problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
Problem 9.6. (20 points) There were $n$ identical-looking coins weighing $x_{1}, x_{2}, \ldots, x_{n}$ grams (the weights of the coins are pairwise distinct positive real numbers), and also weightless stickers with numbers $x_{1}, x_{2}, \ldots, x_{n}$. At night, a lab assistant weighed the coins and labeled them with ... | Answer: 2 yes
Solution. Consider the weights 10, 20, 25, 30, 35, 201, 203, 207 (weights will be measured in grams hereafter). We will perform two checks:
$$
\begin{aligned}
10+20+30 & =25+35 \\
10+25+201 & <30+207
\end{aligned}
$$
First, consider the first weighing. We will prove that if some three coins balance som... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Kolya came up with an entertainment for himself: he rearranges the digits in the number 2015, then places a multiplication sign between any two digits. In this process, none of the resulting two factors should start with zero. Then he calculates the value of this expression. For example: $150 \cdot 2=300$, or $10 \c... | Answer: 1050
Solution. Note three facts (their justification is obvious). First, in the case of the maximum product, the digits in each of the factors are in descending order. That is, 0 should be at the end of one of the factors. Second, it does not matter at the end of which of the two factors to place 0, the produc... | 1050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha - 7 matches... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The numbers $x$ and $y$ are such that $x+y=xy=17$. Find the value of the expression:
$$
\left(x^{2}-17 x\right)\left(y+\frac{17}{y}\right)
$$ | Answer: -289
Solution. $x-17=-y, \frac{17}{y}=x$. The desired expression is:
$$
x(x-17)(y+x)=-x y \cdot 17=-289 .
$$
Criteria.
$(-)$
$(-$.
(-/+) Serious arithmetic error affecting the solution process, or several arithmetic errors.
$(+/ 2)$
(+/-) One arithmetic error in an otherwise correct solution (for exampl... | -289 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 4. Problem $4.1 *$
In the class, 6 students received a grade of 5, 7 received a grade of 4, and 1 received a grade of 3. The teacher told them to form pairs with different grades, where the student with the better grade would explain to the student with the worse grade where they made a mistake.
In how many ways co... | # 4. Problem $4.1 *$
In the class, 6 students received a grade of 5, 7 received a grade of 4, and 1 received a grade of 3. The teacher told them to form pairs with different grades, where the student with the better grade would explain to the student with the worse grade where they made a mistake.
In how many ways co... | 5040 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Kolya came up with an entertainment for himself: he rearranges the digits in the number 2015, then places a multiplication sign between any two digits and calculates the value of the resulting expression. For example: $150 \cdot 2=300$, or $10 \cdot 25=250$. What is the largest number he can get as a result of such ... | Answer: 1050
Solution. Note three facts (their justification is obvious). First, in the case of the maximum product, the digits in each of the factors are in descending order. That is, 0 should be at the end of one of the factors. Second, it does not matter at the end of which of the two factors to place 0, the produc... | 1050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers $x$ and $y$ are such that $x+y=xy=17$. Find the value of the expression:
$$
\left(x^{2}-17 x\right)\left(y+\frac{17}{y}\right)
$$ | Answer: -289
Solution. $x-17=-y, \frac{17}{y}=x$. The desired expression is:
$$
x(x-17)(y+x)=-x y \cdot 17=-289
$$
Criteria.
$(-)$
$(-$.
(-/+) Serious arithmetic error affecting the solution process, or several arithmetic errors.
$(+/ 2)$
(+/-) One arithmetic error in an otherwise correct solution.
(+.) Omissi... | -289 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Problem 2
Angles $A, B, C$ of triangle $A B C$ are $44^{\circ}, 66^{\circ}$, and $70^{\circ}$ respectively. The bisector of angle $A B C$ and the perpendicular bisector of side $A C$ intersect at point $D$. How many degrees is angle $A D C$? (Write the answer without the degree symbol.)
# | # 2. Task 2
Angles $A, B, C$ of triangle $A B C$ are $44^{\circ}, 66^{\circ}$, and $70^{\circ}$ respectively.
The bisector of angle $A B C$ and the perpendicular bisector of side $A C$ intersect at point $D$. How many degrees does angle $A D C$ measure? (Write the answer without the degree symbol.)
## Answer: 114 | 114 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. In the expression
$$
(* *+*)(* *+*)=* * * *
$$
insert digits instead of asterisks so that a correct equality is obtained and no more than 4 different digits are used. (A number cannot start with zero). | Solution. For example, $(90+9)(10+1)=1089$. Or: $(99+1)(10+9)=1900$. Criteria.
(+.) Correct left side.
$(+)$ Correct answer. | 1089 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11-2. Mr. $A$ stood at the point with coordinates $(0,0)$ for an hour. During the same hour, moving uniformly and in a straight line, Mr. $B$ traveled from the point $(22,0)$ to the point $(2,20)$. During this same hour, Miss $C$, also moving uniformly and in a straight line, traveled from the point $(30,4)$ to... | Answer: 53.
Solution. Informally speaking, the problem in this task is not to find a path of calculations leading to the answer; but to find a path to the answer that passes through not too many intermediate calculations. Let's show how to do this.
