Competition · AMC preparation · step 4 of 4

AMC 8 · 2013 · #25

Grade 7 geometry-2d
Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum
arc-measureperimeterarea-circles identify-subproblemscasework ↑ Prerequisites: perimeterarc-measurefraction-arithmetic
📏 Long solution 💡 4 insights 📊 Diagram
Problem
A ball of diameter 4 inches (so radius r = 2) rolls without slipping along a track made of three semicircular arcs with radii R₁ = 100, R₂ = 60, and R₃ = 80 inches. The ball always touches the track. We are asked for the total distance traveled by the ball's center (not the contact point) from A to B.

Pick an answer.

(A)
$238\pi$
(B)
$240\pi$
(C)
$260\pi$
(D)
$280\pi$
(E)
$500\pi$

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole trick is geometric: on a valley the ball's center sweeps a smaller semicircle (radius R - 2), on a hill a larger one (radius R + 2). A quick sketch of the ball sitting in a valley and on top of a hill makes this ± r rule obvious — that's Tool #1 (Draw a Diagram). Then Tool #7 (Identify Subproblems) splits the track into three independent semicircles: compute each center-path length with π × radius, then add. No algebra, no advanced geometry — just one circle fact (semicircle length = π r) applied three times.

1STEP 1

Sketch the valley and hill arcs

Sketch the ball in each case: the center's path radius is R - r in a valley and R + r on a hill.

valley: center radius = R - r, hill: center radius = R + r
2STEP 2

Measure the first valley arc

Arc 1 (R₁ = 100) is a valley, so the center sweeps a semicircle of radius 100 - 2 = 98, giving length 98π.

L₁ = π × (100 - 2) = 98π
3STEP 3

Measure the hill arc

Arc 2 (R₂ = 60) is a hill, so the center rides r above the track: radius 60 + 2 = 62, giving length 62π.

L₂ = π × (60 + 2) = 62π
4STEP 4

Measure the second valley arc

Arc 3 (R₃ = 80) is another valley, so use R - r again: radius 80 - 2 = 78, giving length 78π.

L₃ = π × (80 - 2) = 78π
5STEP 5

Add the three arc lengths

Add the three center-path lengths L₁ + L₂ + L₃ to get the total distance.

L₁ + L₂ + L₃ = 98π + 62π + 78π = (98 + 62 + 78)π = 238π → (A)
Answer
238π
Sanity-check against the track itself. The three semicircle track lengths are 100π + 60π + 80π = 240π. Our answer 238π differs by 2π, which is exactly (-2 + 2 - 2)π — the algebraic sum of ± r corrections (valley, hill, valley) with r = 2. The center-path is 2π shorter than the track because there are two valleys and only one hill, so the -r corrections win by one. Magnitude and sign both check out, and the answer matches choice (A).
💡Key takeaway

Once you see the ± r rule from one quick sketch, this AMC 8 problem is just the Grade 7 circumference formula applied three times!

  • Sketch the valley and hill arcs
  • Measure the first valley arc
  • Measure the hill arc
  • Measure the second valley arc
  • Add the three arc lengths

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