## A Geometry Problem from the 2007 Irish Mathematical Olympiad

January 25, 2009

Problem: Problem 3 from the Twentieth Irish Mathematical Olympiad from Saturday, 12 May 2007. Collected at the 2008 International Mathematical Olympiad, Madrid, Spain, July 2008.

The point $P$ is a fixed point on a circle and $Q$ is a fixed point on a line. The point $R$ is a variable point on the circle and $P$, $Q$, and $R$ are not collinear. The circle through $P$, $Q$, and $R$ meets the line again at $V$. Show that the line $VR$ passes through a fixed point.

Solution: From the official solutions of the Twentieth Irish Mathematical Olympiad from Saturday, 12 May 2007. There are several diagrams possible depending on relative positions of the circle, line and point $P$. In one case, $\angle PRV$ and $\angle PQV$ are equal and in the other case are complementary. $\angle PRV=180-\angle PQV$ as $PQVR$ is a cyclic quadrilateral. $S$ is the intersection of $VR$ and the circle. $T$ is the intersection of the line $PQ$ and the circle. $\angle PRV=\angle STP$. Thus $S$ is a fixed point on the circle and $S$ is on $VR$. Hence $VR$ always passes through $S$. The other case is similar.

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Second solution: Let $X$ be the other point where $PQ$ meets the circle and let $W$ be the other point where the line through $X$ parallel to $QV$ meets the circle, clearly $W$ is a fixed point on the circle. Now $PQVR$ is a cyclic quadrilateral. Therefore $\angle PRV=18{0}^{\circ }-\angle PQV$. Similarly $PXWR$ is a cyclic quadrilateral, so $\angle PRW=18{0}^{\circ }-\angle PXW$. But $\angle PXW=\angle PQV$. Therefore $\angle PRW=\angle PRV$. Hence, $R$, $W$ and $V$ are collinear, as required. (Note that different configurations are possible depending on which side of the line $PQ$ the point $R$ lies. However, the arguments are essentially the same in all cases.)

Commentary: This was a problem that I did not see how to solve, so ultimately I entered the official solution. It was obvious to see the common point $S$ from the Geogebra diagram, but I could see no way to show that it was fixed. I think the key was that I was not prepared to see the cyclic quadrilateral. Note that a cyclic quadrilateral is central to both solutions. Many of these Olympiad problems depend on cyclic quadrilaterals.