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Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,212 |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,214 |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,216 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,218 |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,219 |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,222 |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,225 |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,226 |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 x+1009 q+1008 s=0$ have a solution? | # Solution.
From the condition, it follows that $p^{2}0, s>0)$.
The discriminant of the third equation
$$
\begin{aligned}
& D=(1009 p+1008 r)^{2}-4 \cdot 2017(1009 q+1008 s)=(1009 p)^{2}+2 \cdot 1009 \cdot 1008 p r+(1008 r)^{2}- \\
& -4 \cdot 2017(1009 q+1008 s)<1009^{2} \cdot 4 q+2 \cdot 1009 \cdot 1008 \cdot 4 \sq... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,229 |
3. The length of the side of a square is an integer. On each side of the square, a trapezoid is constructed outside it such that the upper bases of the trapezoids and their lateral sides form a regular dodecagon.
Can the area of this dodecagon be equal to $2 \underbrace{1111 \ldots 11}_{2017 \text { ones }}$? | # Solution.
$$
S_{\text {dodecagon }}=S_{\text {square }}+4 S_{\text {trapezoid }}
$$
Let the side of the square be $a \in \mathbb{N}$. Consider the trapezoid $A B C D$ that is being constr... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,230 | |
4. A circle with center at point $O$ is inscribed in quadrilateral $A B C D$. Segments $O A, O B, O C$ and $O D$ intersect the circle at points $K, L, M$ and $N$ respectively. $P$ is the intersection point of the diagonals of quadrilateral $K L M N$, $Q$ is the midpoint of segment $K L$. Prove that lines $P Q$ and $M N... | # Solution.
$K O$ is the bisector of $\angle L K N, L O$ is the bisector of $\angle K L M$. Therefore,
$\angle K O L=180^{\circ}-\angle L K O-\angle K L O=180^{\circ}-\frac{1}{2}(\angle L K N+\angle K L M)$.
Similarly, $\angle M O N=180^{\circ}-\frac{1}{2}(\angle L M N+\angle K N M)$.
Then $\angle K O L+\angle M O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,232 |
5. In the Emerald City, there is a social network called "Magic Mirror," which is used only by the residents of this city. The total number of friends in the "Magic Mirror" for any two residents of the Emerald City is no less than the total population of the city. The number of residents in the city is greater than 2.
... | # Solution.
a) Let the number of city residents be $N$. Suppose there is a person with fewer than 2 friends in their list. Pair this person with any other person who is not their friend. Together, they have no more than $1 + (N-2) = N-1$ friends, which contradicts the problem's condition.
b) Suppose this is not the c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,233 |
1. A number is called good if it has no identical digits and is divisible by the sum of its digits. Come up with at least two good two-digit numbers that remain good after increasing each of their digits by 1. | # Solution:
For example, the number 10 (it is divisible by $1+0$, and the number 21 is divisible by $2+1$) and the number 70 (it is divisible by $7+0$, and the number 81 is divisible by $8+1$).
## Criteria:
One good number found - 3 points, two numbers - 7 points. | 10,70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,234 |
2. Agronomist Bilbo noticed that if the length of his rectangular field were 20 meters longer, the perimeter of the field would be twice as large. However, if the width of the field were twice as large, the perimeter of the field would be 18 meters larger. What is the area of the field?
# | # Answer: $99 \mu^{2}$.
## Solution:
## Method № 1.
If the width of the field has doubled, then the perimeter has simply increased by twice the width of the field, but according to the problem, this change is 18 meters - which means the width of the field is 9 meters. When the length of the field increases by 20 met... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,235 |
3. The numbers $1,2,3,4,5,6,7,8,9$ are written into the cells of a $3 \times 3$ table. After that, all possible sums of numbers standing in adjacent (by side) cells are written down in a notebook. What is the smallest number of different numbers that could have been written in the notebook? | # Solution:
Consider the number in the central cell of the table. Next to it are 4 different neighbors - they give 4 different sums with the central number, so there are already at least 4 different sums written down. An example where there are exactly 4 of them exists (one is shown on the right, the sums are $8,9,10,... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,236 |
4. In the parliament of Anchuria, some deputies are at odds with each other (and each has at least one enemy), some are friends, and some are indifferent to each other. Friendship, enmity, and indifference are mutual, and for each deputy, the principle applies: "A friend of my enemy is also my enemy." Prove that some d... | # Solution:
Let's take any deputy. By the condition, he has at least one enemy - let's choose any enemy. Of these two conflicting deputies, we will call $A$ the one who has more friends, and the other deputy we will call $B$ (if they have the same number of friends, it doesn't matter which one is $A$ and which one is ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 16,237 |
5. Robot Wall-E prints numbers. If the last printed number is even, then after it, Wall-E will print half of this number, and if it is odd - first add 1001 to the number, divide this sum by two, and print the result. The first number printed was 1. Is it true that eventually among the printed numbers, all natural numbe... | Answer: Incorrect.
