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4. Stones lie in three piles: one has 51 stones, another has 49 stones, and the third has 5 stones. It is allowed to combine any piles into one, as well as to split a pile with an even number of stones into two equal piles. Is it possible to get 105 piles, each with one stone? | 4. Note that if at some point the number of stones in each pile is divisible by an odd number $a$, then in all further piles obtained by allowed actions, the number of stones will also be divisible by $a$ (see facts 5 and 9).
After the first move, three possible arrangements of stones can be obtained: piles of 100 sto... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,938 |
4. In an acute triangle $A B C$, the altitudes $A H_{A}, B H_{B}$, and $C H_{C}$ are drawn. Prove that the triangle with vertices at the intersection points of the altitudes of triangles $A H_{B} H_{C}, B H_{A} H_{C}, C H_{A} H_{B}$ is equal to triangle $H_{A} H_{B} H_{C}$. | 4. Let $H_{1}, H_{2}, H_{3}$ be the orthocenters (points of intersection of the altitudes) of triangles $A H_{B} H_{C}, B H_{A} H_{C}, C H_{A} H_{B}$ respectively, $H$ be the orthocenter of triangle $A B C$, and $M_{1}, M_{2}, M_{3}$ be the midpoints of segments $H_{B} H_{C}, H_{C} H_{A}$, and $H_{A} H_{B}$. We will sh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,939 |
4. Prove that there does not exist a polynomial of degree not less than two with non-negative integer coefficients, the value of which for any prime $p$ is a prime number. | 4. Suppose such a polynomial $Q(x)$ exists. Let $Q(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$, where $a_{0}, a_{1}, \ldots, \ldots, a_{n}$ are non-negative integers, $a_{n} \neq 0, n \geqslant 2$.
If $a_{0}=0$, then $Q(x)=x\left(a_{n} x^{n-1}+a_{n-1} x^{n-2}+\ldots+a_{1}\right)$, hence for a prime $p$ the nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,940 |
6. (Continuation.) Show that there is no way to guarantee Grisha's success in 18 attempts.
## 9 t h g r a d e
##
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6. Let the total points scored by participant $A$ be denoted by $S_{A}$, and their strength coefficient by $F_{A}$. Consider the sum $\sum_{A} S_{A} F_{A}$ (the sum is taken over all participants). We claim that this sum is zero.
Indeed, define the number $r(A, B)$ as follows:
$$
r(A, B)=\left\{\begin{aligned}
1, & \... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,944 |
1. On the island, $2 / 3$ of all men are married and $3 / 5$ of all women are married. What fraction of the island's population is married | 1. Let the number of married couples on the island be $N$ (i.e., $N$ women are married, and $N$ men are married). According to the problem, $N$ married women constitute $3 / 5$ of all women on the island, so there are $5 N / 3$ women on the island. Similarly, married men constitute $2 / 3$ of all men on the island, so ... | \frac{12}{19} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,947 |
1. The row of recruits stood facing the sergeant. On the command "left," some turned left, some turned right, and the rest turned around. Can the sergeant always find a place in the row so that there are an equal number of recruits facing him on both sides? | 1. Let's agree that in the case when the sergeant stands in the formation, the letter $m$ will denote the number of people standing to the left of the sergeant facing him, and the letter $n$ will denote the number of people standing to the right of the sergeant facing him.
Suppose the sergeant first stands at the left... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,948 |
1. The tangents of the angles of a triangle are natural numbers. What can they be equal to? | 1. The first method. Let the angles of the triangle be denoted by $\alpha, \beta, \gamma (\alpha \leqslant \beta \leqslant \gamma)$. Since $\alpha+\beta+\gamma=\pi$, the inequality $02, n>3$. In this case, it would be necessary to increase two angles of the triangle $ABC$, leaving the third unchanged, which is impossib... | (1,2,3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,949 |
2. The square of the sum of the digits of the number $A$ is equal to the sum of the digits of the number $A^{2}$. Find all such two-digit numbers $A$. | 2. Note that $A^{2} \leqslant 99^{2}=9801<9999$. Therefore, the sum of the digits of the number $A^{2}$ is less than $9 \cdot 4=36$. Since it is equal to the square of the sum of the digits of the number $A$, the sum of the digits of the number $A$ is less than $\sqrt{36}=6$.
But if the sum of the digits is less than ... | 10,11,12,13,20,21,30,31,40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,950 |
2. Let $a, b, c$ be the sides of a triangle. Prove the inequality $a^{3}+b^{3}+3 a b c>c^{3}$. | 2. The first method. By the triangle inequality, $a+b>c$. Moreover, for any non-negative $a$ and $b$, the inequality $a^{2}-a b+b^{2} \geqslant 0$ holds. Indeed,
$$
a^{2}-a b+b^{2} \geqslant a^{2}-2 a b+b^{2}=(a-b)^{2} \geqslant 0
$$
Therefore,
$$
(a+b)\left(a^{2}-a b+b^{2}\right) \geqslant c\left(a^{2}-a b+b^{2}\ri... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,951 |
2. For positive numbers $a, b, c$, it is known that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant a+b+c$. Prove that $a+b+c \geqslant 3 a b c$. | 2. Multiplying the inequality
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant a+b+c
$$
by the common denominator, we obtain the equivalent inequality
$$
b c+a c+a b \geqslant(a+b+c) a b c .
$$
Now let's prove the auxiliary inequality (valid for all $a, b$ and $c$)
$$
(a+b+c)^{2} \geqslant 3(a b+b c+a c)
$$
We st... | +b+\geqslant3 | Inequalities | proof | Yes | Yes | olympiads | false | 25,952 |
2. Prove that on the graph of the function $y=x^{3}$, there exists a point $A$, and on the graph of the function $y=x^{3}+|x|+1$, there exists a point $B$, such that the distance $A B$ does not exceed $\frac{1}{100}$. | 2. Let $t>0$. Then the point $B$ with coordinates $(t, c)$, where $c=t^{3}+t+1$, lies on the graph of the function $y=x^{3}+|x|+1$.
The point $A$ with coordinates $(\sqrt[3]{c}, c)$ lies on the graph of the function $y=x^{3}$. We will show that for sufficiently large $t$, the distance between points $A$ and $B$ does n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,953 |
3. Given a circle with diameter $A B$. Another circle with center at $A$ intersects the segment $A B$ at point $C$, such that $A C < \frac{1}{2} A B$. A common tangent to both circles touches the first circle at point $D$. Prove that the line $C D$ is perpendicular to $A B$. | 3. Let $E$ be the point of tangency of the second circle with the common tangent, and $O$ the center of the first circle (Fig. 143). Triangle $A O D$ is isosceles (since $A O$ and $O D$ are radii of the first circle), so $\angle O D A = \angle O A D$.
