AMC12 2011 A

AMC12 2011 A · Q17

AMC12 2011 A · Q17. It mainly tests Triangles (properties), Circle theorems.

Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

半径分别为 $1$、$2$ 和 $3$ 的三个圆两两外切。由这些切点确定的三角形的面积是多少？

(A) \frac{3}{5} \frac{3}{5}

(B) \frac{4}{5} \frac{4}{5}

(D) \frac{6}{5} \frac{6}{5}

(E) \frac{4}{3} \frac{4}{3}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6. The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine, $\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$ $\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$ $\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$ which add up to $4.8$. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.

这三个圆的圆心构成一个 3-4-5 三角形，其面积为 6。由每个圆的圆心与该圆上的两个切点所确定的三个三角形的面积，利用正弦面积公式分别为 $\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$ $\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$ $\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$ 它们相加为 $4.8$。所求面积等于大的 3-4-5 三角形面积减去这三个小三角形面积，即 $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$。

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