On the Total Edge-Length of a Tetrahedron

Abstract

Proof. Let r be the circle with minimum area that encloses the triangle ABC in the plane, and let 0 be its center. (r might not be the circumscribed circle of the triangle ABC.) Then 0 lies either inside the triangle ABC or on a side of ABC. First, consider the case that 0 lies on a side, say, on the side AB. Then AB = 2R and AC + CB > AB. Hence L(ABC) > 4R. Next, suppose that 0 is interior to the triangle ABC. We may put A = (a, 0), B = (-a, 0), C = (xl, yl), y, > 0. Let AC' be a diameter of r, and let C' = (xo, yo). Then, since 0 is interior to AABC, it follows that 0 L(AB C'). Since 0 lies on AC', it follows that L(AB C') > 4R. Hence L(ABC) > 4R.