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 $O{A}_i{A}_j$ with a plus sign or a minus sign, depending on the third vertex $A_k$ of the triangle $A_i{A}_j{A}_k$: if $A_k$ lies on the major arc $A_i{A}_j$ then we have a plus sign; else we have a minus sign (it won't matter if $A_i{A}_j$ is a diameter, because in that case the area of $O{A}_i{A}_j$ is zero).

Therefore, if $A_i{A}_j$ subtend a minor arc of $k\cdot \frac{2\pi }{n}$, $1\le k\le \lfloor n/2\rfloor$, the area of the triangle $O{A}_i{A}_j$ 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 $S_2(\theta )={\sum }_{k=1}^{\lfloor n/2\rfloor }\cos k\theta ⟹S_2^{\prime }(\theta )=-{\sum }_{k=1}^{\lfloor n/2\rfloor }k\sin k\theta$. So we want to compute $\frac{n^2}{2}{S}_1(\theta )+n\cdot S_2^{\prime }(\theta )$.

$$\begin{array}{rl}{S}_2(\theta )+i{S}_1(\theta )& =\sum _{k=1}^{\lfloor n/2\rfloor }\cos k\theta +i\sin k\theta =\sum _{k=1}^{\lfloor n/2\rfloor }{\omega }^k=\omega \cdot \frac{{\omega }^{\lfloor n/2\rfloor }-1}{\omega -1}\\ & ={\omega }^{\lfloor n/2\rfloor /2+1/2}\frac{{\omega }^{\lfloor n/2\rfloor /2}-{\omega }^{-\lfloor n/2\rfloor /2}}{{\omega }^{1/2}-{\omega }^{-1/2}}\\ & =\left(\cos \left(\frac{(\lfloor n/2\rfloor +1)\theta }{2}\right)+i\sin \left(\frac{(\lfloor n/2\rfloor +1)\theta }{2}\right)\right)\cdot \frac{\sin (\lfloor n/2\rfloor \theta /2)}{\sin (\theta /2)},\end{array}$$

$$\begin{array}{rl}{S}_1& =\frac{\sin \left(\frac{(\lfloor n/2\rfloor +1)\theta }{2}\right)\sin \left(\frac{\lfloor n/2\rfloor \theta }{2}\right)}{\sin (\theta /2)}=\frac{\cos (\theta /2)-\cos ((\lfloor n/2\rfloor +1/2)\theta )}{2\sin (\theta /2)}\\ & =\frac{1}{2}\cot (\theta /2)-\frac{\cos ((\lfloor n/2\rfloor +1/2)\theta )}{2\sin (\theta /2)}\\ S_2& =\frac{\cos \left(\frac{(\lfloor n/2\rfloor +1)\theta }{2}\right)\sin \left(\frac{\lfloor n/2\rfloor \theta }{2}\right)}{\sin (\theta /2)}=\frac{\sin (\theta /2)+\sin ((\lfloor n/2\rfloor +1/2)\theta )}{2\sin (\theta /2)}\\ & =\frac{1}{2}+\frac{\sin ((\lfloor n/2\rfloor +1/2)\theta )}{2\sin (\theta /2)}\end{array}$$

$$\begin{array}{rl}{S}_2^{\prime }(\theta )& =\frac{\left(\lfloor \frac{n}{2}\rfloor +\frac{1}{2}\right)\cos \left(\left(\lfloor \frac{n}{2}\rfloor +\frac{1}{2}\right)\theta \right)\sin \left(\frac{\theta }{2}\right)-\frac{1}{2}\cos \left(\frac{\theta }{2}\right)\sin \left(\left(\lfloor \frac{n}{2}\rfloor +\frac{1}{2}\right)\theta \right)}{2{\sin }^2(\theta /2)}\\ & =\frac{\lfloor n/2\rfloor }{2}\frac{\cos ((\lfloor n/2\rfloor +1/2)\theta )}{\sin (\theta /2)}-\frac{1}{4}\frac{\sin (\lfloor n/2\rfloor \theta )}{{\sin }^2(\theta /2)}\end{array}$$

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

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