On Pythagorean Pyramid we tried to build a tetrahedron out of four equal right angled triangles, but the attempt fell flat (pun intended!).

Is it possible to have a tetrahedron built out of right angled triangles, dropping the condition that all triangles be the same? Can you manage to have three equal faces? Or maybe two pairs of equal faces?

Scribbling on some paper I discovered that if I can make a tetrahedron with two pair of equal faces:

two 3 x 4 x 5 triangles, and two 3 x √7 x 4 triangles.

If it makes it easier to see how to put those together, let the four vertices of the tetrahedron be A, B, C, and D.

AC = 5
AB = 3
CD = 3
BD = √7
AD = 4
BC = 4

Then,
ABC = 3 x 4 x 5
ACD = 3 x 4 x 5
ABC = 3 x √7 x 4
BCD = 3 x √7 x 4

note x, in this context, is not meant to indicate multiplication... just to separate the length of the triangle sides.

Posted by Thalamus on 2004-05-03 14:10:53