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3. The circle $\omega$ is circumscribed around an acute-angled triangle $ABC$. The tangent to the circle $\omega$ at point $C$ intersects the line $AB$ at point $K$. The point $M$ is the midpoint of the segment $CK$. The line $BM$ intersects the circle $\omega$ again at point $L$, and the line $KL$ intersects the circl... | Solution. Since $K C$ is tangent to the circle $\omega, \angle M C B=$ $\angle M L C$ and, therefore, triangles $M C B$ and $M L C$ are similar. Hence, $\frac{M B}{M C}=$ $\frac{M C}{M L}$. Consequently, $\frac{M B}{M K}=\frac{M K}{M L}$. This means that triangles $M K B$ and $M L K$ are similar. Thus,
$$
\angle C K N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,873 |
4. The numbers $u$ and $v$ are roots of the quadratic polynomial $x^{2} + a x + b$ with integer coefficients. For any natural number $n$, prove that if $a^{2}$ is divisible by $b$, then $u^{2 n} + v^{2 n}$ is divisible by $b^{n}$. | Solution. We will prove by induction on $n$ that $u^{n}+v^{n}$ is divisible by $b^{[n / 2]}$. The base cases $n=1$ and $n=2$ are easily verified: $u+v$ is divisible by $b^{0}=1$ and $u^{2}+v^{2}=$ $(u+v)^{2}-2 u v=a^{2}-2 b$ is divisible by $b^{1}$. We will establish the transition from $n-2$ and $n-1$ to $n$. Since $u... | proof | Algebra | proof | Yes | Yes | olympiads | false | 7,874 |
5. Given a checkerboard $n \times n$. A rook moves according to standard chess rules (i.e., in one move, the rook can be moved to a cell in the same row or the same column). For which $n$ is it possible to visit all cells of the board exactly once, if immediately after a horizontal move of the rook, a vertical move mus... | Answer: $n$ - even
Solution. Let $n$ be odd. Choose a column in which the rook neither started nor ended its route. Consider the first visit of the rook to this column, which was along a row. Therefore, the next move was along a column, and the rook visited a cell in the same column and then left it. Similarly, on the... | n- | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,875 |
6. Find all pairs of natural numbers $n$ and $k$ for which $(n+1)^{k}=$ $n!+1$. (As usual, $n!$ denotes the product of all natural numbers not exceeding $n$. For example, $4!=1 \cdot 2 \cdot 3 \cdot 4$.) | Answer: $n=k=1, n=2$ and $k=1, n=4$ and $k=2$.
Solution. If $n=1$, then $2^{k}=1!+1=2$, so $k=1$. Let $n \geqslant 2$, then the number $n!+1$ is odd and, therefore, $n+1$ is also odd. This means that $n$ is even. If $n=2$, then $3^{k}=2!+1=3$, so $k=1$. If $n=4$, then $5^{k}=4!+1=25$, so $k=2$. Now let $n \geqslant 6$... | n=k=1,n=2k=1,n=4k=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,876 |
1. (10 points) If two sides of the first triangle are respectively equal to two sides of the second triangle and the angle opposite one of these sides in the first triangle is equal to the angle opposite the corresponding equal side in the second triangle, then these triangles
a) are equal
b) are not equal
c) may be... | Answer: c) and d).
Solution:
The triangles mentioned in the problem can obviously be equal, for example, if they are equilateral. At the same time, equal triangles have equal areas, so answ... | )) | Geometry | MCQ | Yes | Yes | olympiads | false | 7,877 |
2. (10 points) Choose the correct statements, assuming that $A$ is the area of a square with a diagonal of $2 \sqrt{2}$, $B$ is the area of a rectangle whose center and two adjacent vertices have coordinates $(2,2)$, $(4,1)$, and $(4,3)$ respectively, and $C$ is the area of the triangle formed by the coordinate axes an... | Answer: a) and d).
Solution: It is not difficult to see that
- $A=4 \cdot \frac{1}{2} \cdot \sqrt{2} \cdot \sqrt{2}=4$
- the coordinates of the two remaining vertices of the rectangle will be $(0,3)$ and $(0,1)$ - these are the coordinates of the points symmetric to the points $(4,1)$ and $(4,3)$ with respect to the ... | )) | Geometry | MCQ | Yes | Yes | olympiads | false | 7,878 |
3. (20 points) Young marketer Masha was supposed to survey 50 customers in an electronics store throughout the day. However, there were fewer customers in the store that day. What is the maximum number of customers Masha could have surveyed, given that according to her data, 7 of the respondents made an impulse purchas... | Answer: 47.
Solution: Let the number of customers surveyed be $x$. Then, the number of customers who made a purchase under the influence of advertising is $(x-7) \cdot 3 / 4$, and the number of customers who made a purchase on the advice of a sales consultant is $(x-7)/4$. Since the number of customers can only be an ... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,879 |
4. (20 points) The reception hall in the palace of the tritriandiat kingdom consists of points on the plane whose coordinates satisfy the condition $4|x|+5|y| \leqslant 20$. How many identical two-sided parquet tiles, shaped as right triangles with legs of 1 and 5/4, are needed to tile the floor of the hall? Tiling is ... | Answer: 64.
Solution: It is not hard to see that the reception hall is a rhombus with vertices at points $(-5,0),(0,4),(5,0)$ and $(0,-4)$, and each quarter of the hall (bounded by the coordinate axes and one of the sides, i.e., forming a right-angled triangle) is similar to one parquet tile with a similarity ratio of... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,880 |
5. (30 points) The midpoints of sides $B A, A N, N B$ of triangle $N B A$ are denoted by points $L$, E and $D$ respectively, and the point of intersection of the angle bisectors of triangle $N B A$ is denoted by X. $P$ is the point of intersection of lines $B X$ and $E L, V-$ lines $A X$ and $D L$, and at points $T$ an... | # Solution:
Since $E L \| N B$, then $\angle L P B = \angle P B D$, but $\angle P B D = \angle P B L$ by the condition, so triangle $L B P$ is isosceles with $L B = L P$.
Similarly, triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,881 |
6. (30 points) In triangle $K I A$, where side $K I$ is less than side $K A$, the bisector of angle $K$ intersects side $IA$ at point $O$. Let $N$ be the midpoint of $IA$, and $H$ be the foot of the perpendicular dropped from vertex $I$ to segment $KO$. Line $IH$ intersects segment $KN$ at point $Q$. Prove that $OQ$ an... | Solution: Let $I Q$ intersect side $K A$ at point $V$. In triangle $K I V$, the height $K H$ is also the bisector, and therefore the median, so $I H = H V$. Therefore, $H N$ is the midline of triangle $I V A$ and $H N \| A K$.
Let $W$ be the point of intersection of $N H$ and $K I$. We get that $N W$ is the midline of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,882 |
7. (40 points) Among the pensioners of one of the stars in the Tau Ceti system, the following pastime is popular: a board with a grid of 2016 by 2017 is painted in gold and silver in a checkerboard pattern, after which the numbers 0 or 1 are written at the vertices of each cell in such a way that the sum of the numbers... | Answer: 0, 2 or 4.
