Let be a circle of radius . Let be circles of equal radius . Suppose each of the circles touches the circle externally. Also, for , the circle touches externally, and touches externally. Then, which of the following statements is/are TRUE?

Question

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? [2022]

Smart Achievers · Accepted Answer

(3, 4)

Refer to diagram,

In $∆ A O B$

$\sin (\frac{π}{n}) = \frac{r}{R + r}$

$\Rightarrow c o s e c (\frac{π}{n}) = \frac{R}{r} + 1$

$\Rightarrow R = r [c o s e c (\frac{π}{n}) - 1]$

$If n = 4, R = r (\sqrt{2} - 1)$

$If n = 5, R = r (c o s e c \frac{π}{5} - 1)$

$∵$ $c o s e c \frac{π}{5} < c o s e c \frac{π}{6}$

$(c o s e c \frac{π}{5} - 1) < 2 - 1 = 1$ $∴$ $R < r$

$If n = 8, R = r (c o s e c \frac{π}{8} - 1)$ $∴$ $c o s e c \frac{π}{8} > c o s e c \frac{π}{4}$

$(c o s e c \frac{π}{8} - 1) > \sqrt{2} - 1 \Rightarrow R > r (\sqrt{2} - 1)$

$If n = 12, R = r (c o s e c \frac{π}{12} - 1)$

$R = r (\sqrt{2} (\sqrt{3} + 1) - 1), R < \sqrt{2} (\sqrt{3} + 1) r$

Solution

1 If $n = 4$ , then $(\sqrt{2} - 1) r < R$

2 If $n = 5$ , then $r < R$

3 If $n = 8$ , then $(\sqrt{2} - 1) r < R$

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

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Solution

1 If n=4, then (2-1)r<R

2 If n=5, then r<R

3 If n=8, then (2-1)r<R

4 If n=12, then 2(3+1)r>R

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1 If $n = 4$ , then $(\sqrt{2} - 1) r < R$

2 If $n = 5$ , then $r < R$

3 If $n = 8$ , then $(\sqrt{2} - 1) r < R$

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