Menu Close

n-is-a-number-such-that-regular-n-gon-is-possible-with-straightedge-and-compass-only-Write-first-thirty-values-of-n-What-are-other-properties-of-such-numbers-If-values-of-n-are-arranged-in-or




Question Number 3519 by Rasheed Soomro last updated on 14/Dec/15
n is a number such that regular  n−gon is  possible with straightedge and  compass only.  ∗Write first thirty values of n.  ∗What are other properties of such numbers ?  ∗If values of n are arranged in order, what is  the formula for generating Nth number?
$${n}\:{is}\:{a}\:{number}\:{such}\:{that}\:{regular}\:\:{n}−{gon}\:{is} \\ $$$${possible}\:{with}\:{straightedge}\:{and}\:\:{compass}\:{only}. \\ $$$$\ast{Write}\:{first}\:{thirty}\:{values}\:{of}\:{n}. \\ $$$$\ast{What}\:{are}\:{other}\:{properties}\:{of}\:{such}\:{numbers}\:? \\ $$$$\ast{If}\:{values}\:{of}\:{n}\:{are}\:{arranged}\:{in}\:{order},\:{what}\:{is} \\ $$$${the}\:{formula}\:{for}\:{generating}\:{Nth}\:{number}? \\ $$
Commented by prakash jain last updated on 14/Dec/15
A regular n−gon is contructible with ruler  and compass if and only if sine of internal  angle x is a solution of a quadratic with  integer coefficients.
$$\mathrm{A}\:\mathrm{regular}\:{n}−{gon}\:\mathrm{is}\:\mathrm{contructible}\:\mathrm{with}\:\mathrm{ruler} \\ $$$$\mathrm{and}\:\mathrm{compass}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{sine}\:\mathrm{of}\:\mathrm{internal} \\ $$$$\mathrm{angle}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{with} \\ $$$$\mathrm{integer}\:\mathrm{coefficients}. \\ $$
Commented by prakash jain last updated on 14/Dec/15
3−60°−ok, 4−90°−ok,5−108°−ok,6−120°−ok  8−135°−ok,10−8×18°−ok,12−150°−ok  9−140°−not ok−requires drawing of 10°  7−((5×180)/7)−to be checked  11−((9×180)/(11))−to be checked  Each angle can be individual checked. I will  check if a general formula for n can be derived.
$$\mathrm{3}−\mathrm{60}°−\mathrm{ok},\:\mathrm{4}−\mathrm{90}°−\mathrm{ok},\mathrm{5}−\mathrm{108}°−\mathrm{ok},\mathrm{6}−\mathrm{120}°−\mathrm{ok} \\ $$$$\mathrm{8}−\mathrm{135}°−\mathrm{ok},\mathrm{10}−\mathrm{8}×\mathrm{18}°−\mathrm{ok},\mathrm{12}−\mathrm{150}°−\mathrm{ok} \\ $$$$\mathrm{9}−\mathrm{140}°−\mathrm{not}\:\mathrm{ok}−\mathrm{requires}\:\mathrm{drawing}\:\mathrm{of}\:\mathrm{10}° \\ $$$$\mathrm{7}−\frac{\mathrm{5}×\mathrm{180}}{\mathrm{7}}−{to}\:{be}\:{checked} \\ $$$$\mathrm{11}−\frac{\mathrm{9}×\mathrm{180}}{\mathrm{11}}−{to}\:{be}\:{checked} \\ $$$$\mathrm{Each}\:\mathrm{angle}\:\mathrm{can}\:\mathrm{be}\:\mathrm{individual}\:\mathrm{checked}.\:\mathrm{I}\:\mathrm{will} \\ $$$$\mathrm{check}\:\mathrm{if}\:\mathrm{a}\:\mathrm{general}\:\mathrm{formula}\:\mathrm{for}\:{n}\:\mathrm{can}\:\mathrm{be}\:\mathrm{derived}. \\ $$
Commented by prakash jain last updated on 14/Dec/15
If let us p−gon is contructible than n−gon  is contructible if   ((2π)/n)=(1/2^k )((2π)/p)   or n=2^k p  ....(A)  Since we can contruct p−gon, we can draw  (((p−2)π)/p). Hence we can draw ((2π)/p).  since we can bisect an angle we can draw  (1/2^k )∙((2π)/p)⇒we can draw ((2π)/n)⇒we can draw π−((2π)/n)  ⇒we can contruct 2^k ∙p polygon.  given that we can contruct   3 ⇒6,12,24,48,... can be constructed  4⇒4,8,16,32,64,... can be constructed  5⇒5,10,20,... can be contructed.
