Let $f:S\to\mathbb{R}$ be the function from the set of all right triangles into the set of real numbers, defined by $f(\Delta ABC)=\frac{h}{r}$, where $h$ is the height with respect to the hypotenuse and $r$ is the inscribed circle's radius. Find the image, $Im(f)$, of the function.

Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$. [i]Laurențiu Panaitopol[/i]

Let $f:(0;\infty) -- (0;\infty)$ such that $f(x^y)=(f(x))^{f(y)}$. Prove $f(xy)=f(x)f(y)$ and $f(x+y)=f(x)+f(y)$ for all positive real $x,y$.

$ABC$ is a triangle, $G$ its centroid and $A',B',C'$ the midpoints of its sides $BC,CA,AB$, respectively. Prove that if the quadrilateral $AC'GB'$ is cyclic then $AB \cdot CC' = AC \cdot BB'$:

How many positive integers less than 2005 are relatively prime to 1001?

Cauchy the magician has a new card trick. He takes a standard deck(which consists of 52 cards with 13 denominations in each 4 suits) and let Schwartz to shuffle randomly. Schwartz is told to take $ m $ cards not more than $ \frac{1}{3} $ form the top of the deck. Then, Cauchy take $ 18 $ cards one by one from the top of the remaining deck and show it to Schwartz with the second card is placed in front of the first card (from Schwartz view) and so on. He ask Schwartz to memorize the $ m-th $ card when showing the cards. Let it be $ C_1 $. After that, Cauchy places the $ 18 $ cards and the $ m $ cards on the bottom of the deck with the $ m $ cards are placed lower than the $ 18 $ cards. Now, Cauchy distributes and flip the cards on the table from the top of the deck while shouting the numbers $ 10 $ until $ 1 $ with the following operation: a) When a card flipped has the number which is same as the number shouted by Cauchy, stop the distribution and continue with another set.\\ b) When $ 10 $ cards are flipped and none of the cards flipped has the number which is same as the number shouted by Cauchy, take a card from the top of the deck and place it on top of the set with backside(the site which has no value) facing up. Then continue with another set.\\ Cauchy stops when 3 sets of cards are placed. Then, he adds up all the numbers on top of each sets of cards( backside is consider $ 0 $ ). Let $ k $ be the sum. He placed another $ k $ cards to the table from the top of the remaining deck. Finally, he shows the first card on top of the remaining deck to Schwartz. Let it be $ C_2 $. Show that $ C_1 = C_2 $.

A cyclic $n$-gon is divided by non-intersecting (inside the $n$-gon) diagonals to $n-2$ triangles. Each of these triangles is similar to at least one of the remaining ones. For what $n$ this is possible?

In a tetrahedron we know that sum of angles of all vertices is $180^\circ.$ (e.g. for vertex $A$, we have $\angle BAC + \angle CAD + \angle DAB=180^\circ.$) Prove that faces of this tetrahedron are four congruent triangles.

A linear binomial $l(z) = Az + B$ with complex coefficients $A$ and $B$ is given. It is known that the maximal value of $|l(z)|$ on the segment $-1 \leq x \leq 1$ $(y = 0)$ of the real line in the complex plane $z = x + iy$ is equal to $M.$ Prove that for every $z$ \[|l(z)| \leq M \rho,\] where $\rho$ is the sum of distances from the point $P=z$ to the points $Q_1: z = 1$ and $Q_3: z = -1.$

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$? $\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32$ [asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]

Let $(x_1,x_2,\ldots)$ be a sequence of positive real numbers satisfying ${\displaystyle \sum_{n=1}^{\infty}\frac{x_n}{2n-1}=1}$. Prove that $$ \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} \le2. $$ (Proposed by Gerhard J. Woeginger, The Netherlands)

Does there exist an irreducible two variable polynomial $f(x,y)\in \mathbb{Q}[x,y]$ such that it has only four roots $(0,1),(1,0),(0,-1),(-1,0)$ on the unit circle.

The positive integers $ A$, $ B$, $ A \minus{} B$, and $ A \plus{} B$ are all prime numbers. The sum of these four primes is $ \textbf{(A)}\ \text{even} \qquad \textbf{(B)}\ \text{divisible by }3 \qquad \textbf{(C)}\ \text{divisible by }5 \qquad \textbf{(D)}\ \text{divisible by }7 \\ \textbf{(E)}\ \text{prime}$

The number $0$ or $1$ is to be assigned to each of the $n$ vertices of a regular polygon. In how many different ways can this be done (if we consider two assignments that can be obtained one from the other through rotation in the plane of the polygon to be identical)?

