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AMC 8 · 2014 · #25

Grade 7 geometry-2drate-ratio
Archetype Arithmetic Expression Decomposition
area-circlesperimeterrateunit-conversion identify-subproblemsdimensional-analysis ↑ Prerequisites: area-circlesrateunit-conversion
📏 Medium solution 💡 4 insights 📊 Diagram

Problem

A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?

Note: 1 mile = 5280 feet

(A) π11(B) π10(C) π5(D) 2π5(E) 2π3\textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3}

Pick an answer.

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

Understand

Restated: 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?

Givens: Highway length (straight-line) $= 1$ mile $= 5280$ feet; Highway width $= 40$ feet, equal to each semicircle's diameter; Path = repeating semicircles spanning the full width, alternating sides; Riding speed $= 5$ mph; Answer choices: (A) $\tfrac{\pi}{11}$, (B) $\tfrac{\pi}{10}$, (C) $\tfrac{\pi}{5}$, (D) $\tfrac{2\pi}{5}$, (E) $\tfrac{2\pi}{3}$ (hours)

Unknowns: The time in hours Robert spends riding the full $1$-mile stretch

Understand

Restated: 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?

Givens: Highway length (straight-line) $= 1$ mile $= 5280$ feet; Highway width $= 40$ feet, equal to each semicircle's diameter; Path = repeating semicircles spanning the full width, alternating sides; Riding speed $= 5$ mph; Answer choices: (A) $\tfrac{\pi}{11}$, (B) $\tfrac{\pi}{10}$, (C) $\tfrac{\pi}{5}$, (D) $\tfrac{2\pi}{5}$, (E) $\tfrac{2\pi}{3}$ (hours)

Plan

Primary tool: #1 Draw a Diagram

Secondary: #9 Solve an Easier Related Problem, #8 Analyze the Units

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 $\to$ miles before dividing by mph to get hours.

Execute — Answer: B

#1 Draw a Diagram 7.G.B.4 Step 1
  • Draw one semicircle stretched across the $40$-foot-wide highway.
  • Its diameter equals the width, so $d = 40$ ft and the radius is $r = \tfrac{d}{2} = 20$ ft.
  • The next semicircle flips to the other side and starts where this one ends, so each semicircle advances Robert exactly $40$ ft (one diameter) along the highway.
$$d = 40 \text{ ft}, \quad r = \tfrac{40}{2} = 20 \text{ ft}$$

💡 Sketching one tile of a repeating pattern turns a scary mile-long path into a single shape whose size is obvious from the picture.

#9 Solve an Easier Related Problem 7.G.B.4 Step 2
  • Solve the easier problem first: what is the arc length of just one semicircle?
  • A full circle of radius $r$ has circumference $2\pi r$, so a semicircle's arc is half of that, $\pi r$.
  • Plug in $r = 20$.
$$\text{arc of one semicircle} = \pi r = 20\pi \text{ ft}$$

💡 Knowing one tile (one semicircle) is enough — the full path is just many copies of it.

#9 Solve an Easier Related Problem 5.MD.A.1 Step 3
  • Count how many tiles fit in the $1$-mile stretch.
  • The straight-line length is $5280$ ft, and each semicircle advances $40$ ft of straight-line distance, so the number of semicircles is $5280 \div 40 = 132$.
$$N = \dfrac{5280 \text{ ft}}{40 \text{ ft / semicircle}} = 132 \text{ semicircles}$$

💡 Dividing the total length by the length-per-tile is the standard "how many tiles fit" move.

#8 Analyze the Units 5.MD.A.1 Step 4

Multiply arc-per-tile by number-of-tiles to get the total riding distance, then convert feet to miles using $5280$ ft $= 1$ mi so the units match the speed (mph).

$$\text{total} = 132 \times 20\pi = 2640\pi \text{ ft} = \dfrac{2640\pi}{5280} \text{ mi} = \dfrac{\pi}{2} \text{ mi}$$

💡 Converting feet to miles before dividing by mph keeps the units honest — the answer comes out in hours, not in a feet/mph mess.

#8 Analyze the Units 6.RP.A.3 Step 5

Apply $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ with consistent units (miles and mph) to get hours.

$$t = \dfrac{\tfrac{\pi}{2} \text{ mi}}{5 \text{ mph}} = \dfrac{\pi}{10} \text{ hr} \;\Rightarrow\; \textbf{(B)}$$

💡 Distance divided by speed gives time — the rate relationship $d = rt$ used in reverse.

[1] #1 7.G.B.4 Draw one semicircle stretched across the $40$-foot-wide highway. Its diameter eq
[2] #9 7.G.B.4 Solve the easier problem first: what is the arc length of just one semicircle? A
[3] #9 5.MD.A.1 Count how many tiles fit in the $1$-mile stretch. The straight-line length is $5
[4] #8 5.MD.A.1 Multiply arc-per-tile by number-of-tiles to get the total riding distance, then
[5] #8 6.RP.A.3 Apply $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ with consistent unit

Review

Reasonableness: Sanity check the magnitude. Robert's riding distance is $\tfrac{\pi}{2} \approx 1.57$ miles to cover a $1$-mile straight stretch — about $57\%$ longer because of the curving, which matches the fact that a semicircle is $\tfrac{\pi}{2} \approx 1.57$ times longer than its diameter. At $5$ mph, $1.57$ mi takes about $\tfrac{1.57}{5} \approx 0.314$ hr, and $\tfrac{\pi}{10} \approx 0.314$ — perfect match. Also $0.314$ hr is roughly $19$ minutes, a reasonable bike time for $1.5$ miles.

Alternative: Tool #7 (Identify Subproblems) gives a one-line shortcut by working in ratios. Every $40$ ft of straight road costs the rider $20\pi$ ft of arc, so the total path is $\tfrac{20\pi}{40} = \tfrac{\pi}{2}$ times the straight length — i.e. $\tfrac{\pi}{2}$ mile. Then $t = \tfrac{\pi/2}{5} = \tfrac{\pi}{10}$ hr. The ratio $\tfrac{\pi}{2}$ comes from "semicircle arc / diameter", so we never need the count $132$ at all.

CCSS standards used (min grade 7)

  • 7.G.B.4 Know the formulas for area and circumference of a circle; solve problems involving them (Computing the arc length of one semicircle as $\pi r = 20\pi$ ft using the circumference formula $C = 2\pi r$ halved.)
  • 5.MD.A.1 Convert among different-sized standard measurement units within a given system (Counting $5280 \div 40 = 132$ semicircles in $1$ mile and converting the total distance from $2640\pi$ ft to $\tfrac{\pi}{2}$ mi.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Applying $\text{time} = \text{distance} / \text{speed}$ to get $\tfrac{\pi/2}{5} = \tfrac{\pi}{10}$ hour from the total miles and the mph speed.)

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

⭐ 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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