AMC12 2009 A

AMC12 2009 A · Q19

AMC12 2009 A · Q19. It mainly tests Area & perimeter, Polygons.

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were $A$ and $B$, respectively. Each polygon had a side length of $2$. Which of the following is true?

Andrea 在一个正五边形内接一个圆，在该五边形外接一个圆，并计算两圆之间区域的面积。Bethany 对一个正七边形（7 边）做了同样的事。两块区域的面积分别为 $A$ 和 $B$。每个多边形的边长均为 $2$。以下哪项正确？

(A) $A = \frac{25}{49}B$ $A = \frac{25}{49}B$

(B) $A = \frac{5}{7}B$ $A = \frac{5}{7}B$

(D) $A = \frac{7}{5}B$ $A = \frac{7}{5}B$

(E) $A = \frac{49}{25}B$ $A = \frac{49}{25}B$

Answer

Correct choice: (C)

正确答案：（C）

Solution

In any regular polygon with side length $2$, consider the isosceles triangle formed by the center of the polygon $S$ and two consecutive vertices $X$ and $Y$. We are given that $XY=2$. Obviously $SX=SY=r$, where $r$ is the radius of the circumcircle. Let $T$ be the midpoint of $XY$. Then $XT=TY=1$, and $TS=\rho$, where $\rho$ is the radius of the incircle. Applying the Pythagorean theorem on the triangle $STX$, we get that $\rho^2 + 1 = r^2$. Then the area between the circumcircle and the incircle can be computed as $\pi r^2 - \pi \rho^2 = \pi r^2 - \pi (r^2 - 1) = \pi$. Hence $A=\pi$, $B=\pi$, and therefore $\boxed{\textbf{(C) }A=B}$.

在任意边长为 $2$ 的正多边形中，考虑由多边形中心 $S$ 与相邻两个顶点 $X$、$Y$ 构成的等腰三角形。已知 $XY=2$。显然 $SX=SY=r$，其中 $r$ 为外接圆半径。设 $T$ 为 $XY$ 的中点，则 $XT=TY=1$，且 $TS=\rho$，其中 $\rho$ 为内切圆半径。在直角三角形 $STX$ 中应用勾股定理，得 $\rho^2+1=r^2$。两圆之间的面积为 $\pi r^2-\pi\rho^2=\pi r^2-\pi(r^2-1)=\pi$。因此 $A=\pi$，$B=\pi$，从而 $\boxed{\textbf{(C) }A=B}$。

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