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.

 Problem 548: Triangle, Transversal, Complete Quadrilateral, Circumcircles, Circumcenters, Similarity, Concyclic Points
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Level: High School, SAT Prep, College geometry

2 comments:

  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

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  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

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