The altitudes of a triangle $\triangle ABC$ are used to form the sides of a second triangle $\triangle A_1B_1C_1$. The altitudes of $\triangle A_1B_1C_1$ are then used to form the sides of a third triangle $\triangle A_2B_2C_2$. Prove that $\triangle A_2B_2C_2$ is similar to $\triangle ABC$.

The incentre of a triangle is joined by three segments to the three vertices of the triangle, thereby dividing it into three smaller triangles. If one of these three triangles is similar to the original triangle, find its angles. (A Shapovalov)

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac{2}{3}$, determine the length of $YZ$

Circles with radii $1, 2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency? $ \textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3} $

Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $X_A$ lying on the tangent at $H$ to the circumcircle of triangle $BHC$ is such that $AH=AX_A$ and $X_A \not= H$. Points $X_B,X_C$ are defined similarly. Prove that the triangle $X_AX_BX_C$ and the orthotriangle of $ABC$ are similar.

Let $ \overline{AB}$ be a diameter of a circle and $ C$ be a point on $ \overline{AB}$ with $ 2 \cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ \overline{DC} \perp \overline{AB}$ and $ \overline{DE}$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area of $ \triangle ABD$? [asy]unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$ \textbf{(A)} \ \frac {1}{6} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {2}{3}$

Let $ABCD$ be a trapezoid where $AB$ is parallel to $CD.$ Let $P$ be the intersection of diagonal $AC$ and diagonal $BD.$ If the area of triangle $PAB$ is $16,$ and the area of triangle $PCD$ is $25,$ find the area of the trapezoid.

Points $D$ and $E$ are taken on sides $BC$ and $CA$ of a triangle $ BD\equal{}AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of $\angle ACB$ intersects $AD$ and $BE$ at $Q$ and $R$ respectively. Prove that $ \frac{PQ}{PR}\equal{}\frac{AD}{BE}$.

In triangle $ABC$, a point $O$ is taken such that $\angle COA = \angle B + 60^o$, $\angle COB = \angle A + 60^o$, $\angle AOB = \angle C + 60^o$.Prove that if a triangle can be formed from the segments $AO$, $BO$, $CO$, then a triangle can also be formed from the altitudes of triangle $ABC$ and these triangles are similar.

$ABC$ is a triangle. Take $a = BC$ etc as usual. Take points $T_1, T_2$ on the side $AB$ so that $AT_1 = T_1T_2 = T_2B$. Similarly, take points $T_3, T_4$ on the side BC so that $BT_3 = T_3T_4 = T_4C$, and points $T_5, T_6$ on the side $CA$ so that $CT_5 = T_5T_6 = T_6A$. Show that if $a' = BT_5, b' = CT_1, c'=AT_3$, then there is a triangle $A'B'C'$ with sides $a', b', c'$ ($a' = B'C$' etc). In the same way we take points $T_i'$ on the sides of $A'B'C' $ and put $a'' = B'T_6', b'' = C'T_2', c'' = A'T_4'$. Show that there is a triangle $A'' B'' C'' $ with sides $a'' b'' , c''$ and that it is similar to $ABC$. Find $a'' /a$.

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]

It is known that the bisector of the angle $\angle ADC$ of the inscribed quadrilateral $ABCD$ passes through the center of the circle inscribed in the triangle $ABC$. Let $M$ be an arbitrary point of the arc $ABC$ of the circle circumscribed around $ABCD$. Denote by $P$ and $Q$ the centers of the circles inscribed in the triangles $ABM$ and $BCM$. Prove that all triangles $DPQ$ obtained by moving point $M$ are similar to each other. Find the angle $\angle PDQ$ and ratio $BP : PQ$ if $\angle BAC = \alpha$, $\angle BCA = \beta$

Trapezoid $ABCD$ has $\overline{AB} \parallel \overline{CD}$, $BC = CD = 43$, and $\overline{AD} \perp \overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. GIven that $OP = 11$, the length $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m + n$? $\textbf{(A)}\: 65\qquad\textbf{(B)}\: 132\qquad\textbf{(C)}\: 157\qquad\textbf{(D)}\: 194\qquad\textbf{(E)}\: 215$

