Regular 2^k-gon and reflections

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Regular 2^k-gon and reflections

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Source: Iran TST 2012-Third exam-1st day-P1

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#1 h May 15, 2012, 10:34 AM • 7 Y

Y by hadikh, mahanmath, Amir Hossein, mijail, Adventure10, Mango247, and 1 other user

Consider a regular $2^k$ -gon with center $O$ and label its sides clockwise by $l_1,l_2,...,l_{2^k}$ . Reflect $O$ with respect to $l_1$ , then reflect the resulting point with respect to $l_2$ and do this process until the last side. Prove that the distance between the final point and $O$ is less than the perimeter of the $2^k$ -gon.

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#2 h May 22, 2012, 11:52 AM • 5 Y

Y by a00012025, mijail, Adventure10, Mango247, and 1 other user

Suppose, $A_1A_2.....A_{2^k}$ (clockwise) is the regular polygon. $A_iA_{i+1}\equiv l_i$ . Now note that reflections on $A_1A_2,A_2A_3$ is equivalent to the clockwise rotation centered at $A_2$ that maps $A_2A_{2^k}$ to $A_2A_4$ and similar for the next pairs of reflections. So we consider the movement of $A_{2^k}$ in this sequence of rotations centered at $A_2,A_4,....,A_{2^k}$ . Clearly, at the first rotation $A_2A_{2^k}$ maps to $A_2A_4$ and the next rotation maps it to $A_4A_6$ and so on. So at the end it comes back to $A_2A_{2^k}$ again. Now note that in this way we also get that if $A_{2^k}$ maps to $P$ finally. Then $A_{2^k}P$ is equal to the perimeter of $A_2A_4....A_{2^k}$ . Also note that, the total rotation is by $0$ degree. So if image of $O$ is $O'$ after all these rotations, then $OO'=A_{2^k}P$ which is clearly less than the perimeter of the $2^k$ -gon.

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#3 h Apr 23, 2013, 11:32 AM • 2 Y

Y by Adventure10, Mango247

Suppose $z_2$ be reflection of $z_1$ in line $a_1a_2$ . Then it’s easy to see, $z_2-a_1=e^{2i\theta}(\frac{|z_1|^2}{z_1}-\frac {|a_1|^2}{a_1})$ . Where $\theta=arg(a_2-a_1)$ . Now in general, if $z_{n+1}$ is reflection of $z_n$ in line $a_na_{n+1}$ then, $z_{n+1}=a_n+ e^{2i\theta(n)}(\frac{|z_n|^2}{z_n}-\frac {|a_n|^2}{a_n})$ .
Where $\theta(n)=arg(a_{n+1}-a_n)$ .Now take $O=0$ and radius is one.
So take $a_n=e^{i\alpha(n)}$ where $\alpha(n)=\frac {(n-1)\pi}{2^{k-1}}$ .
Assuming $z_n=x_n+iy_n$ for all $n$ we’ve $x_{n+1}-Cos(\alpha(n))=Cos(2\theta(n))(x_n-Cos(\alpha(n)))$ and $y_{n+1}-Cos(\alpha(n))=-Cos(2\theta(n))(y_n-Cos(\alpha(n)))$ .
Now basically see,
$\theta(n)=tan^{-1}(\frac {Sin(\alpha_{n+1})-Sin(\alpha(n))}{Cos(\alpha_{n+1})-Cos(\alpha(n))}$ . And so, $\theta(n)=\frac {\pi+(2n-1)\pi/(2^{k-1})}{2}$ .So $Cos(2\theta(n))=-Cos(\frac {(2n-1)\pi}{2^k})$ .
So now we’ve $x_{n+1}=Cos(\frac {n-1\pi}{2^{k-1}}) -Cos(\frac {(2n-1)\pi}{2^k})(x_n- Cos(\frac {n-1\pi}{2^{k-1}}) )$ . Now taking, $f(n)= Cos(\frac {n-1\pi}{2^{k-1}})-Cos(\frac {n-1\pi}{2^{k-1}})$ and,
$g(n)= Cos(\frac {(n-1)\pi}{2^{k-1}})\times Cos(\frac {(n-1)\pi}{2^{k-1}})$ we’ve $x_{n+1}=f(n)x_n+g(n)$ .
Now, just easy induction gives,
$x_{n+1}=f(n-1)f(n-2).....f(1)g(0)+f(n-1)...f(2)g(1)+...f(n-1)g(n-2)+g(n-1)$ .
Also we get similarly,
$y_{n+1}= f'(n-1)f'(n-2)...f'(1)g(0)+f'(n-1)..f'(2)g(1)+...f'(n-1)g(n-2)-g(n-1)$ ,
where $f'(n)=- Cos(\frac {n-1\pi}{2^{k-1}})-Cos(\frac {n-1\pi}{2^{k-1}})$ .
So now we’ve $x^2_n+y^2_n$ for all $n$ . Using this it’s not difficult to show $x^2_{2^k}+y^2_{2^k}\geq (2^{k+1}Sin(\frac {\pi}{2^{k-1}}))^2$ . ( Hint to prove, induction on $k$ ).

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