Question

如下图所示，优弧 $AD$ 和劣弧 $BC$ 有同一个圆心 $O$ 。并且 $A$ 在 $O$ 和 $B$ 之间， $D$ 在 $O$ 和 $C$ 之间。优弧 $AD$ 、劣弧 $BC$ ，以及两条线段 $AB$ 和 $CD$ 的长度都为 $2\pi$ 。

$O$ 到 $A$ 的距离是多少？

In the figure shown below, major arc $AD$ and minor arc $BC$ have the same center, $O.$ Also, $A$ lies between $O$ and $B,$ and $D$ lies between $O$ and $C.$ Major arc $AD,$ minor arc $BC,$ and each of the two segments $AB$ and $CD$ have length $2\pi.$

What is the distance from $O$ to $A?$

$1$

$1 - \pi + \sqrt{1 + \pi^2}$

$\dfrac{1}{2}\pi$

$\dfrac{1}{2}\sqrt{1 + \pi^2}$

$2$

答案：B

知识点：弧二次方程

难度评级：1530

小提示：

设 $OA = R_1$ ， $OB = R_2$ ， $\angle AOD = \alpha$ ；弧长给出 $R_2\alpha = 2\pi$ 和 $R_1(2\pi - \alpha) = 2\pi$

Let $OA = R_1,$ $OB = R_2,$ and $\angle AOD = \alpha$ ; the arcs give $R_2\alpha = 2\pi$ and $R_1(2\pi - \alpha) = 2\pi$

大提示：

线段给出 $R_2 - R_1 = 2\pi$ ；消去半径可得 $\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0$

The segments give $R_2 - R_1 = 2\pi$ ; eliminating the radii yields $\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0$

解答：

设 $R_1 = OA = OD$ ， $R_2 = OB = OC$ ，并设 $\alpha = \angle AOD = \angle BOC$ （两条射线重合）。劣弧 $BC$ 的长度为 $R_2\alpha = 2\pi$ ，而优弧 $AD$ 是其反向的大弧，所以 $R_1(2\pi - \alpha) = 2\pi$ 。

每条线段 $AB = CD = R_2 - R_1 = 2\pi$ 。

由前两个方程， $R_2 = \dfrac{2\pi}{\alpha}$ ， $R_1 = \dfrac{2\pi}{2\pi - \alpha}$ 。代入 $R_2 - R_1 = 2\pi$ 并两边除以 $2\pi$ ，得到 $\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1\text{，}$ 它可化简为 $\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0$ 。

较大的根大于 $2\pi$ ，不可能是这段劣弧的圆心角，所以 $\alpha = (1+\pi) - \sqrt{1+\pi^2}$ 。于是 $\begin{aligned} R_1 &= \frac{2\pi}{2\pi - \alpha} \\ &= \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} \\ &= 1 - \pi + \sqrt{1+\pi^2} \end{aligned}\text{，}$ 其中最后一步作了有理化（分母乘以 $\sqrt{1+\pi^2} - (\pi-1)$ 等于 $2\pi$ ）。

因此，正确答案是 B。

Let $R_1 = OA = OD$ and $R_2 = OB = OC,$ and let $\alpha = \angle AOD = \angle BOC$ (the rays coincide). The minor arc $BC$ has length $R_2\alpha = 2\pi,$ and the major arc $AD$ is the reflex arc, so $R_1(2\pi - \alpha) = 2\pi.$

Each segment $AB = CD = R_2 - R_1 = 2\pi.$

From the first two equations, $R_2 = \dfrac{2\pi}{\alpha}$ and $R_1 = \dfrac{2\pi}{2\pi - \alpha}.$ Substituting into $R_2 - R_1 = 2\pi$ and dividing by $2\pi$ gives $\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1,$ which simplifies to $\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0.$

The larger root is greater than $2\pi,$ so it cannot be an angle of this minor arc. Thus $\alpha = (1+\pi) - \sqrt{1+\pi^2}.$ Then $\begin{aligned} R_1 &= \frac{2\pi}{2\pi - \alpha} \\ &= \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} \\ &= 1 - \pi + \sqrt{1+\pi^2}, \end{aligned}$ after rationalizing (the denominator times $\sqrt{1+\pi^2} - (\pi-1)$ equals $2\pi$ ).

Thus, the correct answer is B.

2025 AMC 12A 第 10 题

答案：B

难度评级：1530

小提示：

大提示：

解答：

其他年份的第 10 题