problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
89*. Find all pairs of integers \(x\) and \(y\) that satisfy the equation
\[
x^{2} + x = y^{4} + y^{3} + y^{2} + y
\] | 89. Answer: $(0 ;-1),(-1 ;-1),(0 ; 0),(-1 ; 0),(5 ; 2)$, we get
Multiplying both sides of the equation by 4 and adding 1 to each,
$(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right)$. If $y$ is an integer and different from $-1,0,1$ and 2, then $3 y^{2}+4 y+1>0$ a... | (0,-1),(-1,-1),(0,0),(-1,0),(5,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,294 |
90. In a sequence of integers (positive), each term, starting from the third, is equal to the absolute difference of the two preceding ones.
What is the maximum number of terms such a sequence can have if each of its terms does not exceed 1967? | 90. Answer: 2952. We will prove that the length (number of terms) of a sequence satisfying the condition of the problem, where the largest term is the second and equals $n$, does not exceed $d_{n} = [3(n+1) / 2]$, and for the sequence $n-1, n, 1, \ldots, \ldots, 1,1$ the length is exactly $d_{n}$.
We will reason by in... | 2952 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,295 |
91. "THE SUICIDAL KING". On a chessboard of size $1000 \times 1000$, there is a black king and 499 white rooks. Prove that, regardless of the initial arrangement of the pieces, the king can place himself under attack by a white rook, no matter how the whites play. (Moves are made in the same way as in regular chess.) | 91. If the king moves from the bottom left corner of the board to the top right corner along the diagonal, at some point he will inevitably be under attack by a white rook.
To prove this, it is sufficient to note that after the king's first move, all white rooks must be above the third rank and to the right of the thi... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,296 |
92. Three consecutive vertices of a rhombus lie on the sides $A B, B C, C D$ of a given square with side 1. Find the area of the figure filled by the fourth vertices of such rhombi. | 92. Answer: $S=1$.
Let $K, L, N$ be the vertices of the rhombus on the sides $A B, B C$, and $A D$ of the square (Fig. $52, a$). Note that the length $K B$ is equal to the distance from point $M$ to the line $A D$. Therefore, if we fix point $K$, the possible positions of point $M$ fill a certain segment $M_{1} M_{2}$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,297 |
93*. A natural number $k$ has the property that if $n$ is divisible by $k$, then the number written with the same digits as $n$, but in reverse order, is also divisible by $k$. Prove that $k$ is a divisor of 99.
$$
\begin{gathered}
2 \text {-nd International Mathematical Olympiad,} \\
1968 \text { (Leningrad) }
\end{g... | 93. First, note that the number $k$ is coprime with 10. Indeed, there exists a number divisible by $k$ and starting with 1. The reversed number is also divisible by $k$ and ends with 1.
Now, let's take a number starting with the digits 500 and divisible by $k: 500 a b c \ldots z$ (where $a, b, c, \ldots z$ are the dig... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,298 |
94. In an octagon, all angles are equal, and the side lengths are integers. Prove that the opposite sides of the octagon are equal to each other. | 94. Let $a_{1}, a_{2}, \ldots, a_{8}$ be the lengths of the sides of the octagon (Fig. 53). From the equality of all its angles, it immediately follows that its opposite sides are parallel. By extending two pairs of opposite sides $a_{1}$ and $a_{5}, a_{3}$ and $a_{7}$, we obtain a rectangle from which our octagon can ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,299 |
95. Which of the numbers is greater: $31^{11}$ or $17^{14}$? | 95. Answer: $17^{14}>31^{11}$. The proof follows from the obvious inequalities:
$$
17^{14}>16^{14}=2^{56}>2^{56}=32^{11}>31^{11}
$$ | 17^{14}>31^{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,300 |
96. On a grid paper with a cell side of 1 cm, a circle with a radius of $100 \mathrm{~cm}$ is drawn, not passing through the vertices of the cells and not touching the sides of the cells. How many cells can this circle intersect? | 96. Answer: 800 or 799.
One of the directions of the grid lines will be called horizontal, and the other - vertical. A circle with a diameter of 200, not passing through the nodes and not touching the grid lines, intersects 200 horizontal lines and 200 vertical lines, and each of them twice. Thus, the maximum number o... | 800or799 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,301 |
97. Among the students admitted to the University of Peoples' Friendship, exactly 50 people know English, exactly 50 people know French, and exactly 50 people know Spanish. Prove that the students can be divided into 5 (not necessarily equal) groups such that in each group there are exactly 10 people who know English, ... | 97. First, we will prove that it is possible to choose a group of students in which each language is known by exactly two people.
We will denote each student by a set of the initial letters of the names of the languages they know. Thus, student a knows English but not the others, student if knows Spanish and French bu... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,302 |
98. Prove the identity:
$\frac{2}{x^{2}-1}+\frac{4}{x^{2}-4}+\frac{6}{x^{2}-9}+\ldots+\frac{20}{x^{2}-100}=$
$=11\left(\frac{1}{(x-1)(x+10)}+\frac{1}{(x-2)(x+9)}+\ldots+\frac{1}{(x-10)(x+1)}\right)$. | 98. To solve, it is sufficient to use the identities
$$
\frac{2 k}{x^{2}-k^{2}}=\frac{1}{x-k}-\frac{1}{x+k}
$$
and
$$
\frac{11}{(x-11+k)(x+k)}=\frac{1}{x-11+k}-\frac{1}{x+k}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,303 |
99. In a regular $n$-gon ( $n>5$ ), the difference between the largest and smallest diagonals is equal to the side. Find $n$. | 99. Answer: $n=9$.
Let $a_{n}$ be the side length, and $D_{n}$ and $d_{n}$ be the lengths of the largest and smallest diagonals of a regular $n$-gon. For $n=4$ and $n=5$, all diagonals are equal. For $n=6$ and $n=7$, $D_{n}-a_{n}=2 A K=D_{8}-d_{8}$. For $n=9$ (Fig. 55,6), similarly, we get $\angle A B K=30^{\circ}$, s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,304 |
100. The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is formed according to the following rule:
$$
a_{1}=1, a_{2}=a_{1}+\frac{1}{a_{1}}, \ldots, a_{n}=a_{n-1}+\frac{1}{a_{n-1}}
$$
Prove that $a_{100}>14$. | 100. We will immediately prove the two-sided estimate $141$ we have $a_{k}^{2}=a_{k-1}^{2}+2+\frac{1}{a_{k-1}^{2}}$, and $a_{k}>1$. Therefore, $a_{k-1}^{2}+2<a_{k}^{2}<a_{k-1}^{2}+3$.
By summing these inequalities for all $1 \leqslant k \leqslant n$, we get (taking into account that $a_{1}=1$)
$$
2 n-1<a_{n}^{2}<3 n-... | 14<a_{100}<18 | Algebra | proof | Yes | Yes | olympiads | false | 25,305 |
101. Inside an acute-angled triangle $ABC$, a point $O$ is taken, and inside an acute-angled triangle $A' B' C'$, a point $O'$ is taken. Perpendiculars are dropped from point $O$: $OA_1$ to side $BC$, $OB_1$ to side $CA$, and $OC_1$ to side $AB$. Similarly, perpendiculars are dropped from point $O'$: $O'A_1'$ to $B'C'$... | 101. Let's translate $\Delta A^{\prime} B^{\prime} C^{\prime}$ parallel so that point $O^{\prime}$ moves to point $O$. The vertices of the resulting triangle will be denoted, as before, by $A^{\prime}, B^{\prime}$, and $C^{\prime}$.