As is known, the area of a triangle formed by vectors $\overrightarro... | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the expression
$$
(* *+*)(* *+*)=* * * *
$$
insert digits instead of asterisks so that a correct equality is obtained and no more than 4 different digits are used. (A number cannot start with zero). | Solution. For example, $(90+9)(10+1)=1089$. Or: $(99+1)(10+9)=1900$.
Criteria.
(+.) Correct left side.
$(+)$ Correct answer. | 1089 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.6. The altitudes of an acute-angled scalene triangle $ABC$ intersect at point $H$. $I$ is the incenter of triangle $ABC$, $O$ is the circumcenter of triangle $BHC$. It is known that point $I$ lies on the segment $OA$. Find the angle $BAC$. | Answer: $60^{\circ}$
Solution. Note that point $O$ lies on the perpendicular bisector of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $A B C$, and dividing the arc $B C$ in half. Therefore, point $O$ l... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. Point $B$ is the midpoint of segment $A C$. Square $A B D E$ and equilateral triangle $B C F$ are located in the same half-plane relative to line $A C$. Find (in degrees) the measure of the acute angle between lines $C D$ and $A F$. | Answer: $75^{\circ}$.
Solution. Note that points $A D F C$ lie on the same circle with center at point $B$. The arc $F C$ is $60^{\circ}$. Therefore, the angle $F A C$ is $30^{\circ}$ (by the inscribed angle theorem). Triangle $B D C$ is isosceles with a right angle at vertex $B$. Therefore, the angle $D C A$ is $\pi ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. The lines containing the altitudes of the non-isosceles triangle \(ABC\) intersect at point \(H\). \(I\) is the incenter of triangle \(ABC\), \(O\) is the circumcenter of triangle \(BHC\). It is known that point \(I\) lies on the segment \(OA\). Find the angle \(BAC\). | Answer: $60^{\circ}$
Solution. Note that point $O$ lies on the bisector (perpendicular through the midpoint) of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $B A C$, and dividing the arc $B C$ in half.... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.6. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2}, \f... | Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n-1} \subset S_{n}$, then $a_{n-1} \geq a_{n}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$. Furt... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. (15 points) Find the smallest ten-digit natural number, all digits of which are different, such that when all even digits are erased, 97531 remains, and when all odd digits are erased, 02468 remains. | Answer: 9024675318.
Solution. Let $A=9024675318$. Consider an arbitrary number $N=\overline{a_{1} a_{2} \ldots a_{10}}$ that satisfies the conditions of the problem. Clearly, $a_{1}=9$, since $a_{1} \neq 0$. Let $7=a_{k}$ for some $k$. We will consider several cases.
- If $k>6$, then $N=9024687531>A$.
- If $kA$.
- If... | 9024675318 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. (15 points) Given an isosceles triangle $A B C(A B=B C)$. On the sides $A B, B C, A C$ points $K, L, M$ are marked respectively such that $\angle A K M=90^{\circ}, \angle B L K=90^{\circ}$ and $K M=K L$. What is the measure of angle $C M L ?$ | Answer: $90^{\circ}$.
Solution. From the condition, it follows that $\angle M K L + \angle L K B = 90^{\circ} = \angle L K B + \angle L B K$, from which $\angle M K L = \angle L B K$ (Fig. 1). Triangles $A B C$ and $M K L$ are isosceles with equal angles at the vertex, so the base angles are also equal: $\angle K L M ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. (15 points) Several boxes are stored in a warehouse. It is known that there are no more than 60 boxes, and each of them contains either 59 apples or 60 oranges. After a box with a certain number of oranges was brought to the warehouse, the number of fruits in the warehouse became equal. What is the smalles... | Answer: 30.
Solution. Let the number of boxes with oranges be denoted by $n$, the number of boxes with apples by $m$, and the number of oranges in the brought box by $x$. According to the problem, $59 m = 60 n + x$. By moving $59 n$ to the left side, we get $59(m - n) = n + x$. From this, it follows that $n + x$ is a ... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. (15 points) On an island, there live knights who always tell the truth, and liars who always lie. One day, 100 residents of this island lined up, and each of them said one of the following phrases:
- "To the left of me, there are as many liars as there are knights."
- "To the left of me, there are 1 more ... | Answer: 50.
Solution. Suppose there are no fewer than 51 knights in the row. Consider the rightmost of them. To his left, there are at least 50 knights and no more than 49 liars, so he could not have said any of the listed phrases, leading to a contradiction. Therefore, the total number of knights cannot exceed 50.
N... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. On side $AB$ of triangle $ABC$, a point $D$ is taken. In triangle $ADC$, the angle bisectors $AP$ and $CQ$ are drawn. On side $AC$ of triangle $ADC$, a point $R$ is taken such that $PR \perp CQ$. It is known that the angle bisector of angle $D$ of triangle $BCD$ is perpendicular to segment $PB$, $AB=18$, $AP=12$. Fi... | Solution. Let $A B=n, A P=m, A D=c, A C=b, D P=p, C P=q, B D=D P=p, C P=C R=q$.
