Question

Let $M=\left\{(x, y) \in R \times R: x^{2}+y^{2} \leq r^{2}\right\}$, where $r>0$. Consider the geometric progression $a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots .$ Let $S_{0}=0$ and for $n \geq 1$, let $S_{n}$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$, let $C_{n}$ denote the circle with center $\left(S_{n-1}, 0\right)$ and radius $a_{n}$, and $D_{n}$ denote the circle with center $\left(S_{n-1}, S_{n-1}\right)$ and radius $a_{n}$.
Consider $M$ with $r=\frac{1025}{513}$. Let $k$ be the number of all those circles $C_{n}$ that are inside $M$. Let $1$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then

• A. $k+2 l=22$
• B. $2 k +1=26$
• C. $2 k +3 l =34$
• D. $3 k +2 l =40$

Answer 1

Answer: (D)

$C _{1} \rightarrow(0,0), r =1$
$C _{2} \rightarrow(1,0), r =1 / 2$
$C _{3} \rightarrow(3 / 2,0), r =1 / 4$
$\vdots$
$C_{n}\left(2\left(1-\frac{1}{2^{n-1}}\right), 0\right), r=\frac{1}{2^{n-1}}$
$\Rightarrow 2\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2^{n-1}} < r$
$ 2-\frac{1}{2^{n-1}} < r$
$\Rightarrow 2-\frac{1}{2^{n-1}} < \frac{1025}{513}$
$\Rightarrow \frac{1}{2^{n-1}}>\frac{1}{513}$
$\Rightarrow 2^{n-1} < 513 $
$\Rightarrow n-1 \leq 9$
$\Rightarrow n \leq 10 \Rightarrow k=10$
Also no two by $C _{1}, C _{3}, C _{5}, C _{7}, C _{9}$ intersect each other.
And no two of $C _{2}, C _{4}, C _{6}, C _{8}, C _{10}$ intersect each other
For both, we get $1=5$
$\Rightarrow 3 k +2 l =40$