Thought Process

What I noticed is that we need triangles with integer sides where A^2 + (B+C)^2 = d^2, and we can generate these pythagorean triples using the method from Problem 9, Please read that problem to understand better.

Essentially we generate triangles where d = m^2 + n^2, and A or (B+C) = m^2 - n^2 or 2mn, now for the final part of the problem, we need to figure out all the possible combinations of the triangles.

Case 1: A = m^2 - n^2, B + C = 2mn

Case 2: A = 2mn, B + C = m^2 - n^2

Now we note the following:

1. If 2A < B + C => B or C or both are greater than A this is impossible, so there are 0 combinations of this type

2. If A >= B + C => A >= B, C, we must ensure that B >= C

   • We can do a few example such as B+C = 5 => (B,C) = (4,1) or (3,2)

   • B+C = 6 => (B,C) = (5,1) or (4,2) or (3,3)

   • There will be floor((B+C)/2) combinations

3. If B+C >= A

   • You can do a few more examples B+C = 15, A = 8 => (A,B,C) = (8,8,7)

   • B+C = 12, A = 9 => (A,B,C) = (9,9,3) or (9,8,4) or (9,7,5) or (9,6,6) 4 possibilities

   • What I notice is that if B+C is divisible by 2, we can include the solution where B = C as a solution

   • Therefore if B + C is divisble by 2 then there are A - floor((B+C)/2) + 1 combinations

   • Otherwise there are A - floor((B+C)/2) combinations

I just wrote a quick functions combinations which does part 3, used my Primitive Pythagorean Triple generator for Part 2, and the problem is easily solved!