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10.4. A circle is inscribed in a triangle. Prove that the triangle formed by the points of tangency is acute. | Solution: We will prove the lemma. Let a circle with center at point $O$ be inscribed in an angle with vertex $A$ and let it touch the sides of the angle at points $M$ and $N$. Then $\angle M N O=\angle N M O=\frac{\angle N A M}{2}$ - see the figure.
, 31(=2+29), 32(=3+29), 33(=2+31), 34$ $(=5+29)$ or $102(=5+97), 103(=2+101), 104(=31+73), 105(=2+103)$, $106(=47+59)$. We will show that there are no six consecutive semiprime numbers.
Indeed, among any six consecutive numbers... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,681 |
10.6. Ali-baba came to a cave where there was gold and diamonds. Ali-baba had one large bag with him. It is known that a full bag of gold weighs 200 kg, and if the entire bag is filled with diamonds, it will weigh 40 kg (the empty bag weighs nothing). A kilogram of gold costs 20 dinars, and a kilogram of diamonds costs... | # Solution:
Method 1. Let Ali-Baba have placed a set of gold and diamonds that gives him the maximum possible revenue - we will call such a bag optimal. Then either the bag is filled to the brim, or it weighs exactly 100 kg: otherwise, more diamonds could be added to it and the revenue would increase. Consider the fir... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,682 |
11.1. Inside a convex pentagon, a point is marked and connected to all vertices. What is the maximum number of the ten segments drawn (five sides and five segments connecting the marked point to the vertices of the pentagon) that can have a length of 1? (A. Kuznetsov) | Answer: 9 segments.
Solution. First, we prove that all 10 segments cannot have a length of 1. Assume the opposite. Let $A B C D E$ be a pentagon, $O$ be a point inside it, and all 10 drawn segments have a length of 1 (see Fig. 6). Then triangles $O A B, O B C$, $O C D, O D E$, and $O E A$ are equilateral, so $\angle A... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,683 |
11.2. In each cell of a $1001 \times 1001$ table, a 0 or 1 was placed. It turned out that in any column, there are more zeros than ones. Is it necessarily true that there will be two columns such that the number of rows where the intersections with these two columns contain only zeros is greater than the number of rows... | Answer. Yes, definitely.
Solution. We will show that the required condition is satisfied by any two columns of the table. Remove all columns from the table except for the two we are considering. The total number of zeros in these columns is greater than the total number of ones; this means that there are at least 1002... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,684 |
11.3. Given an isosceles triangle \(ABC\), where \(\angle B = 135^\circ\). Let \(M\) be the midpoint of segment \(AC\). Point \(O\) is the center of the circumcircle \(\Omega\) of triangle \(ABC\). Ray \(BM\) intersects the circle \(\Omega\) again at point \(D\). Prove that the center of the circumcircle \(\Gamma\) of ... | Solution. Since $\angle A B C=135^{\circ}$, then $\angle A O C=90^{\circ}$. Since $O M$ is the median in the right triangle $A O C$, we have $O M=A M=M C$. On the extension of the segment $O M$ beyond point $M$, mark point $E$ such that $O M=M E$ (see Fig. 8). Since the quadrilateral $A B C D$ is inscribed, $B M \cdot ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,685 |
11.4. Initially, the numbers $1-\sqrt{2}, \sqrt{2}$, and $1+\sqrt{2}$ were written on the board. Every minute, all three numbers $x, y$, and $z$ on the board are erased, and instead, the numbers $x^{2}+x y+y^{2}$, $y^{2}+y z+z^{2}$, and $z^{2}+x z+x^{2}$ are written on the board. Can all three numbers on the board be r... | Answer. No, they cannot.
Solution. Consider the pairwise differences of the listed numbers. After one operation, the set of differences $x-y, y-z, z-x$ transitions to the set $(y-x)(x+y+z),(z-y)(x+y+z),(x-z)(x+y+z)$. Since at the initial moment these differences were $1-2 \sqrt{2}, -1$, and $2 \sqrt{2}$, at any moment... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,686 |
11.5. Let's call a trapezoid with bases 1 and 3 a "boat," which is obtained by gluing two triangular half-cells to the opposite sides of a unit square. In a $100 \times 100$ square, an invisible boat is placed (it can be rotated, it does not go beyond the boundaries of the square, its middle cell is entirely on one of ... | Answer: 4000 shots.
First solution. We will call a boat horizontal or vertical depending on whether its parallel sides are horizontal or vertical.
First, we will show that 4000 shots are enough. We divide the $100 \times 100$ square into 400 squares of size $5 \times 5$, and in each square, we make 10 shots as shown ... | 4000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,687 |
Problem 11.1. Masha lives in apartment No. 290, which is located in the 4th entrance of a 17-story building. On which floor does Masha live? (The number of apartments is the same in all entrances of the building on all 17 floors; apartment numbers start from 1.$)$ | # Answer: 7.
Solution. Let $x$ be the number of apartments per floor, then there are $17 x$ apartments in each entrance. Thus, in the first three entrances, there are $51 x$ apartments, and in the first four $68 x$.
If $x \geqslant 6$, then in the first three entrances there are at least 306 apartments, so apartment ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,688 |
Task 11.2. On the table, there are 30 coins: 23 ten-ruble coins and 7 five-ruble coins, with 20 of these coins lying heads up and the remaining 10 tails up. What is the smallest $k$ such that among any $k$ randomly selected coins, there will definitely be a ten-ruble coin lying heads up? | Answer: 18.
Solution. If you choose 18 coins, then among them there will be no more than 10 lying heads down, so at least 8 coins will be lying heads up. Among these coins, no more than 7 will be five-ruble coins, so at least one will be a ten-ruble coin, which is what we need.
