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8.3. On the coordinate plane, four lines are constructed, the equations of which have the form $y=k x+b$. All coefficients and free terms are different natural numbers from 1 to 8. Can these 4 lines divide the plane into exactly 8 parts? | Answer: they can.
Solution. Consider, for example, the lines given by the equations: $y=8 x+1, y=7 x+2, y=6 x+3$, and $y=5 x+4$. Each of them passes through the point ( $1 ; 9$ ), so they divide the plane into exactly 8 parts.
There are other examples where all lines pass through the specified point. This condition i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,709 |
8.4. In trapezoid $A B C D$, point $M$ is the midpoint of the lateral side $C D$. Rays $B D$ and $B M$ divide angle $A B C$ into three equal parts. Diagonal $A C$ is the bisector of angle $B A D$. Find the angles of the trapezoid. | Answer: $\angle A=72^{\circ}, \angle B=108^{\circ}, \angle C=54^{\circ}, \angle D=126^{\circ}$.
Solution. Let $\angle A B D=\alpha$, then $\angle A B C=3 \alpha, \angle B A D=180^{\circ}-$ $3 \alpha, \angle B D A=\angle D B C=2 \alpha$ (see Fig. 8.4).
In triangle $B C D$, segment $B M$ is both a bisector and a median... | \angleA=72,\angleB=108,\angleC=54,\angleD=126 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,710 |
8.5. On an island, there live 33 knights, as well as liars and fantasists. Each resident of this island was asked in turn: “How many of you are knights?”. Ten different answers were received, each of which was given by more than one resident. Knights always tell the truth, liars always give a wrong number that has not ... | Answer: Yes.
Solution. Let's call a number $m$ "initial" if it was named, but the previous number $m-1$ was not named. According to the problem, the fantasizers did not name initial numbers. If an initial number was named by a liar, then neither another liar nor a knight could name it again. But each number was named ... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,711 |
8.6. Points $M$ and $N$ are the midpoints of sides $B C$ and $A D$ of quadrilateral $A B C D$. It is known that $\angle B=$ $150^{\circ}, \angle C=90^{\circ}$ and $A B=C D$. Find the angle between the lines $M N$ and $B C$. | Answer: $60^{\circ}$.
Solution. First method. Construct parallelogram $A B M K$ and rectangle $C D L M$ (see Fig. 8.6a). Since $A K\|B C\| L D$ and $A K=B M=$ $M C=L D$, then $A K D L$ is also a parallelogram. Therefore, the midpoint $N$ of its diagonal $A D$ is also the midpoint of diagonal $K L$.
. | Answer. 01.
Solution. The combination 01 occurs 16 times in the tetrad. After it, $7+8=15$ times there is a 0 or 1, and one time there is not. Therefore, one of the combinations 01 stands at the end of the line.
Criteria. Full solution - 7 points. Partial examples of sequences with the correct answer - 1. Only answer... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,713 |
2. On a white sheet of paper, there are the numbers: 1 and 2. The following action is allowed: you can increase one of the numbers on the sheet by the sum of the digits of the other. Can the numbers 1 and 2 be transformed into:
a) 2021 and $2021?$
b) 2022 and $2022?$ | Answer. a) yes; b) no.
Solution. a) We will turn the number 2 into the number 2021 by adding 1 (we will add one to the two 2020 times). Note that the sum of the digits of the number 2021 is $2+0+2+1=5$, therefore, by adding 5 to one 404 times, we will get two numbers 2021 and 2021.
b) The sum of the digits of the num... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,714 |
3. There are 21 different applications installed on the phone. In how many ways can six applications be selected for deletion so that among them are three applications from the following six $T V F T' V' F'$, but none of the pairs $T T', V V', F F'$ are included? | Answer. $8 \cdot C_{15}^{3}=3640$.
Solution. Choose one application from each pair $T T^{\prime}, V V^{\prime}, F F^{\prime}-8$ ways. Thus, out of 21 applications, 6 have already been excluded, leaving 15. Choose 3 applications from the remaining 15, which can be done in $\frac{15 \cdot 14 \cdot 13}{3!}$ (or $C_{15}^{... | 3640 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,715 |
4. Around a circle, $m$ numbers are written in such a way that any two adjacent numbers differ by 1. We call a number strong if both of its neighbors are less than it, and weak if both of its neighbors are greater than it. Let the sum of all strong numbers be $S$, and the sum of all weak numbers be $s$. Prove that $m=2... | Solution. Write down $m$ pairs of adjacent numbers and in each pair subtract the smaller number from the larger one. Add these differences. On the one hand, this sum is equal to $m$, since each difference is 1. On the other hand, each strong number will enter into two differences with "+" signs, each weak number both t... | 2(S-) | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 17,716 |
5. Katya wrote down the set $X=\{1,2,3,4,5,6,7,8,9\}$ in her notebook and decided to split it into two subsets. Prove that no matter how Katya splits the set into two subsets, at least one of the resulting subsets will contain three numbers such that the sum of two of them is twice the third. | Solution. The solution will be based on 5, since the pairs of numbers $(1,9),(2,8),(3,7),(4,6)$ sum up to twice the value of five. Next, we will consider various options, but to simplify this, note the following property: consecutive numbers with the same step immediately satisfy the condition of the problem (from the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,717 |
1. Petya ran down the escalator, counting the steps. Exactly halfway down, he stumbled and the rest of the way he tumbled down (Petya flies three times faster than he runs). How many steps are there on the escalator if Petya counted 20 steps with his feet (i.e., before falling) and 30 steps with his sides (after fallin... | Answer: 80
Solution. Let the escalator have a length of $2 \mathrm{~L}$ (steps), the speed of the escalator be $u$, and Petya runs with a speed of $\mathrm{x}$, and flies with a speed of $3 \mathrm{x}$. Then the time until the fall is $\frac{L}{u+x}$ and during this time Petya will count $\frac{L x}{u+x}$ steps. From ... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,718 |
