1996 AMC 12 第 20 题

Question

在

xy

平面内，从

(0,0)

到

(12,16)

且不进入圆

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

内部的最短路径长度是多少？

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

圆心是两个端点的中点

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

最短的允许路径由两条切线段和两个切点之间的较短圆弧组成

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

每个端点到圆心的距离都是

10

，所以每条切线段的长度为

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

。在切点处的直角三角形中，圆心处的角为

60^\circ

。由于从圆心到两个端点的射线方向相反，两个切点之间的小弧所对圆心角为

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

，所以弧长为

\frac{5\pi}{3}

。总长度为

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

，正确答案是 C。

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.

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