Programming Challenge: a classic trigonometry problem

[Pekis]

Interesting enough, on the crossed ladders Wikipedia page, in order to fold a sheet of paper in thirds if there is already a mark on the half height:

Starting from 1/c = 1/sqrt(a^2 - d^2) + 1/sqrt(b^2 - d^2)
(where d is the distance between the ladders)

Knowing (with Pythagore):
a^2 = d^2 + 1^2 (where 1 is the relative sheet height) => a^2 - d^2 = 1
b^2 = d^2 + (1/2)^2 (where 1/2 is the relative sheet half height) => b^2 -d^2 = 1/4

So
1/c = 1/sqrt(1) + 1/sqrtr(1/4) = 3 => c = 1/3
i.e the intersection point will be at the the third of the relative sheet height, for any relative sheet width !