2. Can the number 2020 be represented as the sum of the squares of six odd numbers? | Answer: No.
Solution. The square of an odd number $2 n+1$ is $4 n^{2}+4 n+1$. The number $n(n+1)$ is even, so the square of an odd number gives a remainder of 1 when divided by 8. Therefore, the sum of 6 squares of odd numbers, when divided by 8, has a remainder of 6. But for 2020, the remainder is 4, which is a contr... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,719 |
3. Solve the equation $\left(101 x^{2}-18 x+1\right)^{2}-121 x^{2} \cdot\left(101 x^{2}-18 x+1\right)+2020 x^{4}=0$ Answer: $\frac{1}{9}, \frac{1}{18}$ | Solution. Let $y=\left(101 x^{2}-18 x+1\right), z=\frac{y}{x^{2}}$. After dividing the equation by (non-zero number: $\mathrm{x}=0$ is not suitable) $x^{4}$, we get $z^{2}-121 z+2020=0$. The roots of this equation can be easily found using Vieta's theorem: $\mathrm{z}=101$ or $\mathrm{z}=20$. In the first case, we get ... | \frac{1}{9},\frac{1}{18} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,720 |
4. In triangle $ABC$, the angle bisectors $\mathrm{AK}$ and $\mathrm{CL}$ and the median $\mathrm{BM}$ are drawn. It turns out that $\mathrm{ML}$ is the bisector of angle AMB, and $\mathrm{MK}$ is the bisector of angle CMB. Find the angles of triangle ABC. | Answer: $30^{\circ}, 30^{\circ}, 120^{\circ}$.
Solution. By the property of the angle bisector, AM:MB=AL:LB=AC:CB. But AC=2AM, so $C B=2 M B$. Similarly, $A B=2 M B$, and thus, triangle $A B C$ is isosceles. But then $B M$ is the altitude, triangle $B M C$ is right-angled, and its leg BM is half of its hypotenuse BC. ... | 30,30,120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,721 |
5. We will call a right-angled triangle elegant if one of its legs is 10 times longer than the other. Is it possible to cut a square into 2020 identical elegant triangles? | Answer: Yes
Solution. An elegant triangle with legs of 10 and 100 can be cut into 100 triangles with legs of 1 and 10 (the cuts are straight lines parallel to its sides). From four such large triangles, we can form a square (its sides will be the hypotenuses of these triangles) with a hole (in the shape of a square wi... | 2020 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,722 |
# 2. CONDITION
The sum of several numbers is equal to 1. Can the sum of their squares be less than $0.01 ?$ | Solution. If, for example, each of the $n$ numbers is equal to $1 / n$, then the sum of these numbers is 1, and the sum of their squares is $1 / n$ and for $n>100$ the sum of the squares will be less than $0.01$.
Answer: it can. | itcan | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 17,724 |
# 5. CONDITION
A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road pass on flat ground, if the tourist's speed is 4 km/h when climbing uphi... | Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4+(9-x) / 6$ hours on this part. Since the tourist walks $2 x / 5$ hours on flat g... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,725 |
# 6. CONDITION
From point $B$ of a square billiard, we launch a ball parallel to the diagonal. Find the set of points on the billiard such that if a second ball is launched simultaneously with the first, at the same speed and in the same direction, they will collide. | Answer. The desired set of points is the combination of three segments, two of which pass through point $B$ and are parallel to the sides of the square, and the third segment is parallel to the chosen diagonal and passes through the point symmetric to point $B$ with respect to this diagonal. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,726 |
1. Buratino left Papa Carlo's house and arrived at the Field of Wonders exactly at 22:00. If his walking speed had been 25% faster, he would have arrived at 21:30. At what time did he leave the house? | Answer: at 19:30.
Solution: If Buratino spent $t$ (hours) on his journey, then with the increased speed, he would have spent 1.25 times less, i.e., $\frac{4}{5} t$. Therefore, he would have saved $\frac{1}{5} t$, which amounted to 30 minutes. Thus, he spent 2.5 hours on the way to the Field of Wonders, and he left hom... | 19:30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,727 |
2. Compare two numbers: $\sqrt{3} \sin 10^{\circ}$ and $\sin 80^{\circ}$. | Answer: the second is greater
Solution: Consider the difference between the first and second numbers:
$$
\begin{aligned}
& \sqrt{3} \sin 10^{\circ}-\sin 80^{\circ}=\sqrt{3} \sin 10^{\circ}-\cos 10^{\circ}=2\left(\frac{\sqrt{3}}{2} \sin 10^{\circ}-\frac{1}{2} \cos 10^{\circ}\right)= \\
& =2\left(\cos 30^{\circ} \sin 1... | theisgreater | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,728 |
3. Do there exist positive numbers $a, b, c$ such that the numbers $d$ and $\sqrt{d}$ are respectively roots of the equations $a x^{2}+b x-c=0$ and $\sqrt{a} x^{2}+\sqrt{b} x-\sqrt{c}=0 ?$ | Answer: No.
Solution: Let such numbers exist, substitute the values of $d$ and $\sqrt{d}$ for $x$ in the equations. Then we have $c=a d^{2}+b d$ and $\sqrt{c}=\sqrt{a} d+\sqrt{b} \sqrt{d}$. In the last equation, both sides are positive, square both sides: $c=(\sqrt{a} d+\sqrt{b} \sqrt{d})^{2}=a d^{2}+2 d \sqrt{a b d}+... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,729 |
4. A three-digit number, all digits of which are different and non-zero, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Find the smallest balanced number. | Answer: 132.
Solution: Let the desired number be of the form $\overline{a b c}$. Then it can be represented as the sum of six different two-digit numbers: $\overline{a b c}=\overline{a b}+\overline{b a}+\overline{a c}+\overline{c a}+\overline{b c}+\overline{c b}$.
From the last equality, we get $100 a+10 b+c=22 a+22 ... | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,730 |
5. Three circles with radii 1, 2, 3 touch each other externally at three points. Find the radius of the circle passing through these three points. | Answer: 1.
Solution: Let $\mathrm{O}_{1}, \mathrm{O}_{2}$ and $\mathrm{O}_{3}$ be the centers of the given circles, K, M, N the points of tangency, such that $\mathrm{O}_{1} \mathrm{~K}=\mathrm{O}_{1} \mathrm{~N}=1, \mathrm{O}_{2} \mathrm{~K}=\mathrm{O}_{2} \mathrm{M}=2$ and $\mathrm{O}_{3} \mathrm{~N}=$ $\mathrm{O}_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,731 |
6. Do there exist 2022 consecutive natural numbers among which there are exactly 22 prime numbers? | Answer: They exist.
Solution: Let $P_{n}$ denote the number of prime numbers among 2022 consecutive numbers from $n$ to $n+2021$. Note that $P_{n+1}$ differs from $P_{n}$ by at most one. It is easy to verify that $P_{1}>22$ (there are already 25 prime numbers in the first hundred: $2,3,5,7, \ldots, 97$). On the other ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,732 |
1. You have a bag of granulated sugar, a balance scale, a 1 kg weight, and paper bags in which you can package the sugar. You need to measure out 50 kg of sugar, using no more than 6 weighings. How can you do this? | Solution. Let $\Gamma$ denote a 1 kg weight. The right and left sides of the equalities correspond to the right and left pans of the balance.
