A set $ S$ of points from the space will be called [b]completely symmetric[/b] if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.

Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.

Let $AA',BB',CC'$ be parallel lines not lying in the same plane. Denote $U$ the intersection of the planes $A'BC,AB'C,ABC'$ and $V$ the intersection of the planes $AB'C',A'BC',A'B'C$. Show that the line $UV$ is parallel with $AA'$.

Given triangle $ABC$ in which $\angle CAB= 30^{\circ}$ and $\angle ACB=60^{\circ}$. On the ray $AB$ a point $D$ is chosen, and on the ray $CB$ a point $E$ is chosen such that $\angle BDE=60^{\circ}$. Lines $AC$ and $DE$ intersect at $F$. Prove that the circumcircle of $AEF$ passes through a fixed point, which is different from $A$ and does not depend on $D$.

Some paper squares of $k$ distinct colors are placed on a rectangular table, with sides parallel to the sides of the table. Suppose that for any $k$ squares of distinct colors, some two of them can be nailed on the table with only one nail. Prove that there is a color such that all squares of that color can be nailed with $2k-2$ nails.

At a meeting of students of the 9th "G" class, it was decided to declare the 9th "G" a presidential republic. Four blocs nominated their candidates for the presidency: “Our Street”, “Our Yard”, “Our House” and “Our Entrance”. When discussing how to select a president four proposals were made. A. “What is there to think about! Have each student place a piece of paper with the name of the candidate they support in the box. Whoever gets the most votes is the president.” B. “No, you can’t do that. If no one gets more than half the votes, a repeat vote must be held, in which the top two from the first vote must participate.” C. “We must choose the one who is better than anyone else. How to do it? Let each person make a list: in the first place on his list he should put the best in his opinion, in the second place - the second, etc. If in most lists B is higher than A, then he is better than B. So, B is better than everyone if he is better than A, better than B and better than D.” D. “Let everyone really make their own list, as V said. For the first place on the list, the candidate receives three points, for the second - 2, for the third - 1 point, and for the fourth - 0. Whoever scores the most points will he’s the president.” As you can see, all four proposed methods are quite democratic. And yet, can it turn out that with method A one candidate wins, with method B another candidate wins, with C a third one, and in option D a fourth one? It is known that there are 29 people in the class, but the applicants (there are four of them) do not participate in the voting. Each student votes strictly according to his list (see speech B). After a heated discussion, option B was adopted. Interestingly, if elections had been held immediately, then after two rounds a representative of the Our Yard bloc would have become president. However, elections were scheduled for a week later. Students belonging to the “Our Yard” bloc, not knowing the true state of affairs, based on the principle “you can’t spoil porridge with butter,” launched a vigorous campaign in support of their candidate. As a result of this agitation, many students did not change their opinion. True, in some lists the position of the representative of “Our Yard” has improved. (All the changes boiled down to the fact that only this candidate’s position improved.) But as a result, another was elected president. How could this happen? [hide=might have typos, here is the original wording] На собрании учеников 9-го «Г» класса было принято решение — объявить 9-й «Г» президентской республикой. Своих кандидатов на пост президента выдвинули четыре блока: «Наша улица», «Наш двор», «Наш дом» и «Наш подъезд». При обсуждении способов выбора президента прозвучало четыре предложения. А. «Чего здесь думать! Пусть каждый ученик опустит в ящик бумажку с фамилией поддерживаемого им претендента. Кто наберет больше голосов, тот и президент.» Б. «Нет, так нельзя. Если никто не наберет больше половины голосов, надо устроить повторное голосование, в котором должны участвовать двое лучших по результатам первого голосования.» В. «Надо выбрать того, кто лучше любого другого. Как это сделать? Пусть каждый человек составит список: на первое место в своем списке он должен поставить самого лучшего по его мнению, на второе — второго и т.д. Если в большинстве списков В стоит выше А, то он лучше В. Значит, В лучше всех, если он лучше А, лучше Б и лучше Г.» Г. «Пусть и в самом деле каждый составит свой список, как сказал В. За первое место в списке кандидат получает три очка, за второе — 2, за третье — 1 очко, а за четвертое — 0. Кто наберет больше всех очков, тот и президент.» Как видим, все четыре предложенных способа вполне демократичны. И все же, может ли получиться так, что при способе А побеждает один кандидат, при способе Б — другой кандидат, при В — третий, ну, а в варианте Г — четвертый? Известно, что в классе 29 человек, но претенденты (их четверо) в голосовании не участвуют. Каждый ученик голосует строго в соответствии со своим списком (см. выступление В). После бурного обсуждения был принят вариант Б. Интересно, что если бы сразу же были проведены выборы, то после двух туров президентом стал бы представитель блока «Наш двор». Однако выборы были назначены на неделю позже. Ученики, входящие в блок «Наш двор», не зная ис тинного положения дел, исходя из принципа «кашу маслом не испортишь», развернули бурную агитацию в поддержку своего кандидата. В результате этой агитации многие ученики никак не изменили своего мнения. Правда, в некоторых списках улучшилось положение представителя «Наш двор». (Все изменения свелись к тому, что улучшилось положение только этого претендента.) Но в результате президентом был избран другой. Как такое могло случиться? [\hide]

