This is not really an advanced problem and only indirectly related to geometry, but I instantly though of it because of its short solution and trick, which is really nice for the introduction to complex numbers.

Let $z_1,z_2,z_3,z_4\in\mathbb{C}$ be points with $|z_1|=|z_2|=|z_3|=|z_4|$ and $z_1+z_2+z_3+z_4=0$, then prove there are two pairs of antipodal points among them.

Consider the polynomial $(z-z_1)(z-z_2)(z-z_3)(z-z_4)$. The cubic coefficient vanishes by proposition, so does the linear one as: \begin{align*} &z_2z_3z_4 +z_1z_3z_4 +z_1z_2z_4 +z_1z_2z_3 =z_1z_2z_3z_4\left( \frac{1}{z_1} +\frac{1}{z_2} +\frac{1}{z_3} +\frac{1}{z_4}\right) \\ =&z_1z_2z_3z_4\left( \frac{z_1^*}{|z_1|^2} +\frac{z_2^*}{|z_2|^2} +\frac{z_3^*}{|z_3|^2} +\frac{z_4^*}{|z_4|^2}\right) =\frac{z_1z_2z_3z_4}{|z_1|^2} (z_1+z_2+z_3+z_4)^*=0. \end{align*} As a result for every root $z$ of the polynomial, $-z$ is also a root.