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3. For what values of the parameter a does the equation $\cos ^{4} 2 x-2(a+2) \cos ^{2} 2 x-(2 a+5)=0$ have at least one solution | Solution: Let's make the substitution $\cos ^{2} 2 x=t, t \in[0 ; 1]$, then the equation will take the form: $t^{2}-2(a+2) t-(2 a+5)=0$. We will solve the quadratic equation with respect to $t$: $t_{1,2}=\frac{2(a+2) \pm \sqrt{4(a+2)^{2}+4(2 a+5)}}{2}=\frac{2 a+4 \pm 2(a+3)}{2} ; t_{1}=-1, t_{2}=2 a+5$.
Notice that $t... | \in[-2.5,-2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,790 |
# 4. Does there exist a tetrahedron, all faces of which are isosceles triangles, and no two of them are equal? | Suppose such a tetrahedron \(ABCD\) exists. First, note that from one vertex, three equal edges cannot emanate. Indeed, if \(AB = AC = AD\), then since among the segments \(BC, BD\), and \(CD\) there are at least two equal (making \(\triangle BCD\) isosceles), among the triangles \(\triangle ABC, \triangle ABD\), and \... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,791 |
5. On a chessboard of size $2015 \times 2015$, dominoes are placed. It is known that in every row and every column there is a cell covered by a domino. What is the minimum number of dominoes required for this? | Solution Let's start filling the 2015 × 2015 square with dominoes starting from the lower left corner (see figure). The first domino will be placed horizontally, covering two vertical lines and one horizontal line. The next domino will be placed vertically, covering two horizontal lines and one vertical line. In total,... | 1344 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,792 |
1. Each of the three addends is 3 less than their sum. What are the addends? Justify your answer. | 1. Correct answer without explanation - 3 points. Full solution - 7 points. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,793 |
1. Instead of the ellipsis signs, insert such numbers so that the expression $\left(x^{2}+\ldots \times x+2\right) \times(x+3)=(x+\ldots) \times\left(x^{2}+\ldots \times x+6\right)$ becomes an identity. | Answer. $\left(x^{2}+3 x+2\right)(x+3)=(x+1)\left(x^{2}+5 x+6\right)$
Solution. Let the unknown coefficients be $a, b$, respectively:
$\left(x^{2}+a x+2\right)(x+3)=(x+b)\left(x^{2}+c x+6\right) \quad$ and bring $\quad$ the polynomials in the left and right parts to the standard form:
$$
x^{3}+(a+3) x^{2}+(3 a+2) x+... | (x^{2}+3x+2)(x+3)=(x+1)(x^{2}+5x+6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,798 |
3. Anya and Danya together weigh 82 kg, Danya and Tanya - 74 kg, Tanya and Vanya - 75 kg, Vanya and Manya - 65 kg, Manya and Anya - 62 kg. Who is the heaviest and how much does he/she weigh? | Answer. Vanya weighs 43 kg.
Solution. Adding the weights given in the condition: $82+74+75+65+62=358$, we get the doubled weight of all the children. That is, all the children together weigh $358 / 2=179$.
Anya, Danya, Tanya, and Vanya together weigh $82+75=157$, so Manya weighs 179 $157=22$.
Similarly, we find that... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,800 |
4. Solve the numerical riddle: TETA+BETA=GAMMA. (Different letters - different digits.) | Answer: $4940+5940=10880$
Solution. Since $A+A$ ends in $A$, then $A=0$. Since $\Gamma$ is the result of carrying over to the next digit, then $\Gamma=1$. Since $A+A$ ends in $A$, then $A=0$. This means there is no carry to the tens place, i.e., $T+T$ ends in $M$, and thus $M$ is even. There is also no carry to the hu... | 4940+5940=10880 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,801 |
5. In triangle $\mathrm{ABC}$, point $\mathrm{M}$ is the midpoint of $\mathrm{AC}$, $\mathrm{MD}$ and $\mathrm{ME}$ are the angle bisectors of triangles $\mathrm{ABM}$ and $\mathrm{CBM}$, respectively. Segments $\mathrm{BM}$ and $\mathrm{DE}$ intersect at point $\mathrm{F}$. Find $\mathrm{MF}$, if $\mathrm{DE}=7$. | # Answer: 3.5
Solution. By the property of the angle bisector from triangles AMB and CMB, we get that $\frac{A D}{B D}=\frac{A M}{B M}$ and $\frac{C E}{B E}=\frac{C M}{B M}$. According to the condition, $A M=C M$, therefore, $\frac{A D}{B D}=\frac{C E}{B E}$, hence, $D E \| A C$ (by the converse of Thales' theorem for... | 3.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,802 |
1. A student passed 31 exams over 5 years of study. Each subsequent year, he passed more exams than the previous year, and in the fifth year, he passed three times as many exams as in the first year. How many exams did he pass in the fourth year? | # Answer: 8.
Solution: Let $a, b, c, d, e$ be the number of exams taken in each year of study. According to the problem, $a+b+c+d+e=31, a<b<c<d<e$.
Replace the numbers $b, c, d, e$ in the equation with definitely not larger values:
$a+(a+1)+(a+2)+(a+3)+3a \leq 31$. We get $7a \leq 25$. Replacing these numbers with d... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,803 |
2. The magnitudes $\alpha$ and $\beta$ of acute angles satisfy the equation
$\sin ^{2} \alpha+\sin ^{2} \beta=\sin (\alpha+\beta)$. Prove that $\alpha+\beta=\frac{\pi}{2}$. | Solution. From the condition, it follows that $\sin \alpha(\sin \alpha-\cos \beta)=\sin \beta(\cos \alpha-\sin \beta)$.
