In a convex pentagon $ABCDE$ the sides have lengths $1,2,3,4,$ and $5$, though not necessarily in that order. Let $F,G,H,$ and $I$ be the midpoints of the sides $AB$, $BC$, $CD$, and $DE$, respectively. Let $X$ be the midpoint of segment $FH$, and $Y$ be the midpoint of segment $GI$. The length of segment $XY$ is an integer. Find all possible values for the length of side $AE$.

Let $b$, $c$ be integer numbers, and define $f(x)=(x+b)^2-c$. i) If $p$ is a prime number such that $c$ is divisible by $p$ but not by $p^{2}$, show that for every integer $n$, $f(n)$ is not divisible by $p^{2}$. ii) Let $q \neq 2$ be a prime divisor of $c$. If $q$ divides $f(n)$ for some integer $n$, show that for every integer $r$ there exists an integer $n'$ such that $f(n')$ is divisible by $qr$.

Find the smallest possible area of an ellipse passing through $(2,0)$, $(0,3)$, $(0,7)$, and $(6,0)$.

Consider a finite binary string $b$ with at least $2017$ ones. Show that one can insert some plus signs in between pairs of digits such that the resulting sum, when performed in base $2$, is equal to a power of two. [i]Proposed by David Stoner

$O$ is the circumcenter of triangle $ABC$. The bisector from $A$ intersects the opposite side in point $P$. Prove that the following is satisfied: $$AP^2 + OA^2 - OP^2 = bc.$$

Given natural numbers $k$ and $n, k \le n, n \ge 3,$ find the set of all values in the interval $(0, \pi)$ that the $k^{th}-$largest among the interior angles of a convex $n$-gon can take.

We are given a chessboard 100 x 100, $k$ barriers (each with length 1), and one ball. We want to put the barriers between the cells of the board and put the ball in some cell, in such way that the ball can get to each possible cell on the board. The only way that the ball can move is by lifting the board so it can go only forward, backward, to the left or to the right. The ball passes all cells on its way until it reaches a barrier or the edge of the board where it stops. What’s the least number of barriers we need so we can achieve that?

In acute triangle $ABC$, $AB > AC$. Let $M$ be the midpoint of side $BC$. The exterior angle bisector of $\widehat{BAC}$ meet ray $BC$ at $P$. Point $K$ and $F$ lie on line $PA$ such that $MF \perp BC$ and $MK \perp PA$. Prove that $BC^2 = 4 PF \cdot AK$. [asy] defaultpen(fontsize(10)); size(7cm); pair A = (4.6,4), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P); draw(K--M--F--P--B--A--C); pair point = I; pair[] p={A,B,C,M,P,F,K}; string s = "A,B,C,M,P,F,K"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy]

[b]8.[/b] Does there exist a Euclidean ring which is properly contained in the field $V$ of real numbers, and whose quotient field is $V$? [b](A.21)[/b]

Let be a natural number $ a\ge 2. $ Prove that for any choice of primes which has the property that none of them divides any of the numbers $ N_n=1+a+a^2+a^3+\cdots +a^{2n} , $ with natural $ n, $ there is another prime not among this choice which doesn't divide any of the numbers $ N_n. $

Let $M$ and $m$ be, respectively, the greatest and the least ten-digit numbers that are rearrangements of the digits $0$ through $9$ such that no two adjacent digits are consecutive. Find $M - m$.

Prove that we can put $ \Omega(\frac1{\epsilon})$ points on surface of a sphere with radius 1 such that distance of each of these points and the plane passing through center and two of other points is at least $ \epsilon$.

A [i]slab[/i] is the set of points strictly between two parallel planes. Prove that a countable sequence of slabs, the sum of whose thicknesses converges, cannot fill space.

In base $10$, the number $2013$ ends in the digit $3$. In base $9$, on the other hand, the same number is written as $(2676)_9$ and ends in the digit $6$. For how many positive integers $b$ does the base-$b$ representation of $2013$ end in the digit $3$? $\textbf{(A) }6\qquad \textbf{(B) }9\qquad \textbf{(C) }13\qquad \textbf{(D) }16\qquad \textbf{(E) }18\qquad$

Suppose that each of $n$ people knows exactly one piece of information and all $n$ pieces are different. Every time person $A$ phones person $B$, $A$ tells $B$ everything he knows, while tells $A$ nothing. What is the minimum of phone calls between pairs of people needed for everyone to know everything?

A function $f$ from the positive integers to the positive integers is called [i]Canadian[/i] if it satisfies $$\gcd\left(f(f(x)), f(x+y)\right)=\gcd(x, y)$$ for all pairs of positive integers $x$ and $y$. Find all positive integers $m$ such that $f(m)=m$ for all Canadian functions $f$.

Given real numbers $a,b,c,d,e>1$. Prove that \[ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge 20 \]

Let $ABC$ be a triangle, and let $\ell$ be the line through $A$ and perpendicular to the line $BC$. The reflection of $\ell$ in the line $AB$ crosses the line through $B$ and perpendicular to $AB$ at $P$. The reflection of $\ell$ in the line $AC$ crosses the line through $C$ and perpendicular to $AC$ at $Q$. Show that the line $PQ$ passes through the orthocenter of the triangle $ABC$. Flavian Georgescu

Given an expression $$x_1 : x_2 : ... : x_n$$ ( $:$ means division). We can put the braces as we want. How many expressions can we obtain?

Let $\{a_i\}$ be a strictly increasing sequence of positive integers. Define the sequence $\{s_k\}$ as $$s_k = \sum_{i=1}^{k}\frac{1}{[a_i,a_{i+1}]},$$ where $[a_i,a_{i+1}]$ is the least commun multiple of $a_i$ and $a_{i+1}$. Show that the sequence $\{s_k\}$ is convergent.

Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$. (Anton Trygub)

Endre wrote $n$ (not necessarily distinct) integers on a paper. Then for each of the $2^n$ subsets, Kelemen wrote their sum on the blackboard. a) For which values of $n$ is it possible that two different $n$-tuples give the same numbers on the blackboard? b) Prove that if Endre only wrote positive integers on the paper and Ferenc only sees the numbers on the blackboard, then he can determine which integers are on the paper.

$(GBR 3)$ A smooth solid consists of a right circular cylinder of height $h$ and base-radius $r$, surmounted by a hemisphere of radius $r$ and center $O.$ The solid stands on a horizontal table. One end of a string is attached to a point on the base. The string is stretched (initially being kept in the vertical plane) over the highest point of the solid and held down at the point $P$ on the hemisphere such that $OP$ makes an angle $\alpha$ with the horizontal. Show that if $\alpha$ is small enough, the string will slacken if slightly displaced and no longer remain in a vertical plane. If then pulled tight through $P$, show that it will cross the common circular section of the hemisphere and cylinder at a point $Q$ such that $\angle SOQ = \phi$, $S$ being where it initially crossed this section, and $\sin \phi = \frac{r \tan \alpha}{h}$.

Triangle $ ABC$ obeys $ AB = 2AC$ and $ \angle{BAC} = 120^{\circ}.$ Points $ P$ and $ Q$ lie on segment $ BC$ such that \begin{eqnarray*} AB^2 + BC \cdot CP = BC^2 \\ 3AC^2 + 2BC \cdot CQ = BC^2 \end{eqnarray*} Find $ \angle{PAQ}$ in degrees.

A rectangle has sides of length $a$ and $36$. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36$. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2$. [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236 [/asy]