## Thursday, July 29, 2010

### Problem 486: Tangent circles, Chord, Arc, Midpoint, Collinearity, Congruence

Geometry Problem
Click the figure below to see the complete problem 486 about Tangent circles, Chord, Arc, Midpoint, Collinearity, Congruence.

Complete Problem 486

Level: High School, SAT Prep, College geometry

1. Denote (XYZ) angle XYZ
Connect BF,AB, BC . Extend GA to intersect DE at M
Note that A, B and C are collinear ( Circle A tangent to circle B at C)
BF perpen. To DE
GM perpen. To DE ( G is the midpoint of arc DE and Tri. DGE isosceles)
So GM // BF and
(GAC)=(FBC) ( corresponding angles)
Both triangles GAC and FBC are isosceles
(BCF)=(ACG) (both angles =1/2*(180- (GAC))

So G, F, C are collinear

We have (GFE)=(GCE) (face 2 equal arcs DG & GE)
Triangle GFE will congruent to Triangle GEC ( case AA)
And we have GE/GC=GF/GE and GE^2=GF.GC
Consider circle B , secant GFC and tangent GH , we have GH^2=GF.GC
So GD=GE=GH

Different method:

Consider geometric inversion transformation with inversion center t G and inversion radius GE= GD
1. Inversion transformation as defined above, line DE will become circle C1 . Circle C1 will become line DE
2. Circle C2 tangent to circle C1 and line DE ( before inversion) will become a circle with center located on line GB and tangent to both circle C1 and line DE . So circle C1 will stay at the same location
3. Point C (tangent point of C1 and C2) will become F (tangent to DE and C2) . C and F are 2 corresponding points of inversion .So G, C, F are collinear
4. In this inversion with inversion radius GE we have GE^2= GC.GF = GH^2 . So GE=GH=GD

Peter Tran
vstran@yahoo.com

2. ang(GFD + DFH + HFC) = 1/2 arc(FC + FH + HC)=1/2∙2π

= π

3. To c.t.e.o

You state that ang(GFD + DFH + HFC) = 1/2 arc(FC + FH + HC)=1/2∙2π

Note that Ang(GFD) = 1/2 arc(FC) only if G, F and C are collinear . We try to prove that G,F,C collinear !!!

Peter Tran

4. To Peter
Ok, thanks

Tangent at G is // to DE
from P475, G, F, C are collinear