Find all points P such that B C P H A + A C P H B + A B P H C is minimum.

Question

Find all points P such that             B      C              P              H        A              +            A      C              P              H        B              +            A      B              P              H        C             is minimum.

Genevieve James · Accepted Answer

Step 1Let:  P      H    B    =  yand altitudes from A, B and C as       h    a   ,       h    b   , and       h    c   for sides a, b and c respectively.It can be shown that:      x          h      a        +      y          h      b        +      z          h      c        =  1Or:      a    x        a          h      a        +      b    y        b          h      b        +      c    z        c          h      c        =  1  a      h    a    =  b      h    b    =  c      h    c    =  2  Swhere S is the area of triangle, so we have to minimize   f  (  x  ,  y  ,  z  )  =  A  =      a    x    +      b    y    +      c    z  subjected to   g  (  x  ,  y  ,  z  )  =  a  x  +  b  y  +  c  z  −  2  SWe use Lagrange multiplier; we have:      L    (  x  ,  y  ,  z  ,  λ  )  =  f  (  x  ,  y  ,  z  )  +  λ  g  (  x  ,  y  ,  z  )Taking partial derivative we get:  −      a          x      2        −      b          y      2        −      c          z      2        +  λ  (  a  +  b  +  c  )  =  0which gives:  a            (        λ  −      1          x      2                  )        +  b            (        λ  −      1          y      2                  )        +  c            (        λ  −      1          z      2                  )        =  0which gives:            (        λ  −      1          x      2                  )        =            (        λ  −      1          y      2                  )        =            (        λ  −      1          z      2                  )        =  0Or:  x  =  y  =  zOr we may put   λ  =      1    x   ,   λ  =      1    y   and   λ  =      1    z   in g(x, y, z) and get:      a    λ    +      b    λ    +      c    λ    =  2  S  ⇒  λ  =            a      +      b      +      c              2      S      putting this in f(x, y, z) we get:  f  (  x  ,  y  ,  z  )  =  A  =            (      a      +      b      +      c              )        2                    2      S      which is minimum of A.

timbreoizy · Accepted Answer

Step 1Consider the areas of the triangles       S    A    =  (  P  B  C  )     ,         S    B    =  (  P  A  C  )     ,         S    C    =  (  P  A  B  )  ,  S  =  (  A  B  C  ) . Note that       S    A    +      S    B    +      S    C    =  S and that the quantity that we seek to minimize is  Q  =            B      C              P              H        A              +            A      C              P              H        B              +            B      A              P              H        C              =            a      2              2              S        A              +            b      2              2              S        B              +            c      2              2              S        C            and the problem can be recast as the minimization of the quantity   Q  (      S    A    ,      S    B    ,      S    C    ) under the constraint   h  (      S    A    ,      S    B    ,      S    C    )  =      S    A    +      S    B    +      S    C    =  S (why can there not be more constraints?) This problem can be solved in various ways but the most elegant one is to use the Cauchy-Schwarz inequality (how?) to conclude that            a      2              2              S        A              +            b      2              2              S        B              +            c      2              2              S        C              ≥            (      a      +      b      +      c              )        2                    2      S      The equality holds only when   P      H    A    =  P      H    B    =  P      H    C   (why?). This is obeyed by the incenter of the triangle. This solution is also unique, because   Q  −  λ  h is convex whenever the areas of the triangles are positive (which is trivially true in the interior of ABC).

Find all points P such that <mfrac> <mrow> B C </mrow> <mrow>

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