Hellenic Mathematical Olympiad

(a) Since $w$ is a root of the equation $z^2 + z + 1 = 0$, we have $w^2 + w + 1 = 0$. Multiplying both parts by $w$:
$$w^3 + w^2 + w = 0 \Leftrightarrow w^3 + \underbrace{w^2 + w + 1}_{0} = 1 \Leftrightarrow w^3 = 1.$$
From the last equation we find $|w| = 1$. Substituting in relation (I) $w^2 = -w - 1$, we find:
$$z_1(-1-w) + z_3w + z_5 = 0 \Leftrightarrow -z_1 - z_1w + z_3w + z_5 = 0 \Leftrightarrow (z_3 - z_1)w = z_1 - z_5.$$
Hence
$$|(z_3 - z_1)w| = |z_1 - z_5| \Leftrightarrow |z_3 - z_1| |w| = |z_1 - z_5| \Leftrightarrow \boxed{|z_3 - z_1| = |z_1 - z_5|} \quad (A).$$
Substituting in relation (I) $w = -w^2 - 1$, we find:
$$z_1 w^2 + z_3(-w^2 - 1) + z_5 = 0 \Leftrightarrow z_1 w^2 - z_3 w^2 - z_3 + z_5 = 0 \Leftrightarrow (z_1 - z_3)w^2 = z_5 - z_3.$$
Hence we have
$$|(z_1 - z_3)w|^2 = |z_5 - z_3|^2 \Leftrightarrow |z_1 - z_3|^2 |w|^2 = |z_5 - z_3|^2 \Leftrightarrow \boxed{|z_3 - z_1| = |z_5 - z_3|} \quad (B).$$
From (A) and (B) we obtain the equalities:
$$|z_1 - z_3| = |z_3 - z_5| = |z_5 - z_1|,$$
that is the triangle $A_1A_3A_5$ is equilateral..

(β) Similarly, using relation (II) we prove that the triangle $A_2A_4A_6$ is equilateral. From a known proposition of Euclidean Geometry we have that $A_1A_2 + A_1A_6 = A_1A_4$, and then using measures of complex numbers we have:

$$|z_1 - z_2| + |z_6 - z_1| = |z_1 - z_4|. \qquad (1)$$
Similarly, from the equality $A_3A_2 + A_3A_4 = A_3A_6$ using measures of complex numbers we get:
$$|z_2 - z_3| + |z_3 - z_4| = |z_3 - z_6|. \qquad (2)$$
Also, from equality $A_5A_4 + A_5A_6 = A_5A_2$ we find:
$$|z_4 - z_5| + |z_5 - z_6| = |z_2 - z_5|. \qquad (3)$$

Summing up by parts the relations (1), (2) and (3) and using the equalities
$$|z_1 - z_4| = |z_3 - z_6| = |z_2 - z_5|$$
we find:
$$|z_1 - z_2| + |z_2 - z_3| + |z_3 - z_4| + |z_4 - z_5| + |z_5 - z_6| + |z_6 - z_1| = \\ 3|z_1 - z_4| = 3|z_2 - z_5| = 3|z_3 - z_6|.$$