一道“简单”的几何题

wayne

呵呵，有道理， 我再想想，改改程序

KeyTo9_Fans

看看1楼的式子用Tex排出来效果如何 第一个式子： $\frac{\sqrt{576+6\sqrt{105-6\root{3}{620+12\sqrt{849}}-6\root{3}{620-12\sqrt{849}}}-6\sqrt{8\root{3}{63332+1860\sqrt{849}}+8\root{3}{63332-1860\sqrt{849}}+70\root{3}{620+12\sqrt{849}}+70\root{3}{620-12\sqrt{849}}+969}+6\sqrt{1938-8\root{3}{63332+1860\sqrt{849}}-8\root{3}{63332-1860\sqrt{849}}-70\root{3}{620+12\sqrt{849}}-70\root{3}{620-12\sqrt{849}}+90\sqrt{105-6\root{3}{620+12\sqrt{849}}-6\root{3}{620-12\sqrt{849}}}}-6\sqrt{210+6\root{3}{620+12\sqrt{849}}+6\root{3}{620-12\sqrt{849}}+6\sqrt{8\root{3}{63332+1860\sqrt{849}}+8\root{3}{63332-1860\sqrt{849}}+70\root{3}{620+12\sqrt{849}}+70\root{3}{620-12\sqrt{849}}+969}}}}{6}$ 第二个式子： $\frac{\sqrt{144-3\sqrt{6\root{3}{37484+12\sqrt{849}}+6\root{3}{37484-12\sqrt{849}}+201}-3\sqrt{402-6\root{3}{37484+12\sqrt{849}}-6\root{3}{37484-12\sqrt{849}}+6\sqrt{8\root{3}{175646564+112452\sqrt{849}}+8\root{3}{175646564-112452\sqrt{849}}-134\root{3}{37484+12\sqrt{849}}-134\root{3}{37484-12\sqrt{849}}+9}}}}{3}$ 还有3楼的式子： $\sqrt{16-\sqrt{2/3\root{3}{37484+12\sqrt{849}}+2/3\root{3}{37484-12\sqrt{849}}+67/3}-\sqrt{134/3-2/3\root{3}{37484+12\sqrt{849}}-2/3\root{3}{37484-12\sqrt{849}}+2\sqrt{2/3\root{3}{-108+12\sqrt{849}}-2/3\root{3}{108+12\sqrt{849}}+1}}}$ 另外，此贴一发，Fans的积分就达到4位数了。

wayne

重新编了程序。还是只寻找正整数解的情况。这次快了好几个数量级，找出x<1000的所有正整数解的情况，1秒钟就能搞定 这是前几个 {x,{a,b,c}}

{40, {{401, 58, 38}}}, {56, {{119, 70, 30}}}, {63, {{663, 87, 55}, {105, 87, 35}}}, {70, {{182, 74, 21}}}, {80, {{802, 116, 76}, {116, 100, 35}}}, {96, {{296, 104, 35}}}, {105, {{273, 175,90}, {175, 119, 40}}}, {112, {{238, 140, 60}, {238, 113, 14}}} {117, {{533, 195, 120}}}, {119, {{7081, 169, 118}}}, {120, {{1203, 174, 114}}}, {126, {{1326, 174, 110}, {210, 174, 70}}}, {140, {{364, 175, 80}, {364, 148, 42}}}, {144, {{219, 156, 44}}}, {160, {{1604, 232, 152}, {232, 200, 70}}}, {165, {{915, 429,275}, {429, 187, 72}}}, {168, {{357, 210, 90}, {210, 182, 45}}}, {189, {{1989, 261, 165}, {315, 261, 105}}}, {192, {{592, 208, 70}}}, {200, {{2005, 290, 190}}}

我之前的那个随机搜索算法搜了十分钟原来是在下图处转圈，前面这么多小的值都给无视了， 。。。

gxqcn

{200, {{2005, 290, 190}}} 等同于 {40, {{401, 58, 38}}}

数学星空

呵,只要x值算出来是前面结果整数倍的均是重复的 例:{40, {{401, 58, 38}}}={80, {{802, 116, 76}}}={120, {{1203, 174, 114}}}={160, {{1604, 232, 152}={200, {{2005, 290, 190}}}.... 当然,编程排除掉这些重复项也比较简单...

