AMC12 2008 B

AMC12 2008 B · Q9

AMC12 2008 B · Q9. It mainly tests Pythagorean theorem, Circle theorems.

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?

点 $A$ 和 $B$ 在半径为 $5$ 的圆上，且 $AB = 6$。点 $C$ 是小弧 $AB$ 的中点。线段 $AC$ 的长度是多少？

(A) $\sqrt{10}$ $\sqrt{10}$

(B) $\frac{7}{2}$ $\frac{7}{2}$

(D) $\sqrt{15}$ $\sqrt{15}$

(E) 4 4

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$. The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.

设 $\alpha$ 为所对弧 $AB$ 的圆心角。由余弦定理， $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$，得 $\cos(\alpha) = 7/25$。由半角公式， $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$。由余弦定理可得 $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$，对应选项 $\boxed{\text{A}}$。

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