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.



set of centers of sphere inscribed in tetrahedron
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.)