On the other hand, if initially on the ... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,689 |
Problem 11.3. The product of positive numbers $a$ and $b$ is 1. It is known that
$$
(3 a+2 b)(3 b+2 a)=295
$$
Find $a+b$. | Answer: 7.
Solution. Expanding the brackets, we get
$$
295=6 a^{2}+6 b^{2}+13 a b=6\left(a^{2}+b^{2}\right)+13
$$
from which $a^{2}+b^{2}=47$. Then
$$
(a+b)^{2}=a^{2}+b^{2}+2 a b=47+2=49=7^{2}
$$
which gives $a+b=7$ (note that $a+b>0$, since $a>0$ and $b>0$). | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,690 |
Problem 11.4. A convex pentagon $A B C D E$ is such that $\angle A B C=128^{\circ}, \angle C D E=9^{\circ}$ $104^{\circ}, A B=B C, A E=E D$. How many degrees does the angle $A D B$ measure?
 | Answer: 26.
Fig. 14: to the solution of problem 11.4
Solution. The angles at the base of the isosceles triangle $ABC$ are equal to $\frac{1}{2}\left(180^{\circ}-128^{\circ}\right)=26^{\circ}... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,691 |
Problem 11.5. For what least natural $n$ can the numbers from 1 to $n$ be arranged in a circle so that each number is either greater than all 40 following it in the clockwise direction, or less than all 30 following it in the clockwise direction? | Answer: 70.
Solution. If $n \leqslant 39$, then for the number $n$ the condition cannot be satisfied: it cannot be greater than the next 40 numbers (since it is not greater than itself), nor can it be less than the next 30 numbers (since it is the largest).
If $40 \leqslant n \leqslant 69$, then for the number 40 the... | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,692 |
Problem 11.6. The polynomial $P(x)$ has all coefficients as non-negative integers. It is known that $P(1)=4$ and $P(5)=152$. What is $P(11) ?$ | Answer: 1454.
Solution. Suppose the degree of the polynomial $P$ is not less than 4, then its leading coefficient is not less than 1. Since all other coefficients of $P(x)$ are non-negative, then $P(5) \geqslant 5^{4}=625$, which contradicts the condition $P(5)=152$.
Therefore, the degree of the polynomial $P$ is not... | 1454 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,693 |
Problem 11.7. The centers of six spheres of radius 1 are located at the vertices of a regular hexagon with side 2. These spheres are internally tangent to a larger sphere $S$ with its center at the center of the hexagon. The sphere $P$ is externally tangent to the six spheres and internally tangent to the sphere $S$. W... | Answer: 1.5.
Solution. Let the centers of the first six spheres be $A, B, \ldots, F$ (in the order in which they form a hexagon), and the plane of the hexagon be $\alpha$; the center of sphere $S$ (i.e., the center of the hexagon) be $O$, and its radius be $r$; the center of sphere $P$ be $Z$, and its desired radius b... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,694 |
Problem 11.8. In a $28 \times 35$ table, some $k$ cells are painted red, another $r-$ in pink, and the remaining $s-$ in blue. It is known that
- $k \geqslant r \geqslant s$
- each boundary cell has at least 2 neighbors of the same color;
- each non-boundary cell has at least 3 neighbors of the same color.
What is th... | Answer: 28.
Solution. From the condition, it is easy to understand that each cell can have no more than one neighbor of a different color.
We will prove that the coloring of the table must be "striped," meaning that either each row or each column is completely colored in one color. To do this, it is sufficient to sho... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,695 |
1. The Big Carousel represents figurines of fairy-tale animals moving in a circle, one after another, on which one can ride. Ten students from classes $6 \mathrm{a}$, $6$ б, and $6$ в, and five students from classes $7 \mathrm{a}$ and $7$ б, were seated on the carousel, occupying all the seats. It turned out that no tw... | Solution.
First, note that if we find four or more sixth-graders from the same class sitting in a row, then, by that, we have found three such sixth-graders.
Suppose no such triplet of sixth-graders was found. This means that all sixth-graders from the same class sit individually or in pairs. According to the problem... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,696 |
2. $A B C D$ is a trapezoid with bases $A D$ and $B C$, such that a circle with diameter $B C$ can be constructed, passing through the midpoints of the diagonals $A C$ and $B D$, and touching $A D$. Find the angles of the trapezoid. | Solution.
First method. Let point $M$ be the midpoint of diagonal $A C$, and point $O$ be the center of the circle and the midpoint of base $B C$. In triangle $A B C$, segment $M O$ is the ... | 30,150,150,30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,697 |
3. Find all solutions to the equation
$$
n^{6}+3 n^{5}+3 n^{4}+2 n^{3}+3 n^{2}+3 n+1=m^{3}
$$
where $m, n$ are integers. | Solution.
Transform the left side of the equation
\[
\begin{gathered}
n^{6}+3 n^{5}+3 n^{4}+n^{3}+n^{3}+3 n^{2}+3 n+1=m^{3} \\
n^{3}\left(n^{3}+3 n^{2}+3 n+1\right)+n^{3}+3 n^{2}+3 n+1=m^{3} \\
n^{3}(n+1)^{3}+(n+1)^{3}=m^{3} \\
\left(n^{3}+1\right)(n+1)^{3}=m^{3}
\end{gathered}
\]
The product of integers on the left... | n=-1,=0orn=0,=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,698 |
4. A natural number $n$ is such that the number $n+1$ is divisible by 8. Prove that the sum of all divisors of the number $n$, including 1 and the number itself, is divisible by 8. | Solution.
