Is it possible to partition any rectangle into congruent isosceles triangles?
How to Partition a Rectangle into Congruent Isosceles Triangles
co.combinatoricsdiscrete geometrymg.metric-geometry
Related Solutions
See http://www.michaelbeeson.com/research/papers/SevenTriangles.pdf and http://www.michaelbeeson.com/research/papers/TriangleTiling1.pdf http://www.michaelbeeson.com/research/papers/TriangleTiling2.pdf http://www.michaelbeeson.com/research/papers/TriangleTiling3.pdf
Beeson conjectures that the cases you list, together with $2n^2$ (a special case of $m^2+n^2$ and $6n^2$ (barycentrically subdivide an equilateral triangle and then split each of the resulting 30-60-90 right triangles into $n^2$ similar triangles) are the only ones possible, without the constraint that the original triangle and the smaller ones be similar. He proves that no subdivision is possible for $n=7$, $11$, $19$, and $23$.
Not an answer. But permit me to draw attention to Robert J. MacG. Dawson's website on congruent sphere tilings,
including this beautiful tiling by triangles:
Best Answer
No. Note that the acute angle of your triangle must divide $\pi/2$ (look at a corner), so there are countably many such triangles (up to similarity), and hence you get only a countable set of possible ratios of sides.