Olympiad problems

AND:

OR:

NO:

Found problems: 2

2019 Philippine TST, 2

In a triangle $ABC$ with circumcircle $\Gamma$, $M$ is the midpoint of $BC$ and point $D$ lies on segment $MC$. Point $G$ lies on ray $\overrightarrow{BC}$ past $C$ such that $\frac{BC}{DC} = \frac{BG}{GC}$, and let $N$ be the midpoint of $DG$. The points $P$, $S$, and $T$ are defined as follows: [list = i] [*] Line $CA$ meets the circumcircle $\Gamma_1$ of triangle $AGD$ again at point $P$. [*] Line $PM$ meets $\Gamma_1$ again at $S$. [*] Line $PN$ meets the line through $A$ that is parallel to $BC$ at $Q$. Line $CQ$ meets $\Gamma$ again at $T$. [/list] Prove that the points $P$, $S$, $T$, and $C$ are concyclic.

2023 Euler Olympiad, Round 2, 4

Tags: geometry , trapezoid , euler , right angle , harmonic bundles

Let $ABCD$ be a trapezoid, with $AD \parallel BC$, let $M$ be the midpoint of $AD$, and let $C_1$ be symmetric point to $C$ with respect to line $BD$. Segment $BM$ meets diagonal $AC$ at point $K$, and ray $C_1K$ meets line $BD$ at point $H$. Prove that $\angle{AHD}$ is a right angle. [i]Proposed by Giorgi Arabidze, Georgia[/i]