Let $\triangle ABC$ have $AB=6$, $BC=7$, and $CA=8$, and denote by $\omega$ its circumcircle. Let $N$ be a point on $\omega$ such that $AN$ is a diameter of $\omega$. Furthermore, let the tangent to $\omega$ at $A$ intersect $BC$ at $T$, and let the second intersection point of $NT$ with $\omega$ be $X$. The length of $\overline{AX}$ can be written in the form $\tfrac m{\sqrt n}$ for positive integers $m$ and $n$, where $n$ is not divisible by the square of any prime. Find $100m+n$. [i]Proposed by David Altizio[/i]

Side $AB$ of triangle $ABC$ has length $8$ inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$, and $E$ is on segment $BC$. Line $AE$ extended bisects angle $FEC$. If $DE$ has length $5$ inches, then the length of $CE$, in inches, is: $\textbf{(A)}\ \dfrac{51}{4} \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ \dfrac{53}{4} \qquad \textbf{(D)}\ \dfrac{40}{3} \qquad \textbf{(E)}\ \dfrac{27}{2} $

A non-right triangle $A_1A_2A_3$ is given. Circles $C_1$ and $C_2$ are tangent at $A_3, C_2$ and $C_3$ are tangent at $A_1$, and $C_3$ and $C_1$ are tangent at $A_2$. Points $O_1,O_2,O_3$ are the centers of $C_1, C_2, C_3$, respectively. Supposing that the triangles $A_1A_2A_3$ and $O_1O_2O_3$ are similar, determine their angles.

Determine whether there is a tetrahedron $ABCD$ with the longest edge of length 1 such that all its faces are similar right triangles with right angles at vertices $B,C.$ If so, determine which edge is the longest, which is the shortest and what is its length.

Given triangle $ABC$ and points $P$. Let $A_1,B_1,C_1$ be the second points of intersection of straight lines $AP, BP, CP$ with the circumscribed circle of $ABC$. Let points $A_2, B_2, C_2$ be symmetric to $A_1,B_1,C_1$ wrt $BC,CA,AB$, respectively. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar. (A. Zaslavsky)

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 $BC$.

Let $ABC$ be a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2 EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$.

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C’$, and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$. Suppose that $C’D=\frac{1}{3}$. What is the perimeter of $\triangle AEC’$? [asy] //Diagram by Samrocksnature pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.62); pair C=(1,0); pair B=(0,0); pair G=(0,0.25); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); [/asy] $\textbf{(A) }2 \qquad \textbf{(B) }1+\frac{2}{3}\sqrt{3} \qquad \textbf{(C) }\frac{13}{6} \qquad \textbf{(D) }1+\frac{3}{4}\sqrt{3} \qquad \textbf{(E) }\frac{7}{3}$

In triangle $ABC$ we have $\angle ACB = 90^o$. The point $M$ is the midpoint of $AB$. The line through $M$ parallel to $BC$ intersects $AC$ in $D$. The midpoint of line segment $CD$ is $E$. The lines $BD$ and $CM$ are perpendicular. (a) Prove that triangles $CME$ and $ABD$ are similar. (b) Prove that $EM$ and $AB$ are perpendicular. [asy] unitsize(1 cm); pair A, B, C, D, E, M; A = (0,0); B = (4,0); C = (2.6,2); M = (A + B)/2; D = (A + C)/2; E = (C + D)/2; draw(A--B--C--cycle); draw(C--M--D--B); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, NW); dot("$E$", E, NW); dot("$M$", M, S); [/asy] [i]Be aware: the figure is not drawn to scale.[/i]

Let $P$ be a point inside triangle $ABC$, and suppose lines $AP$, $BP$, $CP$ meet the circumcircle again at $T$, $S$, $R$ (here $T \neq A$, $S \neq B$, $R \neq C$). Let $U$ be any point in the interior of $PT$. A line through $U$ parallel to $AB$ meets $CR$ at $W$, and the line through $U$ parallel to $AC$ meets $BS$ again at $V$. Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$. Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$.

Let $ABCD$ be a parallelogram of center $O$. Points $M$ and $N$ are the midpoints of $BO$ and $CD$, respectively. Prove that if the triangles $ABC$ and $AMN$ are similar, then $ABCD$ is a square.

Given three points $P$, $A$, $B$ and a circle such that the lines $PA$ and $PB$ are tangent to the circle at the points $A$ and $B$, respectively. A line through the point $P$ intersects that circle at two points $C$ and $D$. Through the point $B$, draw a line parallel to $PA$; let this line intersect the lines $AC$ and $AD$ at the points $E$ and $F$, respectively. Prove that $BE = BF$.