As is known, the tangent is perpendicular to the radius drawn to th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,954 |
3. In a convex quadrilateral $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E, A F$, and $E F$ divide the quadrilateral into 4 triangles, the areas of which are consecutive natural numbers. What is the maximum possible value of the area of triangle $A B D$? | 3. Let the areas of the triangles be $n, n+1, n+2$, $n+3$. Then the area of the quadrilateral $A B C D$ is $4 n+6$. It is easy to see that the area of triangle $B C D$ is four times the area of triangle $E C F$, so this area is at least $4 n$. Therefore,
$$
S_{A B D}=S_{A B C D}-S_{B C D} \leqslant(4 n+6)-4 n=6 .
$$
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,955 |
4. Find all integers $x$ and $y$ that satisfy the equation $x^{4}-2 y^{2}=1$. | 4. If $(x, y)$ is a solution to the equation $x^{4}-2 y^{2}=1$, then $(-x, y),(x,-y)$, and $(-x,-y)$ are also solutions. Therefore, we will only look for non-negative solutions. It is clear that $x$ is an odd number, $x=2 t+1$. Rewrite the equation as
$x^{4}-1=(x-1)(x+1)\left(x^{2}+1\right)=2 t \cdot(2 t+2) \cdot\left... | x=\1,y=0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,958 |
4. Each viewer who bought a ticket for the first row of the cinema took one of the seats in the first row. It turned out that all seats in the first row were occupied, but each viewer was sitting in the wrong seat. The usher can swap neighbors if both are sitting in the wrong seats. Can the usher always manage to seat ... | 4. Let's number all the spectators by their ticket numbers $1, \ldots, n$. For definiteness, let the rightmost seat have number $n$, and the leftmost seat have number 1. We will reduce the problem to the same problem with fewer spectators by seating spectator $n$ in their place.
Suppose the usher acts as follows. He m... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,959 |
4. In an increasing sequence of natural numbers, each number starting from the 2002nd is a divisor of the sum of all previous numbers. Prove that there will be some number in the sequence, starting from which each number is equal to the sum of all previous ones. | 4. Let the $n$-th term of the sequence be denoted by $a_{n}$, and the sum of the first $n$ terms by $S_{n}$. Suppose the quotient of the division of $S_{n-1}$ by $a_{n}$ is $k_{n}$, i.e., $S_{n-1}=a_{n} k_{n}$. By the condition, $k_{n}$ is a natural number for $n \geqslant 2002$.
Since $a_{n+1}>a_{n}$, we get
$$
k_{n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,960 |
5. In triangle $ABC$, medians $AD$ and $BE$ intersect at point $M$. Prove that if angle $AMB$ is a) right; b) acute, then $AC + BC > 3AB$. | 5. The first method. In triangle $A M B$, we draw medians $A F$ and $B G$ and denote their intersection point by $N$ (Fig. 145). It is known that the intersection point of the medians divides them in the ratio $1: 2$. Therefore,
$$
M F=\frac{1}{2} M B=M E
$$
since $M$ is the intersection point of the medians of trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,961 |
5. In the city of Udoyev, mayoral elections proceed as follows. If in a given round of voting no candidate receives more than half of the votes, then a subsequent round is held with the participation of all candidates except the one who received the fewest votes. (No two candidates ever receive the same number of votes... | 5. a) Ostap could not take the last, 2002nd place in the first round, as otherwise he would have been immediately eliminated from the list of candidates. Therefore, $k \leqslant 2001$.
Suppose all candidates in the first round received almost the same number of votes, Ostap took the second-to-last place, and in each s... | 2001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,963 |
5. Let $A A_{1}, B B_{1}, C C_{1}$ be the altitudes of an acute triangle $A B C$; $O_{A}, O_{B}, O_{C}$ be the centers of the inscribed circles of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ respectively; $T_{A}, T_{B}, T_{C}$ be the points of tangency of the inscribed circle of triangle $A B C$ with sides ... | 5. Let $I$ be the center of the inscribed circle of triangle $ABC$, and let $r$ be its radius. Let $\angle BAC = \alpha$ (Fig. 153). We will prove that all sides of the hexagon $T_{A} O_{C} T_{B} O_{A} T_{C} O_{B}$ are equal to $r$.
First, we will prove that $T_{B} O_{A} \perp AB$.
We find
$$
AB_{1} = AB \cos \alpha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,964 |
6*. In a rectangular grid of $m \times n$, each cell can be either alive or dead. Every minute, all living cells die simultaneously, and those dead cells that had an odd number of living neighbors (by side) come to life. Indicate all pairs ( $m, n$ ) for which there exists an initial arrangement of living and dead cell... | 6. Let's first consider the case of a rectangle $1 \times n$. For $n=2$ and $n \geqslant 4$, eternal living configurations in the rectangle $1 \times n$ exist: each of the configurations in Fig. 147 has a period of 2, i.e., every second minute it returns to its initial state. The four cases in the figure correspond to ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,965 |
6. An acute triangle was cut into two (not necessarily triangular) parts by a straight cut, then one of these parts was cut again into two parts, and so on: at each step, any of the existing parts was chosen and cut (along a straight line) into two. After several steps, it turned out that the original triangle had spli... | 6. First, note that all resulting polygons will be convex. We will prove by induction on the number of parts (see fact 24) the following statement: after each cut, there will be a part (polygon) with at least three non-obtuse angles.
Base of induction. Initially, there is one part, and it is a triangle with three acut... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,966 |
6. Can all points of a square and a circle be painted black and white so that the sets of white points of these figures are similar to each other and the sets of black points are also similar to each other (possibly with different similarity coefficients)?
11 c l a s s | 6. Consider the following coloring of a square: inscribe a circle in the square and paint black the points of the square lying outside the circle; inscribe a square with sides parallel to the original square in the obtained circle. Paint white the points of the circle lying outside the "small" square. We will color the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,967 |
1. There are 4 people in the family. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead, Mom's salary is doubled - then by $15 \%$, if Dad's salary is doubled - then by $25 \%$. By what percentage will the family's total income increase if Grandpa's pension is d... | 1. The first method. If Masha's scholarship is doubled, the family income will increase by the amount of this scholarship. Therefore, Masha's scholarship constitutes $5 \%$ of the total income. Similarly, Mom's salary is $15 \%$, and Dad's is $25 \%$. The remaining share $100 \% - 5 \% - 15 \% - 25 \% = 55 \%$ is attri... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,969 |
1. Do there exist natural numbers $a, b$, and $c$ such that each of the equations
$$
\begin{array}{ll}
a x^{2}+b x+c=0, & a x^{2}+b x-c=0 \\
a x^{2}-b x+c=0, & a x^{2}-b x-c=0
\end{array}
$$
has both roots as integers? | 1. The solution can simply consist of checking that the numbers $a=1, b=5, c=6$ indeed satisfy the condition. Let's try to explain how such numbers can be found. First, let's try to simplify our task. Notice, first of all, that the roots of equations differing only in the sign of $b x$ are opposite, so it is sufficient... | =1,b=5,=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,971 |
1. For positive numbers $x, y, z$, the equality
$$
\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}=\frac{x^{2}}{z}+\frac{z^{2}}{y}+\frac{y^{2}}{x}
$$
is satisfied. Prove that at least two of the numbers $x, y, z$ are equal to each other. | 1. Freeing ourselves from the denominator, we bring our equality to the form
$$
x^{3} z-x^{3} y+z^{3} y-z^{3} x+y^{3} x-y^{3} z=0
$$
We factorize the left-hand side:
$$
\begin{gathered}
x^{3} z-x^{3} y+z^{3} y-z^{3} x+y^{3} x-y^{3} z= \\
=x^{3}(z-y)+z^{3}(y-x)+y^{3}(x-z)= \\
=x^{3}((z-x)+(x-y))+z^{3}(y-x)+y^{3}(x-z)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,972 |
2. The product of five numbers is not equal to zero. Each of these numbers was decreased by one, yet their product remained unchanged. Provide an example of such numbers. | 2. We can construct an example simply by taking the first four numbers equal to 2 (we cannot take ones - after subtraction, we would get zeros). Let the fifth number be $x$. For these numbers to satisfy the condition of the problem, it is sufficient that the equality $16 x=x-1$ holds, from which $x=-1 / 15$. | -\frac{1}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,974 |
2. A closed broken line passing through each vertex exactly once is drawn along the edges of a convex polyhedron with 2003 vertices. Prove that in each of the parts into which this broken line divides the surface of the polyhedron, the number of faces with an odd number of sides is odd. | 2. Let's choose any of the resulting parts. Consider the sum $a_{1}+a_{2}+\ldots+a_{n}$, where $a_{i}$ is the number of sides of the $i$-th face.
