Henry Ernest Dudeney/Modern Puzzles/42 - The Puzzle of the Runners/Solution

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Modern Puzzles by Henry Ernest Dudeney: $42$

The Puzzle of the Runners
Two men ran a race round a circular course, going in opposite directions.
Brown was the best runner and gave Tompkins a start of $\tfrac 1 8$ of the distance.
But Brown, with a contempt for his opponent, took things too easily at the beginning,
and when he had run $\tfrac 1 6$ of his distance he met Tompkins,
and saw that his chance of winning the race was very small.
How much faster than he went before must Brown now run in order to tie with his competitor?


Solution

$20 \tfrac 1 4$ times faster.

Probably not technically achievable.


Proof

While Brown has run $\dfrac 1 6$ of the course, or $\dfrac 4 {24}$ of it, Tomkins has already run $\dfrac 5 6 - \dfrac 1 8$, that is $\dfrac {17} {24}$ of it.

Thus Tomkins's pace has up till now been $\dfrac {11} 4$ of Brown's.

Brown now has $\dfrac 5 6$ of the course to run, while Tomkins has $\dfrac 1 6$ left.

Thus Brown must speed up to $5$ times as fast as Tomkins in order to pass the finish line at the same time as Tomkins.

Thus he must go at $5 \times \dfrac {17} 4$, which is $\dfrac {85} 4$ times as fast as he went at first.

Then we note that $\dfrac {85} 4$ times as fast is the same thing as $\dfrac {85} 4 - 1$ times faster.

Hence the correct answer is $\dfrac {81} 4 = 20 \tfrac 1 4$ times faster.

$\blacksquare$


Sources

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