Library / Mongolian National Mathematical Olympiad /13 of 31

Mongolian National Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Mongolia

Let $ABCD$ be a circumscribed quadrilateral with circumcenter $O$ and let $K$ , $L$ , $M$ and $N$ denote the midpoints of the sides $AB$ , $BC$ , $CD$ and $DA$ respectively. Suppose that the lines $KM$ and $LN$ do not pass through $O$ . Let $E := KM \cap LN$ . Point $P$ is chosen on the interval $KM$ to satisfy $\angle KOE = \angle MOP$ and point $Q$ is chosen on the interval $LN$ to satisfy $\angle LOE = \angle NOQ$ . Show that the points $O$ , $P$ and $Q$ are collinear.

(Batzaya G.)

Solution

It suffices to prove that $\angle POQ = 180^\circ$ .

Since $O$ is the circumcenter, the points $K$ , $L$ , $M$ , $N$ are the feet of the perpendiculars from $O$ to the sides of $ABCD$ . Hence $OMDN$ and $OKBL$ are circumscribed. Hence
$\begin{align*} \angle POQ &= \angle POM + \angle MON + \angle NOQ \\ &= \angle KOE + (180^\circ - \angle MDN) + \angle LOE \\ &= \angle KOL + \angle KBL \\ &= 180^\circ. \end{align*}$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.