2025 AMC 12A Problem 10

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.