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11.1. Find the domain and range of the function $\sqrt{1-\cos 2 x+2 \sin x}+\frac{1}{\sqrt{\sin ^{2} x+\sin x}}$. | Answer: The domain of definition $2 k \pi < x < \pi + 2 k \pi \Leftrightarrow \sin x(\sin x+1)>0 \Leftrightarrow$ $\sin x>0$ (since $\sin x \geq-1) \Leftrightarrow 2 k \pi < x < \pi + 2 k \pi (k \in \mathbb{Z})$. Further, investigating the function $y(t)$, we obtain the minimum point $t_{0}=2^{-1 / 4}<\sqrt{2}$ and the... | 2\sqrt[4]{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,502 |
11.3. Can a square be constructed on a coordinate plane with vertices at integer points and with an area, a) equal to $2000;$ b) equal to $2015?$ | Answer: a) can; b) cannot. Solution. a) Since $2000=20^{2} \cdot 5=20^{2}\left(2^{2}+1^{2}\right)$, it is possible to construct a square. Indeed, consider the vertices $A(0 ; 0), B(-20 ; 40)$, $C(20 ; 60), D(40 ; 20)$. The lengths of all sides of the quadrilateral $A B C D$ are $\sqrt{20^{2}+40^{2}}=\sqrt{2000}$, and a... | )can;b)cannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,504 |
8.1 The average age of the teaching staff of a school, consisting of 20 teachers, was 49 years. When another teacher joined the school, the average age became 48 years. How old is the new teacher? | Answer: 28 years old. Solution: Before the new teacher arrived, the total age of the teachers was 49*20=980. Then the total age became 48*21=1008. Therefore, the new teacher is $1008-$ $980=28$ years old. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,506 |
8.2 Given a triangle, all sides of which are less than one. Prove that there exists a containing isosceles triangle, all sides of which are also less than one. | Solution. Consider in triangle $A B C$ the angle that does not exceed 60 degrees (such an angle exists, since the sum of all three angles is 180 degrees). Let, for definiteness, this be angle $A$ and $A B \leq A C$. We mark a point $B_{1}$ on the ray $A B$ such that $A B_{1}=A C$. Then the isosceles triangle $A B_{1} C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,507 |
8.4 On a grid sheet of paper sized $60 \times 70$ cells (horizontally and vertically respectively), Lena drew a coordinate system (the origin is at the center of the sheet, the $x$-axis is horizontal, the $y$-axis is vertical, and the axes are drawn to the edges of the sheet) and plotted the graph $y=0.83 x$. Then Lena... | Answer: 108. Solution: The graph passes through the first and third quadrants. Let's count the number of shaded cells in the first quadrant (in the third quadrant, there will be the same number, since the graph is a straight line passing through the origin, and therefore centrally symmetric). The graph intersects 29 ve... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,509 |
8.2 Let $s(n)$ denote the sum of the digits of a natural number $n$. Solve the equation $n+s(n)=2018$. | Answer: $n=2008$. Solution. Since $n2018-29=1989$, i.e., $n$ is written in the form $\overline{199 x}$ or $\overline{200 x}$ or $\overline{201 x}$ where $x$ is some digit. In the first case, we have the equation $1990+x+19+x=2018$, which gives a non-integer value of $x$. Similarly, in the third case, the equation $2010... | 2008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,511 |
8.3 On the sides $AB$, $BC$, and $AC$ of triangle $ABC$, points $C_{1}$, $A_{1}$, and $B_{1}$ are marked such that $C_{1}$ is the midpoint of $AB$ and $\angle B_{1} C_{1} A_{1} = \angle C$, $\angle C_{1} A_{1} B_{1} = \angle A$, $\angle A_{1} B_{1} C_{1} = \angle B$. Is it necessarily true that points $A_{1}$ and $B_{1... | Answer: Not necessarily. Solution. Consider the following example: let $ABC$ be an isosceles right triangle with the right angle at $C$. From the midpoint of the hypotenuse $C_{1}$, draw two perpendicular lines (at angles to the hypotenuse, different from $45^{\circ}$). Then the points of intersection with the legs (po... | notnecessarily | Geometry | proof | Yes | Yes | olympiads | false | 22,512 |
8.4 a) Given natural numbers $a$ and $b$, such that $3a + b$ and $3b + a$ give the same remainder when divided by 10. Is it true that the numbers $a$ and $b$ themselves give the same remainder when divided by 10? b) Is it true that natural numbers $a, b$, and $c$ give the same remainder when divided by 10, if it is kno... | Answer: a) incorrect; b) correct. Solution. a) For example, consider $a=1$ and $b=6$, then both numbers $3 a+b$ and $3 b+a$ end in 9. b) Let $s=a+b+c$. By reducing each of the numbers $2 a+b, 2 b+c$, and $2 c+a$ by $s$, we get the numbers $a-c, b-a, c-b$ (some of which may be negative), and these numbers have the same ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,513 |
10.2. Find all values of the parameter $a$ for which the equation $|x+a|=\frac{1}{x}$ has exactly two roots. | Answer: $a=-2$. Solution. The intersection of the graphs of the right and left parts can only be in the first
the graph $y=|x+a|$ in the first quadrant represents the line $y=x+a$, which inte... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,515 |
10.3. Plot on the coordinate plane the set of points satisfying the inequality $\sqrt{x^{2}-2 x y}>\sqrt{1-y^{2}}$ | Solution. The given inequality is equivalent to the system:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x ^ { 2 } - 2 x y > 1 - y ^ { 2 } } \\
{ 1 - y ^ { 2 } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-y)^{2}>1 \\
|y| \leq 1
\end{array} \Leftrightarrow\right.\right. \\
& \text { either }\left\... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,516 | |
10.4. The sequence $a_{n}$ is defined by the relations $a_{n+1}=\frac{1}{2}+\frac{a_{n}^{2}}{2} ; a_{1}=\frac{1}{2}$. Prove that $a_{n}$ is monotonically increasing and $a_{n}<2$ for all $n$. | Solution. We will prove the monotonicity of $a_{n}$. We have $a_{n+1}>a_{n} \Leftrightarrow \frac{1}{2}+\frac{a_{n}^{2}}{2}>a_{n} \Leftrightarrow\left(a_{n}-1\right)^{2}>0$. This inequality is strict, since in fact $a_{n}<1$, which we will prove by mathematical induction. Indeed, $a_{1}=\frac{1}{2}<1$, and if the inequ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,517 |
10.5. On the coordinate plane, a family of concentric circles centered at point $M(\sqrt{2} ; \sqrt{3})$ is considered. a) Is there a circle in this family that has two rational points? b) Prove that there exists a circle in this family, inside which (i.e., inside the disk) there are exactly 2014 integer points. (A rat... | Answer: a) will not be found. Solution. a) Suppose, to the contrary, that there exist two rational points $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}\right)$ on the circle of the given family. Then $\left(x_{1}-\sqrt{2}\right)^{2}+\left(y_{1}-\sqrt{3}\right)^{2}=\left(x_{2}-\sqrt{2}\right)^{2}+\left(y... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,518 |
