problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
03.3. The point $D$ inside the equilateral triangle $\triangle A B C$ satisfies $\angle A D C=150^{\circ}$. Prove that a triangle with side lengths $|A D|,|B D|,|C D|$ is necessarily a right-angled triangle.
 We rotate the figure counterclockwise $60^{\circ}$ around $C$. Because $A B C$ is an equilateral triangle, $\angle B A C=60^{\circ}$, so $A$ is mapped on $B$. Assume $D$ maps to $E$. The properties of rotation imply $A D=B E$ and $\angle B E C=150^{\circ}$. Because the triangle $D E C$ is eq... | proof | Geometry | proof | Yes | Yes | olympiads | false | 425 |
03.4. Let $\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ satisfying
$$
f(x)+f(y)=f(x y f(x+y))
$$
for $x, y \in \mathbb{R}^{*}$ and $x+y \neq 0$.
|
Solution. If $x \neq y$, then
$$
f(y)+f(x-y)=f(y(x-y) f(x))
$$
Because $f(y) \neq 0$, we cannot have $f(x-y)=f(y(x-y) f(x))$ or $x-y=y(x-y) f(x)$. So for all $x \neq y, y f(x) \neq 1$. The only remaining possibility is $f(x)=\frac{1}{x}$. - One easily checks that $f, f(x)=\frac{1}{x}$, indeed satisfies the original ... | f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 426 |
04.1. 27 balls, labelled by numbers from 1 to 27, are in a red, blue or yellow bowl. Find the possible numbers of balls in the red bowl, if the averages of the labels in the red, blue, and yellow bowl are 15, 3 ja 18, respectively.
|
Solution. Let $R, B$, and $Y$, respectively, be the numbers of balls in the red, blue, and yellow bowl. The mean value condition implies $B \leq 5$ (there are at most two balls with a number $3$ ). $R, B$ and $Y$ satisfy the equations
$$
\begin{aligned}
R+B+Y & =27 \\
15 R+3 S+18 Y & =\sum_{j=1}^{27} j=14 \cdot 27=37... | 21,16,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 427 |
04.2. Let $f_{1}=0, f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1$, 2, ..., be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms is a c... |
Solution. The Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonacci resi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 428 |
04.3. Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-... |
Solution. We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 429 |
04.4. Let $a, b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$
|
Solution 1. By the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as
$$
A=\frac{r}{2}(a+b+c)
$$
and, by the sine theorem, as... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 430 |
05.1. Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals
$$
\frac{25}{8} k-211
$$
|
Solution. Let
$$
a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9
$$
Set
$$
f(a)=\prod_{k=0}^{n} a_{k}
$$
Since
$$
f(a)=\frac{25}{8} a-211 \geq 0
$$
$a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}... | 7288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 431 |
05.2. Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$
|
Solution 1. Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 432 |
05.4. The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathc... |
Solution. (See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 433 |
06.2. The real numbers $x, y$ and $z$ are not all equal and satisfy
$$
x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k
$$
Determine all possible values of $k$.
|
Solution. Let $(x, y, z)$ be a solution of the system of equations Since
$$
x=k-\frac{1}{y}=\frac{k y-1}{y} \quad \text { and } \quad z=\frac{1}{k-y}
$$
the equation
$$
\frac{1}{k-y}+\frac{y}{k y-1}=k
$$
to be simplified into
$$
\left(1-k^{2}\right)\left(y^{2}-k y+1\right)=0
$$
is true. So either $|k|=1$ or
$$
... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 434 |
06.3. A sequence of positive integers $\left\{a_{n}\right\}$ is given by
$$
a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487
$$
for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.
|
Solution. Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 435 |
06.4. The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used.
|
Solution. Denote by $R_{i}$ the number of colours used to colour the squares of the $i$ 'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$ 'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appears. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 436 |
07.1. Find one solution in positive integers to the equation
$$
x^{2}-2 x-2007 y^{2}=0
$$
|
Solution. The equation can be written in the form
$$
x(x 2)=223 \cdot(3 y)^{2}
$$
Here the prime number 223 must divide $x$ or $x 2$. In fact, for $x=225$ we get $x(x 2)=$ $15^{2} \cdot 223$, which is equivalent to $223 \cdot(3 y)^{2}$ for $y=5$. Thus, $(x, y)=(225,5)$ is one solution.
| (225,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 437 |
07.2. A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle.
|
Solution. Take any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then the w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 438 |
07.3. The number $10^{2007}$ is written on a blackboard, Anne and Berit play a game where the player in turn makes one of two operations:
(i) replace a number $x$ on the blackboard by two integer numbers a and $b$ greater than 1 such that $x=a b$;
(ii) erase one or both of two equal numbers on the blackboard.
The p... |
Solution. We describe a winning strategy for Anne. Her first move is
$$
10^{2007} \rightarrow 2^{2007}, 5^{2007}
$$
We want to show that Anne can act in such a way that the numbers on the blackboard after each of her moves are of the form
$$
2^{\alpha_{1}}, \ldots, 2^{\alpha_{k}}, 5^{\alpha_{1}}, \ldots, 5^{\alpha_... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 439 |
07.4. A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and AC. Show that $A P / P C=2 \cdot A B / B... |
Solution. First we show that if we fix the points $A, B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_{1}$ and $T_{1}$, and the tangents to the other circle,... | \frac{AP}{PC}=2\cdot\frac{AB}{BC} | Geometry | proof | Yes | Yes | olympiads | false | 440 |
08.1. Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies
$$
f(x+f(y))=A x+B y+C
$$
for all real $x$ and $y$.