Using the property of the angle bisector and the formula for the length of the angle bisector, we get $\left\{\begin{array}{l}\frac{c}{b}=\frac{p}{q}, \\ c+p=A B=n, \\ m^{2}=b c-p q,\end{array}\right.$
$\left\{\begin{array}{l}c=\frac{b p}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. ... | Answer: 12 points.
## Solution. FIRST WAY. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? (Enter the year.) Points for the task: 8.
# | # 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? (Enter the year.) Points for the task: 8.
## Answer: 2048 | 2048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Problem 7*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Find ... | # 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8. Problem 8.10
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
# | # 8. Problem 8.10
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without the unit of measurement).
## Answer: 38 | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 9. Problem 9.10
In a convex 10-gon \(A_{1} A_{2} \ldots A_{10}\), all sides and all diagonals connecting vertices every other one (i.e., \(A_{1} A_{3}, A_{2} A_{4}\), etc.) are drawn, except for the side \(A_{1} A_{10}\) and the diagonals \(A_{1} A_{9}, A_{2} A_{10}\).
We call a path leading from \(A_{1}\) to \(A_{... | # 9. Problem 9.10
In a convex 10-gon $A_{1} A_{2} \ldots A_{10}$, all sides and all diagonals connecting vertices every other one (i.e., $A_{1} A_{3}, A_{2} A_{4}$, etc.) are drawn, except for the side $A_{1} A_{10}$ and the diagonals $A_{1} A_{9}, A_{2} A_{10}$.
A path from $A_{1}$ to $A_{10}$ is defined as a non-se... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | # 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? Enter the year. Points for the task: 8.
## Answer: 2048 | # 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? Enter the year. Points for the task: 8.
## Answer: 2048 | 2048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | # 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Each move of a chess knight is a move of one square horizontally and two squares vertically, or vice versa - one square vertically and two squares horizontally. (In the diagram on the right, the knight marked with the letter $\mathrm{K}$ can move to any of the shaded squares in one move.)
A chess knight is placed i... | Answer: 2
Solution. Let's number the cells as shown in Figure 1. The knight can move from any cell to any cell with the same number, and cannot move to other cells.
| $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (based on materials from the Ural Tournaments)
In triangle $ABC$, angle $B$ is $80^{\circ}$. On side $BC$, point $D$ is marked such that $AB = AD = CD$. On side $AB$, point $F$ is marked such that $AF = BD$. On segment $AC$, point $E$ is marked such that $AB = AE$. Find angle $AEF$.
 $n=404$? b) $n=406... | Answer: 202 colors in both parts. Solution:
Example for part a) with 202 colors:
Let's call a 198-block a sequence of 198 consecutive houses, where any pair of houses 99 apart is painted in a unique color (i.e., among all 404 houses, there are no other houses of that color).
Let's call a 2-block a pair of houses pai... | 202 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (Folklore)
There are balls in three boxes. In the first - red ones, in the second - white ones, in the third - both red and white balls. Each box has a label: "red," "white," "mixed," but it is known that none of the labels correspond to the actual contents. Seventh-grader Sergey wants to find out which b... | Answer: Yes. Solution:
Take from the box "mixed":
Red $\Rightarrow$ in the box with mixed balls red $\Rightarrow$ in the box with white balls not white and not red (then mixed) $\Rightarrow$ in the box with red balls white
Criteria: Correct solution: 15 points (+)
Non-obvious and unclear consequence of the form “in... | 432 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
26. The demand and supply functions for a good in the market are linear. Initially, equilibrium in the market was established at a price of $14 per unit of the good and a quantity of 42 units. The government decided to support producers: for this purpose, it buys any quantity of the good that producers are willing to s... | Solution: demand has increased, it is necessary to find a new demand function from two points: $p 1=20$, $q 1=42+12=54 ; \quad p 2=29, \quad q 2=0 . \quad p=a-b q, \quad b=(29-20) / 54=1 / 6 ; \quad a=29$. The new demand function is $p=29-1 / 6 q: p=14, q=90$. The state will purchase 90-42=48 units | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
27. On a perfectly competitive market for a good, the market demand and market supply functions are given respectively as: $Q^{d}=60-14 P$ and $Q^{s}=20+6 P$. It is known that if the amount of labor used by a typical firm is $\mathbf{L}$ units, then the marginal product of labor is $\frac{160}{L^{2}}$. Determine how ma... | Solution: find the equilibrium price of the good $60-14 P=20+6 P, P=2$. Optimal choice of the firm: MPL $x P=w ; \frac{160}{L^{2}} \times 2=5 ; \boldsymbol{L}=\boldsymbol{8}$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. ... | Answer: 12 points.
## Solution. FIRST WAY. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. a) Find at least two different natural numbers $n$, for each of which the number $n^{2}+2015 n$ is a perfect square of a natural number.
b) Find the number of all natural numbers $n$, for each of which the number $n^{2}+$ $2015 n$ is a perfect square of a natural number. | Answer. a) For example $n=496$ and $n=1007^{2}=1014049$. b) 13.
