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# Prove that no equilateral triangle can have all of vertices on lattice points

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

1. mike says:

Should B(T) = B(A) + B(B) – 2V(B) + 2 since our altitude lattice points are on the boundaries of both A and B and thus must be subtracted out when calculating out B(T) ?? of course we add 2 to get the vertex point and a base lattice point. Carrying out the calculations further we arrive at another contradiction that a(A) = I(B) + 1/2 B(B).
Thank you very much for posting these answers; it makes for great feedback for the self learner.
mike

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

• HelZ says:

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

• Long Vo says:

However when Bt= Ba + Bb -2(Vb-1) then the statement becomes true. Therefore, there must be a reason that why the author writes Bt= Ba +Bb-Vb+2.

3. Supreeth says:

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.

• nu creation says:

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

• Anonymous says:

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.

• Opieqhd says:

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.