Let $A_1A_2A_3 \cdots A_{2013}$ be a cyclic $2013$-gon. Prove that for every point $P$ not the circumcenter of the $2013$-gon, there exists a point $Q\neq P$ such that $\frac{A_iP}{A_iQ}$ is constant for $i \in \{1, 2, 3, \cdots, 2013\}$. [i]Proposed by Robin Park[/i]

Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, triangle $EUM$ is similar to triangle: [asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ EFA \qquad \textbf{(B)}\ EFC \qquad \textbf{(C)}\ ABM \qquad \textbf{(D)}\ ABU \qquad \textbf{(E)}\ FMC$

The incircle $\odot (I)$ of $\triangle ABC$ touch $AC$ and $AB$ at $E$ and $F$ respectively. Let $H$ be the foot of the altitude from $A$, if $R \equiv IC \cap AH, \ \ Q \equiv BI \cap AH$ prove that the midpoint of $AH$ lies on the radical axis between $\odot (REC)$ and $\odot (QFB)$ I hope that this is not repost :)

Compute \[\sum_{k=1}^{1007}\left(\cos\left(\dfrac{\pi k}{1007}\right)\right)^{2014}.\]

Square $ ABCD$ has side length $ s$, a circle centered at $ E$ has radius $ r$, and $ r$ and $ s$ are both rational. The circle passes through $ D$, and $ D$ lies on $ \overline{BE}$. Point $ F$ lies on the circle, on the same side of $ \overline{BE}$ as $ A$. Segment $ AF$ is tangent to the circle, and $ AF \equal{} \sqrt {9 \plus{} 5\sqrt {2}}$. What is $ r/s$? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$E$",Ep,E); label("$F$",F,NW);[/asy]$ \textbf{(A) } \frac {1}{2}\qquad \textbf{(B) } \frac {5}{9}\qquad \textbf{(C) } \frac {3}{5}\qquad \textbf{(D) } \frac {5}{3}\qquad \textbf{(E) } \frac {9}{5}$

Let X be a dense set homeomorphic to $\mathbb R^n$ in the compact Hausdorff space Y. Prove that for $n\geq 2$ , $Y \setminus X$ is connected, and for n=1 it consists of at most two components.

An airline operates flights between any two capital cities in the European Union. Each flight has a fixed price which is the same in both directions. Furthermore, the flight prices from any given city are pairwise distinct. Anna and Bella wish to visit each city exactly once, not necessarily starting from the same city. While Anna always takes the cheapest flight from her current city to some city she hasn't visited yet, Bella always continues her tour with the most expensive flight available. Is it true that Bella's tour will surely cost at least as much as Anna's tour? [i](Based on a Soviet problem)[/i]

Points $C,E,D$ and $F$ lie on a circle with centre $O$. Two chords $CD$ and $EF$ intersect at a point $N$. The tangents at $C$ and $D$ intersect at $A$, and the tangents at $E$ and $F$ intersect at $B$. Prove that $ON\perp AB$.

Let $f(x)$ be a polynomial of nonzero degree. Can it happen that for any real number $a$, an even number of real numbers satisfy the equation $f(x) = a$?

If $ a > 1$, then the sum of the real solutions of \[\sqrt{a \minus{} \sqrt{a \plus{} x}} \equal{} x\] is equal to $ \textbf{(A)}\ \sqrt{a} \minus{} 1\qquad \textbf{(B)}\ \frac{\sqrt{a} \minus{} 1}{2}\qquad \textbf{(C)}\ \sqrt{a \minus{} 1}\qquad \textbf{(D)}\ \frac{\sqrt{a \minus{} 1}}{2}\qquad \textbf{(E)}\ \frac{\sqrt{4a \minus{} 3} \minus{} 1}{2}$

If the answer to this problem is $x$, then compute the value of $\tfrac{x^2}{8} +2$. [i]Proposed by Lewis Chen [/i]

Find max number $n$ of numbers of three digits such that : 1. Each has digit sum $9$ 2. No one contains digit $0$ 3. Each $2$ have different unit digits 4. Each $2$ have different decimal digits 5. Each $2$ have different hundreds digits

Consider a tetrahedron $SABC$. The incircle of the triangle $ABC$ has the center $I$ and touches its sides $BC$, $CA$, $AB$ at the points $E$, $F$, $D$, respectively. Let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the points on the segments $SA$, $SB$, $SC$ such that $AA^{\prime}=AD$, $BB^{\prime}=BE$, $CC^{\prime}=CF$, and let $S^{\prime}$ be the point diametrically opposite to the point $S$ on the circumsphere of the tetrahedron $SABC$. Assume that the line $SI$ is an altitude of the tetrahedron $SABC$. Show that $S^{\prime}A^{\prime}=S^{\prime}B^{\prime}=S^{\prime}C^{\prime}$.

