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?

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

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

(B) (B) If $n = 5$, then $r < R$

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Circle

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?

Answer: (C), (D)

From the geometry, the centers of the small circles form a regular n-gon. The relationship is:

2 (R + r) sin \frac{π}{n} = 2 r

\frac{R + r}{r} = csc \frac{π}{n}

(A)

n = 4

R + r = 2 r

, so

R = (2 - 1) r

. The statement says

(2 - 1) r < R

, which is false (equality holds).

(B)

n = 5

csc \frac{π}{5} < csc \frac{π}{6} = 2

, so

R + r < 2 r

, giving

R < r

. The statement says

r < R

, which is false.

(C)

n = 8

csc \frac{π}{8} > csc \frac{π}{4} = 2

, so

R + r > 2 r

, giving

R > (2 - 1) r

. The statement is true. ✓

(D)

n = 12

csc \frac{π}{12} = 2 (3 + 1)

, so

R + r = 2 (3 + 1) r

, giving

R = (2 (3 + 1) - 1) r < 2 (3 + 1) r

. Wait, this gives

2 (3 + 1) r > R

. The statement is true. ✓

Therefore, (C) and (D) are correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options