problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 5
values | source stringclasses 11
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|---|---|---|---|---|---|---|---|---|
97.2. Let $A B C D$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that the areas of the triangles $A B P, B C P, C D P$, and $D A P$ are equal. Show that at least one of the diagonals of the quadrilateral bisects the other diagonal.
 We first assume that $P$ does not lie on the diagonal $A C$ and the line $B P$ meets the diagonal $A C$ at $M$. Let $S$ and $T$ be the feet of the perpendiculars from $A$ and $C$ on the line $B P$. The triangles $A P B$ and $C B P$ have equal area. Thus $A S=C T$. If $S \neq T$, then the righ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
97.3. Let $A, B, C$, and $D$ be four different points in the plane. Three of the line segments $A B, A C, A D, B C, B D$, and $C D$ have length a. The other three have length $b$, where $b>a$. Determine all possible values of the quotient $\frac{b}{a}$.
|
Solution. If the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $B C D$, and
$$
\frac{b}{a}=\frac{b... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
97.4. Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2 x)=2 f(x), f(4 x+1)=4 f(x)+3$, and $f(4 x-1)=2 f(2 x-1)-1$. Show that $f$ is an injection, i.e. if $f(x)=f(y)$, then $x=y$.
|
Solution. If $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \equiv 1 \bmod 4$, then $f(x) \equiv 3 \bmod 4$, and if $x \equiv 3 \bmod 4$, then $f(x) \equiv 1 \bmod 4$. Clearly $f(0)=0, f(1)=3, f(2)=6$, and $f(3)=5$. So at least $f$ restricted to the set $\{0,1,2,3\}$ ia an inj... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
98.1. Determine all functions $f$ defined in the set of rational numbers and taking their values in the same set such that the equation $f(x+y)+f(x-y)=2 f(x)+2 f(y)$ holds for all rational numbers $x$ and $y$.
|
Solution. Insert $x=y=0$ in the equation to obtain $2 f(0)=4 f(0)$, which implies $f(0)=0$. Setting $x=0$, one obtains $f(y)+f(-y)=2 f(y)$ of $f(-y)=f(y)$. Then assume $y=n x$, where $n$ is a positive integer. We obtain
$$
f((n+1) x)=2 f(x)+2 f(n x)-f((n-1) x)
$$
In particular, $f(2 x)=2 f(x)+2 f(x)-f(0)=4 f(x)$ and... | f(x)=^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
98.2. Let $C_{1}$ and $C_{2}$ be two circles intersecting at $A$ and $B$. Let $S$ and $T$ be the centres of $C_{1}$ and $C_{2}$, respectively. Let $P$ be a point on the segment $A B$ such that $|A P| \neq|B P|$ and $P \neq A, P \neq B$. We draw a line perpendicular to $S P$ through $P$ and denote by $C$ and $D$ the po... |
Solution. (See Figure 10.) The power of the point $P$ with respect to the circles $C_{1}$ and $C_{2}$ is $P A \cdot P B=P C \cdot P D=P E \cdot P F$. Since $S P$ is perpendicular to the chord $C D, P$
 For which positive numbers $n$ does there exist a sequence $x_{1}, x_{2}, \ldots, x_{n}$, which contains each of the numbers 1, 2, ..., $n$ exactly once and for which $x_{1}+x_{2}+\cdots+x_{k}$ is divisible by $k$ for each $k=1,2, \ldots, n$ ?
(b) Does there exist an infinite sequence $x_{1}, x_{2}, x_{3}, ... |
Solution. (a) We assume that $x_{1}, \ldots, x_{n}$ is the sequence required in the problem. Then $x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}$. This sum should be divisible by $n$. If $n$ is odd, this is possible, since $\frac{(n+1)}{2}$ is an integer. If, on the other hand, $n=2 m$, then $\frac{n(n+1)}{2}=m(2 m+1)=$ $... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
98.4. Let $n$ be a positive integer. Count the number of numbers $k \in\{0,1,2, \ldots, n\}$ such that $\binom{n}{k}$ is odd. Show that this number is a power of two, i.e. of the form $2^{p}$ for some nonnegative integer $p$.
|
Solution. The number of odd binomial coefficients $\binom{n}{k}$ equals the number of ones on the $n$ :th line of the Pascal Triangle $\bmod 2$ :
(We count the lines so that the uppermost l... | 2^{e(n)} | Combinatorics | proof | Yes | Yes | olympiads | false |
99.1. The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text { if } n \leq 1999 \\ n-5, & \text { if } n>1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$.
