problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8.6. On a circular highway 15 km long, there are 6 different cities $A, B, C, D, E, F$ (not necessarily in that order). Baron Munchausen traveled the route $A-B-C-D-E-A$ (each leg of the journey was made along the shortest arc connecting the specified points) and claims that the lengths of the legs were (in kilometers)... | Answer and justify.
Solution to problem 8.6
Solution: a) They could. The cities are in order (for example, in a clockwise direction) $E-B-D-A-C-$ $F-E$, and the distances are as follows: $E ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,824 |
Problem 8.1
It is known that $\frac{2}{a}-\frac{1}{b}=\frac{1}{a+2 b}$. Prove that $\frac{4}{a^{2}}-\frac{1}{b^{2}}=\frac{1}{a b}$.
## Number of points 7 | Answer:
Solution
$\frac{4}{a^{2}}-\frac{1}{b^{2}}=\left(\frac{2}{a}-\frac{1}{b}\right)\left(\frac{2}{a}+\frac{1}{b}\right)=\frac{1}{a+2 b} \cdot \frac{a+2 b}{a b}=\frac{1}{a b}$
## Remark
Solutions are possible where the given equality is transformed into the provable equality through identical transformations.
# | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,825 |
# Task 8.2
A team consisting of Petya, Vasya, and a one-person scooter is participating in a race. The distance is divided into segments of equal length, the number of which is 42, with a checkpoint at the beginning of each. Petya runs a segment in 9 minutes, Vasya in 11 minutes, and on the scooter, either of them can... | # Answer:
18
Solution
If the time of one becomes less than the time of the other of the boys, then the time of the other will increase, and consequently, the team's time will increase. Therefore, the times of the boys must coincide.
Denoting the number of segments driven by Petya as $x$ and solving the equation:
$... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,826 |
Problem 8.3
In triangle $A B C$, angle $\mathrm{B}$ is equal to $120^{\circ}, A B=2 B C$. The perpendicular bisector of side $A B$ intersects $A C$ at point $D$. Find the ratio $C D: D A$.
## Number of points 7
# | # Answer:
$3: 2$
## Solution
We need the ratio of segments lying on the same straight line. We apply Thales' theorem. The angle $CAB$, one straight line $DM$ is already there (and it passes through the point $D$ we need, which is another confirmation that it is appropriate to adapt to Thales' theorem here). We need ... | 3:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,827 |
# Task 8.4
A rectangle is paved with tiles of sizes $1 \times 4$ and $2 \times 2$. The tiles were removed from the rectangle, and one tile of size $2 \times 2$ was replaced with a tile of size $1 \times 4$. Prove that it is now impossible to pave the rectangle with the new set of tiles.
## Number of points 7
# | # Solution
Consider a coloring with 4 colors such that each $2 \times 2$ tile contains exactly one cell of color 1, and each $1 \times 4$ tile contains either zero or two cells of color 1. Therefore, the parity of the number of $2 \times 2$ tiles must match the parity of the number of cells of color 1, which proves th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,828 |
# Task 8.5
Let $a, b, c$ be natural numbers and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1$. Prove that then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{41}{42}$.
## Number of points 7
# | # Solution
We can assume that $a \leq b \leq c$.
Consider all possible cases.
1) $a=2$. Then $b>2$.
If $b=3$, then $c>6$ and
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{41}{42}$
If $b=4$, then $c>4$ and
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{1}{2}+\frac{1}{4}+\frac... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,829 |
3. A train passes a pole in 5 s. How long (in seconds) will it take for the train and an electric train to pass each other, if the speed of the train is twice the speed of the electric train, and the length of the train is three times the length of the electric train | 3. Let $v_{1}, l_{1}$ be the speed (in m/s) and length (in m) of the train; $v_{2}, l_{2}$ be the speed (in m/s) and length (in m) of the electric train. According to the problem, $v_{2}=\frac{1}{2} v_{1} ; l_{2}=\frac{1}{3} l_{1}$. Knowing that the train passes a pole in 5 seconds, we have $\frac{l_{1}}{v_{1}}=5$. To ... | \frac{40}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,830 |
5. There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row. | 5. It follows from the condition that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no fewer than 3*6=18 chip... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,832 |
