Friday, May 4, 2012

Problem 749: Triangle, Rhombus, Parallel, Harmonic Means, Metric Relations

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

Click the figure below to see the problem 749 details.

Online Geometry Problem 748: Quadrilateral, 90 Degrees, Diagonal, Perpendicular, Metric Relations.

Posted by Antonio Gutierrez at 2:03 PM
Labels: , , ,

3 comments:

  1. Peter TranMay 4, 2012 at 3:58 PM

    Since ABEF is a rhombus so AE is an angle bisector of angle BAC
    We have BE/EC=AB/AC= c/b
    BE/BC= c/(b+c)=x/b => x=b.c/(b+c)

    ReplyDelete
  2. PravinMay 4, 2012 at 4:39 PM

    ADEF is a rhombus.FE = x = DE
    x/b = DE/AC = BE/BC and
    x/c = FE/AB = CE/BC
    Add:
    x/b + x/c = (BE + CE)/BC = BC/BC = 1
    x(1/b + 1/c)= 1
    x = bc /(b+c)
    Also
    x = (1/2)[2/(1/b + 1/c)]:=(1/2)HM(b,c)

    ReplyDelete
  3. c.t.e.oApril 15, 2018 at 11:53 AM

    Tr DBE ~ tr EFC => x / (b+x) = (c-x) / x

    ReplyDelete

Subscribe to: Post Comments (Atom)