November 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Heronian octonions?
by Dan Asimov 19 Nov '11

19 Nov '11

Sorry if I was unclear. Yes, S^3 here is the group of quaternions of norm one. And yes, the multiplication is in the quaternions. But what I left unsaid is that the y in the equation now labeled (*) below is a "pure" quaternion -- i.e., y has real part = 0. The pure quaternions H_0 form a copy of R^3. Then for any unit quaternion p, the map g(p): H_0 -> H_0 defined by g(p)(y) := p y p^(-1) acts as a rotation on the of pure quaternions, and so g(p) belongs to SO(3). --Dan Andy wrote: << I wrote: << On Fri, Nov 18, 2011 at 5:06 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: Interesting subject. Yes, the map f: S^3 x S^3 -> SO(4) via f(p,q)(y) := p y q^(-1), >> So you're doing this multiplication in the quaternions, interpreting S^3 as the quaternions of norm 1, right? << Likewise, g: S^3 -> SO(3) via (*) g(p)(y) := p y p^(-1) >> But this has me baffled; how do you multiply a quaternion by a vector in R^3? >> ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] quarter-Heroic triangles are Heronian (bzzz)
by Warren Smith 19 Nov '11

19 Nov '11

THEOREM: The "Heronian" triangles can also be characterized as the triangles with integer sides and rational area (they then automatically have integer area). SORRY my proof was bogus, but the theorem still appears correct (says computer).

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[math-fun] quarter-Heroic triangles are Heronian
by Warren Smith 19 Nov '11

19 Nov '11

Michael Reid's proof that every Heronian triangle is embeddable in the plane with integer coordinates for vertices [posted here 17 Nov 2011]... actually (more strongly) shows that for every triangle with 1. Integer edge lengths a,b,c 2. Rational area. There is no requirement for integer area. Now the Heron area formula Area=sqrt(s*(s-a)*(s-b)*(s-c)) where 2s=a+b+c shows that any triangle with integer edge lengths a,b,c either has 1. Area that is an integer/4 2. Irrational area. This suggests that a more natural class than the "Heronian" triangles are the "quarter-Heroic" triangles whose area is a quarter-integer. But... a funny thing happens. The only quarter-Heroic triangles actually ARE Heronian, i.e. quarter-integer area implies integer area. Equivalently: every quarter-Heroic triangle has EVEN perimeter so that s automatically is an integer. Equivalently: every primitive quarter-Heroic triangle has two odd sides and one even side. Equivalently: THEOREM: The "Heronian" triangles can also be characterized as the triangles with integer sides and rational area (they then automatically have integer area). PROOF: For a primitive, (a,b,c)=(even,even,even) is impossible (not primitive). (a,b,c)=(even,even,odd) and (a,b,c)=(odd,odd,odd) both are impossible since then we'd have 16*area^2=p*(p-2a)*(p-2b)*(p-2c)=odd*odd*odd*odd=odd. Hence (a,b,c)=(odd,odd,even) which causes area and semiperimeter both to be integer. QED.

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[math-fun] tridragon small constants?
by Bill Gosper 18 Nov '11

18 Nov '11

http://gosper.org/islands.png shows four triadic dragons sharing an endpoint, and fanned out at angular intervals π/6, π/3, and π/2. If we suppose each has unit area, it is a minor puzzle to compute the distance between endpoints (12^(1/4)), and the (slightly greater) diameter, which I haven't bothered to rederive. Perhaps the easiest area derivation is to note that the dragons tile the (infinite) plane in a triangular grid, with 3/2 dragons/triangle. This is a case of using infinity to answer a finite question, and is unrelated to the dragon's infinite perimeter, e.g. Small constant #1: What is the diameter minus endpoint span. Notice the purple Hawaiianoid chain of apparently similar islands formed by the purple dragon minus the blue one, and the (necessarily) congruent white chain between the green and gold dragons. While trying to characterize these Islands, Corey & Julian noticed something peculiar. Define A(ø) := the area of intersection of two coterminous unit area tdragons fanned at angle ø. A(π/3)=0 by virtue of the plane tiling. A(π/6) appears to be some simple function of the area of an island chain. But maybe not an island! On very close inspection, A(π/2) is very slightly positive. This is not a positioning artifact--in the fractal limit, coterminous dragons infinitely entwine, and cannot be slid apart without rotation. This is not visually obvious because they are based on a log spiral with a scale gain of 729 per turn. (kth order dragon ~ (rt3 e^(i pi/6))^k. k=12 makes one turn.) It thus appears that the islands are very slightly dissimilar! Small constant #2: What is A(π/2)? Small constant #3: What is the least positive ϵ for which A(π/2+ϵ)=0? --rwg