2. Find all such prime numbers $p$ and $q$, for which the number $2^{2}+p^{2}+q^{2}$ is also prime.
Note. A prime number is an integer greater than one that is divisible only by one and by itself. | Solution. If the numbers $p$ and $q$ are odd, then the number $2^{2}+p^{2}+q^{2}$ is even and, obviously, greater than 2, and therefore cannot be prime. Thus, among the prime numbers $p$ and $q$, there is an even one, which is equal to 2. Suppose this is the number $p$. It remains to find all primes $q$ for which the n... | p=2,q=3orp=3,q=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,833 |
3. In a convex quadrilateral $A B C D$, the side $C D$ is visible from the midpoint of the side at a right angle. Prove that $A D+B C \geq C D$. | Solution. Let the midpoint of side $AB$ be denoted as $O$, and draw rays $AX$ and $BX$ parallel to lines $OD$ and $OC$ respectively, which intersect at point $X$ (Fig. 1). It is clear from the construction that $\triangle AXB$ is a right triangle. Then $OD$ is the perpendicular bisector of $AX$, and $OC$ is the perpend... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,834 |
4. On equal-armed scales, weights of $1, 2, 3, \ldots, n, n>6$ are placed in such a way that the scales are in equilibrium. Is it always possible to remove 3 weights from the scales in such a way that the scales remain in equilibrium? | Solution. If for some $k>1$ one weight $k+1$ is placed on the first scale pan, and on the second - weights $k$ and 1, then it is sufficient to remove these 3 weights and the requirement is met. If such a situation does not occur, then consider the scale pan where the weight 1 is located, as well as the weight with the ... | Always | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,835 |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,839 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,840 |
Problem 5.4. Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.
![](https://cdn.mathpix.com/cropped/2024_05_06_c49ecf27446af41a8bcdg-09.jpg?height=... | Answer: $A=4, B=1, C=2, D=5, E=6, F=3$.
Solution. According to the condition, $A, D, E$ are different digits, not exceeding 6, the sum of which is 15. If these digits are taken to be the maximum possible, then their sum is $4+5+6=15$. Therefore, $A, D, E$ are $4,5,6$ in some order (if at least one of the digits is no ... | A=4,B=1,C=2,D=5,E=6,F=3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,841 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,844 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,846 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,850 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=$ $90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,851 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,853 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,854 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
This results in 8 rectan... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,855 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,856 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,857 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,858 |
8.2. How many solutions does the equation $(2 x+y)^{2}=2017+x^{2}$ have in integers $x, y$? | Answer: four solutions. Solution. Since 2017 is a prime number, from the factorization $2017=(3 x+y)(x+y)$, the possible systems are:
$$
\left\{\begin{array}{l}
3 x+y=2017 \\
x+y=1
\end{array} ;\left\{\begin{array}{l}
3 x+y=1 \\
x+y=2017
\end{array} ;\left\{\begin{array}{l}
3 x+y=-2017 \\
x+y=-1
\end{array} ;\left\{\b... | four\solutions | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,860 |
8.3. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$. | Answer: $60^{\circ}$. Solution. First, we show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$, and thus, ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,861 |
1. (7 points) The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the $2019-$th term of the sequence. | Solution. Let's find the first few terms of the sequence:
$a_{1}=934 ; a_{2}=16 \cdot 13=208 ; a_{3}=10 \cdot 13=130 ; a_{4}=4 \cdot 13=52 ;$
$a_{5}=7 \cdot 13=91 ; a_{6}=10 \cdot 13=130=a_{3}$.
Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a pe... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,863 |
2. (7 points) Does the equation $x^{4}-4 x^{3}-6 x^{2}-3 x+9=0$ have negative roots? | Solution. $\left(x^{4}-6 x^{2}+9\right)-4 x^{3}-3 x=0$, $\left(x^{2}-3\right)^{2}=4 x^{3}+3 x$ $\left(x^{2}-3\right)^{2}=x\left(4 x^{2}+3\right)$
If $x<0$, then $\left(x^{2}-3\right)^{2} \geq 0, x\left(4 x^{2}+3\right)<0$.
Therefore, the obtained equality will be false for any $x<0$. Hence, there are no negative root... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,864 |
3. (7 points) The function $f$ is such that for any $x>0, y>0$ the equality $f(x y)=f(x)+f(y)$ holds. Find $f(2019)$, if $f\left(\frac{1}{2019}\right)=1$. | Solution. When $y=1 \quad f(x)=f(x)+f(1), f(1)=0$.
When $x=2019 \quad y=\frac{1}{2019} f(1)=f(2019)+f\left(\frac{1}{2019}\right)$,
$f(2019)=f(1)-f\left(\frac{1}{2019}\right)$
$f(2019)=-f\left(\frac{1}{2019}\right)=-1$.
