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        riddles >> putnam exam (pure math) >> Overlapping Circle and Triangle
        
(Message started by: THUDandBLUNDER on Dec 13^th, 2006, 4:14pm)

Title: Overlapping Circle and Triangle
Post by THUDandBLUNDER on Dec 13^th, 2006, 4:14pm

An equilateral triangle of unit side length overlaps a circle of radius r centred at the centre of the triangle. What value of r will minimize the shaded area in the diagram?

Title: Re: Overlapping Circle and Triangle
Post by towr on Dec 14^th, 2006, 3:47am

Possibly, it's [hide] 1/3 [/hide]. But after all the problems I encountered in my calculation, I'm not sure of anything at the moment.

Title: Re: Overlapping Circle and Triangle
Post by Grimbal on Dec 14^th, 2006, 5:47am

I would say the same on the simple principle that the proportion of circle inside the triangle must be 1/2. When it is the case the increase of area outside the triangle matches the decrease of area inside the triangle when the radius increases infinitesimally. Geometrically you find a diameter of 2/3.

Title: Re: Overlapping Circle and Triangle
Post by THUDandBLUNDER on Dec 14^th, 2006, 7:08am

Well done, Grimbal. An astute observation. As usual toiling towr is also right. :)

How about if we replace the triangle with a square?

With a regular n-gon?

Title: Re: Overlapping Circle and Triangle
Post by towr on Dec 14^th, 2006, 8:48am

In general I get [hide]sqrt(2) cot( [pi] /n)/[2 sqrt(cos( [pi] /n) + 1)][/hide]

for larger n it's [hide]approximately n/(2 [pi]) ; or better yet n/(2 [pi]) - 1/( [pi]n) [/hide]

Title: Re: Overlapping Circle and Triangle
Post by THUDandBLUNDER on Dec 14^th, 2006, 12:34pm

That's right, towr!
Slightly simpler is [hide](1/2)*cot(pi/n)sec(pi/2n)[/hide]

How did you get your 1/n(pi) refinement?

Also, the optimal circle has equal arc lengths inside and outside of the n-gon for all n.

Title: Re: Overlapping Circle and Triangle
Post by towr on Dec 14^th, 2006, 12:49pm

on 12/14/06 at 12:34:40, THUDandBLUNDER wrote:

How did you get your 1/n(pi) refinement?

I guessed :P
I plotted the difference and saw a hyperbola, and I tried a few values, pi seemed a nice one to leave it at.

Possibly a taylor series could find a better justified fit.

Title: Re: Overlapping Circle and Triangle
Post by Eigenray on Dec 14^th, 2006, 10:05pm

You would expect r ~ n/(2pi) since the circumference ~ n. In fact,

r = n/(2pi) - 5pi/(48 n) - 293pi³/(11520 n³) - ...

But lets flip it around: fix the radius of the circle at 1, and vary the polygon. Then the inradius is

r = cos(pi/2n) = [1+cos(pi/n)]/2,

and if R is the circumradius, then

rR = [1+sec(pi/n)]/2,

and the area is

A = n[tan(pi/n) + sin(pi/n)]/2.