$n$ Roots of Unity

A regular polygon with $n$ sides can be divided into $n$ isosceles triangles.

${ w }^{ n } =1 , n\in\mathbb{N}$ has $n$ solutions. It's known that one of them is always $1$ .

$1=cis(0)$ . Other solutions to ${ w }^{ n } =1$ can be calculated through $cis(0+\frac { 2\pi }{ n })$ .

Hence, the iscosceles triangles will have 2 sides with lengh 1, and the angle formed by these two sides will be $\frac { 2\pi }{ n }$ .

${ Area }_{ polygon }=n\quad { Area }_{ triangle }$

${ A }_{ triangle }=\frac { b\times h }{ 2 }$

$\cos(\frac { 2\pi }{ n } \div 2)=\frac { h }{ 1 } \quad \wedge \quad \sin(\frac { 2\pi }{ n } \div 2)=\frac { b\div 2 }{ 1 }$

$h=\cos(\frac { \pi }{ n } )$

$b=2\sin(\frac{\pi}{n})$

${ A }_{ triangle }=\frac { 2\sin\left( \frac { \pi }{ n } \right) \cos\left( \frac { \pi }{ n } \right) }{ 2 } =\frac { \sin\left( \frac { 2\pi }{ n } \right) \ }{2 }$

${ A }_{ polygon }=n\frac { \sin\left( \frac { 2\pi }{ n } \right) \ }{2 } \Leftrightarrow3=n\frac { \sin\left( \frac { 2\pi }{ n } \right) \ }{2 }$

Solving the equation will get us that $n=12 \vee n=-12$ . Since $n \in \mathbb{N}$ , $n=12.$

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The only way i found to solve that last equation was graphically. If anyone can solve it or has another solution, please post it :).

$Area \ of\ a\ regular\ polygon= (½)(apothem)(perimeter).\\ Apothem=RCos\dfrac {180} n, \ \ and\ perimeter=n*(side\ length) =n*R*Sin\dfrac {180} n.\\ \therefore\ For\ an\ n-gon\ with\ circumradius, R=1,\\ Area=½*\{2(1*Cos\dfrac {180} n * n*1*Sin\dfrac {180} n)\}=3.\\ \implies\ ½*n*Sin\dfrac {360} n=3.\\ Solving,\ n=12.$

Polygon is formed from isosceles triangles 2 sides equal to 1 and angle x between. There are n of these so area = n×1×1×sin(x)/2 = 3. sin(x)=6/n. ×=360/n. sin(360/n) = 6/n. Sin(30)=1/2 and so n=12