Let $ABCD$ be an irregular tetrahedron, and let $K$ be the center of its inscribed sphere. Let $M$ be the center of the inscribed sphere of $KBCD$. Are $A$, $M$, $K$ necessarily collinear? I have trouble finding a simple counterexample.
(Having a counterexample would settle the generalization of this question to complex values Edit: I found another way around that).
Counterexample.
Irregular tetrahedron $ABCD$ with vertices
\begin{align} A&=(0,0,1) ,\quad B=(1,0,0) ,\quad C=(0,1,0) ,\quad D=(2,0,0) \end{align}
has inscribed sphere with center $K$ and radius \begin{align} r&=\tfrac1{22}\,(5-\sqrt3)\approx 0.1485 , \end{align}
tetrahedron $KBCD$ has inscribed sphere with center $M$ and radius \begin{align} r_m&=\tfrac1k\,(5-\sqrt3)\approx 0.0657 ,\\ k&={22+5\,\sqrt{30}-3\,\sqrt{10}+5\sqrt2-\sqrt6+2\,\sqrt{57-29\,\sqrt3}} . \end{align}
Coordinates of the centers are
\begin{align} K&=\tfrac1{22}(9+7\,\sqrt3, 5-\sqrt3, 5-\sqrt3) \\ &\approx(0.960,0.149,0.149) ,\\ M_x&= \tfrac1k\,\left(9+7\,\sqrt3+5\,\sqrt{30}-3\,\sqrt{10}+4\,\sqrt{57-29\,\sqrt3}\right) ,\\ M_y&= \tfrac1k\,(5-\sqrt3+5\,\sqrt2-\sqrt6) ,\\ M_z&=\tfrac1k\,(5-\sqrt3) ,\\ M&\approx(0.994,0.159,0.0657) , \end{align}
\begin{align} \angle MAK\approx 1.65^\circ ,\\ \angle AKM\approx 154^\circ ,\\ \angle KMA\approx 24.35^\circ . \end{align}