Solution: By adding the sums of the numbers at the vertices of all cells, we get an even number, since the entire board consists of an even number of cells. Since all vertices, except for the corner ones, are counted an even number of times, and the corner ones are counted once, the sum of the numbe... | 0,2or4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,883 |
8. (40 points) At the "Horizon" base, 175 university students arrived. Some of them are acquainted with each other, while others are not. It is known that any six students can be accommodated in two three-person rooms such that all those in the same room are acquainted with each other. What is the minimum number of pai... | Answer: 15050.
Solution: It is clear that each student has no more than three unfamiliar students. If each student has no more than two unfamiliar students, then each is familiar with at least 172 students, and the total number of pairs of familiar students is no less than $175 \times 172 / 2 = 15050$. Suppose there i... | 15050 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,884 |
1. (10 points) 16 children of different heights stood in a circle, all facing the center. Each of them said: “My right neighbor is taller than my left neighbor.” What is the minimum number of children who could be lying?
.
Solution: Let's number the positions of the children in the circle from 1 to 16 counterclockwise. We will divide the children into 2 groups: those standing in positions with odd numbers - the first group, and those standing in positions with even numbers - the second group. Note that the statements of chil... | b)1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 7,885 |
2. (10 points) Vasya has:
a) 2 different volumes from the collected works of A. S. Pushkin; the height of each volume is 30 cm;
b) the collected works of E. V. Tarle in 4 volumes; the height of each volume is 25 cm;
c) a book of lyrical poems, 40 cm tall, published by Vasya himself.
Vasya wants to arrange these book... | Answer: a).
Solution: From the condition of symmetric arrangement, two conclusions can be drawn.
1) To the left and right of Vasya's poems, there should be one volume of Pushkin and two volumes of Tarle.
2) For the volumes of Pushkin, 3 pairs of positions are allowed: $(1,7),(2,6),(3,5)$. The remaining free positions... | 3\cdot2!\cdot4! | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,886 |
3. (10 points) The diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect at point $P$, and a point $M$ is marked on diagonal $A C$ such that $A M = M C$. Mark the correct statements.
a) The distances from point $M$ to the ends of diagonal $B D$ are equal: $B M = M D$.
b) The sum of the areas of triangles $A ... | Answer: b).
Solution: Let's consider each statement.
a) The equality $B M = D M$ is equivalent to the point $M$ lying on the perpendicular bisector of $B D$. The example on the left shows a ... | b) | Geometry | MCQ | Yes | Yes | olympiads | false | 7,887 |
4. (20 points) In the Martian calendar, a year consists of 5882 days, and each month has either 100 or 77 days. How many months are there in the Martian calendar? | Answer: 74.
Solution: Let $x$ be the number of months with 100 days, and $y$ be the number of months with 77 days. According to the problem, $100 x + 77 y = 5882$. It is obvious that $y \leqslant 66$, otherwise $x < 0$. Notice that
$$
x \bmod 11 = 100 x \quad \bmod 11 = 5882 \quad \bmod 11 = 8
$$
Thus, $x = 11 k + 8... | 74 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,888 |
5. (20 points) Professor K., wishing to be known as a wit, plans to tell no fewer than two but no more than three different jokes at each of his lectures. At the same time, the sets of jokes told at different lectures should not coincide. How many lectures in total will Professor K. be able to give if he knows 8 jokes? | Answer: 84.
Solution: The professor can use all possible triplets of anecdotes (the number of which is $C_{8}^{3}=56$), as well as all pairs of anecdotes (which is $C_{8}^{2}=28$). Therefore, the maximum number of sets of anecdotes the professor can use in lectures is $56+28=84$. | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,889 |
6. (20 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different five-letter words can be formed from the letters of the word САМСА? And from the letters of the word ПАСТА? In your answer, indicate the sum of the found numbers. | Answer: 90.
Solution: In the word SAMSA, the letter A appears twice and the letter S appears twice. Therefore, the number of different words will be $\frac{5!}{2!\cdot 2!}=30$. In the word PASTA, only the letter A appears twice. Therefore, the number of different words in this case will be $\frac{5!}{2!}=60$. In total... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,890 |
7. (30 points) On the side $AC$ of triangle $ABC$, where $\angle ACB=45^{\circ}$, a point $K$ is marked such that $AK=2KC$. On the segment $BK$, there is a point $S$ such that $AS \perp BK$ and $\angle AKS=60^{\circ}$. Prove that $AS=BS$.
 The altitudes of an acute, non-isosceles triangle \(ABC\), dropped from vertices \(A\) and \(C\), intersect at point \(H\), and also intersect the angle bisector of \(\angle ABC\) at points \(F\) and \(G\) respectively. Prove that triangle \(FGH\) is isosceles.
 or point $F$ lies between points $B$ and $G$ (right figure).
Let $Q$ and $S$ be the feet of the altitudes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,892 |
9. (30 points) On the side $B C$ of the square $A B C D$, points $E$ and $F$ are marked such that $B E: E C = C F: F B = 1: 2$. On the side $C D$, point $G$ is marked such that $C G: G D = 2: 1$. On the side $A D$, points $H$ and $I$ are marked such that $A I: I D = D H: H A = 1: 2$. Segment $B G$ intersects segments $... | Answer: The area of $G D H L$ is larger.
Solution: Let $S_{A B C D}=36 S, S_{C L G}=x$. Then $S_{C D H}=6 S$ and $S_{G D H L}=6 S-x$, $S_{B C G}=12 S$ and $S_{B C L}=12 S-x$. From the equali... | S_{GDHL}=\frac{42}{11}S,S_{EFKJ}=\frac{36}{11}S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,893 |
10. (40 points) Let $P(n)$ denote the product of the digits of a natural number $n$. For what largest natural number $k$ does there exist a natural number $n>10$ such that
$$
P(n)<P(2 n)<\ldots<P(k n) ?
$$ | # Answer: 9.
Solution: First, let's show that $k<10$. Indeed, for $k \geqslant 10$, the set of numbers includes $P(10 n)$. It equals zero because the number $10 n$ ends in 0. Then $P(n)<0$, which is impossible.
Now let's provide an implementation for $k=9$. We can take $n$ as a number of the form $\underbrace{\overli... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,894 |
11. (40 points) The numbers from 1 to 600 are divided into several groups. It is known that if a group contains more than one number, then the sum of any two numbers in this group is divisible by 6. What is the minimum number of groups that can be formed? | Answer: 202.