1 step: 1 kg (sugar $)=\Gamma$;
2 step: 1 kg $+\Gamma=2$ kg;
3 step: 1 kg $+2 \kappa 2+\Gamma=4$ kg;
4 step: 1 kg $+2 \kappa 2+4 \kappa z+\Gamma=8$ kg;
5 step: 1 kg $+2 \kap... | 32+16+2=50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,733 |
3. The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect at point $O$. It is known that the area of triangle $AOB$ is equal to the area of triangle $COD$, and the area of triangle $AOD$ is equal to the area of triangle $BOC$. Prove that $ABCD$ is a parallelogram. | Solution. From the conditions of the problem, it follows that the areas of triangles $ABC$ and $ADC$ are equal. Therefore, the heights of these triangles, dropped to side $AC$, are equal. This means that the areas and heights of triangles $AOB$ and $COD$, dropped to sides $AO$ and $CO$ respectively, are equal. From thi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,735 |
4. Prove that there exist two infinite sequences of integers $\left\{m_{k}\right\}$ and $\left\{n_{k}\right\}$, such that all numbers in each sequence are distinct, and both sequences $\left\{\sqrt{n_{k}+m_{k}^{2}}\right\}$ and $\left\{\sqrt{n_{k}-m_{k}^{2}}\right\}$ also consist of integers. | Solution. Let $n=5 k^{2}, m=2 k$, where $k=0,1,2, \ldots$
Then $\sqrt{n_{k}+m_{k}^{2}}=\sqrt{5 k^{2}+4 k^{2}}=3 k, \sqrt{n_{k}-m_{k}^{2}}=\sqrt{5 k^{2}-4 k^{2}}=k$, as required.
Remark. There are many other solutions.
Evaluation criteria. Constructing any infinite sequence that satisfies the conditions of the proble... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,736 |
5. All integers from 1 to 100 are written on the board. Mitya and Dima take turns erasing one number at a time until only two numbers remain. If their sum is divisible by 7, Mitya wins; if not, Dima wins. Mitya goes first. Who wins if the players do not make mistakes? | Answer: Dima wins.
Solution. Dima can ensure his victory by adhering to the following strategy. First, he mentally divides all the numbers into pairs: 1 and 100, 2 and 99, 3 and 98, ..., 50 and 51. Next, no matter which number Mitya erases, Dima erases the second number from the same pair. In the end, one of the pairs... | Dimawins | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,737 |
11.1 In a competition of meaningless activity, a participant recorded 2022 numbers in a circle such that each number is equal to the product of its two neighbors. What is the maximum number of different numbers that could have been used? | Solution: If there is a zero among the numbers, then its neighbors are also zeros, so all the recorded numbers will be equal to zero. We will assume that there are no zeros among the numbers. Let $a$ and $b$ be two adjacent numbers. Then on the other side of $a$ stands $a / b$, and on the other side of $b$ stands $b / ... | 337 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,738 |
11.2 It is known that for given numbers a and c, the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}+a x+b \\
x=y^{2}+a y+b
\end{array}\right.
$$
has more than one solution. Prove that $a^{2}>2(a+2 b)-1$. | Solution: The graphs of the equations $y=x^{2}+a x+b$ and $x=y^{2}+a y+b$ are symmetric with respect to the line $x=y$. If the parabola $y=x^{2}+a x+b$ does not intersect the axis of symmetry, then the two graphs lie on opposite sides of it and do not intersect, i.e., the system has no solutions. If the parabola $y=x^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,739 |
11.3 Another participant in the senseless activity competition selected 765 different natural numbers from the sequence $1,2,3, \ldots$, 2023. He claims that the sum of no two of the chosen numbers is divisible by 8. Is the respected participant mistaken? | Solution: Suppose the participant is correct. From the list specified in the condition, there are exactly 253 remainders of each type from 1 to 7 when divided by 8, and 252 numbers with a remainder of 0. Numbers with remainders 0 and 4 can be taken no more than one each. It is impossible to take numbers with remainders... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,740 |
11.4 Another participant in the meaningless activity contest brought with him $N$ unit squares, from which he immediately, in front of the amazed jury, formed a rectangle with sides differing by 9. Not stopping there, the participant then formed a large square from these same $N$ squares, but this time 6 squares were l... | Solution 1: Let $x$ be the smaller side of the rectangle, and $y$ be the side of the square. Then $N=x(x+9)=y^{2}+6$. Multiplying both sides of the equation by 4, we get $2 x(2 x+18)=$ $(x+9)^{2}-81=4 y^{2}+24,(2 x-2 y+9)(2 x+2 y+9)=105$. In this product, the factor $2 x+2 y+9$ is positive and no less than 13, and the ... | 10,22,70,682 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,741 |
11.5 The tangents to the circumscribed circle of the right triangle $A B C$ ( $\angle C=$ $90^{\circ}$ ) at points $A$ and $C$ intersect at point $T$. The rays $A B$ and $T C$ intersect at point $S$. It is known that the areas of triangles $\triangle A C T$ and $\triangle B C S$ are equal. Find the ratio of the areas o... | Solution: Let point $M$ be the midpoint of the hypotenuse $AB$, and point $N$ be the midpoint of the leg $AC$. Note that points $M, N$, and $T$ lie on the same line - the perpendicular bisector of segment $AC$. Let the area of triangle $TNC$ be $S_{1}$, and the area of triangle $AMN$ be $S_{2}$. Then $S(\triangle ANT)=... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,742 |
11.5. The angles of a triangle $\alpha, \beta, \gamma$ satisfy the inequalities $\sin \alpha > \cos \beta, \sin \beta > \cos \gamma, \sin \gamma > \cos \alpha$. Prove that the triangle is acute-angled. (I. Bogdanov) | Solution. Suppose the opposite; let $\gamma \geqslant 90^{\circ}$ for definiteness. Then $\alpha+\beta \leqslant 90^{\circ}$, and the angles $\alpha$ and $\beta$ are acute. Therefore, $0<\beta \leqslant 90^{\circ}-\alpha<90^{\circ}$, from which $\cos \beta \geqslant \cos \left(90^{\circ}-\alpha\right)=\sin \alpha$, whi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,743 |
11.6. At the base of the quadrilateral pyramid $S A B C D$ lies a parallelo-
## XXXVI All-Russian Mathematical Olympiad for Schoolchildren
gram $A B C D$. Prove that for any point $O$ inside the pyramid, the sum of the volumes of tetrahedra $O S A B$ and $O S C D$ is equal to the sum of the volumes of tetrahedra $O S... | Solution. Let $X$ be the point of intersection of ray $S O$ with plane
Fig. 3
 | Answer: It cannot.
