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Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,055 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_d3a... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,056 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,057 |
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 | 18,058 |
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... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,060 |
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 | 18,062 |
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 | 18,064 |
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 | 18,065 |
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 | 18,068 |
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 | 18,071 |
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 | 18,072 |
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?
$ be a polynomial of degree $n \geqslant 2$ with non-negative coefficients, and let $a, b$, and $c$ be the lengths of the sides of some acute triangle. Prove that the numbers $\sqrt[n]{P(a)}, \sqrt[n]{P(b)}$, and $\sqrt[n]{P(c)}$ are also the lengths of the sides of some acute triangle.
(N. Agakhanov, O... | Solution. Let, without loss of generality, $a \geqslant b \geqslant c$; these three positive numbers are the lengths of the sides of an acute triangle if and only if $a^{2}<b^{2}+c^{2}$. This means we need to check that $\sqrt[n]{P(a)^{2}}<\sqrt[n]{P(b)^{2}}+\sqrt[n]{P(c)^{2}}$.
Let $P(x)=p_{n} x^{n}+p_{n-1} x^{n-1}+\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,075 |
10.7. Each cell of a $100 \times 100$ board is colored either black or white, and all cells adjacent to the board's border are black. It turns out that there is no monochromatic $2 \times 2$ square anywhere on the board. Prove that there is a $2 \times 2$ square on the board whose cells are colored in a checkerboard pa... | Solution. Suppose the opposite: there are no monochromatic or checkerboard-colored $2 \times 2$ squares on the board. Consider all segments of the grid that separate two cells of different colors (let's call them separators); let their number be $N$.
In any $2 \times 2$ square, there are either exactly one cell of one... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,076 |
10.8. In a scalene triangle $ABC$ inscribed in a circle with center $O$ and circumscribed around a circle with center $I$, the point $B'$, symmetric to point $B$ with respect to the line $OI$, lies inside the angle $ABI$. Prove that the tangents to the circle circumscribed around triangle $B B' I$, drawn at points $B'$... | The first solution. Let the line $B I$ intersect the circumcircle of triangle $A B C$ again at point $S$. Let the rays $S B^{\prime}$ and $C A$ intersect at point $T$ (see Fig. 2). By the trident lemma, we have $S A = S C = S I$. From the equality $I B = I B^{\prime}$, we get $\angle I B^{\prime} B = \angle I B B^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,077 |
# 2. CONDITION
The messenger needs to run 24 miles. Two-thirds of this distance he ran at an average speed of 8 miles per hour. Can he, by increasing his speed, run the remaining distance so that his average speed for the entire journey equals 12 miles per hour? | Solution. To achieve an average speed of 12 miles per hour for the entire journey, the messenger must run 24 miles in 2 hours. But he has already spent these 2 hours on the first 16 miles.
Answer: cannot. | cannot | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,078 |
# 4. CONDITION
The square of a natural number $a$ when divided by a natural number $n$ gives a remainder of 8. The cube of the number $a$ when divided by $n$ gives a remainder of 25. Find $n$. | Solution. Note that the number $\mathrm{x}=\mathrm{a}^{6}-8^{3}=\left(\mathrm{a}^{2}\right)^{3}-8^{3}=\left(\mathrm{a}^{2}-8\right)\left(\mathrm{a}^{4}+8 \mathrm{a}^{2}+64\right)$ is divisible by $n$. Also note that the number $\mathrm{y}=\mathrm{a}^{6}-25^{2}=\left(\mathrm{a}^{3}\right)^{2}-25^{2}=\left(\mathrm{a}^{3}... | 113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,079 |
11.1. The number $x$ is such that both sums $S=\sin 64 x+\sin 65 x$ and $C=$ $=\cos 64 x+\cos 65 x$ are rational numbers. Prove that in one of these sums, both terms are rational.
