Problem - 2178
Let unit vectors $a$, $b$, and $c$ satisfy $a+b+c=0$, prove the angles between these vectors are all $120^\circ$.
Because $c = -(a+b)$, we find that $a$ and $(b+c)$ collinear. Meanwhile, $|a|=|b|$ implies $c$ is on the bisector of $(a+b)$. Because $a$, $b$ and $c$ can form a closed equilateral triangle, all these included angles are $60^\circ$. Hence the angle between $a$ and $c$ are $120^\circ$. By the same reasoning, the other two angles are $120^\circ$ too.