## Solution. FIRST METHOD.
Let $n^{2}+2015 n=m^{2}$. Denote $d=$ GCD $(n, 2015), n=d \nu, 2015=d r$. Then $\nu$ and $r$ are coprime, and
$$
d^{2} \nu(\nu+r)=m^{2}
$$
Since $\nu$ and $\nu+r$ are coprime, each of them must be a square of a natural number... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) In triangle $A B C$, a square $K L M N$ with side length 1 is inscribed: points $K$ and $L$ lie on side $A C$, points $M$ and $N$ lie on sides $A B$ and $B C$ respectively. The area of the square is half the area of the triangle. Find the length of the height $B H$ of triangle $A B C$. | Answer: 2.
Solution 1:
$A C \cdot B H=2 S_{A B C}=4 S_{K L M N}=4$.
Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute. From the condition, it follows that
$$
\begin{gathered}
S_{A M L}+S_{C K N}+S_{B M N}=S_{K L M N} \\
\frac{A L \cdot 1}{2}+\frac{C K \cdot ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) The rivers Stuka and Turka merge into the river Stukatura at some point. Cities $A$ and $B$ are located on the rivers Stuka and Turka, respectively, with city $A$ being twice as far from the confluence as city $B$. A steamboat takes the same amount of time to travel from $A$ to $B$ via these rivers as it ... | # Solution:
Let the distance from the confluence of the rivers to city $B$ be $S$, the own speed of the steamboat be $x$, and the speeds of Shchuk and Turk be $u$ and $v$ respectively. We form the equation:
$$
\frac{2 S}{x+u}+\frac{S}{x-v}=\frac{S}{x+v}+\frac{2 S}{x-u}
$$
By canceling $S$ and moving the terms relate... | 92 | Algebra | proof | Yes | Yes | olympiads | false |
3. Natural numbers $x_{1}, x_{2}, \ldots, x_{13}$ are such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{13}}=2$. What is the minimum value that the sum of these numbers can take? | Answer: 85
# Examples of answer recording: 14
# | 85 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Between cities A and B, there are three buses: a regular one, which stops in cities V, G, and D (in that order); a fast one, which stops only in city $\Gamma$; and an express, which does not stop anywhere along the way.
All buses travel on different roads at a constant speed and take an integer number of hours to t... | Answer: 17
## Examples of answer notation:
14
# | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of solutions of the equation $x y + 5 x + 7 y = 29$ in integers (i.e., the number of pairs of integers (x, y) that satisfy the given equality). | Answer: 14
## Examples of answer notation:
14
# | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangle $\mathrm{ABCD}$. The length of side $\mathrm{BC}$ is one and a half times less than the length of side $\mathrm{AB}$. Point $\mathrm{K}$ is the midpoint of side $\mathrm{AD}$. Point $\mathrm{L}$ on side $\mathrm{CD}$ is such that $\mathrm{CL}=\mathrm{AK}$. Point $\mathrm{M}$ is the intersection of ... | Answer: 8
## Examples of answer recording: 14 0.5
# | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In a watch repair shop, there is a certain number of electronic watches (more than one), displaying time in a 12-hour format (the number of hours on the watch face changes from 1 to 12). All of them run at the same speed, but show completely different times: the number of hours on the face of any two different watch... | Answer: 11
## Examples of answer notation:
14
# | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. A $10 \times 10$ grid is filled with non-negative numbers. It is known that in each (vertical or horizontal) strip of 1x3, the sum of the numbers is 9. What is the maximum value that the sum of all numbers in the grid can take? | Answer: 306
## Examples of answer recording: 14
# | 306 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Six children are sledding in a train formation down a hill. In how many different ways can they slide down if one of the children, who has peculiarities, believes that it is contraindicated for him to sit in even-numbered positions. | Answer: 360
## Examples of answer recording:
14
# | 360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is a parallelogram with an area of $120 . \mathrm{K}$ is the midpoint of side $\mathrm{AD}, \mathrm{L}$ is the midpoint of $\mathrm{CD}$. Find the area of triangle BKL. | Answer: 45
## Examples of answer notation:
14 | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Natural numbers a and b are such that 5 LCM $(a, b)+2$ GCD $(a, b)=120$. Find the greatest possible value of the number a. | Answer: 20
## Examples of answer notation:
14 | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. (3 points)
In the country of Taxia, everyone pays as many percent in taxes as thousands of tugriks their salary amounts to. What salary is the most advantageous to have?
(Salary is measured in a positive, not necessarily integer number of tugriks) | Answer: 50000 tugriks
Solution:
Let's denote the salary as $x$. Then what remains after the tax deduction is $x-\frac{x^{2}}{100000}$. This is a quadratic trinomial with a negative leading coefficient, it reaches its maximum at the vertex when $x=50000$. | 50000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
For each pair of numbers $\overline{a b b}$ and $\overline{a b b}$, where $a$ and $b$ are different digits, the GCD of these numbers was calculated. Find the greatest of these GCDs.
$\overline{a b b}$ is the standard notation for a number consisting of the digits $a, b$, and $b$ in that exact ... | Solution:
45 is the greatest common divisor (GCD) of the numbers 585 and 855.
Note that \(a\) and \(b\) are not equal to 0, as numbers cannot start with 0.
Suppose there are two numbers for which this GCD is greater than 45. Note that \(\overline{b a b} - \overline{a b b} = 90(b - a)\) is divisible by this GCD. Divi... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. (3 points)
In the country of Taxia, everyone pays as many thousandths of their salary in taxes as tugriks the salary amounts to. What salary is the most advantageous to have?