Since $\angle A^{\prime} O B = \angle A_{1} O B_{1}$ and by the condition $O A^{\prime... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,306 |
102*. Prove that any natural number not exceeding $n!$ can be represented as the sum of no more than $n$ terms, among which no two are the same, and each is a divisor of the number $n!$. | 102. We will prove the statement of the problem by induction on p. For $n=1$, the statement is obviously true. Assume that it is true for $k=n$.
Let $a<(n+1)!$. Divide $a$ by $n+1$ with a remainder: $a=$ $=d(n+1)+r$, where $d \leqslant n!$, $r<n+1$.
By the induction hypothesis, $d=d_{1}+d_{2}+\ldots+d_{l}$, where all... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,307 |
103. In triangle $ABC$, a point $D$ is taken on side $AB$ and a point $E$ on $AC$. It turns out that $DE \parallel BC$, $AD = DE = AC$, and $BD = AE$.
Prove that the length of $BD$ is equal to the length of the side of a regular decagon inscribed in a circle of radius $R = AC$. | 103. By a parallel translation that moves point $D$ to point $B$, we translate triangle $A D E$ to triangle $K B L$, where $K L \| A C$ and, moreover, $K B=L B$ (Fig. 56),
If $\angle K .4 L=\alpha$, then $\angle K L A=\alpha ; \angle B K L=2 \alpha ; \angle B A C=$ $=\angle B C A=2 \alpha$; further, $\triangle A C L=\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,308 |
104. On the edges $AB, AC, AD$ of the tetrahedron $ABCD$, spheres are constructed with these edges as diameters. Prove that these spheres cover the tetrahedron. | 104. From point $A$, drop a perpendicular $A H$ to the plane $B C D$, and from point $H$ - perpendiculars $H K, H L$ and $H M$ to the lines $B C, B D, C D$. Each of the pyramids $A B K L, A C K M, A D M L$ is covered by a corresponding sphere. | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,309 |
105. a) In the cells of a $4 \times 4$ square table, signs “+” and “-” are arranged as shown in Fig. 6. It is allowed to simultaneously change the sign in all cells located in one row, one column, or on a line parallel to one of the diagonals (in particular, in any corner cell). Prove that no matter how many such sign ... | 105. a) It is easy to see that each line parallel to the sides or diagonals of the square intersects an even number of the eight shaded cells shown in Fig. 57. Therefore, the parity of the number of minuses standing in these cells does not change under the specified operations (P13).
b) On an $8 \times 8$ board, a $4 ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,310 |
106. The medians of triangle $ABC$ divide it into six triangles. It turns out that four of the circles inscribed in these triangles are equal. Prove that triangle $ABC$ is equilateral. | 106. The areas of all six triangles into which a given triangle is divided by its medians are equal.
Due to the equality of the radii of the inscribed circles and the formula $S=p r$, it follows that the perimeters of four of such triangles are equal.
Since two of them adjoin one side, say $AB$ (Fig. 58), then $AM=MB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,311 |
107. Prove that the equation
$$
x^{2}+x+1=p y
$$
has solutions in integers $(x, y)$ for infinitely many primes $p$. | 107. The proof of this statement resembles Euclid's proof of the infinitude of the set of prime numbers (P2).
Suppose that the equation $x^{2}+x+1=p y$ has integer solutions $(x, y)$ only for a finite number of prime numbers $p_{1}, p_{2}, \ldots, p_{m}$.
Let $P=p_{1} p_{2} \ldots p_{m}$. Then the number $P^{2}+P+1$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,312 |
109*. It is known that among the numbers $a_{1}, a_{2}, \ldots, a_{n}$, the numbers $1, 1 / 2, 1 / 3, \ldots, 1 / n$ appear exactly once, and among the numbers $b_{1}, b_{2}, \ldots, b_{n}$ - also. It is also known that $a_{1}+b_{1} \geqslant a_{2}+b_{2} \geqslant a_{3}+b_{3} \geqslant \ldots \geqslant a_{n}+$ $+b_{n}$... | 109. For any $m(1 \leqslant m \leqslant n)$ among $m$ pairs ( $a_{k}, b_{k}$ ), $1 \leqslant k \leqslant m$, one of the inequalities $a_{k} \geqslant b_{k}$ or $b_{k} \geqslant a_{k}$ is satisfied for not less than $m / 2$ pairs.
Let, for example, $b_{k} \geqslant a_{k}$ for not less than $m / 2$ pairs. If $b_{l}$ is ... | a_{}+b_{}\leqslant4/ | Inequalities | proof | Yes | Yes | olympiads | false | 25,313 |
110. On the teacher's desk, there are scales. On the scales, there are weights of not necessarily the same weight, on each of which the surnames of one or several students are written. When a student enters the classroom, they move to the other pan of the scales each weight that has their surname written on it. Prove t... | 110. The statement of the problem can be proved by induction on the number of students $n$. However, we will provide another solution that follows from such an almost obvious fact: the sum of $2^{k}$ products $x_{1} x_{2} \ldots x_{k}$, where $\left(x_{1}, x_{2} \ldots, x_{k}\right)$ are all possible sets of numbers +1... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,314 |
111. The city has a plan view of a rectangle divided into cells: $n$ streets are parallel to each other, $m$ others intersect them at right angles. On the city streets - but not at the intersections - there are police officers. Each police officer reports the number of the car passing by, the direction of its movement,... | 111. Answer: the smallest number of police officers is $(m-1)(n-1)$.
If the police officers are placed in the required manner, the entire grid of streets breaks down into some number $k$ of pieces that do not contain closed routes (cycles), otherwise
(n-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,315 |
112. The incircle of triangle $ABC$ touches side $AC$ at point $K$. Prove that the line connecting the midpoint of side $AC$ with the center of the incircle bisects segment $BK$.
untranslated text:
112. Вписанная в треугольник $A B C$ окружность касается стороны $A C$ в точке $K$. Докажите, что прямая, соединяющая се... | 112. Let 2 be the diametrically opposite point to $K$ on the inscribed circle with center $O$, $D$ be the midpoint of segment $A C$, and $E$ be the point of its intersection with line $B L$. It is sufficient to prove that $E D=D K$ (then $D O$ goes along the midline of $\triangle E B K$), i.e., that $A E=K C$,
Note th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,316 |
113. The sequence of numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ satisfies the following conditions:
$$
\begin{aligned}
& a_{1}=0, \quad\left|a_{2}\right|=\left|a_{1}+1\right|, \quad\left|a_{3}\right|=\left|a_{2}+1\right|, \ldots \\
& \ldots,\left|a_{n}\right|=\left|a_{n-1}+1\right|
\end{aligned}
$$
Prove that $\fra... | 113. Let us square all the equalities $a_{1}=0 ; \quad\left|a_{2}\right|=\left|a_{1}+1\right|$. "., $\left|a_{n+1}\right|=\left|a_{n}+1\right|$ and add them. After simplifications, we get $a_{n+1}=2\left(a_{1}+a_{2}+\ldots+a_{n}\right)+n \geqslant 0$, from which $a_{1}+$ $+a_{2}+\ldots+a_{n} \geqslant-n / 2$.
