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Difference between revisions of "1986 IMO Problems/Problem 2"
 
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Now it is clear that the triangle <math>A_1A_2A_3</math> is equilateral.
 
Now it is clear that the triangle <math>A_1A_2A_3</math> is equilateral.
 +
Shen kislay kai
  
 
== See Also == {{IMO box|year=1986|num-b=1|num-a=3}}
 
== See Also == {{IMO box|year=1986|num-b=1|num-a=3}}

Latest revision as of 12:05, 3 September 2024

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

See Also

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