Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let $\Omega$ be a circle of radius $8$ centered at point $O$ , and let $M$ be a point on $\Omega$ . Let $S$ be the set of points $P$ such that $P$ is contained within $\Omega$ , or such that there exists some rectangle $ABCD$ containing $P$ whose center is on $\Omega$ with $AB=4$ , $BC=5$ , and $BC \parallel OM$ . Find the area of $S$ .

Solution

Solution:

Answer: $164+64\pi$

We wish to consider the union of all rectangles $ABCD$ with $AB=4$ , $BC=5$ , and $BC \parallel \overline{OM}$ , with center $X$ on $\Omega$ . Consider translating rectangle $ABCD$ along the radius $XO$ to a rectangle $A'B'C'D'$ now centered at $O$ . It is now clear that every point inside $ABCD$ is a translate of a point in $A'B'C'D'$ , and furthermore, any rectangle $ABCD$ translates along the appropriate radius to the same rectangle $A'B'C'D'$ .

We see that the boundary of this region can be constructed by constructing a quarter-circle at each vertex, then connecting these quarter-circles with tangents to form four rectangular regions. Now, splitting our region into four quarter circles and five rectangles, we compute the desired area to be

$4 \cdot \frac{1}{4}(8)^2 \pi + 2(4 \cdot 8) + 2(5 \cdot 8) + (4 \cdot 5) = 164 + 64\pi$

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