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
$r = \frac{\left( 2^{199} - 1 \right)\sqrt{2}}{2^{198}}$. The number
of all those circles $D_{n}$ that are inside $M$ is
Answer: 199
Solution: $\because\ r = \frac{\left( 2^{199} - 1 \right)\sqrt{2}}{2^{198}}$\
Now, $\sqrt{2}S_{n - 1} + a_{n} < \frac{2^{199} - 1}{2^{198}}\sqrt{2}$\
$${\Rightarrow 2\sqrt{2}\left( 1 - \frac{1}{2^{n - 1}} \right) + \frac{1}{2^{n - 1}} < \frac{2^{199} - 1}{2^{198}}
}{\Rightarrow \frac{2\sqrt{2} - 1}{{2.2}^{n - 2}} > \frac{\sqrt{2}}{2^{198}}
}{\Rightarrow 2^{n - 2} < \left( 2 - \frac{1}{\sqrt{2}} \right)2^{197}
}{\therefore n \leq 199 \Rightarrow n = 199}$$