Sunday, July 1, 2012

Problem 777: Triangle, Distance from the Orthocenter to a Vertex, Circle, Circumradius, Cosine

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

Click the figure below to see the complete problem 777.

Online Geometry Problem 777: Triangle, Distance from the Orthocenter to a Vertex, Circle, Circumradius, Cosine.

6 comments:

  1. http://img839.imageshack.us/img839/7122/problem777.png

    Draw lines per attached sketch
    We have LC//BH and LB//CE
    So BLCH is a parallelogram and
    BH=LC=AL. cos(ALC)= 2.R.Cos(ABC)

    ReplyDelete
  2. HB = BD / sin(BHD)
    = AB*cos(ABC) / sin(ACB)...CDHF concyclic
    = 2R*cos(ABC)...Sine rule

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  3. AC/sin(ABC) = 2R (R is the circumradius)
    The statement is then reduced to prove :
    AC/sin(ABC) = BH/cos(ABC) <=> AC = BH*tanABC
    which is the properties of the altitude

    q.e.d.

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  4. Let M be the midpoint of AC.
    Median BM intersects Euler line OH at centroid G.
    ∆BGH ~ ∆MOG since OM ∥ BH
    So BH/OM = BG/GM = 2 and
    BH = 2OM = 2.OA.cos∠AOM = 2R cos∠ABC
    [since∠AOM=(1/2)∠AOC=(1/2)(2∠ABC)=∠ABC]

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  5. Since < BHD = < C, BD = BH sin <C ...,(1)

    Also AB = 2R sin <C ....(2)

    And BD = AB cos <B ...(3)

    Eliminating AB and BD between these 3 equations

    BH = 2R cos < B

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Alternatively let AO meet the circle at X then prove that BHCX is a parellelogram (refer my proof of Problem 778)

    Then BH = CX = 2R cos <B since < AXC = < B

    ReplyDelete