|
Solution. If $n \geq 2005$, then $f(n)=n-5 \geq 2000$, and the equation $f(n)=1999$ has no solutions. Let $1 \leq k \leq 4$. Then
$$
\begin{gathered}
2000-k=f(2005-k)=f(f(2010-k)) \\
=f(1999-k)=f(f(2004-k))=f(1993-k)
\end{gathered}
$$
Let $k=1$. We obtain three solutions $1999=f(2004)=f(1998)=f(1992)$. Moreover, $19... | 1999=f(6n),ifonlyifn=1,2,\ldots,334 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
99.2. Consider 7-gons inscribed in a circle such that all sides of the 7-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a 7-gon.
|
Solution. It is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\angle A B C=\angle B C D=120^{\circ}$, and arc $B C$ equals $b^{\circ}$, then arcs $A B$ and... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
99.3. The infinite integer plane $\mathbb{Z} \times \mathbb{Z}=\mathbb{Z}^{2}$ consists of all number pairs $(x, y)$, where $x$ and $y$ are integers. Let $a$ and $b$ be non-negative integers. We call any move from a point $(x, y)$ to any of the points $(x \pm a, y \pm b)$ or $(x \pm b, y \pm a) a(a, b)$-knight move. D... |
Solution. If the greatest common divisor of $a$ and $b$ is $d$, only points whose coordinates are multiples of $d$ can be reached by a sequence of $(a, b)$-knight moves starting from the origin. So $d=1$ is a necessary condition for the possibility of reaching every point in the integer plane. In any $(a, b)$-knight m... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
99.4. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers and $n \geq 1$. Show that
$$
\begin{aligned}
& n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) \\
& \quad \geq\left(\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}}\right)\left(n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}
$$
When does e... |
Solution. The inequality of the problem can be written as
$$
\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}} \leq \frac{n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)}{n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}
$$
A small manipulation of the right hand side brings the inequality to the equivalent form
$$
\frac{1}{\... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)
|
Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such t... | 333333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
00.2. The persons $P_{1}, P_{1}, \ldots, P_{n-1}, P_{n}$ sit around a table, in this order, and each one of them has a number of coins. In the start, $P_{1}$ has one coin more than $P_{2}, P_{2}$ has one coin more than $P_{3}$, etc., up to $P_{n-1}$ who has one coin more than $P_{n}$. Now $P_{1}$ gives one coin to $P_... |
Solution. Assume that $P_{n}$ has $m$ coins in the start. Then $P_{n-1}$ has $m+1$ coins, ... and $P_{1}$ has $m+n-1$ coins. In every move a player receives $k$ coins and gives $k+1$ coins away, so her net loss is one coin. After the first round, when $P_{n}$ has given $n$ coins to $P_{1}$, $P_{n}$ has $m-1$ coins, $P... | 6or63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
00.3. In the triangle $A B C$, the bisector of angle $B$ meets $A C$ at $D$ and the bisector of angle $C$ meets $A B$ at $E$. The bisectors meet each other at $O$. Furthermore, $O D=O E$. Prove that either $A B C$ is isosceles or $\angle B A C=60^{\circ}$.
