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楼主 |
发表于 2026-2-20 11:08:08
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简化题目——面积为1的正n(n=3,4,5,...)边形, 将其放入三角形中,
三角形最小面积——{1.00000, 2.00000, 1.78885, 1.50000, 1.74224, 1.70711, 1.58626, 1.69443, 1.68239, 1.61603, 1.67720, 1.67167, 1.62973, 1.66905, 1.66606, 1.63716, 1.66455, 1.66275, 1.64163, 1.66180, 1.66064, 1.64453, 1.66001, 1.65921}
通项公式是这样——Table[Piecewise[{{3 Sqrt[3] Cot[Pi/n]/n, Mod[n, 3] == 0}, {8 Cos[(n + 6 - 4 Mod[n, 3]) Pi/(6 n)] (Cos[Pi/n] + Sin[(n + 6 - 4 Mod[n, 3]) Pi/(6 n)])/(n Sin[2 Pi/n]), Mod[n, 3] ≠ 0}}], {n, 3, 60}]
1.00000, 2.00000, 1.78885, 1.50000, 1.74224, 1.70711, 1.58626, 1.69443, 1.68239, 1.61603, 1.67720, 1.67167, 1.62973, 1.66905, 1.66606, 1.63716, 1.66455, 1.66275, 1.64163, 1.66180, 1.66064, 1.64453, 1.66001, 1.65921, 1.64652, 1.65876,
1.65819, 1.64794, 1.65787, 1.65745, 1.64899, 1.65721, 1.65689, 1.64979, 1.65670, 1.65645, 1.65041, 1.65630, 1.65611, 1.65090, 1.65599, 1.65583, 1.65130, 1.65573, 1.65560, 1.65162, 1.65552, 1.65542, 1.65189, 1.65535, 1.65526, 1.65212,
1.65520, 1.65513, 1.65231, 1.65508, 1.65502, 1.65247, 1.65498, 1.65492, 1.65262, 1.65488, 1.65484, 1.65274, 1.65481, 1.65476, 1.65284, 1.65474, 1.65470, 1.65294, 1.65468, 1.65464, 1.65302, 1.65462, 1.65459, 1.65309, 1.65457, 1.65455,
1.65316, 1.65453, 1.65451, 1.65322, 1.65449, 1.65447, 1.65327, 1.65446, 1.65444, 1.65331, 1.65443, 1.65441, 1.65336, 1.65440, 1.65439, 1.65340, 1.65438, 1.65436, 1.65343, 1.65435, 1.65434, 1.65346, 1.65433, 1.65432, 1.65349, 1.65431}
复杂的数据是这样:
{1,
2,
4/Sqrt[5],
3/2,
8/7 (Cos[\[Pi]/7] + Sin[(3 \[Pi])/14]),
1 + 1/Sqrt[2],
Cot[\[Pi]/9]/Sqrt[3],
2/5 (2 + Sqrt[5]),
4/11 (Cos[\[Pi]/22] + 2 Cos[(5 \[Pi])/22]) Csc[(2 \[Pi])/11],
1/4 (3 + 2 Sqrt[3]),
4/13 Csc[(2 \[Pi])/13] (2 Cos[(3 \[Pi])/26] + Sin[(3 \[Pi])/13]),
4/7 Cot[\[Pi]/7] (Cos[\[Pi]/14] + Sin[\[Pi]/7]),
1/5 Sqrt[3] Cot[\[Pi]/15],
1/8 (Sqrt[2] + 4 Cos[\[Pi]/8]) Csc[\[Pi]/8],
4/17 (Cos[(3 \[Pi])/34] + 2 Cos[(7 \[Pi])/34]) Csc[(2 \[Pi])/17],
Cot[\[Pi]/18]/(2 Sqrt[3]),
4/19 (2 Cos[(5 \[Pi])/38] + Cos[(9 \[Pi])/38]) Csc[(2 \[Pi])/19],
1/5 (3 + Sqrt[5] + Sqrt[5 + 2 Sqrt[5]]),
