Problem of the Week 1124

A Fair Race

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Solution

Let us assume that all runners run at constant speed.

Here is the solution of Jonathan Foster, extended to the case of n runners.

Consider n runners, A(1), A(2), ...., A(n), each of whom gives a 5-meter head start to the next.

A(1) runs 100^{(n − 1)} meters while A(2) runs 95 * 100^{(n − 2)}.

A(2) runs 95 * 100^{(n − 2)} while A(3) runs 95² 100^{(n − 3)}.

A(3) runs 95² 100^{(n − 3)} while A(4) runs 95³ 100^{(n − 4)}.

...

A(n − 1) runs 95^{(n − 2)} 100 while A(n) runs 95^{(n − 1)}.

So A(1) runs 100^{(n − 1)} meters while A(n) runs 95^{(n − 1)}, and so A(1) runs 100 while A(n) runs 95^{(n − 1)}/(100^{(n − 2)}).

Therefore, the first should give the last a head start of 100 − 95^{(n − 1)}/(100^{(n − 2)}), which is 14.2625 meters if n = 4.

The general expression is more simply written as

100 (1 − (95/100)^{(n − 1)})

So if the race distance is D and the fraction given as a head start is p, the answer is D (1 − (1 − p)^{(n − 1)}).

[Back to Problem 1124]

© Copyright 2009 Stan Wagon. Reproduced with permission.