## Solution:
## First method.
Note that if an even number $n$ is not divisible by 7, then the number $n / 2$ is also not divisible by 7. If an odd number $n$ is not divisible by 7, then the number $n + 1001$ (since 1001 is divisible by 7) is also not divisible by 7, and therefore the number $(n + ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,238 |
1. The residents of the Country of Fairy Tales have invited Baba Yaga, Koschei, and Zmey Gorynych as honored guests to their celebration. The residents have learned the following. If Baba Yaga comes to the celebration, then Koschei will not come. If Koschei comes, then Zmey Gorynych will not come. If Zmey Gorynych come... | # Solution
The condition "If Baba Yaga comes to the party, then Koschei will not come" is equivalent to the statement that both Baba Yaga and Koschei come to the party - an impossible event.
By similar conditions, it is also impossible for both Koschei and Zmey Gorynych, or both Zmey Gorynych and Baba Yaga, to come t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 16,239 |
2. In a certain state, there are coins worth 1 ruble, 3 rubles, and 5 rubles. Can a salary of 200 rubles be paid with thirteen coins? Justify your answer.
# | # Solution
Let $x, y, z$ be the number of coins of 1 ruble, 3 rubles, and 5 rubles, respectively. We obtain the system of equations
$$
\left\{\begin{array}{c}
x+y+z=13 \\
x+3 y+5 z=200
\end{array}\right.
$$
From the system, we get the equation, which has no solutions in integers: $2y + 4z = 187$. Therefore, the answ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,240 |
4. It is known that the sum of the squares of two natural numbers is divisible by 7. Is it true that their product is divisible by 49? Justify your answer.
# | # Solution
Consider the set of remainders of the division of a natural number by 7: $\{0,1,2,3,4,5,6\}$. Therefore, the set of remainders of the division of the square of a natural number by 7 consists of the numbers $\{0,1,2,4\}$. Based on the obtained set of remainders, the sum of the squares of natural numbers will... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,242 |
1. A palindrome is a natural number that reads the same from left to right as it does from right to left. Does there exist a five-digit palindrome that is equal to the sum of two four-digit palindromes? | Answer: For example, $6006+5005=11011$.
Other examples are possible: 7777+4444=12221.
Criteria. Answer without an example: 0 points. | 11011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,244 |
2. Which three-digit numbers are more numerous: those in which all digits have the same parity, or those in which adjacent digits have different parity | Answer: equally
Instructions. Method 1. We will consider the numbers of interest by tens. In each ten, the hundreds and units digits are the same (they are of the same parity). There are exactly 10 tens: five of each parity, and thus, from this ten, five numbers will be added to each type. Therefore, the numbers of ea... | 225 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,245 |
3. On a line, seven points A, B, C, D, E, F, G are marked in the given order. It turned out that $A G=23 \mathrm{~cm}, B F=17 \mathrm{~cm}$ and $\mathrm{CE}=9 \mathrm{~cm}$. Find the sum of the lengths of all segments with endpoints at these points. | Answer: 224 cm.
Instructions. There are five pairs of segments that sum up to AG and AG itself. There are three pairs of segments that sum up to BF and BF itself. There is one pair of segments that sum up to CE and CE itself. Therefore, the total length of all segments is $6 \cdot 23 + 4 \cdot 17 + 2 \cdot 9 = 138 + 6... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,246 |
4. Is it possible to arrange balls of seven colors in five boxes standing in a circle so that each box contains three balls of different colors, and no two balls of the same color are in adjacent boxes? | Answer: No.
Instructions. Since there are $15=3 \cdot 5$ balls and seven colors, there will be at least three balls of one color. If there are no more than two balls of each color, then the total number of balls is no more than $2 \cdot 7=14<15$. Balls of the same color cannot be in the same box. But then there are tw... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,247 |
5. There are seven weights weighing $1 g, 2 g, \ldots, 7$ g. They were all placed on the scales so that equilibrium was achieved. Vasya claims that he can always remove three weights, one of which weighs 1 g, such that the equilibrium is maintained. Is Vasya right? | Answer: incorrect.
Instruction. If equilibrium was achieved when weights of $1,6,7$ were placed on one pan and the rest on the other, it would not be possible to remove three weights. If equilibrium is achieved after removing three weights, it would also be achieved if these three weights were left, and the others rem... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,248 |
6. From a paper square $8 \times 8$, p seven-cell corners were cut out. It turned out that it was impossible to cut out any more such corners. For what smallest $n$ is this possible? A seven-cell corner is obtained by cutting out a $3 \times 3$ square (in cells) from a $4 \times 4$ square. | Answer: $\mathrm{n}=3$.
Instructions. Example. The figures show two examples of placing three corners so that no more can be cut out.