(N. Agakhanov) | Solution. Note that the number
$$
\begin{aligned}
& S^{2}+C^{2}=\left(\sin ^{2} 64 x+\cos ^{2} 64 x\right)+\left(\sin ^{2} 65 x+\cos ^{2} 65 x\right)+ \\
&+2(\sin 64 x \sin 65 x+\cos 64 x \cos 65 x)= \\
&=2+2 \cos (65 x-64 x)=2+2 \cos x
\end{aligned}
$$
is rational, from which $\cos x$ is also a rational number. Give... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,080 |
11.2. An acute isosceles triangle \(ABC (AB = AC)\) is inscribed in a circle with center at point \(O\). The rays \(BO\) and \(CO\) intersect the sides \(AC\) and \(AB\) at points \(B'\) and \(C'\) respectively. A line \(\ell\) is drawn through point \(C'\) parallel to the line \(AC\). Prove that the line \(\ell\) is t... | Solution. Let the line $A O$ intersect $\ell$ at point $T$ (see Fig. 3). By symmetry with respect to $A O$, we have $\angle B^{\prime} T O = \angle C^{\prime} T O$. Since $\ell \| A C$, we get $\angle C^{\prime} T O = \angle O A C = \angle O C A$. Therefore, $\angle B^{\prime} T O = \angle B^{\prime} C O$, which means ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,081 |
11.3. On a board, there are $n$ positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ written in a row. Vasya wants to write a number $b_{i} \geqslant a_{i}$ under each number $a_{i}$ such that for any two of the numbers $b_{1}, b_{2}, \ldots, b_{n}$, the ratio of one to the other is an integer. Prove that Vasya can write the... | Solution. We will prove that there exist even numbers $b_{1}, b_{2}, \ldots, b_{n}$, satisfying the following (stronger) conditions:
(1) $b_{i} \geqslant a_{i}$ for all $i \leqslant n$;
(2) $b_{1} b_{2} \ldots b_{n} \leqslant 2^{(n-1) / 2} a_{1} a_{2} \ldots a_{n}$
(3) the ratio of any two of the numbers $b_{i}$ is ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,082 |
11.4. A magician and an assistant have a deck of cards; one side (the "back") of all cards is the same, while the other is painted in one of 2017 colors (the deck contains 1,000,000 cards of each color). The magician and the assistant are going to perform the following trick. The magician leaves the room, and the audie... | Answer. $n=2018$.
Solution. Let $k=2017$.
For $n=k+1$, the trick is easy to arrange. The magician and the assistant number the colors from 1 to $k$. The assistant, seeing the color of the last, $(k+1)$-th card (let its number be $a$), leaves the $a$-th card open. The magician, seeing which numbered card is open, can ... | 2018 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,083 |
1. The number $a$ is 1 greater than the number $b$. Can the numbers $a^{2}$ and $b^{2}$ be equal? | Answer. They can.
Solution. If $a=\frac{1}{2}, b=-\frac{1}{2}$, then $a=b+1$ or $a^{2}=b^{2}$.
Alternatively, the system of equations can be solved: $\left\{\begin{array}{l}a^{2}=b^{2}, \\ a=b+1 .\end{array}\right.$
Grading criteria.
- Correct answer with the numbers $a$ and $b$ specified - 7 points.
- System of eq... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,084 |
2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take for Petya to run down from the ... | Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights 2 seconds longer than it takes his mother to ride the elevator, and 3 flights 2 seconds faster than his mother. Therefore, it takes... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,085 |
3. Plot the graph of the function $y=\frac{x^{2}}{|x|}$. | Answer. See the figure.
Solution. Since $x^{2}=|x|^{2}$, then $y=|x|$, and $x \neq 0$.
We can also, using the definition of the absolute value, obtain that
$$
y=\left\{\begin{array}{l}
x, \text { if } x>0, \\
-x, \text { if } x<0
\end{array} \quad \text { (for } x=0\right. \text { the function is not }
$$
. If there were no more than 4 marked points in each such square, then there would be no more than \(25 \cdot 4 = 100\) points in total, which contradicts the condit... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,087 |
5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,088 |
6. Given a rectangle $A B C D$. Point $M$ is the midpoint of side $A B$, point $K$ is the midpoint of side $B C$. Segments $A K$ and $C M$ intersect at point $E$. How many times smaller is the area of quadrilateral $M B K E$ compared to the area of quadrilateral $A E C D$? | Answer: 4 times.
