> 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.
it's already been pointed out that only non-obtuse triangles can
occur as a "hypoteface". however, we can reduce to that case as
follows.
let K(a, b, c) = sqrt(s(s - a)(s - b)(s - c)) , which is heron's
formula. then 16 K^2 = c^2 (2 a^2 + 2 b^2 - c^2) - (a^2 - b^2)^2 ,
which shows that K(a, b, c) = K(a, b, c') , where
(c')^2 = 2 a^2 + 2 b^2 - c^2 . this corresponds to a simple geometric
transformation: attach two (a, b, c) triangles along their c edge
to make a parallelogram. then divide the parallelogram into two
congruent (a, b, c') triangles using the other diagonal.
the relationship c^2 + (c')^2 = 2 (a^2 + b^2) is the usual
parallelogram law.
it is easy to show that any obtuse triangle can be transformed into
a non-obtuse triangle by a sequence of transformations of this ilk.
(always attach along the edge opposite the obtuse angle; repeat
as needed ... have patience ... you will get there!)
mike
