AoPS - Problem

Source: Cono Sur 2004 #2

Tags: geometry, cono sur

fprosk

18.11.2015 17:22

Given a circle $C$ and a point $P$ on its exterior, two tangents to the circle are drawn through $P$, with $A$ and $B$ being the points of tangency. We take a point $Q$ on the minor arc $AB$ of $C$. Let $M$ be the intersection of $AQ$ with the line perpendicular to $AQ$ that goes through $P$, and let $N$ be the intersection of $BQ$ with the line perpendicular to $BQ$ that goes through $P$. Show that, by varying $Q$ on the minor arc $AB$, all of the lines $MN$ pass through the same point.

Let $D$ be the midpoint of $AB$ and $X$ the midpoint of $DP$. I will show that $DMPN$ is always a parallelogramm, hence $X$ is the common midpoint of $MN$ and $DP$. In particular, $MN$ passes through $X$ which is fixed. Since $ \angle BNP = \angle BDP = 90^\circ$, the quadrilateral $BPND$ is cyclic. Hence, we get $$\angle NDA = \angle NPB = 90^\circ - \angle PBN = 90^\circ - \angle PBQ = 90^\circ - \angle BAQ = 90^\circ - \angle DAQ,$$which implies that $DN \perp AQ$. Since by definition also $PM \perp AQ$, we conclude that $PM \parallel DN$. Analogously we can prove that $DM \parallel PN$, which finishes the proof.