Let G be a circle of radius R > 0. Let G$$_1$$, G$$_2$$, ..., G$$_n$$ be $$n$$ circles of equal radius $$r > 0$$. Suppose each of the $$n$$ circles G$$_1$$, G$$_2$$, ..., G$$_n$$ touches the circle G externally. Also, for $$i = 1, 2, ..., n-1$$, the circle G$$_i$$ touches G$$_{i+1}$$ externally, and G$$_n$$ touches G$$_1$$ externally. Then, which of the following statements is/are TRUE?

Let the centre of the big circle $$G$$ be $$O$$ and its radius be $$R$$.
Each small circle $$G_i$$ has centre $$A_i$$ and radius $$r$$.

Because every $$G_i$$ touches $$G$$ externally, the distance $$OA_i$$ is the sum of the radii: $$OA_i = R + r$$.
Because consecutive circles $$G_i$$ and $$G_{i+1}$$ touch externally, the distance between their centres is $$A_iA_{i+1}=2r$$.

The centres $$A_1,A_2,\ldots ,A_n$$ are therefore the vertices of a regular $$n$$-gon inscribed in the circle with centre $$O$$ and radius $$R+r$$.
Hence the central angle between two consecutive vertices is $$\theta = \dfrac{2\pi}{n}$$.

Consider the isosceles triangle $$OA_iA_{i+1}$$. Applying the cosine rule,

$$\begin{aligned} A_iA_{i+1}^2 &= OA_i^2 + OA_{i+1}^2 - 2(OA_i)(OA_{i+1})\cos\theta \\ (2r)^2 &= (R+r)^2 + (R+r)^2 - 2(R+r)^2\cos\theta \\ 4r^2 &= 2(R+r)^2\bigl(1-\cos\theta\bigr). \end{aligned}$$

Rearranging gives

$$\begin{aligned} (R+r)^2 &= \frac{2r^2}{1-\cos\theta},\\ R+r &= \frac{\sqrt{2}\,r}{\sqrt{1-\cos\theta}},\\ R &= r\!\left(\frac{\sqrt{2}}{\sqrt{1-\cos\theta}} - 1\right).\qquad -(1) \end{aligned}$$

Thus the ratio $$\dfrac{R}{r}$$ depends only on $$n$$ through $$\theta=\dfrac{2\pi}{n}$$.

$$\dfrac{R}{r}= \sqrt{2}-1,\qquad\text{so } R = (\sqrt{2}-1)\,r.$$ Option A claims $$(\sqrt{2}-1)r \lt R$$, but we have equality; therefore Option A is false.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.3090}}-1 \approx \frac{1.4142}{0.8312}-1 \approx 0.698.$$ Hence $$R\approx 0.698\,r,$$ i.e. $$R \lt r.$$ Option B states $$r \lt R$$, which is false.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.7071}}-1 = \frac{1.4142}{\sqrt{0.2929}}-1 = \frac{1.4142}{0.5412}-1 \approx 1.613.$$ Therefore $$R \approx 1.613\,r \gt (\sqrt{2}-1)\,r\;(\approx 0.414\,r).$$ So the inequality $$(\sqrt{2}-1)r \lt R$$ holds. Option C is true.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.8660}}-1 = \frac{1.4142}{\sqrt{0.1340}}-1 = \frac{1.4142}{0.3660}-1 \approx 2.866.$$ Compute $$\sqrt{2}(\sqrt{3}+1) = 1.4142\,(1.732 + 1) = 1.4142 \times 2.732 \approx 3.866.$$ Since $$3.866\,r \gt 2.866\,r = R,$$ we have $$\sqrt{2}(\sqrt{3}+1)r \gt R.$$ Option D is true.

Hence the correct statements are:
Option C (If $$n = 8$$, then $$(\sqrt{2}-1)r \lt R$$)
Option D (If $$n = 12$$, then $$\sqrt{2}(\sqrt{3}+1)r \gt R$$).