2015 AMC 10A Problem 14

P11-15~4 minPrint

Circles Transformations & Symmetry Set up the equation/formula and compute; no special trick needed

The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?

Show solution without answering

Key insight

The disk turns from rolling and also from traveling around the clock: 3 turns per lap. So the arrow is up again after one third of a lap.

Solution

Two different motions turn the arrow. Measure each one as the disk moves some angle around the clock, then add.

Rolling. The clock's radius ( $20$ ) is twice the disk's radius ( $10$ ), so the clock's edge is twice as long as the disk's edge. If the touching point moves a quarter of the way around the clock, the disk rolls along a length equal to half of its own edge. In general, the disk spins twice as much as the angle it moves around the clock, measured against the line from the clock's center to the disk's center.
Traveling around. That line itself turns as the disk goes around. Picture the disk sliding (not rolling) around the clock with the arrow always pointing away from the clock's center: at $3$ o'clock it points right, at $6$ o'clock it points down. So traveling an angle around the clock turns the arrow by that same angle, in the same clockwise direction.
Add them. If the disk moves an angle $\theta$ around the clock, the arrow turns $2\theta + \theta = 3\theta$ in total.
The arrow points straight up again after one full turn, $360^\circ$ . So $3\theta = 360^\circ$ and $\theta = 120^\circ$ .
Each hour on a clock face is $360^\circ \div 12 = 30^\circ$ . So $120^\circ$ is $4$ hours: clockwise from $12$ o'clock, the disk touches at $4$ o'clock.

The answer is $\boxed{\textbf{(C)}\ 4\text{ o'clock}}$ .

Why this works

A coin rolling around the outside of another coin makes one more turn than the edge lengths alone suggest, because its center also circles around. For a disk of radius $r$ rolling around a circle of radius $R$ , the number of turns per lap is $\frac{R + r}{r}$ ; here $\frac{20 + 10}{10} = 3$ . Try it with two equal coins on a table: the rolling coin turns twice, not once, in one lap.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Suppose the tangent point has moved through an angle $\theta$ (in radians) around the clock, so the disk's center has also swung through $\theta$ about the clock's center. Two things rotate the arrow.

First, rolling without slipping: the disk has covered an arc of length $20\theta$ on the clock, which is an arc of $20\theta / 10 = 2\theta$ on its own circumference, so the disk has turned $2\theta$ relative to the direction from the clock's center to the disk's center.

Second, that reference direction itself has turned through $\theta$ (clockwise, the same sense as the rolling).

So the arrow has rotated through $2\theta + \theta = 3\theta$ in total. It next points straight up when $3\theta = 2\pi$ , i.e. $\theta = \frac{2\pi}{3} = 120^\circ$ . Starting from $12$ o'clock and going clockwise $120^\circ$ , which is $4$ hours' worth of $30^\circ$ each, the disk is tangent at $4$ o'clock.

The answer is $\boxed{\textbf{(C)}\ 4\text{ o'clock}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Introduction to Geometry, Chapter 12: Circles and Angles
Introduction to Geometry, Chapter 13: Power of a Point