8.1. There are $n$ kg of cereal ( $n$ - an integer), a balance scale, and one 3-kilogram weight.
a) Prove that if $n$ is not divisible by 3, then in several weighings, it is possible to measure out 1 kg of cereal; b) Is it possible to measure out 1 kg of cereal in three weighings when $n=19$? | Answer: b) it is possible. Solution. a) If a weight is placed on one scale pan and the other is filled with grain to balance the scales, the remaining grain will weigh $n-3$ kg. Let $n=3 k+r$, where $r$ is the remainder of the division of $n$ by 3, and $k$ is the quotient (by condition $r \neq 0$). Therefore, $r$ equal... | proof | Other | proof | Yes | Yes | olympiads | false | 22,519 |
8.3. Initially, the board had $n$ numbers written on it: $1,2, \ldots, n$. It is allowed to erase any two numbers on the board and write down the absolute value of their difference instead. What is the smallest number that can end up on the board after ( $n-1$ ) such operations a) for $n=111$; b) for $n=110$? | Answer: a) 0; b) 1. Solution. a) Using the following 55 operations, we can obtain 56 ones: $\{1,3-2,5-4, \ldots, 111-110\}=\{1,1, \ldots, 1\}$. From these, using 28 operations, we can get 28 zeros: $\{1-1,1-1, \ldots, 1-1\}=\{0,0, \ldots, 0\}$, and then after 27 operations - one 0. b) Note that with any operation, the ... | )0;b)1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,521 |
8.5. Given a convex quadrilateral $A B C D$, where $A B=A D=1, \angle A=80^{\circ}$, $\angle C=140^{\circ}$. Find the length of the diagonal $A C$. | Answer: 1. Solution. We will prove that $A C=1$ by contradiction. If $A C>1$, then in triangle $A B C$ the larger side $A C$ is opposite the larger angle: $\angle B>\angle B C A$. Similarly, for triangle $A D C$ we have $\angle D>\angle D C A$. Adding these inequalities, we get $\angle B+\angle D>\angle C=140^{\circ}$.... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,523 |
7.1. Three positive numbers were changed as follows: the first was increased by $10 \%$, the second by $13 \%$, and the third was decreased by $20 \%$. Will the product of the three resulting numbers be greater or less than the product of the original numbers? | Answer: less. Solution. Let $a ; b ; c$ be the original numbers. Then $1.1 a ; 1.13 b ; 0.8 c$ are the obtained numbers, and their product ( $(1.1 \cdot 1.13 \cdot 0.8$ ) $a b c=0.9944 a b c<a b c$). | 0.9944 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,524 |
7.2. From a three-digit number, the sum of its digits was subtracted and the result was 261. Find the second digit of the original number. | Answer: 7. Solution. Let $\overline{x y z}=100 x+10 y+z$ be the original number. According to the condition, $99 x+9 y=261$, i.e., $11 x+y=29$. Since $y \leq 9$, for $29-y$ to be divisible by 11, we get $y=7$ (and then $x=2$). | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,525 |
7.3. Can a rectangle $100 \times 70$ be cut into three rectangles so that their areas are in the ratio $1: 2: 4$? | Answer: Yes. Solution. See the figure.
 | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,526 |
7.4. a) Prove that the number $\frac{2014^{2}+2016^{2}}{2}$ can be represented as the sum of the squares of two natural numbers. b) Prove a more general fact: the half-sum of the squares of two different even numbers can be represented as the sum of the squares of two natural numbers. | Solution. a) Let $n=$ 1012. Then we have $\frac{(2 n)^{2}+(2 n+2)^{2}}{2}=2 n^{2}+2(n+1)=$ $\left.4 n^{2}+4 n+2=(2 n+1)^{2}+1^{2} . \boldsymbol{\mathbf { \sigma }}\right) \frac{(2 n)^{2}+(2 m)^{2}}{2}=2 n^{2}+2 m^{2}=(n-m)^{2}+(n+m)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,527 |
10.1. Solve the equation $\left(x^{4}+x+1\right)(\sqrt[3]{80}-\sqrt[3]{0.01})=2(\sqrt[3]{5.12}+\sqrt[3]{0.03375})$. | Answer: $x_{1}=0, x_{2}=-1$. Solution. Multiply both sides of the equation by $\sqrt[3]{100}$, we get $\left(x^{4}+x+1\right)(\sqrt[3]{8000}-\sqrt[3]{1})=2(\sqrt[3]{512}+\sqrt[3]{3.375}) \Leftrightarrow 19 \cdot\left(x^{4}+x+1\right)=2(8+1.5) \Leftrightarrow x^{4}+x+1=1 \Leftrightarrow$ $x\left(x^{3}+1\right)=0$. There... | x_{1}=0,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,528 |
10.3. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $\angle B A M = \angle B C A$. Prove that the center of the circumcircle of triangle $A B C$ lies on the line passing through point $B$ and perpendicular to $A M$. | Solution. On the line passing through point $B$ and perpendicular to line $A M$, take such a point $N$ that
$A N = B N$ (it lies on the perpendicular bisector of $A B$). Then
$$
\angle N A B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,530 |
10.4. The sequence of integers $a_{n}$ is defined as follows: $a_{n+1}=a_{n}^{2}-a_{n}+1, a_{1}=100$. Prove that any two distinct terms of the sequence are coprime. | Solution. Let $a_{n}$ and $a_{n+m}$ be two arbitrary terms of the sequence ($m>0$). We will prove that $a_{n+m}$ can be represented as $a_{n+m}=a_{n} \cdot P_{m}\left(a_{n}\right)+1$, where $P_{m}(x)$ is a polynomial with integer coefficients. We will prove this fact by induction. For $m=1$ we have $a_{n+1}=a_{n}^{2}-a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,531 |
7.1. A cyclist was initially traveling at a speed of 20 (km/h). After covering one-third of the distance, he glanced at his watch and decided to increase his speed by 20%. He continued the rest of the journey at the new speed. What was the cyclist's average speed? | Answer: 22.5 km/h. Solution. Let $a$ be the distance, $v=20($ km/h) be the initial speed. Then the new speed is $1.2 v$. The entire journey will take $\frac{a}{3 v}+\frac{2 a}{3 \cdot 1.2 v}=\frac{8}{9} \frac{a}{v}$ (hours). Thus, the average speed is $a: \frac{8}{9} \frac{a}{v}=\frac{9}{8} v=22.5$ (km/h). | 22.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,533 |
7.2. The two adjacent sides of a rectangle are in the ratio 3:7. What is the area of the rectangle if its perimeter is 40 cm? | Answer: $84 \mathrm{~cm}^{2}$. Solution. Let $x$ be the shorter side of the rectangle, then $\frac{7}{3} x$ is the longer side. From the conditions of the problem, we get the equation $2\left(x+\frac{7}{3} x\right)=40$. Hence, $x=6$ (cm), and the area is $x \frac{7}{3} x==84$ $\left(\mathrm{cm}^{2}\right)$. | 84\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,534 |
7.3. Append two digits to the number 2020 so that the resulting six-digit number is divisible by 36. Find all possible solutions. | Answer: 32 or 68. Solution: Note that $36=9 \cdot 4$. Since the sum of the first four digits is 4, by the divisibility rule for 9, the sum of the last two digits of the obtained number can be either 5 or 14. In the first case, the two digits we are looking for must form the number 32 (other options with a sum of 5 are ... | 32or68 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,535 |
