
Re: geometry challenge  annular region
Posted:
Jan 10, 2014 3:58 PM


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 ngon 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.
 jiw

