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?


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Question Number 3519 by Rasheed Soomro last updated on 14/Dec/15

$n i s a n u m b e r s u c h t h a t r e g u l a r n - g o n i s$ $p o s s i b l e w i t h s t r a i g h t e d g e a n d c o m p a s s o n l y .$ $* W r i t e f i r s t t h i r t y v a l u e s o f n .$ $* W h a t a r e o t h e r p r o p e r t i e s o f s u c h n u m b e r s ?$ $* I f v a l u e s o f n a r e a r r a n g e d i n o r d e r, w h a t i s$ $t h e f o r m u l a f o r g e n e r a t i n g N t h n u m b e r ?$
Commented by prakash jain last updated on 14/Dec/15

$A regular n - g o n is contructible with ruler$ $and compass if and only if sine of internal$ $angle x is a solution of a quadratic with$ $integer coefficients .$
Commented by prakash jain last updated on 14/Dec/15

$3 - 60 ° - ok, 4 - 90 ° - ok, 5 - 108 ° - ok, 6 - 120 ° - ok$ $8 - 135 ° - ok, 10 - 8 \times 18 ° - ok, 12 - 150 ° - ok$ $9 - 140 ° - not ok - requires drawing of 10 °$ $7 - \frac{5 \times 180}{7} - t o b e c h e c k e d$ $11 - \frac{9 \times 180}{11} - t o b e c h e c k e d$ $Each angle can be individual checked . I will$ $check if a general formula for n can be derived .$
Commented by prakash jain last updated on 14/Dec/15

$If let us p - gon is contructible than n - g o n$ $is contructible if$ $\frac{2 π}{n} = \frac{1}{2^{k}} \frac{2 π}{p} or n = 2^{k} p . . . . (A)$ $Since we can contruct p - g o n, we can draw$ $\frac{(p - 2) π}{p} . Hence we can draw \frac{2 π}{p} .$ $since we can bisect an angle we can draw$ $\frac{1}{2^{k}} \cdot \frac{2 π}{p} \Rightarrow we can draw \frac{2 π}{n} \Rightarrow we can draw π - \frac{2 π}{n}$ $\Rightarrow we can contruct 2^{k} \cdot p p o l y g o n .$ $g i v e n t h a t w e c a n c o n t r u c t$ $3 \Rightarrow 6, 12, 24, 48, . . . can be constructed$ $4 \Rightarrow 4, 8, 16, 32, 64, . . . can be constructed$ $5 \Rightarrow 5, 10, 20, . . . can be contructed .$
Commented by prakash jain last updated on 14/Dec/15

$Also we can state that if a polygon p$ $c a n n o t be drawn than than n gon such$ $that n = k p cannot be drawn .$ $Assume n - g o n c a n b e d r a w n$ $\frac{2 π}{n} can be drawn \Rightarrow \frac{2 π}{k p} can be drawn$ $\Rightarrow k (\frac{2 π}{k p}) can be drawn b y d r a w n i n \frac{2 π}{k p}$ $m u l i t p l e t i m e s \Rightarrow \frac{2 π}{p} can be drawn \Rightarrow$ $π - \frac{2 π}{p} = \frac{(p - 2) π}{p} can be drawn contradicts$ $that p - g o n cannot be drawn .$ $Hence if p - g o n cannot be drawn than any$ $n - g o n such that n = k p cannot be drawn .$ $Note that it does not mean if p can be$ $drawn k p can be drawn . It is only for$ $exclusion .$
Commented by Rasheed Soomro last updated on 14/Dec/15

$LOT o f TH α NKS!$
	Answered by prakash jain last updated on 14/Dec/15

	$Continuing from comments we now$ $need to find .$ $a . which prime numbers polygons$ $cannot be drawn$ $b . properties of prime number polygons which can be$ $drawn$ $c . properties of composite number which can$ $be drawn .$ $Continue$
	Commented by prakash jain last updated on 15/Dec/15

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