The Bulldozers and the Bee

Problem
Two bulldozers (presumably driven by a pair of insane enemies) are 20 miles apart, heading towards each other at 10 miles per hour, on a collision course. At the same time, a bee takes off from the blade of one bulldozer at 20 miles per hour, towards the other bulldozer. As soon as the bee reaches the other bulldozer, it reverses direction instantaneously and heads off at 20 miles per hour back towards the first bulldozer. It continues to do this until the bulldozers collide, squashing the bee between them and killing her.

The question is: how far does the bee fly before the collision?

Solution
This is frequently asked as a trick question.

The Long Answer
Let $d$ be the total distance the bee travels.

Let $D_1$ be the initial separation of the bulldozers in miles.

Let $d_n$ be the distance the bee travels on each leg of her journey.

Let $d'_n$ be the distance that one of the bulldozers travels during the time the bee travels $d_n$.

Let $D_n$ be the distance the bulldozers are apart at the start of each leg of the journey.

The bee travels twice as fast as each of the bulldozers. So on each leg, $d_n = 2 d'_n$.

Consider the $m$th leg of the journey.

The bee travels $d_m$, and the bulldozers travel $\dfrac {d_m} 2$. These two together equal $D_m$.

Therefore $d_m = \dfrac {2 D_m} 3$, while $d'_m = \dfrac {D_m} 3$.

At the start of leg $m + 1$, both bulldozers have covered the distance $\dfrac {D_m} 3$. So at the start of the second leg, the bulldozers are $D_{m+1} = D_m - \dfrac {2 D_m} 3 = \dfrac {D_m} 3$.

This gives us a recurrence formula: $\displaystyle d_{n+1} = \begin{cases} \frac {2 D_1} 3 & : n = 1 \\ \frac {d_n} 3 & : n > 1 \end{cases}$

It can be seen that the answer can be calculated by Sum of Geometric Progression, and comes out as 20 miles.

The Short Answer
The bulldozers are travelling at 10 mph and are 20 miles apart. Therefore they travel 10 miles each and collide after one hour.

The bee is flying at 20 mph and therefore travels 20 miles in that time.

Comment
The question has been phrased in several different forms.

One many apocryphal tales concerning John von Neumann is that he was asked this question. He instantly gave the answer.

"So you've heard this one then? You solved it the quick way?" he was asked.

"I solved it by summing an infinite geometric progression. There's a quicker way?" was the reply.

The point is that there are (at least) two ways to solve the problem, and they come to the same value.

That is:
 * $\displaystyle 20 \times 2 \sum_{n \mathop \ge 1} \left({\frac 1 3}\right)^n = 20$

Pointless quibbles
Whether a bee can actually fly at 20 miles per hour is doubtful, let alone sustain that speed for a whole hour. I may be completely wrong. This may be completely reasonable.

Even if she could, she could not reverse direction instantaneously. The laws of physics are completely against it.