Solution. Draw segments $MK$ and $AC$. Quadrilateral $MBKE$ consists of triangles $MBK$ and $MKE$, while quadrilateral $AEC D$ consists of triangles $AEC$ and $ACD$. We can reason in different ways.
1st method. Triangles $MBK$ and $ACD$ are right-angled, and the legs of the first are half the length ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,089 |
9.5. On the board, $n>3$ different natural numbers are written, each less than $(n-1)!=1 \cdot 2 \cdot \ldots \cdot(n-1)$. For each pair of these numbers, Seryozha divided the larger by the smaller with a remainder and wrote down the resulting quotient (so, if he divided 100 by 7, he would get $100=14 \cdot 7+2$ and wr... | Solution. Suppose the opposite. Let $a_{1}, a_{2}, \ldots, a_{n}$ be the numbers on the board in ascending order, and let $q_{i}$ be the quotient of the division of $a_{i+1}$ by $a_{i} (i=1,2, \ldots, n-1)$; then $a_{i+1} \geqslant q_{i} a_{i}$. Since all $q_{i}$ are distinct, we have $q_{1} q_{2} \ldots q_{n-1} \geqsl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,090 |
9.6. Is it true that for any three distinct natural numbers $a$, $b$, and $c$, there exists a quadratic trinomial with integer coefficients and a positive leading coefficient that takes the values $a^{3}$, $b^{3}$, and $c^{3}$ at some integer points?
(A. Khryabrov) | Answer. Yes, correct.
Solution. We will show that the quadratic polynomial
$f(x)=(a+b+c) x^{2}-(a b+b c+c a) x+a b c=x^{3}-(x-a)(x-b)(x-c)$ fits. It is clear that its coefficients are integers and the leading coefficient $a+b+c$ is positive. Finally, it is easy to see that $f(a)=a^{3}-0=a^{3}$; similarly, $f(b)=b^{3}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,091 |
9.7. In an isosceles triangle \(ABC\), where \(\angle ACB = 60^\circ\), is inscribed in a circle \(\Omega\). On the bisector of angle \(BAC\), a point \(A'\) is chosen, and on the bisector of angle \(ABC\), a point \(B'\) is chosen such that \(AB' \parallel BC\) and \(BA' \parallel AC\). The line \(A'B'\) intersects \(... | Solution. From the parallelism of lines $A B^{\prime}$ and $B C$, we get that $\angle A B^{\prime} B = \angle C B B^{\prime} = \angle A B B^{\prime}$. Therefore, $A B^{\prime} = A B$. Similarly, $A B = A^{\prime} B$. Let $\angle B A C = 2 \alpha$, $\angle A B C = 2 \beta$. Without loss of generality, let $\alpha > \bet... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,092 |
9.8. Each cell of a $100 \times 100$ board is colored either black or white, and all cells adjacent to the board's border are black. It turns out that there is no monochromatic $2 \times 2$ square anywhere on the board. Prove that there is a $2 \times 2$ square on the board whose cells are colored in a checkerboard pat... | Solution. Suppose the opposite: there are no monochromatic or checkerboard-colored $2 \times 2$ squares on the board. Consider all segments of the grid that separate two cells of different colors (let's call them separators); let their number be $N$.
In any $2 \times 2$ square, there are either exactly one cell of one... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,093 |
10.1. There are 40 pencils of four colors - 10 pencils of each color. They were distributed among 10 children so that each received 4 pencils. What is the smallest number of children that can always be selected to ensure that pencils of all colors are found among them, regardless of the distribution of pencils?
(I. Bo... | 10.1. Answer. 3 boys.