7.4. The sum of ten different natural numbers is greater than 144. Prove that among these ten numbers, there are three numbers whose sum is not less than 54. | Solution. Let $a<b<c$ be the three largest numbers among the given ones. If $a \geq 17$, then $a+b+c \geq$ $17+18+19=54$, and the statement is proved. Now consider the case $a \leq 16$ and assume the opposite of the statement of the problem. Then $a+b+c<54$, and the other seven (smaller) numbers among the given ten sum... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,536 |
7.1. Kolya and Petya exchanged stamps. Before the exchange, Kolya had 5 more stamps than Petya. After Kolya exchanged $24\%$ of his stamps for $20\%$ of Petya's stamps, Kolya had one stamp less than Petya. How many stamps did the boys have before the exchange? | Answer. Petya had 45 stamps, Kolya had 50 stamps. Solution. Let Petya have $x$ stamps before the exchange, then Kolya had $(x+5)$ stamps. After the exchange, Petya had $x-\frac{x}{5}+(x+5) \cdot \frac{6}{25}$, and Kolya had $x+5-(x+5) \cdot \frac{6}{25}+\frac{x}{5}$. Solving the equation $x-\frac{x}{5}+(x+5) \cdot \fra... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,537 |
7.2. For a 92-digit natural number n, the first 90 digits are known: from the 1st to the 10th - ones, from the 11th to the 20th - twos, and so on, from the 81st to the 90th - nines. Find the last two digits of the number n, given that n is divisible by 72. | Answer: 36. Solution: Let the last digits be $x$ and $y$. The number $n$ must be divisible by 9 and 8. The number consisting of the first 90 digits is divisible by 9, since the sum of its digits is divisible by 9. Therefore, the number $\overline{x y}$ is also divisible by 9. In addition, by the divisibility rule for 4... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,538 |
7.3. a) Can the numbers $1,2,3,4,5,6,7,8$ be rearranged so that adjacent numbers differ by either 2 or 3? b) The same problem for one hundred numbers $1,2,3, \ldots, 100$. | Answer. a) It is possible, for example: $1,3,6,8,5,2,4,7$; b) It is possible, for example: $1,3,5,2,4,6,8,10,7,9, \ldots$.
Solution. a) Constructing an example is helped by a graph of possible neighbors. b) The next quintet of numbers $5 k+1$, $5 k+3,5 k+5,5 k+2,5 k+4$ precedes the following quintet $5(k+1)+1,5(k+1)+3... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,539 |
7.4. Given a rectangle that is not a square, where the numerical value of the area is equal to three times the perimeter. Prove that one of the sides of the rectangle is greater than 12. | Solution. Let $a, b$ be the sides of the rectangle. According to the condition, $a b=3(2 a+2 b)$. From this, dividing by $a b$, we get $\frac{6}{a}+\frac{6}{b}=1$. The positive numbers $\frac{6}{a}$ and $\frac{6}{b}$ are not equal to $\frac{1}{2}$ (since $a \neq b$), so one of them is less than $\frac{1}{2}$, and the o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,540 |
7.1. $\quad$ In the 7a class, $60 \%$ of the students are girls. When two boys and one girl were absent due to illness, the percentage of girls present was $62.5 \%$. How many girls and boys are there in the class according to the list? | Answer: 21 girls and 14 boys. Solution. Let there be $d$ girls and $m$ boys in the class. From the conditions of the problem, we have two equations: $\frac{d}{d+m}=0.6$ and $\frac{d-1}{d+m-3}=0.625$. From the first equation, $2 d=3 m$. Substituting $m=\frac{2}{3} d$ into the second equation and solving it, we get $d=21... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,541 |
7.2. There are 11 kg of cereal. How can you measure out 1 kg of cereal using two weighings on a balance scale, if you have one 3 kg weight? | Solution. First weighing: place a weight (3 kg) on one scale pan, and on the other, initially 11 kg of grain, and keep pouring grain onto the first pan until equilibrium is achieved. We get 3 kg (weight) + 4 kg (grain) $=7$ kg (grain) (since $3+x=11-x=>x=4$). Second weighing: from the obtained 4 kg of grain, pour out 3... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,542 |
7.3. Find a six-digit number that, when multiplied by 9, is written with the same digits as the original number but in reverse order? How many such six-digit numbers are there? | Answer: 109989 is the only number. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (whi... | 109989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,543 |
7.4. On the edges of a cube, numbers $1,2, \ldots, 12$ were arranged in some order, and for each face, the sum of the four numbers on its edges was calculated. Prove that there are two faces, on one of which the corresponding sum is greater than 25, and on the other - less than 27. | Solution. Let's calculate the corresponding sum on each face and then add these sums for all six faces. We will get $(1+2+\ldots+12) \cdot 2$ as a result, since in this calculation any edge will be counted twice. Thus, the total sum is 156, and then for at least one face, its sum is not less than $\frac{156}{6}=26$. (I... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,544 |
9.1. We have a set consisting of $n$ weights with masses $1,2, \ldots, n$. Can all the weights be divided into two piles of equal weight if: a) $n=99$, b) $n=98$. | Answer. a) It is possible; b) it is not possible.
Solution. See the solution to problem 7.4. Part b) follows (as in problem 7.4(a)) from the oddness of the total weight of all weights. Part a) (as in problem 7.4(b)) follows from the fact that $99=1+98$, and the number of all consecutive numbers from 2 to 97 is divisib... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,545 |
9.2. Given one hundred numbers: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{100}$. We will compute 98 differences: $a_{1}=1-\frac{1}{3}, a_{2}=\frac{1}{2}-\frac{1}{4}, \ldots$, $a_{98}=\frac{1}{98}-\frac{1}{100}$. What is the sum of all these differences? | Answer: 14651/9900. Solution.
$a_{1}+a_{2}+\ldots+a_{98}=\left(1-\underset{\sim}{\frac{1}{3}}\right)+\left(\frac{1}{2}-\underset{\sim}{\frac{1}{4}}\right)+\left(\frac{1}{3}-\frac{1}{\sim}\right)+\left(\frac{1}{4}-\frac{1}{\sim}\right)+\ldots+\left(\frac{1}{\underline{97}}-\frac{1}{99}\right)+\left(\underline{\underline... | \frac{14651}{9900} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,546 |
9.3. Find the natural number $x$ that satisfies the equation
$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3} .
$$ | Answer. $\quad 2012$.
