## A Geometry Problem from the

2007 Irish Mathematical Olympiad

Steven R. Dunbar

Department of Mathematics

University of Nebraska-Lincoln

Lincoln, Nebraska

www.math.unl.edu/$\sim $sdunbar1

sdunbar1 at unl dot edu

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.

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.