We will show that it is always possible to choose three boys such that they have pencils of all colors. Since there are 10 pencils of each color and each boy received 4 pencils, at least one boy must have received pencils of at least two different colors. It remains to add to him two boys who hav... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,094 |
10.3. Through the center $O$ of the circumcircle of an acute, non-isosceles triangle $ABC$, lines are drawn perpendicular to the sides $AB$ and $AC$. These lines intersect the altitude $AD$ of triangle $ABC$ at points $P$ and $Q$. The point $M$ is the midpoint of side $BC$, and $S$ is the center of the circumcircle of ... | 10.3. Let for definiteness $A B > A C$ (see Fig. 4). Denote by $L$ the midpoint of segment $A B$. Notice that $\angle A O L = \frac{1}{2} \angle A O B = \angle A C B$. Therefore, $\angle B A O = 90^{\circ} - \angle A O L = 90^{\circ} - \angle A C B = \angle C A D$.
The sides of triangle $O P Q$ are perpendicular to th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,096 |
10.4. In each cell of a $100 \times 100$ square, a certain natural number is written. A rectangle, whose sides lie along the grid lines, is called good if the sum of the numbers in all its cells is divisible by 17. It is allowed to simultaneously color all the cells in some good rectangle. It is forbidden to color a ce... | 10.4. Answer. $9744=100^{2}-16^{2}$ cells.
Lemma. Suppose a strip $1 \times k$ is filled with natural numbers. Then in it, one can paint several non-overlapping good rectangles, containing no less than $k-16$ cells.
Proof. Induction on $k$. For $k \leqslant 16$, nothing needs to be painted. Let $k \geqslant 17$. Supp... | 9744=100^{2}-16^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,097 |
11.1. Plot the graph of the function $\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$. | Answer. The graph of the function will be the line $y = 4$.
## Solution.
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,098 |
11.2. The numbers $x$ and $y$ satisfy the inequality $x>y>\frac{3}{x-y}$. Prove that $x^{2}>y^{2}+6$.
# | # Solution.
By adding the inequalities $x>\frac{3}{x-y}$ and $y>\frac{3}{x-y}$, we get that $x+y>\frac{6}{x-y}$.
From the condition $x>y$, it follows that the denominator of the fractions is positive, so we can multiply by it without changing the inequality sign. Then we get: $(x+y)(x-y)>6$, that is, $x^{2}-$ $y^{2}>... | x^{2}-y^{2}>6 | Inequalities | proof | Yes | Yes | olympiads | false | 18,099 |
11.3. The Pyramid of Khufu has a square base, and its lateral faces are equal isosceles triangles. Can the angle of the face at the apex of the pyramid be equal to $95^{\circ} ?$ | Answer: No, it cannot.
Solution.
The lateral surface of the pyramid consists of four equal isosceles triangles. If we cut the lateral surface of the pyramid along the lateral edges and unfold it onto a plane, we will get the figure shown in Fig. 1. In this case, the common point of the triangles is the vertex of the ... | No,itcannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,100 |
11.4. Prove that $n^{2}+n+1$ for any natural $n$:
a) is an odd number;
b) is not a square of any other natural number.
# | # Solution.
a) $n^{2}+n+1=n(n+1)+1$. Since $n(n+1)$ is an even number, then $n(n+1)+1$ will be an odd number;
b) The squares closest to the number $n^{2}+n+1$ are among the natural numbers $n^{2}$ and $(n+1)^{2}$, but $n^{2}<n^{2}+n+1<(n+1)^{2}$.
Since $n^{2}$ and $(n+1)^{2}$ are squares of consecutive natural numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,101 |
11.5. Three different numbers from 1 to 9 are written on the board. In one move, it is allowed to either add 1 to two of the numbers, or subtract 1 from all the numbers. Is it true that by making no more than 30 such moves, it is always possible to achieve that only zeros remain on the board? | Answer: Incorrect.