Note that the number $n$ gives a remainder of 7 when divided by 8. Let $n=a b$. Consider the remainders of $a$ and $b$ when divided by 8 and form a table of the product of remainders (in other words, consider the table of products $a b$ modulo 8).
| | $\mathbf{0}$ | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,699 |
5. Let $a, b, c$ be the sides of a triangle. Prove the inequality
$$
\frac{a^{3}}{c^{3}}+\frac{b^{3}}{c^{3}}+\frac{3 a b}{c^{2}}>1
$$ | Solution.
Rewrite the inequality to an equivalent form
$$
a^{3}+b^{3}+3 a b c>c^{3}
$$
From the triangle inequality, it follows that
$$
a+b>c>0
$$
Multiply both sides of the inequality by the positive number $\left(a^{2}-a b+b^{2}\right)$ (it equals 0 only when $a=b=0$)
$$
(a+b)\left(a^{2}-a b+b^{2}\right)>c\left... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,700 |
1. Can we find four different natural numbers, each of which is not divisible by 2, 3, or 4, but the sum of any two is divisible by 2, the sum of any three is divisible by 3, and the sum of all four is divisible by 4? | Answer: For example, 5, 17, 29, and 41.
Solution. The four numbers mentioned can be written in the form $12k + 5$, where $k$ takes the values 0, $1, 2, 3$. Therefore, the sum of any three numbers
$$
\left(12k_{1} + 5\right) + \left(12k_{2} + 5\right) + \left(12k_{3} + 5\right) = 12\left(k_{1} + k_{2} + k_{3}\right) +... | 5,17,29,41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,701 |
2. Numbers from 1 to 8 were arranged in some order in a circle, and then the sums of adjacent numbers were recorded. Could it be that 8 consecutive numbers (in some order) were obtained? | Answer: no.
Solution. Suppose we have eight consecutive sum numbers
$$
x-3, x-2, x-1, x, x+1, x+2, x+3, x+4 .
$$
The sum of these numbers is $8 x+4$, and it must be twice the sum of the numbers from 1 to 8 (each number participates in exactly two sums). Then
$$
8 x+4=2 \cdot(1+2+\ldots+8)=72, \Longleftrightarrow 2 ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,702 |
3. Find all distinct prime numbers $a, b$ and $c$ such that $a b+b c+c a \geqslant a b c$. | Answer: $a=2, b=3, c=5$ and all sets obtained by permuting these numbers.
Solution. Suppose that $a<b<c$, then $a b<b c$ and $a c<b c$.
If $a \geqslant 3$, then $a b+b c+c a<3 b c \leqslant a b c$, which is a contradiction. The number $a$ is prime, so $a=2$, and then we get $2 b+2 c+b c \geqslant 2 b c$, or $0 \geqsl... | =2,b=3,=5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,703 |
4. At a round table, 30 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each person was asked: "How many knights are among your neighbors?" (Two people are called neighbors of each other if there is no one else sitting between them.) 10 people answered "one,"... | Answer: 22 knights.
Solution. From the condition of the problem, it follows that the number of knights who have two knight-neighbors does not exceed 10; the same can be said about the number of knights who have one knight-neighbor. Therefore, the number of pairs of knight-neighbors does not exceed $(2 \cdot 10 + 1 \cd... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,704 |
5. In triangle $A B C$, angle $A$ is $75^{\circ}$, and angle $C$ is $60^{\circ}$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid off, equal to half of side $A C$. Find angle $B D C$. | Answer: $\angle B D C=45^{\circ}$.
Solution. In this triangle, angle $A B C$ is obviously equal to $180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$. Draw the altitude $A H$ and note that triangle $A H B$ is isosceles, since $\angle B A H=90^{\circ}-45^{\circ}=\angle A B H$ (Fig. 1). Therefore, $A H=B H$. In the right tr... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,705 |
# Problem 10.2
The function $f(x)$ is defined for all real numbers, and for any $x$, the equalities $f(x+3)=f(3-x)$ and $f(x+11)=f(11-x)$ hold. Prove that $f(x)$ is a periodic function.
Number of points 7
# | # Solution
We will prove that the given function has a period of 16. There are different ways to reason about this.
First method.
$f(x+16)=f((x+5)+11)=f(6-x)=f((3-x)+3)=f(x)$.
Second method.
The first equality means that the points with coordinates ( $x+3, f(x+3)$ ) and ( $3-x, f(3-x)$ ) are symmetric with respect... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,707 |
# Problem 10.3
Given a rectangle $A B C D$. Through point $B$, two perpendicular lines are drawn. The first line intersects side $A D$ at point $K$, and the second line intersects the extension of side $C D$ at point $L$. Let $F$ be the intersection of $K L$ and $A C$. Prove that $B F \perp K L$.
## Number of points ... | Solution
Since $\angle A B K = \angle C B L$, triangles $A B K$ and $C B L$ are similar. Therefore, triangles $A B C$ and $K B L$ are also similar, and $\angle B K F = \angle B A F$. Conseque... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,708 |
# Task 10.4
The sequence of numbers $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=1, a_{n+1}=\frac{3 a_{n}}{4}+\frac{1}{a_{n}}(n \geq 1)$
Prove that a) the sequence $\left\{a_{\mathrm{n}}\right\}$ is bounded; b) $\left|a_{1000}-2\right|<2^{-1000}$.
