JEE Advanced 2022 — Paper 2 — MATH-II solution

JEE Advanced2022Paper 2MATH-II

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?

A
If $n = 4$ , then $(2 - 1) r < R$
B
If $n = 5$ , then $r < R$
C
If $n = 8$ , then $(2 - 1) r < R$
D
If $n = 12$ , then $2 (3 + 1) r > R$

Solution

The centres of $G_{1}, \dots, G_{n}$ lie on a circle of radius $R + r$ around the centre of $G$ , separated by angle $\frac{2 π}{n}$ . The condition $G_{i}$ touches $G_{i + 1}$ externally gives $sin \frac{π}{n} = \frac{r}{R + r}$ , so

$R = r (csc \frac{π}{n} - 1)$ .

(A) $n = 4$ : $R = r (2 - 1)$ , so the strict inequality $(2 - 1) r < R$ fails (equality holds).

(B) $n = 5$ : $csc 3 6^{\circ} \approx 1.701$ , so $R \approx 0.701 r < r$ , i.e. $r > R$ , so the claim is false.

(D) $n = 12$ : $csc 1 5^{\circ} = 6 + 2 = 2 (3 + 1)$ , so $R = r [2 (3 + 1) - 1] < 2 (3 + 1) r$ . True.

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