Cells belonging to the same corner are marked with the s... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,249 |
1. Solve the equation in integers:
$(\mathrm{x}-\mathrm{y})^{3}+(\mathrm{y}-\mathrm{z})^{3}+(\mathrm{z}-\mathrm{x})^{3}=30 .(6$ points $)$ | Solution. Transforming the given equation, we get:
$3(x-y)(y-z)(z-x)=30$ or $(x-y)(y-z)(z-x)=10$.
Thus, the integers $(x-y),(y-z),(z-x)$ are divisors of the number 10, and the sum of these divisors is zero. It is not hard to verify that there are no such divisors of the number 10. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,250 |
2. From a three-digit number, the sum of its digits was subtracted. The same operation was performed on the resulting number, and so on, 100 times. Prove that the result will be zero. (6 points) | Solution. Since $\overline{a b c}-(a+b+c)=9 \cdot(11 a+b)$, the first difference is divisible by 9. The sum of its digits is divisible by 9, which means the second, and similarly, all other differences will be divisible by 9.
The sum of the digits of a three-digit number divisible by 9 can be 9, 18, or 27. Therefore, ... | 0 | Number Theory | proof | Yes | Yes | olympiads | false | 16,251 |
3. Do there exist everywhere defined functions $f(x)$ and $g(y)$ such that for any $x$ and $y$, $f(x) \cdot g(y) = x + y - 1$? (6 points) | Solution. Suppose such functions exist. Then for any y
for $x=0: f(0) \cdot g(y)=y-1$,
for $x=1: f(1) \cdot g(y)=y$.
Obviously, $\mathrm{f}(0) \neq 0, \mathrm{f}(1) \neq 0$, hence
$g(y)=\frac{y-1}{f(0)}=\frac{y}{f(1)}$
This equality does not hold for all $\mathrm{y}$ (it is false for $\mathrm{y}=0$), therefore, su... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,252 |
4. Solve the equation
$$
\frac{1}{x}+\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8}-\frac{1}{x+10}+\frac{1}{x+12}+\frac{1}{x+14}=0
$$
S o l u t i o n. Substituting $\mathrm{y}=\mathrm{x}+7$ makes the given equation symmetric:
$$
\frac{1}{y-7}+\frac{1}{y-5}-\frac{1}{y-3}-\frac{1}{y-1}-\frac{1}{y+1}-\frac{1}{... | Answer: $\mathrm{x}=-7, \mathrm{x}=-7 \pm \sqrt{19 \pm 6 \sqrt{5}}$. | -7,-7\\sqrt{19\6\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,253 |
5. In a convex quadrilateral $A B C D$ with internal angles $<180^{\circ}$, point $E$ is the intersection of the diagonals, $F_{1}, F_{2}$ are the areas of triangles $\triangle A B E, \Delta C D E$, and $F$ is the area of quadrilateral $A B C D$. Prove that $\sqrt{F_{1}}+\sqrt{F_{2}} \leq \sqrt{F}$. In what case is equ... | Solution.
We have $F=F_{1}+F_{2}+F_{3}+F_{4}$. The inequality to be proven is equivalent to the inequality
$$
0<\sqrt{F_{1}}+\sqrt{F_{2}} \leq \sqrt{F_{1}+F_{2}+F_{3}+F_{4}} .
$$
After squa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,254 |
10.1. Let's call a natural number an almost square if it equals the product of two consecutive natural numbers. Prove that every almost square can be represented as the quotient of two almost squares.
(V. Senderov) | Solution. Any almost square can be written in the form
$$
n(n+1)=\frac{n(n+2)(n+1)^{2}}{(n+1)(n+2)}=\frac{\left(n^{2}+2 n\right)\left(n^{2}+2 n+1\right)}{(n+1)(n+2)}
$$
In the numerator and denominator of the last fraction, it is also clear that there are almost squares. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,255 |
10.2. Given a parallelogram $A B C D$, where $A B<A C<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find th... | Answer: $60^{\circ}$.
Solution. Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C>A C$, we have $\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A$; thus, $A^{\prime}$ lies on the extension of segment $D C$ beyond poi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,256 |
10.3. At a freestyle cycling competition, there were 100 judges. Each judge ranked all participants from the best to the worst according to their opinion. It turned out that for no three participants $A, B, C$ did there exist three judges, one of whom considered $A$ the best of the three and $B$ the worst, another cons... | Solution. We will construct a graph where the vertices are the participants, and there will be a directed edge from $A$ to $B$ if $A$ is better than $B$ according to at least 51 judges (in this case we will write $A \rightarrow B$). Thus, $A$ and $B$ will not be connected by an edge exactly when each is better than the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,257 |
10.4. Let $S(k)$ denote the sum of the digits of a natural number $k$. A natural number $a$ is called $n$-good if there exists a sequence of natural numbers $a_{0}, a_{1}, \ldots, a_{n}$ such that $a_{n}=a$ and $a_{i+1}=a_{i}-S\left(a_{i}\right)$ for all $i=0,1, \ldots, n-1$. Is it true that for any natural $n$ there e... | Answer. Yes.