## Number of points 7
# | # Solution
Obviously, $a_{n}>0$. By the Cauchy inequality for all $n$
$$
a_{n+1}=\frac{3 a_{n}}{4}+\frac{1}{a_{n}} \geq 2 \sqrt{\frac{3 a_{n}}{4} \cdot \frac{1}{a_{n}}}=\sqrt{3}>1
$$
Consider
$$
a_{n+1}=2+\frac{3 a_{n}}{4}+\frac{1}{a_{n}}-2=2+\frac{3 a_{n}^{2}-8 a_{n}+4}{4 a_{n}}=2+\frac{\left(a_{n}-2\right)\left(3... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,709 |
# Task 10.5
The field is a $41 \times 41$ grid, in one of the cells of which a tank is hidden. The fighter aircraft shoots one cell per shot. If a hit occurs, the tank moves to an adjacent cell along a side; if not, it remains in place. After each shot, the pilot does not know whether a hit occurred. To destroy the ta... | # Answer:
2521 shots
## Solution
## Example.
Let's color the cells in a checkerboard pattern so that the corner cells are black. Suppose the pilot first shoots at all the white cells, then at all the black ones, and then again at all the white ones. If the tank was on a white cell, the pilot will destroy it in the ... | 2521 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,710 |
1. On the board, several different integers are written such that the product of the three smallest of them is 8, and the product of the three largest of them is 27. Can it happen that exactly five numbers are written on the board? | Answer: can.
Solution. Example: $9,3,1,-2,-4$.
Criteria. Answer without example: 0 points. | can | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,711 |
3. Real numbers $a$ and $b$ are such that $a^{5}+b^{5}=3, a^{15}+b^{15}=9$. Find the value of the expression $a^{10}+b^{10}$. | Answer: 5.
Solution. Let $x=a^{5}, y=b^{5}$. Then $x+y=3, x^{3}+y^{3}=9 ;(x+y)\left(x^{2}-x y+y^{2}\right)=9$; $3\left(x^{2}-x y+y^{2}\right)=9 ; x^{2}-x y+y^{2}=3 ; x^{2}+2 x y+y^{2}=9 ; 3 x y=6 ; x y=2 ; x^{2}+y^{2}=9-2 x y=9-$ $2 \cdot 2=5 ; a^{10}+b^{10}=x^{2}+y^{2}=5$
Criteria. Simplification using substitution w... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,713 |
4. In a company of 8 people, each person is acquainted with exactly 6 others. In how many ways can four people be chosen such that any two of them are acquainted? (We assume that if A is acquainted with B, then B is also acquainted with A, and that a person is not acquainted with themselves, as the concept of acquainta... | Answer: 16.
Solution. Each person in the company does not know $8-1-6=1$ person in this company. This means the company can be divided into four pairs of strangers. From each pair, we can only take one person to form a company of four people who are all acquainted with each other. The selection of such a company consi... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,714 |
5. On the side $\mathrm{BC}$ of triangle $\mathrm{ABC}$, a point $\mathrm{P}$ is taken such that $\angle \mathrm{PAB}=45^{\circ}$. The perpendicular bisector of segment $\mathrm{AP}$ intersects side $\mathrm{AC}$ at point $\mathrm{Q}$. It turns out that $\mathrm{PQ} \perp \mathrm{BC}$. Prove that triangle $\mathrm{ABC}... | Solution. Method 1. Let $\mathrm{D}$ be the point of intersection of the perpendicular bisector of segment $\mathrm{AP}$ with side $\mathrm{AB}$. Points on the perpendicular bisector of a segment are equidistant from its endpoints. Therefore, $\mathrm{AD}=\mathrm{DP}$ and $\mathrm{AQ}=\mathrm{QP}$. Triangle $\mathrm{AD... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,715 |
6. Let $a$ and $b$ be natural numbers. Prove that at least one of the numbers: $a, b, a+b$ - equals the difference of the squares of two integers. | Solution. Since $(n+1)^{2}-n^{2}=2 n+1$, any odd number can be represented as the difference of squares of two consecutive numbers. Since $(n+1)^{2}-(n-1)^{2}=4 n$, any number divisible by four is the difference of two squares. It remains to consider the case when a and $b$ are divisible by 2 but not by 4. In this case... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,716 |
11.1. It is known that
$$
\sin x \cos y=\cos x \sin y=\frac{1}{2}
$$
Find $\cos 2 x-\sin 2 y$. | Answer: -1 or 1.
First solution. Multiplying the given equations and multiplying the product by 4, we get $\sin 2 x \sin 2 y=1$. From this, $\sin 2 x=\sin 2 y=1$ or $\sin 2 x=\sin 2 y=-1$ (both cases are possible; it is sufficient to take $x=y=\frac{\pi}{4}$ or $x=y=\frac{3 \pi}{4}$). In both cases, $\cos 2 x=0$. Then... | -1or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,717 |
11.2. The numbers $x$ and $y$ satisfy the inequality $x>y>\frac{2}{x-y}$. Prove that $x^{2}>y^{2}+4$. | Solution. By adding the inequalities $x>\frac{2}{x-y}$ and $y>\frac{2}{x-y}$, we get that $x+y>\frac{4}{x-y}$.
From the condition $x>y$, it follows that the denominator of the fractions is positive, so we can multiply by it without changing the inequality sign. Then we get: $(x+y)(x-y)>4$, that is, $x^{2}-y^{2}>4$. Th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,718 |
11.3. A circle can be described around the base of an $n$-sided pyramid. It is known that the center of this circle is equidistant from all the midpoints of the lateral edges of the pyramid. Prove that the lengths of all the lateral edges of the pyramid are equal. | Solution. Let $O$ be the center of the circle circumscribed around the base $A_{1} A_{2} \ldots A_{n}$ of the pyramid $S A_{1} A_{2} \ldots A_{n}$, and let points $M_{1}, M_{2}, \ldots, M_{n}$ be the midpoints of the lateral edges $S A_{1}, S A_{2}, \ldots, S A_{n}$ of the pyramid $S A_{1} A_{2} \ldots A_{n}$ (see Fig.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,719 |
11.4. Is it true that any number divisible by 6 and greater than 1000 can be represented as
$$
n(n+1)(n+2)(n+3)(n+4)-m(m+1)(m+2)
$$
where \( m \) and \( n \) are natural numbers? | Answer: Incorrect.
