In a triangle $ABC$, let $D$, $E$ and $F$ be the touchpoints of the inscribed circle and the sides $AB$, $BC$ and $CA$. Show that the triangles $DEF$ and $ABC$ are similar if and only if $ABC$ is equilateral.

A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$. Let $I$ be the incenter of $ABC$. Prove that $AI^2 = AD \cdot AE$.

Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.

For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$. [asy] defaultpen(linewidth(0.7)); pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22); pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3); draw(a--b--c--cycle); draw(p--q--r--cycle); label("$A$",a,dir(point--a)); label("$B$",b,dir(point--b)); label("$C$",c,dir(point--c)); label("$P$",p,dir(point--p)); label("$Q$",q,dir(point--q)); label("$R$",r,dir(point--r));[/asy]

Let $ABC$ be an arbitrary triangle. On the perpendicular bisector of $AB$, there is a point $P$ inside of triangle $ABC$. On the sides $BC$ and $CA$, triangles $BQC$ and $CRA$ are placed externally. These triangles satisfy $\vartriangle BPA \sim \vartriangle BQC \sim \vartriangle CRA$. (So $Q$ and $A$ lie on opposite sides of $BC$, and $R$ and $B$ lie on opposite sides of $AC$.) Show that the points $P, Q, C$ and $R$ form a parallelogram.

The side lengths $a,$ $b,$ $c$ of a triangle $ABC$ form an arithmetical progression (such that $b-a=c-b$). The side lengths $a_{1},$ $b_{1},$ $c_{1}$ of a triangle $A_{1}B_{1}C_{1}$ also form an arithmetical progression (with $b_{1}-a_{1}=c_{1}-b_{1}$). [Hereby, $a=BC,$ $b=CA,$ $c=AB, $ $a_{1}=B_{1}C_{1},$ $b_{1}=C_{1}A_{1},$ $c_{1}=A_{1}B_{1}.$] Moreover, we know that $\measuredangle CAB=\measuredangle C_{1}A_{1}B_{1}.$ Show that triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar.

Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$ are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$

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.

Let $k$ be a circle with diameter $AB$. A point $C$ is chosen inside the segment $AB$ and a point $D$ is chosen on $k$ such that $BCD$ is an acute-angled triangle, with circumcentre denoted by $O$. Let $E$ be the intersection of the circle $k$ and the line $BO$ (different from $B$). Show that the triangles $BCD$ and $ECA$ are similar.

Suppose $P$ is an interior point of a triangle $ABC$ such that the ratios \[ \frac{d(A,BC)}{d(P,BC)} , \frac{d(B,CA)}{d(P,CA)} , \frac{d(C,AB)}{d(P,AB)} \] are all equal. Find the common value of these ratios. $d(X,YZ)$ represents the perpendicular distance fro $X$ to the line $YZ$.

Triangles $ABC$ and $A_1B_1C_1$ are similar and differently oriented. On the segment $AA_1$, a point $A'$ is taken such that $AA' / A_1A'= BC / B_1C_1$. We similarly construct $B'$ and $C'$. Prove that $A', B',C'$ lie on one straight line.

Let $ABC$ be a triangle inscribed in circle $\omega$ and $P$ a point in its interior. The lines $AP,BP$ and $CP$ intersect circle $\omega$ for the second time at $D,E$ and $F,$ respectively. If $A',B',C'$ are the reflections of $A,B,C$ with respect to the lines $EF,FD,DE,$ respectively, prove that the triangles $ABC$ and $A'B'C'$ are similar.

Let $ABC$ be a scalene triangle, and let $D$ be a point on side $BC$ satisfying $\angle BAD=\angle DAC$. Suppose that $X$ and $Y$ are points inside $ABC$ such that triangles $ABX$ and $ACY$ are similar and quadrilaterals $ACDX$ and $ABDY$ are cyclic. Let lines $BX$ and $CY$ meet at $S$ and lines $BY$ and $CX$ meet at $T$. Prove that lines $DS$ and $AT$ are parallel. [i]Michael Ren[/i]

A semicircle with diameter $d$ is contained in a square whose sides have length $8$. Given the maximum value of $d$ is $m- \sqrt{n}$, find $m+n$.

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 $\triangle{JHU}$.

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)

In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$ [asy] size(100); pathpen = linewidth(0.7); pointpen = black+linewidth(3); pair A = (0,0), C = (6,0), B = (6,8), D = (2*A+B)/3, E = (2*C+B)/3; D(D("A",A,SW)--D("B",B,NW)--D("C",C,SE)--cycle); D(D("D",D,NW)--D("E",E,plain.E)); D(rightanglemark(D,E,B,16)); D(rightanglemark(A,C,B,16));[/asy] $\mathbf{(A)}\;5\qquad \mathbf{(B)}\;\frac{16}{3}\qquad \mathbf{(C)}\; \frac{20}{3}\qquad \mathbf{(D)}\; \frac{15}{2}\qquad \mathbf{(E)}\; 8$

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.$

We construct similar isosceles triangles on the sides of the triangle $ ABC $: triangle $ APB $ outside the triangle $ ABC $ ($ AP = PB $), triangle $ CQA $ outside the triangle $ ABC $ ($ CQ = QA $), triangle $ CRB $ inside the triangle $ ABC $ ($ CR = RB $). Prove that $ APRQ $ is a parallelogram or that the points $ A, P, R, Q $ lie on a straight line.

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB$, $BC$, and $CD$ are diameters of circle $O$, $N$, and $P$, respectively. Circles $O$, $N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length [asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, dir(0)); label("$P$", P, S); label("$N$", N, S); label("$O$", O, S); label("$E$", E, dir(120)); label("$F$", F, NE); label("$G$", G, dir(100));[/asy] $\textbf {(A) } 20 \qquad \textbf {(B) } 15\sqrt{2} \qquad \textbf {(C) } 24 \qquad \textbf{(D) } 25 \qquad \textbf {(E) } \text{none of these}$

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}$