Solution. For natural numbers $n$ and $k$, let's introduce the notation $f(n) = n - S(n)$ and $f^{k}(n) = \underbrace{f(f(\ldots(n)}_{k} \ldots))$. When the number $n$ is increased by 1, the number $S(n)$ either increases by 1 (if $n$ does not end in 9) or decreases. This means that the function $f$ is no... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,258 |
10.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with? (N. Agakhanov) | Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,259 |
10.7. Four pairwise distinct integers, each with an absolute value greater than a million, are written on the board. It is known that there is no natural number greater than 1 that divides all four of the written numbers. Pete wrote down the six pairwise sums of these numbers in his notebook, divided these six sums int... | Answer. Yes, they could.
First solution. For example, the numbers
$$
\begin{gathered}
x=N^{2}-3 N+1, \quad y=N^{2}-N+1 \\
z=-3 N^{2}+3 N-1, \quad t=N^{2}+N-1
\end{gathered}
$$
where $N$ is a natural number greater than a million.
It is easy to see that these numbers are also greater than a million in absolute value... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,261 |
10.1. In a contest with 5 participants, there were several questions. $N a$ for each question, one of the participants gave a wrong answer, while the others answered correctly. The number of correct answers given by Petya is 10 - this is less than that of any other participant. The number of correct answers given by Va... | Solution: For each question, there were four correct answers, which means the total number of correct answers is a multiple of four. The maximum number of correct answers is $13+3 \cdot 12+10=59$, and the minimum is $-13+3 \cdot 11+10=56$. In the range from 56 to 59, there is only one number that is a multiple of 4: th... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,264 |
10.2. Can the signs $*$ in the expression $\frac{1}{2} * \frac{1}{3} * \frac{1}{4} * \frac{1}{5} * \frac{1}{6} * \frac{1}{7} * \frac{1}{8}$ be replaced with plus and minus signs so that the absolute value of this expression becomes less than $\frac{1}{500}$? | Solution: The least common multiple of all the denominator numbers is 840. If the numerator after converting all fractions to this common denominator is $0, -1$, or $1$, the answer is positive; otherwise, it is negative. The numerators of the fractions after conversion are $420, 280, 210, 168, 140, 120$, and 105. The l... | Itisimpossible | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,265 |
10.3. Prove that for any natural number $n$ and any real $\alpha$, the following double inequality holds:
$$
\frac{2}{3}<\frac{1}{n+\sin ^{2} \alpha}+\frac{1}{n+2 \sin ^{2} \alpha}+\ldots+\frac{1}{n+(2 n+1) \sin ^{2} \alpha} \leqslant 3
$$ | Solution: Let $t=\sin ^{2} \alpha$ (where $t$ is a parameter taking values from the interval $[0 ; 1]$). We need to prove that
$$
\frac{2}{3}\frac{2}{3}
$$
Also note that the estimate 3 is achieved (for $n=1, \alpha=0$), while the estimate $\frac{2}{3}$ is not. Recommendations for checking:
| is in the work | points... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,266 |
10.4. On a plane, three pairwise intersecting lines $l, p, q$ are drawn. Using a compass and a ruler, construct points $A$ and $B$ on lines $l$ and $p$ respectively, such that segment $AB$ is perpendicular to line $q$ and is bisected by this line. Determine how many solutions the problem has depending on the mutual arr... | # Solution:
Analysis. Suppose such points are found. Then under reflection about the line $q$, they transform into each other. Since point $A$ lies on the line $l$, point $B$ must lie on the line $l^{\prime}$, which is the image of the line $l$ under the specified reflection. Additionally, $B$ lies on the line $p$. Th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,267 |
10.5. Let $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find all values that the number $x+y$ can take, and prove that no other values are possible. | Solution: Consider the function $f(x)=x+\sqrt{x^{2}+1}$. Since for any real $x$ we have $\sqrt{x^{2}+1}>\sqrt{x^{2}}=|x| \geqslant -x$, this function is positive everywhere. Let $a>0$. Solving the equation $f(x)=a$, we find that the value $a$ is taken by the function at the unique point $x=\frac{a^{2}-1}{2 a}$. Our equ... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 16,268 |
1. Option 1. To a confectionery factory for cake packaging, 5 rolls of ribbon, each 60 m long, were delivered. How many cuts need to be made to get pieces of ribbon, each 1 m 50 cm long? | Answer: 195.
Solution: From one roll, 40 pieces of ribbon, each 1 m 50 cm long, can be obtained. For this, 39 cuts are needed. Therefore, a total of $5 \cdot 39=195$ cuts are required. | 195 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,270 |
# 2. Option 1
Given a rectangular grid. We will call two cells adjacent if they share a side. Let's count the number of cells that have exactly four adjacent cells. It turned out to be 23. How many cells have exactly three adjacent cells? | Answer: 48.
