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Ring of Circles (Posted on 2013-08-17) Difficulty: 3 of 5

Define a Ring of Circles ROC(n,r,P,a) as a set of
n≥3 circles (each with radius r, each with center on
a circle with center P and radius a, and each
externally tangent to both its neighbors).

With b>a, each of the circles in ROC(6,R,P,b)
is externally tangent to two of the circles in
ROC(6,r,P,a).

What is the ratio b/a?

Extra credit - what is the ratio if 6
is replaced with n in the problem?

See The Solution Submitted by Bractals    
Rating: 4.0000 (1 votes)

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Solution Solution Comment 1 of 1
Let two adjacent circles in ROC(n,r,P,a) have centres at U and V and
touch at M. Let a circle in ROC(n,R,P,b) have its centre at X on PM
produced and touch the line PU produced at T.

In triangles PMU and PTX, sin(180/n) = r/a = R/b.
Let s = sin(180/n) and c = cos(180/n), then r = as and R = bs.

From triangle MXU,         sin(/MXU) = r/(R+r) = a/(a+b),
and from triangle TUX,    sin(/TUX) = R/(R+r) = b/(a+b).

Also, /TUX = /MXU + 180/n, so using a compound angle formula:

            sin(/TUX) = sin(/MXU)cos(180/n) + cos(/MXU)sin(180/n)

            b/(a+b) = ac/(a+b) + s*sqrt(1 – a2/(a+b)2)

which simplifies to b – ac = s*sqrt(b2 + 2ab).

Squaring then produces a quadratic equation in b/a, whose greatest
root (which satisfies b>a) is:

            b/a = (c + s2 + s*sqrt(1 + 2c))/c2

 In particular, when n = 6, s = sin(30) = ½, c = cos(30) = sqrt(3)/2,
which gives:

     b/a = (1/3)(1 + 2*sqrt(3) + 2*sqrt(1 + sqrt(3))) ~ 2.58996..



  Posted by Harry on 2013-08-21 10:46:48
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