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1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with \( n \); if connected, then the numbers \( a + b \) and \( n \) must have a common natural divisor greater than 1. For what smallest \( n \) does such an arrangement exist?
|
Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two three-link cycles - for example, $(a, b, c)$ and $(a, d, c)$. But there is only one such cycle in the picture.
2) $n$ has at least two distinct prime divisors. Suppose $n$ is a power of a prime number $p$. Consider a chain $(a, b, c, d)$ of sequentially connected numbers. By the condition,
$$
a+d=(a+b)-(b+c)+(c+d) \vdots p
$$
Then the numbers $a$ and $d$ are also connected, which means that there is a cycle of length 4 in the picture, which is not there.
Thus, the number $n$ has at least two distinct odd prime divisors, hence $n \geqslant 15$. The arrangement for $n=15$ is shown in the figure.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-f(x)=0$ has exactly three solutions. Find the ordinate of the vertex of the trinomial $f(x)$.
|
# Answer: 0.
Solution. Suppose the leading coefficient of the trinomial is positive. Note that $(f(x))^{3}-f(x)=f(x) \cdot(f(x)-1) \cdot(f(x)+1)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-1$, and fewer roots than the equation $f(x)=1$. It is also clear that no two equations have common roots. Therefore, the equation $f(x)=0$ has exactly one root. Consequently, the ordinate of the vertex of the trinomial $f(x)$ is zero. Similarly, the case where the leading coefficient of the trinomial is negative is also considered.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1?
|
Answer: 45.
Solution. We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 45 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Suppose for definiteness that $a<b$. Then $k^{2} \leqslant a<b<(k+1)^{2}$ and, consequently, $\sqrt{b}-\sqrt{a}<(k+1)-k=1$.
Now let's present 44 numbers, all differences between the square roots of which are at least 1:
\[
\begin{array}{llllll}
1^{2}, & 2^{2}, & 3^{2}, & 4^{2}, & \ldots, & 44^{2} .
\end{array}
\]
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6.
|
Answer: 400.
Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then $c-b=6$, which is impossible. Therefore, in each set of ten, no more than four numbers are selected, and in the first thousand numbers, there are no more than 400, since there are a hundred sets of ten in a thousand.
If we take all the numbers ending in $1, 2, 3$ or 4, then there will be exactly 400, but no difference will be equal to 4, 5 or 6.
|
400
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the maximum number of chess kings that can be placed on a $12 \times 12$ board so that each king attacks exactly one of the others?
|
Answer: 56.
Solution. Note that two kings attack each other if and only if their cells share at least one vertex. For each pair of attacking kings, mark the vertices of the cells they occupy. In this case, no fewer than six vertices are marked for each such pair. Since different pairs of kings mark different vertices (otherwise, some king would attack more than one king), the total number of pairs of kings is no more than $\left[13^{2} / 6\right]=28$, and the number of kings is no more than 56.
The arrangement of 56 kings is shown in the figure.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. From the numbers 1, 2, 3, ..., 2016, $k$ numbers are chosen. What is the smallest $k$ such that among the chosen numbers, there will definitely be two numbers whose difference is greater than 672 and less than 1344?
|
Answer: 674.
Solution. Let $n=672$. Then $2 n=1344$ and $3 n=2016$. Suppose it is possible to choose $674=n+2$ numbers such that no required pair of numbers can be found among them. Let $m$ be the smallest of the chosen numbers. Then the numbers $m+n+1, m+n+2$, $\ldots, m+2 n-1$ are not chosen. Remove them and the number $m$ from the set $\{1, \ldots, 3 n\}$, and denote the remaining set by $E$. Consider the pairs of numbers
\[
\begin{aligned}
& (1, n+2), \quad(2, n+3), \quad \ldots(m-1, m+n) \\
& (m+1, m+2 n), \quad(m+2, m+2 n+1), \quad \ldots, \quad(n+1,3 n) .
\end{aligned}
\]
There are $(m-1)+(n-m+1)=n$ such pairs. Note that the union of the left and right parts of these pairs gives the set $E$. Then any chosen number coincides with the left or right part of one of the pairs. By assumption, there are exactly $n+1$ such numbers, so there will be two of them, say $a$ and $b$, belonging to the same pair. Then their difference is $n+1$ or $2 n-1$. Thus, $a$ and $b$ satisfy the condition of the problem, which is impossible.
If the numbers $1,2,3, \ldots, 673$ are chosen, then it will not be possible to find the required two numbers.
|
674
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a board $2016 \times 2016$. For what smallest $k$ can the cells of the board be colored in $k$ colors such that
1) one of the diagonals is painted in the first color;
2) cells symmetric with respect to this diagonal are painted in the same color;
3) any two cells located in the same row on opposite sides of the cell of the first color are painted in different colors (the cells do not necessarily have to be adjacent to the cell of the first color).
|
Answer: 11.
Solution. Let the cells of the diagonal running from the top left corner to the bottom right corner be painted in the first color. Denote by $C_{i}$ the set of colors in which the cells of the $i$-th row, located to the left of the diagonal of a single color, are painted. We will prove that $C_{i} \neq C_{j}$ for $i<j$. Indeed, the cell located in the $i$-th row and $j$-th column has a color not included in $C_{i}$. But the cell located in the $j$-th row and $i$-th column is painted in the same color, and its color is included in $C_{j}$. Thus, the sets $C_{1}, C_{2}, \ldots, C_{2016}$ are different subsets of the set $\{1,2, \ldots, k\}$. Then there are no more than $2^{k}$ of them. Therefore, $k \geqslant 11$.
By induction, we will show how to paint the board $2^{k} \times 2^{k}$ in $k$ colors as required. This will be sufficient, since by leaving only the rows from 1 to 2016 and the columns from 1 to 2016, we will obtain a $2016 \times 2016$ board with the required coloring.
Base case $k=1$ is obvious, since a $2 \times 2$ board can be painted in one color. Inductive step from $k$ to $k+1$. If we already know how to paint a $2^{k} \times 2^{k}$ board as required, then we will paint a $2^{k+1} \times 2^{k+1}$ board as follows: place two copies of the $2^{k} \times 2^{k}$ board, one in the top left corner and the other in the bottom right corner, and paint the remaining cells in the $(k+1)$-th color.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. At a cactus lovers' meeting, 80 cactus enthusiasts presented their collections, each consisting of cacti of different species. It turned out that no species of cactus is present in all collections at once, but any 15 people have cacti of the same species. What is the smallest total number of cactus species that can be in all the collections?
|
Answer: 16.
Solution. We will show that 16 cacti were possible. Number the cacti from 1 to 16. Let the 1st cactus lover have all cacti except the first; the 2nd cactus lover have all except the second cactus; the 15th cactus lover have all except the fifteenth cactus; and the cactus lovers from the 16th to the 80th have all cacti except the sixteenth. Then, any 15 cactus lovers will have a common cactus.
Now, we will establish that they must have more than 15 cacti. Suppose the opposite: let there be a total of $k \leqslant 15$ cacti. Number the cacti from 1 to $k$. For the cactus with number $i$, there exists a cactus lover $A_{i}$ who does not have it. But then, for the cactus lovers $A_{1}, A_{2}, \ldots, A_{k}$, there is no cactus that all of them have. And, even more so, there is no such cactus if we add a few more cactus lovers to them so that their number becomes 15. Contradiction.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have?
|
Answer: 2.
Solution. Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. Then the equation $f(x)=0$ has exactly one root. Therefore, the equation $f(x)=1$ has exactly two roots, and the equation $f(x)=-1$ has no roots. Thus, the equation $(f(x))^{2}-1=(f(x)+1)(f(x)-1)=0$ has two roots. Similarly, the case when the leading coefficient of the polynomial is negative is considered.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$?
|
Answer: 13.
Solution. We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there are 13 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Let's assume for definiteness that $a<b$. Then $k^{3} \leqslant a<b<(k+1)^{3}$ and, consequently, $0<\sqrt[3]{b}-\sqrt[3]{a}<(k+1)-k=1$.
Now let's present 12 numbers, all differences between the cube roots of which are at least 1:
\[
1^{3}, \quad 2^{3}, \quad 3^{3}, \quad 4^{3}, \quad \ldots, \quad 12^{3} .
\]
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The cells of a $9 \times 9$ board are painted in black and white in a checkerboard pattern. How many ways are there to place 9 rooks on the same-colored cells of the board so that they do not attack each other? (A rook attacks any cell that is in the same row or column as itself.)
|
Answer: $4!5!=2880$.
Solution. Let the upper left corner of the board be black for definiteness. Note that black cells come in two types: cells with both coordinates even, and cells with both coordinates odd. If a rook is on a black cell with both even coordinates, then all black cells it attacks also have even coordinates. Similarly, a rook standing on a black cell with both odd coordinates attacks only black cells with both odd coordinates. Therefore, we need to count the number of ways to place 9 rooks on two boards: $4 \times 4$ and $5 \times 5$ (a board consisting only of black cells with both even coordinates, and a board consisting only of black cells with both odd coordinates). For the first board, this number is 4!, and for the second, it is 5!. Therefore, the total number of ways to place the rooks on black cells is $4!5!=2880$.
Now let's show that it is impossible to place 9 rooks on white cells in the required manner. Consider the cells whose first coordinate is odd and the second is even. No more than four rooks can be placed on such cells, as they are contained in four rows. Similarly, no more than four rooks can be placed on white cells whose first coordinate is even and the second is odd, as they are contained in four columns.
|
2880
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table?
|
Answer: 28.
Solution. Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they would be telling the truth). Also, both of them cannot be knights, since they see each other. Therefore, one of them is a knight, and the other is a liar. It remains to note that the situation where two knights sit next to each other, and all the others are liars, is possible.
|
28
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the cells of a $2015 \times n$ table, non-negative numbers are arranged such that in each row and each column there is at least one positive number. It is known that if a cell contains a positive number, then the sum of all numbers in its column is equal to the sum of all numbers in its row. For which $n$ is this possible?
|
Answer: $n=2015$.
Solution. We will prove by induction on $m+n$ that for an $m \times n$ table, the specified arrangement is possible only when $m=n$. For $m+n=2$, the statement is obvious. Consider an $m \times n$ table. Let the sum of the numbers in the first row be $a$. Consider all rows and columns whose sum of numbers is also $a$. If these are all the rows and all the columns, then the sum of all numbers in the table, calculated by rows, is $m a$, and calculated by columns, is $n a$. Therefore, $m=n$. If these are not all the rows and all the columns, then consider any row with a different sum. Then at the intersection of this row with a column with sum $a$, there is a zero. The same conclusion is true for a column with a sum different from $a$. Now, let's strike out from the table all rows and columns whose sum of numbers is different from $a$. After striking out, there cannot be any zero rows and columns, since from the rows and columns that remained, only zeros were struck out. Repeating the reasoning with the calculation of the sum of the numbers in the table, we understand that there will be an equal number of rows and columns left (let there be $k$ of each). Now, let's strike out from the original table all rows and columns whose sum of numbers is equal to $a$. We will get a table of size $(m-k) \times (n-k)$, satisfying the condition of the problem. Then, by the inductive hypothesis, $m-k=n-k$. Therefore, $m=n$.
|
2015
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green so that in each $2 \times 2$ square there are two blue and two green cells?
|
Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider the first row of the table. Suppose there are no adjacent cells of the same color in this row. Then the color of the first cell in the second row uniquely determines the colors of the other cells in the second row: they either all match the colors of the cells in the first row or are opposite to them. Therefore, the second row can be painted in two ways. In any of these ways, there are also no adjacent cells of the same color. Similarly, the color of the first cell in the third row uniquely determines its coloring. Thus, the third row can also be painted in two ways, and so on. Therefore, the number of ways to paint the table with such a fixed first row is exactly $2^{9}$. The number of ways to paint the table so that there are no adjacent cells of the same color in the first row is $2^{10}$.
Suppose there are two adjacent cells of the same color in the first row. Let's assume for definiteness that they are green. Then there must be two blue cells under them. Then the colors of the cells in the second row are uniquely determined. Therefore, there must be two green cells under the two adjacent blue cells. Thus, the colors of the cells in the third row are also uniquely determined, and so on. Therefore, the first row with adjacent cells of the same color uniquely determines the coloring of the entire table. The number of ways to paint the first row in two colors is $2^{10}$, but in two of them there are no adjacent cells of the same color. Therefore, the number of colorings of the table with such first rows is $2^{10}-2$, and the total number of colorings is $2^{10}+\left(2^{10}-2\right)=2^{11}-2=2046$.
|
2046
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!" was heard exactly 230 times. How many times was the phrase "You are a knight!" heard?
|
Answer: 526.
Solution. Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knight pairs is $\frac{230}{2}=115=r \ell$. Since $r \ell=115=5 \cdot 23$ and $r, \ell \geqslant 2$, either $r=5$ and $\ell=23$, or $r=23$ and $\ell=5$. In either case, the number of knight-knight and liar-liar pairs is $\frac{5 \cdot 4}{2}+\frac{23 \cdot 22}{2}=263$. Therefore, the phrase "You are a knight!" was said 526 times.
|
526
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In an $8 \times 8$ table, natural numbers are arranged. The numbers in cells symmetric with respect to both diagonals of the table are equal. It is known that the sum of all numbers in the table is 1000, and the sum of the numbers on the diagonals is 200. For what smallest number $M$ can we assert that the sum of the numbers in each row does not exceed $M$?
|
Answer: $M=288$.
Solution. Consider the upper half of the table. Let the numbers be arranged as shown in the figure. Due to symmetry, the marked numbers completely determine the placement of the remaining numbers in the table.
| $a_{1}$ | $b_{1}$ | $b_{2}$ | $b_{3}$ | $c_{3}$ | $c_{2}$ | $c_{1}$ | $d_{1}$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | $a_{2}$ | $b_{4}$ | $b_{5}$ | $c_{5}$ | $c_{4}$ | $d_{2}$ | |
| | | $a_{3}$ | $b_{6}$ | $c_{6}$ | $d_{3}$ | | |
| | | | $a_{4}$ | $d_{4}$ | | | |
Let
$$
\begin{array}{ll}
A=a_{1}+a_{2}+a_{3}+a_{4}, & B=b_{1}+b_{2}+b_{3}+b_{4}+b_{5}+b_{6} \\
D=d_{1}+d_{2}+d_{3}+d_{4}, & C=c_{1}+c_{2}+c_{3}+c_{4}+c_{5}+c_{6}
\end{array}
$$
By the condition, $2(A+D)=200$ and $2(A+D)+4(B+C)=1000$. Therefore, $A+D=100$ and $B+C=200$. Since the numbers are natural, $a_{i}+d_{j} \leqslant A+D-6=94$, and the sum of any three numbers $b_{i}$ and any three numbers $c_{j}$ does not exceed $B+C-6=194$. It remains to note that in each row there is one number of the form $a_{i}$ and $d_{j}$, and three numbers of the form $b_{i}$ and $c_{j}$, so their sum does not exceed $94+194=288$. Therefore, $M=288$ fits. We will show that smaller $M$ do not work. Consider the table
| 47 | 95 | 1 | 1 | 1 | 1 | 95 | 47 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 95 | 1 | 1 | 1 | 1 | 1 | 1 | 95 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 95 | 1 | 1 | 1 | 1 | 1 | 1 | 95 |
| 47 | 95 | 1 | 1 | 1 | 1 | 95 | 47 |
The sum of the numbers in it is $8 \cdot 95+4 \cdot 47+52=1000$, on the diagonals $4 \cdot 47+12=200$, and in the first row $-2 \cdot 47+2 \cdot 95+4=288$.
|
288
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$.
|
Answer: 128
Solution. By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cdot b^{6}+c \cdot c^{6}+d \cdot d^{6} \leqslant 2\left(a^{6}+b^{6}+c^{6}+d^{6}\right)=2 \cdot 64=128
$$
Equality is achieved when $a=2, b=c=d=0$. Therefore, $a^{7}+b^{7}+c^{7}+d^{7}$ does not exceed 128 and can be equal to it.
|
128
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all natural numbers $n$ such that the number $2^{n}+n^{2}+25$ is a cube of a prime number.
|
Answer: $n=6$
Solution. Let $2^{n}+n^{2}+25=p^{3}$ for some prime number $p$. Since $p>3$, $p$ is an odd prime number. Then $n$ is an even number, and $2^{n}$ gives a remainder of 1 when divided by three. If $n$ is not divisible by three, then $n^{2}$ gives a remainder of 1 when divided by three, and then $2^{n}+n^{2}+25$ is divisible by three, which is impossible. Therefore, $n$ is divisible by six and can be written as $n=6 k$. Consequently, $2^{n}+n^{2}+25=64^{k}+36 k^{2}+25>\left(4^{k}\right)^{3}$. Therefore,
$$
64^{k}+36 k^{2}+25 \geqslant\left(4^{k}+1\right)^{3}=64^{k}+3 \cdot 16^{k}+3 \cdot 4^{k}+1 .
