Tuesday, May 29, 2012

Problem 757: Equilateral Triangle, Circumcircle, Chords, a Half of Harmonic Mean

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

Click the figure below to see the problem 757 details.

Online Geometry Problem 757: Equilateral Triangle, Circumcircle, Chords, a Half of Harmonic Mean.

6 comments:

  1. Denote S(XYZ)= area of triangle XYZ
    We have S(ADB)=S(ADE)+S(DEB)
    Or ½.b.c.sin(120)=1/2.b.x.sin(60)+1/2.c.x sin(60)
    Note that sin(60)=sin(120)
    So b.c=b.x+c.x => x=b.c/(b+c)

    ReplyDelete
  2. Tr.ADE///Tr.CDB since /_CDB=/_ADE both=60 and /_DAE=/_DCB (same sector); hence x/b = c/CD --(1)
    By Ptolemy or by Problem 256, CD=b+c --(2)
    Combine (1)&(2) to get, x/b=c/(b+c) or 1/x=1/b+1/c

    ReplyDelete
  3. By similar triangles (ADE ~ BEC)&(BDE ~ AEC),
    x/b = BE/BC, x/c = AE/AC
    Since ABC is equiliteral, (BE + AE = AC = BC)
    Then x/b + x/c = BE/BC + AE / AC = 1
    The remaining could easily done by rearranging of terms
    Q.E.D.

    ReplyDelete
  4. By Sine Rule,
    x/b + x/c
    = sin DAE / sin AED + sin DBE / sin DEB
    = sin DCB / sin BEC + sin DCA / sin AEC
    = BE/AC + EC/AC
    = (BE + EC) / AC
    = BC/AC
    = 1
    etc

    ReplyDelete
  5. A pure geometric solution:

    Let be P on ray AD such that DP=DB. As ADBC is cyclic, x=bc(b+c) QED

    Video solution is coming, Go-Greetings :)

    ReplyDelete
    Replies
    1. WTF, what happened with my post?!?!!?

      i will complete this: (...) cylic, triangles ADE and APB are similar, then (...)

      Go-greetings!

      Delete