Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$. Let $z_1$, $z_2$, $z_3$ be distinct complex numbers such that $|z_1| = |z_2| = |z_3| > 0$. If $z_1 + z_2z_3$, $z_2 + z_3z_1$, and $z_3 + z_1z_2$ are real numbers, prove that $z_1z_2z_3 = 1$.
I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).
I tried several approaches: The first was to substitute $a + ib$, $c + id$, and $f + ig$ for $z_1$, $z_2$, and $z_3$, respectively, and bash. I used $(1+z_1)(1+z_2)(1+z_3)$, which didn't really get me anywhere.
After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

We have $\dfrac{z}{|z|}+\dfrac{|z|}{z} \in \mathbb{R}$ when $z \ne 0$
So we have $\dfrac{z_1}{|z_1|}+\dfrac{|z_1|}{z_1} \in \mathbb{R}$ and $z_1+z_2z_3 \in \mathbb{R} \Rightarrow \dfrac{z_1}{|z_1|}+\dfrac{z_2z_3}{|z_1|} \in \mathbb{R}$
Subtracting we get $\dfrac{z_1z_2z_3-|z_1|^2}{z|z_1|} \in \mathbb{R} \Rightarrow z_1z_2z_3=|z_1|^2$
Hence $|z_1z_2z_3|=|z_1|^3=|z_1|^2 \Rightarrow |z_1|=1$ and the conclusion follows
