Author |
Topic: Circumscribing Circles (Read 565 times) |
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
Circumscribing Circles
« on: Nov 26th, 2008, 8:51am » |
 Quote  Modify
|
Draw a circle of radius 1 and circumscribe it with an equilateral triangle. Now draw the circumscribing circle of the triangle and then circumscribe this circle with a square. Continue in this fashion, drawing a circumscribing n-gon. then its circumscribing circle, and then the (n+1)-gon which circumscribes the circle.
Do the circumscribing circles increase without limit?
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Circumscribing Circles
« Reply #1 on: Nov 26th, 2008, 9:20am » |
Quote Modify
|
The size of the circles, prodk=3..inf 1/cos(pi/(2k)), seems to converge, but I don't know how to make sure.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: Circumscribing Circles
« Reply #2 on: Nov 28th, 2008, 10:56am » |
Quote Modify
|
on Nov 26th, 2008, 9:20am, towr wrote:| The size of the circles, prodk=3..inf 1/cos(pi/(2k)), seems to converge, but I don't know how to make sure. |
|
Yes, it converges, even with 2k replaced by k. 
Can you give an answer to a few decimal places?
|
|
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Circumscribing Circles
« Reply #3 on: Nov 28th, 2008, 11:50am » |
Quote Modify
|
on Nov 28th, 2008, 10:56am, ThudanBlunder wrote:| Yes, it converges, even with 2k replaced by k. |
|
Ah, yes, I meant 2 /(2k); I was supposed to be dividing the whole circle in 2k angles, not just half.
Quote:| Can you give an answer to a few decimal places? |
|
Seems to be 8.700 Of course the precision leaves to be desired, since any small rounding errors at the machine-word level explode in multiplications.
|
| « Last Edit: Nov 28th, 2008, 11:51am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Circumscribing Circles
« Reply #5 on: Nov 28th, 2008, 11:56am » |
Quote Modify
|
| hidden: | log[1/cos(/n)] = 2/(2n2) + O(1/n4),
so the sum coverges.
Moreover, for any finite N, we can find both
 n  N 1/cos(/n) and  n>N 2/(2n2)
in closed form. Taking N=100 gives 8.70001. But if we take the first 4 terms in the Taylor series, then N=100 gives 8.7000366252081943, accurate to 15 digits.
r = 8.7000366252081945032224098591130... |
Edit: Ah, this is more or less equivalent to formula 7 at Mathworld I guess, once one knows the Taylor series of log sec.
|
| « Last Edit: Nov 28th, 2008, 12:10pm by Eigenray » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print