Finding largest small polygons using symbolic computations

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1 Finding largest small polygons using symbolic computations by Dragutin Svrtan Faculty of Science, Department of Mathematics, Bijenička Zagreb, Croatia, Abstract A small polygon is a convex polygon of unit diameter. The problem of determining the largest area of small n gons was already studied by Reinhardt in 19. He showed that for n odd, the obvious configuration is optimal (regular n gon, for n odd, has maximal area). For n even this is not the case. For n = 6 the largest area F 6, a plane hexagon of unit diameter can have, satisfies a 10 th degree irreducible polynomial over the integers (famous Graham s largest small hexagon JCT(1975)). This problem needed factorizations 1

2 of a polynomial of degree 40 with coefficients having 15 to 5 digits. Graham introduced the diameter graph of a polygon by joining only those vertices which are at maximal distance. For n = 6 there are 10 possible graphs. The case n = 8 is much more complicated (it has 31 diameter graphs and in 00. C. Audet, P. Hansen, F. Messine solved approximately a difficult global optimization with 10 variables and 0 constraints (Hansen s little octagon)). In 007., Foster and Szabo proved Graham s conjecture that the diameter graphs are cycles with a pending edge. In this talk we reprove Graham & Johnson s computations with a polynomial of degree 14 instead of 40. Moreover, under the axial symmetry conjecture we obtain, for the first time, explicit equations for F 8 (of degree 4) and F 10 (of degree 15) by intriguing symbolic computations (involving even 800 digit numbers).

3 1 Small pentagons (x, y ) = (x 1 cos α, y 1 sin α), (x 3, y 3 ) = ( x 1 +cos β, y 1 sin β) 3

4 Area equation: A 5 = 1 x 3 y 3 x 1 y x 1 y 1 x 1 y x 1 y 1 x y S = 1 (x 3y 1 y 3 x 1 ) + x 1 y 1 1 (x 1y + y x 1 ) S Constraint: = y +y 3 x 1 + x 3 x y 1 + x 1 y 1 S = y 1 sin α+y 1 sin β x 1 + x 1+cos β=x 1 +cos α y 1 + x 1 y 1 S = 1 (sin α + sin β)x (cos α + cos β)y 1 x 1 y 1 S (x 3 x ) + (y 3 y ) = 1 ( x 1 + cos α + cos β) + (sin α sin β) = 1 4x 1 4x 1 (cos α + cos β) + cos(α + β) + 1 = 0 We fix (x 1, y 1 ), x 1 + y1 = 1 (1 x 1 1 ). Instead of coordinates (α, β), we introduce new coordinates (K, L) so that K = α + β, 4

5 L = α β. Then the area equation becomes ( A 5 = x 1 sin K + y 1 cos K ) cos L x 1y 1 S with the constraint: 4x 1 8x 1 cos K cos L + cos K + 1 = 0 By considering (α, β) with K = α + β fixed we see that the constraint implies that cos L is also fixed, so the area is maximized if the only varying term (x 1 sin K cos L ) in A 5 is maximized, which implies L = α β = 0. Thus the area maximizing small pentagon should be axially symmetric. From the constraint we get the equation for K (corresponding to the choice (x 1, y 1 ), x 1 + y1 = 1): 4x 1 8x 1 cos K + cos K + 1 = 0 5

6 which, by using cos K = cos K 1, implies The area equation cos K = x A 5 = x 1 sin K + y 1(cos K x 1) S = x 1 1 (x ) + y 1 S = x 1 1 (x ) + 1 x 1 S This equation can be rationalized by using the substitution x 1 =. One obtains implicit equation 1 t 1+t 1 Eq 5 := (16S + 5)t t 7 + 4(16S + 35)S St 5 + (48S 61)S 4 + 3St 3 + 4(4S 1)(4S + 1)t 3St + 16S 3 with the discriminant discr(eq 5, t) = 7 (56S 4 400S +15)(65536S S S 883). 6

7 The largest small pentagon equation is the A 5 = 56S 4 400S + 15 (The second factor has only two real solutions ± which should be discarded.) 7

8 Small hexagons 8

9 Area equation: A 6 = 1 x 3 y 3 x 1 y x 1 y 1 x 1 y x 1 y 1 x y + x 1(1 y 1 ) S = 1 (x 3y 1 y 3 x 1 ) + x 1 y 1 1 (x 1y + y x 1 ) + x 1 x 1 y 1 S = y +y 3 x 1 + x 3 x y 1 + x 1 S = 1 (sin α + sin β)x (cos α + cos β)y 1 + x 1 (1 y 1 ) S = sin K cos L x 1 + cos K cos L y 1 + x 1 (1 y 1 ) S, (K = α + β, L = α β) Constraint: (x 3 x ) + (y 3 y ) = 1 ( x 1 + cos α + cos β) + (sin α sin β) = 1 4x 1 4x 1 (cos α + cos β) + cos(α + β) + 1 = 0 We fix (x 1, y 1 ), x 1 + y1 = 1 (1 x 1 1 ). Instead of coordinates (α, β), we introduce new coordinates (K, L) so that K = α + β, 9

