Radius Relations for Externally Tangent Circles Array

To determine the correct statements, we analyze the geometric arrangement of the given circles.

Let $O$ be the center of the main circle $G$ of radius $R$. Let $O_1, O_2, \dots, O_n$ be the centers of the $n$ identical circles $G_1, G_2, \dots, G_n$, each having radius $r$.

Since each circle $G_i$ touches $G$ externally, the distance from the origin $O$ to each center $O_i$ is: $O O_i = R + r$

Since each circle $G_i$ touches $G_{i+1}$ (and $G_n$ touches $G_1$) externally, the distance between adjacent centers is: $O_i O_{i+1} = r + r = 2r$

The $n$ centers $O_1, O_2, \dots, O_n$ form a regular $n$-gon centered at $O$. The central angle subtended by any adjacent pair of centers $O_i O_{i+1}$ at $O$ is: $\theta = \frac{2\pi}{n}$

By considering the isosceles triangle $\triangle O O_i O_{i+1}$ with equal sides $O O_i = O O_{i+1} = R + r$ and base $O_i O_{i+1} = 2r$, the perpendicular from $O$ to $O_i O_{i+1}$ bisects the central angle $\theta$. Thus: $\sin\left(\frac{\pi}{n}\right) = \frac{r}{R + r}$

Rearranging this formula gives: $R + r = \frac{r}{\sin\left(\frac{\pi}{n}\right)} \implies R = r \left( \frac{1}{\sin\left(\frac{\pi}{n}\right)} - 1 \right)$

Now, we evaluate each of the given options:

Option (A): For $n = 4$

$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ Substituting this into the relation: $R + r = \sqrt{2}r \implies R = (\sqrt{2} - 1)r$ Thus, $(\sqrt{2} - 1)r = R$. The strict inequality $(\sqrt{2} - 1)r < R$ is false. Therefore, Option (A) is INCORRECT.

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Option (B): For $n = 5$

Since $\frac{\pi}{5} = 36^\circ > 30^\circ = \frac{\pi}{6}$, we have: $\sin\left(\frac{\pi}{5}\right) > \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ $\frac{r}{R + r} > \frac{1}{2} \implies 2r > R + r \implies r > R$ This contradicts the statement $r < R$. Therefore, Option (B) is INCORRECT.

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Option (C): For $n = 8$

Since $\frac{\pi}{8} < \frac{\pi}{4}$, we have: $\sin\left(\frac{\pi}{8}\right) < \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ $\frac{r}{R + r} < \frac{1}{\sqrt{2}} \implies \sqrt{2}r < R + r \implies (\sqrt{2} - 1)r < R$ Therefore, Option (C) is CORRECT.

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Option (D): For $n = 12$

We know that: $\sin\left(\frac{\pi}{12}\right) = \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

Substituting this value into $R + r = \frac{r}{\sin\left(\frac{\pi}{12}\right)}$: $R + r = r \cdot \frac{2\sqrt{2}}{\sqrt{3} - 1} = r \frac{2\sqrt{2}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \sqrt{2}(\sqrt{3} + 1)r$

Since $r > 0$, we get: $\sqrt{2}(\sqrt{3} + 1)r = R + r > R$ Therefore, Option (D) is CORRECT.

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Conclusion:

The correct statements are (C) and (D).