Rich Schroeppel wrote:
If we have a right tetrahedron, where one of the corners matches a cube's corner, there are three leg-faces and one hypote-face. It turns out that the square of the hypoteface area is equal to the sum of the squares of the three legface areas. (I don't know why, or if it works in more dimensions.) Any triangle can be a hypoteface, and the lengths of the X,Y,Z legs are easy to calculate: If the hypoteface edges are A,B,C, then X2+Y2 = A2, etc. and X2 = (A2+B2-C2)/2. The legface areas are XY/2 etc., the area squares are X2Y2/4, and it's not hard to check that the sum of squares of the three legface areas matches Hero^2.
I thought of that too ... but is it true that "any triangle can be a hypoteface"? I think it only works for acute-angled ones. If your corners are (p,0,0), (0,q,0), (0,0,r) then the squared side-lengths are p^2+q^2 etc., so a^2+b^2-c^2 = 2r^2 > 0. What am I missing? -- g