Greek Anthology Book XIV: 7. - Problem

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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.


Let $t$ be the number of hours it takes to fill a jar.

Let $r, l, m, f$ be the flow rate in numbers of jars per day of (respectively) right eye, left eye, mouth and foot.

In $t$ hours, the various contributions of each of the spouts is:

Right eye: $r t$
Left eye: $l t$
Right foot: $f t$
Mouth: $m t$

So for the total contribution to be $1$ jar, we have:

\(\ds \paren {r + l + f + m} t\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac 1 {r + l + f + m}\)

There are $24$ hours in a day.

Hence, we have:

\(\ds r\) \(=\) \(\ds \frac 1 {2 \times 24}\) that is, $1$ jar in $2$ days
\(\ds l\) \(=\) \(\ds \frac 1 {3 \times 24}\) that is, $1$ jar in $3$ days
\(\ds f\) \(=\) \(\ds \frac 1 {4 \times 24}\) that is, $1$ jar in $4$ days
\(\ds m\) \(=\) \(\ds \frac 1 6\) $1$ jar in $6$ hours

and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {r + l + f + m}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\dfrac 1 {2 \times 24} + \dfrac 1 {3 \times 24} + \dfrac 1 {4 \times 24} + \dfrac 1 6}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac {12 \times 24} {6 + 4 + 3 + 48}\) multiplying top and bottom by $12 \times 24$
\(\ds \) \(=\) \(\ds \dfrac {288} {61}\)
\(\ds \) \(=\) \(\ds 4 \frac {44} {61}\)

So the jar will be filled in $4 \frac {44} {61}$ hours, or $4$ hours, $43$ minutes and $17$ seconds, approximately.



This is the prototype of a whole class of problems of this type which schoolchildren have been set over the centuries, for example:

$A$ and $B$ together can do a piece of work in $6$ days.
$B$ and $C$ together can do it in $20$ days.
$C$ and $A$ together can do it in $7 \frac 1 2$ days.

How many days will each require to do the job separately?

Historical Note

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.