Including the 2 newest states, let alone what the next one might be (Puerto Rico? England? Canada? Afghanistan? Iraq?) would cause the east coast states to be so scrunched up that the flag would need to be 500 feet across to show them, unless the chosen "centres" were located way outside the "states" borders. We can simplify the problem enormously by merely omitting stars for all states that went for Bushie Jr. Naturally, Florida gets a star. Or not. Steve Gray ----- Original Message ----- From: "Jon Perry" <perry@globalnet.co.uk> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Saturday, September 06, 2003 1:56 AM Subject: RE: [math-fun] more 51 star designs

How about taking a snapshot of the 51 states, select a representive centre for each state, transform the curved image onto a flat image, and let each centre become a star?

Can we reverse the question - how many states should the USA have in order to produce the most aesthetic flag? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Shel Kaphan Sent: 09 September 2003 00:32 To: math-fun Subject: Re: [math-fun] more 51 star designs * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * This is the last flag we'll need no matter how many new states, because the "ellipsis" can be interpreted as "etc." _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

Assuming that a rectangular design is desirable and assuming that symmetry is desirable. One way to get this symmetry with 51 stars is to locate one of the stars in the middle of the rectangle and arrange the other stars symmetrically around it. Some of these have been posted here. A second way to approach the problem is to see it as a packing problem of packing circles into a rectangle. Here the desire may not be to get the most dense packing, but a packing that also displays symmetry. One way to do this is to generates a series of packings based on 4 rows of 8 stars and 3 rows of 7 stars less 2 stars. Here are some of these possibilities: 1. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 2. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 3. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * etc. These all have a rectangular convex hull. We could also pack the stars into a non rectangular box. To get something like. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Which to me at least, seems aesthetically pleasing. This is also a variation on the 4 rows of 8 and 3 of 7 where the 2 extras have been removed from the top and bottom rows. Making 5 rows of 7 and 2 of 8. This is my favorite so far. On a waving flag, the arrangement would look rectangular and not have the holes in it that the other suggestions above have. One could also take the following arrangement: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * and streatch the outside rows proportionally in order to get a rectangular convex hull that doesn't touch the endpoints of the 2nd 4th and 6th rows. Regards Otto otto@olympus.net