October 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Distance ratio question
by N. J. A. Sloane 25 Oct '08

25 Oct '08

Ron Hardin and I did do some work on it, but the most extensive was done in the following papers: MR1600759 (99b:52040) Graham, R. L.; Lubachevsky, B. D.; Nurmela, K. J.; Östergård, P. R. J. Dense packings of congruent circles in a circle. Discrete Math. 181 (1998), no. 1-3, 139--154. (Reviewer: A. Florian) 52C15 (05B40) MR1455513 (98f:52021) Lubachevsky, B. D.; Graham, R. L. Curved hexagonal packings of equal disks in a circle. Discrete Comput. Geom. 18 (1997), no. 2, 179--194. 52C15 (05B40) MR1450127 (98a:52028) Lubachevsky, B. D.; Graham, R. L. Dense packings of $3k(k+1)+1$ equal disks in a circle for $k=1,2,3,4$, and $5$. Computing and combinatorics (Xi'an, 1995), 303--312, Lecture Notes in Comput. Sci., 959, Springer, Berlin, 1995. 52C15 (05B40) etc. Neil >By the way, Kevin Buzzard & I just realized that >this M/m problem is (trivially) equivalent to the >"Penny Problem": -- > > Given n non-overlapping* unit disks in the plane, > what is the smallest R = R(n) for which they can > be arranged inside of a disk of radius R ? > >Neil, did you do any numerical work on this question? >

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Re: [math-fun] Distance ratio question
by Dan Asimov 25 Oct '08

25 Oct '08

<< On Saturday 25 October 2008, Dan Asimov wrote: [SNIP: algebra] > which seems to agree with Tom's but is > twice what Gareth wrote. > > So, Tom & I win, two to one. Ha. The *reason* why I showed my working was because I'm appallingly good at making errors when doing that sort of calculation. >> And the reason I showed mine is I wanted to prove that at least once -- if only on my second try -- I could get something right. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Distance ratio question
by Dan Asimov 25 Oct '08

25 Oct '08

Suprise! I screwed up again. Oh, well. Let's see if I can avoid screwing up the 2nd time and get something like what Gareth & Tom did: Of course, the right thing to do is look at at circle packed with a triangular array of points. WLOG say the circle is radius 1. Then M = max dist. ~ 2. As for m = min dist., let that be the side of each triangle in the array. Then each triangle's area is m^2 sqrt(3)/4, while the circle's is pi, so the number of triangles F ~ 4pi/(m^2 sqrt(3)). Here n is the number V of vertices, and each vertex (if the array is oriented so some edges are horizontal) can be made to correspond to 2 triangles -- the ones directly above & below it. Thus F ~ 2V = 2n. Thus 2n ~ 4pi / (m^2 sqrt(3)) and so 1/m ~ sqrt( n sqrt(3) / (2pi) ), so M/m ~ 2/m ~ sqrt(2 sqrt(3)/pi) * sqrt(n), which seems to agree with Tom's but is twice what Gareth wrote. So, Tom & I win, two to one. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Pfaffians
by Eugene Salamin 24 Oct '08

24 Oct '08

The determinant of a matrix is a polynomial in its elements. For an antisymmetric matrix of odd order, the determinant is zero, while for an antisymmetric matrix of even order, this polynomial factors into the square of another polynomial, called the Pfaffian. Does anyone know of a simple, easy to follow, proof that the determinant factors as asserted? Gene

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Re: [math-fun] tetrahedron volume
by Eugene Salamin 24 Oct '08

24 Oct '08

----- Original Message ---- From: "rwg(a)sdf.lonestar.org" <rwg(a)sdf.lonestar.org> To: math-fun <math-fun(a)mailman.xmission.com> Cc: math-fun <math-fun(a)mailman.xmission.com> Sent: Monday, October 20, 2008 3:37:08 AM Subject: [math-fun] tetraroller volume True or False Quickie: The volume of a tetrahedron is determined by the areas of its faces. _______________________________________________ False. There exists a face area preserving continuous deformation that alters the volume. Start with a regular tetrahedron. Perform a dilatation along the mutual perpendicular of two opposite edges, while simultaneously performing a dilatation in the transverse plane so as to preserve the face areas. This is possible because all four faces are inclined with respect to that mutual perpendicular by the same angle. The volume is not preserved, in particular the volume goes to zero as the tetrahedron is squeezed flat. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] Andrew Gleason
by Michael Reid 22 Oct '08

