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Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,174 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,176 |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,177 |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,180 |
Problem 9.7. On the coordinate plane, there is a point $A$. On the $O x$ axis, there is a point $B$, and on the $O y$ axis, there is a point $C$.
It is known that the equations of the lines $A B, B C, A C$ in some order have the form $y=a x+4$, $y=2 x+b$ and $y=\frac{a}{2} x+8$ for some real numbers $a$ and $b$.
Find... | Answer: 13 or 20.
Solution. Let the graph of the function $y=a x+4$ be denoted by $l_{1}$, the function $y=2 x+b-$ by $l_{2}$, and the function $y=\frac{a}{2} x+8-$ by $l_{3}$.
From the condition, it follows that $l_{1}, l_{2}, l_{3}$ are pairwise intersecting lines (each of the points $A, B, C$ belongs to some two o... | 13or20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,182 |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,184 |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,185 |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
=1$, that is, $A+B=90^{\circ}$.
Second method. Adding the original equalities, we get $\sin \left(A+45^{\circ}\right)+\sin \left(B+45^{\circ}\right)=2$, from which $\sin \left(A+45^{\circ}\right)=\si... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,188 |
11.3. Solve the equation $20[x]-14\{x\}=2014$ ([x] - the integer part of the number $x$, i.e., the greatest integer not exceeding $x,\{x\}$ - the fractional part of the number $x$ : $\{x\}=x-[x]$). | Answer: $x=101 \frac{3}{7}$.
From the inequality $0 \leq\{x\}<1$ it follows that $0 \leq 14\{x\}<14,0 \leq 20[x]-2014<14$. Adding 2014 and dividing by 20, we get $\frac{2014}{20} \leq[x]<\frac{2028}{20}$ or $100.7 \leq[x]<101.4$. Thus, $[x]=101,\{x\}=\frac{20 \cdot 101-2014}{14}=\frac{3}{7}$ and $x=101+\frac{3}{7}$.
... | 101\frac{3}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,189 |
1. In each cell of a $3 \times 3$ table, an irrational number was written. Could it happen that the sum of the numbers in each row and each column is a rational number? | 1. Answer. Yes. Solution. For example, the following arrangement works
| $1-\sqrt{2}$ | $2 \sqrt{2}$ | $2-\sqrt{2}$ |
| :--- | :--- | :--- |
| $3-\sqrt{2}$ | $-\sqrt{2}$ | $2 \sqrt{2}$ |
| $2 \sqrt{2}$ | $-\sqrt{2}$ | $-\sqrt{2}$ |
Grading criteria. Any correct arrangement - 7 points, in all other cases $-\mathbf{0}$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,194 |
2. Two parks with a total area of 110 hectares are divided into the same number of plots, and in each park, the plots have the same area, but differ from those in the other park. If the first park were divided into plots of the same area as the second, it would result in 75 plots. If the second park were divided into p... | 2. Answer. 50 and 60 hectares. Solution. Let x be the area of the first park. According to the condition, we will construct a table (for the new division):
| Park | In the new division | | |
| :---: | :---: | :---: | :---: |
| | Area of the park | Number of plots | Area of the plot |
| First | x | 75 | x/75 |
| Sec... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,195 |
3. At the base of the triangular pyramid $\mathrm{SABC}$ lies an acute isosceles triangle $\mathrm{ABC}$ ( $\mathrm{AC}=\mathrm{BC}$ ). $\mathrm{SH}$ is the height of the pyramid, and $\mathrm{H}$ is the point of intersection of the altitudes of triangle $\mathrm{ABC}$. Prove that the heights of the pyramid, drawn from... | 3. Solution. Let AM and $\mathrm{BN}$ be the altitudes of triangle $\mathrm{ABC}$. From the theorem of three perpendiculars, it follows that BC is perpendicular to SM, and thus to the plane ASM. Similarly, BN is perpendicular to SN, and thus to the plane BSN. This means that the height of the pyramid $\mathrm{AH}_{1}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,196 |
4. Given the functions $f(x)=x^{2}+4 x+3$ and $g(x)=x^{2}+2 x-1$. Find all integer solutions to the equation $f(g(f(x)))=g(f(g(x)))$. | 4. Answer. $x=-2$. Solution. Let's represent the functions as $f(x)=x^{2}+4 x+3=(x+2)^{2}-1$ and $g(x)=x^{2}+2 x-1=(x+1)^{2}-2 . \quad$ Then $\quad f(g(x))=\left((x+1)^{2}-2+2\right)^{2}-1=(x+1)^{4}-1$, $g(f(x))=\left((x+2)^{2}-1+1\right)^{2}-2=(x+2)^{4}-2$. Performing similar operations, we get $g(f(g(x)))=(x+1)^{8}-2... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,197 |
5. In triangle ABC, angle C is equal to $60^{\circ}$, points I and O are the centers of the inscribed and circumscribed circles of triangle ABC, respectively. Prove that the segment connecting point I and the midpoint of arc $\mathrm{AB}$ is equal to the radius of this circumscribed circle? | 5. Solution. Let $M$ be the midpoint of arc $AB$.
The central angle $\angle A O B=2 \angle A C B=120^{\circ}$ and the angle bisector of $\angle C$ passes through $I$ and $M$. Note that triangles $AOM$ and $BOM$ are equilateral. Express the angle $\angle A I B=180^{\circ}-(\angle B A C+\angle A B C) / 2=$ $=180^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,198 |
6. Robot Petya displays three three-digit numbers on the screen every minute, which sum up to 2019. Robot Vasya swaps the first and last digits in each of these numbers and then adds the resulting numbers. What is the largest sum Vasya can obtain? | 6. Answer. 2118. Solution. Let the original numbers be $\overline{a_{1} b_{1} c_{1}}, \overline{a_{2} b_{2} c_{2}}, \overline{a_{3} b_{3} c_{3}}$, then $100\left(a_{1}+a_{2}+a_{3}\right)+10\left(b_{1}+b_{2}+b_{3}\right)+\left(c_{1}+c_{2}+c_{3}\right)=2019$. Let $S=100\left(c_{1}+c_{2}+c_{3}\right)+10\left(b_{1}+b_{2}+b... | 2118 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,199 |
11.1. Solving the equation $x^{4}-3 x^{3}-2 x^{2}-4 x+1=0$, Vasya obtained exactly one negative root. Could Vasya be wrong? | Answer: Yes, I made a mistake.
