Not Sangaku

Here are the dimensions of all $5$ triangles with their incircle diameters and areas

$\left(a,b,c,d,A \right)=$

$\left(225,351,360,162,37908 \right)$

$\left(360,480,600,240,86400 \right) \;\;$ (Recognize this?)

$\left(416,480,448,256,86016 \right)$

$\left(600,806,950,408,240312 \right)$

$\left(799,1222,1175,564,450636 \right)$

where $a$ is the smallest side, and then proceeding clockwise, with $d$ being the diameter and $A$ being the area.

If the areas and sides are reals, there'd be an infinity of solutions. However, restricting all areas and sides to integers points to an unique solution. Hence, it's not a Sangaku problem, as it involves number theory.

If any integer diameter $d$ satisfies for integer $k,m,n$ the following relation

$d=2k\left( mn-{ k }^{ 2 } \right)$

then there is an Heronian triangle associated with $k,m,n$ for which $d$ is the diameter of its incircle. Given any integer diameter $d$ , there can be multiple but not infinitely many such Heronian triangles. For integer $k,m,n$ , the sides of the Heronian triangle would be

$m({n}^{2}+{k}^{2})$
$n({m}^{2}+{k}^{2})$
$(m+n)(mn-{k}^{2})$

That Pythagorean right triangle in the center, as an hunch, is a great starting point for cracking this math puzzle. Given that the ratio of the sides "look like" $3:4:5$ , and given the incircle diameter of $240$ , it's an easy step to work out $3$ of the internal edges $360,480,600$ . Then work from there, by finding which combinations of $k,m,n$ would yield a given diameter and a given side, for each of the other $3$ triangles. Start by finding the bottom $2$ and see if their bases are colinear. In fact, at that point, the problem is already solved.

It is curious how close the following "intuitive" but incorrect solution is to the correct one.

$\sqrt { { 162 }^{ 2 }+{ 240 }^{ 2 }+{ 256 }^{ 2 }+{ 408 }^{ 2 } } =562$