Solution: For $k=0,1, \ldots, 5$, let $G_{k}$ be the set of numbers from 1 to 600 that give a remainder of $k$ when divided by 6. Suppose a number $a$ from $G_{k}$ is included in some group. If another number $b$ is also in this group, then it belongs to $G_{6-k}$, otherwise $a+b$ is not divisible by 6. S... | 202 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,895 |
12. (40 points) Five boys played a word game: each of them wrote down 7 different words. It turned out that each boy had exactly 2 words that were not found in any of the other boys' lists. What is the maximum number of different words that the boys could have written in total? | Answer: 22.
Solution: In total, $5 \times 7=35$ words were written. Since each boy wrote exactly 2 words that did not appear in any of the other boys' writings, there were a total of $5 \times 2=10$ such unique words. From the remaining 25 words (repeated), each was written at least twice. Therefore, there are no more... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,896 |
13. (40 points) In a store where all items cost a whole number of rubles, two special offers are in effect:
1) A customer who buys at least three items can choose one item as a gift (free of charge), the cost of which does not exceed the minimum cost of the paid items;
2) A customer who buys exactly one item for no les... | Answer: 504.
Solution: Let $S$ denote the total cost of the four items that interested the customer, and let $X$ be the minimum possible cost of the items. Being interested in exactly four items means that the customer can take advantage of either offer 1 or offer 2, but not both.
Let the items cost $s_{1}, s_{2}, s_... | 504 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,897 |
14. (40 points) Prove that among the three numbers $777^{2021} \cdot 999^{2021}-1, 999^{2021} \cdot 555^{2021}-1$ and $555^{2021} \cdot 777^{2021}-1$ only one is divisible by 4. | Solution: Let $m$ and $n$ be odd natural numbers. If they give the same remainder when divided by 4, then either $m n \bmod 4=1$, or $m n \bmod 4=9 \bmod 4=1$. In either case, $m n-1$ is divisible by 4. If $m$ and $n$ give different remainders when divided by 4, then $m n \bmod 4=3$, and $m n-1$ is not divisible by 4.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,898 |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,902 |
8. (40 points) $B$ In the 8-A class, there are $n$ students ( $n \geq 2$ ). For them, clubs are organized, each of which is attended by at least two students. Any two clubs that have at least two common students differ in the number of participants. Prove that the number of clubs is no more than $(n-1)^{2}$. | Solution: It is sufficient to prove that for each $2 \leq k \leq n$ there are no more than $\frac{n(n-1)}{k(k-1)}$ circles attended by exactly $k$ people. Indeed, since there cannot be fewer than 2 or more than $n$ people in a circle, the number of circles does not exceed
$$
n(n-1)\left(\frac{1}{1 \cdot 2}+\frac{1}{2 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,903 |
1. Prove that for any natural $n$ there exist such integers $x$ and $y$ that $x^{2}+y^{2}-2017$ is divisible by $n$. | Solution. As $x$ and $y$, one can take the numbers 44 and 9. They are suitable for all $n$, since $44^{2}+9^{2}-2017=0$. | 44^{2}+9^{2}-2017=0 | Number Theory | proof | Yes | Yes | olympiads | false | 7,904 |
4. Given an isosceles triangle $A B C$ with base $B C$. Point $M$ is the midpoint of side $A C$, point $P$ is the midpoint of $A M$, and point $Q$ is marked on side $A B$ such that $A Q=3 B Q$. Prove that $B P+M Q>A C$.
. Consider the time interval during which each runner will have run an integer number of laps (for example, 29! hours). Since the stand arc constitutes $\frac{1}{10}$ of the track length, each runner will ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,908 |
6. On the board, there are 10000 odd numbers that are not divisible by 5. Prove that it is possible to choose several numbers such that their sum ends in 1379. (If one number is chosen, then it is equal to the sum.) | Solution. Let's check two statements first.
1) If a number $m$ is coprime with 10000, and numbers $p$ and $q$ give different remainders when divided by 10000, then numbers $pm$ and $qm$ also give different remainders. Indeed, $pm - qm = (p - q)m$. The first factor does not divide 10000, and the second is coprime with ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,909 |
1. Prove that for any natural $n$ there exist such integers $x$ and $y$ that $x^{2}+y^{2}-2018$ is divisible by $n$. | Solution. As $x$ and $y$, one can take the numbers 43 and 13. They are suitable for all $n$, since $43^{2}+13^{2}-2018=0$. | 43^{2}+13^{2}-2018=0 | Number Theory | proof | Yes | Yes | olympiads | false | 7,910 |
2. Kostya "combined" the addition table and the multiplication table of natural numbers: he constructed a table whose rows and columns correspond to natural numbers starting from 3, and filled in all the cells: in the cell at the intersection of the $r$-th row and $s$-th column, he placed the number $r s - (s + r)$.
!... | Solution. If the number $n+1$ is not prime, then it can be factored into a product of two factors, both greater than 1. Let's write them as $r-1$ and $s-1$, where $r$ and $s$ are natural numbers, $r, s \geqslant 3$. Then
$$
n=n+1-1=(r-1)(s-1)-1=r s-s-r
$$
which means the number $n$ is contained in the table. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,911 |
4. Given an isosceles triangle $A B C$ with base $B C$. Point $M$ is the midpoint of side $A B$, point $Q$ is the midpoint of $A M$, and point $P$ is marked on side $A C$ such that $A P=3 P C$. Prove that $P Q+C M>A B$.
. Consider the time interval during which each runner will have run an integer number of laps (for example, 29! hours). Since the stand arc constitutes $\frac{1}{20}$ of the track length, each runner will ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,914 |
6. On the board, there are 1000 odd numbers that are not divisible by 5. Prove that it is possible to choose several numbers such that their sum ends in 713. (If one number is chosen, then it is equal to the sum.) | Solution. Let's check two statements first.
1) If a number $m$ is coprime with 10000, and numbers $p$ and $q$ give different remainders when divided by 10000, then numbers $pm$ and $qm$ also give different remainders. Indeed, $pm - qm = (p - q)m$. The first factor does not divide 10000, and the second is coprime with ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,915 |
1. (10 points) On the board, all three-digit numbers divisible by 5 were written, where the hundreds digit is greater than the tens digit, and the tens digit is greater than the units digit. There turned out to be $A$ such numbers. Then, all three-digit numbers divisible by 5 were written, where the hundreds digit is l... | Answer: a)
Solution: Let's denote the three-digit number as $\overline{x y z}$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. According to the problem, the numbers we are looking for are divisible by 5, which means $z$ is either 0 or 5.
Let's determine what $A$ is. If $z=0$, we a... | ) | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,916 |
2. (10 points) Vasya has 9 different books by Arkady and Boris Strugatsky, each containing one work by the authors. Vasya wants to arrange these books on a shelf so that: a) the novels "The Beetle in the Anthill" and "Waves Extinguish the Wind" (in any order) are next to each other; b) the novellas "Anxiety" and "A Tal... | Answer: a).
