Question

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

Solution:

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 $\sin^2(\angle BEA) = \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} = \frac{27}{31},$ and $m + n = 27 + 31 = 58.$

2014 AIME I Problem 10

Solution:

Problem 10 in Other Years