Date: Thu, 26 Oct 95 14:43:55 CDT From: rusin (Dave Rusin) To: surgent@imap2.asu.edu Subject: Re: Formula for area of a nonagon? In article you write: >Is there a formula that gives the area of a nonagon (regular 9-sided >polygon)? i.e. given the length of a side and an apothem, can one >calculate the area of the nonagon? Better yet, does one exist that >completely avoids trig? My answers are yes, eh?, and maybe, respectively. The area is 18 times the area of a "slice", which is a right triangle with a 20-degree angle. Thus the total area is (9/2)ab, where a is the distance from the center of the enclosing circle to the midpoint of a side, and b is the length of a side. The relationship of these two is that tan(20*pi/180)=(b/2)/a. I don't know what an apothem is, but assuming the usual use of "apo" from greek, this ought to be the radius of the enclosing circle, which is sqrt(a^2+(b/2)^2). You can use trig to rewrite this one too. I'm inclined to say there is no non-trigonometric formula which involves just one of these parameters, since the 20 degree angle is known not to be constructible with compass and straightedge, so there should be no "legal" way to dispense with one of them. However, there are ways of trisecting an angle with _marked_ straightedges, so I suppose you can draw a nice diagram starting only with a (say) and get a rectangle with the same area as the nonagon. dave >Email me at surgent@imap2.asu.edu Your newsreader affixed your address as ADGNR@asuvm.inre.asu.edu; perhaps you can get your site administrator to change that if you don't like it. ============================================================================== Date: Thu, 26 Oct 1995 13:05:06 -0700 (MST) From: surgent@IMAP2.ASU.EDU Subject: Re: Formula for area of a nonagon? To: Dave Rusin Dave- Thanks for your input- I was hoping to avoid trig, but I guess it's not possible. This has been a problem that always bugged me- If we can construct a 40 dgree angle, we should be able to find an exact value for sin 40, etc. I have a list of exact values for sin and cos of all angles of multiple 3, starting with 3, 6, 9, etc., up through 45. If I could get a sin 40, then I could use the trig identities to find sin 1, sin 2, etc. This has, as far as I know, no social value, but it does entertain me sometimes... Scott Surgent, ASU math ============================================================================== Date: Thu, 26 Oct 95 16:38:43 CDT From: rusin (Dave Rusin) To: surgent@IMAP2.ASU.EDU Subject: Re: Formula for area of a nonagon? Well, you _can_ get sin(40) etc., just not by using square roots and so on. You know sin(3x)=sin(2x+x)=2sinxcos^2 + (cos^2x-sin^x)sinx = 2 S (1-S^2)+ (1-2S^2)S = 3S-4S^3 (S=sin x), so for example since sin(120)=sqrt(3)/2 we know that S=sin(40) is the root of the following cubic: 4S^3 - 3S + sqrt(3)/2 = 0. Well, there's a formula for the roots to a cubic. I don't remember it off hand but it must be in CRC Handbooks and abstract algebra books; it's s something like: the roots of x^3+px+q=0 are (get ready) cuberoot(-q/2 + sqrt(q^2/4+p^3/27)) + cuberoot(-q/2 - sqrt(q^2/4+p^3/27)). So there, you have a formula for the sine of 40. There is a little problem here -- this formula should also give the sines of 40+360*n/3 for any n by the same reasoning. Well, no problem -- a cubic has three roots. You get all the roots by taking all the cube roots above (recall cube root of a can be r*w where r is any fixed cube root and w is any of the three cube roots of 1). But this suggests that you need to invoke complex numbers to get your answers, rather an unexpected development. Oh, you can try to get around that by sticking to the real parts, but you'll find these look like trigonometric formulae in general -- I'll bet a nickel it simplifies in this particular cubic to simply S= ... sin(40) ! So then what have you gained? One last response of course is to declare sin(40) to be a root of the above cubic and then solve the cubic numerically, say by Newton's method, but that lacks a certain elegance. Otherwise, you're SOL. dave ============================================================================== Date: Fri, 27 Oct 1995 11:24:58 -0700 (MST) From: surgent@IMAP2.ASU.EDU Subject: Re: Formula for area of a nonagon? To: Dave Rusin Dave- I've actually tried to calculate sin 20 using that sin^3 trig identity you mentioned. But I end up with a negative under the radical that is in turn inside the cube root radical. Very messy. I understand there are limitations to the general cubic formula, and I think I'm violating them... There are proofs out there that show the non-constructability of the septa- and nonagons. I have a feeling that I am trying to find something that is simply impossible. Oh well. It is an entertaining problem. I'm even looking at some series solutions. But it's elegance I'm after. I've never liked "decimal expansions", simply because I wanted to know where they came from! I was able to come up with an exact value fr sin 3, which is quite long, but involves nothing more than the 4 basic operations and the use of radicals. Your help is most appreciated. Again, this isn't earthshattering stuff I'm seeking; just a hobby-like problem that I've been bandying about for some time now. Let me know what other ideas you encounter. Scott Surgent