AMC 10 · 2006 · #15

Grade 7 geometry-2d
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed
rateratio-proportioncircle-circumference identify-subproblems ↑ Prerequisites: rate
📏 Medium solution 💡 3 insights
Problem
Odell and Kershaw run for 30 minutes on a circular track, starting on the same radial line. Odell runs clockwise at 250 m/min in the inner lane, radius 50 meters. Kershaw runs counterclockwise at 300 m/min in the outer lane, radius 60 meters. Because the lanes differ, they count as passing each other whenever they lie on the same radial line. How many times after the start do they pass each other?

Pick an answer.

(A)
29
(B)
42
(C)
45
(D)
47
(E)
50

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Analyze the Units

The runners are on different-sized circles, so comparing their meters is misleading — what lines them up is how far around the circle each has gone, i.e. laps. Tool #8 (Analyze the Units) converts each speed in meters-per-minute into minutes-per-lap by dividing by the circumference 2π r; this is the move that exposes the hidden coincidence — both take the same time per lap. Tool #1 (Draw a Diagram) reframes 'pass each other' as 'same radial line', so lanes stop mattering. Tool #4 (Introduce a Variable) names one lap-time T and turns opposite-direction motion into a fixed meeting interval. Finally, because the count comes out just below a whole number, Tool #3 (Eliminate Possibilities) settles the tempting 48-vs-47 rounding trap against the answer list.

1STEP 1

Reframe passing as same angle

Different lanes never collide — passing just means sharing a radial line, the same fraction of a lap, so only the angle matters.

pass ⇔ same angle ⇔ same fraction of a lap
2STEP 2

Find each runner's time per lap

Odell's lap is 100π m at 250 m/min and Kershaw's is 120π m at 300 m/min, so both lap in T = 0.4π minutes.

T_Odell = 100π/250 = 0.4π, T_Kershaw = 120π/300 = 0.4π → T = 0.4π min/lap
3STEP 3

Turn opposite directions into a meeting interval

Running opposite ways closes the gap twice as fast, so they line up every half lap-time — once every 0.2π minutes.

2 · t/T = 1, 2, 3, … → t = T/2, T, 3T/2, … → interval = T/2 = 0.2π min
4STEP 4

Count the passings in 30 minutes

30 ÷ 0.2π = 150/π ≈ 47.75, and the 48th meeting falls at ≈ 30.2 min, so there are 47 passings — (D).

30/0.2π = 150/π ≈ 47.75 → 47 passings → (D)
Answer
47
Sanity-check the size: about one passing every 0.2π ≈ 0.63 minutes over 30 minutes gives roughly 30/0.63 ≈ 48, so an answer in the high-40s is right, ruling out 29 and 42. The exact value 150/π ≈ 47.75 is not a whole number, so it must round down to the last completed passing, 47; the boundary check confirms the 48th meeting falls at ≈ 30.2 min, just after the whistle. The trap answer (E) 50 comes from rounding 47.75 up or from mishandling the half-lap interval, and (C) 45/(B) 42 come from comparing meters instead of laps.
💡Key takeaway

Different lanes only change lap length, so turn each speed into minutes-per-lap; here both runners lap in the same time, they meet every half-lap since they run opposite ways, and 30 ÷ 0.2π ≈ 47.75 rounds down to 47 real passings.

  • Reframe passing as same angle
  • Find each runner's time per lap
  • Turn opposite directions into a meeting interval
  • Count the passings in 30 minutes

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