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.