Solution. See problem 8.4. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,547 |
9.4. a) Prove that a square can be divided into 40 smaller squares.
b) Can it be achieved that in the desired division, the sides of the smaller squares take only two values \(a < b\), differing by no more than 25% (i.e., \(\frac{b}{a} \leq 1.25\))? | Answer. b) Yes. Solution. a) We can provide various partition options. For example, first, we divide the square into 25 smaller squares with a side length of $1 / 5$ (without loss of generality, we consider the original square to be a unit square), then we take 5 of these smaller squares and divide each of them into 4 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,548 |
9.5. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$. | Answer. $\quad 60^{\circ}$. Solution. First, we will show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$.... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,549 |
7.1. Append one digit to the left and one digit to the right of the number 2016 so that the resulting six-digit number is divisible by 72 (provide all solutions). | Answer: 920160 and 120168. Solution. The desired number must be divisible by $72=8 \cdot 9$. For divisibility by 8, the three-digit number formed by the last digits must be divisible by 8. Since 16 is already divisible by 8, we need to append 0 or 8 on the right. For divisibility by 9, the sum of the digits must be div... | 920160120168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,550 |
7.2. A father had three sons, and he left them 9 ares of land as an inheritance - a rectangle measuring 25 m $\times 36$ m. The brothers decided to divide the land into three rectangular plots - three ares for each brother. How many options are there for the division (in terms of the length and width of the plots), and... | Answer: 4 options; the smallest length is 49 m in the option of dividing into a plot of $25 \times 12$ and two plots of $12.5 \times 24$. Solution. Let $ABCD$ be the original rectangle; $AB=25, BC=36$. Since it has 4 vertices and 3 plots, two vertices must belong to one plot. First, consider the case where the vertices... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,551 |
7.3. The numbers $a, b$, and $c$ satisfy the relation $\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}$. Find $a+b+c$, given that $b \neq c$. | Answer: 0. Solution. From the equality $\frac{a+b}{c}=\frac{a+c}{b}$, we get $a(b-c)=(c-b)(c+b)$. Dividing this equality by $b-c \neq 0$, we obtain $b+c=-a$. Therefore, $a+b+c=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,552 |
7.4. How many zeros does the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with, where $s(n)$ denotes the sum of the digits of the natural number $n$? | Answer: .19 zeros. Solution. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a digit sum of either 5, 10, or 15. There are 6 numbers with a digit sum of 5: these are $5, 14, 23, 32, 41, 50$. There are 9 numbers with a digit sum of 10: these are $19, 28, \... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,553 |
8.1. In triangle $ABC$, the bisector $BM$ is drawn. Prove that $AM < AB$ and $MC < BC$. | Solution. By the property of the exterior angle $\angle A M B=\angle M B C+\angle B C A>\angle M B C=\angle M B A$. Therefore, in triangle $A B M$, opposite the larger angle lies the longer side: $A B>A M$. Similarly, $M C<B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,554 |
8.2. A father had three sons, and he left them 9 ares of land as an inheritance - a rectangle measuring 25 m $\times 36$ m. The brothers decided to divide the land into three rectangular plots - three ares for each brother. How many options are there for the division (in terms of the length and width of the plots), and... | Answer: 4 options; the shortest length is 49 m in the option of dividing into a plot of $25 \times 12$ and two plots of $12.5 \times 24$. Solution. See problem 7.2 | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,555 |
2. Let $s(n)$ denote the sum of the digits (in decimal notation) of a natural number $n$. Find all natural $n$ for which $n+s(n)=2011$. | Answer: 1991. Hint. Since $n<2011$, then $s(n) \leq 2+9+9+9=29$. Therefore, $n=2011-$ $s(n) \geq 1982$. Since the numbers $n=2011$ and $n=2010$ obviously do not fit, the first three digits of the number $n$ can be one of three possibilities: 198, or 199, or 200. Let the last (fourth) digit of the number $n$ be $x$. The... | 1991 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,558 |
4. In triangle $A B C$, angle $A$ is equal to $60^{\circ}$, the distances from vertices $B$ and $C$ to the center of the inscribed circle of triangle $A B C$ are 3 and 4, respectively. Find the radius of the circle circumscribed around triangle $A B C$. | Answer. $\sqrt{\frac{37}{3}}$ Indication. Let $O$ be the center of the inscribed circle. Then $\angle O B C+\angle B C O$ $=\frac{1}{2}(\angle B+\angle C)=\frac{1}{2} \cdot 120^{\circ}=60^{\circ}$, and therefore $\angle B O C=180^{\circ}-60^{\circ}=120^{\circ}$. By the cosine theorem, we get $B C^{2}=B O^{2}+O C^{2}-2 ... | \sqrt{\frac{37}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,559 |
5. The unit square in the first quadrant of the coordinate plane ( $0 \leq x, y \leq 1$ ) is divided into squares with side length $2 \cdot 10^{-4}$. How many nodes of this partition (inside the unit square) lie on the parabola $y=x^{2}$? | Answer: 49. Note. The nodes of the partition have coordinates of the form ( $i / 5000, j / 5000$ ), where $i$, $j=1,2, \ldots, 4999$. The condition that a given node lies on the parabola is $\frac{j}{5000}=\left(\frac{i}{5000}\right)^{2}$, i.e., $i^{2}=j \cdot 5^{4} \cdot 2^{3}$. Therefore, the number $i$ must have the... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,560 |
9.2. The bisectors of the external angles $B$ and $C$ of triangle $ABC$ intersect at point $M$. a) Can angle $BMC$ be obtuse? b) Find angle $BAC$ if it is known that $\angle BMC = \frac{\angle BAM}{2}$. | Answer: a) cannot; b) $120^{\circ}$. Hint. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle B M C=180^{\circ}-\left(\frac{180^{\circ}-\beta}{2}+\frac{180^{\circ}-\gamma}{2}\right)=\frac{\beta+\gamma}{2}=90^{\circ}-\frac{\alpha}{2}$. a) Therefore, angle $B M C$ cannot be obtuse. b) from the equation ... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,561 |
9.4. For a quadrilateral $A B C D$ inscribed in a circle of radius $R$, the following holds: $A B^{2}+B C^{2}+C D^{2}+A D^{2}=8 R^{2}$. Can we assert that at least one of the diagonals of $A B C D$ is a diameter of the circle? | Answer: No. Hint. It is sufficient to consider a quadrilateral with mutually perpendicular diagonals, which are not diameters. This solution can be reached by introducing the angular magnitudes of the arcs $\alpha, \beta, \gamma, \delta$, into which the circle is divided, and then
