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2025 AMC 12A 第 10 题

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10.

如下图所示，优弧 ADAD 和劣弧 BCBC 有同一个圆心 OO。并且 AA 在 OO 和 BB 之间，DD 在 OO 和 CC 之间。优弧 ADAD、劣弧 BCBC，以及两条线段 ABAB 和 CDCD 的长度都为 2π2\pi。
OO 到 AA 的距离是多少？
In the figure shown below, major arc ADAD and minor arc BCBC have the same center, O.O. Also, AA lies between OO and B,B, and DD lies between OO and C.C. Major arc AD,AD, minor arc BC,BC, and each of the two segments ABAB and CDCD have length 2π.2\pi.
What is the distance from OO to A?A?
11
1−π+1+π21 - \pi + \sqrt{1 + \pi^2}
12π\dfrac{1}{2}\pi
121+π2\dfrac{1}{2}\sqrt{1 + \pi^2}
22
答案：B
知识点：弧二次方程
难度评级：1530
小提示：
设 OA=R1OA = R_1，OB=R2OB = R_2，∠AOD=α\angle AOD = \alpha；弧长给出 R2α=2πR_2\alpha = 2\pi 和 R1(2π−α)=2πR_1(2\pi - \alpha) = 2\pi
Let OA=R1,OA = R_1, OB=R2,OB = R_2, and ∠AOD=α\angle AOD = \alpha; the arcs give R2α=2πR_2\alpha = 2\pi and R1(2π−α)=2πR_1(2\pi - \alpha) = 2\pi
大提示：
线段给出 R2−R1=2πR_2 - R_1 = 2\pi；消去半径可得 α2−2(1+π)α+2π=0\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0
The segments give R2−R1=2πR_2 - R_1 = 2\pi; eliminating the radii yields α2−2(1+π)α+2π=0\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0
解答：
设 R1=OA=ODR_1 = OA = OD，R2=OB=OCR_2 = OB = OC，并设 α=∠AOD=∠BOC\alpha = \angle AOD = \angle BOC（两条射线重合）。劣弧 BCBC 的长度为 R2α=2πR_2\alpha = 2\pi，而优弧 ADAD 是其反向的大弧，所以 R1(2π−α)=2πR_1(2\pi - \alpha) = 2\pi。
每条线段 AB=CD=R2−R1=2πAB = CD = R_2 - R_1 = 2\pi。
由前两个方程，R2=2παR_2 = \dfrac{2\pi}{\alpha}，R1=2π2π−αR_1 = \dfrac{2\pi}{2\pi - \alpha}。代入 R2−R1=2πR_2 - R_1 = 2\pi 并两边除以 2π2\pi，得到 1α−12π−α=1，\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1\text{，} 它可化简为 α2−2(1+π)α+2π=0\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0。
较大的根大于 2π2\pi，不可能是这段劣弧的圆心角，所以 α=(1+π)−1+π2\alpha = (1+\pi) - \sqrt{1+\pi^2}。于是 R1=2π2π−α=2ππ−1+1+π2=1−π+1+π2， \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 1+π2−(π−1)\sqrt{1+\pi^2} - (\pi-1) 等于 2π2\pi）。
因此，正确答案是 B。
Let R1=OA=ODR_1 = OA = OD and R2=OB=OC,R_2 = OB = OC, and let α=∠AOD=∠BOC\alpha = \angle AOD = \angle BOC (the rays coincide). The minor arc BCBC has length R2α=2π,R_2\alpha = 2\pi, and the major arc ADAD is the reflex arc, so R1(2π−α)=2π.R_1(2\pi - \alpha) = 2\pi.
Each segment AB=CD=R2−R1=2π.AB = CD = R_2 - R_1 = 2\pi.
From the first two equations, R2=2παR_2 = \dfrac{2\pi}{\alpha} and R1=2π2π−α.R_1 = \dfrac{2\pi}{2\pi - \alpha}. Substituting into R2−R1=2πR_2 - R_1 = 2\pi and dividing by 2π2\pi gives 1α−12π−α=1,\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1, which simplifies to α2−2(1+π)α+2π=0.\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0.
The larger root is greater than 2π,2\pi, so it cannot be an angle of this minor arc. Thus α=(1+π)−1+π2.\alpha = (1+\pi) - \sqrt{1+\pi^2}. Then R1=2π2π−α=2ππ−1+1+π2=1−π+1+π2, \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 1+π2−(π−1)\sqrt{1+\pi^2} - (\pi-1) equals 2π2\pi).
Thus, the correct answer is B.

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