题目184：包含原点的三角形

欧拉计划

Triangles containing the origin

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?

题目：

考虑满足如下条件的点 (x,y) 的集合 I_r ：以原点为圆心，半径为r形成的圆内的整数坐标点，即 x² + y² <r²。

如果半径为 2，I₂ 包括了 (0,0)， (1,0)， (1,1)，(0,1)， (-1,1)， (-1,0)，(-1,-1)，(0,-1) 和 (1,-1) 九个点。由这些点形成的三角形中，总共有八个三角形把原点包含其中。下面展示了其中两个，别的则是这两个的旋转得到的。

如果半径为 3，则圆内整数坐标点形成的三角形中，有 360 个把原点围在内部，半径为 5 时，则有 10600 个。

有多少个顶点在 I₁₀₅ 中的三角形将原点包含其中？