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Re: [math-fun] Heronian octonions?
by Dan Asimov 18 Nov '11

18 Nov '11

Interesting subject. Yes, the map f: S^3 x S^3 -> SO(4) via f(p,q)(y) := p y q^(-1), for all y in R^4, is a homomorphic double covering of SO(4). (where SO(n) = rotation group of R^n; S^3 = group of unit quaternions.) But dim(SO(8)) = 28, so there are not enough dimensions in the domain of the corresponding map F: S^7 x S^7 -> SO(8) for it to be onto. (And it's easy to see why it's not a homomorphism.) Likewise, g: S^3 -> SO(3) via g(p)(y) := p y p^(-1) is also a homomorphic double covering, but the corresponding map G: S^7 -> SO(7) again can't be onto, since dim(SO(7)) = 21 (and again it clearly can't be a homomorphism). These things are well-known and probably explained nicely in The Octonions by Conway & Smith, QUESTION: Are there nice descriptions of the images F(S^7 x S^7) \sub SO(8), and G(S^7) \sub SO(7) ??? --Dan ---------- Warren wrote: << . . . There is a nice characterization of 2D, 3D, and 4D (and even 5D) rotations using complex #s & quaternions. But there is no such nice characterization using octonions. . . . >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Heronian octonions?
by Warren Smith 18 Nov '11

18 Nov '11

>And finally and even more wildly --- are we eventually going to discover octonions lurking around the corner somewhere in this discussion? --I doubt it. There is a nice characterization of 2D, 3D, and 4D (and even 5D) rotations using complex #s & quaternions. But there is no such nice characterization using octonions. I think this proof technique is only going to work up to 3D (if there).

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[math-fun] An alternative way to count the rationals?
by Bill Gosper 18 Nov '11

18 Nov '11

JZucker> Well, the Calkin-Wilf tree is still the best, of course. Thank you, Joshua! I had somehow lived in total ignorance of Newman's amazing, useful, and tiny invention: NestList[1/(2*Floor[#] - # + 1) &, 0, 69] 1 1 3 2 1 4 3 5 2 5 3 1 5 4 7 {0, 1, -, 2, -, -, -, 3, -, -, -, -, -, -, -, 4, -, -, -, -, 2 3 2 3 4 3 5 2 5 3 4 5 4 7 3 3 8 5 7 2 7 5 8 3 7 4 1 6 5 9 4 11 -, -, -, -, -, -, -, -, -, -, -, 5, -, -, -, -, --, --, 8 5 7 2 7 5 8 3 7 4 5 6 5 9 4 11 7 7 10 3 11 8 13 5 12 7 9 2 9 7 12 5 --, --, --, --, --, --, --, --, -, -, -, -, --, --, --, 10 3 11 8 13 5 12 7 9 2 9 7 12 5 13 13 8 11 3 10 7 11 4 9 5 1 7 6 11 5 --, --, --, --, --, --, --, -, -, -, 6, -, -, --, --, --, ...} 8 11 3 10 7 11 4 9 5 6 7 6 11 5 14 (http://oeis.org/A002487) Julian derived the reverse sequence: NestList[2*Ceiling[1/#] - 1 - 1/# &, 14/9, 9] 14 5 11 6 7 1 5 9 4 {--, --, --, --, -, -, 6, -, -, -} 9 14 5 11 6 7 6 5 9 and found the waiting time from 355/113 to 0 was 67107848. I.e., intermediate swell was >>355. But the real surprise came with plotting the forward sequence starting with Sqrt[-1] in a neighborhood of the origin--the Ford Pinto circles <http:gosper.org/fordpinto.png>! (Complete with some particulate emissions.) (Using Mma's convention: Floor[z]==Floor[Re[z]]+I*Floor[Im[z]].) The reverse sequence plots the left half (mirror image). --rwg rwg> Duh, the original argument was < -1, so the z/(z-1) transformation gives a single, convergent series: 10 3 1/12 1/12 13 1 5 2 3 5 sqrt(6) 17 23 hyper_2f1(--, --, -, --------) = --------------------- 24 24 6 2 2 1/6 17 23 3125 Duh, all you need is z<1/2.

