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.
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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.
$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.2.2" open="on"]American Mathematical Contest 2017, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" open="off" _builder_version="4.2.2" inline_fonts="Abhaya Libre"]
Therefore, The required area = Area of $\triangle UXY$ - $2 \times$ Area of the sector $SXR$.
[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$
And the are of sector $SXR= \frac{2\pi}{3}$
ANS : $4\sqrt{3}-\frac{4\pi}{3}$[/et_pb_tab][et_pb_tab title="Formulas Used " _builder_version="4.2.2"]Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle]
Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle][/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" title_level="h2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]