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Correction Re: [math-fun] Hero[n]'s formula
by dasimov＠earthlink.net 26 Apr '06

26 Apr '06

I wrote: << More generally the shape of an arbitrary right p-simplex (p-tetrahedron) is made by an affine hyperplane in R^p that intersects each positive coordinate axis. Again, the p-otenuse has one right corner at the origin of the posiive orthant, and the rest of them must be "p-acute" -- defined by requiring that each dihedral angle at the corner is acute. >> but should have worded the last sentence as: << Again the p-otenuse has one right corner at the origin of the positive orthant, and the rest of them must be "p-acute" -- defined by requiring that each angle between two edges emanating from this corner is acute. >> --Dan

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Re: [math-fun] Hero[n]'s formula
by dasimov＠earthlink.net 26 Apr '06

26 Apr '06

Gareth wrote: << Rich Schroeppel wrote: > . . . Any triangle can be a hypoteface . . . I thought of that too ... but is it true that "any triangle can be a hypoteface"? I think it only works for acute-angled ones. . . . >> Definitely only works for acute triangles: Consider how a third plane can intersect a right dihedral angle. More generally the shape of an arbitrary right p-simplex (p-tetrahedron) is made by an affine hyperplane in R^p that intersects each positive coordinate axis. Again, the p-otenuse has one right corner at the origin of the posiive orthant, and the rest of them must be "p-acute" -- defined by requiring that each dihedral angle at the corner is acute. --Dan

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RE: [math-fun] Hero[n]'s formula
by Schroeppel, Richard 26 Apr '06

26 Apr '06

Somewhat more seriously: If we have a right tetrahedron, where one of the corners matches a cube's corner, there are three leg-faces and one hypote-face. It turns out that the square of the hypoteface area is equal to the sum of the squares of the three legface areas. (I don't know why, or if it works in more dimensions.) Any triangle can be a hypoteface, and the lengths of the X,Y,Z legs are easy to calculate: If the hypoteface edges are A,B,C, then X2+Y2 = A2, etc. and X2 = (A2+B2-C2)/2. The legface areas are XY/2 etc., the area squares are X2Y2/4, and it's not hard to check that the sum of squares of the three legface areas matches Hero^2. Rich -----Original Message----- From: Schroeppel, Richard Sent: Wednesday, April 26, 2006 1:26 PM To: 'math-fun' Cc: Schroeppel, Richard; 'rcs(a)cs.arizona.edu' Subject: RE: [math-fun] Hero[n]'s formula Let's see now. If I wanted the volume of a tetrahedron, I could start by adding up the areas of three faces and subtracting the fourth. That would give me four terms to multiply together, for a total degree of 8. Maybe throw in the sum of all the areas, for degree 10. Since I want a degree 3 answer, I'll raise it to the 3/10 power. What's wrong with this picture? Rich PS: Doyle's web page looks like a lot of fun. I haven't looked beyond the Hero's formula discussion with JHC, but the titles look intriguing. --R -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Richard Guy Sent: Wednesday, April 26, 2006 12:48 PM To: math-fun Subject: Re: [math-fun] Hero[n]'s formula Here is a heuristic, not a proof. But is very suggestive, and may have influenced people two thousand or more years ago. Note that one doesn't need to think about imaginary numbers --- merely that the factors vanish as the triangle degenerates. Leave out the square roots until the dimension has got too big and needs reducing. R. What is a triangle? a + b > c, etc. Might expect area to contain factors sqrt(b+c-a) sqrt(c+a-b) sqrt(b+c-a) To get dimension up to 2 we need another factor, one that is symmetric in a,b,c. sqrt(a+b+c) -- and we're home within a constant, which an easy case --- say 3,4,5 --- shows to be 1/4. Note also: Conway's `extraversion' gives equal weight to the 4 quantities s s-a s-b s-c On Wed, 26 Apr 2006, Eugene Salamin wrote: > --- Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote: > >> As all munsters know, area^2 = s(s-a)(s-b)(s-c) = xyz(x+y+z). >> Is there a really good proof of this? The best one I've been able to >> cook up[1] says: let t,u,v = tan A/2 etc.; then xt=r (the inradius) >> etc., so t:u:v = 1/x:1/y:1/z, and (tangent-sum formula, angle-sum of >> triangle) uv+vt+tu=1, so t = 1/x / sqrt(1/yz+1/zx+1/xy), so r^2 = >> 1/(1/yz+1/zx+1/xy) = xyz/(x+y+z) and as rs=area we're done. But this >> feels a bit too complicated. Is there perhaps a neat proof in 4 or 6 >> dimensions that looks at the cartesian product of the triangle (or >> some tetrahedron with the triangle as base) with itself? >> >> [1] Presumably it's been known for centuries. >> >> g > > Adjoin two congruent triangles along their c edge to obtain a > parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). > Using the law of cosines, A^2 can be expressed as a polynomial > P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So > > P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c). > > Q is of degree 1, and since P is symmetric, so is Q. Then Q is a > multiple of a+b+c. The fixed constant can be found by consdering e.g. > an equilateral triangle. > > Gene _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Fwd: Re: [math-fun] Hero[n]'s formula
by Henry Baker 26 Apr '06