Solution. Let $a$ and $b$ be the sides of the rectangle. The total number of cells that have exactly 4 neighbors by side is $(a-2)(b-2)$, and on the other hand, there are 23 such cells. Since 23 is a prime number, the numbers $a-2$ and $b-2$ are equal to the numbers 1 and 23 in some order. The number of ce... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,271 |
# 3. Variant 1
Three rectangles A, B, and C are drawn on the sheet (see figure).
Rectangles A and B have the same width, while rectangles B and C have the same length (width - top to bottom... | Answer: 20.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 3 cm, the area increases by $3 b$. Therefore, $3 b=12, b=4$. The area of rectangle B is larger than that of A by $24 \mathrm{~cm}^{2}$, so the length of rectangle B is $24: 3=8$ cm. Therefore, the length ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,272 |
# 4. Variant 1
The train was moving from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 15 minutes. After that, the driver had to increase the train's speed by 4 times to arrive at point B on schedule. How many minutes does the train travel from poi... | Answer: 40.
Solution. Let's represent the train's path as a segment divided into 8 parts. Denote the intervals it travels in equal time with arcs.
Since the train stopped for 15 minutes, it... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,273 |
# 5. Option 1.
An apple, three pears, and two bananas together weigh 920 g; two apples, four bananas, and five pears together weigh 1 kg 710 g. How many grams does a pear weigh? | Answer: 130
Solution: From the first condition, it follows that two apples, four bananas, and six pears together weigh 1 kg 840 g. Therefore, a pear weighs $1840-1710=130$ g. | 130 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,274 |
# 6. Variant 1
A doll maker makes one doll in 1 hour 45 minutes. After every three dolls made, the master has to rest for half an hour. Ten dolls are needed for gifts. At what time the next day (specify hours and minutes) will the order be completed if the master started making dolls at 10:00 AM and worked through the... | Answer: 5:00.
Solution: The master takes 105 minutes to make one doll. Then ten dolls are made in 1050 minutes. During this time, the master will have to rest 3 times for 30 minutes each, which is 90 minutes. In total, the entire order will take 1140 minutes. This is exactly 19 hours.
## Variant 2
A doll maker takes... | 5:00 | Other | math-word-problem | Yes | Yes | olympiads | false | 16,275 |
# 7. Option 1
Diligent Masha wrote down in a row all natural numbers from 372 to 506 inclusive. Then she calculated two sums: first, the sum of all odd numbers in this row, and then the sum of all even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get? | Answer: 439.
Solution. Let's divide all numbers from 372 to 506, except 506, into pairs such that each even number corresponds to the next odd number (this is possible because the last number is 505). In each pair, the odd number will be 1 greater than the even one. There will be (505 - 371) : 2 = 67 pairs in the rang... | 439 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,276 |
# 8. Variant 1
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 11 sticks, and laid out a pattern inside the triangle such that the triangle was divided into smaller triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle wi... | Answer: 198.
Solution (1st method). All the sticks can be divided into those that go horizontally, at an angle to the right, and at an angle to the left:
In the first such row, there is 1 s... | 198 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,277 |
11.1. Answer. $\frac{1}{2021} \cdot\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2021}\right)>\frac{1}{2022} \cdot\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$. | Solution. Let $\frac{1}{2021} \cdot\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2021}\right)=x, \frac{1}{2022} \cdot\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)=y$, $a=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2021}$. Then $x=\frac{a}{2021}, y=\frac{1}{2022}\left(a+\frac{1}{2022}\right)$. Therefore, ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,278 |
11.3. Answer. $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$. | Solution.
Let $O$ be the common center of the given circles. From the condition, it follows that $BO$ and $CO$ are the bisectors of angles $ABC$ and $BCD$ ($O$ is the incenter) and, moreover... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,280 |
10.6. On the board, three natural numbers were written: two ten-digit numbers $a$ and $b$, as well as their sum $a+b$. What is the maximum number of odd digits that could have been written on the board?
(I. Bogdanov, P. Kozhevnikov) | Answer: 30.
Solution: Note that the number $a+b$ has no more than 11 digits, so in total, no more than 31 digits are written on the board. At the same time, all three numbers $a, b, a+b$ cannot be odd simultaneously. Therefore, one of their last three digits is even, which means that there are no more than 30 odd digi... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,283 |
10.7. Vasya wrote natural numbers from 1 to 81 in the cells of a $9 \times 9$ table (each cell contains one number, all numbers are distinct). It turned out that any two numbers differing by 3 are in adjacent cells. Is it true that there will necessarily be two corner cells, the difference of the numbers in which is di... | Answer. Correct.
Solution. Consider the remainders of the numbers in the four corner cells when divided by 3. By the Pigeonhole Principle, at least two of these numbers, $x$ and $y$, will have the same remainder, meaning that the difference $y-x$ is divisible by 3. Without loss of generality, assume $x < y$.
