AMC12 2025 A

AMC12 2025 A · Q10

AMC12 2025 A · Q10. It mainly tests Quadratic equations, Fractions.

In the figure shown below, major arc $\widehat{AD}$ and minor arc $\widehat{BC}$ have the same center, $O$. Also, $A$ lies between $O$ and $B$, and $D$ lies between $O$ and $C$. Major arc $\widehat{AD}$, minor arc $\widehat{BC}$, and each of the two segments $\overline{AB}$ and $\overline{CD}$ have length $2\pi$. What is the distance from $O$ to $A$?

如图所示，大弧$\widehat{AD}$和小弧$\widehat{BC}$有相同的圆心$O$。此外，$A$位于$O$和$B$之间，$D$位于$O$和$C$之间。大弧$\widehat{AD}$、小弧$\widehat{BC}$以及线段$\overline{AB}$和$\overline{CD}$的长度均为$2\pi$。求$O$到$A$的距离。

(A) 1 1

(B) 1 - \pi + \sqrt{\pi^{2} + 1} 1 - \pi + \sqrt{\pi^{2} + 1}

(D) \frac{\sqrt{\pi^{2} + 1}}{2} \frac{\sqrt{\pi^{2} + 1}}{2}

(E) 2 2

Answer

Correct choice: (B)

正确答案：（B）

Solution

The ratio between the radius and the arc length is constant. For the inner circle, the radius, which we will denote as $r$, has a corresponding arc length of $2\pi r - 2\pi$ (the circumference minus the major arc length). For the outer circle, the radius, which is $r + 2\pi$, has a corresponding arc length of $2\pi$. We therefore write the equation \[\frac{r}{2\pi r - 2\pi} = \frac{r+2\pi}{2\pi},\] which simplifies to \[r = (r+2\pi)(r-1)\] \[0 = r^2 + (2\pi-2)r - 2\pi.\] Applying the Quadratic Formula, we get that $r =\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$

半径与弧长的比值为常数。对于内圆，半径记为$r$，其对应弧长为$2\pi r - 2\pi$（周长减去大弧长）。对于外圆，半径为$r + 2\pi$，对应弧长为$2\pi$。因此写方程 \[\frac{r}{2\pi r - 2\pi} = \frac{r+2\pi}{2\pi},\] 化简为 \[r = (r+2\pi)(r-1)\] \[0 = r^2 + (2\pi-2)r - 2\pi.\] 应用二次求根公式，得$r=\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$

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