Subject: Mathematics — Circles
Question: Let
$M = \left\{ (x,y)\mathbb{\in R \times R:}x^{2} + y^{2} \leq r^{2} \right\}$
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
progressions. For $n \geq 1,$let $C_{n}$ denote the circle with center
$(S_{n - 1},0)$ and radius, $a_{n}$ and $D_{n}$ denote the circle with
center $(S_{n - 1},S_{n - 1})$ and $a_{n}$.
Consider $M$ with $r = \frac{1025}{513}$. Let $k$ be the number of all
those circles $C_{n}$ that are inside $M$. Let $l$ be the maximum
possible number of circles among these $k$ circles such that no two
circles intersect. Then
Answer: $3k + 2l = 40$
Solution: $\because a_{n} = \frac{1}{2^{n - 1}}$\
$${S_{n} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + \ldots + \frac{1}{2^{n - 1}}
}{= 2\left( 1 - \frac{1}{2^{n}} \right) = 2 - \frac{1}{2^{n - 1}}
}$$For circles $C_{n}$ to inside $M$\
$${S_{n - 1} + a_{n} < \frac{1025}{513}
}{\Rightarrow \ 2 - \frac{1}{2^{n - 2}} + \frac{1}{2^{n - 1}} < \frac{1025}{513}
}{\Rightarrow \ 1 - \frac{1}{2^{n}} < \frac{1025}{1026} = 1 - \frac{1}{1026}
}{\Rightarrow 2^{n} < 2026 \Rightarrow n \leq 10 \Rightarrow k = 10
}$$Also $l = 5$\
$$3k + 2l = 30 + 10 = 40$$