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Harvard-MIT Math Tournament

Geometry Difficulty 4.1 AIME Find the answer United States

Problem:
A circle of radius $3$ crosses the center of a square of side length $2$ . Find the difference between the areas of the nonoverlapping portions of the figures.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Call the area of the square $s$ , the area of the circle $c$ , and the area of the overlapping portion $x$ . The area of the circle not overlapped by the square is $c - x$ and the area of the square not overlapped by the circle is $s - x$ , so the difference between these two is $(c - x) - (s - x) = c - s = 9\pi - 4$ .

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