## Generating Primitive isosceles Heronian triangles from PPT

Pythagorean triples:

$m, \; n$   are positive integers, with   $m \; > \; n$.   The integers

$a \; = \; m^2 \; - \; n^2$
$b \; = \; 2 \, m \, n$
$c \; = \; m^2 \; + \; n^2$

form a Pythagorean triple.

Perimeter   $P \; = \; m^2 + n^2 + 2 \, m \, n + m^2 - n^2 \; = \; 2 \, m^2 + 2 \, m \, n$
Area   $A \; = \; m \, n \, (m^2 - n^2)$

(1)   $(c, \; c, \; 2 \, a)$   —->   $m^2 + n^2, \; m^2 + n^2, \; 2 \,(m^2 - n^2)$
(2)   $(c, \; c, \; 2 \, b)$   —->   $m^2 + n^2, \; m^2 + n^2, \; 4 \, m \, n$

It’s clear that these triples satisfy the triangle inequality.

(1)

Let’s compute the area of the triangle

$(m^2 + n^2) \; + \; (m^2 + n^2) \; + \; 2 \,(m^2 - n^2) \; = \; 4 \, m^2$
$(m^2 + n^2) \; + \; (m^2 + n^2) \; - \; 2 \,(m^2 - n^2) \; = \; 4 \, n^2$
$(m^2 + n^2) \; - \; (m^2 + n^2) \; + \; 2 \,(m^2 - n^2) \; = \; 2 \,(m^2 - n^2)$
$-(m^2 + n^2) \; + \; (m^2 + n^2) \; + \; 2 \,(m^2 - n^2) \; = \; 2 \,(m^2 - n^2)$

$4 \, A_1 \; = \; \sqrt { \, (4 \, m^2) \,(4 \, n^2) \: 2 \,(m^2 - n^2) \: 2 \,(m^2 - n^2) \, }$
$4 \, A_1 \; = \; \sqrt { \, (16 (m n)^2 \: 4(m^2 - n^2)^2 \, }$
$4 \, A_1 \; = \; 8 \, m \, n \, (m^2 - n^2)$

$A_1 \; = \; 2 \, m \, n \, (m^2 - n^2)$

we find that it’s an integer.

(2)

Let’s compute the area of the triangle

$(m^2 + n^2) \; + \; (m^2 + n^2) \; + \; 4 \, m \, n \; = \; 2(m + n)^2$
$(m^2 + n^2) \; + \; (m^2 + n^2) \; - \; 4 \, m \, n \; = \; 2(m - n)^2$
$(m^2 + n^2) \; - \; (m^2 + n^2) \; + \; 4 \, m \, n \; = \; 4 \, m \, n$
$-(m^2 + n^2) \; + \; (m^2 + n^2) \; + \; 4 \, m \, n \; = \; 4 \, m \, n$

$4 \, A_2 \; = \; \sqrt { \, 2 \,(m + n)^2 \: 2(m - n)^2 \: (4 \, m \, n) \: (4 \, m \, n) \, }$
$4 \, A_2 \; = \; \sqrt { \, 4 \,(m^2 - n^2)^2 \: (4 \, m \, n) \: (4 \, m \, n) \, }$
$4 \, A_2 \; = \; 8 \, m \, n \, (m^2 - n^2)$

$A_2 \; = \; 2 \, m \, n \, (m^2 - n^2)$

we find that it’s an integer.