2010 AMC 12B Problems/Problem 18

Revision as of 01:34, 27 December 2017 by Drunkenninja (talk | contribs) (Solution 2 (Geometric))


A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

$\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}$


Solution 1 (Complex Numbers)

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] has magnitude less than or equal to $1$. Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.

Solution 2 (Geometric)

The first frog hop doesn't matter because no matter where the frog hops, it lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.

No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how $\frac{1}{2}x^2$ and $x^2$ are tangent. The area ratio of the two circles is \[\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}\].

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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