AMC 10 · 2002 · #24

Grade 8 geometry-2d
thirty-sixty-ninety-trianglearc-measurerate identify-subproblems ↑ Prerequisites: ratethirty-sixty-ninety-triangle
📏 Long solution 💡 3 insights
Problem
A Ferris wheel is a circle of radius 20 feet that spins at one full turn per minute. A rider starts at the very bottom. Find how many seconds pass before the rider is 10 feet higher than the bottom.

Pick an answer.

(A)
5
(B)
6
(C)
7.5
(D)
10
(E)
15

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The question mixes a spinning circle with a straight-up height, so Tool #1 (Draw a Diagram) is the anchor: sketch the wheel as a circle, mark the center, the bottom point, and the rider. That picture turns 'height above the bottom' into a length inside a right triangle. Tool #7 (Identify Subproblems) splits the work into two clean pieces — first find the angle the wheel has turned through, then convert that angle into time. Tool #4 (Introduce a Variable) names the rider's sideways distance so the Pythagorean theorem can pin the point down. Finally Tool #8 (Analyze the Units) carries the angle over to seconds, since a constant one-turn-per-minute rate means an equal share of the angle is an equal share of the 60 seconds.

1STEP 1

Draw the wheel and place the points

Put the bottom at B=(0,0). The radius is 20, so the center is O=(0,20), and the rider's target sits at P=(x,10).

OB = OP = 20, B=(0,0), O=(0,20), P=(x,10)
2STEP 2

Find where the rider is

P is one radius from O, so x² + 10² = 20². Then x² = 300 and the rider sits at x = 10√(3).

x² + 10² = 20² → x² = 300 → x = 10√(3)
3STEP 3

Measure the angle turned

The chord BP = √(300 + 100) = 20 matches both radii, so triangle OBP is equilateral and the wheel has turned ∠ BOP = 60°.

BP = √(300+100) = 20 = OB = OP → △ OBP equilateral → ∠ BOP = 60°
4STEP 4

Turn the angle into seconds

60° is 60/360 = 1/6 of a turn, and at a steady rate that is 1/6 of the 60 seconds: 10 seconds, choice (D).

60°/360° = 1/6, 1/6 × 60 s = 10 s (D)
Answer
10
Check the size of the answer. Half the wheel (from bottom all the way to the top, a rise of 40 feet) is 180° and would take 30 seconds. A rise of only 10 feet is far less than halfway up, so the time should be much less than 30 seconds, and 10 seconds fits. The traps make sense too: 15 (E) is a quarter turn, which would lift the rider to the center height of 20 feet, not 10; 5 (A) and 6 (B) are too small for a 60° sweep. Only (D) 10 matches the 60° the geometry forces.
💡Key takeaway

Turn 'how high' into an angle at the center of the wheel, then trade that angle for time — the same fraction of the circle is the same fraction of the minute.

  • Draw the wheel and place the points
  • Find where the rider is
  • Measure the angle turned
  • Turn the angle into seconds