Question

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

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

$15$

$17$

$19$

$21$

$23$

答案：C

知识点：变换扇形面积分割

难度评级：2520

解答：

设 $A', B', C', D'$ 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 $AP = \sqrt{2}$ 且 $PC = AC - AP = \sqrt{6},$ 所以 $A$ 和 $C$ 处的扇形面积分别为 $\tfrac{\pi}{2}$ 和 $\tfrac{3\pi}{2}.$ 若 $H$ 是 $AA',$ 的中点，则 $PH=AH=1$ 且 $HB=\sqrt3,$ 所以 $\triangle BPH$ 是 $30$ - $60$ - $90^\circ$ 三角形，并且 $PB=2.$ 因此沿 $BC$ 的两个 $60^\circ$ 扇形面积各为 $2\pi/3.$ 以 $PH$ 为高的两个三角形贡献 $\sqrt3-1,$ 另一对全等三角形的高为 $\sqrt3$ ，贡献 $3-\sqrt3.$ 所以四个三角形共贡献 $2.$

总面积为 $\begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered}$ 因此 $a + b + c = 10 + 6 + 3 = 19.$

因此，正确答案是 C。

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

Since $AP = \sqrt{2}$ and $PC = AC - AP = \sqrt{6},$ the sectors at $A$ and $C$ have areas $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$ If $H$ is the midpoint of $AA',$ then $PH=AH=1$ and $HB=\sqrt3,$ so $\triangle BPH$ is a $30$ - $60$ - $90^\circ$ triangle and $PB=2.$ Hence the two $60^\circ$ sectors along $BC$ each have area $2\pi/3.$ The two triangles with altitude $PH$ contribute $\sqrt3-1,$ and the other congruent pair has altitude $\sqrt3$ and contributes $3-\sqrt3.$ Thus the four triangles contribute $2.$

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

Thus, the correct answer is C.

2013 AMC 12A 第 23 题

答案：C

难度评级：2520

解答：

其他年份的第 23 题