$$\mathrm{If}\:\mathrm{let}\:\mathrm{us}\:{p}−\mathrm{gon}\:\mathrm{is}\:\mathrm{contructible}\:\mathrm{than}\:{n}−{gon} \\ $$$$\mathrm{is}\:\mathrm{contructible}\:\mathrm{if}\: \\ $$$$\frac{\mathrm{2}\pi}{{n}}=\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\frac{\mathrm{2}\pi}{{p}}\:\:\:\mathrm{or}\:{n}=\mathrm{2}^{{k}} {p}\:\:….\left({A}\right) \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{can}\:\mathrm{contruct}\:{p}−{gon},\:\mathrm{we}\:\mathrm{can}\:\mathrm{draw} \\ $$$$\frac{\left({p}−\mathrm{2}\right)\pi}{{p}}.\:\mathrm{Hence}\:\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\frac{\mathrm{2}\pi}{{p}}. \\ $$$$\mathrm{since}\:\mathrm{we}\:\mathrm{can}\:\mathrm{bisect}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{we}\:\mathrm{can}\:\mathrm{draw} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\centerdot\frac{\mathrm{2}\pi}{{p}}\Rightarrow\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\frac{\mathrm{2}\pi}{{n}}\Rightarrow\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\pi−\frac{\mathrm{2}\pi}{{n}} \\ $$$$\Rightarrow\mathrm{we}\:\mathrm{can}\:\mathrm{contruct}\:\mathrm{2}^{{k}} \centerdot{p}\:{polygon}. \\ $$$${given}\:{that}\:{we}\:{can}\:{contruct}\: \\ $$$$\mathrm{3}\:\Rightarrow\mathrm{6},\mathrm{12},\mathrm{24},\mathrm{48},…\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{4}\Rightarrow\mathrm{4},\mathrm{8},\mathrm{16},\mathrm{32},\mathrm{64},…\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{5}\Rightarrow\mathrm{5},\mathrm{10},\mathrm{20},…\:\mathrm{can}\:\mathrm{be}\:\mathrm{contructed}. \\ $$
Commented by prakash jain last updated on 14/Dec/15
Also we can state that if a polygon p  cannot be drawn than than n gon such  that n=kp cannot be drawn.  Assume n−gon can be drawn  ((2π)/n) can be drawn⇒((2π)/(kp)) can be drawn  ⇒k(((2π)/(kp))) can be drawn by drawnin ((2π)/(kp))   mulitple times⇒((2π)/p) can be drawn⇒  π−((2π)/p) = (((p−2)π)/p) can be drawn contradicts  that p−gon cannot be drawn.  Hence if p−gon cannot be drawn than any  n−gon such that n=kp cannot be drawn.  Note that it does not mean if p can be  drawn kp can be drawn. It is only for  exclusion.
$$\mathrm{Also}\:\mathrm{we}\:\mathrm{can}\:\mathrm{state}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{polygon}\:{p} \\ $$$${cannot}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{than}\:\mathrm{than}\:{n}\:\mathrm{gon}\:\mathrm{such} \\ $$$$\mathrm{that}\:{n}={kp}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}. \\ $$$$\mathrm{Assume}\:{n}−{gon}\:{can}\:{be}\:{drawn} \\ $$$$\frac{\mathrm{2}\pi}{{n}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\Rightarrow\frac{\mathrm{2}\pi}{{kp}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn} \\ $$$$\Rightarrow{k}\left(\frac{\mathrm{2}\pi}{{kp}}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:{by}\:{drawnin}\:\frac{\mathrm{2}\pi}{{kp}}\: \\ $$$${mulitple}\:{times}\Rightarrow\frac{\mathrm{2}\pi}{{p}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\Rightarrow \\ $$$$\pi−\frac{\mathrm{2}\pi}{{p}}\:=\:\frac{\left({p}−\mathrm{2}\right)\pi}{{p}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{contradicts} \\ $$$$\mathrm{that}\:{p}−{gon}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}. \\ $$$$\mathrm{Hence}\:\mathrm{if}\:{p}−{gon}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{than}\:\mathrm{any} \\ $$$${n}−{gon}\:\mathrm{such}\:\mathrm{that}\:{n}={kp}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}. \\ $$$$\mathrm{Note}\:\mathrm{that}\:\mathrm{it}\:\mathrm{does}\:\mathrm{not}\:\mathrm{mean}\:\mathrm{if}\:{p}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{drawn}\:{kp}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{only}\:\mathrm{for} \\ $$$$\mathrm{exclusion}. \\ $$
Commented by Rasheed Soomro last updated on 14/Dec/15
LOT    of   THαNKS !
