Inscribed Regular Polygons

Consider a regular $n$ -sided polygon, then the central angle is $\dfrac {2\pi}n$ , the outer radius $r_o$ is the circumradius of the polygon and the inner radius $r_i$ is the apothem or inradius of the polygon. The ratio $\dfrac {r_i}{r_o} = \cos\left(\dfrac \pi n \right)$ .

Label the radii of the five circles $r_1$ to $r_5$ from outermost to innermost. We have $\dfrac {r_2}{r_1} = \cos \dfrac \pi 3$ , $\dfrac {r_3}{r_2} = \cos \dfrac \pi 4$ , $\dfrac {r_4}{r_3} = \cos \dfrac \pi 5$ , and $\dfrac {r_5}{r_4} = \cos \dfrac \pi 6$ . Then we have:

$\begin{aligned} x & = \frac {r_5}{r_1} = \cos \frac \pi 3 \cos \frac \pi 4 \cos \frac \pi 5 \cos \frac \pi 6 \\ & = \frac 12 \cdot \frac 1{\sqrt 2} \cdot \frac {1+\sqrt 5}4 \cdot \frac {\sqrt 3}2 \\ & = \frac {\sqrt 3+\sqrt{15}}{16\sqrt 2} = \frac {\sqrt 6 + \sqrt{30}}{32} \approx 0.247709854 \\ \implies \lfloor 10^5 x \rfloor & = \boxed {24770} \end{aligned}$

When $n \to \infty$ , $\displaystyle x = \prod_{n=3}^\infty \cos \dfrac \pi n$ . Since $\sum_{n=3}^\infty \ln \left(\cos \dfrac \pi n \right) \approx -2.16333$ (according the Wolfram Alpha, the series converges by comparison test -- hope that @Pi Han Goh can prove this), $x \approx e^{-2.16333} \approx 0.114942$ .

The ratio of the inradius, $r$ , to the circumradius, $R$ , of a regular n-gon is

$\hspace{4cm}\dfrac{r}{R} = \cos\dfrac{\pi}{n}$

So the required ratio of our sequence of four polygons can be expressed becomes

$\hspace{4cm}x = \cos\dfrac{\pi}{3}\cdot\cos\dfrac{\pi}{4}\cdot\cos\dfrac{\pi}{5}\cdot\cos\dfrac{\pi}{6} = \dfrac{\sqrt6+\sqrt{30}}{32} \approx 0.2477098$

Trigonometry identities courtesy of Wolfram Alpha . And the requested result is $\boxed{24770}$ .