Tuesday, July 9, 2013

Problem 901: Intersecting Circles, Common External Tangent, Secant, Circumcircle, Concyclic Points, Cyclic Quadrilateral

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 901.

Online Geometry Problem 901: Intersecting Circles, Common External Tangent, Secant, Circumcircle, Concyclic Points, Cyclic Quadrilateral. GeoGebra, HTML5 Animation for iPad and more tablets

7 comments:

  1. Observe that
    (1) ∠HAD = ∠DJB
    (2) ∠HDA = ∠CDA + ∠HDC = ∠AED + ∠DBJ = ∠ABJ
    So ∆HAD~∆HJB, with power of cirlce, H-D-J collinear.
    Hence,
    ∠HJB = ∠DJB = ∠FAB = ∠FCB
    by the converse of ext.∠ of cylic quad, Q.E.D.

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  2. Can you explain why ∠HDC = ∠DBJ? Thanks.

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    Replies
    1. Sorry i miss a small step in my previous solution,
      Extend CD to point X, X is on the other side of C,
      ∠HDC = ∠JDX = ∠DEJ = ∠DBJ

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    2. Note to William Fung
      Your statement ∠HDC = ∠JDX is true only if H, D, J are collinear.

      Peter Tran

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  3. http://img853.imageshack.us/img853/4403/fso.png

    Let HD cut circle O1 at J’
    Since H is on radical line AB of circles O and O1
    So We have HC.HF=HA.HB=HD.HJ’ => F,C,D,J’ is cocyclic
    ∠(CFJ’) supplement to ∠(CDJ’)
    ∠(CDJ’) supplement to ∠(HDC) and ∠(HDC)= ∠(DEJ’)
    So ∠(DEJ’)= ∠(CFJ) => G,F,E,J’ is cocyclic => J’ coincide to J
    The rest of my solution is similar to W. Fung solution

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  4. It seems that the past few problems could give enough hints on how to solve #738. Any thoughts?

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