Let $G$ be a circle of radius $R>0$. Let $G_1,G_2,\dots ,G_n$ be $n$ circles of equal radius $r>0$. Suppose each of the $n$ circles $G_1,G_2,\dots ,G_n$ touches the circle $G$ externally. Also, for $i=1,2,\dots ,n-1$, 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 ?

The correct statements are: If $n=8$, then $\left(\sqrt{2}-1\right)r<R$ and If $n=12$, then $\sqrt{2}\left(\sqrt{3}+1\right)r>R$.

Step 1: Understand the Geometric Relationship
Let $O$ be the center of circle $G$ of radius $R$, and let $O_1,O_2,\dots ,O_n$ be the centers of the $n$ equal circles $G_1,G_2,\dots ,G_n$ of radius $r$.
Since each small circle touches the central circle $G$ externally, the distance from $O$ to the center of any small circle is:

$O{O}_i=R+r$

Since adjacent circles $G_i$ and $G_{i+1}$ touch each other externally, the distance between their centers is:

$O_i{O}_{i+1}=r+r=2r$

Step 2: Apply Trigonometry
The $n$ centers $O_1,O_2,\dots ,O_{\mathrm{n}}$ form a regular polygon with $n$ sides. The angle subtended by each side at the central point $O$ is:

$\angle O_{i}O{O}_{i+1}=\frac{2\pi }{n}$

Dropping a perpendicular from $O$ to the segment $O_i{O}_{i+1}$ bisects both the angle and the segment. In the resulting right-angled triangle:

$sin\left(\frac{\pi }{n}\right)=\frac{r}{R+r}$

Rearranging to express $R$ in terms of $r$:

$R+r=\frac{r}{sin\left(\pi /n\right)}$

$R=r\left(\frac{1}{sin\left(\pi /n\right)}-1\right)$

Step 3: Analyze for $n=8$
For $n=8$, we have:

$sin\left(\frac{\pi }{8}\right)<sin\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$

Taking the reciprocal:

$\frac{1}{sin\left(\pi /8\right)}>\sqrt{2}$

Subtracting $1$ from both sides:

$\frac{1}{sin\left(\pi /8\right)}-1>\sqrt{2}-1$

Multiplying by $r$:

$R>\left(\sqrt{2}-1\right)r$

which is equivalent to:

$\left(\sqrt{2}-1\right)r<R$

Thus, the statement for $n=8$ is TRUE.

Step 4: Analyze for $n=12$
For $n=12$:

$sin\left(\frac{\pi }{12}\right)=sin\left(15°\right)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Therefore:

$\frac{1}{sin\left(\pi /12\right)}=\frac{2\sqrt{2}}{\sqrt{3}-1}=\sqrt{2}\left(\sqrt{3}+1\right)$

Substituting this into our expression for $R$:

$R=r\left(\sqrt{2}\left(\sqrt{3}+1\right)-1\right)$

Since $\sqrt{2}\left(\sqrt{3}+1\right)-1<\sqrt{2}\left(\sqrt{3}+1\right)$, it follows that:

$R<\sqrt{2}\left(\sqrt{3}+1\right)r$

which is equivalent to:

$\sqrt{2}\left(\sqrt{3}+1\right)r>R$

Thus, the statement for $n=12$ is TRUE.