## Thursday, December 16, 2010

### Problem 548: Triangle, Transversal, Complete Quadrilateral, Circumcircles, Circumcenters, Similarity, Concyclic Points

Geometry Problem
Click the figure below to see the complete problem 548 about Triangle, Transversal, Complete Quadrilateral, Circumcircles, Circumcenters, Similarity, Concyclic Points.

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Level: High School, SAT Prep, College geometry

1. http://img839.imageshack.us/img839/4444/problem548.png

1 C1, C2, C3 and C4 meet at a point G : See solution of problem 547
2 O4O2 perpendicular to DG and O2O3 perpen, to GE so m(DGE) =m(B)
And m(O4O2O3)=m(B)
Similarly m(O2O4O3)=m(A) so triangle ABC similar to triangle O4O2O3 .

3 We have O1O2 perpen. To BG and O1O4 perpen to AG
So angle( O4O1O2) supplement to angle (BGA)
But angle(BGA)= angle(C) =angle (O4O3O2)
So 4 points O1,O2,O3 ,O4 concyclic.

We have angle (BGA)= angle (DGF) both angle supplement to angle A
angle (BCF)= angle (DBG) both angle supplement to angle( ACG)
So tri. GBD similar to tri. GCF .
Tri. GCF is the image of Tri. GBD in the rotation +dilation transformation with center at G
Since O2, O3 is the center of circumcircles of tri. GBD and GCF , O3 will be the image of O2 in this transformation and angle (O2GO3)= angle (BGC) = supplement to angle (A) or angle (O2O4O3)
points O1, O2, O3, O4 and G are concyclic

Peter Tran

2. (1)
Let C2 intersect C3 again at G
DE subtends <B at G,
EF subtends <C at G.
So DF subtends <(Pi - A)at G & < A at A;
A,F,G,D are concyclic; C4 passes through G
(2)
O2O3 perpendicular GE,
O3O4 perpendicular GF,
O2O4 perpendicular GD.
<O2O3O4 = <EGF = <C,
<O2O4O3 = Pi - <DGF = <A,
<O3O2O4 = <DGE = <B.
So triangles ABC, O3O2O4 are similar
(3)
O1O2 perp BG,
O1O3 perp CG,
So <O2O1O3 = <A = <O2O4O3;
O1,O2,O3,O4 are concyclic