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2015 AMC 10A Problem 14

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2015 AMC 10A Problem 14AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 3 of 5 · about 4 min · position P11-15

The diagram below shows the circular face of a clock with radius 2020 cm and a circular disk with radius 1010 cm externally tangent to the clock face at 1212 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?

figure

  1. A)

    2o′clock\mathrm{2 o'clock}

  2. B)

    3o′clock\mathrm{3 o'clock}

  3. C)

    4o′clock\mathrm{4 o'clock}

  4. D)

    6o′clock\mathrm{6 o'clock}

  5. E)

    8o′clock\mathrm{8 o'clock}

Answer

C

Key insight

A disk rolling outside a circle of triple its radius spins 3 times per lap, so the arrow is upright again after one third of a lap: 4 o'clock.

Solution

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

First, rolling without slipping: the disk has covered an arc of length 20θ20\theta on the clock, which is an arc of 20θ/10=2θ20\theta / 10 = 2\theta on its own circumference, so the disk has turned 2θ2\theta relative to the direction from the clock's center to the disk's center.

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

So the arrow has rotated through 2θ+θ=3θ2\theta + \theta = 3\theta in total. It next points straight up when 3θ=2π3\theta = 2\pi, i.e. θ=2π3=120∘\theta = \frac{2\pi}{3} = 120^\circ. Starting from 1212 o'clock and going clockwise 120∘120^\circ, which is 44 hours' worth of 30∘30^\circ each, the disk is tangent at 44 o'clock.

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

Why this works

A disk of radius rr rolling around the outside of a circle of radius RR makes R+rr\frac{R + r}{r} full turns per lap, not Rr\frac{R}{r}: the extra turn comes from the disk's center orbiting the fixed circle. Here 20+1010=3\frac{20 + 10}{10} = 3 turns per lap, so one turn takes a third of a lap. Whenever a rolling object also orbits, add the orbital angle to the rolling angle.

Alternative approach

Test the answer choices against the rule "the disk spins 33 times per lap": one spin is one third of the way around, at 44 o'clock; 66 o'clock would be 1.51.5 spins and 33 o'clock only 0.750.75 spins.

The trap

Matching arc lengths only (disk spins 20/10 = 2 times per lap) and answering 6 o'clock, forgetting that the disk's center also travels around the clock.

Common mistakes

  • Matching arc lengths only (disk spins 20/10=220/10 = 2 times per lap) and answering 66 o'clock, forgetting that the disk's center also travels around the clock.
  • Converting 120∘120^\circ to the clock incorrectly (each hour mark is 30∘30^\circ, so 120∘120^\circ is four hours, not two).

Techniques

Set up the equation/formula and compute; no special trick needed