(Salary is measured in a positive, not necessarily integer number of tugriks) | Answer: 500 tugriks
Solution:
Let the salary be $x$. Then what remains after the tax deduction is $x-\frac{x^{2}}{1000}$. This is a quadratic trinomial with a negative leading coefficient, it reaches its maximum at the vertex when $x=500$. | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
In triangle $A B C$, the median $B M$ is drawn, in triangle $M C B$ - the median $B N$, in triangle $B N A$ - the median $N K$. It turned out that $N K \perp B M$. Find $A C: B C$.
Answer: 2
# | # Solution:
Let $\vec{b}=\overrightarrow{C B}$ and $\vec{a}=\overrightarrow{C A}$.
$\overrightarrow{N K}=\overrightarrow{C K}-\overrightarrow{C N}=\frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}}{4}=\frac{\vec{a}+2 \vec{b}}{4}$.
$\overrightarrow{B M}=\overrightarrow{C M}-\overrightarrow{C B}=\frac{\vec{a}}{2}-\vec{b}$
$0=\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
For each pair of numbers $\overline{a b b}$ and $\overline{a b a}$, where $a$ and $b$ are different digits, the GCD of these numbers was calculated. Find the greatest of these GCDs.
$\overline{a a b}$ - standard notation for a number consisting of digits $a, a$ and $b$ in exactly that order.
... | Solution:
18 is the greatest common divisor (GCD) of the numbers 828 and 882.
Suppose there are two numbers for which this GCD is greater than 18. Note that $\overline{a a b}-\overline{a b a}=9(a-b)$ is divisible by this GCD. $|a-b| \leqslant 9$ so for the GCD to be greater than 9, it must contain factors from both t... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On side $AB$ of triangle $ABC$, point $D$ is marked, and on side $BC$, point $E$ is marked. Segments $CD$ and $AE$ intersect at point $L$. It turns out that $CL = CE$, $AL = DL$. The angle $ACE$ is 24 degrees. Find the degree measure of angle $ABC$. | Answer: 27
## Examples of answer notation:
45
# | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In the store, they sell bags of apples weighing 3 kg (one bag costs 20 rubles), bags of pears weighing 4 kg (one bag costs 35 rubles), and bags of plums weighing 5 kg (one bag costs 50 rubles). Anya has 155 rubles, what is the maximum number of kg of fruit she can buy? | Answer: 22
## Examples of answer notation:
11
# | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 2. (3 points)
In a certain company, the $20 \%$ most useful employees perform $80 \%$ of the work. What is the smallest percentage of work that the $40 \%$ most useful employees can perform?
We will consider an employee more useful if they perform more work. | Answer: $85 \%$
## Solution:
$40 \%$ of the most useful employees are divided into $20 \%$ who perform $80 \%$ of the work, and the next $20 \%$ who are part of the $80 \%$ performing the remaining $20 \%$. These second $20 \%$ make up a quarter of the $80 \%$. Since they are the most useful among these $80 \%$, they... | 85 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
30 people are standing in a row, each of them is either a knight, who always tells the truth, or a liar, who always lies. They were numbered from left to right, after which each person with an odd number said: "All people with higher numbers than mine are liars," and each person with an even nu... | # Answer: 28
## Solution:
We will prove that under the given conditions, there must be exactly two knights, and all others are liars.
Consider the people with odd numbers. If a person with an odd number $n$ is telling the truth, then the person with the odd number $n+2$ must also be telling the truth, since all peop... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
Three runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.
At some point, the first runner meets the second. After 15 minutes, the second runner meets the third for the first time. Anothe... | # Answer: 80
## Solution:
Let the first runner meet the second, then after $a$ minutes the second runner meets the third for the first time, and after another $b$ minutes the third runner meets the first for the first time.
