Probability of frog jumping

Question

I'm trying to understand the solution given to this problem on AOPS site:

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 (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 $\dfrac{1}{4}$.

..........

I only understand $|a+b+c|$. I don't know where other three terms come from. Also why only one of these has magnitude less than or equal to $1$?

Thank you.

Answer 1

Where these terms come from: the idea is to modify the problem to get an equivalent experiment.

• Simply following the word problem, we could choose jumps $a,b,c$ uniformly from the unit circle and then end up at $a+b+c$.

• Instead, we could choose $b$ in two steps: first choose an angle $\theta \in [0,\pi)$, then flip a coin to let $b$ be one of $e^{i\theta}$ or $-e^{i\theta}$. This is also uniform on the unit circle. We could do the same thing for $c$.

• If, using the second method, we choose the angles for $b$ and $c$, but wait to choose the sign until the end, then we are flipping two coins to choose between $\{a+b+c, a+b+(-c), a+(-b)+c, a+(-b)+(-c)\}$.

Why exactly one of these is in the unit circle: I can't promise this is the best argument, but here's one. We'll show that at most one of these points is in the unit circle, then show that at least one is in the unit circle.

The four points are the vertices of a rhombus with sides parallel to $2b$ and $2c$, which is centered at $a$. Here's a typical picture:

Assume none of the corners are on the boundary of the unit circle, which holds with probability $1$. Then two adjacent corners can't both be in the circle, because they're two units apart. Two opposite corners can't both be in the circle, because the midpoint between them (which is $a$) would also be in the circle. So at most one of these four points is in the unit circle.

On the other hand, from $a$, the four directions $b+c, b-c, -b+c, -b-c$ are perpendicular, so two of them point into the circle; without loss of generality, say that it's $b+c$ and $b-c$. Now consider the following two right triangles:

1. The right triangle with vertices $a$, $a+b+c$, and $a+b-c$.
2. The right triangle inscribed in the unit circle with one vertex at $a$ and the two sides being the two chords from $a$ going in the same directions as $b+c$ and $b-c$.

Both right triangles have hypotenuse $2$: the first one because $|(a+b+c)-(a+b-c)| = |2c|=2$, and the second one because its hypotenuse is a diameter of the unit circle. So the first right triangle has a side that's shorter than the corresponding side of the second one. This means one of its vertices lies on the side of the second right triangle, and is in the interior of the circle, QED.