# Aristotle’s Tetrahedral Blunder

Aristotle falsely believed that congruent regular tetrahedra tile space:

Among surfaces it is agreed that there are three figures which fill the place that contain them–the triangle, the square and the hexagon: among solids only two, the pyramid and the cube. -De Caelo

(Here the pyramid means the regular tetrahedron.) Even more remarkable than Aristotle’s false belief is that “it took 1800 years for Aristotle’s error to be resolved,” according to the fascinating history documented by Lagarias and Zong. In partial defense of Aristotle, they write that “of course, at the time of Aristotle, methods of geometric measurement and computation were more limited and computers were not available!”

Here we prove that tetrahedra do not tile space, using only math that Aristotle might have understood. This means it took 1800 years more than it should have to set the record straight. Aristotle’s referees should have caught his error before publication!

We limit our proof to the case when the tetrahedra are arranged face-to-face, but a small adaptation of our argument, which we leave to the reader, shows that no tiling is possible even without the face-to-face constraint.

Let us assume for a contradiction that congruent tetrahedra tile space. By placing tetrahedra face-to-face, they are also edge-to-edge and vertex-to-vertex. All the tetrahedra around a fixed vertex V form a Platonic solid with n equilateral triangular faces, where n is the number of tetrahedra meeting at the center V. There are only three Platonic solids with triangular faces: the tetrahedron itself, the octahedron, and the icosahedron. We can immediately rule out the tetrahedron and the octahedron because the angle from the center V to two adjacent vertices of the solid is not an acute angle. (The angle would have to be 60 degrees if part of a tetrahedron tiling). Thus, the tiling must consist of n=20 tetrahedra around the center V, forming an icosahedron. a diameter of the icosahedron and one face of a tetrahedron around V

To complete the proof we rule out the icosahedron. Consider a plane containing the diameter (the black vertical line) of the icosahedron, and one triangular face as shown. If we let each edge be 1 unit in length, then the diameter is 2, and the vertex V’ has height 1/2 above the center. By symmetry, the icosahedron has five vertices at height 1/2 and five vertices at height -1/2 below center (as shown below).  The distance between the planes at heights +1/2 and -1/2 is 1.  There is a unique point (the orthogonal projection, shown as a blue dot in the figure to the left) on the plane of height -1/2 at distance 1 from V’. This leads immediately to a contradiction: following the two edges on the tetrahedron from V’ to V1 and V2, we find that there are at least two such points! In the icosahedron, the vertex V’ at height 1/2 cannot have distance 1 from two distinct points V1, V2 at height -1/2.

Acknowledgement: The tetrahelix image is from a post by Brian Hayes