+a^{2} b^{2}(a-b) \geqslant 4+a^{2} b^{2}(a-b)$. Since $a^{3}>b^{3}$, we have $a>b$, which means $a^{2} b^{2}(a-b) \geqslant 0$. Therefore, $2\left(a^{2}+b^{2}\right) \geqslant 4$, from which the ... | ^{2}+b^{2}\geqslant2 | Inequalities | proof | Yes | Yes | olympiads | false | 13,105 |
9.7. On the side $AC$ of triangle $ABC$, an arbitrary point $D$ is marked. Let $E$ and $F$ be the points symmetric to point $D$ with respect to the bisectors of angles $A$ and $C$ respectively. Prove that the midpoint of segment $EF$ lies on the line $A_{0} C_{0}$, where $A_{0}$ and $C_{0}$ are the points of tangency o... | Solution. Let $B_{0}$ be the point of tangency of the inscribed circle with side $A C$. We can assume that point $D$ lies on the segment $A B_{0}$. Points $A_{0}$ and $C_{0}$ are symmetric to point $B_{0}$ with respect to the bisectors of angles $C$ and $A$ respectively. Therefore, point $E$ lies on the segment $A C_{0... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,106 |
9.8. Given an $n \times n$ square. Initially, its cells are painted in white and black in a checkerboard pattern, with at least one of the corner cells being black. In one move, it is allowed to simultaneously repaint the four cells in some $2 \times 2$ square according to the following rule: each white cell is repaint... | Answer. For all $n$ that are multiples of three.
Solution. Suppose we managed to repaint the cells as required by the conditions of the problem. We will call cells of the first type those that were initially white, and cells of the second type those that were black. Note that if a cell is repainted three times, it doe... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,107 |
Problem 10.1. (a) (2 points) A natural number $n$ is less than 120. What is the largest remainder that the number 209 can give when divided by $n$?
(b) (2 points) A natural number $n$ is less than 90. What is the largest remainder that the number 209 can give when divided by $n$? | Answer: (a) 104. (b) 69.
Solution. Let $209=n k+r$, where $k-$ is the quotient, and $r$ is the remainder of the division. Since $rn k+r=209=n k+r>r k+r=r(k+1)$, hence
$$
k+1>\frac{209}{n} \quad \text { and } \quad r\frac{209}{119}$, i.e., $k \geqslant 1$. Then $r\frac{209}{89}$, i.e., $k \geqslant 2$. Then $r<\frac{2... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,108 |
Problem 10.2. A real number $a$ is such that of the two equations
$$
5+|x-2|=a \text { and } 7-|2 x+6|=a
$$
one has exactly one solution, and the other has exactly two solutions. What can $a$ be? List all possible options. | Answer: 5, 7.
Solution. Consider the first equation, equivalent to the equation $|x-2|=a-5$. If $a-5>0$, then it has exactly two roots $x=$ $2 \pm(a-5)$. If $a-5=0$, i.e., $a=5$, then it has exactly one root $x=2$.
Now consider the second equation, equivalent to the equation $|2 x+6|=7-a$. If $7-a>0$, then it has exa... | 5,7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,109 |
Problem 10.3. On the board, natural numbers $a, b, c, d$ are written. It is known that among the six sums
$$
a+b, \quad b+c, c+d, d+a, a+c, b+d
$$
three are equal to 23, and the other three are equal to 34.
(a) (1 point) What is the value of $a+b+c+d$?
(b) (3 points) What is the smallest of the numbers $a, b, c, d$... | Answer: (a) 57. (b) 6.
Solution. (a) Let's add all 6 sums $a+b, b+c, c+d, d+a, a+c, b+d$. Since three of them are equal to 23 and the other three are equal to 34, we get $23 \cdot 3 + 34 \cdot 3$. On the other hand, we get $3(a+b+c+d)$. Therefore,
$$
a+b+c+d=\frac{23 \cdot 3 + 34 \cdot 3}{3}=57
$$
(b) Suppose that a... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,110 |
Problem 10.4. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and the quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=17$?
(b) (2 points) How many degrees does the angle $K M D$ measure, if... | Answer: (a) 8.5. (b) 48.
Solution. (a) Note that KBMD is a cyclic trapezoid, so it is isosceles, i.e., $M D=K B$. Therefore,
$$
M D=K B=\frac{B C}{2}=\frac{A D}{2}=8.5
$$
Fig. 7: to the so... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,111 |
Problem 10.5. One winter day, 43 children were throwing snowballs at each other. Each of them threw exactly one snowball at someone else. It is known that:
- the first threw a snowball at the one who threw a snowball at the second,
- the second threw a snowball at the one who threw a snowball at the third,
- the forty... | # Answer: 24.
Solution. First, note that not only did each throw exactly one snowball, but each was hit by exactly one snowball. Indeed, from the phrase "the first threw a snowball at the one who threw a snowball at the second," it follows that someone threw a snowball at the second; similarly, it is established that ... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,112 |
Problem 10.6. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14, 26, 182$. (b) 24.
Solution. Since $a^{3}+p^{3}=(a+p)\left(a^{2}-a p+p^{2}\right)$, and $a^{2}-p^{2}=(a+p)(a-p)$, the condition of divisibility is equivalent to $a^{2}-a p+p^{2}=a(a-p)+p^{2}$ being divisible by $a-p$. Note that $a(a-p)$ is divisible by $a-p$, so $p^{2}$ must be divis... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,113 |
Problem 10.7. At one meal, Karlson can eat no more than 5 kg of jam. If he opens a new jar of jam, he must eat it completely during this meal. (Karlson will not open a new jar if he has to eat more than 5 kg of jam together with what he has just eaten.)
Little Boy has several jars of raspberry jam weighing a total of ... | Answer: 12.
Solution. We will prove that in 12 meals, Karlson will always be able to eat all the jam.