$$
If $k \geqslant 2$, then $3 \cdot 4^{k}>25$ and $3 \cdot 16^{k} \geqslant 36 k^{2}$ (the latter follows, for example, from the easily verifiable by induction inequality $4^{k} \geqslant 6 k$). Thus, $k=1$ and $n=6$. It remains to verify that such $n$ works: $2^{6}+6^{2}+25=125=5^{3}$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $a$ for which the quadratic trinomials $x^{2}+2 x+a$ and $x^{2}+a x+2=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial.
|
Answer: $a=-4$
Solution. If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $2^{2}-2 a=a^{2}-2 \cdot 2$. Thus, we need to solve the equation $a^{2}+2 a-8=0$. Its roots are $a=-4$ and $a=2$. The first solution is valid, while the second solution is not valid, since the quadratic polynomial $x^{2}+2 x+2$ does not have roots.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $a$ for which the quadratic trinomials $x^{2}-6 x+4 a$ and $x^{2}+a x+6=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial.
|
Answer: $a=-12$
If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $6^{2}-8 a=a^{2}-2 \cdot 6$. Thus, we need to solve the equation $a^{2}+8 a-48=0$. Its roots are $a=-12$ and $a=4$. The first solution is valid, while the second solution is not valid, since the quadratic polynomial $x^{2}-6 x+16$ does not have roots.
|
-12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. On an island, there live liars and knights, a total of 2021 people. Knights always tell the truth, and liars always lie. Every resident of the island knows whether each person is a knight or a liar. One fine day, all the residents of the island lined up. After that, each resident of the island stated: "The number of liars standing to the right of me is greater than the number of knights standing to the left of me." How many knights are there on the island? Provide all possible options and prove that there are no others.
|
Answer: 1010
Solution. The rightmost islander cannot be telling the truth, since there is no one to the left of him, and in particular, there are no liars. Therefore, he is a liar. The leftmost islander cannot be lying, since there is at least one liar to the left of him, and there is no one to the right of him, and in particular, there are no knights. Therefore, he is a knight. Temporarily remove these two from the row. For all the others, the number of liars to the right and the number of knights to the left has decreased by one, so the truthfulness and falsity of their statements have not changed. Therefore, by repeating the same reasoning, we establish that the rightmost islander is a liar, and the leftmost is a knight. Remove the two outermost islanders again, and continue this process until only one resident remains. He is lying and therefore is a liar. Thus, in the middle of the row stands a liar, all to the right of him are also liars, and all to the left of him are knights. Therefore, there are 1010 knights.
|
1010
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree grows. In your answer, write the integer part of the obtained number. (The integer part of a number is the greatest integer not exceeding the given number.)
|
Answer: 40.
Solution. Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is $100 \mathrm{~m} \times 100 \mathrm{~m}=10000 \mathrm{~m}^{2}$. Thus, one tree grows on $\frac{10000 \mathrm{M}^{2}}{250}=40 \mathrm{M}^{2}$.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exactly at 12 o'clock before the start of the five-minute correct movement interval?
|
Answer: 28.
Solution. In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Since $252 \cdot 2=60 \cdot 8+24$, the hand will travel 8 full circles, plus 24 minutes, and then 5 minutes in the correct direction and 1 minute in the opposite direction. In total, $24+5-1=28$.
|
28
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (50 points) Two multiplication examples of two natural numbers were written on the board, and in one of the examples, the numbers were the same. The troublemaker Petya erased both the original numbers and almost all the digits of the multiplication results, leaving only the last three digits: ... 689 and ... 759. Can the excellent student Misha determine which of these three digits belong to the number that resulted from the multiplication of the same numbers?
|
Answer: It could be 689.
First solution. Statement: at least one of the two last digits of the square of a natural number is necessarily even. Then the second number is eliminated. Example for the first number: $133 \cdot 133=17689$.
Proof of the statement: the statement is true for even numbers, let's consider the squares of odd numbers. Let the second-to-last digit of the original number be $n$, the last digit be $m$, where $m \in\{1,3,5,7,9\}$, and all other digits of the number be $k$. Then
$$
\left[\begin{array}{l}
(100 k+10 n+1)^{2}=100 \cdot(\ldots)+10 \cdot 2 n+1 \\
(100 k+10 n+3)^{2}=100 \cdot(\ldots)+10 \cdot 6 n+9 \\
(100 k+10 n+5)^{2}=100 \cdot(\ldots)+10 \cdot 2+5 \\
(100 k+10 n+7)^{2}=100 \cdot(\ldots)+10 \cdot(4 n+4)+9 \\
(100 k+10 n+9)^{2}=100 \cdot(\ldots)+10 \cdot(8 n+8)+1
\end{array}\right.
$$
In all cases, the second-to-last digit of the square is even.
Second solution. We will base our reasoning on the following statement: the square of a natural number is either divisible by 4 or gives a remainder of 1 when divided by 8. Indeed, if the original number is even, then its square is divisible by 4, and if the original number is odd, then it can be represented as $8 q+r$, where $r \in\{1,3,5,7\}$. Then $(8 q+$ $+r)^{2}=8 \cdot(\ldots)+1$. Having the last three digits of the number in front of us, we can check the divisibility rule by 4 or find the remainder when the number is divided by 8.
|
689
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green such that in each $2 \times 2$ square there are two blue and two green cells?
|
Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider the first row of the table. Suppose there are no adjacent cells of the same color in this row. Then the color of the first cell in the second row uniquely determines the colors of the other cells in the second row: they either all match the colors of the cells in the first row or are opposite to them. Therefore, the second row can be painted in two ways. In any of these ways, there are also no adjacent cells of the same color. Similarly, the color of the first cell in the third row uniquely determines its coloring. Thus, the third row can also be painted in two ways, and so on. Therefore, the number of ways to paint the table with such a fixed first row is exactly $2^{9}$. The number of ways to paint the table so that there are no adjacent cells of the same color in the first row is $2^{10}$.
Suppose there are two adjacent cells of the same color in the first row. Let's assume for definiteness that they are green. Then there must be two blue cells under them. Then the colors of the cells in the second row are uniquely determined. Therefore, there must be two green cells under the two adjacent blue cells. Thus, the colors of the cells in the third row are also uniquely determined, and so on. Therefore, the first row with adjacent cells of the same color uniquely determines the coloring of the entire table. The number of ways to paint the first row in two colors is $2^{10}$, but in two of them there are no adjacent cells of the same color. Therefore, the number of colorings of the table with such first rows is $2^{10}-2$, and the total number of colorings is $2^{10}+\left(2^{10}-2\right)=2^{11}-2=2046$.
|
2046
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many numbers of the form $\overline{a b a b c}$ are divisible by 5?
|
Answer: 180.
Solution. Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot start with zero, and the value of $b$ can be anything.
Thus, we get that the total number of such numbers is $2 \cdot 9 \cdot 10=180$.
|
180
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (20 points) Nезнayka came up with a password for his email consisting of five characters. Deciding to check the reliability of this password, he calculated all possible combinations that can be formed from these five characters. In the end, he got 20 different combinations. Is a password with this number of combinations possible, or did Nезнayka make a mistake in his calculations? If it is possible, provide an example of a suitable password.
|
For example, error.
Solution. The maximum number of different combinations that can be formed from 5 symbols equals $5!(n!=1 \cdot 2 \cdots n$ - the factorial of number $n$, i.e., the product of all natural numbers from 1 to $n$ inclusive). This number is obtained when all 5 symbols are distinct.
Suppose among the given 5 symbols there are $k \leqslant 5$ identical ones. Consider an arbitrary combination of these symbols. Clearly, swapping any two identical symbols in the considered combination while keeping the other symbols in their places does not yield a new combination. There can be $k!$ such permutations among the identical symbols. Thus, each combination of symbols that are not identical repeats $k!$ times. Therefore, the total number of different combinations is $5!/ k!$. Similarly, if in addition to $k$ identical symbols there are also $n$ other identical symbols, then the total number of different combinations is $5!/(k!\cdot n!)$.
It is not hard to see that 20 different combinations are obtained when $k=3, n=1$.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Masha has 1000 beads in 50 different colors, 20 beads of each color. What is the smallest $n$ such that for any way of making a necklace from all the beads, one can choose $n$ consecutive beads, among which there are beads of 25 different colors?
|
Answer: $n=462$.
First solution. Let's call a segment of the necklace of length $m$ a set of $m$ consecutive beads. If the beads in the necklace are arranged in groups of 20 of the same color, then a segment of length 461 cannot contain more than 24 different colors. Therefore, $n \geqslant 462$. Consider a segment of the necklace of length 462. Suppose that in this segment, 25 beads of different colors do not appear. Number the beads of the necklace counterclockwise so that the first beads are those of the chosen segment. Let $m$ be the number of the first bead of a color that is not present in the segment (say, yellow). We will show that the segment of the necklace from $m-461$ to $m$ contains 25 colors. Indeed, the yellow color is present in it. In the remaining part of the segment, the yellow color cannot appear by construction. But there are beads of at least 24 colors in it, since $461 > 23 \cdot 20$.
Second solution. Let's call a segment of the necklace 462 consecutive beads. We will show that in some segment, there will be 25 different colors. Suppose the opposite. Then in any segment, there are no more than 24 different colors. Consider all possible pairs consisting of a segment of the necklace and some color present in it. In each segment, there are no more than 24 different colors, and the total number of segments of the necklace is 1000. Therefore, the number of pairs does not exceed $24 \cdot 1000 = 24000$. Now consider some specific color (say, blue). There are 20 blue beads in total, and between the farthest blue beads $A$ and $B$, there are always at least 18 beads (exactly 18 only when all blue beads are consecutive). Thus, there are at least 19 segments containing $B$ and not containing $A$. The bead $A$ itself is included in 462 segments. Therefore, blue beads are contained in at least 481 segments. These arguments are valid for any color. Therefore, the total number of pairs is at least $481 \cdot 50 = 24050 > 24000$, which is impossible.
|
462
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips?
|
Answer: 4801.
Solution. Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total number of chips must be at least $99^{2}-5000=4801$.
Now, place chips in the cells of the expanded table where the product of the coordinates is divisible by four. In this case, each of the 625 squares $4 \times 4$ will have exactly 8 chips. Moreover, the added row and column will be completely filled with chips. Therefore, in the original table, there will be $625 \cdot 8-199=4801$ chips.
|
4801
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Boy Tolya loves to swim. When he comes to one grandmother's cottage, he swims in the Volkhov and floats downstream from one beach to another in 18 minutes. On the way back, it takes him exactly 1 hour. When he came to another grandmother's cottage, he swam the same distance downstream in the Luga River in 20 minutes. How much time will it take him to return back?
|
Answer: 45 minutes.
Solution. Let the distance between the beaches be $x$ km. Then in Volkhov, Tolya swims downstream at a speed of $x / 18$ km/min, and upstream at a speed of $x / 60$ km/min. Therefore, Tolya's own speed is $\frac{1}{2}(x / 60 + x / 18) = 13 x / 360$ km/min. In Luga, Tolya swims downstream at a speed of $x / 20$ km/min. Thus, the speed of the current in Luga is $x / 20 - 13 x / 360 = 5 x / 360$ km/min. Therefore, Tolya's speed against the current in Luga is $13 x / 360 - 5 x / 360 = x / 45$ km/min. Hence, Tolya will swim $x$ km against the current in Luga in 45 minutes.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The cells of a $20 \times 20$ table are painted in $n$ colors, and there are cells of each color. In each row and each column of the table, no more than six different colors are used. What is the largest $n$ for which this is possible?
|
Answer: 101.
Solution. Suppose there is a coloring with 102 colors. Then there are two rows that together use at least 12 colors (otherwise, the total number of colors does not exceed $11+5 \cdot 18=101$). Let's assume for definiteness that these are the first and second rows. By the condition, no more than six different colors are present in each row. Therefore, exactly six colors are used in both the first and second rows, and all these colors are different. We will call the colors used in the first two rows dark, and the rest light. Now consider the columns of the table. In each of them, there are no more than four light colors, since the two top cells are painted in different dark colors. Then the total number of light colors used in the table does not exceed $4 \cdot 20=80$ and 12 dark colors. Thus, the total number of colors does not exceed $80+12=92<102$, which is impossible.
A coloring with 101 colors is given below. Empty cells are painted in the 100th color.
| 0 | 1 | 2 | 3 | 4 | | | | | | | | | | | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 5 | 6 | 7 | 8 | 9 | | | | | | | | | | | | | | |
| | | 10 | 11 | 12 | 13 | 14 | | | | | | | | | | | | | |
| | | | 15 | 16 | 17 | 18 | 19 | | | | | | | | | | | | |
| $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ |
| | | | | | | | | | | | | | | | 75 | 76 | 77 | 78 | 79 |
| 80 | | | | | | | | | | | | | | | | | 81 | 82 | 83 | 84 |
| 85 | 86 | | | | | | | | | | | | | | | | 87 | 88 | 89 |
| 90 | 91 | 92 | | | | | | | | | | | | | | | | | 93 | 94 |
| 95 | 96 | 97 | 98 | | | | | | | | | | | | | | | | | 99 |
|
101
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (10 points) We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013?
|
Answer: 2.
Solution. Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first and second digits in the month's notation. Since there are 12 months in a year, $\mathrm{M}_{1} \leqslant 1$, hence $\mathrm{M}_{1}=0$. Moreover, no month has more than 31 days, and of the remaining digits, only 2 is suitable for $Д_{1}$. The digits 4 and 5 can be placed in positions $Д_{2}$ and $\mathrm{M}_{2}$ in any order, giving a valid date.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (10 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different six-letter words can be formed from the letters of the word СКАЛКА? And how many seven-letter words from the letters of the word ТЕФТЕЛЬ? In your answer, indicate the quotient of dividing the larger of the found numbers by the smaller.
#
|
# Answer: 7.
Solution. The word СКАЛКА has six letters, but the letter А and the letter K each appear twice. Therefore, the number of different words will be $\frac{6!}{2!\cdot 2!}$. In
the word ТЕФТЕЛЬ, there are seven letters, and the letters $\mathrm{T}$ and $\mathrm{E}$ each appear twice, so the number of different words that can be formed from the letters of this word will be greater and will be equal to $\frac{7!}{2!2!}$. It is easy to see that the quotient of dividing the larger by the smaller will be equal to 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (20 points) Several positive numbers are written on the board. The sum of the five largest of them is 0.29 of the sum of all the numbers, and the sum of the five smallest is 0.26 of the sum of all the numbers. How many numbers are written on the board?
|
Answer: 18.
Solution. Let the total number of numbers be $k+10$. From the condition, it is clear that $k>0$. We can assume that the sum of all numbers is 1 (otherwise, we divide each number by this sum, and the condition of the problem remains). We order the numbers in ascending order:
$$
a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5} \leqslant a_{5+1} \ldots \leqslant a_{5+k} \leqslant a_{6+k} \leqslant a_{7+k} \leqslant a_{8+k} \leqslant a_{9+k} \leqslant a_{10+k}
$$
Then
$$
0.26=a_{1}+\ldots+a_{5} \leqslant 5 a_{5} \leqslant 5 a_{6} \Longrightarrow a_{6} \leqslant 0.26 / 5
$$
Moreover,
$$
0.29=a_{k+6}+\ldots+a_{k+10} \geqslant 5 a_{k+6} \geqslant 5 a_{k+5} \Longrightarrow a_{k+5} \geqslant 0.26 / 5
$$
Let $s=a_{6}+\ldots+a_{k+5}$. Note that $s=1-0.26-0.29=0.45$. Therefore,
$$
0.45=s \geqslant k \cdot a_{6} \geqslant \frac{0.26 k}{5}, \quad 0.45=s \leqslant k \cdot a_{k+5} \leqslant \frac{0.29 k}{5}
$$
Thus,
$$
\frac{0.45 \cdot 5}{0.29} \leqslant k \leqslant \frac{0.45 \cdot 5}{0.26} \Longleftrightarrow 7.75<\frac{225}{29} \leqslant k \leqslant \frac{225}{26}<8.65
$$
The only integer $k$ that satisfies this double inequality is 8. Therefore, the total number of numbers written on the board is $8+10=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) Students' written work is evaluated on a two-point grading system, i.e., a work will either be credited if it is done well, or not credited if it is done poorly. The works are first checked by a neural network, which gives an incorrect answer in $10 \%$ of cases, and then all works deemed uncredited are rechecked manually by experts, who do not make mistakes. The neural network can both incorrectly classify a good work as uncredited and a bad work as credited. It is known that among all the works submitted by students, $20 \%$ are actually bad. What is the smallest percentage of bad works that can be among those that experts recheck after the neural network's selection? Provide the integer part of the number.
|
Answer: 66.
