Greek Anthology Book XIV: Metrodorus: 131/Historical Note

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Historical Note on Metrodorus' Arithmetical Epigram no. $131$

Open me and I, a spout with abundant flow, will fill the present cistern in four hours;
the one on my right requires four more hours to fill it,
and the third twice as much.
But if you bid them both join me in pouring forth a stream of water, we will fill it in a small part of the day.


In W.R. Paton's $1918$ translation of The Greek Anthology Book XIV, he gives the answer as $2 \frac 2 {11}$ hours.

The discrepancy between this and the $2 \frac 2 7$ reported in the calculation is explained by the imprecision of the wording of the epigram.


If we take:

the third twice as much

to mean:

the third spout takes twice as much time to fill a cistern as the second spout

then we deduce that the third spout takes $16$ hours (that is, twice four plus four hours) to fill the cistern.

Hence we arrive at the solution $2 \frac 2 7$ hours.


However, if we interpret:

the third twice as much

to mean:

the third spout takes twice as much longer (that is, eight hours) than the first spout than the second one does to fill a cistern

that would mean the third spout takes $4$ hours plus twice $4$ hours, that is $12$ hours, to fill a cistern.


If this assumption is made, then $2 \frac 2 {11}$ is correct.


Proof

Thus we have:

\(\ds a\) \(=\) \(\ds \frac 1 4\) that is, $1$ cistern in $4$ hours
\(\ds b\) \(=\) \(\ds \frac 1 8\) that is, $1$ cistern in $4 + 4$ hours
\(\ds c\) \(=\) \(\ds \frac 1 {12}\) that is, $1$ cistern in $4 + 2 \times 4$ hours


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\dfrac 1 4 + \dfrac 1 8 + \dfrac 1 {12} }\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac {24} {6 + 3 + 2}\) multiplying top and bottom by $24$
\(\ds \) \(=\) \(\ds \frac {24} {11}\)
\(\ds \) \(=\) \(\ds 2 \frac 2 {11}\)

So the cistern will be filled in $2 \frac 2 {11}$ hours.


Sources

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