A = ( x 1 , y 1 ), B = ( x 2 , y 2 ), and C = ( x 3 , y 3 ) are three (distinct) non-collinear points in the Cartesian plane, and a = | B C ¯ | , b = | A C ¯ | , and c = | A B ¯ | . The incenter of the triangle is ( a x 1 + b x 2 + c x 3 a + b + c , a y 1 + b y 2 + c y 3 a + b + c ) .The x-coordinate of the incenter is a "weighted average" of the x-coordinates of the vertices of the given triangle, and the y-coordinate of the incenter is the same "weighted average" of the y-coordinates of the same vertices. I am requesting an explanation for this statement.

Question

A  =  (      x      1    ,      y      1    ),   B  =  (      x      2    ,      y      2    ), and   C  =  (      x      3    ,      y      3    ) are three (distinct) non-collinear points in the Cartesian plane, and   a  =      |                  B        C            ¯        |  ,   b  =      |                  A        C            ¯        |  , and   c  =      |                  A        B            ¯        |  . The incenter of the triangle is      (                  a                  x                  1                +        b                  x                  2                +        c                  x                  3                            a        +        b        +        c              ,                       a                  y                  1                +        b                  y                  2                +        c                  y                  3                            a        +        b        +        c              )    .The x-coordinate of the incenter is a &quot;weighted average&quot; of the x-coordinates of the vertices of the given triangle, and the y-coordinate of the incenter is the same &quot;weighted average&quot; of the y-coordinates of the same vertices. I am requesting an explanation for this statement.

britspears523jp · Accepted Answer

The bisector of angle   A intersects side   B  C at a point       A    ′  , and according to angle bisector theorem we have:       A    ′    B  :      A    ′    C  =  c  :  b. It follows that       A    ′   is a weighted average of   B and   C, , with weights given by the lengths of the opposite sides:      A    ′    =      b          b      +      c        B  +      c          b      +      c        C  ,and of course we have analogous expressions for the similarly defined points       B    ′   and       C    ′  .The incenter   I of   A  B  C is the intersection of   A      A    ′  ,   B      B    ′   and   C      C    ′  . It is then hardly surprising that it turns out to be the weighted average of   A,   B and   C. For instance: as   I belongs to segment   A      A    ′   we can write:  I  =  (  1  −  t  )  A  +  t      A    ′    =  (  1  −  t  )  A  +            t      b              b      +      c        B  +            t      c              b      +      c        C  ,for some   t  ∈  [  0  ,  1  ]. But the expression for   I must be symmetric when exchanging   A,   B, and   C among them, and it is easy to verify that   t  =            b      +      c              a      +      b      +      c       does the trick, leading to your formula for   I:  I  =      a          a      +      b      +      c        A  +      b          a      +      b      +      c        B  +      c          a      +      b      +      c        C  .

A = ( x 1 </msub> , y 1 </msub> ) , B

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