Solution: Let the numbers 1, 8, and 9 be initially written on the board, and after several moves, they all become zeros. Let's examine how the differences between the second and the first number, as well as between the third and the first number, change. Initially, these differences are 7 and 8. Whe... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,102 |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,105 |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,106 |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
 in the squares so that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,110 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on line $A L$. Carefully finding the intersection point of these lines ... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,111 |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,113 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,114 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,115 |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,116 |
Problem 9.7. Through points $A(0 ; 14)$ and $B(0 ; 4)$, two parallel lines are drawn. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,118 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,119 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
)$.
Note. An answer without specifying the sequence of operations - 0 points. | 17=32-(16-(8-4-2-1)) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,121 |
10.2. Find all quadratic trinomials $f(x)$ for which the equality $f(2 x+1)=4 x^{2}+14 x+7$ holds. | Answer: $f(x)=x^{2}+5 x+1$.
First solution. Let $t=2 x+1$, then $x=1 / 2(t-1)$. Therefore, $f(t)=(t-1)^{2}+7(t-1)+7=t^{2}+5 t+1$.
Second: $4 x^{2}+14 x+7=\left(4 x^{2}+4 x+1\right)+10 x+5+1=(2 x+1)^{2}+5(2 x+1)+1$.
Third: quadratic trinomials are equal if their corresponding coefficients are equal. Denote $f(x)=a x^... | f(x)=x^{2}+5x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,122 |
10.3. What is the maximum number of digits that a natural number can have, where all digits are different, and it is divisible by each of its digits? | Answer: 7 digits.
Evaluation. There are 10 digits in total. The number cannot contain the digit 0, so there are no more than 9. If all 9, then the digit 5 must be at the end of the number (divisibility rule for 5), but in this case, the number must also be divisible by 2. Therefore, the digit 5 is not in this number. ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,123 |
10.4. The diagonals of trapezoid $ABCD$ intersect at point $O$. The circumcircles of triangles $AOB$ and $COD$ intersect at point $M$ on the base $AD$. Prove that triangle $BMC$ is isosceles. | First solution. Angles $A O B$ and $C O D$ are equal as
vertical angles, angles $A O B$ and $A M B$, $C O D$ and $C M D$ are equal as inscribed angles subtending the same arc, angles $A M B$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,124 |
2. In triangle $A B C$, the median $B M$ is twice as short as side $A B$ and forms an angle of $40^{\circ}$ with it. Find the angle $A B C$. | Solution. Extend median $B M$ beyond point $M$ by the same length to get point $D$ (see figure). Since $A B = B D$, triangle $A B D$ is isosceles. Therefore, $\angle B A D = \angle B D A = (180^{\circ} - 40^{\circ}) : 2 = 70^{\circ}$.
Quadrilateral $A B C D$ is a parallelogram because its diagonals bisect each other. ... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,125 |
1. Cut a cross made of 5 cells into 5 pieces that can be assembled into a square. | 1. Answer: as shown in the figure.
Solution: Cut off the triangles protruding beyond the square and insert them into the empty spaces.
Instructions for checking:
There are other solutions... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,126 |
3. A five-digit number is called humped if its middle digit is greater than the others, and valley-like if this digit is smaller than the others. Which are more numerous - humped or valley-like numbers? | 3. Answer: ravine.
Solution: There is a one-to-one correspondence between ravine numbers not starting with 9 and all humpback numbers, given by the formula: $x \leftrightarrow 99999-x$.
Ravine numbers starting with 9 are extra.
## Instructions for checking:
Students can simply count the number of both types of numb... | ravine | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,128 |
4. A hexagon is formed from 3 yellow and 3 blue sticks, such that the colors of the sticks alternate along its perimeter. From any three consecutive sticks, a triangle can be formed. Prove that a triangle can also be formed from the sticks of one of the colors. | 4. Proof: Let the sticks lying in a circle be: $a, b, c, d, e, f$. (a, c, e - blue).
For simplicity, let their lengths be denoted by the same letters.
Let, for definiteness, $a$ be the largest blue stick, and the largest yellow stick lies next to $a$. Let this be $b$.
We have, $d+c>b$ (by condition).
