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 楼主|
发表于 2013-8-25 11:07:51
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| 突然发现楼上的问题与http://bbs.emath.ac.cn/forum.php ... page%3D6&page=3 中的两个问题有点关系:
 
 对于双椭圆(内接和外切)的五边形ABCDE(五边依次为$a,b,c,d,e$),存在的条件?
 内接椭圆为$x^2/m^2+y^2/n^2=1$
 $(m*cos(t)-m*cos(t+(2*pi)/5))^2+(n*sin(t)-n*sin(t+(2*pi)/5))^2=a^2$
 $(m*cos(t+(2*pi)/5)-m*cos(t+(4*pi)/5))^2+(n*sin(t+(2*pi)/5)-n*sin(t+(4*pi)/5))^2=b^2 $
 $(m*cos(t+(4*pi)/5)-m*cos(t+(6*pi)/5))^2+(n*sin(t+(4*pi)/5)-n*sin(t+(6*pi)/5))^2=c^2 $
 $(m*cos(t+(6*pi)/5)-m*cos(t+(8*pi)/5))^2+(n*sin(t+(6*pi)/5)-n*sin(t+(8*pi)/5))^2=d^2$
 $(m*cos(t+(8*pi)/5)-m*cos(t+2*pi))^2+(n*sin(t+(8*pi)/5)-n*sin(t+2*pi))^2=e^2$
 化简为:
 $c^2*m^2+c^2*n^2+(5*m^4)/2-(5*n^4)/2-(sqrt(5)+1)/2*m^2*b^2-m^2*a^2+(sqrt(5)+1)/2*n^2*b^2+n^2*a^2=0$
 
 $7.184447626*d^2*m^2*n^2-11.62468045*m^2*b^2*n^2-11.62468045*m^2*a^2*n^2-3.592223813*d^2*m^4-$
 $3.592223813*d^2*n^4+5.550291030*m^4*n^2+5.550291030*m^2*n^4+5.812340223*m^4*b^2+5.812340224*m^4*a^2+$
 $5.812340223*n^4*b^2+5.812340224*n^4*a^2-5.550291030*m^6-5.550291030*n^6=0 $
 
 $7.184447624*m^2*b^2*n^2+11.62468045*m^2*a^2*n^2+7.184447626*e^2*m^2*n^2-8.980559532*m^4*n^2-8.980559533*m^2*n^4-3.592223812*m^4*b^2-$$5.812340224*m^4*a^2-3.592223812*n^4*b^2-5.812340224*n^4*a^2-3.592223813*e^2*m^4-3.592223813*e^2*n^4+8.980559532*m^6+8.980559532*n^6=0$
 
 有谁能给出简化的最终结果??
 
 $ (a^2-c^2)/(d^2-e^2)=(sqrt(5)-1)/2$
 
 
 
 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
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