Greek Anthology Book XIV: Metrodorus: 133

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Arithmetical Epigram of Metrodorus

What a fine stream do these two river-gods and beautiful Bacchus pour into the bowl.
The current of the streams of all is not the same.
Nile flowing alone will fill it up in a day, so much water does he spout from his paps,
and the thyrsus of Bacchus, sending forth wine, will fill it in three days,
and thy horn, Achelous, in two days.
Now run all together and you will fill it in a few hours.


Solution

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

Let $a, b, c$ be the flow rate in numbers of bowls per hour of (respectively) Nile, Bacchus and Achelous.


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 $1$ bowl, 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, $1$ bowl in $1$ day
\(\ds b\) \(=\) \(\ds \frac 1 3\) that is, $1$ bowl in $3$ days
\(\ds c\) \(=\) \(\ds \frac 1 2\) that is, $1$ bowl in $2$ days


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {1 + \dfrac 1 3 + \dfrac 1 2}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac 6 {6 + 2 + 3}\) multiplying top and bottom by $6$
\(\ds \) \(=\) \(\ds \frac 6 {11}\)

So the bowl will be filled in $\dfrac 6 {11}$ of a day.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources

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