# The Bulldozers and the Bee

## Contents

## 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: $\displaystyle d_{n+1} = \begin{cases} \dfrac {2 D_1} 3 & : n = 1 \\ \dfrac {d_n} 3 & : n > 1 \end{cases}$

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

$\blacksquare$

### The Short Answer

The bulldozers are travelling at $10$ mph and are $20$ miles apart.

Therefore they travel $10$ miles each and collide after $1$ hour.

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

$\blacksquare$

## 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.

## Historical Note

The problem of **The Bulldozers and the Bee** 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:

- $\displaystyle 20 \times 2 \sum_{n \mathop \ge 1} \paren {\frac 1 3}^n = 20$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.1$: The bulldozers and the bee