10 L = α β. Then the area equation becomes ( A 6 = x 1 sin K + y 1 cos K ) cos L + x 1(1 y 1 ) S with the constraint: 4x 1 8x 1 cos K cos L + cos K + 1 = 0 By considering (α, β) with K = α + β fixed we see that the constraint implies that cos L is also fixed, so the area is maximized if the only varying term (x 1 sin K cos L ) in A 6 is maximized, which implies L = α β = 0. Thus the area maximizing small hexagon should be axially symmetric. From the constraint we get the equation for K (corresponding to the choice (x 1, y 1 ), x 1 + y1 = 1): 4x 1 8x 1 cos K + cos K + 1 = 0 10

11 which, by using cos K = cos K 1, implies The area equation cos K = x A 6 = x 1 sin K + y 1 cos K + x 1(1 y 1 ) S = x 1 1 ( 1 + x ( 1) + 1 x 1) 1 x 1 + x 1 S ( 1 + The function F (x) = ( 1 x) 1 x + x 1 ( 1 + x) was studied by S. C. Johnson and R. L. Graham as SIGSAM Problem #7 and it has a maximum at x , where it attains the value y This value is the root of an irreducible polynomial of 10 th degree over the integers and it was obtained in four steps: (1) F (x) = 0 and clear the radicals. The result is a 10 th degree polynomial P (x) which has a root x )

12 () In the equation y = F (x) clear the radicals in order to get a polynomial Q(x, y) such that Q(x, y) = 0 when y = F (x). Q has degree four in y. (3) Eliminate x from P and Q. The resultant R(y) has degree 40 with coefficients having 15 to 5 digits. (4) Factor R(y) into 4 irreducible factors each of degree 10. Only one factor has a root at so it is the desired minimal polynomial: 4096y y y y y y y y y 78488y Graham and Johnson wrote: It would be of considerable interest if this problem could be solved by some other method which avoided the factorization. (Bell Labs ALTRAN resultant ARES step (3), reduction mod various primes, Hensel lemma step (4).) 1

13 Now we give ultimately simple method to study the famous Graham s equation which avoids factorization. OUR APPROACH: We use any of the two rational parameterizations corresponding to the two radicals in A 6 and then we take the discriminant (which corresponds to Lagrange multipliers): i) The substitution x 1 = 1 t 1+t 1 leads to the implicit equation Q 1 = [ (t + 1) S + (3t 1)(t + 1) ] +4t (5t +1)(t 3). Then discrim(q 1, t) 60 = (4096S S S S S S S S S 78488y )(S 1) 4. ii) The substitution x 1 = 1 t 1+t leads also to a degree 14 equation but with the extraneous factor given by (S 1) (S + 1). 13

14 3 Largest symmetric small heptagon Area equation: A 7 = (x x 1 )(y 1 y ) + ((x x 1 ) + (x 1 x + x 4 ))(y 1 y + y 4 (y 1 y ))+ +(x 1 x + x 4 + x 1 )(y y 4 ) S = = (x x 1 )y 1 + (3x 1 x + x 4 )y + ( x 1 + x )y 4 S Constraints: eq 1 := x 1 + y 1 1, eq := x + y 1 eq 3 := x 4 + y 4 1, eq 4 := x 1 x + x 4 1 By solving for x 4 and y 4 from A 7, eq 4 and plugging into eq 3 we obtain the implicit equation for S: ( ( x1 ) ) Eq = ( x 1 + x ) + x [S (x x 1 )y 1 (3x 1 x + x 4 )y ] = 0 14

15 Then by rational parametrization x 1 = 1 t, y 1+t 1 = t, x 1+t = 1 s, y 1+s s = s we obtain 1+s Eq = 4(Ss 4 t 4 + s 3 t 4 + Ss 4 t + Ss t 4 + 4s 4 t 6s 3 t 4s t 3 + 5st 4 + Ss Ss t + St 4 7s 3 + 4s t + st 4t 3 + Ss + St 3s + S) (s t 3s + 3t 1) (s t + 5s 3t + 1)(3s t s + 7t + 3) Now we proceed with computation of some iterated discriminants: i) discrim(eq, s) = 64 (t + 1) 6 P 40 P 8 where the factor P 40 has degree 40 in t. (The factor P 8 of degree 8 in t has discrim(p 8, t) = 60 (56S 6 7) ), hence it should be discarded). ii) discrim(eq, t) = 56 (s + 1) 7 P 4 Q 8 where the factor P 4 has degree 4 in s. The factor Q 8 of degree 8 in s has discrim(q 8, s) = 60 (56S 7) (S + 1) (S 1), hence it should be discarded. 15

