>>11513165Let's go to cases. First, since the point a is special, we'll count every triangle including a as an endpoint. Then we'll count all triangles not including a.
Since every other point is collinear with a, we need only establish that a given pair of points are collinear with each other, and that the three points are not all collinear on the same line. By 2), the two points must not have the same index number, and by 3) and 4), they must share a letter. On the three lines b, c and d, two of four points may be chosen. The number of triangles including a as an endpoint is thus .
We must now consider triangles not including a as an endpoint. Each such triangle's endpoints must be pairwise collinear, but not all three collinear together on the same line.
By 2)-4), any two of such a triangle's endpoints must have either a letter in common, or a number in common. This leads to subcases.
Suppose that a first point is chosen, and that a second point is then chosen sharing a letter with the first. Then by 3), the third point to be chosen must not share that same letter with the first and second points, and it must be collinear with both the first and second points. That is, by 2) it must share a number with one of the points, and by 3) it must yet share a letter with the other point. But this letter must not be the same with the first and second points. So one of the points must have two letters.
Similarly, suppose that a first point is chosen, and that a second point is then chosen sharing a number with the first. By 2), the third point to be chosen must not share that same number with the first and second points, and it must be collinear with both the first and second points. That is, by 3) it must share a letter with one of the points, and by 3) again it must yet share another letter with the other point, the two letters being different. So again one of the points must have two letters.