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[math-fun] Maths and Sport: Countdown to the Olympic Games
by Henry Baker 18 Nov '11

18 Nov '11

FYI -- http://sport.maths.org/content/ Maths and Sport: Countdown to the Games 'Maths and Sport: Countdown to the Games' is an exciting new project from the award-winning Millennium Mathematics Project at the University of Cambridge. The 2012 Olympic and Paralympic Games offer an extraordinary and exciting opportunity to explore mathematics and science at work in real-world contexts. To celebrate London 2012 we're therefore developing free online mathematical articles and activities exploring maths and science through the Olympic and Paralympic Games, aimed at a wide audience from school students and their teachers to members of the public. We'll continue to publish new content throughout the run-up to the 2012 Games. Featured article - Leaning into 2012 Leaning into 2012 This article explores the Velodrome, the first of the London 2012 venues to be completed. With its sweeping curved roof and beautiful cedar clad exterior the Velodrome is a stunning building. But what the athletes are most excited about is the elegant wooden cycle track enclosed inside. How does its geometry contribute to speed? Featured activity - Track Design Track design You are designing and marking out the running track for a new Olympic stadium. Where should the staggered starting positions be marked for the 200m and 400m events? This activity is aimed at GCSE maths students (Key Stage 4).

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[math-fun] Heronian primitive triangles -- single formula describing all
by Warren Smith 18 Nov '11

18 Nov '11

I also point out that (assuming Reid proof valid) by combining my Buchholz-->nonprimitive coordinates formula, and the Gaussian GCD, we can actually write just a single formula in terms of n,m,k, rational operations, and Gaussian GCD, giving coordinates for every PRIMITIVE Heronian triangle's vertices A,B,C. That is quite cool. Here are the nonprimitive coordinate formulas: xC = m*(n-k)*(n+k); yC = 2*k*m*n; xA = n*(m-k)*(m+k) + xC; yA = 0; xB = 0; yB = 0; and now the primitivization as Gaussian integers is 0, xA/G, and (xC+yC*i)/G where G=GaussianGCD(xA, xC+yC*i).

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[math-fun] Heronian Triangles and Gaussian Integer GCD
by Warren Smith 18 Nov '11

