Integer-sided triangles with equal perimeters and equal areas What is a way to find the side-lengths of non-congruent triangles whose perimeters are equal, and whose areas are equal?
(I have posted an answer below. This shows how answers to my question are related to answers to a question posted elsewhere on math.SE.)
 A: Find some sets of three integers $\{x_i, y_i, z_i\}$, each with sum equal to $s$ and product equal to $p$, as shown, for example, in S. Dolan's answer here. Then the side-lengths of the $i$th triangle are $\{a_i, b_i, c_i\}=\{s-x_i, s-y_i, s-z_i\}$. The triangles have semi-perimeter $s$ and area $\Delta=\sqrt{sp}$. As a bonus, seeing as a triangle's inradius is $r=\Delta/s$, the triangles have equal inradii, too.
If you select only sets where $sp$ is a square, then the area will be an integer and the triangles will be Heronian.
A: Consider the triangles,
$(a,a,2b)=(p^2+4q),(p^2+4q),(8q)$
$(d,d,2f)=(4p+q^2)(4p+q^2),(8p)$
Since perimeter's are equal we get the condition, 
$(p+q)=8$  ----(1)
Using the value $p=(8-q)$ we also have the semi perimeter, 
$s=p^2-8p+64$
We have the area formula for triangle $(a,b,c)$, 
$A^2=s(s-a)(s-b)(s-c)$
So we get,
$(s-a)=4(8-p)$
$(s-2b)=p^2$
$(s-d)=(4p)$
$(s-2f)=(8-p)^2$
Hence common Area is,
$A^2=(p^2-8p+64)(4)^2(p)^2(8-p)^2$  ----(2)
To make the (RHS) of (2) a square the equations (1) & (2) 
are both satisfied at $(p,q)=(3,5)$
Hence we have two triangles:
$(a,a,2b)=(29,29,40)$ &  
$(d,d,2f)=(37,37,24)$
their, perimeter=$98$ & area=$420$
Below is the link to paper written by Andrew Bremner & 
also link to article by Seiji Tomita
https://www.emis.de//journals/AMI/2006/ami2006-bremner.pdf
http://www.maroon.dti.ne.jp/fermat/dioph312e.html
