Using a compass of fixed opening and straightedge, what is the shortest way to find the centroid of $10$ points?

Question

This question is actually mentioned in the OEIS sequence $A157650$. After reviewing the initial terms, I believe that for $n=1..9$, the centroid of $n$ points can be found with the following step counts: $$0, 4, 9, 12, 19, 21, 28, 28, 36$$

Here I need to claim that a compass of fixed opening is a compass whose distance is fixed, and which draws circles of a predetermined and constant, but arbitrary radius.

When I attempted to find the centroid of $10$ points, I failed to accomplish it within $39$ steps (Divide all points into $5$ groups of $2$ points each, find the centroid of each group, and then make the centroid of these five midpoints. This requires $5×4+19=39$ steps). But $A157650(10) = 38$. Does anyone know how to construct the centroid of $10$ points with $38$ steps?

See the Figure 1 and Figure 2 for constructing the centroid of $5$ points.

Figure 1:

Figure 2:

Answer 1

This is not a complete answer, but an idea as to how to systematize the calculation. After exploring the problem by hand, it's hard to see how you can beat $39$ steps. But to what extent can we rule out a more efficient solution? One idea is to allow only certain natural operations using the fixed-opening ("rusty") compass on the set of centroids generated so far. If those operations are sufficient to reproduce all the earlier entries in the series, but can't find the centroid of ten points in less than $39$ steps, then at least we can say that a qualitatively new idea would be needed to do so.

Starting with $N$ points (each of which is its own centroid), the proposed operations are these:

• A: Circle an existing centroid ($1$ step) (once a centroid is already circled, this has no further effect).
• B: Given circled centroids of a disjoint pair of point sets $A$ and $B$, where $|A|=|B|$, form the centroid of $A\cup B$ ($2$ steps).
• C: Given centroids of a disjoint pair of point sets $A$ and $B$, and of a different disjoint pair of point sets $C$ and $D$, where $A\cup B = C\cup D$, form the centroid of $A\cup B$ ($2$ steps).

Operation B finds the midpoint of the line segment connecting two (circled) centroids, by drawing the line connecting the intersection points of the two surrounding circles and intersecting it with the line connecting the two centroids. This midpoint is, in fact, the centroid of the combined point set, provided that the initial point sets are disjoint and have equal size.

Operation C is the application of the Archimedes Lemma: since the centroid of $A\cup B=C\cup D$ lies on the line connecting the centroids of $A$ and $B$, and also on the line connecting the centroids of $C$ and $D$, we can find it by drawing the two lines and taking their intersection.

The state after any operation will be a set of centroids, each of which is either circled or uncircled. If we represent centroids by strings (e.g., $abcd$ if uncircled, or $[abcd]$ if circled), then examples of the three operations are the following:

• A: $abcd \rightarrow [abcd]$
• B: $[abc], [def] \rightarrow abcdef$
• C: $(ab, [cd]), ([a], bcd) \rightarrow abcd$.

When searching for a derivation of the centroid of $N$ labeled points, we will treat as equivalent all states that differ only through permutations of the labels. For instance, then, the centroid of $3$ points can be minimally derived as follows:

• 1: $a \rightarrow [a]$
• 2: $b \rightarrow [b]$
• 3: $c \rightarrow [c]$
• 4,5: $[a], [b] \rightarrow ab$
• 6,7: $[a], [c] \rightarrow ac$
• 8,9: $(ab, c), (ac, b) \rightarrow abc.$