3 circle

Geometry Level 2

The arc of circle A A is tangent to the circle B B and the circle C C . Let the radius of the circle A A be r r and the radius of the circle A A equal to 4 4 times the circle radius B B and the radius of Circle C C Is s s . Find the value of r s \large \frac { r }{ s } .


The answer is 9.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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2 solutions

This is a well know problem from Sangaku Circle . You might also want to see Ford Circle . Similar problem with this .

The formula is 1 r left + 1 r right = 1 r middle \dfrac{1}{\sqrt{r_{\text{left}}}} + \dfrac{1}{\sqrt{r_{\text{right}}}} = \dfrac{1}{\sqrt{r_{\text{middle}}}}

So 1 r + 1 1 4 r = 1 s 3 r = 1 s 3 s = r 9 s = r r s = 9 \begin{aligned} \dfrac{1}{\sqrt{r}} + \dfrac{1}{\sqrt{\frac{1}{4}r}}&= \dfrac{1}{\sqrt{s}}\\ \dfrac{3}{\sqrt{r}} &= \frac{1}{\sqrt{s}}\\ 3 \sqrt{s} &= \sqrt{r}\\ 9s&=r\\ \dfrac{r}{s} &= \color{#D61F06}{\boxed{9}} \\ \end{aligned}

If you want the proof, Wildan's solution has it.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

We can simplify this problem with r = 4 k r = 4k for radius of circle A A so that the radius of circle B B is k k .

( A ′ B ) 2 = ( 5 k ) 2 − ( 3 k ) 2 (A'B)^2 = (5k)^2 - (3k)^2

( A ′ B ) 2 = 25 k 2 − 9 2 (A'B)^2 = 25k^2 - 9^2

( A ′ B ) 2 = 16 k 2 (A'B)^2 = 16k^2

A ′ B = 4 k A'B = 4k

( E C ) 2 = ( A C ) 2 − ( A E ) 2 ( E C ) 2 = ( 4 k + s ) 2 − ( 4 k − s ) 2 ( E C ) 2 = ( 4 k + s + 4 k − s ) ( 4 k + s − 4 k + s ) ( E C ) 2 = 8 k ⋅ 2 s ( E C ) 2 = 16 k s E C = 4 k s { (EC) }^{ 2 }={ (AC) }^{ 2 }-{ (AE) }^{ 2 }\\ { (EC) }^{ 2 }={ (4k+s) }^{ 2 }-{ (4k-s) }^{ 2 }\\ { (EC) }^{ 2 }=(4k+s+4k-s)(4k+s-4k+s)\\ { (EC) }^{ 2 }=8k\cdot 2s\\ { (EC) }^{ 2 }=16ks\\ EC=4\sqrt { ks }

( F C ) 2 = ( B C ) 2 − ( F B ) 2 ( F C ) 2 = ( k + s ) 2 − ( k − s ) 2 ( F C ) 2 = ( k + s + k − s ) ( k + s − k + s ) ( F C ) 2 = 2 k ⋅ 2 s ( F C ) 2 = 4 k s F C = 2 k s { (FC) }^{ 2 }={ (BC) }^{ 2 }-{ (FB) }^{ 2 }\\ { (FC) }^{ 2 }={ (k+s) }^{ 2 }-{ (k-s) }^{ 2 }\\ { (FC) }^{ 2 }=(k+s+k-s)(k+s-k+s)\\ { (FC) }^{ 2 }=2k\cdot 2s\\ { (FC) }^{ 2 }=4ks\\ FC=2\sqrt { ks }

E F = E C + F C 4 k = 4 k s + 2 k s 4 k = 6 k s 16 k 2 = 36 k s 16 k = 36 s 36 16 = k s 36 4 = 4 k s 9 1 = r s EF=EC+FC\\ 4k=4\sqrt { ks } +2\sqrt { ks } \\ 4k=6\sqrt { ks } \\ 16{ k }^{ 2 }=36ks\\ 16k=36s\\ \frac { 36 }{ 16 } =\frac { k }{ s } \\ \frac { 36 }{ 4 } =\frac { 4k }{ s } \\ \frac { 9 }{ 1 } =\frac { r }{ s }

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