Let $ ABC$ be a triangle with $ \angle BAC \equal{} 90^\circ$. A circle is tangent to the sides $ AB$ and $ AC$ at $ X$ and $ Y$ respectively, such that the points on the circle diametrically opposite $ X$ and $ Y$ both lie on the side $ BC$. Given that $ AB \equal{} 6$, find the area of the portion of the circle that lies outside the triangle. [asy]import olympiad; import math; import graph; unitsize(20mm); defaultpen(fontsize(8pt)); pair A = (0,0); pair B = A + right; pair C = A + up; pair O = (1/3, 1/3); pair Xprime = (1/3,2/3); pair Yprime = (2/3,1/3); fill(Arc(O,1/3,0,90)--Xprime--Yprime--cycle,0.7*white); draw(A--B--C--cycle); draw(Circle(O, 1/3)); draw((0,1/3)--(2/3,1/3)); draw((1/3,0)--(1/3,2/3)); label("$A$",A, SW); label("$B$",B, down); label("$C$",C, left); label("$X$",(1/3,0), down); label("$Y$",(0,1/3), left);[/asy]

Pinocchio claims that he can take some non-right-angled triangles , all of which are similar to one another and some of which may be congruent to one another, and put them together to form a rectangle. Is Pinocchio lying? (A Fedotov)

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

Let $ABC$ be a 90-degree triangle with hypotenuse $BC$. Let $D$ and $E$ distinct points on segment $BC$ and $P, Q$ be the foot of the perpendicular from $D$ to $AB$ and $E$ to $AC$, respectively. $DP$ and $EQ$ intersect at $R$. Lines $CR$ and $AB$ intersect at $M$ and lines $BR$ and $AC$ intersect at $N$. Prove that $MN \parallel BC$ if and only if $BD=CE$.

Two different points $A$ and $B$ and a circle $\omega$ that passes through $A$ and $B$ are given. $P$ is a variable point on $\omega$ (different from $A$ and $B$). $M$ is a point such that $MP$ is the bisector of the angle $\angle{APB}$ ($M$ lies outside of $\omega$) and $MP=AP+BP$. Find the geometrical locus of $M$.

A pyramid has a square base with sides of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube? $ \textbf{(A)}\ 5\sqrt{2}-7 \qquad \textbf{(B)}\ 7-4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9} $

$A_0B_0C_0$ and $A_1B_1C_1$ are acute-angled triangles. Describe, and prove, how to construct the triangle $ABC$ with the largest possible area which is circumscribed about $A_0B_0C_0$ (so $BC$ contains $B_0, CA$ contains $B_0$, and $AB$ contains $C_0$) and similar to $A_1B_1C_1.$

In the figure, $ABCD$ is a square of side length 1. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? [asy] unitsize(150); pair A,B,C,D,E,F,G,H,J,K; A=(1,0); B=(0,0); C=(0,1); D=(1,1); draw(A--B--C--D--A); E=(2-sqrt(3),0); F=(2-sqrt(3),1); draw(E--F); G=(1,sqrt(3)/2); H=(2.5-sqrt(3),1); K=(2-sqrt(3),1-sqrt(3)/2); J=(0.5,0); draw(G--H--K--J--G); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$K$",K,W); label("$J$",J,S); [/asy] $ \textbf{(A) }\dfrac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\dfrac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\dfrac{\sqrt{3}}{6}\qquad\textbf{(E) }1-\dfrac{\sqrt{2}}{2} $

On the sides of a triangle $ABC$ right angled at $A$ three points $D, E$ and $F$ (respectively $BC, AC$ and $AB$) are chosen so that the quadrilateral $AFDE$ is a square. If $x$ is the length of the side of the square, show that \[\frac{1}{x}=\frac{1}{AB}+\frac{1}{AC}\]

Let $ABC$ be an acute triangle, and let $M$ and $N$ be two points on the line $AC$ such that the vectors $MN$ and $AC$ are identical. Let $X$ be the orthogonal projection of $M$ on $BC$, and let $Y$ be the orthogonal projection of $N$ on $AB$. Finally, let $H$ be the orthocenter of triangle $ABC$. Show that the points $B$, $X$, $H$, $Y$ lie on one circle.

Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.

$ABC$ is a triangle. $A' , B' , C'$ are points on the segments $BC, CA, AB$ respectively. $\angle B' A' C' = \angle A$ , $\frac{AC'}{C'B} = \frac{BA' }{A' C} = \frac{CB'}{B'A}$. Show that $ABC$ and $A'B'C'$ are similar.

