AMC10 2003 A

AMC10 2003 A · Q17

AMC10 2003 A · Q17. It mainly tests Triangles (properties), Circle theorems.

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

一个正三角形的周长（英寸数）等于其外接圆面积（平方英寸数）。求该圆的半径（英寸）。

(A) $\frac{3\sqrt{2}}{\pi}$ $\frac{3\sqrt{2}}{\pi}$

(B) $\frac{3\sqrt{3}}{\pi}$ $\frac{3\sqrt{3}}{\pi}$

(D) $\frac{6}{\pi}$ $\frac{6}{\pi}$

(E) $\sqrt{3\pi}$ $\sqrt{3\pi}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let the triangle have vertices $A$, $B$, and $C$, let $O$ be the center of the circle, and let $D$ be the midpoint of $\overline{BC}$. Triangle $COD$ is a 30-60-90 degree triangle. If $r$ is the radius of the circle, then the sides of $\triangle COD$ are $r$, $r/2$, and $r\sqrt{3}/2$. The perimeter of $\triangle ABC$ is $6\left(r\sqrt{3}/2\right)=3r\sqrt{3}$, and the area of the circle is $\pi r^2$. Thus $3r\sqrt{3}=\pi r^2$, and $r=(3\sqrt{3})/\pi$.

（B）设三角形的顶点为 $A$、$B$、$C$，$O$ 为圆心，$D$ 为线段 $\overline{BC}$ 的中点。三角形 $COD$ 是一个 $30^\circ\!-\!60^\circ\!-\!90^\circ$ 三角形。若 $r$ 为圆的半径，则 $\triangle COD$ 的三边分别为 $r$、$r/2$、$r\sqrt{3}/2$。$\triangle ABC$ 的周长为 $6\left(r\sqrt{3}/2\right)=3r\sqrt{3}$，圆的面积为 $\pi r^2$。因此 $3r\sqrt{3}=\pi r^2$，从而 $r=(3\sqrt{3})/\pi$。

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