How many triangles formed by three vertices of a regular $17$-gon are obtuse?
As an extension, how many triangles formed by three vertices of a regular $n$-gon are obtuse?
$\endgroup$1 Answer
$\begingroup$Clearly it's $n$ times the number of obtuse triangles with the obtuse vertex fixed at a specific point.
The angles of a regular $n$-gon have size $\frac{(n-2)\pi}n$. There are $n{-}3$ lines dividing this angle into $n{-}2$ equal parts of size $\frac\pi n$ (see the inscribed angle theorem). This means a triangle is obtuse if its remaining vertices are more than $\frac n2$ segments apart, as $\frac n2\frac\pi n=\frac\pi2$.
Therefore setting $k=\!\!\left\lfloor\frac n2\right\rfloor\!{+}1$, the answer is $$n\cdot((n{-}k{-}1)+(n{-k}-2)+\ldots+1)=\frac{n(n-k)(n-k-1)}2=\frac{n\left(\left\lceil\frac n2\right\rceil\!{-}1\right)\!\left(\left\lceil\frac n2\right\rceil\!{-}2\right)}2.$$
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