Sunday, June 24, 2012

Problem 773: Intersecting Circles, Midpoint, Secant line, Bisection

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

Click the figure below to see the complete problem 773.

Online Geometry Problem 773: Intersecting Circles, Midpoint, Secant line, Bisection.



    Draw new lines per attached sketch
    We have ∠BDA=1/2*∠BQA= ∠BQO
    And ∠BCA=1/2*∠BOA =∠BOQ
    ∆BOQ similar to ∆BCD …. ( case SAS)
    ∆ BCD is the image ∆BOQ in the spiral similarity geometry transformation with following details.
    center = point B , angle of rotation θ= ∠OBC=∠QBD , ratio of similarity= BC/BO=BD/BQ
    Since M and E are midpoints of OQ and CD , so E is the image of M in this geometry transformation.
    So we have ∆BME similar to ∆BOC => MB=ME and E is the point of intersection of CD and circle M

  2. Please draw a figure
    Let F be diametrically opposite to A w.r.t. cirle (M).
    Join EF. So ∠AEF = 90°, E is the foot of ⊥ from F on CD.
    Let O', M', Q' be the feet of the ⊥s from O, M, Q on CD.
    So CO' = O'A, AM' = M'E, AQ' = Q'D
    Join OF, QF.
    OAQF is a parallelogram since diagonals OQ, AF bisect each other.
    OA = FQ and OF = AQ
    So O'A = EQ' and O'E = AQ'
    Now CE = CA + AE = 2O'A + AE
    = 2EQ' + AE = EQ' + (EQ' + AE)
    = EQ' + AQ' = EQ' + Q'D = ED as required.

  3. Ref: For figure of Problem 773, please go to the following link:

  4. Let OC=r and QD=R
    This is true, if and only if
    EA*EC=ED*EA or
    EO^2-r^2=R^2-EQ^2 or
    R^2+r^2=EO^2+EQ^2 or
    AO^2+AQ^2=EO^2+EQ^2 or
    2(OQ^2)+(AM^2)/2=2(OQ^2)+(EM^2)2 or
    But this is true


  6. Let X,Y, and Z be the feet of the perpendiculars from O,M, and Q to CD. Then XA=(1/2)AC and AZ=(1/2)AD, so XZ=(1/2)CD, so XY=YZ=(1/4)CD, so XY=XA+AY=(1/2)AC+(1/2)AE=(1/2)CE=(1/4)CD, so CE=(1/2)CD.

  7. (O ,r) ,(Q,R) The two circles
    EA.EC= EO^2 - r^2, EA.ΕD=R^2-EQ^2
    EA.EC- EA.ΕD= EO^2+ EQ^2-( r^2+ R^2)=2ΕΜ^2+2ΟΜ^2-2ΑΜ^2-2ΟΜ^2=0(by Theorem median in triangles ΟΕQ,OAQ, and because ΑΜ=ΕΜ ).So, EA.EC- EA.ΕD=0 ,EA.EC=EA.ΕD, EC=ED