1999 Bay Area Mathematical Olympiad

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 $AE\u2215EB=CF\u2215FD$. Let $P$ be the point on the segment such that $PE\u2215PF=AB\u2215CD$. Prove that the ratio between the areas of $\u25b3APD$ and $\u25b3BPC$ does not depend on the choice of $E$ and $F$.

Solution: If $AD$ and $BC$ are parallel, then $ABCD$ is a trapezoid. Because $ABCD$ is a cyclic quadrilateral, $\angle BAD+\angle DCB=18{0}^{\circ}$. By the parallelism, $\angle BCD$ is equal to the angle supplementary to $\angle ADC$. Then $\angle ADC=\angle BAD$. Then the trapezoid $ABCD$ is isosceles, that is, $AB=CD$. Then $PE=PF$, and $P$ is the midpoint of $\overline{EF}$. Let $Q$ be the foot of the altitude of $\u25b3APD$ to $\overline{AD}$ and $R$ be the foot of the altitude of $\u25b3BPC$. Then $PR=PQ$. Hence the ratio between the areas of $\u25b3APD$ and $\u25b3BPC$ depends only on the ratio $AD\u2215BC$.

Assume $\stackrel{\u20e1}{AD}$ and $\stackrel{\u20e1}{BC}$ are not parallel. Let the intersection point be $G$. Because $ABCD$ is a cyclic quadrilateral, $\angle DAB+\angle DCB=18{0}^{\circ}$. Also, $\angle DCG+\angle DCB=18{0}^{\circ}$, hence $\angle DAB=\angle DCG$. Since the angle at $G$ is in common, $\u25b3GAB$ is similar to $\u25b3GCD$, with similarity ratio $AB\u2215CD$. From the given information, $AB\u2215CD=AE\u2215CF=PE\u2215PF$. The points $E$ and $F$ are corresponding points in $\u25b3GAB$ and $\u25b3GCD$ so that $\u25b3GAE$ is similar to $\u25b3GCF$ with similarity ratio $AB\u2215CD=AE\u2215CF=PE\u2215PF$. Thus, $GE\u2215GF=PE\u2215PF$. Hence, $GP$ is the angle bisector of $\angle AGB$. Therefore, $d\left(P,\overline{AG}\right)=d\left(P,\overline{BC}\right)$. That is, the altitudes of the two triangles in question, $\u25b3APD$ and $\u25b3BPC$. Then the ratio of the areas of the two triangles depends only on the ratio of their bases $AD\u2215BC$, 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 $\overline{AB}$ and $\overline{BC}$ to intersect in point $G$ is a good idea to create some triangles involving the ratio $AB\u2215CD$. 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 $\u25b3AGB$ and $\u25b3CGD$ and therefore $\u25b3AGE$ and $\u25b3CGF$ 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 $AE\u2215EB$ 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 $AB\u2215CD$ algebraically and then enter $P$ as a convex combination of $E$ and $F$.