16 Now we compute the greatest common divisor of 1 := discrim(p 40, t) and 1 := discrim(p 4, s). We obtain gcd( 1, 1 ) = P 6 P 44 where P 6 = 4096S S S and P 44 has degree 44 in S and has the largest real root which is smaller than 1. So the minimal polynomial for the area of a largest small heptagon is P 6, and it is approximately S = 7 8 sin π 7 cos π and P 6 is the minimal polynomial for the area of a regular 7 gon (of diameter 1). 16

17 4 Largest symmetric small octagon For the area equation A 8 for symmetric small octagon we have A 8 = A 7 + x 1 (1 y 1 ) Then we have new equation: ( ( x1 ) ) Eq = ( x 1 + x ) + x [S (x x 1 )y 1 (3x 1 x + x 4 )y x 1 (1 y 1 )] which, with substitution x 1 = 1 t,..., y 1+t 1 = s, gives the similar 1 s equation Eq = Now we have: i) discrim(eq, s) = 64 t 4 (t + 1) 34 P 3 P 4, where the factor P 3 has degree 3 in t. (The factor P 4 of degree 4 in t has 17

18 discrim(p 4, t) = 18 (S 1)(8S S S 8S 7)(4S 8S + 7) with only real roots ± ). ii) discrim(eq, t) = 68 (s + 4s + 5)(Ss + s + S s 1) 4 (s + 1) 34 Q 3, where Q 3 has degree 3 in s. Now we compute the greatest common divisor of 1 = discrim(p 3, t)/ 891 and 1 = discrim(q 3, s)/ 969 and we get explicitly the desired equation F 8 = gcd( 1, 1 ) = S for the area of the largest small octagon (Hansen s octagon). 18

19 Remark. We also managed to get the area equation for the largest symmetric nonagon as follows: 4096S S S = 0. Its second largest real root is 9 8 sin π 9 cos π which corresponds to a regular nonagon (in accordance with Reinhardt 19 result). 19

20 5 Addendum 5.1 Graham s result R = S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S factor(r) = ( 4096S S S S S S S S S 3416S + 41 ) ( 4096S S S S S S S S S 78488S ) ( 819S S S 8 58S S S S S S + 340S 8179) ( 819S S S S S S S S S 48556S ) 0

21 5. A8eq A8eq := S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S

22 5.3 Sizes of coefficients map(length, coef f s(op(1, op(3, Delta3)), S)); {180, 183, 187, 1830, 1834, 1836, 1840, 1843, 1846, 1849, 185, 1855, 1857, 1860, 1863, 1866, 1869, 187, 1874, 1876, 1880, 188, 1885, 1888, 1890, 189, 1896, 1898, 1901, 1903, 1905, 1908, 1911, 1913, 1915, 1918, 190, 19, 194, 197, 1930, 193, 1934, 1936, 1938, 1941, 1943, 1945, 1947, 1949, 1950, 195, 1954, 1956, 1958, 1960, 1961, 1963, 1965, 1966, 1968, 1970, 197, 1973, 1975, 1977, 1979, 1980, 198, 1984, 1985, 1987, 1988, 1989, 1991, 199, 1994, 1996, 1997, 1999, 000, 001, 003, 005, 007, 009, 010, 01, 013, 015, 016, 018, 019, 01, 0, 03, 04, 06, 08, 030, 031, 033, 034, 035, 036, 038, 040, 04, 043, 045, 047, 048, 049, 051, 05, 054, 055, 056, 057, 059, 061, 06, 064, 065, 066, 067, 069, 071, 073, 074, 076, 077, 079, 080, 08, 083, 085, 086, 088, 089, 090, 09, 093, 095, 096, 097, 099, 100, 101, 103, 104, 106, 107, 108, 109, 111, 11, 113, 115, 116, 117, 118, 119, 11, 1, 13, 14, 15, 16, 18, 19, 130, 131, 13, 134, 136, 137, 139, 140, 141, 143, 144, 145, 147, 148, 149, 150, 15, 153, 154, 155, 157, 158, 159, 160, 16, 163, 164, 165, 166, 168, 169, 170, 171, 17, 174, 175, 176, 177, 178, 179, 181, 18, 183, 184, 185, 186, 187, 188, 189, 191, 19, 193, 194, 195, 196, 197, 198, 199, 00, 0, 03, 04, 05, 06, 07, 09, 10, 11, 1, 13, 14, 16, 17, 18, 19, 0, 1,, 4, 5, 6, 7, 8, 9, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 5, 53, 54, 56, 57, 58, 59, 60, 61, 6, 63, 64, 65, 66, 67, 68, 69, 70, 71, 7, 73, 74, 75, 76, 77, 78, 79, 80, 81, 8, 84, 85, 86, 87, 88, 89, 90, 91, 9, 93, 94, 95, 96, 97, 98, 99, 300, 301, 30, 303, 304, 305, 306, 307, 308, 309, 310, 31, 313, 314, 315, 316, 317, 318, 319, 30, 31, 3, 33, 34, 35, 36, 37, 38, 39, 330, 331, 33, 333, 334, 335, 336, 337, 338, 339, 340, 341, 34, 343, 344, 345, 346, 347, 348, 349, 350, 351, 35, 353, 354, 355, 356, 357, 358, 359, 360, 361, 36, 363, 364, 365, 366, 367, 368, 369, 370, 371, 37, 373, 374, 375, 376, 377, 378, 379, 380, 381, 38, 383, 384, 385, 386, 387, 388, 389, 390, 391, 39, 393, 394, 395, 396, 397, 398, 399, 400, 401, 40, 403, 404, 405, 406, 407, 408, 409, 410, 411,

