Wednesday, July 18, 2012

Problem 785: Equilateral Triangle, Parallel to a side, Perpendicular to a side

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

Click the figure below to see the complete problem 785.

Online Geometry Problem 785: Equilateral Triangle, Parallel to a side, Perpendicular to a side.

3 comments:

  1. Let K be the mid-point of AC, such that BK is the altitude.

    Note that triangles ADE and EFC are both equilateral,
    and also DBFE is a parallelogram, so
    AE = DE = BF
    EC = EF = DB

    Thus
    a + d + f
    = d + EC + AE + f
    = d + DB + BF + f
    = GB + BH
    = 2*MK + 2*KN
    = 2*(MK + KN)
    = 2x

    ReplyDelete
  2. Let AE = b, then AD = b, EC = a-b, FC = a-b
    Then
    AM = ADcos60 = (b-d)/2
    NC = CFcos60 = (a-b-f)/2
    by considering the length of AC = AM + MN + NC,
    a = (b-d)/2 + x + (a-b-f)/2
    x = (a+d+f)/2
    q.e.d.

    ReplyDelete
  3. Triangles ADE and CFE are equilateral
    So DE + EF = AE + EC = a
    Draw DJ, FK each perpendicular to AC
    x = ME + EN = (MJ + JE)+(EK + KN)
    = [(GD + DE) + (EF + FH)]cos 60°
    = [GD + (DE + EF) + FH]/2
    = (d + a + f)/2

    ReplyDelete