What's the smallest number of similar but non-congruent triangles you can use to subdivide a square ?|
i just went at it starting with a = 1/2 and going from there,
and things seemed so good but when i worked out the length of b i got something fairly off.
So then i came up with some factor x s.t. a * x + b * x = 1, which gave me say a' and then worked thru to get the length of b', but my total length for the side of the square ended up being 1/((1+(125/253)^2)^3) + (125/253) = ~1.01338216 so either i made a math mistake (highly possible) or a different design is needed.
various solutions are here: http://www.mathforum.org/epigone/geometry-puzzles/lordramblum/2s9m48lnllde@legacy.
20060509: i just re-read the preceding paragraph after like two years and realized a huge flaw:
we don't want X s.t. aX + bX = 1, we want X s.t. aX + b(1-X) = 1. not going to work it thru right now tho.