Bailee Short

Bailee Short

Answered

2022-06-24

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.

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Answer & Explanation

britspears523jp

britspears523jp

Expert

2022-06-25Added 28 answers

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 .

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