If the yellow ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,129 |
5. The numbers $1,2,3, \ldots, 99$ are written on the board. Petya and Vasya are playing a game, with Petya starting. Each move involves erasing three numbers that sum to 150. The player who cannot make a move loses. Which player can win, regardless of how the opponent plays? | # 5. Answer: Petya.
Solution: His first move is to erase the number 50 and any 2 numbers that sum to 100, for example, 1 and 99. He then divides all the remaining numbers into pairs with a sum of 100: (2,98), (3,97), ..., (49,51).
Vasya cannot erase two numbers from the same pair in one move, as their sum is 100, and... | Petya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,130 |
9.1. Solve the equation:
$$
\left(x^{2}-20\right)^{2}+\left(x^{2}-19\right)^{2}=2019
$$ | Solution. Let $x^{2}-20=y$; then $y^{2}+y-1009=0(D=4037)$ and $y=\frac{-1 \pm \sqrt{4037}}{2}$. For the smaller root, the reverse substitution leads to the condition $x^{2}<0$ (no solutions), the larger root gives Answer: $x= \pm \sqrt{\frac{39+\sqrt{4037}}{2}}$. | \\sqrt{\frac{39+\sqrt{4037}}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,131 |
9.2. Troublesome ninth-graders Vova and Dima tore up the school wall newspaper that criticized their behavior, diligence, and speech culture. Moreover, each piece of the newspaper that fell into Vova's hands he tore into 7 pieces, while Dima only tore into 4 (he was criticized less). Later, the school cleaner collected... | Solution. In all operations, the number of newspaper scraps increases by 6 or by 3, so the remainder of the division by 3 of the total number of scraps at any moment is preserved. At the beginning of the process (when the newspaper was whole), this remainder was 1. The number 2019 gives a zero remainder when divided by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,132 |
9.3. It is known that $a d > b c$, and also $\frac{a}{b} > \frac{c}{d}$. Prove that then
$$
\frac{a}{b} > \frac{a+c}{b+d} > \frac{c}{d}
$$ | Solution. The second of the given inequalities can be rewritten as $\frac{a d-b c}{b d}>0$, from which, taking into account the first, $b d>0$, then $b d+b^{2}>0$ and $b d+d^{2}>0$. The left part of the double inequality can be rewritten as $\frac{a d-b c}{b d+b^{2}}>0-$ which is true. Similarly, the right part of the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,133 |
9.4. From point A, tangents are drawn to a circle with center at point $O$: $B$ and $C$ are the points of tangency. Point $M$ is the midpoint of segment $AO$. Prove that the circle circumscribed around triangle $AMC$ is tangent to line $AC$. | Solution. $z$ - center of the circle passing through points $A, M, b$ - intersection of the perpendicular bisectors of $AM$ and $BM$; $D, E$ - the feet of the perpendiculars; since $M$ is the midpoint of the hypotenuse of the right triangle $ABO$, then $AM = BM$, $\angle AZD = \angle DZM = \angle MZE = \angle EZB, \ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,134 |
9.5. In the parliament of the island state of Promenade-and-Tornado, 2019 indigenous inhabitants were elected, who are divided into knights and liars: knights always tell the truth, liars always lie. At the first meeting, 2016 of them sat in the parliamentary seats arranged in the hall in a rectangle of $42 \times 48$,... | Solution. If two liars are adjacent in the hall, then the entire hall is filled with only liars, which corresponds to the maximum number of liars in the hall. For the minimum number of liars, each liar is adjacent only to knights. The rectangle $42 \times 48$ can be tiled with 224 squares of $3 \times 3$. The minimum n... | 227 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,135 |
1. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, 626 is a palindrome, while 2015 is not). Represent the number 2015 as the sum of two palindromes. | Answer: $2015=1551+464$
Comment. To find the solution, one could reason as follows:
Since 2002 does not work, the larger addend must be of the form $\overline{1 \text { AА1 }}$. Then the second addend must end in 4, as it equals 2015 - $\overline{1 \text { AА1 }}$, i.e., it has the form $\overline{4 \mathrm{~B} 4}$. ... | 2015=1551+464 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,136 |
2. A fraction in its simplest form was written on the board. Petya decreased its numerator by 1 and its denominator by 2. Vasya, on the other hand, increased the numerator by 1 and left the denominator unchanged. It turned out that the boys ended up with the same value. What exactly could their result be? | Answer. 1.