We say that two triangles are oriented similarly if they are similar and have the same orientation. Prove that if $ALT, ARM, ORT, $ and $ULM$ are four triangles which are oriented similarly, then $A$ is the midpoint of the line segment $OU.$

Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0<x<\frac{\pi}2$. The length of the hypotenuse is $\text{(A)} \ \frac 43 \qquad \text{(B)} \ \frac 32 \qquad \text{(C)} \ \frac{3\sqrt{5}}{5} \qquad \text{(D)} \ \frac{2\sqrt{5}}{3} \qquad \text{(E)} \ \text{not uniquely determined}$

a) The triangle $ABC$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the triangle $A_1B_1C_1$. The corresponding segments $[AB]$ and $[A_1B_1]$ intersect in the point $C_2, [BC]$ and $[B_1C_1]$ -- $A_2, [AC]$ and $[A_1C_1]$ -- $B_2$. Prove that the triangle $A_2B_2C_2$ is similar to the triangle $ABC$. b) The quadrangle $ABCD$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the quadrangle $A_1B_1C_1D_1$. Prove that the points of intersection of the corresponding lines ( $(AB$) and $(A_1B_1), (BC)$ and $(B_1C_1), (CD)$ and $(C_1D_1), (DA)$ and $(D_1A_1)$ ) are the vertices of the parallelogram.

In a convex quadrilateral $ABCD, \angle A < 90^o, \angle B < 90^o$ and $AB > CD$. Points $P$ and $Q$ are on the segments $BC$ and $AD$ respectively. Suppose the triangles $APD$ and $BQC$ are similar. Prove that $AB$ is parallel to $CD$.

Given a triangle $ABC$ and points $P$ and $Q$. It is known that the triangles formed by the projections $P$ and $Q$ on the sides of $ABC$ are similar (vertices lying on the same sides of the original triangle correspond to each other). Prove that line $PQ$ passes through the center of the circumscribed circle of triangle $ABC$. (A. Zaslavsky)

We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$. a) Prove that the triangles $ABC$ and $AF E$ are similar. b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.

In $ \bigtriangleup ABC $, $ AB = 86 $, and $ AC = 97 $. A circle with center $ A $ and radius $ AB $ intersects $ \overline{BC} $ at points $ B $ and $ X $. Moreover $ \overline{BX} $ and $ \overline{CX} $ have integer lengths. What is $ BC $? $ \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 $

The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$ respectively. Let the line $AD$ intersect this incircle of triangle $ABC$ at a point $X$ (apart from $D$). Assume that this point $X$ is the midpoint of the segment $AD$, this means, $AX = XD$. Let the line $BX$ meet the incircle of triangle $ABC$ at a point $Y$ (apart from $X$), and let the line $CX$ meet the incircle of triangle $ABC$ at a point $Z$ (apart from $X$). Show that $EY = FZ$.

A paper cup has a base that is a circle with radius $r$, a top that is a circle with radius $2r$, and sides that connect the two circles with straight line segments as shown below. This cup has height $h$ and volume $V$. A second cup that is exactly the same shape as the first is held upright inside the first cup so that its base is a distance of $\tfrac{h}2$ from the base of the first cup. The volume of liquid that will t inside the first cup and outside the second cup can be written $\tfrac{m}{n}\cdot V$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] pair s = (10,1); draw(ellipse((0,0),4,1)^^ellipse((0,-6),2,.5)); fill((3,-6)--(-3,-6)--(0,-2.1)--cycle,white); draw((4,0)--(2,-6)^^(-4,0)--(-2,-6)); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-2.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); pair s = (10,-2); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-4.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); //darn :([/asy]

Let $N_5$ be the answer to problem 5. Triangle $JHU$ satisfies $JH=N_5$ and $JU=6$. Point $X$ lies on $\overline{HU}$ such that $\overline{JX}$ is an altitude of $\triangle{JHU}$, point $Y$ is the midpoint of $\overline{JU}$, and $\overline{JX}$ and $\overline{HY}$ intersect at $Z$. Assume that $\triangle{HZX}$ is similar to $\triangle{JZY}$ (in this vertex order). Compute the area of $\triangle{JHU}$.

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 $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

The parallel sides of a trapezoid are $ 3$ and $ 9$. The non-parallel sides are $ 4$ and $ 6$. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is: [asy]defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair B = (2.25,0); pair C = (2,1); pair D = (1,1); pair E = waypoint(A--D,0.25); pair F = waypoint(B--C,0.25); draw(A--B--C--D--cycle); draw(E--F); label("6",midpoint(A--D),NW); label("3",midpoint(C--D),N); label("4",midpoint(C--B),NE); label("9",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 4: 3\qquad \textbf{(B)}\ 3: 2\qquad \textbf{(C)}\ 4: 1\qquad \textbf{(D)}\ 3: 1\qquad \textbf{(E)}\ 6: 1$