26 Apr '06

>If you "unroll" the triangle by breaking it into three triangles, >where the bases of the triangles are the original 3 sides, and >the altitudes of the three triangles are all equal to the radius >of the incenter, then the area is the sum of the 3 smaller >triangles, and yet this area is _independent_ of everything but >the perimeter and the radius of the incenter: > >A = A_1 + A_2 + A_3 > = p*r, where r = radius of the incircle This should be (1/2) p * r, of course.

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RE: [math-fun] Hero[n]'s formula
by Schroeppel, Richard 26 Apr '06

26 Apr '06

Let's see now. If I wanted the volume of a tetrahedron, I could start by adding up the areas of three faces and subtracting the fourth. That would give me four terms to multiply together, for a total degree of 8. Maybe throw in the sum of all the areas, for degree 10. Since I want a degree 3 answer, I'll raise it to the 3/10 power. What's wrong with this picture? Rich PS: Doyle's web page looks like a lot of fun. I haven't looked beyond the Hero's formula discussion with JHC, but the titles look intriguing. --R -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Richard Guy Sent: Wednesday, April 26, 2006 12:48 PM To: math-fun Subject: Re: [math-fun] Hero[n]'s formula Here is a heuristic, not a proof. But is very suggestive, and may have influenced people two thousand or more years ago. Note that one doesn't need to think about imaginary numbers --- merely that the factors vanish as the triangle degenerates. Leave out the square roots until the dimension has got too big and needs reducing. R. What is a triangle? a + b > c, etc. Might expect area to contain factors sqrt(b+c-a) sqrt(c+a-b) sqrt(b+c-a) To get dimension up to 2 we need another factor, one that is symmetric in a,b,c. sqrt(a+b+c) -- and we're home within a constant, which an easy case --- say 3,4,5 --- shows to be 1/4. Note also: Conway's `extraversion' gives equal weight to the 4 quantities s s-a s-b s-c On Wed, 26 Apr 2006, Eugene Salamin wrote: > --- Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote: > >> As all munsters know, area^2 = s(s-a)(s-b)(s-c) = xyz(x+y+z). >> Is there a really good proof of this? The best one I've been able to >> cook up[1] says: let t,u,v = tan A/2 etc.; then xt=r (the inradius) >> etc., so t:u:v = 1/x:1/y:1/z, and (tangent-sum formula, angle-sum of >> triangle) uv+vt+tu=1, so t = 1/x / sqrt(1/yz+1/zx+1/xy), so r^2 = >> 1/(1/yz+1/zx+1/xy) = xyz/(x+y+z) and as rs=area we're done. But this >> feels a bit too complicated. Is there perhaps a neat proof in 4 or 6 >> dimensions that looks at the cartesian product of the triangle (or >> some tetrahedron with the triangle as base) with itself? >> >> [1] Presumably it's been known for centuries. >> >> g > > Adjoin two congruent triangles along their c edge to obtain a > parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). > Using the law of cosines, A^2 can be expressed as a polynomial > P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So > > P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c). > > Q is of degree 1, and since P is symmetric, so is Q. Then Q is a > multiple of a+b+c. The fixed constant can be found by consdering e.g. > an equilateral triangle. > > Gene _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] A problem for the shower
by David Wolfe 26 Apr '06

26 Apr '06

A question struck me in the shower this morning: Suppose you would like a piece of soap which dissolves at a uniform rate. I.e., as it gets smaller, the soap bar's surface area remains constant until it vanishes. I have convinced myself that one should be able to construct such a bar of soap with arbitrarily small holes. As the soap gets smaller, the holes are exposed to increase the surface area. Need it have holes?