Solution: Transform the equation: $x^{4}-3 x^{3}-2 x^{2}-4 x+1=0 \Leftrightarrow\left(x^{2}-1\right)^{2}=x\left(3 x^{2}+4\right)$.
If $x<0$, the left side is positive, while the right side is negative for all values of $x$;
only the correct answer or incorrect solution - 0 points. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,200 |
11.2. A line with a positive slope passes through the point (0, 2020) and intersects the parabola $y=x^{2}$ at two points with integer coordinates. What values can the slope take? List all possible options and explain why there are no others. | Answer: $81,192,399,501,1008,2019$
## Solution:
The equation of the line is $y=kx+b$. Since the line passes through the point $(0,2020)$, then $b=2020$.
The points of intersection of the parabola and the line are the integer and distinct roots of the equation $x^{2}=$ $kx+2020$, or $x^{2}-kx-2020=0$. According to Vi... | 81,192,399,501,1008,2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,201 |
11.3. Given a triangle $A B C$. It is known that $\angle B=60^{\circ}, \angle C=75^{\circ}$. On side $B C$ as the hypotenuse, an isosceles right triangle $B D C$ is constructed inside triangle $A B C$. What is the measure of angle $D A C$? | Answer: $30^{0}$.
Solution 1: From the problem statement, it follows that $\angle A=45^{0}$. Draw a circle with center $\mathrm{M}$ and radius $\mathrm{MB}=\mathrm{MC}$. Since $\angle \mathrm{BDC}=90^{\circ}$, the major arc BC is seen at an angle of $45^{0}$. Therefore, vertex A lies on this circle. This means that tr... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,202 |
11.4. Prove the inequality $\left(1+x+x^{2}+\cdots+x^{2020}\right) \cdot\left(1+x^{2020}\right) \geq 4040 x^{2020}$, where $x \geq 0$.
# | # Solution:
Let's expand the brackets and group the terms:
$$
\begin{gathered}
\left(1+x+x^{2}+\cdots+x^{2020}\right) \cdot\left(1+x^{2020}\right)=1+x+x^{2}+\cdots+x^{2020}+x^{2020}+x^{2021}+\cdots+x^{4040}= \\
\left(1+x^{4040}\right)+\left(x+x^{4039}\right)+\left(x^{1}+x^{4038}\right)+\cdots+\left(x^{2020}+x^{2020}\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,203 |
11.5. In a mathematics class, there are 36 students. Exactly one of them recently became a winner of a mathematics olympiad. Each of his classmates has exactly five friends in common with him. Prove that there is a student in the class with an odd number of friends.
# | # Solution
Let it not be so, and each student in the class has an even number of friends. We will divide the class into three groups. In the first group, there will be only the winner, in the second group - his friends (an even number of them), and in the third group - everyone else (an odd number of them).
Each stud... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,204 |
9.1. At the market, a dealer has many carpets. He agrees to exchange a carpet of size $a \times b$ for either a carpet of size $\frac{1}{a} \times \frac{1}{b}$, or two carpets of sizes $c \times b$ and $\frac{a}{c} \times b$ (in each such exchange, the customer can choose the number $c$ themselves). A traveler told a s... | Answer: It's a lie.
Solution. Let's call a carpet large if all its sides are greater than 1, and small if all its sides are less than 1. Thus, initially, the traveler had one large carpet. We will prove that the total number of large and small carpets does not decrease; from this it follows that the described situatio... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 15,205 |
9.2. Circle $\omega$ touches the sides of angle $B A C$ at points $B$ and $C$. Line $\ell$ intersects segments $A B$ and $A C$ at points $K$ and $L$ respectively. Circle $\omega$ intersects $\ell$ at points $P$ and $Q$. Points $S$ and $T$ are chosen on segment $B C$ such that $K S \| A C$ and $L T \| A B$. Prove that p... | Solution. If $\ell \| B C$, the statement is obvious due to symmetry with respect to the perpendicular bisector of $B C$.
Now let the lines $\ell$ and $B C$ intersect at point $X$ (see Fig. 1). From the parallelism, we get $\frac{X B}{X T}=\frac{X K}{X L}=\frac{X S}{X C}$, hence $X T \cdot X S=X B \cdot X C$. Since po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,206 |
9.3. Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then, for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be $N-2$. What values cou... | Answer: $N=3$.
Solution. Note immediately that $d_{s+1-i}=N / d_{i}$ for all $i=$ $=1,2, \ldots, s$.
The number $d_{i+1}-d_{i}$ is divisible by the GCD $\left(d_{i}, d_{i+1}\right)$, so the GCD $\left(d_{i}, d_{i+1}\right) \leqslant d_{i+1}-d_{i}$. For $i=1, \ldots, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)-$ GCD $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,207 |
9.4. From a square sheet of grid paper $100 \times 100$, 1950 two-cell rectangles were cut out along the grid lines. Prove that from the remaining part, a four-cell figure of the form $\square$ - possibly rotated - can be cut out along the grid lines. (If such a figure is already among the remaining parts, it is consid... | The first solution. Imagine that the dominoes (rectangles $1 \times 2$) have not yet been cut out, and we will cut them out one by one. At each moment of the process, we will call the price of an uncut cell the number of its uncut side neighbors, decreased by 2 (for example, the price of a non-corner cell lying on the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,208 |
1. In the zoo, there are 10 elephants and large balance scales. It is known that if any four elephants stand on the left pan of the scales, and any three on the right, the left pan will outweigh. Five elephants stood on the left pan and four on the right. Will the left pan definitely outweigh? | Answer: Not necessarily.
Solution. We will provide a counterexample. Let five elephants weigh 7 tons each, and another five weigh 9 tons each. Then, any four elephants on the left pan together weigh no less than $4 \cdot 7 = 28$ tons, while any three elephants on the right pan weigh no more than $3 \cdot 9 = 27$ tons,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,210 |
2. Two-digit numbers are written on the board. Each number is composite, but any two numbers are coprime. What is the maximum number of numbers that can be written? | Answer: four numbers.