If $\sin \alpha>\cos \beta$ and $\cos \alpha>\sin \beta$, then $1=\sin ^{2} \alpha+\cos ^{2} \alpha>\sin ^{2} \beta+\cos ^{2} \beta=1-$ contradiction. Similarly, a contradiction arises if the signs i... | \alpha+\beta=\frac{\pi}{2} | Algebra | proof | Yes | Yes | olympiads | false | 16,804 |
3. The village of knights and liars on the map has the shape of a $9 \times 9$ grid, with one person living in each cell - a knight or a liar. Knights always tell the truth, while liars always lie. Cells are considered neighbors if they share a side or a corner. Each resident said: “Among my neighbors, there is an odd ... | Solution. Divide the board into nine $3 \times 3$ squares. We will prove that in each such square there is an odd number of liars. Consider a resident from the central cell. If he is a knight, then among his neighbors there is an odd number of liars, and thus in the entire square there is an odd number of liars. If he ... | odd | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,805 |
4. A circle with center at point $O$ is circumscribed around quadrilateral $A B C D$. The diagonals of the quadrilateral are perpendicular. Find the length of side $B C$, if the distance from point $O$ to side $A D$ is 1. | Answer: $B C=2$
Solution. Let $O E \perp A D$, then $O E=1$. Draw a line through point $A$ perpendicular to $A D$, which intersects the circle at point $M$. Then $D M$ is a diameter and $D O=O M$. Since $\angle D B A=\angle D M A$ (as inscribed angles subtending the same arc), $\angle M D A=90^{\circ}-\angle D M A$ an... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,806 |
# Task 9.1
There are 2005 coins on the table. Two players play the following game: they take turns; on a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. Who will win with correct play?
... | # Answer:
The first player wins.
## Solution
We will describe the strategy for the first player. On the first move, he should take 85 coins from the table. On each subsequent move, if the second player takes x coins, then the first player should take $101 - x$ coins (he can always do this because if x is an even num... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,810 |
# Task 9.2
Factorize $x^{4}+2021 x^{2}+2020 x+2021$.
## Number of points 7
# | # Solution
Group as follows
$\left(x^{4}+x^{2}+1\right)+2020\left(x^{2}+x+1\right)$
Next,
$\left(x^{4}+x^{2}+1\right)=x^{4}+2 x^{2}+1-x^{2}=\left(x^{2}+1\right)^{2}-x^{2}=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$
So it is divisible by $x^{2}+x+1$
Therefore,
$x^{4}+2021 x^{2}+2020 x+2021=$
$=\left(x^{2}+x+1\... | (x^{2}+x+1)(x^{2}-x+2021) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,811 |
# Problem 9.3
In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
## Number of points 7 | Answer:
$30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
Figure a
!(m+1)!=(n+m)!$
Number of points 7 | Answer:
$\{(2 ; 4),(4 ; 2)\}$
## Solution
Obviously, $\mathrm{n}>1$ and $\mathrm{m}>1$ (since when $\mathrm{n}=1$ we get $2(m+1)!=(1+m)!$ and similarly when $\mathrm{m}=1$).
Divide the equation by $(n+1)!$, we get $1 \cdot 2 \cdot 3 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$
Here on the left side there are $\... | (2,4),(4,2) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,813 |
# Task 9.5
In the class, there are 30 students: excellent students, average students, and poor students. Excellent students always answer questions correctly, poor students always make mistakes, and average students answer the questions given to them strictly in turn, alternating between correct and incorrect answers.... | # Answer:
20 C-students
Solution. Let $\mathrm{a}$ be the number of excellent students, $\mathrm{b}$ be the number of poor students, $\mathrm{c}$ be the number of C-students who answered the first question incorrectly, answered the second question correctly, and answered the third question incorrectly (we will call t... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,814 |
7.2. In the album, a checkered rectangle $3 \times 7$ is drawn. Igor the robot was asked to trace all the lines with a marker, and it took him 26 minutes (the robot draws lines at a constant speed). How many minutes will it take him to trace all the lines of a checkered square $5 \times 5$? | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = 1 / 2$ minutes on the side of one cell.
The rectangle... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,816 |
7.3. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | Answer: 8.
Solution: Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,817 |
7.4. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer. 443 words.
Solution. One word is required to pronounce 29 numbers: $1,2,3,4,5,6,7,8,9,10,11,12$, $13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \cdot 7... | 443 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,818 |
7.5. In a large square hall, two square carpets were brought, the side of one carpet being twice the side of the other. When they were placed in opposite corners of the hall, they covered $4 \mathrm{~m}^{2}$ in two layers, and when they were placed in adjacent corners, they covered 14 m². What are the dimensions of the... | Answer. $19 \times 19 \mathrm{~m}^{2}$.
Solution. In the first case, the intersection of the carpets is a square with an area of 4 m² (left figure), so the length of the side of this square is 2 m. In the second case, the intersection is a rectangle, one side of which is also 2 m (right figure). Therefore, the other s... | 19\times19\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,819 |
11.4. Plane $\alpha$ intersects the edges $A B, B C, C D$ and $D A$ of the tetrahedron $A B C D$ at points $K, L, M$ and $N$ respectively. It turns out that the dihedral angles $\angle(K L A, K L M)$, $\angle(L M B, L M N)$, $\angle(M N C, M N K)$ and $\angle(N K D, N K L)$ are equal. (Here, $\angle(P Q R, P Q S)$ deno... | Solution. Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ be the projections of vertices $A, B, C, D$ respectively onto the plane $\alpha$. Let $X$ be an arbitrary point on the extension of segment $K L$ beyond point $K$. Then we have $\angle(K X A, K X N)=\angle(K L A, K L M)$ and $\angle(K N A, K N X) = \angle(N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,820 |
6.1. In a large greed pill, there are 11 g of antimatter, in a medium pill -1.1 g, and in a small pill -0.11 g. The doctor prescribed Robin-Bobin to consume exactly 20.13 g of antimatter. Can Robin-Bobin follow the doctor's prescription by eating at least one pill of each type? (If he can, explain how; if not, explain ... | Answer: will be able to.
Solution. Let's list all possible examples of sets of pills.
1) 1 large, 1 medium, 73 small; 2) 1 large, 2 medium, 63 small;
2) 1 large, 3 medium, 53 small; 4) 1 large, 4 medium, 43 small;
3) 1 large, 5 medium, 33 small; 6) 1 large, 6 medium, 23 small;
4) 1 large, 7 medium, 13 small; 8) 1 lar... | willbeableto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,821 |
6.2. Second-graders Kolya, Vasya, Misha, Stepa, and Grisha took turns correctly solving five multiplication table problems. Each subsequent boy received an answer one and a half times greater than the previous one. What numbers did Stepa multiply? | Answer: 6 and 9.