Answer. $f(2019)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,865 |
4. (7 points) From a sheet of graph paper, a polygon without holes was cut out along the grid lines. It is known that it can be cut into $2 \times 1$ rectangles. Prove that it has at least one side of even length. | Solution. Let's color the cells in a checkerboard pattern. Since the polygon can be cut into $2 \times 1$ rectangles, the number of white cells is equal to the number of black cells. Therefore, the total number of sides of white cells is equal to the number of sides of black cells. Since inside the figure, each side of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,866 |
5. (7 points) On side $AB$ of triangle $ABC$, a point $K$ is marked, and on side $AC$ - a point $M$. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find the angle $BAC$.
... | # Solution.
Add the area of triangle $B P K$ to both sides of the equality $S_{\text {AKPM }}=S_{\text {BPC }}$:
$$
\begin{aligned}
& S_{A K P M}+S_{B P K}=S_{B P C}+S_{B P K}, \\
& S_{A B ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,867 |
7.1 Four princesses each thought of a two-digit number, and Ivan thought of a four-digit number. After they wrote down their numbers in a row in some order, the result was 132040530321. Find Ivan's number.
Answer: 5303 | Solution: Let's go through the options. Option 1320 is not suitable because the remaining part of the long number is divided into segments of two adjacent digits: $40,53,03,21$, and the segment 03 is impossible as it is not a two-digit number. Option 3204 is impossible due to the invalid segment 05 (or the segment 1 fr... | 5303 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,868 |
7.3 It is known that for some natural number $n$ each of the numbers $3 n-1$ and $n-10$ is divisible by a prime number p. Find the number p. | Answer: 29
Solution. The number $3 n-1-3(n-10)=29$ is divisible by p. Since $29$ is a prime number, then $\mathrm{p}=29$. | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,869 |
7.4 In the class, there are more than 30 people but less than 40. Any boy is friends with three girls, and any girl is friends with five boys. How many people are in the class? | Answer: 32
Solution. Let $\mathrm{m}$ be the number of boys, $\mathrm{d}$ be the number of girls, and $\mathrm{r}$ be the number of friendly pairs "boy-girl". According to the problem, $\mathrm{r}=3 \mathrm{~m}$ and $\mathrm{r}=5 \mathrm{~d}$. Therefore, r is divisible by 3 and 5, and thus by 15: $\mathrm{r}=15 \mathr... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,870 |
7.5 In the cells of a $3 \times 3$ square, Petya placed the numbers $1,2,3, \ldots, 9$ (each one only once), and then calculated the sums in each row, each column, and each of the two diagonals. Could these 8 sums be equal to $13,14, \ldots, 20 ?$ | Answer: No
Solution. Suppose they could. Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$ be the sums of the three rows. The sum $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}$ is the sum of all numbers in the square, i.e., $1+2+3+\ldots+9=45$. Similarly, if $\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}$ are th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,871 |
7.1. In the first grade, there are more than 10 but fewer than 15 children. On New Year's morning, Father Frost came to them with a bag of candies and distributed them equally to everyone, and found that he had 9 candies left. How many candies did each child receive if there were initially 207 candies in the bag? | Solution. 207-9=198 candies were given out by Father Frost. 198=18$\cdot$11 - factorization taking into account that one of the factors meets the condition of being between 10 and 15. Therefore, 11 children received 18 candies each.
Answer: 18 candies.
## Criteria:
7 points - complete solution;
6 points - factoriza... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,872 |
7.2. Find the GCD of all numbers obtained by all possible permutations of the digits of the number 202120222023
Solution. By the divisibility rule, all these numbers are divisible by 9 (the sum of the digits is 18). A sufficient condition to prove that there are no other numbers is that the difference between any two ... | Answer: 9.
## Criteria
7 points - complete solution;
3 points - GCD found, but not proven that there can be no other divisors. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,873 |
7.4. Several knights and liars are standing in a circle, all of different heights. Each of them said the phrase: "I am taller than exactly one of my neighbors." Could there have been exactly 2023 liars among those standing in the circle? (Knights always tell the truth, while liars always lie). | Solution. The "dwarfs" - those who are shorter than both of their neighbors, and the "giants" - those who are taller than both of their neighbors, lie. Dwarfs and giants alternate in the circle, so together they form an even number.
Answer: No.