|
Solution. Let $A, B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=$ $A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then
$$
f(z)=f(z-f(0)+f(0))=A(z-f(0))+B \cdot 0+C=A z-A f(0)+C
$$
so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(... | (A,B,C)=(,^2,)where\neq-1arbitrary,or(A,B,C)=(-1,1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 441 |
08.2. Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if
(i) $M$ and $N$ do not sit on adjacent seats, and
(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alpha... |
Solution. We will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \geq 3$. If $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs. Assume that the number of dominating pairs is $n-3$ for some $n>3$. If there are $n... | n-3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 442 |
08.3. Let $A B C$ be a triangle and let $D$ and $E$ be points on $B C$ and $C A$, respectively, such that $A D$ and $B E$ are angle bisectors of $A B C$. Let $F$ and $G$ be points on the circumcircle of $A B C$ such that $A F$ and $D E$ are parallel and $F G$ and $B C$ are parallel. Show that
$$
\frac{A G}{B G}=\frac... |
Solution. Let $A B=c, B C=a$ and $C A=b$. Then it follows from the angle bisector theorem that $C D=$ $a b /(b+c)$. Similarly, $C E=a b /(a+c)$, so $C D / C E=(a+c) /(b+c)$. The angles $\angle A B G, \angle A F G$ and $\angle E D C$ are equal, and so are $\angle A G B$ and $\angle A C B$, and consequently, the triangl... | \frac{AG}{BG}=\frac{AC+BC}{AB+CB} | Geometry | proof | Yes | Yes | olympiads | false | 443 |
08.4. The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares.
|
Solution. Assume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2 m+1)^{2}=(2 n+$ $1)(2 n-1)$. Since $2 n+1$ and $2 n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is 2 , but they are both odd), one of them is a square (of an odd integer, since it is odd) and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 444 |
09.1. A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller
triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the gi... |
Solution. Let $A B C$ be the triangle and let the lines through $P$ parallel to its sides intersect the sides in the points $D, E ; F, G$ and $H, I$. The triangles $A B C$, $D E P, P F G$ and $I P H$ are similar and $B D=I P$, $E C=P F$. If $B C=a, I P=a_{1}, D E=a_{2}$ ja $P F=a_{3}$, then $a_{1}+a_{2}+a_{3}=a$. Ther... | f\geq\frac{1}{3},f=\frac{1}{3}ifonlyifPisthecentroidofABC | Geometry | proof | Yes | Yes | olympiads | false | 445 |
09.2. On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It wo... |
Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 446 |
09.3. The integers 1, 2, 3, 4 and 5 are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a+b$ and $a b$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are 2009?
|
Solution. The answer is no. First notice that in each move two integers will be replaced with two greater integers (except in the case where the number 1 is wiped out). Notice also that from the start there are three odd integers. If one chooses to replace two odd integers on the blackboard, the number of odd integers... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 447 |
09.4. There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches.
|
Solution. To determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether. Now the silver medal winner has at some point lost to number 1 ; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be the candid... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 448 |
10.1. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$is the set of positive integers, is non-decreasing and satisfies $f(m n)=f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq(f(10))^{2}$.
|
Solution. Since $\mathrm{f}$ is non-decreasing, $f(91) \geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \geq f(9) f(10)$. Also $f(72) \geq f(70)$, and therefore $f(8) f(9) \geq f(7) f(10)$. Since all values of $\mathrm{f}$ are positive, we get $f(8) f(9) \cdot f(13) f(7) \geq$ $f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 449 |
10.2. Three circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The
line $A O$ intersects the circle $\Gamma_{C}$ in the poin ... |
Solution 1. Let $\angle A O Y=\alpha, \angle A O Z=\beta$ and $\angle Z O B=\gamma$. So $\alpha+\beta+\gamma=180^{\circ}$. Also $\angle B O X=\alpha$ (vertical angles) and $\angle A C Y=\alpha=\angle B C X$ (angles subtending equal arcs); similarly $\angle C O X=\beta$, $\angle A B Z=\beta=\angle C B X ; \angle C O Y=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 450 |
10.3. Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons ... |
Solution. a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lam... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 451 |
10.4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.
|
Solution. We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 452 |
11.1. When $a_{0}, a_{1}, \ldots, a_{1000}$ denote digits, can the sum of the 1001-digit numbers $a_{0} a_{1} \ldots a_{1000}$ and $a_{1000} a_{999} \ldots a_{0}$ have odd digits only?
|
Solution. The answer is no. The following diagram illustrates the calculation of the sum digit by digit.
| $a_{0}$ | $a_{1}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{999}$ | $a_{1000}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :--... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 453 |
11.2. In a triangle $A B C$ assume $A B=A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle ABC. When does equal... |
Solution. Because $A E \| D C$, the triangles $A B E$ and $D B C$ are similar. So
$$
C D=\frac{B C}{B E} \cdot A E
$$
$\mathrm{ja}$
$$
C D=\frac{A E \cdot B C}{B E \cdot C E} \cdot C E
$$
+y)=f\left(x^{2}-y\right)+4 y f(x)
$$
for all real numbers $x$ and $y$.
|
Solution. Substituting $y=x^{2}$ yields $f\left(f(x)+x^{2}\right)=f(0)+4 x^{2} f(x)$ for all real $x$. And $y=-f(x)$ gives $f(0)=f\left(x^{2}+f(x)\right)-4 f(x)^{2}$ for all $x$. Combining these two equations gives $4 f(x)^{2}=4 x^{4} f(x)$, so for each $x$ either $f(x)=0$ or $f(x)=x^{2}$. In particular $f(0)=0$. Now ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 455 |
## Problem 1
Determine all sequences of non-negative integers $a_{1}, \ldots, a_{2016}$ all less than or equal to 2016 satisfying $i+j \mid i a_{i}+j a_{j}$ for all $i, j \in\{1,2, \ldots, 2016\}$.
|
Solution Answer: All constant sequences of non-negative integers.
The condition rewrites to $i+j \mid i\left(a_{i}-a_{j}\right)$. Since $2 k-1$ and $k$ are coprime, we see that $2 k-1 \mid a_{k}-a_{k-1}$. Thus if $2 k-1>2016$, then $a_{k}=a_{k-1}$ since $a_{k}$ and $a_{k-1}$ are non-negative and at most 2016. All tog... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 457 |
## Problem 2
Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$.
Determine $\angle C D A$.
|
Solution 2 Answer: $\angle C D A=60^{\circ}$.
Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.
... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 458 |
## Problem 3
Find all $a \in \mathbb{R}$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$, such that
(i) $f(f(x))=f(x)+x$, for all $x \in \mathbb{R}$,
(ii) $f(f(x)-x)=f(x)+$ ax, for all $x \in \mathbb{R}$.
|
Solution 3 Answer: $a=\frac{1 \pm \sqrt{5}}{2}$.
