Wednesday, March 5, 2014

Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, 90 Degree, Common External Tangent, Rectangle, Midpoint of Arc

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

Click the figure below to view the complete problem 989

Online Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, 90 Degree, Common External Tangent, Rectangle, Midpoint of Arc

4 comments:

  1. http://s25.postimg.org/vsrf9vrnj/Prob_989.png

    Let M, L, N, O are points shown on the sketch
    1. We have MG⊥GH and NH⊥GH
    Since LG=LB=LH=> L is the midpoint of GH
    ∠(GBH)= ½(∠GMB+∠BNH)= 90
    And ∠ (MLN)=1/2(∠GLB)+ ∠BLH)=90
    In right triangle MLN , BL^2=BM.BN
    In right triangle ADC , DB^2=BA.BC= 4. BM.BN=> L is the midpoint of BD
    And DGBH is a parallelogram with right angle ∠GBH => DGBH is a rectangle
    2. Connect AG and CH
    We have ∠ (AGB=∠ (BGD)= 90 => A,G,D are collinear
    Triangles AMG and AOD are isosceles with common angle ∠GAM
    So ∠ (AGM)= ∠ (ADO) => OD//MG => OD ⊥ EF
    So D is the midpoint of arc EF

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  2. I have a comment on the second part.
    DO || HN, and they are perpendicular to EF.
    So DO bisects both chord EF and arc EF.

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    Replies
    1. Thank you for your comment on my solution.
      Peter

      Delete
  3. Because <GAC=<FHC, AGHC is cyclic. Let AG and HC meet at D'. <D'AC+<D'CA=90, so D' must be on big circle. Furthermore, because D'G*D'A=D'H*D'C from cyclic property of AGHC, D' must also be on radical axis of 2 small circles, which is their common tangent. Therefore, D' coincides with D. It is easy to see that GDHB has 3 right angles from Thales theorem in each circle and must therefore be rectangle.
    <GAC=<FHC=arcDF/2+arcFC/2=arcDE+arcFC/2.
    This means that arcDF=arcDE so D is midpoint of arcEF.

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