# Triangles with Congruent Corresponding Sides that Cannot fold into a Tetrahedron

I've been trying to find, without much success, 4 triangles whose corresponding sides are congruent that cannot be folded into a tetrahedron.

Anyone has any clue how to approach this problem?

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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 – Gerhard Paseman Feb 28 '12 at 22:36
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 – Gerhard Paseman Feb 28 '12 at 22:43
@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. – drum Feb 28 '12 at 22:54
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 – Gerhard Paseman Feb 28 '12 at 23:14

What do you mean by "corresponding sides"? If what you mean that you have a gluing diagram which is consistent, just take your triangles $ABC, ABD, ACD, BCD$ in such a way that the angles at $A$ in all three triangles sharing that vertex is $5\pi/6,$ and otherwise the three triangles with vertex at $A$ are isosceles (so the other two angles are $\pi/10$) the triangle $BCD$ is equilateral. Notice that these triangles do not glue into a tetrahedron, since the total angle at $A$ is greater than $2\pi$ (since $15/6 > 2$).