## Criteria:
7 points - complete solution;
2 points - noted that around... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,875 |
7.5. On December 31, 2011, Yevgeny Alexandrovich's age coincided with the sum of the digits of his birth year. How old was Yevgeny Alexandrovich on December 31, 2014? Prove the uniqueness of the answer.
Solution. The maximum sum of the digits of the birth year can be $1+9+9+9=28$, and the minimum - 2. Therefore, Y.A. ... | Answer: 23 years.
## Criteria:
7 points - complete solution;
4 points - equations are correctly set up;
3 points - the earliest and latest year of birth are estimated;
2 points - noted that in each decade there is no more than one suitable year. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,876 |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,879 |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,880 |
Problem 5.5. From a $6 \times 6$ square grid, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
 in the squares so that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. The top-left corner cell of the $5 \times 5$ table will be called $A$, and the bottom-right corner c... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,884 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to the line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on the line $A L$. Carefully finding the intersection point of thes... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,885 |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,887 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,888 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,889 |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,890 |
Problem 9.7. Two parallel lines are drawn through points $A(0 ; 14)$ and $B(0 ; 4)$. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L-... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,892 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,893 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
Solution. The number of pins Marfa takes out on the $n$-th step is expressed by the linear function $2+k \cdot(n-1)$. Therefore, the inequalities $2+9 \cdot k>45$ and $2+14 \cdot k<90$ hold. From this, we have $\frac{43}{9}<k<\frac{44}{7}$, which gives the natural values $k=5$ and $k=6$. | 5,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,896 |
3.1. A tangent $FK$ is drawn to the circumcircle $K$ of triangle $FDC$, and $\angle KFC=58^{\circ}$. Points $K$ and $D$ lie on opposite sides of line $FC$, as shown in the figure. Find the acute angle between the angle bisectors of $\angle CFD$ and $\angle FCD$. Express your answer in degrees.
$\angle KFC$ is the angle between the chord and the tangent, which is equal to half the arc $FC$, just like the inscribed angle $\angle FDC$ ... | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,897 |
4.1. Among fifty consecutive natural numbers, exactly 8 are divisible by 7 without a remainder. What is the remainder when the eleventh number in the sequence is divided by 7?
# | # Answer: 3
Solution. The closest numbers that give the same remainder when divided by 7 differ by 7. Therefore, among fifty consecutive numbers, 6 groups of numbers with the same remainder are formed, each containing 7 numbers, and only one group containing 8 numbers, with the first and last numbers among them. Thus,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,898 |
5.1. For real numbers $a$ and $b$, it is known that $a b=5, \frac{1}{a^{2}}+\frac{1}{b^{2}}=0.6$. Write down all possible values of $a+b$. | Answer: $5, -5$ (All answers)
Solution. Transform the given conditions: $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{(a+b)^{2}-2 a b}{a^{2} b^{2}}=\frac{(a+b)^{2}-10}{25}=0.6$. Then $(a+b)^{2}=0.6 \cdot 25+10=25$. Therefore, the desired value is 5 or -5.
This situation is possible because we... | 5,-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,899 |
6.1. Four chess players - Ivanov, Petrov, Vasiliev, and Kuznetsov - played a round-robin tournament (each played one game against each other). A win earns 1 point, a draw earns 0.5 points each. It turned out that the player who took first place had 3 points, and the player who took last place had 0.5 points. How many v... | Answer: 36
Solution. First, let's determine what quantities of points the four participants can have under the given conditions. Since the one who took first place has 3 points, he won against everyone, and the one who took second place could not have scored more than 2, as he lost to the first. There were a total of ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,900 |
7.1. People stand in a circle - liars who always lie, and ryiars who always tell the truth. And each of them said that among the people standing next to them, there are as many liars as ryiars. How many people are there in total if there are 48 ryiars? | Answer: 72
Solution. Let's denote a knight as $\mathrm{P}$, and a liar as L. Notice that each knight stands between a knight and a liar, otherwise he would have told a lie. Therefore, knights stand in groups of two. Liars cannot stand in groups larger than one, as in such a case, the liar standing at the edge of the g... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,901 |