Color th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,284 |
10.8. Point $M$ is the midpoint of the base $A C$ of isosceles triangle $A B C$. On the extensions of segments $A C$ and $B C$ beyond point $C$, points $D$ and $K$ are marked respectively such that $B C = C D$ and $C M = C K$. Prove that the circumcircles of triangles $A B D$ and $M C K$ are tangent.
(A. Kuznetsov) | Solution. Draw the bisector $m$ of angle $B C D$. By construction, $B$ and $D$, as well as $M$ and $K$, are symmetric with respect to $m$.
From symmetry, $B M$ and $D K$ intersect at point $X$, which lies on $m$. Since $X M \perp C M$, then $X K \perp C K$, which means $X$ lies on the circle $(C M K)$, and $C X$ is th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,285 |
10.9. A magician and an assistant are going to perform the following trick. They have $n \geqslant 3$ cards numbered $1,2, \ldots, n$, and a row of $n$ cells the size of a card. The back sides of all cards are indistinguishable. The audience places cards 1 and 2 on any two spots; the assistant, seeing this, places the ... | Answer. For all $n$.
Solution. Let's present one of the possible agreements between the magician and the assistant. Mark the leftmost position with a * (the magician will flip the card at this position). Number all other positions from 1 to $n-1$; the magician and the assistant will consider these positions to be arra... | Foralln | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,286 |
10.10. Vitya wrote down $n$ different natural numbers in his notebook. For each pair of numbers from the notebook, he wrote their least common multiple on the board. Could it happen for some $n>100$ that $\frac{n(n-1)}{2}$ numbers on the board are (in some order) consecutive terms of a non-constant arithmetic progressi... | # Answer. No.
First solution. Let's call a set of $n$ numbers in the notebook $k r a$ beautiful if it results in the required set of least common multiples. Suppose a beautiful set of $n>100$ numbers exists. Choose from all such sets the set with the smallest sum of numbers.
Notice that if the difference of the resul... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,287 |
1. From two settlements, which are 60 km apart, two cars set off in the same direction at the same time. The first car, traveling at a speed of 90 km/h, caught up with the second car after three hours. What is the speed of the second car? | Solution. The closing speed of the cars is $60: 3=20(\kappa m / h)$, so the speed of the second car is $90-20=70(km / h)$.
Answer: 70 km $/$ h.
Comment. The correct answer was obtained through correct reasoning - 7 points. The closing speed was correctly found, but the wrong answer was given - 3 points. | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,289 |
2. Vera, Nadya, and Lyuba have dresses of three different colors - pink, purple, and turquoise, and their hats are also of these three colors. Only Vera has the same color for her dress and hat, Nadya's are not pink, and Lyuba has a purple hat. Indicate the colors of the dress and hat for each girl. | Solution. Since Luba's hat is purple, and Nadya's is not pink, Nadya's must be turquoise, which means Vera's is pink. Then Vera's dress is also pink. Nadya's dress and hat are of different colors, so her dress is purple, and Luba's is turquoise.
Answer: Vera has a pink dress and a pink hat, Nadya has a purple dress an... | Vera: | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,290 |
3. For tasks a) and b), given the numbers $A$ and $B$, determine whether there exists a natural number that, when divided by $A$, gives a remainder of 1, and when divided by $B$, gives a remainder of 2: a) $A=6, B=8$; b) $A=7, B=9$. | Solution. a) If such a natural number $C$ exists, then on the one hand, when divided by 6 it gives a remainder of 1, which means $C$ is an odd number, and on the other hand, when divided by 8 it gives a remainder of 2, which means $C$ is an even number. This contradiction implies that such a natural number $C$ does not... | )no;b)yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,291 |
4. A square grid consisting of 64 cells is cut along the cell edges into several pieces. All pieces must be squares, not necessarily of the same size. Indicate some ways of such cutting into 10, 11, and 12 pieces. | Solution. The ways of cutting can be different. For example, the answer is:
Comment. It is clear that there can be several ways to cut even into the same number of parts. At least one way o... | yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,292 |
11.2. In space, three segments $A_{1} A_{2}, B_{1} B_{2}$, and $C_{1} C_{2}$ are given, not lying in the same plane and intersecting at one point $P$. Denote by $O_{i j k}$ the center of the sphere passing through the points $A_{i}, B_{j}$, $C_{k}$, and $P$. Prove that the lines $O_{111} O_{222}, O_{112} O_{221}, O_{12... | Solution. For any segment $XY$, the perpendicular bisector of this segment is called the plane that is perpendicular to it and passes through its midpoint, i.e., the geometric locus of points equidistant from $X$ and $Y$.
