A regular polygon [closed]

In a regular polygon : $ABCDEFGHIJKL$ that has 12 sides .How to find $$\frac{AB}{AF} + \frac{AF}{AB}$$

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Unsourced, unmotivated cut'n'paste problems with no sign of effort on the part of the poster don't go over so well here. Please edit in accord. –  Gerry Myerson Jun 27 at 9:01

closed as off-topic by O.L., Micah, Adriano, Dominic Michaelis, Danny Cheuk Jul 24 at 4:38

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Inscribe the Dodecagon into a Circle. Now, in $\triangle AGF$, $\angle AFG = 90, \angle AGF = 75$. Because, any inscribed angle over a circle diameter is 90 and each triangle in a Dodecagon is 150, for example $\angle HGF = 150$.

Now, $\frac{AF}{GF}$ = tan (75)

So,$\frac{AF}{AB}$ = tan (75); [AB = GF].

So, the answer is: $tan(75) + \frac{1}{tan(75)} = 4$.

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For a polygon {2p}, one can find the odd chords by adding successive chords of {p}, so eg

$AB = 0+1 = 1$, $AD = 1+\sqrt{3}$, $AF = \sqrt{3}+2$. The even chords belong to an inscribed hexagon, the edge being the shortchord AC of the dodecagon. One can find this here by multiplying $AD$ by $\sqrt{2}$, and then dividing by the diameter of the hexagon $2$.

It then comes to find the value of $2+\sqrt{3}$ and $1/(2+\sqrt{3})$, which gives $4$.

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Position the regular polygon at $(0,0)$ in the coordinate plane. Let the $A$ be at $(r,0)$. Then $|AB|=r(2-\sqrt{3})$. The coordinate of $F$ will be $(r\cos\frac{5\pi}{6},r\sin\frac{5\pi}{6})=(-r\frac{\sqrt{3}}{2},\frac{r}{2})$. $|AF|=\sqrt{(r+r\frac{\sqrt{3}}{2})^2+(-\frac{r}{2})^2}=r\sqrt{2+\sqrt{3}}$
Therefore, $$\displaystyle{\frac{|AB|}{|AF|}=\sqrt{\frac{2-{\sqrt{3}}}{2+\sqrt{3}}}=2-\sqrt{3}}$$
$$\displaystyle{\frac{|AF|}{|AB|}=2+\sqrt{3}}$$
$$\displaystyle{\frac{|AB|}{|AF|}+\frac{|AF|}{|AB|}=4}$$