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Topic: Kissing Circles (Read 2255 times) 

ThudnBlunder
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Kissing Circles
« on: Aug 23^{rd}, 2010, 5:00am » 
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If the radii of four mutually kissing circles are in geometric progression, find exact possible values for the common ratio.


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towr
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Re: Kissing Circles
« Reply #1 on: Aug 23^{rd}, 2010, 5:48am » 
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You can use the formula at http://mathworld.wolfram.com/SoddyCircles.html which gives the radius of 4 touching circles. Then you only need to impose the constraints they're in geometric progression; which gives you a six degree polynomial to solve. Wolframalpha is happy enough to solve it (giving 1/2+sqrt(5)/2+sqrt(1/2 (1+sqrt(5))) and it's inverse), but I'll think a bit more about whether there's a nice way to do it by hand.

« Last Edit: Aug 23^{rd}, 2010, 5:49am by towr » 
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TenaliRaman
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Re: Kissing Circles
« Reply #2 on: Aug 23^{rd}, 2010, 2:03pm » 
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on Aug 23^{rd}, 2010, 5:48am, towr wrote:1/2+sqrt(5)/2+sqrt(1/2 (1+sqrt(5))) and it's inverse), 
 That is phi + sqrt(phi) right? It's beautiful!  AI

« Last Edit: Aug 23^{rd}, 2010, 2:04pm by TenaliRaman » 
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towr
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Re: Kissing Circles
« Reply #3 on: Aug 23^{rd}, 2010, 2:50pm » 
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Yup, it is. But I still don't see a nice way to derive it. The Soddy circles formula plus geometric constraint gives 2 (x^{6} + x^{4} + x^{2} + 1) = (x^{3} + x^{2} + x + 1)^{2} Which simplifies to (x^{2} + 1) (x^{4}  2 x^{3}  2 x^{2}  2 x + 1) = 0 We can dismiss (x^{2} + 1) as a source for real solutions, so then we have x^{4}  2 x^{3}  2 x^{2}  2 x + 1 = 0 And then I'm stuck letting wolframalpha finishing it off. Any ideas? [edit] We can use that if x is a solution, then so is 1/x Expand (xa)(x1/a)(xb)(x1/b), the coefficients have to be equal to the ones of x^{4}  2 x^{3}  2 x^{2}  2 x + 1, so we get  1/a  a  1/b  b = 2 2 + 1/(a b) + a/b + b/a + a b = 2 Take a' = a+1/a b' = b+1/b Then a'+b'=2 a'*b'= 4 a'*(2a') = 4 Which gives a' = 1 + sqrt(5) b' = 1  sqrt(5) (or we can exchange a' and b') For real a,b, we'd have a',b' >= 2, so we only need to look at a+1/a = 1 + sqrt(5) So, then a+1/a = 2phi a^{2}  2phi a + 1 = 0 a = (2phi +/ sqrt(4 phi^{2}  4))/2 = phi +/ sqrt(phi) [/edit]

« Last Edit: Aug 23^{rd}, 2010, 3:35pm by towr » 
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ThudnBlunder
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Re: Kissing Circles
« Reply #4 on: Aug 23^{rd}, 2010, 3:53pm » 
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on Aug 23^{rd}, 2010, 2:03pm, TenaliRaman wrote: That is phi + sqrt(phi) right?  AI 
 Yes. Here is a painless method for solving the polynomial. To paraphrase Snoddy, "The sum of the squares of the radii equals half the square of their sum." Algebraically, (1 + r + r^{2} + r^{3})^{2} = 2(1 + r^{2} + r^{4} + r^{6}) Factoring out 1 + r^{2}, we have r^{4}  2r^{3}  2r^{2}  2r + 1 = 0 Because the GP is either increasing or decreasing, if r is a solution then 1/r is also a solution. So we should expect a factor of the form r^{2}  kr + 1, where k = r + (1/r), thus getting r^{4}  2r^{3}  2r^{2}  2r + 1 = (r^{2}  kr + 1)[r^{2} + (k  2)r + 1] Comparing coefficients, k^{2}  2k  3 = 1 and so k = 1 5 Choosing the positive root, and leaving the negative root for the anally retentive to write a song about, we have r^{2}  (1 + 5)r + 1 = 0 and finally, r = There are two solutions because the 4th circle can be either internally or externally tangent to the other three.

« Last Edit: May 31^{st}, 2011, 3:08pm by ThudnBlunder » 
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Noke Lieu
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Re: Kissing Circles
« Reply #5 on: Sep 29^{th}, 2010, 9:23pm » 
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1 minus the square root of five, You're part of an answer I derive. Though a number you are I can't go that far. Oh, how will our love survive? ...now for some black coffee...


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