Let the first and second runners meet at point $A$, the second and third at point $B$, and th... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
Natural numbers $a, b, c$ are such that $\operatorname{GCD}(\operatorname{LCM}(a, b), c) \cdot \operatorname{LCM}(\operatorname{GCD}(a, b), c)=200$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(a, b), c)$ can take? | Answer: 10
## Solution:
Note that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$200=2^{3} \cdot 5^{2}... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The three heights of a tetrahedron are three, four, and four times greater than the radius of its inscribed sphere, respectively. How many times greater is the fourth height than the radius of the inscribed sphere? | Answer: 6
## Examples of answer notations:
14
$1 / 4$
# | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The AC-2016 calculator can perform two operations: taking the cube root and calculating the tangent. Initially, the number $2^{-243}$ was entered into the calculator. What is the minimum number of operations required to obtain a number greater than 1? | Answer: 7
## Examples of answer notation:
## 10 | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the country, there are 9 cities, some of which are connected by postal flights. To send a letter from one city to another, you need to stick as many stamps on it as the number of flights required (the route with the fewest flights is used). It is known that even if two cities are not connected by a flight, it is ... | Answer: 240
## Examples of answer recording:
1000
# | 240 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Dima took the fractional-linear function $\frac{a x+2 b}{c x+2 d}$, where $a, b, c, d-$ are positive numbers, and added it to all the remaining 23 functions that result from it by permuting the numbers $a, b, c, d$. Find the root of the sum of all these functions, independent of the numbers $a, b, c, d$. | Answer: -1
## Examples of answer notation:
14
$1 / 4$
$-1.4$
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the last digit of the integer part of the number $(\sqrt{37}+\sqrt{35})^{2016}$. | Answer: 1
## Examples of answer recording:
## 0
# | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A cube is circumscribed around a sphere of radius 1. From one of the centers of the cube's faces, vectors are drawn to all other centers of the faces and vertices. The scalar products of each pair of different vectors were calculated, a total of 78. What is the sum of these scalar products? | Answer: 76
## 10th grade
# | 76 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For the function $\mathrm{f}(\mathrm{x})$, the condition $\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))+3 \mathrm{f}(\mathrm{f}(\mathrm{x}))+9 \mathrm{f}(\mathrm{x})+27 \mathrm{x}=0$ is satisfied. Find $\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(2))))$. | Answer: 162
## Examples of how to write answers:
14
$1 / 4$
$-0.25$
# | 162 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
Petya wrote the number 11234567 on the board, and then all the numbers obtained from it by rearranging the digits, in ascending order. What was the position of the number $46753211$? | Answer: 12240
## Solution:
The number of a number is the quantity of numbers that are listed no later than it. The numbers listed no later than 46753211 are:
(1) Numbers starting with 1. In this case, the remaining digits can be arranged in 7! = 5040 ways.
(2) Numbers starting with 2 or 3. For each variant of the f... | 12240 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
Circles $O_{1}, O_{2}$, and $O_{3}$ are located inside circle $O_{4}$ with radius 6, touching it internally, and touching each other externally. Moreover, circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$. Find the radius of circle $O_{3}$.
# | # Answer: 2
## Solution:
We will solve the problem in general for all cases, specifically proving that if the radius of circle $O_{4}$ is $R$, then the radius of circle $O_{3}$ is $\frac{R}{3}$.
Since circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$ and are internally tangent to it, for each of ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
In a $7 \times 7$ table, some cells are black, and the rest are white. In each white cell, the total number of black cells on the same row or column is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take?
# | # Answer: 168
## Solution:
The number in the white cell consists of two addends: a "horizontal" and a "vertical" one. Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the ... | 168 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a regular hexagon $A B C D E F$, with side $10 \sqrt[4]{27}$. Find the area of the union of triangles ACE and BDF.
 | Answer: 900
## Examples of how to write answers:
14
$1 / 4$
1.4
# | 900 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given triangle $\mathrm{ABC}: \mathrm{BK}, \mathrm{CL}$ - angle bisectors, M - the point of their intersection. It turns out that triangle $\mathrm{AMC}$ is isosceles, one of whose angles is 150 degrees. Find what the perimeter of triangle $\mathrm{ABC}$ can be, if it is known that $\mathrm{BK}=4-2 \sqrt{3}$. | Answer: 4
## Examples of how to write answers:
14
$1 / 4$
0.25
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of solutions in natural numbers for the equation $(x-4)^{2}-35=(y-3)^{2}$. | Answer: 3
## Examples of answer notation:
14
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At the hitmen convention, 1000 participants gathered, each receiving a registration number from 1 to 1000. By the end of the convention, it turned out that all hitmen, except number 1, were killed. It is known that each hitman could only kill hitmen with higher numbers, and the number of his victims could not exceed... | Answer: 10
## Examples of answer notation:
5
# | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In the country of Aviania, there are 40 cities, some of which are connected by two-way flights. Moreover, between any two cities, there is only one reasonable air route (i.e., a route where the same flight is not used in different directions).
For each city, the air distance to the capital was calculated. It is cal... | Answer: 780.
## Examples of answer recording:
45 | 780 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a regular 11-gon. In how many ways can 3 diagonals be drawn in it so that every two of them intersect (inside the 11-gon)? | Answer: 462
## Examples of answer recording:
45
## Problem 6 (3 points). | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Anya and Kolya were collecting apples. It turned out that Anya collected as many apples as the percentage of the total apples collected by Kolya, and vice versa, Kolya collected as many apples as the percentage of the total apples collected by Anya. How many apples did Anya collect and how many did Kolya ... | Answer: 50 apples each.
Solution: If we add the two given conditions, it turns out that the total number of apples equals the total percentage. Therefore, Anya and Kolya collected a total of 100 apples. If one of them had collected more apples, they would have also collected more percentages. On the other hand, the co... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) Point $C$ is located on the segment $A E$. On one side of the line $A E$, points $B$ and $D$ are marked such that $A B C$ is an equilateral triangle, and $C D E$ is an isosceles right triangle with a right angle at $D$. It turns out that triangle $B C D$ is isosceles with base $B C$. Find the angle $A D E... | # Answer: $105^{\circ}$
Solution: The acute angles of triangle $C D E$ are each $45^{\circ}$, and the angles of triangle $A B C$ are each $60^{\circ}$. Therefore, angle $B C D$ is $75^{\circ}$. Consequently, $\angle C B D=75^{\circ}, \angle C D B=30^{\circ}$.