Solution. We will represent the students' works in the diagram below.
Neural network errors can:
a) classify all 10% of good works as bad;
b) classify part of the good works as bad and part of the bad works as good;
c) classify all 10% of bad works as good.
Let's analyze the general case. Let \( x \) be the percentage of all works that are incorrectly classified as good (these are the bad works that will not be reviewed by experts); \( 0 \leqslant x \leqslant 10 \). Then the experts will receive for rechecking (in percentages of all works): bad works that the neural network did not misclassify - these are \((20-x) \% \);
good works that were incorrectly classified as bad - these are \((10-x) \% \).
Thus, the percentage of bad works to be rechecked is
\[
\frac{(20-x) \times 100}{(20-x)+(10-x)} = \frac{(20-x) \times 100}{30-2x} = 50 + \frac{500}{30-2x}
\]
This function is increasing over its entire domain, so under the conditions of the problem, it takes the smallest value when \( x = 0 \). This value is \( 66 \frac{2}{3} \), and its integer part is 66.
|
66
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (20 points) Let a sequence of non-negative integers be given
$$
k, k+1, k+2, \ldots, k+n
$$
Find the smallest $k$, for which the sum of all numbers in the sequence is equal to 100.
#
|
# Answer: 9.
Solution. The given sequence of numbers is an arithmetic progression with $n+1$ terms. Its sum is $\frac{(n+1)(2 k+n)}{2}$. Therefore, the condition can be rewritten as $(n+1)(2 k+n)=200$, where $n$ is a non-negative integer. It is clear that $k$ decreases as $n$ increases. Thus, we need to find the largest $n$ for which $k$ is an integer. Note that
$$
200=(n+1)(2 k+n)>n^{2} \Longrightarrow n<\sqrt{200} \Longrightarrow n \leqslant 14
$$
Moreover, $200=2^{3} \cdot 5^{2}$. Let's consider two cases.
1) $n$ is even. The largest $n \leqslant 14$ for which 200 is divisible by $n+1$ is 4, from which $k=\frac{40-4}{2}=18$.
2) $n$ is odd. Then $n$ can only be $1, 3, 7$ (otherwise 200 is not divisible by $n+1$). For $n=7$, we get $k=9$. This will be the smallest.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
|
Answer: $17^{\circ}$.
Solution. Since in the right triangle $ABC$ the angle $BAC=17^{\circ}$, the angle $BCA=73^{\circ}$. Note that triangle $QMC$ is isosceles with base $MC$, as by condition $QM=QC$. Similarly, triangle $PMA$ is isosceles with base $MA$. Therefore, the angles at the bases of these triangles are equal. From this, we get that $\angle QMP=180^{\circ}-\angle PMA-\angle QMC=180^{\circ}-17^{\circ}-73^{\circ}=90^{\circ}$.
Let points $O$ and $R$ be the feet of the perpendiculars dropped from points $Q$ and $P$ to the line $AC$, respectively. It is easy to notice that triangles $QOM$ and $MRP$ are similar by two angles. Since $CM=MA$, from the isosceles nature of triangles $QCM$ and $PMA$, it follows that
$$
MO=\frac{1}{2} CM=\frac{1}{2} MA=MR
$$
Let $x=MO$. From the right triangles $QOM$ and $PMR$, we get that $MQ=\frac{x}{\cos 73^{\circ}}$ and $MP=\frac{x}{\cos 17^{\circ}}$ respectively. Thus, for triangle $MQP$ we have
$$
\operatorname{tg} \angle PQM=\frac{MP}{MQ}=\frac{x}{\cos 17^{\circ}} \cdot \frac{\cos 73^{\circ}}{x}=\frac{\cos \left(\pi / 2-17^{\circ}\right)}{\cos 17^{\circ}}=\operatorname{tg} 17^{\circ}
$$
Since in the right triangle $MQP$ the angle $PQM$ is acute, from the last equality it follows that $\angle PQM=17^{\circ}$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. (40 points) Let's call the tail of a natural number any number that is obtained from it by discarding one or several of its first digits. For example, 234, 34, and 4 are tails of the number 1234. Does there exist a six-digit number without zeros in its decimal representation that is divisible by each of its tails?
|
Answer: Yes (721875 fits).
Solution. Suppose the required number exists. Let's write it as $A=$ $=\overline{a_{5} a_{4} \ldots a_{0}}$. Then $A$ is divisible by its five-digit tail, that is, by $\overline{a_{4} \ldots a_{0}}$. Therefore, $\overline{a_{4} \ldots a_{0}}$ divides the difference between $A$ and its tail, which is $a_{5} \cdot 10^{5}$. Thus, we need to check the five-digit divisors of $a_{5} \cdot 10^{5}$ without zeros in their decimal representation. Any divisor of the number $a_{5} \cdot 10^{5}$ is the product of $a_{5}$ and some powers of five and two. Since this product should not end in zero, it cannot contain both two and five simultaneously. If we take only twos, we get a number not exceeding $a_{5} \cdot 2^{5} \leqslant 9 \cdot 2^{5}=288$, which is not a five-digit number. Therefore, we need the number $a_{5} \cdot 5^{n}$, where $a_{5}$ is odd. Note that $n=5$, otherwise $a_{5} \cdot 5^{n} \leqslant 9 \cdot 625=5625$, and this number is also not five-digit. Finally, $a_{5}>3$, since the number $3 \cdot 5^{5}=9375$ is four-digit.
Considering the constraints, we get three possible options: $5 \cdot 5^{5}$, $7 \cdot 5^{5}, 9 \cdot 5^{5}$. The second one gives the number 721875, which fits:
$$
\begin{gathered}
721875 / 21875=33 ; 721875 / 1875=385 ; 721875 / 875=825 ; \\
721875 / 75=9625 ; 721875 / 5=144375
\end{gathered}
$$
It is not difficult to check that the other two options do not work. However, this is not necessary, as the answer has already been obtained.
|
721875
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (15 points) In a bookstore, there is a rule for "summing" discounts: "if different types of discounts apply to an item, they are applied sequentially one after another." For example, if two discounts A% and B% apply to an item, the first discount is applied to the original price, and the second discount is applied to the result of the first discount. Currently, two discounts are in effect in the store: "Autumn" at $25 \%$ and "Random" - a non-zero discount of a whole number of percent, determined at the moment of purchase. Someone with a loyalty card, which gives a $20 \%$ discount on all items, bought a book for 69 rubles, the original price of which was 230 rubles. What was the size of the "Random" discount?
|
Answer: $50 \%$.
Solution. Note that the sequence of applying discounts is irrelevant, since applying a discount of A\% is equivalent to multiplying the cost of the item by ( $1-\frac{A}{100}$ ). Let $R$ be the magnitude of the "random" discount. Then, as a result of "summing" the three discounts, we get
$$
230 \cdot\left(1-\frac{25}{100}\right) \cdot\left(1-\frac{20}{100}\right) \cdot\left(1-\frac{R}{100}\right)=69 \Longleftrightarrow R=50
$$
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers.
|
Answer: 36.
Solution. In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 words. Indeed, for the letters K and Ш, there are $4 \cdot 3=12$ positions, and we write the letters А in the remaining places. Thus, in total, we get $24+12=36$ words.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (35 points) Journalists have found out that
a) in the lineage of Tsar Pafnuty, all descendants are male: the tsar himself had 2 sons, 60 of his descendants also had 2 sons each, and 20 had 1 son each;
b) in the lineage of Tsar Zinovy, all descendants are female: the tsar himself had 4 daughters, 35 of his descendants had 3 daughters each, and another 35 had 1 daughter each.
The rest of the descendants of both tsars had no children. Who had more descendants?
|
Answer: At Zinovy.
Solution. We need to find the total number of children in the lineage of Pafnuty, including the children of the king himself. It is equal to $60 \cdot 2+20 \cdot 1+2=142$. The total number of children in the lineage of King Zinovy is $4+35 \cdot 3+35 \cdot 1=144$.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (50 points) A $5 \times 5$ square of cells was cut into several pieces of different areas, each consisting of an integer number of cells. What is the maximum number of pieces that could result from such a cutting?
|
Answer: 6.
Solution. We will show that there cannot be more than 6 parts. Indeed, the total area of 7 parts cannot be less than $1+2+3+4+5+6+7=28$, which exceeds the area of the square.
Now we will show how a $5 \times 5$ square can be cut into 6 parts of different sizes:
Thus, the areas of the pieces will be $2, 8, 3, 7, 1,$ and 4.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (50 points) From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property?
|
Answer: 40.
Solution. Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this group. Suppose there is another number $C$, and $j=C \bmod 5$. Then the numbers $i+j$ and $5-i+j$ are divisible by 5, as well as their difference, which is $2 i-5$, which is impossible for $i \neq 0$.
Now consider the alternative case, where all numbers in the group are divisible by 5. In the given range, there are exactly 40 numbers that are multiples of 5, and all these numbers can be included in the group simultaneously.
|
40
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (50 points) Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true?
|
Answer: 6.
Solution. Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let's provide an example where exactly 6 statements are true. Suppose the knights with odd numbers guessed the numbers from 1 to 6, and those with even numbers guessed the numbers from 7 to 12 (in any order). Then all the knights with even numbers are telling the truth.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On a $5 \times 7$ grid, 9 cells are marked. We will call a pair of cells with a common side interesting if exactly one cell in the pair is marked. What is the maximum number of interesting pairs that can be?
|
Answer: 35.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, it is no more than 3. Then the total number of interesting pairs does not exceed $9 \cdot 4 = 36$. At the same time, if there are two adjacent marked cells, the interesting pair containing them is counted twice. Note that among the 9 cells in the $3 \times 5$ rectangle, there are necessarily two adjacent cells. Therefore, among the marked cells, there is either a boundary cell or two adjacent cells. Thus, the total number of interesting pairs does not exceed 35. An example of a marking with 35 interesting pairs is provided below.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$
|
Answer: 64.
Solution. By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \frac{b+c}{2 \sqrt{b c}} \cdot \frac{c+d}{2 \sqrt{c d}} \cdot \frac{d+a}{2 \sqrt{d a}} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$.
|
64
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Quadrilateral $A B C D$ is inscribed in a circle. A tangent line $\ell$ is drawn to this circle at point $C$. Circle $\omega$ passes through points $A$ and $B$ and is tangent to line $\ell$ at point $P$. Line $P B$ intersects segment $C D$ at point $Q$. Find the ratio $\frac{B C}{C Q}$, given that $B$ is the point of tangency of circle $\omega$.
|
Answer: 1.
Solution. The angle between the tangent $BD$ and the chord $AB$ of the circle $\omega$ is equal to the inscribed angle that subtends $AB$, so $\angle APB = \angle ABD = \angle ACD$. Therefore, the quadrilateral $APCQ$ is cyclic, from which $\angle CQB = \angle CAP$. Applying the tangent-secant theorem twice more, we also get the equalities $\angle BCP = \angle BAC$ and $\angle BPC = \angle BAP$. Then,
$$
\angle CBQ = 180^{\circ} - \angle CBP = \angle BCP + \angle BPC = \angle BAC + \angle BAP = \angle CAP = \angle CQB
$$
Thus, triangle $BCQ$ is isosceles, from which $BC = CQ$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On a $7 \times 7$ checkerboard, 14 cells are marked. We will call a pair of cells with a common side interesting if at least one cell in the pair is marked. What is the maximum number of interesting pairs that can be?
|
# Answer: 55.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, no more than 3. Then the total number of interesting pairs does not exceed $14 \cdot 4 = 56$. At the same time, if there are two adjacent marked cells, the interesting pair containing them is counted twice. Note that among the 14 cells in a $5 \times 5$ square, there are necessarily two adjacent ones. Therefore, among the marked cells, there is either a boundary cell or two adjacent ones. Thus, the total number of interesting pairs does not exceed 55. An example of a marking with 55 interesting pairs is provided below.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$
|
Answer: 64.
Solution. By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdot b c}=64 \cdot \frac{a^{2}+b^{2}}{2 a b} \cdot \frac{b^{2}+c^{2}}{2 b c} \cdot \frac{c^{2}+d^{2}}{2 c d} \cdot \frac{d^{2}+a^{2}}{2 d a} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$.
|
64
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A circle $\omega$ is circumscribed around triangle $A B C$. Tangents to the circle, drawn at points $A$ and $B$, intersect at point $K$. Point $M$ is the midpoint of side $A C$. A line passing through point $K$ parallel to $A C$ intersects side $B C$ at point $L$. Find the angle $A M L$.
|
Answer: $90^{\circ}$.
Solution. Let $\alpha=\angle A C B$. The angle between the tangent $A K$ and the chord $A B$ of the circle $\omega$ is equal to the inscribed angle that subtends $A B$, hence $\alpha=\angle B A K=\angle A B K$. Since
$A C \| K L$, we get $\angle B L K=\angle A C B=\angle B A K$. Therefore, the quadrilateral $A L B K$ is cyclic. Then
$$
\angle A L C=180^{\circ}-\angle A L B=\angle A K B=180^{\circ}-2 \alpha \quad \text { and } \quad \angle L C A=\alpha=180^{\circ}-\angle A L C-\alpha=\angle L A C .
$$
Thus, triangle $A L C$ is isosceles, and its median $L M$ is also an altitude. Therefore, $\angle A M L=90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 20, 40, and 40, and the radius of the base of the cone is 21. Find the height of the cone.
|
Answer: 28.
Solution. Let $O$ be the center of the base of the cone, $h$ be its height, and $2 \alpha$ be the angle of inclination of the cone's generators to the table. Consider the section passing through the axis of symmetry of the cone and the center of one of the spheres (see the left figure). Let $K$ be the point of contact of the sphere with the table, and $R$ be the radius of the sphere. Then
$$
O K=R \cdot \operatorname{tg} \alpha+21
$$
Let $A, B, C$ be the points of contact of the spheres with the table (see the right figure). From the condition of the spheres touching, $B C=80$ and
$$
A B=A C=\sqrt{(40+20)^{2}-(40-20)^{2}}=40 \sqrt{2}
$$
Thus, the height $A D$ of the triangle $A B C$ is the perpendicular bisector of $B C$. From (*) it follows that the point $O$ is equidistant from $B$ and $C$, i.e., it lies on $A D$, and $O A=20 \operatorname{tg} \alpha+21$. Note that
$$
B D=40, \quad A D=\sqrt{A B^{2}-B D^{2}}=40, \quad O D=A D-O A=19-20 \operatorname{tg} \alpha
$$
The equality (*) gives $O B=40 \operatorname{tg} \alpha+21$, and by the Pythagorean theorem
$O B^{2}=O D^{2}+B D^{2} \Longleftrightarrow(40 \operatorname{tg} \alpha+21)^{2}=(19-20 \operatorname{tg} \alpha)^{2}+40^{2} \Longleftrightarrow 30 \operatorname{tg}^{2} \alpha+61 \operatorname{tg} \alpha-38=0 \Longleftrightarrow \operatorname{tg} \alpha=\frac{1}{2}$.
From this,
$$
\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{4}{3} \quad \text { and } \quad h=21 \cdot \operatorname{tg} 2 \alpha=28
$$
## Variant 3
|
28
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Each cell of a $5 \times 6$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of red cells is not less than the number of blue cells and not less than the number of yellow cells, and in each column of the table, the number of blue cells is not less than the number of red cells and not less than the number of yellow cells. How many yellow cells can there be in such a table? Provide an example of a corresponding coloring.
|
Answer: 6.
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ |
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ |
Solution. Since in each row there are no fewer red cells than blue ones, there are no fewer of them in the entire table. Then the number of red and blue cells is the same in each column. Indeed, if in one of the columns there are fewer red cells than blue ones, then there will be fewer of them in the entire table, since in the other columns there are no more blue ones. In addition, the number of blue cells in each column is no less than $\frac{5}{3}$, that is, it is at least 2. Therefore, the table contains at least 12 blue cells and as many red ones. Therefore, there are no more than 6 yellow cells in the table. On the other hand, in each column the total number of red and blue cells is even, so there is at least one yellow cell in it. Therefore, the table contains no fewer than 6 yellow cells. An example of coloring with six yellow cells is shown in the figure.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BC$ intersects circle $\omega_{1}$ again at point $P$, and line $AD$ intersects circle $\omega_{2}$ again at point $Q$. It is known that points $P, Q$, and $E$ are distinct. Find the angle $PEQ$.
|
Answer: $180^{\circ}$.
