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 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 \dots . (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, \dots can be constructed

4 \Rightarrow 4, 8, 16, 32, 64, \dots can be constructed

5 \Rightarrow 5, 10, 20, \dots 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

\boldsymbol LOT o f \boldsymbol TH α \boldsymbol 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 .