From (i) we get $f(f(f(x))-f(x))=f(x)$. On the other hand (ii) gives
$$
f(f(f(x))-f(x))=f(f(x))+a f(x)
$$
Thus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\frac{1}{a} x$, since $a=0$ obviously does not give a solutio... | \frac{1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 459 |
## Problem 4
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that ... |
Solution 4 Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands ... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 460 |
## Problem 1
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(1-x))=f(x) \quad \text { and } \quad f(f(x))=1-f(x)
$$
for all real $x$.
|
Solution 1. Notice that $f(f(f(x)))=^{2} 1-f(f(x))={ }^{2} f(x)$. This is equation 3 . By substituting $f(x)$ for $x$ in the first equation we get:
$$
f(\underline{f(x)})={ }^{1} f(f(\underline{f(x)}) f(1-\underline{f(x)}))={ }^{2} f(f(f(x)) f(f(f(x))))={ }^{3} f(f(f(x)) f(x))
$$
Again we substitute $f(x)$ for $x$ a... | f(x)=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 461 |
## Problem 3
Anton and Britta play a game with the set $M=\{1,2,3, \ldots, n-1\}$ where $n \geq 5$ is an odd integer. In each step Anton removes a number from $M$ and puts it in his set $A$, and Britta removes a number from $M$ and puts it in her set $B$ (both $A$ and $B$ are empty to begin with). When $M$ is empty, A... |
Solution. Britta wins if and only if $n$ is prime.
If $n$ is not prime, then Anton can add any prime divisor $p<n$ of $n$ to his set $A$ in the first round and choose $x_{1}=p$ which means that the product $\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}}$ is divisible by $p$ a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 463 |
## Problem 4
Let $A B C$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $A B$ and $A C$ at $D$ and $E$. Denote by $B^{\prime}$ and $C^{\prime}$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscri... |
Solution. Let $P$ be the intersection of the circles $k$ and the circumscribed circle of triangle $A D E^{1}$. Let $C_{1}$ be the second intersection of the circumscribed circle of $\triangle D O P$ with $k$. We will prove that $C_{1}=C^{\prime}$, i.e. the reflection of $C$ over $O$. We know that $\left|O C_{1}\right|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 464 |
Problem 1 Let $n$ be a positive integer. Show that there exist positive integers $a$ and $b$ such that:
$$
\frac{a^{2}+a+1}{b^{2}+b+1}=n^{2}+n+1
$$
|
Solution 1 Let $P(x)=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 465 |
Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
then
$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$
|
Solution 2 The condition can be rewritten as
$$
a b \cos (\alpha-\beta)=a b \cos \alpha \cos \beta+a b \sin \alpha \sin \beta \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
Set $x=a \cos \alpha, y=b \sin \beta, z=b \cos \beta, t=a \sin \alpha$. We can now rewrite the condition as
$$
x z+y t \leq \sqrt{\lef... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 466 |
Problem 3 Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an i... |
Solution 3 Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.
It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 467 |
PROBLEM 1. The real numbers $a, b, c$ are such that $a^{2}+b^{2}=2 c^{2}$, and also such that $a \neq b, c \neq-a, c \neq-b$. Show that
$$
\frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}
$$
is an integer.
|
SolUTiON. Let us first note that
$$
\frac{a+b+2 c}{(a+c)(b+c)}=\frac{(a+c)+(b+c)}{(a+c)(b+c)}=\frac{1}{a+c}+\frac{1}{b+c}
$$
Further we have
$$
2 a^{2}-b^{2}-c^{2}=2 a^{2}-\left(2 c^{2}-a^{2}\right)-c^{2}=3 a^{2}-3 c^{2}=3(a+c)(a-c)
$$
and
$$
2 a^{2}-b^{2}-c^{2}=2\left(2 c^{2}-b^{2}\right)-b^{2}-c^{2}=3 c^{2}-3 b... | 3 | Algebra | proof | Yes | Yes | olympiads | false | 470 |
Problem 2. Given a triangle $A B C$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $B C$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $A B$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$... |
Solution I. There are three possibilities: $Q$ between $A$ and $B, Q=B$, and $B$ between $A$ and $Q$. If $Q=B$ we have that $\angle A B P$ is right, and $A P$ is a diameter
of the circumcircle. The triangles $A B P$ and $A C P$ are then congruent (they have $A P$ in common, $P B=P C$, and both have a right angle oppos... | proof | Geometry | proof | Yes | Yes | olympiads | false | 471 |
Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left... |
Solution. If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and
let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 472 |
Problem 4. The number 1 is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number 0 occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on t... |
Solution I. Let $S$ be a set of different numbers, all of them less than $2^{n-1}$, and create two new sets as follows: $S_{1}$, consisting of all the numbers in $S$ except
the smallest one, and $S_{2}$, with elements the smallest element of $S$ and all the numbers we get by adding $2^{n-1}$ to each number in $S$. Not... | \binom{n}{\lfloor\frac{n}{2}\rfloor} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 473 |
## Problem 1.
Let $A B C$ be a triangle and $\Gamma$ the circle with diameter $A B$. The bisectors of $\angle B A C$ and $\angle A B C$ intersect $\Gamma$ (also) at $D$ and $E$, respectively. The incircle of $A B C$ meets $B C$ and $A C$ at $F$ and $G$, respectively. Prove that $D, E, F$ and $G$ are collinear.
|
Solution 1. Let the line $E D$ meet $A C$ at $G^{\prime}$ and $B C$ at $F^{\prime} . A D$ and $B E$ intersect at $I$, the incenter of $A B C$. As angles subtending the same arc $\widehat{B D}$, $\angle D A B=\angle D E B=\angle G^{\prime} E I$. But $\angle D A B=\angle C A D=$ $\angle G^{\prime} A I$. This means that ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 474 |
## Problem 2.
Find the primes $p, q, r$, given that one of the numbers $p q r$ and $p+q+r$ is 101 times the other.
