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.

Posted by Antonio Gutierrez at 8:26 AM
Labels: , , ,

5 comments:

  1. Jacob HaJuly 18, 2012 at 5:44 PM

    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. W FungJuly 18, 2012 at 7:03 PM

    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. PravinJuly 18, 2012 at 8:02 PM

    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
  4. Sumith PeirisNovember 4, 2017 at 1:05 AM

    Let AG = p so that AM = p/2
    Let CH = q so that CN = q/2

    Since AED and CEF are equilateral
    p+d+q+f = a ....(1)
    p/2 + d + q/2+f = x
    So 2x = p+2d+q+2f....(2)

    Comparing (1) and (2) the result follows easily

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Peter TranNovember 7, 2017 at 11:06 AM

    https://photos.app.goo.gl/hq74dAKX1uzHKGqG3
    define points G1, D1, F1 and H1 per attached sketch
    we have DB=EF=FC
    so DB+BF=FC+BF=a
    GG1=d/2
    HH1=f/2
    So x= GG1+DD1+FF1+HH1= DG/2+DB/2+BF/2+FH/2
    X= (a+d+f)/2

    ReplyDelete

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