Greek Anthology Book XIV: Metrodorus: 144

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Arithmetical Epigram of Metrodorus

$A$. How heavy is the base I have to stand on together with myself!
$B$. And my base together together with myself weighs the same number of talents.
$A$. But I alone weigh twice as much as your base.
$B$: And I alone weigh three times the weight of yours.


Solution

Presumably $A$ and $B$ are talking statues.


Let $a$ talents be the weight of $A$.

Let $b$ talents be the weight of $B$.

Let $c$ talents be the weight of $A$'s base.

Let $d$ talents be the weight of $B$'s base.


We have:

\(\text {(1)}: \quad\) \(\ds a + c\) \(=\) \(\ds b + d\)
\(\text {(2)}: \quad\) \(\ds a\) \(=\) \(\ds 2 d\)
\(\text {(3)}: \quad\) \(\ds b\) \(=\) \(\ds 3 c\)

We have $3$ simultaneous equations in $4$ unknowns.

Hence we cannot calculate the actual weights, but the proportions only.


So:

\(\ds 2 d + c\) \(=\) \(\ds 3 c + d\) substituting from $(2)$ and $(3)$ into $(1)$
\(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds 2 c\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 4 c\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 4 c\)


So, taking the base of $A$ as a unit, we have:

$A$ weighs $4$ times its base
$B$ weighs $3$ times the base of $A$
the base of $B$ weighs twice the base of $B$.

$\blacksquare$


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:

From these data not the actual weights but the proportions alone can be determined.
The statue $A$ was a third part heavier than $B$, and $B$ only weighed $\dfrac 3 4$ of the statue $A$.
The base of $B$ weighed thrice as much as the base of $A$.


This appears to be incorrect.


Sources

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