Each edge of the polyhedron that the broken line does not pass through is counted twice in this sum, and therefore the parity of the sum does not depend on the number of suc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,975 |
3. Is it possible to color some cells of an $8 \times 8$ board so that in any $3 \times 3$ square there are exactly 5 colored cells, and in each $2 \times 4$ rectangle (vertical or horizontal) there are exactly 4 colored cells? | 3. Suppose such a coloring is possible.
First, note that an $8 \times 8$ board can be cut into eight $4 \times 2$ rectangles, so there are exactly $8 \times 4 = 32$ shaded cells on it.
Now, cut the board into four $3 \times 3$ squares, three $4 \times 2$ rectangles, and a corner $2 \times 2$ square (Fig. 154). In the... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,977 |
3. The store has three floors, and movement between them is only possible by elevator. A study of floor visitation showed that from the start of the working day until the store closes:
1) of the customers entering the elevator on the second floor, half go to the first floor, and half - to the third;
2) among the custom... | 3. The first conversation. Suppose that during the day, $x$ customers entered the elevator on the first floor, $-y$ on the second floor, and $-z$ on the third floor. Note that the number of customers who exited the elevator on each floor during the day is equal to the number of customers who entered on that same floor ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,978 |
3. Let $P(x)$ be a polynomial with leading coefficient 1, and let the sequence of integers $a_{1}, a_{2}, a_{3}, \ldots$ be such that $P\left(a_{1}\right)=0, P\left(a_{2}\right)=a_{1}, P\left(a_{3}\right)=a_{2}$, and so on. The numbers in the sequence do not repeat. What degree can $P(x)$ have? | 3. If the degree of the polynomial $P(x)$ is 0, then $P(x)=1$ (the leading coefficient is 1). But then $P\left(a_{1}\right) \neq 0$. If the degree of the polynomial is 1, for example, the polynomial $P(x)=x-1$ and the sequence $1,2,3, \ldots\left(a_{n}=n\right)$ satisfy the condition of the problem.
Let's prove that t... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,979 |
3. Given a cyclic quadrilateral $A B C D$. Points $P$ and $Q$ are symmetric to point $C$ with respect to the lines $A B$ and $A D$ respectively. Prove that the line $P Q$ passes through the orthocenter (the point of intersection of the altitudes) $H$ of triangle $A B D$.
| 3. Let $U$ and $V$ be the points symmetric to $H$ with respect to $AB$ and $AD$ respectively, $X$ be the intersection point of lines $UC$ and $AB$, and $Y$ be the intersection point of lines $VC$ and $AD$. First, note that points $U$ and $V$ lie on the circumcircle of the quadrilateral (Fig. 166). To prove this, observ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,980 |
4. In triangle $A B C$, points $X$ and $Y$ are taken on sides $A C$ and $B C$ such that $\angle A B X=\angle Y A C$, $\angle A Y B=\angle B X C$, and $X C=Y B$. Find the angles of triangle $A B C$. | 4. For the external angles $B X C$ and $A Y B$ of triangles $A B X$ and $C A Y$, we can write the equalities
$$
\begin{aligned}
& \angle B X C=\angle A B X+\angle B A X \\
& \angle A Y B=\angle Y A C+\angle Y C A
\end{aligned}
$$
a)
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,983 |
4. Around the perimeter of a round cake with a diameter of $n / \pi$ meters, there are $n$ cherries. If at the ends of some arc there are cherries, then the number of other cherries on this arc is less than the length of the arc in meters. Prove that the cake can be cut into $n$ equal sectors such that each piece conta... | 4. Let's take some point on the circle as the starting point. Let $a_{i}$ be the length of the arc from the starting point to the $i$-th cherry in a clockwise direction. Consider the numbers $b_{i}=a_{i}-i$.
Lemma. $\left|b_{m}-b_{k}\right|<1$ for any $k < m$. We need to prove that $b_{m}-b_{k}<1$ and $b_{k}-b_{m}<1$.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,984 |
5. In the country, there are 15 cities, some of which are connected by air routes belonging to three airlines. It is known that even if any one of the airlines ceases operations, it will still be possible to travel from any city to any other (possibly with layovers), using the flights of the remaining two airlines. Wha... | 5. We will prove that fewer than 21 airlines are not sufficient. First, note that if 15 cities are connected by airlines in such a way that one can travel from any city to any other, then there are no fewer than 14 airlines. Indeed, starting from an arbitrary city, we will try to visit all the others, and visiting each... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,985 |
5. A right triangle $ABC$ with hypotenuse $AB$ is inscribed in a circle. Let $K$ be the midpoint of the arc $BC$ not containing point $A$, $N$ be the midpoint of segment $AC$, and $M$ be the point of intersection of ray $KN$ with the circle. Tangents to the circle at points $A$ and $C$ intersect at point $E$. Prove tha... | 5. Let point $O$ be the center of the circle (Fig. 160). Since triangle $ABC$ is a right triangle, point $O$ coincides with the midpoint of the hypotenuse $AB$ (see fact 14). Angle $NOK$ is a right angle: indeed, the midline $NO$ of triangle $ABC$ is parallel to its side $BC$, and the line $OK$ contains the height of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,986 |
5. In the country, there are several cities connected by roads with one-way and two-way traffic. It is known that from any city to any other, there is exactly one way to travel without passing through the same city twice. Prove that the country can be divided into three provinces in such a way that no road connects two... | 5. Let's show how to divide a country into provinces so that the condition of the problem is met. If the country has only one city, the problem is trivial: we can take two empty provinces.
We will call several cities of the country an autonomous region if from each city of this autonomous region, you can travel to any... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,987 |
5. For a convex polyhedron, the internal dihedral angle at each edge is acute. How many faces can the polyhedron have? | 5. For each of the faces, consider the vector of the external normal, i.e., a vector perpendicular to this face and directed outward from the polyhedron.
