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

flag	asked Nov 6 2010 at 0:04 John Iskra 193●1●6

Joseph O'Rourke · Accepted Answer · 2010-11-07 00:42:12Z UTC

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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.

edited Nov 7 2010 at 0:42

Joseph O'Rourke

answered Nov 6 2010 at 15:07

Fedor Petrov
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	Brilliant! $\mbox{}$ – Joseph O'Rourke Nov 6 2010 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 2010 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 2010 at 21:56
	Got it. Thank you! That is a really nice proof. – John Iskra Nov 9 2010 at 2:31

Tony Huynh · Answer 2 · 2010-11-06 00:17:57Z UTC

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If the length divided by the width is rational, then yes. Just partition the rectangle into congruent squares and cut each square along a diagonal.

answered Nov 6 2010 at 0:17

Tony Huynh
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Is this a necessary condition as well? – Jeremy West Nov 6 2010 at 0:22

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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 2010 at 1:04

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Yeah. Thank you to those who have posted so far: I did know, though, that the answer is 'yes' in the cases listed above. In fact, relaxing any of the hypotheses seems to give a positive answer. That is, if the triangles need not be isosceles, if they need not be congruent etc. I was asked this question without the congruency condition by a sophomore student I have. It seems like, potentially, a good research project, but I didn't know whether it was 1. known 2. trivial or 3. too hard. This is outside my area of specialization, so I thought I'd post here to get some sense of things. – John Iskra Nov 6 2010 at 1:37

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If you knew all that, you could have saved people some work by mentioning it in your original statement. – Gerry Myerson Nov 6 2010 at 1:53

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@Gerry: some was equal to about epsilon here, so no worries as far as I am concerned. @unknown: There is nothing wrong with extremely long questions on MO. One reason why I like MO is that I can learn a lot from just reading questions, and in this sense longer questions are better than short ones. Concerning your points {1,2,3}: I am not sure whether the answer is known, but combining Theo's comment with my answer gives a complete characterization. If one can thrash out the details of Theo's comment, I think it would be neither trivial (2) or too hard (3). Probably 2.4. – Tony Huynh Nov 6 2010 at 12:01

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Partitioning a Rectangle

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