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Topic: geometry challenge -- annular region
Replies: 9   Last Post: Jan 14, 2014 10:30 AM

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James Waldby

Posts: 308
Registered: 1/27/11
Re: geometry challenge -- annular region
Posted: Jan 10, 2014 3:58 PM
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On Fri, 10 Jan 2014 12:36:51 -0800, konyberg wrote:
> On Friday, January 10, 2014 4:19:51 AM UTC+1, quasi wrote:
>> Here's a geometry challenge (elementary, but perhaps a little
>> rough) ...
>> No proof needed, but an exact answer is required (not just a
>> numerical estimate) together with theoretically based heuristic
>> reasoning to support the result.

>> Everything takes place in R^2.
>> Let n be an integer with n >= 3.
>> Let r be a real number with sin(Pi/n) < r < 1.
>> Let P_1, ..., P_n be the vertices of a regular n-gon inscribed
>> in the standard unit circle.

>> For i = 1, ..., n, let D_i be a disk of radius r centered at P_i.
>> Let a(n,r) be the largest possible area for an annulus centered
>> at the origin and contained in the union of the disks
>> D_1, ..., D_n.

>> Let b(n,r) = n*(Pi*(r^2))
>> Let f(n,r) = a(n,r)/b(n,r)

>> Question:
>> What is the least upper bound of the range of f?

> a(r,n); is it equal to the difference between the circle spanned by
> the circles of radius 1 and the circles spanned by sin(Pi/n)? This
> is not clear to me.

a(r,n) is the area of the largest annulus covered by D_1...D_n.
Call that largest annulus L(r,n), and suppose the inner circle
of L is P(r,n) and the outer circle of L is Q(r,n). Since r<1,
there is an open area in the center of the coverage of D_1...D_n,
hence 1 > radius(P) > 0. Also, 1+r > radius(Q) >= 1.


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