Solution. First solution. Substituting $x=1234$ into both quadratic polynomials and equating them, we get $1234^{2} \cdot b+1234 \cdot c+a=$ $=1234^{2} \cdot c+1234 \cdot a+b$, or, after moving all terms to the left side, $\left(1234^{2}-1\right) b+\left(1234-1234^{2}\right) c+(1-1234) a=0$. Dividin... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,745 |
11.8. In the cells of a $100 \times 100$ square, the numbers $1,2, \ldots, 10000$ were placed, each exactly once; at the same time, numbers differing by 1 were placed in cells adjacent by side. After this, the distances between the centers of each pair of cells, the numbers in which differ by exactly 5000, were calcula... | Answer: $50 \sqrt{2}$.
(I. Bogdanov)
Solution. We will number the rows (from bottom to top) and columns (from left to right) of the square with numbers from 1 to 100; we will denote a cell by a pair of the numbers of its row and column. We will call the distance between cells the distance between their centers. Cells... | 50\sqrt{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,746 |
7.1. Each student in class 7B ate the same number of chocolate bars during math lessons over the week. Nine of them together ate fewer than 288 chocolate bars in a week, while ten of them together ate more than 300 chocolate bars. How many chocolate bars did each student in class 7B eat? Explain your answer. | Answer: 31 chocolates each
Solution 1: Since ten students ate more than 300 chocolates, nine students ate more than (310:10)$\cdot$9 = 270 chocolates. It is also known that nine students together ate less than 288 chocolates. The only number divisible by 9 in the range from 271 to 287 is 279. Therefore, nine students ... | 31 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 17,747 |
7.2. Find the largest natural number in which each digit, starting from the third, is equal to the sum of all the previous digits of the number. | Answer: 101248.
Solution: Let the first digit of the number be $a$, the second digit be $b$. Then the third digit is $(a+b)$, the fourth digit is $(2a+2b)$, the fifth digit is $(4a+4b)$, and the sixth digit is $(8a+8b)$. There cannot be a seventh digit, because if it exists, it would be equal to $(16a+16b)$, but this ... | 101248 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,748 |
7.3. Given an equilateral triangle $\mathrm{ABC}$. On the sides $\mathrm{AB}$ and $\mathrm{BC}$, isosceles right triangles ABP and BCQ are constructed externally with right angles $\angle \mathrm{ABP}$ and $\angle \mathrm{BCQ}$. Find the angle $\angle \mathrm{PAQ}$. | Answer: 90.
Solution: Consider triangle ACQ. It is isosceles because $\mathrm{AC}=\mathrm{BC}$ (by the condition of the equilateral triangle $\mathrm{ABC}$), and $\mathrm{BC}=\mathrm{CQ}$ (by the condition of the isosceles triangle BCQ). The angle at the vertex $\angle A C Q=60 \circ+90 \circ=150$ . Then $\angle \math... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,749 |
7.4. For the municipal stage of the All-Russian School Students' Olympiad, 12 classrooms of school $\mathrm{N}$ were used (all classrooms are on the same floor and are in a row).
It turned out that the number of students in adjacent classrooms differed by one. Could 245 students have come to participate in the Olympia... | Answer: It could not.
Solution: The number of students in two adjacent classrooms differs by 1, so the total number of students in these two classrooms is odd. There are six such pairs of classrooms. Therefore, the total number of students who participated in the Olympiad is the sum of six odd numbers, which is an eve... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,750 |
7.5. What is the minimum number of cells that need to be marked on a 5 by 5 board so that among the marked cells there are no adjacent ones (having a common side or a common vertex), and adding any one cell to these would violate the first condition? | Answer: 4 cells.
Solution: Estimation. Divide the board into four parts (see fig.). In each of them, a cell must be marked, otherwise the black cell contained in it can be added.
Example. The four black cells in the figure satisfy both conditions.
Criteria: Answer only - 0 points. Estimation only - 5 points. Example... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,751 |
5. If the scales showed the exact weight, the sum of the results of the first three weighings would give twice the total weight of the three portfolios. However, the scales are inaccurate, and the total error over the three weighings is no more than 1.5 kg. Therefore, the three portfolios together weigh no less than $\... | Answer: Colin - 3.5 kg, Petya - 3 kg, Vasya - 2 kg. | Colin-3.5\,Petya-3\,Vasya-2\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,752 |
1. Arrange the natural numbers from 1 to 101 in a row so that the difference between any two adjacent numbers is equal to 2 or 5. | Answer: $101,99,97, \ldots, 9,4,2,7,5,3,1,6,8,10, \ldots, 100$.
Remark. Other examples are possible.
Criteria. If the solution is incorrect - 0 points.
If any correct example is provided - 7 points.
