```

wu :: forums
(http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)

riddles >> medium >> Kissing Circles
(Message started by: ThudanBlunder on Aug 23rd, 2010, 5:00am)

```

Title: Kissing Circles
Post by ThudanBlunder on Aug 23rd, 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 23rd, 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 23rd, 2010, 2:03pm

on 08/23/10 at 05:48:35, towr wrote:
 [hide]1/2+sqrt(5)/2+sqrt(1/2 (1+sqrt(5))) and it's inverse), [/hide]

That is [hide]phi + sqrt(phi)[/hide] right? It's beautiful!

-- AI

Title: Re: Kissing Circles
Post by towr on Aug 23rd, 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 (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?
[/hide]

[hide]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)[/hide]
[/edit]

Title: Re: Kissing Circles
Post by ThudanBlunder on Aug 23rd, 2010, 3:53pm

on 08/23/10 at 14:03:45, TenaliRaman wrote:
 That is [hide]phi + sqrt(phi)[/hide] right?-- AI

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 + 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 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
r2 - (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 29th, 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...