15
$\begingroup$

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

Improve this question
$\endgroup$

2 Answers 2

Reset to default
24
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Brilliant! $\mbox{}$ $\endgroup$
    – Joseph O'Rourke
    Commented Nov 6, 2010 at 21:03
  • $\begingroup$ 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! $\endgroup$
    – John Iskra
    Commented Nov 8, 2010 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – Fedor Petrov
    Commented Nov 8, 2010 at 21:56
  • $\begingroup$ Got it. Thank you! That is a really nice proof. $\endgroup$
    – John Iskra
    Commented Nov 9, 2010 at 2:31
9
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • 4
    $\begingroup$ Is this a necessary condition as well? $\endgroup$
    – Jeremy West
    Commented Nov 6, 2010 at 0:22
  • 4
    $\begingroup$ 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. $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 6, 2010 at 1:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – John Iskra
    Commented Nov 6, 2010 at 1:37
  • 5
    $\begingroup$ If you knew all that, you could have saved people some work by mentioning it in your original statement. $\endgroup$
    – Gerry Myerson
    Commented Nov 6, 2010 at 1:53
  • 2
    $\begingroup$ @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. $\endgroup$
    – Tony Huynh
    Commented Nov 6, 2010 at 12:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .