Maximally symmetric Heron simplices ====Warren=D=Smith===Dec=2011====== A "Heron" simplex is one with all edge lengths, triangle-areas, ..., and volume, all integer. Heron simplices "don't want to exist" because many quantities need to be integer that "want" to be irrational algebraic numbers. Thus when searching for Heron simplices, symmetry is your friend. For a symmetric Heron simplex, the search is simpler (fewer parameters describe the simplex) and also there are fewer quantities that we need to force "against their will" to be integers (since the symmetry causes many of them to automatically be equal). TRIANGLES: Obviously a Heron triangle cannot be equilateral. (Area would then be irrational.) The most symmetric it can be is isoceles, i.e. 2-symmetric. And there are an infinity of isoceles Heron triangles, as you can see by gluing together two identical mirrored pythagorean right triangles at a common leg. Result: legs=m*m+n*n, base=2*(m*m-n*n)=2*(m-n)*(m+n) or legs=m*m+n*n, base=4*m*n are two 2-parameter families of isoceles Heron triangles, showing the number of such triangles with diameter<N is at least of order N. I elsewhere proved that the number of (general) Heron triangles with diameter<N is of order N^(1+o(1)) so we see that the isoceles ones constitute a goodly fraction (perhaps even a nonzero constant fraction?) of all Heron triangles. TETRAHEDRA: It is not hard to see that the most symmetric a Heron tetrahedron could hope to be is 8-symmetric. This would be achieved if it had 2 edge lengths a and b with "the a's forming a quadrilateral and the b's its two diagonals." However, apparently no 8-symmetric Heron tetrahedra exist... or at least, computer says there are none with diameter<500000... Let's prove that: In order for one to exist, it would be necessary and sufficient for a and b to obey: a=odd b=even 2<=b<=2*a-2 a>=3 (2*a*a - b*b)/2 = SquareInteger <==> 2*a*a-b*b = EvenSquare ((2*a+b)*(2*a-b))/4 = SquareInteger <==> 4*a*a-b*b=EvenSquare Can that happen? By Euclid's characterization of Pythagorean triples http://en.wikipedia.org/wiki/Pythagorean_triple we have from the last condition that for any primitive solution (a,b) that b=4*m*n and a=m*m+n*n with m,n of opposite parity wlog [or anyway with m and having different numbers of 2s in their factorizations] and hence from the penultimate condition we'd see 2*(m*m+n*n) = 16*m*m*n*n + evensquare which is impossible since the right hand side contains an even number of 2s in its factorization and the left hand side contains an odd number. Hence THEOREM: An 8-symmetric Heron tetrahedron cannot exist! [A different proof of this nonexistence theorem was also given, shortly before his table 2, by Ralph Heiner Buchholz: Perfect Pyramids, Bulletin Australian Mathematical Society 45 (1992) 353-368, and Buchholz also classifies all possible "types" of tetrahedra...] Given that an 8-symmetric Heron tetrahedron cannot exist, it is not hard to see that the most symmetric a Heron tetrahedron could hope to be is 4-symmetric. One way this can be achieved by having 2 edge lengths a, b, c with "the a's forming a quadrilateral and b and c its two diagonals." I constructed a 2-parameter infinity of this kind of Heron tetrahedra previously, showing at least order N^(1/14) of primitive 4-symmetric Heron tetrahedra exist with diameter<N. The number of 4-symmetric Heron tetrahedra with diameter<=16384 is 25, which is 6.3% of the 394 general Heron tetrahedra in that range. It would not surprise me if the 4-symmetric Heron tetrahedra constitute a nonzero constant fraction of all Heron tetrahedra; indeed perhaps they and tetrahedra derived from them constitute asymptotically a 100% fraction of all Heron tetrahedra. I made the following CENSUS: diameter #of [a,c,c,c,c,b]-type primitive Heron tets with diam<that ---------------------------------------------------------------------- 1024 0 2048 5 4096 8 8192 14 16384 25 32768 46 65536 57 131072 103 and the count seems to be growing proportionally to diam^0.7 roughly. A different kind of 4-symmetric tetrahedron has 4 faces, all of which are congruent Heron triangles with sides a,b,c. (Fred Lunnon calls these "isohedral" although I think that is a bad name. That bad name has already been used to mean something else in the world of tilings.) There are known to be an infinity of this other kind of 4-symmetric Heron tetrahedron also. [Theorem 1 of Buchholz shows this using elliptic curves -- getting an exceedingly weak lower bound of order lnln(Diameter) on the count -- but apparently, unknown to Buchholtz, this infinitude via the same proof was actually already understood by E.Haentzschel: Sitzungsber. Berlin Math. Gesell. 12 (1913) 101-108 and 17 (1918) 37-39 following up on groundwork by R. G:untsche: Sitzungsber. Berlin Math. Gesell. 6 (1907) 38-53 and Archiv Math. Phys.(3) 11 (1907) 38-53.] The isohedral Heron tetrahedra seem far less numerous than the [a,c,c,c,c,b] type. Here is a CENSUS: diameter #of [a,b,c,c,b,a]-type primitive Heron tets with diam<that ---------------------------------------------------------------------- 256 1 512 1 1024 2 2048 3 4096 6 8192 7 16384 10 32768 12 and the count seems to be growing proportionally to diam^0.6 very roughly. All the [a,b,c,c,b,a]-type primitive Heron tetrahera with diam<32768: a b c ------------------------ 203 195 148 888 875 533 1804 1479 1183 2431 2296 2175 2873 2748 1825 3111 2639 2180 5512 5215 1887 8484 6625 6409 11275 10136 8619 14637 10933 10100 19695 16448 13073 32708 31493 24525 4D-SIMPLICES: Considering permutations of the vertices... There would be one and only one way a 4-simplex could have an order-5 symmetry (or any multiple of 5): if it had 2 squared edge lengths a and b with the a's forming a pentagon and the b's the "inscribed pentagram." However, the 4-volume of such a beast for Heronity would need to be integer, which would be equivalent to demanding 5*(b^2-3*b*a+a^2)^2 = square which is impossible since the "square" would contain an odd number of 5s in its factorization. Hence THEOREM: a 5-symmetric Heron 4-simplex is impossible! Given that, the most symmetric a Heron 4-simplex could hope to be is if at least one vertex is uninvolved in the symmetry, so it is really a lower-dimensional symmetry. One possible way to build a 4-symmetric 4-simplex would be to make one 3D face be an isohedral Heron tetrahedron with edge lengths a,b,c; then join all 4 of its vertices to a 5th vertex via 4 equal edges of length d. (If d were odd this would require a,b,c all to be even.) I did a computer search up to diameter 263000 for this kind of 4-simplex (Heron tetrahedra of type [a,b,c,c,b,a]+dddd) and found nothing. Here is a Cayley matrix for a different possible 4-simplex (squared edge lengths as entries) which also would be 4-symmetric if it existed. [0 a a b b 1] [a 0 c e e 1] [a c 0 e e 1] [b e e 0 d 1] [b e e d 0 1] [1 1 1 1 1 0] And my computer undertook a search for those up to diameter 133000 and found nothing. Based on heuristic "probability" arguments I suspect there are no 4D Heron simplices. However, I'd be a lot more convinced of that if anybody carried out an exhaustive search up to diameter 10^9, which might now be feasible with a good implementation of my N^(1+o(1))-time exhaustive generation algorithm. But it is not trivial to implement that algorithm and there are many games you can play within that framework to "prune the search." -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)