JEE Challenger
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MathHardAdv 2021 • Paper 2

Number of Circles Dn Contained Within Region M

Comprehension Passage

Let M={(x,y)∈R×R:x2+y2≤r2},M = \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \leq r^2\}, where r>0r > 0. Consider the geometric progression given by an=12n−1,n=1,2,3,…a_n = \frac{1}{2^{n-1}}, n = 1, 2, 3, \dots. Let S0=0S_0 = 0 and, for n≥1n \geq 1, let SnS_n represent the sum of the first nn terms of this sequence. For n≥1n \geq 1, let CnC_n denote the circle with center at (Sn−1,0)(S_{n-1}, 0) and radius ana_n, and DnD_n denote the circle with center at (Sn−1,Sn−1)(S_{n-1}, S_{n-1}) and radius ana_n.

Suppose region MM has radius r=(2199−1)22198r = \frac{(2^{199}-1)\sqrt{2}}{2^{198}}. The total number of circles DnD_n that are contained entirely inside MM equals

Options

A

198

B

199

Correct
C

200

D

201

Topics & Concepts

Conic SectionsCirclesGeometric ProgressionSum of Series

Step-by-Step Solution

To determine the number of circles DnD_n contained entirely inside the region MM, we first establish the mathematical condition for a circle to lie inside a centered disk.

1. Definition of Parameters

Given:

  • The sequence terms are an=12n−1=21−na_n = \frac{1}{2^{n-1}} = 2^{1-n} for n≥1n \geq 1.

  • The partial sum S0=0S_0 = 0, and for n≥1n \geq 1: Sn=∑k=1nak=1−(12)n1−12=2(1−2−n)=2−21−nS_n = \sum_{k=1}^n a_k = \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 2\left(1 - 2^{-n}\right) = 2 - 2^{1-n} Therefore, for the center coordinates of circle DnD_n, we have: Sn−1=2−22−nfor n≥1S_{n-1} = 2 - 2^{2-n} \quad \text{for } n \geq 1

  • The circle DnD_n has center (Sn−1,Sn−1)(S_{n-1}, S_{n-1}) and radius an=21−na_n = 2^{1-n}.

  • The region MM is defined by x2+y2≤r2x^2 + y^2 \leq r^2, which is a disk centered at the origin (0,0)(0,0) with radius: r=(2199−1)22198=2(2−12198)=22−22198r = \frac{(2^{199}-1)\sqrt{2}}{2^{198}} = \sqrt{2}\left(2 - \frac{1}{2^{198}}\right) = 2\sqrt{2} - \frac{\sqrt{2}}{2^{198}}

2. Condition for Circle DnD_n to lie inside Region MM

A circle with center (x0,y0)(x_0, y_0) and radius RR is entirely contained within the region x2+y2≤r2x^2 + y^2 \leq r^2 if and only if the sum of the distance from the origin to its center and its radius is less than or equal to rr: x02+y02+R≤r\sqrt{x_0^2 + y_0^2} + R \leq r

Substituting the center (Sn−1,Sn−1)(S_{n-1}, S_{n-1}) and radius ana_n: Sn−12+Sn−12+an≤r  ⟹  2Sn−1+an≤r\sqrt{S_{n-1}^2 + S_{n-1}^2} + a_n \leq r \implies \sqrt{2}S_{n-1} + a_n \leq r

3. Solving the Inequality

Substitute Sn−1S_{n-1} and ana_n into the inequality: LHS=2(2−22−n)+21−n=22−22⋅21−n+21−n=22−21−n(22−1)\text{LHS} = \sqrt{2}\left(2 - 2^{2-n}\right) + 2^{1-n} = 2\sqrt{2} - 2\sqrt{2} \cdot 2^{1-n} + 2^{1-n} = 2\sqrt{2} - 2^{1-n}\left(2\sqrt{2} - 1\right)

Set this less than or equal to rr: 22−22−12n−1≤22−221982\sqrt{2} - \frac{2\sqrt{2} - 1}{2^{n-1}} \leq 2\sqrt{2} - \frac{\sqrt{2}}{2^{198}}

Subtracting 222\sqrt{2} from both sides and multiplying by −1-1: 22−12n−1≥22198\frac{2\sqrt{2} - 1}{2^{n-1}} \geq \frac{\sqrt{2}}{2^{198}}

Rearranging for 2n−12^{n-1}: 2n−1≤2198⋅(22−12)=2198(2−12)2^{n-1} \leq 2^{198} \cdot \left(\frac{2\sqrt{2} - 1}{\sqrt{2}}\right) = 2^{198} \left(2 - \frac{1}{\sqrt{2}}\right)

4. Finding the Number of Integer Solutions for nn

Now we approximate 2−122 - \frac{1}{\sqrt{2}}: 12≈0.707  ⟹  2−12≈1.293\frac{1}{\sqrt{2}} \approx 0.707 \implies 2 - \frac{1}{\sqrt{2}} \approx 1.293

Since 1<2−12<21 < 2 - \frac{1}{\sqrt{2}} < 2, we have: 2198<2198(2−12)<21992^{198} < 2^{198}\left(2 - \frac{1}{\sqrt{2}}\right) < 2^{199}

Taking the base-2 logarithm: 198<log⁡2[2198(2−12)]<199198 < \log_2\left[2^{198}\left(2 - \frac{1}{\sqrt{2}}\right)\right] < 199

Thus, the largest integer value for n−1n-1 satisfying the inequality is 198198: n−1≤198  ⟹  n≤199n - 1 \leq 198 \implies n \leq 199

Since n≥1n \geq 1, the possible values for nn are n∈{1,2,3,…,199}n \in \{1, 2, 3, \dots, 199\}.

Therefore, the total number of circles DnD_n contained entirely inside region MM is 199.

Correct Option: (B)

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