椭圆内接三角形各边长平方和及倒数和的极值问题
我们已经得到了有关外椭圆 \(\frac{x^2}{m^2}+\frac{y^2}{n^2}=1\)内接凸N边形各边长和的最大值结论：即当N边形内切一个与外椭圆共焦点的内椭圆\(\frac{x^2}{m^2-t}+\frac{y^2}{n^2-t}=1\)时，各边长和取最大值，且为定值
具体见：https://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=3740&fromuid=1455
我们现在讨论
1.外椭圆 \(\frac{x^2}{m^2}+\frac{y^2}{n^2}=1\)内接三角形\(\triangle ABC\) 各边长的平方和的极值条件？
记\(BC=a,AC=b,AB=c\),求\(a^2+b^2+c^2\)的最大值条件？
2.外椭圆 \(\frac{x^2}{m^2}+\frac{y^2}{n^2}=1\)内接三角形\(\triangle ABC\) 各边长的倒数和的极值条件？
记\(BC=a,AC=b,AB=c\),求\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)的最小值条件？
我们利用对下式求导：
\(a^2+b^2+c^2=m^2(\cos(\beta)-\cos(\gamma))^2+n^2(\sin(\beta)-\sin(\gamma))^2+m^2(\cos(\alpha)-\cos(\gamma))^2+n^2(\sin(\alpha)-\sin(\gamma))^2+
m^2(\cos(\alpha)-\cos(\beta))^2+n^2(\sin(\alpha)-\sin(\beta))^2\)
可以得到：
\(2m^2(\cos(\gamma)-\cos(\alpha))\sin(\alpha)-2n^2(\sin(\gamma)-\sin(\alpha))\cos(\alpha)-2m^2(\cos(\alpha)-\cos(\beta))\sin(\alpha)+2n^2(\sin(\alpha)-\sin(\beta))\cos(\alpha)-k=0\)
\(-2m^2(\cos(\beta)-\cos(\gamma))\sin(\beta)+2n^2(\sin(\beta)-\sin(\gamma))\cos(\beta)+2m^2(\cos(\alpha)-\cos(\beta))\sin(\beta)-2n^2(\sin(\alpha)-\sin(\beta))\cos(\beta)-k=0\)
\(2m^2(\cos(\beta)-\cos(\gamma))\sin(\gamma)-2n^2(\sin(\beta)-\sin(\gamma))\cos(\gamma)-2m^2(\cos(\gamma)-\cos(\alpha))\sin(\gamma)+2n^2(\sin(\gamma)-\sin(\alpha))\cos(\gamma)-k=0\)
通过数值计算，得到的结果见下图~~取\(m=5,n=3\)
经过数值计算分析得到
对于内接椭圆三角形各边长平方和取最大值，我们有如下结论：
1.当\(n\leq m\leq \frac{2}{\sqrt{3}}n\)时
\(\triangle ABC\) 取等腰三角形，且顶点A位于短轴顶点,\(b=c\)时，
三角形各边长的平方和取最大值：\(a^2+b^2+c^2=\frac{18m^4}{3m^2-n^2}\)
且\(A,B[\frac{m(-p^2+1)}{p^2+1}, \frac{2np}{p^2+1}],C[\frac{-m(-p^2+1)}{p^2+1}, \frac{2np}{p^2+1}]\)
\(n^2p^2+6m^2p-2n^2p+n^2=0\)
2.当\(\frac{2}{\sqrt{3}}n \leq m\)时
\(\triangle ABC\) 退化为一条直线（长轴）时，即\(a=0,b=2m,c=2m\)时，
三角形各边长的平方和取最大值：\(a^2+b^2+c^2=8m^2\)
例：\(m=5,n=3\) 时 \(a=0,b=10,c=10\) 对于对于内接椭圆三角形各边长倒数和取最小值，我们有如下结论：
1.当\(n\leq m\)时，
\(\triangle ABC\) 取等腰三角形，且顶点A位于短轴顶点,\(b=c\)时，
三角形各边长的倒数和取最小值：\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=L\)
\(A,B[\frac{m(-p^2+1)}{p^2+1},\frac{2np}{p^2+1}], C[\frac{-m(-p^2+1)}{p^2+1},\frac{2np}{p^2+1}]\)
\(-4m^2n^4p^{10}+(-16m^4n^2-8m^2n^4)p^9+(-15m^6-61m^4n^2+15m^2n^4+n^6)p^8+(-90m^6-58m^4n^2+26m^2n^4-6n^6)p^7+
(-225m^6+61m^4n^2-11m^2n^4+15n^6)p^6+(-300m^6+148m^4n^2-36m^2n^4-20n^6)p^5+(-225m^6+61m^4n^2-11m^2n^4+15n^6)p^4+(-90m^6-58m^4n^2+26m^2n^4-6n^6)p^3+(-15m^6-61m^4n^2+15m^2n^4+n^6)p^2+(-16m^4n^2-8m^2n^4)p-4m^2n^4=0\)
\(65536L^{10}m^{16}n^6+(-184320m^{16}n^4-581632m^{14}n^6+536576m^{12}n^8)L^8+(283392m^{16}n^2+1391616m^{14}n^4-327168m^{12}n^6-2196480m^{10}n^8+967424m^8n^{10})L^6+(-54000m^{16}-2430432m^{14}n^2+1136880m^{12}n^4+3554752m^{10}n^6-1755408m^8n^8-512992m^6n^{10}-240m^4n^{12})L^4+(459000m^{14}+3736800m^{12}n^2-10910520m^{10}n^4+9341504m^8n^6-2702008m^6n^8+75104m^4n^{10}+120m^2n^{12})L^2-759375m^{12}+1316250m^{10}n^2-374625m^8n^4-162900m^6n^6-18465m^4n^8-870m^2n^{10}-15n^{12}=0\)
2.当\(m\leq n\)时，
\(\triangle ABC\) 取等腰三角形，且顶点A位于短轴顶点,\(b=c\)时，
三角形各边长的倒数和取最小值：\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=L\)
\(A,B[\frac{m(-p^2+1)}{p^2+1},\frac{2np}{p^2+1}], C[\frac{m(-p^2+1)}{p^2+1},\frac{-2np}{p^2+1}]\)
\(m^6p^{10}+(-2m^6+3m^4n^2)p^8+(m^6-6m^4n^2+3m^2n^4)p^6+(-61m^4n^2+58m^2n^4-15n^6)p^4+(-61m^2n^4+30n^6)p^2-15n^6=0\)
\(65536L^{10}m^6n^{16}+(536576m^8n^{12}-581632m^6n^{14}-184320m^4n^{16})L^8+(967424m^{10}n^8-2196480m^8n^{10}-327168m^6n^{12}+1391616m^4n^{14}+283392m^2n^{16})L^6+(-240m^{12}n^4-512992m^{10}n^6-1755408m^8n^8+3554752m^6n^{10}+1136880m^4n^{12}-2430432m^2n^{14}-54000n^{16})L^4+(120m^{12}n^2+75104m^{10}n^4-2702008m^8n^6+9341504m^6n^8-10910520m^4n^{10}+3736800m^2n^{12}+459000n^{14})L^2-15m^{12}-870m^{10}n^2-18465m^8n^4-162900m^6n^6-374625m^4n^8+1316250m^2n^{10}-759375n^{12}=0\)
例如：\(m=5,n=3\) 时 \(p=1.369602407,a=9.524982860,b=6.164238702,c=6.164238702,L=0.4294391346\)
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