Solution: Let's write down the solution for the general case when there are $N$ different books. Vasya needs to arrange in a row $N-2$ objects: $N-4$ books not mentioned in points a) and b), and two pairs from a) and b). This can be done in $(N-2)!$ ways. In addition, within each pair, the books can be swa... | 4\cdot7! | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,917 |
4. (20 points) In math class, each dwarf needs to find a three-digit number such that when 198 is added to it, the result is a number with the same digits but in reverse order. For what maximum number of dwarfs could all the numbers they find be different? | Answer: 70.
Solution: By the Pigeonhole Principle, the maximum number of gnomes is equal to the number of numbers that satisfy the condition of the problem.
Let the three-digit number be denoted as $\overline{x y z}$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. Since the number... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,919 |
6. (30 points) On the side $N A$ of triangle $N B A$, points $Q$ and $F$ are marked such that $N Q = F A = N A / 4$. A point $L$ is chosen on the segment $Q F$. Lines through points $Q$ and $F$ parallel to $B L$ intersect sides $N B$ and $A B$ at points $D$ and $K$ respectively. Is it true that the sum of the areas of ... | Answer: Yes, correct.
## Solution:
Since $L B \| Q D$, the areas of triangles $L Q D$ and $B Q D$ are equal - these triangles have the base $D Q$, and their heights are equal to the distanc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,921 |
7. (30 points) Before the lesson, the teacher wrote the following problem on the board: «In a circle of radius $r$, an isosceles triangle is inscribed, the sum of the lengths of the base and the height of which is equal to the diameter of the circle. Find the height of the triangle.» However, the underachiever Vasya se... | Answer: No, he will not be able to.
Solution: The length of side $a$ of the triangle cannot exceed the diameter of the circle, and the height $h$ of the inscribed triangle is less than the diameter of the circumscribed circle. Therefore, the sum of the lengths of the base and the height is less than $4 r$. Since $\pi>... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,922 |
8. (30 points) D student Vasya dreamed that the following statement is true: if in triangle $ABC$ the median $CC_1$, drawn to side $AB$, is greater than the median $AA_1$, drawn to side $BC$, then $\angle CAB$ is less than $\angle BCA$. Excellent student Petya believes that this statement is incorrect. Determine who is... | Answer: Petya is right.
Solution: First, let's prove the auxiliary statement: the larger side of a triangle corresponds to the smaller median. Consider triangle $KMN$, where $MN > MK$. Let t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,923 |
9. (40 points) On an island, there live only 50 knights, who always tell the truth, and 15 commoners, who can either tell the truth or lie. A scatterbrained professor, who came to the island to give a lecture, forgot what color hat he was wearing. How many of the local residents should the professor ask about the color... | Answer: 31.
Solution: Since the professor will only ask about the color of the hat, to accurately determine the color of the hat, it is necessary to survey more knights than commoners. In the worst case, among those surveyed, there could be all the commoners on the island, i.e., 15 people; therefore, it is necessary t... | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,924 |
1. (10 points) In a $3 \times 5$ table, all numbers from 1 to 15 were arranged such that the sum of the numbers in any column of three cells is divisible by 3. Then some of them were erased and replaced with numbers $a, b, c$:
| 11 | $a$ | 3 | 8 | 10 |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 9 | 13 | 15 | 12 |... | Answer: $a=1, b=14$.
Solution: For the sum of the numbers in each column to be divisible by three, it is necessary that 1) $c$ is divisible by three; 2) $a$ gives a remainder of $1$ when divided by three; 3) $b$ gives a remainder of $2$ when divided by three. Therefore, the fourth option does not fit.
According to th... | =1,b=14 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 7,925 |
2. (20 points) How many even 100-digit numbers exist, each digit of whose decimal representation is one of the digits 0, 1, or 3? | Answer: $2 \cdot 3^{98}$.
Solution: According to the condition, the number must be even, so the least significant digit can only be the digit 0; for the hundredth
place, we have two options - the digit 1 or the digit 3; for the other places (from the second to the ninety-ninth) - three options - any of the digits $1,2... | 2\cdot3^{98} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,926 |
3. (20 points) How many numbers from the sequence
20142015, 201402015, 2014002015, 20140002015, 201400002015, ... are perfect squares? | Answer: None.
Solution: The sum of the digits of any number in this sequence is 15. The number 15 is divisible by 3, but not by 9. Therefore, there cannot be any perfect squares in the sequence. | None | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,927 |
4. (20 points) Arrange the numbers $\Psi, \Omega, \Theta$ in non-increasing order, if
$$
\begin{aligned}
& \Psi=\frac{1}{2} \cdot(1+2-3-4+5+6-7-8+\ldots-2012) \\
& \Omega=1-2+3-4+\ldots-2014 \\
& \Theta=1-3+5-7+\ldots-2015
\end{aligned}
$$ | Answer: $\Theta, \Omega, \Psi$.
Solution:
$$
\begin{gathered}
\Psi=\frac{1}{2} \cdot(-4) \cdot 503=-1006 \\
\Omega=(-1) \cdot 1007=-1007 \\
\Theta=(-2) \cdot 504=-1008
\end{gathered}
$$ | \Theta,\Omega,\Psi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,928 |
5. (30 points) In the royal dining hall, there are three tables, and three identical pies are served. For lunch, the king invited six princes to his table. At the second table, one can seat from 12 to 18 courtiers, and at the third table, from 10 to 20 knights. Each pie is cut into equal pieces according to the number ... | Answer: Both guests and courtiers will be 14 people on that day.
First solution: Let the number of courtiers at the table be $a$, and the number of knights be $-b$, then the dining rule can be written as $\frac{1}{a} + \frac{1}{b} = \frac{1}{7} \Leftrightarrow \frac{1}{b} = \frac{1}{7} - \frac{1}{a}$. Maximizing the v... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,929 |
6. (40 points) On a circle with center $O$, points $A$ and $B$ are taken such that the angle $A O B$ is $60^{\circ}$. From an arbitrary point $R$ on the smaller arc $A B$, segments $R X$ and $R Y$ are drawn such that point $X$ lies on segment $O A$ and point $Y$ lies on segment $O B$. It turns out that the angle $R X O... | First solution: Let $r_{1}$ be the radius of the circle centered at point $O$. Since $\angle R X O + \angle R Y O = 180^{\circ}$, the quadrilateral $R X O Y$ is cyclic; denote the radius of its circumscribed circle by $r_{2}$. The diagonal $O R$ equals $r_{1}$ and subtends an arc on which the inscribed angle $R X O$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,930 |
7. (40 points) $Q L$ is the bisector of triangle $P Q R$, and $M$ is the center of the circumcircle of triangle $P Q L$. It turned out that points $M$ and $L$ are symmetric with respect to $P Q$. Find the angles of triangle $P Q L$.
---
The text has been translated while preserving the original formatting and line br... | Answer: $\angle P L Q=120^{\circ}, \angle L P Q=\angle P Q L=30^{\circ}$.
