Saturday, September 21, 2013

Geometry Problem 926: Two Equilateral Triangles, Midpoint, Perpendicular, Metric Relations

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 926.

Online Geometry Problem 926: Two Equilateral Triangles, Midpoint, Perpendicular, Metric Relations.

Posted by Antonio Gutierrez at 4:37 PM
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4 comments:

  1. EditorSeptember 21, 2013 at 5:29 PM

    Let P on BC be the feet of A-altitude for tr ABC, then by the spiral similarity that carries ABC into DFE, we have <PAG= <FBC, so APGB is cyclic, therefore <AGB=<APB=90°.

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  2. Jacob HA (EMK2000)September 21, 2013 at 5:39 PM

    Let T(M,θ,k) be a transformation that rotate a point P, with angle θ clockwise about M, with a scale factor k. i.e. ∠PMP'=θ and P'M=k×PM.

    Consider T(M,90°,√3), then
    A→B
    D→F
    ∠AMD = ∠BMF = 90°−∠BMD

    Thus ΔAMD→ΔBMF. Hence, AD→BF.
    i.e. AD⊥BF and BF = AD×√3.

    ReplyDelete
  3. Peter TranSeptember 21, 2013 at 7:41 PM

    Triangles AMD and BMF are similar ..( case SAS)
    and MB/MA=MF/MD= sqrt(3)
    So BMF is the image of AMD in the hemothethy and rotation transformation center M, ratio= sqrt(3) , angle of rotation=90
    in this transformation correspondent angles are preserved
    so AD ⊥BF and BF=AD.sqrt(3)

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  4. Sumith PeirisNovember 2, 2016 at 6:00 AM

    If AB = 2b and DF = 2a then,

    DM = a and AM = b
    BM =√3b and FM = √3a

    So triangles AMD ≈ BMF since < AMD = < BMF and AM/BM = DM/MF

    Hence < MAD = MBG and so AMGB is concyclic and < AGB = 90

    Further BF/AD = MF/MD = √3

    Sumith Peiris
    Moratuwa
    Sri Lanka

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