Question

Let $G$ be a circle of radius $R>0$. Let $G _1, G _2, \ldots, G _{ n }$ be $n$ circles of equal radius $r>0$. Suppose each of the $n$ circles $G _1, G _2, \ldots, G _{ n }$ touches the circle $G$ externally. Also, for $i =1,2, \ldots, n -1$, 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?

• A. If \(n =4\), then \((\sqrt{2}-1) r < R\)
• B. If \(n =5\), then \(r < R\)
• C. If \(n =8\), then \((\sqrt{2}-1) r < R\)
• D. If \(n =12\), then \(\sqrt{2}(\sqrt{3}+1) r > R\)

Answer 1

The Correct Option is D

To solve this problem, let's first analyze the configuration of circles. We have a larger circle \(G\) with radius \(R\), and \(n\) smaller circles, each with radius \(r\), arranged in such a way that each smaller circle touches the larger circle from outside, and each pair of adjacent smaller circles touch each other externally.

The centers of the smaller circles \(G_1, G_2, \ldots, G_n\) and the center of the larger circle \(G\) form a regular \(n\)-gon inscribed in a larger circle with radius \(R + r\). The distance between any two adjacent centers of the smaller circles is \(2r\), and this is equal to the side of the \(n\)-gon.

For a regular \(n\)-gon inscribed in a circle of radius \(R + r\), the side length \(s\) is given by:

\(s = 2(R + r) \sin\left(\frac{\pi}{n}\right)\)

Equating this to \(2r\), we have:

\(2r = 2(R + r) \sin\left(\frac{\pi}{n}\right)\)

Solving for \(R\), we get:

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

Now, let's check each option:

1. If \(n = 4\):
2. If \(n = 5\):
3. If \(n = 8\):
4. If \(n = 12\):

Based on these calculations, the correct answer is:

If \(n = 12\), then \(\sqrt{2}(\sqrt{3}+1)r > R\).