AMC10 2013 A

AMC10 2013 A · Q20

AMC10 2013 A · Q20. It mainly tests Area & perimeter, Transformations.

A unit square is rotated 45° about its center. What is the area of the region swept out by the interior of the square?

一个单位正方形绕其中心旋转45°。正方形内部扫过的区域面积是多少？

(A) $1 - \sqrt{2}/2 + \pi/4$ $1 - \sqrt{2}/2 + \pi/4$

(B) $1/2 + \pi/4$ $1/2 + \pi/4$

(D) $\sqrt{2}/2 + \pi/4$ $\sqrt{2}/2 + \pi/4$

(E) $1 + \sqrt{2}/4 + \pi/8$ $1 + \sqrt{2}/4 + \pi/8$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $O$ be the center of unit square $ABCD$, let $A$ and $B$ be rotated to points $A'$ and $B'$, and let $OA'$ and $A'B'$ intersect $AB$ at $E$ and $F$, respectively. Then one quarter of the region swept out by the interior of the square consists of the $45^\circ$ sector $AOA'$ with radius $\frac{\sqrt{2}}{2}$, isosceles right triangle $OEB$ with leg length $\frac{1}{2}$, and isosceles right triangle $AEF$ with leg length $\frac{\sqrt{2}-1}{2}$. Thus the area of the region is $$ 4\left(\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\pi}{8}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}-1}{2}\right)^2\right)=2-\sqrt{2}+\frac{\pi}{4}. $$

答案（C）：设 $O$ 为单位正方形 $ABCD$ 的中心，将 $A$ 与 $B$ 旋转到点 $A'$ 与 $B'$，并令 $OA'$ 与 $A'B'$ 分别在 $AB$ 上交于 $E$ 与 $F$。则正方形内部扫过区域的四分之一由以下部分组成：半径为 $\frac{\sqrt{2}}{2}$ 的 $45^\circ$ 扇形 $AOA'$、直角等腰三角形 $OEB$（直角边长为 $\frac{1}{2}$）、以及直角等腰三角形 $AEF$（直角边长为 $\frac{\sqrt{2}-1}{2}$）。因此该区域面积为 $$ 4\left(\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\pi}{8}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}-1}{2}\right)^2\right)=2-\sqrt{2}+\frac{\pi}{4}. $$

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