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2013 AMC 12A 第 23 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

23.

ABCDABCD 是一个边长为 3+1\sqrt{3} + 1 的正方形。点 PP 在 AC‾\overline{AC} 上，且 AP=2AP = \sqrt{2}。将正方形 ABCDABCD 围成的区域绕中心 PP 逆时针旋转 90∘90^\circ，扫过区域的面积为 1c(aπ+b)\dfrac{1}{c}(a\pi + b)，其中 aa，bb 和 cc 为正整数且 gcd⁡(a,b,c)=1\gcd(a, b, c) = 1。求 a+b+ca + b + c。

ABCDABCD is a square of side length 3+1.\sqrt{3} + 1. Point PP is on AC‾\overline{AC} such that AP=2.AP = \sqrt{2}. The square region bounded by ABCDABCD is rotated 90∘90^\circ counterclockwise with center P,P, sweeping out a region whose area is 1c(aπ+b),\dfrac{1}{c}(a\pi + b), where a,a, b,b, and cc are positive integers and gcd⁡(a,b,c)=1.\gcd(a, b, c) = 1. What is a+b+c?a + b + c?

1515

1717

1919

2121

2323

答案：C
知识点：变换扇形面积分割
难度评级：2520
小提示：

跟踪像点 A′,B′,C′,D′A', B', C', D'；扫过区域由四个圆扇形和四个三角形组成

Track the images A′,B′,C′,D′;A', B', C', D'; the swept region is four circular sectors plus four triangles

大提示：

AP=2AP = \sqrt{2} 且 PC=6PC = \sqrt{6} 给出扇形半径，而 △BPH\triangle BPH 是 3030-6060-9090 度三角形。

AP=2AP = \sqrt{2} and PC=6PC = \sqrt{6} give the sector radii, and △BPH\triangle BPH is a 3030–6060–9090 triangle

解答：

设 A′,B′,C′,D′A', B', C', D' 为各顶点旋转后的像。扫过的区域可分解为四个圆扇形和四个三角形。

因为 AP=2AP = \sqrt{2} 且 PC=AC−AP=6PC = AC - AP = \sqrt{6}，所以 AA 和 CC 处的扇形面积分别为 π2\tfrac{\pi}{2} 和 3π2\tfrac{3\pi}{2}。若 HH 是 AA′AA' 的中点，则 PH=AH=1PH=AH=1 且 HB=3HB=\sqrt3，所以 △BPH\triangle BPH 是 3030-6060-90∘90^\circ 三角形，并且 PB=2PB=2。因此沿 BCBC 的两个 60∘60^\circ 扇形面积各为 2π3\frac{2\pi}{3}。以 PHPH 为高的两个三角形贡献 3−1\sqrt3-1，另一对全等三角形的高为 3\sqrt3，贡献 3−33-\sqrt3。所以四个三角形共贡献 22。

总面积为 π2+3π2+2⋅2π3+2=10π+63， \begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3} \end{gathered}\text{，}因此 a+b+c=10+6+3=19a + b + c = 10 + 6 + 3 = 19。

因此，正确答案是 C。

Let A′,B′,C′,D′A', B', C', D' be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since AP=2AP = \sqrt{2} and PC=AC−AP=6,PC = AC - AP = \sqrt{6}, the sectors at AA and CC have areas π2\tfrac{\pi}{2} and 3π2.\tfrac{3\pi}{2}. If HH is the midpoint of AA′,AA', then PH=AH=1PH=AH=1 and HB=3,HB=\sqrt3, so △BPH\triangle BPH is a 3030-6060-90∘90^\circ triangle and PB=2.PB=2. Hence the two 60∘60^\circ sectors along BCBC each have area 2π3.\frac{2\pi}{3}. The two triangles with altitude PHPH contribute 3−1,\sqrt3-1, and the other congruent pair has altitude 3\sqrt3 and contributes 3−3.3-\sqrt3. Thus the four triangles contribute 2.2.

The total area is π2+3π2+2⋅2π3+2=10π+63, \begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered} so a+b+c=10+6+3=19.a + b + c = 10 + 6 + 3 = 19.

Thus, the correct answer is C.

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