All points of the form $O_{1 j k}$ lie in the perpendicular bisector $\alpha_{1}$ of the segment ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,294 |
11.3. On a grid sheet of size $100 \times 100$, several non-overlapping cardboard isosceles right triangles with a leg of 1 were placed; each triangle occupies exactly half of one of the cells. It turned out that each unit segment of the grid (including boundary segments) is covered by exactly one leg of a triangle. Fi... | Answer. $49 \cdot 50=2450$ cells
Solution. Let $n=50$. We will call a triangle upper if it is located above the line containing its horizontal leg, and lower otherwise. Number the horizontal lines of the grid from bottom to top with numbers from 0 to $2n$.
Denote by $u_{k}$ (respectively $d_{k}$) the number of segmen... | 2450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,295 |
# 11.4. In coordinate space, all planes with equations $x \pm y \pm$ $\pm z=n$ (for all integers $n$) were drawn. They divided the space into tetrahedra and octahedra. Let the point $\left(x_{0}, y_{0}, z_{0}\right)$ with rational coordinates not lie on any of the drawn planes. Prove that there exists a natural number ... | Lemma. Let rational numbers $a, b$, and $c$ be such that $a + b + c$ is not an integer. Then there exists a natural number $k$ such that the numbers $ka, kb$, and $kc$ are not integers, and $12$. This means there exists the smallest natural number $k$ for which $f(k) > 1$ (then $f(k-1) \leqslant 1$). We will show that ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,296 |
8.3. The numbers $a, b, c$ satisfy the relation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$. Find $(a+b)(b+c)(a+c)$. | Answer: 0. Solution: Move $\frac{1}{a}$ to the right side, we get $\frac{b+c}{b c}=\frac{-(b+c)}{a(a+b+c)}$. If $b+c \neq 0$, then we will have (multiplying by the denominator)
$$
a^{2}+a b+a c+b c=0 \Leftrightarrow a(a+b)+c(a+b)=0 \Leftrightarrow(a+b)(a+c)=0
$$
Thus, in any case $(a+b)(b+c)(a+c)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,299 |
8.4. In trapezoid $ABCD$ with bases $AD$ and $BC$, the diagonals intersect at point $M$. It turns out that $AB = DM$ and $\angle ABD = \angle CBD$. Prove that a) $\angle BAD > 60^{\circ}$; b) $AB > BC$. | Solution. a). From the conditions of the problem, it follows that $A B=A D$, since $\angle D B A=\angle D B C=\angle B D A$ (since $A D \| B C$) and thus, $\triangle A B D$ is isosceles. Let $\beta=\angle A B D=\angle B D A=\angle D B C, \alpha=\angle B A D$. Then $\alpha>\beta$, since in triangle $A B D$ the side $B D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,300 |
8.5. Kolya drew 10 segments and marked all their intersection points in red. Counting the red points, he noticed the following property: on each segment, there are three red points. a) Provide an example of the arrangement of 10 segments with this property. b) What is the maximum number of red points for 10 segments wi... | Answer: b) 15. Solution. a) See the example in the figure. As another example, you can take two copies of the right part of the figure from problem 7.5. b) The example given in the figure shows that 15 red points can be obtained. Let's prove that this is the maximum possible number. Number all 10 segments and write dow... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,301 |
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 | 16,303 |
Problem 4.5. Hooligan Dima laid out a structure 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 diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | 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 | 16,304 |
Problem 4.7. Schoolchildren Alexander, Boris, Sergey, Daria, and Elena attended choir classes from Monday to Friday. It is known that:
- every day, exactly three of the schoolchildren were present, and exactly two were absent
- no one was absent for two consecutive days, and no one was present for three consecutive da... | Answer: Alexander, Boris, Daria.
Solution. Since no one was present for three consecutive days, the maximum number of days one can attend is 4. Four classes a week are possible only if Wednesday is skipped. Since it is not allowed to skip two days in a row, the minimum number of days one can attend is 2. Two classes a... | Alexer,Boris,Daria | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,305 |
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. 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.
Add the distance from Asey to Bory to the obtained sum: $32+8=40$. Now all three branches (to Gala - twice) and the main road w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,306 |
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 | 16,307 |
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 | 16,308 |
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 | 16,309 |
Problem 6.3. In the cells of a $4 \times 4$ table, the numbers $1,2,3,4$ are arranged such that
- each number appears in each row and each column;
- in all four parts shown in the figure, the sums of the numbers are equal.
Determine in which cells the twos are located based on the two numbers in the figure.
![](http... | Answer: In row $A$, the two is in column 2, in row $B-1$, in row $C-4$, in row $D-3$.
Solution. First, let's find what the sum of the numbers in each of the parts into which the board is divided is. In each column, the numbers from 1 to 4 appear once. Their sum is $1+2+3+4=10$, and the sum of all numbers on the board ... | A2,B1,C4,D3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,310 |
Problem 6.4. Misha made himself a homemade dartboard during the summer at the cottage. The round board is divided by circles into several sectors - darts can be thrown into it. Points are awarded for hitting a sector as indicated on the board.