Solution: Let segments $M L$ and $P Q$ intersect at point $H$. According to the problem, $M L \perp P Q$ and $M H=H L... | \anglePLQ=120,\angleLPQ=\anglePQL=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,931 |
1. (10 points) In workshop $B$, there are beads of two shapes - "cubes" and "pyramids". The "cubes" are green and blue, and the "pyramids" are red and blue. Traditions require that in any jewelry, adjacent beads must be of different shapes and different colors. The jeweler made a necklace in the form of a ring, using a... | Answer: b) and c).
Solution: We will prove that the solution to the problem is numbers of the form $6+2k$, where $k$ is a non-negative integer. Let's agree to denote green and blue cubes as $\mathrm{K}$ and $\mathrm{K}$, and red and blue pyramids as П and П.
Firstly, note that the number of beads is even. Indeed, cub... | 6+2k | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,932 |
2. (10 points) It is known that one of four coins is counterfeit and differs in weight from the genuine ones. It is required to determine which coin is counterfeit using a balance scale without weights. Which of the following statements are true?
a) The counterfeit coin can be determined in 2 weighings.
b) The counte... | Answer: a)
Solution: Let's choose two of the available coins and weigh them (one on each pan of the balance). This way, we can determine which pair contains the counterfeit coin (if the balance is even, the counterfeit coin is among the remaining two coins; otherwise, it is among the chosen ones).
Now, take one coin ... | ) | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 7,933 |
3. (20 points) During the draw before the math marathon, team captains were asked to name the smallest possible sum of the digits in the decimal representation of the number $n+1$, given that the sum of the digits of the number $n$ is 2017. What was the answer given by the captain of the team that won the draw? | Answer: 2.
Solution: First, we show that the answer is not less than 2. If the sum of the digits of the number $n+1$ is 1, then $n+1=10 \ldots 0$, and the decimal representation of $n$ consists entirely of nines. Then the number $n$ is divisible by 9, and the sum of its digits, therefore, is also. But this is impossib... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,934 |
4. (20 points) Before the math lesson, the teacher wrote nine consecutive numbers on the board, but the duty students accidentally erased one of them. When the lesson began, it turned out that the sum of the remaining eight numbers is 1703. Which number did the duty students erase? | Answer: 214.
Solution: Let the average of the original numbers be $a$. Then these numbers can be written in a symmetric form:
$$
a-4, a-3, a-2, a-1, a, a+1, a+2, a+3, a+4
$$
The erased number has the form $a+b$, where $-4 \leqslant b \leqslant 4$, and the sum of the remaining numbers is $9a - (a+b) = 8a - b$. On the... | 214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,935 |
5. (20 points) Alexei came up with the following game. First, he chooses a number $x$ such that $2017 \leqslant x \leqslant 2117$. Then he checks if $x$ is divisible by 3, 5, 7, 9, and 11 without a remainder. If $x$ is divisible by 3, Alexei awards the number 3 points, if by 5 - then 5 points, ..., if by 11 - then 11 p... | Answer: $2079=11 \cdot 9 \cdot 7 \cdot 3$.
Solution: Note that the divisibility of $x$ by 9 immediately gives 12 points, as it also implies divisibility by 3. If the number $x$ is not divisible by 11, it will score no more than $9+7+5+3=24$ points, and if it is not divisible by 9, it will score no more than $11+7+5+3=... | 2079 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,936 |
6. (30 points) In triangle $K I A$, angles $K$ and $I$ are equal to $30^{\circ}$. On the line passing through point $K$ perpendicular to side $K I$, a point $N$ is marked such that $A N$ is equal to $K I$. Find the measure of angle $KAN$. | Answer: $90^{\circ}$ or $30^{\circ}$.
Solution: Let $K I=2 a$, and point $H$ be the foot of the perpendicular from vertex $A$ to the
line $K N$. According to the condition, $\triangle K I A... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,937 |
7. (30 points) In the convex quadrilateral FIDO, the opposite sides FI and DO are equal to each other and greater than side DI. It is known that $\angle F I O = \angle D I O$. Prove that $F O$ is greater than $D I$.
# | # Solution:
Mark a point $E$ on the segment $F I$ such that $I E = I D$ (this is possible since $F I > D I$). Triangles $I D O$ and $I E O$ are congruent by two sides and the included angle (... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,938 |
8. (40 points) The shaman of the Sunlovers tribe calculates every midnight whether the coming day will be lucky: according to the tribe's beliefs, the day with number D from the beginning of time will be lucky if the number
$$
\left(D^{2}+4\right)\left(R^{2}+4\right)-2 D\left(R^{2}+4\right)-2 R\left(D^{2}+4\right)
$$
... | Answer: No, they do not exist.
Solution: It is easy to notice that the question of the problem is equivalent to the following: "Do there exist such natural numbers $D \geqslant R$ that the value of the expression given in the condition is negative?". Let's transform the minuend:
$$
\left(D^{2}+4\right)\left(R^{2}+4\r... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,939 |
9. (40 points) Given the following numbers: 20172017, 20172018, 20172019, 20172020, and 20172021. Is there a number among them that is coprime with all the others? If so, which one? | Answer: $20172017, 20172019$.
Solution: The given numbers are five consecutive natural numbers. The even numbers 20172018 and 20172020 do not fit.
Adjacent numbers are always coprime:
$$
n=m q, \quad n+1=m p, \quad 1=m(p-q) \Rightarrow m=1
$$
Adjacent odd numbers are also always coprime:
$$
2 k+1=m q, \quad 2 k+3=... | 20172017,20172019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,940 |
1. In a tennis tournament, there are $2^{n}+4$ schoolchildren participating, where $n-$ is a natural number greater than 4. A win earns 1 point, and a loss earns 0 points. Before each round, pairs are formed by drawing lots from participants who have the same number of points (those who do not find a pair skip the roun... | Answer: $C_{n}^{3}+1$.
Solution. First, let's prove two statements.
1) In a tournament with $2^{n}$ participants, there will never be a player without a pair, and by the end of the tournament, for any $k \in\{0, \ldots, n\}$, $C_{n}^{k}$ students will have $k$ points. We will use induction on $n$. For $n=1$, everythi... | C_{n}^{3}+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,941 |
2. Find the minimum value of the expression for $a, b>0$
$$
\frac{|6 a-4 b|+|3(a+b \sqrt{3})+2(a \sqrt{3}-b)|}{\sqrt{a^{2}+b^{2}}}
$$ | Answer: $\sqrt{39}$.
Solution 1. The expression to be minimized is $2\left(d_{1}+d_{2}\right)$, where $d_{1}, d_{2}$ are the distances from point $A(3,2)$ to the lines $\ell_{1}, \ell_{2}$, given by the equations
$$
\ell_{1}: a x-b y=0 \quad \text { and } \quad \ell_{2}:(a+b \sqrt{3}) x+(a \sqrt{3}-b) y=0 .
$$
These... | \sqrt{39} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,942 |
3. Inside the circle $\omega$ are located intersecting at points $K$ and $L$ circles $\omega_{1}$ and $\omega_{2}$, touching the circle $\omega$ at points $M$ and $N$. It turned out that points $K, M$, and $N$ lie on the same line. Find the radius of the circle $\omega$, if the radii of the circles $\omega_{1}$ and $\o... | Answer: 8.
