题目184：包含原点的三角形
Triangles containing the originConsider the set Ir 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 Ir ：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？
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