Tuesday, November 23, 2010

Fox 316

3 comments:

  1. Let x=BP.
    Area(PQR)=0.5*PR*PQ*sin(<RPQ)
    But
    PR=2*x*sin(B/2), PQ=2([BC]-x)²cos(C/2)
    and
    <RPQ=180º-<RPB-<QPC=180-(180-B)/2-(180-C)/2=(B+C)/2=Pi/2-A/2
    then
    Area(PQR)=2*x*([BC]-x)*sin(B/2)*sin(C/2)*cos(A/2)
    Then maximum area occurs when x=BC/2 and it is
    Maximum(A(PQR))=0.5*[BC]^2*sin(B/2)*sin(C/2)*cos(A/2)
    Then apply identities:
    2*sin(u)*sin(v)=cos(u-v)-cos(u+v)
    2*cos(u)*cos(v)=cos(u+v)+cos(u-v).
    and A,B,C relations to get the result.

    César Lozada
    Venezuela

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  2. Please read
    PQ=2*[BC]-x)*cos(C/2)

    instead of
    PQ=2([BC]-x)²cos(C/2)

    Thanks
    César Lozada
    Venezuela

    ReplyDelete
  3. César, thank you for confirming the result!

    ReplyDelete