We will distribute the jars into piles according to the following algorithm. In each pile (starting with the first, then the second, and so on), we will place jars one by one until the pile contains more than 5 kg of... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,114 |
Problem 10.8. Given a triangle $A B C$. Let point $I$ be the center of its inscribed circle, and points $P$ and $Q$ be the midpoints of sides $A B$ and $A C$ respectively. It turns out that $\angle P I Q + \angle B I C = 180^{\circ}$. Find the length of segment $B C$, if $A B = 20$ and $A C = 14$.
Fig. 8: to... | \frac{34}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,115 |
Variant 10.4.1. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=17$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $\ang... | Answer: (a) 8.5. (b) 48. | 8.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,119 |
Variant 10.4.2. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K = B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D = 19$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $... | Answer: (a) 9.5. (b) 42. | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,120 |
Variant 10.4.3. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=15$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $\ang... | Answer: (a) 7.5. (b) 39. | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,121 |
Variant 10.4.4. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=13$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $\ang... | Answer: (a) 6.5. (b) 36. | 6.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,122 |
Variant 10.6.1. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14,26,182$. (b) 24. | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,123 |
Variant 10.6.2. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 17)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 18. | Answer: (a) any of the numbers $18,34,306$. (b) 21. | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,124 |
Variant 10.6.3. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 19)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 24. | Answer: (a) any of the numbers $20,38,380$. (b) 27. | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,125 |
Variant 10.6.4. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 11)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 16. | Answer: (a) any of the numbers $12, 22, 132$. (b) 18. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,126 |
9.1. Several consecutive natural numbers are written on the board. Exactly $52 \%$ of them are even. How many even numbers are written on the board | # Answer: 13.
Solution. Since the natural numbers written down are consecutive, even and odd numbers alternate. According to the condition, there are more even numbers, which means the sequence starts and ends with even numbers.
First method. Let $n$ be the number of even numbers, then the number of odd numbers is $(... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,127 |
9.2. The graph of the function $y=x^{2}+a x+b$ is shown in the figure. It is known that the line $A B$ is perpendicular to the line $y=x$. Find the length of the segment $O C$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: 1.
Solution. Since $y(0)=b$, then $B(0 ; b)$. Now let's find the length of the segment $O A$.
First method. Since the line $A B$ is perpendicular to the line $y=x$, it is parallel to the line $y=-x$. Moreover, this line passes through the point $B(0 ; b)$. Therefore,
\), a circle with center \(O\) is inscribed, touching side \(AB\) at point \(E\). On the extension of side \(AC\) beyond point \(A\), a point \(D\) is chosen such that \(AD = \frac{1}{2} AC\). Prove that lines \(DE\) and \(AO\) are parallel. | Solution. Let $M$ be the point of tangency of the circle with side $A C$ (see Fig. 9.3). Since triangle $A B C$ is isosceles, $M$ is the midpoint of $A C$. Thus, $D A=A M=M C$.
On the other hand, $A E=A M$ (segments of tangents drawn from the same point to a circle). From the equality $A E=A M=A D$, it follows that tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,129 |
9.4. In a square table of size $100 \times 100$, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded? | Answer: 198.
Solution. Example. We will color all the cells of one row and all the cells of one column, except for their common cell. In this case, the condition of the problem is satisfied, and exactly 198 cells are colored.
Estimate. We will prove that no more than 198 cells could have been colored in the required ... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,130 |
9.5. The heights $A D$ and $B E$ of an acute-angled triangle $A B C$ intersect at point $H$. The circumcircle of triangle $A B H$ intersects sides $A C$ and $B C$ at points $F$ and $G$ respectively. Find $F G$, if $D E=$ $=5$ cm. | Answer: 10 cm.
Solution. Let $\angle H B F=\alpha$ (see Fig. 9.5). Then $\angle F A H=\angle H B F=\alpha$ (inscribed angles subtending the same arc). From the right triangle $A D C: \angle C=90^{\circ}-\alpha$, and from the right triangle $E C B: \angle E B C=90^{\circ}-\angle C=\alpha$.
Thus, $B E$ is the height an... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,131 |
9.6. Twenty-five coins are arranged into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. Each time a group is divided into two, the product of the number of coins in the two resultin... | Answer: 300.
Solution. First method. Let's represent the coins as points and connect each pair of points with a segment. We will get $\frac{25(25-1)}{2}$ $=300$ segments. With each division of one group of coins into two, we will erase all segments connecting points corresponding to coins that end up in different grou... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,132 |
5. Option 1.
Petya is the oldest child in the family. He has two sisters, Anya and Katya, and a brother Vasya. Petya calculated that Anya and Katya together are 19 years old, and Anya and Vasya together are 14 years old. Determine how old Katya is, if it is known that the two youngest children together are 7 years old... | Solution. Pete cannot be among the two younger children, and among the remaining children, the unconsidered pair is Katya and Vasya, so together they are 7 years old. If you add Katya's age to Anya's age, it will be 5 years more than if you add Vasya's age to Anya's age, so Anya is 5 years older than Vasya. But togethe... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,133 |
11.1 Prove that of the two numbers $a=|\cos x-\sin x|, b=|\cos x+\sin x|$, at least one is not less than 1. | Solution: For example, $a^{2}+b^{2}=\cos ^{2} x-2 \cos x \sin x+\sin ^{2} x+\cos ^{2} x+2 \cos x \sin x+\sin ^{2} x=$ $=2\left(\cos ^{2} x+\sin ^{2} x\right)=2$.
Therefore, among the numbers $a \geq 0$ and $b \geq 0$, there is one not less than 1. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,134 |
11.2 Point $\mathrm{D}$ is on side $\mathrm{AC}$ of triangle $\mathrm{ABC}$. Given that $\angle \mathrm{BDC}=\angle \mathrm{ABC}$. The bisectors of angles $\mathrm{ABC}$ and $\mathrm{BDC}$ intersect at point $\mathrm{K}$. Prove that $\mathrm{AK}=\mathrm{BK}$. | Solution: See fig.