Solution 1. Since $\angle B C O=\angle P C O_{1}$, the isosceles triangles $B O C$ and $P O_{1} C$ are similar, from which $\angle B O C=\angle P O_{1} C$. Similarly, it can be verified that $\angle A O D=\angle Q O_{2} D$. Then the segments $O_{1} P$ and $O_{2} Q$ are parallel to the line $A B$ and, therefore, to each other. Hence, $\angle P O_{1} E=\angle Q O_{2} E$. Thus, the isosceles triangles $P O_{1} E$ and $Q O_{2} E$ are similar, from which $\angle O_{1} E P=\angle O_{2} E Q$. Therefore, the point $E$ lies on the segment $P Q$, which gives the answer.
Solution 2. Note a useful property of tangent circles: if a secant $U V$ passes through the point $T$ of tangency of two circles, then the inscribed angles subtended by the arcs it intercepts are equal. Indeed, since an inscribed angle is equal to the angle between the tangent and the secant, the equalities $\angle U U_{1} T=\angle U T Y=\angle X T V=\angle V V_{1} T$ hold (see the left figure below).
Let the intersection of $A D$ and $B C$ be $R$, and the second intersection of $A C$ with the circle $\omega_{1}$ be $S$. Let $\angle O O_{1} O_{2}=2 \alpha$ and $\angle O O_{2} O_{1}=2 \beta$. From the isosceles triangles $O_{1} C E$ and $O_{2} D E$, we get that
$$
\angle O_{1} E C=90^{\circ}-\alpha, \quad \angle O_{2} E D=90^{\circ}-\beta \quad \text { and } \quad \angle C E D=180^{\circ}-\angle O_{1} E C-\angle O_{2} E D=\alpha+\beta
$$
On the other hand, from the sum of the angles of triangle $O O_{1} O_{2}$, we find that $\angle C O D=180^{\circ}-2 \alpha-2 \beta$. The inscribed angle $\angle C A D$ is half the central angle $\angle C O D$. Then
$$
\angle C A D=90^{\circ}-\alpha-\beta, \quad \angle A C R=\angle A C B=90^{\circ}, \quad \text { from which } \quad \angle C R D=\alpha+\beta
$$
Thus, $\angle C R D=\angle C E D$. From the cyclic quadrilateral $P E C S$, it follows that
$$
\angle P E C=180^{\circ}-\angle P S C=180^{\circ}-\angle C A B=90^{\circ}+\angle R B A
$$
Moreover, $\angle D E Q=\angle D B A=90^{\circ}-\angle R A B$. Now we can find the angle $P E Q$:
$$
\begin{aligned}
& \angle P E Q=\angle P E C+\angle C E D-\angle D E Q=\angle P E C+\angle C R D-\angle D E Q= \\
& =\left(90^{\circ}+\angle R B A\right)+\left(180^{\circ}-\angle R A B-\angle R B A\right)-\left(90^{\circ}-\angle R A B\right)=180^{\circ}
\end{aligned}
$$
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Each cell of a $5 \times 5$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of yellow cells is not less than the number of red cells and not less than the number of blue cells, and in each column of the table, the number of red cells is not less than the number of yellow cells and not less than the number of blue cells. How many blue cells can there be in such a table? Provide an example of a corresponding coloring.
|
Answer: 5.
| $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ |
Solution. Since in each row the number of yellow cells is not less than the number of red cells, the same is true for the entire table. Therefore, the number of yellow and red cells is the same in each column. Indeed, if in one of the columns the number of yellow cells is less than the number of red cells, then there will be fewer of them in the entire table, since in the other columns their number is not greater than the number of red cells. Moreover, the number of red cells in each column is not less than $\frac{5}{3}$, that is, it is at least 2. Thus, the table contains at least 10 red cells and the same number of yellow cells. Therefore, there are no more than 5 blue cells in the table. On the other hand, in each column the total number of yellow and red cells is even, so there is at least one blue cell in each column. Thus, the table contains no fewer than 5 blue cells. An example of coloring with five blue cells is shown in the figure.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BD$ intersects circle $\omega_{2}$ again at point $P$, and line $AC$ intersects circle $\omega_{1}$ again at point $Q$. Find the angle $PEQ$.
|
Answer: $180^{\circ}$.
Solution 1. Since $\angle A C O=\angle Q C O_{1}$, the isosceles triangles $A O C$ and $Q O_{1} C$ are similar, from which $\angle A O C=\angle Q O_{1} C$. Similarly, it can be verified that $\angle B O D=\angle P O_{2} D$. Then the segments $O_{1} Q$ and $O_{2} P$ are parallel to the line $A B$ and, therefore, to each other. Hence, $\angle O_{1} Q E=\angle O_{2} P E$, which is equivalent to $\angle O_{1} E Q=\angle O_{2} E P$. This means that point $E$ lies on the segment $P Q$, from which $\angle P E Q=180^{\circ}$.
Solution 2. Note a useful property of tangent circles: if a secant $U V$ passes through the point $T$ of tangency of two circles, then the inscribed angles subtending the arcs cut by it are equal. Indeed, since the inscribed angle is equal to the angle between the tangent and the secant, the equalities $\angle U U_{1} T=\angle U T Y=\angle X T V=\angle V V_{1} T$ hold (see the left figure below).
Let the intersection of $A D$ and $B C$ be $R$. Let $\angle O O_{1} O_{2}=2 \alpha$ and $\angle O O_{2} O_{1}=2 \beta$. Then, from the isosceles triangles $O_{1} C E$ and $O_{2} D E$, we get that
$$
\angle O_{1} E C=90^{\circ}-\alpha, \quad \angle O_{2} E D=90^{\circ}-\beta \quad \text { and } \quad \angle C E D=180^{\circ}-\angle O_{1} E C-\angle O_{2} E D=\alpha+\beta .
$$
On the other hand, from the sum of the angles of triangle $O O_{1} O_{2}$, we find that $\angle C O D=180^{\circ}-2 \alpha-2 \beta$. The inscribed angle $\angle C A D$ is half of the central angle $\angle C O D$. Then
$$
\angle D A C=90^{\circ}-\alpha-\beta, \quad \angle A C R=\angle A C B=90^{\circ}, \quad \text { from which } \angle D R C=\alpha+\beta .
$$
Thus, $\angle D R C=\angle D E C$. By the proven statement,
$$
\angle P E D=\angle B A D=\angle R A B \quad \text { and } \quad \angle C E Q=\angle A B C=\angle R B A .
$$
Now we can find the angle $P E Q$ :
$$
\begin{aligned}
\angle P E Q & =\angle P E D+\angle D E C+\angle C E Q=\angle P E D+\angle D R C+\angle C E Q= \\
& =\angle P E D+180^{\circ}-\angle R A B-\angle R B A+\angle C E Q=180^{\circ} .
\end{aligned}
$$
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table.
|
Answer: 48.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, as each of them is counted twice - in the row and in the column. Then
$$
S \geqslant \frac{1}{2}(5+6+\ldots+13+14)=\frac{1}{2} \cdot 95=47 \frac{1}{2}, \quad \text { that is } \quad S \geqslant 48 .
$$
An example for $S=48$ is given below.
| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 2 | 2 |
| 1 | 1 | 2 | 3 | 3 |
| 1 | 2 | 2 | 3 | 3 |
| 2 | 3 | 3 | 3 | 4 |
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{d}\right)^{3}+\left(d+\frac{1}{a}\right)^{3}
$$
|
Answer: 32.
Solution 1. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(2 \sqrt{\frac{a}{b}}\right)^{3}+\left(2 \sqrt{\frac{b}{c}}\right)^{3}+\left(2 \sqrt{\frac{c}{d}}\right)^{3}+\left(2 \sqrt{\frac{d}{a}}\right)^{3} \geqslant 32\left(\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot \frac{d}{a}\right)^{3 / 8}=32
$$
Equality is achieved when $a=b=c=d=1$.
Solution 2. We will use the inequality $x^{3}+y^{3} \geqslant \frac{1}{4}(x+y)^{3}$, which is true for $x, y>0$. Applying it three times, we get
$$
x^{3}+y^{3}+z^{3}+w^{3} \geqslant \frac{1}{4}\left((x+y)^{3}+(z+w)^{3}\right) \geqslant \frac{1}{16}(x+y+z+w)^{3} \quad \text { for } \quad x, y, z, w>0
$$
Then, by the Cauchy inequality for means,
$$
A \geqslant \frac{1}{16}\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{d}+d+\frac{1}{a}\right)^{3} \geqslant \frac{1}{16}\left(8 \sqrt[8]{a \cdot \frac{1}{b} \cdot b \cdot \frac{1}{c} \cdot c \cdot \frac{1}{d} \cdot d \cdot \frac{1}{a}}\right)^{3}=\frac{8^{3}}{16}=32
$$
Equality is achieved when $a=b=c=d=1$.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given an acute-angled triangle $A B C$. A circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Tangents drawn to the circle at points $D$ and $E$ intersect at point $K$. Find the angle between the lines $A K$ and $B C$.
|
Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle D B E, \beta=\angle B C D, \gamma=\angle C B E, O$ - the intersection point of segments $C D$ and $B E$. The angle between the tangent $D K$ and the chord $D E$ is equal to the inscribed angle that subtends
$$
\angle D E K=\angle E D K=\alpha
$$
Moreover, $\angle C D E=\gamma$. Due to the cyclic nature of quadrilateral $B D E C$
$$
\angle A E K=\angle A E D-\alpha=\angle C B D-\alpha=\gamma \quad \text { and, similarly } \quad \angle A D K=\beta .
$$
Since $\alpha+\beta+\gamma=\angle A D C=90^{\circ}$, we get
$$
\angle D A E=180^{\circ}-2 \alpha-\beta-\gamma=90^{\circ}-\alpha=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=\frac{1}{2} \angle D K E
$$
Note that a circle $\omega$ can be circumscribed around quadrilateral $A D O E$. Then the angle $D A E$ is inscribed in $\omega$, and point $K$ lies on the perpendicular bisector of chord $B E$, i.e., on the diameter of $\omega$. Therefore, $K$ is the center of $\omega$, from which
$$
\angle D A K=\angle A D K=\beta=90^{\circ}-\alpha-\gamma=90^{\circ}-\angle C B D
$$
Thus, $A K \perp B C$.
Solution 2. Let
$$
\alpha=\angle D B E, \quad \beta=\angle B C D, \gamma=\angle C B E, \varphi=\angle D A K, \quad \psi=\angle E A K
$$
The angle between the tangent $D K$ and the chord $D E$ is equal to the inscribed angle that subtends $D E$, hence
$$
\angle D E K=\angle E D K=\alpha
$$
Moreover, $\angle C D E=\gamma$. Due to the cyclic nature of quadrilateral $B D E C$
$$
\angle A E K=\angle A E D-\alpha=\angle C B D-\alpha=\gamma \quad \text { and, similarly } \quad \angle A D K=\beta .
$$
Since $\alpha+\beta+\gamma=\angle A D C=90^{\circ}$, we get
$$
\gamma=90^{\circ}-\alpha-\beta, \quad \psi=180^{\circ}-2 \alpha-\beta-\gamma-\varphi=90^{\circ}-\alpha-\varphi
$$
The Law of Sines for triangles $D A K$ and $E A K$ gives
$$
\frac{A K}{\sin \beta}=\frac{D K}{\sin \varphi}, \quad \frac{A K}{\sin \gamma}=\frac{E K}{\sin \psi}
$$
Since $D K=E K$, we can eliminate the sides from these equations. We get
$$
\begin{aligned}
& \sin \gamma \sin \varphi=\sin \beta \sin \psi \Longleftrightarrow \cos (\alpha+\beta) \sin \varphi=\sin \beta \cos (\alpha+\varphi) \Longleftrightarrow \\
& \Longleftrightarrow \sin (\alpha+\beta+\varphi)+\sin (\varphi-\alpha-\beta)=\sin (\alpha+\beta+\varphi)+\sin (\beta-\alpha-\varphi) \Longleftrightarrow 2 \sin (\varphi-\beta) \cos \alpha=0
\end{aligned}
$$
Then
$$
\angle D A K=\varphi=\beta=90^{\circ}-\alpha-\gamma=90^{\circ}-\angle C B D
$$
from which $A K \perp B C$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table.
|
Answer: 43.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, since each of them is counted twice - in the row and in the column. Then
$$
S \geqslant \frac{1}{2}(4+5+\ldots+12+13)=\frac{1}{2} \cdot 85=42 \frac{1}{2}, \quad \text { that is } \quad S \geqslant 43
$$
An example for $S=43$ is given below.
| 1 | 1 | 1 | 1 | 1 | 2 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 2 | 2 | 3 |
| 1 | 1 | 2 | 3 | 3 | 2 |
| 1 | 2 | 2 | 2 | 3 | 4 |
|
43
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$
|
Answer: 32.
Solution. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{d a}}\right)^{3}+\left(\frac{2 d}{\sqrt{a b}}\right)^{3} \geqslant 32\left(\frac{a^{2}}{b c} \cdot \frac{b^{2}}{c d} \cdot \frac{c^{2}}{d a} \cdot \frac{d^{2}}{a b}\right)^{3 / 8}=32
$$
Equality is achieved when $a=b=c=d=1$.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Point $M$ is the midpoint of side $AB$ of triangle $ABC$. A circle $\omega_{1}$ is drawn through points $A$ and $M$, tangent to line $AC$, and a circle $\omega_{2}$ is drawn through points $B$ and $M$, tangent to line $BC$. Circles $\omega_{1}$ and $\omega_{2}$ intersect again at point $D$. Point $E$ lies inside triangle $ABC$ and is symmetric to point $D$ with respect to line $AB$. Find the angle $CEM$.
|
Answer: $180^{\circ}$
Solution 1. The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $\angle C A B = \angle A D M$. Similar reasoning for $\omega_{2}$ gives $\angle A B C = \angle B D M$. Extend triangle $A D B$ to parallelogram $A D B K$ (see the right figure). Triangles $A C B$ and $D A K$ are similar by two angles. Point $M$ is the midpoint of segments $A B$ and $D K$, so triangles $A C M$ and $D A M$ are also similar (by angle and proportional sides). Therefore, $\angle A M C = \angle A M D$. Moreover, from the symmetry of points $D$ and $E$ with respect to $A B$, it follows that $\angle A M E = \angle A M D$. Thus, point $E$ lies on segment $C M$.
Solution 2. By the properties of tangents, $\angle A D M = \angle C A M$ and $\angle M D B = \angle C B M$. Therefore,
$$
\angle A D B = \angle A D M + \angle M D B = \angle C A M + \angle C B M = 180^{\circ} - \angle A C B
$$
and quadrilateral $A C B D$ is cyclic. Then
$$
\angle B D C = \angle B A C = \angle M D A \quad \text{and} \quad \angle B C D = \angle B A D = \angle M A D.
$$
Thus, triangles $B D C$ and $M D A$ are similar by two angles, so
$$
\frac{B D}{M D} = \frac{B C}{M A} = \frac{C B}{M B}
$$
Therefore, triangles $B D M$ and $C B M$ are similar by angle and ratio of sides, and in particular, angles $\angle C M B$ and $\angle D M B$ are equal. Due to the symmetry of points $D$ and $E$ with respect to $A B$, we get
$$
\angle E M B = \angle D M B = \angle C M B
$$
Thus, point $E$ lies on line $C M$, from which $\angle C E M = 180^{\circ}$.
