Thursday, June 7, 2012

Problem 761: Scalene Triangle, 60 degrees, Euler Line, Equilateral Triangle

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.
Click the figure below to see the problem 761 details.

Online Geometry Problem 761: Scalene Triangle with an angle of 60 degrees, Euler Line, Equilateral Triangle.

2 comments:

  1. http://img835.imageshack.us/img835/6848/problem761.png

    Draw additional lines per attached sketch
    Note that ∆ABN is equilateral and BG/BM=BQ/BS= 2/3 => QG⊥BQ
    And HG=2.GO and HQ=2.QR …. (Property of Euler line)
    OR=RQ. √3=1/3 . HR. √3
    So tan(x)= OR/HR= 1/√3 => x=30
    And triangle ADE is equilateral

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  2. Let X and Y be the midpoints of AC, AB
    Let L, M be the feet of the altitudes from B, C
    1/2 = cos 60° = AL/AB = AL/c. So AL = c/2 = AY
    1/2 = cos 60° = AM/AC = AM/b. So AM = b/2 = AX
    Follows LX = AX - AL = (b - c)/2 ; MY = AM - AY = (b - c)/2
    So LX = MY = d say
    Next ∆ACM is right angled at M and ∠ACM = 30°
    (i.e.)∠LCH = 30°
    So 1/2 = sin 30° = HL/CH, HL = CH/2 = OY by a well known property
    Similar to HL = OY, we have HM = OX
    For convenience denote OX = x, OY = y
    Next HL ∥ OX implies LE/XE = HL/OX = y/x, 1 + LX/XE = y/x,
    LX/XE = (y - x)/x,
    (y - x).XE = x. LX = x.d
    Similarly
    HM ∥ OY implies DY/DM = OY/HM = y/x, 1 + MY/DM = y/x,
    MY/DM = (y - x)/x,
    (y - x).DM = x. MY = x.d
    Thus XE = DM
    Hence finally,
    AD = AM + MD = b/2 + MD = AX + XE = AE and so ∆ADE is equilateral
    (minor typos, if any may be excepted)

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