Lattice Equilateral Triangle

Problem: In two dimensions, a lattice polygon is a polygon in a Cartesian coordinate plane such that the two coordinates of each vertex are integers. In three dimensions, a lattice polyhedron is a polyhedron such that the three coordinates of each vertex are integers. (a) Prove that a lattice triangle cannot be an equilateral triangle. (b) Is it possible for a lattice tetrahedron to be a regular tetrahedron? Solution: The solution to (a) depends on the fact that the area of a lattice triangle must be a number of the form n/2, where n is an integer. (See below for three different proofs of this fact.) 3 2 But the area of an equilateral triangle of side s is s , and if the triangle is a lattice 4 triangle, s 2 is an integer, so the area is not of the form n/2. The answer to (b) is "yes", and an example is given by the tetrahedron with vertices A = (0, 0, 0), B = (0, 1, 1), C = (1, 0, 1), and D = (1, 1, 0). Direct computation shows AB = AC = AD = BC = BD = CD = √2. Proofs of fact about lattice triangles: (1) The most elementary proof comes from creating R, the smallest bounding rectangle of the lattice triangle T that has sides parallel to the axes. The area of T is the area of R (an integer), less the area of (at most) 3 right lattice triangles. The area of a lattice right triangle is clearly an integer or a half-integer. (2) The surveyor's formula for the area of an n-gon in the case n = 3 asserts the area of 1 3 triangle is ∑ ( xi yi +1 − yi xi +1 ) where ( x4 , y4 ) ≡ ( x1 , y1 ) . If each coordinate is an integer, 2 i =1 the area must be one-half an integer. For the surveyor's formula, see http://www.maa.org/pubs/Calc_articles/ma063.pdf. (3) The fact that the area of a lattice polygon is one-half an integer is an immediate consequence of Pick's Theorem (http://en.wikipedia.org/wiki/Pick's_theorem).