Coffin Question: Triangle in grids

Official solution: Suppose one of the vertices is at the origin. Now suppose the coordinates of the other two vertices are at (a,b) and (c,d).

If all the numbers are divisible by two, we can reduce the triangle by half. This we can assume that one of the numbers is not divisible by two. Suppose a is odd.

If b is odd, then $a^2+b^2$ is of the form $4k+ 2$ . As $a^2+b^2 = c^2+d^2$ we must conclude that both c and d are odd. The problem is that the square of the length of the third side is $(a-c)^2+(b-d)^2$ , which is divisible by 4. So the triangle can't be equilateral.

if b is even, then $a^2+b^2$ is of the form $4k+1$ . As $a^2+b^2 = c^2+d^2$ we must conclude that c & d have different parity. Hence $a-c$ and $b-d$ must have the same parity and, correspondingly, the square of the length of the third side again is even. Contradiction.

Solution 2:

Observe that the angle any segment between two grid vertices make with either axis has to have a rational tangent. In addition, as $tan(a+b) = \frac{\tan{a} + \tan{b}}{1-\tan{a}\tan{b}}$

the tangent of the sum or difference of the two angles with rational tangents is rational. Therefore any non-right-angle given by the grid points has rational tangents. Since $\tan 60$ (degrees) = $sqrt{3}$ is irrational, the equilateral triangle cannot have all three vertices on a square grid