Brilliant Logo Triangles

Let $\{A,B,C\}$ be a set of three points, as described in the problem. Then $A$ and $B$ lie on some great circle (say $\gamma_C$ ), $A$ and $C$ lie on some great circle (say $\gamma_B$ ), and $B$ and $C$ lie on some great circle (say $\gamma_A$ ). So, we can count the number of possible sets $\{A,B,C\}$ by counting the number of ways to choose three great circles.

There are 15 great circles, so there are $\binom{15}{3} = 455$ ways to choose three of them. However, we must exclude the cases where all three great circles pass through the same pair of antipodal points.

There are 12 points that lie on exactly five great circles, 20 points that lie on exactly three great circles, and 30 points that lie on exactly two great circles.

Among the 12 points that lie on exactly five great circles, there are six pairs of antipodal points. For each such pair of antipodal points, there are $\binom{5}{3} = 10$ ways to choose three great circles that go through the points. Among the 20 points that lie on exactly three great circles, there are ten pairs of antipodal points. For each such pair of antipodal points, there is $\binom{3}{3} = 1$ way to choose three great circles that go through the points. (We can ignore the 30 points that lie on exactly two great circles.) This gives us a total of $455 - 6 \cdot 10 - 10 \cdot 1 = 385$ viable ways to choose three great circles. Call them $\gamma_A$ , $\gamma_B$ , and $\gamma_C$ .

Then there are two ways to choose the point $A$ (from the two intersections of $\gamma_B$ and $\gamma_C$ ), two ways to choose the point $B$ , and two ways to choose the point $C$ , so the number of sets $\{A,B,C\}$ is $8 \cdot 385 = 3080$ .

There are $3$ kinds of points on the spherical disdyakis triacontahedron, those that are at the intersections of $2$ , $3$ , and $5$ lines. Hence, all triangles can be categorized by such points as vertices, and counted separately. Below is the list of the number of triangles of each, and the grand total.

$\left\{ 2,2,2 \right\} \qquad 040$
$\left\{ 2,2,3 \right\} \qquad 000$
$\left\{ 2,2,5 \right\} \qquad 000$
$\left\{ 2,3,3 \right\} \qquad 000$
$\left\{ 2,3,5 \right\} \qquad 960$
$\left\{ 2,5,5 \right\} \qquad 480$
$\left\{ 3,3,3 \right\} \qquad 000$
$\left\{ 3,3,5 \right\} \qquad 480$
$\left\{ 3,5,5 \right\} \qquad 960$
$\left\{ 5,5,5 \right\} \qquad 160$

$Total\quad \quad \quad 3080$

As an example, for $\left\{ 5,5,5 \right\}$ triangles, if we disregard condition 3), and noting that every $\left\{ 5\right\}$ point lies on a great circle with any other $\left\{ 5\right\}$ point, then we can compute the total number of triangles

$\dfrac { (12)(11)(10) }{ (3)(2)(1) } =220$

However, there are $6$ pairs of $\left\{ 5\right\}$ points that are antipodal, so that for each such pair, there are $10$ triangles (sets of $3$ points, technically) that all fail to meet condition 3), so that the net is

$220-60=160$

As another example, there are no $\left\{ 3,3,3 \right\}$ triangles, which can be determined by inspection of the stereographic projection.

Messy casework is (not) the way to go.

(BTW, if you add up all the numbers, you get twice the correct answer, because this is my scratch paper for the first wrong answer that I submitted)