Geometry Problem 2 from the
1999 Bay Area Mathematical Olympiad

Steven R. Dunbar
Department of Mathematics
Univesity of Nebraska-Lincoln
Lincoln, Nebraska www.math.unl.edu/ sdunbar1
sdunbar1 at unl dot edu

January 28, 2009

Problem: Let ABCD be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let E and F be variable points on the sides AB and CD, respectively, such that AEEB = CFFD. Let P be the point on the segment such that PEPF = ABCD. Prove that the ratio between the areas of APD and BPC does not depend on the choice of E and F.

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Solution: If AD and BC are parallel, then ABCD is a trapezoid. Because ABCD is a cyclic quadrilateral, BAD + DCB = 180. By the parallelism, BCD is equal to the angle supplementary to ADC. Then ADC = BAD. Then the trapezoid ABCD is isosceles, that is, AB = CD. Then PE = PF, and P is the midpoint of EF¯. Let Q be the foot of the altitude of APD to AD¯ and R be the foot of the altitude of BPC. Then PR = PQ. Hence the ratio between the areas of APD and BPC depends only on the ratio ADBC.

Assume AD and BC are not parallel. Let the intersection point be G. Because ABCD is a cyclic quadrilateral, DAB + DCB = 180. Also, DCG + DCB = 180, hence DAB = DCG. Since the angle at G is in common, GAB is similar to GCD, with similarity ratio ABCD. From the given information, ABCD = AECF = PEPF. The points E and F are corresponding points in GAB and GCD so that GAE is similar to GCF with similarity ratio ABCD = AECF = PEPF. Thus, GEGF = PEPF. Hence, GP is the angle bisector of AGB. Therefore, d(P,AG¯) = d(P,BC¯). That is, the altitudes of the two triangles in question, APD and BPC. Then the ratio of the areas of the two triangles depends only on the ratio of their bases ADBC, and is independent of the choice of E and F.

Commentary: Zuming Feng told me that this was an IMO Short List problem. A little searching found it on the MathLinks Forum, along with some solutions as 1998 IMO Shortlist Geometry Problem 2. The proof presented here is an adaptation of the proof by Yimin Ge, posted on the MathLinks Forum, June 28, 2006.

There’s no obvious similarity in this problem initially, but the statement in terms of ratios suggests that there ought to be some similar figures around. Extending AB¯ and BC¯ to intersect in point G is a good idea to create some triangles involving the ratio ABCD. Using given information, the property of opposite angles in a cyclic quadrilateral establishes the similarity. Of course, the creation of the auxiliary point G assumes that the lines are not parallel, dividing the problem into cases. However, the non-parallel case is easy. So this problem is another example of solving a geometry problem by judiciously adding in points and lines to illuminate special relationships. It’s a little harder to see in this case, because the created AGB and CGD and therefore AGE and CGF are reversed in orientation.

From the point of view of GeoGebra this was an interesting problem since the entry of points E, F and P was easier with algebraic entry rather than geometric entry. Also, since the ratio AEEB is variable, it is natural to use a slider to represent it. Once the ratio is set, then it easy to set E as a convex combination of points A and B. Likewise, represent F as a convex combination of C and D. Finally, it is easy to measure the ratio ABCD algebraically and then enter P as a convex combination of E and F.