Lattice Equilateral Triangle V2

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) This proof could be taught to middle school students. Call a line segment that is parallel to one of the axes an orthosegment. Consider three basic cases (See pictures below). 1 Case I: The triangle has two orthosegments, and the area is ab , where a and b are the 2 legs of the right triangle, hence integers. 1 Case II: The triangle has exactly one orthosegment, and the area is bh , where b (the 2 base) is the length of the orthosegment and h is the altitude on the orthosegment, both integers. Case III: The triangle has no orthosegments. Let R be the bounding box, the smallest bounding rectangle of the lattice triangle T that has sides parallel to the axes. Clearly the area of R is an integer, so it suffices to show that the area of the complement of the triangle with respect to the bounding box is of the form n/2. Case IIIa: All 3 vertices of the triangle are on distinct sides of the bounding box. The complement consists of the union of three right triangles with legs parallel to the axes, so the area of the complement is the sum of areas of the triangles, which are of the form n/2. Case IIIb: One vertex of the triangle (call it P) is in the interior of the bounding box and the other two are on diagonally opposite corners. Label the bounding box ABCD, so that the triangle is ACP with P and D on the same side of the line AC. Drop perpendiculars from P to AD and CD. This divides the complement into a rectangle R' with sides parallel

to the axes (and integral area) and three right triangles with legs parallel to the axes and area of the form n/2.

Case 1: Two orthosegments

Case 2a: One orthosegment, interior altitude.

Case 2b: One orthosegment, exterior altitude.

II

I

III

II

I IV

III

Case 3a: No orthosegments, all vertices on bounding box.

Case 3b: No orthosegments, one vertex not on bounding box.

An alternative proof in the same vein involves introducing at most two trapezoids and at most two right triangles, and expressing the area of the given triangle using set union and difference. A typical case looks line this: C A

D

B

E

Using the convention that ( P1... Pn ) is the area of the polygon P1... Pn we have (ABC) = (ADEC) – (ADB) – (BCE). Since each polygon on the right side is of the form n/2, so is (ABC). We leave it to the reader to complete the proof for other possible triangle configurations.

(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).