15

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

Martin Sleziak
  • 4,608
John Iskra
  • 404

2 Answers2

24

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.

Joseph O'Rourke
  • 149,182
  • 34
  • 342
  • 933
Fedor Petrov
  • 102,548
  • Brilliant! $\mbox{}$ – Joseph O'Rourke Nov 06 '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 08 '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 08 '10 at 21:56
  • Got it. Thank you! That is a really nice proof. – John Iskra Nov 09 '10 at 02:31
9

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.

Tony Huynh
  • 31,500
  • 4
    Is this a necessary condition as well? – Jeremy West Nov 06 '10 at 00:22
  • Not if you allow infinitely many partitions. For example, consider the infinite partition of a Golden Rectangle into infinitely many squares. Gerhard "Ask Me About System Design" Paseman, 2010.11.05 – Gerhard Paseman Nov 06 '10 at 00:58
  • 4
    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 06 '10 at 01:04
  • @Gerhard: OP requires that all triangles in the partition be congruent. So your square partition of the golden rectangle leads to a "triangle partition" with infinitely small triangles. Surely I can do any rectangle if I'm allowed infinitely small triangles, by approximating the ratio of sides by rationals. – Theo Johnson-Freyd Nov 06 '10 at 01:05
  • 1
    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 06 '10 at 01:37
  • 5
    If you knew all that, you could have saved people some work by mentioning it in your original statement. – Gerry Myerson Nov 06 '10 at 01:53
  • 1
    Sorry. I asked the question I wanted answered, not other related questions. Requiring that in general here, might make some questions very very very.... very long. :) – John Iskra Nov 06 '10 at 03:39
  • 2
    @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 06 '10 at 12:01