$$
A B^{2}+B C^{2}+C D^{2}+A D^{2}=4 ... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 22,562 |
9.5. Let $s(n)$ denote the sum of the digits of a natural number $n$. How many zeros does the number equal to the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with? | Answer: 19 zeros. Hint. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a sum of digits of either 5, 10, or 15. There are 6 numbers with a sum of 5: these are $5,14,23,32,41,50$. There are 9 numbers with a sum of digits 10: these are $19,28, \ldots, 91$. ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,563 |
9.2. Petya says to Kolya: «If you think of a quadratic trinomial that has roots and tell me only the leading coefficient and the distance between the roots, then I will guess the y-coordinate of the vertex on its graph». Kolya believes that Petya is wrong: after all, to define a quadratic trinomial, you need to know th... | Answer: Pete. Solution. Let $a$ be the leading coefficient, $d=\left|x_{2}-x_{1}\right|$ be the distance between the roots. The result follows from the relations $a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)=$ $a\left(\left(x-\frac{x_{1}+x_{2}}{2}\right)^{2}+x_{1} x_{2}-\left(\frac{x_{1}+x_{2}}{2}\right)^{2}... | -(\frac{}{2})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,565 |
9.4. Given a scalene triangle, the lengths of whose sides form an arithmetic progression. Prove that this triangle has two angles less than $60^{\circ}$. | Solution. Let the lengths of the sides be $a-d, a, a+d$, where $d>0$. Clearly, opposite the smaller side $a-d$ lies the smaller angle in the triangle, i.e., this angle is less than $180^{\circ} / 3=60^{\circ}$. We will show that the angle $\alpha$ opposite the side $a$ is also less than $60^{\circ}$. From the cosine th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,567 |
9.5. There are $n$ weights, each weighing an integer number of grams, and their total weight is 100 grams. Is it true that the weights can always be divided into two groups to balance the scales if a) $n=50 ;$ b) $n=51$? | Answer: a) incorrect; b) correct. Solution. a) If we take one weight of 51 grams and 49 weights of 1 gram each, then it is impossible to divide them into two parts of equal weight. b) Let $a_{1}, a_{2}, \ldots, a_{51}$ be the weights of the weights. Consider a regular 100-gon $A_{1} A_{2} \ldots A_{100}$ and mark 51 ve... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,568 |
7.1. In a four-digit number, Petya erased the first digit and obtained a three-digit number. Then he divided the original number by this three-digit number and got a quotient of 3, with a remainder of 8. What is the original number? (Find all possible numbers). | Answer: 1496 or 2996. Solution. Let $x$ be the first digit, $y$ be the three-digit number obtained after crossing out the first digit. Then $1000 x+y=3 y+8$, i.e., $500 x=y+4$. From this, considering the inequalities $0<y<1000$, we get that $x$ is either 1 or 2. Then, respectively, $y=496$ or $y=996$. | 1496or2996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,569 |
7.2. In quadrilateral $A B C D$, where $A B=C D$, diagonal $A C$ is drawn. Prove that if angle $A C B$ is obtuse, then angle $A D C$ is acute. | Solution. Suppose, for the sake of contradiction, that angle $D$ is not acute. Then in $\triangle A C D$ we have $A C>C D$. But in $\triangle A B C$, opposite the obtuse angle $A C B$ lies the larger side: $A B>A C$. Thus, $C D<A C<A B$. We have reached a contradiction with the condition $C D=A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,570 |
7.3. Kolya has seven antique coins: four identical doubloons and three identical crowns. He has forgotten their exact weight, but remembers that a doubloon weighs either 5 or 6 grams, and a crown weighs either 7 or 8 grams. Can he determine the exact weight of the coins using two weighings on a balance scale without we... | Answer: He can. Solution. First weighing: put all four doubloons on the left pan and all three crowns on the right pan. Thus, the weight of the coins on the left pan is 20 or 24 grams, and on the right - 21 or 24 grams. If the scales are in balance, then the weight of the coins is unambiguously determined: the doubloon... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,571 |
7.4. A rectangular grid with a cell side of 1 cm and an area of $2021 \mathrm{~cm}^{2}$ is cut into four rectangular pieces by two perpendicular cuts along the grid lines. Prove that at least one of the pieces has an area of at least $528 \mathrm{~cm}^{2}$. | Solution. The prime divisors of the number 2021 are 43 and 47, and $2021=43 \cdot 47$. Therefore, the integer sides $a$ and $b$ of the original rectangle can be either 1) $a=2021, b=1$, or 2) $a=47, b=43$. However, it is obvious that case 1) $a=2021, b=1$ is impossible, since such a rectangle cannot be cut into square ... | 528 | Number Theory | proof | Yes | Yes | olympiads | false | 22,572 |
7.5. Along a circle, 25 numbers: $1,2, \ldots, 25$ were written in some order. Could it be that any two adjacent numbers differ either by 10 or by a multiple (an integer) of each other? | Answer: it could not. Solution. Suppose, to the contrary, that it is possible to arrange the numbers, and consider the three largest prime numbers less than 25, namely 17, 19, and 23. Let $n$ be any of these three numbers. Since $n+10>25$ and $2n>25$, the two numbers adjacent to $n$ on the circle can only be $n-10$ and... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,573 |
10.1. Solve the equation $\left|x^{2}-100\right|=10 x+90$. | Answer. $x_{1}=5+\sqrt{215}, x_{2}=-5+\sqrt{35}$. Solution. If $x^{2}-100 \geq 0$, i.e., under the condition $|x| \geq 10$, we have the equation $x^{2}-10 x-190=0, x=5 \pm \sqrt{215}$. The root $5+\sqrt{215}$ satisfies the condition $|x| \geq 10$, while the root $5-\sqrt{215}$ does not. If $|x|<10$, then we have the eq... | x_{1}=5+\sqrt{215},x_{2}=-5+\sqrt{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,574 |
10.2. Given a triangular pyramid $S A B C$ with mutually perpendicular lateral edges $S A, S B, S C$. Prove that $\triangle A B C$ is acute. | Solution. Let the lengths of the lateral edges be denoted by $a, b, c$. Then the squares of the sides of $\triangle A B C$ by the Pythagorean theorem are $a^{2}+b^{2}, b^{2}+c^{2}, a^{2}+c^{2}$. Therefore, the sum of the squares of any two sides of the base is greater than the square of the third side, which means that... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,575 |
10.1. Graph the solution set of the system of equations $\left\{\begin{array}{l}x-2 y=1 \\ x^{3}-8 y^{3}-6 x y=1\end{array}\right.$ on the coordinate plane. | Answer. The set of solutions of the system is the line $y=\frac{x-1}{2}$. Solution. The second equation is a consequence of the first, since
$$
\begin{aligned}
& x^{3}-8 y^{3}-6 x y=(x-2 y)\left(x^{2}+2 x y+4 y^{2}\right)-6 x y= \\