Solution. Let $O, O_{1}, O_{2}$ be the centers of the circles $\omega, \omega_{1}, \omega_{2}$, respectively. The radii $O M$ and $O_{1} M$ of the circles $\omega$ and $\omega_{1... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,943 |
5. In the country of Alphia, there are 150 cities, some of which are connected by express trains that do not stop at intermediate stations. It is known that any four cities can be divided into two pairs such that there is an express train running between the cities of each pair. What is the minimum number of pairs of c... | Answer: 11025.
Solution. Suppose that some city (let's call it Alfsk) is connected by express trains to no more than 146 cities. Then a quartet of cities, consisting of Alfsk and any three with which it is not connected, does not satisfy the problem's condition, since Alfsk cannot be paired with any of the three remai... | 11025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,945 |
6. A cone with a base radius of 2 and a slant height of c contains three spheres of radius r. They touch each other (externally), the lateral surface of the cone, and the first two spheres touch the base of the cone. Find the maximum value of $r$. | Answer: $\sqrt{3}-1$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A$ be the point of tangency of the first two, $PO$ be the height of the cone, and $B$ be the projec... | \sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,946 |
1. In an arm wrestling tournament, there are $2^{n}+6$ athletes participating, where $n$ is a natural number greater than 7. One point is awarded for a win, and zero points for a loss. Before each round, pairs are formed by drawing lots among participants with the same number of points (those who do not find a pair ski... | Answer: $35 \cdot 2^{n-7}+2$.
Solution. First, we prove two statements.
1) In a tournament with $2^{n}$ participants, after the $m$-th round, the number of athletes with $k$ points will be $2^{n-m} \cdot C_{m}^{k}$, where $m \leqslant n$ and $k \in\{0, \ldots, m\}$. Let $f(m, k)$ be the number of participants who hav... | 35\cdot2^{n-7}+2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,947 |
2. Find the maximum value of the expression for $a, b>0$
$$
\frac{|4 a-10 b|+|2(a-b \sqrt{3})-5(a \sqrt{3}+b)|}{\sqrt{a^{2}+b^{2}}}
$$ | # Answer: $2 \sqrt{87}$.
Solution 1. The expression to be maximized is $2\left(d_{1}+d_{2}\right)$, where $d_{1}, d_{2}$ are the distances from point $A(2,5)$ to the lines $\ell_{1}, \ell_{2}$, given by the equations
$$
\ell_{1}: a x-b y=0 \quad \text { and } \quad \ell_{2}:(a-b \sqrt{3}) x-(a \sqrt{3}+b) y=0
$$
The... | 2\sqrt{87} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,948 |
3. Inside the circle $\omega$ is a circle $\omega_{1}$ that is tangent to it at point $K$. The circle $\omega_{2}$ is tangent to the circle $\omega_{1}$ at point $L$ and intersects the circle $\omega$ at points $M$ and $N$. It turns out that points $K, L$, and $M$ lie on the same line. Find the radius of the circle $\o... | Answer: 11.
Solution. Let $O, O_{1}, O_{2}$ be the centers of the circles $\omega, \omega_{1}, \omega_{2}$, respectively. The radii $O K$ and $O_{1} K$ of the circles $\omega$ and $\omega_{... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,949 |
5. In the country of Betia, there are 125 cities, some of which are connected by express trains that do not stop at intermediate stations. It is known that any four cities can be toured in a circle in some order. What is the minimum number of pairs of cities connected by express trains? | Answer: 7688.
Solution. Suppose that some city (let's call it Betsk) is connected by express trains to no more than 122 cities. Take a quartet of cities consisting of Betsk and any three other cities, two of which Betsk is not connected to. This quartet cannot be traveled in a circle, otherwise Betsk must be connected... | 7688 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,951 |
6. Three spheres of radius $r$ are placed inside a cone with height 4 and slant height 8. They touch each other (externally), the lateral surface of the cone, and the first two spheres touch the base of the cone. Find the maximum value of $r$. | Answer: $\frac{12}{5+2 \sqrt{3}}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A$ be the point of tangency of the first two, $PO$ be the height of the cone, and $B$ b... | \frac{12}{5+2\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,952 |
1. In an arm wrestling tournament, 510 athletes are participating. 1 point is awarded for a win, and 0 points for a loss. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with a difference of no more than 1 point in their ... | # Answer: 9.
Solution. Let there be $N=2 m+k$ athletes in the leading group before the next round, with $2 m$ of them meeting each other, and $k$ meeting participants who have one point less. Note that if athletes with $n$ and $n+1$ points meet, they will end up with $n$ and $n+2$ points regardless of the outcome. The... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,953 |
3. An isosceles trapezoid $A B C D$ with bases $A B$ and $D C$ has a circle inscribed in it with the center at point $O$. Find the area of the trapezoid if $O B=b$ and $O C=c$. | Answer: $2 b c$.
Solution. Let $r$ be the radius of the circle, $K$ be the point of tangency of the circle with side $B C$, and $M N$ be the height of the trapezoid passing through point $O... | 2bc | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,955 |
5. In the village of Sosnovka, there are 240 residents, some of whom are acquainted with each other, while others are not. It is known that any five residents can be seated at a round table such that each of them is acquainted with both of their neighbors. What is the minimum number of pairs of acquaintances that can e... | Answer: 28440.
Solution. Suppose that some resident (let's call him Petya) is not acquainted with at least three other residents. Choose the following five people: Petya, the three residents he is not acquainted with, and one more arbitrary person. They cannot be seated around a round table in the required manner, sin... | 28440 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,957 |
6. A cone, whose base diameter equals its slant height, contains three identical spheres that touch each other externally. Two of the spheres touch the lateral surface and the base of the cone. The third sphere touches the lateral surface of the cone at a point lying in the same plane as the centers of the spheres. Fin... | Answer: $\frac{5}{4}+\sqrt{3}$.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A$ be the point of tangency of the first two, $PO$ be the height of the cone, $B$ be the ... | \frac{5}{4}+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,958 |
1. In a tennis tournament, 254 schoolchildren are participating. A win earns 1 point, a loss earns 0 points. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with the same number of points compete, but in one of the pairs,... | Answer: 56.
Solution. Let $f(m, k)$ denote the number of students who have $k$ points after $m$ rounds. If $f(m, k)$ is even for any $k \in \{0, \ldots, m\}$, then participants with different numbers of points will not meet in the $(m+1)$-th round (otherwise, there would be at least two such meetings). Also, note that... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,959 |
2. Find the minimum value of the expression for $a, b \geqslant 0$
$$
\frac{|a-3 b-2|+|3 a-b|}{\sqrt{a^{2}+(b+1)^{2}}}
$$ | Answer: 2.