$1^{\circ}$. Let's prove that the angle between any two external normals is obtuse or straight. Suppose this is not the case, and there exist two faces $\Gamma_{1}$ ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,988 |
6. Borya thought of an integer greater than 100. Kira names an integer greater than 1. If Borya's number is divisible by this number, Kira wins; otherwise, Borya subtracts the named number from his, and Kira names the next number. She is not allowed to repeat numbers that have been named before. If Borya's number becom... | 6. Let's try playing as Kira. The most natural first move is to name the number "2". If we didn't guess and Borya's number was odd, then we will name "3". If we miss again, at least Borya's number will become even. Unfortunately, we can no longer name "2", so we need to act more cleverly.
To understand which of Borya'... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,989 |
6. 100 prisoners were placed in jail. The guard told them:
“I will give you the evening to talk to each other, and then I will place each of you in separate cells, and you will not be able to communicate anymore. Sometimes I will take one of you to a room that has a lamp (which is initially off). When leaving the room... | 6. Let's consider one of the possible strategies for the prisoners. We will choose one of the prisoners (we will call him the "counter," and the rest of the prisoners - "ordinary"). He will count the prisoners who have visited the room as follows. Initially, the number of counted prisoners is 0. Then, if, upon entering... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,990 |
6. Given an infinite sequence of polynomials $P_{1}(x), P_{2}(x), \ldots$ Does there always exist a finite set of functions $f_{1}(x), f_{2}(x), \ldots, f_{N}(x)$, compositions of which can express any of them (for example, $P_{1}(x)=$ $\left.=f_{2}\left(f_{1}\left(f_{2}(x)\right)\right)\right) ?$
## 11th grade | 6. Let, for example,
$$
\begin{aligned}
& f(x)=\pi+\operatorname{arctg} x \\
& g(x)=x+\pi
\end{aligned}
$$
$h(x)$ is a function that on the interval $\left(-\frac{\pi}{2}+\pi n ; \frac{\pi}{2}+\pi n\right)$ equals $P_{n}(\operatorname{tg} x)$ (i.e., the graph of the polynomial $P_{n}$ is "compressed" into the interva... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,991 |
6. On the shore of a round island called Somewhere, there are 20 villages, each inhabited by 20 wrestlers. A tournament was held where each wrestler faced all wrestlers from all other villages. Village A is considered stronger than Village B if at least $k$ matches between wrestlers from these villages end with a victo... | 6. Let's provide an example showing that the described situation is possible when $k \leqslant 290$. We will order all the wrestlers by strength and renumber them in ascending order of strength (the first being the weakest). We will call the 210 weakest wrestlers novices, and the 190 strongest - masters. In particular,... | 290 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,992 |
7. Given the equality
$$
\left(a^{m_{1}}-1\right) \ldots\left(a^{m_{n}}-1\right)=\left(a^{k_{1}}+1\right) \ldots\left(a^{k_{l}}+1\right)
$$
where $a, n, l$ and all exponents are natural numbers, and $a>1$. Find all possible values of the number $a$. | 7. The values $a=2$ and $a=3$ are possible: for example, $2^{2}-1=2+1, (3-1)^{2}=3+1$. Suppose that for some $a>3$ the required equality holds:
$$
A=\left(a^{m_{1}}-1\right) \ldots\left(a^{m_{n}}-1\right)=\left(a^{k_{1}}+1\right) \ldots\left(a^{k_{l}}+1\right)
$$
Lemma 1. Any $m_{i}$ and $a-1$ are powers of two.
Pro... | =2=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,993 |
1. For the quadratic equation $x^{2} + p x + q = 0$, the coefficients $p$ and $q$ were increased by one. This operation was repeated four times. Provide an example of such an initial equation so that each of the five resulting equations has integer roots. | 1. It is not difficult to check that the roots of the equation $x^{2}+3 x+2=0$ are -1 and -2. After increasing the coefficients by one, we get the equation $x^{2}+4 x+3=0$ with roots -1 and -3, then the equation with roots -1 and -4, then - with roots -1 and -5, and finally, the equation with roots -1 and -6.
To verif... | x^{2}+3x+2=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,994 |
1. The stock price of the company "Roga i Kopыта" increases or decreases by 17 percent every day at 12:00 (the stock price is not rounded). Can the stock price take the same value twice? | 1. Note that when the stock price increases, it is multiplied by $\frac{117}{100}$, and when it decreases, it is multiplied by $\frac{83}{100}$. Therefore, after $k$ increases and $l$ decreases, the stock price will be multiplied by
$$
\left(\frac{117}{100}\right)^{k}\left(\frac{83}{100}\right)^{l}
$$
If this number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,995 |
1. An arithmetic progression consists of integers. The sum of the first $n$ terms of this progression is a power of two. Prove that $n$ is also a power of two. | 1. Let $a$ and $b$ be the first and $n$-th terms of the progression, and $S$ be the sum of the first $n$ terms. Then
$$
S=\frac{a+b}{2} n
$$
Thus, $2 S$ is divisible by $n$. Since $2 S$ is a power of two, $n$ must also be a power of two.
Comment. We used the following statement: any divisor of a power of two is itse... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,996 |
1. Prove that any quadratic trinomial can be represented as the sum of two quadratic trinomials with zero discriminants. | 1. The first method. Consider the quadratic trinomial $f(x)=a x^{2}+b x+c$. We will complete the square, for which we denote $t=x+\frac{b}{2 a}$ and $D=b^{2}-4 a c$. Then
$$
a x^{2}+b x+c=a\left(t^{2}-\frac{D}{4 a^{2}}\right)
$$
For $D \leqslant 0$, let $p=\frac{\sqrt{-D}}{2 a}$. Then the desired representation is
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,997 |
2. For the quadratic equation $x^{2} + p x + q = 0$, the coefficients $p$ and $q$ were increased by one. This operation was repeated nine times. Could it be that each of the ten resulting equations has integer roots? | 2. For example, the equation $x^{2}+3 x+2=0$ works. Indeed, using the converse of Vieta's theorem, it is easy to understand that the roots of the equation $x^{2}+(q+1) x+q=0$ are the numbers -1 and $-q$. See also the solution to problem 1 for grade 8 and the comments on it. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,999 |
2. Does there exist a tetrahedron, all of whose faces are equal right-angled triangles? | 2. Suppose there exists a tetrahedron \(ABCD\) (Fig. 174). Let \(AB\) be the hypotenuse of triangle \(ABC\). Then \(AB\) is also the hypotenuse of triangle \(ABD\). It is not hard to see that in this case, \(CD\) is the hypotenuse of triangles \(ACD\) and \(BCD\). The midpoint of the hypotenuse \(CD\) is denoted by \(M... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,000 |
3. A billiard table has the shape of a polygon (not necessarily convex), in which adjacent sides are perpendicular to each other. The vertices of this polygon are pockets, into which the ball remains if it falls. A ball is shot from a vertex with an (internal) angle of $90^{\circ}$, which reflects off the sides (edges ... | 3. Proof by contradiction. Suppose the ball returned to the initial vertex. Denote this vertex by $A$, and the points of reflection by $A_{1}, \ldots, A_{n}$. Choose one of the sides of the billiard table and call lines parallel to it vertical, and lines perpendicular to it horizontal. Then each side of the billiard ta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,003 |
3. Let's call numbers of the form $\sqrt{a+b \sqrt{2}}$, where $a$ and $b$ are integers not equal to zero, white numbers. Similarly, let's call numbers of the form $\sqrt{c+d \sqrt{7}}$, where $c$ and $d$ are integers not equal to zero, black numbers. Can a black number equal the sum of several white numbers? | 3. For example,
$$
\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}=\sqrt{6+2 \sqrt{7}}
$$
This equality is easily verified by squaring both sides.
