\section*{Problem 4 - 061244}

Given a natural number \(n>3\). Let \(V=P_{1} P_{2} \ldots P_{n}\) be a regular \(n\)-sided polygon in the plane. Determine the total number of distinct obtuse triangles \(\Delta P_{k} P_{l} P_{m}\) (where \(P_{k}, P_{l}, P_{m}\) are vertices of \(V\))!
}

We distinguish the cases \(n\) even and \(n\) odd.

Let \(P_{k}, P_{l}\) and \(P_{m}\) be the vertices of the triangle in a clockwise direction. First, we note that the triangle \(P_{k} P_{l} P_{m}\) has a right angle at \(P_{l}\) if and only if \(P_{l}\) lies on the semicircle over the segment between \(P_{k}\) and \(P_{m}\) (Thales' theorem).

We now require that the desired obtuse angle is at \(P_{l}\). This is obviously the case when \(P_{l}\) and \(P_{k}\) lie on an arc of a circle that is less than half the circumference of the regular \(n\)-gon. First, we fix an arbitrary point \(P_{k}\). For this, we have \(n\) possibilities in both cases.

1. Case: Let \(n\) be even.

We now look for the two missing points with the above requirement. If we count \(\frac{n}{2}\) points from \(P_{k}\) and set the point \(P_{m}\), then every triangle at \(P_{m}\) has a right angle. We can therefore choose the two missing points for the obtuse triangle from \(\frac{n}{2}-1\).

2. Case: Let \(n\) be odd.

If we draw the diameter through \(P_{k}\), then in this case exactly \(\frac{n-1}{2}\) vertices lie on each semicircle. From these, we can again choose 2 arbitrarily.

Let \(A(n)\) be the total number of obtuse triangles, we thus obtain:

\[
A(n)= \begin{cases}n\left(\frac{n}{2}-1\right) & , n \text { even } \\ n\left(\frac{n-1}{2}\right) & , n \text { odd }\end{cases}
\]

\section*{Solution by a Member of Matheplanet}

Alternative solution:

If \(\triangle P_{k} P_{l} P_{m}\) is obtuse, then exactly one angle is greater than \(90^{\circ}\); let this be at \(P_{l}\). The arc from \(P_{k}\) to \(P_{m}\), which contains \(P_{l}\), is smaller than half the circumference. On the other hand, any three points that lie on an arc shorter than half the circumference form an obtuse triangle. Therefore, the number we are looking for is equal to the number of ways to choose three points in this manner.

For the selection of the "first" point \(P_{k}\), there are \(n\) possibilities, as all \(n\) points are considered. For each selection of \(P_{k}\), then for \(P_{l}\) and \(P_{m}\), any selection of 2 points from the points \(P_{k+1}, P_{k+2}, \ldots ; P_{k+\lfloor(n-1) / 2\rfloor}\) can be used, where we set \(P_{n+i} \equiv P_{i}(i=1, \ldots, n)\) in this list. Thus,

\[
\begin{aligned}
n \cdot\left(\left\lfloor\frac{n-1}{2}\right\rfloor\right) & =n \cdot \frac{1}{2}\left\lfloor\frac{n-1}{2}\right\rfloor\left(\left\lfloor\frac{n-1}{2}\right\rfloor-1\right)=\frac{1}{2} n\left\lfloor\frac{n-1}{2}\right\rfloor\left\lfloor\frac{n-3}{2}\right\rfloor \\
& = \begin{cases}\frac{1}{8} n(n-1)(n-3) & \text { for } n \text { odd } \\
\frac{1}{8} n(n-2)(n-4) & \text { for } n \text { even }\end{cases}
\end{aligned}
\]
is the total number of all possibilities.

Adapted from [12]