Solution 1. The minimized expression is $d_{1}+d_{2}$, where $d_{1}, d_{2}$ are the distances from points $A(1,3)$ and $B(3,1)$ to the line $\ell$ defined by the equation $a x-(b+1) y+1=0$. It can be rewritten as
$y=\frac{a}{b+1} x+\frac{1}{b+1}$ or $y=p x+q$, where $p \geqslant 0, q \in(0,1]$. We can assum... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,960 |
3. Points $K, L$ and $M$ are the midpoints of sides $AB, BC$ and $CD$ of parallelogram $ABCD$. It turned out that quadrilaterals $KBLM$ and $BCDK$ are cyclic. Find the ratio $AC: AD$.
---
The translation is provided as requested, maintaining the original formatting and structure. | Answer: 2.
Solution. Let $N$ be the point of intersection of the diagonals of the parallelogram. Quadrilateral $K B L M$ is an inscribed trapezoid, so $K B = L M$. Multiplying this equality... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,961 |
5. In the village of Beryozovka, there are 200 residents, some of whom are acquainted with each other, while others are not. It is known that any six residents can be seated at a round table such that each of them is acquainted with both of their neighbors. What is the minimum number of pairs of acquainted residents in... | Answer: 19600.
Solution. Suppose that some resident (let's call him Vasya) is not acquainted with at least four other residents. We will select the following six people: Vasya, four residents he is not acquainted with, and one more arbitrary person. They cannot be seated around a round table in the required manner, si... | 19600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,963 |
6. In a cone, three spheres of radius $\sqrt{24}$ are placed, each touching the other two externally. Two of the spheres touch the lateral surface and the base of the cone. The third sphere touches the lateral surface of the cone at a point lying in the same plane as the centers of the spheres. Find the radius of the b... | Answer: $7+4 \sqrt{3}+2 \sqrt{6}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A$ be the point of tangency of the first two, $PO$ be the height of the cone, $B$ be th... | 7+4\sqrt{3}+2\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,964 |
1. In a tennis tournament, 1152 schoolchildren are participating. 1 point is awarded for a win, and 0 points for a loss. Before each round, pairs are formed by drawing lots among participants with the same number of points (those who do not find a pair are awarded a point without playing). A player is eliminated after ... | Answer: 14.
Solution. Note that $1152=1024+128$. Let's prove two statements first.
1) In a tournament with $2^{n}$ participants, for any $m \in\{1, \ldots, n\}$, after the $m$-th round, there will be $2^{n-m}$ participants without any losses and $m \cdot 2^{n-m}$ participants with one loss. We will use induction on $... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,965 |
3. Inside triangle $ABC$, a point $P$ is chosen such that $AP=BP$ and $CP=AC$. Find $\angle CBP$, given that $\angle BAC = 2 \angle ABC$.
---
Here is the translation of the provided text, maintaining the original formatting and structure. | Answer: $30^{\circ}$.
Solution 1. Draw the perpendicular bisector of side $A B$. Obviously, it will pass through point $P$. Let $C^{\prime}$ be the point symmetric to $C$ with respect to th... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,967 |
4. Solve the equation $2^{x}+6^{y}+5=15^{z}$ in natural numbers. | Answer: $(2,1,1),(2,3,2)$.
Solution. The right side of the equation is divisible by 3, and the left side gives the same remainder when divided by 3 as $(-1)^{x}+2$. Therefore, $x$ is even and, in particular, $x \geqslant 2$. We can write the equation as $2^{x}+6^{y}=15^{z}-5$ and consider two cases.
1) $z$ is odd. Th... | (2,1,1),(2,3,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,968 |
5. A health camp was visited by 175 schoolchildren. Some of the children know each other, while others do not. It is known that any six schoolchildren can be accommodated in two three-person rooms so that all schoolchildren in the same room know each other. What is the minimum number of pairs of schoolchildren who coul... | Answer: 15050.
Solution. Suppose that some student (let's call him Vasya) is not acquainted with at least four students. We will select the following six children: Vasya, four students he is not acquainted with, and one more arbitrary student. It is impossible to seat them in two rooms, since out of this six, Vasya is... | 15050 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,969 |
6. Four spheres are placed inside a cone, each touching each other (externally) and the lateral surface of the cone. Three of the spheres have a radius of 3 and also touch the base of the cone. Find the radius of the fourth sphere if the angle between the slant height and the base of the cone is $\frac{\pi}{3}$. | Answer: $9-4 \sqrt{2}$.
Solution. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the spheres, $r$ be the radius of the fourth sphere, $BO$ be the height of the cone, and $D$ be the proje... | 9-4\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,970 |
1. In a qualification arm wrestling tournament, 896 athletes participate. A victory earns 1 point, a defeat earns 0 points. Before each round, pairs are formed by drawing lots among participants with an equal number of points (those who do not find a pair are awarded a point without playing). After the second defeat, a... | Answer: 10.
Solution. Note that $896=1024-128$. First, let's prove an auxiliary statement.
In a tournament with $2^{n}$ participants, for any $m \in\{1, \ldots, n\}$, after the $m$-th round, there will be $2^{n-m}$ athletes without any losses and $m \cdot 2^{n-m}$ athletes with one loss. We will use induction on $m$.... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,971 |
3. Given a triangle $ABC$ with the largest side $BC$. The bisector of its angle $C$ intersects the altitudes $AA_{1}$ and $BB_{1}$ at points $P$ and $Q$ respectively, and the circumcircle of $ABC$ at point $L$. Find $\angle ACB$, if it is known that $AP=LQ$.
| Answer: $60^{\circ}$.
Solution. Let $\alpha=\angle B C L, \beta=\angle A L C, \gamma=\angle B L C$ (see the figure). We will prove the equality of triangles $A L P$ and $B L Q$. Note that $A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,973 |
4. Solve the equation $2^{x}+5^{y}+19=30^{z}$ in natural numbers. | Answer: $(8,4,2)$.
Solution. The right-hand side of the equation is divisible by 5, and the left-hand side gives the same remainder when divided by 5 as $2^{x}-1$. Therefore, $x$ is a multiple of 4, and in particular, $x \geqslant 4$. Moreover, the right-hand side is divisible by 3, and the left-hand side gives the sa... | (8,4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,974 |
5. In a health camp, 225 schoolchildren are resting. Some of the children know each other, while others do not. It is known that among any six schoolchildren, there are three non-intersecting pairs who know each other. What is the minimum number of pairs of schoolchildren who could be resting in the camp? | Answer: 24750.
Solution. Suppose that some student (let's call him Vasya) is not acquainted with at least five other students. Then Vasya and the five students he is not acquainted with form a group of six that does not satisfy the condition of the problem. Therefore, each student can have no more than four unknowns. ... | 24750 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,975 |
6. A cone, whose height and radius of the base are both 7, contains four identical spheres. Each of them touches two others (externally), the base, and the lateral surface of the cone. A fifth sphere touches the lateral surface of the cone and all the identical spheres (externally). Find the radius of the fifth sphere. | Answer: $2 \sqrt{2}-1$.
