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Thread: Three circles [or 4]

20061016, 02:23 #1
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Three circles [or 4]
Notwithstanding the accuracy of the drawing...
There are three circles with radii of 6, 7, and 8. The three circles are tangent to one another. In the area formed by the three larger circles, there is a 4th circle that is also tangent to the other three. What is the radius of the smaller, shaded 4th circle? Ricky

20061016, 10:31 #2
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Re: Three circles [or 4]
Using a combination of equatiions and brute force: <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">~ 1.070064</span hide>

20061016, 17:26 #3
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Re: Three circles [or 4]
We have yet to work this one out for ourselves. I am assuming (for the moment) that your answer is correct <img src=/S/smile.gif border=0 alt=smile width=15 height=15>. So, please enjoy your reward of 1/2 pound of yummy Belgian choccy. <img src=/S/yum.gif border=0 alt=yum width=15 height=15>
I hope we're on the right track... <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">Did your solution begin by forming a triangle whose sides were formed by connecting the three centerpoints of the larger circles?</span hide> Ricky

20061016, 19:08 #4
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Re: Three circles [or 4]
Yep!

20061018, 20:51 #5
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Re: Three circles [or 4]
OK
I have been playing with this on and (mostly) off for a few days now and I was thinking that this was solved using Descartes' Theorem or four kissing circles.
The general terms I have gathered are:
X =(Kone+Ktwo+Kthree)Jerry

20061018, 21:20 #6
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Re: Three circles [or 4]
I placed the centres of the circles on a coordinate grid. The centres of the large circles can be easily computed; that of the small circle is unknown, so call it (x, y). You can then translate the requirement of the puzzle into an algebraic equation. I worked this out on paper. I then used Excel's goal seeker to solve this equation  that was the brute force.
Descartes' Theorem is indeed the 'official solution of the puzzle. With radii of 6, 7 and 8, the curvatures are 1/6, 1/7 and 1/8. The equation evaluates to
k = (1/6+1/7+1/8) +/ 2*SQRT(1/6*1/7+1/6*1/8+1/7*1/8) = 73/168 +/ 1/2
The minus variant results in k = 11/168 or r = 168/11. This is the radius of the 'outer' circle touching the three.
The plus variant results in k = 157/168 or r = 168/157. This is the radius of the 'inner' circle touching the three, i.e. the shaded circle.

20061018, 21:20 #7
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Re: Three circles [or 4]
<img src=/S/blush.gif border=0 alt=blush width=15 height=15> I see where I went wrong now, but I now concur with you <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>
Jerry

20061019, 05:07 #8
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Re: Three circles [or 4]
I pulled out a cad programs, drew the three circles as shown, and then the 4th in the middle.
Using the "properties" function, I found the radius of the small circle, and then calculated the area as requested.
For the record, I got an answer only slightly higher than that of Hans. .004 or so highter.Christopher Baldrey

20061019, 12:12 #9
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Re: Three circles [or 4]
Chris
This a fun java page with it in
Soddy Circles ...enjoyJerry

20061020, 11:25 #10
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Re: Three circles [or 4]
thats how i would have done it <img src=/S/laugh.gif border=0 alt=laugh width=15 height=15>
in catia v4: curve2 > circle > multitangent (click the 3 circles), then just analyze the new circle <img src=/S/thumbup.gif border=0 alt=thumbup width=15 height=15><img src=/w3timages/blueline.gif width=33% height=2>
<big>John</big>

20061020, 18:24 #11
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Re: Three circles [or 4]
Computers (once in a while) do actually make things easier. It's not just a rumour anymore!
Christopher Baldrey