Solution. Note that the product of five consecutive natural numbers $n(n+1)(n+2)(n+3)(n+4)$ is divisible by 5. Let's consider several cases.
If $m$ has a remainder of 0, 3, or 4 when divided by 5, then $m(m+1)(m+2)$ is divisible by 5.
If $m$ has a remainder of 1 when divided by 5, then $m(m+1)(m+2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,720 |
11.5. Can one choose a number $n \geqslant 3$ and fill an $n \times (n+3)$ table (with $n$ rows and $n+3$ columns) with distinct natural numbers from 1 to $n(n+3)$ in such a way that in each row there are three numbers, one of which is the product of the other two? | Answer: No.
Solution: Suppose the table could be filled in the required manner. Consider in each row the three numbers: two multipliers and their product. Mark the smallest multiplier in each row. Note that a multiplier cannot equal 1, because in this case, the row would contain identical numbers. Since there are $n$ ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,721 |
11.1. In each cell of a $6 \times 6$ table, numbers are written. All the numbers in the top row and all the numbers in the left column are the same. Each of the other numbers in the table is equal to the sum of the numbers written in the two adjacent cells - the cell to the left and the cell above. What number can be w... | Answer: 8. Let $a$ be the number written in the corner cell at the top left. Then the table is filled in sequentially. For the bottom right corner, we get the relation $252 a = 2016$, from which it follows that $a = 8$.
Comment. An answer with a chain of calculations - 7 points. An answer obtained using combinatorial ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,722 |
11.2. Find the set of values of the function $f(x)=\sin ^{2} x+\cos x$. | Answer: the set of values of the function is the interval $[-1 ; 5 / 4]$.
On the one hand, $f(x)=\sin ^{2} x+\cos x \geq 0-1=-1$ and this estimate is achieved $f(\pi)=-1$.
On the other hand, $f(x)=1+\cos x-\cos ^{2} x=5 / 4-(1 / 2-\cos x)^{2} \leq 5 / 4$. This estimate is achieved $f(\pi / 3)=5 / 4$.
Thus, the funct... | [-1;\frac{5}{4}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,723 |
11.4. A set of composite numbers from the set $\{1,2,3,4, \ldots, 2016\}$ is called good if any two numbers in this set do not have common divisors (other than 1). What is the maximum number of numbers that a good set can have? | Answer: 14 numbers.
Example. $2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$. All the presented numbers are composite, as they are squares of natural numbers greater than 1. It is also clear that different prime numbers do not have common divisors greater th... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,724 |
7.1. Write the number 2013 several times in a row so that the resulting number is divisible by 9. Explain your answer. | Answer: for example, 201320132013.
Solution. We will provide several ways to justify this.
First method. The sum of the digits of the written number is $(2+0+1+3) \cdot 3=18$, so it is divisible by 9 (by the divisibility rule for 9).
Second method. The sum of the digits of the number 2013 is divisible by 3, so if we... | 201320132013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,725 |
7.2. The height of the room is 3 meters. During its renovation, it was found that more paint is used for each wall than for the floor. Can the area of the floor of this room be more than 10 square meters? Explain your answer. | Answer: No, it cannot.
Solution. We will provide several ways to justify this.
First method. Since more paint is used for each wall than for the floor, the area of the floor is less than the area of each of the walls. The area of the floor is the product of the room's width and its length, while the area of one of th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,726 |
7.3. Yesterday, Sasha cooked soup and put in too little salt, so he had to add more salt. Today, he put in twice as much salt, but still had to add more salt, though only half the amount he added yesterday. By what factor does Sasha need to increase today's portion of salt so that he doesn't have to add more salt tomor... | Answer: 1.5 times.
Solution. We will present two methods of solving: "arithmetic" and "algebraic."
First method. Doubling the amount of salt compensated for half of yesterday's addition, so the initial amount of salt that Sasha put in yesterday is one-third of the required amount. Therefore, the initial amount of sal... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,727 |
7.5. In the sum $+1+3+9+27+81+243+729$, any addends can be crossed out and the signs before some of the remaining numbers can be changed from “+” to “-”. Masha wants to use this method to first obtain an expression whose value is 1, then, starting over, obtain an expression whose value is 2, then (starting over again) ... | Answer: up to the number 1093 (inclusive).
Solution. The number 1 is obtained by erasing all addends except the first. Then Masha can get the numbers $2=-1+3, 3=+3$, and $4=+1+3$.
We will show that by adding the addend 9, one can obtain any integer from $5=9-4$ to $13=9+4$. Indeed, $5=-1-3+9; 6=-3+9; 7=+1-3+9; 8=-1+9... | 1093 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,729 |
1. On January 1, 2013, a little boy was given a bag of chocolate candies, containing 300 candies. Each day, the little boy ate one candy. On Sundays, Karlson would fly over, and the little boy would treat him to a couple of candies. How many candies did Karlson eat? (January 1, 2013, was a Tuesday).
2. Petya can swap ... | 1. Answer: 66.
Solution. In a full week, Little One eats 7 candies (one each day), and Karlson eats 2 (on Sunday), so together they eat 9 candies. Dividing 300 by 9 with a remainder, we get the quotient 33 and the remainder 3 (33$\cdot$9+3=300). In the last incomplete week, Little One will eat 3 candies: on Wednesday,... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,730 |
3. Three statements are made about a four-digit number: "The digits 1, 4, and 5 appear in the number", "The digits 1, 5, and 9 appear in the number", "The digits 7, 8, and 9 appear in the number". Exactly two of these statements are true. Which digits appear in the number? | 3. Answer: $1,4,5,9$.
