Greek Anthology Book XIV: Metrodorus: 135

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Arithmetical Epigram of Metrodorus

We three Loves stand here pouring out water for the bath, sending streams into the fair-flowing tank.
I on the right, from my long-winged feet, fill it full in the sixth part of a day;
I on the left, from my jar, fill it in four hours;
and I in the middle, from my bow, in just half a day.
Tell me in what a short time we should fill it, pouring water from wings, bow and jar all at once.


Solution

First note that the day in this context is taken to be $12$ hours long.


Let $t$ be the number of days it takes to fill the tank.

Let $a, b, c$ be the flow rate in numbers of tanks per day of (respectively) the Loves on the right, the left and the middle.


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

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


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


We have:

\(\ds a\) \(=\) \(\ds 6\) that is, $1$ tank in $\dfrac 1 6$ of a day
\(\ds b\) \(=\) \(\ds \frac {12} 4 = 3\) that is, $1$ tank in $4$ hours which is $\dfrac 4 {12}$ of a day
\(\ds c\) \(=\) \(\ds 2\) that is, $1$ tank in $\dfrac 1 2$ days


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {6 + 3 + 2}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac 1 {11}\)

So the tank will be filled in $\dfrac 1 {11}$ of a day.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources

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