|
Solution. (See Figure 11.) Consider the triangles $A O E$ and $A O D$. They have two equal pairs of sides and the angles facing one of these pairs are equal. Then either $A O E$ and $A O D$ are congruent or $\angle A E O=180^{\circ}-\angle A D O$. In the first case, $\angle B E O=\angle C D O$, and
=0, f(1)=1$, and
$$
\frac{1}{2} \leq \frac{f(z)-f(y)}{f(y)-f(x)} \leq 2
$$
for all $0 \leq x<y<z \leq 1$ with $z-y=y-x$. Prove that
$$
\frac{1}{7} \leq f\left(\frac{1}{3}\right) \leq \frac{4}{7}
$$
|
Solution. We set $f\left(\frac{1}{3}\right)=a$ and $f\left(\frac{2}{3}\right)=b$. Applying the inequality of the problem for $x=\frac{1}{3}, y=\frac{2}{3}$ and $z=1$, as well as for $x=0, y=\frac{1}{3}$, and $z=\frac{2}{3}$, we obtain
$$
\frac{1}{2} \leq \frac{1-b}{b-a} \leq 2, \quad \frac{1}{2} \leq \frac{b-a}{a} \l... | \frac{1}{7}\leq\leq\frac{4}{7} | Inequalities | proof | Yes | Yes | olympiads | false |
01.1. Let $A$ be a finite collection of squares in the coordinate plane such that the vertices of all squares that belong to $A$ are $(m, n),(m+1, n),(m, n+1)$, and $(m+1, n+1)$ for some integers $m$ and $n$. Show that there exists a subcollection $B$ of $A$ such that $B$ contains at least $25 \%$ of the squares in $A... |
Solution. Divide the plane into two sets by painting the strips of squares parallel to the $y$ axis alternately red and green. Denote the sets of red and green squares by $R$ and $G$, respectively. Of the sets $A \cap R$ and $A \cap G$ at least one contains at least one half of the squares in $A$. Denote this set by $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
01.2. Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(... |
Solution. Let $g(6 x)=f(x)$. Then $g$ is bounded, and
$$
\begin{gathered}
g(t+2)=f\left(\frac{t}{6}+\frac{1}{3}\right), \quad g(t+3)=f\left(\frac{t}{6}+\frac{1}{2}\right) \\
g(t+5)=f\left(\frac{t}{6}+\frac{5}{6}\right), \quad g(t+2)+g(t+3)=g(t)+g(t+5) \\
g(t+5)-g(t+3)=g(t+2)-g(t)
\end{gathered}
$$
for all real numbe... | proof | Algebra | proof | Yes | Yes | olympiads | false |
01.3. Determine the number of real roots of the equation
$$
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0
$$
|
Solution. Write
$$
\begin{gathered}
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2} \\
=x(x-1)\left(x^{6}+2 x^{4}+3 x^{2}+4\right)+\frac{5}{2}
\end{gathered}
$$
If $x(x-1) \geq 0$, i.e. $x \leq 0$ or $x \geq 1$, the equation has no roots. If $0x(x-1)=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4} \ge... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
01.4. Let $A B C D E F$ be a convex hexagon, in which each of the diagonals $A D, B E$, and $C F$ divides the hexagon in two quadrilaterals of equal area. Show that $A D, B E$, and $C F$ are concurrent.
 Denote the area of a figure by $|\cdot|$. Let $A D$ and $B E$ intersect at $P, A D$ and $C F$ at $Q$, and $B E$ and $C F$ at $R$. Assume that $P, Q$, and $R$ are different. We may assume that $P$ lies between $B$ and $R$, and $Q$ lies between $C$ and $R$. Both $|A B P|$ and $|D E P|$ differ ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
02.1. The trapezium $A B C D$, where $A B$ and $C D$ are parallel and $A D<C D$, is inscribed in the circle c. Let DP be a chord of the circle, parallel to AC. Assume that the tangent to $c$ at $D$ meets the line $A B$ at $E$ and that $P B$ and $D C$ meet at $Q$. Show that $E Q=A C$.
|
Solution. (See Figure 13.) since $A D<C D, \angle P D C=\angle D C A<\angle D A C$. This implies that arc $C P$ is smaller than arc $C D$, and $P$ lies on that arc $C D$ which does not include $A$ and $B$. We show that the triangles $A D E$ and $C B Q$ are congruent. As a trapezium inscribed in a circle, $A B C D$ is ... | AC=EQ | Geometry | proof | Yes | Yes | olympiads | false |
02.2. In two bowls there are in total $N$ balls, numbered from 1 to $N$. One ball is moved from one of the bowls to the other. The average of the numbers in the bowls is increased in both of the bowls by the same amount, $x$. Determine the largest possible value of $x$.