Solution. First method. Each boy received an answer $\frac{3}{2}$ times greater than the previous one. Therefore, Grisha received an answer $\frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2}=\frac{81}{16}$ times greater than Kolya. Since all the numbers received by the boys are integer... | 6\times9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,822 |
6.4. All inhabitants of the island are either knights who always tell the truth, or liars who always lie. A traveler met five islanders. In response to his question, "How many of you are knights?", the first answered: "None!", and two others answered: "One". What did the others answer? | Answer: The others answered: "Two".
Solution. The knight could not have answered "None," as this would have been a lie. Therefore, the first one is a liar. The other two answered the same, so they are either both knights or both liars. But if they were knights, there would be at least two knights on the island, and th... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,824 |
6.5. The sheikh distributed his treasures into nine bags: the first bag contained 1 kg, the second - 2 kg, the third - 3 kg, and so on, the ninth - 9 kg. The treacherous vizier stole part of the treasures from one of the bags. How can the sheikh determine in two weighings on a balance scale without weights which one ex... | Solution. We will present two possible methods of solving.
First method. I weighing. Divide the bags into groups of three bags each so that the total weights of the bags in some two groups are equal. For example, $1+3+7, 2+4+5$ and $6+8+9$. Weigh two groups of bags, in which the weight should be the same. If the scale... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,825 |
1. Ivan and Petr are running in the same direction on circular tracks with a common center, and initially, they are at the minimum distance from each other. Ivan completes one full circle every 20 seconds, while Petr completes one full circle every 28 seconds. After what least amount of time will they be at the maximum... | Answer: 35 seconds.
Solution. Ivan and Petr will be at the minimum distance from each other at the starting points after the LCM $(20,28)=140$ seconds. In this time, Ivan will complete 7 laps, and Petr will complete 5 laps relative to the starting point. Consider this movement in a reference frame where Petr is statio... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,826 |
3. Find all functions $f$, defined on the set of real numbers and taking real values, such that for any real $x$ and $y$ the equality $f(x y)=f(x) f(y)+2 x y$ holds. | Answer: There are no such functions.
Solution. Substitute 1 for x and y. Then $\mathrm{f}(1)=\mathrm{f}(1)^{2}+2$. Thus, $f(1)=a$ is a root of the quadratic equation $a^{2}-a+2=0$. Its discriminant is $1^{2}-4 \cdot 2=-7<0$. The equation has no roots, and therefore there is no such function.
Criteria. Answer without ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,828 |
4. Rational numbers $a, b$ and $с$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$.
---
Translation:
4. Rational numbers $a, b$ and $c$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=2 c^{2}$, or $(a+b)^{2}=3 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=3 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 16,829 |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that $\mathrm{CA} \cdot \mathrm{CB}=\mathrm{AB} \cdot \mathrm{B}^{2}$. Prove that the area of triangle ABI is equal to the area of quadrilateral CDIE. | Solution. Let $S(C D I E)=S_{1}, S(A B I)=S_{2}, S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio of the adjacent sides, we have
$\left(S_{1}+S_{4}\right) /\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,830 |
6. In one company, among any 9 people, there are two people who know each other. Prove that in this company, there will be a group of eight people such that each of the others knows someone from this group. | Solution. Consider the largest group $G$ of people who are pairwise unfamiliar with each other. In this group, there are no more than eight people, otherwise, there would be nine people among them, none of whom are familiar with each other, which contradicts the condition. Since this is the largest group, every other p... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,831 |
7.1. Find the number of three-digit numbers for which the second digit is less than the third by 3. | 7.1. The first digit of the number can be chosen in 9 ways (it can be any digit from 1 to 9), the second digit in 7 ways (it can be any digit from 0 to 6), and the third digit is uniquely determined. We get $9 \cdot 7 \cdot 1=63$ three-digit numbers that satisfy the condition of the problem.
Answer: 63 | 63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,832 |
7.2. Vasya left the village of Vasilki and walked to the village of Romashki. At the same time, Roma left the village of Romashki for the village of Vasilki. The distance between the villages is 36 km, Vasya's speed is 5 km/h, and Roma's speed is 4 km/h. At the same time as Vasya, Dima left Vasilki on a bicycle at a sp... | 7.2. Dima the cyclist was traveling at a speed of $5+4=9$ km/h for as long as it took Vasya and Roma to meet. Both of them walked 9 km in an hour. Therefore, the total time spent on the journey was $36: 9=4$ hours. In this time, the cyclist traveled $9 \cdot 4=36$ km. Answer: 36 km. | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,833 |
7.3. A plot of $80 \times 50$ meters is allocated for gardens and is fenced on the outside. How should 5 straight fences of the same length be installed inside the plot to divide it into 5 rectangular plots of equal area? | 7.3. One of the possible solutions is shown in the figure. The fences have a length of 40 m, their ends are marked with bold dots (the horizontal segment is composed of two fences).
 | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,834 |
7.4. The monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? | 7.4. Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed each monkey no more than one tangerine, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. We will show how to satisfy 45 monkeys:
5 monkeys eat a pear, a banana, ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,835 |
7.5. Zya decided to buy a crumblik. In the store, they also sold kryambliks. Zya bought a kryamblik and received coupons worth $50\%$ of the cost of the purchased kryamblik. With these coupons, he was able to pay $20\%$ of the cost of the crumblik. After paying the remaining amount, he bought the crumblik as well. By w... | 7.5. From the condition of the problem, it follows that $50 \%$ of the cost of a kryamblik is equal to $20 \%$ of the cost of a krumblik. This means that the kryamblik constitutes $40 \%$ of the cost of the krumblik. Zya paid the full cost of the kryamblik and the remaining $80 \%$ of the cost of the krumblik. Thus, he... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,836 |
10.1. What is the sum of the digits of the number $A=100^{40}-100^{30}+100^{20}-100^{10}+1$? | # Answer: 361.
Solution. The number is the sum of three numbers: a number composed of 20 nines followed by 60 zeros, a number composed of 20 nines followed by 20 zeros, and finally the number 1. All nines and the one fall on the zeros of the other addends, so there is no carry-over, and the answer is $180+180+1=361$.
