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楼主 |
发表于 2014-2-15 00:26:52
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对于椭圆\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)\)内接等角\(N(\geqslant 5)\)边形,我们得到如下结论:
当椭圆\(\frac{a}{b}\in [\frac{1}{x_0},x_0]\)时,仅存在四个等角\(N(\geqslant 5)\)边形,且这4个等角形对称分布在椭圆左右或者上下两侧。
例\(N=5\)时
\(x_0=\frac{\sqrt{25+20\sqrt{5}}}{5}\)
设等角五边形各个顶点坐标依次为\((0,-b),(x_1,y_1),(x_2,y_2),(-x_2,y_2),(-x_1,y_1)\)[另一个关于X轴对称],设各边长分别为\(L_1,L_2,L_3\),总长为\(L\),面积为\(S\)
\( (-a^2k_1^2-b^2)x_1+2a^2bk_1=0 \)
\( (-a^2k_1^2-b^2)y_1^2-2b^3y_1+a^2b^2k_1^2-b^4=0 \)
\( (a^4k_1^2k_2^2+a^2b^2k_1^2+a^2b^2k_2^2+b^4)x_2+2a^4bk_1^2k_2-2a^4bk_1k_2^2+2a^2b^3k_1-2a^2k_2b^3=0 \)
\( (a^4k_1^2k_2^2+a^2b^2k_1^2+a^2b^2k_2^2+b^4)y_2+a^4k_2^2k_1^2b-a^2k_1^2b^3+4a^2b^3k_1k_2-a^2b^3k_2^2+b^5=0 \)
\(L_1^2-x_1^2-(b+y_1)^2=0\)
\(L_2^2-(x_1-x_2)^2-(y_1-y_2)^2=0\)
\(L_3=2x_2\)
\(L-2(L_1+L_2+x_2)=0\)
\(-51200a^8b^4n(a^2+b^2)(5a^6t-65a^4b^2t+79a^2b^4t-43b^6t+25a^6-85a^4b^2+235a^2b^4-103b^6)+1024a^6b^3(5a^4+10a^2b^2+b^4)(25a^6n^2t-25a^4b^2n^2t+75a^2b^4n^2t-19b^6n^2t+50a^6n^2-100a^4b^2n^2+170a^2b^4n^2-80b^6n^2+50a^6t+100a^4b^2t+250a^2b^4t-40b^6t-
125a^6-275a^4b^2-375a^2b^4+15b^6)L+320a^4b^2n(a^2+b^2)(5a^4+10a^2b^2+b^4)^2(3a^2t-b^2t+5a^2-15b^2)L^2-16a^2b(5a^4+10a^2b^2+b^4)^3(
2a^2n^2t+5a^2n^2-b^2n^2+5a^2t-b^2t-10a^2)L^3+n(5a^4+10a^2b^2+b^4)^4L^4=0\)
\(-n(5a^4+10a^2b^2+b^4)^2(20a^4t+4a^2b^2t-45a^4-10a^2b^2-b^4)S+20b^2a^4(325a^8t+580a^6b^2t+430a^4b^4t+68a^2b^6t+5b^8t-725a^8-1300a^6b^2-990a^4b^4-180a^2b^6-5b^8)=0\)
注:\(t=\sqrt{5},n=\sqrt{5-2\sqrt{5}},k_1=\sqrt{5-2\sqrt{5}},k_2=-\sqrt{5+2\sqrt{5}}\) |
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