Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 638.
Wednesday, July 20, 2011
Problem 638: Arbelos and Inscribed Circle, Concyclic Points
Subscribe to:
Post Comments (Atom)
Thanks Henkie for the note (problem 638).
ReplyDeletehttp://img714.imageshack.us/img714/1382/problem638.png
ReplyDelete1. Let O, I, J and P are centers of circles ( see picture)
We have ( FAC)= 90-1/2*(FJA) ( tri. FAJ is isosceles)
(FEC)=(FEP)+(CEP)
(CEP)=90+1/2*(EOC) ( Tri. COE is isosceles)
And (FEP)=1/2*(OPJ) ( Tri. FPE is isosceles)
So (FAC)+(FEC)= 90-1/2*(FJA)+1/2*(OPJ)+90+1/2*(EOC)
But (FJA)=(OPJ)+(EOC) so (FAC)+(FEC)= 180
Quadrilateral AFEC is cyclic
2. Let GH cut AB at Q.
Note that Q is the center of dilation transformation Dil(Q,OG/IH) to transform circle I to circle O
Q also be center of Inversion transformation Inv(Q, QC^2) to transform circle I to Circle O
in this inversion, Circle J stay the same. Circle P tangent to circles I, O, J will stay the same after inversion.
So QC=QF and Q, D, E are collinear and QD.QE= QC^2 .
QH become perpendicular bisector of FC and H become center of circle (AFEC)
3. In the inversion Inv(Q, QC^2) , points (A,F,E,C) become points (B, F, D,C) so quadrilateral BFDC is cyclic.
Note that GO and GQ are perpend. Bisectors of CB and CF . So G will be center of circle (BFDC)
Peter Tran