AMC 8 · 2013 · #25
Grade 7 geometry-2dProblem
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 inches, inches, and 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.
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
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) $500\pi$
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
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 $= R + r$).
💡 This is the whole problem in one picture. Once you see the $\pm r$ rule, the rest is arithmetic.
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 $\pi \rho$.
💡 Subproblem 1: just one arc, with the smaller radius because it's a valley.
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 $= 60 + 2 = 62$.
💡 Subproblem 2: same circle formula, but $+r$ instead of $-r$ because it's a hill.
7.G.B.4 Step 4 - Arc 3 ($R_3 = 80$) is another valley, so use $R - r$ again.
- Path radius $= 80 - 2 = 78$.
💡 Subproblem 3: valley, so subtract $r$ again.
4.NBT.B.4 Step 5 - Add the three pieces.
- The total distance the center travels is $L_1 + L_2 + L_3$.
💡 Final combine step: three nonnegative whole-number coefficients added, then a single $\pi$ factored out.
7.G.B.4 Sketch the ball sitting in a valley arc and on top of a hill arc. In both cases 7.G.B.4 Arc 1 ($R_1 = 100$) is a valley, so the center sweeps a semicircle of radius $10 7.G.B.4 Arc 2 ($R_2 = 60$) is a hill (it bulges upward), so the center rides $2$ in abov 7.G.B.4 Arc 3 ($R_3 = 80$) is another valley, so use $R - r$ again. Path radius $= 80 - 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.4Know 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.4Fluently 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!
⭐ 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!