Let $ABCDEF$ be an inscribed hexagon with $$AB.CD.EF=BC.DE.FA$$ Let $B_1$ be the reflection point of $B$ with respect to $AC$ and $D_1$ be the reflection point of $D$ with respect to $CE,$ and finally let $F_1$ be the reflection point of $F$ with respect to $AE.$ Prove that $\triangle B_1D_1F_1\sim BDF.$

Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on $\left[AB\right]$. $Q$ is the point on $BC$ such that $PQ$ is perpendicular to $AB$. $R$ is the point on $AC$ such that $QR$ is perpendicular to $BC$. And $S$ is the point on $AB$ such that $RS$ is perpendicular to $AC$. $Q'$ is the point on $BC$ such that $PQ'$ is perpendicular to $BC$. $R'$ is the point on $AC$ such that $Q'R'$ is perpendicular to $AC$. And $S'$ is the point on $AB$ such that $R'S'$ is perpendicular to $AB$. Determine $\frac{|PB|}{|AB|}$ if $S=S'$.

$C$ is a point on the semicircle diameter $AB$, between $A$ and $B$. $D$ is the foot of the perpendicular from $C$ to $AB$. The circle $K_1$ is the incircle of $ABC$, the circle $K_2$ touches $CD,DA$ and the semicircle, the circle $K_3$ touches $CD,DB$ and the semicircle. Prove that $K_1,K_2$ and $K_3$ have another common tangent apart from $AB$.

In an acute triangle $\bigtriangleup ABC$, $AB<AC<BC$, and $A_1,B_1,C_1$ are the projections of $A,B,C$ to the corresponding sides. Let the reflection of $B_1$ wrt $CC_1$ be $Q$, and the reflection of $C_1$ wrt $BB_1$ be $P$. Prove that the circumcirle of $A_1PQ$ passes through the midpoint of $BC$.

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$. [asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]

Let $ABC$ be an acute triangle. Points $B'$ and $C'$ are located on the interior of sides $AB$ and $AC$, respectively. Let $M$ denote the second intersection of the circumcircles of triangles $ABC$ and $AB'C'$, while let $N$ denote the second intersection of the circumcircles of triangles $ABC'$ and $AB'C$. Reflect $M$ across lines $AB$ and $AC$, and let $l$ denote the line through the reflections. a) Prove that the line through $M$ perpendicular to $AM$, the line $AK$, and $l$ are either concurrent or all parallel. b) Show that if the three lines are concurrent at $S$, then triangles $SBC'$ and $SCB'$ have equal areas. [i]Proposed by Áron Bán-Szabó, Budapest[/i]

Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$. Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.

Point $P$ is taken in the interior of the triangle $ABC$, so that \[\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).\] Let $L$ be the foot of the angle bisector of $\angle B$. The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$. Prove that $QB$ is the angle bisector of $\angle AQC$. [i]Proposed by S. Berlov[/i]

Let $ABC$ be a triangle. Prove that there exists a unique point $P$ for which one can find points $D$, $E$ and $F$ such that the quadrilaterals $APBF$, $BPCD$, $CPAE$, $EPFA$, $FPDB$, and $DPEC$ are all parallelograms. [i]Proposed by Lewis Chen[/i]

Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \angle CAD$ and $ \angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.

Let $ABC$ be an acute angled triangle. Let $D,E,F$ be points on $BC, CA, AB$ such that $AD$ is the median, $BE$ is the internal bisector and $CF$ is the altitude. Suppose that $\angle FDE=\angle C, \angle DEF=\angle A$ and $\angle EFD=\angle B.$ Show that $ABC$ is equilateral.

Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible? [asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy] $\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Let squares be constructed on the sides $BC,CA,AB$ of a triangle $ABC$, all to the outside of the triangle, and let $A_1,B_1, C_1$ be their centers. Starting from the triangle $A_1B_1C_1$ one analogously obtains a triangle $A_2B_2C_2$. If $S, S_1, S_2$ denote the areas of triangles$ ABC,A_1B_1C_1,A_2B_2C_2$, respectively, prove that $S = 8S_1 - 4S_2.$

The figure shows a triangular pool table with sides $a$, $b$ and $c$. Located at point $S$ on $c$ a sphere - which can be assumed as a point. After kick-off, as indicated in the figure, it runs through as a result of reflections to $a, b, a, b$ and $c$ (in $S$) always the same track. The reflection occurs according to law of reflection. Characterize entilrely all triangles $ABC$, which allow such an orbit, and determine the locus of $S$. [img]https://cdn.artofproblemsolving.com/attachments/5/b/7662943e5b9ad321226e0c5f5daa3c4ac9faaa.png[/img]

(A.Zaslavsky, 9--10) The reflections of diagonal $ BD$ of a quadrilateral $ ABCD$ in the bisectors of angles $ B$ and $ D$ pass through the midpoint of diagonal $ AC$. Prove that the reflections of diagonal $ AC$ in the bisectors of angles $ A$ and $ C$ pass through the midpoint of diagonal $ BD$ (There was an error in published condition of this problem).

On the real number line, paint red all points that correspond to integers of the form $81x+100y$, where $x$ and $y$ are positive integers. Paint the remaining integer point blue. Find a point $P$ on the line such that, for every integer point $T$, the reflection of $T$ with respect to $P$ is an integer point of a different colour than $T$.

Let $ABCD$ be a cyclic quadrilateral. The lines $BC$ and $AD$ meet at a point $P$. Let $Q$ be the point on the line $BP$, different from $B$, such that $PQ=BP$. Consider the parallelograms $CAQR$ and $DBCS$. Prove that the points $C,Q,R,S$ lie on a circle.

Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

In $\triangle ABC$, $\angle ABC=90^{\circ}$. Points $D,G$ lie on side $AC$. Points $E, F$ lie on segment $BD$, such that $AE \perp BD $ and $GF \perp BD$. Show that if $BE=EF$, then $\angle ABG=\angle DFC$.

Let $I$ be the incentre of a triangle $ABC$ and $\Gamma_a$ be the excircle opposite $A$ touching $BC$ at $D$. If $ID$ meets $\Gamma_a$ again at $S$, prove that $DS$ bisects $\angle BSC$.

In $\triangle ABC$, $AB>AC$. In the circumcircle $(O)$ of $\triangle ABC$, $M$ is the midpoint of arc $BAC$. The incircle $(I)$ of $\triangle ABC$ touches $BC$ at $D$, the line through $D$ parallel to $AI$ intersects $(I)$ again at $P$. Prove that $AP$ and $IM$ intersect at a point on $(O)$.

Let $ABC$ be an acute triangle inscribed in a circle centered at $O$. Let $M$ be a point on the small arc $AB$ of the triangle's circumcircle. The perpendicular dropped from $M$ on the ray $OA$ intersects the sides $AB$ and $AC$ at the points $K$ and $L$, respectively. Similarly, the perpendicular dropped from $M$ on the ray $OB$ intersects the sides $AB$ and $BC$ at $N$ and $P$, respectively. Assume that $KL=MN$. Find the size of the angle $\angle{MLP}$ in terms of the angles of the triangle $ABC$.