July 2003 - mathfun@mailman.xmission.com

Re: [math-fun] inscribed polygon area
by reid＠math.arizona.edu 30 Jul '03

30 Jul '03

> << > just by chance, i was working on something related to this the other day. > the formula for the circumradius of a convex cyclic quadrilateral, > with side lengths a , b , c and d is: > > /------------------------------------------------------------------ > / (ab + cd) (ac + bd) (ad + bc) > R = \ / ------------------------------------------------------------------ > \/ (-a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d) > > the formula for the circumradius of a triangle can be obtained by > putting d = 0 into this formula. > >> > > This is a nice formula! glad you like it; i thought it was pretty neat, too. perhaps i should have said that i derived this formula for what i was working on. i had not seen it before, though i can't imagine it's really "new". the symmetry in the variables, as you note, means that the circumradius is independent of the order of the sides. by exploiting this properly, it is possible to give a nice formula for the diagonal of a cyclic quadrilateral. i'll leave this as a cute puzzle for you. mike

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Re: [math-fun] Heron, Brahmagupta, Robbins, etc.
by asimovd＠aol.com 30 Jul '03

30 Jul '03

Henry writes: << In yesterday's (Tue, Jul 29) Wall Street Journal (!), a front page article talked about David P. Robbins's search for a generalization of Heron's & Brahmagupta's formulae for the area of polygons inscribed in circles, when given only their edge lengths. The article indicated that Dr. Robbins was using only pencil & paper, which I found a bit odd, considering that there are a number of symbolic algebra systems that I would imagine would be quite useful for this problem. >> I wasn't able to access the article, but given the above information I don't see what the puzzle is. As long as a polygon inscribed in a circle is non-self-intersecting and contains the center of the circle in its interior, its area is the sum of the areas of all triangles T_E formed by each edge E and the center of the circle. Assuming radius = 1, and that E = edgelength, the area A_E of T_E is given by A_E = E * sqrt( 1 - (E/2)^2 ) / 2. (If the center of the circle is not in the interior of the polygon and/or the polygon is self-intersecting, then the polygon's area is a signed sum of these triangles' areas.) What am I missing? --Dan

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Re: [math-fun] inscribed polygon area
by asimovd＠aol.com 30 Jul '03

30 Jul '03

Mike Reid writes: << just by chance, i was working on something related to this the other day. the formula for the circumradius of a convex cyclic quadrilateral, with side lengths a , b , c and d is: /------------------------------------------------------------------ / (ab + cd) (ac + bd) (ad + bc) R = \ / ------------------------------------------------------------------ \/ (-a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d) the formula for the circumradius of a triangle can be obtained by putting d = 0 into this formula. >> This is a nice formula! Not having thought about this before, what I find particularly surprising is that the quadrilateral circumradius formula is independent of the cyclic order of the edge-lengths. Is this easy to see a priori ? Is it also true for convex polygons of > 4 sides ? Wait -- consider any 3 consecutive (ccw) vertices A,B,C of a convex n-gon. If the triangle ABC is cut from the n-gon and re-glued so that they now appear in ccw order C,B,A, then it's easy to see that doesn't change the circumcircle (by its symmetry about the perpendicular bisector of AC). Since all permutations of the cyclic ordeer of the edges are generated by ones of this sort, that shows the circumcircle and circumradius are functions purely of the multiset of edge lengths. I think. (If true, I never realized this independence before.) --Dan

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Re: [math-fun] inscribed polygon area
by reid＠math.arizona.edu 30 Jul '03

30 Jul '03

> I think one way of expressing the hard problem is > "given a set of edge-lengths {Ei}, determine the radius of the circumscribing circle". > > IIRC, the formula for the circumradius of a triangle contains Hero's area formula > as a term. just by chance, i was working on something related to this the other day. the formula for the circumradius of a convex cyclic quadrilateral, with side lengths a , b , c and d is: /------------------------------------------------------------------ / (ab + cd) (ac + bd) (ad + bc) R = \ / ------------------------------------------------------------------ \/ (-a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d) the formula for the circumradius of a triangle can be obtained by putting d = 0 into this formula. mike

