382 pickover/pickover.06.p
Description
This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.
382 pickover/pickover.06.p
Title: Cliff Puzzle 6: Star Chambers
From: cliff@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself so I can cite you appropriately if you provide unique
information. PLEASE ALSO directly mail me a copy of your response in
addition to any responding you do in the newsgroup. I will assume it is
OK to describe your answer in any article or publication I may write in
the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
As many of you probably know, 5-sided stars produced by drawing a
continuous line with your pencil can nest inside each other. (One star
can sit inside the pentagon produced by the larger star. Each of the
5 points of the small star coincide with the 5 points of the
internal pentagon of the large star.)
Start with a five sided star formed with 5 line segments, each 1 inch
long. Continually nest stars so that the assembly of stars gets bigger
and bigger.
Questions:
1. How many nestings N are required to make star N
have an edge-length equal to the diameter of the sun (4.5E9 feet)?
2. How many nestings N are required to make the cumulative length
of lines of all the nested stars equal to the diameter of the sun?
pickover/pickover.06.s
-------------------------
Cliff Pickover,
So here I am, waiting to see if one of my long Grobner basis
calculations is going to finish before the machine goes down.
This is a good time to read news, and I came across this trivial
problem in rec.games.puzzles. I'm not sure why I'm responding,
perhaps the hour, or perhaps curiousity to see what will come
of this, but I could have done this the day in high school
when I learned how to compute cos(pi/5). The ratio between
side lengths of successive pentagrams is r = (3+sqrt(5))/2
= 1 + golden ratio = 2.618... . The smallest N for which
r^N > 5.48e10 (slightly more accurate value for sun's diameter
in inches) is 26, with r^26 = 7.37e10. The smallest N for which
5[r^(N+1)-1]/(r-1) > 5.48e10 is 24, with 5(r^25 - 1)/(r-1) = 8.70e10.
This seems too trivial to post, but do with this response as you like.
Bob Holt
-------------------------
I just started reading 'rec.puzzles', so have just seen this one and
the one before (#5)... and to be honest I'm not sure why you put this one
out, since it is pretty straightforward.
>Start with a five sided star formed with 5 line segments, each 1 inch
>long. Continually nest stars so that the assembly of stars gets bigger
>and bigger.
The analytical (and general) answer to this problem comes from the
basic relationship of a "chord" of a regular pentagon, which is defined
as follows:
_=*=_
_=/ / \=_
_=/ | \=_
_=/ | \=_
* / *
| | <-- "chord" |
\ | /
| / |
\ | /
| / |
*-------------*
compared to the length of one of the sides is the golden ratio, i.e.
_
1 + \/5
--------- .
2
It can then be derived that the length of the "chord" (i.e. segment
length) of the next bigger Star compared to the length of the "chord"
of its incribed Star is the square of the golden ratio, or the golden
ratio plus one, same thing.
>Questions:
>1. How many nestings N are required to make star N
>have an edge-length equal to the diameter of the sun (4.5E9 feet)?
Back-of-envelope calculations as follows:
4.5E9 * 12 = total of 5.22E10 inches. ratio of Star sizes approx. 2.618.
The best answer is 27 nested Stars, although it produces a Star with
a "chord" length of 7.366E10 inches, i.e. a bit bigger. The first, and
smallest Star, is assumed to be the one with "chord" length of 1 inch.
>2. How many nestings N are required to make the cumulative length
>of lines of all the nested stars equal to the diameter of the sun?
This is just the sum of a geometric sequence with the ratio being
the golden ratio squared (or the golden ratio plus one).
_
/ 1 + \/5 \ 2
So, S = 1 inch, and S = S | --------- |
0 n n-1 \ 2 /
The sum is just the standard geometric summation, which I can't remember
offhand, but the contributing terms in the sum (other than the last term),
are less than one 1.6th of the total (by conincidence the inverse of the
golden ratio ;-). This means that the 25th Star (term) is the determining
factor, and in this case is the answer with a total length of 8.694E10
inches amoung all of them, and 5.373E10 inches for just the sum of the
segments of the 25th Star (again, counting the first one as side length
of 1 inch, or sum of 5 inches).
Well, that's it, I guess. Sorry to be so exhaustive, but I like to
use analytical methods to be sure I have the right answer.
My .signature explains most of what you need to know. What I mean
by "Honorary Grad Student" is that I have been taking Grad math classes
since my sophomore year, and for all intensive purposes might as well
be one. My Snail-mail address is 1521 S.W. 66th Ave., Portland, OR
97225. As to info about myself... I love learning about things, and
mathematics and consequently computers tend to be a great focus.
Anyway, if you have any more puzzles, pass them along... I am still
pondering on that sequence (puzzle #5) that you posted.
Thanks for your time.
Erich
--
"I haven't lost my mind; I know exactly where it is."
/ -- Erich Stefan Boleyn -- \ --=> *Mad Genius wanna-be* <=--
{ Honorary Grad. Student (Math) } Internet E-mail: <erich@gemini.mth.pdx.edu>
\ Portland State University / WARNING: INTERESTED AND EXCITABLE
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