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2010 AMC 12B Problem 18

Problem 18 of 25IntermediateCounting & Probability

A frog makes 33 jumps, each exactly 11 meter long. The directions of the jumps are chosen independently and at random. What is the probability that the frog’s final position is no more than 11 meter from its starting position?

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Solution

This is a continuous (geometric) probability. Anchor the second jump from P=(0,0)P=(0,0) to Q=(1,0),Q=(1,0), and let α,β\alpha,\beta be the directions of the first and third jumps, so the start is A=(cosα,sinα)A=(\cos\alpha,\sin\alpha) and the end is B=(1+cosβ,sinβ).B=(1+\cos\beta,\sin\beta). Taking 0απ0\le\alpha\le\pi and 0β2π,0\le\beta\le2\pi, the requirement AB1AB\le1 holds exactly when αβπ.\alpha\le\beta\le\pi. In the αβ\alpha\beta-rectangle of area 2π2,2\pi^2, the favorable region is a triangle of area π22,\tfrac{\pi^2}{2}, so the probability is π222π2=14.\dfrac{\frac{\pi^2}{2}}{2\pi^2}=\dfrac14. Thus, the correct answer is C.

More practice

Concepts: geometric probability · random walk

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.