Start with a circle of some radius - call it Ro.

Inscribe a triangle inside this circle.  Now inscribe a new circle inside the triangle.  Inside 
the new circle, inscribe a square.  Inside the square, inscribe another circle.  Inside that 
circle, inscribe a pentagon.  Etc.  Etc.

The question is:  does this progression converge to a nonzero radius?  Or does it just keep 
shrinking to zero?

Here's my stab at the problem:

The radius of the nth circle goes like:

Rn = Ro * product[m=3,n] ( cos(pi/m) )

So as n goes to infinity:

Rinf = Ro * product[m=3,inf] ( cos(pi/m) )

If we take Ro = 1, then the computer shows a
convergence to Rinf = 0.11499878... 

To get a cheap approximation to this number, I took
the natural logarithm of both sides of the Rinf
equation:

ln(Rinf) = ln(Ro) + sum[m=3,inf] ( ln(cos(pi/m)) )

Let's be lazy bastards and say that ln(x) = x-1 if
we're really close to x=1.  Let's also set Ro=1 to get
rid of that term.  Now we have:

ln(Rinf) = sum[m=3,inf] ( cos(pi/m) - 1 )

Let's get even cheaper and say that cos(pi/m) = 1 -
0.5*(pi/m)^2.  We're left with:

ln(Rinf) = sum[m=3,inf] ( -0.5*(pi/m)^2 )

That's just a harmonic series of power 2.  
sum[m=1,inf] ( m^-2 ) = pi^2 / 6, which is zeta(2). 
Three cheers for the Riemann Zeta Function.  

Our approximation is now:

ln(Rinf) = -0.5*(pi^2)*( (pi^2)/6 - 1.0 - 0.25)

We have to subtract out the m=1 and m=2 terms from the
sum because we started with a triangle at m=3.  That
leaves:

ln(Rinf) = -1.9489...

Rinf = 0.1424...

That's not too far from Rinf = 0.115...

Is there a closed form for sum( ln(cos(pi/m)) )?  Anyone?