["Circumcenter of mass"](https://en.wikipedia.org/wiki/Circumcenter_of_mass)
is a natural generalization of circumcenter to non-cyclic polygons.
CCM(P) can be defined as the weighted average of the circumcenters
of the triangles in any triangulation of P, the weights being proportional
to the triangle areas.
It's a surprising fact that that the resulting point is independent of the triangulation chosen.

There is, apparently, an analogous generalization of incenter to non-tangential polygons.
I've verified this experimentally, but I've been unable to find any literature on it.

Call this generalization "incenter of mass", ICM(P), for convex polygons P.

I don't know a simple formula for ICM(P),
but we can compute it for any convex polygon P recursively,
by taking as an ansatz the assumed property that ICM(P)
will be *some* (mysteriously) weighted average of the ICMs of its parts, and that it will be the same regardless of which partitioning is chosen.

The appropriate kind of partitioning here is dual, in a sense, to the kind of partitioning
down to a triangulation which is done during the computation of CCM.
That is, partitioning is done at a chosen pair of *sides*, rather than a pair of *vertices*, of P,
and repeated partitioning eventually arrives at a fully refined set of side-triples, rather than vertex-triples (triangles).  We can see immediately that this makes sense when the polygon is tangential-- all side-triples will agree on a common point which is the incenter of P.

The recursive algorithm proceeds as follows.
For the recursion, we extend the definition of ICM to apply to not only convex polygons,
but also to any subsequence S of the cyclic sequence of sides (directed line segments)
of a convex polygon P.

- For subsequence S consisting of only 3 segments, find $\mathrm{ICM}(S)$ directly as for an incenter of a triangle: that is, find the center of the unique circle having those three sides (directed lines) as tangents, being careful to choose sidedness so that it's in the interior of the original P.  That is, it's the unique point equidistant (using signed distance) from the three lines.

- For subsequence S having more than 3 sides, subdivide into two smaller problems, in two different ways:

    (1) Cut the cyclic list of sides $S=[s_0,...s_{n-1}]$ into two (overlapping) parts,
        at any two non-adjacent entries in it; let's say $s_0$ and $s_2$ for definiteness.
        Recursively compute the two points $\mathrm{ICM}([s_0,s_1,s_2])$ and $\mathrm{ICM}([s_2,s_3,...,s_{n-1},s_0])$.
        We know the desired $\mathrm{ICM}(S)$ will be some weighted average
        of those two points (i.e. it will lie on the line
        through those two points) but we don't know the weights yet.

    (2) Cut $S$ into two overlapping parts
        at a *different* pair of non-adjacent sides; this time at, say, $s_1$ and $s_3$.
        Recursively compute the two points $\mathrm{ICM}([s_1,s_2,s_3])$ and $\mathrm{ICM}([s_3,...,s_{n-1},s_0,s_1])$.
        Again, we know $\mathrm{ICM}(S)$ will lie on the line through these two points.

    (3) The point $\mathrm{ICM}(S)$ can now be pinpointed, as the intersection of the two lines
        computed in (1) and (2).

The following picture shows an example of the recursive step in the case that P is a quadrilateral:
the four circles are computed as the "incenters" of the four side-triples;
each of the circle centers is joined to its opposite by a line segment,
and then ICM(P) is computed as the intersection of the two segments.
(And the weights are still mysterious!  I.e. how do we predict how far along each line segment that intersection point will be, as a fraction of the segment length?  I don't see it.)

[![the recursive step in the case of a quadrilateral][1]][1]

I haven't proved this, but I've verified experimentally that it works: that is,
the algorithm described above gives a consistent answer,
independent of which subdivision is chosen at each stage in the recursion.
In other words, surprisingly, ICM(P) is well-defined, just as (surprisingly) CCM(P) is.

My questions:
- Is this a known result?
- Is there a simple formula for ICM(P), as a weighted average of the incenters of the side-triples corresponding to triangles in any given triangulation of the dual of P?  (I anticipate something of the flavor of CCM, whose weights are proportional to the triangle areas of a triangulation of P, but I haven't been able to figure out exactly what it is; the above diagram doesn't seem to reveal any obvious formula.)
- Is there a duality relationship between ICM and CCM?  E.g. perhaps ICM(P) can be computed
  as some simple function of CCM(P') for some polygon P' derived from P.
- Any other insights about how ICM relates to other properties?

  [1]: https://i.sstatic.net/JfSB61r2.png