Let $\angle \mathrm{ABC}=2 \beta$. Then $\angle \mathrm{ABK}=\beta$, $\angle \mathrm{CDK}=\beta, \angle \mathrm{ADK}=180^{\circ}-\angle \mathrm{CDK}=180^{\circ}-\beta$. Consider the quadrilateral АDKB. In it, the sum of the opposite angles АВК and АDK is $\beta+\left(180^{\circ}-\beta\right)=180^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,135 |
11.4 The infinite sequence of numbers $a_{1}, a_{2}, a_{3}, \ldots$ is formed according to the rule: $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}+\frac{1}{a_{n}}}$ for $n=1,2, \ldots$. Prove that some interval of length 1 contains more than a thousand terms of this sequence. | Solution. It is clear that the sequence is increasing: $a_{1}\frac{1}{a_{n}\left(a_{n+1}+a_{n+1}\right)}=\frac{1}{2 a_{n} a_{n+1}}
\end{aligned}
$
Using (2), it is easy to show that the sequence increases without bound, but it is sufficient for us to establish that, starting from some number, all members of the sequen... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,136 |
9.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$. (N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=mp$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number, ... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,140 |
9.6. Trapezoid $A B C D$ with bases $A B$ and $C D$ is inscribed in circle $\Omega$. Circle $\omega$ passes through points $C, D$ and intersects segments $C A, C B$ at points $A_{1}, B_{1}$ respectively. Points $A_{2}$ and $B_{2}$ are symmetric to points $A_{1}$ and $B_{1}$ with respect to the midpoints of segments $C ... | The first solution. The statement of the problem is equivalent to the equality $C A_{2} \cdot C A = C B_{2} \cdot C B$. Since $A A_{1} = C A_{2}$ and $B B_{1} = C B_{2}$, it is sufficient to prove that $A A_{1} \cdot A C = B B_{1} \cdot B C$.
Let $D_{1}$ be the second point of intersection of $\omega$ with $A D$ (see ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,141 |
9.7. In the Republic of Mathematicians, a number $\alpha>2$ was chosen and coins of 1 ruble and $\alpha^{k}$ rubles for each natural $k$ were issued. At the same time, $\alpha$ was chosen so that the denominations of all coins, except the smallest one, are irrational. Could it be that any amount in a natural number of ... | Answer. It could.
Solution. We will show that the mathematicians could have chosen the number $\alpha=\frac{-1+\sqrt{29}}{2}$; this number is a root of the equation $\alpha^{2}+\alpha=7$. Clearly, $\alpha>2$. It is not hard to see that for natural $m$ we have $(2 \alpha)^{m}=a_{m}+b_{m} \sqrt{29}$, where $a_{m}$ and $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,142 |
9.8. In a state, there are $n$ cities, and an express train runs between each pair of them (in both directions). For any express train, the ticket prices for the "outbound" and "return" trips are the same, and for any different express trains, these prices are different. Prove that a traveler can choose an initial city... | First solution. Remove all express trains, and then start launching them back one by one in ascending order of price (i.e., the first one launched is the cheapest, the second one is the cheapest of the remaining, and so on). At each moment, in each city, we will write the maximum number of express trains that can be se... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,143 |
10.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$.
(N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=m p$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number,... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,144 |
10.6. Point $M$ is the midpoint of side $AC$ of triangle $ABC$. Points $P$ and $Q$ are chosen on segments $AM$ and $CM$ respectively such that $PQ = AC / 2$. The circumcircle of triangle $ABQ$ intersects side $BC$ at point $X \neq B$, and the circumcircle of triangle $BCP$ intersects side $AB$ at point $Y \neq B$. Prov... | The first solution. From the inscribed quadrilaterals $BCPY$ and $BAQX$, it follows that $\angle APY = \angle ABC = \angle CQX$. Let the line passing through $M$ parallel to $QX$ intersect the line $BC$ at point $K$, and the line passing through $M$ parallel to $PY$ intersect the line $AB$ at point $L$ (see Fig. 3). Th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,145 |
10.7. In the Republic of Mathematicians, a number $\alpha>2$ was chosen and coins with denominations of 1 ruble, as well as $\alpha^{k}$ rubles for each natural number $k$, were issued. In this case, $\alpha$ was chosen such that the denominations of all coins, except the smallest one, are irrational. Could it be that ... | Answer. It could.
Solution. We will show that the mathematicians could have chosen the number $\alpha=\frac{-1+\sqrt{29}}{2}$; this number is a root of the equation $\alpha^{2}+\alpha=7$. Clearly, $\alpha>2$. It is not hard to see that for natural $m$ we have $(2 \alpha)^{m}=a_{m}+b_{m} \sqrt{29}$, where $a_{m}$ and $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,146 |
10.8. On the plane, there are $n$ convex pairwise intersecting $k$-gons. Any of them can be transformed into any other by a homothety with a positive coefficient. Prove that there is a point on the plane that belongs to at least $1+\frac{n-1}{2 k}$ of these $k$-gons.
(A. Akopyan) | Lemma. Let $P$ and $P^{\prime}$ be intersecting convex polygons that are homothetic with a positive coefficient. Then one of the vertices of one of them lies in the other.
Proof. If one of the polygons is completely inside the other, the statement is obvious. Otherwise, there exists a side $AB$ of polygon $P$ that int... | 1+\frac{n-1}{2k} | Geometry | proof | Yes | Yes | olympiads | false | 13,147 |
11.5. A natural number $n$ is called good if each of its natural divisors, increased by 1, is a divisor of the number $n+1$. Find all good natural numbers.
(S. Berlov) | Answer. One and all odd prime numbers.
