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?
Best Answer
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:
Subtracting these $10+840+10+10+10+10=890$ triples leaves exactly $10,000$ triangles.