RADII

$\begin{aligned} A & = 2 \prod_{k=3}^\infty \left(1-\tan^4 \frac \pi {2^n} \right) = 2 \prod_{k=3}^\infty \left(1-\tan^2 \frac \pi {2^k} \right) \left(1+\tan^2 \frac \pi {2^k} \right) \\ & = 2 \prod_{k=3}^\infty \frac {1-\tan^2 \frac \pi {2^k}}{1+\tan^2 \frac \pi {2^k}} \left(1+\tan^2 \frac \pi {2^k} \right)^2 = 2 \prod_{k=3}^\infty \cos \frac \pi{2^{n-1}} \sec^4 \frac \pi {2^k} \\ & = 2 \prod_{k=3}^\infty \frac {\cos \frac \pi{2^{k-1}}}{\cos^4 \frac \pi {2^k}} = 2\cos \frac \pi 4 \prod_{k=3}^\infty \frac 1{\cos^3 \frac \pi {2^k}} = \prod_{k=3}^\infty \frac {\sqrt 2}{\cos^3 \frac \pi {2^k}} & \small \blue{\text{See proof }\prod_{k=2}^n \cos \frac \pi{2^k} = \frac 1{2^{n-1}\sin \frac \pi{2^n}}} \\ & = \lim_{n \to \infty} \sqrt 2 \cos^3 \frac \pi 4 \left( \prod_{k=2}^n \frac 1{\cos \frac \pi{2^k}} \right)^3 = \frac 12 \left(\lim_{n \to \infty} 2^{n-1} \sin \frac \pi {2^n} \right)^3 \\ & = \frac 12 \left(\lim_{n \to \infty} \frac { \pi \sin \frac \pi {2^n}}{2 \cdot \frac \pi{2^n}} \right)^3 = \frac {\pi^3}{16} \end{aligned}$

Therefore $A = \frac {\pi^2}{16} = \pi R^2$ , $\implies R = \frac \pi 4$ and the radius of the circle with area $R$ is $r = \frac 12 = \log_2 (\sqrt 2) = \log_2 \sec \frac \pi 4 = \boxed{\log_2 \sec R}$ .

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Proof: Let us prove by induction the claim $\displaystyle P(n) = \prod_{k=2}^n \cos \frac \pi{2^k} = \frac 1{2^{n-1} \sin \frac \pi {2^n}}$ for $n \ge 2$ . For $n=2$ , $P(2) = \cos \dfrac \pi 4 = \dfrac 1{\sqrt 2} = \dfrac 1{2 \sin \frac \pi 4}$ . The claim is true for $n=2$ . Assuming that the claim is true for $n$ , then:

$\begin{aligned} P(n+1) & = P(n) \cos \frac \pi{2^{n+1}} = \frac {\cos \frac \pi{2^{n+1}}}{2^{n-1}\sin \frac \pi {2^n}} = \frac {\cos \frac \pi{2^{n+1}}}{2^n \sin \frac \pi {2^{n+1}}\cos \frac \pi{2^{n+1}}} = \frac 1{2^n \sin \frac \pi {2^{n+1}}} \end{aligned}$

The claim is also true of $n+1$ and therefore is true for all $n \ge 2$ .