Solution 3. Let
$$
\alpha = \angle C A B, \quad \beta = \angle A B C, \varphi = \angle A M D, \psi = \angle A M C
$$
The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $\angle C A B = \angle A D M$. Similar reasoning for $\omega_{2}$ gives $\angle A B C = \angle B D M$. Note that
$$
\angle D A M = 180^{\circ} - \alpha - \varphi \quad \angle D M B = 180^{\circ} - \beta - (180^{\circ} - \varphi) = \varphi - \beta
$$
Then the Law of Sines for triangles $A D M$ and $D M B$ gives
$$
\frac{A M}{\sin \alpha} = \frac{D M}{\sin (\alpha + \varphi)}, \quad \frac{B M}{\sin \beta} = \frac{D M}{\sin (\varphi - \beta)}
$$
Since $A M = B M$, we can eliminate the sides from these equations. We get
$$
\frac{\sin \beta}{\sin \alpha} = \frac{\sin (\varphi - \beta)}{\sin (\alpha + \varphi)}
$$
Similar reasoning for triangles $A C M$ and $C M B$ gives
$$
\frac{\sin \beta}{\sin \alpha} = \frac{\sin (\psi - \beta)}{\sin (\alpha + \psi)}
$$
Eliminating the left side of these relations, we get
$$
\begin{aligned}
\sin (\varphi - \beta) \sin (\psi + \alpha) = & \sin (\psi - \beta) \sin (\varphi + \alpha) \Longleftrightarrow \cos (\varphi - \beta - \alpha - \psi) - \cos (\varphi - \beta + \alpha + \psi) = \\
& = \cos (\psi - \beta - \alpha - \varphi) - \cos (\varphi - \beta + \alpha + \psi) \Longleftrightarrow 2 \sin (\alpha + \beta) \sin (\varphi - \psi) = 0
\end{aligned}
$$
from which $\varphi = \psi$. Moreover, from the symmetry of points $D$ and $E$ with respect to $A B$, it follows that $\angle A M E = \varphi$. Thus, point $E$ lies on segment $C M$.
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. What is the minimum number of cells that need to be marked in a $7 \times 7$ table so that every vertical or horizontal strip $1 \times 4$ contains at least one marked cell?
|
Answer: 12.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row and $m$ axial cells in the central column. In rows and columns that do not contain axial cells, there should be at least two marked cells. In total, in such rows, there will be at least $2 \cdot(2 n-2-m)=4(n-1)-2 m$ marked cells, and in columns - at least $4(n-1)-2 k$ cells. At the same time, each cell will be counted no more than twice. Taking into account the axial cells, we get that there are at least
$$
\frac{1}{2}(4(n-1)-2 m+4(n-1)-2 k)+k+m=4(n-1) \text { cells. }
$$
The value $4(n-1)$ is achieved if all cells of the central row and column, except for the center of the board, are marked. Assuming $n=4$, we get the answer.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given numbers $x, y, z \in\left[0, \frac{\pi}{2}\right]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$
|
Answer: 1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is positive and does not depend on $x$, while the argument of the second cosine lies in $\left[0, \frac{\pi}{2}\right]$, since $\frac{\pi}{2} \geqslant \frac{z+y}{2} \geqslant x$. Therefore, the right-hand side will be the smallest when $x=0$. In this case,
$$
A=\cos y+\cos z+\cos (y-z)=\cos z+2 \cos \frac{z}{2} \cdot \cos \left(y-\frac{z}{2}\right)
$$
Notice that $-\frac{z}{2} \leqslant y-\frac{z}{2} \leqslant \frac{z}{2}$, hence $\cos \left(y-\frac{z}{2}\right) \geqslant \cos \frac{z}{2}$. Therefore,
$$
A \geqslant \cos z+2 \cos ^{2} \frac{z}{2}=2 \cos z+1 \geqslant 1
$$
Equality is achieved when $x=0, y=z=\frac{\pi}{2}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. What is the minimum number of cells that need to be marked in a $9 \times 9$ table so that each vertical or horizontal strip $1 \times 5$ contains at least one marked cell?
|
Answer: 16.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row and $m$ axial cells in the central column. In rows and columns that do not contain axial cells, there should be at least two marked cells. In total, in such rows, there will be at least $2 \cdot(2 n-2-m)=4(n-1)-2 m$ marked cells, and in columns - at least $4(n-1)-2 k$ cells. At the same time, each cell will be counted no more than twice. Taking into account the axial cells, we get that there are at least
$$
\frac{1}{2}(4(n-1)-2 m+4(n-1)-2 k)+k+m=4(n-1) \text { cells. }
$$
The value $4(n-1)$ is achieved if all cells of the central row and column, except for the center of the board, are marked. Assuming $n=5$, we get the answer.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given numbers $x, y, z \in [0, \pi]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$
|
Answer: -1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is non-negative and does not depend on $x$, while the argument of the second cosine lies in the interval $[0, \pi]$, since $\pi \geqslant \frac{z+y}{2} \geqslant x$. Therefore, the right-hand side will be the smallest when $x=0$. In this case,
$$
A=\cos y+\cos z+\cos (y-z)=\cos z+2 \cos \frac{z}{2} \cdot \cos \left(y-\frac{z}{2}\right)
$$
Notice that $-\frac{\pi}{2} \leqslant-\frac{z}{2} \leqslant y-\frac{z}{2} \leqslant \frac{z}{2} \leqslant \frac{\pi}{2}$, hence
$$
A=\cos z+2 \cos \frac{z}{2} \cdot \cos \left(y-\frac{z}{2}\right) \geqslant \cos z \geqslant-1
$$
Equality is achieved when $x=0, y=z=\pi$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ again at point $X$ and circle $\omega_{2}$ again at point $Y$. Point $M$ is the midpoint of segment $XY$. Find the angle $OMX$.
|
Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A C$, i.e., $\angle A X C=\angle B A C=\alpha$. Similarly, we get $\angle A Y B=\alpha$. Extend quadrilateral $C X Y B$ to form triangle $X Y Z$ (see figure). Note that
$$
\angle B O C=2 \angle B A C=2 \alpha=\angle Z X Y+\angle Z Y X=180^{\circ}-\angle X Z Y
$$
which means quadrilateral $O B Z C$ is cyclic. Then
$$
\angle O Z B=\angle O C B=\angle O B C=\angle O Z C
$$
Therefore, ray $Z O$ is the angle bisector of isosceles triangle $X Z Y$, which is also the median and altitude. Thus, point $M$ lies on ray $Z O$ and $\angle O M X=90^{\circ}$.
Solution 2. Let line $X Y$ intersect circle $\omega$ again at point $K$. Since line $A C$ is tangent to circle $\omega_{2}$,
$$
\angle B Y K=\angle B Y A=\angle B A C
$$
Angle $Y B A$ subtends arc $Y A$ of circle $\omega_{2}$, not containing point $B$. Therefore, it is equal to the angle between chord $Y A$ and the tangent to $\omega_{2}$ at point $A$, i.e., angle $X A C$. Considering the cyclic quadrilateral $A B C K$, we get
$$
\angle Y B A=\angle X A C=\angle K A C=\angle K B C
$$
which implies $\angle Y B K=\angle A B C$. Therefore, triangles $Y B K$ and $A B C$ are similar by two angles. Then
$$
\frac{Y K}{A C}=\frac{B K}{B C} \Longleftrightarrow Y K=\frac{A C \cdot B K}{B C}
$$
Similarly, it can be verified that triangles $C A X$ and $C B K$ are similar and $\frac{A X}{B K}=\frac{A C}{B C}$. Thus,
$$
A X=\frac{A C \cdot B K}{B C}=Y K
$$
From this, it follows that point $M$ is the midpoint of segment $A K$. Note that point $O$ is the circumcenter of triangle $A K B$ and, therefore, lies on the perpendicular bisector of side $A K$. Hence, $\angle O M X=\angle O M K=90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible.
|
Answer: 220000001.
Solution. Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a_{7}+a_{k} a_{8} \quad \text { for any } \quad k \in\{0, \ldots, 8\}
$$
Clearly, $b_{0}=a_{0} a_{8} \bmod 10$. Since $10^{17}>m n \geqslant a_{0} a_{8} \cdot 10^{16}$, we get $a_{0} a_{8} \leqslant 9$, which means $b_{0}=a_{0} a_{8}$. Suppose that for some $k0$, we get $a_{0}=1$ and $a_{k}=0$ for $k=1, \ldots, 6$. The number $m=220000001$ clearly satisfies the condition of the problem.
|
220000001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a $3 \times 3$ table, 9 numbers are arranged such that the six products of these numbers in the rows and columns of the table are all different. What is the maximum number of these numbers that can equal one?
|
Answer: 5.
Solution. Let's call the index of the table the total number of its rows and columns consisting of ones. According to the condition, the index does not exceed 1. Let $n$ be an element of the table different from 1. Then in one row or one column with $n$ there is another number not equal to 1 (otherwise the products in the row and column containing $n$ would be equal to $n$). Therefore, non-unit elements occur in pairs. There are at least two such pairs, otherwise the index of the table would be no less than 3. If there are exactly two pairs, then they do not intersect, otherwise the index of the table is 2. Thus, the table contains at least 4 numbers different from 1, and the number of ones does not exceed 5. An example of a table with 5 ones is given below.
| 1 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 2 | 3 |
| 5 | 7 | 1 |
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Point $O$ is the center of the circumcircle of triangle $ABC$. On the circumcircle of triangle $BOC$, outside triangle $ABC$, point $X$ is chosen. On the rays $XB$ and $XC$ beyond points $B$ and $C$, points $Y$ and $Z$ are chosen respectively such that $XY = XZ$. The circumcircle of triangle $ABY$ intersects side $AC$ at point $T$. Find the angle $\angle Y T Z$.
|
Answer: $180^{\circ}$.
Solution. Note that $\angle B Y T=\angle B A T$ as inscribed angles subtending the same arc. Since quadrilateral $B O C X$ is cyclic, we get
$$
180^{\circ}-\angle B X C=\angle B O C=2 \angle B A C=2 \angle B Y T
$$
On the other hand, triangle $Y X Z$ is isosceles, so $180^{\circ}-\angle Y X Z=2 \angle X Y Z$. Therefore, $\angle X Y T=\angle X Y Z$, which means point $T$ lies on segment $Y Z$.
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a $3 \times 4$ table, 12 numbers are arranged such that all seven sums of these numbers in the rows and columns of the table are distinct. What is the maximum number of numbers in this table that can be zero?
|
Answer: 8.
Solution. Let's call the index of the table the total number of its zero rows and columns. According to the condition, the index does not exceed 1. Let $n$ be a non-zero element of the table. Then in the same row or in the same column with $n$ there is another non-zero number (otherwise the sums in the row and column containing $n$ are equal to $n$). Therefore, non-zero elements occur in pairs. There are at least two such pairs, otherwise the index of the table is not less than 4. If there are exactly two pairs, then they do not intersect, otherwise the index of the table is 3. Thus, the table contains at least 4 non-zero numbers, and the number of zeros does not exceed 8. An example of a table with 8 zeros is given below.
| 0 | 0 | 0 | 0 |
| :--- | :--- | :--- | :--- |
| 1 | 2 | 0 | 0 |
| 0 | 0 | 4 | 5 |
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given numbers $x, y, z \in [0, \pi]$. Find the maximum value of the expression
$$
A=\sin (x-y)+\sin (y-z)+\sin (z-x)
$$
|
Answer: 2.
Solution 1. We can assume that $x \leqslant y$ and $x \leqslant z$, since the expression $A$ does not change under cyclic permutation of the variables. Notice that
$$
\sin (x-y)+\sin (z-x)=2 \sin \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The argument of the sine on the right-hand side lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and the cosine in $[0, \pi]$, since $\pi \geqslant \frac{z+y}{2} \geqslant x$. Consider two cases.
1) $z>y$. The right-hand side will be maximized at the maximum possible $x$, i.e., when $x=y$, and the value of $A$ will be zero.
2) $z \leqslant y$. For fixed $y$ and $z$, the right-hand side will reach its maximum when $x=0$. In this case,
$$
A=\sin z-\sin y+\sin (y-z)=-\sin y+2 \sin \frac{y}{2} \cdot \cos \left(\frac{y}{2}-z\right) \leqslant 2 \sin \frac{y}{2}-\sin y \leqslant 2
$$
Thus, $A \leqslant 2$. Equality is achieved when $x=0, y=\pi, z=\frac{\pi}{2}$.
Solution 2. Let $a=x-y, b=y-z, c=z-x$. The sum of the numbers $a, b, c$ is zero, so at least one of them is non-positive. Suppose, for example, $c \leqslant 0$. Then $c \in[-\pi, 0]$, hence $\sin c \leqslant 0$. Therefore,
$$
A=\sin a+\sin b+\sin c \leqslant \sin a+\sin b \leqslant 2
$$
Equality is achieved when $x=0, y=\pi, z=\frac{\pi}{2}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Point $O$ is the center of the circumcircle of triangle $A B C$. Points $Q$ and $R$ are chosen on sides $A B$ and $B C$ respectively. Line $Q R$ intersects the circumcircle of triangle $A B R$ again at point $P$ and intersects the circumcircle of triangle $B C Q$ again at point $S$. Lines $A P$ and $C S$ intersect at point $K$. Find the angle between lines $K O$ and $Q R$.
|
Answer: $90^{\circ}$.
Solution. Note that $\angle C S Q=\angle C B Q$ and $\angle A P R=\angle A B R$ as inscribed angles subtending the same arc. Therefore, $\angle K S P=\angle K P S$, which means triangle $P K S$ is isosceles. Since
$$
180^{\circ}-\angle A K C=2 \angle K S P=2 \angle A B C=\angle A O C
$$
quadrilateral $A O C K$ is cyclic. Since triangle $A O C$ is isosceles, we have
$$
\angle A K O=\angle A C O=\angle C A O=\angle C K O.
$$
Therefore, line $K O$ is the angle bisector of isosceles triangle $P K S$, and thus also its altitude. Hence, $K O \perp Q R$.
different decimal digits. This product is a six-digit number, its extreme digits are equal, and between them are two pairs of identical adjacent digits. What is written on the board?
Answer: $633556=847 \cdot 748$.
Solution. Since the problem does not change when I and K are swapped, we will assume $I \geqslant K \cdot 10000 > m \cdot 100000$.
From this, $d \geqslant m$, and equality is not possible, otherwise $I \cdot K \vdots 11$. Additionally, the inequality $p < (I+1)(K+1) \cdot 10000$ holds, from which we get
$$
m < d \leqslant \left[\frac{I \cdot K + I + K + 1}{10}\right] = m + \left[\frac{d + I + K + 1}{10}\right]
$$
From the condition, it follows that $p$ is divisible by 11. Then $\overline{KC I}$ or $\overline{I C K}$ is divisible by 11, which is equivalent to $I + K - C \vdots 11$. Therefore, either $C = I + K$ or $C = I + K - 11$. Let's consider these cases.
1) Suppose $C = I + K$. Since $d + I + K + 1 = d + 1 + C < 20$, from (*) we get $d = m + 1$, which means $I \cdot K = 11m + 1$. Additionally, $I + K \leqslant 9$, from which $I \cdot K \leqslant 20$. Therefore, $I \cdot K = 12$, and the pair (I, K) is $(2,6)$ or $(3,4)$. Since $682 \cdot 286 = 195052$ and $473 \cdot 374 = 176902$, these cases do not fit.
2) Suppose $C = I + K - 11$. Since $d + I + K + 1 \leqslant 27 < 30$, from (*) we get $d = m + 1$ or $d = m + 2$, from which $I \cdot K \bmod 11$ is 1 or 2. Additionally, $I + K \geqslant 11$, from which $I \cdot K \geqslant 18$. Therefore, $I \cdot K$ can take values $24, 35, 45, 56$, and the pair (I, K) can be $(3,8), (5,7), (5,9), (7,8)$. Since
$$
308 \cdot 803 = 247324, \quad 715 \cdot 517 = 369655, \quad 935 \cdot 539 = 503965 \quad 847 \cdot 748 = 633556
$$
only the last case fits.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. (40 points) For the quadratic function $p(x)=a x^{2}+b x+c$, for some integers $n$, the equality $p(n)=p\left(n^{2}\right)$ holds. Provide an example of the function $p(x)$ for which the number of such numbers $n$ is the largest. What is this largest number of numbers $n$?
|
Answer: The maximum number of numbers $n$ is 4. An example of a function: $p(x)=x^{2}-6 x+1$.
Solution: Since $0=0^{2}$ and $1=1^{2}$, for any function $p(x)$ there are at least two such numbers $n$.
Write the equality $p(n)=p\left(n^{2}\right)$ and transform it:
$$
a n^{2}+b n+c=a n^{4}+b n^{2}+c \Leftrightarrow a n^{2}\left(n^{2}-1\right)=b n(1-n)
$$
Further, assuming that $n \neq 0$ and $n \neq 1$, we get that $b=-a n(n+1)$.
