Prove that an equilateral triangle cannot have all of vertices on lattice points, i.e., points such that both are integers.

*Proof.*Suppose there exists such an equilateral triangle . Then,

for two disjoint, congruent right triangles . Since the vertices of are at lattice points, we know the altitude from the vertex to the base must pass through lattice points (where is the height of ). Therefore, denoting the lattice points on this altitude by , we have

Since is a polygon with lattice point vertices we know by the previous exercise, that . Further, by Exercise #2 of this section, we know . So,

But, , so,

But, , so this is a contradiction. Therefore, cannot have its vertices at lattice points and be equilateral

Union of borders of A and B crosses the altitude twice. Shouldn’t Bt = Ba + Bb – 2(Vb – 2) ?

I think it would be: Bt = Ba + Bb – 2(Vb – 1)

Will this be a valid alternate proof?

Because the triangle is equilateral with all of its coordinates being lattice points, its height is going to be sqrt(3)*a/4 (where a is the length of one of its sides). This means that the area, 0.5bh is irrational; but Pick’s formula always gives you a rational result. Hence, contradiction.

If a is sqrt(3) the area you get is rational

Area would actually be (a^2)*sqrt(3). But (a^2) is an integer since vertices are lattice points. Now we need the result that sqrt3 is irrational and a rational times an irrational is irrational.

Formula for area of equilateral triangle with side length ‘a’ is: (sqrt(3)*a*a)/4;

On a side note:

Since, vertices are lattice points, say (p, q) and (r, s), then side length can be calculated by distance formula:

side_length = sqrt( (p-r)^2 + (q-s)^2) );

Then we have side_length of form sqrt(z), where z is a positive integer, and hence, side_length squared is ‘z’, an integer. Consequently, area calculated using above formula would be: (sqrt(3)*side_length*side_length)/4 = (sqrt(3)*z)/4, which is clearly irrational.