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.

See also:
Complete Problem 486

Level: High School, SAT Prep, College geometry

4 comments:

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