Wednesday, July 20, 2011

Problem 638: Arbelos and Inscribed Circle, Concyclic Points

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 638.

Geometry Problem 638: Arbelos and Inscribed Circle, Concyclic Points.

Posted by Antonio Gutierrez at 8:02 AM
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5 comments:

  1. Antonio GutierrezJuly 21, 2011 at 2:11 PM

    Thanks Henkie for the note (problem 638).

    ReplyDelete
  2. Peter TranAugust 12, 2011 at 11:06 PM

    http://img714.imageshack.us/img714/1382/problem638.png
    1. 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

    ReplyDelete
    Replies
    1. IchungchenOctober 5, 2021 at 3:49 AM

      excuse me, can you tell me why "QH become perpendicular bisector of FC"? thank you very much.

      Delete
    2. Peter TranOctober 8, 2021 at 1:23 PM

      Thank you for your comment . The correct inversion will be I(C, CA.CB) not I(Q, QC^2). See solution to the problem 1168 and sketch below for details.

      https://photos.app.goo.gl/3PmVBiGuTH1skZUJ9

      Delete
  3. W FungOctober 7, 2012 at 7:24 PM

    Set up inversion transformation I(C,r), for any r.

    The line ACB is invariant as line A'C'B'.
    Then circle AC and AB would become two parrallel lines perpendicular to A'B' passing through A' and B' respectively.
    Circle AB becomes a circle P with diameter A'B', touching the two parrallel lines.
    Now circle DEF becomes a circle Q touching P at F' and also the two parrllel lines at D' and E'. (obviously, P and Q are having the same radius)
    A'D'E'B' is now a rectangle and F' lies on the diagonal of B'D'.
    Hence B',F',D' collinear and B,F,C,D concyclic.
    (Symmetric for A,C,E,F)

    Again from the inversion image,
    ∠CFA =∠C'A'F'=∠C'B'F'=∠CFB.
    So CF is the angle bisector of the 90 degree ∠AFB, so ∠CFB = 45 and ∠CGB is 90.
    Yet, G must lies on the perpendicular bisector of BC, so G is the midpoint of semicircle BC.
    (Symmetric for H)

    Q.E.D.

    ReplyDelete

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