Difference between revisions of "2020 INMO Problems/Problem 5"
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(a) Show that <math>3, 4, 6</math> are frameable. | (a) Show that <math>3, 4, 6</math> are frameable. | ||
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(b) Show that any integer <math>n \geqslant 7</math> is not frameable. | (b) Show that any integer <math>n \geqslant 7</math> is not frameable. | ||
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(c) Determine whether <math>5</math> is frameable. | (c) Determine whether <math>5</math> is frameable. | ||
Latest revision as of 02:12, 22 September 2023
Problem
Infinitely many equidistant parallel lines are drawn in the plane. A positive integer is called frameable if it is possible to draw a regular polygon with sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.
(a) Show that are frameable.
(b) Show that any integer is not frameable.
(c) Determine whether is frameable.
Solution
At first of all suppose 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 where is a real constant .
[*] also suppose two consecutive lines are one unit apart to each other .
[*] Suppose are n ()points of n sided regular polygon . Suppose for all . Where is a complex number with
Also denote the parallel lines as iff
So we can see that .
So, .
.
From get ,
both are rational number .
So, from get is also a rational number .
is rational.
We get and similarly we get is rational for .
So are .
We get .
So , is not .
.
we always get a polynomial ()with integer cofficient such that . In general and its a Monic polynomial.
It can be done by simple induction .
We can get some examples for .
[*] .
[*] and .
Note that .
.
Here . So our claim is proved .
Now .
If is rational then has a rational root then ,
and so,
for we have only two chice that is .
So ,for the above is possible. So, for any are. ~Trishan