The Bulldozers and the Bee/The Long Answer

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?

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

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

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.13$: Example