Solution. One of the three statements can be false. 1) If the statement "The digits 1, 4, and 5 appear in the number" is false, then the statements "The digits 1, 5, and 9 appear in the number" and "The digits 7, 8, and 9 appear in the number" are true, and the digits in the four-digit number are... | 1,4,5,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,731 |
4. On the table, there are four stacks of identical coins consisting of 5 coins, 6 coins, 7 coins, and 19 coins. In one of these stacks, one coin has been replaced with a coin of a different weight, which looks identical to the others. How can you use a balance scale without weights to find two stacks in which all the ... | 4. Solution. Take a coin from the stack of 7 coins and place it in the stack of 5 coins. This will result in two stacks of 6 coins each. Place these on the different pans of the scale. If equilibrium is achieved, the sought stacks are the ones with 5 and 7 coins. Otherwise, the sought stacks are the ones with 6 and 19 ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,732 |
5. A $5 \times 5$ square is to be cut into two types of rectangles: $1 \times 4$ and $1 \times 3$. How many rectangles can result from the cutting? Justify your answer. | 5. Answer: 7.
Solution. Example:
If there are no more than 6 rectangles, then they occupy no more than $6 \cdot 4 = 24$ cells, while there are 25 cells. Therefore, there must be at least 7 r... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,733 |
1. 50 students from fifth to ninth grade published a total of 60 photos on Instagram, each not less than one. All students of the same grade (same parallel) published an equal number of photos, while students of different grades (different parallels) published a different number. How many students published only one ph... | Solution. Let each of the students publish one photo first, during which 50 photos out of 60 will be published. There are 10 photos left to be published, which we will call additional photos. We have a total of five different classes (parallels), and these 10 additional photos must be published by different students fr... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,734 |
2. Find the value of the expression
$$
2^{2}+4^{2}+6^{2}+\ldots+2018^{2}+2020^{2}-1^{2}-3^{2}-5^{2}-\ldots-2017^{2}-2019^{2}
$$ | # Solution.
$$
\begin{aligned}
& 2020^{2}-2019^{2}+2018^{2}-2017^{2}+\ldots+6^{2}-5^{2}+4^{2}-3^{2}+2^{2}-1^{2}= \\
& =(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+\ldots+ \\
& +(6-5)(6+5)+(4-3)(4+3)+(2-1)(2+1)= \\
& =2020+2019+2018+2017+\ldots+6+5+4+3+2+1= \\
& =\frac{2020+1}{2} \cdot 2020=2041210
\end{aligned}
$$
... | 2041210 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,735 |
3. $a$ and $b$ are positive numbers, and $a+b=2$. Prove the inequality
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1} \geq 1
$$ | Solution. Let's write two equalities.
$$
\frac{1}{a^{2}+1}=\frac{a^{2}+1-a^{2}}{a^{2}+1}=1-\frac{a^{2}}{a^{2}+1}
$$
Similarly,
$$
\frac{1}{b^{2}+1}=1-\frac{b^{2}}{b^{2}+1}
$$
Rewrite the inequality in an equivalent form.
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1} \geq 1
$$
$$
\begin{gathered}
1-\frac{a^{2}}{a^{2}+1}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,736 |
4. In an acute triangle $ABC$, side $AB$ is the diameter of a circle that intersects the sides $AC$ and $BC$ at points $P$ and $Q$ respectively. The tangents to the circle at points $P$ and $Q$ intersect at point $F$. Prove that the lines $CF$ and $AB$ are perpendicular.
 x+m-1=0
$$
the smallest? | 3. $\mathrm{m}=3$.
Note that the discriminant $\mathrm{D}$ of the equation is equal to
$$
\mathrm{D}=(\mathrm{m}+1)^{2}-4(\mathrm{~m}-1)=(\mathrm{m}-1)^{2}+4>0
$$
therefore, the equation has two roots for any $\mathrm{m}$. Let's represent the sum of the squares of the roots $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ as
... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,739 |
4. In a convex quadrilateral $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}, \mathrm{M}$ is the point of intersection of the diagonals, $\mathrm{K}$ is the point of intersection of the angle bisectors of angles A and D. Prove that points A, M, K, and D lie on the same circle. | 4. It is sufficient to prove the equality of angles $\angle \mathrm{AMD}=\angle \mathrm{AKD}$. Let's perform the calculations.
$\angle \mathrm{AKD}=180^{\circ}-1 / 2(\angle \mathrm{A}+\angle \mathrm{D})=1 / 2(\angle \mathrm{B}+\angle \mathrm{C})$.
On the other hand, triangles $\mathrm{ABC}$ and $\mathrm{BCD}$ are iso... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,740 |
5. In an internet community of 20 participants, there are friends. Any two participants A and B are either friends themselves or there exists a chain
$\mathrm{A}-\mathrm{C}_{1}-\ldots-\mathrm{C}_{\mathrm{k}}-\mathrm{B}$
where adjacent participants in the chain are friends, that is, $\mathrm{C}_{1}$ is a friend of $\m... | 5. For a chain of friends $\mathrm{A}-\mathrm{C}_{1}-\mathrm{C}_{2}-\ldots-\mathrm{C}_{\mathrm{k}}-\mathrm{B}$, its length is the number $\mathrm{k}+1$. We will call a chain of friends A - $\mathrm{C}_{1}-\mathrm{C}_{2}-\ldots-\mathrm{C}_{\mathrm{k}}-\mathrm{B}$, whose length is greater than 1, closed if A is friends w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,741 |
1.1. Arina wrote down a number and the number of the month of her birthday, multiplied them and got 248. In which month was Arina born? Write the number of the month in your answer.
 | Answer: 8
Solution. $248=2 * 2 * 2 * 31$. Obviously, 31 is the number, then 8 is the month number. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,742 |
3.1. A number was multiplied by its first digit and the result was 494, by its second digit - 988, by its third digit - 1729. Find this number. | Answer: 247
Solution. It can be noticed that all three obtained numbers are divisible by 247, from which we get the answer. | 247 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,744 |
4.1. To the fraction $\frac{1}{6}$, some fraction was added, and the result turned out to be a proper fraction with a denominator less than 6. What is the largest fraction that could have been added?
she 7. What is the largest fraction that could have been added?
