Question

圆心为 $O$ 的圆面积为 $156\pi$ 。 $\triangle ABC$ 是等边三角形， $\overline{BC}$ 是圆的一条弦， $OA = 4\sqrt{3}$ ，且点 $O$ 在 $\triangle ABC$ 外。求三角形 $ABC$ 的边长。

A circle with center $O$ has area $156\pi.$ Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3},$ and point $O$ is outside $\triangle ABC.$ What is the side length of $\triangle ABC?$

$2\sqrt{3}$

$6$

$4\sqrt{3}$

$12$

$18$

答案：B

知识点：弦等边三角形勾股定理

难度评级：1860

解答：

参考下图：

圆半径满足 $BO = \sqrt{156}$ 。

延长 $\overline{AO}$ 与 $\overline{BC}$ 交于 $X$ 。设 $s$ 为 $\triangle ABC$ 的边长。

于是且 $BX = \dfrac{s}{2}$ $AX = \dfrac{s\sqrt3}{2}.$

$\triangle OXB$ 是直角三角形，所以 $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

化简得 $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

由于 $s$ 为正， $s = 6$ 。

所以正确答案是 B。

Consider the following diagram:

Using the formula for the area of a circle, we have that $BO = \sqrt{156}$ since it is a radius.

Extend $\overline{AO}$ to intersect $\overline{BC}$ at $X.$ Let $s$ be the side length of $\triangle ABC.$

Then we have that $BX = \dfrac{s}{2}$ and $AX = \dfrac{s\sqrt3}{2}.$

We have that $\triangle OXB$ is right, which means that we can apply the Pythagorean Theorem. This gives us $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

Simplifying, we get $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

Since $s$ is positive, we must have that $s = 6.$

Thus, B is the correct answer.

2010 AMC 10B 第 19 题

答案：B

难度评级：1860

解答：

其他年份的第 19 题