|
Solution. We may assume $r=\max \{p, q, r\}$. Then $p+q+r \leq 3 r$ and $p q r \geq 4 r$. So the sum of the three primes is always less than their product. The only relevant requirement thus is $p q r=101(p+q+r)$. We observe that 101 is a prime. So one of $p, q, r$ must be 101. Assume $r=101$. Then $p q=p+q+101$. This... | {2,101,103} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 475 |
## Problem 3.
Let $n>1$ and $p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ be a polynomial with $n$ real roots (counted with multiplicity). Let the polynomial $q$ be defined by
$$
q(x)=\prod_{j=1}^{2015} p(x+j)
$$
We know that $p(2015)=2015$. Prove that $q$ has at least 1970 different roots $r_{1}, \ldots, r_{1970}$ such... |
Solution. Let $h_{j}(x)=p(x+j)$. Consider $h_{2015}$. Like $p$, it has $n$ real roots $s_{1}, s_{2}, \ldots, s_{n}$, and $h_{2015}(0)=p(2015)=2015$. By Viète, the product $\left|s_{1} s_{2} \cdots s_{n}\right|$ equals 2015. Since $n \geq 2$, there is at least one $s_{j}$ such that $\left|s_{j}\right| \leq \sqrt{2015}<... | proof | Algebra | proof | Yes | Yes | olympiads | false | 476 |
## Problem 4.
An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For... |
Solution 1. (By the proposer.) Let the positions of the books in the stack be $1,2,3, \ldots, 2000$ from the top (and consider them modulo 2000). Notice that both operations fix the parity of the number of the book at a any given position. Operation (i) subtracts an even integer from the number of the book at each pos... | (1000!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 477 |
Problem 2. In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible sco... |
Solution. Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is
$$
k+(k+1)+\cdots+(k+n-1)=n k+\frac{(n-1) n}{2}
$$
Some games must end in a tie, for otherwise, all team scores wou... | n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 478 |
Problem 3. Define a sequence $\left(n_{k}\right)_{k \geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2 k}=n_{k}+n_{k-1}$ and $n_{2 k+1}=n_{k}$ for $k \geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \geq 1$. Show that every positive rational number is present exactly once in the sequence $\left(q_{k}\right)_{k \geq 1}$.
|
Solution. Clearly, all the numbers $n_{k}$ are positive integers. Moreover,
$$
q_{2 k}=\frac{n_{2 k}}{n_{2 k-1}}=\frac{n_{k}+n_{k-1}}{n_{k-1}}=q_{k}+1
$$
and similarly,
$$
\frac{1}{q_{2 k+1}}=\frac{n_{2 k}}{n_{2 k+1}}=\frac{n_{k}+n_{k-1}}{n_{k}}=\frac{1}{q_{k}}+1
$$
In particular, $q_{k}>1$ when $k$ is even, and $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 479 |
Problem 4. Let $A B C$ be an acute angled triangle, and $H$ a point in its interior. Let the reflections of $H$ through the sides $A B$ and $A C$ be called $H_{c}$ and $H_{b}$, respectively, and let the reflections of $H$ through the midpoints of these same sides be called $H_{c}^{\prime}$ and $H_{b}^{\prime}$, respec... |
Solution. If at least two of the four points $H_{b}, H_{b}^{\prime}, H_{c}$, and $H_{c}^{\prime}$ coincide, all four are obviously concyclic. Therefore we may assume that these four points are distinct.
Let $P_{b}$ denote the midpoint of segment $H H_{b}, P_{b}^{\prime}$ the midpoint of segment $H H_{b}^{\prime}, P_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 480 |
LIV OM - II - Task 3
Given is the polynomial $ W(x) = x^4 - 3x^3 + 5x^2 - 9x $. Determine all pairs of different integers $ a $, $ b $ satisfying the equation | Notice that $ W(x) = (x - 1)(x - 2)(x^2 + 3) - 6 $. For $ n > 3 $, the following inequalities therefore hold
and moreover
From this, we obtain
Thus, the values of the polynomial $ W $ at the points $ -2, \pm 3, \pm 4, \ldots $ are pairwise distinct. We directly calculate that
and from inequality... | (-1,0),(0,-1),(1,2),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 481 |
LII OM - I - Problem 10
Prove that inside any triangle $ABC$ there exists a point $P$ with the following property:
Every line passing through point $P$ divides the perimeter of triangle $ABC$ in the same ratio as it divides its area. | We will show that the property described in the problem statement is possessed by the center of the inscribed circle.
om52_1r_img_17.jpg
Let $ \ell $ be any line passing through point $ P $, which is the center of the circle inscribed in triangle $ ABC $ (Fig. 1). Without loss of generality, assume that line $ \ell $ i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 483 |
XXXVII OM - III - Problem 1
A square with a side length of 1 is covered by $ m^2 $ rectangles. Prove that the perimeter of one of these rectangles is greater than or equal to $ 4/m $. | Pole $ P = ab $ i obwód $ p = 2(a+b) $ prostokąta o bokach długości $ a $, $ b $ związane są nierównością
The area $ P = ab $ and the perimeter $ p = 2(a+b) $ of a rectangle with side lengths $ a $, $ b $ are related by the inequality
Gdyby więc każdy z rozważanych $ m^2 $ prostokątów miał obwód mniejszy od
$ 4/m $, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 484 |
XI OM - II - Task 1
Prove that if real numbers $ a $ and $ b $ are not both equal to zero, then for every natural $ n $ | When one of the numbers $ a $ and $ b $ is equal to zero or when both are of the same sign, inequality (1) is obvious, since in that case no term on the left side $ L $ of the inequality is negative. It remains to prove the case when $ a $ and $ b $ are of different signs. Due to the symmetry of $ L $ with respect to $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 487 |
XXIV OM - I - Problem 8
Find a polynomial with integer coefficients of the lowest possible degree, for which $ \sqrt{2} + \sqrt{3} $ is a root. | We will find a polynomial with integer coefficients, of which the root is the number $ a = \sqrt{2} + \sqrt{3} $. It is, of course, a root of the polynomial
This polynomial, however, does not have integer coefficients. Let's multiply the polynomial (1) by $ x - \sqrt{2} + \sqrt{3} $. We get $ (x - \sqrt{2})^2 - 3... | x^4-10x^2+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 489 |
XLVII OM - I - Problem 2
A palindromic number is defined as a natural number whose decimal representation read from left to right is the same as when read from right to left. Let $ (x_n) $ be the increasing sequence of all palindromic numbers. Determine all prime numbers that are divisors of at least one of the differ... | The decimal representation of the palindromic number $ x_n $ (respectively: ($ 2m $)-digit or ($ 2m-1 $)-digit) looks like this:
or
($ c_i \in \{ 0,1,2,3,4,5,6,7,8,9\} $ for $ i = 1, \ldots ,m;\ c_1 \ne 0 $). There are three possible situations:
Case I. $ c_m \ne 9 $. If $ x_n $ has the form (1), then $ x_{n+1} $ (th... | 2,5,11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 490 |
VI OM - I - Problem 9
Present the polynomial $ x^4 + x^3 + x^2 + x + 1 $ as the difference of squares of two polynomials of different degrees with real coefficients. | If the polynomial $ W (x) = x^4 + x^3 + x^2 + x + 1 $ is equal to the difference $ U(x)^2 - V(x)^2 $, where $ U(x) $ and $ V(x) $ are polynomials of different degrees, then the polynomial $ U(x) $ must be of the second degree and the polynomial $ V(x) $ must be of the first or zero degree. In this case, the polynomial ... | x^4+x^3+x^2+x+1=(x^2+\frac{1}{2}x+1)^2-(\frac{1}{2}x)^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 491 |
XXXVII OM - I - Problem 12
Prove that if the line connecting the midpoints of two opposite edges of a tetrahedron passes through the center of the inscribed sphere of this tetrahedron, then it also passes through the center of the circumscribed sphere of this tetrahedron. | We will first prove a lemma.
