Consider the set I_r of points (x,y) with integer co-ordinates in the interior of the circle with radius r, centered at the origin, i.e. x² + y² < r².

For a radius of 2, I₂ contains the nine points (0,0), (1,0), (1,1), (0,1), (-1,1), (-1,0), (-1,-1), (0,-1) and (1,-1). There are eight triangles having all three vertices in I₂ which contain the origin in the interior. Two of them are shown below, the others are obtained from these by rotation.

For a radius of 3, there are 360 triangles containing the origin in the interior and having all vertices in I₃ and for I₅ the number is 10600.

How many triangles are there containing the origin in the interior and having all three vertices in I₁₀₅?