For each function $p(x)$, the value $-\frac{b}{a}$ is known and fixed. Therefore, the values of $n$ of interest can be obtained as solutions to the quadratic equation $n^{2}+n+\frac{b}{a}=0$. Its roots are $n_{1,2}=\frac{-1 \pm \sqrt{D}}{2}$, where $D=1-4 \frac{b}{a}$. It is not difficult to see that the roots of the equation will be integers only if the numerator of the fraction is even, which means the discriminant must be a perfect square of some odd number; and in this case, we will have two integer solutions to the equation; in all other cases, there will be no integer roots.
The relationship between $n_{1}$ and $n_{2}$ follows from Vieta's theorem: $n_{1}+n_{2}=-1$, i.e., $n_{2}=-1-n_{1}$.
Thus, the set of numbers for which the condition of the problem is satisfied is the set $\left\{0,1, n_{1},-1-n_{1}\right\}$. Now let's determine whether there are such $n_{1}$ that this set consists of four numbers. It is easy to see that the conditions
$$
\left\{\begin{array} { l }
{ n _ { 1 } \neq 0 , } \\
{ n _ { 1 } \neq 1 , } \\
{ - 1 - n _ { 1 } \neq 0 , } \\
{ - 1 - n _ { 1 } \neq 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
n_{1} \neq 0 \\
n_{1} \neq 1 \\
n_{1} \neq-1 \\
n_{1} \neq-2
\end{array}\right.\right.
$$
Obviously, such $n_{1}$ exist, and therefore, the maximum number of numbers $n$ for which the condition of the problem is satisfied is 4. To construct an example, let $n_{1}=2, a=1$. We get that $n_{2}=-1-n_{1}=-3$ and $b=-a n_{1}\left(n_{1}+1\right)=-1 \cdot 2 \cdot 3=-6$. Let $c=1$. Then we get that $p(x)=x^{2}-6 x+1$. Let's check: $p(0)=p\left(0^{2}\right)=1$, $p(1)=p\left(1^{2}\right)=-4, p(2)=p\left(2^{2}\right)=-7, p(-3)=p\left((-3)^{2}\right)=28$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There turned out to be $B$ such numbers. What is $100 B+A$?
|
Answer: 413.
Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 to 9, giving us 9 numbers. And if $y=5$, then $x$ can take values only from 6 to 9, giving us another 4 numbers. In total, we get 13 numbers.
Let's calculate $B$. Here, there can only be the case where $y=5$, because $x$ cannot be less than 0. So, $x$ can take values from 1 to 4. Therefore, $B=4$. From this, we get that $100 B+A=400+13=413$.
|
413
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself).
|
Answer: 9.
Solution: The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The position of zero can be chosen in three ways, as it should not be the first. Then, in three ways, we choose the position of 1, and on the two remaining places, we put the twos. In the end, we get $3 \cdot 3=9$ possible numbers.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (20 points) The pensioners of one of the planets in the Alpha Centauri system enjoy spending their free time solving cybrow solitaire puzzles: they choose natural numbers from a certain interval $[A, B]$ in such a way that the sum of any two of the chosen numbers is not divisible by a given number $N$. Last week, the newspaper "Alphacentauri Panorama" offered its readers a solitaire puzzle with the values $A=1353, B=2134, N=11$. What is the maximum number of numbers that can be a solution to such a solitaire puzzle?
|
Answer: 356.
Solution: For $k=0,1, \ldots, 10$, let $I_{k}$ be the set of all numbers in $[A, B)$ that give a remainder of $k$ when divided by 11. Since $A$ and $B$ are multiples of 11, all sets $I_{k}$ contain an equal number of elements. Therefore, all numbers in $[A, B)$ that are not multiples of 11 can be paired as $(x, y)$, where $x \in I_{k}, y \in I_{11-k}$ for some $k$ from $\{1, \ldots, 5\}$. The number of such pairs is $\frac{2134-1353}{11} \cdot 5=355$. Clearly, from each pair, at most one number can be included in the final set. Additionally, this set can contain at most one number that is a multiple of 11. Thus, the solution includes no more than 356 numbers.
Now, we will show that a set of 356 numbers is possible. Include $A$ and all numbers that give an odd remainder when divided by 11 in this set. Take two distinct numbers $x$ and $y$ from this set, different from each other and from $A$. Then $(x \bmod 11) + (y \bmod 11)$ is an even number between 2 and 18. Therefore, it is not divisible by 11, and thus $x+y$ is also not divisible by 11. Clearly, $A+x$ is also not
divisible by 11. This means that such a set satisfies the problem's conditions. It remains to note that this set contains $\frac{2134-1353}{11} \cdot 5 + 1 = 356$ numbers.
|
356
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (20 points) Before the geometry lesson, the teacher wrote on the board the values of all the angles (in degrees) of a certain convex polygon. However, the duty students erased one of the written numbers. When the lesson began, it turned out that the sum of the remaining numbers was 1703. What number did the duty students erase?
|
Answer: 97.
Solution: Let the polygon have $n$ vertices. Since the $n$-gon is convex, each of its angles is less than $180^{\circ}$, and the sum of all angles is $(n-2) \cdot 180^{\circ}$. Therefore, the sum of all angles of the polygon minus one lies in the interval from $180(n-3)$ to $180(n-2)$. Then
$$
180(n-3)<1703<180(n-2), \quad \text { hence } \quad n=\left[\frac{1703}{180}\right]+3=12 \text {. }
$$
Therefore, the missing angle is $180 \cdot 10-1703=97$ degrees.
|
97
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. All three-digit numbers from 100 to 999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \) could this have happened?
|
Answer: $k=5$.
Solution. An example of a five-digit number that satisfies the condition of the problem is 22923. Indeed, Kostya could underline the fragment «22923» at the junction of the numbers 229 and 230, and Andrey - at the junction of the numbers 922 and 923.
Now we will show that it is not possible to choose a six-digit fragment. First, let's note two obvious facts.
1) If the middle digits of two consecutive three-digit numbers are different, then the smaller number ends in 9, and the larger one - in 0.
2) If the leading digits of two consecutive three-digit numbers are different, then the smaller number ends in 99, and the larger one - in 00.
Suppose the boys underlined a six-digit fragment «abcdef». It can be highlighted in the string in one of three ways, as shown in the diagram:
(here $i, j, k$ are three-digit numbers, and asterisks denote some digits). Note that Kostya and Andrey used different methods, otherwise they would have underlined the same block. Suppose Kostya chose his block using the third method. Then $b$ and $e$ are the last digits of the consecutive numbers $k$ and $k+1$, so they are different. If Andrey used the second configuration, then the numbers $j+1$ and $j+2$ have different leading digits. By 2), the number $j+2$ ends in 00, and in particular, $f=0$. If Andrey used the first configuration, then the numbers $i$ and $i+1$ have different middle digits, and $f=0$ by 1). But $f \neq 0$, since the number $k+2$ is a three-digit number. The case where the boys used the first two configurations is handled similarly.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. All four-digit numbers from 1000 to 9999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \) could this have happened?
|
Answer: $k=7$.
Solution. An example of a seven-digit number that satisfies the condition of the problem is 2229223. Indeed, Kostya could underline the fragment «2229223» at the junction of the numbers 2229 and 2230, and Andrey - at the junction of the numbers 9222 and 9223.
Now we will show that it is impossible to choose an eight-digit fragment. First, note the obvious fact: if the second-to-last digits of two consecutive four-digit numbers are different, then the smaller number ends in 9, and the larger one ends in 0. Suppose the boys underlined an eight-digit fragment «abcdefgh». It can be highlighted in the string in one of four ways, as shown in the diagram:
$$
\underbrace{a b c d}_{i} \underbrace{e f g h}_{i+1} \underbrace{* a b c}_{j} \underbrace{d e f g}_{j+1} \underbrace{h * * *}_{j+2} \underbrace{* * a b}_{k} \underbrace{c d e f}_{k+1} \underbrace{g h * *}_{k+2} \underbrace{* * * a}_{m} \underbrace{b c d e}_{m+1} \underbrace{f g h *}_{m+2}
$$
(here $i, j, k, m$ are four-digit numbers, and asterisks denote some digits). Note that Kostya and Andrey used different methods, otherwise they would have underlined the same block. Suppose Kostya chose his block in the first way, and Andrey in the second. Then $c$ and $g$ are the last digits of adjacent numbers $j$ and $j+1$, so they are different. Therefore, the numbers $i$ and $i+1$ have different second-to-last digits. According to the above remark, the number $i+1$ ends in 0, that is, $h=0$. But this is impossible, since the number $j+2$ is a four-digit number. Other configurations are analyzed similarly.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest path to the beacon. For example, if a beacon is located in room $A$ (in chess notation, this is room a4), and the robot is in room $B$ (this is room g4), the distance between them is 7 links. The robot has the maze diagram and the beacons are marked on it. For what minimum number of beacons can the robot, when in any room, uniquely determine its location by sound?
a) Indicate how such a number of beacons can be placed. Why will the robot be able to determine its room by them?
b) Prove that it is impossible to manage with a smaller number of beacons.
Fig. 1: Maze
|
Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
Fig. 3: Distances to beacon a5
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 2): part $\mathcal{K}$ is room $K$ and two long dead-end corridors that lead from it to the left and right, part $\mathcal{L}$ is room $L$ and two dead-end corridors that lead from it upwards and to the right, part $\mathcal{N}$ is room $N$ and two dead-end corridors that lead from it to the left and downwards. If there are no more than 2 beacons, then one of these parts will not have a beacon. For example, let this be part $\mathcal{K}$, signals from beacons will reach this part via the edge entering room $K$ from below. Then a robot in room a3 will hear the same signals as in room b3, and therefore will not be able to distinguish which of these two rooms it is in. Similarly, indistinguishable rooms can be found in other parts if they do not contain a beacon.
a) Example. Let's check that beacons in rooms a1, d3, a5 will allow the robot to orient itself in this maze. Here, it is easier to perform a case-by-case check than to come up with general considerations. In Fig. 3, we marked the distances from all rooms to beacon a5. As we can see, rooms with distances from 0 to 5 are uniquely defined. Rooms with a distance of 6 are at different distances from beacon d3. Similarly, rooms with distances of $8, 9, 10, 11$. As for the four rooms with a distance of 7, two of them, v2 and v4, are at distances of 3 and 7 from beacon d3 and are thus uniquely defined by the distances to a5 and to d3. The other two rooms, a2 and b1, are at a distance of 5 from d3, but they are at different distances from a1.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the maze diagram in Fig. 7, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest path to the beacon. For example, if a beacon is in room $A$ (in chess notation, this is room a4), and the robot is in room $B$ (this is room g4), the distance between them is 5 links. The robot has the maze diagram and the beacons are marked on it. For what minimum number of beacons can the robot, when in any room, uniquely determine its location by sound?
a) Indicate how such a number of beacons can be placed. Why will the robot be able to determine its room by them?
b) Prove that it is impossible to manage with fewer beacons.
Fig. 7: Maze
|
Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 8): part $\mathcal{K}$ is room $K$, the long dead-end corridor that exits from it to the left, and the dead-end corridor below, part $\mathcal{L}$ is room $L$ and the two dead-end corridors that exit from it upwards and to the right, part $\mathcal{N}$ is room $N$ and the two dead-end corridors that exit from it to the left and below. If there are no more than 2 beacons, then one of these parts will not have a beacon. For example, let this be part $\mathcal{K}$, signals from the beacons will reach this part via the link entering room $K$ from the right. Then a robot, being in room a5, will hear the same signals as in room b4, and therefore will not be able to distinguish,
Fig. 8: Parts of the maze
Fig. 9: Distances to beacon b3
which of these two rooms it is in. Similarly, indistinguishable rooms can be found in other parts if they do not contain a beacon.
a) Example. Let's check that beacons in rooms a1, b3, d4 will allow the robot to orient itself in this maze. Here, it is easier to perform a case-by-case analysis rather than come up with general considerations. In Fig. 9, we marked the distances from all rooms to beacon b3. As we can see, rooms with distances $0,1,2,3,4,8$ and 12 are uniquely defined. Rooms with a distance of 5 are at different distances from beacon a1. Similarly, rooms with a distance of 6 and 7. There are three rooms with a distance of 9: one of them - b2 is at a distance of 2 from beacon a1, and the other two are at a distance of 4 from a1, but they have different distances to d4. Two rooms with a distance of 10 - d2 and g3 - are at distances of 4 and 2 from beacon d4, and the other three - b1, a2, and g1 - are at a distance of 6 from d4, but at different distances from a1. Finally, two rooms with a distance of 11 - d3 and g4 - are at distances of 5 and 1 from beacon d4, and the other two - a1 and d1 - are at a distance of 7 from d4, but at different distances from a1.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the maze diagram in Fig. 13, each segment (link) is a corridor, and each circle is a small room. Some rooms have beacons that hum, each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest path to the beacon. For example, if a beacon is in room $A$ (in chess notation, this is room a4), and the robot is in room $B$ (this is room g4), the distance between them is 5 links. The robot has the maze diagram and the beacons are marked on it. For what minimum number of beacons can the robot, when in any room, uniquely determine its location by sound?
a) Indicate how such a number of beacons can be placed. Why will the robot be able to determine its room by them?
b) Prove that it is impossible to manage with fewer beacons.
Fig. 13: Maze
|
Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
## Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 14): part $\mathcal{K}$ is room $K$ and the dead-end corridors that lead from it to the right and down, part $\mathcal{L}$ is room $L$ and the two dead-end corridors that lead from it up and to the right, part $\mathcal{N}$ is room $N$ and the two dead-end corridors that lead from it to the left and down. If there are no more than 2 beacons, then one of these parts will not have a beacon. For example, let this be part $\mathcal{K}$, signals from beacons will reach this part via the link entering room $K$ from the right. Then the robot, being in room a3, will hear the same signals as in room b4, and therefore will not be able to distinguish which of these two rooms it is in. Similarly, indistinguishable rooms can be found in other parts if they do not contain a beacon.
a) Example. Let's check that beacons in rooms a1, b3, d4 will allow the robot to orient itself in this maze. Here, it is easier to perform a case-by-case analysis than to come up with some general considerations. In Figure 15, we marked the distances from all rooms to beacon b3. As we can see, rooms with distances $0,1,2,5$ and 8 are uniquely defined. Rooms with a distance of 3 are at different distances from beacon a1. Similarly, rooms with distances of 4, 6, and 7. There are three rooms with a distance of 9: one of them -62- is at a distance of 2 from beacon a1, while the other two are at a distance of 4 from a1, but they have different
Fig. 14: Parts of the maze
Fig. 15: Distances to beacon b3
distances to d4. Two rooms with a distance of 10 - g1 and g3 - are at a distance of 5 from beacon a1, but they differ in their distance to d4, while another one - b1 - is at a distance of 1 from a1 and thus differs from the previous two. Two rooms with a distance of 11 - d3 and g4 - are at distances of 1 and 3 from beacon d4, while the other two - a1 and d1 - are at a distance of 7 from d4, but at different distances from a1. Finally, rooms with a distance of 12 differ in their distance to d4.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In each cell of a $100 \times 100$ table, a natural number is written. In each row, there are at least 10 different numbers, and in any four consecutive rows, there are no more than 15 different numbers. What is the maximum number of different numbers that can be in the table?
|
Answer: 175
Solution. In one line, there are no less than 10 different numbers, so in the next three lines together, there appear no more than 5 new numbers. Therefore, the first four lines contain no more than 15 different numbers, and each of the following three lines adds no more than 5 new numbers, making the total number of numbers no more than $15+32 \cdot 5=175$.
Let's provide an example with 175 numbers. We will number the lines from 1 to 100. In the first line, we will place numbers from 1 to 10, and in the lines with numbers from $3k-1$ to $3k+1$, we will place numbers 1 to 5 and numbers from $5k+6$ to $5k+10$. Then, in each line, there will be 5 unique numbers and numbers from 1 to 5, i.e., exactly 10 different numbers, and in every four lines, there will be exactly 15 different numbers. Thus, in the table, there will be numbers from 1 to $5 \cdot 33 + 10 = 175$.
Remark. It can also be proven by induction that the number of different numbers in the table does not exceed 175. Specifically, it can be proven that in any $3n+1$ consecutive lines, there are no more than $5(n+2)$ different numbers. The base case $n=1$ is true by the condition. We will establish the transition from $n$ to $n+1$. Consider $3n+4$ consecutive lines. Suppose the fourth line from the bottom contains $k \geqslant 10$ different numbers. Then, in the three lowest lines, there are no more than $15-k$ different numbers. And in the remaining $3n+1$ lines, by the inductive hypothesis, there are no more than $5(n+2)$ numbers. Therefore, the total number of different numbers will be more than $5(n+2)+15-k=5(n+5)-k \leqslant 5(n+3)$.
|
175
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find all primes $p$ for which the numbers $p+1$ and $p^{2}+1$ are double the squares of natural numbers.