22 Oct '08

i was saddened to hear about andy gleason's passing. i took several classes from him last century, and each ranks very high on the list of classes in which i learned the most. i feel very fortunate to have had him as a teacher. i specifically remember, in his classical geometry course, he showed how desargues' theorem/axiom follows directly from the incidence axioms in 3 dimensional projective space, but not so in 2 dimensional space. he then remarked that if you can embed a 2 dimensional space into a 3 dimensional space, it sometimes gives you added information! indeed, there are projective planes that are "non-desarguesian" and therefore cannot be embedded in 3 dimensional projective space. this idea stuck with me and i've used it several times. one application is the following. the "radical axis" of two circles is the locus of all points in the euclidean plane that have equal "power" with respect to each circle. (the "power" of a point with respect to a circle is the square of the length of a tangent to the circle from that point; for points inside the circle, this must be modified to, say, the negative of the product of the two lengths of the segments into which the point divides a chord of the circle.) for two circles that intersect, it is easy to show that their radical axis is the line through the two points of intersection (or common tangent, if they intersect tangentially). for two circles (non-concentric) that do not intersect, their radical axis is still a line. i used gleason's idea to give a nice proof of this (it works better in the case that neither circle contains the other; otherwise one needs to handle the concentric case separately). i also recall, somewhat vaguely, that gleason's title was professor of mathematicks [sic] and natural philosophy. the obituary suggests that they have modernized(?) the spelling in the meantime. do anyone else remember this? or am i mis-remembering? i also note that ethan bolker, who is quoted in the boston globe article, is probably the same e. bolker listed in the references R recently sent for UPINT problem C5 "sums determining members of a set". mike > Whit Diffie passed along the news that Andrew Gleason has died. --Rich > > <url is split> > http://www.boston.com/bostonglobe/obituaries/articles/2008/10/20/ > andrew_gleason_helped_solve_vexing_geometry_problem/ >

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[math-fun] Fw: Andrew Gleason
by Dan Asimov 22 Oct '08

22 Oct '08

Rich wrote: << >Whit Diffie passed along the news that Andrew Gleason has died. --Rich > ><url is split> >http://www.boston.com/bostonglobe/obituaries/articles/2008/10/20/ andrew_gleason_helped_solve_vexing_geometry_problem/ >> Here's another one with links: < http://www.math.harvard.edu/history/gleason/index.html >. I think Gleason was most famous for solving (with Deane Montgomery and Leo Zippin) Hilbert's 5th Problem, which asks whether, given a topological manifold M with a topological group structure on it (i.e., the map MxM -> M via (x,y) |-> x/y is continuous), there is always a real analytic atlas on M for which the group operations are real analytic. (At least that's how I remember it.) (Before that, he was a Putnam Fellow three times.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Andrew Gleason
by rcs＠xmission.com 21 Oct '08

21 Oct '08

Whit Diffie passed along the news that Andrew Gleason has died. --Rich <url is split> http://www.boston.com/bostonglobe/obituaries/articles/2008/10/20/ andrew_gleason_helped_solve_vexing_geometry_problem/

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Re: [math-fun] tetrahedron volume
by Eugene Salamin 21 Oct '08

21 Oct '08

----- Original Message ---- From: Stephen Gray <stevebg(a)roadrunner.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Monday, October 20, 2008 1:32:32 PM Subject: Re: [math-fun] tetrahedron volume More simply, if a vertex moves parallel to the opposite face, volume is preserved. Move a vertex in that manner so that it goes way off to one side. When the moving vertex is no longer above a point on the opposite side, further motion will increase the altitudes of the 3 changing triangles, so their areas increase also. QED Steve Gray ________________________________ This shows that the areas can change while the volume remains fixed. But RWG was asking if it is possible to change the volume while holding the four face areas fixed. One possible way to complete your proof is to scale the tetrahedron. But it is not clear that such a scaling is possible that preserves all four face areas. Gene -----Original Message----- From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, October 20, 2008 12:18 PM To: math-fun Subject: Re: [math-fun] tetrahedron volume ----- Original Message ---- From: "rwg(a)sdf.lonestar.org" <rwg(a)sdf.lonestar.org> To: math-fun <math-fun(a)mailman.xmission.com> Cc: math-fun <math-fun(a)mailman.xmission.com> Sent: Monday, October 20, 2008 3:37:08 AM Subject: [math-fun] tetraroller volume True or False Quickie: The volume of a tetrahedron is determined by the areas of its faces. _______________________________________________ False. There exists a face area preserving continuous deformation that alters the volume. Start with a regular tetrahedron. Perform a dilatation along the mutual perpendicular of two opposite edges, while simultaneously performing a dilatation in the transverse plane so as to preserve the face areas. This is possible because all four faces are inclined with respect to that mutual perpendicular by the same angle. The volume is not preserved, in particular the volume goes to zero as the tetrahedron is squeezed flat. Gene _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] tetrahedron volume
by Dan Asimov 21 Oct '08

21 Oct '08

Rich asks these tetrahedron questions: << (a) Given the areas of the four faces, what's the maximum volume? (b) What extra information is needed (beyond the face areas) to determine the volume? One or two edges? One of the lines connecting centers of opposite edges? Recall that altitude + opposite face determines the volume. Also that three face areas of a right tetrahedron determine the volume. So maybe we should be thinking about angles, or solid angles. >> Some other worthy questions are: (c) What are the possible 4 face areas of a tetrahedron? (To avoid extraneous info, one could assume the total surface area is constant, and the face areas given in non-decreasing order -- leading to the quotient of a regular 3-simplex by S_4 -- before finding which subset of this answers the question.) (d) What are the possible 4 solid angles at the corners of a tetrahedron? (Taking the exterior solid angle may be more convenient, since these add up to 4pi.) (e) What are the possible lengths of the 6 edges of a tetrahedron? (This would amount to labeling the 1-skeleton of a tetrahedron with lengths that can be realized. Rich once showed that satisfying the triangle inequalities for each face is not sufficient, as may have been mentioned here in the past.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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