Question

一个半径为 $1$ 的圆盘与一个半径为 $5$ 的圆盘外切。设 $A$ 为两圆盘的切点， $C$ 为小圆盘的圆心， $E$ 为大圆盘的圆心。保持大圆盘不动，让小圆盘沿大圆盘外侧滚动，直到小圆盘转过 $360^\circ$ 。也就是说，若小圆盘圆心移动到点 $D$ ，而小圆盘上原本位于 $A$ 的点现在移动到点 $B$ ，则 $\overline{AC}$ 平行于 $\overline{BD}$ 。于是 $\sin^2(\angle BEA) = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

A disk with radius $1$ is externally tangent to a disk with radius $5.$ Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ.$ That is, if the center of the smaller disk has moved to the point $D,$ and the point on the smaller disk that began at $A$ has now moved to point $B,$ then $\overline{AC}$ is parallel to $\overline{BD}.$ Then $\sin^2(\angle BEA) = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：58

知识点：圆变换坐标几何三角学

难度评级：2920

解答：

将 $E$ 放在原点，并取 $C = (6, 0)$ ，于是 $A = (5, 0)$ 。半径为 $1$ 的圆沿半径为 $5$ 的固定圆外侧无滑动滚动时，若其圆心绕 $E$ 扫过角 $\varphi$ ，滚动接触会使圆盘相对于圆心连线转过 $5\varphi$ ，而这条圆心连线本身的转动又增加 $\varphi$ ，所以圆盘在固定坐标系中共转过 $6\varphi$ 。转过 $360^\circ$ 因此意味着 $\varphi = 60^\circ$ ，所以 $D = 6(\cos 60^\circ, \sin 60^\circ)$ $= (3, 3\sqrt{3})$ 。

完整转过 $360^\circ$ 后，圆盘回到原来的朝向，因此从圆心到标记点的向量不变： $B = D + (A - C)$ $= (3 - 1, 3\sqrt{3})$ $= (2, 3\sqrt{3})$ 。（特别地， $\overline{BD}$ 平行于 $\overline{AC}$ ，正如题目所述。）

射线 $EA$ 是正 $x$ 轴，所以 $\begin{aligned} \sin^2(\angle BEA) &= \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} \\ &= \frac{27}{31}, \end{aligned}$ 因而 $m + n = 27 + 31 = 58$ 。

Place $E$ at the origin with $C = (6, 0),$ so $A = (5, 0).$ When a circle of radius $1$ rolls without slipping outside a fixed circle of radius $5$ and its center sweeps an angle $\varphi$ about $E,$ the rolling contact turns the disk through $5\varphi$ relative to the line of centers, and the revolution of that line adds $\varphi$ more, so the disk turns $6\varphi$ in the ground frame. Turning through $360^\circ$ therefore means $\varphi = 60^\circ,$ so $D = 6(\cos 60^\circ, \sin 60^\circ)$ $= (3, 3\sqrt{3}).$

Having turned through a full $360^\circ,$ the disk is back in its original orientation, so the vector from its center to the marked point is unchanged: $B = D + (A - C)$ $= (3 - 1, 3\sqrt{3})$ $= (2, 3\sqrt{3}).$ (In particular $\overline{BD}$ is parallel to $\overline{AC},$ as the problem states.)

The ray $EA$ is the positive $x$ -axis, so $\begin{aligned} \sin^2(\angle BEA) &= \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} \\ &= \frac{27}{31}, \end{aligned}$ and $m + n = 27 + 31 = 58.$

2014 AIME I 第 10 题

答案：58

难度评级：2920

解答：

其他年份的第 10 题