Two circles $K_{1}$ and $K_{2}$, centered at $O_{1}$ and $O_{2}$ with radii $1$ and $\sqrt{2}$ respectively, intersect at $A$ and $B$. Let $C$ be a point on $K_{2}$ such that the midpoint of $AC$ lies on $K_{1}$. Find the length of the segment $AC$ if $O_{1}O_{2}=2$

Let $\triangle ABC$ be an acute triangle, and let $D$ be the projection of $A$ on $BC$. Let $M,N$ be the midpoints of $AB$ and $AC$ respectively. Let $\Gamma_1$ and $\Gamma_2$ be the circumcircles of $\triangle BDM$ and $\triangle CDN$ respectively, and let $K$ be the other intersection point of $\Gamma_1$ and $\Gamma_2$. Let $P$ be an arbitrary point on $BC$ and $E,F$ are on $AC$ and $AB$ respectively such that $PEAF$ is a parallelogram. Prove that if $MN$ is a common tangent line of $\Gamma_1$ and $\Gamma_2$, then $K,E,A,F$ are concyclic.

Prove that $\frac 12 \cdot \sqrt{4\sin^2 36^{\circ} - 1}=\cos 72^\circ$.

Triangle $ \triangle ABC$ is given. Points $ D$ i $ E$ are on line $ AB$ such that $ D \minus{} A \minus{} B \minus{} E, AD \equal{} AC$ and $ BE \equal{} BC$. Bisector of internal angles at $ A$ and $ B$ intersect $ BC,AC$ at $ P$ and $ Q$, and circumcircle of $ ABC$ at $ M$ and $ N$. Line which connects $ A$ with center of circumcircle of $ BME$ and line which connects $ B$ and center of circumcircle of $ AND$ intersect at $ X$. Prove that $ CX \perp PQ$.

Assume every side length of a triangle $ABC$ is more than $2$ and two of its angles are given by $\angle ABC = 57^\circ$ and $ACB = 63^\circ$. Point $P$ is chosen on side $BC$ with $BP:PC = 2:1$. Points $M,N$ are chosen on sides $AB$ and $AC$, respectively so that $BM = 2$ and $CN = 1$. Let $Q$ be the point on segment $MN$ for which $MQ:QN = 2:1$. Find the value of $PQ$. Your answer must be in simplest form.

The straight line $ \overline{AB}$ is divided at $ C$ so that $ AC\equal{}3CB$. Circles are described on $ \overline{AC}$ and $ \overline{CB}$ as diameters and a common tangent meets $ AB$ produced at $ D$. Then $ BD$ equals: $ \textbf{(A)}\ \text{diameter of the smaller circle} \\ \textbf{(B)}\ \text{radius of the smaller circle} \\ \textbf{(C)}\ \text{radius of the larger circle} \\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

A right triangle $ABC$ with hypotenuse $AB$ has side $AC = 15$. Altitude $CH$ divides $AB$ into segments $AH$ And $HB$, with $HB = 16$. The area of $\triangle ABC$ is: [asy] size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); pair A = origin, H = (5,0), B = (13,0), C = (5,6.5); draw(C--A--B--C--H^^rightanglemark(C,H,B,16)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,N); label("$H$",H,S); label("$15$",C/2,NW); label("$16$",(H+B)/2,S); [/asy] $\textbf{(A) }120\qquad \textbf{(B) }144\qquad \textbf{(C) }150\qquad \textbf{(D) }216\qquad \textbf{(E) }144\sqrt5$

Circles $Q_1$ and $Q_2$ are tangent to each other externally at $T$. Points $A$ and $E$ are on $Q_1$, lines $AB$ and $DE$ are tangent to $Q_2$ at $B$ and $D$, respectively, lines $AE$ and $BD$ meet at point $P$. Prove that (1) $\frac{AB}{AT}=\frac{ED}{ET}$; (2) $\angle ATP + \angle ETP = 180^{\circ}$. [asy]import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-6,xmax=5.94,ymin=-3.19,ymax=3.43; pair Q_1=(-2.5,-0.5), T=(-1.5,-0.5), Q_2=(0.5,-0.5), A=(-2.09,0.41), B=(-0.42,1.28), D=(-0.2,-2.37), P=(-0.52,2.96); D(CR(Q_1,1)); D(CR(Q_2,2)); D(A--B); D((-3.13,-1.27)--D); D(P--(-3.13,-1.27)); D(P--D); D(T--(-3.13,-1.27)); D(T--A); D(T--P); D(Q_1); MP("Q_1",(-2.46,-0.44),NE*lsf); D(T); MP("T",(-1.46,-0.44),NE*lsf); D(Q_2); MP("Q_2",(0.54,-0.44),NE*lsf); D(A); MP("A",(-2.22,0.58),NE*lsf); D(B); MP("B",(-0.35,1.45),NE*lsf); D((-3.13,-1.27)); MP("E",(-3.52,-1.62),NE*lsf); D(D); MP("D",(-0.17,-2.31),NE*lsf); D(P); MP("P",(-0.47,3.02),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

The diagonals of a convex quadrilateral dissect it into four similar triangles. Prove that this quadrilateral can also be dissected into two congruent triangles.

