椭圆内接N边形的最大面积

数学星空

对于$N=3$邓寿才利用代数不等式证明了
设椭圆内接三角形的三个顶点坐标分别为
$A(m*cos(t_1),n*sin(t_1)),B(m*cos(t_2),n*sin(t_2)),C(m*cos(t_3),n*sin(t_3))$
则：AB直线方程 $n*(sin(t_2)-sin(t_1))*x-m*(cos(t_2)-cos(t_1))*y+m*n*sin(t_1-t_2)=0$
$|AB|=sqrt(m^2*(cos(t_1)-cos(t_2))^2+n^2*(sin(t_1)-sin(t_2))^2)$
三角形面积(设$t=(t_1+t_2)/2$)
$S=1/2*m*n*|sin(t_1-t_2)+sin(t_2-t_3)+sin(t_3-t_1)|$
$<=m*n*|sin(t)*(1-cos(t))|$
$=sqrt((1/3*(m*n)^2*(3+3*cos(t))*(1-cos(t))^3))$
$<=sqrt(1/3*(m*n)^2*(((3+3*cos(t))+3*(1-cos(t)))/4)^4)$
$=(3*sqrt(3)*m*n)/4$
仅当$t_2=t_1+(2*pi)/3,t_3=t_2+(2*pi)/3$ 时取等号

数学星空

一般地： 我们可以仿上面的方法得到下面不等式 设$t_k $为正实数，$k=1..n, t_k<=t_(k+1),t_(n+1)=t_1$则有 $|sum_(i=1)^n sin(t_(k+1)-t_k)|<=n*sin((2*pi)/n)$ 仅当$t_(k+1)=t_k+(2*pi)/n$时取等号

kastin

数学星空 发表于 2012-4-26 20:28
在http://bbs.emath.ac.cn/thread-3740-1-1.html中我们解决了椭圆内接N边形的最大周长问题。
现在我们来讨 ...

设一平面上有一个圆，将该平面与水平面倾斜theta角度，记dS为圆域内面积微分，dS'为相应的射影，那么有dS=cos(theta)*dS'，因此积分可知圆的面积与其射影（椭圆）的面积相差一个比例常数cos(theta)。同理，圆内接多边形面积与其射影也是这个关系。

由于圆内接正n边形的面积是最大的，所以，其射影对应的多边形是对应椭圆的内接多边形中面积最大的一个。该椭圆的离心率为e=sin(tehta)。

数学星空

对于双椭圆内接和外切五边形问题，即28#方程，我们进一步化简得到如下结果：

1. \[\cos(2t)=\frac{-5c^2+5m^2+5n^2-\sqrt{5}c^2}{5(m-n)(m+n)}\]

2. \[e=\frac{\sqrt{-2\sqrt{5}a^2+2\sqrt{5}b^2+2a^2-2b^2+4c^2}}{2}\]

3. \[d=\frac{\sqrt{ 2\sqrt{5}b^2-2\sqrt{5}c^2+4a^2-2b^2+2c^2}}{2}\]

4. \(a^4-2a^2b^2-a^2c^2+a^2e^2+b^4+b^2c^2-b^2e^2-c^4+2c^2e^2-e^4=0\)

5. \(a^4-a^2b^2+a^2c^2-2a^2d^2-b^4+2b^2c^2+b^2d^2-c^4-c^2d^2+d^4=0\)

6. \((a^2+ab+2ac-b^2+bc+c^2)(a^2-ab+2ac-b^2-bc+c^2)(a^2-ab-2ac-b^2+bc+c^2)(a^2+ab-2ac-b^2-bc+c^2)+(20a^6-30a^4b^2-20a^4c^2+10a^2b^4-20a^2b^2c^2-20a^2c^4+10b^6+10b^4c^2-30b^2c^4+20c^6)n+(-30a^4+70a^2b^2+140a^2c^2-45b^4+70b^2c^2-30c^4)n^2+(-100a^2-50b^2-100c^2)n^3+125n^4=0\)

7. \((a^2+ab+2ac-b^2+bc+c^2)(a^2-ab+2ac-b^2-bc+c^2)(a^2-ab-2ac-b^2+bc+c^2)(a^2+ab-2ac-b^2-bc+c^2)+(20a^6-30a^4b^2-20a^4c^2+10a^2b^4-20a^2b^2c^2-20a^2c^4+10b^6+10b^4c^2-30b^2c^4+20c^6)m+(-30a^4+70a^2b^2+140a^2c^2-45b^4+70b^2c^2-30c^4)m^2+(-100a^2-50b^2-100c^2)m^3+125m^4=0\)

