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

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

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

    \[ T = A \cup B \]

for two disjoint, congruent right triangles A, B. Since the vertices of T are at lattice points, we know the altitude from the vertex to the base must pass through h lattice points (where h is the height of T). Therefore, denoting the lattice points on this altitude by V_B = h+1, we have

    \[ B_T = B_A + B_B -V_B + 2, \qquad I_T = I_A + I_B +V_B - 2. \]

Since T is a polygon with lattice point vertices we know by the previous exercise, that a(T) = I_T + \frac{1}{2} B_T -1. Further, by Exercise #2 of this section, we know a(T) = \frac{1}{2} bh. So,

    \begin{align*}  &&I_T + \frac{1}{2}B_T - 1 &= (I_A + I_B + V_B - 2) + \frac{1}{2} (B_A + B_B - V_B+2) \\ \implies && I_T + \frac{1}{2}B_T - 1 &= 2I_A + B_A - 2 + \frac{1}{2} V_B &(B_A = B_B, I_A = I_B)\\ \implies && I_T + \frac{1}{2} B_T - 1 &= 2I_A + B_A - 2 + \frac{1}{2}(h+1) &(V_B = h+1) \\ \implies && I_T + \frac{1}{2} B_t - 1 &= 2(a(A)) + \frac{1}{2}(h+1)  \end{align*}

But, \frac{1}{2} a(T) = a(A) = a(B), so,

    \[ I_T + \frac{1}{2} B_T - 1 &= a(T) + \frac{1}{2}(h+1) \qquad \implies \qquad a(T) = a(T) + \frac{1}{2}(h+1). \]

But, h > 0, so this is a contradiction. Therefore, T cannot have its vertices at lattice points and be equilateral. \qquad \blacksquare


  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.

  2. 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.

    • 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.

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