wayne

有道理 只保留非平凡解，即a,b,c互质的解，x<1000，总共有74组： {40, {{401, 58, 38}}} {56, {{119, 70, 30}}} {63, {{663, 87, 55}, {105, 87, 35}}} {70, {{182, 74, 21}}} {80, {{116, 100, 35}}} {96, {{296, 104, 35}}} {105, {{273, 175, 90}, {175, 119, 40}}} {112, {{238, 113, 14}}} {117, {{533, 195, 120}}} {119, {{7081, 169, 118}}} {140, {{364, 175, 80}}} {144, {{219, 156, 44}}} {165, {{915, 429, 275}, {429, 187, 72}}} {168, {{210, 182, 45}}} {224, {{1799, 476, 340}}} {225, {{1695, 255, 112}}} {252, {{1148, 273, 96}, {420, 273, 80}}} {260, {{1700, 676, 455}}} {264, {{814, 275, 70}, {561, 286, 90}}} {275, {{715, 365, 176}}} {300, {{500, 375, 144}}} {315, {{1435, 357, 150}}} {364, {{1300, 455, 224}}} {385, {{1375, 407, 120}}} {400, {{1625, 500, 252}}} {408, {{442, 425, 70}}} {448, {{3152, 848, 585}}} {455, {{7969, 481, 153}, {2137, 679, 406}}} {460, {{1196, 667, 336}}} {462, {{2562, 770, 495}}} {483, {{1725, 667, 360}}} {495, {{3729, 825, 560}, {825, 583, 210}}} {496, {{2015, 1054, 630}}} {520, {{1105, 754, 350}, {650, 533, 90}}} {525, {{1875, 875, 504}}} {528, {{1628, 660, 315}, {1122, 822, 385}}} {552, {{1702, 1173, 630}}} {555, {{4181, 1443, 1008}}} {576, {{1224, 776, 351}}} {585, {{4407, 689, 336}, {689, 663, 168}}} {595, {{3637, 757, 414}, {2125, 1547, 840}, {1547, 1099, 561}}} {600, {{870, 845, 306}, {750, 625, 126}}} {612, {{1887, 1020, 560}}} {616, {{4334, 1309, 910}, {1166, 770, 315}}} {630, {{6630, 1306, 975}}} {637, {{28987, 763, 414}}} {651, {{6851, 1085, 770}}} {660, {{4972, 1199, 832}}} {663, {{1937, 1105, 595}}} {672, {{4728, 840, 455}, {2353, 1022, 574}}} {675, {{9125, 1125, 819}}} {680, {{986, 697, 126}}} {693, {{3843, 707, 135}}} {728, {{2422, 1547, 858}, {1547, 1378, 630}}} {735, {{2625, 959, 495}}} {741, {{4845, 1235, 819}}} {759, {{1495, 809, 230}}} {760, {{3838, 950, 495}, {1976, 1615, 800}}} {765, {{1989, 901, 378}}} {775, {{9703, 2015, 1560}}} {792, {{6558, 2442, 1705}, {2442, 1067, 546}}} {800, {{4040, 1160, 693}}} {819, {{15981, 3731, 2964}, {15981, 1869, 1520}, {5355, 981, 490}}} {825, {{1625, 1375, 616}}} {880, {{1276, 884, 77}, {1100, 979, 260}}} {918, {{23418, 2682, 2275}}} {920, {{4646, 943, 198}}} {931, {{22819, 3325, 2800}}} {936, {{2886, 2210, 1155}, {1560, 975, 224}}} {945, {{21273, 2639, 2208}, {1233, 1071, 308}}} {952, {{1802, 1073, 374}}} {975, {{1455, 1105, 351}}} {980, {{4949, 2548, 1584}}} {992, {{7967, 1240, 680}}}

数学星空

wayne 能否把你编写的Mathematica程序贴上来,让我们也学习学习...

KeyTo9_Fans

3楼说BC的长可以表示成 $\sqrt{16-\sqrt(y)-\sqrt{67-y-sqrt{z}}}$ 其中y和z的值确实是那两个三次方程的实根。 可是，$67-y-sqrt{z}<0$啊，再开平方不就成虚数了么？ 你可能会说最后结果是实数，可是虚部不为0的数开平方是得不到实数的啊。 是不是你最后一步算错了？ 12楼抄了你的式子，发现和1楼的式子对不上。 当年也得到过{56, {119, 70, 30}}。 参见 http://tieba.baidu.com/f?kz=74349506 不过现在才知道原来有那么多整数解。 谢谢wayne了。

wayne

1. cds1 = DeleteCases[Table[t = Divisors[ii^2]; tt = (Length[t] - 1)/2;
2. data = Select[Transpose[{t, Reverse[t]}][[1 ;; tt]], EvenQ[#[[2]] - #[[1]]] &];
3. If[Length[data] >= 2, {ii, Table[Total@jj/2, {jj, data}]}, {}], {ii, 2, 500}], {}];

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1. cds2 = Table[{ii[[1]], tmp = Sqrt[Subsets[ii[[2]], {2}]^2 - ii[[1]]^2];
2. Flatten /@ Thread@{Subsets[ii[[2]], {2}], Map[1/Total[#] &, 1/tmp]}}, {ii, cds1}];

复制代码

1. ans = DeleteCases[Table[{ii[[1]], Select[ii[[2]], And @@ IntegerQ /@ # &]}, {ii, cds2}], {_, {}}];ans1 = DeleteCases[Table[{ii[[1]], Select[ii[[2]], GCD @@ # == 1 &]}, {ii, ans}], {_, {}}]; Table[{"x=", Framed[ii[[1]], RoundingRadius -> 15, FrameMargins -> 5, Background -> Green], "\t(a,b,c)=", Sequence @@ Row /@ Map[Framed, ii[[2]], {-1}]}, {ii, ans1}] // Grid

复制代码

wayne

18# KeyTo9_Fans 原来你提出这题，这蓄谋已久啊~~