Another ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,317 |
114. Given a convex quadrilateral $A B C D$, the lengths of all sides and diagonals of which are rational. Let $O$ be the point of intersection of the diagonals. Prove that the length of the segment $A O$ is a rational number.
3rd All-Union Olympiad, 1969 (Kiev)
| racc | 1st day | 2nd |
| :---: | :---: | :---: |
| |... | 114. Let $\angle A B C=\beta, \angle A B D=\beta_{1} ; \angle D B C=\beta_{2}$. Applying the Law of Sines to triangles $A O B$ and $B O C$, we get
$$
\frac{A O}{O C}=\frac{A O}{A B} \cdot \frac{B C}{O C} \cdot \frac{A B}{B C}=\frac{A B}{B C} \cdot \frac{\sin \beta_{1}}{\sin \beta_{2}}
$$
Since $A B$ and $B C$ are rat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,318 |
115. On the base $AD$ of the trapezoid $ABCD$, there is a point $E$ such that the perimeters of triangles $ABE$, $BCE$, and $CDE$ are equal. Prove that $BC = AD / 2$. | 115. On the line $B C$, choose a point $C_{1}$ such that $A B C_{1} E$ is a parallelogram. Then the perimeters of triangles $A B E$, $B E C_{1}$, and $B E C$ are equal. From this, it follows that points $C$ and $C_{1}$ coincide. Therefore, $B C = A E$. Similarly, it can be proved that $B C = E D$. | BC=AD/2 | Geometry | proof | Yes | Yes | olympiads | false | 25,319 |
116. In the center of a field shaped as a square, there is a wolf, and at the vertices of the square, there are four dogs. The wolf can run throughout the field, while the dogs can only run along the sides of the square. It is known that the wolf can outrun a dog, but two dogs can outrun the wolf. The maximum speed of ... | 116. Let v be the maximum speed of the wolf. Draw two lines through the point where the wolf is located, parallel to the diagonals of the square. These lines intersect the contour of the square at points $C_{1}, C_{2}, C_{3}, C_{4}$.
Since the speed of movement of each of the points $C_{1}, C_{2}, C_{3}$, and $C_{4}$ ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,320 |
117. Given a finite sequence of zeros and ones, possessing two properties:
a) if in some arbitrary place in the sequence you select 5 digits in a row and in any other place also select 5 digits in a row, then these quintets will be different (these quintets can overlap, for example 0110101);
## $\checkmark$
b) if yo... | 117. Let abcd be the last four digits. In the sequence, there must be a quintet abcdo (otherwise, 0 could be added while preserving property a), which contradicts b), and abcd1. Therefore, the quartet abcd will appear in the sequence three times, and since neither 0 nor 1 can stand before abcd more than once, one of th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,321 |
118. $a, b, c$ and $d$ are positive numbers. Prove that among the inequalities
$$
\begin{gathered}
a+b<c+d \\
(a+b)(c+d)<a b+c d \\
(a+b) c d<a b(c+d)
\end{gathered}
$$
there is at least one that is false. | 118. Multiplying the first and second inequalities, we get $(a+b)^{2}(a+b)^{2} c d \geqslant 4 a b c d$, from which $a b+c d>4 c d$, i.e., $a b>3 c d$.
Thus, simultaneously $a b>3 c d$ and $c d>3 a b$, which is impossible. 119. Answer: $a=5$.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,322 |
120. Given a natural number $n$. Let us write down all fractions of the form $\frac{1}{p q}$, where $p$ and $q$ are coprime, $0 < p < q \leq n$. Prove that the sum of all such fractions is $1 / 2$. | 120. The approved task can be proven by induction.
For $n=1$ it is valid: $\frac{1}{1 \cdot 1}=1$. When transitioning from $n-1$ to $n$, we need to discard all fractions $\frac{1}{p q}$ with coprime $p$ and $q$, for which $p<q, p+q=n, p$ and $q$ are coprime, and add all fractions of the form $\frac{1}{p n}$, where $p<... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,323 |
121*. In space, $n$ points are arranged such that any three of them form a triangle, one of whose angles is greater than $120^{\circ}$. Prove that these points can be labeled as $A_{1}$, $A_{2}, \ldots, A_{n}$ in such a way that each of the angles $A_{i} A_{j} A_{k}$, where $1 \leqslant i<j<k \leqslant n$, will be grea... | 121. Let's choose from our system of points two, the distance between which is the greatest: let, for example, these be points $A$ and $B$.
We will prove that each of the angles $X A Y$ (and $X B Y$), where $X$ and $Y$ are some of the given points, is less than $120^{\circ}$. Indeed, since in triangles $A X B$ and $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,324 |
122. Four different three-digit integers, starting with the same digit, have the property that their sum is divisible by three of them without a remainder. Find these numbers. | 122. Answer: $108,135,180,117$.
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the unknown numbers, $S$ be their sum, and $a$ be the first digit of each of them. It is clear that $100 a \leqslant x_{i}<100(a+1)$ $(i=1,2,3,4)$.
Using these inequalities, we get $x_{t}+300 a \leqslant S<$ l $<x_{i}+300(a+1)$, from which
$$
1+\fra... | 108,135,180,117 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,325 |
123. In a certain state, the airline system is arranged in such a way that any city is connected by air routes to no more than three other cities, and from any city to any other, one can travel with no more than one layover.
What is the maximum number of cities that can be in this state? | 123. Answer: 10 cities.
From any city $A$, one can reach no more than three cities, and from each of them no more than two (excluding $A$). Thus, the total number of cities is no more than $1+3+3 \cdot 2=$ $=10$.
The example in Fig. 60 shows that the required system of air routes in a state with ten cities exists.
$... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,326 |
124. Consider a convex pentagon with all sides of equal length.
a) Prove that there exists a point inside it, lying on the longest diagonal, from which all sides are visible at angles not exceeding a right angle.
b) Prove that the circles constructed on its sides as diameters do not cover the entire pentagon. | 124. a) If $K$ is the midpoint of the largest diagonal $AD$ of a convex pentagon $ABCDE$, all sides of which are equal to $a$, then $\angle AKE = \angle EKD = 90^{\circ}$. Since $AC \leq AD$, it follows that $\angle BAC > \angle DAE$; therefore, $\angle BAK > \angle KAE$. From this, it follows that points $A$ and $B$ l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,327 |
125. On the board, the equation is written
$$
x^{3}+\ldots+x^{2}+\ldots+x+\ldots=0
$$
Two players play a game. The first player places an integer, different from zero (positive or negative), in any of the empty spaces. Then the second player places an integer in one of the remaining spaces. Finally, the first player ... | 125. If the first player puts -1 before $x$ in the first degree, and on his second move he places the opposite number of what the second player placed on the last remaining spot, then the polynomial will be of the form $x^{3}-a x^{2}-x+a=$ $=\left(x^{2}-1\right)(x-a)$. The roots of this polynomial are $-1,1, a$-integer... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,328 |
126*. In the country's football championship, 20 teams are participating. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | 126. Answer: 90 games.
