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Topic: Kissing Circles (Read 2405 times) |
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ThudnBlunder
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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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towr
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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), |
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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
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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
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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
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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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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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