image under a rotation center through an angle of 120°

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orl

3647 posts

orl

#1 h Nov 11, 2005, 8:18 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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.

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ZetaX

7579 posts

ZetaX

#2 h Jan 25, 2006, 7:20 PM • 5 Y

Y by Sskkrr, Adventure10, Mango247, Mango247, and 1 other user

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.

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archimedes1

1482 posts

archimedes1

#3 h Jun 27, 2008, 1:10 AM • 2 Y

Y by Adventure10, Mango247

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

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FOURRIER

727 posts

FOURRIER

#4 h Jun 27, 2008, 12:12 PM • 2 Y

Y by Adventure10, Mango247

There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf

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archimedes1

1482 posts

archimedes1

#5 h Jun 27, 2008, 4:23 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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.
Synthetic

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FOURRIER

727 posts

FOURRIER

#6 h Jun 28, 2008, 2:24 PM • 3 Y

Y by Vietnamisalwaysinmyheart, Adventure10, Mango247

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

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sqing

41881 posts

sqing

#7 h Nov 19, 2018, 4:07 AM • 1 Y

Y by Adventure10

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.

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Sskkrr

1 post

Sskkrr

#8 h Jan 3, 2020, 6:40 AM • 2 Y

Y by Adventure10, Mango247

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

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HamstPan38825

8857 posts

HamstPan38825

#9 h Dec 12, 2022, 6:40 PM

Y by

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.

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AshAuktober

997 posts

AshAuktober

#10 h Mar 6, 2024, 8:18 AM

Y by

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. $\square$

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