Solution. Evaluation. Since any two of the written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of some of them must be at leas... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,211 |
3. In an acute-angled triangle $A B C$, altitudes $A D$ and $C E$ are drawn. Points $M$ and $N$ are the feet of the perpendiculars dropped from points $A$ and $C$ to the line $D E$, respectively. Prove that $M E=D N$. | Solution. Since $\angle A D C=\angle A E C$, the quadrilateral $A E D C$ is cyclic. We can reason in different ways from here.
First method. By the property of a cyclic quadrilateral, $\angle N D C=\angle B A C=\alpha, \angle M E A=\angle B C A=\gamma$ (see Fig. 9.3a). Then, using the right triangles $A M E$ and $A E ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,212 |
4. Which is greater: $\sqrt{2016}+\sqrt{2015+\sqrt{2016}}$ or $\sqrt{2015}+\sqrt{2016+\sqrt{2015}}$? | Answer: $\sqrt{2016}+\sqrt{2015+\sqrt{2016}}>\sqrt{2015}+\sqrt{2016+\sqrt{2015}}$.
Solution. First method. Let $a=\sqrt{2016}+\sqrt{2015+\sqrt{2016}}>0, b=\sqrt{2015}+\sqrt{2016+\sqrt{2015}}>$ 0. Then $a^{2}-b^{2}=(\sqrt{2016}+\sqrt{2015+\sqrt{2016}})^{2}-(\sqrt{2015}+\sqrt{2016+\sqrt{2015}})^{2}=\sqrt{2016}-\sqrt{201... | \sqrt{2016}+\sqrt{2015+\sqrt{2016}}>\sqrt{2015}+\sqrt{2016+\sqrt{2015}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,213 |
5. Hermann and Chekalin laid out 13 different cards on the table. Each card can lie in one of two positions: face up or face down. The players must take turns flipping one card at a time. The player who, after their move, repeats any previous situation (including the initial one) loses. Chekalin made the first move. Wh... | Answer. Chekalinshchik.
Solution. Chekalinshchik's winning strategy is to flip the same card every time (for example, the Queen of Spades). All possible positions can be divided into pairs that differ only in the position of the Queen of Spades. If, in response to Chekalinshchik's move, Hermann also flips the Queen of... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,214 |
6. The altitudes of an acute-angled, non-isosceles triangle \(ABC\) intersect at point \(H\). \(O\) is the center of the circumcircle of triangle \(BHC\). The center \(I\) of the inscribed circle of triangle \(ABC\) lies on the segment \(OA\). Find the angle \(BAC\). | Answer: $60^{\circ}$.
Solution. From the problem statement, it follows that point O lies at the intersection of the angle bisector of angle $A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, point O lies on this circle and is the midpoint of arc $BC$ (se... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,215 |
11.1. Given two five-digit numbers without the digits 0 and 1 in their notation. The absolute value of their difference is a four-digit number \( S \). It is known that if each digit of one of the original numbers is decreased by 1, then the absolute value of the difference becomes 10002. What values can the number \( ... | Answer: 1109.
Solution: Let $A$ and $B$ be the two given numbers, and $C$ be the number obtained from $B$ by decreasing each of its digits by 1, that is, $C = B - 11111$. If $A10000 > B - A$ (this number is four-digit), then $C > B$. This is a contradiction. Therefore, $A > C$. Also, the case $A > B$ is impossible (th... | 1109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,216 |
11.2 The difference of an increasing arithmetic progression consisting of natural numbers is a natural number ending in 2019. Can three consecutive terms of this progression be squares of natural numbers? | Answer: They cannot.
Solution: Suppose that three consecutive terms of the progression are perfect squares. Denote them as $a^{2}<b^{2}<c^{2}$. Let the common difference of the progression be $d=2 k+1$. Then $c^{2}-a^{2}=2 d=4 k+2$. But $c^{2}-a^{2}=(c-a)(c+a)$, and both factors have the same parity (they differ by $2... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,217 |
11.3. On a wooden wall, the vertices of triangle $A C E$ were marked. Nails were hammered perpendicularly into the wall so that the parts of the nails protruding outward had lengths: $A B=1, C D=2, E F=4$ ( $B, D, F$ - the heads of the nails). Could the distances between the heads of the nails be $B D=\sqrt{2}, D F=\sq... | Answer: Could not.
Solution: Suppose the distances between the hats are as given in the condition. From the rectangular trapezoid $A B D C$, we calculate the length of the lateral side $A C$: $A C^{2}=B D^{2}-(C D-A B)^{2}=2-1=1$, i.e., $A C=1$. Similarly, we find $C E$ and $E A$: $C E=\sqrt{5-(4-2)^{2}}=1, E A=\sqrt{... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,218 |
11.4. It is known that $x+0.5>y^{2}+z^{2}$. Prove that $x+y+z>-1$. | Solution. Add $y+z$ to both sides of the given inequality. We get: $x+y+z+0.5>y^{2}+y+z^{2}+z=(y+0.5)^{2}+(z+0.5)^{2}-0.5$. Therefore, $x+y+z>(y+0.5)^{2}+(z+0.5)^{2}-1 \geq-1$, which is what we needed. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,219 |
11.5. In each of the 320 boxes, there are either 6, 11, or 15 balls, and all three types of boxes are present. Is it true that one can choose several boxes in which the total number of balls is exactly 1001? | Answer. Correct.
Solution. Note that $6 \cdot 11 \cdot 15=990$. Choose one box with 11 balls. We will show that from the remaining 319 boxes, we can select several in which the total number of balls is exactly 990. We will show that these 990 balls can be obtained by selecting boxes of one type. Indeed, if this were n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,220 |
1. What is the greatest value that the sum of the cosines of all angles of an isosceles triangle can take? | Solution. Let $\alpha$ be the acute angle of an isosceles triangle, $0 < \alpha < \pi / 2$.
$2 \cos \alpha + \cos (\pi - 2 \alpha) = -2 \cos^2 \alpha + 2 \cos \alpha + 1$. When $\cos \alpha = 1 / 2$, this quadratic trinomial takes the maximum value, which is 1.5.
Comment. An unjustified answer is scored 0 points. The... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,221 |
2. In volleyball competitions, where there are no ties, 5 teams participate. All teams played against each other. The team that took 1st place won all their matches, and the teams that took 2nd and 3rd place each won exactly two matches. In the case of equal points, the position is determined by the result of the match... | Solution. Let's denote a team by a point. If team A won against team B, we draw an arrow from A to B.
