Consider M with r = ( 2 199 - 1 ) 2 2 198 . The number of all those circles D n that are inside M is [2021]

Question

Let $M = {(x, y) \in ℝ \times ℝ : x^{2} + y^{2} \leq r^{2}}$ 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 $n \geq 1$ , 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 $(S_{n - 1}, 0)$ and radius $a_{n}$ , and $D_{n}$ denote the circle with center $(S_{n - 1}, S_{n - 1})$ and radius $a_{n}$ . [2021]

Q. Consider $M$ with $r = \frac{(2^{199} - 1) \sqrt{2}}{2^{198}}$ . The number of all those circles $D_{n}$ that are inside $M$ is

Smart Achievers · Accepted Answer

(2)

$∵$ $r = \frac{(2^{199} - 1) \sqrt{2}}{2^{198}}$

$Now, \sqrt{2} S_{n - 1} + a_{n} < \frac{2^{199} - 1}{2^{198}} \sqrt{2}$

$\Rightarrow 2 \sqrt{2} (1 - \frac{1}{2^{n - 1}}) + \frac{1}{2^{n - 1}} < \frac{2^{199} - 1}{2^{198}} \sqrt{2}$

$\Rightarrow \frac{2 \sqrt{2} - 1}{2 \cdot 2^{n - 2}} > \frac{\sqrt{2}}{2^{198}}$

$\Rightarrow 2^{n - 2} < (2 - \frac{1}{\sqrt{2}}) 2^{197} \Rightarrow n \leq 199 \Rightarrow n = 199$

Solution

1 198

2 199

3 200

4 201

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