Školjka

%V0 The vertices $D,E,F$ of an equilateral triangle lie on the sides $BC,CA,AB$ respectively of a triangle $ABC.$ If $a,b,c$ are the respective lengths of these sides, and $S$ the area of $ABC,$ prove that $$DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}.$$

%V0 Given is a convex polygon $P$ with $n$ vertices. Triangle whose vertices lie on vertices of $P$ is called good if all its sides are equal in length. Prove that there are at most $\frac {2n}{3}$ good triangles. Author: unknown author, Ukraine

%V0 Given an acute triangle $ABC$ with $\angle B > \angle C$. Point $I$ is the incenter, and $R$ the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $AD$ such that $AK = 2R$, and $D$ separates $A$ and $K$. Lines $DI$ and $KI$ meet sides $AC$ and $BC$ at $E,F$ respectively. Let $IE = IF$. Prove that $\angle B\leq 3\angle C$. Author: Davoud Vakili, Iran

%V0 Point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\omega$ be the incircle of triangle $CPD$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $APD$ and $BPC$ at points $K$ and $L$, respectively. Let lines $AC$ and $BD$ meet at $E$, and let lines $AK$ and $BL$ meet at $F$. Prove that points $E$, $I$, and $F$ are collinear. Author: Waldemar Pompe, Poland

%V0 Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_1$ and $O_2$ are circumcenters and points $H_1$ and $H_2$ are orthocenters of triangles $ABP$ and $CDP$, respectively. Denote the midpoints of segments $O_1H_1$ and $O_2H_2$ by $E_1$ and $E_2$, respectively. Prove that the perpendicular from $E_1$ on $CD$, the perpendicular from $E_2$ on $AB$ and the lines $H_1H_2$ are concurrent. Proposed by Ukraine

%V0 Let $ABC$ be a triangle with incenter $I$ and let $X$, $Y$ and $Z$ be the incenters of the triangles $BIC$, $CIA$ and $AIB$, respectively. Let the triangle $XYZ$ be equilateral. Prove that $ABC$ is equilateral too. Proposed by Mirsaleh Bahavarnia, Iran