Misha threw 8 darts 3 times. The second time he scored twice as many points... | Answer: 48.
Solution. The smallest possible score that can be achieved with eight darts is $3 \cdot 8=24$. Then, the second time, Misha scored no less than $24 \cdot 2=48$ points, and the third time, no less than $48 \cdot 1.5=72$.
On the other hand, $72=9 \cdot 8$ is the highest possible score that can be achieved w... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,311 |
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 | 16,312 |
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 | 16,314 |
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 | 16,315 |
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 | 16,316 |
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 | 16,317 |
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 | 16,318 |
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 $L M$, given that $A K=4, B L=31, M C=3$.
. Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,319 |
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 | 16,321 |
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 | 16,322 |
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 | 16,323 |
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 $D Z$.
$ (where $t$ is in hours), the car is in motion for exactly half of this time. If the car moves a... | Itcould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,325 |
11.2 Let $x \in \mathbf{R}$. Prove that the numbers $x+\sqrt{3}$ and $x^{3}+5 \sqrt{3}$ cannot both be rational at the same time. | Solution. Suppose the opposite, and let the numbers $a=x+\sqrt{3}$ and $b=x^{3}+5 \sqrt{3}$ be rational. Then $x=a-\sqrt{3}, x^{3}=a^{3}-3 \sqrt{3} a^{2}+9 a-3 \sqrt{3}, b=a^{3}+9 a+\sqrt{3}\left(-3 a^{2}+2\right)$. The number $\sqrt{3}\left(-3 a^{2}+2\right)=b-a^{3}-9 a$ is rational, the number $-3 a^{2}+2$ is also ra... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,326 |
11.3 Given three pairwise perpendicular lines. The distance between any two of them is a. Find the area of the parallelogram, two vertices of which are located on one line, and the other two - on the other two lines. | Solution. Consider a cube $ABCD A_{1} B_{1} C_{1} D_{1} \mathrm{c}$ with edge $a$. We can assume that the given pairwise perpendicular skew lines are precisely the lines $AD, A_{1}B_{1}$, and $CC_{1}$ - see the figure. We can also assume that we are considering a parallelogram $KLMN$ with two vertices $K$ and $L$ lying... | ^{2}\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,327 |
11.4 The pirates, led by John Silver on Treasure Island, found Billy Bones' chest, which contained 40 coins worth 1 ducat each and 40 coins worth 5 ducats each. John Silver has not yet decided how to divide this money among all the pirates (he does not want to take anything for himself). For what maximum number of pira... | Solution. We will show that if there are no more than 11 pirates, Silver can divide the coins as he wishes. Indeed, let the $i$-th pirate need to receive $S_{i}$ coins. $S_{i}=5 x_{i}+a_{i}$, for some integers $x_{i}$ and $a_{i}$ (where $a_{i}$ is the remainder of the division of $S_{i}$ by 5). Note that the sum of all... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,328 |
11.5 The equation $(x+a)(x+b)=-9$ has a root $x_{0}=ab$ where $a, b<0$. Prove that $a+b<-6$. | Solution. Method 1. By the condition $(a b+a)(a b+b)=-9$, which is equivalent to the equation $a b(a b+a+b+1)=-9$. Let $a b=t, a+b=p$. Then $t^{2}+p t+t+9=0$, from which $p=-t-1-\frac{9}{t}$. The number $t$ is positive, as the product of two negative numbers. Then $p=-7-\left(\sqrt{t}-\frac{3}{\sqrt{t}}\right)^{2} \leq... | +b\leqslant-7 | Algebra | proof | Yes | Yes | olympiads | false | 16,329 |
11.6 Squares $ABCD$ and $A_{1}B_{1}C_{1}D_{1}$ (vertices are named in a clockwise direction) lie in the same plane and coincide at vertices $C$ and $B_{1}$ (there are no other common points between these squares). Points $O$ and $O_{1}$ are the centers of the squares. Prove that the line $OO_{1}$ intersects segments $A... | Solution. Consider a $90^{\circ}$ rotation with the center at point $C=B_{1}$. Under such a transformation, point $B$ will move to point $D$, point $A_{1}$ to point $C_{1}$, and segment $B A_{1}$ to segment $C_{1} D$. Therefore, segments $B A_{1}$ and $C_{1} D$ are equal and perpendicular.
![](https://cdn.mathpix.com/... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,330 |
10.2 Find any three vectors with a zero sum such that subtracting the third vector from the sum of any two vectors results in a vector of length 1. | 10.2 three vectors of length $1 / 2$, directed from the center of an equilateral triangle to its vertices.
The algebraic sum $\mathbf{a}+\mathbf{b}-\mathbf{c}$ is obtained from the zero sum of vectors a, b, and c by subtracting twice the vector c. Therefore, the length of each vector should be equal to $1 / 2$. The th... | threevectorsoflength1/2,directedfromthecenterofanequilateraltriangletoitsvertices. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,331 |
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