& =x^{2}+2 x y+4 y^{2}-6 x y=(x-2 y)^{2}=1
\end{aligned}
$$
Therefore, the system of t... | \frac{x-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,578 |
10.3. Does there exist an irrational number $x \in [0.3 ; 0.4]$, such that $x(x+1)(x+2)$ is an integer? | Answer. It exists. Let $P(x)=x(x+1)(x+2)-1$ and consider the equation $P(x)=0$. Since $P(0.3)=-0.103$, and $P(0.4)=0.344$ - values of different signs, the equation has a root on the interval $(0.3 ; 0.4)$. Denote this root by $x_{0}$ and show that $x_{0}$ is an irrational number. If $x_{0}=\frac{p}{q}$, where $p, q-$ a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,580 |
10.4. Prove that for any acute-angled triangle $A B C$, a triangular pyramid $S A B C$ with mutually perpendicular lateral edges $S A, S B, S C$ can be constructed. | Solution. Let $a=BC, b=AC, c=AB$. We will show that in a space with a rectangular coordinate system, we can mark points $A'(x, 0,0), B'(0, y, 0)$, and $C'(0,0, z)$ on the coordinate axes such that $A'B'=AB, B'C'=BC$, and $A'C'=AC$. Indeed, we have a system of three equations: $x^2 + y^2 = c^2$, $y^2 + z^2 = a^2$, and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,581 |
10.2. On a chessboard, 8 rooks were placed so that they do not attack each other. Prove that in any rectangular cell area of size $4 \times 5$ (cells) there is at least one rook. | Solution. Assume the opposite, and let's say, for definiteness, the rectangle is located in five horizontals and four verticals. Then any rook is among the other three horizontals or among the other four verticals (or simultaneously, i.e., at their intersection). But in three horizontals, there are no more than three r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,582 |
10.3. Find all quadratic trinomials $P(x)=x^{2}+b x+c$ such that $P(x)$ has integer roots, and the sum of its coefficients (i.e., $1+b+c$) is 10. | Answer. $(x-2)(x-11),(x-3)(x-6), x(x+9),(x+4)(x+1)$. Solution. See problem 9.3. | (x-2)(x-11),(x-3)(x-6),x(x+9),(x+4)(x+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,583 |
10.4. For a right triangle $ABC$, the length of the hypotenuse $AB$ and the leg $AC$ satisfy the inequalities $100<AB<101$ and $99<AC<100$. Prove that $\triangle ABC$ can be divided into several triangles, each of which has a side of length 1, and there exists a division in which there are no more than 21 such triangle... | Solution. By the Pythagorean theorem $CB=\sqrt{AB^{2}-AC^{2}}<\sqrt{101^{2}-99^{2}}=20$. From point $B$ along the leg $CB$, we sequentially lay out unit segments $1=BK_{1}=K_{1}K_{2}=\ldots=K_{n-1}K_{n}$ such that $CK_{n}<1$ (where $n$ is the integer part of the length of $CB$, i.e., $n<20$). If the length of $CB$ is a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,584 |
10.2. Prove that in any Pythagorean triangle there is a side whose length is divisible by 5 (a Pythagorean triangle is a right triangle with integer sides). | Solution. Note that the remainders of the squares of integers when divided by 5 can be either 0, 1, or 4. If we assume, for the sake of contradiction, that the lengths of the sides $a, b, c$ of a Pythagorean triangle are not divisible by 5, then from the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, we obtain a contradictio... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,586 |
10.3. 14 tennis players participated in a round-robin tournament (each player played one match against each other). Prove that there exist three players such that each of the remaining 11 players lost at least one match to one of these three. (There are no ties in tennis). | Solution. First, we will show that there is a player who has won at least seven matches. Indeed, otherwise, the total number of wins by all players would be no more than $14 \cdot 6 = 84$. However, the total number of wins is equal to the number of all matches played, which is $(14 \cdot 13) / 2 = 91$ - a contradiction... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,587 |
10.2. In triangle $A B C$, angle $B$ is equal to $60^{\circ}$. Points $M$ and $N$ are marked on sides $A B$ and $B C$ respectively. It turns out that $A M=M N=N C$. Prove that the intersection point of segments $C M$ and $A N$ coincides with the center of the circle circumscribed around $\triangle A B C$. | Solution. Let $O$ be the intersection point of $C M$ and $A N$. Denote $\angle M A O=\alpha, \angle N C O=\beta$. Then, in the isosceles triangles $A M N$ and $M N C$, we have: $\angle M N O=\alpha, \angle N M O=\beta$. By the property of the exterior angle $\angle M O A$ in triangles $M N O$ and $A O C$, we get: $\ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,590 |
10.3. The sum $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{45}$ is represented as a fraction with the denominator $45!=1 \cdot 2 \cdots 45$. How many zeros (in decimal notation) does the numerator of this fraction end with? | Answer: 8 zeros. Solution. The numerator of the fraction is the sum of numbers of the form $1 \cdot 2 \cdot \ldots \cdot(k-1)(k+1) \cdot \ldots .45$ (the product lacks one of the natural numbers from 1 to 45). Let's denote such a term as $c_{k}$. Note that $45!=5^{10} \cdot 2^{n} \cdot \boldsymbol{p}$, where $\boldsymb... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,591 |
10.1. Given the equation $x^{3}+5 y=y^{3}+5 x$. Do there exist solutions to this equation for a) natural numbers $x \neq y$? ; b) positive real numbers $x \neq y$? | Answer: a) do not exist; b) exist. Solution. a) See item a) of problem 9.1. b) Let's take, for example, $y=2 x$, and substitute it into the last equation: $x^{2}+2 x^{2}+4 x^{2}=5$. Then $x=\sqrt{\frac{5}{7}}$ and $y=2 \sqrt{\frac{5}{7}}$ satisfy the equation. | \sqrt{\frac{5}{7}},2\sqrt{\frac{5}{7}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,593 |
10.3. Given a quadrilateral $A B C D$, in which a circle can be inscribed. Prove that the two circles inscribed in triangles $A B C$ and $A D C$ touch the diagonal $A C$ at the same point. | Solution. Let $A B=a, B C=b, C D=c, A D=d, A C=e$, and let the circle inscribed in triangle $A B C$ touch $A C$ at point $M$. Then, from the equality of tangent segments, we get that $A M=\frac{a+e-b}{2}$. Similarly, for the circle inscribed in triangle $A C D$ and touching $A C$ at point $N$, we get: $A N=\frac{d+e-c}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,595 |
10.4. a) Prove the inequality $\sqrt{n+1}+2 \sqrt{n}<\sqrt{9 n+3}$ for all natural numbers $n$; b) does there exist a natural number $n$ such that $[\sqrt{n+1}+2 \sqrt{n}]<[\sqrt{9 n+3}]$? ([a] — the integer part of the number $a$). | Answer: b) does not exist. Solution.. a) After squaring both sides and separating the square root, we get the equivalent true inequality $2 \sqrt{n(n+1)}<2 n+1 \Leftrightarrow 4 n(n+1)<4 n^{2}+4 n+1 \Leftrightarrow 0<1$. b) Suppose, for the sake of contradiction, that such an $n$ exists. Then for some natural number $k... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,596 |
