Question

Let $P_1$ be a regular polygon of $n$ sides with side length $x$ and let $P_{k+1}$ be the regular polygon of $n$ sides formed by joining the mid points of consecutive sides of $P_k$. Also perimeters of the $P_i$ 's form an infinite geometric progression. If $n=6$ and $x=\frac{1}{2+\sqrt{3}}$, then find the sum of this infinite geometric progression.

Answer 1

Answer: 12

The sum of the perimeters, then is
$S=nx+nx\cos \theta +nx{\cos }^2\theta +\dots \dots \dots =\frac{nx}{1-\cos \theta }\left(\text{ where }\theta =\frac{\pi }{n}\right)$
$\text{ Put }n=6,\frac{6x}{1-\cos \frac{\pi }{6}}=\frac{12x}{2-\sqrt{3}}=12$
$\text{ (where }x\text{ is }\frac{1}{2+\sqrt{3}}=2-\sqrt{3}).$
$\cos (\pi -2\theta )=\frac{\frac{x^2}{4}+\frac{x^2}{4}-x^{\prime 2}}{2\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)}$
$\frac{-x^2}{2}\cos 2\theta =\frac{x^2}{2}-x^{\prime 2}$
$x^{\prime 2}=\frac{x^2}{2}(1+\cos 2\theta )=x^2{\cos }^2\theta$
$\therefore x^{\prime }=x\cos \theta .$