Let $a_0=\pi /2$, and let $a_n=\sin (a_{n-1})$ for $n\ge 1$. Determine whether \[ \sum_{n=1}^{\infty}a_n^2 \] converges.

Talithia throws a party on the fifth Saturday of every month that has five Saturdays. That is, if a month has five Saturdays, Talithia has a party on the fifth Saturday of that month, and if a month has four Saturdays, then Talithia does not have a party that month. Given that January $1$, $2010$ was a Friday, compute the number of parties Talithia will have in $2010$.

Addison and Emerson are playing a card game with three rounds. Addison has the cards $1, 3$, and $5$, and Emerson has the cards $2, 4$, and $6$. In advance of the game, both designate each one of their cards to be played for either round one, two, or three. Cards cannot be played for multiple rounds. In each round, both show each other their designated card for that round, and the person with the higher-numbered card wins the round. The person who wins the most rounds wins the game. Let $m/n$ be the probability that Emerson wins, where $m$ and $n$ are relatively prime positive integers. Find $m +n$. [i]Proposed by Ada Tsui[/i]

Let $\triangle ABC$ be a right triangle with $\angle A = 90^\circ$ and $AB = 1.$ Let $x$ be the length that $AC$ must be so that the perpendicular bisector of $AC$ is tangent to the incircle of $\triangle ABC.$ Let $y$ be the length that $BC$ must be so that the perpendicular bisector of $BC$ is tangent to the incircle of $\triangle ABC.$ (Note that $x$ and $y$ arise in different triangles.) Then $x+y=\tfrac{m}{n}$ for positive integers $m, n$ with $m,n$ in simplest form. Compute $m + n.$

The football tournament was played in one round. $3$ points were given for a win, $1$ point for a draw, and $0$ points for a loss. Could it be that the first place team under the old scoring system (win - $2$ points, draw - $1$ point, loss - $0$) would be last?

In the country some mathematicians know each other and any division of them into two sets contain 2 friends from different sets.It is known that if you put any set of four or more mathematicians at a round table so that any two neighbours know each other , then at the table there are two friends not sitting next to each other.We denote by $c_i $ the number of sets of $i$ pairwise familiar mathematicians(by saying "familiar" it means know each other).Prove that $c_1-c_2+c_3-c_4+...=1$

A triangle $ ABC$ has integer sides, $ \angle A\equal{}2 \angle B$ and $ \angle C>90^{\circ}$. Find the minimum possible perimeter of this triangle.

Does there exist a non-negative integer n, such that the first four digits of n! is 1993?

A pentagon $ABCDE$ inscribed in a circle for which $BC<CD$ and $AB<DE$ is the base of a pyramid with vertex $S$. If $AS$ is the longest edge starting from $S$, prove that $BS>CS$.

You have $\text{5}$ different strings with weights tied at various point, all hanging from the ceiling, and reaching down to the floor. The string is released at the top, allowing the weights to fall. Which one will create a regular, uniform beating sound as the weights hit the floor? [asy] size(300); // (A) bar picture bar; draw(bar,(0,0)--(0,42)); for (int i=0;i<43;i+=2) { draw(bar,(-2,i)--(-3,i)); } add(bar); picture ball; filldraw(ball,circle((0,0),0.5),gray); add(ball); add(shift(12*up)*ball); add(shift(22*up)*ball); add(shift(30*up)*ball); add(shift(36*up)*ball); add(shift(40*up)*ball); add(shift(42*up)*ball); label(scale(0.75)*"(A)",(-1,0),2*S); // (B) bar add(shift(15*right)*bar); add(shift(15*right)*ball); add(shift(6*up)*shift(15*right)*ball); add(shift(12*up)*shift(15*right)*ball); add(shift(18*up)*shift(15*right)*ball); add(shift(24*up)*shift(15*right)*ball); add(shift(30*up)*shift(15*right)*ball); add(shift(36*up)*shift(15*right)*ball); add(shift(42*up)*shift(15*right)*ball); label(scale(0.75)*"(B)",(14,0),2*S); // (C) bar add(shift(30*right)*bar); add(shift(30*right)*ball); add(shift(2*up)*shift(30*right)*ball); add(shift(6*up)*shift(30*right)*ball); add(shift(12*up)*shift(30*right)*ball); add(shift(20*up)*shift(30*right)*ball); add(shift(30*up)*shift(30*right)*ball); add(shift(42*up)*shift(30*right)*ball); label(scale(0.75)*"(C)",(29,0),2*S); // (D) bar add(shift(45*right)*bar); add(shift(45*right)*ball); add(shift(2*up)*shift(45*right)*ball); add(shift(8*up)*shift(45*right)*ball); add(shift(18*up)*shift(45*right)*ball); add(shift(32*up)*shift(45*right)*ball); label(scale(0.75)*"(D)",(44,0),2*S); // (E) bar add(shift(60*right)*bar); add(shift(60*right)*ball); add(shift(2*up)*shift(60*right)*ball); add(shift(10*up)*shift(60*right)*ball); add(shift(28*up)*shift(60*right)*ball); label(scale(0.75)*"(E)",(59,0),2*S); [/asy]