Let among any three teams, there be two that have already played against each other. Choose a team $A$ that has played the least number of games $-k$. Each of the $k$ teams that have already played with $A$, as well as team $A$ itself, has played no fewer than $k$ games. From the $(19-k)$ teams ... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,329 |
127. $h_{k}$ - apothem of a regular $k$-gon inscribed in a circle of radius $R$. Prove that
$$
(n+1) h_{n+1}-n h_{n}>R
$$ | 127. The inequality of the problem is equivalent to the relations
$$
(n+1) \cos \frac{\pi}{n+1}-n \cos \frac{\pi}{n}>1
$$
or
$$
n\left(\cos \frac{\pi}{n+1}-\cos \frac{\pi}{n}\right)>1-\cos \frac{\pi}{n+1}
$$
i.e.,
$$
n \sin \frac{\pi}{2 n(n+1)} \cdot \sin \frac{\pi(2 n+1)}{2 n(n+1)}>\sin ^{2} \frac{\pi}{2(n+1)}
$$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,330 |
128*. Prove that for any positive numbers $a_{1}, a_{2}, \ldots, a_{n}$, the inequality
$$
\begin{aligned}
& \frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+\ldots \\
& \quad \ldots+\frac{a_{n-2}}{a_{n-1}+a_{n}}+\frac{a_{n-1}}{a_{n}+a_{1}}+\frac{a_{n}}{a_{1}+a_{2}}>\frac{n}{4}
\end{aligned}
$$
4th All-Union Olymp... | 128. Let $a_{i_{1}}=\max _{1 \leqslant i \leqslant n} a_{i}$. Choose $a_{i_{2}}$ - the largest of the numbers in the denominator of the fraction with numerator $a_{i_{1}}, a_{i_{3}}$ - the largest of the numbers in the denominator of the fraction with numerator $a_{i_{2}}$, and generally $a_{i_{k}}$ the largest of the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,331 |
129. Given a circle, its diameter $A B$ and a point $C$ on this diameter. Construct two points $X$ and $Y$ on the circle, symmetric with respect to the diameter $A B$, such that the line $YC$ is perpendicular to the line $X A$. | 129. Suppose the problem is solved. Since line $C Y$ is perpendicular to line $A X$, then $C Y \| B X$. Let $K$ be the intersection point of segments $A B$ and $X Y$. It is clear that $\triangle K B X = \triangle C K Y$. Therefore, $C K = K B$.
To construct, it is sufficient to draw a perpendicular through the midpoin... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,332 |
130. Prove that if the product of three positive numbers is 1, and the sum of these numbers is strictly greater than the sum of their reciprocals, then exactly one of these numbers is greater than 1. | 130. Let $x, y, \frac{1}{x y}$ be these numbers. If $x+y+\frac{1}{x y}>\frac{1}{x}+$ $+\frac{1}{y}+x y$, then after transformations we get $(x-1)(y-1) \times$ $\times\left(\frac{1}{x y}-1\right)>0$, from which it is clear that exactly one of the factors must be positive. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,333 |
131. How many sides in a convex polygon can be equal in length to the longest diagonal? | 131. Answer: no more than two. An example of a polygon with two sides equal to the largest diagonal is shown in Fig. 61.
Suppose there are more than two such sides. We select two of them, $A B$ and $C D$, which do not share any vertices (this is possible, [^3] since a polygon with diagonals is not a triangle). Then at... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,334 |
132. The digits of a certain seventeen-digit number are written in reverse order. The resulting number is added to the original. Prove that at least one of the digits of their sum will be even. | 132. Assume the opposite. Write down two numbers one under the other and add them column by column. Since the last digit of the sum is odd, the sum of the first digits must also be odd. Therefore, a one does not carry over from the previous column. This means that the sum of the digits in the second column, and consequ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,335 |
133. a) The castle has the shape (in plan) of an equilateral triangle with a side of 100 meters. It is divided into 100 triangular halls. All the walls of the halls have the same length - 10 meters, and a door is made in the middle of each wall between the halls. Prove that if a person wants to walk through the castle,... | 133. We will color the triangles in two colors as shown in Fig. $62, a_{2}$
The number of black triangles is $k$ more than the number of white triangles. Therefore, the total number of white triangles is $\frac{1}{2}\left(k^{2}-k\right)$, and the number of black triangles is $\frac{1}{2}\left(k^{2}+k\right)$. In the c... | k^{2}-k+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,336 |
134. Five segments are such that any three of them can form a triangle. Prove that at least one of these triangles is acute. | 134. Let $a \leqslant b \leqslant c \leqslant d \leqslant e-$ be the lengths of given segments. Suppose that no three of these segments can form an acute triangle. Then, since in a non-acute triangle the square of the length of the largest side is greater than or equal to the sum of the squares of the lengths of the ot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,337 |
137*. Prove that from any two hundred integers, one can select a hundred numbers, the sum of which is divisible by a hundred.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 137. We will prove that from any 199 (integer) numbers, one can choose 100 whose sum is divisible by 100. For this, we will only need the following facts: from any 3 numbers, one can choose 2 of the same parity (this is obvious); from any 9 numbers, one can choose 5 whose sum is divisible by 5.
Let $P_{m}$ denote the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,340 |
138. In triangle $ABC$, a line $MO$ is drawn through the midpoint $M$ of side $BC$ and the center $O$ of the inscribed circle of the triangle. This line intersects the altitude $AH$ at point $E$. Prove that the segment $AE$ is equal to the radius of the inscribed circle. | 138. First solution. Let $P$ be the point of tangency of the inscribed circle with side $BC$, and $PQ$ the diameter of the inscribed circle. In the solution to problem 112, it was proven that $AQ \parallel MO$, so $AEOQ$ is a parallelogram, hence $OQ = AE = r$.
Second solution. Let $a, b, c$ be the lengths of the side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,341 |
139. Prove that for each natural number $k$ there are infinitely many natural numbers $t$, not containing zeros in their decimal representation, such that $t$ and $k t$ have the same sum of digits. | 139. If the number $k$ is written as $k=\overline{a_{n} a_{n-1} \ldots a_{0}}$ and $t$ is a number consisting of $m>n$ nines: $t=\underbrace{99 \ldots 9}_{m}=10^{m}-1$, then
$$
k t=a_{n} a_{n-1} \ldots\left(a_{0}-1\right) \underbrace{99 \ldots 9\left(9-a_{n}\right)\left(9-a_{n-1}\right) \ldots\left(9-a_{1}\right)\left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,342 |
140. Two equal rectangles are positioned such that their contours intersect at eight points. Prove that the area of the common part of these rectangles is greater than half the area of each of them. | 140. Let $a$ and $b$ be the lengths of the sides of each rectangle. If the contours of the rectangles intersect at eight points, then on each side of any of them, there are exactly two points of intersection with the adjacent sides of the other. If there are fewer than two points of intersection on any side, then there... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,343 |
141. Cards have all five-digit numbers from 11111 to 99999 written on them. Then these cards are placed in one chain in a random order. Prove that the resulting 444,445-digit number cannot be a power of two. | 141. We will prove that any such number is divisible by $11111 \mathrm{n}$, and therefore is not a power of two. To do this, note that $10^{5}$ when divided by 11111 leaves a remainder of $1: 10^{5}=9 \cdot 1111+1$. Therefore, any of the obtained numbers gives the same remainder when divided by 11111 as the sum of all ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,344 |
142*. All natural numbers, in decimal notation of which there are no more than $n$ digits, are divided into two groups. The first group includes all numbers with an odd sum of digits, and the second group includes all numbers with an even sum of digits. Prove that if $1 \leqslant k<n$, then the sum of the $k$-th powers... | 142. Let $D_{1}$ be the set of 10 digits $\{0,1,2, \ldots, 9\}$; $A_{1}=\{0,2, \ldots, 8\}$ - the set of even digits, $B_{1}=\{1,3, \ldots, 9\}$ - the set of odd digits. In general, for any $n$, denote by $D_{n}$ the set of all numbers with no more than $n$ digits, $A_{n}$ and $B_{n}$ - its subsets of numbers with even... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,345 |
143*. The vertices of a regular $n$-gon are colored with several colors (each with one color) such that points of the same color are vertices of a regular polygon. Prove that among these polygons, there will be two equal ones.