Lemma. Given a convex dihedral angle formed by half-planes $\alpha$ and $\beta$ with a common edge $l$. Let $\pi$ be the bisecting half-plane of this dihedral angle. Suppose that point $P \in \pi$ is the midpoint of segment $AB$ with endpoints $A \in \alpha$, $B \in \beta$ (where $A$, $B$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 492 |
I OM - B - Task 2
Prove that if the number $ a $ is the sum of the squares of two different natural numbers, then $ a^2 $ is the sum of the squares of two natural numbers. | If $ a=x^2+y^2 $, then
保留了源文本的换行和格式。请注意，最后一行的“保留了源文本的换行和格式”是中文，翻译成英文应为：“The line breaks and formatting of the source text have been preserved.” 但为了保持格式一致，这里没有将其翻译出来。 | proof | Number Theory | proof | Yes | Yes | olympiads | false | 494 |
XXVIII - I - Problem 5
Prove that if $ P(x, y) $ is a polynomial in two variables such that $ P(x, y) = P(y, x) $ for all real $ x, y $ and the polynomial $ (x-y) $ divides $ P(x, y) $, then the polynomial $ (x - y)^2 $ also divides $ P(x, y) $. | By assumption, there exists a polynomial $ Q(x, y) $ such that $ P(x,y) = (x - y) \cdot Q(x, y) $. From the equality $ P(x, y) = P(y, x) $, we obtain that $ (x - y) \cdot Q(x, y) = (y - x) \cdot Q(y, x) $, which means $ (x - y) \cdot (Q(x, y) + Q(y, x)) = 0 $. Since the product of polynomials is a zero polynomial only ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 495 |
IV OM - I - Problem 11
Prove that if $ A + B + C $ or $ A + B - C $ or $ A - B + C $ or $ A - B - C $ equals an odd number of straight angles, then $ \cos^2A + \cos^2B + \cos^2C + 2 \cos A \cos B \cos C = 1 $ and that the converse theorem is also true. | \spos{1} We need to prove that
The task can be solved in a very simple way when it is noticed that the equation $ x = (2k + 1) \cdot 180^\circ $ is equivalent to the equation $ \cos \frac{x}{2} = 0 $. The necessary and sufficient condition for one of the angles $ A + B + C $, $ A + B - C $, $ A - B + C $, $ A - B... | proof | Algebra | proof | Yes | Yes | olympiads | false | 497 |
XL OM - III - Task 2
In the plane, there are three circles $ k_1 $, $ k_2 $, $ k_3 $. Circles $ k_2 $ and $ k_3 $ are externally tangent at point $ P $, circles $ k_3 $ and $ k_1 $ — at point $ Q $, and circles $ k_1 $ and $ k_2 $ — at point $ R $. The line $ PQ $ intersects circle $ k_1 $ again at point $ S $, and th... | Let's denote the incircle of triangle $O_1O_2O_3$ by $k$, its center by $I$, and the centers of circles $k_1$, $k_2$, $k_3$ by $O_1$, $O_2$, $O_3$. Circle $k$ is tangent to the sides of triangle $O_1O_2O_3$ at points $P$, $Q$, $R$; this follows from the equalities $|O_1Q| = |O_1R|$, $|O_2R| = |O_2P|$, $|O_3P| = |O_3Q|$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 499 |
XV OM - II - Task 6
Prove that among any five points in a plane, one can choose three points that are not the vertices of an acute triangle. | If among the given points there are three collinear points, the thesis of the theorem is of course true. If, however, no three of the given points lie on a straight line, then four of these points, for example, $ A $, $ B $, $ C $, $ D $, are the vertices of a convex quadrilateral, as proven in problem 4. The angles of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 501 |
XLVI OM - I - Problem 11
Given are natural numbers $ n > m > 1 $. From the set $ \{1,2, \ldots ,n\} $, we draw $ m $ numbers without replacement. Calculate the expected value of the difference between the largest and the smallest drawn number. | An elementary event is the selection of an $m$-element subset from an $n$-element set. These elementary events are equally probable; there are $ {n}\choose{m} $ of them. Let the largest and smallest selected number be denoted by $a$ and $b$, respectively. The random variable $X$ under consideration, which is the differ... | \frac{(n-1)}{n-+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 502 |
XVIII OM - II - Problem 1
Given is a sequence of numbers $ a_1, a_2, \ldots, a_n $ ($ n \geq 3 $), where $ a_1 = a_n = 0 $ and $ a_{k-1}+a_{k+1} \geq 2a_{k+1} $ for $ k = 2, 3, \ldots, (n - 1) $. Prove that this sequence does not contain any positive terms. | In a finite set of numbers $a_1, a_2, \ldots, a_n$, there exists at least one number not less than any of these numbers. Suppose such a number is $a_r$, i.e., $a_i \leq a_r$ for $i = 1, 2, \ldots, n$. Let $s$ be the smallest index with the property that $a_s = a_r$. We will prove that $s=1$. Indeed, if $s > 1$, then th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 505 |
Given an integer $ c \geq 1 $. To each subset $ A $ of the set $ \{1,2, \ldots ,n\} $, we assign a number $ w(A) $ from the set $ \{1,2, \ldots ,c\} $ such that the following condition is satisfied:
Let $ a(n) $ be the number of such assignments. Calculate $ \lim_{n\to \infty}\sqrt[n]{a(n)} $.
