2010
Problem - 2259
Let $A (x_1, y_1)$, $B (x_2, y_2)$, and $C (x_3, y_3)$ be three points on the unit circle, and $$x_1 + x_2 + x_3 = y_1+y_2+y_3=0$$ Prove $$x_1^2 +x_2^2+x_3^2=y_1^2+y_2^2+y_3^2=\frac{3}{2}$$
Given $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=0$, it is easy to see that $\angle{AOB}=\angle{BOC}=\angle{COA}=120^\circ$.
Let $A=(\cos\alpha,\sin\alpha)$. Then $B = (\cos(\alpha+120^\circ),\sin(\alpha+120^\circ))$ and $C = (\cos(\alpha-120^\circ),\sin(\alpha-120^\circ))$. Then,
\begin{align*}
&x_1^2 + x_2^2 + x_3^2 \\
=\quad&\cos^2\alpha + \cos^2(\alpha+120^\circ)+\cos^2(\alpha - 120^\circ)\\
=\quad&\frac{1+\cos 2\alpha}{2} + \frac{1+\cos(2\alpha + 240^\circ)}{2} + \frac{1+\cos(2\alpha - 240^\circ)}{2}\\
=\quad&\frac{3}{2} + \frac{1}{2}\cdot(\cos 2\alpha + \cos(2\alpha + 240^\circ) + \cos(2\alpha - 240^\circ))\\
=\quad&\frac{3}{2} + \frac{1}{2}\cdot(\cos 2\alpha + 2\cos 2\alpha \cos 240^\circ )\\
=\quad&\frac{3}{2}
\end{align*}
Clearly, $x_1^2+x_2^2 + x_3^2 + y_1^2+y_2^2+y_3^2 = 3$, therefore $y_1^2+y_2^2+y_3^2=\frac{3}{2}$ too.