Wednesday, August 10, 2011

Problem 656: Triangle, Cevian, Concurrency

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Mohak Jain.

Click the figure below to see the complete problem 656.

 Online Geometry Problem 656: Triangle, Cevian, Concurrency.



    Let ( FAP)= α1 , (PAE)= α2 (QBD)=β1, (QBF)=β2, (RCE)=ϒ1, (RCD)=ϒ2
    Let E1 and F1 are the projection of F and E over AP ( see picture)
    We have sin(α1)/sin(α2)=(FF1/AF)/(EE1/AE)= (PF/AF)/(PE/AE) =(PF/PE)/(AE/AF) (1)
    Similarly we also have
    sin(β1)/sin(β2)=(QD/QF)/(BF/BD) (2)
    sin(ϒ1)/sin(ϒ2)=(RE/RD)/(DC/EC) (3)
    Multiply expressions (1) x (2) x(3) side by side we have
    sin(α1)/sin(α2) x sin(β1)/sin(β2) x sin(ϒ1)/sin(ϒ2) = Numerator /Denumerator where
    Numerator=(PF/PE).( QD/QF).( RE/RD)
    Denumerator=( AE/AF).( BF/BD).( DC/EC)
    Since DP, FR and EQ are concurrent, value of numerator =1 per Ceva’s theorem
    Since AD, CF and BE are concurrent, value of denumerator = 1 per Ceva’s theorem
    So sin(α1)/sin(α2) x sin(β1)/sin(β2) x sin(ϒ1)/sin(ϒ2) =1
    And AP, CR and BQ will concurrent per inverse of Ceva’s theorem
    Note that Ceva’s theorem can be state using ratio of segments or ratio of sine of angles.
    Peter Tran

  2. Denote : A = {a}, B = {b}, C = {c} be projective points in 1-dimensional subspace.
    Since D,E,F lies on BC,CA,AB, we can have F = {a+b}, E = {c+a}, D = {b+c}.
    Similarly, P = {(c+a)+(a+b)} = {2a+b+c}.
    Symmetrically, Q = {a+2b+c} , R = {a+b+2c}.
    Now projective line
    AP = {(a) X (2a+b+c)} = {(a)X(b+c)}
    BQ = {(b) X (a+2b+c)} = {(b)X(a+c)}
    CR = {(c) X (a+b+2c)} = {(c)X(a+b)}
    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.