How can one guess such an equality? We will look for a black number equal to the sum of two white numbers. Experience with radicals suggests taking conjugate white numbers:
$$
\sqrt... | \sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}=\sqrt{6+2\sqrt{7}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,004 |
3. Prove that for any natural number $d$ there exists a natural number $n$ divisible by it, in the decimal representation of which one can erase some non-zero digit so that the resulting number is also divisible by $d$.
| 3. First method. The number $n$ can be written in the form $n=10^{k}(10 a+b)+c$, where $0 \leqslant c < d$. The number $10^{k} a+10^{k-1}$ can be divided by $d$ with a remainder: $10^{k} a+10^{k-1}=d q+r, 0 \leqslant r < d$. If $r=0$, then $10^{k} a+c=d q$ is divisible by $d$.
Second method (proposed by school student... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,005 |
4. The stock price of the company "Roga i Kopыта" increases or decreases by $n$ percent every day at $12^{00}$, where $n$ is a fixed positive integer less than 100 (the stock price is not rounded). Does there exist an $n$ for which the stock price can take the same value twice? | 4. First method. Note that when the stock price increases, it is multiplied by $1+\frac{n}{100}$, and when it decreases, it is multiplied by $1-\frac{n}{100}$. Therefore, after $k$ increases and $l$ decreases, the stock price will be multiplied by
$$
\left(1+\frac{n}{100}\right)^{k}\left(1-\frac{n}{100}\right)^{l}
$$
... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,006 |
4*. Let $l_{a}, l_{b}$ and $l_{c}$ be the lengths of the angle bisectors of angles $A, B$ and $C$ of triangle $A B C$, and $m_{a}, m_{b}$ and $m_{c}$ be the lengths of the corresponding medians. Prove that
$$
\frac{l_{a}}{m_{a}}+\frac{l_{b}}{m_{b}}+\frac{l_{c}}{m_{c}}>1
$$ | 4. Let $a, b$, and $c$ be the lengths of the sides of triangle $ABC$. Without loss of generality, we can assume that $a \leqslant b \leqslant c$. Let $I$ be the point of intersection of the angle bisectors of triangle $ABC$ (Fig. 173). Then
 Seven convex polygons were cut out of cardboard and placed on a table such that any six of them can be nailed to the table with two nails, but all seven cannot. Provide an example of such polygons and their arrangement. (Polygons can overlap.)
b)* Eight convex polygons were cut out of cardboard and placed on a t... | 5. a) Consider a convex heptagon \(A B C D E F G\) (for example, a regular one). The seven sought polygons are the triangles defined by pairs of adjacent sides of the heptagon (i.e., triangles \(A B C, B C D\), \(C D E, D E F, E F G, F G A\), and \(G A B\), see Fig. 169, a). We will prove that they satisfy the conditio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,010 |
5. We will call a natural number permitted if it has no more than 20 different prime divisors. Initially, there is a pile of 2004! (i.e., $1 \cdot 2 \cdot \ldots \cdot 2004$)
stones. Two players take turns removing a permitted number of stones from the pile (possibly a different number each time). The player who takes... | 5. Let $l$ be the product of the first 21 prime numbers. Note that $l$ is the smallest non-divisible number. It is not hard to check that the 21st prime number is $71<2004$. Therefore, 2004! is divisible by $l$.
Consider the number $m$, obtained after the first player's first move. Since the number of stones taken in ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,011 |
5. The radius of the circumscribed circle of triangle $ABC$ is equal to the radius of the circle that touches side $AB$ at point $C'$ and the extensions of the other two sides at points $A'$ and $B'$. Prove that the center of the circumscribed circle of triangle $ABC$ coincides with the orthocenter (the point of inters... | 5. The first method. Let $O$ and $O^{\prime}$ be the centers of the circumcircle and the excircle mentioned in the problem statement (see Fact 16). We will assume that $A^{\prime}$ lies on the ray $C B$, and $B^{\prime}$ lies on the ray $C A$ (Fig. $175, a$). It is sufficient to prove that $O A^{\prime} \perp B^{\prime... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,012 |
5. For given natural numbers $k_{0}<k_{1}<k_{2}$, determine the smallest number of roots on the interval $[0 ; 2 \pi)$ that the equation of the form
$$
\sin k_{0} x+A_{1} \sin k_{1} x+A_{2} \sin k_{2} x=0
$$
where $A_{1}, A_{2} \in \mathbb{R}$, can have. | 5. Let $N(F)$ denote the number of zeros of the function $F$ on the half-interval $[0 ; 2 \pi)$, i.e., the number of values of the argument $x \in [0 ; 2 \pi)$ for which $F(x)=0$. Then, for $A_{1}=A_{2}=0$, we have $N\left(\sin k_{0} x\right)=2 k_{0}$. Indeed, the zeros are the numbers $x_{n}=\frac{\pi n}{k_{0}}$, wher... | 2k_{0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,013 |
6. An extrasensory perception (ESP) practitioner has a deck of 36 cards face down in front of him (four suits, nine cards of each suit). He names the suit of the top card, after which the card is revealed to him. Then he names the suit of the next card, and so on. The task of the ESP practitioner is to guess the suit a... | 6. a) Note that guessing 18 cards is not difficult. Indeed, the first two backs can "encode" the suit of the second card (by associating each suit with one of the four possible arrangements of the two backs), the next two backs can encode the suit of the fourth card, and so on.
When only two cards remain in the deck, ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,015 |
1. Find at least one integer solution to the equation
$$
a^{2} b^{2}+a^{2}+b^{2}+1=2005
$$ | 1. To solve the problem, it is sufficient to verify that the numbers $a=2, b=20$ satisfy our equation.
Let's explain how to find all integer solutions to this equation (although this was not required for solving the problem). Factorize the left-hand side of the equation:
$$
a^{2} b^{2}+a^{2}+b^{2}+1=a^{2}\left(b^{2}+... | =\2,b=\20or=\20,b=\2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,018 |
1. The discriminants of three given quadratic trinomials are 1, 4, and 9. Prove that it is possible to choose one root from each of them such that their sum equals the sum of the remaining roots. | 1. From the formula for the roots of a quadratic equation, it follows that if $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $x^{2}+p x+q$, with $x_{1} \geqslant x_{2}$, and $D$ is the discriminant of this trinomial, then $x_{1}-x_{2}=\sqrt{D}$. Indeed,
$$
x_{1}-x_{2}=\frac{-b+\sqrt{D}}{2}-\frac{-b-\sqrt... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,019 |
1. Does there exist a flat quadrilateral in which the tangents of all interior angles are equal?
Comment. A rectangle does not work, as the tangents of its angles are not defined. | 1. At first glance, it seems that the answer is negative, since the only quadrilateral with all angles equal is a rectangle. However, from the equality of tangents, it does not follow that the angles are equal, but only that the difference of the angles is a multiple of $\pi$.
