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2011 AMC 10B Problem 12

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2011 AMC 10B Problem 12AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 2 of 5 · about 3 min · position P11-15

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of 66 meters, and it takes her 3636 seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?

  1. A)

    π3\frac{\pi}{3}

  2. B)

    2π3\frac{2\pi}{3}

  3. C)

    π\pi

  4. D)

    4π3\frac{4\pi}{3}

  5. E)

    5π3\frac{5\pi}{3}

Answer

A

Key insight

The straight sides are the same length on both edges; only the two semicircles differ, and their radii differ by 6, adding exactly 12 pi meters.

Solution

Let the straight sides have length LL and the inner semicircles radius rr. The outer semicircles then have radius r+6r + 6.

Inner edge length: 2L+2πr2L + 2\pi r (two straight sides plus two semicircles, which together form a full circle).
Outer edge length: 2L+2π(r+6)2L + 2\pi(r + 6).

The difference is 2π(r+6)−2πr=12π2\pi(r + 6) - 2\pi r = 12\pi meters, independent of LL and rr.

Walking 12π12\pi extra meters takes 3636 extra seconds at constant speed vv, so
v=12π36=π3 m/s.v = \frac{12\pi}{36} = \frac{\pi}{3}\ \text{m/s}.

The answer is (A) π3\boxed{\textbf{(A)}\ \frac{\pi}{3}}.

Why this works

Circumference is linear in radius, so widening a circular lane by ww adds 2πw2\pi w regardless of the original size, and straight segments contribute nothing to the difference. Write both perimeters with the same unknowns and subtract; the unknowns cancel by design.

The trap

Introducing unknowns for the straight length and inner radius and getting stuck, or using 6 as the difference of diameters (giving 6 pi).

Common mistakes

  • Introducing unknowns for the straight length and inner radius and getting stuck, or using 66 as the difference of diameters (giving 6π6\pi).
  • Counting only one semicircle's growth (6π6\pi extra) instead of both ends.

Techniques

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