October 2003 - mathfun@mailman.xmission.com

Re: [math-fun] cyclic quadrilaterals
by Dan Hoey 27 Oct '03

27 Oct '03

John Conway <conway(a)math.princeton.edu> writes: > ... Alternatively, let AEPF be any circle through A and P, and > then BFPD and CEPD be the circumcircles of BFP and CEP > (which will automatically intersect at a point D on BC). > This is all part of "the Miquel theory" of the triangle. JHC By considering the point at infinity, I understand this to mean that if we have points 01234567 in cocircular quartets 0123, 4567, 0145, 2367, and 0246, then quartet 1357 must also be cocircular. I didn't know that. Are all such patterns known? Say, given a mapping of a hypergraph to the plane, where vertices are mapped to points, and the images of the vertices of each hyperedge are cocircular, can we tell what other subsets of vertices are forced to map to cocircular points? Dan

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[math-fun] Re: cyclic quadrilaterals
by James Propp 27 Oct '03

27 Oct '03

John Conway writes: > Given P, there's a 1-parameter family. You can get them all by >dropping "isoclines" from P to the sides - ie., lines such that >angles AFP, BDP, CEP are all equal. Alternatively, let AEPF >be any circle through A and P, and then BFPD and CEPD be >the circumcircles of BFP and CEP (which will automatically >intersect at a point DE on BC). Thanks, John. Since this gives us 3 real degrees of freedom in total, I wonder if we get a nice parametrization of the full set of configurations (for A,B,C fixed) by specifying the three lengths AP, BP, and CP (or perhaps the three lengths DP, EP, and FP, or perhaps the lengths AD, BE, and CF, or ...). What I ideally want is a set of three parameters such that the distances between all the points can be expressed as rational functions of the parameters and the original lengths AB, AC, BC. Is this too much to hope for? Jim

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Re: [math-fun] RE: Prentice Hall strikes again
by Dan Hoey 27 Oct '03

27 Oct '03

Eugene Salamin <gene_salamin(a)yahoo.com> maunders again: > What is needed to to do away with the obstacles to free choice. What is needed is to do away with that obstacle to the discussion of mathematics. Does he really think repetition of this rant is going to sell anyone on it? Dan

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Re: [math-fun] cyclic quadrilaterals
by reid＠math.arizona.edu 27 Oct '03

27 Oct '03

> Given a generic triangle ABC, in "how many ways" can you choose an interior > point P and points D, E, and F on sides BC, AC, and AB (respectively) so > that quadrilaterals AEPF, BFPD, and CDPE are cyclic? That is to say, how > many degrees of freedom do you have? Is there a nice way to parametrize there are three degrees of freedom. the internal point P can be anywhere inside the triangle. if P is fixed, there is still another degree of freedom, as JHC notes. a nice way to find all such configurations is as follows. draw the triangle ABC on a sheet of paper. draw a point P , with three rays emanating from it, on a transparency. the three rays, PQ , PR and PS should have angles between them so that angle QPR [resp. RPS , SPQ ] is the supplement of angle A [resp. B , C ] of the triangle. (we also need that the orientation of these three rays is the same (clockwise v. counterclockwise) as the orientation of the vertices A , B , C .) the configurations are obtained by placing the transparency over the paper such that P is inside the triangle, and setting D [resp. E , F ] to be the points where ray PQ intersects BC [resp. PR intersects CA , PS intersects AB ]. two degrees of freedom are moving P around inside the triangle, and the third is rotation about P . to see that the three quadrilaterals are cyclic, note that the opposite angles are supplementary, by construction of the angles between the rays. i'll leave the task of finding an explicit parametrization and formulae for the lengths for others. mike

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[math-fun] cyclic quadrilaterals
by James Propp 27 Oct '03

