Tuesday, July 10, 2012

Problem 781: Triangle, Orthocenter, Circumcenter, Circle, Vertex, Altitude, Midpoint, Perpendicular

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

Click the figure below to see the complete problem 781.

Online Geometry Problem 781: Triangle, Orthocenter, Circumcenter, Circle, Vertex, Altitude, Midpoint, Perpendicular.

Posted by Antonio Gutierrez at 5:25 PM
Labels: , , , , , , , ,

4 comments:

  1. PravinJuly 13, 2012 at 11:11 PM

    A trigonometric Proof: (sketch)
    Let∠DMB=θ
    In ΔOBM:R/cosθ = R cosB/sin(90°–θ–C+A)
    = R cosB/cos(C–A+θ),
    cosB cos θ = cos(C-A)cosθ – sin(C–A)sinθ,
    (*) tanθ = 2cosC. cosA/sin(C–A)
    In ΔBDM: DM/cosA = R cosB/ sin(θ+90°–A)
    = Rcos B/cos(A – θ)
    In ΔBEM: ME/cosC = RcosB /sin(θ–(90°–C))
    = - Rcos B /cos(C+θ)
    It suffices to prove:
    cosA cos(C+θ)+cosC.cos(A-θ)= 0 (or)
    cosθ(2cos C.cos A)= sinθ.sin(C – A)
    This follows from (*)
    Hence DM = ME

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  2. AnonymousJuly 16, 2012 at 4:35 AM

    Let N be the midpoint of AB. Then NM || AH and ON || CH so angle(ONM)=angle(C1HA)=B.
    But ONDM is inscrided, then angle(ONM)=angle(ODM)=B.
    The same is true for angle OEM. So DO=DE.

    ReplyDelete
    Replies
    1. Antonio GutierrezJuly 16, 2012 at 5:57 AM

      So DO=EO, therefore OM is the perpendicular bisector of DE and M is the midpoint of DE.

      Delete
  3. Peter TranJuly 16, 2012 at 5:10 PM

    Anonymous

    You got an excellent solution. !
    attached is the sketch per your solution
    http://img9.imageshack.us/img9/3192/problem781.png

    Peter Tran

    ReplyDelete

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