23 41, 413, 414, 415, 416, 417, 418, 419, 40, 41, 4, 43, 44, 45, 46, 47, 48, 49, 430, 431, 43, 433, 434, 435, 436, 437, 438, 439, 440, 441, 44, 443, 444, 445, 446, 447, 448, 449, 450, 451, 45, 453, 454, 455, 456, 457, 458, 459, 460, 461, 46, 463, 464, 465, 466, 467, 468, 469, 470, 471, 47, 473, 474, 475, 476, 477, 478, 479, 480, 481, 48, 483, 484, 485, 486, 487, 488, 489, 490, 491, 49, 493, 494, 495, 496, 497, 498, 499, 500, 501, 50, 503, 504, 505, 506, 507, 508, 509, 510, 511, 51, 513, 514, 515, 516, 517, 518, 519, 50, 51, 5, 53, 54, 55, 56, 57, 58, 59, 530, 531, 53, 533, 534, 535, 536, 537, 538, 539, 540, 541, 54, 543, 544, 545, 546, 547, 548, 549, 550, 551, 55, 553, 554, 555, 556, 557, 558, 559, 560, 561, 56, 563, 564, 565, 566, 567, 568, 569, 570, 571, 57, 573, 574, 575, 576, 577, 578, 579, 580, 581, 58, 583, 584, 585, 586, 587, 588, 589, 590, 591, 59, 593, 594, 595, 596, 597, 598, 599, 600, 601, 60, 603, 604, 605, 606, 607, 608, 609, 610, 611, 61, 613, 614, 615, 616, 617, 618, 619, 60, 61, 6, 63, 64, 65, 66, 67, 68, 69, 630, 631, 63, 633, 634, 635, 636, 637, 638, 639, 640, 641, 64, 643, 644, 645, 646, 647, 648, 649, 650, 651, 65, 653, 654, 655, 656, 657, 658, 659, 660, 661, 66, 663, 664, 665, 666, 667, 668, 669, 670, 671, 67, 673, 674, 675, 676, 677, 678, 679, 680, 681, 68, 683, 684, 685, 686, 687, 688, 689, 690, 691, 69, 693, 694, 695, 696, 697, 698, 699, 700, 701, 70, 703, 704, 705, 706, 707, 708, 709, 710, 711, 71, 713, 714, 715, 716, 717, 718, 719, 70, 71, 7, 73, 74, 75, 76, 77, 78, 79, 730, 731, 73, 733, 734, 735, 736, 737, 738, 739, 740, 741, 74, 743, 744, 745, 746, 747, 748, 749, 750, 751, 75, 753, 754, 755, 756, 757, 758, 759, 760, 761, 76, 763, 764, 765, 766, 767, 768, 769, 770, 771, 77, 773, 774, 775, 776, 777, 778, 779, 780, 781, 78, 783, 784, 785, 786, 787, 788, 789, 790, 791, 79, 793, 794, 795, 796, 797, 798, 799, 800, 801, 80, 803, 804, 805, 806, 807, 808, 809, 810, 811, 81, 813, 814, 815, 816, 817, 818, 819, 80, 81, 8, 83, 84, 85, 86, 87, 88, 89, 830, 831, 83, 833, 834, 835, 836, 837, 838, 839, 840, 841, 84, 843, 844, 845, 846, 847, 848, 849, 850, 851, 85, 853, 854, 855, 856, 857, 858, 859, 860, 861, 86, 863, 864, 865, 866, 867, 868, 869, 870, 871, 87, 873, 874, 875, 876, 877, 878, 879, 880, 881, 88, 883, 884, 885, 886, 887, 888, 889, 890, 891, 89, 893} 3

24 5.4 Aeq10 := op(, Delta3); Aeq10 := op(, Delta3); S S S S S S S S S S S S

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