A [i]diagonal line[/i] of a (not necessarily convex) polygon with at least four sides is any line through two non-adjacent vertices of that polygon. Determine all polygons with at least four sides satisfying the following condition: The reflexion of each vertex in each diagonal line lies inside or on the boundary of the polygon. [i]The Problem Selection Committee[/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 $CF$.

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 $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

Let $ABC$ be a triangle with $CA>BC>AB$. Let $O$ and $H$ be the circumcentre and orthocentre of triangle $ABC$ respectively. Denote by $D$ and $E$ the midpoints of the arcs $AB$ and $AC$ of the circumcircle of triangle $ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ about $AB$ and $E'$ the reflection of $E$ about $AC$. Prove that $O,H,D',E'$ are concylic if and only if $A,D',E'$ are collinear.

Let $ABC$ be a triangle and its bisector at $A$ cuts its circumcircle at $D.$ Let $I$ be the incenter of triangle $ABC,$ $M$ be the midpoint of $BC,$ $P$ is the symmetric to $I$ with respect to $M$ (Assuming $P$ is in the circumcircle). Extend $DP$ until it cuts the circumcircle again at $N.$ Prove that among segments $AN, BN, CN$, there is a segment that is the sum of the other two.

We say a finite set $S$ of points in the plane is [i]very[/i] if for every point $X$ in $S$, there exists an inversion with center $X$ mapping every point in $S$ other than $X$ to another point in $S$ (possibly the same point). (a) Fix an integer $n$. Prove that if $n \ge 2$, then any line segment $\overline{AB}$ contains a unique very set $S$ of size $n$ such that $A, B \in S$. (b) Find the largest possible size of a very set not contained in any line. (Here, an [i]inversion[/i] with center $O$ and radius $r$ sends every point $P$ other than $O$ to the point $P'$ along ray $OP$ such that $OP\cdot OP' = r^2$.) [i]Proposed by Sammy Luo[/i]

Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.

In a acute triangle $\triangle ABC$, denote $D, E$ as the foot of the perpendicular from $B$ to $AC$ and $C$ to $AB$. Denote the reflection of $E$ with respect to $AC, BC$ as $S, T$. The circumcircle of $\triangle CST$ hits $AC$ at point $X (\not= C)$. Denote the circumcenter of $\triangle CST$ as $O$. Prove that $XO \perp DE$.

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color. Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?

A triangle $ABC$ is given together with an arbitrary circle $\omega$. Let $\alpha$ be the reflection of $\omega$ with respect to $A$, $\beta$ the reflection of $\omega$ with respect to $B$, and $\gamma$ the reflection of $\omega$ with respect to $C$. It is known that the circles $\alpha, \beta, \gamma$ don't intersect each other. Let $P$ be the meeting point of the two internal common tangents to $\beta, \gamma$ (see picture). Similarly, $Q$ is the meeting point of the internal common tangents of $\alpha, \gamma$, and $R$ is the meeting point of the internal common tangents of $\alpha, \beta$. Prove that the triangles $PQR, ABC$ are congruent.

Let $A$ be point in the exterior of the circle $\mathcal C$. Two lines passing through $A$ intersect the circle $\mathcal C$ in points $B$ and $C$ (with $B$ between $A$ and $C$) respectively in $D$ and $E$ (with $D$ between $A$ and $E$). The parallel from $D$ to $BC$ intersects the second time the circle $\mathcal C$ in $F$. Let $G$ be the second point of intersection between the circle $\mathcal C$ and the line $AF$ and $M$ the point in which the lines $AB$ and $EG$ intersect. Prove that \[ \frac 1{AM} = \frac 1{AB} + \frac 1{AC}. \]

Two circles intersect at two points $A$ and $B$. A line $\ell$ which passes through the point $A$ meets the two circles again at the points $C$ and $D$, respectively. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ (which do not contain the point $A$) on the respective circles. Let $K$ be the midpoint of the segment $CD$. Prove that $\measuredangle MKN = 90^{\circ}$.

Let $H$ be the orthocentre of an acute triangle $ABC$. The line through $A$ perpendicular to $AC$ and the line through $B$ perpendicular to $BC$ intersect in $D$. The circle with centre $C$ through $H$ intersects the circumcircle of triangle $ABC$ in the points $E$ and $F$. Prove that $|DE| = |DF| = |AB|$.

In a scalene triangle $ABC$, $D$ is the foot of the altitude through $A$, $E$ is the intersection of $AC$ with the bisector of $\angle ABC$ and $F$ is a point on $AB$. Let $O$ the circumcenter of $ABC$ and $X=AD\cap BE$, $Y=BE\cap CF$, $Z=CF \cap AD$. If $XYZ$ is an equilateral triangle, prove that one of the triangles $OXY$, $OYZ$, $OZX$ must be equilateral.

Acute-angled triangle $ABC$ is inscribed into circle $\Omega$. Lines tangent to $\Omega$ at $B$ and $C$ intersect at $P$. Points $D$ and $E$ are on $AB$ and $AC$ such that $PD$ and $PE$ are perpendicular to $AB$ and $AC$ respectively. Prove that the orthocentre of triangle $ADE$ is the midpoint of $BC$.

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Concentric circles $\Omega_1$ and $\Omega_2$ with radii $1$ and $100$, respectively, are drawn with center $O$. Points $A$ and $B$ are chosen independently at random on the circumferences of $\Omega_1$ and $\Omega_2$, respectively. Denote by $\ell$ the tangent line to $\Omega_1$ passing through $A$, and denote by $P$ the reflection of $B$ across $\ell$. Compute the expected value of $OP^2$. [i]Proposed by Lewis Chen[/i]

$ ABCD$ is a quadrilateral. $ A'BCD'$ is the reflection of $ ABCD$ in $ BC,$ $ A''B'CD'$ is the reflection of $ A'BCD'$ in $ CD'$ and $ A''B''C'D'$ is the reflection of $ A''B'CD'$ in $ D'A''.$ Show that; if the lines $ AA''$ and $ BB''$ are parallel, then ABCD is a cyclic quadrilateral.

Let $ABCD$ an inscribed quadrilateral and $r$ and $s$ the reflections of the straight line through $A$ and $B$ over the inner angle bisectors of angles $\angle{CAD}$ and $\angle{CBD}$, respectively. Let $P$ the point of intersection of $r$ and $s$ and let $O$ the circumcentre of $ABCD$. Prove that $OP \perp CD$.

Let $ ABC$ be a triangle and let $ I$ be the incenter. $ Ia$ $ Ib$ and $ Ic$ are the excenters opposite to points $ A$ $ B$ and $ C$ respectively. Let $ La$ be the line joining the orthocenters of triangles $ IBC$ and $ IaBC$. Define $ Lb$ and $ Lc$ in the same way. Prove that $ La$ $ Lb$ and $ Lc$ are concurrent. Daniel

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]

Suppose a circle $C$ of radius $\sqrt2$ touches the $Y$ -axis at the origin $(0, 0)$. A ray of light $L$, parallel to the $X$-axis, reflects on a point $P$ on the circumference of $C$, and after reflection, the reflected ray $L'$ becomes parallel to the $Y$ -axis. Find the distance between the ray $L$ and the $X$-axis.

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 $CF$.