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

## 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:

$\dfrac p q = t$

 $\ds \dfrac 1 2 \paren {t + \dfrac 1 t}$ $=$ $\ds 1.54$ $\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 $2291$.