The number of outgoing arrows equals the number of wins. In total, 10 arrows can be draw... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,222 |
3. The lengths of the diagonals of a rhombus and the length of its side form a geometric progression. Find the sine of the angle between the side of the rhombus and its larger diagonal, given that it is greater than $1 / 2$. | Solution. Let $2 b$ be the major diagonal, $2 a$ be the minor diagonal, and $c$ be the side of the rhombus. The sine of the required angle is $\frac{a}{c}>\frac{1}{2}$, that is, $c<2 a<2 b$. Thus, the order of the terms of the geometric progression is determined. By the property of a geometric progression, $(2 a)^{2}=2... | \sqrt{\frac{\sqrt{17}-1}{8}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,223 |
4. The product of all natural numbers from 1 to $\mathrm{n}$ is denoted by $n$! (read as “ $n$ - factorial”). Which of the numbers is greater: 200! or $100^{200}$? | Solution. We will prove that $200!<100^{200}$, for which we will divide the factors on the left into quartets of the form $k(101-k)(100+k)(201-k)$ for $k=1,2, \ldots, 50$ and compare each quartet with $100^{4}$: $\left(k^{2}-101 k\right)\left(k^{2}-101 k-20100\right)<$ $100^{4}$. The variable $t=\left(k^{2}-101 k\right... | 200!<100^{200} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,224 |
5. In a convex quadrilateral $A B C D$, points $P$ and $mathrm{Q}$ are located on sides $A B$ and $C D$ respectively. It is known that $A Q\|C P, B Q\| D P$, $A B \perp B C$ and $C D \perp D P$. Prove that $A B \perp A D$. | Solution. Since $A Q \| C P$, the corresponding angles $B A Q$ and $B P C$ are equal (secant $A B$). Since by the condition $\angle P B C=\angle C D P=90^{\circ}$, the quadrilateral $B P D C$ is inscribed in a circle (with diameter $P C$). The inscribed angles $B P C$ and $B D C$ in this circle are equal (they subtend ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,225 |
1. In the forest, two gnomes lived - Senya and Vanya. Senya lied on Mondays, Tuesdays, and Wednesdays, while Vanya lied on Tuesdays, Thursdays, and Saturdays. On all other days, they only told the truth. One day, upon meeting, one of them was asked: "What is your name?"
- Senya, - he replied.
- And what day of the wee... | Solution. The first gnome cannot be Senya, and can only be Vanya, and on the day of the conversation, he was lying. The second gnome was Senya, and he was also lying. Both gnomes could only lie on Tuesday.
Answer: the first - Vanya, the second - Senya, the conversation took place on Tuesday. | thefirst-Vanya,the-Senya,theconversationtookplaceonTuesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,226 |
3. In three piles, there are 11, 7, and 6 pins respectively. It is allowed to transfer from one pile to another as many pins as are already in the other pile. How can you equalize the number of pins in all piles in three moves? | Answer: $(11,7,6)-(4,14,6)-(4,8,12)-(8,8,8)$. | (8,8,8) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,228 |
4. Little One gave a big box of candies to Karlson. Karlson ate all the candies in three days. On the first day, he ate 0.2 of the entire box and 16 more candies. On the second day, he ate -0.3 of the remainder and 20 more candies. On the third day, he ate -0.75 of the remainder and the last 30 candies. How many candie... | Solution. Let $x$ be the number of candies in the box. On the first day, $(0.2x + 16)$ candies were eaten; on the second and third days, $(0.8x - 16)$ candies were eaten. On the second day, $(0.3(0.8x - 16) + 20) = (0.24x + 15.2)$ candies were eaten; on the third day, $(0.56x - 31.2)$ candies remained. Since on the thi... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,229 |
5. 109 apples are distributed into bags. In some bags, there are $x$ apples, and in others, there are 3 apples. Find all possible values of $x$ if the total number of bags is 20 (there should be no empty bags). | Solution. If there were 3 apples in each package, there would be a total of 60 apples, but in fact, there were 49 more apples. This means that the "extra" apples need to be evenly distributed among some of the packages. Since $49=7 \cdot 7=49 \cdot 1$ and there are 20 packages in total, either 7 packages contain 7 "ext... | 10or52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,230 |
1. The coefficients $a, b, c$ of the quadratic trinomial $a x^{2}+b x+c$ are natural numbers not exceeding 2019. Can such a trinomial have two distinct roots that differ by less than 0.01? | Solution. Answer: For example, $1001 x^{2}+2001 x+1000$. The roots of the quadratic equation $a x^{2}+b x+c=0$ differ by the value $\frac{\sqrt{b^{2}-4 a c}}{a}$. For example, let's make $\sqrt{b^{2}-4 a c}=1$. Notice that $(2 n+1)^{2}=4 n(n+1)+1$. Then, for $a=n+1$, $b=2 n+1$, and $c=n$, we have $\frac{b^{2}-4 a c}{a}... | 1001x^{2}+2001x+1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,231 |
2. Solve the equation in natural numbers $x, y, z$ (where $x!$ denotes the factorial of $x$)
$$
\frac{x+y}{z}=\frac{x!+y!}{z!}
$$ | Solution. Answer: $(a, b, c)$, where each of $a, b, c$ is independently 1 or $2 ;(a, a, a)$ for $a \geqslant 3$.
Rewrite the equation as $(x+y)(z-1)!=x!+y!$ !
