Problem of the Week 1242Pythagoras Leaves FlatlandA "right tetrahedron" is a tetrahedron having three right angles meeting at a vertex V. Think of it as sitting in the first octant of 3space with V at the origin and the three faces containing V lying on the xy plane, the xz plane, and the yz plane. These three faces are called the "legs" and the fourth face is the "hypotenuse." De Gua's theorem says that the sum of the squares of the areas of the three legs equals the square of the area of the hypotenuse. Show that there is a tetrahedron that satisfies the conclusion of de Gua's theorem, but is not a right tetrahedron. That is, there are three faces with areas that, when squared and summed, equal the squared area of the fourth face, but the tetrahedron is not a right tetrahedron. Source: I saw this result in the very nice A Mathematical Space Odyssey by Claudi Alsina and Roger B. Nelsen. (MAA; Dolciani Expositions #50). Of course, the Pythagoras formula in the plane fails for any nonright triangle. The book has many tidbits that were new to me, such as de Gua's theorem and Problem 1242, but also interesting observations about, for example:
