Vector radius with triangle and inscribed circle

Question

Can you give me hints on how to solve this one: Knowing the radius-vectors r1, r2 and r3, of vertex of triangle, find radius vector of center of circle inscribed inside that triangle. So basically I should somehow represent that radius vector of center of that circle with r1, r2 and r3? What should I use to express in terms of them?

Answer 1

Observe first of all that if $|B-A|=|C-A|$ then vector $(B-A)+(C-A)$ is parallel to the bisector of $\angle BAC$. For a generic triangle $ABC$ we can then construct a vector $v_A$ parallel to the bisector of $\angle BAC$ as follows: $$ v_A={B-A\over|B-A|}+{C-A\over|C-A|}. $$ Any point $P$ on that bisector can be then represented as $P=A+tv_A$, where $t$ is any real number. In the same way, you can construct $v_B$ and represent any point $Q$ on the bisector of $\angle ABC$ as $Q=B+sv_B$, where $s$ is any real number.

The incenter you are after is the intersection of those two bisectors. To find it, just solve for $t$ and $s$ the equation: $A+tv_A=B+sv_B$.

Answer 2

Let $a = |\mathbf{r_2} - \mathbf{r_3}|, b = |\mathbf{r_3} - \mathbf{r_1}|, c = |\mathbf{r_1} - \mathbf{r_2}|$. Then the position vector $\mathbf{r}$ of the incenter is given by \begin{align*} \mathbf{r} = \frac{a\mathbf{r_1} +b\mathbf{r_2}+c\mathbf{r_3}}{a+b+c} \end{align*}