1996 AMC 12 Problem 20

Question

In the

xy

-plane, what is the length of the shortest path from

(0,0)

to

(12,16)

that does not go inside the circle

(x-6)^2+(y-8)^2=25?

10\sqrt3

10\sqrt5

10\sqrt3+\frac{5\pi}{3}

\frac{40\sqrt3}{3}

10+5\pi

The circle’s center is the midpoint of the two endpoints

The shortest permitted path consists of two tangent segments and the shorter arc between their tangency points

Each endpoint is

10

units from the circle’s center, so each tangent segment has length

\sqrt{10^2-5^2}=5\sqrt3.

In the right triangle at a tangency point, the angle at the center is

60^\circ.

Since the endpoint rays are opposite, the intervening minor arc subtends

180^\circ-2(60^\circ)=60^\circ,

so its length is

\frac{5\pi}{3}.

The total is

10\sqrt3+\frac{5\pi}{3},

and the correct answer is C.

Problem 20 in other years