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Sep 23, 2022 at 20:08 comment added Oscar Lanzi I think the question is obtuse, really.
Apr 7, 2019 at 6:14 history edited Martin Sleziak
removed the deprecated (geometry) tag - see the tag info: https://mathoverflow.net/tags/geometry/info; if there are some other geometry-related tags which are suitable, please use some of them instead
May 28, 2012 at 6:33 answer added Carl Sebeny timeline score: 0
Feb 28, 2012 at 23:14 comment added Gerhard Paseman If the 6 sides were lengthened to 1 from 0.99, you could tile the equilateral triangle with the three isosceles triangles, and still not get a tetrahedron. If I did not mess up, the areas of the three isosceles triangles do not add up to the area of the equilateral triangle, which they must exceed if the four triangles are to form a tetrahedron. This is like Igor's example, which I did not see until after posting. Gerhard "Ask Me About System Design" Paseman, 2012.02.28
Feb 28, 2012 at 22:57 vote accept drum
Feb 28, 2012 at 22:54 comment added drum @Gerhard Paseman, would you explain to me why does your solution work? I can't understand why wouldn't these triangles fold into a tetrahedron.
Feb 28, 2012 at 22:43 comment added Gerhard Paseman I am going to guess that an example of what you want is an equilateral triangle of side sqrt(3) with 3 isosceles triangle each having a base of sqrt(3) and the other sides of length 0.99. However, this example is so apparent that I hope you mean something else. Gerhard "Ask Me About System Design" Paseman, 2012.02.28
Feb 28, 2012 at 22:36 comment added Gerhard Paseman Your question is unclear to me. In spite of that, my suggestion is to unfold every tetrahedron to see which 4-tuples of triangles are allowed, and then avoid picking from that set. Gerhard "Ask Me About System Design" Paseman, 2012.02.28
Feb 28, 2012 at 22:34 answer added Igor Rivin timeline score: 1
Feb 28, 2012 at 22:22 history asked drum CC BY-SA 3.0