$$\boldsymbol{\mathcal{LOT}}\:\:\:\:{of}\:\:\:\boldsymbol{\mathcal{TH}}\alpha\boldsymbol{\mathcal{NKS}}\:! \\ $$
Answered by prakash jain last updated on 14/Dec/15
Continuing from comments we now  need to find.  a. which prime numbers polygons       cannot be drawn  b. properties of prime number polygons which can be       drawn  c. properties of composite number which can       be drawn.  Continue
$$\mathrm{Continuing}\:\mathrm{from}\:\mathrm{comments}\:\mathrm{we}\:\mathrm{now} \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{find}. \\ $$$$\mathrm{a}.\:\mathrm{which}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{polygons}\: \\ $$$$\:\:\:\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn} \\ $$$$\mathrm{b}.\:\mathrm{properties}\:\mathrm{of}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{polygons}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be} \\ $$$$\:\:\:\:\:\mathrm{drawn} \\ $$$$\mathrm{c}.\:\mathrm{properties}\:\mathrm{of}\:\mathrm{composite}\:\mathrm{number}\:\mathrm{which}\:\mathrm{can} \\ $$$$\:\:\:\:\:\mathrm{be}\:\mathrm{drawn}. \\ $$$$\mathrm{Continue} \\ $$
Commented by prakash jain last updated on 15/Dec/15
Adding answer (without proof) for completness.  Prime number polygon can be constructed  iff p is of the form 2^2^n  −1.  Composite number n polygon can be constructed  if n takes the form =2^k p_1 ∙p_2 ∙..  where p_i  is of the form 2^2^n  −1.  Note that composite numbers factor can  include each p_i  only once.  3=2^2^1  −1⇒construtible.  7 not constructible  9=3×3 not constructible.
$$\mathrm{Adding}\:\mathrm{answer}\:\left(\mathrm{without}\:\mathrm{proof}\right)\:\mathrm{for}\:\mathrm{completness}. \\ $$$$\mathrm{Prime}\:\mathrm{number}\:\mathrm{polygon}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{iff}\:{p}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{2}^{\mathrm{2}^{{n}} } −\mathrm{1}. \\ $$$$\mathrm{Composite}\:\mathrm{number}\:{n}\:\mathrm{polygon}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{if}\:{n}\:\mathrm{takes}\:\mathrm{the}\:\mathrm{form}\:=\mathrm{2}^{{k}} {p}_{\mathrm{1}} \centerdot{p}_{\mathrm{2}} \centerdot.. \\ $$$$\mathrm{where}\:{p}_{{i}} \:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{2}^{\mathrm{2}^{{n}} } −\mathrm{1}. \\ $$$$\mathrm{Note}\:\mathrm{that}\:\mathrm{composite}\:\mathrm{numbers}\:\mathrm{factor}\:\mathrm{can} \\ $$$$\mathrm{include}\:\mathrm{each}\:{p}_{{i}} \:\mathrm{only}\:\mathrm{once}. \\ $$$$\mathrm{3}=\mathrm{2}^{\mathrm{2}^{\mathrm{1}} } −\mathrm{1}\Rightarrow\mathrm{construtible}. \\ $$$$\mathrm{7}\:\mathrm{not}\:\mathrm{constructible} \\ $$$$\mathrm{9}=\mathrm{3}×\mathrm{3}\:\mathrm{not}\:\mathrm{constructible}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *