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.

## Wednesday, August 10, 2011

### Problem 656: Triangle, Cevian, Concurrency

Labels:
cevian,
concurrent,
line,
triangulo

Subscribe to:
Post Comments (Atom)

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

ReplyDeleteLet ( 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

Denote : A = {a}, B = {b}, C = {c} be projective points in 1-dimensional subspace.

ReplyDeleteSince 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.

http://www.youtube.com/watch?v=4GheODmMYHA

ReplyDelete