2006 AMC 12A 第 13 题

Question

如图，一个边长为

3

、

4

、

5

的直角三角形，其三个顶点是三个两两外切圆的圆心。这些圆的面积之和是多少？

The vertices of a

3

–

4

–

5

right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of these circles?

12\pi

\dfrac{25\pi}{2}

13\pi

\dfrac{27\pi}{2}

14\pi

相邻两个顶点处圆的半径之和等于它们之间的边长。

The radii at two adjacent vertices sum to the side length between them

解

r + s = 3,\ r + t = 4,\ s + t = 5

。

Solve

r + s = 3,\ r + t = 4,\ s + t = 5

若

r, s, t

是三个顶点处圆的半径，则

r + s = 3,\ r + t = 4,\ s + t = 5

。三式相加得

r + s + t = 6

，所以

r = 1,\ s = 2,\ t = 3

。

面积之和为

\pi(1^2 + 2^2 + 3^2) = 14\pi

。

因此，正确答案是 E。

If

r, s, t

are the radii at the vertices, then

r + s = 3,\ r + t = 4,\ s + t = 5.

Adding all three gives

r + s + t = 6,

so

r = 1,\ s = 2,\ t = 3.

The sum of the areas is

\pi(1^2 + 2^2 + 3^2) = 14\pi.

Thus, the correct answer is E.

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