... | 361 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,837 |
10.2. The set $M$ consists of the products of pairs of consecutive natural numbers: $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots$ Prove that the sum of some two elements of the set $M$ is equal to $2^{2021}$. | Solution. Consider the sum of two consecutive products: $S=(n-1) n+$ $n(n+1)=2 n^{2}$. Therefore, if $n^{2}=2^{2020}\left(n=2^{1010}\right)$, then $S=2^{2021}$. | 2^{2021} | Number Theory | proof | Yes | Yes | olympiads | false | 16,838 |
10.3. Given three quadratic trinomials $f(x)=a x^{2}+b x+c, g(x)=b x^{2}+c x+a, h(x)=c x^{2}+$ $a x+b$, where $a, b, c$ are distinct non-zero real numbers. From them, three equations were formed: $f(x)=g(x), f(x)=h(x), g(x)=h(x)$. Find the product of all roots of these three equations, given that each of them has two d... | Answer. 1.
Solution. Since it is known that all equations have roots, we can use Vieta's theorem. Then the product of all roots will be equal to $\frac{c-a}{a-b} \cdot \frac{a-b}{b-c} \cdot \frac{b-c}{c-a}=1$.
Comment. Correct answer without justification - 0 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,839 |
10.4. On the extension of side $AC$ of triangle $ABC$ beyond point $C$, a point $D$ is chosen. Let $S_{1}$ be the circumcircle of triangle $ABD$, and $S_{2}$ be the circumcircle of triangle $CBD$. The tangent to circle $S_{1}$ at point $A$ and the tangent to circle $S_{2}$ at point $C$ intersect at point $P$. Prove tha... | Solution. By the property of the angle between a tangent and a chord, angle $B A P$ is equal to angle $B D A$. Similarly, angle $B C P$ is equal to angle $B D C$, which is equal to angle $B D A$. Therefore, angles $B A P$ and $B C P$, subtending segment $B P$, are equal. This means that quadrilateral $A P B C$ can be i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,840 |
10.5. Given the "skeleton" of a $10 \times 10$ grid (that is, the set of vertical and horizontal segments dividing the square into unit squares, including the boundary of the square). This skeleton is divided into corners (consisting of two unit segments) and segments of length 2 (also consisting of two unit segments).... | Answer: It could not.
Solution: Consider coloring the segments in two colors: all vertical segments will be painted white, and all horizontal segments - black. Then, in each corner, there will be exactly one white and one black segment, while in a segment of length 2, there will be two segments of the same color. Note... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,841 |
10.6. For a natural number $n$, denote by $S_{n}$ the least common multiple of all numbers $1,2, \ldots, n$. Does there exist a natural number $m$ such that $S_{m+1}=4 S_{m}$?
(A. Kuznetsov) | Answer. No.
Solution. Suppose the opposite. Let $S_{m+1}$ be divisible by $2^{s}$ but not by $2^{s+1}$; then $s \geqslant 2$. This means that among the numbers $1,2, \ldots, m+1$ there is a number $a$ that is divisible by $2^{s}$. But then the number $a / 2$ does not exceed $m$ and is divisible by $2^{s-1}$; hence, $S... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,842 |
10.7. Petya took some three-digit natural numbers $a_{0}, a_{1}, \ldots, a_{9}$ and wrote the following equation on the board:
$$
a_{9} x^{9}+a_{8} x^{8}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}=*
$$
Prove that Vasya can write a certain 30-digit natural number instead of the asterisk so that the resulting equation has an int... | Solution. Let $\overline{x_{i} y_{i} z_{i}}$ be the decimal representation of a three-digit number $a_{i}$. Substituting $x = 1000$ into the left side of the equation gives $a_{9} \cdot 1000^{9} + a_{8} \cdot 1000^{8} + \ldots + a_{1} \cdot 1000 + a_{0} = \overline{x_{9} y_{9} z_{9} \underbrace{0000 \ldots 0}} + \overl... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,843 |
10.8. The bisector of angle $A$ of parallelogram $A B C D$ intersects side $B C$ at point $K$. A point $L$ is chosen on side $A B$ such that
$A L=C K$. Segments $A K$ and $C L$ intersect at point $M$. On the extension of segment $A D$ beyond point $D$, point $N$ is marked. It is known that quadrilateral $A L M N$ is c... | First solution. Since $AM$ is the bisector of angle $LAN$, segments $LM$ and $MN$ are equal as chords subtending equal arcs (see Fig. 3). Now it is sufficient to prove that $CM = LM$ (then $CM = LM = MN$, so $CNL$ is a right triangle, and $NM$ is its median drawn from the right angle).
Since $\angle BKA = \angle NAK =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,844 |
10.9. Given a natural number $k$. Along the road, there are $n$ poles at equal intervals. Misha painted them in $k$ colors and for each pair of monochromatic poles, between which there are no other poles of the same color, he calculated the distance between them. All these distances turned out to be different. For what... | Answer. $3 k-1$.
Solution. Let's number the poles from 1 to $n$ along the road and assume the distance between adjacent poles is 1. A pair of poles of the same color, between which there are no other poles of the same color, will be called good.
Estimate. Suppose $n$ poles are painted such that the condition of the p... | 3k-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,845 |
8.2. In triangle $A B C$, angle $A B C$ is 120 degrees, and points $K$ and $M$ are marked on side $A C$ such that $A K=A B$ and $C M = C B$. A perpendicular $K H$ is drawn from point $K$ to line $B M$. Find the ratio $B K : K H$. | Answer: $2: 1$.
Comments. Only the correct answer - $\underline{0 \text { points. }}$.
Solution. Since $A K=A B$ and $C M=C B$, then angle $K B M=$ angle $K B A+$ angle $M B C-$ angle $A B C=\left(180^{\circ}-\right.$ angle $\left.C A B\right) / 2+$ $\left(180^{\circ}-\right.$ angle $\left.A C B\right) / 2-120^{\circ... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,847 |
8.3. Toshi is traveling from point A to point B via point C. From A to C, Toshi travels at an average speed of 75 km/h, and from C to B, Toshi travels at an average speed of 145 km/h. The entire journey from A to B took Toshi 4 hours and 48 minutes. The next day, Toshi travels back at an average speed of 100 km/h. The ... | Answer: 290 km.