1) Let $z>x$ and $z>y$. Then $(x+y)(z-1)!=x!+y!\leqslant 2(z-1)!$ ! From this, $x+y \leqslant 2$ and $x=y=1$. In this case, $(z-1)!=1$ and $z=1$ or $z=2$. Thus... | (,b,),whereeachof,b,isindependently1or2;(,,)for\geqslant3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,232 |
4. We will call a natural number a "rare number" if it cannot be represented as the product of more than four integer factors, each greater than one. Seven consecutive three-digit numbers were written in a row so that a string of 21 digits was formed. Prove that some six consecutive digits in this string, taken in the ... | Solution. Consider six digits pertaining to two consecutive three-digit numbers. The number $A$, represented by these six digits, has the form $1001 x+1$ (where $x$ is the smaller of the two three-digit numbers). The number $A$ in three out of six cases (when $x$ is even) is not divisible by 2, it is also not divisible... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,234 |
5. The figure "archer" on a grid board attacks along a ray - along cells upwards, downwards, to the right, or to the left (exactly one of the four directions; the directions for different archers are independent). What is the maximum number of non-attacking archers that can be placed on an $8 \times 8$ chessboard? | Solution. Answer: 28. We will prove that it is impossible to place more than 28 archers. Consider any arrangement with the maximum possible number of archers. Perform the following two operations sequentially:
1) turn those archers who are standing at the edge of the board so that they shoot "outward" (this will not "... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,235 |
7.2. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. In the last convocation, 2020 deputies were elected to the parliament. At the first plenary session ... | Solution. Note that the statement was made by more than half of the deputies. If all those who made the statement are knights, then their statements turn out to be false; if all those who made the statement are liars, then their statements turn out to be true - both are impossible. Therefore, among those who made the s... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,237 |
7.3. A trading organization wholesale purchased exotic fruits, the moisture content of which was $99\%$ of their mass. After delivering the fruits to the market, the moisture content dropped to $98\%$. By what percentage should the trading organization increase the retail price of the fruits (the price at which it will... | Solution. In 100 kg of fresh fruits, there was 99 kg of water and 1 kg of solid mass. After drying, 1 kg of solid mass constituted $2 \%$ of the mass of 50 kg, i.e., every 100 kg of fresh fruits dried down to 50 kg, i.e., by half, and the retail price, compared to the wholesale price, should be increased by two times.
... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,238 |
7.4. Find the smallest natural number ending in the digit 6 (on the right) which, when this digit is moved to the beginning (to the left), increases exactly four times. | Solution. Multiplication "in column" is performed from the end, so we can start the process of multiplication by sequentially finding all the digits: *****6 ( $6 \times 4=24$ - "4 we write, 2 in mind" - the 4 written under the line is the 4th digit that stood before 6; by writing 4 in the top line before 6, we will con... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,239 |
7.5. The following numbers are written on the wall: $1,3,4,6,8,9,11,12,16$. Four of these numbers were written by Vova, four numbers were written by Dima, and one number was simply the house number of the local police officer. The police officer found out that the sum of the numbers written by Vova is three times the s... | Solution. The sum of all numbers on the wall is $1+3+4+6+8+9+11+12+16=70$. Let $n$ be the sum of the numbers written by Dima, then $3n$ is the sum of the numbers written by Vova, $4n$ is the sum of the numbers written by both, $70-4n-$ is the house number of the district police officer, which when divided by 4 leaves a... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,240 |
1. Masha left the house for school. A few minutes later, Vanya ran out of the same house to school. He overtook Masha at one-third of the way, and when he arrived at school, Masha still had half of the way left to go. How many times faster does Vanya run compared to how Masha walks? | Solution. At one third of the way, Masha and Vanya were at the same time. After that, Vanya ran $2 / 3$ of the way, while Masha walked $1 / 2 - 1 / 3 = 1 / 6$ of the way in the same time. This means that he runs $2 / 3$ : $1 / 6 = 4$ times more in the same amount of time than Masha. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,241 |
2. At a mathematics olympiad, each of the 11 seventh-grade students solved 3 problems. It is known that for any two of them, there is a problem that one of them solved and the other did not. Prove that they were offered no fewer than 6 problems. | Solution. If there are no more than 5 problems, then there are no more than 10 different sets of 3 problems. Therefore, there will be two seventh-graders who have solved the same set of problems. This contradicts the condition.
Comment. If the case for 5 problems is considered and it is not stated that for a smaller n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,242 |
4. Two cards have four different digits written on them - one on each side of the card. Can it be so that every two-digit number that can be formed from these cards is prime? (Digits cannot be flipped upside down, i.e., you cannot turn a 6 into a 9 or vice versa.) | Solution. All two-digit numbers ending in $0, 2, 4, 6$ or 8 are even, and those ending in 5 are multiples of five. Therefore, such numbers cannot be prime, and it makes no sense to write these digits on cards. The remaining digits are 1, 3, 7, and 9. If the digits 3 and 9 are written on different cards, then they can f... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,243 |
5. What is the smallest identical number of pencils that need to be placed in each of 6 boxes so that any 4 boxes contain pencils of any of the 26 pre-specified colors (there are enough pencils available)? Prove that fewer is impossible. | Solution. Let's assume that we have fewer than 3 pencils of some color. Then, if we take 4 boxes in which such pencils are not present (and such boxes can be found, since there are no more than $2 y x$ pencils of that color), the condition of the problem will not be met. This means that there are at least 3 pencils of ... | 78 | Combinatorics | proof | Yes | Yes | olympiads | false | 15,244 |
1. Real numbers $a, b, c$ are such that $a+1 / b=9, b+1 / c=10$, $c+1 / a=11$. Find the value of the expression $a b c+1 /(a b c)$. | Answer: 960.
Sketch of the solution. By multiplying the equations, expanding the brackets, and grouping, we get: $a b c + 1/(a b c) + a + 1/b + b + 1/c + c + 1/a = 990$. From this, $a b c + 1/(a b c) = 990 - 9 - 10 - 11 = 960$. | 960 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,245 |
2. In a football tournament, 17 teams participate, and each team plays against each other exactly once. A team earns 3 points for a win. For a draw, 1 point is awarded. The losing team gets no points. What is the maximum number of teams that can accumulate exactly 10 points? | Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 10 points each. The total points include all points scored by these teams in matches against each other (at least 2) and possibly in matches against other teams: $10 n \geq 2(n-1) n / 2$. Hence, $n \leq 11$.
Example: All eleven teams pl... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,246 |
3. Provide an example of three different integers, one of which is equal to the difference of the other two, and another is the quotient of the other two. | Answer: the numbers $2, -2, -4$.
The numbers fit because $2 = (-2) - (-4); -2 = -4 / 2$.
Remark. These numbers can be found by solving in integers one of the solutions of the system $\left\{\begin{array}{c}a=b-c \\ b=\frac{c}{a}\end{array}\right.$ or a similar one.
Criteria. Correct triplet without specifying how it ... | 2,-2,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,247 |
4. Given three non-zero real numbers $a, b, c$ such that the equations: $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ each have two roots. How many of the roots of these equations can be negative? | Answer: 2.
Sketch of the solution.
If the numbers $a, b, c$ are replaced by their opposites, then the "new" equations will have the same set of roots as the original ones. There are two possible cases: the numbers $a, b, c$ are of the same sign; among them, there are both positive and negative numbers.
