Let $ABC$ be a triangle such that $AC$ is its shortest side. A point $P$ is inside it and satisfies that $BP = AC$. Let $R$ be the midpoint of $BC$ and let $M$ be the midpoint of $AP$. Let $E$ be the intersection of $BP$ and $AC$. Prove that the bisector of angle $\angle BE A$ is perpendicular to segment $MR$.

Let $n\ge 2$ be an integer. $a_1,a_2,\dotsc,a_n$ are arbitrarily chosen positive integers with $(a_1,a_2,\dotsc,a_n)=1$. Let $A=a_1+a_2+\dotsb+a_n$ and $(A,a_i)=d_i$. Let $(a_2,a_3,\dotsc,a_n)=D_1, (a_1,a_3,\dotsc,a_n)=D_2,\dotsc, (a_1,a_2,\dotsc,a_{n-1})=D_n$. Find the minimum of $\prod\limits_{i=1}^n\dfrac{A-a_i}{d_iD_i}$

Eddy is traveling to England and needs to exchange USD to GBP (US dollars to British pounds). The current exchange rate is $1.3$ USD for $1$ GBP. He exchanges $x$ USD to GBP and while in England, uses $\tfrac{x}{2}$ GBP. When he returns, the value of the British pound has changed so that $1$ GBP equals $\alpha$ USD. After exchanging all his remaining GBP, he notes that he has $\tfrac{x}{2}$ USD left. What is $\alpha$?

A sequence of positive integers $a_1, a_2, \ldots$ satisfies $a_k + a_l = a_m + a_n$ for all positive integers $k,l,m,n$ satisfying $kl = mn$. Prove that if $p$ divides $q$ then $a_p \le a_q$.

$\triangle ABC$ is a triangle such that $\angle A = 60^{\circ}$. The incenter of $\triangle ABC$ is $I$. $AI$ intersects with $BC$ at $D$, $BI$ intersects with $CA$ at $E$, and $CI$ intersects with $AB$ at $F$, respectively. Also, the circumcircle of $\triangle DEF$ is $\omega$. The tangential line of $\omega$ at $E$ and $F$ intersects at $T$. Show that $\angle BTC \ge 60^{\circ}$

Equilateral $ \triangle ABC$ has side length $ 2$, $ M$ is the midpoint of $ \overline{AC}$, and $ C$ is the midpoint of $ \overline{BD}$. What is the area of $ \triangle CDM$? [asy]size(200);defaultpen(linewidth(.8pt)+fontsize(8pt)); pair B = (0,0); pair A = 2*dir(60); pair C = (2,0); pair D = (4,0); pair M = midpoint(A--C); label("$A$",A,NW);label("$B$",B,SW);label("$C$",C, SE);label("$M$",M,NE);label("$D$",D,SE); draw(A--B--C--cycle); draw(C--D--M--cycle);[/asy]$ \textbf{(A)}\ \frac {\sqrt {2}}{2}\qquad \textbf{(B)}\ \frac {3}{4}\qquad \textbf{(C)}\ \frac {\sqrt {3}}{2}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ \sqrt {2}$

A $2$-kg rock is suspended by a massless string from one end of a uniform $1$-meter measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the $0.20$-meter mark? [asy] size(250); draw((0,0)--(7.5,0)--(7.5,0.2)--(0,0.2)--cycle); draw((1.5,0)--(1.5,0.2)); draw((3,0)--(3,0.2)); draw((4.5,0)--(4.5,0.2)); draw((6,0)--(6,0.2)); filldraw((1.5,0)--(1.2,-1.25)--(1.8,-1.25)--cycle, gray(.8)); draw((0,0)--(0,-0.4)); filldraw((0,-0.4)--(-0.05,-0.4)--(-0.1,-0.375)--(-0.2,-0.375)--(-0.3,-0.4)--(-0.3,-0.45)--(-0.4,-0.6)--(-0.35,-0.7)--(-0.15,-0.75)--(-0.1,-0.825)--(0.1,-0.84)--(0.15,-0.8)--(0.15,-0.75)--(0.25,-0.7)--(0.25,-0.55)--(0.2,-0.4)--(0.1,-0.35)--cycle, gray(.4)); [/asy] $ \textbf {(A) } 0.20 \, \text{kg} \qquad \textbf {(B) } 1.00 \, \text{kg} \qquad \textbf {(C) } 1.33 \, \text{kg} \qquad \textbf {(D) } 2.00 \, \text{kg} \qquad \textbf {(E) } 3.00 \, \text{kg} $

In the Cartesian plane, each point with integer coordinates is colored either red, green, or blue. It is possible to form right isosceles triangles ($45^\circ - 90^\circ - 45^\circ$) using colored points as vertices. Prove that regardless of how the coloring is done, there always exists a right isosceles triangle such that all its vertices are either the same color or all different colors.

A circle passing through vertices $A$ and $B$ of triangle $ABC$ intersects the sides $AC$ and $BC$ again at points $P$ and $Q$, respectively. Given that the median from vertex $C$ bisect the arc $PQ$ of the circle. Prove that $ABC$ is an isosceles triangle.

For each integer $N>1$, there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by $N$. If $69,90,$ and $125$ are congruent in one such system, then in that same system, $81$ is congruent to $\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }8$

How many positive whole numbers less than $100$ are divisible by $3$, but not by $2$?

A cube with the edge of length $n$ is divided onto $n^3$ unit ones. Let us choose some of them and draw three lines parallel to the edges through their centres. What is the least possible number of the chosen small cubes necessary to make those lines cross all the smaller cubes? a) Find the answer for the small $n$ ($n = 2,3,4$). b) Try to find the answer for $n = 10$. c) If You can not solve the general problem, try to estimate that value from the upper and lower side. d) Note, that You can reformulate the problem in such a way: Consider all the triples $(x_1,x_2,x_3)$, where $x_i$ can be one of the integers $1,2,...,n$. What is the minimal number of the triples necessary to provide the property: [i]for each of the triples there exist the chosen one, that differs only in one coordinate. [/i] Try to find the answer for the situation with more than three coordinates, for example, with four.

$\textbf{(Lucas Numbers)}$ The Lucas numbers are defined by $L_0 = 2$, $L_1 = 1$, and $L_{n+2} = L_{n+1} + L_n$ for every $n \ge 0$. There are $N$ integers $1 \le n \le 2016$ such that $L_n$ contains the digit $1$. Estimate $N$. An estimate of $E$ earns $\left\lfloor 20 - 2|N-E| \right\rfloor$ or $0$ points, whichever is greater.

Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P + \lambda Q$ is the square of a polynomial. [i]Alison Miller[/i]

Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point. [i](Kritesh Dhakal, Nepal)[/i]

Pentagon $ABCDE$ is inscribed in a circle such that $ACDE$ is a square with area $12$. What is the largest possible area of pentagon $ABCDE$? $\text{(A) }9+3\sqrt{2}\qquad\text{(B) }13\qquad\text{(C) }12+\sqrt{2}\qquad\text{(D) }14\qquad\text{(E) }12+\sqrt{6}-\sqrt{3}$

Let $ABC$ be a triangle with $AB > AC$. Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.