#
|
# Answer: $p=7$
First solution. Let $p+1=2 x^{2}$ and $p^{2}+1=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Therefore, either $y-x$ or $y+x$ is divisible by $p$. From the inequality $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note that $\frac{y^{2}}{x^{2}}=\frac{p^{2}+1}{p+1}>\frac{p^{2}-1}{p+1}=p-1$. Then if $p \geqslant 11$, $y^{2}>12 x^{2}$, hence $y>3 x$ and, thus, $2(y-x)>y+x \geqslant p$. Therefore, $2(y-x)(y+x)>p^{2}>p(p+1)$, which is also impossible. Thus, it remains to consider the cases $p=3, p=5$, and $p=7$. In the first two cases, $p+1$ is not a doubled square, while $p=7$ fits.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In each cell of a $75 \times 75$ table, a natural number is written. In each row, there are at least 15 different numbers, and in any three consecutive rows, there are no more than 25 different numbers. What is the maximum number of different numbers that can be in the table?
|
# Answer: 385
Solution. In one line, there are no less than 15 different numbers, so in the next two lines together, there appear no more than 10 new numbers. Therefore, the first three lines contain no more than 25 different numbers, and each of the following two lines adds no more than 10 new numbers, making the total number of numbers no more than $25 + 36 \cdot 10 = 385$.
Let's provide an example with 385 numbers. We will number the lines from 1 to 75. In the first line, we will place numbers from 1 to 15, and in the line with number $k$, we will place numbers from 1 to 10 and numbers from $5k + 6$ to $5k + 10$. Then, in each line, there will be 5 unique numbers and numbers from 1 to 10, i.e., exactly 15 different numbers, and in every three lines, there will be exactly 25 different numbers.
Remark. We can also prove that the number of different numbers in the table does not exceed 385 by induction. Specifically, we can prove that in any $2n + 1$ consecutive lines, there are no more than $5(2n + 3)$ different numbers. The base case $n=1$ is true by the problem statement. We will establish the transition from $n$ to $n+1$. Consider $2n + 3$ consecutive lines. Suppose the third line from the bottom contains $k \geq 15$ different numbers. Then, in the two lowest lines, there are no more than $25 - k$ different numbers. And in the remaining $2n + 1$ lines, by the inductive hypothesis, there are no more than $5(2n + 3)$ numbers. Therefore, the total number of different numbers will be more than $5(2n + 3) + 25 - k = 5(2n + 8) - k \leq 5(2n + 5)$.
|
385
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find all primes $p$ for which the numbers $p+7$ and $p^{2}+7$ are double the squares of natural numbers.
|
Answer: $p=11$
First solution. Let $p+7=2 x^{2}$ and $p^{2}+7=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Since $p-$ is odd, $p \geqslant 3$ and $2 p^{2}>p^{2}+7$. Therefore, $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note that $\frac{y^{2}}{x^{2}}=\frac{p^{2}+7}{p+7}>\frac{p^{2}-49}{p+7}=p-7$. Then if $p \geqslant 17$, $y^{2}>10 x^{2}$, from which $y>3 x$ and, thus, $2(y-x)>y+x \geqslant p$. Therefore, $2(y-x)(y+x)>p^{2}>p(p+1)$, which is also impossible. Thus, it remains to consider the cases $p=3, p=5, p=7, p=11$ and $p=13$. In the last, as well as in the first three of them, $p+7$ is not a doubled square, while $p=11$ fits.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are dropped. A point $T$ is chosen on the plane such that the lines $T A$ and $T B$ are tangents to the circumcircle of triangle $A B C$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the line $A_{1} B_{1}$ intersects the line $C C_{1}$ at point $K$, and the line passing through point $C_{1}$ parallel to $O K$ intersects the segment $C O$ at point $L$. Find the angle $\angle C L A_{1}$.
|
Answer: $90^{\circ}$
Solution. Let $\angle B A C=\alpha$ and $\angle A C B=\gamma$. Triangles $A O T$ and $B O T$ are congruent by three sides, and the inscribed angle $\angle A C B$ subtends the same arc as the central angle $\angle A O B$, so $\gamma=\angle A C B=\frac{1}{2} \angle A O B=\angle T O B$. Therefore, $\frac{O B}{O T}=\cos \gamma$. Similarly, $\angle B O C=2 \angle B A C$ and from the isosceles property of triangle $B O C$, it follows that $\angle B C O=90^{\circ}-\frac{1}{2} \angle B O C=90^{\circ}-\angle B A C$. Since $\angle A B_{1} B=90^{\circ}=\angle A A_{1} B$, quadrilateral $A B_{1} A_{1} B$ is cyclic. Therefore, $\angle B A C=\angle C A_{1} B_{1}$. Thus, $\angle B C O=90^{\circ}-\angle B A C=90^{\circ}-\angle C A_{1} B_{1}$ and lines $A_{1} B_{1}$ and $C O$ are perpendicular. Therefore, lines $C O$ and $T K$ are parallel, and quadrilateral $C O T K$ is a parallelogram, in particular, $C K=O T$. Then, by Thales' theorem, $\frac{C L}{C C_{1}}=\frac{C O}{C K}=\frac{O B}{O T}=\cos \gamma$. Therefore,
$$
C L=C C_{1} \cos \gamma=A C \sin \alpha \cos \gamma=C A_{1} \sin \alpha=C A_{1} \sin \angle C A_{1} B_{1} .
$$
Thus, point $L$ is the foot of the perpendicular dropped from $C$ to line $A_{1} B_{1}$, meaning $\angle C L A_{1}=90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. What is the minimum number of cells that need to be marked in a $15 \times 15$ table so that each vertical or horizontal strip $1 \times 10$ contains at least one marked cell.
|
Answer: 20
Solution. Let's cut the $15 \times 15$ table without the central $5 \times 5$ square into 20 rectangles of $1 \times 10$ (see the left figure). Therefore, we will need to mark at least 20 cells. An example with 20 cells: all cells of two parallel diagonals of length 10 are marked (see the right figure).
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. What is the minimum number of cells that need to be marked in a $20 \times 20$ table so that each vertical or horizontal strip of $1 \times 12$ contains at least one marked cell.
|
Answer: 32
Solution. Let's cut the $20 \times 20$ table without the central $4 \times 4$ square into 32 rectangles of $1 \times 12$ (see the left figure). Therefore, we will need to mark at least 32 cells. An example with 32 cells: all cells of three parallel diagonals of lengths 4, 16, and 12 are marked (see the right figure).
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On the leg $AC$ of the right triangle $ABC$ with hypotenuse $AB$, a point $P$ is marked. Point $D$ is the foot of the perpendicular dropped from vertex $A$ to the line $BP$, and point $E$ is the foot of the perpendicular dropped from point $P$ to the side $AB$. On the plane, a point $T$ is chosen such that the lines $TA$ and $TP$ are tangents to the circumcircle of triangle $PAB$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the line $DE$ intersects the line $BC$ at point $Q$, and the line passing through point $C$ parallel to $OQ$ intersects the segment $BO$ at point $K$. Find the angle $\angle OKE$.
|
Answer: $90^{\circ}$
Solution. Let $\angle A B P=\varphi$ and $\angle C B P=\psi$. Triangles $A O T$ and $P O T$ are equal by three sides, and the inscribed angle $\angle A B P$ subtends the same arc as the central angle $\angle A O P$, so $\gamma=\angle A B P=\frac{1}{2} \angle A O P=\angle A O T$. Therefore, $\frac{O A}{O T}=\cos \varphi$. The central angle $\angle A O B$ subtends the smaller arc $A B$, while the inscribed angle $\angle A P B$ subtends the larger arc $A B$, so $\angle A P B+\frac{1}{2} \angle A O B=180^{\circ}$. From the isosceles triangle $A O B$, we have
$$
\angle A B O=90^{\circ}-\frac{1}{2} \angle A O B=90^{\circ}-\left(180^{\circ}-\angle A P B\right)=\angle A P B-90^{\circ}=\angle P B C=\psi
$$
Since $\angle A D P=90^{\circ}=\angle A E P$, quadrilateral $A D P E$ is cyclic, in particular, $\angle E A P=\angle E D P=\angle B D E$. Therefore,
$$
\angle B D E+\angle D B O=\angle B D E+\angle D B A+\angle A B O=\angle E A P+\angle A O T+\angle B A O=90^{\circ}
$$
Thus, lines $D E$ and $B O$ are perpendicular, and lines $B O$ and $Q T$ are parallel. Points $O$ and $T$ lie on the perpendicular bisector of segment $A P$, so lines $A P$ and $O T$ are perpendicular. Therefore, lines $B C$ and $O T$ are parallel, and quadrilateral $B O T Q$ is a parallelogram, in particular, $B Q=O T$. By Thales' theorem for parallel lines $C K$ and $O Q$, we have $\frac{B K}{B C}=\frac{B O}{B Q}=\frac{O A}{O T}=\cos \varphi$. Therefore,
$$
B K=B C \cos \varphi=B P \cos \psi \cos \varphi=B E \cos \psi=B E \cos \angle E B K
$$
Thus, point $K$ is the foot of the perpendicular dropped from $E$ to line $B O$, meaning $\angle B K E=90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The sum of positive numbers $a, b, c$ and $d$ is not less than 8. Find the minimum value of the expression
$\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}$.
|
Answer: 1
First solution. By the inequality of means for four numbers, we have
$$
\begin{aligned}
\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{a+b}{16}+\frac{a+c}{16} & +\frac{a+d}{16} \geqslant \\
& \geqslant 4 \sqrt[4]{\frac{a^{4}}{(a+b)(a+c)(a+d)} \cdot \frac{a+b}{16} \cdot \frac{a+c}{16} \cdot \frac{a+d}{16}}=\frac{a}{2}
\end{aligned}
$$
Therefore,
$$
\frac{a^{4}}{(a+b)(a+c)(a+d)} \geqslant \frac{a}{2}-\left(\frac{a+b}{16}+\frac{a+c}{16}+\frac{a+d}{16}\right)=\frac{5 a}{16}-\frac{b+c+d}{16}
$$
Summing this inequality with three similar ones, we get
$$
\begin{aligned}
& \frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)} \geqslant \\
& \geqslant\left(\frac{5 a}{16}-\frac{b+c+d}{16}\right)+\left(\frac{5 b}{16}-\frac{c+d+a}{16}\right)+\left(\frac{5 c}{16}-\frac{d+a+b}{16}\right)+\left(\frac{5 d}{16}-\frac{a+b+c}{16}\right)= \\
& =\frac{2(a+b+c+d)}{16} \geqslant 1 .
\end{aligned}
$$
If $a=b=c=d=2$, then the sum of the fractions in the problem condition is 1, so the minimum value of the expression is 1.
Second solution. For brevity, let
$$
K=\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}
$$
By the Cauchy-Bunyakovsky inequality for the sets of numbers
$$
\frac{a^{4}}{(a+b)(a+c)(a+d)}, \frac{b^{4}}{(b+c)(b+d)(b+a)}, \frac{c^{4}}{(c+d)(c+a)(c+b)}, \frac{d^{4}}{(d+a)(d+b)(d+c)} \text { and }
$$
we have
$$
\begin{aligned}
& (2(a+b+c+d)) K=((a+d)+(b+a)+(c+b)+(d+c)) K \geqslant \\
& \quad \geqslant\left(\frac{a^{2}}{\sqrt{(a+b)(a+c)}}+\frac{b^{2}}{\sqrt{(b+c)(b+d)}}+\frac{c^{2}}{\sqrt{(c+d)(c+a)}}+\frac{d^{2}}{\sqrt{(d+a)(d+b)}}\right)^{2}
\end{aligned}
$$
Let the last expression in parentheses be denoted by $L$ and estimate it by the Cauchy-Bunyakovsky inequality for the sets
$$
\begin{aligned}
& \frac{a^{2}}{\sqrt{(a+b)(a+c)}}, \quad \frac{b^{2}}{\sqrt{(b+c)(b+d)}}, \quad \frac{c^{2}}{\sqrt{(c+d)(c+a)}}, \quad \frac{d^{2}}{\sqrt{(d+a)(d+b)}} \text { and } \\
& \sqrt{(a+b)(a+c)}, \quad \sqrt{(b+c)(b+d)}, \quad \sqrt{(c+d)(c+a)}, \quad \sqrt{(d+a)(d+b)} .
\end{aligned}
$$
Then $L M \geqslant(a+b+c+d)^{2}$, where
$$
M=\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+d)}+\sqrt{(c+d)(c+a)}+\sqrt{(d+a)(d+b)}
$$
Thus,
$$
K \geqslant \frac{L^{2}}{2(a+b+c+d)} \geqslant \frac{(a+b+c+d)^{4}}{2(a+b+c+d) M^{2}}=\frac{(a+b+c+d)^{3}}{2 M^{2}}
$$
Finally, by the inequality of means for two numbers $\sqrt{(x+y)(x+z)} \leqslant \frac{(x+y)+(x+z)}{2}$, therefore
$$
M \leqslant \frac{2 a+b+c}{2}+\frac{2 b+c+d}{2}+\frac{2 c+d+a}{2}+\frac{2 d+a+b}{2}=2(a+b+c+d)
$$
Thus,
$$
K \geqslant \frac{1}{2} \cdot \frac{(a+b+c+d)^{3}}{(2(a+b+c+d))^{2}}=\frac{a+b+c+d}{8} \geqslant 1
$$
If $a=b=c=d=2$, then $K=1$, so the minimum value of $K$ is 1.
Third solution. For brevity, let
$$
K=\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}
$$
By the Cauchy-Bunyakovsky inequality for the sets of numbers
$$
\begin{aligned}
& \frac{a^{4}}{(a+b)(a+c)(a+d)}, \frac{b^{4}}{(b+c)(b+d)(b+a)}, \frac{c^{4}}{(c+d)(c+a)(c+b)}, \frac{d^{4}}{(d+a)(d+b)(d+c)} \text { and } \\
& (a+c)(a+d), \quad(b+d)(b+a), \quad(c+a)(c+b), \quad(d+b)(d+c)
\end{aligned}
$$
we have
$$
\begin{aligned}
(a+b+c+d)^{2} K & =((a+c)(a+d)+(b+d)(b+a)+(c+a)(c+b)+(d+b)(d+c)) K \geqslant \\
& \geqslant\left(\frac{a^{2}}{\sqrt{a+b}}+\frac{b^{2}}{\sqrt{b+c}}+\frac{c^{2}}{\sqrt{c+d}}+\frac{d^{2}}{\sqrt{d+a}}\right)^{2}
\end{aligned}
$$
Let the last expression in parentheses be denoted by $L$ and estimate it by the Cauchy-Bunyakovsky inequality for the sets
$$
\frac{a^{2}}{\sqrt{a+b}}, \quad \frac{b^{2}}{\sqrt{b+c}}, \quad \frac{c^{2}}{\sqrt{c+d}}, \quad \frac{d^{2}}{\sqrt{d+a}} \text { and } \sqrt{a+b}, \sqrt{b+c}, \sqrt{c+d}, \sqrt{d+a}
$$
Then $L M \geqslant(a+b+c+d)^{2}$, where $M=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}$. Thus,
$$
K \geqslant \frac{L^{2}}{(a+b+c+d)^{2}} \geqslant \frac{1}{(a+b+c+d)^{2}} \cdot \frac{(a+b+c+d)^{4}}{M^{2}}=\left(\frac{a+b+c+d}{M}\right)^{2}
$$
By the inequality of means for two numbers $2 \sqrt{x+y} \leqslant \frac{x+y+4}{2}$, therefore
$$
M \leqslant \frac{a+b+4}{4}+\frac{b+c+4}{4}+\frac{c+d+4}{4}+\frac{d+a+4}{4}=\frac{2 s+16}{4}
$$
where $s=a+b+c+d \geqslant 8$. Thus, we have proved that
$$
K \geqslant\left(\frac{s}{M}\right)^{2} \geqslant\left(\frac{4 s}{2 s+16}\right)^{2}
$$
It remains to note that $4 s=2 s+2 s \geqslant 2 s+16$, so the last expression is not less than 1. If $a=b=c=d=2$, then $K=1$, so the minimum value of $K$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and triangle $A P D$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. A perpendicular is drawn from point $E$ to line $A C$, and a perpendicular is drawn from point $F$ to line $B D$, these perpendiculars intersect at point $Q$. Find the angle between the lines $P Q$ and $B C$.
