Bailee Short

2022-06-24

$A=\left({x}_{1},{y}_{1}\right)$, $B=\left({x}_{2},{y}_{2}\right)$, and $C=\left({x}_{3},{y}_{3}\right)$ are three (distinct) non-collinear points in the Cartesian plane, and $a=|\overline{BC}|$, $b=|\overline{AC}|$, and $c=|\overline{AB}|$. The incenter of the triangle is

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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Expert

The bisector of angle $A$ intersects side $BC$ at a point ${A}^{\prime }$, and according to angle bisector theorem we have: ${A}^{\prime }B:{A}^{\prime }C=c:b$. It follows that ${A}^{\prime }$ is a weighted average of $B$ and $C$, , with weights given by the lengths of the opposite sides:
${A}^{\prime }=\frac{b}{b+c}B+\frac{c}{b+c}C,$
and of course we have analogous expressions for the similarly defined points ${B}^{\prime }$ and ${C}^{\prime }$.
The incenter $I$ of $ABC$ is the intersection of $A{A}^{\prime }$, $B{B}^{\prime }$ and $C{C}^{\prime }$. 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 $A{A}^{\prime }$ we can write:
$I=\left(1-t\right)A+t{A}^{\prime }=\left(1-t\right)A+\frac{tb}{b+c}B+\frac{tc}{b+c}C,$
for some $t\in \left[0,1\right]$. But the expression for $I$ must be symmetric when exchanging $A$, $B$, and $C$ among them, and it is easy to verify that $t=\frac{b+c}{a+b+c}$ does the trick, leading to your formula for $I$:
$I=\frac{a}{a+b+c}A+\frac{b}{a+b+c}B+\frac{c}{a+b+c}C.$

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