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RE: [math-fun] arctan: update sieving and a fun relation
by Schroeppel, Richard 25 Apr '06

25 Apr '06

A potential problem with dropping the priority queues is the amount of random memory access. If your problem won't fit in the memory cache, things slow down a lot. There's a way of organizing priority queues that drops most of the (heavy) reorganization overhead. The key idea is to use a larger base like 256, instead of 2. Bins come in a few sizes: the stuff to be processed in the immediate future (about 256 cells), the less-than immediate future (256 cells, each holding unsorted values in a width-256 range), and the far future (256 cells, each holding a width-65536 range). You spend the bulk of your time doing whatever processing is required for individual values. When a low-level 256-range is completed, you open up the next higher level cell and spread out its contents into the low-level 256 array. There's never any sorting, and only one or two levels of rebinning into smaller bins. You can tinker with the bin-size ratios at each level to optimize your problem; there's nothing sacred about 256. The advantage of this scheme is that it's compatible with cache memory, and will even work tolerably if there are disk files involved. One interesting gimmick with divisor lists is that the insertion of a next-number-with-this-divisor item, (item: N+D, divisor D) into slot N+D can always be at the top of the list. When the time comes to process N+D, the list of divisors will be in sorted (increasing) order already. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Joerg Arndt Sent: Tuesday, April 25, 2006 2:00 AM To: math-fun Subject: [math-fun] arctan: update sieving and a fun relation Sieving (finding all X<=M so that X^2+1 factors into primes <= 761): First version: factor X^2+1 and test smoothness. takes 11,000 cycles per test. Improvements (gcd trick) accelerate method but still not good enough. Second version: adding logarithms (as described in earlier mail) and use priority queues. Takes 310 cycles. Tedious optimizations bring it down to 240 cycles. Inspection of the machine code shows the reschedule of the binary heaps eats most of the cycles. Third version: same but _discard_ priority queues. Careful but straightforward optimization bring method down to just over 11 cycles per test! Analyzing memory throughput and intructions involved: both reasonably close to peak performance. Any further speedup would need completely new ideas. <clip>

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[math-fun] arctan: update sieving and a fun relation
by Joerg Arndt 25 Apr '06

25 Apr '06

Sieving (finding all X<=M so that X^2+1 factors into primes <= 761): First version: factor X^2+1 and test smoothness. takes 11,000 cycles per test. Improvements (gcd trick) accelerate method but still not good enough. Second version: adding logarithms (as described in earlier mail) and use priority queues. Takes 310 cycles. Tedious optimizations bring it down to 240 cycles. Inspection of the machine code shows the reschedule of the binary heaps eats most of the cycles. Third version: same but _discard_ priority queues. Careful but straightforward optimization bring method down to just over 11 cycles per test! Analyzing memory throughput and intructions involved: both reasonably close to peak performance. Any further speedup would need completely new ideas. Curious arccot relation (write [X] for arccot(X)): If [X] was exactly 1/X we'd have [X] = [X+1] + [X*(X+1)] (and trivially [N*X] = [N*(X+1)] + [N*X*(X+1)]) But [X] = 1/X -1/(3*X^3) +- ... and we have [K*N] = [(K+1)*N] + [K*(K+1)*N] - [Q(K,N)] (*) where Q(K,N):=K^2*(K+1)^2*N^3+(K^2+K+1)*N The relation depends on the factorization of the argument X. Contrast the well known [X] = [X+D] + [X+(X^2+1)/D] which depends on the factorization of X^2+1. Note that relation (*) gives different expressions for [K*N], [N*K], [1*(N*K)], and [(N*K)*1] Is relation (*) already known? Worth to be added on one of Eric's pages? The arccot connectedness of Q(K,N) is reflected in Q(K,N)^2 + 1 == ((N*K)^2+1) * (((K+1)*N)^2+1) * ((K*(K+1)*N)^2+1)

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[math-fun] OBKR/UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO/TWTQSJQSSEKZZWATJKLUDIAWINFBNYP/VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
by dasimov＠earthlink.net 25 Apr '06

25 Apr '06

http://www.nytimes.com/2006/04/22/us/22puzzle.html

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[math-fun] Function puzzle
by franktaw＠netscape.net 24 Apr '06

24 Apr '06

Find a strictly increasing sequence of positive integers a(k) such that (a(0) + a(1)*x + a(2)*x^2 + a(3)*x^3 + ...) * (a(0) - a(1)*x + a(2)*x^2 - a(3)*x^3 + ...) = 1. Note: there are infinitely many solutions. One with a radius of convergence of 1 would be preferred. (No radius of convergence larger than 1 is possible with the given conditions.) Franklin T. Adams-Watters ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com

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