Solution. Let x km be the distance between B and C, and y km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
 each set contains at least one element;
2) any two sets have at least one common element;
3) no two sets are identical.
Prove that from the original five elements, it is possible to form one more (fifteenth) set that, together with th... | Solution. If among the 14 specified sets there is one consisting of a single element, then all the other sets include this element. There are a total of $2^{4}=16$ sets (including the empty set) without this element. By adding this element to them, we get 16 suitable sets. If among the 14 sets there are no single-eleme... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,850 |
1.1. Vasya thought of a natural number greater than 99 but less than 1000. The sum of the first and last digits of this number is 1, and the product of the first and second digits is 4. What number did Vasya think of? | Answer: 140
Solution. Note that the sum of 1 can only be obtained by adding the digits 1 and 0, and the first digit of the number cannot be 0, so it is 1, from which the last digit is 0. Since the product of the first and second digits is 4, we get that the second digit is 4, and the answer is the number 140. | 140 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,851 |
4.1. In a $4 \times 4$ square, each of the 16 cells was painted either black or white. Then, in each of the nine $2 \times 2$ squares that can be identified within this square, the number of black cells was counted. The resulting numbers were 0, 2, 2, 3, 3, 4, 4, 4, 4. How many black cells can there be in the large squ... | Answer: 11
Solution. Note that squares with 0 white cells and 4 white cells cannot intersect. If a square with 0 white cells is not in a corner, then it does not intersect with more than three other squares (the square can be in the center or adjacent to the middle of a side). Let the white square be in the bottom-lef... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,858 |
5.1. Mother gives pocket money to her children: 1 ruble to Anya, 2 rubles to Borya, 3 rubles to Vitya, then 4 rubles to Anya, 5 rubles to Borya and so on until she gives 202 rubles to Anya, and 203 rubles to Borya. By how many rubles will Anya receive more than Vitya? | # Answer: 68
Solution. Note that Anya will receive money one more time than Vitya. If we remove the first ruble, then each subsequent time Anya receives one more ruble than Vitya. Thus, it remains to determine how many such times there were. Apart from 1 ruble and 2 rubles, all the "moves" by mom break down into tripl... | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,859 |
7.1. In a hotel, rooms are arranged in a row in order from 1 to 10000. Masha and Alina checked into the hotel in two different rooms. The sum of the room numbers they are staying in is 2022, and the sum of the room numbers of all the rooms between them is 3033. In which room is Masha staying, if her room has a lower nu... | Answer: 1009
Solution. If there is exactly one room between Masha's and Alina's rooms, then it is room number 3033, but then Masha's and Alina's rooms are 3032 and 3034, and their sum is not 2022.
Consider room B, whose number is 1 more than Masha's room. Also consider room V, whose number is 1 less than Alina's room... | 1009 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,861 |
8.1. In a batch, there are 90000 boxes of vaccine weighing 3300 grams each, and 5000 small boxes weighing 200 grams each. What is the minimum number of temperature-controlled containers needed if, according to the new rules, no more than 100 kilograms can be placed in each? | Answer: 3000
Solution. Note that no more than 30 boxes of vaccines can be placed in one container. Thus, less than $90000 / 30=3000$ containers would not be enough. Also, note that if a container holds 30 boxes, there is still room for 5 small boxes, so all these small boxes will fit into the free spaces. | 3000 | Other | math-word-problem | Yes | Yes | olympiads | false | 16,862 |
9.1. Ivan wanted to buy nails. In one store, where 100 g of nails cost 180 rubles, he could not buy the necessary amount because he was short 1430 rubles. Then he went to another store, where 100 g cost 120 rubles. He bought the required amount and received 490 rubles in change. How many kilograms of nails did Ivan buy... | Answer: 3.2 kg.
According to the problem, the price difference is 60 rubles per 100 g, or 600 rubles per kilogram. If this is multiplied by the mass of the nails in kg, we get $1430+490=1920$. Therefore, Ivan bought $1920: 600=3.2$ (kg). | 3.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,863 |
9.3. For the coefficients $a, b, c$ and $d$ of the two quadratic trinomials $x^{2}+b x+c$ and $x^{2}+$ $a x+d$, it is known that $0<a<b<c<d$. Can these trinomials have a common root? | Answer: No, they cannot.
Since the coefficients of both trinomials are positive, their roots (if they exist) are negative. The common root $x_{0}$ of these trinomials is a root of their difference, that is, $x_{0}(b-a)=d-c$. From the condition, it follows that $d-c>0$ and $b-a>0$, so $x_{0}>0$. Contradiction. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,864 |
9.5. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number? | Answer: 4 numbers. Example: $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13$, $1+7+9=17, 3+7+9=19$ are prime. Evaluation. Note that among five natural numbers, it is always possible to choose three whose sum is a composite number. Consider the remainders of these five numbers when divided by 3. If there are three id... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,865 |
7.1. Ten guides were leading ten tour groups through the halls of the Hermitage. The groups did not necessarily have equal numbers, but on average, there were 9 people in each group. When one of the groups finished their tour, the average number of tourists per group decreased to 8 people. How many people were in the g... | Answer: 18.
Solution: The total number of excursionists initially was $9 \cdot 10=90$. After one of the groups finished the excursion, $8 \cdot 9=72$ excursionists remained. Therefore, the group that finished the excursion had $90-72=18$ excursionists.
Comment: A correct answer without justification - 0 points. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,867 |
7.2. In the school gym, there is one arm wrestling table. The physical education teacher organized a school tournament. He calls any two participants who have not yet met each other for a match. There are no draws. If a participant loses twice, they are eliminated from the tournament. After 29 matches were held, all pa... | Answer: 16.
Solution: Each participant is eliminated after exactly two losses. In the situation where two "finalists" remain, the total number of losses is 29. If $n$ people have been eliminated from the tournament, then they have collectively suffered $2 n$ losses, while the "finalists" could have $0(0+0), 1(0+1)$ or... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,868 |
7.3. Can a square grid $35 \times 35$ be cut into rectangular grid pieces such that the length and width of each differ by 3? | Answer: No.