The triangle was divided into five triangles similar to it. Is it true that the original triangle is right-angled? (S. Markelov)

Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$, respectively. Place points $X$ and $Y$ on line $BC$ such that $\angle HM_BX = \angle HM_CY = 90^{\circ}$. Prove that triangles $OXY$ and $HBC$ are similar.

A triangle can be cut into three similar triangles. Prove that it can be cut into any number of triangles similar to each other.

A trapezium has parallel sides of length equal to $a$ and $b$ ($a <b$), and the distance between the parallel sides is the altitude $h$. The extensions of the non-parallel lines intersect at a point that is a vertex of two triangles that have as sides the parallel sides of the trapezium. Express the areas of the triangles as functions of $a,b$ and $h$.

Consider a circular cone with vertex $ V$, and let $ ABC$ be a triangle inscribed in the base of the cone, such that $ AB$ is a diameter and $ AC \equal{} BC$. Let $ L$ be a point on $ BV$ such that the volume of the cone is 4 times the volume of the tetrahedron $ ABCL$. Find the value of $ BL/LV$.

Let $ABCD$ be a parallelogram and points $E,F$ be on its exterior. If triangles $BCF$ and $DEC$ are similar, i.e. $\triangle BCF \sim \triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.

Let $ABC$ be a triangle with $b\neq c$. Points $D$ is the midpoint of $BC$ and let $E$ be the foot of angle $A$ bisector. In the exterior of the triangle we construct the similar triangles $AMB$ and $ANC$ . Prove: a) $MN\bot AD \Longleftrightarrow MA \bot AB$ b) $MN\bot AE \Longleftrightarrow M,A,N$ are colinear.

Circle $ \Gamma_{1},$ with radius $ r,$ is internally tangent to circle $ \Gamma_{2}$ at $ S.$ Chord $ AB$ of $ \Gamma_{2}$ is tangent to $ \Gamma_{1}$ at $ C.$ Let $ M$ be the midpoint of arc $ AB$ (not containing $ S$), and let $ N$ be the foot of the perpendicular from $ M$ to line $ AB.$ Prove that $ AC\cdot CB\equal{}2r\cdot MN.$

Given a regular tetrahedron $OABC$. Take points $P,\ Q,\ R$ on the sides $OA,\ OB,\ OC$ respectively. Note that $P,\ Q,\ R$ are different from the vertices of the tetrahedron $OABC$. If $\triangle{PQR}$ is an equilateral triangle, then prove that three sides $PQ,\ QR,\ RP$ are pararell to three sides $AB,\ BC,\ CA$ respectively. 30 points

The angles of the triangle $ABC$ satisfy $\angle A / \angle C = \angle B / \angle A = 2$. The incenter is $O. K, L$ are the excenters of the excircles opposite $B$ and $A$ respectively. Show that triangles $ABC$ and $OKL$ are similar.

Find all integers $n\geq 3$ such that there exists a convex $n$-gon $A_1A_2\dots A_n$ which satisfies the following conditions: - All interior angles of the polygon are equal - Not all sides of the polygon are equal - There exists a triangle $T$ and a point $O$ inside the polygon such that the $n$ triangles $OA_1A_2,\ OA_2A_3,\ \dots,\ OA_{n-1}A_n,\ OA_nA_1$ are all similar to $T$, not necessarily in the same vertex order.

Assume every side length of a triangle $ABC$ is more than $2$ and two of its angles are given by $\angle ABC = 57^\circ$ and $ACB = 63^\circ$. Point $P$ is chosen on side $BC$ with $BP:PC = 2:1$. Points $M,N$ are chosen on sides $AB$ and $AC$, respectively so that $BM = 2$ and $CN = 1$. Let $Q$ be the point on segment $MN$ for which $MQ:QN = 2:1$. Find the value of $PQ$. Your answer must be in simplest form.

Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD}$, respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}$. The length of a side of this smaller square is $\displaystyle \frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c$.

Let $ABC$ be an acute triangle and let $A_1$, $B_1$, $C_1$ be points on the sides $BC, CA$ and $AB$, respectively. Show that the triangles $ABC$ and $A_1B_1C_1$ are similar ($\angle A = \angle A_1, \angle B = \angle B_1,\angle C = \angle C_1$) if and only if the orthocentre of the triangle $A_1B_1C_1$ and the circumcentre of the triangle $ABC$ coincide.