All of the polygons in the figure below are regular. Prove that $ABCD$ is an isosceles trapezoid. Proposed by Mahdi Etesamifard - Iran

In an isosceles triangle $ABC$ with $AB = AC$ and $\angle A = 30^o$, points $L$ and $M$ lie on the sides $AB$ and $AC$, respectively such that $AL = CM$. Point $K$ lies on $AB$ such that $\angle AMK = 45^o$. If $\angle LMC = 75^o$, prove that $KM +ML = BC$. Proposed by Mahdi Etesamifard - Iran

Let $ABCD$ be a square with side length $1$. How many points $P$ inside the square (not on its sides) have the property that the square can be cut into $10$ triangles of equal area such that all of them have $P$ as a vertex? Proposed by Josef Tkadlec - Czech Republic

Let $ABCD$ be a convex quadrilateral. Let $E$ be the intersection of its diagonals. Suppose that $CD = BC = BE$. Prove that $AD + DC\ge AB$. Proposed by Dominik Burek - Poland

A polygon is decomposed into triangles by drawing some non-intersecting interior diagonals in such a way that for every pair of triangles of the triangulation sharing a common side, the sum of the angles opposite to this common side is greater than $180^o$. a) Prove that this polygon is convex. b) Prove that the circumcircle of every triangle used in the decomposition contains the entire polygon. Proposed by Morteza Saghafian - Iran

Points $M$ and $N$ are the midpoints of sides $AB$ and $BC$ of the square $ABCD$. According to the fgure, we have drawn a regular hexagon and a regular $12$-gon. The points $P, Q$ and $R$ are the centers of these three polygons. Prove that $PQRS$ is a cyclic quadrilateral. Proposed by Mahdi Etesamifard - Iran

A convex hexagon $ABCDEF$ with an interior point $P$ is given. Assume that $BCEF$ is a square and both $ABP$ and $PCD$ are right isosceles triangles with right angles at $B$ and $C$, respectively. Lines $AF$ and $DE$ intersect at $G$. Prove that $GP$ is perpendicular to $BC$. Proposed by Patrik Bak - Slovakia

Let $\omega$ be the circumcircle of the triangle $ABC$ with $\angle B = 3\angle C$. The internal angle bisector of $\angle A$, intersects $\omega$ and $BC$ at $M$ and $D$, respectively. Point $E$ lies on the extension of the line $MC$ from $M$ such that $ME$ is equal to the radius of $\omega$. Prove that circumcircles of triangles $ACE$ and $BDM$ are tangent. Proposed by Mehran Talaei - Iran

Let $ABC$ be a triangle and $P$ be the midpoint of arc $BAC$ of circumcircle of triangle $ABC$ with orthocenter $H$. Let $Q, S$ be points such that $HAPQ$ and $SACQ$ are parallelograms. Let $T$ be the midpoint of $AQ$, and $R$ be the intersection point of the lines $SQ$ and $PB$. Prove that $AB$, $SH$ and $TR$ are concurrent. Proposed by Dominik Burek - Poland

There are $n$ points in the plane such that at least $99\%$ of quadrilaterals with vertices from these points are convex. Can we find a convex polygon in the plane having at least $90\%$ of the points as vertices? Proposed by Morteza Saghafian - Iran

We are given an acute triangle $ABC$. The angle bisector of $\angle BAC$ cuts $BC$ at $P$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, so that $BC \parallel DE$. Points $K$ and $L$ lie on segments $PD$ and $PE$, respectively, so that points $A$, $D$, $E$, $K$, $L$ are concyclic. Prove that points $B$, $C$, $K$, $L$ are also concyclic. Proposed by Patrik Bak, Slovakia

Let ${I}$ be the incenter of $\triangle {ABC}$ and ${BX}$, ${CY}$ are its two angle bisectors. ${M}$ is the midpoint of arc $\overset{\frown}{BAC}$. It is known that $MXIY$ are concyclic. Prove that the area of quadrilateral $MBIC$ is equal to that of pentagon $BXIYC$. Proposed by Dominik Burek - Poland

There are several discs whose radii are no more that $1$, and whose centers all lie on a segment with length ${l}$. Prove that the union of all the discs has a perimeter not exceeding $4l+8$. Proposed by Morteza Saghafian - Iran

Let $ABC$ be a triangle with bisectors $BE$ and $CF$ meet at $I$. Let $D$ be the projection of $I$ on the $BC$. Let M and $N$ be the orthocenters of triangles $AIF$ and $AIE$, respectively. Lines $EM$ and $FN$ meet at $P.$ Let $X$ be the midpoint of $BC$. Let $Y$ be the point lying on the line $AD$ such that $XY \perp IP$. Prove that line $AI$ bisects the segment $XY$. Proposed by Tran Quang Hung - Vietnam

In triangle $ABC$ points $M$ and $N$ are the midpoints of sides $AC$ and $AB$, respectively and $D$ is the projection of $A$ into $BC$. Point $O$ is the circumcenter of $ABC$ and circumcircles of $BOC$, $DMN$ intersect at points $R, T$. Lines $DT$, $DR$ intersect line $MN$ at $E$ and $F$, respectively. Lines $CT$, $BR$ intersect at $K$. A point $P$ lies on $KD$ such that $PK$ is the angle bisector of $\angle BPC$. Prove that the circumcircles of $ART$ and $PEF$ are tangent. Proposed by Mehran Talaei - Iran