# Heron of Alexandria/Problems/Pythagorean Triangle

## Problem

In a Pythagorean triangle, the sum of the area and the perimeter is $280$.
Find the sides and the area.

## Solution

The Pythagorean triangle has sides $20$, $21$ and $29$.

Its area is $210$.

## Proof

There is no simple way to solve this except by trial and error.

First we note that:

the area of the $20-21-29$ Pythagorean triangle is $\dfrac {20 \times 21} 2 = 210$

and we have that:

$210 + 20 + 21 + 29 = 280$

The sequence of solutions of the Pythagorean equation can be tabulated as follows:

$\begin{array} {r r | r r | r r r | c} m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 5 & \text{Primitive} \\ \hline 3 & 1 & 9 & 1 & 6 & 8 & 10 \\ 3 & 2 & 9 & 4 & 12 & 5 & 13 & \text{Primitive} \\ \hline 4 & 1 & 16 & 1 & 8 & 15 & 17 & \text{Primitive} \\ 4 & 2 & 16 & 4 & 16 & 12 & 20 \\ 4 & 3 & 16 & 9 & 24 & 7 & 25 & \text{Primitive} \\ \hline 5 & 1 & 25 & 1 & 10 & 24 & 26 \\ 5 & 2 & 25 & 4 & 20 & 21 & 29 & \text{Primitive} \\ 5 & 3 & 25 & 9 & 30 & 16 & 34 \\ 5 & 4 & 25 & 16 & 40 & 9 & 41 & \text{Primitive} \\ \hline 6 & 1 & 36 & 1 & 12 & 35 & 37 & \text{Primitive} \\ 6 & 2 & 36 & 4 & 24 & 32 & 40 \\ 6 & 3 & 36 & 9 & 36 & 27 & 45 \\ 6 & 4 & 36 & 16 & 48 & 20 & 52 \\ 6 & 5 & 36 & 25 & 60 & 11 & 61 & \text{Primitive} \\ \hline 7 & 1 & 49 & 1 & 14 & 48 & 50 \\ 7 & 2 & 49 & 4 & 28 & 45 & 53 & \text{Primitive} \\ 7 & 3 & 49 & 9 & 42 & 40 & 58 \\ 7 & 4 & 49 & 16 & 56 & 33 & 65 & \text{Primitive} \\ 7 & 5 & 49 & 25 & 70 & 24 & 74 \\ 7 & 6 & 49 & 36 & 84 & 13 & 85 & \text{Primitive} \\ \hline \end{array}$

from which it can soon be seen that the $20-21-29$ triangle is the one required.

$\blacksquare$