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 $a=\left|\overline{BC}\right|$, $b=\left|\overline{AC}\right|$, and $c=\left|\overline{AB}\right|$. The incenter of the triangle is

$(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\text{}\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}).$

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.

Answer & Explanation

britspears523jp

Expert

2022-06-25Added 28 answers

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:

$I=(1-t)A+t{A}^{\prime}=(1-t)A+\frac{tb}{b+c}B+\frac{tc}{b+c}C,$

for some $t\in [0,1]$. But the expression for $I$ must be symmetric when exchanging $A$, $B$, and $C$ among them, and it is easy to verify that $t=\frac{b+c}{a+b+c}$ does the trick, leading to your formula for $I$:

$I=\frac{a}{a+b+c}A+\frac{b}{a+b+c}B+\frac{c}{a+b+c}C.$

Most Popular Questions