

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

Title: Re: Kissing Circles Post by towr on Aug 23^{rd}, 2010, 5:48am [hide]You can use the formula at [/hide][hide]http://mathworld.wolfram.com/SoddyCircles.html[/hide] (http://mathworld.wolfram.com/SoddyCircles.html)[hide] 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.[/hide] 

Title: Re: Kissing Circles Post by TenaliRaman on Aug 23^{rd}, 2010, 2:03pm on 08/23/10 at 05:48:35, towr wrote:
That is [hide]phi + sqrt(phi)[/hide] right? It's beautiful!  AI 

Title: Re: Kissing Circles Post by towr on Aug 23^{rd}, 2010, 2:50pm Yup, it is. But I still don't see a nice way to derive it. [hide] 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? [/hide] [edit] [hide]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)[/hide] [/edit] 

Title: Re: Kissing Circles Post by ThudanBlunder on Aug 23^{rd}, 2010, 3:53pm on 08/23/10 at 14:03:45, TenaliRaman wrote:
Yes. Here is a painless method for solving the polynomial. [hide]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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 Choosing the positive root, and leaving the negative root for the anally retentive to write a song about, we have r^{2}  (1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5)r + 1 = 0 and finally, r = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif There are [/hide]two solutions (http://www.gogeometry.com/geometry/soddy_descartes_circles.htm)[hide] because the 4th circle can be either internally or externally tangent to the other three.[/hide] 

Title: Re: Kissing Circles Post by Noke Lieu on Sep 29^{th}, 2010, 9:23pm 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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