Incenter-of-mass of a polygon

Question

"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.)

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?

Answer 1

Given convex polygon P, the generalized incenter ICM(P) can be computed simply as follows.

1. Choose any "dual triangulation" of P-- that is, n-2 side-triples of P, corresponding to a triangulation of the (structural) dual of P. Any such triangulation will do (since the answer turns out to be triangulation-independent), but we get the least-tangled-looking picture if we use the dual-triangulation corresponding to the n-2 internal nodes of the medial axis of P, as shown in the following picture. Each of the n-2 chosen side-triples has an incircle, with an incenter. ICM(P) will be some weighted average of those n-2 incenters. We need to figure out the weights.

2. Define P' to be the tangential polygon having the unit circle as its incircle, whose angles and side directions are the same as those of P.

3. Define Q to be the reciprocal polygon of P' about the unit circle; that is, Q is the cyclic polygon whose vertices are the points where the sides of P' tangentially meet the unit circle. Alternatively, we can construct Q directly from P: its vertices, on the unit circle, are the outward unit normals of the sides of P.

4. Triangulate Q, into n-2 triangles corresponding to the n-2 chosen side-triples of P. The areas of these n-2 triangles are the desired weights, so the answer is:

$$ \mathrm{ICM(P)} = {{\Sigma\ (\mathrm{area\ of\ triangle}_i) * \mathrm{incenter}_i} \over {\Sigma\ \mathrm{area\ of\ triangle}_i}} $$

We can get a sense of the weighting by drawing each little triangle of Q with its centroid positioned at the point in P that its weight is being applied to; then ICM(P) is the overall area centroid of the union of the little blue triangles in this picture of P.

(Note that the precise scale of the little triangles in this picture doesn't matter; e.g. we could have drawn them all twice as big, or 1/3 as big; what matters is only their centroids and their area proportions.)

Observations:

• In the case that P is a tangential polygon (i.e. it has an incircle), all of the n-2 incenters are the same, and so the result is just the incenter of P, as required.
• The result is, magically, triangulation-independent. I haven't included a proof of this.

Two more examples:

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