First case. D... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,248 |
6. Several cells on a $14 \times 14$ board are marked. It is known that no two marked cells are in the same row and the same column, and also, that a knight can, starting from some marked cell, visit all marked cells in several jumps, visiting each exactly once. What is the maximum possible number of marked cells? | Answer: 13.
Since there is no more than one marked cell (field) in each row, there are no more than 14 marked cells. Suppose there are 14 marked cells. Number the rows and columns from 1 to 14 from bottom to top and from left to right, and color the cells in a checkerboard pattern, where the sum of the row and column ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,249 |
7.1. In the expression $5 * 4 * 3 * 2 * 1=0$, replace the asterisks with the four arithmetic operation signs (using each sign exactly once) to make the equation true. No other signs, including parentheses, can be used. | Solution: Answer: $5-4 \cdot 3: 2+1=0$.
Note: The given example is not the only possible one. A complete solution requires only one (any) correct example.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correctly replaced asterisks | 7 points |
| Solutions that do not meet the condition (... | 5-4\cdot3:2+1=0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,250 |
7.2. Petya and Borya each made five shots at one target, hitting "5" once, "7" twice, "8" once, "9" twice, "10" twice, "11" once, and "12" once. With his last four shots, Petya scored seven times more points than with his first shot. It is known that both Petya and Borya hit "10". Who hit "12"? Justify your answer. | Solution: We have that with the last four shots, Petya scored no more than $12+11+10+9=42$, which means his first shot was no more than $42: 7=6$. Therefore, his first shot hit the "five," and the last four shots in total scored 35. During this, there was exactly one "ten," so the remaining three shots scored 25. This ... | Borya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,251 |
7.3. ("We choose, we are chosen"). Three boys - Kolya, Petya, and Yura are in love with three girls: Tanya, Zina, and Galia. However, although Tanya, Zina, and Galia are also in love with Kolya, Petya, and Yura, the love turned out to be unrequited. Kolya loves a girl who is in love with a boy who loves Tanya. Petya lo... | Solution: Since it is a love without reciprocation, all attachments follow the cycle Y1 - D1 - Y2 - D2 - Y3 - D3 - Y1 (where Y and D represent the young man and the young woman, respectively). Without loss of generality, let Y1 be Kolya. Then D2 is Tanya. If now Y3 is Petya, then D1 is Zina, and Y2 is Yura. In this cas... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,252 |
7.4. Lena has two triangles and a quadrilateral, and Vasya has two triangles (different from Lena's) and a pentagon (all figures are cut out of a sheet of plywood). Both claim that they can form both a $4 \times 6$ rectangle and a $3 \times 8$ rectangle from their figures.
a) Could it be that Lena is right?
b) Could ... | Solution: a) Lena can be right. b) Vasya can also be right. Possible constructions are shown in the figure.
Lena's construction
 Provide an example of such a triangle and the segment drawn in it.
b) Determine what values the angles of such a trian... | Solution: Let's immediately solve part b). After drawing one segment in the triangle, either 2 or 3 triangles will be formed. The 3 triangles case only occurs if the segment connects a vertex to a point on the opposite side. Since we need a right, acute, and obtuse triangle, this last case must be the one that applies.... | 90,30,60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,254 |
7.6. Natural numbers $m$ and $n$ are such that $88 m=81 n$. Prove that the number $m+n$ is composite. | Solution: The number $88(m+n)=88 m+88 n=81 n+88 n=169 n$ is divisible by $169=13^{2}$. Since the numbers 88 and 13 are coprime, the number $m+n$ is divisible by $13^{2}$, making it a composite number.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 po... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,255 |
1. Solve the equation $\sqrt{\sin ^{2} x+\lg ^{2} x-1}=\sin x+\lg x-1$. | Answer: $10, \pi / 2+2 n \pi, n-$ an integer, non-negative.
Instructions. By squaring and factoring, we get $\sin x=1$ or $\lg x=1$. From the first equation, we have $x=\pi / 2+2 n \pi, n-$ an integer. From the second, $x=10$. Since the logarithm is defined only for positive numbers, $n$ is non-negative. It is easy to... | 10,\pi/2+2n\pi,n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,256 |
5. Find all functions f defined on the set of real numbers such that $\mathrm{f}(\mathrm{xy})+\mathrm{f}(\mathrm{xz}) \geq 1+\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{yz})$ for all real $\mathrm{x}, \mathrm{y}, \mathrm{z}$. | Answer: $\mathrm{f}(\mathrm{x})=1$ (a constant function, all values of which are equal to 1).
Instructions. Let $\mathrm{x}=\mathrm{y}=\mathrm{z}=1$ and we get $\mathrm{f}(1)+\mathrm{f}(1) \geq 1+\mathrm{f}(1) \mathrm{f}(1)$, or $(f(1)-1)^{2} \leq 0$ and thus, $f(1)=1$.
Let $\mathrm{x}=\mathrm{y}=\mathrm{z}=0$ and we... | \mathrm{f}(\mathrm{x})=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,257 |
# 8.1. Can a cube be cut into 71 smaller cubes (of any non-zero size)? | Answer: Yes, it is possible.
Solution: Let's divide each edge of the cube in half and cut the cube into 8 smaller cubes. Now, take one of these smaller cubes and divide it into 8 even smaller cubes. One cube disappears, but 8 new ones appear, increasing the total number of parts by \(8-1=7\) to 15. Repeating this oper... | 71 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,258 |
8.2. Two cyclists, Andrey and Boris, are riding at a constant and identical speed along a straight highway in the same direction, so that the distance between them remains constant. There is a turnoff to a village ahead. At some point in time, the distance from Andrey to the turnoff was equal to the square of the dista... | Answer: 2 or 0 km.
Solution: Let the first mentioned distances be $a$ and $b$, then $a=b^{2}$. When each of them had traveled another kilometer, the remaining distance to the turn was $a-1$ and $b-1$ km, respectively, so $a-1=3(b-1)$, which means $b^{2}-1=3(b-1),(b-1)(b+1)=3(b-1)$, from which $b=1$ or $b=2$. When $b=1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,259 |
8.3. In triangle $A B C$ with angle $C$ equal to $30^{\circ}$, median $A D$ is drawn. Angle $A D B$ is equal to $45^{\circ}$. Find angle $B A D$.
