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?

A simple way of getting a right angled tetrahedron is:
consider ABC (the right angle is at A) the base of the tetrahedron, and pick a point D such that angles DCA and DCB are right angles (in other words, DC is perpendicular to the ABC plane). DA will then be perpendicular to AB, and we get our fourth right angle.

If we make AB=AC=CD=1, then we get two 1-1-√2 triangles (ABC, CDA) and two 1-√2-√3 triangles (DCB, DAB). We can also get all faces to be different if we start with a 3-4-5 triangle and take DC=6.

Posted by e.g. on 2004-05-04 11:09:48