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