Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:
Let $ABCD$ be an isosceles trapezoid such that $AB = 17$ , $BC = DA = 25$ , and $CD = 31$ . Points $P$ and $Q$ are selected on sides $AD$ and $BC$ , respectively, such that $AP = CQ$ and $PQ = 25$ . Suppose that the circle with diameter $PQ$ intersects the sides $AB$ and $CD$ at four points which are vertices of a convex quadrilateral. Compute the area of this quadrilateral.

Solution

Solution:
Let the midpoint of $PQ$ be $M$ ; note that $M$ lies on the midline of $ABCD$ . Let $B'$ and $C'$ be a translate of $BC$ (parallel to $AB$ and $CD$ ) so that $M$ is the midpoint of $B'$ and $C'$ . Since $MB' = MC' = 25/2 = MP = MQ$ , $B'$ and $C'$ are one of the four intersections of the circle with diameter $PQ$ and the sides $AB$ and $CD$ . We may also define $A'$ and $D'$ similarly and get that they are also among the four points.

It follows that the desired quadrilateral is $B'D'C'A'$ , which is a rectangle with height equal to the height of $ABCD$ (which is $24$ ), and width equal to $\frac{1}{2}(31-17) = 7$ . Thus the area is $24 \cdot 7 = 168$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.