Tuesday, September 30, 2014

Geometry Problem 1048: Circles, Tangent, Perpendicular, Diameter, Angle Bisector

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 1048.

Online Math: Geometry Problem 1048: Circles, Tangent, Perpendicular, Diameter, Angle Bisector, Polya's Mind Map.

6 comments:

  1. Hi Antonio,

    I would like to add the folllowings:
    The point B is the center of the inscribed circle of the triangle DFH and the smallest circles passing through B and touching the three sides are Archimedean.

    Hiroshi

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  2. As far as I know, the two Archimedean circles touching the sides DF and DH are new.

    Hiroshi

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  3. Hi, Hiroshi
    Please explain why B is the center of the inscribed circle of triangle DFH concretely.

    ReplyDelete
  4. http://s25.postimg.org/ag5sakzpb/pro_1048.png

    Connect lines per attached sketch
    We have GM=GB=GN => MBN is a right triangle
    In right triangles AEC and O1GO2 we have GB^2=BO1.BO2 and BE^2=BA.BC
    So BG= ½ BE and MBNE is a rectangle
    We have ∠ (BNE)= ∠ (BNC)=90 =>E, N, C are collinear
    Similarly A,M,E are collinear
    Note that ∠ (BCE)= Arc(AF)+Arc(FE)
    And ∠ (EMN)= Arc(AF)+Arc(EH)
    But ∠ (EMN)= ∠ (BNM)= ∠ (BCE) => Arc( FE)=Arc(EH)=> DE is an angle bisector of angle FDH

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  5. By theorem from previous problem, AMNC is concyclic. Therefore, AM and NC must meet on radical axis DE of O1 and O2 at point P. <APC must be 90 because <MAB+<NCB=90, so P is also on big circle. P must coincide with E as result.
    <ACN=<AMF=<ACF+<EDH, but also
    <ACN=(<ADF=<ACF)+<FDE.
    Therefore <ACF+<EDH=<ACF+<FDE, so <EDH=<FDE

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  6. Let radius of O1 = a and that of O2 = b
    Let MG = GB = GN = p and let GB = q

    So 2a X 2b = (p+q)^2
    But MN^2 = 4p^2 = (a+b)^2 - (b-a)^2 = 4ab

    So p+q = 2p hence p=q hence BMEN is a rectangle, AME, CNE are collinear and ADCE is a kite.
    So < BGN = < MAD hence AMGD is cyclic and so < BMG = < ADG = < HFE + < FEM = = < HDE + < ADF ==>. < FDB = < HDB

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