27 Oct '03

Given a generic triangle ABC, in "how many ways" can you choose an interior point P and points D, E, and F on sides BC, AC, and AB (respectively) so that quadrilaterals AEPF, BFPD, and CDPE are cyclic? That is to say, how many degrees of freedom do you have? Is there a nice way to parametrize the set of such configurations (with A,B,C fixed), so as to express the lengths of segments AP, BP, CP, DP, EP, and FP in terms of the lengths of segments AB, AC, and BC together with the extra parameter(s)? B / \ / \ F D / P \ / \ A-----E-----C Jim Propp University of Wisconsin Madison, WI

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[math-fun] Re: Erdos-Strauss conjecture
by Dean Hickerson 27 Oct '03

27 Oct '03

Jon Perry (perry(a)globalnet.co.uk) wrote: > Here's what I have; > > The problem (4/n=1/a+1/b+1/c for all n>1 with a,b,c integers) has easy > solutions for n=0,2,3mod4. > > If n=4k+1, then consider; > > 4/(4k+1) - 1/a - 1/b = 1/c > > Rearranging gives; > > c=(4k+1)ab/[4ab-(4k+1)(a+b)] > > wlog, let a=x(4k+1), as this is required for c to be an integer: If you mean that x must be an integer, this isn't true. For example, consider 4/5 = 1/2 + 1/5 + 1/10: k=1, a=2, b=5, c=10, x=2/5. > Simplifying; > > c=(4k+1)xb/[4bx-4kx-x-b] > > Again, b=yx; If you mean that y must be an integer, this isn't true. For example, consider 4/29 = 1/87 + 1/8 + 1/696: k=7, a=87, b=8, c=696, x=3, y=8/3. > c=(4k+1)xy/[4xy-4k-1-y] > > For c to be an integer, [1] 4xy-y=z(4k+1), and then; If you mean that z must be an integer, this isn't true. For example, consider 4/5 = 1/5 + 1/2 + 1/10: k=1, a=5, b=2, c=10, x=1, y=2, z=6/5. Dean Hickerson dean(a)math.ucdavis.edu

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[math-fun] ESC
by Jon Perry 27 Oct '03

27 Oct '03

I havbe been asked to define the Erdos-Strauss conjecture - a reference to it can be found at MathWorld; http://mathworld.wolfram.com/Erdos-StrausConjecture.html This page covers it in greater depth; http://math.uindy.edu/swett/esc.htm Jon Perry perry(a)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

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[math-fun] greatest integer
by Mike Stay 26 Oct '03

26 Oct '03

Are there any useful definitions of greatest integer function for algebraic integers? For the Gaussian integers, it's pretty straightforward because it uses the Euclidean norm--find the nearest integer to a complex x with norm less than magnitude(x). It's unique except for some points on diagonals where there are two. What about Z[sqrt(-3)]? I guess one could talk about integers with norm less than magnitude(x), but the magnitude of x is using a Euclidean norm while Z[sqrt(3)] uses a hyperbolic norm, N(a+b sqrt(-3)) = a^2-3b^2. (I think I got that right.) When the greatest integer isn't unique, can one show there are only a finite number of possibilities? -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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[math-fun] Boston talks on geometric sculpture
by Tom Knight 26 Oct '03

26 Oct '03

Those of you in the Cambridge area may be interested in these talks:

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[math-fun] Erdos-Strauss conjecture
by Jon Perry 26 Oct '03

26 Oct '03

Here's what I have; The problem (4/n=1/a+1/b+1/c for all n>1 with a,b,c integers) has easy solutions for n=0,2,3mod4. If n=4k+1, then consider; 4/(4k+1) - 1/a - 1/b = 1/c Rearranging gives; c=(4k+1)ab/[4ab-(4k+1)(a+b)] wlog, let a=x(4k+1), as this is required for c to be an integer: Simplifying; c=(4k+1)xb/[4bx-4kx-x-b] Again, b=yx; c=(4k+1)xy/[4xy-4k-1-y] For c to be an integer, [1] 4xy-y=z(4k+1), and then; [2] c=xy/(z-1) So solutions to ESC for n=4k+1 are given by [1] and [2]. But here is where I get stuck. Jon Perry perry(a)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

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