Solution. It is clear that $n=1$ satisfies the condition. Also, all odd primes satisfy it: if $n=p$, then its divisors increased by 1 are 2 and $p+1$; both of them divide $p+1$. On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisi... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,148 |
11.6. The sphere $\omega$ passes through the vertex $S$ of the pyramid $SABC$ and intersects the edges $SA, SB$, and $SC$ again at points $A_1, B_1$, and $C_1$ respectively. The sphere $\Omega$, circumscribed around the pyramid $SABC$, intersects $\omega$ along a circle lying in a plane parallel to the plane $(ABC)$. P... | The first solution. The statement of the problem is equivalent to the equality $S A_{2} \cdot S A = S B_{2} \cdot S B = S C_{2} \cdot S C$. Therefore, due to the equalities $A A_{1} = S A_{2}$ and two similar ones, it is sufficient to prove that $A A_{1} \cdot A S = B B_{1} \cdot B S = C C_{1} \cdot C S$.
Let $\ell$ b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,149 |
11.7. Initially, the polynomials $x^{3}-3 x^{2}+5$ and $x^{2}-4 x$ are written on the board. If the polynomials $f(x)$ and $g(x)$ are already written on the board, it is allowed to add to the board the polynomials $f(x) \pm g(x), f(x) g(x), f(g(x))$ and $c f(x)$, where $c$ is an arbitrary (not necessarily integer) cons... | Answer: It cannot.
First solution. Let $f(x)$ and $g(x)$ be two polynomials, and for some point $x_{0}$, the equalities $f^{\prime}\left(x_{0}\right)=0$ and $g^{\prime}\left(x_{0}\right)=0$ hold. Then, obviously, $(f \pm g)^{\prime}\left(x_{0}\right)=0$ and $c f^{\prime}\left(x_{0}\right)=0$. Also, $(f g)^{\prime}\lef... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,150 |
11.8. Two players are playing a card game. They have a deck of $n$ pairwise distinct cards. For any two cards in the deck, it is known which one beats the other (however, if $A$ beats $B$, and $B$ beats $C$, it can happen that $C$ beats $A$). The deck is distributed between the players in an arbitrary manner. On each t... | Solution. Let's list all possible situations that can occur in the game (i.e., all possible pairs of decks of the participants). We will call a situation final if all the cards are with one player; critical if one of the players has exactly one card; and regular if both players have at least two cards. We will draw an ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,151 |
10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$ | Answer: 0.
First solution. From $\sin x = \operatorname{tg} y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$ or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,152 |
10.2. Given the expression $A=x y+y z+z x$, where $x, y, z$ are integers. If the number $x$ is increased by 1, and the numbers $y$ and $z$ are decreased by 2, then the value of the expression $A$ does not change. Prove that the number $(-1) \cdot A-$ is the square of an integer. | Solution. By the condition $x y+y z+z x=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)$, from which $4 x+y+z=0$, or $x=-\frac{y+z}{4}$. Then $(-1) \cdot A=-(x y+y z+z x)=-y z-x(y+z)=$ $-y z+\frac{y+z}{4}(y+z)=\frac{(y+z)^{2}-4 y z}{4}=\frac{(y-z)^{2}}{4}=\left(\frac{y-z}{2}\right)^{2}$. Since $y+z=-4 x$, then $y$ and $z$ have the sa... | (-1)\cdotA=(\frac{y-z}{2})^2 | Algebra | proof | Yes | Yes | olympiads | false | 13,153 |
10.3. The numbers $x$ and $y$ satisfy the inequalities $x^{7}>y^{6}$ and $y^{7}>x^{6}$. Prove that $x+y>2$. | Solution. We will prove that $x>1$ and $y>1$, from which the required statement will follow. The right-hand sides of the inequalities are non-negative, so both numbers $x$ and $y$ are positive. Multiplying the given inequalities and canceling by the positive number $x^{6} y^{6}$, we get: $x y>1$. This means that at lea... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,154 |
10.4. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), have settled in these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupi... | Answer: 12 knights.
Solution: Note that for each knight, at least one of their neighbors must be a liar. We will show that there must be no fewer than 4 liars (thus showing that there are no more than 12 knights). Suppose there are no more than 3 liars, then there will be a "vertical row" of rooms where only knights l... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,155 |
10.5. In an acute-angled triangle $A B C$, the altitude $C H$ is drawn. Point $P$ is symmetric to point $A$ with respect to the line $B C$. The line $C H$ intersects the circumcircle of triangle $A C P$ again at point $K$. The line $K P$ intersects the segment $A B$ at point $M$. Prove that $A C = C M$. | Solution. Since $CH$ is the height of triangle $ACM$, the equality $AC = CM$ is equivalent to $CH$ being the perpendicular bisector of segment $AM$, which means that $KH$ is the perpendicular bisector of segment $AM$. This is equivalent to $AK = KM$, which means $\angle AKH = \angle MKH$, or $\angle AKC = \alpha = \ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,156 |
1. First, we will calculate the number of ways to choose three points out of the 60 available. The first point can be chosen in 60 ways, the second in 59 ways, and the third in 58 ways; as a result, we get $205320=60 \cdot 59 \cdot 58$ options. Since the selections of vertices ABC, ACB, CBA, CAB, BAC, and BCA all give ... | Answer: 34190
Recommendations for evaluating solutions: to determine the number of triplets of points, one can use the combination formula $C_{60}^{3}=\frac{60!}{3!57!}=34220$. | 34190 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,157 |
2. First method: Find the digits $a, b, c$ such that the number $\overline{387 a b c}$ is divisible by 5, 6, and 7. Any even number divisible by 5 ends in 0, so $c=0$. For divisibility by 3, it is necessary that $3+8+7+a+b=18+a+b$ is divisible by 3. Since 18 is divisible by 3, $a+b$ must be divisible by 3. Let's take t... | Answer: 387030, 387240, 387450, 387660, 387870.
