Greek Anthology Book XIV: Metrodorus: 132

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Arithmetical Epigram of Metrodorus

This is Polyphemus the brazen Cyclops, and as if on him someone made an eye, a mouth, and a hand, connecting them with pipes.
He looks quite as if he were dripping water and seems also to be spouting it from his mouth.
None of the spouts are irregular;
that from his hand when running will fill the cistern in three days only,
that from his eye in one day,
and his mouth in two-fifths of a day.
Who will tell me the time it takes when all three are running?


Solution

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

Let $a, b, c$ be the flow rate in numbers of cisterns per hour of (respectively) the eye, the mouth and the hand.


In $t$ hours, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.

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


\(\ds \paren {a + b + c} t\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)


We have:

\(\ds a\) \(=\) \(\ds 1\) that is, $1$ cistern in $1$ day
\(\ds b\) \(=\) \(\ds \frac 5 2\) that is, $1$ cistern in $\dfrac 2 5$ of a day
\(\ds c\) \(=\) \(\ds \frac 1 3\) that is, $1$ cistern in $3$ days


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {1 + \dfrac 5 2 + \dfrac 1 3}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac 6 {6 + 15 + 2}\) multiplying top and bottom by $6$
\(\ds \) \(=\) \(\ds \frac 6 {23}\)

So the cistern will be filled in $\dfrac 6 {23}$ of a day.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources

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