Partitioning a Rectangle

Is it possible to partition any rectangle into congruent isosceles triangles?

-

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.

-
Brilliant! $\mbox{}$ –  Joseph O'Rourke Nov 6 '10 at 21:03
You seem to be saying that there is a one-to-one correspondence between classes of similar triangles and classes of rectangles of sides of a given ratio. Perhaps I'm misunderstanding your argument, but if I'm not, I don't see why that has to be true. Can you elaborate a bit? Many thanks! –  John Iskra Nov 8 '10 at 20:43
Not one-to-one, but if we fix triangle (with sides a and b), then both sides of rectangle are linear combinations of a,b with integer coefficients. So, there are at most countably many of them for fixed a,b. –  Fedor Petrov Nov 8 '10 at 21:56
Got it. Thank you! That is a really nice proof. –  John Iskra Nov 9 '10 at 2:31
The argument for necessity would run something like this: 1. By looking at a corner, convince yourself that the triangles must be right (i.e. a half-square). (Some angle on the triangle must divide $90^\circ$; then analyze by cases which angle it can be.) 2. Using irrationally of $\sqrt2$, argue that for any rectangle tiled with half-squares, all short (say) sides must be parallel/perpendicular to each other (as opposed to at $45^\circ$). 3. Whether all short sides are parallel to the sides of the rectangle or perpendicular, you still win. –  Theo Johnson-Freyd Nov 6 '10 at 1:04