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.
 A: Let $r = |z_1| = |z_2| = |z_3| > 0$ be the common modulus and $P = z_1z_2z_3$. Then
$$
 z_1 + z_2 z_3 = \overline{z_1} + \overline{z_2} \cdot \overline {z_3} = \frac{r^2}{z_1} + \frac{r^4}{z_2 z_3} \, .
$$
Multiplication with $z_1$ gives
$$
z_1^2 + P = r^2 + \frac{r^4 z_1^2}{P} 
\implies z_1^2 \left( 1 - \frac{r^4}{P}\right) = r^2 - P \, .
$$
The last identity holds for $z_2$ and $z_3$ as well.
It follows that $P = r^4$ because otherwise $z_1^2=z_2^2=z_3^2$, which is not possible for distinct $z_1, z_2, z_3$.
So we have $z_1z_2z_3 = r^4$. Taking the absolute value implies that $r=1$.
A: My solution relies heavily on the fact that the two sides of a parallelogram are equal if and only if the diagonals of the said parallelogram are orthogonal (when the parallelogram is a rhombus). On the complex plane this can be reformulated as follows: two complex numbers share the same magnitude if and only if their sum and difference vectors are orthogonal to each other.

I will first prove that the common length of the three complex numbers is $=1$. I will show that $|z_3|=1$, the others are done in the same way.
Consider the known real numbers $u=z_1+z_2z_3$ and $v=z_2+z_1z_3$. We have that
$$
u+v=(z_1+z_2)(1+z_3)
$$
and
$$
u-v=(z_1-z_2)(1-z_3)
$$
are also both real. Let us denote $\phi_k=\arg z_k, k=1,2,3$. Because $|z_1|=|z_2|$, the parallelogram with sides $z_1$ and $z_2$ is a rhombus. From this (or by a number of other means) it follows that modulo an integer multiple of $\pi$ we have
$$
\arg(z_1+z_2)\equiv\frac{\phi_1+\phi_2}2
$$
and
$$
\arg(z_1-z_2)\equiv\frac{\pi+\phi_1+\phi_2}2.
$$
In other words, the arguments of these two numbers differ from each other by $\pm\pi/2$.
As $\arg(u+v)\equiv \arg(u-v)\equiv0\pmod\pi$, the arguments of
$1-z_3$ and $1+z_3$ must also differ from each other by $\pm\pi/2$. Therefore the diagonals $1\pm z_3$ of the parallelogram with sides $1$ and $z_3$ are orthogonal, making the said parallelogram a rhombus. Therefore $|z_3|=1$ as claimed.
Another consequence of $|z_3|=1$ is that
$$
\arg(1-z_3)=\frac{\pi+\phi_3}2.
$$
Putting these pieces together we get (modulo $\pi$)
$$
0\equiv\arg(u-v)=\arg(z_1-z_2)+\arg(1-z_3)=\frac{2\pi+\phi_1+\phi_2+\phi_3}2\equiv\frac{\phi_1+\phi_2+\phi_3}2.
$$
So $\phi_1+\phi_2+\phi_3$ is an integer multiple of $2\pi$ and the claim follows.
A: Alternative approach:
$x$ and $y$ say hello from kludgiville.
Given that $$(z_1\overline{z_1}) = (z_2\overline{z_2}) = (z_3\overline{z_3}). \tag1$$
Let $(x_k + iy_k)$ denote $z_k : k \in \{1,2,3\}.$
From the constraints:
$(x_1 + iy_1) + (x_2 + iy_2)(x_3 + iy_3) = a \in \mathbb{R}.$ 
$(x_2 + iy_2) + (x_1 + iy_1)(x_3 + iy_3) = b \in \mathbb{R}.$ 
$(x_3 + iy_3) + (x_1 + iy_1)(x_2 + iy_2) = c \in \mathbb{R}.$
Let $s_1 = (a-z_1), s_2 = (b-z_2), s_3 = (c-z_3).$ 
Then $$(z_1 z_2 z_3) = (s_1 z_1) = (s_2 z_2) = (s_3 z_3).\tag2$$
Therefore: 
$s_1 = (a - z_1) = a - (x_1 + iy_1) = a + (x_1 - iy_1) - 2x_1 = [a + \overline{z} - 2x_1] \implies $ 
$$(s_1 z_1) = (az_1) + (z_1\overline{z_1}) - 2x_1z_1 = (z_1\overline{z_1}) + z_1(a - 2x_1).\tag3$$
Similarly: 
$$(s_2 z_2) = (z_2\overline{z_2}) + z_2(b - 2x_2).$$
$$(s_3 z_3) = (z_3\overline{z_3}) + z_3(c - 2x_3).$$
Using equations (1) and (2) above, this implies that 
$$z_1(a - 2x_1) = z_2(b - 2x_2) = z_3(c - 2x_3).$$
Suppose that $(a - 2x_1) \neq 0.$
$$\text{Then} ~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k$$
$$\text{and} ~z_1 = kz_2 ~\text{and} |z_1| = |z_2|.$$
The only way that this is possible is if $z_1 = z_2$.
This is a contradiction, since it violates the constraint that $z_1, z_2, z_3$ are all distinct.

Edit
The above analysis is definitely the right approach, but contains a flaw.  See the comment/analysis of JyrkiLahtonen immediately following this answer.  Applying his comment/analysis, the explicit remedy is as follows:
Exploring the hypothesis that $(a - 2x_1) \neq 0$, and keeping in mind that none of $z_1,z_2,z_3$ can $= 0$ [else, by equation (1) above, they all $= 0$ and are therefore not distinct], you have that
$$~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_1$$
$$\text{and} ~\frac{c - 2x_3}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_2.$$
This implies that $~z_1 = k_1z_2 = k_2z_3,~~$ with
$~~|z_1| = |z_2| = |z_3|.$
This implies that $~|k_1| = 1 = |k_2|.$
Since, in this hypothetical $~~k_1, k_2 \in \mathbb{R},$ 
this implies that $~k_1,k_2~$ are both $~\in \{+1, -1\}.$
In order for $z_2$ and $z_3$ to both be distinct from $z_1$, it is necessary for $k_1 = -1 = k_2.$  However, in that case, you have that $z_2 = z_3$.
Therefore, the assumption that $(a-2x_1) \neq 0$ implies that $z_1, z_2, z_3$ can not all be distinct.  Therefore, one can conclude that $(a-2x_1) = 0.$

Therefore, from equation (3) above,
$$(z_1 z_2 z_3) = (s_1 z_1) = z_1 \overline{z_1} \implies
(z_2 z_3) = \overline{z_1}. \tag4 $$
Using equation (1) above, this implies that
$$|z_1|^2 = |z_2 z_3| = |\overline{z_1}| = |z_1| \implies |z_1| = 1.$$
Therefore, by equation (4) above, $$(z_1 z_2 z_3) = 1.$$
A: 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
A: Gosh... well, I just looked at the three-way symmetry, noted all points must lie on a circle around the origin, and deduced $z_1 = 1$, $z_2 = e^{2 i \pi/3}$ and $z_3 = e^{-2 i \pi/3}$, and confirmed these worked.

I suppose one could work through the algebra more rigorously, but on an exam... I think this is fine!