It remains to come up with a quadrilatera... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,020 |
1. The numbers $a$ and $b$ are such that the first equation of the system
$$
\left\{\begin{array}{l}
\sin x + a = b x \\
\cos x = b
\end{array}\right.
$$
has exactly two solutions. Prove that the system has at least one solution. | 1. Option $A$. By the condition, the function $y=\sin x+a-b x$ is zero at exactly two points. Let's denote them as $x_{1}$ and $x_{2}, x_{1}1)$. We will show that on the intervals $\left(-\infty ; x_{1}\right)$ and $\left(x_{2} ;+\infty\right)$, the function has different signs. Without loss of generality, we can assum... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,021 |
2. A square sheet of checkered paper $8 \times 8$ was folded several times along the grid lines so that a $1 \times 1$ square was obtained. It was then cut along a segment connecting the midpoints of two opposite sides of the square. Into how many pieces could the square have split as a result? | 2. Let the cut be vertical (the case of a horizontal cut is analogous). Draw vertical segments in all $1 \times 1$ squares, connecting the midpoints of opposite sides. Notice that when folding along the grid lines, these
$. Let's add to these numbers their sum. The resulting set of numbers $(1,2,3,6)$ also has the property that the s... | 3\cdot2^{2003} | Number Theory | proof | Yes | Yes | olympiads | false | 26,023 |
2. On the graph of a polynomial with integer coefficients, two points with integer coordinates are marked. Prove that if the distance between them is an integer, then the segment connecting them is parallel to the x-axis. | 2. Let the marked points have coordinates $\left(x_{1}, P\left(x_{1}\right)\right)$ and $\left(x_{2}, P\left(x_{2}\right)\right)$, where
Fig. 188 $x_{1} \neq x_{2}$. By shifting the origin ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,024 |
2. The sum of the absolute values of the terms of a finite arithmetic progression is 100. If all its terms are increased by 1 or all its terms are increased by 2, then in both cases the sum of the absolute values of the terms of the resulting progression will also be equal to 100. What values can the quantity $n^{2} d$... | 2. Let's denote the sum of the absolute values of the terms of an arithmetic progression by $S(S=100$ in version A and $S=250$ in version B). We will show that $n^{2} d=4 S$, if the given progression $a_{1}, a_{2}, \ldots, a_{n}$ is increasing (for a decreasing progression, the solution is analogous and $n^{2} d=-4 S$)... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,025 |
3. The altitudes $A A^{\prime}$ and $B B^{\prime}$ of triangle $A B C$ intersect at point $H$. Points $X$ and $Y$ are the midpoints of segments $A B$ and $C H$ respectively. Prove that the lines $X Y$ and $A^{\prime} B^{\prime}$ are perpendicular. | 3. Since $A A^{\prime}$ and $B B^{\prime}$ are altitudes, triangles $A A^{\prime} B$, $A B^{\prime} B$, $C A^{\prime} H$, and $C B^{\prime} H$ are right triangles (Fig. 181). The median of a right triangle drawn to the hypotenuse is equal to half of the hypotenuse (see fact 14), so $X A^{\prime} = \frac{1}{2} A B = X B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,026 |
3. The circle $\omega_{1}$ passes through the center of the circle $\omega_{2}$. From a point $C$, lying on $\omega_{1}$, tangents to $\omega_{2}$ are drawn, intersecting $\omega_{1}$ again at points $A$ and $B$. Prove that the segment $A B$ is perpendicular to the line passing through the centers of the circles. | 3. Let $O_{2}$ be the center of the circle $\omega_{2}$, and $E$ and $F$ be the points of tangency of the lines $A C$ and $B C$ with the circle $\omega_{2}$ (Fig. 184). The right triangles $O_{2} C E$ and $O_{2} C F$ are equal by the leg and hypotenuse, so $\angle O_{2} C A = \angle O_{2} C B$. Since these angles are i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,027 |
3. On the sides of triangle $A B C$, squares $A B B_{1} A_{2}, B C C_{1} B_{2}$, and $C A A_{1} C_{2}$ are constructed outwardly. On segments $A_{1} A_{2}$ and $B_{1} B_{2}$, squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$ are also constructed outwardly from $\triangle A A_{1} A_{2}$ and $\triangle B B_... | 3. The first method. Since $A B=B B_{1}, B C=B B_{2}$ and $\angle B_{1} B B_{2}=\pi-\angle A B C$, then $S_{A B C}=S_{B B_{1} B_{2}}$ (Fig. 189). Similarly, $S_{B_{1} A_{2} A_{3}}=S_{A A_{1} A_{2}}=S_{A B C}=S_{B B_{1} B_{2}}=S_{B_{1} A_{2} B_{4}}$.
Thus, triangles $B_{1} A_{2} A_{3}$ and $B_{1} A_{2} B_{4}$ have a co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,028 |
3. A board of size $2005 \times 2005$ is divided into unit square cells. Some of the cells on the board are numbered in some order with the numbers $1,2, \ldots$ such that within a distance of less than 10 from any unnumbered cell, there is a numbered cell. Prove that there will be two cells within a distance of less t... | 3. Variant A. Consider an arbitrary square grid with a side length of 105, which we will call the large square. Divide it into 25 squares of size $21 \times 21$, which we will call lalylis.
The center of a cell with a number, located at a distance of less than 10 from the center of some small square, must be in this s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,029 |
4. Around a circle, 2005 natural numbers are arranged. Prove that there exist two adjacent numbers such that after their removal, the remaining numbers cannot be divided into two groups with equal sums. | 4. Let it not be so, that is, after removing any two adjacent numbers, the remaining numbers can be divided into two groups with equal sums. Suppose first that all numbers are even (see fact 23). Divide all numbers by two. Note that the specified property will be preserved after division, i.e., again, after removing an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,030 |
4. The constructor consists of a set of rectangular parallelepipeds. All of them can be placed in one box, also having the shape of a rectangular parallelepiped. In a defective set, one of the dimensions of each parallelepiped turned out to be less than the standard. Can the dimension of the box, in which the set fits,... | 4. Let's describe the simplest possible counterexample. Suppose two blocks of size $1 \times 2 \times 3$ are placed in a box of size $2 \times 2 \times 3$. Slightly reduce (for example, by 0.1) one of the sides of length 3 for one block, and one of the sides of length 2 for the other. Since one side of the second block... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,032 |
4. With a convex quadrilateral $A B C D$, the following operation is performed: one of the given vertices is replaced by a point symmetric to this vertex with respect to the perpendicular bisector of the diagonal (of which it is not an endpoint), denoting the new point by the same letter. This operation is sequentially... | 4. Let $O$ be the point of intersection of the perpendicular bisectors of the diagonals. Note that this point remains fixed when any number of operations are applied to the quadrilateral. Let the angles formed by the sides of the quadrilateral and the segments $A O, B O$, $C O, D O$ be denoted by $\alpha_{1}, \alpha_{2... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,033 |
5. Divide the circle into several equal parts so that the center of the circle does not lie on the boundary of at least one of them. | 5. Divide the circle with center at point $O$ into six equal parts by points $A, B, C, D, E$, and $F$. Triangle $OAB$ is equilateral because $OA = OB$ and $\angle AOB = \frac{360^\circ}{6} = 60^\circ$. Similarly, triangles $OBC, OCD, ODE, OEF$, and $OFA$ are equilateral.
