Talk:2010 AMC 12B Problems/Problem 18

Solution No. 2 is wrong due to the incorrect assumption that after two jumps, the frog ends up at a random point (with probability distributed by area) on a disk of radius 2 centered at the starting point.

Proof that this is in fact false:

Suppose the frog ends up at a random point on a disk of radius 2 after 2 jumps. Consider the probability of the frog ending within 1m of the origin after two jumps. By our assumption, the frog will end up somewhere on $x^2 + y^2 < 4$. We are looking for the probability that $x^2 + y^2 < 1$ which is just $.25$. Lets calculate this same probability a different way. Fix the first jump to be directly right. The second jump results in a point on the circle $(x-1)^2+y^2 = 1$. Note that in order for the jump to land the frog within 1 of the origin, we must have that $x<.5$ after the jump; in other words, there is an angle range for the second jump of 120 degrees which results in the frog returning to within one of the origin. $120 / 360 = 1/3$ is therefore the probability that the frog will end within one of the origin. Since we have shown that $1/4 = 1/3$, we have a contradiction.

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