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.

1. http://img9.imageshack.us/img9/390/problem656.png

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.