tag:blogger.com,1999:blog-6933544261975483399.post6874727391653727205..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 656: Triangle, Cevian, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-9672675719900419872013-02-23T18:01:13.379-08:002013-02-23T18:01:13.379-08:00http://www.youtube.com/watch?v=4GheODmMYHAhttp://www.youtube.com/watch?v=4GheODmMYHAAnonymoushttps://www.blogger.com/profile/05208522114695973019noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70055831720288338872013-02-21T22:05:49.705-08:002013-02-21T22:05:49.705-08:00Denote : A = {a}, B = {b}, C = {c} be projective p...Denote : A = {a}, B = {b}, C = {c} be projective points in 1-dimensional subspace.<br />Since D,E,F lies on BC,CA,AB, we can have F = {a+b}, E = {c+a}, D = {b+c}.<br />Similarly, P = {(c+a)+(a+b)} = {2a+b+c}.<br />Symmetrically, Q = {a+2b+c} , R = {a+b+2c}.<br />Now projective line <br />AP = {(a) X (2a+b+c)} = {(a)X(b+c)}<br />BQ = {(b) X (a+2b+c)} = {(b)X(a+c)}<br />CR = {(c) X (a+b+2c)} = {(c)X(a+b)}<br />Since (a)X(b+c) + (b)X(a+c) + (c)X(a+b) = 0, projective lines AP,BQ,CR are linearly dependent and hence concurrent.W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78524621684330456222011-08-10T14:47:48.943-07:002011-08-10T14:47:48.943-07:00http://img9.imageshack.us/img9/390/problem656.png
...http://img9.imageshack.us/img9/390/problem656.png<br /><br />Let ( FAP)= α1 , (PAE)= α2 (QBD)=β1, (QBF)=β2, (RCE)=ϒ1, (RCD)=ϒ2<br />Let E1 and F1 are the projection of F and E over AP ( see picture)<br />We have sin(α1)/sin(α2)=(FF1/AF)/(EE1/AE)= (PF/AF)/(PE/AE) =(PF/PE)/(AE/AF) (1)<br />Similarly we also have <br />sin(β1)/sin(β2)=(QD/QF)/(BF/BD) (2)<br />sin(ϒ1)/sin(ϒ2)=(RE/RD)/(DC/EC) (3)<br />Multiply expressions (1) x (2) x(3) side by side we have<br />sin(α1)/sin(α2) x sin(β1)/sin(β2) x sin(ϒ1)/sin(ϒ2) = Numerator /Denumerator where<br />Numerator=(PF/PE).( QD/QF).( RE/RD)<br />Denumerator=( AE/AF).( BF/BD).( DC/EC)<br />Since DP, FR and EQ are concurrent, value of numerator =1 per Ceva’s theorem<br />Since AD, CF and BE are concurrent, value of denumerator = 1 per Ceva’s theorem<br />So sin(α1)/sin(α2) x sin(β1)/sin(β2) x sin(ϒ1)/sin(ϒ2) =1<br />And AP, CR and BQ will concurrent per inverse of Ceva’s theorem <br />Note that Ceva’s theorem can be state using ratio of segments or ratio of sine of angles.<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com