Evaluation recommendations: Students may find only one or a few solutions. If only one solution is found, the task is scored 3 points; if several but not all solutions are found - 5 points. Full solution - 7 points. | 387030,387240,387450,387660,387870 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,158 |
1. We took 2015 consecutive natural numbers, divisible by 2015. Can their sum be the 2015th power of some natural number?
# | # Solution.
Let the first of the given numbers be denoted as \( a \). Then the second number is \( a + 2015 \), the third number is \( a + 2015 \cdot 2 \), and so on. The last number is \( a + 2015 \cdot 2014 \). These numbers form an arithmetic progression with a common difference of 2015. Their sum \( S \) is
\[
\f... | 2015^{2015} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,159 |
3. On the left half of the board, the number 21 is written, and on the right half, the number 8 is written. It is allowed to take an arbitrary number \(a\) from the left half of the board and an arbitrary number \(b\) from the right half of the board, compute the numbers \(ab\), \(a^3 + b^3\), and write \(ab\) on the l... | # Answer: No Solution.
Note that the number $21: 7$, and 8 gives a remainder of 1. Therefore, after the first operation, the number $a b: 7$, and the number $a^{3}+b^{3}$ gives a remainder of 1 when divided by 7. Thus, we get that on the left board we will write a number divisible by 7, and on the right board a numbe... | No | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,161 |
4. Two circles touch each other internally at point $K$. A line intersects the larger circle at points $A$ and $B$, and the smaller circle at points $C$ and $D$. Point $C$ lies between $A$ and $D$. Prove that $\angle A K C = \angle B K D$. | # Solution.
Let $l$ be the common tangent to the circles at point $K$, $l \cap (A B) = M$. $\angle B K D = \angle D K M - \angle B K M$
$\angle B K M$ is the angle between the tangent and th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,162 |
5. In a regular 9-gon, all sides and diagonals are painted either red or blue. It turned out that there are no three vertices of the 9-gon connected by segments forming a red triangle. Prove that there will be 4 vertices of the 9-gon forming a quadrilateral, all sides and diagonals of which are painted blue.
# | # Solution.
Suppose there exists a vertex $A$ from which 4 or more red segments emerge: $A B, A C, A D, A E$. Since there cannot be any red triangles, the vertices $B, C, D,$ and $E$ are connected by blue segments. The desired quadrilateral has been found.
Consider the case where fewer than 4 red segments emerge from... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,163 |
1. It is known that the quadratic function $f(x)=x^{2}+a x+b$ has zeros $x_{1}$, $x_{2}$ and $f(2014)=f(2016)$. Find $\frac{x_{1}+x_{2}}{2}$. | # Answer: 2015
## First Solution.
Since the graph of the quadratic function $f(x)$ is symmetric with respect to the line $x=-a / 2$ and $f(2014)=f(2016), f\left(x_{1}\right)=f\left(x_{2}\right)=0$, then
$-\frac{a}{2}=\frac{x_{1}+x_{2}}{2}=\frac{2014+2016}{2}=2015$.
## Second Solution.
Since $f(2014)=f(2016)$, then... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,164 |
3. On the left half of the board, the number 21 is written, and on the right half, the number 15 is written. It is allowed to take an arbitrary number $a$ from the left half of the board and an arbitrary number $b$ from the right half of the board, compute the numbers $a b, a^{3}+b^{3}$, and write $a b$ on the left and... | # Answer: No First solution.
Notice that the number $21 \vdots 7$, while 15 gives a remainder of 1. This means that after the first operation, the number $a b \vdots 7$, and the number $a^{3}+b^{3}$ gives a remainder of 1 when divided by 7. Thus, we get that on the left board we will write a number divisible by 7, an... | No | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,166 |
4. In triangle $A B C$ with bisector $A E$, the circumcircles of triangles $A E B, A E C$ intersect side $A C$ at point $N$, and side $A B$ at point $M$ respectively. Prove the equality of the lengths of segments $AM$ and $AN$. | Solution. Two secants are drawn from point $B$ to the second circle, so $B E \cdot B C = B M \cdot B A$. Similarly, $C E \cdot C B = C N \cdot C A$. Dividing the second equality by the first, we get $\frac{C E}{B E} = \frac{C N}{B M} \cdot \frac{C A}{B A}$.
Considering that the angle bisector divides the opposite side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,167 |
5. It is known that on a chessboard, 8 rooks can be placed so that they do not attack each other. Schoolboy Pete does not like the chessboard coloring, and he colored the board in 32 colors, so that there are exactly two cells of each color. Will he now be able to place 8 rooks so that they do not attack each other and... | # Solution.
Note that the total number of ways to place 8 rooks on the board so that they do not attack each other is $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 8$! (on the first vertical there are 8 ways, on the second 7, and so on).
Now let's count the number of ways to place 8 rooks so that some ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,168 |
3. In a square grid of cells, some cells are painted black. It turned out that no black cell shares a side with more than one other black cell? What is the maximum number of cells that could be painted black?
