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

7 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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  4. Draw a circle with diameter BD. It meets (AB) in X and (BC) in Y.

    <BXD=<BYD=90 deg by Thales' theorem.
    <BXA=<BYC=90 deg by Thales' theorem.
    So X lies on AD and Y lies on BD.
    Thales again: <XDY = <ADC = 90 deg.
    Now note that <BXY = 90 deg - <BDX = <BAX.
    So XY is tangent to the (AB) at X and similarly tangent to (BC) at Y. So X=G and Y=H.

    By similarity the tangent to the (AC) at D is parallel to GH. Let M be the midpoint of {AC), then MD is perpendicular to EF, hence bisects EF and bisects angle EMF, and D is midpoint of arc EF.

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  5. Let GH, BD cut at X. Let O be the centre of AC and let OD cut GH at Y.
    Let AB = 2a, BC = 2b and let GH = 2t,

    Now XG = XB = XH = t. Also GF^2 = (a-b)^2 + (a+b)^2 = 4ab = 4t^2 = BD^2 hence BX = FX = DX = GX = t. It follows that BFDG is a rectangle

    If < DAO = p, < DOB = p = < BXH
    Therefore OYXB is concyclic and so OY is perpendicular to EG
    It follows that EY = YF and hence ED = DF which is the result we need

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. https://photos.app.goo.gl/Ps4iad2ZZNznvwP17

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