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Problem 922

AMC 12 late, AIME early

Geometry Difficulty 4.7 Prove it Annual Harvard-MIT November Tournament · United States

Let $C$ be the circle of radius $12$ centered at $(0,0)$ . What is the length of the shortest path in the plane between $(8 \sqrt{3}, 0)$ and $(0,12 \sqrt{2})$ that does not pass through the interior of $C$ ?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:
$12 + 4 \sqrt{3} + \pi$

The shortest path consists of a tangent to the circle, a circular arc, and then another tangent. The first tangent, from $(8 \sqrt{3}, 0)$ to the circle, has length $4 \sqrt{3}$ , because it is a leg of a $30$ - $60$ - $90$ right triangle. The $15^{\circ}$ arc has length $\frac{15}{360}(24\pi)$ , or $\pi$ , and the final tangent, to $(0,12 \sqrt{2})$ , has length $12$ .

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