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.

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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