Rollin' Stones Pt. 2

Geometry Level 5

If the radii of the red circle and the blue circle as shown above are 1 and 4 respectively, find the sum of the reciprocals of the radii of the seven green circles.

The answer can be expressed as a b \dfrac{a}{b} where a a and b b are coprime positive integers. Find a + b a+b .

Note : All circles that seem to touch with each other and/or the line are actually tangent with them. For example, the red circle, the blue circle, and the largest green circle are tangent to each other and to the line.


The answer is 213.

Digvijay Singh by Digvijay Singh , 21, India

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2 solutions

Digvijay Singh
Feb 21, 2016

If two circles are mutually tangent to each other and a line

Then, R 3 = R 1 R 2 ( R 1 + R 2 ) 2 \large R_3=\frac{R_1 R_2}{(\sqrt{R_1}+\sqrt{R_2})^2}

This formula is derived from the fact that 1 R 3 = 1 R 1 + 1 R 2 \frac{1}{\sqrt{R_3}}=\frac{1}{\sqrt{R_1}}+\frac{1}{\sqrt{R_2}}

For four mutually tangent circles like

R 4 = R 1 R 2 + R 2 R 3 + R 1 R 3 − 2 R 1 R 2 R 3 ( R 1 + R 2 + R 3 ) ( R 1 R 2 ) 2 + ( R 2 R 3 ) 2 + ( R 1 R 3 ) 2 − 2 R 1 R 2 R 3 ( R 1 + R 2 + R 3 ) R 1 R 2 R 3 R_4=\frac{R_1 R_2+R_2 R_3 +R_1 R_3-2\sqrt{R_1 R_2 R_3(R_1+R_2+R_3)}}{(R_1 R_2)^2+(R_2 R_3)^2 +(R_1 R_3)^2-2R_1 R_2 R_3(R_1+R_2+R_3)} R_1 R_2 R_3 .

This formula is derived from the fact that

2 ( ( 1 R 1 ) 2 + ( 1 R 2 ) 2 + ( 1 R 3 ) 2 + ( 1 R 4 ) 2 ) = ( 1 R 1 + 1 R 2 + 1 R 3 + 1 R 4 ) 2 2((\frac{1}{R_1})^2+(\frac{1}{R_2})^2+(\frac{1}{R_3})^2+(\frac{1}{R_4})^2)=(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4})^2

Applying these formulae a few times gives the radii of the green circles 4 9 , 4 25 , 1 4 , 1 7 , 4 49 , 4 25 , 4 57 \frac{4}{9},\frac{4}{25},\frac{1}{4},\frac{1}{7},\frac{4}{49},\frac{4}{25},\frac{4}{57} .

Here is the exact figure showing the radii of the green circles:

And their reciprocals add up to 209 4 \frac{209}{4} .

9 Helpful 0 Interesting 1 Brilliant 0 Confused

Really nice sketch that gives insight of the problem.

Niranjan Khanderia - 5 years, 3 months ago

This type questions in which book ? Please tell me [email protected]

Kamalsingh Prajapati - 2 years, 8 months ago

Descartes Formula for reciprocal of radii of four mutually tangential circle W h e r e 1 r i = k i i s : − \ k 4 = k 1 + k 2 + k 3 + 2 ∗ k 1 ∗ k 2 + k 2 ∗ k 3 + k 3 ∗ k 1 . If the third is a st. line, it is obvious k 3 = 0. In our case, we have four groups. Applying the above formula, we get :- Where ~~ \dfrac 1 r_i=k_i~~is:-~~\ k_4=k_1+k_2+k_3+2*\sqrt{k_1*k_2+k_2*k_3+k_3*k_1}.\\ \text{ If the third is a st. line, it is obvious } k_3=0.\\ \text{In our case, we have four groups. Applying the above formula, we get :-}

( A ) B i g G r e e n ∣ k 1 = 1 , ∣ k 2 = 1 4 , ∣ k 3 = 0 , . . . ∣ ∴ K b i g = 9 4 ∣ (A) \ Big\ Green \ |\ k_1=1, \ |\ k_2=\frac 1 4,\ |\ k_3=0,...\ |\ \therefore K_{big}=\frac 9 4\ | \\

( B ) W i t h R e d R = 1 ⟹ k 1 = 1 , ∣ k 3 = 0 ∣ . . . . . . . . . M i d d l e . . . . . . . . . . ∣ k 2 = 9 4 . . . . . ∣ ∴ K M d R = 25 4 ∣ . . . . . . . . . S m a l l , . . . . . . . . . . ∣ k 2 = 25 4 . . . . . ∣ ∴ K S R = 49 4 ∣ (B) \ With\ Red\ \ R=1\ \ \implies ~~k_1=1,\ |\ k_3=0 \ |\\ ......... Middle .......... |\ \ k_2=\frac 9 4\ ..... |\ \therefore K_{Md_R}=\frac{25} 4 \ | \\ ......... Small ,.......... |\ \ k_2=\frac{25} 4\ ..... |\ \therefore K_{S_R}=\frac {49} 4 \ | \\

( C ) W i t h B l u e R = 4 ⟹ k 1 = 1 4 , ∣ k 3 = 0 ∣ . . . . . . . . . M i d d l e . . . . . . . . . . ∣ k 2 = 9 4 . . . . . ∣ ∴ K M d B = 4 ∣ . . . . . . . . . S m a l l , . . . . . . . . . . ∣ k 2 = 4 . . . . . . ∣ ∴ K S B = 25 4 ∣ (C) \ With\ Blue \ \ R=4\ \ \implies ~~k_1=\frac 1 4,\ |\ k_3=0 \ |\\ ......... Middle .......... |\ \ k_2=\frac 9 4\ ..... |\ \therefore K_{Md_B}= 4 \ | \\ ......... Small ,.......... |\ \ k_2= 4\ ...... |\ \therefore K_{S_B}=\frac {25} 4 \ | \\

( D ) W i t h R e d R = 1 a n d B l u e R = 4 ⟹ k 1 = 1 ∣ k 3 = 1 4 , ∣ . . . . . . . . . M i d d l e . . . . . . . . . . ∣ k 2 = 9 4 . . . . . ∣ ∴ K M d b o t h = 7 ∣ . . . . . . . . . S m a l l , . . . . . . . . . . ∣ k 2 = 7 . . . . . . ∣ ∴ K S b o t h = 57 4 ∣ (D) \ With\ Red \ \ R=1\ \ and \ \ Blue \ \ R=4 \ \ \implies ~~k_1=1 \ |k_3=\frac 1 4,\ |\\ ......... Middle .......... |\ \ k_2=\frac 9 4\ ..... |\ \therefore K_{Md_both}= 7 \ | \\ ......... Small ,.......... |\ \ k_2= 7\ ...... |\ \therefore K_{S_both}=\frac {57} 4 \ | \\

Adding all the seven Ks, we get 209 4 . a + b = 213 \dfrac{209} 4. ~~~a+b= \Huge \color{#D61F06}{\boxed {~213~}}

Referance:- link text

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