The Fly and the Trains Problem



The problem is this: Two trains are on the same line, 60 miles apart, heading towards each other, each traveling at 30 mph. A fly that can travel at 60mph leaves one engine flying towards the other. Upon reaching the other engine, it instantaneously turns around, and heads back to the other engine. This is repeated until the two trains crash and the fly is anihilated at the same time.

Question: How far does the fly travel before it is "splatted"?

The interesting story: According to several persons who were present, this problem was presented to the famous mathematician John von Neumann at a cocktail party. After thinking about it for a while, he answered 60 miles. When asked how he did that he replied "I summed the series."

What series?

In the diagram above, the two engines are d1 miles apart at the time the fly leaves the leftmost engine which is traveling at sL mph; the fly is traveling at sF mph. The righthand engine is traveling at sR mph.

The distance traveled by the fly to meet the righthand engine is:

d2 = d 1 - (d2 /sF) x sR

where (d2 /sF) is the time that it takes the fly to travel between the two engines.

That is d2 = d1 / (1 + sR / sF)

Thus if the initial distance between the two engines is d0, then the distance traveled in the first segment of the fly's flight is:

d0 / (1 + sR / sF),

and on the second leg, d0 / (1 + sR / sF) / (1 + sL / sF),

and so on. But in general sR = sL. Let the engine speeds be sT

Let r = (1 + sT / sF).

So the total distance traveled by the fly is:

d0 (1/r + 1/r2+ 1/r3+ 1/r4+ 1/r5 + ...)

Now in the problem given to von Neumann, d0 = 60, and r = 1.5 Probably, von Neuman recognized that in the limit
(1/1.5 + 1/1.52+ 1/1.5r3+ 1/1.54+ 1/1.55 + ... ) sums to 1.

But, he had to work out the mathematics in his head!

A simpler solution: Can you think of a simpler way of coming up with the solution? Click here to check your solution.

Last updated 2002/05/30
© J.A.N. Lee, 2002.