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[math-fun] inscribed polygon area
by Richard Schroeppel 30 Jul '03

30 Jul '03

> What am I missing? > Assuming radius = 1, I think one way of expressing the hard problem is "given a set of edge-lengths {Ei}, determine the radius of the circumscribing circle". IIRC, the formula for the circumradius of a triangle contains Hero's area formula as a term. Rich rcs(a)cs.arizona.edu

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[math-fun] Heron, Brahmagupta, Robbins, etc.
by Henry Baker 30 Jul '03

30 Jul '03

In yesterday's (Tue, Jul 29) Wall Street Journal (!), a front page article talked about David P. Robbins's search for a generalization of Heron's & Brahmagupta's formulae for the area of polygons inscribed in circles, when given only their edge lengths. The article indicated that Dr. Robbins was using only pencil & paper, which I found a bit odd, considering that there are a number of symbolic algebra systems that I would imagine would be quite useful for this problem. I searched the web, but the only publications I could find by Robbins on this subject were: Robbins D P Areas of polygons inscribed in a circle, Discrete Comput. Geom., 12 (1994), 223--236 David P. Robbins. Areas of polygons inscribed in a circle. Amer. Math. Monthly, 102(6):523-530, 1995 Apparently, he has found generalizations to pentagons and hexagons, but 7-sided figures currently has him stumped. The following paper talks about these papers: www.geocities.com/teufel_pi/papers/cyclic.ps.gz Dr. Robbins works here: http://www.idaccr.org, and this is his email: david.robbins(a)idaccr.org I could not find a personal web page for Dr. Robbins.

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[math-fun] Re: Kepler conjecture & Flyspeck
by Richard Schroeppel 28 Jul '03

28 Jul '03

I guess I'm a bit puzzled that some people reject computer proofs. I think they're just complicated calculations, and should be subjected to the same tests as other long calculations: Double checking, the code available for review, sometimes programmed twice by independent teams, etc. This is the same process as the Mersenne folks use, and it seems a reasonable approach. The chance of some sort of error is small but nonzero, but that's also true for long human proofs. The number of people who've checked every detail of the classification of simple groups must be tiny. I've personally checked that M13 = 8191 and M17 = 131071 are prime, and I could do M31 if pressed. But for M11213, I depend on the machine. Similarly, when I claim that 2^(1/5) generates a Euclidean number field, my proof depends on a computer check of 100,000 boxes. This in turn depends on my program, the Lisp interpreter, the editor, the bignum package, the particular machine not having hiccuped, etc. If the result is important enough, someone will do an independent check. --- There's a considerable body of machine verified proofs, from the basic axioms, of some pretty complex results: The Prime Number Theorem has been done. The Kepler Group might be able to use some of this material for a start. The issue of the reviewers giving up could be the subject of another long message. We need a way to give reviewers some credit for a thankless job. Rich rcs(a)cs.arizona.edu

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[math-fun] kepler conjecture, flyspeck
by Thane Plambeck 27 Jul '03

27 Jul '03

Good article in Nature on Kepler conjecture and Flyspeck: *********** Does the proof stack up? Think peer review takes too long? One mathematician has waited four years to have his paper reviewed, only to hear that the exhausted reviewers can't be certain whether his proof his correct. George Szpiro investigates... ************ (PDF) http://www.qxmail.com/mathematics/halesflyspeck.pdf

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[math-fun] new software for combinatorial games
by Richard Schroeppel 21 Jul '03

21 Jul '03

Several folks here play with these things. :-) Rich ---------------------- Date: Mon, 21 Jul 2003 09:49:31 -0500 From: David Wolfe <wolfe(a)gac.edu> Subject: New combinatorial games software Reply-To: wolfe(a)gustavus.edu This is a message to everyone who has sent me an inquiry in the last decade about combinatorial game theory and/or, more specifically, my software toolkit for analyzing combinatorial games. I'm happy to report that Aaron Siegel at Berkeley has written a new Combinatorial Game Suite which is a dramatic improvement over my toolkit. In particular, (1) The Suite is designed to be more flexible and extensible. For instance, you can implement the rules to a new game by writing a plug-in without making any changes to source code. (2) The Suite is written in Java making it portable. (It's been text on Windows 2000/XP, Linux and Solaris.) (3) Aaron has provided additional functionality that were not part of my toolkit distribution, such as support for loopy games and thermographs. (4) The Suite's has an easy to use graphical user interface. If you are still interested in doing investigations in combinatorial games, I highly recommend using his software. It's available at http://cgsuite.sourceforge.net/ (It's GNU public license open-source software.) David

