Question

Let $M=\left\{(x,y)\in R\times R:x^2+y^2\le r^2\right\}$, where $r>0$. Consider the geometric progression $a_n=\frac{1}{2^{n-1}},n=1,2,3,\dots .$ Let $S_0=0$ and for $n\ge 1$, let $S_n$ denote the sum of the first $n$ terms of this progression. For $n\ge 1$, 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. $k+2l=22$
• B. $2k+1=26$
• C. $2k+3l=34$
• D. $3k+2l=40$

Answer 1

Answer: (D)

$C_1\to (0,0),r=1$
$C_2\to (1,0),r=1/2$
$C_3\to (3/2,0),r=1/4$
$⋮$
$C_n\left(2\left(1-\frac{1}{2^{n-1}}\right),0\right),r=\frac{1}{2^{n-1}}$
$\Rightarrow 2\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2^{n-1}}<r$
$2-\frac{1}{2^{n-1}}<r$
$\Rightarrow 2-\frac{1}{2^{n-1}}<\frac{1025}{513}$
$\Rightarrow \frac{1}{2^{n-1}}>\frac{1}{513}$
$\Rightarrow 2^{n-1}<513$
$\Rightarrow n-1\le 9$
$\Rightarrow n\le 10\Rightarrow k=10$
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 $1=5$
$\Rightarrow 3k+2l=40$