Thursday, February 2, 2017

Geometry Problem 1312: Triangle, 60 Degrees, Circle, Circumcenter, Orthocenter, Altitude, Distance, Perpendicular, Measurement

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

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Geometry Problem 1312: Triangle, 60 Degrees, Circle, Circumcenter, Orthocenter, Altitude, Distance, Perpendicular, Measurement.

Posted by Antonio Gutierrez at 7:37 PM
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7 comments:

  1. APOSTOLIS MANOLOUDISFebruary 3, 2017 at 7:09 AM

    Problem 1312
    The extension of OH intersects AB and BC in K,L respectively .According to the problem 761 the triangle BKL is equilateral .Is <AOC=2<ABC=120, <AHC=90+<ABC/2=120.So
    AOHC is cyclic.Draw CPM//KL (M in AB, P in AH).Is <AMC=120=<AOC or AMOHC is
    , so AH-HC=AH-HP=AP=12.Draw MN perpendicular AP. Is OD=MN=x, MA=2x and
    triangle MAN from pythagorean theorem 4x^2=x^2+36 or x=2√(3 ).
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. AnonymousFebruary 3, 2017 at 9:08 AM

    Extend AH to meet BC at E
    The perpendicular from O to BC cuts BC in half and say meet at P.
    So OPED is a rectangle with PE = OD = x -------(1)
    let AB = c and consider the 30-60-90 triangle ABE
    we have BE = c/2 and AE = Sqrt(3)c/2 -------(2)
    From (1) and (2) EC = PC-PE = BP-PE = BE-PE-PE = c-4x/2 --------(3)
    Now consider the 30-60-90 triangle ECH
    we have
    HE = (c-4x)/2Sqrt(3)
    CH = (c-4x)/Sqrt(3) --------(4)

    Given AH-CH = 12
    => (AE-HE)-CH = 12
    => Sqrt(3)c/2-(c-4x)/2Sqrt(3)- (c-4x)/Sqrt(3) = 12 (from (2) and (4))
    => 3c-c+8x-2c+4x = 12*2*Sqrt(3)
    => x = 2Sqrt(3)

    ReplyDelete
  3. Sumith PeirisFebruary 3, 2017 at 9:47 AM

    Let R be the circumradius.

    Since < AOC = < AHC = 120, AOHC is concyclic.
    So < OHD = 30 and OH = 2x

    Applying Ptolemy,
    2x.b + R. CH = R. AH

    But R = b/sqrt3 considering Tr. AOC.

    So x = (AH - CH).R/b = (12 / 2)/sqrt3
    x = 2.sqrt3

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. c.t.e.o.February 3, 2017 at 10:29 AM

    Extend AH to P (P on circle) => ΔHPC equilateral
    Extend CH to M (M on circle) => ΔABC ≡ ΔAMC so if we draw perpendicular to
    CM and AB we get x inradius of AHG (G on AB)
    AD = 1/2 AP => AD = DH+HP = DH+HC (1) But AH-HC = 12 (2)
    From (1) and (2) DH = 6
    From ΔODH and pythagore theorem x = 2√3

    ReplyDelete
  5. Peter TranFebruary 3, 2017 at 3:35 PM

    https://goo.gl/photos/Mo9Wudh9iSaMpX4S8

    Let c= AH, a= HC
    Let AH cut circle ABC at E
    Note that BC is the perpen. Bisector of HE => HCE is equilateral
    So HE=HC=a
    D is the midpoint of AE so AD= ½(a+c)
    AOK is 30-60-90 triangle so AO= AC/sqrt(3)
    In triangle AHC we have angle AHC= 120 and AC^2= a^2+c^2+a.c
    In triangle AOD => x^2= AO^2-AD^2
    X^2= 1/3(a^2+c^2+a.c)-1/4(a+c)^2
    After simplification we get 12.x^2= (a-c)^2= 144
    So x=2.sqrt(3)

    ReplyDelete
  6. AnonymousFebruary 3, 2017 at 10:00 PM

    Sumith Peiris, Moratuwa, Sri Lanka
    - This solution is EXCELLENT!

    ReplyDelete

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