Edit: Thanks @below, this stuff is wrong. The issue is that I defined value incorrectly, I should have defined it as the number of others which form an isosceles triangle with the two lines, where the other triangle is the base of the triangle. Then, the maximum value of any two lines is $2$, as pointed out by math90 . We claim that Mario's maximum number of coins is $4 \cdot \binom{20}{2} = 760.$ For a pair of lines, define their value to be the number of other lines which form an isosceles triangle with them. Then, observe that Mario's wealth is exactly equal to the sum of values of all pairs of lines (by an easy double-counting argument). It's also clear that each pair of lines has value at most $4$. Therefore, this directly implies the upper bound already. Now, to obtain this bound, consider some lines with arguments $72i + 45k$ degrees, for $0 \le i \le 4, 0 \le k \le 3,$ and no three lines meeting at a common point. This is easily checked to obtain the maximum number of $760$. $\square$

Above: I got $2\cdot\binom{20}{2}=380$. Each pair of lines has a value at most $2$, since we have two options for the third line. The third line is parallel to the internal/external angle bisector.

math90 wrote: Above: I got $2\cdot\binom{20}{2}=380$. Each pair of lines has a value at most $2$, since we have two options for the third line. The third line is parallel to the internal/external angle bisector. Yeah!I also

math90 wrote: Above: I got $2\cdot\binom{20}{2}=380$. Each pair of lines has a value at most $2$, since we have two options for the third line. The third line is parallel to the internal/external angle bisector. Are you taking into account the fact that equilateral triangles have value $3$?

Above: Yes, I take. The number of points of each triangle is the number of pairs of equal edges.