Let $\Omega$ be a unit circle and $A$ be a point on $\Omega$. An angle $0 < \theta < 180^\circ$ is chosen uniformly at random, and $\Omega$ is rotated $\theta$ degrees clockwise about $A$. What is the expected area swept by this rotation?

Estimate the value of $e^f$ , where $f = e^e$ .

To each pair of nonzero real numbers $a$ and $b$ a real number $a*b$ is assigned so that $a*(b*c) = (a*b)c$ and $a*a = 1$ for all $a,b,c$. Solve the equation $x*36 = 216$.

The Fibonacci sequence $f_n$ is defined by $f_1=f_2=1$ and $f_{n+2}=f_{n+1}+f_n$ for $n\in\mathbb N$. (a) Show that $f_{1005}$ is divisible by $10$. (b) Show that $f_{1005}$ is not divisible by $100$.

$x_i$ are non-negative reals. $x_1 + x_2 + ...+ x_n = s$. Show that $x_1x_2 + x_2x_3 + ... + x_{n-1}x_n \le \frac{s^2}{4}$.

There are three kinds of things, which are designated respectively by the words (stripped of all common meaning) [i]notes[/i], [i]staves [/i], and [i]heads[/i]. There can be a certain relationship between a note and a head, which is expressed by the saying: they match. Also, a note and a head can match and two different staves can match. Given are the following axioms: (a) If a note and a head each match the same stave, then they match, (b) If two different notes both match with stave B, and also both match with head V, then B and V match, (c) If two staves match, then there is a note that matches both, (d) If a note and a stave are given, then there is a head that matches both. Prove the following theorem, denoting the axiom you apply by its letter. If three staves that differ from each other, each one matches every other, and no note matches any of the three staves, then there is a head that matches all three staves. [hide=original wording] Er zijn drie soorten van dingen, die respectievelijk worden aangeduid met de (van alle gangbare betekenis ontdane) woorden noten, balken en vellen. Tussen een noot en een vel kan een zekere betrekking bestaan die uitgedrukt wordt door de zegswijze: zij passen bij elkaar. Ook kunnen een noot en een vel bij elkaar passen en twee verschillende balken kunnen bij elkaar passen. Gegeven zijn de volgende axioma’s: (a) Als een noot en een vel elk passen bij de zelfde balk, dan passen zij bij elkaar; (b) Als tw’ee verschillende noten beide passen bij balk b, en ook passen bij het vel v, dan passen b en v bij elkaar; (c) Als twee balken bij elkaar passen, dan is er een noot die bij beiden past; (d) Als een noot en een balk zijn gegeven, dan is er een vel dat bij beiden past. Bewijs de volgende stelling en geef daarbij telkens door zijn letter het axioma aan dat U toepast. Als van drie onderling verschillende balken elke past bij elke andere en er geen noot bij de drie balken past, dan is er een vel dat bij alle drie de balken past.[/hide]

There are three eight-digit positive integers which are equal to the sum of the eighth powers of their digits. Given that two of the numbers are $24678051$ and $88593477$, compute the third number. [i]Proposed by Vincent Huang[/i]

Let $f(x)=x^2-2$ and let $f^{(n)}(x)$ denote the $n$-th iteration of $f$. Let $H=\{x:f^{(100)}(x)\leq -1\}$. Find the length of $H$ (the sum of the lengths of the intervals of $H$).

A subset $U \subset R$ is open if for any $x \in U$, there exist real numbers $a, b$ such that $x \in (a, b) \subset U$. Suppose $S \subset R$ has the property that any open set intersecting $(0, 1)$ also intersects $S$. Let $T$ be a countable collection of open sets containing $S$. Prove that the intersection of all of the sets of $T$ is not a countable subset of $R$. (A set $\Gamma$ is countable if there exists a bijective function $f : \Gamma \to Z$.)

Let $P$ be an arbitrary point on the incircle $k$ of triangle $ABC$ with center $I$, different from the points of tangency with its sides. The tangent to $k$ at $P$ intersects the lines $BC$, $AC$, $AB$ at points $A_0$, $B_0$, $C_0$, respectively. The lines through $A_0$, $B_0$, $C_0$, parallel to the bisectors of the angles $\angle BAC$, $\angle ABC$, $\angle ACB$, form a triangle $\Delta$. Prove that the line $PI$ is tangent to the circumcircle of $\Delta$.

For a finite set of naturals $(C)$, the product of its elements is going to be noted $P(C)$. We are going to define $P (\phi) = 1$. Calculate the value of the expression $$\sum_{C \subseteq \{1,2,...,n\}} \frac{1}{P(C)}$$