wu :: forums
wu :: forums - Kissing Circles

Welcome, Guest. Please Login or Register.
Sep 23rd, 2020, 12:16pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   medium
(Moderators: ThudnBlunder, william wu, Eigenray, Grimbal, Icarus, towr, SMQ)
   Kissing Circles
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Kissing Circles  (Read 2259 times)
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Kissing Circles  
« on: Aug 23rd, 2010, 5:00am »
Quote Quote Modify Modify

If the radii of four mutually kissing circles are in geometric progression, find exact possible values for the common ratio.
IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
Posts: 13723
Re: Kissing Circles  
« Reply #1 on: Aug 23rd, 2010, 5:48am »
Quote Quote Modify Modify

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 23rd, 2010, 5:49am by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
TenaliRaman
Uberpuzzler
*****



I am no special. I am only passionately curious.

   


Gender: male
Posts: 1001
Re: Kissing Circles  
« Reply #2 on: Aug 23rd, 2010, 2:03pm »
Quote Quote Modify Modify

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
« Last Edit: Aug 23rd, 2010, 2:04pm by TenaliRaman » IP Logged

Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
Posts: 13723
Re: Kissing Circles  
« Reply #3 on: Aug 23rd, 2010, 2:50pm »
Quote Quote Modify Modify

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]
« Last Edit: Aug 23rd, 2010, 3:35pm by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: Kissing Circles  
« Reply #4 on: Aug 23rd, 2010, 3:53pm »
Quote Quote Modify Modify

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.
 
« Last Edit: May 31st, 2011, 3:08pm by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
Noke Lieu
Uberpuzzler
*****



pen... paper... let's go! (and bit of plastic)

   
WWW

Gender: male
Posts: 1884
Re: Kissing Circles  
« Reply #5 on: Sep 29th, 2010, 9:23pm »
Quote Quote Modify Modify

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...
IP Logged

a shade of wit and the art of farce.
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright 2000-2004 Yet another Bulletin Board