 symmetric if, upon rotation by an angle $2 \pi k / m$ (for some $k < m$), it transforms into itself. The sum of the vectors in such a system is clearly zero, as it does not change upon rotation by an angle $2 \pi / m \le... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,346 |
144. Prove that for any natural $n$ there exists a number composed of the digits 1 and 2, divisible by $2^{n}$. | 144. This can be proven by induction. For $n=1$, the number 2 should be taken. If $A=2^{n} \cdot B-n$ is an $n$-digit number divisible by $2^{n}$, then one of the numbers $2 \cdot 10^{n}+A$ or $1 \cdot 10^{n}+A$ is divisible by $2^{n+1}$, because one of the numbers $5^{n}+B$ or $2 \cdot 5^{n}+B$ is even.
$\nabla$ The ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,347 |
146. a) Two players are playing a game. The first player writes down two rows of 10 numbers each, such that the following rule is satisfied: if the number $b$ is written under the number $a$, and the number $d$ is written under the number $c$, then $a+d=b+c$. The second player, knowing this rule, wants to determine all... | 146. a) Answer: 11 questions. Knowing all 10 numbers in one row and one in another, it is easy to restore all the other numbers. If, however, only 10 questions are asked, it is impossible to uniquely restore the numbers; on the one hand, in each column, at least one number must be known—otherwise, an arbitrary number $... | +n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,349 |
147*. In a square with a side of 1, several circles are placed, the diameter of each of which is less than 0.001. The distance between any two points of any two circles is not equal to 0.001. Prove that the total area covered by the circles does not exceed 0.34. | 147. Consider the figures $F_{1}$ and $F_{2}$, obtained from the given figure $F$ (a union of circles) by translations along vectors of length 0.001, forming an angle of $60^{\circ}$ with each other.
The three figures $F, F_{1}, F_{2}$ do not overlap and lie within a square with side length 1.001. Therefore, the area ... | 0.34 | Geometry | proof | Yes | Yes | olympiads | false | 25,350 |
148*. Three vessels contain a whole number of liters of water each. It is allowed to pour from any vessel into any other as much water as the latter already contains. Prove that by several such pourings one can empty one of the vessels. (The vessels are large enough: each can hold all the available water.)
翻译结果如下：
14... | 148. Let the vessels $A, B$, and $C$ contain $a, b$, and $c$ liters of water, respectively, with $0<a \leqslant b \leqslant c$. It is sufficient to achieve, through several pourings, that one of the vessels will contain less than $a$ liters of water (by repeating this procedure, we can reduce the amount of water in one... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,351 |
149. Prove that if for numbers $p_{1}, p_{2}, q_{1}, q_{2}$ the inequality $\left(q_{1}-q_{2}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-\right.$ - $\left.p_{2} q_{1}\right)<0$ holds, then the quadratic trinomials $x^{2}+p_{1} x+q_{1}$ and $x^{2}+p_{2} x+q_{2}$ have real roots and between the two roots of eac... | 149. The condition that the roots of the quadratic polynomials $f_{1}(x)=x^{2}+p_{1} x+q_{1}$ and $\mathrm{f}_{2}(x)=x^{2}+p_{2} x+q_{2}$ are real and interlaced is equivalent to the graphs $y=f_{1}(x)$ and $y=f_{2}(x)$ intersecting at a point $(x_{0}, y_{0})$ lying below the x-axis: $y_{0}<0$. Solving the equation $f_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,352 |
150. Projections of a body onto two planes - circles. Prove that these circles have equal radii. | 150. If the planes are parallel, this is obvious. Let them not be parallel. Then the projection of the body onto the line $l$ of intersection of the planes can be obtained as the projection onto $l$ of each of the projections of this body onto one and the other plane. (Everywhere we are talking about orthogonal project... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,353 |
151. Several numbers are written in a circle. If for some four consecutive numbers $a, b, c, d$ it turns out that $(a-d)(b-c)<0$, then the numbers $b$ and $c$ can be swapped. Prove that such an operation can be performed only a finite number of times. | 151. As a result of the sorting operation, the sum $S$ of the pairwise products of adjacent numbers increases (in this sum $a b+$ $+b c+c d$ is replaced by $a c+c b+b d$, and $a b+c d<a c+b d_{\text {n }}$ if $(a-d)(b-c)<0$.) However, the sum $S$ can take only a finite number of different values (P13). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,354 |
152. a) Prove that a line dividing a given triangle into two polygons of equal area and equal perimeter passes through the center of the circle inscribed in the triangle.
b) Prove an analogous statement for an arbitrary polygon in which a circle can be inscribed.
c) Prove that all lines dividing both the area and the... | 152. Let's immediately solve the general problem b). Suppose a line $l$ intersects the contour of a circumscribed polygon at points $R$ and $Q$ and divides its perimeter in half. Then the broken line consisting of two segments $O R$ and $O Q$, where $O$ is the center of the circle, divides its area in half; this is pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,355 |
154. a) At vertex $A_{1}$ of a regular 12-sided polygon $A_{1} A_{2} \ldots A_{12}$, there is a minus sign, while the others have plus signs. It is allowed to simultaneously change the sign to the opposite in any six consecutive vertices of the polygon. Prove that it is impossible to achieve, after several such operati... | 154. a) Let's divide all 12 vertices into 6 pairs of opposite vertices: $A_{1} A_{7}, A_{2} A_{8}, A_{3} A_{9}, \ldots, A_{6} A_{12}$. With each operation, only one vertex in each pair changes its sign. Therefore, in pairs $A_{2} A_{8}, \ldots, A_{6} A_{12}$, after the $(2 k-1)$-th operation, the signs will be differen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,357 |
155*. On an infinite sheet of graph paper, $N$ cells are painted black. Prove that it is possible to cut out a finite number of squares from the sheet such that the following two conditions are met:
1) all black cells will lie within the cut-out squares
2) in any cut-out square, the area of black cells will constitute... | 155. We can act by the "binary search" method. Cut out a large square $K_{0}$ of $2^{n} \times 2^{n}$ cells from the plane, containing all the black cells and at least four times as many white cells; then the area of the black cells is less than $1 / 5$ of the area of the square $K_{0}$. Cut $K_{0}$ into four squares $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,358 |
156. A cube with edge length $n$ is divided into $n^{3}$ unit cubes. We select several cubes and draw three lines through the center of each, parallel to the edges. How many cubes can be chosen at a minimum so that the lines drawn through them cross all the cubes?
a) Provide the answer for small values of $n$: for $n=... | 156. Answer: the smallest number $A_{n}$ of cubes satisfying the condition is
$$
A_{n}= \begin{cases}n^{2} / 2, & \text { if } n \text { is even, } \\ \left(n^{2}+1\right) / 2, & \text { if } n \text { is odd. }\end{cases}
$$
In particular, $A_{2}=2, A_{3}=5, A_{4}=8, A_{5}=12, A_{10}=50$.
The proof that fewer cubes... | A_{n}=\begin{cases}n^{2}/2,&\text{if}n\text{iseven,}\\(n^{2}+1)/2,&\text{if}n\text{isodd.}\end{cases} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,359 |
157. a) Consider the function $f(x, y)=x^{2}+x y+$ $+y^{2}$. Prove that for any point $(x, y)$ there exist integers $(m, n)$ such that $f(x-m, y-n)=$ $=(x-m)^{2}+(x-m)(y-n)+(y-n)^{2} \leqslant 1 / 2$.
b) * Denote by $\bar{f}(x, y)$ the minimum of the numbers $f(x-m, y-n)$ for all integers $m$ and $n$. The statement of... | 157. For each point $(x, y)$ on the plane, we need to find a point $(m, n)$ on the integer lattice for which the value $f_{a}(x-m, y-n)$ is as small as possible, and to find an upper estimate of this minimum $\vec{f}_{a}(x, y)=\min _{m, n} f_{a}(x-m, y-n)$. It is convenient to call the quantity $\left.\sqrt{f_{a}\left(... | f_{}\leqslant\frac{1}{+2} | Algebra | proof | Yes | Yes | olympiads | false | 25,360 |
159. In rectangle $A B C D$, point $M$ is the midpoint of side $A D$, and point $N$ is the midpoint of side $B C$. On the extension of segment $D C$ beyond point $D$, point $P$ is taken. Denote the intersection point of lines $P M$ and $A C$ as $Q$. Prove that $\angle Q N M = \angle M N P$. | 159. Let $R$ be the point of intersection of the lines $Q N$ and $C D$, and $O$ the center of the rectangle $A B C D$. From $O M=O N$ it follows that $P C = C R$, and therefore triangle $P N R$ is isosceles (with $N C$ being both the median and the altitude of this triangle).
The statement of the problem follows from ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,362 |
161. Find the greatest integer $x$ such that the number
$$
4^{27}+4^{1000}+4^{x}
$$
is a perfect square. | 161. Answer: $x=1972$. Since $4^{27}+4^{1000}+4^{x}=$ $=2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)$, the expression in parentheses will be a perfect square when $2 x-54=2 \cdot 1945$, i.e., when $x=1972$. If $x>1972$, then $2^{2(x-27)}<1+2 \cdot 2^{1945}+2^{2(x-27)}<\left(2^{x-27}+1\right)^{2}$, i.e., the given n... | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,364 |
162. a) Let $a, m, n$ be natural numbers, $a>1$. Prove that if $a^{m}+1$ is divisible by $a^{n}+1$, then $m$ is divisible by $n$.
b) Let $a, b, m, n$ be natural numbers, where $a$ is coprime with $b$ and $a>1$. Prove that if $a^{m}+b^{m}$ is divisible by $a^{n}+b^{n}$, then $m$ is divisible by $n$. | 162. First, let's prove the following two statements.
a) if \(a^{k}+b^{k}\) is divisible by \(a^{n}+b^{n}\), then \(a^{k-n}-b^{k-n}\) is also divisible by \(a^{n}+b^{n}\);
b) if \(a^{t}-b^{t}\) is divisible by \(a^{n}+b^{n}\), then \(a^{l-n}+b^{t-n}\) is also divisible by \(a^{n}+b^{n}\).
These statements easily fol... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,365 |
163. A triangular table is constructed according to the following rule: the top row contains a natural number $a$, and below each number $k$ on the left, $k^{2}$ is written, and on the right, the number $k+1$ is written. For example, when $\boldsymbol{a}=2$, the table shown in Fig. 9 is obtained. Prove that in each row... | 163. Suppose that in some rows of the table, the same numbers appear. Let $n$ be the number of the highest such row, and $p$ and $q$ be the equal numbers in this row.
Since there are no equal numbers in the $(n-1)$-th row, $p$ and $q$ are obtained from numbers $r$ and $s$ in this row through different operations: let ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,366 |
164*. Given several squares, the sum of whose areas is 1. Prove that they can be placed without overlapping in a square of area 2. | 164. Arrange the squares in descending order of their sides and start placing them from the largest $x \times x$, from left to right along the bottom side of the square area $\sqrt{2} \times \sqrt{2}$ (Fig. 67); as soon as the next square does not fit - it protrudes beyond the right side of the area, - we draw a horizo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,367 |
165. $O$ is the intersection point of the diagonals of a convex quadrilateral $A B C D$. Prove that the line passing through the points of intersection of the medians of triangles $A O B$ and $C O D$ is perpendicular to the line passing through the points of intersection of the altitudes of triangles $B O C$ and $A O D... | 165. Points $K, L, M, N$ - the intersections of the altitudes of triangles $A O B, B O C, C O D$ and $D O A$ - are the vertices of a parallelogram, two sides of which lie on lines passing through points $A$ and $C$ perpendicular to $B D$, and the other two - on lines passing through points $B$ and $D$ perpendicular to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,368 |
166. Each of the nine lines divides the square into two quadrilaterals, the areas of which are in the ratio $2: 3$. Prove that at least three of these nine lines pass through one point. | 166. Each line dividing a square into two trapezoids (or rectangles) with an area ratio of $2: 3$, divides the median line of the square, running along the midlines of these trapezoids, in the same ratio; this follows from the formula: the area of a trapezoid is equal to the product of the height and the length of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,369 |
169. Let $x, y$ be positive numbers, $s$ be the smallest of the numbers $x, y+\frac{1}{x}, \frac{1}{y}$. Find the greatest possible value of $s$. For which $x$ and $y$ is it achieved? | 169. Answer: the greatest possible value of $s$ is $\sqrt{2}$, which is achieved when $x=\sqrt{2}, y=\sqrt{2} / 2$.
We need to determine the greatest $s$ for which the inequalities $x \geqslant s, y+1 / x \geqslant s$, $1 / y \geqslant s$ can hold simultaneously (at least one of them must be an equality). It is clear ... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,371 |
170. A point $O$ inside a convex polygon forms an isosceles triangle with every pair of its vertices. Prove that this point is equidistant from the vertices of the polygon. | 170. For triangle $ABC$, it is easy to prove: if a point lies inside this triangle and triangles $AOB_1$, $BOC$, $COA$ are isosceles, then at least two of the angles $\angle AOB$, $\angle BOC$, $\angle COA$ are greater than $120^{\circ}$. Let these angles be $\angle AOB$ and $\angle BOC$. Then $AO = BO = OC$. Using thi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,372 |
171. Is it possible to arrange the digits $0,1,2$ in the cells of a $100 \times 100$ grid of squared paper in such a way that in every $3 \times 4$ rectangle of cells there are three zeros, four ones, and five twos? | 171. Answer: no.
