Greek Anthology Book XIV: Metrodorus: 131

From ProofWiki
Jump to navigation Jump to search

Arithmetical Epigram of Metrodorus

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.


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 first, second and third spouts.


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:

\(\ds a\) \(=\) \(\ds \frac 1 4\) that is, $1$ cistern in $4$ hours
\(\ds b\) \(=\) \(\ds \frac 1 8\) that is, $1$ cistern in $4 + 4$ hours
\(\ds c\) \(=\) \(\ds \frac 1 {16}\) that is, $1$ cistern in $2 \times {\paren 4 + 4}$ hours


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\dfrac 1 4 + \dfrac 1 8 + \dfrac 1 {16} }\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac {16} {4 + 2 + 1}\) multiplying top and bottom by $16$
\(\ds \) \(=\) \(\ds \frac {16} 7\)
\(\ds \) \(=\) \(\ds 2 \frac 2 7\)

So the cistern will be filled in $2 \frac 2 7$ hours.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Historical Note

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.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Greek_Anthology_Book_XIV:_Metrodorus:_131&oldid=480712"