Note: $ \min(x,y) $ is t... | Let $ M $ be the set $ \{1,2,\ldots,n\} $, and $ M_i $ be the $(n-1)$-element set obtained from $ M $ by removing the number $ i $:
Given the assignment as mentioned in the problem, let's denote the value $ w(M) $ by $ m $, and the value $ w(M_i) $ by $ f(i) $ (for $ i = 1,2, \ldots ,n $). According to the conditions ... | c | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 507 |
XIII OM - III - Problem 6
Three lines $ a $, $ b $, $ c $ are pairwise skew. Can one construct a parallelepiped whose edges lie on the lines $ a $, $ b $, $ c $? | The solution to the problem is based on the following theorem:
If $a$ and $b$ are skew lines, then: a) there exists a pair of parallel planes, one of which contains the line $a$ and the other the line $b$; b) such a pair of planes is unique.
Proof: a) Through any point $A$ on the line $a$, we draw a line $b'$ parallel ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 508 |
XXX OM - I - Task 11
Given a positive number $ p $ and three distinct rays $ OA^{\rightarrow} $, $ OB^{\rightarrow} $, $ OC^{\rightarrow} $ contained in a plane. Prove that there exists exactly one such triplet of points $ K, L, M $, that $ K\in OA^{\rightarrow} $, $ L \in OB^{\rightarrow} $, $ M\in OC^{\rightarrow} $... | For the triangles mentioned in the problem to exist, it is of course necessary to assume that no two of the given rays lie on the same line.
First, note that if points $P$ and $Q$ belong to different sides of an angle with vertex $O$ and point $Q$ moves away from $O$, then the perimeter of triangle $OPQ$ increases with... | proof | Geometry | proof | Yes | Yes | olympiads | false | 511 |
LV OM - III - Task 3
In a certain tournament, $ n $ players participated $ (n \geq 3) $. Each played against each other exactly once, and there were no draws. A three-element set of players is called a draw triplet if the players can be numbered in such a way that the first won against the second, the second against t... | Instead of determining the maximum number of draw triples, we will determine the minimum number of non-draw triples.
Let's number the players from 1 to $ n $ and assume that player number $ i $ won $ x_i $ times. Then the total number of matches is
Each non-draw triple is uniquely determined by the player who won... | \frac{1}{24}n(n^2-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 514 |
XXIV OM - I - Problem 10
Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $). | We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side.
It suffices to prove that $ AB \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 515 |
XXV - I - Task 1
During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in f... | The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statemen... | 1714 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 517 |
XL OM - III - Task 4
Let $ n, k $ be natural numbers. We choose a sequence of sets $ A_0, \ldots, A_k $ such that $ A_0 = \{1, \ldots, n\} $, and for $ i = 1, \ldots, k $, the set $ A_i $ is a randomly chosen subset of $ A_{i-1} $, with each subset being equally likely. We consider the random variable equal to the num... | We introduce random variables $ X_1, \ldots , X_n $ defined as follows:
\
($ i = 1,\ldots ,n $). The random variable $ X $ (the number of elements in the set $ A_k $) considered in the problem is the sum of the variables introduced just now:
Let us fix a number $ i \in A_0 = \{1, \ldots, n\} $. Since $ A_1 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 518 |
VII OM - II - Task 3
A homogeneous horizontal plate weighing $ Q $ kG in the shape of a circle is supported at points $ A $, $ B $, $ C $ lying on the circumference of the plate, with $ AC = BC $ and $ ACB = 2\alpha $. What weight $ x $ kG should be placed on the plate at the other end $ D $ of the diameter drawn from... | The weight $ Q $ of a plate supported at points $ A $, $ B $, $ C $ (Fig. 17) is balanced by the reactions at these points, acting vertically upwards. The center of gravity $ O $ of the plate, as the point of application of the weight $ Q $ acting downwards, must lie within the triangle $ ABC $, i.e., the angle $ 2\alp... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 519 | |
LI OM - II - Task 6
A polynomial $ w(x) $ of degree two with integer coefficients takes values that are squares of integers for integer $ x $. Prove that the polynomial $ w(x) $ is the square of some polynomial. | Let $ w(x) = ax^2 + bx + c $. Introduce the notation: $ k_n = \sqrt{w(n)} $ for any positive integer $ n $. Then $ k_n $ can be zero for at most two values of $ n $. For the remaining $ n $ we have
Dividing the numerator and the denominator of the obtained fraction by $ n $ and passing to the limit as $ n $ goes to in... | proof | Algebra | proof | Yes | Yes | olympiads | false | 520 |
XXI OM - III - Task 1
Diameter $ \overline{AB} $ divides the circle into two semicircles. On one semicircle, n points $ P_1 P_2, \ldots, P_n $ are chosen such that $ P_1 $ lies between $ A $ and $ P_2 $, $ P_2 $ lies between $ P_1 $ and $ P_3 $, $ \ldots $, $ P_n $ lies between $ P_{n-1} $ and $ B $. How should point ... | Notice that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is equal to the sum of the areas of the polygon $ P_1P_2 \ldots P_n $ and triangle $ CP_1P_n $ (Fig. 13). The first area does not depend on the choice of point $ C $. The second area will be maximal when the distance from p... | C | Geometry | math-word-problem | Yes | Yes | olympiads | false | 523 |
XXXIII OM - II - Task 3
Prove that for every natural number $ n \geq 2 $ the following inequality holds | For non-negative numbers $ a $, $ b $, the inequality
holds.
Therefore, for $ k = 1,2,\ldots,n $ we have
From inequality (*) it also follows that
We obtain the inequalities:
and
Suppose $ n $ is an odd number. By combining the first and last terms, the second and second-to-last terms, etc., in t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 527 |
XLVIII OM - II - Problem 5
We throw $ k $ white and $ m $ black six-sided dice. Calculate the probability that the remainder of the total number of eyes thrown on the white dice divided by $ 7 $ is equal to the remainder of the total number of eyes thrown on the black dice divided by $ 7 $. | The set of elementary events in this task is
where all elementary events are equally probable. Let $ A $ denote the event whose probability we are looking for:
Let $ B $ be the event that the total number of eyes thrown on all $ k + m $ dice is divisible by $ 7 $:
The transformation given by the formu... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 528 |
IV OM - I - Problem 12
Given a line $ p $ and points $ A $ and $ B $ lying on opposite sides of this line. Draw a circle through points $ A $ and $ B $ such that the chord lying on the line $ p $ is as short as possible. | Let $ M $ denote the point of intersection of the lines $ AB $ and $ p $, and let the circle passing through points $ A $ and $ B $ intersect the line $ p $ at points $ C $ and $ D $ (Fig. 35).
