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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.

$$\mathrm{Adding}\:\mathrm{answer}\:\left(\mathrm{without}\:\mathrm{proof}\right)\:\mathrm{for}\:\mathrm{completness}. \\ $$$$\mathrm{Prime}\:\mathrm{number}\:\mathrm{polygon}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{iff}\:{p}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{2}^{\mathrm{2}^{{n}} } −\mathrm{1}. \\ $$$$\mathrm{Composite}\:\mathrm{number}\:{n}\:\mathrm{polygon}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed} \\ $$$$\mathrm{if}\:{n}\:\mathrm{takes}\:\mathrm{the}\:\mathrm{form}\:=\mathrm{2}^{{k}} {p}_{\mathrm{1}} \centerdot{p}_{\mathrm{2}} \centerdot.. \\ $$$$\mathrm{where}\:{p}_{{i}} \:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{2}^{\mathrm{2}^{{n}} } −\mathrm{1}. \\ $$$$\mathrm{Note}\:\mathrm{that}\:\mathrm{composite}\:\mathrm{numbers}\:\mathrm{factor}\:\mathrm{can} \\ $$$$\mathrm{include}\:\mathrm{each}\:{p}_{{i}} \:\mathrm{only}\:\mathrm{once}. \\ $$$$\mathrm{3}=\mathrm{2}^{\mathrm{2}^{\mathrm{1}} } −\mathrm{1}\Rightarrow\mathrm{construtible}. \\ $$$$\mathrm{7}\:\mathrm{not}\:\mathrm{constructible} \\ $$$$\mathrm{9}=\mathrm{3}×\mathrm{3}\:\mathrm{not}\:\mathrm{constructible}. \\ $$

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