If the reasoning is correct but there is a computational error - 3 points. If the solution is correct - 7 points. | 101,99,97,\ldots,9,4,2,7,5,3,1,6,8,10,\ldots,100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,753 |
2. How many five-digit numbers exist that are not divisible by 1000, and have the first, third, and last digits even?
Otvet: 9960. | Solution. The first digit of the number can be any of the four (2, 4, 6, or 8), the second and fourth can be any of ten each, and the third and fifth, if we abandon the condition "not divisible by a thousand," can be any of five (0, 2, 4, 6, or 8). Therefore, there are $4 \times 10 \times 5 \times 10 \times 5=10000$ fi... | 9960 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,754 |
3. Solve the equation in integers: $x^{2}+y^{2}=x+y+2$.
Answer: ( $-1 ; 1),(-1 ; 0),(0 ;-1),(0 ; 2),(1 ;-1),(1 ; 2),(2 ; 1),(2 ; 0$). | Solution. The equation is equivalent to the following: $x^{2}-x=-y^{2}+y+2$. The function $f(x) = x^{2}-x=(x-0.5)^{2}-0.25$ takes values in the interval $[-0.25 ;+\infty)$, $g(y)=-y^{2}+y+2=-(y-0.5)^{2}+2.25-$ on the interval $[2.25 ;-\infty)$, that is, they have only three common integer values: $0, 1, 2$. Since $x$ a... | (-1,1),(-1,0),(0,-1),(0,2),(1,-1),(1,2),(2,1),(2,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,755 |
4. In the inscribed quadrilateral $ABCD$, the following equalities are satisfied: $AB=AC$ and $BC=CD$. We marked point $P$ - the midpoint of the arc $CD$ of the circumscribed circle, not containing point $A$, and point $Q$ - the intersection point of the diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are p... | Solution. Let the measures of angles $\angle B A C$ and $\angle D A C$ be $\alpha$, and the measure of angle $\angle A B D$ be $\beta$. Then the angles $\angle D B C$ and $\angle C D B$ are also equal to $\alpha$, and the angle $\angle A C D$ is equal to $\beta$, as inscribed angles subtending the same arcs. Due to the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,756 |
5. In a white $10 \times 10$ square, on the first move, a $1 \times 1$ cell rectangle is painted, on the second move - a $1 \times 2$ cell rectangle, on the third - $1 \times 3$ and so on, as long as it is possible to do so. After what minimum number of moves could this process end? (Cells cannot be painted over again.... | # Answer: after 6 moves.
Solution. Evaluation. On the board, 16 rectangles of size $1 \times 6$ can be highlighted, of which a maximum of $1+2+3+4+5=15$ will contain colored cells, meaning that a sixth move is always possible. Example. The process is illustrated in the diagram, showing a scenario where a seventh move ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,757 |
2. Solve the equation in natural numbers $n!+3 n+8=k^{2} \quad(n!=1 \times 2 \times 3 \times$ $\ldots \times n)$. | # Solution
For any natural number $n \geq 3 \quad n!+3n+8$ gives a remainder of 2 when divided by 3. A perfect square does not give a remainder of 2 when divided by 3.
Indeed,
$(3m)^{2}=9m^{2}$ - is divisible by 3,
$(3m \pm 1)^{2}=9m^{2} \pm 6m+1$ - gives a remainder of 1 when divided by 3.
Thus, the equation $n!+... | k=4,n=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,759 |
4. Is the number $1+2^{5^{2017}}$ prime? | # Solution
$5^{2017}$ is divisible by 5, so we can write this number as $5 n$, where $n \in N$
We get the expression $1+2^{5 n}$
$1+2^{5 n}=\left(2^{n}+1\right)\left(2^{4 n}-2^{3 n}+2^{2 n}-2^{n}+1\right)$
Since $2^{4 n}-2^{3 n}>1,2^{2 n}-2^{n}>1$, therefore, $2^{4 n}-2^{3 n}+2^{2 n}-2^{n}+1>3$, and $\left(2^{n}+1\... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,761 |
5. Can a $50 \times 50$ square be cut into strips of $1 \times 4$? | Solution
Let's introduce a coloring (Fig. 3).
Fig. 3
Each strip of size $1 \times 4$ covers exactly 2 white and 2 black cells. Therefore, if the board can be cut, the number of white and b... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,762 |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,763 |
# Task No. 6.3
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
 balls. Each is painted in some color. If you take out any three balls from the box, there will definitely be at least one red and at least one blue among them. How many balls can be in the box | Answer: 4. Solution. In the box, there are no more than two red and blue balls (otherwise, it would be possible to draw three red or three blue balls) and no more than one ball of other colors (otherwise, it would be possible to draw one blue or red ball and two balls of other colors). Therefore, there are no more than... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,765 |
9.3. On 5 cards, natural numbers from 1 to 10 are written (each exactly once, one on each side). It is known that on each card, one of the numbers divides the other. The cards are laid out on the table so that the numbers on the top sides are visible, while the numbers on the bottom sides are not. Can the numbers on th... | Answer: Yes, it is possible. If the card has a seven, then on the reverse side there must be a one, because there are no other numbers among the numbers from 1 to 10 that divide 7 or are divisible by 7. On the reverse side of the card with a five, there can only be 1 or 10, but one is already taken, so it must be ten. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,766 |
9.5. A chip is placed in the central cell of a $7 \times 7$ square. Two players take turns moving the chip to an adjacent cell by side. The first player can move the chip in the same direction as the second player's previous move or turn left, while the second player can move the chip in the same direction as the first... | Answer: No, it cannot. It is clear that only the player who places the chip in a corner on their turn can win. Let's color the cells of the board in a checkerboard pattern. With each move, the color of the cell where the chip is placed changes. Since all the corner cells of the $7 \times 7$ board are the same color as ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,767 |
2. According to the condition, we write the equation where $x$ is the given number: $x^{2}+15=y^{2}$, $y$ is an integer. We transform the equation and solve it in integers. $\mathrm{y}^{2}-\mathrm{x}^{2}=15$, $(\mathrm{y}-\mathrm{x})(\mathrm{y}+\mathrm{x})=15$. From this, we obtain eight systems of equations:
1), 2) $... | Answer: there are such numbers: $7, -7, 1, -1$. | 7,-7,1,-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,769 |
11.1. Given polynomials $P(x)$ and $Q(x)$ of degree ten, with leading coefficients equal to 1. It is known that the equation $P(x)=Q(x)$ has no real roots. Prove that the equation $P(x+1)=Q(x-1)$ has at least one real root. (I. Bogdanov) | Solution. Let $P(x)=x^{10}+p_{9} x^{9}+\ldots+p_{0}$ and $Q(x)=x^{10}+$ $+q_{9} x^{9}+\ldots+q_{0}$. Then the polynomial $P(x)-Q(x)=\left(p_{9}-q_{9}\right) x^{9}+\ldots+$ $+\left(p_{0}-q_{0}\right)$ has no real roots; but, if $p_{9} \neq q_{9}$, then the degree of this polynomial is odd, and it has a root. Therefore, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,771 |
11.2. The inscribed and exscribed spheres of the triangular pyramid $A B C D$ touch its face $B C D$ at different points $X$ and $Y$. Prove that triangle $A X Y$ is obtuse. (The exscribed sphere of the pyramid touches one of its faces and also the planes of the other faces outside these faces.)
