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# Test of Mathematics Solution Subjective 170 - Infinite Circles  This is a Test of Mathematics Solution Subjective 170 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

Let ${C_n}$ be an infinite sequence of circles lying in the positive quadrant of the $XY$-plane, with strictly decreasing radii and satisfying the following conditions. Each $C_n$ touches both the $X$-axis and the $Y$-axis. Further, for all $n\geq 1$, the circle $C_{n+1}$ touches the circle $C_n$ externally. If $C_1$ has radius $10\: cm$, then show that the sum of the areas of all these circles is $\frac{25\pi}{3\sqrt{2}-4} \: cm^2$.

## Solution

Consider the following diagram where the Green line segment is $R_n$, the radius of the $n^{th}$ circle, and the Yellow line segment is $R_{n+1}$. As we are told about the symmetricity of the figure in the problem we can say that:

$\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n$

$=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)$

$=> R_{n+1}= (3-2\sqrt{2})R_n$

Let's say $=> R_{n+1}= \alpha.R_n$.

Now the total sum of the areas of the circles is:

$(\pi R_1^2 + \pi R_2^2 + \cdots ) = \pi (R_1^2 + R_2^2 + R_3^2 + \cdots )$

Now as $R_{n+1}= \alpha.R_n$, we can say that:

$\pi (R_1^2 + R_2^2 + R_3^2 + \cdots ) = \pi (R_1^2 + \alpha^2 R_1^2 + \alpha^4 R_1^2 + \cdots ) = \pi \frac{R_1^2}{1-\alpha^2}$ as $\alpha^2 < 1$.

Substituting the value of $\alpha = 3-2\sqrt{2}$ and $R_1 = 10 \: cm$we have,

Sum = $\frac{25\pi}{3\sqrt{2}-4} \: cm^2$.

Hence Proved. This is a Test of Mathematics Solution Subjective 170 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

Let ${C_n}$ be an infinite sequence of circles lying in the positive quadrant of the $XY$-plane, with strictly decreasing radii and satisfying the following conditions. Each $C_n$ touches both the $X$-axis and the $Y$-axis. Further, for all $n\geq 1$, the circle $C_{n+1}$ touches the circle $C_n$ externally. If $C_1$ has radius $10\: cm$, then show that the sum of the areas of all these circles is $\frac{25\pi}{3\sqrt{2}-4} \: cm^2$.

## Solution

Consider the following diagram where the Green line segment is $R_n$, the radius of the $n^{th}$ circle, and the Yellow line segment is $R_{n+1}$. As we are told about the symmetricity of the figure in the problem we can say that:

$\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n$

$=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)$

$=> R_{n+1}= (3-2\sqrt{2})R_n$

Let's say $=> R_{n+1}= \alpha.R_n$.

Now the total sum of the areas of the circles is:

$(\pi R_1^2 + \pi R_2^2 + \cdots ) = \pi (R_1^2 + R_2^2 + R_3^2 + \cdots )$

Now as $R_{n+1}= \alpha.R_n$, we can say that:

$\pi (R_1^2 + R_2^2 + R_3^2 + \cdots ) = \pi (R_1^2 + \alpha^2 R_1^2 + \alpha^4 R_1^2 + \cdots ) = \pi \frac{R_1^2}{1-\alpha^2}$ as $\alpha^2 < 1$.

Substituting the value of $\alpha = 3-2\sqrt{2}$ and $R_1 = 10 \: cm$we have,

Sum = $\frac{25\pi}{3\sqrt{2}-4} \: cm^2$.

Hence Proved.

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