Answer: $\frac{29}{42}$ | \frac{29}{42} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,747 |
5.1. It is known that in the past chess tournament, in each round all players were paired, the loser was eliminated (there were no draws). It is known that the winner played 6 games. How many participants in the tournament won at least 2 games more than they lost? | Answer: 8
Solution. Since the winner played 6 games, there were a total of $2^{6}=64$ players. The win-loss ratio for those eliminated in the 1st round is -1; for those who lost in the 2nd round, it is 0; for those who lost in the 3rd round, it is +1; for those who lost in the 4th round and beyond, it is at least +2. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,748 |
6.1. In a basketball team, there is a certain number of players. The coach added up all their heights and divided by the number of players (let's call this the average height), getting 190 cm. After the first game, the coach removed Nikolai, who is 197 cm tall, and replaced him with Petr, who is 181 cm tall, after whic... | Answer: 8
Solution. The change in the total height of the players by $197-181=16$ cm leads to a change in the average height of the team by $190-188=2$ cm, so the team has $\frac{16}{2}=8$ people. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,749 |
7.1. On the counter lie 10 weights with masses $n, n+1, \ldots, n+9$. The seller took one of them, after which the total weight of all the remaining ones was 1457. Which weight did the seller take? | # Answer: 158
Solution. The total weight of all 10 weights is $10 n+45$. Suppose the weight taken has a weight of $n+x$, where $x<10$. Then
$10 n+45-n-x=1457$.
$9 n+45-x=1457$.
We can see that 1457, when divided by 9, gives a remainder of 8, so $x=1$. Then
$9 n+45-1=1457$
$9 n=1413$.
$n=157$.
$n+x=158$. | 158 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,750 |
8.1. In a circle, 58 balls of two colors - red and blue - are arranged. It is known that the number of triples of consecutive balls, among which there are more red ones, is the same as the number of triples with a majority of blue ones. What is the smallest number of red balls that could be present?
$ is equal to $0.8 n (100\% - 20\% = 80\%$ of $n)$. Therefore, $m = 2n$, and girls make up $\frac{n}{m+n... | Answer: $33 \frac{1}{3} \%$. | 33\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,754 |
11.2. Let's call a natural number curious if after multiplying it by 9 it is written with the same digits but in reverse order. Prove that the set of curious numbers is infinite. | Solution. Reasoning similarly to how it is done in solving problem 7.4, consider the numbers $N_{k}=1099 \ldots 989$ (here between the digits 0 and 8 there are $k$ nines). It is easy to verify directly that all numbers $N_{k}$ for any natural $k$ are curious. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,756 |
11.3. Solve the inequality $\sqrt{x}-\sqrt{1-x}+8 \sin x<8$. | Answer: $0 \leq x \leq 1$. Solution. The domain of the inequality is the interval $[0 ; 1]$. On $[0 ; 1]$, the function $f(x)=\sqrt{x}-\sqrt{1-x}+8 \sin x$ is monotonically increasing. Let's check the value at the right end of the interval, specifically, we will show that $f(1)<8$. Indeed, $1+8 \sin 1<1+8 \sin \pi / 3$... | 0\leqx\leq1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 13,757 |
11.4. a) Given a rectangular parallelepiped with a volume of 2017 and integer coordinates of vertices in a Cartesian coordinate system in space. Find the diagonal of the parallelepiped, given that its edges are parallel to the coordinate axes. b) Does there exist a rectangular parallelepiped with a volume of 2017 and i... | Answer: a) $\sqrt{2017^{2}+2}$, b) it exists. Solution. a) Let the dimensions of the parallelepiped be $a \times b \times c$. Then we have $a b c=2017$, and due to the simplicity of the number 2017, we get that the edges $a, b, c$ are (in some order) $2017,1,1$, so the diagonal is $\sqrt{2017^{2}+1^{2}+1^{2}}$. b) 2017... | \sqrt{4035} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,758 |
1. In a $7 \times 7$ table (see figure), the elements of each row and each column form arithmetic progressions. What is the number $x$ in the central cell?
 | Answer: 111.
Solution. The middle term of an arithmetic progression is equal to the half-sum of the extreme terms, so in the middle cell of the first row, the number is $(3+143) / 2=73$, and in the middle cell of the last row, the number is $(82+216) / 2=149$.
$ points. According to the problem, the ratio of the number of points scored by girls to the number of points scored by boys is 1:4. Therefore, the gir... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,761 |
3. On a $6 \times 6$ board, 6 chips are placed such that each horizontal and vertical row contains exactly one chip. Can the remaining part of the board be tiled with $1 \times 2$ rectangles (in any orientation)? | Answer: No.
Solution. We will color the board in a checkerboard pattern and number the rows and columns. If the remaining part of the board can be tiled with $1 \times 2$ rectangles, then it contains an equal number of black and white cells, so 3 tokens are on black cells, and 3 on white cells. Note that for a white c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,762 |
4. Through the vertex $B$ of the rectangle $A B C D$, two perpendicular lines are drawn. One line intersects the side $A D$ at point $K$, and the other line intersects the extension of side $C D$ at point $L$. The lines $K L$ and $A C$ intersect at point $F$. Prove that the quadrilateral $A B F K$ can be inscribed in a... | Solution. Since $\angle A B K=\angle A B L-\angle K B L=\angle A B L-90^{\circ}=\angle A B L-$ $\angle A B C=\angle C B L$, the right triangles $A B K$ and $C B L$ are similar. From this, $\frac{A B}{B C}=\frac{B K}{B L}$, then $\frac{A B}{B K}=\frac{B C}{B L}$ and the right triangles $A B C$ and $K B L$ are also simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,763 |
5. Let $f(x)=x^{3}+a x^{2}+b x+c$, where $a, b$ and $c$ are distinct non-zero integers. For which $a, b$ and $c$ do the equalities $f(a)=a^{3}$ and $f(b)=b^{3}$ hold? | Answer: $a=-2, b=4, c=16$.
