AMC 8 · 2014 · #25

Grade 7 geometry-2drate-ratio
Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum
area-circlesperimeterrateunit-conversion identify-subproblemsdimensional-analysis ↑ Prerequisites: area-circlesrateunit-conversion
📏 Medium solution 💡 4 insights 📊 Diagram
Problem
A straight 1-mile stretch of highway is 40 feet wide and closed to cars. Robert rides his bike along a path made of equal semicircles whose diameters span the full 40-foot width, repeating end-to-end down the closed mile. If he rides at 5 mph, how many hours does it take to cover the 1-mile stretch?

Pick an answer.

(A)
$\frac{\pi}{11}$
(B)
$\frac{\pi}{10}$
(C)
$\frac{\pi}{5}$
(D)
$\frac{2\pi}{5}$
(E)
$\frac{2\pi}{3}$

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Diagram) is the unlock: sketch one semicircle stretched across the 40-ft width, and two facts pop out — its diameter is 40 ft, and it advances the rider 40 ft along the highway. Tool #9 (Easier Problem) then handles the repetition: solve one semicircle (its arc length and its straight-line advance), and the full path is just "copy that tile N times". Tool #8 (Analyze the Units) is the bookkeeping at the end — the arc length comes out in feet but the speed is in mph, so we must convert feet → miles before dividing by mph to get hours.

1STEP 1

Draw one semicircle across the 40-ft highway; its diameter is the width, so radius r = 20 ft, and each tile advances one diameter forward.

d = 40 ft, r = 402\frac{40}{2} = 20 ft
2STEP 2

Solve one tile first: a full circle's circumference is 2π r, so one semicircle's arc is half of that, π r = 20π ft.

arc of one semicircle = π r = 20π ft
3STEP 3

The straight length is 5280 ft and each tile advances 40 ft, so the path holds 132 semicircles (5280 ÷ 40).

N = (5280 ft)/(40 ft / semicircle) = 132 semicircles
4STEP 4

Multiply arc by count: 132 × 20π = 2640π ft, then convert with 5280 ft = 1 mi to get π2\frac{π}{2} mi.

total = 132 × 20π = 2640π ft = 2640π5280\frac{2640π}{5280} mi = π2\frac{π}{2} mi
5STEP 5

Divide distance by speed: time = (π2\frac{π}{2} mi) ÷ (5 mph) = π10\frac{π}{10} hr, choice (B).

t = (π2\frac{π}{2} mi)/(5 mph) = π10\frac{π}{10} hr → (B)
Answer
π10\frac{π}{10}
Sanity check the magnitude. Robert's riding distance is π2\frac{π}{2} ≈ 1.57 miles to cover a 1-mile straight stretch — about 57% longer because of the curving, which matches the fact that a semicircle is π2\frac{π}{2} ≈ 1.57 times longer than its diameter. At 5 mph, 1.57 mi takes about 1.575\frac{1.57}{5} ≈ 0.314 hr, and π10\frac{π}{10} ≈ 0.314 — perfect match. Also 0.314 hr is roughly 19 minutes, a reasonable bike time for 1.5 miles.
💡Key takeaway

Draw one semicircle and the whole mile becomes easy — Grade 7 circle circumference plus a simple distance-divided-by-speed is all you need!

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