# | # Answer: $30^{\circ}$.
Solution. Draw the height $B H$ (see the figure). In the right triangle $B H C$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B H=\frac{B C}{2}=B D$. The angle $H B C$ is $180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$. Triangle $B H D$ is isosceles with an angle of $60^{\circ}$, so i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,260 |
8.4. It is known that for real numbers $a$ and $b$ the following equalities hold:
$$
a^{3}-3 a b^{2}=11, \quad b^{3}-3 a^{2} b=2
$$
What values can the expression $a^{2}+b^{2}$ take? | Answer: 5.
Solution. We have
$$
\left(a^{2}+b^{2}\right)^{3}=a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}=\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2}=11^{2}+2^{2}=125
$$
From this, $a^{2}+b^{2}=5$.
Comment. A correct and justified solution - 7 points. The correct answer obtained based on a simple e... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,261 |
8.5. What is the maximum number of members that can be in a sequence of non-zero integers, for which the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative? | Answer: 16.
Solution: Estimation. Assume there are no fewer than 17 numbers. Construct a table with 7 columns and 11 rows, in which the first 17 numbers are arranged.
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{2}$ | $a_{3}$ |... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,262 |
1. Compare the numbers $\frac{2010}{2011}$ and $\frac{20102010}{20112011}$. | 1. Since $\frac{20102010}{20112011}=\frac{2010 \cdot 10001}{2011 \cdot 10001}=\frac{2010}{2011}$, the numbers will be equal. | \frac{2010}{2011}=\frac{20102010}{20112011} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,263 |
2. On the New Vasyuki Currency Exchange, 11 tugriks are exchanged for 14 dinars, 22 rupees for 21 dinars, 10 rupees for 3 thalers, and 5 crowns for 2 thalers. How many tugriks can be exchanged for 13 crowns? | 2. For 5 crowns, they give two thalers, which means for one crown, they give $2 / 5$ thalers. Similarly, for one thaler, they give 10/3 rupees, for one rupee - 21/22 dinars, for one dinar - 11/14 tugriks. Therefore, for one crown, they give $2 / 5$ thalers, or (2/5) $\cdot$ (10/3) rupees, or (2/5) $\cdot$ (10/3) $\cdot... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,264 |
3. Seryozha cut a square birthday cake weighing 900 g with two straight cuts parallel to one pair of sides, and two cuts parallel to the other pair of sides, into 9 rectangular pieces. Prove that Petya can choose such three pieces of cake that do not share any sides and their total weight is not less than 300 g. | 3. Consider the triples of pieces marked with the same numbers in the figure. The total weight of the pieces of at least one triple is not less than 300g; otherwise, the weight of the cake would be less than $300 \times 3 = 900$ g.
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 3 | 1 | 2 |
| 2 | 3 | 1 | | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,265 |
4. A merchant accidentally mixed candies of the 1st grade (at 3 rubles per pound) and candies of the 2nd grade (at 2 rubles per pound). At what price should this mixture be sold to receive the same amount of money, given that initially the total cost of all candies of the 1st grade was equal to the total cost of all ca... | 4. Let's denote by a the original cost of all candies of the 1st type. Then the total revenue from unsold candies of the 1st and 2nd type would be 2a rubles. In this case, the merchant would have $\frac{a}{3}$ pounds of candies of the 1st type, and $\frac{\text { a }}{2}$ pounds of candies of the 2nd type. Thus, for th... | 2.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,266 |
5. Every day in November, Winnie-the-Pooh visited either Piglet, or Rabbit, or Eeyore. For any two consecutive days, he visited Piglet at least once, and for any three consecutive days, he visited Rabbit at least once. How many times could he have visited Eeyore during this time? (November has 30 days. Winnie-the-Pooh ... | 5. Suppose Winnie-the-Pooh visited Eeyore on some day in November, but not on the 1st or the 30th. Since over any two consecutive days he visited Piglet at least once, he must have visited Piglet the day before visiting Eeyore, and he must also have visited Piglet the day after visiting Eeyore. This means that over the... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,267 |
1. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,268 |
3. (7 points) In triangle $ABC$, angle bisectors are drawn from vertices $A$ and $B$, and a median is drawn from vertex $C$. It turns out that the points of their pairwise intersections form a right triangle. Find the angles of triangle $ABC$.
| Solution. Let $M$ be the point of intersection of the angle bisectors of triangle $ABC$, and its median $CO$ intersects the drawn angle bisectors at points $K$ and $L$ (see figure).
Since $\a... | 90,60,30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,270 |
4. (7 points) Petya was adding two natural numbers and by mistake appended an extra digit to the end of one of them. As a result, instead of getting the correct answer of 12345, he got the sum of 44444. What numbers was he adding?
# | # Solution.
Let Pete need to add numbers $x$ and $y$. When he appended a digit $c$ to one of them (for definiteness, to $x$), the number became $10 x + c$. Thus, we obtain the system of equations:
$$
\left\{\begin{array}{c}
x + y = 12345 \\
(10 x + c) + y = 44444
\end{array}\right.
$$
Subtracting the first equation ... | 35668779 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,271 |
5. (7 points) Compare the result of the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 98 \cdot 99$ and the number $50^{99}$.
# | # Solution.
Since $\sqrt{a b}<\frac{a+b}{2}$, then $a b<\left(\frac{a+b}{2}\right)^{2}$. To use this property, let's rewrite the product as follows:
$$
\begin{aligned}
& 1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 98 \cdot 99=1 \cdot 99 \cdot 2 \cdot 98 \cdot 3 \cdot 97 \cdot \ldots \cdot 49 \cdot 51 \cdot 50 . \\
&... | 1\cdot2\cdot3\cdot4\cdot\ldots\cdot98\cdot99<50^{99} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,272 |
10.1. Given ten positive numbers, any two of which are distinct. Prove that among them there are either three numbers whose product is greater than the product of any two of the remaining, or three numbers whose product is greater than the product of any four of the remaining. (A. Gолованов) | First solution. Take any 5 of the given numbers: $a, b, c, d, e$. If $a b c > d e$, then the statement of the problem is true. If $d e \geqslant a b c$, take two more numbers $f$ and $g$. Let's say $f > g$. Then $d e f > a b c g$; hence, the statement of the problem is true in this case as well.