Solution. Let the fraction $\frac{a}{b}$ have been written. Then Petya obtained $\frac{a-1}{b-2}$, and Vasya obtained $\frac{a+1}{b}$. Since they got the same result, $\frac{a-1}{b-2}=\frac{a+1}{b}$, from which $b-a=1$. Therefore, the original fraction had the form $\frac{a}{a+1}$. And Petya obtained the fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,137 |
3. Dima was supposed to arrive at the station at 18:00. By this time, his father was supposed to pick him up in a car. However, Dima managed to catch an earlier train and arrived at the station at 17:05. He didn't wait for his father and started walking towards him. On the way, they met, Dima got into the car, and they... | Answer: 6 km/h
Solution. Dima arrived home 10 minutes earlier, during which time the car would have traveled the distance Dima walked twice. Therefore, on the way to the station, the father saved 5 minutes and met Dima at 17:55. This means Dima walked the distance from the station to the meeting point in 50 minutes, s... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,138 |
8. "Did you answer the previous questions honestly?"
40 gnomes answered "yes" to the first question, 50 to the second, 70 to the third, and 100 to the fourth. How many honest gnomes are there in the underground kingdom? | Answer: 40 honest gnomes.
## Solution.
On the 4th question, both an honest and a liar will answer "yes," so there are 100 gnomes in the underground kingdom.
An honest gnome will answer "yes" to one of the first three questions and "no" to two. A liar, on the other hand, will answer "yes" to two of the first three qu... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,139 |
5. In triangle $ABC$, the median from vertex $A$ is perpendicular to the bisector of angle $B$, and the median from vertex $B$ is perpendicular to the bisector of angle $A$. It is known that side $AB=1$. Find the perimeter of triangle $ABC$. | Answer: 5.
Solution. Let $A M$ be the median drawn from vertex $A$. Then, in triangle $A B M$, the bisector of angle $B$ is perpendicular to side $A M$, i.e., the bisector is also an altitude. Therefore, this triangle is isosceles, $A B = B M = 1$. Hence, $B C = 2 B M = 2$. Similarly, from the second condition, we get... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,140 |
6. There are three vessels with volumes of 3 liters, 4 liters, and 5 liters, without any markings, a water tap, a sink, and 3 liters of syrup in the smallest vessel. Can you, using pourings, obtain 6 liters of a water-syrup mixture such that the amount of water is equal to the amount of syrup in each vessel? | Solution.
For example, as follows (see the table below, c - syrup, w - water, f - final mixture).
| |  | 4-liter container | 5-liter container |
| :---: | :---: | :---: | :---: |
| Pour the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,141 |
2. In a row, 10 natural numbers are written in ascending order such that each number, except the first, is divisible by one of the preceding numbers. The first number is not equal to 1, and the sum of all 10 numbers is 275. Restore these numbers.
Let the first number be $x$, then each of the remaining 9 numbers must b... | Answer: $5,10,15,20,25,30,35,40,45,50$.
Criteria: correct answer without explanation - 2 points. | 5,10,15,20,25,30,35,40,45,50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,143 |
5. What is the maximum number of sides a polygon can have if each of its angles is either $172^{\circ}$ or $173^{\circ}$?
Let the number of angles with a degree measure of $172^{\circ}$ be $a$, and those with $173^{\circ}-b$. Then the sum of all angles of the polygon will be $172a + 173b$. On the other hand, the sum o... | Answer: 51.
Criteria: correct answer without explanation - 2 points. | 51 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,144 |
11.1. Prove that there exist at least 2020 different positive integers $\mathrm{n}$ such that the number $\mathrm{n}+0.25$ is the square of some rational number. | 11.1. Prove that there exist at least 2020 different positive integers $\mathrm{n}$ such that the number $\mathrm{n}+0.25$ is the square of some rational number.
