Timeline for Partitioning a Rectangle into Congruent Isosceles Triangles
Current License: CC BY-SA 2.5
9 events
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| Nov 6, 2010 at 12:01 | comment | added | Tony Huynh | @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. | |
| Nov 6, 2010 at 3:39 | comment | added | John Iskra | 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. :) | |
| Nov 6, 2010 at 1:53 | comment | added | Gerry Myerson | If you knew all that, you could have saved people some work by mentioning it in your original statement. | |
| Nov 6, 2010 at 1:37 | comment | added | John Iskra | 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. | |
| Nov 6, 2010 at 1:05 | comment | added | Theo Johnson-Freyd | @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. | |
| Nov 6, 2010 at 1:04 | comment | added | Theo Johnson-Freyd | 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. | |
| Nov 6, 2010 at 0:58 | comment | added | Gerhard Paseman | 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 | |
| Nov 6, 2010 at 0:22 | comment | added | Jeremy West | Is this a necessary condition as well? | |
| Nov 6, 2010 at 0:17 | history | answered | Tony Huynh | CC BY-SA 2.5 |