|
Solution. Consider the situation before the ball is moved from urn one to urn two. Let the number of balls in urn one be $n$, and let the sum of numbers in the balls in that urn be $a$. The number of balls in urn two is $m$ and the sum of numbers $b$. If $q$ is the number written in the ball which was moved, the condi... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
02.3. Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different.. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
$i=1,2, \ldots, n$, are equal, then the products
$$... |
Solution. Let $P(x)=\left(x+b_{1}\right)\left(x+b_{2}\right) \cdots\left(x+b_{n}\right)$. Let $P\left(a_{1}\right)=P\left(a_{2}\right)=\ldots=P\left(a_{n}\right)=d$. Thus $a_{1}, a_{2}, \ldots, a_{n}$ are the roots of the $n$ :th degree polynomial equation $P(x)-d=0$. Then $P(x)-d=c\left(x-a_{1}\right)\left(x-a_{2}\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false |
02.4. Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by 11. They only look at nine-digit numbers consisting of all the digits 1, 2, .., 9. Anna claims that the probability of such a number to be a multiple of 11 is exactly 1/11. Eva has... |
Solution. We write the numbers in consideration, $n=a_{0}+10 a_{1}+10^{2} a_{2}+\cdots+10^{8} a_{8}$, in the form
$$
\begin{gathered}
a_{0}+(11-1) a_{1}+(99+1) a_{2}+(1001-1) a_{3} \\
+(9999+1) a_{4}+(100001-1) a_{5}+(999999+1) a_{6} \\
\quad+(10000001-1) a_{7}+(99999999+1) a_{8} \\
=\left(a_{0}-a_{1}+a_{2}-a_{3}+a_{... | \frac{11}{126}<\frac{1}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
03.1. Stones are placed on the squares of a chessboard having 10 rows and 14 columns. There is an odd number of stones on each row and each column. The squares are coloured black and white in the usual fashion. Show that the number of stones on black squares is even. Note that there can be more than one stone on a squ... |
Solution. Changing the order of rows or columns does not influence the number of stones on a row, on a column or on black squares. Thus we can order the rows and columns in such a way that the $5 \times 7$ rectangles in the upper left and lower right corner are black and the other two $5 \times 7$ rectangles are white... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
03.2. Find all triples of integers $(x, y, z)$ satisfying
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003
$$
|
Solution. It is a well-known fact (which can be rediscovered e.g. by noticing that the left hand side is a polynomial in $x$ having $-(y+z)$ as a zero) that
$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
&=(x+y+z) \frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{2}
\end{aligne... | (668,668,667) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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
$$
\begin{aligned}
R+B+Y & =27 \\
15 R+3 S+18 Y & =\sum_{j=1}^{27} j=14 \cdot 27=37... | 21,16,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
04.2. Let $f_{1}=0, f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1$, 2, ..., be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms is a c... |
Solution. The Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonacci resi... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
04.3. Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\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 |
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 |
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 |
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
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
05.4. The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathc... |
Solution. (See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are e... | proof | Geometry | proof | Yes | Yes | olympiads | false |
06.2. The real numbers $x, y$ and $z$ are not all equal and satisfy
$$
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 |
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 |
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 |
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 |
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 |
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
$$
2^{\alpha_{1}}, \ldots, 2^{\alpha_{k}}, 5^{\alpha_{1}}, \ldots, 5^{\alpha_... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
07.4. A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and AC. Show that $A P / P C=2 \cdot A B / B... |
Solution. First we show that if we fix the points $A, B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_{1}$ and $T_{1}$, and the tangents to the other circle,... | \frac{AP}{PC}=2\cdot\frac{AB}{BC} | Geometry | proof | Yes | Yes | olympiads | false |
08.1. Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies
$$
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
$$
so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(... | (A,B,C)=(,^2,)where\neq-1arbitrary,or(A,B,C)=(-1,1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
08.2. Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if
(i) $M$ and $N$ do not sit on adjacent seats, and
(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alpha... |
Solution. We will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \geq 3$. If $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs. Assume that the number of dominating pairs is $n-3$ for some $n>3$. If there are $n... | n-3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
08.3. Let $A B C$ be a triangle and let $D$ and $E$ be points on $B C$ and $C A$, respectively, such that $A D$ and $B E$ are angle bisectors of $A B C$. Let $F$ and $G$ be points on the circumcircle of $A B C$ such that $A F$ and $D E$ are parallel and $F G$ and $B C$ are parallel. Show that
$$
\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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
11.4. Show that for any integer $n \geq 2$ the sum of the fractions $\frac{1}{a b}$, where $a$ and $b$ are relatively prime positive integers such that $an$, equals $\frac{1}{2}$.
|
Solution. We prove this by induction. First observe that the statement holds for $n=2$, because $a=1$ and $b=2$ are the only numbers which satisfy the conditions. Next we show that increasing $n$ by 1 does not change the sum, so it remains equal to $\frac{1}{2}$. To that end it suffices to show that the sum of the ter... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
## 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 |
## 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 |
## 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)
$$
Thus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\frac{1}{a} x$, since $a=0$ obviously does not give a solutio... | \frac{1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem 4
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that ... |
Solution 4 Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands ... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem 1
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
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 |
## Problem 2
In Wonderland, the towns are connected by roads, and whenever there is a direct road between two towns there is also a route between these two towns that does not use that road. (There is at most one direct road between any two towns.) The Queen
of Hearts ordered the Spades to provide a list of all "even"... |
Solution. The answer is $\frac{1}{2} n x$.