In a square of cells, some cells are painted black. It turned out that no black cell shares a side with more t... | Solution. An example of properly coloring 8 squares is shown in Fig. 1a. Suppose more than 8 were colored, then at least one of the four $2 \times 2$ squares would have at least three cells colored (Fig. 1b). But then at least one of them would share adjacent sides with two other black cells. This contradiction complet... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,170 |
4. On the diagonal $B D$ of parallelogram $A B C D$, points $P$ and $Q$ are chosen such that $B P=P Q=Q D$. Lines $\boldsymbol{C P}$ and $\boldsymbol{C} \boldsymbol{Q}$ intersect sides $\boldsymbol{A B}$ and $\boldsymbol{A D}$ at points $M$ and $N$, respectively. Find the ratio $M N: B D$. | Solution. Draw the diagonal $AC$, then the intersection point of the diagonals $O$ - divides both diagonals in half (Fig. 2). Then for $\triangle ABC$, the segment $BO$ is a median, and the point $P$ satisfies the condition $BP: PO=2: 1$, which means $P$ is the centroid of $\triangle ABC$. Therefore, $CM$ is also a med... | MN:BD=1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,171 |
5. For the function $f(x)=a x^{2}+b x+c, a \neq 0$ the condition $5 a+b+2 c=0$ is satisfied. Prove that the equation $f(x)=0$ has two roots. | Solution. The given condition can be rewritten in the following form:
$$
5 a+b+2 c=0 \Leftrightarrow a-b+c+4 a+2 b+c=0
$$
And the last equality can be written as $f(-1)+f(2)=0$. Thus, either $f(-1)=f(2)=0$ and the parabola has two real roots, or these values have different signs, from which it also follows that it ha... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,172 |
1. Three bees collect nectar from 88 garden flowers. Each flower was visited by at least one bee. Each bee visited exactly 54 flowers. We will call a flower sweet if it was visited by all three bees, and bitter if it was visited by exactly one. Which of these 88 flowers are more: sweet or bitter, and by how many? | Solution. Let among the 88 garden flowers, $s$ be sweet and $g$ be bitter. Then the number of flowers visited by two bees is exactly $88-s-g$. On one hand, the total number of bee landings on the flowers is $3 \cdot 54=162$ (each bee visited 54 flowers), and on the other hand, it is $3s + 2(88-s-g) + g = s - g + 176$. ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,173 |
2. Find all pairs of natural numbers $a$ and $b$ that satisfy the condition: $\operatorname{LCM}(a, b) + \operatorname{GCD}(a, b) = ab / 5$ and prove that there are no other such pairs. | Solution. We will use the formula $\operatorname{LCM}(a, b) \cdot \operatorname{GCD}(a, b)=a b$, as well as the fact that the least common multiple of numbers is divisible by their greatest common divisor. Let, for brevity, $\operatorname{GCD}(a, b)=x, \operatorname{LCM}(a, b)=k x$ (here $k-$ is some natural number), t... | (6,30),(10,10),(30,6) | Number Theory | proof | Yes | Yes | olympiads | false | 13,174 |
3. An infinite lattice is composed of equilateral triangles with a side length of 1 cm. Each edge of the lattice is painted one of three colors. Is it true that there will always be a vertex from which a snail can crawl 100 cm along edges of the same color without passing over the same edge twice? | Solution. Incorrect. One of the possible counterexamples is shown in the figure. It is based on tiling the plane with hexagons, which can be constructed on a triangular grid. A similar idea can be approached by starting from the coloring of a chessboard, to which small areas of a third color are added, located at the n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,175 |
4. A numerical sequence is defined by the relation: $a_{0}=1, a_{k+1}=a_{k}+\frac{1}{a_{k}}$. Prove that $a_{2021}>60$. | Solution. From the relations in the condition, all $a_{k}$ are positive. Squaring the recurrence relation, we get the estimate
$$
a_{k+1}^{2}=\left(a_{k}+\frac{1}{a_{k}}\right)^{2}=a_{k}^{2}+2+\frac{1}{a_{k}^{2}}>a_{k}^{2}+2
$$
We then apply this inequality several times in the right-hand side:
$$
a_{k+1}^{2}>a_{k}^... | a_{2021}>\sqrt{4043}>60 | Algebra | proof | Yes | Yes | olympiads | false | 13,176 |
5. Find the area of a triangle inscribed in a parallelogram, given that the remaining part of the parallelogram consists of three triangles of unit area. | Solution. Let $ABCD$ be a parallelogram. To inscribe a triangle in it so that the remaining part of the parallelogram represents three triangles, there is only one way: by placing one of the vertices of the triangle at a vertex of the parallelogram (for example, at $C$), and the other two vertices on the sides adjacent... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,177 |
1. Can the number 2017 be represented as the sum of four addends, each of which in decimal notation uses only one digit, and in the notation of which there are different numbers of digits? | Solution. The required representation: 2017=1111+888+11+7.
Criteria. Any correct representation: 7 points. | 2017=1111+888+11+7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,178 |
2. Seventeen aborigines, representing several tribes, stood in a circle. They tell the truth to their fellow tribesmen and lie to representatives of other tribes. Could it happen that each of them said to the neighbor on the right: "My neighbor on the left is from a different tribe"? | Answer: it could not.
Solution. Each aborigine has exactly one of his neighbors as his tribesman. If both neighbors are tribesmen, he must tell the truth, but he lies. If both neighbors are from other tribes, he must lie, but he tells the truth. Therefore, all aborigines must be divided into pairs of neighboring tribe... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,179 |
3. Given three numbers. If each of them is increased by 1, then their product also increases by 1. If all the original numbers are increased by 2, then their product also increases by 2.
Find these numbers. | Answer: $-1,-1,-1$.
## Solution.
Let the required numbers be $\mathrm{a}, \mathrm{b}, \mathrm{c}$. Then $(\mathrm{a}+1)(\mathrm{b}+1)(\mathrm{c}+1)=\mathrm{abc}+1$ (1) and $(\mathrm{a}+2)(\mathrm{b}+2)(\mathrm{c}+2)=\mathrm{abc}+2(2)$. After expanding the brackets and simplifying, we get $\mathrm{a}+\mathrm{b}+\mathr... | -1,-1,-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,180 |
4. In a right isosceles triangle $\mathrm{ABC}$
$(\mathrm{AC}=\mathrm{BC})$ points $\mathrm{D}$ and $\mathrm{F}$ are the midpoints of segments $\mathrm{AB}$ and $\mathrm{BC}$
respectively. On the ray DC
point $\mathrm{E}$ is marked such that
$\mathrm{AF}=\mathrm{FE}$. Find the angles of triangle
AFE. | Answer: $45^{\circ}, 45^{\circ}, 90^{\circ}$.
Solution. Triangle CDB is a right isosceles triangle.
