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2008 AMC 8

Problem 12

Problem 12 · 2008 AMC 8 Medium
Algebra & Patterns geometric-sequence

A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 23 of the height it reached in the previous bounce. On which bounce will it rise to a height less than 0.5 meters?

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Answer: C — 5th bounce.
Show hints
Hint 1 of 2
Each bounce just keeps two-thirds of the last height — so don't restart from 3 each time, multiply the height you already have by 2/3.
Still stuck? Show hint 2 →
Hint 2 of 2
Walk the heights down one bounce at a time and stop the moment you cross below 0.5; with only a few bounces, listing beats any formula.
Show solution
Approach: walk the heights down by ×2/3
  1. Bounce 1 reaches 2. Then keep two-thirds each time: 2 → 4/3 ≈ 1.33 (b2) → 8/9 ≈ 0.89 (b3) → 16/27 ≈ 0.59 (b4) → 32/81 ≈ 0.40 (b5).
  2. Bounce 4 is still above 0.5; bounce 5 is the first below 0.5, so the answer is 5.
  3. Why this transfers: a sequence where each term is a fixed multiple of the last is geometric — for "when does it cross a threshold" just iterate; for far-off terms use height = 2·(2/3)n−1.
Another way — closed-form check:
  1. Measuring from the drop, height after bounce n is 3·(2/3)n.
  2. Bounce 4: 16/27 ≈ 0.59 > 0.5; bounce 5: 32/81 ≈ 0.40 < 0.5 ⇒ 5.
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