Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A) + \cos(3B) + \cos(3C) = 1$. Two sides of the triangle have lengths $10$ and $13$. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}$. Find $m$.

Let $\triangle{ABC}$ be an acute-angled triangle and let $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, respectively, such that \[\angle{AFE}=\angle{BFD}\mbox{,}\quad\angle{BDF}=\angle{CDE}\quad\mbox{and}\quad\angle{CED}=\angle{AEF}\mbox{.}\] Prove that $D$, $E$ and $F$ are the feet of the perpendiculars through $A$, $B$ and $C$ on $BC$, $CA$ and $AB$, respectively. [i](Swiss Mathematical Olympiad 2011, Final round, problem 2)[/i]

A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$. [asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]

A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance of $10$ m from the point where the sphere touches the ground. At the same instant a meter stick (held vertically with one end on the ground) casts a shadow of length $2$ m. What is the radius of the sphere in meters? (Assume the sun's rays are parallel and the meter stick is a line segment.) $ \textbf{(A)}\ \frac{5}{2}\qquad\textbf{(B)}\ 9 - 4\sqrt{5}\qquad\textbf{(C)}\ 8\sqrt{10} - 23\qquad\textbf{(D)}\ 6 - \sqrt{15}\qquad\textbf{(E)}\ 10\sqrt{5} - 20 $

In quadrilateral $ABCD$, we have $AB = 5$, $BC = 6$, $CD = 5$, $DA = 4$, and $\angle ABC = 90^\circ$. Let $AC$ and $BD$ meet at $E$. Compute $\dfrac{BE}{ED}$.

Let $A_1A_2A_3A_4$ be a quadrilateral with no pair of parallel sides. For each $i=1, 2, 3, 4$, define $\omega_1$ to be the circle touching the quadrilateral externally, and which is tangent to the lines $A_{i-1}A_i, A_iA_{i+1}$ and $A_{i+1}A_{i+2}$ (indices are considered modulo $4$ so $A_0=A_4, A_5=A_1$ and $A_6=A_2$). Let $T_i$ be the point of tangency of $\omega_i$ with the side $A_iA_{i+1}$. Prove that the lines $A_1A_2, A_3A_4$ and $T_2T_4$ are concurrent if and only if the lines $A_2A_3, A_4A_1$ and $T_1T_3$ are concurrent. [i]Pavel Kozhevnikov, Russia[/i]

Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

Let $A_1A_2A_3A_4$ be a quadrilateral with no pair of parallel sides. For each $i=1, 2, 3, 4$, define $\omega_1$ to be the circle touching the quadrilateral externally, and which is tangent to the lines $A_{i-1}A_i, A_iA_{i+1}$ and $A_{i+1}A_{i+2}$ (indices are considered modulo $4$ so $A_0=A_4, A_5=A_1$ and $A_6=A_2$). Let $T_i$ be the point of tangency of $\omega_i$ with the side $A_iA_{i+1}$. Prove that the lines $A_1A_2, A_3A_4$ and $T_2T_4$ are concurrent if and only if the lines $A_2A_3, A_4A_1$ and $T_1T_3$ are concurrent. [i]Pavel Kozhevnikov, Russia[/i]

In triangle $ ABC$, side $ AC$ and the perpendicular bisector of $ BC$ meet in point $ D$, and $ BD$ bisects $ \angle ABC$. If $ AD \equal{} 9$ and $ DC \equal{} 7$, what is the area of triangle $ ABD$? $ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 14\sqrt5 \qquad \textbf{(E)}\ 28\sqrt5$

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.65)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(Arc(O, r, 45, 360-17.0312)); draw(A--B--C);dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,SE); [/asy]

In triangle $ABC$, angles $A$ and $B$ measure 60 degrees and 45 degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24.$ The area of triangle $ABC$ can be written in the form $a+b\sqrt{c},$ where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.

Given an acute triangle $ABC$, let $l_a$ be the line passing $A$ and perpendicular to $AB$, $l_b$ be the line passing $B$ and perpendicular to $BC$, and $l_c$ be the line passing $C$ and perpendicular to $CA$. Let $D$ be the intersection of $l_b$ and $l_c$, $E$ be the intersection of $l_c$ and $l_a$, and $F$ be the intersection of $l_a$ and $l_b$. Prove that the area of the triangle $DEF$ is at least three times of the area of $ABC$.

Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \[ \frac{\cot \gamma}{\cot \alpha+\cot \beta}. \]

$D, \: E , \: F$ are points on the sides $AB, \: BC, \: CA,$ respectively, of a triangle $ABC$ such that $AD=AF, \: BD=BE,$ and $DE=DF.$ Let $I$ be the incenter of the triangle $ABC,$ and let $K$ be the point of intersection of the line $BI$ and the tangent line through $A$ to the circumcircle of the triangle $ABI.$ Show that $AK=EK$ if $AK=AD.$

Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that \[ \frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. \] When does equality hold?