Let there be $n$ students and $m$ clubs. The students joined the clubs so that the following is true: - For all students $x$, you can choose some clubs such that $x$ is the only student who joined all of the chosen clubs. Let the number of clubs each student joined be $a_1,a_2,...,a_m$. Prove that $$a_1!(m - a_1)! + a_2!(m - a_2)! + ... + a_n!(m -a_n)! \le m!$$

Consider two triangles, $ABC$ and $DEF$, and any point $O$. We take any point $X$ in $\vartriangle ABC$ and any point $Y$ in $\vartriangle DEF$ and draw a parallelogram $OXY Z$. Prove that the locus of all possible points $Z$ form a polygon. How many sides can it have? Prove that its perimeter is equal to the sum of perimeters of the original triangles.

Fernanda decided to decorate a square blanket with a ribbon and buttons, placing a button in the center of each square where the ribbon passes and forming the design indicated in the figure. If Fernanda sews the first button in the shaded square on line $0$, on which line does she sew the $2007$th button? [img]https://cdn.artofproblemsolving.com/attachments/2/9/0c9c85ec6448ee3f6f363c8f4bcdd5209f53f6.png[/img]

An equilateral triangle $ABC$ is inscribed in a circle $\Omega$ and has incircle $\omega$. Points $P$ and $Q$ are in segments $AC$ and $AB$, respectively, such that $PQ$ is tangent to $\omega$. The circle $\Omega_B$ has center $P$ and radius $PB$ and the circle $\Omega_C$ is defined similarly. Prove that $\Omega$, $\Omega_B$ and $\Omega_C$ have a common point.

Solve the system $\begin{cases} 10x_1 + 3x_2 + 4x_3 + x_4 + x_5 = 0 \\ 11x_2 + 2x_3 + 2x_4 + 3x_5 + x_6 = 0 \\ 15x_3 + 4x_4 + 5x_5 + 4x_6 + x_7 = 0 \\ 2x_1 + x_2 - 3x_3 + 12x_4 - 3x_5 + x_6 + x_7 = 0 \\ 6x_1 - 5x_2 + 3x_3 - x_4 + 17x_5 + x_6 = 0 \\ 3x_1 + 2x_2 - 3x_3 + 4x_4 + x_5 - 16x_6 + 2x_7 = 0\\ 4x_1 - 8x_2 + x_3 + x_4 + 3x_5 + 19x_7 = 0 \end{cases}$

In triangle $ABC$ the bisector of angle $A$, the perpendicular to side $AB$ from its midpoint, and the altitude from vertex $B$, intersect in the same point. Prove that the bisector of angle $A$, the perpendicular to side $AC$ from its midpoint, and the altitude from vertex $C$ also intersect in the same point.

$3m$ balls numbered $1, 1, 1, 2, 2, 2, 3, 3, 3, \ldots, m, m, m$ are distributed into $8$ boxes so that any two boxes contain identical balls. Find the minimal possible value of $m$.

An ant lies on each corner of a $20 \times 23$ rectangle. Each second, each ant independently and randomly chooses to move one unit vertically or horizontally away from its corner. After $10$ seconds, find the expected area of the convex quadrilateral whose vertices are the positions of the ants.

Let $ABC$ be a triangle with $\angle ABC = 60^{\circ}$. Find the angles of the triangle if $\angle BHI = 60^{\circ}$, where $H$ and $I$ are the orthocenter and incenter of $ABC$