Draw an arc of a circle with center at point $A... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,034 |
5. On a circle, $n$ digits are arranged, none of which are 0. Senya and Zhenya copy $n-1$ digits into their notebooks, reading them clockwise. It turns out that although they started from different places, the $(n-1)$-digit numbers they wrote down are the same. Prove that the circle can be cut into several arcs such th... | 5. Since Sanya and Zhenya ended up with the same numbers, each of the digits $1,2, \ldots, 9$ appears the same number of times in Sanya's and Zhenya's numbers. Therefore, the digits that do not appear in Sanya's and Zhenya's numbers are the same. Indeed, let the digit that Sanya did not write be $i$ and it appears $n_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,035 |
5. Given the sequence $a_{n}=1+2^{n}+\ldots+5^{n}$. Do there exist 5 consecutive terms of this sequence that are divisible by 2005? | 5. The first method. First, we prove that for any \( m \) not divisible by 5, \( m^4 - 1 \) is divisible by 5 (this statement is a particular case of Fermat's Little Theorem, see the comment to problem 6 for 11th grade of the 1995 Olympiad).
We have \( m^4 - 1 = (m-1)(m+1)(m^2 + 1) \). The possible remainders when \( ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,036 |
5. To some natural number, two two-digit numbers were appended sequentially to the right. The resulting number turned out to be equal to the cube of the sum of the three original numbers. Find all possible triples of the original numbers. | 5. Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x=a+b+c$. According to the condition, the numbers $a, b, c$ and $x$ satisfy the equation $10^{4} a+100 b+c=x^{3}$ (see fact 11). Therefore,
$$
x^{3}=10^{4} a+100 b+c44$;
2) $x=45(x-1=44), 45^{3}=91125, a=9, b=11, c... | 9,11,25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,037 |
6. On a plane, there are 2005 points (no three of which lie on the same line). Each pair of points is connected by a segment. Tiger and Donkey play the following game. Donkey marks each segment with one of the digits, and then Tiger marks each point with one of the digits. Donkey wins if there are two points marked wit... | 6. First, let's outline the solution if there are only 8 points, and only three digits can be used: 0, 1, and 2 (note that \(8 = 2^3\)). The Donkey should mark the segments as shown in Fig. 183 (16 segments in the figure are not shown; the Donkey should mark them with the digit 2).
. Let the intersection of the circumcircles of triangles $B_{1} C_{2} A_{1}$ and $C_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,039 |
6. In space, there are 200 points. Each pair of them is connected by a segment, and the segments do not intersect each other. Two players have $k$ colors of paint at their disposal. The first player paints each segment in one of the $k$ colors, then the second player paints each point in one of the same colors. If ther... | 6. a) We will prove by induction (see fact 24) the following statement: if the number of colors is $k$, and the number of points is not less than $2^{k}$, then the first player can guarantee a win.
For $k=1$, the statement is obvious. Suppose it is proven for $k-1$ colors; we will prove it for $k$ colors. Choose an ar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 26,040 |
6. On a rectangular sheet of paper, a circle is drawn, inside which Misha mentally selects $n$ points, and Kolya tries to guess them. In one attempt, Kolya indicates one point on the sheet (inside or outside the circle), and Misha tells Kolya the distance from it to the nearest unguessed point. If this distance turns o... | 6. Let's first show how to guess one point in $2 n+1$ attempts. Suppose the line $l$ does not intersect the circle given in the problem's condition; draw a segment of this line and mark $n+1$ points $X_{0}, \ldots, X_{n}$ on it. Find out the distance from each of these points to the nearest hidden point. By the Pigeonh... | (n+1)^{2}-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 26,041 |
1. Given two circles $S_{1}$ and $S_{\mathrm{z}}$ and a line $l$. Draw a line parallel to $l$ such that the distance between the points of intersection of this line with the circles $S_{1}$ and $S_{\mathrm{z}}$ has a given magnitude $a$. | 1. Let's translate the circle $S_{1}$ parallel to the line $l$ by a distance $a$ to the position $S_{1}^{\prime}$; let $A^{\prime}$ and $B^{\prime}$ be the points of intersection of $S_{1}^{\prime}$ with the circle $S_{2}$ (see Fig. 107). The line passing through $A^{\prime}$ and $B^{\prime}$ parallel to $l$ will be th... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,042 |
4. Prove that if the midline $M N$ of quadrilateral $A B C D$ (where $M$ is the midpoint of side $A D$ and $N$ is the midpoint of side $B C$) is equal to half the sum of the bases $A B$ and $C D$, then the quadrilateral is a trapezoid. | 4. Let's parallel translate sides $AB$ and $DC$ of quadrilateral $ABCD$ to new positions $MB'$ and $MC'$ (Fig. 110). The resulting quadrilaterals $AMBB'$ and $DMCC'$ will be parallelograms, and therefore, $BB' \| AM$ and $BB' = AM$, $CC' \| DM$ and $CC' = DM$. But $AM = MD$ ($M$ is the midpoint of side $AD$); thus, seg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,045 |
9. Through the common point $A$ of circles $S_{1}$ and $S_{2}$, draw a line $l$ such that
a) circles $S_{1}$ and $S_{2}$ cut off equal chords on $l$;
b) the difference of the chords cut off on $l$ by circles $S_{1}$ and $S_{2}$ has a given length $a$.