|
Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of intersection of the lines $P Q$ and $B C$. Since the angles $\angle B A C$ and $\angle B D C$ subtend the same arc, they are equal. Therefore, the right triangles $A E^{\prime} E$ and $B F^{\prime} F$ are similar. Then, $\frac{E E^{\prime}}{F F^{\prime}}=\frac{A E}{D F}=\frac{B E}{C F}$, the last equality because points $E$ and $F$ are the midpoints of segments $A B$ and $C D$. Moreover, $\angle B E E^{\prime}=\angle C F F^{\prime}$, so triangles $B E E^{\prime}$ and $C F F^{\prime}$ are similar, and in particular, $\angle A B E^{\prime}=\angle D C F^{\prime}$. The angles $\angle A B D$ and $\angle A C D$ subtend the same arc and are therefore equal. Thus, $\angle E^{\prime} B F^{\prime}=\angle A B D-\angle A B E^{\prime}=\angle A C D-\angle D C F^{\prime}=\angle E^{\prime} C F^{\prime}$. Therefore, the quadrilateral $B E^{\prime} F^{\prime} C$ is cyclic and
$$
\angle P B T=\angle C B F^{\prime}=\angle C E^{\prime} F^{\prime}=\angle P E F^{\prime}=\angle P Q F^{\prime}
$$
(the last equality of angles follows from the cyclic nature of the quadrilateral $P E^{\prime} Q F^{\prime}$). It remains to calculate the angles:
$$
\angle P T B=180^{\circ}-\angle P B T-\angle B P T=180^{\circ}-\angle P Q F^{\prime}-\angle Q P F^{\prime}=90^{\circ}
$$
Second solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $K$ and $L$ be the midpoints of segments $A P$ and $D P$, and $T$ be the point of intersection of the lines $P Q$ and $B C$. Let $O$ be the center of the circumcircle of triangle $A P D$, then $O K$ and $O L$ are the perpendicular bisectors of segments $A P$ and $D P$, respectively. Let $\angle B A P=\varphi$ for brevity. Then $\angle C D P=\varphi$, since the angles $\angle B A P$ and $\angle C D P$ subtend the same arc. Therefore, triangles $B A P$ and $C D P$ are similar by two angles and, hence,
$$
\frac{A P}{D P}=\frac{A B}{C D}=\frac{A E}{D F}=\frac{A E \cos \varphi}{D F \cos \varphi}=\frac{A E^{\prime}}{D F^{\prime}}
$$
Thus,
$$
\frac{P E^{\prime}}{P F^{\prime}}=\frac{A P-A E^{\prime}}{D P-D F^{\prime}}=\frac{A P}{D P}=\frac{P K}{P L}
$$
Let the lines $K O$ and $P Q$ intersect at point $K^{\prime}$, and the lines $L O$ and $P Q$ intersect at point $L^{\prime}$. Then triangles $P E^{\prime} Q$ and $P K K^{\prime}$ are similar with the ratio $\frac{P E^{\prime}}{P K}=\frac{P F^{\prime}}{P L}$, and with the same ratio of similarity, triangles $P F^{\prime} Q$ and $P L L^{\prime}$ are similar. Therefore, $P K^{\prime}=P L^{\prime}$ and, hence, points $K^{\prime}, L^{\prime}$, and $O$ coincide. Thus, the line $P Q$ passes through the point $O$. It remains to calculate the angles:
$\angle P T B=180^{\circ}-\angle C B D-\angle B P T=180^{\circ}-\angle C A D-\angle O P D=180^{\circ}-\angle P O L-\angle O P D=90^{\circ}$.
In the penultimate equality, it was used that the central angle $\angle P O D$ is equal to twice the inscribed angle $\angle C A D$ on one side and to twice the angle $\angle P O L$ on the other side.
Third solution. Draw a line $\ell$ from point $P$ perpendicular to side $B C$, let $T$ be its point of intersection with $B C$. Draw lines through points $A$ and $D$ perpendicular to diagonals $A C$ and $B D$, respectively. Let $S$ be their point of intersection. Since $\angle P A S=90^{\circ}=\angle P D S$, points $A, P, D$, and $S$ lie on a circle with diameter $P S$. Then (the equality $\angle D A P=\angle D B C$ follows from the cyclic nature of quadrilateral $A B C D$)
$$
\angle D P S=\angle D A S=90^{\circ}-\angle D A P=90^{\circ}-\angle D B C=\angle B P T
$$
Therefore, point $S$ lies on line $\ell$.
Let $H$ be the orthocenter of triangle $P B C$. Clearly, it also lies on line $\ell$. Then the lines $A S, B H$, and $E Q$ are parallel, and since $A E=E B$, line $E Q$ is the midline of trapezoid $A B H S$. Therefore, $E Q$ intersects line $\ell$ at the midpoint of segment $S H$. Similarly, line $F Q$ also intersects line $\ell$ at the midpoint of segment $S H$. But then lines $E Q, F Q$, and $\ell$ intersect at the midpoint of segment $S H$ and this point is point $Q$. Thus, $Q$ also lies on line $\ell$. In particular, lines $P Q$ and $B C$ intersect at a right angle.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and the angle $A P B$ is obtuse. Points $E$ and $F$ are the midpoints of sides $A D$ and $B C$ respectively. A perpendicular is drawn from point $E$ to the line $A C$, and a perpendicular is drawn from point $F$ to the line $B D$, these perpendiculars intersect at point $Q$. Find the angle between the lines $P Q$ and $C D$.
|
Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $B P$, respectively, and let $T$ be the point of intersection of the lines $P Q$ and $C D$. Since the angles $\angle C A D$ and $\angle C B D$ subtend the same arc, they are equal. Therefore, the right triangles $A E^{\prime} E$ and $D F^{\prime} F$ are similar. Then, $\frac{E E^{\prime}}{F F^{\prime}}=\frac{A E}{B F}=\frac{D E}{C F}$, the last equality because points $E$ and $F$ are the midpoints of segments $A D$ and $B C$. Moreover, $\angle D E E^{\prime}=\angle C F F^{\prime}$, so triangles $D E E^{\prime}$ and $C F F^{\prime}$ are similar, and in particular, $\angle A D E^{\prime}=\angle B C F^{\prime}$. The angles $\angle A C B$ and $\angle A D B$ subtend the same arc and are therefore equal. Thus, $\angle E^{\prime} B F^{\prime}=\angle A B D-\angle A B E^{\prime}=\angle A C D-\angle D C F^{\prime}=\angle E^{\prime} C F^{\prime}$. Therefore, the quadrilateral $B E^{\prime} F^{\prime} C$ is cyclic and
$$
\angle T D P=\angle C D F^{\prime}=\angle C E^{\prime} F^{\prime}=\angle P E^{\prime} F^{\prime}=\angle P Q F^{\prime}
$$
(the last equality of angles follows from the cyclic nature of the quadrilateral $P E^{\prime} Q F^{\prime}$). It remains to calculate the angles:
$$
\angle P T D=180^{\circ}-\angle T D P-\angle D P T=180^{\circ}-\angle P Q F^{\prime}-\angle Q P F^{\prime}=90^{\circ}
$$
Second solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $B P$, respectively, and let $K$ and $L$ be the midpoints of segments $A P$ and $B P$, and $T$ be the point of intersection of the lines $P Q$ and $C D$. Let $O$ be the center of the circumcircle of triangle $A P B$, then $O K$ and $O L$ are the perpendicular bisectors of segments $A P$ and $B P$, respectively. Let $\angle D A P=\varphi$ for brevity. Then $\angle C B P=\varphi$, since the angles $\angle D A P$ and $\angle C B P$ subtend the same arc. Therefore, triangles $C B P$ and $D A P$ are similar by two angles and, hence,
$$
\frac{A P}{B P}=\frac{A D}{B C}=\frac{A E}{B F}=\frac{A E \cos \varphi}{B F \cos \varphi}=\frac{A E^{\prime}}{B F^{\prime}}
$$
Thus,
$$
\frac{P E^{\prime}}{P F^{\prime}}=\frac{A P-A E^{\prime}}{B P-B F^{\prime}}=\frac{A P}{B P}=\frac{P K}{P L}
$$
Let the lines $K O$ and $P Q$ intersect at point $K^{\prime}$, and the lines $L O$ and $P Q$ intersect at point $L^{\prime}$. Then triangles $P E^{\prime} Q$ and $P K K^{\prime}$ are similar with the ratio $\frac{P E^{\prime}}{P K}=\frac{P F^{\prime}}{P L}$, and with the same ratio of similarity, triangles $P F^{\prime} Q$ and $P L L^{\prime}$ are similar. Therefore, $P K^{\prime}=P L^{\prime}$ and, hence, points $K^{\prime}, L^{\prime}$, and $O$ coincide. Thus, the line $P Q$ passes through the point $O$. It remains to calculate the angles:
$\angle P T C=180^{\circ}-\angle A C B-\angle C P T=180^{\circ}-\angle A B D-\angle A P O=180^{\circ}-\angle P O K-\angle A P O=90^{\circ}$.
In the penultimate equality, it was used that the central angle $\angle A O P$ is equal to twice the inscribed angle $\angle A B D$ on one side and to twice the angle $\angle P O K$ on the other side.
Third solution. Draw a line $\ell$ from point $P$ perpendicular to side $C D$, let $T$ be its point of intersection with $C D$. Draw lines through points $A$ and $B$ perpendicular to diagonals $A C$ and $B D$, respectively. Let $S$ be their point of intersection. Since $\angle P A S=90^{\circ}=\angle P B S$, points $A, P, B$, and $S$ lie on a circle with diameter $P S$. Then (the equality $\angle A B P=\angle A C D$ follows from the cyclic nature of quadrilateral $A B C D$)
$$
\angle A P S=\angle A B S=90^{\circ}-\angle A B P=90^{\circ}-\angle A C D=\angle C P T .
$$
Therefore, point $S$ lies on line $\ell$.
Let $H$ be the orthocenter of triangle $C P D$. Clearly, it also lies on line $\ell$. Then the lines $A S, D H$, and $E Q$ are parallel, and since $A E=E D$, line $E Q$ is the midline of trapezoid $A D H S$. Therefore, $E Q$ intersects line $\ell$ at the midpoint of segment $S H$. Similarly, line $F Q$ also intersects line $\ell$ at the midpoint of segment $S H$. But then lines $E Q, F Q$, and $\ell$ intersect at the midpoint of segment $S H$ and this point is point $Q$. Therefore, $Q$ also lies on line $\ell$. In particular, lines $P Q$ and $C D$ intersect at a right angle.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
|
Answer: $17^{\circ}$.
Solution. Since in the right triangle $ABC$ the angle $BAC=17^{\circ}$, the angle $BCA=73^{\circ}$. Note that triangle $QMC$ is isosceles with base $MC$, as by condition $QM=QC$. Similarly, triangle $PMA$ is isosceles with base $MA$. Therefore, the angles at the bases of these triangles are equal. From this, we get that $\angle QMP=180^{\circ}-\angle PMA-\angle QMC=180^{\circ}-17^{\circ}-73^{\circ}=90^{\circ}$.
Let points $O$ and $R$ be the feet of the perpendiculars dropped from points $Q$ and $P$ to the line $AC$, respectively. It is easy to notice that triangles $QOM$ and $MRP$ are similar by two angles. Since $CM=MA$, from the isosceles nature of triangles $QCM$ and $PMA$, it follows that
$$
MO=\frac{1}{2} CM=\frac{1}{2} MA=MR
$$
Let $x=MO$. From the right triangles $QOM$ and $PMR$, we get that $MQ=\frac{x}{\cos 73^{\circ}}$ and $MP=\frac{x}{\cos 17^{\circ}}$ respectively. Thus, for triangle $MQP$ we have
$$
\operatorname{tg} \angle PQM=\frac{MP}{MQ}=\frac{x}{\cos 17^{\circ}} \cdot \frac{\cos 73^{\circ}}{x}=\frac{\cos \left(\pi / 2-17^{\circ}\right)}{\cos 17^{\circ}}=\operatorname{tg} 17^{\circ}
$$
Since in the right triangle $MQP$ the angle $PQM$ is acute, from the last equality it follows that $\angle PQM=17^{\circ}$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (20 points) At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will be admitted to the evening program.
|
Answer: 6.
Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other two will be admitted to the daytime department. The number of ways to choose 2 out of 4 applicants to be admitted to the daytime department is $C_{4}^{2}=6$.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (30 points) On the sides $B C$ and $A C$ of the isosceles triangle $A B C (A B = A C)$, points $D$ and $E$ were found respectively such that $A E = A D, \angle E D C = 18^{\circ}$. Find the measure of the angle $\angle B A D$.
|
Answer: $36^{\circ}$.
Solution. Denote the angles as indicated in the figure. The angle $A D C$ is an exterior angle for triangle $A D B$; hence, $\beta+18^{\circ}=\alpha+x$. Angles $A B C$ and $A C B$ are the base angles of isosceles triangle $A B C$; hence, $\alpha=\frac{180^{\circ}-(x+y)}{2}$. Angles $A D C$ and $A E D$ are the base angles of isosceles triangle $A D E$; hence, $\beta=\frac{180^{\circ}-y}{2}$. Substitute these expressions into the relation for angle $A D C$:
$$
\frac{180^{\circ}-y}{2}+18^{\circ}=\frac{180^{\circ}-(x+y)}{2}+x
$$
Simplifying the expression, we get $18^{\circ}=\frac{x}{2}$, which means $x=36^{\circ}$.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take?
|
Answer: 6.
Solution. Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b\left(c^{2}+3\right)}{b+c}=b(c+a), \quad \frac{c\left(a^{2}+3\right)}{c+a}=c(a+b)
$$
Thus,
$$
\begin{gathered}
\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}=a(b+c)+b(c+a)+c(a+b)= \\
\quad=a b+a c+b c+b a+c a+c b=2(a b+b c+c a)=2 \times 3=6
\end{gathered}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $2 f(2 x-3)-f(3 x+1)$ also has exactly one root. Find the root of the trinomial $f(x)$.
|
Answer: -11.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=2 f(2 x-3)-f(3 x+1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ and, hence, $f(x)=x^{2}+b x+\frac{b^{2}}{4}$. Then,
\[
\begin{aligned}
g(x) & =2 f(2 x-3)-f(3 x+1)=2(2 x-3)^{2}+2 b(2 x-3)+\frac{b^{2}}{2}-\left((3 x+1)^{2}+b(3 x+1)+\frac{b^{2}}{4}\right)= \\
& =-x^{2}+(b-30) x+\left(17-7 b+\frac{b^{2}}{4}\right)
\end{aligned}
\]
Therefore,
\[
0=(b-30)^{2}+4\left(17-7 b+\frac{b^{2}}{4}\right)=2 b^{2}-88 b+968=2(b-22)^{2}
\]
Thus, the discriminant of the polynomial $g(x)$ is 0 only when $b=22$, which means $f(x)=$ $x^{2}+22 x+121$ and its only root is -11.
|
-11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In the cells of an $11 \times 11$ square, zeros and ones are arranged in such a way that in any figure of four cells in the shape of $\square$, the sum of the numbers is odd. (The figure can be rotated and flipped). What is the minimum number of ones that can be in such an arrangement?
|
Answer: 25
Solution. Place 25 figures in the square without any common cells (see the left figure). Each of them contains at least one unit, so the total number of units is no less than 25. A suitable arrangement of 25 units: in all cells with even coordinates (see the right figure).
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $f(3 x+2)-2 f(2 x-1)$ also has exactly one root. Find the root of the trinomial $f(x)$.
|
Answer: -7.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=f(3 x+2)-2 f(2 x-1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ and, hence, $f(x)=x^{2}+b x+\frac{b^{2}}{4}$. Then
\[
\begin{aligned}
g(x) & =f(3 x+2)-2 f(2 x-1)=(3 x+3)^{2}+b(3 x+2)+\frac{b^{2}}{2}-\left((2 x-1)^{2}+b(2 x-1)+\frac{b^{2}}{4}\right)= \\
& =x^{2}+(20-b) x+\left(2+4 b-\frac{b^{2}}{4}\right)
\end{aligned}
\]
Therefore,
\[
0=(20-b)^{2}-4\left(2+4 b-\frac{b^{2}}{4}\right)=2 b^{2}-56 b+392=2(b-14)^{2}
\]
Thus, the discriminant of the polynomial $g(x)$ is 0 only when $b=14$, which means $f(x)=$ $x^{2}+14 x+49$ and its only root is -7.
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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