Sunday, July 20, 2014

Geometry Problem 1031: Trapezoid, Intersecting Circles, Common Chord, Tangent Line, Midpoint, Parallel Lines

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1031.

Online Math: Geometry Problem 1031: Trapezoid, Intersecting Circles, Common Chord, Tangent Line, Midpoint, Parallel Lines

3 comments:

  1. See sketch for location of S, L, N
    We have SE^2=SA.SB and SF^2=SC.SD
    SB/SA=SC/SD
    1. Calculate p^2= (SE.SD)^2= SA.SB.SD^2
    And p1^2=(SA.SF)^2= SC.SD.SA^2
    (p/p1)^2= (SB/SA).(SD/SC)= 1
    So p=p1 => quadrilateral AFED is cyclic
    Similarly quadrilateral BFEC is cyclic

    2. We have ∠(FAE)= ∠(FDE)= ∠(BFC)= ∠ BEC) => AE//FC and FD//BE
    And FNEL is a parallelogram
    We also have NE.NA=NF.ND => N will be on radical line GH of circles O and Q
    Similarly L will be on GH
    M is the intersection of 2 diagonals of a parallelogram FNEL so M is the midpoint of EF.

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  2. below is the sketch of problem 1031
    http://s25.postimg.org/xlwtf6fjz/pro_1031.png
    Peter Tran

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