Let $\vartriangle ABC$ be an equilateral triangle with side $1$. The points $P$, $Q$, $R$ are chosen on the sides of the segments $AB$, $BC$, $AC$ respectively in such a way that $$\frac{AP}{PB}=\frac{BQ}{QC}=\frac{CR}{RA}=\frac25.$$ Find the area of triangle $PQR$. [img]https://cdn.artofproblemsolving.com/attachments/8/4/6184d66bd3ae23db29a93eeef241c46ae0ad44.png[/img]

In a triangle $ABC$, the bisector of the angle at $A$ of a triangle $ABC$ intersects the segment $BC$ and the circumcircle of $ABC$ at points $A_1$ and $A_2$, respectively. Points $B_1,B_2,C_1,C_2$ are analogously defined. Prove that $$\frac{A_1A_2}{BA_2+CA_2}+\frac{B_1B_2}{CB_2+AB_2}+\frac{C_1C_2}{AC_2+BC_2}\ge\frac34.$$

Given is a triangle $ABC$ with side lengths $a, b, c$ ($a = \overline{BC}$, $b = \overline{CA}$, $c = \overline{AB}$) and area $F$. The side $AB$ is extended beyond $A$ by a and beyond $B$ by $b$. Correspondingly, $BC$ is extended beyond $B$ and $C$ by $b$ and $c$, respectively. Eventually $CA$ is extended beyond $C$ and $A$ by $c$ and $a$, respectively. Connecting the outer endpoints of the extensions , a hexagon if formed with area $G$. Prove that $\frac{G}{F}>13$.

The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$, is $15$, and the sum of the squares of the terms of this series is $45$. The first term of the series is $\textbf{(A) }12\qquad\textbf{(B) }10\qquad\textbf{(C) }5\qquad\textbf{(D) }3\qquad \textbf{(E) }2$

Let there be two circles. Find all points $M$ such that there exist two points, one on each circle such that $M$ is their midpoint.

The lines $AD , BE$ and $CF$ intersect in $S$ within a triangle $ABC$ . It is given that $AS: DS = 3: 2$ and $BS: ES = 4: 3$ . Determine the ratio $CS: FS$ . [asy] unitsize (1 cm); pair A, B, C, D, E, F, S; A = (0,0); B = (5,0); C = (1,4); S = (14*A + 15*B + 6*C)/35; D = extension(A,S,B,C); E = extension(B,S,C,A); F = extension(C,S,A,B); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, NE); dot("$E$", E, W); dot("$F$", F, dir(270)); dot("$S$", S, NE); [/asy]

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, \[ \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}. \]

As illustrated below, we can dissect every triangle $ABC$ into four pieces so that piece 1 is a triangle similar to the original triangle, while the other three pieces can be assembled into a triangle also similar to the original triangle. Determine the ratios of the sizes of the three triangles and verify that the construction works. [asy] import olympiad;size(350);defaultpen(linewidth(0.7)+fontsize(10)); path p=origin--(13,0)--(9,8)--cycle; path p2=rotate(180)*p, p3=shift(-26,0)*scale(2)*p, p4=shift(-27,-24)*scale(3)*p, p1=shift(-53,-24)*scale(4)*p; pair A=(-53,-24), B=(-8,16), C=(12,-24), D=(-17,8), E=(-1,-24), F=origin, G=(-13,0), H=(-9,-8); label("1", centroid(A,D,E)); label("2", centroid(F,G,H)); label("3", (-10,6)); label("4", (0,-15)); draw(p2^^p3^^p4); filldraw(p1, white, black); pair point = centroid(F,G,H); label("$\mathbf{A}$", A, dir(point--A)); label("$\mathbf{B}$", B, dir(point--B)); label("$\mathbf{C}$", C, dir(point--C)); label("$\mathbf{D}$", D, dir(point--D)); label("$\mathbf{E}$", E, dir(point--E)); label("$\mathbf{F}$", F, dir(point--F)); label("$\mathbf{G}$", G, dir(point--G)); label("$\mathbf{H}$", H, dir(point--H)); real x=90; draw(shift(x)*p1); label("1", centroid(shift(x)*A,shift(x)*D,shift(x)*E)); draw(shift(130,0)*p4); draw(shift(130,0)*shift(-27,-24)*p); draw(shift(130,0)*shift(-1,-24)*p3); label("2", shift(130,0)*shift(-27,-24)*centroid(F,(9,8),(13,0))); label("3", shift(130,0)*shift(-1,-24)*(-10,6)); label("4", shift(130,0)*(0,-15)); label("Piece 2 rotated $180^\circ$", (130,10));[/asy]

An infinite sequence of positive real numbers $a_1,a_2,a_3,\dots$ is called [i]territorial[/i] if for all positive integers $i,j$ with $i<j$, we have $|a_i-a_j|\ge\tfrac1j$. Can we find a territorial sequence $a_1,a_2,a_3,\dots$ for which there exists a real number $c$ with $a_i<c$ for all $i$?

$T$ is a point on the plane of triangle $ABC$ such that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Prove that the lines symmetric to $AT, BT$ and $CT$ with respect to $BC, CA$ and $AB$, respectively, are concurrent.

The side lengths of a trapezoid are $\sqrt[4]{3}, \sqrt[4]{3}, \sqrt[4]{3}$, and $2 \cdot \sqrt[4]{3}$. Its area is the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$.

Amy painted a dart board over a square clock face using the "hour positions" as boundaries. [See figure.] If $t$ is the area of one of the eight triangular regions such as that between $12$ o'clock and $1$ o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between $1$ o'clock and $2$ o'clock, then $\frac{q}{t}=$ [asy] size((80)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)--(.9,0)--(3.1,4)--(.9,4)--(3.1,0)--(2,0)--(2,4)); draw((0,3.1)--(4,.9)--(4,3.1)--(0,.9)--(0,2)--(4,2)); [/asy] $ \textbf{(A)}\ 2\sqrt{3}-2 \qquad\textbf{(B)}\ \frac{3}{2} \qquad\textbf{(C)}\ \frac{\sqrt{5}+1}{2} \qquad\textbf{(D)}\ \sqrt{3} \qquad\textbf{(E)}\ 2 $

In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.

Let $ABCD$ be a parallelogram and $H$ the Orthocentre of $\triangle{ABC}$. The line parallel to $AB$ through $H$ intersects $BC$ at $P$ and $AD$ at $Q$ while the line parallel to $BC$ through $H$ intersects $AB$ at $R$ and $CD$ at $S$. Show that $P$, $Q$, $R$ and $S$ are concyclic. [i](Swiss Mathematical Olympiad 2011, Final round, problem 8)[/i]

Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$. Prove that if $\frac{H_BC}{AC} = \frac{H_CA}{AB}$, then the line symmetric to $BC$ with respect to line $H_BH_C$ is tangent to the circumscribed circle of triangle $H_BH_CA$. [i](Proposed by Mykhailo Bondarenko)[/i]

In triangle $ABC$, $AB=13,$ $BC=15$ and $CA=17.$ Point $D$ is on $\overline{AB},$ $E$ is on $\overline{BC},$ and $F$ is on $\overline{CA}.$ Let $AD=p\cdot AB,$ $BE=q\cdot BC,$ and $CF=r\cdot CA,$ where $p,$ $q,$ and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5.$ The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$