Greek Anthology Book XIV: Metrodorus: 144/Historical Note

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Historical Note on Metrodorus' Arithmetical Epigram no. $144$

$A$. 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.


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.


Proof

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.


Let us assume the solution given by W.R. Paton.

Let us assume that $B$'s first statement is true.

$B$. And my base together together with myself weighs the same number of talents.


We have:

\(\ds a + c\) \(=\) \(\ds b + d\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {4 b} 3 + c\) \(=\) \(\ds b + d\) The statue $A$ was a third part heavier than $B$
\(\ds \leadsto \ \ \) \(\ds \dfrac {4 b} 3 + c\) \(=\) \(\ds b + 3 c\) The base of $B$ weighed thrice as much as the base of $A$.
\(\ds \leadsto \ \ \) \(\ds \dfrac {4 b} 3 - b\) \(=\) \(\ds 3 c - c\)
\(\ds \leadsto \ \ \) \(\ds \dfrac b 3\) \(=\) \(\ds 2 c\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 6 c\)


That is, $B$ weighed $6$ times as much as the base of $A$, which deviates from the initial statement of the problem:

$B$: And I alone weigh three times the weight of yours.


Sources

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