Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $C$ be a circle with center at the origin $O$ of a system of rectangular coordinates, and let $M O N$ be the quarter circle of $C$ in the first quadrant. Let $P Q$ be an arc of $C$ of fixed length that lies in the arc $M N$ . Let $K$ and $L$ be the feet of the perpendiculars from $P$ and $Q$ to $O N$ , and let $V$ and $W$ be the feet of the perpendiculars from $P$ and $Q$ to $O M$ , respectively. Let $A$ be the area of trapezoid $P K L Q$ and $B$ the area of trapezoid $P V W Q$ . Prove that $A+B$ does not depend on where $\operatorname{arc} P Q$ is chosen.

Solution

Solution:

Draw $O P$ , $O U$ , and $O Q$ , and note that $A+B$ is the area of rectangle $P V W U$ , plus the area of rectangle $U K L Q$ , plus twice the area of the triangle $P U Q$ . But the area of triangle $P O U$ is half the area of rectangle $P V W U$ , and the area of triangle $U O Q$ is half the area of rectangle $U K L Q$ , so putting this together $A+B$ is twice the area of triangle $P O Q$ , which depends only on the fixed length of arc $P Q$ .

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