Solution. The equalities $f(a)=a^{3}$ and $f(b)=b^{3}$ give us the equalities $a^{3}+a b+c=0$ and $a b^{2}+b^{2}+$ $c=0$. Subtracting one from the other, we get $(a+1) b^{2}-a b-a^{3}=0$. This expression can be factored as
$$
(a+1) b^{2}-a b-a^{3}=(b-a)\left((a+1) b+a^{2}\right)=0
$$
Since... | =-2,b=4,=16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,764 |
1. A flask contains 40 liters of milk. How can you transfer 6 liters of milk from it into a 10-liter bucket using a 9-liter and a 5-liter bucket? | Solution. We fill the 5-liter bucket with milk and pour it into the bucket. Then we fill the 5-liter bucket again, pour it into the 9-liter bucket, fill the 5-liter bucket once more, and pour the missing 4 liters into the 9-liter bucket. Then, exactly 1 liter remains in the 5-liter bucket, which we pour into the bucket... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,765 |
3. Find all natural $\boldsymbol{n}$ such that the value of the expression $\frac{2 n^{2}-2}{n^{3}-n}$ is a natural number. | Solution. $\frac{2 n^{2}-2}{n^{3}-n}=\frac{2(n-1)(n+1)}{n(n-1)(n+1)}$. For $n \neq \pm 1$, this expression is identically equal to $\frac{2}{n}$. Since $n \neq 1$, the value of the last expression is a natural number only when $n=2$.
Answer: $n=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,767 |
2. (7 points) From Monday to Wednesday, the gnome eats semolina porridge for breakfast, from Thursday to Saturday - rice porridge, and on Sunday he makes himself an omelet. On even-numbered days of the month, the gnome tells the truth, and on odd-numbered days - lies. On which of the first ten days of August 2016 could... | Answer. On Tuesday, August 2, on Wednesday, August 3, on Friday, August 5, on Monday, August 8
Solution. If today is Sunday, Monday, or Tuesday, then tomorrow the gnome eats porridge and the statement turns out to be true. This means that the gnome could have said the given statement only if such a day falls on an eve... | OnTuesday,August2,onWednesday,August3,onFriday,August5,onMonday,August8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,769 |
3. (7 points) The number 20 is written on the board. In one move, it is allowed to either double the number or erase its last digit. Is it possible to get the number 25 in several moves? | Answer. Possible.
Solution. The number 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the required chain of transformations can look like this:
$$
20 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \rightarrow 128 \rightarrow 256 \... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,770 |
5. The grandmother has three grandchildren. If a grandchild finished the first grade, the grandmother gave them one book; if they finished the second grade, she gave them two books; if the third grade, then three books, and so on. The books received as gifts over the years were placed on one shelf. Currently, there are... | Answer. Sixth grade.
Solution. From the condition of the problem, it follows that if the grandson has finished the second grade, then there are $1+2=3$ of his books on the shelf, if he has finished the third grade, then $1+2+3=6$ of his books, and so on. For convenience, let's make a table.
| What grade the grandson ... | Sixth\grade | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,772 |
Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,776 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_22f... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,777 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,778 |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,779 |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,781 |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,783 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,785 |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,786 |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,789 |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,792 |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,793 |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
$ there exists an increasing linear function $\mathrm{g}(\mathrm{x})$ such that $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{g}(\mathrm{x}))$. | Solution. Let $\mathrm{f}(\mathrm{x})=\mathrm{kx}+\ell$. The increase of $\mathrm{f}(\mathrm{x})$ means that $\mathrm{k}>0$. Let $\mathrm{g}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$. We need to find suitable a and b. We have: $\mathrm{g}(\mathrm{g}(\mathrm{x}))=\mathrm{ag}(\mathrm{x})+\mathrm{b}=\mathrm{a}(\mathrm{ax}+\math... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,795 |
9.2 Two triangles are given. The sum of two angles of the first is equal to some angle of the second. The sum of another pair of angles of the first is also equal to some angle of the second. Prove that the first triangle is isosceles. | Solution. Let $\alpha, \beta, \gamma$ be the angles of the first triangle. Let $\alpha+\beta$ be an angle of the second triangle. Another pair of angles from the first triangle, say, $\alpha$ and $\gamma$ gives the sum $\alpha+\gamma$. Suppose that $\alpha+\gamma$ is another angle of the second triangle, different from... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,796 |
9.3 On the extension of side $A D$ of the inscribed quadrilateral $A B C D$ beyond point $D$, point $\mathrm{E}$ is marked such that $\mathrm{AC}=\mathrm{CE}$ and $\angle \mathrm{BDC}=\angle \mathrm{DEC}$. Prove that $\mathrm{AB}=\mathrm{DE}$. | Solution. See fig.
Let $\angle \mathrm{DEC}=\varphi$. Then $\angle \mathrm{DAC}=\varphi$, since triangle ACE is isosceles by condition. Angle BAC is also equal to $\varphi$ (inscribed angles BAC and $\mathrm{BDC}$ subtend the same arc). Moreover, $\angle \mathrm{ABC}=\angle \mathrm{CDE}$, since both of them, together ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,797 |
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