A paper triangle with the angles $20^o$, $20^o$ and $140^o$ is cut into two triangles by the bisector of one of its angles. Then one of these triangles is cut into two by its bisector, and so on. Prove that it is impossible to get a triangle similar to the initial one. (AI Galochkin)

Triangle $A_0B_0C_0$ is given in the plane. Consider all triangles $ABC$ such that: (i) The lines $AB,BC,CA$ pass through $C_0,A_0,B_0$, respectvely, (ii) The triangles $ABC$ and $A_0B_0C_0$ are similar. Find the possible positions of the circumcenter of triangle $ABC$.

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 bisectors of angles $A$, $B$, and $C$ of triangle $ABC$ meet for the second time its circumcircle at points $A_1$, $B_1$, $C_1$ respectively. Let $A_2$, $B_2$, $C_2$ be the midpoints of segments $AA_1$, $BB_1$, $CC_1$ respectively. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar.

Let $ABC$ be an acute triangle with ${AB\neq AC}$. The incircle ${\omega}$ of the triangle touches the sides ${BC, CA}$ and ${AB}$ at ${D, E}$ and ${F}$, respectively. The perpendicular line erected at ${C}$ onto ${BC}$ meets ${EF}$ at ${M}$, and similarly the perpendicular line erected at ${B}$ onto ${BC}$ meets ${EF}$ at ${N}$. The line ${DM}$ meets ${\omega}$ again in ${P}$, and the line ${DN}$ meets ${\omega}$ again at ${Q}$. Prove that ${DP=DQ}$. Ruben Dario & Leo Giugiuc (Romania)

Let $ABC$ and $A'B'C'$ be similar triangles with different orientation such that their orthocenters coincide. Show that lines $AA′, BB′, CC′ are concurrent or parallel.

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] $\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\dfrac{1}{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 6\dfrac{1}{4}$

A concrete sewer pipe fitting is shaped like a cylinder with diameter $48$ with a cone on top. A cylindrical hole of diameter $30$ is bored all the way through the center of the fitting as shown. The cylindrical portion has height $60$ while the conical top portion has height $20$. Find $N$ such that the volume of the concrete is $N \pi$. [asy] import three; size(250); defaultpen(linewidth(0.7)+fontsize(10)); pen dashes = linewidth(0.7) + linetype("2 2"); currentprojection = orthographic(0,-15,5); draw(circle((0,0,0), 15),dashes); draw(circle((0,0,80), 15)); draw(scale3(24)*((-1,0,0)..(0,-1,0)..(1,0,0))); draw(shift((0,0,60))*scale3(24)*((-1,0,0)..(0,-1,0)..(1,0,0))); draw((-24,0,0)--(-24,0,60)--(-15,0,80)); draw((24,0,0)--(24,0,60)--(15,0,80)); draw((-15,0,0)--(-15,0,80),dashes); draw((15,0,0)--(15,0,80),dashes); draw("48", (-24,0,-20)--(24,0,-20)); draw((-15,0,-20)--(-15,0,-17)); draw((15,0,-20)--(15,0,-17)); label("30", (0,0,-15)); draw("60", (50,0,0)--(50,0,60)); draw("20", (50,0,60)--(50,0,80)); draw((50,0,60)--(47,0,60));[/asy]

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is $\text{(A)}\ \frac1{2m+1}\qquad\text{(B)}\ m \qquad\text{(C)}\ 1-m\qquad\text{(D)}\ \frac1{4m} \qquad\text{(E)}\ \frac1{8m^2}$

A triangle $ABC$ is inscribed in a circle $G$. Points $M$ and $N$ are the midpoints of the arcs $BC$ and $AC$ respectively, and $D$ is an arbitrary point on the arc $AB$ (not containing $C$). Points $I_1$ and $I_2$ are the incenters of the triangles $ADC$ and $BDC$, respectively. If the circumcircle of triangle $DI_1I_2$ meets $G$ again at $P$, prove that triangles $PNI_1$ and $PMI_2$ are similar.

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)

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 $EY = FZ$.

Let $I$ be the incenter of a right-angled triangle $ABC$, and $M$ be the midpoint of hypothenuse $AB$. The tangent to the circumcircle of $ABC$ at $C$ meets the line passing through $I$ and parallel to $AB$ at point $P$. Let $H$ be the orthocenter of triangle $PAB$. Prove that lines $CH$ and $PM$ meet at the incircle of triangle $ABC$.

$A_0B_0C_0$ 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 $A_1B_1C_1.$

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

Let $\triangle{PQR}$ be a right triangle with $PQ=90$, $PR=120$, and $QR=150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$. Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$. The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt{10n}$. What is $n$?

Given is a triangle $ABC$ with incircle $\omega$, tangent to $BC, CA, AB$ at $D, E, F$. The perpendicular from $B$ to $BC$ meets $EF$ at $M$, and the perpendicular from $C$ to $BC$ meets $EF$ at $N$. Let $DM$ and $DN$ meet $\omega$ at $P$ and $Q$. Prove that $DP=DQ$.

Chords $ AB$ and $ CD$ of a circle intersect at a point $ E$ inside the circle. Let $ M$ be an interior point of the segment $ EB$. The tangent line at $ E$ to the circle through $ D$, $ E$, and $ M$ intersects the lines $ BC$ and $ AC$ at $ F$ and $ G$, respectively. If \[ \frac {AM}{AB} \equal{} t, \] find $\frac {EG}{EF}$ in terms of $ t$.

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 $\angle BAC = \alpha$, $\angle BCA = \beta$

Let $L$ be the length of the altitude to the hypotenuse of a right triangle with legs $5$ and $12$. Find the least integer greater than $L$.

Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$? [asy] size(200); defaultpen(linewidth(.6pt)+fontsize(12pt)); dotfactor=4; draw((0,0)--(0,2)); draw((0,0)--(1,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1)); draw((0,0)--(1,2)); draw((0,2)--(2,1)); draw((0,2)--(2,2)); draw((2,1)--(2,2)); label("$A$",(0,2),NW); label("$B$",(1,2),N); label("$C$",(4/5,1.55),W); dot((0,2)); dot((1,2)); dot((4/5,1.6)); dot((2,1)); dot((0,0)); [/asy] $ \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt2}{4} $

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 $O$ 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 $ABC$.