Solution: Suppose such a cutting is possible. Since the length and width of the rectangle differ by 3, one of these dimensions must be an even number. Therefore, the area of each rectangle in the cutting will be an even number. This means that the sum of the areas of the rectangles in the cutting will also... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,869 |
7.4. Four pirates divided a treasure of 100 coins. It is known that among them, there are exactly two liars (who always lie) and exactly two knights (who always tell the truth).
They said:
First pirate: “We divided the coins equally.”
Second pirate: “Everyone has a different number of coins, but each got at least 15... | Answer: 40 coins
Solution. Note that the first and fourth pirates could not have told the truth simultaneously.
If the first pirate is a knight, then everyone received 25 coins (such a situation is possible). In this case, the maximum number is 25.
If the fourth pirate is a knight, then each received no more than 35... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,870 |
7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers? | Answer: 50.
Solution: Note that two consecutive numbers cannot both be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number before them (counterclockwise) is also a natural number. Since it is greater than the product of these two natural numbers, it is greater than ea... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,871 |
1. Dima is driving on a straight highway from point A to point B. From point B, a traffic jam is moving towards Dima, and the length of the jam is increasing at a speed of $v$ km/h. The speed of the car in the traffic jam is 10 km/h, and outside the jam - 60 km/h. The navigator in the car shows at any moment how much t... | Answer (in both cases): 12 km/h. Let the time between two identical GPS readings be $t$ hours. During this time, the distance between the car and the traffic jam decreased by $(60+v) t$ km, so the time to cover this distance (from the GPS's perspective) decreased by $(60+v) t / 60$ hours. On the other hand, the traffic... | 12\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,872 |
2. Sergey wrote down the numbers from 500 to 1499 in a row in some order. Under each number, except the leftmost one, he wrote the GCD of that number and its left neighbor, obtaining a second row of 999 numbers. Then he applied the same rule to get a third row of 998 numbers, from it a fourth row of 997 numbers, and so... | Answer: 501 (in the second variant 10001).
Let's prove that all numbers in the 501st row are already equal to 1. Indeed, if there is a number $d \neq 1$ in it, then in the 500th row there are 2 numbers divisible by $d$, in the 449th row - 3 such numbers, ..., in the 1st row there are 501 such numbers. But no number $d... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,873 |
3. In a regular 20-gon, four consecutive vertices $A, B, C$ and $D$ are marked. Inside it, a point $E$ is chosen such that $A E=D E$ and $\angle B E C=2 \angle C E D$. Find the angle $A E B$. | Answer: $39^{\circ}$ (in the $2^{nd}$ variant: $36^{\circ}$).
Note that $ABCD$ is an isosceles trapezoid with angles $\angle ABC = \angle DBC = 180^{\circ} \cdot 18 / 20 = 162^{\circ}$. Point E lies on the perpendicular bisector of the base $AC$, and therefore, triangle $BEC$ is isosceles. Draw the height $EH$ in it, ... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,874 |
4. Given various polynomials $f(x)$ and $g(x)$ of degree 3. It turns out that the polynomials $f(f(x))$ and $(g(x))^{3}$ are equal, as well as the polynomials $f(g(x))$ and $(f(x))^{3}$. Additionally, $f(0)=1$. Find all such pairs of polynomials $f, g$. | Answer: $f(x)=-x^{2}+3 x^{2}-3 x+1=(1-x)^{3}, g(x)=(x-1)^{3}+1$. (In the 2nd variant: $f(x)=-2-(x+2)^{3}$, $\left.g(x)=(x+2)^{3}.\right)$
Let $f(x)=a x^{3}+b x^{2}+c x+1$. By the condition, $a f^{3}(x)+b f^{2}(x)+c f(x)+1=g^{3}(x) u a g^{3}(x)+b g^{2}(x)+c g(x)+1=$ $f^{3}(x)$. Subtract the second equality from the fir... | f(x)=(1-x)^3,(x)=(x-1)^3+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,875 |
5. The exam consists of $N \geqslant 3000$ questions. Each of the 31 students has learned exactly 3000 of them, and every question is known by at least 29 students. Before the exam, the teacher openly laid out all the question cards in a circle. He asked the students to point to one of the questions and explained that ... | Answer: $N=3100$.
Each student does not know $N-3000$ questions and thus can mentally mark exactly that many cards which do not suit them as the initial one. Together, they can indicate no more than $31(N-3000)$ different cards. If $31(N-3000)<N$, then the students can indicate a card that suits everyone. This inequal... | 3100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,876 |
1. In a $3 \times 3$ table, numbers are written as shown in the figure. In one move, it is allowed to choose three cells in the shape of a three-cell corner and decrease the number in each of them by 1. Show how to use such operations to make a table where all cells contain zeros.
=183 \cdot 13^{2013}=61 \cdot 3 \cdot 13^{2013}
$$
Thus, the given sum is divisible by 61.
## Grading Criteria.
- Correct solution - 7 points.
- The number $13^{2013}$ is f... | Yes,itisdivisible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,878 |
3. Given two equations $a x^{2}+b x+c=0$ and $c x^{2}+b x+a=0$, where all coefficients are non-zero. It turns out that they have a common root. Is it true that $a=c$? | Answer. No, it is not correct.
Solution. It is enough to provide an example of two such equations. For instance, the equations $x^{2}-3 x+2=0$ and $2 x^{2}-3 x+1=0$ have a common root $x=1$.
Comment. One can indicate general properties of such equations. Let $x=t$ be a common root, meaning that $a t^{2}+b t+c=0$ and ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,879 |
4. In a certain school, every tenth-grader either always tells the truth or always lies. The principal called several tenth-graders to his office and asked each of them about each of the others, whether they are a truth-teller or a liar. In total, 44 answers of "truth-teller" and 28 answers of "liar" were received. How... | Answer: 16 or 56.