Proof: We reformulate the problem in terms of graph theory with the towns being vertices and the roads being edges of a graph $G=(V, E)$. The given information implies that every edge $e \in E$ is part of a cycle. The subgraphs to be counted are those with every valence even... | \frac{1}{2}nx | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
## 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 |
## 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 |
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
$$
|
Solution 1 Let $P(x)=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.
| proof | Number Theory | proof | Yes | Yes | olympiads | false |
Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
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 |
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 |
## Problem 4
A game is played on an $n \times n$ chessboard. At the beginning there are 99 stones on each square. Two players $A$ and $B$ take turns, where in each turn the player chooses either a row or a column and removes one stone from each square in the chosen row or column. They are only allowed to choose a row ... | ## Solution 1
It is easily seen that both $f(x)=x$ and $f \equiv 0$ solve the equation; we shall show that there are no other solutions.
Setting $x=y=0$ gives $f(0)=0$; if only $y=0$ we get $f\left(x^{2}\right)=(f(x))^{2}$, for all admissible $x$. For $x=1$ we now get $f(1)=0$, or $f(1)=1$.
Case 1. $f(1)=0$ : We hav... | notfound | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
Problem 4 Let $n$ be an integer with $n \geq 3$ and assume that $2 n$ vertices of a regular $(4 n+1)$-gon are coloured. Show that there must exist three of the coloured vertices forming an isosceles triangle.
|
Solution 1 a) Notice that $\left\{2019!\cdot 1^{2019!}, 2019!\cdot 2^{2019!}, \ldots, 2019!\cdot 2019^{2019!}\right\}$ is such a set. Observe that if all the elements are divisible by 2019! then the arithmetic means will be integer for all the subsets. Also, if $A$ is a set such that the geometric means are integer fo... | proof | Combinatorics | proof | Yes | Incomplete | olympiads | false |
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
$$
2 a^{2}-b^{2}-c^{2}=2\left(2 c^{2}-b^{2}\right)-b^{2}-c^{2}=3 c^{2}-3 b... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 2. Given a triangle $A B C$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $B C$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $A B$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$... |
Solution I. There are three possibilities: $Q$ between $A$ and $B, Q=B$, and $B$ between $A$ and $Q$. If $Q=B$ we have that $\angle A B P$ is right, and $A P$ is a diameter
of the circumcircle. The triangles $A B P$ and $A C P$ are then congruent (they have $A P$ in common, $P B=P C$, and both have a right angle oppos... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
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 |
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 |
## 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 |
## 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 |
## 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)
$$
We know that $p(2015)=2015$. Prove that $q$ has at least 1970 different roots $r_{1}, \ldots, r_{1970}$ such... |
Solution. Let $h_{j}(x)=p(x+j)$. Consider $h_{2015}$. Like $p$, it has $n$ real roots $s_{1}, s_{2}, \ldots, s_{n}$, and $h_{2015}(0)=p(2015)=2015$. By Viète, the product $\left|s_{1} s_{2} \cdots s_{n}\right|$ equals 2015. Since $n \geq 2$, there is at least one $s_{j}$ such that $\left|s_{j}\right| \leq \sqrt{2015}<... | proof | Algebra | proof | Yes | Yes | olympiads | false |
## Problem 4.
An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For... |
Solution 1. (By the proposer.) Let the positions of the books in the stack be $1,2,3, \ldots, 2000$ from the top (and consider them modulo 2000). Notice that both operations fix the parity of the number of the book at a any given position. Operation (i) subtracts an even integer from the number of the book at each pos... | (1000!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible sco... |
Solution. Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is
$$
k+(k+1)+\cdots+(k+n-1)=n k+\frac{(n-1) n}{2}
$$
Some games must end in a tie, for otherwise, all team scores wou... | n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Define a sequence $\left(n_{k}\right)_{k \geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2 k}=n_{k}+n_{k-1}$ and $n_{2 k+1}=n_{k}$ for $k \geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \geq 1$. Show that every positive rational number is present exactly once in the sequence $\left(q_{k}\right)_{k \geq 1}$.
|
Solution. Clearly, all the numbers $n_{k}$ are positive integers. Moreover,
$$
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 |
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 |
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 |
XXI OM - II - Task 5
Given is the polynomial $ P(x) = \frac{1}{2} - \frac{1}{3}x + \frac{1}{6}x^2 $. Let $ Q(x) = \sum_{k=0}^{m} b_k x^k $ be a polynomial defined by the formula
Calculate $ \sum_{k=0}^m |b_k| $. | We will use the following obvious facts:
(1) The product of polynomials with positive coefficients is a polynomial with positive coefficients.