利用2,3可以得到30#的结果

\[\frac{a^2-c^2}{d^2-e^2}=\frac{\sqrt{5}-1}{2}\]

数学星空

楼上的结论，还可以写成：

\[\cos(2t)=\frac{(m^2+n^2)-(1+\frac{1}{\sqrt{5}})c^2}{m^2-n^2}\]

\[a^2-e^2=(\sin(\frac{2\pi}{5})-\sin(\frac{\pi}{5})+\cos(\frac{\pi}{10}))(m^2-n^2)\sin(2t)\]

\[b^2-d^2=(-2\cos(\frac{3\pi}{10})+\sin(\frac{2\pi}{5})-\sin(\frac{\pi}{5}))(m^2-n^2)\sin(2t)\]

\[\frac{a^2-e^2}{b^2-d^2}=-\frac{\sqrt{5}-1}{2}\]

数学星空

对于双椭圆内接和外切七边形问题，得到如下结果：

\[a_1^2-a_7^2= (2\cos(\frac{3\pi}{14})-\cos(\frac{\pi}{14}))(m^2-n^2)\sin(2t)\]

\[a_2^2-a_6^2= (3\cos(\frac{5\pi}{14})-\cos(\frac{\pi}{14}))(m^2-n^2)\sin(2t)\]

\[a_3^2-a_5^2= (\cos(\frac{5\pi}{14})+\cos(\frac{3\pi}{14})-2\cos(\frac{\pi}{14}))(m^2-n^2)\sin(2t)\]

\[\cos(2t) = \frac{m^2+n^2-a_4^2g_3}{m^2-n^2}  ，  \frac{1}{g_3}= 1-\cos(\frac{2\pi}{7})\]

数学星空

对于双椭圆内接和外切九边形问题，得到如下结果：

\[a_1^2-a_9^2= (\cos(\frac{5\pi}{18})+\sin(\frac{2\pi}{9})-\sin(\frac{4\pi}{9}))(m^2-n^2)\sin(2t)\]

\[a_2^2-a_8^2= (\sqrt{3}-\sin(\frac{4\pi}{9}))(m^2-n^2)\sin(2t)\]

\[a_3^2-a_7^2=\frac{1}{2} (\cos(\frac{4\pi}{9})\sqrt{3}-\sin(\frac{4\pi}{9})-2\sin(\frac{\pi}{9})-\cos(\frac{\pi}{18}))(m^2-n^2)\sin(2t)\]

\[a_4^2-a_6^2=\frac{1}{2} (-\cos(\frac{7\pi}{18})\sqrt{3}-\cos(\frac{7\pi}{18})+\sqrt{3}+2\sin(\frac{2\pi}{9})-\cos(\frac{\pi}{9}))(m^2-n^2)\sin(2t)\]

\[\cos(2t) = \frac{m^2+n^2-a_5^2g_4}{m^2-n^2}  ，  \frac{1}{g_4}= 1-\cos(\frac{2\pi}{9})\]

数学星空

对于双椭圆内接和外切十一边形问题，得到如下结果：

\[a_1^2-a_{11}^2= (\cos(\frac{7\pi}{22})+\sin(\frac{2\pi}{11})-\sin(\frac{4\pi}{11}))(m^2-n^2)\sin(2t)\]

\[a_2^2-a_{10}^2= (-\sin(\frac{3\pi}{11})+2\sin(\frac{5\pi}{11})-\sin(\frac{4\pi}{11}))(m^2-n^2)\sin(2t)\]

\[a_3^2-a_9^2= (-\sin(\frac{3\pi}{11})+2\sin(\frac{9\pi}{22})+\sin(\frac{\pi}{11}))(m^2-n^2)\sin(2t)\]

\[a_4^2-a_8^2= (-\sin(\frac{3\pi}{11})+\sin(\frac{5\pi}{22})+\sin(\frac{\pi}{11}))(m^2-n^2)\sin(2t)\]

\[a_5^2-a_7^2= (-2\sin(\frac{4\pi}{11})+\sin(\frac{2\pi}{11})+\sin(\frac{5\pi}{11}))(m^2-n^2)\sin(2t)\]

\[\cos(2t) = \frac{m^2+n^2-a_6^2g_5}{m^2-n^2}  ，  \frac{1}{g_5}= 1-\cos(\frac{2\pi}{11})\]

