Construction based on circumcenter and incenter

Question

Construct a triangle given the exact location of its circumcenter and its incenter, and the position of its angle bisector (including its direction), but not its length.

I tried to consider the angles between the points.

Answer 1

There is not a unique triangle with that information.

General argument: A triangle has $6$ degrees of freedom: $x$ and $y$ coordinates for each vertex. Since the given angle bisector goes through the given incenter, you have only $5$ pieces of information: the two centers ($2$ each) and the direction of the angle bisector ($1$).

Specific example: Consider an equilateral triangle. The circumcenter and incenter are identical, with the angle bisector going through both. Expand the triangle with respect to the circumcenter/incenter, and none of your given information changes, despite the changes in all the vertices.

Possible modification: Would it be acceptable to find just one of the triangles that have the given circumcenter, incenter, and angle bisector? This could probably be done by fixing one more datum, such as the location of the vertex on the angle bisector.

Answer 2

Given: Incenter $(I)$, its radius $r$, circumcenter $(O)$ and slope $m$ of line $t$ (angle bisector); ?: translation point; $R$ by the Euler's relation, equation of the circle ($μ$) with center $O$ and radius $R$; Points $A$ and $Z$: intersection of the circle ($μ$) with the line $t$; circle ($ψ$) with center $Z$ and radius $IZ$; Points $B$ and $C$: intersection of the circles $ψ$ and $μ$; translation point.

$\begin{array}{} I=(X_{I},Y_{I}) & O=(X_{O},Y_{O}) \end{array}$

$\begin{array}{} \text{translation point} & k=X_{I} & h=Y_{I} \end{array}$

$\begin{array}{} I=(0,0) & O=(a=X_{O}-k,b=Y_{O}-h) \end{array}$

$\begin{array}{} \text{Euler's relation} & IO=\sqrt{a^2+b^2}=\sqrt{R(R-2r)} & R=r+\sqrt{a^2+b^2+r^2} \end{array}$

$\left\{ \begin{array}{} (x-a)^2+(y-b)^2=R^2 \\ y=m·x \end{array} \right.\ $

$\begin{array}{} x_{A}=\frac{a+b·m+\sqrt{(1+m^2)R^2-(a·m-b)^2}}{1+m^2} & y_{A}=m·x_{A} \\ x_{Z}=\frac{a+b·m-\sqrt{(1+m^2)R^2-(a·m-b)^2}}{1+m^2} & y_{Z}=m·x_{Z} \\ \end{array}$

$\begin{array}{} \text{Eq.1 (Ψ)} & x^2+y^2-2x_{Z}·x-2m·x_{Z}·y=0 \\ \text{Eq.2 (μ)} & x^2+y^2-2a·x+2b·y+a^2+b^2-R^2=0 \\ \end{array}$

$\begin{array}{} \text{Eq.2 - Eq.1} & y=p·x+q & \text{where:} \\ p=\frac{x_{Z}-a}{b-m·x_{Z}} & q=\frac{a^2+b^2-R^2}{2(b-m·x_{Z})} \\ \end{array}$

replacing in Eq.2, solving, we get

$\begin{array}{} x_{B}=\frac{a+bp-pq+\sqrt{(1+p^2)R^2-(ap-b+q)^2}}{1+p^2} & y_{B}=p·x_{B}+q \\ x_{C}=\frac{a+bp-pq-\sqrt{(1+p^2)R^2-(ap-b+q)^2}}{1+p^2} & y_{C}=p·x_{C}+q \\ \end{array}$

$\begin{array}{} X_{A}=x_{A}+k & Y_{A}=y_{A}+h \\ X_{B}=x_{B}+k & Y_{B}=y_{B}+h \\ X_{C}=x_{C}+k & Y_{C}=y_{C}+h \\ \end{array}$