Babylonian Mathematics/Examples/Pythagorean Triangle whose Side Ratio is 1.54

From ProofWiki
Jump to navigation Jump to search

Example of Babylonian Mathematics

Consider a Pythagorean triangle whose hypotenuse and one leg are in the ratio $1.54 : 1$.

What are the lengths of that hypotenuse and that leg?


Solution

The lengths in question are $829$ and $540$.


Proof

Let $a$, $b$ and $c$ be positive integers such that $a^2 + b^2 = c^2$ and such that $1.54 \times a = c$.

Without loss of generality, suppose $a$ is even.

From Solutions of Pythagorean Equation, there exist positive integers $p$ and $q$ such that:

\(\ds a\) \(=\) \(\ds 2 p q\)
\(\ds b\) \(=\) \(\ds p^2 - q^2\)
\(\ds c\) \(=\) \(\ds p^2 + q^2\)

Hence it follows that:

$\dfrac c a = \dfrac 1 2 \paren {\dfrac p q + \dfrac q p}$

The Babylonians would then consult the various standard tables of reciprocals which they used for multiplication.

Without these tables, we set:

$\dfrac p q = t$

and solve the quadratic equation:

\(\ds \dfrac 1 2 \paren {t + \dfrac 1 t}\) \(=\) \(\ds 1.54\)
\(\ds \leadsto \ \ \) \(\ds t^2 + 1\) \(=\) \(\ds 3.08 t\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \pm 2.711 \text { or } 0.369\)

We can discard $0.369$ because we are after $p > q$.

Hence:

\(\ds \dfrac p q\) \(=\) \(\ds \dfrac {27} {10}\) as a rough approximation
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds 27\)
\(\ds q\) \(=\) \(\ds 10\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 2 p q\)
\(\ds \) \(=\) \(\ds 540\)
\(\ds c\) \(=\) \(\ds p^2 + q^2\)
\(\ds \) \(=\) \(\ds 829\)
\(\ds \leadsto \ \ \) \(\ds \dfrac c a\) \(=\) \(\ds 1.535\)

which is what is found in the original Babylonian clay tablet.

$\blacksquare$


Historical Note

This result appears in Plimpton $\mathit { 322 }$ as figures $3452$ and $2291$.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Babylonian_Mathematics/Examples/Pythagorean_Triangle_whose_Side_Ratio_is_1.54&oldid=478818"