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Topic: Kissing Circles (Read 2390 times) |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
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Kissing Circles
« on: Aug 23rd, 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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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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towr
wu::riddles Moderator Uberpuzzler
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Re: Kissing Circles
« Reply #1 on: Aug 23rd, 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.
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« Last Edit: Aug 23rd, 2010, 5:49am by towr » |
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TenaliRaman
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Re: Kissing Circles
« Reply #2 on: Aug 23rd, 2010, 2:03pm » |
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on Aug 23rd, 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
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« Last Edit: Aug 23rd, 2010, 2:04pm by TenaliRaman » |
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towr
wu::riddles Moderator Uberpuzzler
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Re: Kissing Circles
« Reply #3 on: Aug 23rd, 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 (x6 + x4 + x2 + 1) = (x3 + x2 + x + 1)2 Which simplifies to (x2 + 1) (x4 - 2 x3 - 2 x2 - 2 x + 1) = 0 We can dismiss (x2 + 1) as a source for real solutions, so then we have x4 - 2 x3 - 2 x2 - 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 (x-a)(x-1/a)(x-b)(x-1/b), the coefficients have to be equal to the ones of x4 - 2 x3 - 2 x2 - 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'*(2-a') = -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 a2 - 2phi a + 1 = 0 a = (2phi +/- sqrt(4 phi2 - 4))/2 = phi +/- sqrt(phi) [/edit]
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« Last Edit: Aug 23rd, 2010, 3:35pm by towr » |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
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Re: Kissing Circles
« Reply #4 on: Aug 23rd, 2010, 3:53pm » |
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on Aug 23rd, 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 + r2 + r3)2 = 2(1 + r2 + r4 + r6) Factoring out 1 + r2, we have r4 - 2r3 - 2r2 - 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 r2 - kr + 1, where k = r + (1/r), thus getting r4 - 2r3 - 2r2 - 2r + 1 = (r2 - kr + 1)[r2 + (k - 2)r + 1] Comparing coefficients, k2 - 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 r2 - (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.
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« Last Edit: May 31st, 2011, 3:08pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Noke Lieu
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Re: Kissing Circles
« Reply #5 on: Sep 29th, 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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