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
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 \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).
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
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 \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).
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.
