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/ 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 V R 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, PRV and PQV are equal and in the other case are complementary. PRV = 180 PQV as PQV R is a cyclic quadrilateral. S is the intersection of V R and the circle. T is the intersection of the line PQ and the circle. PRV = STP. Thus S is a fixed point on the circle and S is on V R. Hence V R 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 PQV R is a cyclic quadrilateral. Therefore PRV = 180PQV . Similarly PXWR is a cyclic quadrilateral, so PRW = 180PXW. But PXW = PQV . Therefore PRW = 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.