Brazilian Math Olympiad

First consider a triangle $ABC$ and its circumcenter $O$. Then the area of $ABC$ is $\frac{R^2}{2}(\sin 2\angle A + \sin 2\angle B + \sin 2\angle C)$. Notice that if $\angle B > 90^\circ$ then $\sin 2\angle B < 0$.

So the sum is equal to the sum of the areas of triangles $OA_iA_j$ with a plus sign or a minus sign, depending on the third vertex $A_k$ of the triangle $A_iA_jA_k$: if $A_k$ lies on the major arc $A_iA_j$ then we have a plus sign; else we have a minus sign (it won't matter if $A_iA_j$ is a diameter, because in that case the area of $OA_iA_j$ is zero).

Therefore, if $A_iA_j$ subtend a minor arc of $k \cdot \frac{2\pi}{n}$, $1 \le k \le \lfloor n/2 \rfloor$, the area of the triangle $OA_iA_j$ appears with a minus sign $k-1$ times and with a plus sign $n - (k-1) - 2 = n-k-1$ times. So it contributes with the sum $n-k-1 - (k-1) = n-2k$ times.

$$S = \frac{n}{2} \sum_{k=1}^{\lfloor n/2 \rfloor} (n-2k) \sin k\theta = \frac{n^2}{2} \sum_{k=1}^{\lfloor n/2 \rfloor} \sin k\theta - n \sum_{k=1}^{\lfloor n/2 \rfloor} k \sin k\theta$$

Consider the sums $S_1(\theta) = \sum_{k=1}^{\lfloor n/2 \rfloor} \sin k\theta$ and $S_2(\theta) = \sum_{k=1}^{\lfloor n/2 \rfloor} \cos k\theta \implies S_2'(\theta) = -\sum_{k=1}^{\lfloor n/2 \rfloor} k \sin k\theta$. So we want to compute $\frac{n^2}{2} S_1(\theta) + n \cdot S_2'(\theta)$.

$$\begin{align*} S_2(\theta) + iS_1(\theta) &= \sum_{k=1}^{\lfloor n/2 \rfloor} \cos k\theta + i \sin k\theta = \sum_{k=1}^{\lfloor n/2 \rfloor} \omega^k = \omega \cdot \frac{\omega^{\lfloor n/2 \rfloor} - 1}{\omega - 1} \\ &= \omega^{\lfloor n/2 \rfloor/2+1/2} \frac{\omega^{\lfloor n/2 \rfloor/2} - \omega^{-\lfloor n/2 \rfloor/2}}{\omega^{1/2} - \omega^{-1/2}} \\ &= \left( \cos \left( \frac{(\lfloor n/2 \rfloor + 1)\theta}{2} \right) + i \sin \left( \frac{(\lfloor n/2 \rfloor + 1)\theta}{2} \right) \right) \cdot \frac{\sin(\lfloor n/2 \rfloor \theta/2)}{\sin(\theta/2)}, \end{align*}$$

$$\begin{aligned} S_1 &= \frac{\sin\left(\frac{(\lfloor n/2 \rfloor + 1) \theta}{2}\right) \sin\left(\frac{\lfloor n/2 \rfloor \theta}{2}\right)}{\sin(\theta/2)} = \frac{\cos(\theta/2) - \cos((\lfloor n/2 \rfloor + 1/2)\theta)}{2 \sin(\theta/2)} \\ &= \frac{1}{2} \cot(\theta/2) - \frac{\cos((\lfloor n/2 \rfloor + 1/2)\theta)}{2 \sin(\theta/2)} \\[1.5em] S_2 &= \frac{\cos\left(\frac{(\lfloor n/2 \rfloor + 1) \theta}{2}\right) \sin\left(\frac{\lfloor n/2 \rfloor \theta}{2}\right)}{\sin(\theta/2)} = \frac{\sin(\theta/2) + \sin((\lfloor n/2 \rfloor + 1/2)\theta)}{2 \sin(\theta/2)} \\ &= \frac{1}{2} + \frac{\sin((\lfloor n/2 \rfloor + 1/2)\theta)}{2 \sin(\theta/2)} \end{aligned}$$

$$\begin{aligned} S_2'(\theta) &= \frac{\left(\lfloor \frac{n}{2} \rfloor + \frac{1}{2}\right) \cos\left(\left(\lfloor \frac{n}{2} \rfloor + \frac{1}{2}\right) \theta\right) \sin\left(\frac{\theta}{2}\right) - \frac{1}{2} \cos\left(\frac{\theta}{2}\right) \sin\left(\left(\lfloor \frac{n}{2} \rfloor + \frac{1}{2}\right) \theta\right)}{2 \sin^2(\theta/2)} \\ &= \frac{\lfloor n/2 \rfloor}{2} \frac{\cos((\lfloor n/2 \rfloor + 1/2)\theta)}{\sin(\theta/2)} - \frac{1}{4} \frac{\sin(\lfloor n/2 \rfloor \theta)}{\sin^2(\theta/2)} \end{aligned}$$

So the required sum is
$$\frac{n^2}{2} \left( \frac{1}{2} \cot(\theta/2) - \frac{\cos((\lfloor n/2 \rfloor + 1/2)\theta)}{2 \sin(\theta/2)} \right) + n \left( \frac{\lfloor n/2 \rfloor}{2} \frac{\cos((\lfloor n/2 \rfloor + 1/2)\theta)}{\sin(\theta/2)} - \frac{1}{4} \frac{\sin(\lfloor n/2 \rfloor \theta)}{\sin^2(\theta/2)} \right)$$

If $n$ is even, $\lfloor n/2 \rfloor = n/2$ and substituting $\theta = \frac{2\pi}{n}$ the sum simplifies to
$$\frac{n^2}{4} \cot \frac{\pi}{n}$$
If $n$ is odd, $\lfloor n/2 \rfloor = (n-1)/2$ and substituting $\theta = \frac{2\pi}{n}$ the sum also simplifies to
$$\frac{n^2}{4} \cot \frac{\pi}{n}$$

$$\frac{n^2}{4} \cot \frac{\pi}{n}$$