Coordinates for intersection point with inscribed circle

Question

I've been thinking about this for quite a while, so it would be great if someone could help me out here.

Given the vertices of a triangle, expressed with coordinates in the cartesian plane, what are the coordinates of the points where a triangle side touches the circle? I googled and found an expression for the incenter, but is there any decent way for these touching points? I tried myself, but it turned in to an inhuman equations which nor me or WolframAlpha could solve.

Answer 1

The three line segments from the incenter to the sides will be perpendicular to the respective sides, because the incircle is tangent to the triangle at these points.

So, given the expression for the incenter, the slope of the line segment from incenter to side will be the inverse reciprocal of the slope of the side.

Now you can solve for the intersection because you have two points for each side, and point and slope for the incenter line segment.

Answer 2

Suppose you know the incenter $I$ and $A,B$ are two vertices and $P=(x_0,y_0)$ the touching point at $AB$.

So the line $PI$ is perpendicular to $AB$ then

$$m_{PI}=-\frac{1}{m_{AB}}$$

Now you can get the expression of the line $PI$

$$\frac{y-y_0}{x-x_0}=-\frac{1}{m_{AB}}$$

And you also know that $P \in AB$ and is easy to get the expression of the line $AB$. Now just make $P$ to be the intersection between $PI$ and $AB$.

Answer 3

Since the inscribed circle is tangent to all three sides of the triangle, the tangent points on each pair of sides are equidistant from those two sides’ common vertex. This fact leads to a fairly simple solution.

Let $A$, $B$ and $C$ be the vertices of the triangle and let $a$, $b$ and $c$ be the lengths of the respective sides opposite them. We then have the system of equations $$\begin{align}r_A+r_B&=c\\r_B+r_C&=b\\r_C+r_B&=a\end{align}$$ for the three distances. This has the solution $$\begin{align}r_A&=\frac12(b+c-a)\\r_B&=\frac12(a+c-b)\\r_C&=\frac12(a+b-c).\end{align}$$ The point of tangency between $A$ and $B$ is then ${r_BA+r_AB\over r_A+r_B}$ and similarly for the other two sides.

Answer 4

$\def\Pp#1#2#3{\frac1{#1_t-I_c}&=-\frac1{#1-I_c}+\frac1{#2-I_c}+\frac1{#3-I_c}}$

Considering the coordinates $A,B,C$ of the vertices of $\triangle ABC$ as complex numbers, $|B-C|=a$, $|C-A|=b$, $|A-B|=c$, the incenter is found as $$I_c=\frac{a\,A+b\,B+c\,C}{a+b+c},$$ and the relationship between the tangential points $A_t,\ B_t,\ C_t$ and the coordinates $A,B,C$ and $I_c$ is presented in a neat system

\begin{align} \Pp ABC ,\\ \Pp BCA ,\\ \Pp CAB . \end{align}

Deduced from: Liang-Shin Hahn. Complex Numbers and Geometry, 1994 , p. 101.