Greek Anthology Book XIV: 7. - Problem/Historical Note

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Historical Note on Problem $7$, Book $\text {XIV}$ of The Greek Anthology

I am a brazen lion; my spouts are my two eyes, my mouth, and the flat of my right foot.
My right eye fills a jar in two days,
my left eye in three,
and my foot in four.
My mouth is capable of filling it in six hours;
tell me how long all four together will take to fill it.


In W.R. Paton's $1918$ translation of The Greek Anthology Book XIV, he gives:

The scholia propose several, two of which, by not counting fractions, reach the result of four hours; but the strict sum is $3 \frac {33} {37}$ hours.


The above is correct if it is assumed there are $12$ hours in a day.

It would also need to be assumed that, in order to fulfil the conditions of the statement of the problem, the spouts are turned off at night.


This problem was apparently first presented by Heron of Alexandria in his Metrika.

It was still being taught in classrooms up until the middle of the $20$th Century, and was considered the epitome of "useless" mathematics.

This, as David Wells points out in his Curious and Interesting Puzzles of $1992$, is a shame, because the idea behind it is far from useless.


Proof

Assuming a $12$-hour day, the following can be calculated:

\(\ds r\) \(=\) \(\ds \frac 1 {2 \times 12}\) that is, $1$ jar in $2$ days
\(\ds l\) \(=\) \(\ds \frac 1 {3 \times 12}\) that is, $1$ jar in $3$ days
\(\ds f\) \(=\) \(\ds \frac 1 {4 \times 12}\) that is, $1$ jar in $4$ days
\(\ds m\) \(=\) \(\ds \frac 1 6\) $1$ jar in $6$ hours


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {r + l + f + m}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\dfrac 1 {2 \times 12} + \dfrac 1 {3 \times 12} + \dfrac 1 {4 \times 12} + \dfrac 1 6}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac {12 \times 12} {6 + 4 + 3 + 24}\) multiplying top and bottom by $12 \times 12$
\(\ds \) \(=\) \(\ds \dfrac {144} {37}\)
\(\ds \) \(=\) \(\ds 3 \frac {33} {37}\)

So the jar will be filled in $3 \frac {33} {37}$ hours.

$\blacksquare$


Sources

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