Let G be a circle of radius R0. Let G_1, G_2 | vedclass

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(D) Let the centers of the $n$ circles $G_i$ form a regular polygon with side length $2r$. The distance from the center of $G$ to the center of any $G

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$C, D$

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(D) Let the centers of the $n$ circles $G_i$ form a regular polygon with side length $2r$. The distance from the center of $G$ to the center of any $G_i$ is $R+r$. Using the triangle formed by the center of $G$ and the centers of two adjacent circles $G_i$ and $G_{i+1}$,we have: $\sin(\frac{\pi}{n}) = \frac{r}{R+r}$ $\frac{R+r}{r} = \operatorname{cosec}(\frac{\pi}{n}) \implies R = r(\operatorname{cosec}(\frac{\pi}{n}) - 1)$. $(A)$ For $n=4$,$R = r(\operatorname{cosec}(\frac{\pi}{4}) - 1) = r(\sqrt{2}-1)$. Thus,$(\sqrt{2}-1)r = R$. The statement $(\sqrt{2}-1)r $(B)$ For $n=5$,$R = r(\operatorname{cosec}(\frac{\pi}{5}) - 1)$. Since $\operatorname{cosec}(\frac{\pi}{5}) \approx 1.701$,$R \approx 0.701r$,so $r > R$. The statement $r $(C)$ For $n=8$,$R = r(\operatorname{cosec}(\frac{\pi}{8}) - 1)$. Since $\operatorname{cosec}(\frac{\pi}{8}) > \operatorname{cosec}(\frac{\pi}{4}) = \sqrt{2}$,we have $R > r(\sqrt{2}-1)$,which means $(\sqrt{2}-1)r $(D)$ For $n=12$,$R = r(\operatorname{cosec}(\frac{\pi}{12}) - 1)$. We know $\operatorname{cosec}(\frac{\pi}{12}) = \sqrt{2}(\sqrt{3}+1) \approx 3.86$. Thus $R = r(\sqrt{2}(\sqrt{3}+1) - 1)$. Clearly,$\sqrt{2}(\sqrt{3}+1)r > R$. This is $TRUE$. Therefore,the correct statements are $C$ and $D$.

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