Find all the triangles in a dissection of a decagon On page $97$ of Robin Wilson's "Four Colors Suffice", the following puzzle appears:

[P]rove that, if all the angular points of a regular decagon are joined, and all the sides and diagonals produced indefinitely, the number of triangles so formed will be $10,000$.

It is stated that the puzzle, due to one James Maurice Wilson, is intended to "require ingenuity rather than knowledge" for it solution.  I haven't solved the problem, but I think I can proved that $10,000$ is too big.

We have $5$ lines extending the diagonals, and $10$ lines extending the sides.  There are $5$ points ($2$ red, $2$ gray, and $1$ white) on each of the former. There are $8$ points on each of the latter ($2$ each colored green, red, blue, and gray.)
There is $1$ white point, and $10$ points of each of the other four colors.  At most there is one triangle for every set of $3$ non-collinear points:  $$\binom{41}3-10\binom83-5\binom53=10,050$$
Each green point is adjacent to red points, which are in turn adjacent to a common blue point.  The four points are the vertices of a kite-like figure, but if we choose any $3$ of them, there is no triangle, because the diagonals of the kite don't appear.  This eliminates $10\binom43=40$ triangles.
Similarly, each of the red points is adjacent to two blue points and a gray point, forming a kite with one diagonal.  Two of the $4$ choices of $3$ these of these $4$ give a triangle, but the $2$ choices including both blue points do not.  This eliminates another $20$ triangles, so we're already below $10,000$, and there are many other choices of $3$ non-collinear points that don't work either.
Is the stated answer incorrect, or am I missing something?
 A: Well, your solution misses some of the triangles formed by lines that aren't either sides of the decagon or diagonals connecting opposite vertices.
However, I feel like I also have a solution, and my solution proves that $10\,000$ is a bit too small...

The lines we draw are in $10$ equivalence classes of parallel lines:

*

*$5$ equivalence classes containing $5$ parallel lines each, parallel to one of the sides of the decagon. These also include diagonals connecting vertices of the decagon that are $3$ or $5$ steps apart.

*$5$ equivalence classes containing $4$ parallel lines each. These include diagonals connecting vertices of the decagon that are $2$ or $4$ steps apart.

If we choose $3$ different lines from three different equivalence classes, they will form a triangle. This gives us
$$
   \underbrace{\binom 53 \cdot 5^3}_{\text{3 lines of first type}} + \underbrace{\binom 52 \cdot 5^2}_{\text{2 lines of first type}} \cdot \underbrace{\binom 51 \cdot 4}_{\text{1 line of second type}} + \\ \underbrace{\binom 51 \cdot 5}_{\text{1 line of first type}} \cdot \underbrace{\binom 52 \cdot 4^2}_{\text{2 lines of second type}} + \underbrace{\binom 53 \cdot 4^3}_{\text{3 lines of second type}} = 10\,890 
$$
triangles.
(With generating functions, we can also get this number as the coefficient of $x^3$ in $(1+5x)^5 (1 + 4x)^5$. Here, $1+5x$ represents the number of ways we can choose $0$ or $1$ lines from an equivalence class of the first type, and $1+4x$ represents the number of ways we can choose $0$ or $1$ lines from an equivalence class of the second type. We multiply these together, and take the coefficient of $x^3$ to find cases where we choose $3$ lines total.)
A: I can justify the count of exactly $10,000$ triangles. Going off of Misha Lavrov's answer, there are $10,890$ ways to select three mutually non-parallel lines in the diagram. However, some of these triples of lines will intersect in a point, so these must be subtracted to correct the count. Namely,

*

*There are $\binom{5}3=10$ triples of lines which intersect in the center of the decagon.


*For each vertex, there are $9$ lines meeting at the vertex, resulting in $10\cdot \binom{9}3=840$ triples.


*For each of the red points in your diagram, there are three lines meeting there, resulting in $10\cdot \binom{3}3=10$ triples.


*Numbering the vertices $v_1,\dots,v_{10}$, then the lines through $\{v_1,v_6\}$, $\{v_3,v_5\}$, and $\{v_7,v_9\}$ all intersect at the same point. Taking all three rotations of this gives $10$ more triples.


*Similarly to the last point, there are $10$ rotations of each of the following triples, which meet inside the decagon:

*

*$\{v_1,v_6\},\{v_5,v_8\}$ and $\{v_4,v_7\}$.

*$\{v_1,v_6\},\{v_3,v_7\}$ and $\{v_5,v_9\}$.



Subtracting these $10+840+10+10+10+10=890$ triples leaves exactly $10,000$ triangles.
