2011 AMC 12A 第 17 题

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17.

半径为 1122, 和 33 的三个圆两两外切。由这些切点确定的三角形面积是多少?

Circles with radii 1,1, 2,2, and 33 are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

35\dfrac{3}{5}

45\dfrac{4}{5}

11

65\dfrac{6}{5}

43\dfrac{4}{3}

答案:D
知识点:相切圆坐标几何鞋带公式
难度评级:1920
解答:

圆心之间的距离等于半径和:3344, 和 55, 形成一个直角三角形,直角在半径为 11 的圆心处。把该圆心放在 (0,0)(0,0), 半径为 22 的圆心放在 (3,0)(3,0), 半径为 33 的圆心放在 (0,4)(0,4)

切点在线段上,距离等于相应半径:(1,0)(1, 0)(0,1)(0, 1), 以及斜边上的 (3,0)+2(3,4)5=(95,85)(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right)

由鞋带公式,面积为 121(185)+0+95(01)=12125=65. \begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}

因此,正确答案是 D

The centers are separated by the sums of radii: 3,3, 4,4, and 5,5, a right triangle with the right angle at the radius-11 center. Place that center at (0,0),(0,0), the radius-22 center at (3,0),(3,0), and the radius-33 center at (0,4).(0,4).

The tangency points lie on the segments at distances equal to the radii: (1,0),(1, 0), (0,1),(0, 1), and on the hypotenuse at (3,0)+2(3,4)5=(95,85).(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).

By the shoelace formula the area is 121(185)+0+95(01)=12125=65. \begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}

Thus, the correct answer is D.

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