Solution. Let $O_{1}, O_{2}, O_{3}, O_{4}, O_{5}$ be the centers of the spheres, $r$ and $x$ be the radii of the first and fifth spheres, respectively, $B O$ be the height of the cone, and $D$ be the projection of $O_{1}$ onto $B O$. In the diagram, part of the cross-section of the cone by the ... | 2\sqrt{2}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,976 |
1. In a tennis tournament, 512 schoolchildren are participating. 1 point is awarded for a win, and 0 points for a loss. Before each round, pairs are formed by lottery among participants with the same number of points (those who do not find a pair are awarded a point without playing). The tournament ends as soon as a so... | Answer: 84.
Solution. We will show that in a tournament with $2^{n}$ participants, no one will score points without playing, and for any $k \in\{0, \ldots, n\}$, exactly $C_{n}^{k}$ participants will end up with $k$ points. We will use induction on $n$. For $n=1$, this is obvious. Suppose that for some $n$ both statem... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,977 |
2. Find the minimum value of the expression for $a, b>0$
$$
\frac{(2 a+2 a b-b(b+1))^{2}+\left(b-4 a^{2}+2 a(b+1)\right)^{2}}{4 a^{2}+b^{2}}
$$ | Answer: 1.
Solution 1. The minimized expression is the sum of the squares of the distances from point $A(1,2 a)$ to the lines $\ell_{1}$ and $\ell_{2}$, given by the equations
$$
\ell_{1}: 2 a x+b y-b(b+1)=0 \quad \text { and } \quad \ell_{2}: b x-2 a y+2 a(b+1)=0
$$
These lines are perpendicular and intersect at po... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,978 |
3. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $A P: P B = B Q: Q C = 2: 1$. Find $\angle A K B$, where $K$ is the intersection point of segments $A Q$ and $C P$. | Answer: $90^{\circ}$.
Solution 1. Let $B H$ be the height and median of triangle $A B C$. Draw a line through vertex $B$ parallel to $A C$ and denote the point of its intersection with line... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,979 |
4. Solve the equation $2^{x}+3^{y}+1=6^{z}$ in natural numbers.
| Answer: $(1,1,1),(3,3,2),(5,1,2)$.
Solution. The right-hand side of the equation is divisible by 3, and the left-hand side gives the same remainder when divided by 3 as $(-1)^{x}+1$. Therefore, $x$ is odd. Let's consider two possible cases.
1) $x=1$. Note that either $y=z=1$, or $y \geqslant 2$ and $z \geqslant 2$. T... | (1,1,1),(3,3,2),(5,1,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,980 |
5. The government has decided to privatize civil aviation. For each pair of 202 cities in the country, the airline connecting them is sold to one of the private airlines. The mandatory condition for the sale is as follows: each airline must ensure the possibility of flying from any city to any other (possibly with seve... | Answer: 101.
Solution. First, we prove that for $k \in \{1, \ldots, 100\}$, the remainders of the numbers
$$
0, k, 2k, 3k, \ldots, 99k, 100k
$$
when divided by the prime number 101 are distinct. Indeed, if for some $0 \leqslant a < b \leqslant 100$ the numbers $ak$ and $bk$ give the same remainder when divided by 10... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,981 |
6. The region is bounded by two cones with a common base, the height of which is 4, and the radius of the base is 3. Three spheres are placed in the region, each touching the others externally. Two of the spheres are identical and touch both cones, while the third touches the boundary of the region. What is the minimum... | # Answer: $\frac{27}{35}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $BO$ be the height of one of the cones, $r$ and $x$ be the radii of the first two and the thir... | \frac{27}{35} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,982 |
1. In an arm wrestling tournament, $2^{n}$ athletes participate, where $n$ is a natural number greater than 7. One point is awarded for a win, and zero points for a loss. Before each round, pairs are formed by drawing lots among participants with the same number of points (those who do not find a pair are simply given ... | Answer: 8.
Solution. Let $f(m, k)$ be the number of participants who have scored $k$ points after $m$ rounds. We will prove by induction on $m$ that
$$
f(m, k)=2^{n-m} \cdot C_{m}^{k}, \quad \text { where } 0 \leqslant k \leqslant m \leqslant 2^{n}
$$
If $m=0$, then $k=0$, and $f(0,0)=2^{n}$. We will now perform the... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,983 |
2. Find the minimum value of the expression for $a, b>0$
$$
\frac{(3 a b-6 b+a(1-a))^{2}+\left(9 b^{2}+2 a+3 b(1-a)\right)^{2}}{a^{2}+9 b^{2}}
$$ | # Answer: 4.
Solution 1. The minimized expression is the sum of the squares of the distances from point $A(3 b,-2)$ to the lines $\ell_{1}$ and $\ell_{2}$, given by the equations
$$
\ell_{1}: a x+3 b y+a(1-a)=0 \quad \text { and } \quad \ell_{2}: 3 b x-a y+3 b(1-a)=0
$$
These lines are perpendicular and intersect at... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,984 |
3. On the hypotenuse $A B$ of an isosceles right triangle $A B C$, points $K$ and $L$ are marked such that $A K: K L: L B=1: 2: \sqrt{3}$. Find $\angle K C L$. | Answer: $45^{\circ}$.
Solution 1. Let $A K=1$. Then $K L=2$ and $L B=\sqrt{3}$, so $A B=3+\sqrt{3}$ and $A C=B C=\frac{3+\sqrt{3}}{\sqrt{2}}$. Therefore,
$$
A C \cdot B C=6+3 \sqrt{3}=3(2+\... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,985 |
4. Solve the equation $2^{x}+3^{y}+3=10^{z}$ in natural numbers. | Answer: $(2,1,1),(4,4,2)$.
Solution. The left side of the equation, when divided by 3, gives the same remainder as $(-1)^{x}$, while the remainder of the right side for any $z$ is 1. Therefore, $x$ is even and, in particular, $x \geqslant 2$. If $z=1$, then the equation is satisfied by $x=2, y=1$ and only they. Let $z... | (2,1,1),(4,4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,986 |
5. The government has decided to privatize civil aviation. For each pair of the country's 127 cities, the connecting airline is sold to one of the private airlines. Each airline must make all the purchased air routes one-way, but in such a way as to ensure the possibility of flying from any city to any other (possibly ... | # Answer: 63.
Solution. For integers $k$ and $n$, let $N_{k, n}$ denote the remainder when $k \cdot n$ is divided by the prime number 127. First, we prove that if $k$ is not divisible by 127, then the numbers $N_{k, 0}, \ldots, N_{k, 126}$ are distinct. Indeed, if for some $0 \leqslant a < b \leqslant 126$ the numbers... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,987 |
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