Problem: Let $A_1A_2\dots A_{100}$ be a regular $100$-gon with a circumcircle of diameter $1$. Let $M$ be the midpoint of the minor arc $A_1A_2$. Find $$ \sum^{100}_{i=1} |MA_i|^2. $$
My attempt: Let the points $A_k$ be of the form $$ \frac12 e^{2\pi ki/100} = \frac12 \cos\Bigl(\frac{2\pi k}{100}\Bigr) + \frac12 i\sin\Bigl(\frac{2 \pi k}{100}\Bigr) $$ in the complex plane, where $k = 1, 2, \dots, 100$. Then $M$ is clearly $\frac12 \cos\bigl(\frac{3\pi}{100}\bigr) + \frac12 i \sin\bigl(\frac{3\pi}{100}\bigr)$. Hence by the Pythagorean Theorem the sum equals $$ \begin{multline} \sum^{100}_{k = 1} \Biggl( \biggl( \frac12 \cos\Bigl(\frac{2\pi k}{100}\Bigr) - \frac12 \cos\Bigl(\frac{3\pi}{100}\Bigr) \biggr)^2 + \biggl( \frac12 \sin\Bigl(\frac{2 \pi k}{100}\Bigr) - \frac12 \sin\Bigl(\frac{3\pi}{100}\Bigr) \biggr)^2 \Biggr) \\ = \frac12 \sum^{100}_{k=1} \biggl( 1 - \cos\Bigl(\frac{2\pi k}{100}\Bigr) \cos\Bigl(\frac{3\pi}{100}\Bigr) - \sin\Bigl(\frac{2\pi k}{100}\Bigr) \cos\Bigl(\frac{3\pi}{100}\Bigr) \biggr) \end{multline} $$ which equals $$ \frac12 ( 100 - 0 - 0 ) = 50. $$
The explanation for the last step is that $\sum \cos = 0$ and $\sum \sin = 0$ because you can do pairing, e.g., for $\cos$ you can pair $k = 1, 49$, $k = 2, 48$, and then their sum is zero.
Is this solution correct? I am not sure since I submitted my answer to a website but it said it was wrong.
Draw the segments $\overline{MA_1}$ and $\overline{MA_{51}}$. Because $A_1$ and $A_{51}$ are diametrically opposite points, the angle between these two segments is a right angle. So by the Pythagorean Theorem the squared lengths satisfy
$MA_1^2+MA_{51}^2=1.$
There are $50$ such diametrically opposite pairs constituting the vertices of the $100$-gon, so the sum of the squares of all $100$ distances is the sum of $50$ unities, thus $50$ itself.