On the plane is give a broken line $ABCD$ in which $AB = BC = CD = 1$, and $AD$ is not equal to $1$. The positions of $B$ and $C$ are fixed but $A$ and $D$ change their positions in turn according to the following rule (preserving the distance rules given): the point $A$ is reflected with respect to the line $BD$, then $D$ is reflected with respect to the line $AC$ (in which $A$ occupies its new position), then $A$ is reflected with respect to the line $BD$ ($D$ occupying its new position), $D$ is reflected with respect to the line $AC$, and so on. Prove that after several steps $A$ and $D$ coincide with their initial positions. (M Kontzewich)

Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.

Let $O$ be the circumcenter and $I$ be the incenter of an acute triangle $ABC$ with $m(\widehat{B}) \neq m(\widehat{C})$. Let $D$, $E$, $F$ be the midpoints of the sides $[BC]$, $[CA]$, $[AB]$, respectively. Let $T$ be the foot of perpendicular from $I$ to $[AB]$. Let $P$ be the circumcenter of the triangle $DEF$ and $Q$ be the midpoint of $[OI]$. If $A$, $P$, $Q$ are collinear, prove that \[\dfrac{|AO|}{|OD|}-\dfrac{|BC|}{|AT|}=4.\]

Let $K_a,K_b,K_c$ with centres $O_a,O_b,O_c$ be the excircles of a triangle $ABC$, touching the interiors of the sides $BC,CA,AB$ at points $T_a,T_b,T_c$ respectively. Prove that the lines $O_aT_a,O_bT_b,O_cT_c$ are concurrent in a point $P$ for which $PO_a=PO_b=PO_c=2R$ holds, where $R$ denotes the circumradius of $ABC$. Also prove that the circumcentre $O$ of $ABC$ is the midpoint of the segment $PI$, where $I$ is the incentre of $ABC$.

Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.

A pool table has the shape of a triangle whose angles are in a rational ratio. A ball positioned at an interior point of the table is hit by a stick. The ball reflects from the sides of the triangle according to the law of reflection. Prove that the ball will move only along a finite number of segments. (It is assumed that the ball does not reach the vertices of the triangle.)

Let the coordinate planes have the reflection property. A ray falls onto one of them. How does the final direction of the ray after reflecting from all three coordinate planes depend on its initial direction?

Two circles $C_1$ and $C_2$ with the respective radii $r_1$ and $r_2$ intersect in $A$ and $B.$ A variable line $r$ through $B$ meets $C_1$ and $C_2$ again at $P_r$ and $Q_r$ respectively. Prove that there exists a point $M,$ depending only on $C_1$ and $C_2,$ such that the perpendicular bisector of each segment $P_rQ_r$ passes through $M.$

Let $ABC$ be a given triangle with circumcenter $O$ and orthocenter $H$. Let $D, E$ and $F$ be the feet of the perpendiculars from $A, B$ and $C$ to the opposite sides, respectively. Let $A'$ be the reflection of $A$ with respect to $EF$. Prove that $HOA'D$ is a cyclic quadrilateral. [i]Proposed by Imad Uddin Ahmad Hasin[/i]

In an acute-angled triangle $ABC$ the angle at $B$ is greater than $45^\circ$. Points $D,E, F$ are the feet of the altitudes from $A,B,C$ respectively, and $K$ is the point on segment $AF$ such that $\angle DKF = \angle KEF$. (a) Show that such a point $K$ always exists. (b) Prove that $KD^2 = FD^2 + AF \cdot BF$.

