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 . We are looking for the probability that which is just . 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 . Note that in order for the jump to land the frog within 1 of the origin, we must have that 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. is therefore the probability that the frog will end within one of the origin. Since we have shown that , we have a contradiction.