(2) The sum of the coefficients of a polynomial $ f(x) $ is equal to $ f(1) $.
(3) The absolute values of the corresponding coefficients of the polynomials $ f(x) $ and $ f(-x) ... | 1 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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 |
XXIII OM - III - Problem 3
Prove that there exists a polynomial $ P(x) $ with integer coefficients such that for all $ x $ in the interval $ \langle \frac{1}{10}, \frac{9}{10}\rangle $ the inequality holds. | Consider the polynomial $ f_n(x) = \displaystyle \frac{1}{2} [(2x - 1)^n + 1] $. It obviously has integer coefficients. Moreover, for $ x \in \left< \displaystyle \frac{1}{10};\ \displaystyle \frac{9}{10} \right> $ we have
$ \displaystyle -\frac{4}{5} \leq 2x-1 \leq \frac{4}{5} $ and therefore $ \left| f_n(x) - \displa... | proof | Algebra | proof | Incomplete | Yes | olympiads | false |
X OM - III - Task 2
In an equilateral triangle $ ABC $, a point $ O $ is chosen and perpendiculars $ OM $, $ ON $, $ OP $ are dropped to the sides $ BC $, $ CA $, $ AB $, respectively. Prove that the sum of the segments $ AP $, $ BM $, $ CN $ does not depend on the position of point $ O $. | The theorem we need to prove can be reduced to a known theorem (cf. Seventh Mathematical Olympiad, problem 12), that the sum of the distances from any point inside an equilateral triangle to its vertices equals the height of the triangle.
Let us draw through the vertices $A$, $B$, $C$ of the given triangle lines perpen... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
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 |
XXXIV OM - I - Problem 3
The numbers $ x_1, x_2 $ are roots of the equation $ x^2 + ax + b = 0 $, with integer coefficients, and $ |x_1| < 1 $. Prove that $ \lim_{n\to \infty} \{x_2^n\} = 0 $, where $ \{t\} $ is the number in the interval $ \langle -\frac{1}{2}, \frac{1}{2}\rangle $ such that $ t + \{t\} $ is an integ... | We will first prove two lemmas.
Lemma 1. The perimeter of the union of two externally tangent circles is equal to the perimeter of the circle circumscribed around this union, that is, a circle to which the given circles are internally tangent and whose center is collinear with their centers.
Proof. The perimeter of the... | proof | Algebra | proof | Yes | Incomplete | olympiads | false |
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 |
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 |
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 |
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 |
XIX OM - I - Problem 3
On two skew lines lie segments $ AB $ and $ CD $, whose midpoints are points $ M $ and $ N $, respectively. Prove that | Let $ P $ be the midpoint of segment $ AC $ (Fig. 2). Point $ P $ cannot lie on line $ MN $, because in that case lines $ AB $ and $ CD $ would lie in the plane determined by lines $ AC $ and $ MN $, so they would not be skew. In such a case
In triangle $ ABC $, segment $ MP $ connecting the midpoints of sides $ ... | proof | Geometry | proof | Yes | Incomplete | olympiads | false |
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 |
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 |
XLIII OM - I - Problem 5
Prove that if $ n \geq 2 $ is a natural number, then the number
is irrational. | Let the investigated number be denoted by $x_n$. It is a root of the polynomial
When $n$ changes, we thus have a sequence of polynomials satisfying the recursive relationship
From the definition (1), it is clear that the coefficients of each polynomial $P_n(x)$ are integers, and the coefficient of the highest power o... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
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 |
XVIII OM - III - Task 2
Prove that if points $ A_1, B_1, C_1 $ lying on the sides $ BC, CA, AB $ of triangle $ ABC $ are the orthogonal projections of some point $ P $ of the triangle onto these sides, then | From the right-angled triangles $ APB_1 $, $ BPC_1 $, $ CPA_1 $ (Fig. 11) we obtain the respective equalities
a from the right-angled triangles $ APC_1 $, $ BPA_1 $, $ CPB_1 $ the respective equalities
The sum of the right-hand sides of the equalities (2) is equal to the sum of the right-hand sides of the equalities ... | proof | Geometry | proof | Incomplete | Yes | olympiads | false |
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 |
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