Let $\Gamma$ be the circumcircle of an acute triangle $ABC.$ The perpendicular to $AB$ from $C$ meets $AB$ at $D$ and $\Gamma$ again at $E.$ The bisector of angle $C$ meets $AB$ at $F$ and $\Gamma$ again at $G.$ The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I.$ Prove that $AI = EB.$

The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^{\circ}$ around the point $(1,5)$ and then reflected about the line $y=-x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b-a$? $\textbf{(A) }1 \qquad \textbf{(B) }3 \qquad \textbf{(C) }5 \qquad \textbf{(D) }7 \qquad \textbf{(E) }9$

Let $ABC$ be an acute triangle with $\angle BAC=30^{\circ}$. The internal and external angle bisectors of $\angle ABC$ meet the line $AC$ at $B_1$ and $B_2$, respectively, and the internal and external angle bisectors of $\angle ACB$ meet the line $AB$ at $C_1$ and $C_2$, respectively. Suppose that the circles with diameters $B_1B_2$ and $C_1C_2$ meet inside the triangle $ABC$ at point $P$. Prove that $\angle BPC=90^{\circ}$ .

Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$. [i]Proposed by N. Beluhov[/i]

In a triangle $ABC$, $M$ and $N$ are the respective midpoints of the sides $AC$ and $AB$, and $P$ is the point of intersection of $BM$ and $CN$. Prove that, if it is possible to inscribe a circle in the quadrilateral $AMPN$, then the triangle $ABC$ is isosceles.

Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold. (Here we denote $XY$ the length of the line segment $XY$.)

Two circles $\Omega$ and $\Gamma$ are internally tangent at the point $B$. The chord $AC$ of $\Gamma$ is tangent to $\Omega$ at the point $L$, and the segments $AB$ and $BC$ intersect $\Omega$ at the points $M$ and $N$. Let $M_1$ and $N_1$ be the reflections of $M$ and $N$ about the line $BL$; and let $M_2$ and $N_2$ be the reflections of $M$ and $N$ about the line $AC$. The lines $M_1M_2$ and $N_1N_2$ intersect at the point $K$. Prove that the lines $BK$ and $AC$ are perpendicular. [i](M. Karpuk)[/i]

Consider two polygons $ P$ and $ Q$. We want to cut $ P$ into some smaller polygons and put them together in such a way to obtain $ Q$. We can translate the pieces but we can not rotate them or reflect them. We call $ P,Q$ equivalent if and only if we can obtain $ Q$ from $ P$(which is obviously an equivalence relation). [img]http://i3.tinypic.com/4lrb43k.png[/img] a) Let $ P,Q$ be two rectangles with the same area(their sides are not necessarily parallel). Prove that $ P$ and $ Q$ are equivalent. b) Prove that if two triangles are not translation of each other, they are not equivalent. c) Find a necessary and sufficient condition for polygons $ P,Q$ to be equivalent.

The circle $\Gamma$ and the line $\ell$ have no common points. Let $AB$ be the diameter of $\Gamma$ perpendicular to $\ell$, with $B$ closer to $\ell$ than $A$. An arbitrary point $C\not= A$, $B$ is chosen on $\Gamma$. The line $AC$ intersects $\ell$ at $D$. The line $DE$ is tangent to $\Gamma$ at $E$, with $B$ and $E$ on the same side of $AC$. Let $BE$ intersect $\ell$ at $F$, and let $AF$ intersect $\Gamma$ at $G\not= A$. Let $H$ be the reflection of $G$ in $AB$. Show that $F,C$, and $H$ are collinear.

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Let $ABC$ be a triangle. Let $C^{\prime}$ and $A^{\prime}$ be the reflections of its vertices $C$ and $A$, respectively, in the altitude of triangle $ABC$ issuing from $B$. The perpendicular to the line $BA^{\prime}$ through the point $C^{\prime}$ intersects the line $BC$ at $U$; the perpendicular to the line $BC^{\prime}$ through the point $A^{\prime}$ intersects the line $BA$ at $V$. Prove that $UV \parallel CA$. Darij

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

In the triangle $ABC$ we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, and $\angle MCB=55^\circ$, $\angle NBC=80^\circ$. Find $\angle NMC$. [i]St.Petersburg folklore[/i]