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Test of Mathematics Solution Subjective 170 - Infinite Circles

Test of Mathematics at the 10+2 Level

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}.

circle

 

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 \: cmwe have,

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

Hence Proved.

Test of Mathematics at the 10+2 Level

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}.

circle

 

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 \: cmwe have,

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

Hence Proved.

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