(V. Shmarov) | The first solution. Let $X$ be the point of tangency of the plane $(BCD)$ with the inscribed sphere. Let the homothety with center at point $A$, which maps the exsphere to the inscribed sphere, map point $Y$ to some point $Z$ on the inscribed sphere. This homothety maps the plane $(BCD)$ to a plane parallel to $(BCD)$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,772 |
11.3. Find all natural $k$ such that the product of the first $k$ odd prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first).
(V. Senderov) | Answer. Such $k$ does not exist.
Solution. Let $n \geqslant 2$, and $3=p_{1}p_{k}$, otherwise the left side of the equality (*) would be divisible by $q$, which is impossible. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, then we can replace $n$ with $t$, and ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,773 |
11.4. On each of 2013 cards, a number is written, and all these 2013 numbers are distinct. The cards are face down. In one move, it is allowed to point to ten cards, and in response, one of the numbers written on them will be reported (it is unknown which one). For what largest $t$ can it be guaranteed to find $t$ card... | Answer. $t=1986=2013-27$.
Solution. 1. First, we will show that it is impossible to guess 1987 cards. Number the cards $A_{1}, \ldots, A_{2013}$; we will demonstrate how to arrange the answers so that none of the numbers on the cards $A_{1}, \ldots, A_{27}$ can be determined.
For each $i=1, \ldots, 9$, combine the ca... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,774 |
11.1. Three natural numbers are such that the last digit of the sum of any two of them is the last digit of the third number. The product of these three numbers was written on the board, and then everything except the last three digits of this product was erased. What three digits could have remained on the board? Find... | Answer: 000, 250, 500, or 750.
Solution: Let \(a, b, c\) be the given numbers. According to the problem, the numbers \(a+b-c\), \(b+c-a\), and \(c+a-b\) are divisible by 10. Therefore, their sum, which is \(a+b+c\), is also divisible by 10. On the other hand, from the equality \(a+b+c = (a+b-c) + 2c\) and the problem'... | 000,250,500,750 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,775 |
11.2. Let $P(x)$ and $Q(x)$ be reduced quadratic trinomials, each having two distinct roots. It turns out that the sum of two numbers obtained by substituting the roots of the trinomial $P(x)$ into the trinomial $Q(x)$ is equal to the sum of two numbers obtained by substituting the roots of the trinomial $Q(x)$ into th... | Solution. Let $a_{1}$ and $a_{2}$ be the roots of the quadratic polynomial $P(x)$, and $b_{1}$ and $b_{2}$ be the roots of the quadratic polynomial $Q(x)$; then $P(x)=\left(x-a_{1}\right)\left(x-a_{2}\right)$ and $Q(x)=$ $=\left(x-b_{1}\right)\left(x-b_{2}\right)$. Therefore, the condition of the problem takes the form... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,776 |
11.3. Can the set of all natural numbers be partitioned into non-intersecting finite subsets $A_{1}, A_{2}, A_{3}, \ldots$ such that for any natural number $k$, the sum of all numbers in the subset $A_{k}$ equals $k+2013$? (R. Zhenodarov) | Answer: No.
First solution. Suppose the desired partition exists. Let's call a set $A_{k}$ large if it contains more than one element. We will prove by induction on $n$ that there are at least $n$ large sets. For $n=1$, consider the set $A_{k_{1}}$ containing the number 1; the sum of the numbers in it is $k_{1}+2013 >... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,777 |
11.4. In the circle $\Omega$, an acute-angled triangle $ABC$ is inscribed, where $AB > BC$. Let $P$ and $Q$ be the midpoints of the smaller and larger arcs $AC$ of the circle $\Omega$, respectively. Let $M$ be the foot of the perpendicular dropped from point $Q$ to the segment $AB$. Prove that the circle circumscribed ... | First solution. Let $S$ be the midpoint of $BP$, and $O$ be the center of the circle $\Omega$. Then $O$ is the midpoint of the segment $PQ$, and $S$ is the projection of $O$ onto $BP$. Note that $QA = QC$, since $Q$ is the midpoint of the arc $AC$. The isosceles triangles $AQC$ and $POC$ are similar because $\angle QAC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,778 |
# 1. Option 1.
A confectionery factory received 5 rolls of ribbon, each 50 m long, for packaging cakes. How many cuts need to be made to get pieces of ribbon 2 m long? | Answer: 120.
Solution. From one roll, 25 pieces of ribbon, each 2 m long, can be obtained. For this, 24 cuts are needed. Therefore, a total of $5 \cdot 24=120$ cuts are required. | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,779 |
# 2. Option 1.
Vasya took six cards with the numbers $513, 23, 5, 4, 46, 7$ written on them. He wants to arrange them in a row so that the resulting ten-digit number is the smallest possible. Write this number. | Answer: 2344651357.
Solution. If you place the card "23" at the beginning, the ten-digit number will start with the digit 2, and if you place another card, it will start with a larger digit. Therefore, you should start with the card "23". Next, the following digit should be as small as possible, so the second card sho... | 2344651357 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,780 |
# 3. Option 1.
In a box, there are chips. Tolya and Kolya were asked how many chips are in the box. Tolya answered: “Less than 7”, and Kolya answered: “Less than 5”. How many chips can be in the box if it is known that one of the answers is correct? Find all the options. In the answer, write their sum. | Answer: 11.
Solution: If there are 7 or more chips in the box, then both boys are lying. If there are 4 or fewer chips in the box, then both boys are telling the truth. If there are 5 or 6 chips in the box, then Tolya is telling the truth, and Kolya is lying. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,781 |
# 4. Option 1.
In a box, there are red, blue, and green pencils. It is known that if you remove two red pencils, the number of pencils of all colors will be equal. And if after that you add 10 red pencils to the box, the number of red pencils will be exactly half of all the pencils. How many pencils were in the box in... | Answer: 32.
Solution. Before adding pencils, it could be considered that there were as many red ones as green ones. Then, the 10 added pencils are the number of blue ones. So, at the moment of adding, there were 10 red, blue, and green pencils each.
# | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,782 |
# 5. Option 1.
The chocolate bar has the shape of a square with a side of 100 cm, divided into pieces with a side of 1 cm. The sweet tooth ate all the pieces along each of the four sides, as shown in the figure. How many pieces did the sweet tooth eat in total?
. The chocolate bar consists of $100 \times 100=10000$ pieces in total. If 2 rows of pieces are removed from each side, a square with a side length of 96 cm remains. Thus, the sweet tooth will eat $10000-96 \cdot 96=784$ pieces.
. Find the area of this flower bed (in square meters).
 | Answer: 10.
Solution. The area of the flower bed can be calculated as the difference between the area of the plot and the sum of the areas of the two triangular parts not occupied by the flower bed:
 as the number itself). The sum of 99 identical remainders is divisible by 3, which means the sum of any 99 natural numbers with the s... | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,788 |
3. The book is sewn from 12 identical booklets. Each booklet consists of several double sheets nested within each other. The booklets of the book are sewn sequentially one after another. All pages of the book are numbered, starting from 1. The sum of the numbers of four pages of one of the double sheets in the fourth b... | Solution. Let the book have $x$ pages, then each notebook has $\frac{x}{12}$ pages, the first three notebooks have $\frac{3 x}{12}=\frac{x}{4}$ pages, and the first four notebooks have $\frac{4 x}{12}=\frac{x}{3}$ pages. The numbers of the first and second pages of the fourth notebook will be $\frac{x}{4}+1$ and $\frac... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,789 |
4. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are marked such that $AM = BN$ and quadrilateral $AMNC$ is cyclic. Let $BL$ be the angle bisector of triangle $ABC$. Prove that lines $ML$ and $BC$ are parallel.
$ that satisfy the equation
$$
x^{2}+x y+y^{2}=x+20
$$ | Solution. Consider this equation as a quadratic in terms of the variable $x$: $x^{2}+(y-1) x+\left(y^{2}-20\right)=0$. Its discriminant $D=(y-1)^{2}-4\left(y^{2}-20\right)=81-2 y-3 y^{2}$ must be non-negative, that is, $3 y^{2}+2 y-81 \leqslant 0$. This inequality holds for $y \in\left[\frac{-1-2 \sqrt{61}}{3} ; \frac{... | (1;-5),(5;-5),(-4;0),(5;0),(-4;4),(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,791 |
# 7.1. Condition:
A carpenter took a wooden square and cut out 4 smaller equal squares from it, the area of each of which was $9 \%$ of the area of the larger one. The remaining area of the original square was $256 \mathrm{~cm}^{2}$.
 and their opposites (these are negative).
Remark. There are other examples. 1004 ones, 2 threes, and 1010 negative ones. Another example can be obtained by replacing all numbers with their opposites.
## Grading criteria.
Any correct example:... | possible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,798 |
2. Find all primes $p$ such that $a^{3} b-a b^{3}$ is divisible by $p$ for any integers $a$ and $b$. | Answer: 2 and 3.
Solution. Let \( a = 2 \) and \( b = 1 \), then \( a^3 b - a b^3 = 8 - 2 = 6 = 2 \cdot 3 \).
Therefore, there can be no other prime divisors besides 2 and 3.
We will prove that 2 and 3 are suitable.
Factorize this expression:
\( a^3 b - a b^3 = a b (a^2 - b^2) = a b (a - b)(a + b) \).
We will prov... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,799 |
3. In triangle $\mathrm{ABC}$, $\angle \mathrm{A}=60^{\circ}$. Points $\mathrm{M}, \mathrm{N}$, and $\mathrm{K}$ are taken on sides $\mathrm{BC}, \mathrm{AC}$, and $\mathrm{AB}$, respectively. It turns out that $\mathrm{BK}=\mathrm{KM}=\mathrm{MN}=\mathrm{NC}$ and $\mathrm{AN=2AK}$. Prove that a) $\mathrm{KN} \perp \ma... | Solution. a) In triangle $A K N$, draw the median KO to side AN. We get $\mathrm{AK}=\mathrm{AO}=\mathrm{ON}$. Since $\angle \mathrm{A}=60^{\circ}$, triangle ACO is equilateral, and $\angle \mathrm{AKO}=\angle \mathrm{AOK}=60^{\circ}$. Triangle $\mathrm{KON}$ is isosceles (KO=ON) and, therefore, $\angle \mathrm{OKN}=\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,800 |
4. Numbers $a$ and $b$. Prove that $a^{2}+b^{2}-a-b-ab+0.25 \geq 0$. For which $a$ and $b$ is equality achieved? | Answer: equality is achieved when ( $a=0$ and $b=0.5$ ) or $(a=0.5$ and $b=0)$.
Solution. Let's complete the square on the left side: $a^{2}+b^{2}-a-b-ab+0.25=(a+b-0.5)^{2}-3ab=(a+b-0.5)^{2}+(-3ab) \geq 0$
: four $3 \times 4$ and a $1 \times 1$ square. If only three cells are shaded, there will be a white rectangle of 12 cells. How to shade 4 cells is shown in the following figure:
+(y+t)}{2}+\frac{4}{(x+z)(y+t)} \geqslant 3
$$
and let $x+z=a, y+t=b$. We need to prove that $\frac{a+b}{2}+\frac{4}{a b} \geqslant 3$. By the AM-GM inequality, $\frac{a+b}{2}+\frac{4}{a b} \geqslant \sqrt{a b}+\frac{4}{a b}$, so it suffices to prove that $\sqrt{a b}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,808 |
3. Quadrilateral $ABCD$ is inscribed in a circle, and the tangents at points $B$ and $D$ intersect at point $K$, which lies on the line $AC$. A line parallel to $KB$ intersects the lines $BA$, $BD$, and $BC$ at points $P$, $Q$, and $R$ respectively. Prove that $PQ = QR$. | Solution. Since $\triangle K A B$ is similar to $\triangle K B C$, we have $\frac{A B}{B C}=\frac{K B}{K C}$. Similarly, $\frac{A D}{D C}=$ $\frac{K D}{K C}$. Considering that $K B=K D$, we get $\frac{A B}{B C}=\frac{A D}{D C}$ or $\frac{A D}{A B}=\frac{D C}{B C}$. Next,
$$
\frac{P Q}{B Q}=\frac{\sin P B Q}{\sin B P Q... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,809 |
4. In a set of natural numbers not exceeding 30, 22 numbers are marked. Prove that there is a marked number that is equal to the sum of some three other marked numbers. | Solution. Let $a$ be the smallest marked number, $b$ the second largest marked number, and $x_{1}, x_{2}, \ldots, x_{20}$ the other marked numbers. Let's assume for definiteness that $a < b < x_{1} < x_{2} < \ldots < x_{20}$. There are no marked numbers less than $a$. Consider sums of the form $a+b+x_{i}$ (20 sums). Si... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,810 |
5. For a table boxing tournament in Yoshkar-Ola, 52 fighters have gathered. It is known that all have different levels of strength and in the game, the stronger one always defeats the weaker one, with one exception: the weakest fighter is an inconvenient opponent for the strongest and always defeats him. The real stren... | Solution. Answer: they can.
