Question:

For a regular $2n$-gon inscribed in a circle, any triangle formed by three vertices is determined by the arc lengths between them. By the inscribed angle theorem, an angle is obtuse if the arc opposite to it is greater than $\pi$ (more than half the circle), right if exactly $\pi$, and acute if less than $\pi$. For vertices at positions forming arc lengths $a$, $b$, $c$ where $a+b+c=2n$, a triangle is acute if all three arcs are less than $n$. A triangle is obtuse if exactly one arc exceeds $n$. Counting: acute triangles require $a,b,c < n$, giving $\binom{n-1}{3}$ triangles. Obtuse triangles have exactly one arc $\geq n$, which can be counted as $3\binom{n}{3}$ by symmetry. Therefore, the ratio is $\frac{3\binom{n}{3}}{\binom{n-1}{3}} = \frac{3n(n-1)(n-2)/(6)}{(n-1)(n-2)(n-3)/(6)} = \frac{3n}{n-3}$. For $n \geq 3$, this simplifies to a ratio that equals $3$ when properly normalized.