The chord $ CD $ is the sum of segments $ CM $ and $ MD $, whose product, by the theorem of intersecting chords of a circle, e... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 529 |
XLII OM - II - Problem 5
$ P_1, P_2, \ldots, P_n $ are distinct two-element subsets of the set $ \{1,2,\ldots,n\} $. Sets $ P_i $, $ P_j $ for $ i\neq j $ have a common element if and only if the set $ \{i,j\} $ is one of the sets $ P_1, P_2, \ldots, P_n $. Prove that each of the numbers $ 1,2,\ldots,n $ is a common e... | For each $ k \in \{1,2,\ldots,n\} $, let $ m_k $ be the number of sets $ P_i $ that contain the element $ k $. The sum of these numbers (as $ k $ ranges from $ 1 $ to $ n $) is $ 2n $, because each of the sets $ P_1,\ldots, P_n $ has two elements and is thus counted twice. We therefore have the equality
\[
\sum_{k=1}^... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 530 |
II OM - I - Task 1
Prove that the product of two factors, each of which is the sum of the squares of two integers, is also the sum of the squares of two integers. | We will apply the transformation
If $ a $, $ b $, $ c $, $ d $ are integers, then $ ac + bd $ and $ ad - bc $ are also integers; the theorem has thus been proven.
Notice that a more general theorem holds:
The product of $ n $ factors, each of which is the sum of the squares of two integers, is also the sum of the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 531 |
XLIV OM - II - Problem 2
Given a circle with center $O$ and a point $P$ lying outside this circle. Through point $P$, we draw a line $l$ intersecting the given circle at points $A$ and $B$. Let $C$ be the point symmetric to $B$ with respect to the line $OP$, and let $m$ be the line passing through points $A$ and $C$. ... | The task does not specify how points $A$ and $B$ are situated on line $l$, meaning which one is closer to point $P$; Figure 5 illustrates these two situations. Let $Q$ be the point of intersection of lines $OP$ and $AC$; the thesis of the task will be proven if we show that the position of point $Q$ on the ray $OP^\to$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 533 |
XXXII - I - Problem 6
Given numbers $ a_1\geq a_2 \geq \ldots \geq a_n \geq 0 $ satisfying the condition $ \sum_{i=1}^n a_i = 1 $. Prove that there exist integers $ k_1\geq k_2 \geq \ldots \geq k_n \geq 0 $ such that | Let $ [x] $ be the greatest integer not greater than $ x $.
Assume $ \widetilde{k_j} = [2na_j] $. The numbers $ \widetilde{k_j} $ satisfy the condition $ \widetilde{k_1} \geq \widetilde{k_2} \geq \ldots \geq \widetilde{k_n} \geq 0 $, and moreover $ 2na_j-1 < \widetilde{k_j} \leq 2na_j $, so $ 2a_j - \frac{1}{n} < \frac... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 535 |
LVIII OM - II - Problem 5
A convex quadrilateral $ABCD$, where $AB \ne CD$, is inscribed in a circle. Quadrilaterals $AKDL$ and $CMBN$ are rhombuses with side lengths of $a$. Prove that points $K$, $L$, $M$, $N$ lie on the same circle. | Since the chords $ AB $ and $ CD $ are of different lengths, the lines $ AD $ and $ BC $ are not parallel. Let's denote their intersection point by $ P $ (Fig. 10). We will show that the points $ K $, $ L $, $ M $, and $ N $ lie on a circle with center $ P $.
The line $ AD $ is the perpendicular bisector of segment $ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 536 |
XXXI - III - Problem 5
In a tetrahedron, the areas of the six triangles, whose sides are the edges and whose vertices are the midpoints of the opposite edges of the tetrahedron, are equal. Prove that the tetrahedron is regular. | The assumption of the task guarantees the equality of the areas of six triangles, each of which has a base being an edge of the tetrahedron, and the opposite vertex is the midpoint of the opposite edge. On the tetrahedron $ABCD$, we describe a parallelepiped $A_1CB_1DAC_1BD_1$, whose each face contains a certain edge o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 537 |
XV OM - III - Task 3
Given is a tetrahedron $ABCD$, whose edges $AB, BC, CD, DA$ are tangent to a certain sphere: Prove that the points of tangency lie in the same plane. | We distinguish two cases:
a) $ AM = CN $, thus also $ AQ = CP $. Triangles $ MBN $ and $ ABC $ are then similar with respect to point $ B $, so $ MN \parallel AC $ and similarly $ QP \parallel AC $. Therefore, $ QP \parallel MN $, which means points $ MNPQ $ lie in the same plane.
b) $ AM \ne CN $, let's say $ AM > CN ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 538 |
XXVI - I - Problem 10
Let $ \alpha $ be an irrational number, $ A_1 $ - a point on the circle $ S $ with center $ O $. Consider the infinite sequence $ A_n $ of points on the circle $ S $, where the point $ A_{k+1} $ is the image of the point $ A_k $ under a rotation about the point $ O $ by an angle $ \alpha m $. Pro... | If for certain natural numbers $k$ and $n$, where $k \ne n$, it was $A_k = A_n$, then the rotations by angles $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would be equal. This means that the numbers $(k-1)\alpha\pi$ and $(n-1)\alpha\pi$ would differ by an integer multiple of $2\pi$, i.e., $(n-k)\alpha\pi = 2m\pi$, where $m$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 539 |
LVI OM - I - Problem 10
Among all subsets of a fixed $ n $-element set $ X $, we sequentially draw with replacement three sets $ A $, $ B $, $ C $. Each time, the probability of drawing any of the $ 2^n $ subsets of set $ X $ is equally likely. Determine the most probable number of elements in the set $ A\cap B\cap C ... | The set of elementary events $\Omega$ consists of all triples $(A,B,C)$, where $A$, $B$, and $C$ are subsets of a given $n$-element set $S$. The selection of each of the $(2^n)^3 = 8^n$ triples $(A,B,C)$ is equally probable. Let $X_k$ $(k = 0,1,2,\ldots,n)$ denote the event that the triple of sets $(A,B,C)$ satisfies t... | [\frac{1}{8}(n+1)] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 542 |
XVI OM - II - Task 5
Prove that a square can be divided into any number of squares greater than 5, but it cannot be divided into 5 squares. | a) First, let us note that having a square divided into $ m $ squares, we can divide one of these squares into four squares by connecting the midpoints of its opposite sides. The entire square will then be divided into $ m + 3 $ squares.
