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 with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A ball of diameter $4$ inches (so radius $r = 2$) rolls without slipping along a track made of three semicircular arcs with radii $R_1 = 100$, $R_2 = 60$, and $R_3 = 80$ 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 $B$.

Givens: Ball diameter $= 4$ in, so ball radius $r = 2$ in; Track = three semicircular arcs in sequence with radii $R_1 = 100$, $R_2 = 60$, $R_3 = 80$ in; From the figure: arc 1 is a valley (concave up), arc 2 is a hill (concave down), arc 3 is a valley; The ball stays in contact with the track and does not slip; Answer choices: (A) $238\pi$, (B) $240\pi$, (C) $260\pi$, (D) $280\pi$, (E) $500\pi$

Unknowns: The total arc length traced by the *center* of the ball as it rolls from $A$ to $B$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

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 $\pm r$ 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 $\pi \times \text{radius}$, then add. No algebra, no advanced geometry — just one circle fact ($\text{semicircle length} = \pi r$) applied three times.

Execute — Answer: A

#1 Draw a Diagram 7.G.B.4 Step 1

Sketch the ball sitting in a valley arc and on top of a hill arc.
In both cases the ball's center is exactly $r = 2$ in from the contact point, along the radial line through the track's curvature center.
In a valley the center is pulled *toward* the curvature center (path radius $= R - r$).
On a hill the center is pushed *away* from it (path radius $= R + r$).

$$\text{valley: center radius} = R - r, \quad \text{hill: center radius} = R + r$$

💡 This is the whole problem in one picture. Once you see the $\pm r$ rule, the rest is arithmetic.

#7 Identify Subproblems 7.G.B.4 Step 2

Arc 1 ($R_1 = 100$) is a valley, so the center sweeps a semicircle of radius $100 - 2 = 98$.
The length of a semicircle of radius $\rho$ is $\pi \rho$.

$$L_1 = \pi \times (100 - 2) = 98\pi$$

💡 Subproblem 1: just one arc, with the smaller radius because it's a valley.

#7 Identify Subproblems 7.G.B.4 Step 3

Arc 2 ($R_2 = 60$) is a hill (it bulges upward), so the center rides $2$ in above the track.
Path radius $= 60 + 2 = 62$.

$$L_2 = \pi \times (60 + 2) = 62\pi$$

💡 Subproblem 2: same circle formula, but $+r$ instead of $-r$ because it's a hill.

#7 Identify Subproblems 7.G.B.4 Step 4

Arc 3 ($R_3 = 80$) is another valley, so use $R - r$ again.
Path radius $= 80 - 2 = 78$.

$$L_3 = \pi \times (80 - 2) = 78\pi$$

💡 Subproblem 3: valley, so subtract $r$ again.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Add the three pieces.
The total distance the center travels is $L_1 + L_2 + L_3$.

$$L_1 + L_2 + L_3 = 98\pi + 62\pi + 78\pi = (98 + 62 + 78)\pi = 238\pi \;\Rightarrow\; \textbf{(A)}$$

💡 Final combine step: three nonnegative whole-number coefficients added, then a single $\pi$ factored out.

[1] #1 7.G.B.4 Sketch the ball sitting in a valley arc and on top of a hill arc. In both cases

[2] #7 7.G.B.4 Arc 1 ($R_1 = 100$) is a valley, so the center sweeps a semicircle of radius $10

[3] #7 7.G.B.4 Arc 2 ($R_2 = 60$) is a hill (it bulges upward), so the center rides $2$ in abov

[4] #7 7.G.B.4 Arc 3 ($R_3 = 80$) is another valley, so use $R - r$ again. Path radius $= 80 -

[5] #7 4.NBT.B.4 Add the three pieces. The total distance the center travels is $L_1 + L_2 + L_3$

Review

Reasonableness: Sanity-check against the track itself. The three semicircle track lengths are $100\pi + 60\pi + 80\pi = 240\pi$. Our answer $238\pi$ differs by $2\pi$, which is exactly $(-2 + 2 - 2)\pi$ — the algebraic sum of $\pm r$ corrections (valley, hill, valley) with $r = 2$. The center-path is $2\pi$ *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).

Alternative: Tool #3 (Eliminate Possibilities). The naive answer 'just use the track length' gives $100\pi + 60\pi + 80\pi = 240\pi$ — that's choice (B), the classic trap. Any correct answer must differ from $240\pi$ by an integer multiple of $\pi \cdot r = 2\pi$ (one correction per arc). Choices (C) $260\pi$, (D) $280\pi$, (E) $500\pi$ are too far off in the wrong direction to come from $\pm 2$ corrections, so they're eliminated. That leaves (A) $238\pi$ versus (B) $240\pi$, and the $\pm r$ analysis above picks (A).

CCSS standards used (min grade 7)

7.G.B.4 Know the formulas for area and circumference of a circle (Using $\text{(semicircle length)} = \pi \times \text{radius}$ for each of the three arcs traced by the ball's center.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing the three adjusted radii ($100 - 2$, $60 + 2$, $80 - 2$) and summing $98 + 62 + 78 = 238$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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