## No Equilateral Triangles on a Chess Board

Are the centers of any three squares on a chess board exactly
equidistant? To show that the answer is no, assume an equilateral
triangle whose vertices all have integer coordinates. Let the
coordinates of one vertex be the origin (0,0) and the coordinates
of the other two vertices be integer pairs (a,b) and (c,d). Then
we have the equilateral condition
a^2 + b^2 = c^2 + d^2 = (a-c)^2 + (b-d)^2 = Q
where Q is the square of the edge length. Expanding the right hand
expression and rearranging terms gives
(a^2 + b^2) + (c^2 + d^2) - 2(ac+bd) = Q
so we have 2(ac+bd) = Q. Now we can form Q^2 in two different ways
Q^2 = 4(ac+db)^2 = (a^2+b^2)(c^2+d^2)
Expanding both sides gives
4(ac)^2 + 8(abcd) + 4(bd)^2 = (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2
Subtracting (ac)^2 + 2(abcd) + (bd)^2 from both sides gives
3(ac+db)^2 = (ad-bc)^2
which is impossible in integers, because a square cannot equal 3 times
a square (due to unique factorization of integers).
Another form of this argument is to observe that the distance between
any two lattice points is the square root of an integer (by Pythagoras's
Theorem), so the edge length of a lattice triangle must be of the form
E = sqrt(k) for some integer k. Also, we know the "altitude" of an
equilateral triangle with edge E is
sqrt[E^2 - (E/2)^2] = E sqrt(3)/2
so the area of the triangle is E^2 sqrt(3)/2 = k sqrt(3)/2, which
is irrational, basically for the same reason as in the previous
proof, i.e., if we square both sides and clear the fractions we have
an expression asserting that a square equals 3 times a square, which
is impossible due to unique factorization of integers. Now we can
assert that the equilateral triangle cannot have all its vertices
as lattice points, because any triangle whose vertices have integer
coordinates must have rational area, as is clear from the area
formula
(x2-x3)(y1-y2) - (x1-x2)(y2-y3)
A = -------------------------------
2
For more on this formula, see Net Area and Green's Theorem.

Return to MathPages Main Menu