Consider a circle which cuts an ellipse in precisely four points. Let A, B, C and D respectively correspond to the
eccentric angles of these four points.
Show that A + B + C + D is an even multiple of pi radians.
I try to resolve this problem but first of all i must study!
I found at
http://mathworld.wolfram.com/Ellipse.html
this
Let four points on an ellipse with axes parallel to the coordinate axes have angular coordinates. Such points are concyclic when
s1s2s3+s1s2s4+s1s3s4+s2s3s4-(s1+s2+s3+s4)==0 (31)
where the intermediate variable si=tan(ti/2)
I will prove now that this relation is same with
A+B+C+D = 2*k*pi
Divide by 2 i found
A/2+B/2+C/2+D/2 = k*pi
or
A/2+B/2 = k*pi - (C/2+D/2)
using tangent function i found
tan(A/2+B/2) = tan(k*pi-(C/2+D/2))
so
(tan(A/2)+tan(B/2)) / (1-tan(A/2)*tan(B/2)) = - (tan(C/2)+tan(D/2)) / (1-tan(C/2)*tan(D/2))
With the notation
s1=tan(A/2), s2=tan(B/2), s3=tan(C/2), s4=tan(D/2)
(s1+s2)/(1-s1*s2) = -(s3+s4)/(1-s3*s4)
and after calculus i found the above relation (31)
Edited on September 29, 2007, 12:46 pm