Second solution. Order... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,273 |
10.2. Given a convex hexagon $A B C D E F$. It is known that $\angle F A E = \angle B D C$, and the quadrilaterals $A B D F$ and $A C D E$ are cyclic. Prove that the lines $B F$ and $C E$ are parallel. | Solution. Let $K-$ be the intersection point of segments $A E$ and $B F$. Since quadrilaterals $A B D F$ and $A C D E$ are cyclic, we have $\angle A F B = \angle A D B$ and $\angle A D C = \angle A E C$. From this and the condition of the problem, we get $\angle A K B = \angle A F B + \angle F A E = \angle A D B + \ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,274 |
10.3. The sequence of numbers $a_{1}, a_{2}, \ldots$ is defined by the conditions $a_{1}=1$, $a_{2}=143$ and $a_{n+1}=5 \cdot \frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$ for all $n \geqslant 2$. Prove that all terms of the sequence are integers.
(M. Murashkin) | The first solution. The number $a_{3}=5 \cdot 72=360$ is an integer. For $n \geqslant 4$, the following equalities will hold:
$$
a_{n}=5 \cdot \frac{a_{1}+\ldots+a_{n-1}}{n-1} \quad \text { and } \quad a_{n-1}=5 \cdot \frac{a_{1}+\ldots+a_{n-2}}{n-2}
$$
from which it follows that
$$
a_{n}=\frac{5}{n-1}\left(\frac{n-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,275 |
10.4. On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint of exa... | Answer. 1.
First solution. We will prove by induction on $N$ that there is one more even matching than odd. For $N=1$, the statement is obvious: there is only one matching, and it is even. Now we will prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Let the marked points be $A_{1}, A_{2}, ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,276 |
10.5. Given a convex pentagon. Petya wrote down in his notebook the values of the sines of all its angles, and Vasya - the values of the cosines of all its angles. It turned out that among the numbers written down by Petya, there are no four distinct ones. Can all the numbers written down by Vasya be distinct?
(N. Aga... | Answer: No.
Solution: Suppose the opposite; then all the angles of the pentagon are different numbers from the interval $(0, \pi)$. Immediately note that then Petya will not find three equal numbers, because there are no three different angles in this interval with equal sines.
Therefore, Petya must have two pairs of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,277 |
10.6. Petya chose a natural number $a>1$ and wrote down fifteen numbers $1+a, 1+a^{2}, 1+a^{3}, \ldots, 1+a^{15}$ on the board. Then he erased several numbers so that any two remaining numbers are coprime. What is the maximum number of numbers that could remain on the board?
(O. Podlipsky) | # Answer. 4 numbers.
Solution. First, we will show that there cannot be more than four such numbers. Note that if $k$ is odd, then the number $1+a^{n k}=1^{k}+\left(a^{n}\right)^{k}$ is divisible by $1+a^{n}$. Next, each of the numbers $1,2, \ldots, 15$ has one of the forms $k, 2 k, 4 k, 8 k$, where $k$ is odd. Thus, ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,278 |
10.7. Given an $n \times n$ square. Initially, its cells are painted in white and black in a checkerboard pattern, with at least one of the corner cells being black. In one move, it is allowed to simultaneously repaint the four cells in some $2 \times 2$ square according to the following rule: each white cell is painte... | Answer. For all $n$ that are multiples of three.
Solution. Suppose we managed to repaint the cells as required by the conditions of the problem. We will call cells of the first type those that were initially white, and cells of the second type those that were black. Note that if a cell is repainted three times, it doe... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,279 |
10.8. In trapezoid $ABCD$, the lateral side $CD$ is perpendicular to the bases, $O$ is the point of intersection of the diagonals. On the circumscribed circle of triangle $OCD$, a point $S$ is taken, diametrically opposite to point $O$. Prove that $\angle BSC = \angle ASD$.
(V. Shmarov) | Solution. Since $SO$ is the diameter, then $\angle SCA = \angle SCO = \angle SDO = \angle SDB = 90^{\circ}$. To solve the problem, it is sufficient to prove the similarity of the right triangles $SCA$ and $SDB$. Indeed, from the similarity, it follows that the angles $\angle CSA = \angle DSB$, from which $\angle BSC = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,280 |
8.1. Is the number $10^{2021}-2021$ divisible by 9? | Answer: No.
Solution. Let's write the number as $10^{2021}-1-2020$. The number $10^{2021}-1$ is of the form 99...9 and is divisible by 9, while the number 2020 is not. Therefore, their difference is not divisible by 9. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,281 |
8.2. It is known that $k \neq b$ and the graphs of the functions $y=k x+k, y=b x+b$ and $y=a x+c$ intersect at one point. Does this mean that $a=c$? | Answer: Yes.
Solution. It is clear that the point $(-1,0)$ belongs to the first two lines. Since $k \neq b$, these lines are different, so they have no other common points. Therefore, the point $(-1,0)$ also belongs to the graph of the third function, that is, $0=-a+c$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,282 | |
8.3. In triangle $A B C$, the median $A M$ is perpendicular to the bisector $B D$. Find the perimeter of the triangle, given that $A B=1$, and the lengths of all sides are integers. | Answer: 5.
Solution. Let $O$ be the point of intersection of the median $A M$ and the bisector $B D$. Triangles $A B O$ and $M B O$ are congruent (by the common side $B O$ and the two adjacent angles). Therefore, $A B = B M = 1$ and $C M = 1$, since $A M$ is a median. Thus, $A B = 1, B C = 2$. By the triangle inequali... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,283 |
8.4. Masha and Katya wove wreaths from wildflowers: dandelions, cornflowers, and daisies. The total number of flowers in the wreaths of both girls turned out to be 70, with dandelions making up $5 / 9$ of Masha's wreath, and daisies making up 7/17 of Katya's wreath. How many cornflowers are in each girl's wreath, if th... | Answer: Masha has 2 cornflowers in her wreath, while Katya has no cornflowers.
Solution. From the condition, it follows that Masha has a total of $9 n$ flowers in her wreath, and Katya has $-17 k$, where $n$ and $k$ are non-negative integers. Then, in the girls' wreaths, there are $5 n$ daisies and $7 k$ chamomiles. S... | Masha\has\2\cornflowers,\Katya\has\0\cornflowers | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,284 |
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