Solution. If $n$ tenth-graders are called, then $n(n-1)=44+28=72$ answers are given, from which $n=9$. Let $t$ be the number of truth-tellers and $(9-t)$ be the number of liars among these 9 students. The answer "liar" can only be given by a liar about a truth-teller and a truth-teller about a liar, ... | 16or56 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,880 |
5. Can two angle bisectors of a triangle divide it into four parts of equal area | Answer: No, they cannot.
Method 1. Suppose this is possible, i.e., the angle bisectors $A D$ and $B E$ of triangle $A B C$ divide it into four parts of equal area. Let $I$ be the point of intersection of the specified bisectors. The equal-area triangles $A I B$ and $A I E$ have a common height drawn from vertex $A$, s... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,881 |
# 6. Does there exist a natural number that is a multiple of 2015, the sum of whose digits is 2015? | Answer. Exists.
Solution. It is sufficient to provide one example of such a number. We will show a couple of ways to obtain such examples.
Example 1. Notice that $10075=2015 \cdot 5$, and the sum of the digits of the number 10075 is 13.
Then the number $\underbrace{1007510075 \ldots 10075}$ is divisible by 2015, and... | Exists | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,882 |
11.1. Each leg of a right triangle was increased by one. Could its hypotenuse have increased by more than $\sqrt{2}$? | # Answer. Could not.
(C. Vomchonkov)
Solution. First solution. Let the lengths of the legs of the original right triangle be $x$ and $y$. Then its hypotenuse had a length of $\sqrt{x^{2}+y^{2}}$, and after increasing the legs, it became $\sqrt{(x+1)^{2}+(y+1)^{2}}$. Suppose the hypotenuse increases by more than $\sqr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,883 |
11.2. In a row of 2009 weights, the weight of each weight is an integer number of grams and does not exceed 1 kg. The weights of any two adjacent weights differ by exactly 1 g, and the total weight of all the weights in grams is an even number. Prove that the weights can be divided into two piles such that the sums of ... | Solution. It is clear that the weights of all weights placed in odd positions have the same parity, while the weights of all other weights have a different parity. Since the total weight is even, the 1005 weights in odd positions have even weights.
Place the first weight on the left pan of the scales (let its weight b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,884 |
11.3. Quadrilateral $ABCD$ is inscribed in a circle with diameter $AC$. Points $K$ and $M$ are the projections of vertices $A$ and $C$ respectively onto the line $BD$. A line through point $K$ parallel to $BC$ intersects $AC$ at point $P$. Prove that angle $KPM$ is a right angle.
(T. Emelyanova) | Solution. First solution. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$. Suppose, for definiteness, that point $K$ lies on the segment $B E$. Let the line passing through $K$ parallel to $P M$ intersect $A C$ at point $N$ (see Fig. 6). Then $\triangle N K E \sim \triangle P M E$ (since their sid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,885 |
11.4. We will call a triple of natural numbers $(a, b, c)$ square if they form an arithmetic progression (in that order), the number $b$ is coprime with each of the numbers $a$ and $c$, and the number $a b c$ is a perfect square. Prove that for any square triple, there exists another square triple that shares at least ... | Solution. If $b=1$, then $a=c=1$, and another triplet can be chosen as $(1,25,49)$. If $b \neq 1$, then from the mutual simplicity of the difference of the progression $d$ cannot be zero. Then, without loss of generality, $d=b-a=c-b>0$.
Since $b$ is coprime with both $a$ and $c$, it is coprime with $ac$. Furthermore, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,886 |
1. In Grandfather Frost's magical forest, cedars grow one and a half times taller than firs and grow for 9 hours. Firs grow for 2 hours. Grandfather Frost planted cedar seeds at 12 o'clock and fir seeds at 14 o'clock. At what time were the trees of the same height? (The trees grow uniformly for the specified number of ... | Solution: Let the cedars grow to a height of 9 meters, then the firs - 6 meters. In 1 hour, cedars grow by 1 meter, and firs - by 3 meters. At 2 PM, the height of the cedars was 2 meters, and at 3 PM both cedars and firs reached a height of 3 meters. At 4 PM, the firs stopped growing and at 6 PM the cedars caught up wi... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,887 |
2. In a right triangle $\mathrm{ABC}$, from the right angle $\mathrm{B}$, the altitude $\mathrm{BH}$, the angle bisector $\mathrm{BL}$, and the median $\mathrm{BM}$ are drawn. Prove that $\angle \mathrm{LBM} = \angle \mathrm{HBL}$. | Solution. Let $\angle \mathrm{CAB} \leq 45^{\circ}$. Denote $\angle \mathrm{CAB}=\alpha$. Then $\angle \mathrm{ACB}=90^{\circ}-\alpha$.
1) In the right triangle $\triangle \mathrm{BHC}$, $\angle \mathrm{CBH}=90^{\circ}-\angle \mathrm{HCB}=90^{\circ}-(90^{\circ}-\alpha)=\alpha$. $\angle \mathrm{CBL}=45^{\circ}$ (BL is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,888 |
3. Motorcyclists Vasya and Petya are riding with constant speeds on a circular track 1 km long. Vasya noticed that Petya overtakes him every 2 minutes. Then he doubled his speed and now he himself overtakes Petya every 2 minutes. What were the initial speeds of the motorcyclists? Answer: 1000 and 1500 meters per minute... | Solution. Let Vasya's initial speed be Vv m/min, and Petya's speed be Vp m/min. Then in the first case $1000=2 \cdot(V p-V v)$, and in the second case - 1000=2$\cdot$(2Vv - Vp). Adding these equations, we get $2000=2 \cdot$ Vv. Therefore, Vv $=1000$ m/min, and then Vp=1500m/min. | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,889 |
4. Ten different natural numbers are such that the product of any 5 of them is even, and the sum of all 10 numbers is odd. What is their smallest possible sum | Answer: 65.
Solution. The product of 5 odd numbers is odd $\Rightarrow$ among the given ten numbers, there are no more than 4 odd numbers. 4 odd numbers are also impossible, since then the sum of all numbers would be even, while according to the condition, it is odd. Therefore, there are a maximum of 3 odd numbers. If... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,890 |
5. There are candies in five bags. In different bags, there are different numbers of candies, and in total, there are 21 candies. It is known that candies from any two bags can be distributed into the three remaining bags so that the number of candies in these three bags becomes equal. Prove that there is a bag with ex... | Solution. Suppose there is a bag A, which contains 8 or more candies. Then, according to the condition, candies from other bags D, E can be distributed into bags A, B, C so that the number of candies in them becomes equal. Since the number of candies in A has not decreased, there are at least 8 candies in each bag, mea... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,891 |
1. Kostya walked in the park for a long time: he left the central point of the park and, walking along the paths, returned to the central point (possibly passing through the central point several times). On the diagram, the forks are marked with bold dots. When Kostya arrived at a fork, he continued moving without turn... | Answer: No. On all paths leading from the inner circle to the outer circle,
Kostya walked a total of $10+15+20+20=65$ times. However, he should have crossed from the inner circle to the outer... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,892 |
11.1. Solve the inequality: $\sqrt{(x-2)^{2}\left(x-x^{2}\right)}<\sqrt{4 x-1-\left(x^{2}-3 x\right)^{2}}$. | 11.1. Answer: $x=2$.
