AMC 12 2010B Problem Help #18 Can someone explain this solution?
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? 
Solution 1
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 1/4. 
 A: Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 \pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $\frac{\pi}{4\pi} = \frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $\frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