In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin \left( \angle DAE \right)$ can be expressed in the form $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$.

In quadrilateral $ ABCD$, it is given that $ \angle A \equal{} 120^\circ$, angles $ B$ and $ D$ are right angles, $ AB \equal{} 13$, and $ AD \equal{} 46$. Then $ AC \equal{}$ $ \textbf{(A)}\ 60 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 65 \qquad \textbf{(E)}\ 72$

Let $ ABC$ be a triangle inscribed in a circle of radius $ R$, and let $ P$ be a point in the interior of triangle $ ABC$. Prove that \[ \frac {PA}{BC^{2}} \plus{} \frac {PB}{CA^{2}} \plus{} \frac {PC}{AB^{2}}\ge \frac {1}{R}. \] [i]Alternative formulation:[/i] If $ ABC$ is a triangle with sidelengths $ BC\equal{}a$, $ CA\equal{}b$, $ AB\equal{}c$ and circumradius $ R$, and $ P$ is a point inside the triangle $ ABC$, then prove that $ \frac {PA}{a^{2}} \plus{} \frac {PB}{b^{2}} \plus{} \frac {PC}{c^{2}}\ge \frac {1}{R}$.

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$

In this figure $\angle RFS = \angle FDR$, $FD = 4$ inches, $DR = 6$ inches, $FR = 5$ inches, $FS = 7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is: [asy] import olympiad; pair F,R,S,D; F=origin; R=5*dir(aCos(9/16)); S=(7.5,0); D=4*dir(aCos(9/16)+aCos(1/8)); label("$F$",F,SW);label("$R$",R,N); label("$S$",S,SE); label("$D$",D,W); label("$7\frac{1}{2}$",(F+S)/2.5,SE); label("$4$",midpoint(F--D),SW); label("$5$",midpoint(F--R),W); label("$6$",midpoint(D--R),N); draw(F--D--R--F--S--R); markscalefactor=0.1; draw(anglemark(S,F,R)); draw(anglemark(F,D,R)); //Credit to throwaway1489 for the diagram[/asy] $\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\dfrac{1}{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 6\dfrac{1}{4}$

The diagram below shows the regular hexagon $BCEGHJ$ surrounded by the rectangle $ADFI$. Let $\theta$ be the measure of the acute angle between the side $\overline{EG}$ of the hexagon and the diagonal of the rectangle $\overline{AF}$. There are relatively prime positive integers $m$ and $n$ so that $\sin^2\theta  = \tfrac{m}{n}$. Find $m + n$. [asy] import graph; size(3.2cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((-1,3)--(-1,2)--(-0.13,1.5)--(0.73,2)--(0.73,3)--(-0.13,3.5)--cycle); draw((-1,3)--(-1,2)); draw((-1,2)--(-0.13,1.5)); draw((-0.13,1.5)--(0.73,2)); draw((0.73,2)--(0.73,3)); draw((0.73,3)--(-0.13,3.5)); draw((-0.13,3.5)--(-1,3)); draw((-1,3.5)--(0.73,3.5)); draw((0.73,3.5)--(0.73,1.5)); draw((-1,1.5)--(0.73,1.5)); draw((-1,3.5)--(-1,1.5)); label("$ A $",(-1.4,3.9),SE*labelscalefactor); label("$ B $",(-1.4,3.28),SE*labelscalefactor); label("$ C $",(-1.4,2.29),SE*labelscalefactor); label("$ D $",(-1.4,1.45),SE*labelscalefactor); label("$ E $",(-0.3,1.4),SE*labelscalefactor); label("$ F $",(0.8,1.45),SE*labelscalefactor); label("$ G $",(0.8,2.24),SE*labelscalefactor); label("$ H $",(0.8,3.26),SE*labelscalefactor); label("$ I $",(0.8,3.9),SE*labelscalefactor); label("$ J $",(-0.25,3.9),SE*labelscalefactor); [/asy]

Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

Let $O$ be the center of a circle of radius $26$, and let $A$, $B$ be two distinct points on the circle, with $M$ being the midpoint of $AB$. Consider point $C$ for which $CO=34$ and $\angle COM=15^\circ$. Let $N$ be the midpoint of $CO$. Suppose that $\angle ACB=90^\circ$. Find $MN$.

The incenter of $\triangle A_1A_2A_3$ is $I$, and the center of the $A_i$-excircle is $J_i$ ($i=1,2,3$). Let $B_i$ be the intersection point of side $A_{i+1}A_{i+2}$ and the bisector of $\angle A_{i+1}IA_{i+2}$ ($A_{i+3}:=A_i$ $\forall i$). Prove that the three lines $B_iJ_i$ are concurrent.