The altitudes of triangles $A B C$ and $B C D$, dropped to... | 105 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Given the puzzle: AB + BC + DE = FGH. Different letters represent different digits, no digit is equal to 9, and a number cannot start with 0. Find the smallest possible value of FGH. | Answer: 108.
## Solution:
The sum of all the digits at our disposal is 36. Therefore, AB $+\mathrm{B} \Gamma+\mathrm{DE}+$ ZHI is divisible by 9. But AB + BG + DE + ZHI $=2$ ZHI, so ZHI is divisible by 9.
The smallest three-digit number divisible by 9 is 108. Since, for example, $108=25+36+$ 47, this number fits our... | 108 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (5 points) In the cells of a $5 \times 7$ table, the numbers 1, 2, and 3 are arranged such that in any $2 \times 2$ square, there are all three different numbers. What is the smallest value that the sum of the numbers in the entire table can take? | Answer: 55
Solution:
Example:
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 3 | 2 | 3 | 2 | 3 | 2 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 3 | 2 | 3 | 2 | 3 | 2 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Evaluation
Consider the left upper $2 \times 2$ square of the $5 \times 7$ rectangle... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | # Problem 5. (3 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | 1040 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (3 points)
Natural numbers $a$ and $b$ are such that $2a + 3b = \operatorname{LCM}(a, b)$. What values can the number $\frac{\operatorname{LCM}(a, b)}{a}$ take? List all possible options in ascending or descending order, separated by commas. If there are no solutions, write the number 0. | Answer: 0 (no solutions)
## Solution:
Let $d=\operatorname{GCD}(a, b)$. Then $a=x d, b=y d, \operatorname{LCM}(a, b)=x y d$, where $x$ and $y$ are coprime. We get $2 x d+3 y d=x y d$. Dividing by $d$ yields $2 x+3 y=x y$.
The right side of this equation is divisible by $x$, so $2 x+3 y$ is divisible by $x$, which me... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
Inside a pentagon, 1000 points were marked and the pentagon was divided into triangles such that each of the marked points became a vertex of at least one of them. What is the smallest number of triangles that could have resulted?
# | # Answer: 1003
## Solution:
The sum of the angles of a pentagon is $540^{\circ}$. The sum of the angles at each internal point is either $180^{\circ}$, if it lies on the side of one of the triangles and is the vertex of several others, or $360^{\circ}$, if angles of triangles are located on all sides. That is, the su... | 1003 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $ABCD$ with base $AD$, the diagonals are the angle bisectors of $\angle B$ and $\angle C=110^{\circ}$. Find the degree measure of $\angle BAC$.
# | # Solution:
$\angle B D A=\angle C B D$ as alternate interior angles, and $\angle C B D=\angle A B D$, since $B D$ is the bisector. Therefore, triangle $B D A$ is isosceles, that is, $A B=A D$. Similarly, $C D=A D$, so our trapezoid is isosceles (equilateral) and $\angle B=\angle C=110^{\circ}$.
$$
\angle B A C=180^{... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
In an $8 \times 10$ table (8 rows, 10 columns), some cells are black, and the rest are white. In each white cell, the total number of black cells in the same row is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take... | Answer: 200
## Solution:
If a row contains $x$ black cells and $10-x$ white cells, then the sum of the horizontal terms in this row is $x(10-x)$. This value is maximized when $x=5$ and equals 25. This can be understood as $x(10-x)-25=-x^{2}+10 x-25=-(x-5)^{2}$. Therefore, the maximum sum of all numbers is $8 \cdot 25... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) On the side $AC$ of triangle $ABC$, a circle with a radius of 10 cm is constructed with $AC$ as its diameter. This circle intersects sides $AB$ and $BC$ at points $X$ and $Y$ respectively. Find $AX \cdot AB + CY \cdot BC$. | Answer: 400
Solution: Let point $M$ be the midpoint of side $AC$, which is also the center of the circle. Then $AB \cdot BX = BC \cdot BY = (BM + 10)(BM - 10)$. Therefore, $BX = \frac{BM^2 - 100}{AB} = \frac{2AB^2 + 2BC^2 - 400 - 400}{4AB}$. Accordingly, $AX = AB - \frac{2AB^2 + 2BC^2 - 800}{4AB} = \frac{2AB^2 - 2BC^2... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (5 points) A $10 \times 10$ table is filled with zeros. In one operation, the smallest number in the table (if there are several, any one is chosen) is found, and one is added to it, as well as to all numbers in the cells adjacent to it by side or corner. What is the largest number that can appear in one of the cell... | Answer: 20.
Solution: Let's call the $n$-th phase a sequence of consecutive operations applied to numbers equal to $n$. If such operations did not occur, we will say that this phase consists of zero operations. Everything starts with the zero phase.
