Estonian Mathematical Olympiad

Let $n=4k$. For each $i=1,2,\dots ,n$, denote ${\alpha }_i=i\cdot \frac{360^{\circ }}{n}$ and $x_i={\sin }^2{\alpha }_i$. We pair the terms as $(x_1,x_{k+1}),(x_2,x_{k+2}),\dots ,(x_k,x_{2k})$ and $(x_{2k+1},x_{3k+1}),(x_{2k+2},x_{3k+2}),\dots ,(x_{3k},x_{4k})$. Since $${\alpha }_{i+k}=(i+k)\cdot \frac{360^{\circ }}{n}=i\cdot \frac{360^{\circ }}{n}+k\cdot \frac{360^{\circ }}{n}={\alpha }_i+90^{\circ },$$ we have $${\sin }^2{\alpha }_{i+k}={\sin }^2({\alpha }_i+90^{\circ })={\sin }^2(90^{\circ }-(-{\alpha }_i))={\cos }^2(-{\alpha }_i)={\cos }^2{\alpha }_i.$$ Therefore $x_i+x_{i+k}={\sin }^2{\alpha }_i+{\cos }^2{\alpha }_i=1$. Thus, the sum of the numbers of each pair is $1$. Since we have $\frac{n}{2}$ pairs, the sum of all numbers is $\frac{n}{2}$.

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Alternative solution.

Let $n=4k$. Note that $${\sin }^2\left(i\cdot \frac{360^{\circ }}{n}\right)=\frac{1-\cos \left(2i\cdot \frac{360^{\circ }}{n}\right)}{2}=\frac{1}{2}-\frac{1}{2}\cos \left(2i\cdot \frac{360^{\circ }}{n}\right).$$ For each $i=1,2,\dots ,n$, denote ${\alpha }_i=i\cdot \frac{360^{\circ }}{n}$ and $y_i=\cos 2{\alpha }_i$; then the equation to be proven takes the form $\frac{1}{2}n-\frac{1}{2}(y_1+y_2+\cdots +y_n)=\frac{1}{2}n$, which is equivalent to $y_1+y_2+\cdots +y_n=0$. We are going to prove this equality in the rest. We pair the terms as $(y_1,y_{k+1}),(y_2,y_{k+2}),\dots ,(y_k,y_{2k})$ and $(y_{2k+1},y_{3k+1}),(y_{2k+2},y_{3k+2}),\dots ,(y_{3k},y_{4k})$ and prove the angle equality ${\alpha }_{i+k}={\alpha }_i+90^{\circ }$ as in Solution 1. Next, we see that $$\cos 2({\alpha }_i+90^{\circ })=\cos (2{\alpha }_i+180^{\circ })=-\cos 2{\alpha }_i,$$ from which it follows that $y_i+y_{i+k}=\cos 2{\alpha }_i-\cos 2{\alpha }_i=0$. Since the sum of the numbers of each pair is $0$, the sum of all numbers is also $0$.