2013 OIM Problems/Problem 2

Problem

Let $X$ , $Y$ be the ends of a diameter of a circle $\Gamma$ and $N$ the the midpoint of one of the $XY$ arcs of $\Gamma$ . Let $A$ and $B$ be two points on the segment $XY$ . The straight lines $NA$ and $NB$ cut $\Gamma$ again at points $C$ and $D$ , respectively. The tangents to $\Gamma$ in $C$ and $D$ intersect at $P$ . Let $M$ be the point of intersection of segment $XY$ with segment $NP$ . Show that $M$ is the midpoint of segment $AB$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

This proof won't use the fact that XY is a diameter of $\Gamma$ and will prove it for every chord XY.

Let $M'$ be the midpoint of $CD$ and $S = AB \cap NM'$ . We observe that $NM' \equiv NS$ is the median and $NM$ is the symmedian of $\triangle NCD$ , hence $\angle CNM \equiv \angle ANM = \angle DNM' \equiv \angle BNS$ .

Therefore, it suffices to show that $NS$ is symmedian of $\triangle ABN$ , which is equivalent to $AB$ and $CD$ being antiparallel, in other words, we only need to prove that $ACDB$ is cyclic:

$\angle NCD = \frac {\overarc (DN)}{2} = \frac {\overarc (DY + YN)}{2} = \frac {\overarc (DY + NX)}{2} = \angle XYN + \angle DNY \equiv \angle BYN + \angle YNB = \angle ABN$ , where $\overarc PQ$ stands for the arc $PQ$ , which ends the problem

-zuat.e

Page

Toolbox

Search

2013 OIM Problems/Problem 2

Problem

Solution