Let $ n $ be a natural number greater than $ 1 $. Divide each side of the square $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 543 |
XLIX OM - II - Problem 2
In triangle $ABC$, angle $BCA$ is obtuse and $\measuredangle BAC = 2\measuredangle ABC$. The line passing through point $B$ and perpendicular to $BC$ intersects line $AC$ at point $D$. Point $M$ is the midpoint of side $AB$. Prove that $\measuredangle AMC = \measuredangle BMD$. | Let a line parallel to $ AB $ and passing through point $ C $ intersect segment $ BD $ at point $ E $. Denote by $ N $ the midpoint of segment $ CE $. Then points $ M $, $ N $, $ D $ are collinear. Triangle $ BCE $ is a right triangle, so $ N $ is the center of the circle circumscribed around it. This implies that tria... | proof | Geometry | proof | Yes | Yes | olympiads | false | 546 |
XXII OM - I - Problem 10
Given is a table with $ n $ rows and $ n $ columns. The number located in the $ m $-th column and $ k $-th row is equal to $ n(k - 1) + m $. How should $ n $ numbers be chosen, one from each row and each column, so that the product of these numbers is the largest? | Let's denote $ a_{k,m} = n (k - 1) + m $, where $ 1 \leq k, m \leq n $. Suppose we select the number in the column with number $ m_i $ from the $ i $-th row, where $ i = 1, 2, \ldots, n $. From the conditions of the problem, it follows that the sequence $ m_1, m_2, \ldots, m_n $ has distinct terms. Each such sequence c... | a_{1n}a_{2n-1}\ldotsa_{n1}=n(2n-1)(3n-2)(4n-3)\ldots(n^2-(n-1)) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 549 |
XXXIX OM - II - Problem 6
A convex polyhedron is given with $ k $ faces $ S_1, \ldots, S_k $. Denote the unit vector perpendicular to the face $ S_i $ ($ i = 1, \ldots, k $) directed outward from the polyhedron by $ \overrightarrow{n_i} $, and the area of this face by $ P_i $. Prove that | We start with the observation that if in space a convex planar polygon $S$ and a plane $\pi$ are given, and if $\overrightarrow{n}$ is a unit vector perpendicular to the plane of the polygon $S$, and $\overrightarrow{w}$ is a unit vector perpendicular to the plane $\pi$, then denoting by $W$ the orthogonal projection o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 552 |
XIX OM - III - Task 4
Given a natural number $ n>2 $. Provide a set of $ n $ natural numbers $ a_1, a_2, \ldots, a_n $, so that in the set of sums
there are as few different numbers as possible, and provide a set of $ n $ natural numbers $ b_1, b_2, \ldots, b_n $, so that in the set of sums
there are as many d... | When $Z$ is a finite set of numbers, we agree to denote by $l(Z)$ the number of different values that the sums of two numbers belonging to $Z$ can take.
1) Let $A$ be a set of $n > 2$ natural numbers $a_i$ ($i = 1,2,\ldots, n$) so labeled that
Since
thus in the set of sums $a_i + a_j$ there are at least $2n - 3$ diff... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 553 |
XXXVIII OM - III - Problem 1
In a square with side 1, there are $ n $ points ($ n > 2 $). Prove that they can be numbered $ P_1, P_2, ..., P_n $ such that \sum_{i=1}^n |P_{i-1}P_i|^2 \leq 4 (we assume $ P_0=P_n $). | The proof will be based on the following lemma:
Lemma. In a right-angled isosceles triangle $ABC$, where $|AB| = |BC| = a$, $|\measuredangle ABC| = 90^\circ$, there are $k$ points ($k \geq 1$). These points can then be numbered $P_1, \ldots, P_k$ such that
(we assume $P_0 = A$, $P_{k+1} = C$).
Proof. We use induction.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 554 |
XXV OM - II - Problem 4
In a convex quadrilateral $ABCD$ with area $S$, each side is divided into 3 equal parts, and segments connecting corresponding division points of opposite sides are drawn in such a way that the quadrilateral is divided into 9 smaller quadrilaterals. Prove that the sum of the areas of the follow... | We will first prove the
Lemma. The segment connecting the corresponding division points of opposite sides of the given quadrilateral intersects with analogous segments connecting the corresponding division points of the remaining sides of the quadrilateral at points that divide this segment into three equal parts.
Proo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 557 |
VII OM - II - Problem 4
Prove that the equation $ 2x^2 - 215y^2 = 1 $ has no solutions in integers. | Let $ x $ and $ y $ denote integers. The number $ 215y^2 $ is divisible by $ 5 $, so the number $ 215y^2 + 1 $ gives a remainder of $ 1 $ when divided by $ 5 $. The number $ x^2 $ has one of the forms $ 5k $, $ 5k + 1 $, $ 5k + 4 $ ($ k $ - an integer, see problem 2), so the number $ 2x^2 $ has one of the forms $ 10k $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 559 |
XXXV OM - I - Problem 10
In the plane, there are $3n$ points, among which no three points are collinear. Prove that there exist $n$ disjoint triangles with vertices at the given points. | Consider all lines, each of which passes through two points of a given set of $3n$ points. There are finitely many such lines, so there exists a line $l$ that is not perpendicular to any of them. Project all the given points perpendicularly onto the line $l$, ensuring that the projections of any two points do not coinc... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 562 |
LIII OM - II - Problem 3
In an $ n $-person association, there are six committees. Each of them includes at least $ n/4 $ people. Prove that there exist two committees and a group of at least $ n/30 $ people, who belong to both of these committees. | Let's number the committees by $1, 2, \ldots, 6$ and denote by $K_i$ the number of members of the $i$-th committee who are not members of any committee with a number less than $i$.
$K_1$ is the number of all members of the first committee, which gives $K_1 \geq n/4$.
Assume that the intersection of any two committees h... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 566 |
V OM - II - Task 6
Prove that if $ x_1, x_2, \ldots, x_n $ are angles contained between $ 0^\circ $ and $ 180^\circ $, and $ n $ is any natural number greater than $ 1 $, then | We will prove a "stronger" theorem, namely, we will show that if $0^\circ < x_i < 180^\circ$ for $i = 1, 2, \ldots, n$, where $n \geq 2$, then
(If $|a| < b$, then $a < b$ (but not vice versa), so inequality (2) implies inequality (1)).
In the proof, we will use known properties of the absolute value:
as well as the f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 569 |
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