Let the strongest fighter be $S$, and the weakest $W$. Divide the fighters into 13 quartets. In each quartet, conduct the following matches: by pairing the quartet, determine the winners in the pairs (let $a>b$ and $c>d$), then determine the winner in the pair of winners (let $a>c$). Fighte... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,811 |
1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2015$ and $a \neq c$ (the numbers $a, b, c, d$ are not given). | 1. Answer: 1007.5.
First solution. Expanding the brackets, we get $a b-(a+b) x=c d-(c+d) x$. Substituting $d=2015-a$ and $b=2015-c$, we obtain
$a(2015-c)-(a+2015-c) x=c(2015-a)-(c+2015-a) x . \quad \begin{aligned} & \text { combining } \text { and } \\ & \text { similar terms, }\end{aligned}$
we get $(a-c) 2015=2(a-... | 1007.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,812 |
3. What is the minimum number of 3-cell corners that need to be painted in a $5 \times 5$ square so that no more corners can be painted? (Painted corners should not overlap.) | # 3. Answer. 4.
Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked on the diagram. Since none of these corners can be painted, at least one cell in each of them must be painted. Note that one corner cannot paint cells of two marked corners... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,814 |
4. We will call a number greater than 25 semi-prime if it is the sum of some two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? | 4. Answer. 5.
Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers \(2n+1, 2n+3\), and \(2n+5\), greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers \(2n-1, 2n+1\), and \(2n+3\) are prim... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,815 |
10.1. One degree on the Celsius scale is equal to 1.8 degrees on the Fahrenheit scale, while $0^{\circ}$ Celsius corresponds to $32^{\circ}$ Fahrenheit. Can a temperature be expressed by the same number of degrees on both the Celsius and Fahrenheit scales? | Answer: Yes, it can.
From the condition, it follows that the temperature in Fahrenheit is expressed through the temperature in Celsius as follows: $T_{F}=1.8 T_{C}+32^{\circ}$. If $T_{F}=T_{C}$, then $0.8 T_{C}+32=0$, that is, $T_{C}=-40$.
Remarks. An answer without explanations - 0 points; an answer with explanation... | -40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,817 |
10.2. Each face of a rectangular parallelepiped $3 \times 4 \times 5$ is divided into unit squares. Is it possible to inscribe a number in each square so that the sum of the numbers in each cellular ring of width 1, encircling the parallelepiped, is equal to 120? | Answer: Yes, it is possible.
Example. In all the squares (two) of the $3 \times 4$ faces, we write the number 5, in all the squares (two) of the $3 \times 5$ faces, we write the number 8, and in all the squares (two) of the $4 \times 5$ faces, we write the number 9.
Verification: The sum of the numbers in each cell r... | Yes,itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,818 |
8.1. The monthly pension of football fan Ivan Ivanovich is
$$
\frac{149^{6}-199^{3}}{149^{4}+199^{2}+199 \cdot 149^{2}}
$$
rubles, and the cost of a ticket to a World Cup match is 22000 rubles. Will Ivan Ivanovich's pension for one month be enough to buy one ticket? Justify your answer. | Solution: Let $149^{2}=a, 199=b$. Then Ivan Ivanovich's monthly pension (in rubles) is
$$
\frac{a^{3}-b^{3}}{a^{2}+b^{2}+a \cdot b}=\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{a^{2}+a b+b^{2}}=a-b=149^{2}-199=22002
$$
Thus, his pension will be enough for one ticket.
Answer: It will be enough.
Recommendations for check... | 22002 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,819 |
8.2. One of the angles of a triangle is $120^{\circ}$ more than another. Prove that the bisector drawn from the vertex of the third angle is twice the height drawn from this angle. | Solution: Let in triangle $ABC$ angle $C$ be greater than angle $A$ by $120^{\circ}$, and $BD$ and $BE$ be the altitude and the angle bisector, respectively (see the figure). Let angle $A$ be denoted by $\alpha$. Then $\angle C = 120^{\circ} + \alpha$. Since the sum of the interior angles of a triangle is $180^{\circ}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,820 |
8.3. Let \(a\) and \(b\) be different non-zero numbers, and \(a^{2} + \frac{1}{b} = b^{2} + \frac{1}{a}\). Prove that at least one of the numbers \(a\) or \(b\) is negative. | Solution: We will carry out equivalent transformations (for $a \neq b$).
$$
\begin{gathered}
a^{2}+\frac{1}{b}=b^{2}+\frac{1}{a}, \quad a^{2}-b^{2}=\frac{1}{a}-\frac{1}{b} \\
(a-b)(a+b)=\frac{b-a}{a b}, \quad a+b=-\frac{1}{a b}
\end{gathered}
$$
If both numbers $a$ and $b$ are positive, then the number on the left si... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,821 |
8.4. Can a square be cut into several convex pentagons (not necessarily equal)? Justify your answer. | Solution: One example of the cutting is shown in the figure.
To the solution of problem 8.4
Answer: Yes.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,822 |
8.5. The first two digits of a natural four-digit number are either each less than 5, or each greater than 5. The same can be said about the last two digits. How many such numbers are there? Justify your answer. | Solution: There are $4^{2}=16$ two-digit numbers where both digits are greater than 5, and there are $4 \cdot 5=20$ (the first digit is not zero) where both digits are less than 5. In total, there are $16+20=36$ such numbers. This is the conclusion about the first two digits of the number. The last two digits will prov... | 1476 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,823 |
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