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Difference between revisions of "2020 INMO Problems/Problem 5"

(Created page with "==Problem== Infinitely many equidistant parallel lines are drawn in the plane. A positive integer <math>n \geqslant 3</math> is called frameable if it is possible to draw a r...")
 
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for  <math>|\frac{p}{q}|</math> we have only two chice that is <math>|\cos \frac{2\pi}{n}| =1,\frac{1}{2}</math>.
 
for  <math>|\frac{p}{q}|</math> we have only two chice that is <math>|\cos \frac{2\pi}{n}| =1,\frac{1}{2}</math>.
  
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So ,for <math>n=2,0,3,6</math> the above is possible. So, for any <math>n\ge 7</math> are<math>\textbf{ not Frameable}</math>.
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So ,for <math>n=2,0,3,6</math> the above is possible. So, for any <math>n\ge 7</math> are<math>\textbf{ not Frameable}</math>. ~Trishan

Revision as of 13:02, 5 November 2020

Problem

Infinitely many equidistant parallel lines are drawn in the plane. A positive integer $n \geqslant 3$ is called frameable if it is possible to draw a regular polygon with $n$ sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.

(a) Show that $3, 4, 6$ are frameable. (b) Show that any integer $n \geqslant 7$ is not frameable. (c) Determine whether $5$ is frameable.

Solution

At first of all suppose $\Omega$ be the plane in complex plane consists all equidistant parallel lines . [list] [*] Suppose the parallel lines are parallel to the real axis. This is because we always get a transformation $T(ax+by-c)\to e +si$ where $s$ is a real constant .

[*] also suppose two consecutive lines are one unit apart to each other .


[*] Suppose $P_0,\cdots ,P_{n-1}$ are n ($n>2$)points of n sided regular polygon . Suppose $P_i =z .\zeta ^i$ for all $i \in \{0,1,2,\cdots ,n -1\}$. Where $z$ is a complex number with $z=x+yi$

Also denote the parallel lines as $l_i$ iff $P_i \in l_i$

So we can see that $l_k \equiv \ Im ({P_k})=y \cos \frac{2\pi k}{n} +x \sin \frac{2\pi k}{n}$ .

So, $l_0 \equiv y \cdots (1)$ .

$l_1 \equiv y \cos \frac{2\pi }{n} +x \sin \frac{2\pi }{n} \cdots (2)$

$l_{n-1} \equiv y \cos \frac{2\pi }{n} -x \sin \frac{2\pi }{n} \cdots (3)$.

From $(2),(3)$ get ,

$x\sin \frac{2\pi}{n} ,y \cos \frac{2\pi}{n}$ both are rational number .

So, from $(2),(1)$ get $y$ is also a rational number .

$\implies \cos \frac{2\pi}{n}$ is rational.

$\textbf{ Part (a)}$

We get $\cos \frac{2\pi }{3} =\frac{-1}{2} \in \mathbb{Q}$ and similarly we get $\cos \frac{2\pi}{n}$ is rational for $n=4,6$ .

So $n=3,4,6$ are $\textbf{Frameable}$ .

$\underline{\text{Part (c)}}$

We get $\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{2} \in (\mathbb{R}- \mathbb{Q})$.

So ,$n=5$ is not $\textbf{Frameable}$.

$\underline{\textbf{ Part (b)}}$.

$\textbf{Claim:}$ 

we always get a polynomial $P_n$ ($n\ge 1$)with integer cofficient such that $P_n(2\cos t) = 2 \cos nt$. In general $\deg (P_n)=n$ and its a Monic polynomial.

$\textbf{Proof.}$

It can be done by simple induction .

We can get some examples for $n=1,2$.


[*] $P_1(x)=x$ .

[*] $P_2(x)=x^2-2$ and $P_2(2 \cos t) =2 \cos 2t$.


Note that $2\cos (kx) (2\cos x) - 2\cos ((k-1) x)$.

$\implies \boxed{ P_{k+1} (x)= x P_k(x) -P_{k-1}(x)}$.

Here $\deg (P_{k+1} ) =k+1$. So our claim is proved $\blacksquare$.

Now $P_n(2\cos \frac{2\pi}{n})=2\cos (2\pi)=2$.

If $2\cos \frac{2\pi} {n} =\frac{p}{q}$ is rational then $P_n-2$ has a rational root then ,

$| p| \mid 2$ and $|q| \mid 1$ so,

for $|\frac{p}{q}|$ we have only two chice that is $|\cos \frac{2\pi}{n}| =1,\frac{1}{2}$.

So ,for $n=2,0,3,6$ the above is possible. So, for any $n\ge 7$ are$\textbf{ not Frameable}$. ~Trishan

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