8.1. Parallelogram $A B C D$ was folded along the diagonal $B D$ so that vertex $C$ remained in place, and vertex $A$ took the position $A^{\prime}$. Segments $B C$ and $A^{\prime} D$ intersected at point $K$, and $B K: K C=3: 2$. Find the area of triangle $A^{\prime} K C$, if the area of parallelogram $A B C D$ is 27. | Solution. As a result of the action described in the condition, $\triangle A B D$ has undergone a symmetry with respect to the line $B D$ and has transformed into $\triangle A^{\prime} B D$. Then $A^{\prime} D = A D = B C$, $A^{\prime} B = A B = C D$. From this, $\triangle A^{\prime} B D = \triangle C D B$ by three sid... | 3.6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,902 |
1. For what values of $x$ and $y$ is the equality $x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3}$ true? | Solution. Expanding the brackets and grouping the terms by powers of the variable $x$, we obtain a quadratic equation in $x$ of the following form: $x^{2}-y x+\left(y^{2}-y+\frac{1}{3}\right)^{2}=0$. Then $D=y^{2}-4 y^{2}+4 y-\frac{4}{3}=-3 y^{2}+4 y-\frac{4}{3}=-3\left(y-\frac{2}{3}\right)^{2}$. The quadratic equation... | \frac{1}{3},\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,903 |
2. Among 81 coins, there is one counterfeit (lighter) coin. How to find it using no more than four weighings. | Solution. Divide the coins into three piles of 27 coins each. Weigh the first and second piles. If the scales are in balance, the counterfeit coin is in the third pile. If the scales are not in balance, the counterfeit coin is in the lighter pile. After this, divide the pile of 27 coins (which contains the counterfeit ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,904 |
3. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase? | Solution. Let the merchant pay $x$ rubles for the salt during the first purchase in Tver. Then he sold it in Moscow for $x+100$ rubles. The second time, he spent $x+100$ rubles in Tver and received $x+100+120=x+220$ rubles in Moscow. Since the ratio of Moscow and Tver prices did not change, we can set up the proportion... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,905 |
4. In a triangle, two medians are perpendicular to each other, their lengths are 18 cm and 24 cm. Calculate the area of the triangle. | Solution. $S=\frac{1}{2} A C \cdot B M$ (see the figure). By the property of medians, we have $A O=\frac{2}{3} \cdot A E=16, O C=\frac{2}{3} \cdot C D=12$. From $\triangle A O C$, we find $A C=\sqrt{A E^{2}+O C^{2}}=20$. Let $D F \perp A C, O K \perp A C$,
 \cdot(3!\cdot 4!) \cdot \ldots \cdot(19!\cdot 20!)= \\
= & (1!\cdot 1!\cdot 2) \cdot(3!\cdot 3!\cdot 4) \cdot(5!\cdot 5!\cdot 6) \cdot \ldots \cdot(17!\cdot 17!\cdot 18) \cdot(19!\cdot 19!\cdot 20)= \\
= & (1!)^{2} \... | 10! | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,907 |
3. We will build a sequence of numbers in the following way. We will place the number 7 in the first position, and after each number, we will place the sum of the digits of its square, increased by one. For example, the second position will hold the number 14, since $7^{2}=49$, and $4+9+1=14$. The third position will h... | Solution Let's continue finding the first few terms of the sequence: $7 ; 14 ; 17 ; 20 ; 5 ; 8 ; 11 ; 5 ; \ldots$ - the number 5 is repeated. This means the sequence has a period of length 3: the numbers 5; 8; 11 will repeat subsequently. The number 8 is in the sixth position, so for any $\mathrm{k}>0$, the number 8 wi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,908 |
4. Thirty students from five classes came up with 40 tasks, and students from one class came up with the same number of tasks, while students from different classes came up with a different number. How many students came up with one task? | Solution: We will choose 5 students, one from each class. Each of them came up with a different number of problems. Therefore, the total number of problems they came up with is no less than $1+2+3+4+5=$ 15. The remaining 25 students came up with no more than $40-15=25$ problems. It is clear that each of them came up wi... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,909 |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and there are no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as show... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,913 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,914 |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,915 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Let's add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
To the obtained sum, add the distance from Asey to Bory: $32+8=40$. Now all three branches (to Gala - twice) and the main... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,916 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,917 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,918 |
Problem 5.7. After the World Hockey Championship, three journalists wrote an article about the German national team - each for their newspaper.
- The first wrote: “The German national team scored more than 10 but fewer than 17 goals throughout the championship.”
- The second: “The German national team scored more than... | Answer: $11,12,14,16$ or 17.
Solution. Let's consider all possible cases.
If no more than 10 goals were scored, then for an odd number of goals, one statement would be true, and for an even number, none would be true. This does not fit.
If 11 goals were scored, then exactly two statements are true. This fits.
If th... | 11,12,14,16,17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,919 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,920 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,923 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,925 |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,926 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,927 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,928 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,929 |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,930 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,932 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,933 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,934 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
(y+s)=(x+t)(y+t)$
$x y+x s+y s+s^{2}=x y+x t+y t+t^{2}$,
$x s-x t+y s-y t+s^{2}-t^{2}=0$
$x(s-t)+y(s-y)+(s+t)(s-t)=0$,
$(s-t)(x+y+s+t)=0$.
Since $\mathrm{s} \neq \mathrm{t}$, then $\mathrm{x}+\mathrm{y}+\mathrm{s}+\mathrm{t}=0$. | x+y++=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,936 |
9.2 Points A, B, C, D are consecutive vertices of a regular n-gon. What is n if $\angle \mathrm{ACD}=120^{\circ}$? | Solution. We consider the given n-gon as inscribed.
Each small arc between adjacent vertices is $\frac{1}{\mathrm{n}} \cdot 360^{\circ}$. The angle $\mathrm{ACD}$ is an inscribed angle and s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,937 |
9.3 Let a, b, c be non-negative integers such that $28a + 30b + 31c = 365$. Prove that $a + b + c = 12$. | Solution. If it were $\mathrm{a}+\mathrm{b}+\mathrm{c} \leq 11$, then we would have $28 \mathrm{a}+30 \mathrm{~b}+31 \mathrm{c} \leq 31(\mathrm{a}+\mathrm{b}+\mathrm{c}) \leq 31 \cdot 11 < 365$. Therefore, $\mathrm{a}+\mathrm{b}+\mathrm{c} \geq 12$. We will show that the inequality $\mathrm{a}+\mathrm{b}+\mathrm{c} \ge... | 12 | Number Theory | proof | Yes | Yes | olympiads | false | 17,938 |
9.4 Prove that if $\mathrm{m}$ and $\mathrm{n}$ are natural numbers, $\mathrm{m}<\mathrm{n}$, then
$$
m^{2}+\sqrt{m^{2}+m}<n^{2}-\sqrt{n^{2}-n}
$$ | Solution. We need to prove the inequality
$$
n^{2}-m^{2}>\sqrt{n^{2}-n}+\sqrt{m^{2}+m}
$$
Note that the condition $m>n$ with natural $m$ and $n$ gives the inequality $\mathrm{n}-\mathrm{m} \geq 1$. Therefore, $\mathrm{n}^{2}-\mathrm{n} \geq \mathrm{m}^{2}+m$ (indeed, $\mathrm{n}^{2}-\mathrm{m}^{2}-(\mathrm{n}+\mathrm... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,939 |
9.5 Petya and Vasya take turns breaking a stick: first Petya - into two parts (possibly unequal), then Vasya - one of the resulting parts into two, Petya - one of the three parts into two, and so on. The winner is the one who, after their move, can choose four sticks from all the available parts whose lengths form an a... | Solution. The first player, i.e., Petya, wins by acting as follows. His first move is to break the stick in half - resulting in two sticks, each of length, say, $2 \ell$. The second move is Vasya's, he will break one of the sticks of length $2 \ell$ into two - of lengths $\ell-\mathrm{p}$ and $\ell+\mathrm{p}$. The nex... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,940 |
Problem 7.1. In class 7 "B", there are more than 3 but less than 15 children. On New Year's, Santa Claus came to them with a bag containing 195 candies. After distributing the candies equally among all the children in the class, Santa Claus found that there were 8 candies left in the bag. How many candies did each chil... | Answer: 17.
Solution. The children received $195-8=187=11 \cdot 17$ candies. This number must be divisible by the number of children in the class, which is more than 3 and less than 15. Therefore, there are 11 children in the class, and each of them received $187: 11=17$ candies. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,941 |
Problem 7.3. In the example with fractions, some two-digit natural numbers have been replaced by the letters $A$ and $B$.
$$
\frac{A-5}{A}+\frac{4}{B}=1
$$
(a) (2 points) What is the smallest value that $A$ can take?
(b) (2 points) What is the largest value that $B$ can take? | # Answer:
(a) (2 points) 15.
(b) (2 points) 76.
Solution. According to the condition,
\[
1=\frac{A-5}{A}+\frac{4}{B}=1-\frac{5}{A}+\frac{4}{B}
\]
from which we obtain \(\frac{A}{5}=\frac{B}{4}\) and \(4A=5B\). It follows that for some integer \(k\), the equalities \(A=5k\) and \(B=4k\) hold.
Since \(B=4k \geqslan... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,943 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.