AMC12 2006 A

AMC12 2006 A · Q13

AMC12 2006 A · Q13. It mainly tests Systems of equations, Circle theorems.

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 the three circles?

一个 $3-4-5$ 直角三角形的顶点是三个两两外切的圆的圆心，如图所示。三个圆的面积之和是多少？

(A) $12\pi$ $12\pi$

(B) $\frac{25\pi}{2}$ $\frac{25\pi}{2}$

(D) $\frac{27\pi}{2}$ $\frac{27\pi}{2}$

(E) $14\pi$ $14\pi$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let the radius of the smallest circle be $r_A$, the radius of the second largest circle be $r_B$, and the radius of the largest circle be $r_C$. \[r_A + r_B = 3\] \[r_A + r_C = 4\] \[r_ B + r_C = 5\] Adding up all these equations and then dividing both sides by 2, we get, \[r_A + r_B + r_C = 6\] Then, we get $r_A = 1$, $r_B = 2$, and $r_C = 3$ Then we get $1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}$

设最小圆的半径为 $r_A$，第二大圆的半径为 $r_B$，最大圆的半径为 $r_C$。 \[r_A + r_B = 3\] \[r_A + r_C = 4\] \[r_ B + r_C = 5\] 将这三个方程相加并将两边同时除以 2，得到 \[r_A + r_B + r_C = 6\] 于是 $r_A = 1$，$r_B = 2$，$r_C = 3$。因此面积之和为 $1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}$。

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