Proof. It is sufficient to note that for any positive integer $\mathrm{m}$, the number $\mathrm{n}=\mathrm{m}^{2}+\mathrm{m}$ will satisfy the condition of t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,145 |
11.2. Prove that if the sum of pairwise distinct positive numbers $x, y, z, t$ is 1, then at least one of the numbers $\sqrt{x}+\sqrt{y}, \sqrt{x}+\sqrt{z}, \sqrt{x}+t, \sqrt{y}+\sqrt{z}, \sqrt{y}+\sqrt{t}, \sqrt{z}+\sqrt{t}$ is greater than 1. | 11.2. Prove that if the sum of pairwise distinct positive numbers $x, y, z, t$ is 1, then at least one of the numbers $\sqrt{x}+\sqrt{y}, \sqrt{x}+\sqrt{z}, \sqrt{x}+t, \sqrt{y}+\sqrt{z}, \sqrt{y}+\sqrt{t}, \sqrt{z}+\sqrt{t}$ is greater than 1.
Proof. Without loss of generality, we can assume that $x>y>z>t$. It is suf... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,146 |
11.3. Prove that the fraction $\frac{m(n+1)+1}{m(n+1)-n}$ is irreducible for all natural values of $n$ and $m$. | 11.3. Prove that the fraction $\frac{m(n+1)+1}{m(n+1)-n}$ is irreducible for all natural values of $n$ and $m$.
Proof. Suppose this is not the case. Then the difference between the numerator and the denominator, which is $n+1$, is divisible by their common divisor $d>1$. Then $1=(m(n+1)+1)-m(n+1)$ is divisible by $d$.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,147 |
11.4. In a convex quadrilateral $A B C T$, $A B=B C$ and $A T=T C$. A point $P$ is marked on the diagonal $B T$. A line passing through $P$ parallel to $BC$ intersects the line $A T$ at point $M$. A line passing through $P$ parallel to $CT$ intersects the line $AB$ at point $K$. Prove that triangles $PTK$ and $PBM$ hav... | 11.4. In a convex quadrilateral $A B C T$, $A B=B C$ and $A T=T C$. A point $P$ is marked on the diagonal $B T$. A line passing through $P$ parallel to $BC$ intersects the line $AT$ at point $M$. A line passing through $P$ parallel to $CT$ intersects the line $AB$ at point $K$. Prove that triangles $PTK$ and $PBM$ have... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,148 |
11.5 The numbers $1,2,3, \ldots, 46$ are divided into three groups. Prove that there will be at least one group in which there are two numbers whose difference is the square of some integer. | 11.5 The numbers $1,2,3, \ldots, 46$ are divided into three groups. Prove that in at least one of the groups, there will be two numbers whose difference is a perfect square.
Proof. Suppose that in each group, the difference between any two numbers is not a perfect square, i.e., not equal to the numbers $1,4,9,16,25,36... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,149 |
Problem 4.1. Four arithmetic examples were written on the board. Vera erased one plus sign, one minus sign, one multiplication sign, one division sign, and four equal signs.
Instead of identical signs, she wrote the same letters, and for different signs, different letters. Restore the examples.
![](https://cdn.mathpi... | Answer: $\mathrm{a} 4 \mathrm{~b} 5 \mathrm{c} 2 \mathrm{~d} 1 \mathrm{e} 3$.
Solution. Notice that the letter $B$ appears in all examples. It turns out that the letter $B$ must be replaced by the equals sign:
$$
\begin{aligned}
& 4 A 2=2 \\
& 8=4 C 2 \\
& 2 D 3=5 \\
& 4=5 E 1
\end{aligned}
$$
Now we will look at th... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,150 | |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,152 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,153 |
Problem 5.4. Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,157 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,159 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,163 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=$ $90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. Drop a perpendicular from point $D$ in the isosceles triangle $ABD$, let $H$ be its foot (Fig. 3). Since this triangle is acute-... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,164 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,166 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,167 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles of $... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,168 |
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