18 Nov '11

Re Michael Reid's proof, I had earlier sent Lunnon email explaining a proof idea which I believe coincides with Reid's proof idea (I did not have a proof, but I did have this underlying idea), I now excerpt from our correspondence: WDS-->Lunnon: If your triangle vertices are (0,0) (x1, y1) and (x2, y2) all integers, then regard the points as COMPLEX NUMBERS and take G = gcd( x1+y1*i, x2+y2*i ) as a Gaussian integer. We then may divide all 3 vertices by G. This does both a scaling and a rotation, and it scales it down by the most it possibly can (no?). Lunnon-->WDS: Aren't you assuming the result that you're trying to prove here? WDS-->Lunnon: --no, I begin with an integer embedding that I already gave formulas for every coordinate. The only problem is it just might not be primitive. To make it primitive we divide by a real GCD; to find the smallest lattice version we divide by a complex GCD. Lunnon then said, and I agreed, that we ought to check Reid's proof by computer before agreeing with it. About that, here is some computer output I made: Program for messing with Heronian Triangles. Warren D Smith Nov 2011 n=2 m=3 k=1 a=15 b=20 c=25 G=5 semip=30 area=150 xC=9 yC=12 xA=25 GIgcd=3+4*i norm=5 n=2 m=8 k=1 a=40 b=130 c=150 G=10 semip=160 area=2400 xC=24 yC=32 xA=150 GIgcd=6+8*i norm=10 n=2 m=13 k=1 a=65 b=340 c=375 G=5 semip=390 area=9750 xC=39 yC=52 xA=375 GIgcd=3+4*i norm=5 n=2 m=18 k=1 a=90 b=650 c=700 G=10 semip=720 area=25200 xC=54 yC=72 xA=700 GIgcd=6+8*i norm=10 n=3 m=7 k=1 a=70 b=150 c=200 G=10 semip=210 area=4200 xC=56 yC=42 xA=200 GIgcd=8+6*i norm=10 n=3 m=10 k=2 a=130 b=312 c=338 G=26 semip=390 area=20280 xC=50 yC=120 xA=338 GIgcd=10+24*i norm=26 n=3 m=12 k=1 a=120 b=435 c=525 G=15 semip=540 area=18900 xC=96 yC=72 xA=525 GIgcd=12+9*i norm=15 n=4 m=11 k=2 a=220 b=500 c=600 G=20 semip=660 area=52800 xC=132 yC=176 xA=600 GIgcd=12+16*i norm=20 n=4 m=11 k=3 a=275 b=520 c=525 G=5 semip=660 area=69300 xC=77 yC=264 xA=525 GIgcd=4+3*i norm=5 n=4 m=13 k=1 a=221 b=680 c=867 G=17 semip=884 area=45084 xC=195 yC=104 xA=867 GIgcd=15+8*i norm=17 n=5 m=8 k=1 a=208 b=325 c=507 G=13 semip=520 area=20280 xC=192 yC=80 xA=507 GIgcd=12+5*i norm=13 n=6 m=9 k=2 a=360 b=510 c=750 G=30 semip=810 area=81000 xC=288 yC=216 xA=750 GIgcd=24+18*i norm=30 n=7 m=8 k=1 a=400 b=455 c=825 G=5 semip=840 area=46200 xC=384 yC=112 xA=825 GIgcd=3+4*i norm=5 n=7 m=8 k=4 a=520 b=560 c=600 G=40 semip=840 area=134400 xC=264 yC=448 xA=600 GIgcd=32+24*i norm=40 To explain the lines of the table, the Buchholz parameters are n,m,k yielding the (not primitive) triangle with sides a,b,c. All Buchholz triangles with semiperim<1000 should be listed that do not have a trivially integer-embeddable primitivization. G is the integer GCD of (a,b,c). The coordinates of the naive integer embedding of the vertices A,B,C which I had derived in a previous post are A=(xA,0), B=(0,0), and C=(xC,yC). FInally, GIgcd is the Gaussian-integer GCD of (xA, xC+yC*i) and norm is its absolute value (modulus). You will notice an amazing coincidence of the G and norm columns and the amazing fact norm is always an integer. Lunnon now wonders: "What next? On to tetrahedra?" Well, maybe... I point out that in 4 dimensions, the "Hurwitz integers" are the quaternions A+B*i+C*j+D*k such that either (A,B,C,D) or (A+1/2, B+1/2, C+1/2, D+1/2) all are integers. http://en.wikipedia.org/wiki/Hurwitz_quaternion These enjoy a Euclidean-like GCD algorithm, albeit be careful since quaternion multiplication is noncommutative, Any 3D rotation+scaling can be expressed using quaternions as follows: Let (x,y,z), the 3-vector being rotated, be regarded as X=0+x*i+y*j+z*k and perform X --> Q*X*Qconjugate. [Also any 4D rotation+scaling can be expressed using quaternions as follows: X --> Q*X*R.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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