Write $x= A_1-A_2 , y= A_2-A_3 , z= A_3-A_1$. The mappings are (with $\zeta$ being the 'first' third root of unity): $P \mapsto \zeta (P-A_1)+A_1$ $P \mapsto \zeta (P-A_2)+A_2$ $P \mapsto \zeta (P-A_3)+A_3$ and using them in a row gives (easy to calculate out): $P \mapsto P + \zeta^2 x + \zeta y + z$ Using the chain of mappings $1986= 3 \cdot 662$ times is the same as using that mapping $662$ times, thus the mapping $P \mapsto P+ 662(\zeta^2 x + \zeta y + z)$ Now the last shall equal $P$, so $\zeta^2 x + \zeta y + z = 0$, which means (by subtracting $x+y+z=0$) that $(\zeta^2-1) x +(\zeta-1) y =0 \iff x= \zeta y$. Similary, $z=\zeta x$, thus $|x|=|y|=|z|$, thus the triangle is equilateral.

Does anyone have a synthetic solution or know if a nice one exists?

There's one here, but in french p.20 http://www.animath.fr/cours/deho_geo/deho_geo.pdf

FOURRIER wrote: There's one here, but in french p.20 http://www.animath.fr/cours/deho_geo/deho_geo.pdf Thank you very much! Luckily "Math French" is a little similar to "math English." So I think I have understood the solution, and this is not a literal translation from that file.

I think the solution found by Zetax is shorter and better (I also did like him).

orl wrote: Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral. Professor Chang Gengzhe, one of the propositional person of this problem, died in Beijing on November 18, 2018.

FOURRIER wrote: There's one here, but in french p.20 http://www.animath.fr/cours/deho_geo/deho_geo.pdf Sir can you please repost this link because this link is dead when you open this there is no solution of above problem.kindly give me other link for this solution

Let $a_1, a_2, a_3, p_0$ be the complex numbers corresponding to their points in the complex plane, and set $\omega = e^{2i \pi/3}$. We can compute via a series of rotations that$$p_3 = p_0 - a_1 + \omega^2 a_1 + \omega a_2 - \omega a_3 - \omega^2 a_2 + a_3.$$In other words, we must have$$a_3-a_1+\omega^2(a_1-a_2) + \omega(a_2-a_3) = 0$$in order to have $P_{1986} = P_0$ because $3 \mid 1986$. Now, consider the polynomial$$x^2(a_1-a_2) + x(a_2 - a_3) + (a_3 - a_1) = 0.$$It has roots $\omega$ and $1$, so by Vieta's,$$\frac{a_3 - a_1}{a_1 - a_2} = \omega,$$which implies that $a_1, a_2, a_3$ are the vertices of an equilateral triangle, as required.

Observe that by the theory of rotation composition, we have that $P_{k+3}$ is a translation of $P_k \forall k \ge 0$. But since $P_{1986}$ is reached by performing $662$ such translations, and $P_0 = P_{1986}$, we have that $P_k = P_{k+3} \forall k \ge 0$. Now construct an equilateral triangle $A_1A_2B$ such that $B$ is on the same side of line $A_1A_2$ as $A_3$. Then clearly if $P_0 = B$, we also have $P_2 = B$.Therefore $P_2 = P_3 = B$, but this is only possible if $B = A_3$. Since $\Delta A_1A_2B$ is equilateral by definition, we are done. $\square$