9.1. Which of the numbers is greater: $\sqrt{100^{2}-99}+\sqrt{99^{2}-100}$ or $\sqrt{100^{2}-100}+\sqrt{99^{2}-99}$ ? | Answer: the second number is greater. Solution. Let $a=100, b=99, A=\sqrt{a^{2}-b}+\sqrt{b^{2}-a}$, $B=\sqrt{a^{2}-a}+\sqrt{b^{2}-b}$. We have equivalent inequalities (after squaring and eliminating like terms) $A>0$ (since $\left.a \neq b\right)$. | the\\\is\greater | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,597 |
9.4. Given a pentagon with sides (in some order) $1 ; 2 ; 5 ; 6 ; 7$. Prove that a circle cannot be inscribed in this pentagon. | Solution. Note the following fact: if a circle can be inscribed in a pentagon, then the sum of any two non-adjacent sides of the pentagon is less than the sum of the three remaining sides. This fact is proven analogously to the known property of a quadrilateral (one needs to add the corresponding segments into which th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,600 |
8.1. Two rectangles with horizontal and vertical sides are given. The horizontal side of the second rectangle is $11 \%$ longer than the horizontal side of the first, and the vertical side of the second rectangle is $10 \%$ shorter than the vertical side of the first. a) Which rectangle has a larger area and by what pe... | Answer: a) the area of the second rectangle is less than the area of the first by $0.1\%$, b) the ratio of the horizontal side to the vertical side is 290/109 ≈ 22.66. Solution. Let $a$ and $b$ be the horizontal and vertical sides of the first rectangle, respectively, then the sides of the second rectangle are $1.11a$ ... | )0.1,b)\frac{290}{109}\approx2.66 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,602 |
8.2. a) Agent 007 wants to encrypt his number using two natural numbers $m$ and $n$ such that $0.07=\frac{1}{m}+\frac{1}{n}$. Can he do this? b) Can his colleague, Agent 013, encrypt his number in a similar way? | Answer: a) will be able to; b) will be able to. The solution follows from the equalities: $0.07=0.05+0.02=\frac{1}{20}+\frac{1}{50}$ and $0.13=\frac{25}{200}+\frac{1}{200}=\frac{1}{8}+\frac{1}{200}$ | )\frac{1}{20}+\frac{1}{50};b)\frac{1}{8}+\frac{1}{200} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,603 |
8.3. From points $A$ and $B$, two cyclists set off towards each other at the same time. They traveled at constant speeds. After meeting, the first cyclist took 40 minutes to reach point $B$, and the second cyclist took one and a half hours to reach point $A$. Find the time from the start of the journey until they met a... | Answer. Time until the meeting - 1 hour. The speed of the first cyclist is 1.5 times greater than the speed of the second. Solution. Let $v_{1}, v_{2}$ be the speeds of the cyclists, $t$ - the time until the meeting. Then the first cyclist traveled the distance $\mathrm{v}_{1} t$ before the meeting, and the second - th... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,604 |
8.4. In triangle $A B C$, angle $A$ is three times smaller than angle $C$, and side $B C$ is half the length of side $A B$. Find the angles of triangle $A B C$. | Answer. Angles $A, B$ and $C$ are equal to $30^{\circ}, 60^{\circ}, 90^{\circ}$ respectively. Solution. Let $B C=a, \angle A=\alpha$, then $A B=2 a$ and $\angle C=3 \alpha$. Take a point $M$ on side $A B$ such that $\angle A C M=\alpha$, then $\angle B M C=2 \alpha$ by the property of the exterior angle. Thus, $\angle ... | 30,60,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,605 |
8.5. Can a checkerboard square of size $n \times n$ (cells) with a corner cell cut out be cut into dominoes (rectangles $2 \times 1$) such that the number of horizontal and vertical dominoes is the same, if a) $n=101$; b) $n=99$? | Answer. a) It is possible; b) it is not possible. Solution. a) Let the left lower cell be cut off. Fill the lower horizontal (without the cut-off cell) with 50 horizontal dominoes, and the left vertical - with 50 vertical dominoes. There will remain a square $100 \times 100$, which can be divided into an even number of... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,606 |
11.2. Find all parameters $b$, for which the system of equations $\left\{\begin{array}{l}x^{2}-2 x+y^{2}=0 \\ a x+y=a b\end{array}\right.$ has a solution for any $a$. | Answer: $b \in[0 ; 2]$. Solution. The first equation of the system represents a circle $(x-1)^{2}+y^{2}=1$, its center is at the point $(1 ; 0)$, and the radius is 1. The second equation is the equation of a line with a slope of (-a). Note that this line passes through the point $M$ with coordinates $(b ; 0)$. Therefor... | b\in[0;2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,608 |
11.4. a) Investigate the function $y=\frac{\sqrt{x^{2}+1}+x-1}{\sqrt{x^{2}+1}+x+1}$ for evenness (oddness). b) Find the domain and range of this function. | Answer: a) the function is odd; b) domain of definition ( $-\infty, \infty$ ), range of values $(-1 ; 1)$. Solution. See problem 10.4. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,610 |
11.5. The sequence $a_{n}$ is defined as follows: $a_{n+2}=3 a_{n+1}-2 a_{n}, a_{1}=1$, $a_{2}=\frac{10}{9}$. Prove that $a_{n}$ takes integer values for an infinite set of indices $n$. | Solution. Rewrite the expression for $a_{n+2}$ as $a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_{n}\right)$. Therefore, for $n \geq 1$, we will have $a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_{n}\right)=4\left(a_{n}-a_{n-1}\right)=\ldots=2^{n}\left(a_{2}-a_{1}\right)=\frac{1}{9} \cdot 2^{n}$. Then $a_{n+2}=a_{1}+\left(a_{2}-a_{1}\right)+\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,611 |
8.1. Prove that for any natural $n$ the number $n^{3}+9 n^{2}+27 n+35$ is composite. | Solution. Let's represent the expression as $(n+3)^{3}+2^{3}=(n+5)\left((n+3)^{2}-2(n+3)+4\right)=$ $(n+5)\left(n^{2}+4 n+7\right)$, and each factor is obviously greater than one. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,612 |
8.2. Does the equation $x^{9}=2013 y^{10}$ have a solution in natural numbers $x, y$? | Answer: Yes. Solution. We will look for a solution in the form of powers of the number 2013, i.e., $x=2013^{m}, y=2013^{n}$. Then $2013^{9 m}=2013^{1+10 n}$, i.e., $9 m-1=10 n$. We can take $m=9$ (the smallest natural $m$ for which the last digit of the number $9 m$ is 1), then $n=8$. | 2013^{9},2013^{8} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,613 |
8.3. Inside the given triangle $A B C$, construct (using a compass and straightedge) a point $M$ such that the areas of triangles $A B M$, $B C M$, and $C A M$ are equal. | Solution. Let $B K$ be the median and $O$ the point of intersection of the medians in triangle $A B C$. Then, by the property of medians, $O K=\frac{1}{3} B K$. Therefore, the height in triangle $A O C$, dropped to $A C$, is three times smaller than the height of triangle $A B C$ (this follows from the similarity of th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,614 |