$(CD-$ is the median and altitude
of the right isosceles
triangle $\mathrm{ACB}$, and ... | 45,45,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,181 |
5. In the tournament, 15 volleyball teams are playing, and each team plays against all other teams only once. Since there are no draws in volleyball, there is a winner in each match. A team is considered to have performed well if it loses no more than two matches. Find the maximum possible number of teams that performe... | Answer: 5.
Solution.
Evaluation. If the number of teams that played well is not less than 6, then consider six of them. They could have lost no more than $6 \times 2=12$ matches. But the games between them amounted to $6 \times 5 / 2=15$. Thus, they lost no fewer than 15 matches in total. Contradiction.
Example. Pla... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,182 |
3. Let there be $x$ seventh-graders participating in the tournament, who together scored $n$ points. Then the number of eighth-graders participating in the tournament is $10 * x$ people, and the total points they scored is $4.5 * n$ points. Therefore, a total of $11 * x$ students participated in the tournament, and the... | Answer: 1 student from 7th grade participated in the tournament and scored 10 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,183 |
1. Solve the equation: $(x-2016)(x-2017)(x-2018)=(x-2017)(x-2018)(x-2019)$. | Answer: 2017 and 2018.
## Solution:
## First Method.
Move everything to the left side of the equation and transform:
$(x-2016)(x-2017)(x-2018)-(x-2017)(x-2018)(x-2019)=0$
$(x-2017)(x-2018) \cdot((x-2016)-(x-2019))=0$
$(x-2017)(x-2018) \cdot 3=0$
The obtained equation has two roots: 2017 and 2018.
## Second Meth... | 20172018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,186 |
2. Find all solutions to the equation: $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$. | Answer: the interval $\left[\frac{1}{2} ; 1\right]$
## Solution:
Let's find the domain of admissible values of the variable $x$. From the non-negativity of the expressions under the square roots, we have: $2 x-1 \geq 0$ (i.e., $x \geq \frac{1}{2}$) and $x-\sqrt{2 x-1} \geq 0$, from which it immediately follows that $... | [\frac{1}{2};1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,187 |
3. The inscribed quadrilateral $ABCD$ has no parallel sides. The rays $BA$ and $CD$ intersect at point $E$, and the rays $AD$ and $BC$ intersect at point $F$. Prove that if $AE=CF$, then $DE=DF$. | # Solution:
## First Method.
Let $\angle E A D=\alpha$. Then $\angle B A D=180^{\circ}-\alpha$, and since $A B C D$ is a cyclic quadrilateral, by the sum of opposite angles: $\angle B C D=180^{\circ}-\angle B A D=180^{\circ}-\left(180^{\circ}-\alpha\right)=\alpha$, and $\angle F C D=180^{\circ}-\alpha$.
By the Law o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,188 |
4. The archipelago consists of 2018 islands, some of which are connected by ferries (each ferry runs from one island to another and back), and from each island, ferries run to at least 6 other islands. Prove that it is possible to organize a circular trip around the islands of the archipelago, allowing you to visit at ... | # Solution:
Let's choose any island in the archipelago and move from it to another island, from that island to another one, and so on, until we reach an island from which there is nowhere to go (all flights lead to previously visited islands). We get a route like this: $1 \rightarrow 2 \rightarrow 3 \rightarrow \ldots... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,189 |
5. Is it possible to color all natural numbers either red or blue so that in any infinite arithmetic progression consisting of natural numbers, there are numbers of both colors? (To color the numbers means to assign each number one of two colors: either red or blue.) Justify your answer.
# | # Answer: it is possible.
## Solution:
For example, the numbers can be colored as follows: first (in 1 step) color the numbers 1 - red, 2 - blue, then (in 2 steps) 3 and 4 - red, 5 and 6 - blue, then (in 3 steps) $7,8,9$ - red, $10,11,12$ - blue, and so on - on each ($k$-th) step, color the next $k$ consecutive numbe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,190 |
1. Four children are collecting pretty pebbles. Kolya had 70 pebbles, Anya had 30, Petya had 21, and Denis had 45. Could it be that, as a result of trading (exchanging one for one, or two, or three), they all ended up with an equal number of pebbles? | Answer: No.
Solution. The total number of pebbles, which is $70+30+21+45=166$, will not change as a result of the exchange. However, it is not divisible by 4.
Comment. An answer without justification - 0 points. A correct answer obtained by considering an example without a general proof - 1 point, if several examples... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,191 |
3. Vasya has talking cockroaches of two types at home: black and brown. Cockroaches of one color always tell the truth, while cockroaches of the other color can say anything. Vasya placed a cockroach on each cell of a $4 \times 4$ board, and each cockroach claimed that all its side neighbors are brown. Prove that there... | Solution. Assume the opposite: there are 9 black and 7 brown cockroaches on the board. If black cockroaches only told the truth, they could not say "Among my neighbors - only brown ones," because the board can be represented as consisting of eight "dominoes" and at least one domino is completely occupied by black cockr... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 13,193 |
4. Dima recorded his digital password in code: P, AR, AK, RU, SS. The key to the code: different digits are represented by different letters, and the same digits by the same letters, two letters together represent a two-digit number, numbers are separated by commas, the difference between any two adjacent numbers is th... | Answer: $5,12,19,26,33$.
Solution. All these numbers can be determined if we know the first number and the difference $d$ between two consecutive numbers. The first digits of the second and third numbers coincide, meaning they are in the same decade, and their difference, equal to $d$, does not exceed 9. Therefore, by... | 5,12,19,26,33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,194 |
5. Masha talked a lot on the phone with her friends, and the fully charged battery ran out exactly after 24 hours. It is known that the charge lasts for 5 hours of talking or 150 hours of standby. How long did Masha talk to her friends | Answer: $126 / 29$ hours.