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[math-fun] Re: designing a baseball cover
by Henry Baker 21 Jul '03

21 Jul '03

Thanks to Steven Finch for forwarding this pointer to the online article. I re-read the article, and I think I discovered something about baseballs that seams not to have been noticed by the author Richard Thompson, and possible not by other authors he references. Thompson first rotates the baseball into a preferred position, wherein the projection of the seam onto the xy plane is a "C" -- the seam is symmetrical, so that the "C" is double-covered by the seam projection, except for the two ends of the "C". The "C" is, of course, symmetric about the x axis. The "C" is pictured inside the circle projected by the baseball itself. Thompson also notices that due to the symmetry required of the covers, the projected "C" extends equally far on either side of the y axis. Thompson then notices that on particular baseballs he examines, the arms of the "C" _to the right of the y axis_ are _straight lines_! Unfortunately, he doesn't plumb the consequences of this fact, but we will. The straightness of these lines imply that the seam can be embedded not just in the sphere of the baseball, but also in a (non-regular) tetrahedron, and in fact, the seam is the intersection of the sphere with this tetrahedron! This tetrahedron is semiregular, in that 4 edges have one length, and the remaining two (opposite) edges have another (shorter) length -- i.e., all faces are isosceles. The two opposite edges do not intersect the ball, while all four of the other edges intersect the ball at the vertices of a square in the yz plane. In other words, the extreme y values of the "C" indicate the points of intersection of the non-opposite edges with the ball, and also indicate the places where the seam starts following a different face of the tetrahedron. The limiting case of this embedding occurs when the tetrahedron is regular, in which case the ball also intersects the middle of the two opposite edges, and each of the two pieces of leather have zero thickness in their middle. By the way, what do you call the sphere that intersects the middles of each of the edges of a regular tetrahedron? FYI -- The left side of the "C" is an ellipse. There appears to be only one free parameter -- the angle of the arms of the "C" (or the ratio of the edges of the tetrahedron). I dont know if/when the "C" (or the seam) is smooth. At 02:38 PM 7/5/03 -0400, Steven Finch wrote: >Hello! > >The particular web files you mentioned are still available online (zipped >up, however) at: > >http://www.mathcad.com/Library/LibraryContent/baseball.zip > >I don't know why my former employer chose to make these interesting files >less visible. >Best wishes, > >Steve Finch > >>From: Henry Baker <hbaker1(a)pipeline.com> >>Reply-To: math-fun <math-fun(a)mailman.xmission.com> >>To: "Bernie Cosell" <bernie(a)fantasyfarm.com> >>CC: math-fun <math-fun(a)mailman.xmission.com> >>Subject: Re: [math-fun] NY State math exam flapment >>Date: Sat, 05 Jul 2003 09:08:50 -0700 >> >>The original article (Thompson, R.B. 1998. Designing a baseball cover. >>College Mathematics Journal 29(January):48) was pretty good. Unfortunately, >>it does not appear to be online anymore, since the widely referenced mathsoft >>site is gone. The author apparently works for the U. of Arizona, but he doesn't >>have it online either. >> >>Perhaps someone on this list can get Dr. Thompson to post it on his web site? >> >>At 11:01 AM 7/4/03 -0400, Bernie Cosell wrote: >> >On 4 Jul 2003 at 7:25, Henry Baker wrote: >> > >> >> I hope that you know enough about baseball to appreciate the >> >> elegance of the baseball's seam: >> >> >> >> http://www.sciencenews.org/sn_arc98/4_11_98/mathland.htm >> > >> >What property of the 'seam' problem for the baseball precludes just an >> >equator for the seam as a possible solution? Is there some "minimization >> >of necessary stretch" or other criterion that drives the more complicated >> >shape? >> > >> > /Bernie\

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