1/7 Sqrt[3] Cot[\[Pi]/21],
2/11 (2 Cos[(3 \[Pi])/22] + Cos[(5 \[Pi])/22]) Csc[\[Pi]/11],
4/23 (Cos[(5 \[Pi])/46] + 2 Cos[(9 \[Pi])/46]) Csc[(2 \[Pi])/23],
1/8 Sqrt[3] Cot[\[Pi]/24],
4/25 (2 Cos[(7 \[Pi])/50] + Cos[(11 \[Pi])/50]) Csc[(2 \[Pi])/25],
2/13 (Cos[(3 \[Pi])/26] + 2 Cos[(5 \[Pi])/26]) Csc[\[Pi]/13],
Cot[\[Pi]/27]/(3 Sqrt[3]),
1/7 (2 Cos[\[Pi]/7] + Cos[(3 \[Pi])/14]) Csc[\[Pi]/14],
4/29 (Cos[(7 \[Pi])/58] + 2 Cos[(11 \[Pi])/58]) Csc[(2 \[Pi])/29],
1/10 Sqrt[3] Cot[\[Pi]/30],
4/31 (2 Cos[(9 \[Pi])/62] + Cos[(13 \[Pi])/62]) Csc[(2 \[Pi])/31],
1/8 (Cos[\[Pi]/8] + 2 Cos[(3 \[Pi])/16]) Csc[\[Pi]/16],
1/11 Sqrt[3] Cot[\[Pi]/33],
2/17 (2 Cos[(5 \[Pi])/34] + Cos[(7 \[Pi])/34]) Csc[\[Pi]/17],
4/35 (Cos[(9 \[Pi])/70] + 2 Cos[(13 \[Pi])/70]) Csc[(2 \[Pi])/35],
Cot[\[Pi]/36]/(4 Sqrt[3]),
4/37 (2 Cos[(11 \[Pi])/74] + Cos[(15 \[Pi])/74]) Csc[(2 \[Pi])/37],
2/19 (Cos[(5 \[Pi])/38] + 2 Cos[(7 \[Pi])/38]) Csc[\[Pi]/19],
1/13 Sqrt[3] Cot[\[Pi]/39],
1/40 (1 + Sqrt[5] + 8 Cos[(3 \[Pi])/20]) Csc[\[Pi]/20],
4/41 (Cos[(11 \[Pi])/82] + 2 Cos[(15 \[Pi])/82]) Csc[(2 \[Pi])/41],
1/14 Sqrt[3] Cot[\[Pi]/42],
4/43 (2 Cos[(13 \[Pi])/86] + Cos[(17 \[Pi])/86]) Csc[(2 \[Pi])/43],
1/11 (Cos[(3 \[Pi])/22] + 2 Cos[(2 \[Pi])/11]) Csc[\[Pi]/22],
Cot[\[Pi]/45]/(5 Sqrt[3]),
2/23 (2 Cos[(7 \[Pi])/46] + Cos[(9 \[Pi])/46]) Csc[\[Pi]/23],
4/47 (Cos[(13 \[Pi])/94] + 2 Cos[(17 \[Pi])/94]) Csc[(2 \[Pi])/47],
1/16 Sqrt[3] Cot[\[Pi]/48],
4/49 (2 Cos[(15 \[Pi])/98] + Cos[(19 \[Pi])/98]) Csc[(2 \[Pi])/49],
2/25 (Cos[(7 \[Pi])/50] + 2 Cos[(9 \[Pi])/50]) Csc[\[Pi]/25],
1/17 Sqrt[3] Cot[\[Pi]/51],
1/13 (2 Cos[(2 \[Pi])/13] + Cos[(5 \[Pi])/26]) Csc[\[Pi]/26],
4/53 (Cos[(15 \[Pi])/106] + 2 Cos[(19 \[Pi])/106]) Csc[(2 \[Pi])/53],
Cot[\[Pi]/54]/(6 Sqrt[3]),
4/55 (2 Cos[(17 \[Pi])/110] + Cos[(21 \[Pi])/110]) Csc[(2 \[Pi])/55],
1/14 (Cos[\[Pi]/7] + 2 Cos[(5 \[Pi])/28]) Csc[\[Pi]/28],
1/19 Sqrt[3] Cot[\[Pi]/57],
2/29 (2 Cos[(9 \[Pi])/58] + Cos[(11 \[Pi])/58]) Csc[\[Pi]/29],
4/59 (Cos[(17 \[Pi])/118] + 2 Cos[(21 \[Pi])/118]) Csc[(2 \[Pi])/59],
1/20 Sqrt[3] Cot[\[Pi]/60]}
问题1——来个反例。
问题2——提供资料。
问题3——通项公式。
谢谢各位!新年快乐!!! |
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