Let’s say we have 3 pythagorean triangles (integer sides and hypotenuse) all with the same hypotenuse. Is it possible for the areas of two of those triangles to sum to the area of the third triangle? If so, does anyone have an example? I’ve looked all over and done calculations myself, but I haven’t come up with much yet. I know they all lie on the same circle, I’m thinking that may be part of some geometric solution.
My calculations yielded the equation: $wz(z^2 - w^2)[(m^2+n^2)^2][(p^2+q^2)^2] + mn(m^2-n^2)[(w^2 + z^2)^2][(p^2+q^2)^2] = pq(p^2 - q^2)[(w^2 + z^2)^2][(m^2 + n^2)^2]\quad$ where $\quad z > w, m > n,\,$ and $\, p > q\,$ & none of the variables or terms equal 0 or each other (some variables might equal each other, but certain sets of equal variables have shown to be a problem). Any solutions/ parametrizations to this would give such a set of triangles. This is based on the equation $u*(1-u^2)/(1+u^2)^2 + v*(1-v^2)/(1+v^2)^2 = t*(1-t^2)/(1+t^2)^2$ where are variables are less than one.
Thank you!
Ir is possible but probably requires the brute force of a program if such are to be found at all. Here is a list of $\,13\,$ Pythagorean triples with identical radii and their areas. To the right are the sums of each row area and the elements below them. We can see that sums get close here but none are an exact match in this case.
To find other possibles, we must begin with $\,C$-values that have multiple triples as indicated by the number or prime factors. There are $\,2^{n-1}$ primitive triples where $\,n\,$ is the number of prime factors of C. There are then imprimitives which are multiples of triples derived from factors of $C$. Start with these $\,C$-values or combinations of them and good luck in programming.