Let be two natural numbers $ b>a>0 $ and a function $ f:\mathbb{R}\longrightarrow\mathbb{R} $ having the following property. $$ f\left( x^2+ay\right)\ge f\left( x^2+by\right) ,\quad\forall x,y\in\mathbb{R} $$ [b]a)[/b] Show that $ f(s)\le f(0)\le f(t) , $ for any real numbers $ s<0<t. $ [b]b)[/b] Prove that $ f $ is constant on the interval $ (0,\infty ) . $ [b]c)[/b] Give an example of a non-monotone such function.

$ABCD$ is a scalene tetrahedron and let $G$ be its baricentre. A plane $\alpha$ passes through $G$ such that it intersects neither the interior of $\Delta BCD$ nor its perimeter. Prove that $$\textnormal{dist}(A,\alpha)=\textnormal{dist}(B,\alpha)+\textnormal{dist}(C,\alpha)+\textnormal{dist}(D,\alpha).$$ [i] (Adapted from folklore)[/i]

An ant is walking on the edges of an icosahedron of side length $1$. Compute the length of the longest path that the ant can take if it never travels over the same edge twice, but is allowed to revisit vertices. [center]<see attached>[/center]

Let $a_1, a_2, . . . , a_n$ be real numbers. Prove $\sqrt[3]{a_1^3+ a_2^3+ . . . + a_n^3} \le \sqrt{a_1^2+ a_2^2+ . . . + a_n^2} $ (1) When does equality hold in (1)?

a) Factorize $xy - x - y + 1$. b) Prove that if integers $a$ and $b$ satisfy $ |a + b| > |1 + ab|$, then $ab = 0$.

There are 60 empty boxes $B_1,\ldots,B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1\leq k\leq 59$ and splits the boxes into the two groups $B_1,\ldots,B_k$ and $B_{k+1},\ldots,B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning. [i]Czech Republic[/i]

Consider a charged capacitor made with two square plates of side length $L$, uniformly charged, and separated by a very small distance $d$. The EMF across the capacitor is $\xi$. One of the plates is now rotated by a very small angle $\theta$ to the original axis of the capacitor. Find an expression for the difference in charge between the two plates of the capacitor, in terms of (if necessary) $d$, $\theta$, $\xi$, and $L$. Also, approximate your expression by transforming it to algebraic form: i.e. without any non-algebraic functions. For example, logarithms and trigonometric functions are considered non-algebraic. Assume $ d << L $ and $ \theta \approx 0 $. $\emph{Hint}$: You may assume that $ \frac {\theta L}{d} $ is also very small. [i]Problem proposed by Trung Phan[/i] [hide="Clarification"] There are two possible ways to rotate the capacitor. Both were equally scored but this is what was meant: [asy]size(6cm); real h = 7; real w = 2; draw((-w,0)--(-w,h)); draw((0,0)--(0,h), dashed); draw((0,0)--h*dir(64)); draw(arc((0,0),2,64,90)); label("$\theta$", 2*dir(77), dir(77)); [/asy] [/hide]

Let $\mathbb X$ be the set of all bijective functions from the set $S=\{1,2,\cdots, n\}$ to itself. For each $f\in \mathbb X,$ define \[T_f(j)=\left\{\begin{aligned} 1, \ \ \ & \text{if} \ \ f^{(12)}(j)=j,\\ 0, \ \ \ & \text{otherwise}\end{aligned}\right.\] Determine $\sum_{f\in\mathbb X}\sum_{j=1}^nT_{f}(j).$ (Here $f^{(k)}(x)=f(f^{(k-1)}(x))$ for all $k\geq 2.$)

Let $ y$ be the function of $ x$ satisfying the differential equation $ y'' \minus{} y \equal{} 2\sin x$. (1) Let $ y \equal{} e^xu \minus{} \sin x$, find the differential equation with which the function $ u$ with respect to $ x$ satisfies. (2) If $ y(0) \equal{} 3,\ y'(0) \equal{} 0$, then determine $ y$.

In a non-isosceles triangle $ABC$ the centre of the incircle is denoted by $I$. The other intersection point of the angle bisector of $\angle BAC$ and the circumcircle of $\vartriangle ABC$ is $D$. The line through $I$ perpendicular to $AD$ intersects $BC$ in $F$. The midpoint of the circle arc $BC$ on which $A$ lies, is denoted by $M$. The other intersection point of the line $MI$ and the circle through $B, I$ and $C$, is denoted by $N$. Prove that $FN$ is tangent to the circle through $B, I$ and $C$.

Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the splitting line of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^{\circ}$. Find the perimeter of $\triangle ABC$.

Three circles $\omega_1,\omega_2,\omega_3$ are tangent to line $l$ at points $A,B,C$ ($B$ lies between $A,C$) and $\omega_2$ is externally tangent to the other two. Let $X,Y$ be the intersection points of $\omega_2$ with the other common external tangent of $\omega_1,\omega_3$. The perpendicular line through $B$ to $l$ meets $\omega_2$ again at $Z$. Prove that the circle with diameter $AC$ touches $ZX,ZY$. [i]Proposed by Iman Maghsoudi - Siamak Ahmadpour[/i]