Problem of the Week 1242

Pythagoras Leaves Flatland

A "right tetrahedron" is a tetrahedron having three right angles meeting at a vertex V. Think of it as sitting in the first octant of 3-space with V at the origin and the three faces containing V lying on the x-y plane, the x-z plane, and the y-z 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 non-right 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:

  • The universal recycling symbol: Is it a standard Möbius strip, or a strip with three half-turns? Are the three pieces that make up the symbol congruent, or not?
  • The Szilassi polyhedron (described by Wikipedia and MathWorld). This is a polyhedron with 7 polygonal faces. Each face is a hexagon. Each face touches all 6 of the other faces. Topologically, the polyhedron is a torus. So this object shows how the complete graph can embed on a torus in a geometrically simple way: via polygonal faces. Shapeways offers a colored model for sale. I just got mine; very nice. Apparently, computers played a role in the discovery of this remarkable polyhedron.

    [View the solution]

16 June 2017