26th 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 $w$: $$w^3+w^2+w=0\iff w^3+\underset{0}{\underbrace{w^2+w+1}}=1\iff w^3=1.$$ From the last equation we find $|w|=1$. Substituting in relation (I) $w^2=-w-1$, we find: $$z_1(-1-w)+z_{3}w+z_5=0\iff -z_1-z_{1}w+z_{3}w+z_5=0\iff (z_3-z_1)w=z_1-z_5.$$ Hence $$|(z_3-z_1)w|=|z_1-z_5|\iff |z_3-z_1||w|=|z_1-z_5|\iff \boxed{|z_3-z_1|=|z_1-z_5|}(A).$$ Substituting in relation (I) $w=-w^2-1$, we find: $$z_1{w}^2+z_3(-w^2-1)+z_5=0\iff z_1{w}^2-z_3{w}^2-z_3+z_5=0\iff (z_1-z_3)w^2=z_5-z_3.$$ Hence we have $$|(z_1-z_3)w{|}^2=|z_5-z_3{|}^2\iff |z_1-z_3{|}^2|w{|}^2=|z_5-z_3{|}^2\iff \boxed{|z_3-z_1|=|z_5-z_3|}(B).$$ From (A) and (B) we obtain the equalities: $$|z_1-z_3|=|z_3-z_5|=|z_5-z_1|,$$ that is the triangle $A_1{A}_3{A}_5$ is equilateral..

(β) Similarly, using relation (II) we prove that the triangle $A_2{A}_4{A}_6$ is equilateral. From a known proposition of Euclidean Geometry we have that $A_1{A}_2+A_1{A}_6=A_1{A}_4$, and then using measures of complex numbers we have:

$$|z_1-z_2|+|z_6-z_1|=|z_1-z_4|.(1)$$ Similarly, from the equality $A_3{A}_2+A_3{A}_4=A_3{A}_6$ using measures of complex numbers we get: $$|z_2-z_3|+|z_3-z_4|=|z_3-z_6|.(2)$$ Also, from equality $A_5{A}_4+A_5{A}_6=A_5{A}_2$ we find: $$|z_4-z_5|+|z_5-z_6|=|z_2-z_5|.(3)$$ 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: $$\begin{gathered} |z_1-z_2|+|z_2-z_3|+|z_3-z_4|+|z_4-z_5|+|z_5-z_6|+|z_6-z_1|= \\ 3|z_1-z_4|=3|z_2-z_5|=3|z_3-z_6|. \end{gathered}$$