数学星空

http://bbs.emath.ac.cn/forum.php ... 05&fromuid=1455

中给出了同心双椭圆内接外切k边形的条件或者说封闭指数为N的条件为：\(k=3\sim 10\),

\(k=3\)时

\(an+bm-mn=0\)

\(k=4\)时

\(a^2n^2+b^2m^2-m^2n^2=0\)

\(k=5\)时

\(a^3n^3-a^2bmn^2+a^2mn^3-ab^2m^2n+2abm^2n^2-am^2n^3+b^3m^3+b^2m^3n-bm^3n^2-m^3n^3=0\)

\(k=6\)时

\(a^4n^4-2a^2b^2m^2n^2+2a^2m^2n^4+b^4m^4+2b^2m^4n^2-3m^4n^4=0\)

\(k=7\)时

\(a^6n^6+2a^5bmn^5-2a^5mn^6-a^4b^2m^2n^4+2a^4bm^2n^5-a^4m^2n^6-4a^3b^3m^3n^3+4a^3m^3n^6-a^2b^4m^4n^2+2a^2b^2m^4n^4-a^2m^4n^6+2ab^5m^5n+2ab^4m^5n^2-2abm^5n^5-2am^5n^6+b^6m^6-2b^5m^6n-b^4m^6n^2+4b^3m^6n^3-b^2m^6n^4-2bm^6n^5+m^6n^6=0\)

\(k=8\)时

\(a^8n^8+4a^6b^2m^2n^6-4a^6m^2n^8-10a^4b^4m^4n^4+4a^4b^2m^4n^6+6a^4m^4n^8+4a^2b^6m^6n^2+4a^2b^4m^6n^4-4a^2b^2m^6n^6-4a^2m^6n^8+b^8m^8-4b^6m^8n^2+6b^4m^8n^4-4b^2m^8n^6+m^8n^8=0\)

\(k=9\)时

\(a^9n^9-3a^8bmn^8+3a^8mn^9+8a^6b^3m^3n^6+4a^6b^2m^3n^7-4a^6bm^3n^8-8a^6m^3n^9-6a^5b^4m^4n^5+12a^5b^2m^4n^7-6a^5m^4n^9-6a^4b^5m^5n^4-14a^4b^4m^5n^5-8a^4b^3m^5n^6+8a^4b^2m^5n^7+14a^4bm^5n^8+6a^4m^5n^9+8a^3b^6m^6n^3-8a^3b^4m^6n^5-8a^3b^2m^6n^7+8a^3m^6n^9+4a^2b^6m^7n^3+12a^2b^5m^7n^4+8a^2b^4m^7n^5-8a^2b^3m^7n^6-12a^2b^2m^7n^7-4a^2bm^7n^8-3ab^8m^8n-4ab^6m^8n^3+14ab^4m^8n^5-4ab^2m^8n^7-3am^8n^9+b^9m^9+3b^8m^9n-8b^6m^9n^3-6b^5m^9n^4+6b^4m^9n^5+8b^3m^9n^6-3bm^9n^8-m^9n^9=0\)

\(k=10\)时

\(a^{12}n^{12}-6a^{10}b^2m^2n^{10}+6a^{10}m^2n^{12}+15a^8b^4m^4n^8+14a^8b^2m^4n^{10}-29a^8m^4n^{12}-20a^6b^6m^6n^6-20a^6b^4m^6n^8+4a^6b^2m^6n^{10}+36a^6m^6n^{12}+15a^4b^8m^8n^4-20a^4b^6m^8n^6+50a^4b^4m^8n^8-36a^4b^2m^8n^{10}-9a^4m^8n^{12}-6a^2b^{10}m^{10}n^2+14a^2b^8m^{10}n^4+4a^2b^6m^{10}n^6-36a^2b^4m^{10}n^8+34a^2b^2m^{10}n^{10}-10a^2m^{10}n^{12}+b^{12}m^{12}+6b^{10}m^{12}n^2-29b^8m^{12}n^4+36b^6m^{12}n^6-9b^4m^{12}n^8-10b^2m^{12}n^{10}+5m^{12}n^{12}=0\)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

根据12#结论：\(k\)边形内接于双椭圆的最大面积时，仅当内外椭圆共心相似成立，即\(a=m\cos(\frac{\pi}{k}),b=n\cos(\frac{\pi}{k})\)

有兴趣可以代入上面各个方程检验是否成立！

倪举鹏

面积问题好处理些  将椭圆与内多边形投影在某个平面上 椭圆的投影是圆  多边形的投影是圆内正多边形
算算椭圆内多边形周长最长的情况