The left side of the original inequality is defined at $x=2$ and. At the point $x=2$, the inequality is true. We will prove that there are no solutions on the interval $[0 ; 1]$. For this, we will square both sides and bring the inequality to the form $x^{2}(1-x)<-1$. The last inequality is not sa... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,893 |
11.2. The base of the pyramid $S A B C D$ is a convex quadrilateral $A B C D$ such that $B C \cdot A D = B D \cdot A C$. It turns out that $\angle A D S = \angle B D S$ and $\angle A C S = \angle B C S$. Prove that the plane $S A B$ is perpendicular to the plane of the base.
. The maximum absolute value of the difference is 19, the minimum is -1, and there are 19 different values. However, the difference of 19 can be obtained in o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,895 |
11.4. On the graph of the function $y=\frac{1}{x}$, two points $A$ and $B$ with positive abscissas are selected. Points $H_{A}$ and $H_{B}$ are the bases of the perpendiculars dropped from these points to the x-axis, and $O$ is the origin. Prove that the area of the figure bounded by the lines $O A, O B$ and the arc $A... | 11.4. We can assume that the abscissa of point $A$ is less than the abscissa of point $B$. Let $K$ be the intersection point of segments $A H_{A}$ and $O B$. Since $O H_{A} \cdot A H_{A} = O H_{B} \cdot B H_{B} = 1$, the areas of triangles $O A H_{A}$ and $O B H_{B}$ are equal, and thus the areas of triangle $O A K$ an... | proof | Calculus | proof | Yes | Yes | olympiads | false | 16,896 |
11.5. We have the number 1. Petya (starting the game) and Vasya take turns performing the following operations on the current number: in one move, they multiply it by one of the numbers $2,3,4 \ldots 10$, and then add one of the numbers 1, 2, 3... 10 to it. The winner is the one who first gets a number not less than 10... | # 11.5. Answer: Petya
Let's use retrospective analysis (analysis from the end). The player who has a number from 99 to 999 on their turn wins, the player who has 49, ..., 98 loses, the player who has 4, ..., 48 wins, the player who has 2, 3 loses, and the player who has 1 wins. Thus, the player who starts the game, Pe... | Petya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,897 |
1. All horizontal and vertical distances between adjacent points are equal to 1. What is the area of the triangle with vertices at the black points? | Answer. 1.
Solution. The area of the triangle can be found, for example, by subtracting from half the area of the square the area of the square and the areas of two right triangles. We get $S=10-4-2-3=1$. The figure can be divided in other ways.
Comment. An answer without justification - 0 points. A partition that al... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,898 |
2. There were 1009 gnomes with 2017 cards, numbered from 1 to 2017. Ori had one card, and each of the other gnomes had two. All the gnomes knew only the numbers on their own cards. Each gnome, except Ori, said: "I am sure that I cannot give Ori any of my cards so that the sum of the numbers on his two cards would be 20... | Answer: 1009
Solution: If one of the gnomes has a card with the number 1, then he must be sure that Ori does not have a card with the number 2017. This can only be certain if the card with the number 2017 is also in the hands of this gnome. Similarly, the gnome with the card that has the number 2 has another card with... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,899 |
3. Cut a $6 \times 6$ square (see the figure) along the grid lines into 4 equal parts so that each part contains all the digits. Parts are considered equal if they can be perfectly superimposed on each other by rotation only.
| 2 | 9 | 9 | 8 | 7 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 2 | 4 | 4 | 3 | 6... | Solution. See the figure
| 2 | 9 | 9 | 8 | 7 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 2 | 4 | 4 | 3 | 6 |
| 7 | 5 | 1 | 1 | 5 | 6 |
| 8 | 5 | 1 | 1 | 5 | 9 |
| 8 | 3 | 4 | 4 | 2 | 8 |
| 3 | 9 | 6 | 7 | 7 | 2 |
Comment. A correct example is provided - 7 points. A correct figure is sufficient justificati... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,900 |
4. Vanya wrote down a four-digit number, subtracted a two-digit number from it, multiplied the result by a two-digit number, divided by the sum of two single-digit numbers, added a single-digit number, and then divided the result by the sum of three single-digit numbers. To write all the numbers, he used only one digit... | Answer: 2017; any digit.
Solution. Let the digit be $a$. We get $\left(\frac{(\overline{a a a a}-\overline{a \bar{a}} \cdot \cdot \overline{a a}}{a+a}+a\right):(a+a+a)$. Then
$$
\overline{a a a a}-\overline{a a}=\overline{a a 00}=a \cdot 1100 \Rightarrow \frac{\bar{a}}{a a}=a \cdot 11
$$
The numerator of the fractio... | 2017 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,901 |
5. The giants were prepared 813 burgers, among which are cheeseburgers, hamburgers, fishburgers, and chickenburgers. If three of them start eating cheeseburgers, then in that time two giants will eat all the hamburgers. If five take on eating hamburgers, then in that time six giants will eat all the fishburgers. If sev... | Answer: 252 fishburgers, 36 chickenburgers, 210 hamburgers, and 315 cheeseburgers.
Solution: Let $a, b, c$, and $d$ be the quantities of cheeseburgers, hamburgers, fishburgers, and chickenburgers, respectively. According to the problem, $a + b + c + d = 813$. The statement that while three people eat cheeseburgers, tw... | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,902 |
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 | 16,906 |
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_564... | 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 | 16,907 |
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 | 16,908 |
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. How many emeralds are there in total 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 | 16,909 |
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 | 16,911 |
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?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all the moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will ... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,912 |
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