Suppose $A_1A_2 . . . A_{20}$ is a $20-$sided regular polygon.
How many non-isosceles (scalene) triangles
can be formed whose vertices are among the vertices of the polygon but whose sides are not
the sides of the polygon?
I could'nt find a cute answer to this problem. My answer is different from those at other websites. Not to mention that the answers are also different at different websites.
Best Answer
We first count all nondegenerate triangles containing no edge $A_iA_{i+1}$, and then subtract the isosceles triangles among these.
Put the first vertex $v_1$ at $A_0=A_{20}$. If you put $v_2$ at $A_2$ or $A_{18}$ then there are $15$ allowed $A_i$ left for $v_3$. If you put $v_2$ at an $A_i$ with $3\leq i\leq 17$ then there are $2\cdot 3=6$ forbidden $A_i$ for $v_3$. It follows that you can choose $(v_1,v_2,v_3)$ in $$20\cdot2\cdot 15+20\cdot 15\cdot 14=4800$$ ways, giving rise to ${1\over6}\cdot4800=800$ different triangles.
The tip of an isoscles triangle can be chosen in $20$ ways, and then the base in $8$ ways. Since there is no equilateral triangle possible it follows that there are $160$ isosceles triangles.
The total number of admissible triangles therefore is $640$.