Consider a circle $k$ with center $S$ and radius $r$. Denote $\mathsf M$ the set of all triangles with incircle $k$ such that the largest inner angle is twice bigger than the smallest one. For a triangle $\mathcal T\in\mathsf M$ denote its vertices $A,B,C$ in way that $SA\ge SB\ge SC$. Find the locus of points $\{B\mid\mathcal T\in\mathsf M\}$.

Let a semicircle $AB$ be given and let $X$ be an inner point of the arc. Consider a point $Y$ on ray $XA$ such that $XY=XB$. Find the locus of all points $Y$ when $X$ moves on the arc $AB$ (excluding the endpoints).

A segment $AB$ is given in (Euclidean) plane. Consider all triangles $XYZ$ such, that $X$ is an inner point of $AB$, triangles $XBY$ and $XZA$ are similar (in this order of vertices), and points $A, B, Y, Z$ lie on a circle in this order. Find the locus of midpoints of all such segments $YZ$. (Day 1, 2nd problem authors: Michal Rolínek, Jaroslav Švrček)

A segment $AB$ is given in (Euclidean) plane. Consider all triangles $XYZ$ such, that $X$ is an inner point of $AB$, triangles $XBY$ and $XZA$ are similar (in this order of vertices), and points $A, B, Y, Z$ lie on a circle in this order. Find the locus of midpoints of all such segments $YZ$. (Day 1, 2nd problem authors: Michal Rolínek, Jaroslav Švrček)

Let $ABCD$ be a convex quadrilateral. Determine all points $M$, which lie inside $ABCD$, such that the areas of $ABCM$ and $AMCD$ are equal.

Let $O$ denote the origin and let $\gamma$ be the circle with center $(1,0)$ and radius $1$ in the Cartesian system of coordinates. Let $\lambda$ be a real number from the interval $(0,2)$, and let the line $x=\lambda$ intersect the circle $\gamma$ at points $P$ and $Q$. The lines $OP$ and $OQ$ intersect the line $x=2-\lambda$ at the points $P'$ and $Q'$, respectively. Let $\mathcal G$ denote the locus of such points $P'$ and $Q'$ as $\lambda$ varies over the interval $(0,2)$. Prove that there exist points $R$ and $S$ different from the origin in the plane such that for every $A\in \mathcal G$ there exists a point $A'$ on line $OA$ satisfying \[ A'R^2=(A'S-OS)^2=A'A\cdot A'O.\] [i]Proposed by: Áron Bán-Szabó, Budapest[/i]