Problem 9 b) is clearly a generalization of problem 6 a).[^3] | 9. a) Construct the circle $S_{2}^{\prime}$, symmetric to $S_{2}$ with respect to point $A$ (Fig. 117, a). Let $P^{\prime}$ be the point of intersection of circles $S_{1}$ and $S_{2}^{\prime}$, different from $A$. The line $P^{\prime} A$ will be the desired one, as the point $P$ of intersection of this line with circle... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,050 |
12. a) Let $O_{1}, O_{2}, \ldots, O_{n}$ (where $n$ is even) be points on a plane and $AB$ be an arbitrary segment; the segment $A_{1}B_{1}$ is symmetric to $AB$ with respect to $O_{1}$, $A_{2}B_{2}$ is symmetric to $A_{1}B_{1}$ with respect to $O_{2}$, $A_{3}B_{3}$ is symmetric to $A_{2}B_{2}$ with respect to $O_{3}$,... | 12. a) The segment $A_{n} B_{n}$ is obtained from the segment $A B$ after $n$ consecutive symmetries with respect to points $O_{1}$, $O_{2}, \ldots, O_{n}$ ($n$ is even). The sum of symmetries with centers $O_{1}$ and $O_{2}$ is a parallel translation; the sum of symmetries with centers $\hat{O}_{\mathbf{3}}$ and $O_{4... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,053 |
15. Given two lines $l_{1}$ and $l_{2}$, a point $A$, and an angle $\alpha$. Construct a circle with center $A$, on which $l_{1}$ and $l_{2}$ intercept an arc equal to $\alpha$ in angular measure. | 15. Let's rotate the line $l_{1}$ around point $A$ by an angle $\alpha$ to the position $l_{1}^{\prime}$. Let $M$ be the point of intersection of $l_{1}^{\prime}$ with the line $l_{2}$ (Fig. 121). The circle with center at point $A$, passing through point $M$, will be the required one, since the point $M^{\prime}$ of i... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,056 |
16. Construct an equilateral triangle, the vertices of which lie on three given parallel lines or three given concentric circles. | 16. Let $ABC$ be the desired triangle, the vertices of which are located on the given lines $l_{1}, l_{2}$, and $l_{3}$ (Fig. 122). We rotate the line $l_{2}$ around the point $A$ by an angle of $60^{\circ}$ in the direction from $B$ to $C$; then the point $B$ will move to the point $C$.
. Rotate segment $B D$ around the center $O$ of circle $S$ by an angle $\alpha$,
it will go into segment $B^{\prime} C$, forming an angle... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,058 |
20. a) On the sides of an arbitrary triangle $ABC$, outside it, equilateral triangles are constructed. Prove that the centers
Fig. 39. of these triangles $O_{1}, O_{2}, O_{3}$ themselves fo... | 20. a) Consider the successive rotations with centers $O_{1}, O_{2}, O_{3}$ by an angle of $120^{\circ}$ (see Fig. 29 in the text). The first of these rotations maps $A$ to $B$, the second maps $-B$ to $C$, and finally, the third maps $-C$ to $A$.
Thus, point $A$ is a fixed point of the sum of the three considered rot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,061 |
21. On the sides of an arbitrary triangle $A B C$, equilateral triangles $B C A_{1}$,
Fig. 30. $A C B_{1}$ and $A B C_{1}$ are constructed such that vertices $A_{1}$ and $A$, $B_{1}$ and $... | 21. Sequential rotations in the same direction with centers $A_{1}, B_{1}$, and $M$ and angles of rotation $60^{\circ}, 60^{\circ}$, and $240^{\circ}$ translate point $B$ into itself (see drawing 30 in the text). Therefore, the sum of these three rotations represents an identity transformation, and consequently, the su... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,062 |
22. On the sides of an arbitrary parallelogram $A B C D$, outside it, squares are constructed. Prove that their centers $M_{1}$, $M_{2}, M_{3}, M_{4}$ themselves are vertices of a square (Fig. 31).
Will the statement of the problem remain valid if the squares are located on the same side of the sides of the parallelog... | 22. The sum of four rotations with centers at $M_{1}, M_{2}, \mathcal{M}_{3}$, and $M_{4}$ by $90^{\circ}$ each translates the vertex $A$ of a parallelogram into itself. This is only possible if the sum of these rotations represents an identity transformation (compare with the solutions of problems 20 and 21). The sum ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,063 |
23. Given a convex quadrilateral $ABCD$.
Fig. 32.
Show that if the vertices $O_{1}$ and $O_{3}$ of the isosceles right triangles $ABO_{1}$ and $CDO_{3}$ coincide with each other, then the... | 23. Consider four rotations with centers $A, B, C$ and $D$ and angles of rotation equal to $90^{\circ}$ (see Fig. 32 in the text). The sum of the first two rotations represents a symmetry relative to the point $O_{1}$; the sum of the last two rotations represents a symmetry relative to the point $O_{3}$, which coincide... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,064 |
24. a) Given a line $M N$ and two points $A$ and $B$ on the same side of it. Find a point $X$ on the line $M N$ such that the segments $A X$ and $B X$ form equal angles with this line.
b) Given a line $M N$ and two circles $S_{1}$ and $S_{2}$ on the same side of it. Find a point $X$ on the line $M N$ such that the tan... | 24. a) Suppose point $X$ is found, i.e., $\angle A X M = \angle B X N$ (Fig. $128, a$). Let $B^{\prime}$ be the point symmetric to $B$ with respect to the line $M N$; then $\angle B^{\prime} X N = \angle B X N = \angle A X M$. Therefore, points $A, X, B^{\prime}$ lie on the same line. From this, it follows that $X$ can... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,065 |
26. a) Construct a triangle, given the base $A B=a$, the height $h$ dropped to the base, and the difference $\gamma$ of the angles at the base.
b) Construct a triangle, given two sides $A C$ and $B C$ and the difference $\gamma$ of the angles adjacent to the third side. | 26. a) Suppose the problem is solved. Draw a line $MN$ through vertex $C$, parallel to $AC$, and construct point $B'$, symmetric to $B$ with respect to line $MN$ (Fig. 133). Denote the angles at the base $AB$ as $\alpha$ and $\beta$ (we will assume that $\alpha > \beta$), then
$$
\begin{gathered}
\angle ACN = 180^{\ci... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,067 |
29. a) A billiard ball reflects off a straight rail of the billiard table in such a way that the two lines along which it moved before and after the impact are equally inclined to the
Fig. ... | 29. a) Suppose the problem is solved, i.e., points $X_{1}$, $X_{2}, \ldots, X_{n}$ are found on $l_{1}, l_{1}, \ldots, l_{n}$ such that $A X_{1} X_{1} \ldots X_{n} B^{\prime}$ is the path of a billiard ball (in Fig. 137, the case when $n=3$ is shown). It is easy to see that the point $X_{n}$ is the intersection of the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,070 |
30. a) Given a line $l$ and points $A$ and $B$ on the same side of it. Find a point $X$ on the line $l$ such that the sum of the distances $A X + X B$ has a given value $a$.
b) Given a line $l$ and points $A$ and $B$ on opposite sides of it. Find a point $X$ on the line $l$ such that the difference of the distances $A ... | 30. a) Suppose the problem is solved. Let's draw a circle $S_{1}$ with center at point $A$ and radius $a$, and a circle $S_{2}$ with center at point $X$ and radius $X B$ (Fig. $139, a$). Obviously, these two circles touch at a point lying on the line $A X$. Since circle $S_{2}$ passes through point $B$,
 Prove that the three points symmetric to the orthocenter $H$ of an arbitrary triangle $ABC$ with respect to its sides lie on the circumcircle of triangle $ABC$.
b) Construct triangle $ABC$ given three points $H_{1}$, $H_{2}$, $H_{3}$, which are symmetric to the orthocenter of the triangle with respect to its si... | 31. a) Let point $H_{1}$ be symmetric to point $H$ with respect to side $BC$ (Fig. 140). Denote by $P, Q, R$ the feet of the altitudes.
We have $\angle B H_{1} C = \angle B H C$ (since $\tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,072 |
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