$a$
We will present one of the shortest ways to derive a contradiction. We will say that two figures are filled the same if they contain the same number of zeros, ones, an... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,373 |
172. The sum of positive numbers $x_{1}, x_{2}, \ldots, x_{n}$ is 1. Let $s$ be the largest of the numbers
$$
\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \ldots, \frac{x_{n}}{1+x_{1}+x_{2}+\ldots+x_{n}}
$$
Find the smallest possible value of $s$. For which values of $x_{1}, x_{2}, \ldots, x_{n}$ is it achieve... | 172. The smallest value of $s$, equal to $1-2^{-1 / n}$, is achieved when $x_{k}=2^{\mathrm{k} / n}\left(1-2^{-1 / n}\right)$.
Let $y_{0}=1, y_{k}=1+x_{1}+\ldots+x_{k}(1 \leqslant k \leqslant n)$. Then $y_{n}=2, x_{k}=y_{k}-y_{k-1}$, if all these numbers do not exceed $s$, i.e.
$$
\frac{x_{k}}{y_{k}}=\frac{y_{k}-y_{k... | 1-2^{-1/n} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 25,374 |
174. At the trial, 14 coins were presented as physical evidence. The expert found that coins from 1st to 7th are counterfeit, and from 8th to 14th are genuine. The court, however, only knows that counterfeit coins weigh the same, genuine coins weigh the same, and that counterfeit coins are lighter than genuine ones. Th... | 174. Let's present the solution to part b). The expert should conduct the following three weighings.
$1^{\circ}$. The expert places the 1st coin on the left pan and the 8th coin on the right pan. Since the right pan outweighs, the court sees that the 1st coin is counterfeit and the 8th coin is genuine.
$2^{\circ}$. T... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,376 |
175. Prove that a nine-digit number, in the notation of which all digits except zero are used, and which ends with the digit 5, cannot be a perfect square of an integer. | 175. Suppose there exists a perfect square $D=A^{2}$ satisfying the conditions of the problem. The number $A$, obviously, ends in the digit 5, i.e., $A=10 a+5$. Therefore, $D=100 a(a+1)+25$ ends in 25. The number $a(a+1)$ ends in either 2, 6, or 0. Therefore, the third digit from the right of the number $D$ is 6. Thus,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,377 |
176. Given $n$ points, $n>4$. Prove that it is possible to connect them with arrows so that from any point to any other point it is possible to get by following either one arrow or two (any two points can be connected by an arrow in only one direction; it is possible to follow an arrow only in the direction indicated o... | 176. The problem is solved by induction. For $n=3, n=5$ and $n=6$ the required systems of points are shown in Fig. $73, a-c$, while Fig. 73,2 shows one way to transform a system of $n$ points $A_{1}, A_{2}, \ldots, A_{n}$, connected by arrows as needed, into the required system with $n+2$ points $A_{1}$, $A_{2}, \ldots... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,378 |
177. Given an angle with vertex $O$ and a circle tangent to its sides at points $A$ and $B$. From point $A$, a ray is drawn parallel to $OB$, intersecting the circle at point $C$. Segment $OC$ intersects the circle at point $E$, and lines $AE$ and $OB$ intersect at point $K$. Prove that $OK=KB$. | 177. Let $P$ and $Q$ be points
Fig. 74 shows the points of intersection of the tangent to the given circle at point $C$ with the sides of the angle. Since $A P = P C$, $\triangle A P C$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,379 |
178. Real numbers \(a, b, c\) are such that for all numbers \(x\) satisfying the condition \(-1 \leqslant x \leqslant 1\), the inequality
\[
\left|a x^{2}+b x+c\right| \leqslant 1
\]
is satisfied. Prove that for these values of \(x\), the inequality
\[
\left|c x^{2}+b x+a\right| \leqslant 2
\]
is also satisfied. | 178. Let $f(x)=a x^{2}+b x+c, g(x)=c x^{2}+b x+a$. Since $f(1)=g(1)$ and $f(-1)=g(-1)$, then $|g(1)| \leqslant 1$ and $|g(-1)| \leqslant 1$, in addition, $|c|=|f(0)| \leqslant 1$.
Suppose there exists a point $x$ for which $|g(x)|>2$. Then the vertex of the parabola $y=g(x)$ is the point with coordinates ( $x_{0}, g\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,380 |
179. The Tennis Federation has assigned qualification numbers to all its tennis players: the strongest player gets the first number, the next strongest gets the second number, and so on. It is known that in matches where the qualification numbers differ by more than 2, the player with the lower number always wins. A to... | 179. Answer! The highest possible number of the winner is 20.
Since a tennis player with number $k$ can lose (not counting stronger ones) only to the $(k+1)$-th and $(k+2)$-th tennis players, the number of the strongest winner after each round cannot increase by more than 2. Thus, the number of the winner of the entir... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,381 |
180. The quadratic trinomial \( f(x) = ax^2 + bx + c \) is such that the equation \( f(x) = x \) has no real roots. Prove that the equation \( f(f(x)) = x \) also has no real roots. | 180. If the equation $f(x)=x$ has no roots, then either $f(x)>x$ for all $x$ (if $a>0$), or $f(x)<x$, or $f(f(x))<f(x)<x$, which means that the equation $f(f(x))=x$ has no roots.
$\nabla$ The statement of the problem is true not only for a quadratic trinomial, but also for any continuous function. | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,382 |
181. On an infinite grid of white paper, $n$ cells are painted black. At moments $t=1,2, \ldots$ all cells on the sheet are simultaneously repainted according to the following rule: each cell $K$ acquires the color that the majority of three cells had in the previous moment: the cell $K$ itself and its neighbors to the... | 181. a) Note that if we add several more black cells to the given set of black cells, then after recoloring, only additional black cells may appear. Add cells to the initial set $M$ so that a black square of $m \times m$ cells is formed.
After $2 m-1$ steps, nothing will remain of the square (and thus of $M$).
b) Let... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,383 |
182. On the sides of an acute-angled triangle $A B C$, three similar acute-angled triangles $A C_{1} B$, $B A_{1} C$, and $C B_{1} A$ are constructed outwardly (such that $\angle A B_{1} C=\angle A B C_{1}=\angle A_{1} B C$; $\left.\angle B A_{1} C=\angle B A C_{1}=\angle B_{1} A C\right)$.
a) Prove that the circumcir... | 182. Let $D$ be the point of intersection of $A A_{1}$ and $B B_{1}$. Since $\angle A_{1} C A = \angle B_{1} C B$ (Fig. 76) and $A_{1} C : B C = A C : B_{1} C$, triangles $A_{1} C A$ and $B_{1} C B$ are similar, therefore, $\angle D B C = \angle D A_{1} C$. Hence, points $B, D, C, A_{1}$ lie on the same circle. It is a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,384 |
183. $N$ people are not acquainted with each other. Some of them need to be introduced to each other so that no three people have the same number of acquaintances. Prove that this can be done for any $N$. | 183. This problem can be solved by the method of mathematical induction. Such a solution is also possible.
Let us number $N$ people with numbers from 1 to $N$ and introduce person with number $i$ to person with number $j$ if $|i-j| \leqslant$
$\leqslant N / 2$. It is easy to see that with this method of introduction, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,385 |
184. The King traversed an $8 \times 8$ chessboard, visiting each square exactly once and returning to the starting square on the last move (the King moves according to the usual rules). When his path was drawn, connecting the centers of the squares he sequentially passed through with line segments, a closed broken lin... | 184. a) See Fig. 77,6.
b) Let's highlight a border of 28 edge squares on the board. The king visited each of them during his tour. We will number these squares in the order the king visited them. The entire
.