Solution. In one hour of conversation, the battery discharges by $1 / 5$, and in one hour of standby, it discharges by $1 / 150$ of its capacity. Let $x$ be the number of hours needed for the battery to be fully discharged in 24 hours. Then, in standby mode, it will be $24-x$ hours, and the e... | \frac{126}{29} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,195 |
3. Solve the equation $\sin x + \cos x + \sin x \cos x = 2$. | Answer: this equation has no roots.
Solution. The equation $(\sin x-\cos x)^{2}+(\sin x-1)^{2}+(\cos x-1)^{2}=0$ has no solutions because $\sin x$ and $\cos x$ cannot both equal one simultaneously. After expanding the brackets and simplifications, we get the equation $\sin x+\cos x+\sin x \cos x=2$.
Second method. Let... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,198 |
4. Find all functions $f$, defined on the set of real numbers and taking real values, such that for any real $x, y$ and $z$ the equality $f(x y z)=f(x) f(y) f(z)-6 x y z$ holds. | Answer: $f(x)=2 x$.
Solution. Let $\mathrm{f}(1)=\mathrm{a}$. Substitute $\mathrm{y}=\mathrm{z}=1$ into the equation and we get $f(x)=a^{2} f(x)-6 x$. Therefore, $\left(a^{2}-1\right) f(x)=6 x$.
If $\mathrm{a}^{2}-1 \neq 0$, then $\mathrm{f}(\mathrm{x})=\mathrm{kx}$ for some $\mathrm{k}$. If $\mathrm{a}^{2}-1=0$, the... | f(x)=2x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,199 |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find the greatest possible value of angle ACB. | Answer: $60^{\circ}$.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,200 |
6. In one company, among any 11 people, there are two people who know each other. Prove that in this company, there will be a group of ten people such that each of the others knows someone in this group. | Solution. Consider the largest group $G$ of people who are pairwise unfamiliar with each other. In this group, there are no more than ten people; otherwise, there would be eleven people among them, none of whom are familiar with each other, which contradicts the condition. Since this is the largest group, every other p... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,201 |
2. (7 points) In Sun City, they exchange a honey cake for 6 pretzels, and for 9 pretzels, they give 4 doughnuts. How many doughnuts will they give for 3 honey cakes? Explain your answer. | Answer: 8.
Solution.
If for one cookie you get 6 pretzels, then for 3 cookies you will get $3 \times 6=18$ pretzels. 18 pretzels is 2 times 9 pretzels. Therefore, for them, you will get 2 times 4 gingerbread cookies, i.e., 8 gingerbread cookies.
## Grading Criteria.
- Any correct and justified solution - 7 points.
... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,202 |
4. (7 points) To get from the trunk to any leaf of the tree shown in the picture, you need to turn either left or right at each fork. For example, to get to the leaf with the letter A, you need to go this way: ппплп (the letter п means a turn to the right at a fork, the letter л means a turn to the left).
 lppl
b) see figure
## Grading Criteria.
- Correct answers for both parts of the task - 7 points.
- Correct answer only for part b) - 5 points
- Correct answer on... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,204 |
5. (7 points) Vanya, Tanya, and Olya have 12 identical balls: some yellow, some blue, and some red. They distributed the balls into three identical bags, 4 balls in each.
Vanya said: "Look, there are no three identical balls in any bag!" Tanya said: "That's right. But there are also no three different balls in any bag... | Answer. Yes, definitely.
Solution. Each package contains balls of different colors, otherwise Vanya would be wrong. But there cannot be balls of three different colors in any package, otherwise Tanya would be wrong. Therefore, each package contains balls of exactly two colors: 2 balls of one color and 2 balls of anoth... | Yes,definitely | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,205 |
6. Given an acute-angled triangle $A B C$. The height $A A_{1}$ is extended beyond vertex $A$ to segment $A A_{2}=B C$. The height $C_{1}$ is extended beyond vertex $C$ to segment $C_{2}=A B$. Find the angles of triangle $A_{2} B C_{2}$. (R. Zhenodarov) | Answer. $\angle A_{2} B C_{2}=\angle 90^{\circ}, \angle B A_{2} C_{2}=\angle B C_{2} A_{2}=45^{\circ}$.
Solution. Triangles $A B A_{2}$ and $C C_{2} B$ are equal by two sides and the angle between them ( $A B=C C_{2}, A A_{2}=B C$, and the angles
 | Answer. It could not. Solution. Suppose the opposite: the products ab, bc, and ca end with two-digit numbers $n, n+1$, and $n+2$, respectively. Among these three consecutive numbers, there must be an odd one, meaning that the product of some two of the numbers $a, b$, and $c$ is odd. This implies that at least two of t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,207 |
8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of a four-cell figure
$\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). A placement is called unsuccessful if the sum of the numbe... | Answer: 36. Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and in the central cell be \(e\); denote the sum of all these five numbers by \(S\). Then, accor... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,208 |
2. On the day when Dima's brother and sister were congratulating him on his birthday, Dima said: "Look how interesting, I am now twice as old as my brother and three times as old as my sister!" - "And your average age is 11 years," - added Dad. How old did Dima turn? | Answer: 18 years.
Solution. The first method. According to the problem, we can form an equation. Let Dima's age be $x$ years, then his sister's age is $x / 3$, and his brother's age is $-x / 2 ;(x+x / 3+x / 2): 3=11$. After solving this equation, we get that $x=18$. Dima is 18 years old.
It will be useful to provide ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,210 |
3. Once, a detective had to interrogate three witnesses of a robbery: John White, Sam Gray, and Bob Black. John insisted that all of Sam's statements were lies, while Sam kept repeating that Bob was lying. Bob, for his part, tried to convince the detective not to believe either White or, especially, Gray. The detective... | Answer: Sam Gray.
Solution. From the condition of the problem, it is clear that the statements of each of the witnesses were made regarding the statements of the other two witnesses. Let's consider the statement of Bob Black. If what he says is true, then Sam Gray and John White are lying. But from the fact that John ... | SamGray | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,211 |
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