set of centers of sphere inscribed in tetrahedron Having a sphere and three diffrent point $A,B,C$ on this sphere. Find set of all centers of spheres inscribed in a tetrahedron $ABCD$, where $D$ is some point on the given sphere. The problem reduced to 2-dimensions is trivial it's just sum of two arcs of some circle, but in 3-dimensions the set is not so simple. Checking in geogebra it's not sum of some parts of sphere. I don't know what this set looks and how it's described.
 A: A little algebra shows that, for $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$ on the sphere, this surface is an irreducible algebraic surface of degree 5 that is singular at $A$, $B$, and $C$ but is otherwise smooth (even on the plane at infinity). In fact, the hypersurface is defined by
$$
\begin{aligned}
0 &= 2s_1^5-9s_1^3s_2+9s_1^2s_3+9s_1s_2^2-27s_2s_3\\
&\quad -7s_1^4+18s_1^2s_2+18s_1s_3-9s_2^2\\
&\quad\quad +8s_1^3-9s_1s_2-27s_3
-2s_1^2
-2s_1
+1
\end{aligned}
$$
where $s_1=x+y+z$, $s_2=xy+yz+zx$, and $s_3=xyz$.  (Written out in terms of $x$, $y$, and $z$, the polynomial on the right hand side has 56 terms; as Anton suspected, it does not appear to be very nice.)  
The singularities at $A$, $B$, and $C$ are cubic (i.e., the polynomial, when expanded about each of these points, has lowest nonvanishing terms of order $3$), so one suspects that this surface, once the singularities have been resolved, might be a known algebraic surface.  In fact, the projectivization of the tangent cone at each of these three points is a nonsingular cubic curve.  (I have not computed the j-invariant, though.)
A: For $A,B,C=$ $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ (blue below) on a unit sphere $S$,
the surface is a sort of triangular tea bag with corners at $A,B,C$.
Below are two views of $100$ random tetrahedron incenters (red)
on the surface,
corresponding to random points
$D$ (not shown) uniformly distributed on $S$.

      


A typical $D$ (green), the determined tetrahedron, 
the inscribed sphere and its (red) center, are shown below.

                  


A: Here is a depiction of Robert Bryant's surface defined by the $56$-term polynomial
he details:

          


Note one component is the "triangular tea bag" discernable in 
my empirical investigation.
Here is the polynomial:
$$
2 x^5+x^4 y+x^4 z-7 x^4+2 x^3
   y^2+4 x^3 y z-10 x^3 y+2
   x^3 z^2-10 x^3 z+8 x^3+2
   x^2 y^3-12 x^2 y^2 z-15 x^2
   y^2-12 x^2 y z^2+6 x^2 y
   z+15 x^2 y+2 x^2 z^3-15 x^2
   z^2+15 x^2 z-2 x^2+x y^4+4
   x y^3 z-10 x y^3-12 x y^2
   z^2+6 x y^2 z+15 x y^2+4 x
   y z^3+6 x y z^2-6 x y z-4 x
   y+x z^4-10 x z^3+15 x z^2-4
   x z-2 x+2 y^5+y^4 z-7 y^4+2
   y^3 z^2-10 y^3 z+8 y^3+2
   y^2 z^3-15 y^2 z^2+15 y^2
   z-2 y^2+y z^4-10 y z^3+15 y
   z^2-4 y z-2 y+2 z^5-7 z^4+8
   z^3-2 z^2-2 z+1$$
The plot above restricts $(x,y,z)$ to lie on or in the sphere.