Consider together with the given polygon $M$ its symmetric counterpart $M^{\prime}$ relative to point $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,388 |
187. Prove that if \(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\) are positive numbers, then
\[
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)^{2} & \geqslant \\
& \geqslant 4\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}+x_{5} x_{1}\right)
\end{aligned}
\] | 187. If $x_{2} \geqslant x_{1}$, then the inequality follows from the identity
$$
\begin{gathered}
\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)^{2}- \\
-\left(x_{1}-x_{2}+x_{3}-x_{4}+x_{5}\right)^{2}= \\
=4\left(x_{1}+x_{3}+x_{5}\right)\left(x_{2}+x_{4}\right)= \\
=4\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,389 |
189. Numbers are written on cards, each of which is either "+1" or "-1". It is allowed to point to three cards and ask: "What is the product of the numbers on these cards?" (the numbers themselves are not told to us). What is the smallest number of such questions that need to be asked to find out the product of the num... | 189. Answer: a) 10 questions; b) 11; c) 12; d) 50.
a) We divide 30 numbers into 10 triples and find the products in these triples. It is clear that fewer questions will not suffice, as each number must be part of some triple.
b) The product of the first seven numbers is found by multiplying $a_{1} a_{2} a_{3}, a_{1} ... | 10,11,12,50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,391 |
190. Among the numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers, find the smallest in absolute value. Prove that the found number is indeed the smallest. | 190. Answer: $11=36-5^{2}$.
The last digit of the number $36^{h}=6^{2 k}$ is 6, the last digit of the number $5^{t}$ is 5. Therefore, the number $\left|6^{2 k}-5^{t}\right|$ ends in either 1 (if $6^{2 k}>5^{t}$) or 9 (if $6^{2 k}<5^{t}$).
The equation $6^{2 k}-5^{t}=1$ is impossible, because then it would be $5^{l}=\... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,392 |
191. a) Each side of a convex hexagon is greater than 1. Will there always be a diagonal longer than 2?
b) In a convex hexagon $A B C D E F$, the lengths of the diagonals $A D, B E$, and $C F$ are greater than 2. Will there always be a side longer than 1? | 191. a) Answer: not always. Example: if on each side of an equilateral triangle with side 2 isosceles triangles with height $1 / 10$ are constructed, then all sides of the resulting hexagon will be greater than 1, while all diagonals will not exceed 2.
b) Answer: always: Among the three diagonals $A D, B E$, and $F C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,393 |
192. Given two circles of radii $R$ and $r$, touching each other externally. Various trapezoids $A B C D$ are constructed such that each circle touches both lateral sides and one of the bases of the trapezoid. Find the smallest possible length of the lateral side $A B$. | 192. From the similarity of triangles $O_{1} K_{1} P$ and $P K_{2} O_{2}$ (notations are clear from Fig. 79), we obtain $K_{1} p \cdot P K_{2}=R r$, and from the similarity of
Fig. 79 $A_{1... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,394 | |
193*. On the plane, there are $n$ vectors, each of length 1. The sum of all $n$ vectors is the zero vector. Prove that the vectors can be numbered in such a way that for all $k=1,2, \ldots$ ..., n the following condition is satisfied: the sum of the first $k$ vectors has a length of no more than 2. | 193. Let's lay out all vectors from some point $O$.
We will prove that if $k$ vectors have already been chosen with a sum of $s=\overrightarrow{O S}$ (Fig. 80) $|s| \leqslant 1$, then from the remaining vectors, one can either choose a single vector $a$ such that $|s+a| \leqslant 1$, or
.
Substituting into the condition sequentially: $x=y=z=1$, then $x=y=0, z=1$, and finally $x=1, y=-1, z=0$, we get the system $|a+b+c|=1,|a|+|b|+|c|=1,|a-b|+$ $+|b-c|+|c-a|=2$. Since $|a+b+c|=|a|+|b|+$ $+|c|$, all numbers $a,... | two\of\the\\,\b,\\\0,\the\third\is\\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,396 |
195. Given a square $A B C D$. Points $P$ and $Q$ lie on sides $A B$ and $B C$ respectively, such that $B P = B Q$. Let $H$ be the foot of the perpendicular dropped from point $B$ to segment $P C$. Prove that angle $D H Q$ is a right angle. | 195. Let $F$ be the point of intersection of the lines $B H$ and $A D$ (Fig. 81). From the equality of triangles $A B F$ and $B P C$ (by the leg and acute angle), it follows that $A F = B P = B Q$ and $F D = C Q$; therefore, $Q C D F$ is a rectangle.
The circle circumscribed around it passes through the point $H$ ($F ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,397 |
197. Find all natural numbers $n$ and $k$ such that $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits. | 197. Answer: $n=k$ and $k=1,8$ or 9.
If $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits, then $10^{k-1} \leqslant n^{n}<10^{4}$ and $10^{n-1} \leqslant k^{k}<10^{n}$. Let, for definiteness, $n \geqslant k$. Then $n^{n}<10^{n}$, i.e., $n<10$ and $k<10$. It remains to check that $2^{2}<$ $<10,3^{3}<100,4^{4}<10^{3}, ... | n=kk=1,8or9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,399 |
198. On the legs $CA$ and $CB$ of an isosceles right triangle $ABC$, points $D$ and $E$ are chosen respectively such that $CD = CE$. The extensions of the perpendiculars dropped from points $D$ and $C$ to the line $AE$ intersect the hypotenuse $AB$ at points $K$ and $L$ respectively. Prove that $KL = LB$. | 198. Rotate triangle $A B C$ around vertex $C$ by $90^{\circ}$ so that point $A$ moves to $B$. Then $E$ will move to point $F$ on line $A C$, for which $F B\|C L\| D K$ and $F C=C D$, therefore $B L=$ $=L K$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,400 |
199. On an $8 \times 8$ chessboard, two players are playing a game of "cat and mouse." The first player has one token - the mouse, and the second player has several tokens - the cats. All tokens move the same way: one cell to the right, left, up, or down. If the mouse reaches the edge of the board, it can jump off the ... | 199. a) Answer: it is possible. Regardless of the mouse's position, the cats should be placed so that the mouse is on a segment between them, parallel to one of the diagonals of the board, and in response to any move of the mouse, the cats should be moved so that the mouse remains between them on a line parallel to the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,401 |
200. a) Prove that the numbers $1,2,3, \ldots, 32$ can be arranged in such an order that for no two numbers their half-sum equals any of the numbers placed between them.
b) Can the numbers $1,2,3, \ldots, 100$ be arranged in such an order that for no two numbers their half-sum equals any of the numbers placed between ... | 200. a) To achieve the required arrangement, in one half of the line, we will write even numbers, and in the other half - odd numbers. In this case, the half-sum of any two numbers from different halves will not be an integer and therefore will not be contained between them. Then, we will perform a similar procedure wi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,402 |
201. Find all three-digit numbers $A$ that have the following property: the arithmetic mean of all numbers obtained from the number $A$ by different permutations of its digits is equal to $A$. | 201. Answer: 15 numbers ( $111,222, \ldots, 999,407,518,629,370$, $481,592$).
Let $\overline{a b c}$ be the desired number. The condition of the problem means that
$$
\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}=6 \overline{a b c}
$$
from which $222(a+b+c)=6(1... | 111,222,\ldots,999,407,518,629,370,481,592 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,403 |
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