JEE Advanced 2022 Paper 2 Q12 Mathematics Coordinate Geometry Circles Hard

JEE Advanced 2022 Paper 2 · Q12 · Circles

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?

  1. A. If $n=4$, then $(\sqrt{2}-1)r < R$
  2. B. If $n=5$, then $r < R$
  3. C. If $n=8$, then $(\sqrt{2}-1)r < R$
  4. D. If $n=12$, then $\sqrt{2}(\sqrt{3}+1)r > R$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:C, D

Solution

The centres of the $n$ small circles lie on a circle of radius $R+r$ centred at $G$'s centre, with consecutive centres separated by angle $2\pi/n$ and distance $2r$ (since they touch externally). Hence $\sin(\pi/n) = \dfrac{r}{R+r}$, giving $R = r\left(\dfrac{1}{\sin(\pi/n)}-1\right) = r(\csc(\pi/n)-1)$. (A) $n=4$: $\csc(\pi/4)-1=\sqrt{2}-1$, so $R=(\sqrt 2-1)r$, NOT $>(\sqrt 2-1)r$, so (A) false. (B) $n=5$: $\csc 36^\circ-1\approx 1.701-1=0.701<1$, so $R \sqrt 2-1\approx 0.414$, so $R > (\sqrt 2-1)r$, i.e. $(\sqrt 2-1)r < R$, true. (D) $n=12$: $\csc 15^\circ - 1 = \dfrac{1}{\sin 15^\circ}-1\approx 3.864-1=2.864$. Check $\sqrt 2(\sqrt 3+1)\approx 1.414\cdot 2.732 \approx 3.864$, so $\sqrt 2(\sqrt 3+1)r > R$ (since $R\approx 2.864 r$ and $3.864 r > 2.864 r$), true. Hence (C), (D).

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