1986 IMO Problems/Problem 2

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 $P_1,P_2,P_3,...$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^\circ$ clockwise for $k=0,1,2,...$ . Prove that if $P_{1986}=P_0$ , then the triangle $A_1A_2A_3$ is equilateral.

Solution

Consider the triangle and the points on the complex plane. Without loss of generality, let $A_1=0$ , $A_2=1$ , and $A_3=a$ for some complex number $a$ . Then, a rotation about $A_1$ of $120^\circ$ sends point $P$ to point $e^{\frac{4i\pi}{3}}P$ . For $A_2$ , the rotation sends $P$ to $e^{\frac{4i\pi}{3}}(P-1)+1$ and for $A_3$ the rotation sends $P$ to $e^{\frac{4i\pi}{3}}(P-a)+a$ . Thus the result of all three rotations sends $P$ to

$e^{\frac{4i\pi}{3}}(e^{\frac{4i\pi}{3}}(e^{\frac{4i\pi}{3}}P-1)+1-a)+a$

$=P-e^{\frac{2i\pi}{3}}+e^{\frac{4i\pi}{3}}-ae^{\frac{4i\pi}{3}}+a$

$=P-i\sqrt{3}-ae^{\frac{4i\pi}{3}}+a$

Since the transformation $P\to P-i\sqrt{3}-ae^{\frac{4i\pi}{3}}+a$ occurs $1986/3=662$ times, to obtain $P_{1986}$ . But, we have $P_{1986}=P_0$ and so we have

$P_0=P_0+662(-i\sqrt{3}-ae^{\frac{4i\pi}{3}}+a)$

$\Rightarrow -i\sqrt{3}-ae^{\frac{4i\pi}{3}}+a=0$

$\Rightarrow a(-e^{\frac{4i\pi}{3}}+1)=i\sqrt{3}$

$\Rightarrow a=\frac{i\sqrt{3}}{-e^{\frac{4i\pi}{3}}+1}$

$\Rightarrow a=\frac{1}{2}+i\frac{\sqrt{3}}{2}$

Now it is clear that the triangle $A_1A_2A_3$ is equilateral. Shen kislay kai

1986 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1986 IMO Problems/Problem 2

Solution

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