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Circles and Polygons (Posted on 2010-09-05) Difficulty: 3 of 5
A unit circle is inscribed in an equilateral triangle around which is a circumscribed circle, which is inscribed in a square. The process continues as circle, "N gon", circle, "N+1 gon", circle ... to infinity.
All the polygons are regular.
Is the final circle finite, and if so, what is it's radius?

No Solution Yet Submitted by Larry    
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Solution TMI (Spoiler) Comment 1 of 1

In researching the ratio of the circumradius to the inradius for a polygon, I was directed to a site that actually gave away the numeric answer to this puzzle (Wolfram MathWorld: Polygon Circumscribing). So I'll just try to add more precision:

The ratio of circumradius to inradius is:

R/r = 1/cos(pi/n), where n is the number of sides of the polygon.

The desired answer is then the infinite continued product:

Pi{n=3 to inf}sec(pi/n)

But this converges slowly. I stopped the following program:

10   point 25
20   R=1
30   for N=3 to 100000000
40     R=R/cos(#pi/N):if N @ 1000000=0 then print N;R:print
50   next

when it reached R for a 38-million-gon. The last two printed results were:

 37000000  8.7000354648580204580258852546389132977140497012373785478176534839011
 38000000  8.7000354953935489694478082618686095072775991541917901862466964171263

which were not even good to 7 significant figures compared to the Wolfram article.

But that article also offers an accelerated convergence product:

B = Pi{n=3 to inf}((1 - pi^2/(2*n^2) + pi^4/(24*n^4))*sec(pi/n))

which is then multiplied by a complicated constant, for which see the Wolfram article.

The convergence is still rather slow. Here's the program:

   10   point 25
   20   Coef=sqrt(6)/12*#pi^4*(1-#pi^2/2+#pi^4/24)*(1-#pi^2/8+#pi^4/384)/sin(#pi
   30   R=Coef
   40   for N=3 to 200000
   50       R=R*(1-#pi^2/(2*N^2)+#pi^4/(24*N^4))/cos(#pi/N)
   60       if N @ 100=0 then print N,R:print
   70   next

This time after "only" 199,900 and 200,000 iterations respectively, the following are the last two printed successive approximations:

 199900          8.7000366252081945032224098518343955651383396052547892540453533
 200000          8.7000366252081945032224098518525738558260586371560081778797694

Note, of course, that the n values no longer refer to the n-gons as they did in the first program, as the convergence is accelerated.

So, judging from where the values are still changing, the best we can approximate the value is 8.7000366252081945032224098518..., where that 8 at the "end" would be rounded up to 9 if the approximation were terminated there. It agrees with the Wolfram approximation given as 8.700036625....

Edited on September 5, 2010, 2:39 pm
  Posted by Charlie on 2010-09-05 14:38:52

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