AMC10 2010 B

AMC10 2010 B · Q16

AMC10 2010 B · Q16. It mainly tests Pythagorean theorem, Circle theorems.

A square of side length 1 and a circle of radius $\sqrt{3}/3$ share the same center. What is the area inside the circle, but outside the square?

边长为 1 的正方形和半径为 $\sqrt{3}/3$ 的圆共享同一个中心。圆内、正方形外的面积是多少？

(A) $\pi/3-1$ $\pi/3-1$

(B) $2\pi/9-\sqrt{3}/3$ $2\pi/9-\sqrt{3}/3$

(D) 1/4 1/4

(E) $2\pi/9$ $2\pi/9$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $O$ be the common center of the circle and the square. Let $M$ be the midpoint of a side of the square and $P$ and $Q$ be the vertices of the square on the side containing $M$. Since $$ OM^2=\left(\frac12\right)^2<\left(\frac{\sqrt3}{3}\right)^2<\left(\frac{\sqrt2}{2}\right)^2=OP^2=OQ^2, $$ the midpoint of each side is inside the circle and the vertices of the square are outside the circle. Therefore the circle intersects the square in two points along each side. Let $A$ and $B$ be the intersection points of the circle with $\overline{PQ}$. Then $M$ is also the midpoint of $\overline{AB}$ and $\triangle OMA$ is a right triangle. By the Pythagorean Theorem $AM=\frac{1}{2\sqrt3}$, so $\triangle OMA$ is a $30\text{--}60\text{--}90^\circ$ right triangle. Then $\angle AOB=60^\circ$, and the area of the sector corresponding to $\angle AOB$ is $$ \frac16\cdot\pi\cdot\left(\frac{\sqrt3}{3}\right)^2=\frac{\pi}{18}. $$ The area of $\triangle AOB$ is $$ 2\cdot\frac12\cdot\frac12\cdot\frac{1}{2\sqrt3}=\frac{\sqrt3}{12}. $$ The area outside the square but inside the circle is $$ 4\cdot\left(\frac{\pi}{18}-\frac{\sqrt3}{12}\right)=\frac{2\pi}{9}-\frac{\sqrt3}{3}. $$

解答（B）：设 $O$ 为圆与正方形的公共中心。设 $M$ 为正方形某边的中点，$P$ 与 $Q$ 为包含 $M$ 的那条边上的两个顶点。因为 $$ OM^2=\left(\frac12\right)^2<\left(\frac{\sqrt3}{3}\right)^2<\left(\frac{\sqrt2}{2}\right)^2=OP^2=OQ^2, $$ 所以每条边的中点在圆内，而正方形的顶点在圆外。因此圆与正方形在每条边上有两个交点。设 $A$、$B$ 为圆与线段 $\overline{PQ}$ 的交点。则 $M$ 也是 $\overline{AB}$ 的中点，且 $\triangle OMA$ 为直角三角形。由勾股定理 $AM=\frac{1}{2\sqrt3}$，所以 $\triangle OMA$ 是一个 $30\text{--}60\text{--}90^\circ$ 的直角三角形。于是 $\angle AOB=60^\circ$，对应扇形面积为 $$ \frac16\cdot\pi\cdot\left(\frac{\sqrt3}{3}\right)^2=\frac{\pi}{18}. $$ $\triangle AOB$ 的面积为 $$ 2\cdot\frac12\cdot\frac12\cdot\frac{1}{2\sqrt3}=\frac{\sqrt3}{12}. $$ 圆内而正方形外的面积为 $$ 4\cdot\left(\frac{\pi}{18}-\frac{\sqrt3}{12}\right)=\frac{2\pi}{9}-\frac{\sqrt3}{3}. $$

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