During one phase, no number can increase by more than 4. Indeed, if ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) On the side $B C$ of triangle $A B C$ as a diameter, a circle with a radius of 20 cm is constructed. This circle intersects sides $A B$ and $A C$ at points $X$ and $Y$ respectively. Find $B X \cdot A B + C Y \cdot A C$. | Answer: 1600
Solution: Let point $M$ be the midpoint of side $B C$, which is also the center of the circle. Then $A B \cdot A X = A C \cdot A Y = (A M + 20)(A M - 20)$. Therefore, $A X = \frac{A M^{2} - 400}{A B} = \frac{2 A B^{2} + 2 A C^{2} - 1600 - 1600}{4 A B}$. Accordingly, $B X = A B - \frac{2 A B^{2} + 2 A C^{2... | 1600 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (5 points) A $7 \times 7$ table is filled with zeros. In one operation, the smallest number in the table (if there are several, any one is chosen) is found, and one is added to it, as well as to all numbers in the cells adjacent to it by side or corner. What is the largest number that can appear in one of the cells ... | Solution: Let's call the $n$-th phase a sequence of consecutive operations applied to numbers equal to $n$. If such operations did not occur, we will say that this phase consists of zero operations. Everything starts with the zero phase.
During one phase, no number can increase by more than 4. Indeed, if we increased ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $x^{2}+4 x \sin y-4 \cos ^{2} y$. | Answer: -4.
Solution: Add and subtract $4 \sin ^{2} y$. We get the expression $x^{2}+4 x \sin y+4 \sin ^{2} y-4=(x+2 \sin y)^{2}-4$. It is clear that this expression has a minimum value of -4.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{100}$ be natural numbers greater than 1 (not necessarily distinct). In a $100 \times 100$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{4}$ is written. Find the smallest possible... | Answer: -10000
Solution: $\log _{x_{k}} \frac{x_{i}}{4}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 4. Add the minuend to the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this loga... | -10000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $x^{2}-6 x \sin y-9 \cos ^{2} y$. | Answer: -9.
Solution: Add and subtract $9 \sin ^{2} y$. We get the expression $x^{2}-6 x \sin y+9 \sin ^{2} y-9=(x-3 \sin y)^{2}-9$. It is clear that this expression has a minimum value of -9.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $4 x^{2}+4 x \sin y-\cos ^{2} y$. | Answer: -1.
Solution: Add and subtract $\sin ^{2} y$. We get the expression $4 x^{2}+4 x \sin y+\sin ^{2} y-1=(x+2 \sin y)^{2}-1$. It is clear that this expression has a minimum value of -1.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and f... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{200}$ be natural numbers greater than 2 (not necessarily distinct). In a $200 \times 200$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{9}$ is written. Find the smallest possible... | Answer: -40000
Solution: $\log _{x_{k}} \frac{x_{i}}{4}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 9. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this lo... | -40000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{100}$ be natural numbers greater than 1 (not necessarily distinct). In an $80 \times 80$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{16}$ is written. Find the smallest possible... | Answer: -19200
Solution: $\log _{x_{k}} \frac{x_{i}}{16}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 16. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this ... | -19200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have no roots? | Answer: 6
Solution:
Consider, for example, the coefficients 1000, 1001, 1002. The square of any of them is obviously less than the quadruple product of the other two, so all discriminants will be negative. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 10 are inscribed in the same circle such that point $A_{1}$ lies on the arc $B C$, and point $B_{1}$ lies on the arc $A C$. Find $A A_{1}^{2}+B C_{1}^{2}+C B_{1}^{2}$. | Answer: 200
Solution:
Notice that the arcs $A B_{1}, B A_{1}$, and $C C_{1}$ are equal. Let their degree measure be $2 \alpha$. Then the lengths of the arcs $A C_{1}=B B_{1}=C A_{1}=120^{\circ}-2 \alpha$.
By the Law of Sines in triangle $A C A_{1}$, we get $\frac{A A_{1}}{\sin A C A_{1}}=$ $\frac{A C}{A A_{1} C}$, f... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (4 points) For February 23, the boy Zhenya was given a chocolate bar of size $3 \times 3$, on each piece of which a unique picture is drawn, with each picture appearing only once. In each move, Zhenya can bite off one piece that has no more than three common sides with other pieces that have not yet been eaten. In h... | Solution:
If Zhenya had no restrictions, he could eat the pieces in any order, and he would have $9!=362880$ ways.
Not suitable are the ways in which the central cell is eaten before any of the four adjacent to it. Since for each of these five cells, there is an equal number of options where it is earlier than the ot... | 290304 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have two distinct roots | Answer: 6
Solution: Let's take, for example, the coefficients, $-5,1,2$.
If the number -5 is the leading coefficient or the constant term, the equation obviously has two roots of different signs.
For the case when -5 is the second coefficient, let's calculate the discriminant: $5^{2}-1 \cdot 2=23>0$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 12 are inscribed in circle $S$ such that point $A$ lies on arc $B_{1} C_{1}$, and point $B$ lies on arc $A_{1} B_{1}$. Find $A A_{1}^{2}+B B_{1}^{2}+C C_{1}^{2}$. | Answer: 288
Solution:
Notice that the arcs $A B_{1}, B A_{1}$, and $C C_{1}$ are equal. Let their degree measure be $2 \alpha$. Then the lengths of the arcs $A C_{1}=B B_{1}=C A_{1}=120^{\circ}-2 \alpha$.
By the Law of Sines in triangle $A C A_{1}$, we get $\frac{A A_{1}}{\sin A C A_{1}}=$ $\frac{A C}{A A_{1} C}$, f... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.