8.4. A row contains 100 numbers. Is it necessarily true that the sum of all these numbers is positive if it is known that a) the sum of any seven numbers is positive? b) the sum of any seven consecutive numbers is positive? | Answer: a) necessarily; b) no. See problem 7.4. | )necessarily;b)no | Other | math-word-problem | Yes | Yes | olympiads | false | 22,615 |
10.1. Solve the equation $|x+1|=x^{2}+3 x+1$. | Answer: $x_{1}=0, x_{2}=-2-\sqrt{2}$. Solution. If $x \geq-1$, then we have $x^{2}+2 x=0 \Rightarrow x=0$ (the root $x=-2$ does not satisfy the condition $x \geq-1$ ). If $x<-1$, then we have $x^{2}+4 x+2=0 \Rightarrow x=-2-\sqrt{2}$ (the root $-2+\sqrt{2}$ does not satisfy the condition $x<-1$ ). | x_{1}=0,x_{2}=-2-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,617 |
10.3. Given a parallelogram $A B C D$. Can we assert that $A B C D$ is a rectangle if it is known that a) the radii of the inscribed circles of triangles $A B C$ and $A B D$ are equal? b) the radii of the circumscribed circles of triangles $A B C$ and $A B D$ are equal | Answer: a) yes; b) yes. Solution. a) Let $r_{1}, r_{2}$ be the radii of the inscribed circles of triangles $A B C$ and $A B D$ respectively, $p_{1}, p_{2}$ be their semiperimeters. From the formula $p r=S$ it follows that $p_{1} r_{1}=p_{2} r_{2}$, since the areas of triangles $A B C$ and $A B D$ are equal (they are eq... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,619 |
10.4. Find the smallest natural number $n$ for which a) $n!$ is divisible by 2016; b) $n!$ is divisible by $2016^{10}$. (Recall that $\left.n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n.\right)$ | Answer: a) $n=8 ;$ b) $n=63$. Solution. See problem 9.4 . | )8;b)63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,620 |
7.1. There are 19 kg of cereal. Can 1 kg be measured with three weighings on a balance scale if there is one 3 kg weight? | Answer: It is possible. Solution. The first weighing can yield 8 kg of cereal. If a weight is placed on one (left) pan of the scales and, by pouring cereal from the right pan to the left, the scales are balanced. Indeed, from the equation $3+x=19-x$ we get $x=8$, i.e., there will be 8 kg of cereal on the pan with the w... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,621 |
7.2. A rectangle was cut into four rectangles with two perpendicular cuts. It is known that the perimeter of three of them is expressed as an integer. Is it necessarily true that the perimeter of the fourth rectangle will also be an integer? | Answer: necessarily. Solution. Let $A B C D$ be the original rectangle and $P$ be its perimeter. Among the four small rectangles, there are two rectangles containing opposite vertices of the original rectangle. For definiteness, let these be vertices $A$ and $C$, and let the perimeters of the corresponding rectangles b... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,622 |
7.3. In the test in $7 \mathrm{a}$, there were three more girls than boys. The test results (on a five-point scale) showed that there were 6 more fours than fives, and threes were twice as many as fours. Prove that someone received a two or a one. | Solution. Let $n-$ be the number of fives, then the number of fours was $n+6$, and the number of threes $2(n+6)$. The total number of positive grades is $n+n+6+2(n+6)=4n+18$, i.e., an even number. On the other hand, the number of students is odd (it equals $2m+3$, where $m$ is the number of boys. Since the number of st... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 22,623 |
7.4. In a $4 \times 4$ square (cells), crosses were placed in eight cells. Is it necessarily true that in some row or some column there will be exactly two crosses? | Answer: not necessarily. Solution. See example.
 | notnecessarily | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,624 |
7.5. On a chessboard, the centers of some cells are marked in such a way that no triangle with marked vertices is a right triangle. What is the maximum number of points that could have been marked? | Answer: 14 points. Solution. Let's call a marked point vertical if there are no other marked points on its vertical line. Similarly, we define a horizontal point. Note that any marked point is either vertical or horizontal (or both at the same time). Indeed, if the marked point $A$ had another point $B$ on its vertical... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,625 |
8.3. Given a triangle $A B C$, where $\angle B=45^{\circ}, \angle C=89^{\circ}$. Point $M$ is the midpoint of $A B$. Prove that $M B<M C<C B$. | Solution. We will consider $AB$ as the base of the triangle. Since $\angle A = 180^{\circ} - 89^{\circ} - 45^{\circ} = 46^{\circ} > \angle B = 45^{\circ}$, then $BC > AC$. Therefore, the vertex $C$ projects onto the base $AB$ at a point $N$ between $A$ and $M$ (since the larger side has the larger projection). Then the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,628 |
8.4. a) Prove that the number $\frac{2015^{2}+2017^{2}}{2}$ can be represented as the sum of the squares of two natural numbers. b) Prove a more general fact: the half-sum of the squares of two different odd numbers can be represented as the sum of the squares of two natural numbers. | Solution. a) Let $2 k-1=2015$. Then we have $\frac{(2 k-1)^{2}+(2 k+1)^{2}}{2}=4 k^{2}+1$. b) We have $\frac{(2 k+1)^{2}+(2 n+1)^{2}}{2}=2 k^{2}+2 n^{2}+2 k+2 n+1=(k+n+1)^{2}+(k-n)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,629 |
# Task 2. (10 points)
In a right triangle $A B C$ (angle $C$ is right), medians $A M$ and $B N$ are drawn, the lengths of which are 19 and 22, respectively. Find the length of the hypotenuse of this triangle. | # Solution.
Let $AC = 2x$ and $BC = 2y$. By the Pythagorean theorem, we have
$$
\left\{\begin{array}{c}
(2 x)^{2}+y^{2}=19^{2} \\
x^{2}+(2 y)^{2}=22^{2}
\end{array} \Rightarrow 5 x^{2}+5 y^... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,631 |
# Task 3. (12 points)
Accountants, managers, and economists of the bank are sitting at a round table. When the director asked the accountants to raise their hands if an economist was sitting next to them, 20 people raised their hands. And when the director asked the managers to raise their hands if an economist was si... | # Solution
Let's call a group of economists several (possibly one) economists sitting next to each other, with representatives of other professions sitting to their left and right. In this case, if there is no manager or accountant next to whom two economists are sitting, then each person raised their hand no more tha... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,632 |
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