The Mohr-Mascheroni theorem is one of the most interesting results concerning ruler and compass constructions (see [1-3] for a simple proof). It asserts that, as long as the objects we want to construct are points, the full power of the Euclidean tools is in fact not needed and we have the following:

Theorem 1 (Mohr-Mascheroni). Any point that can be constructed with a ruler and compass can be constructed with a compass alone.

In fact, the result has been strengthened in several ways. First, the construction can be performed in such a way that all circles have centers on an arbitrarily chosen segment (see [4,5]). In addition, as noted in [6] based on [7], the theorem can actually be generalized in a different direction by replacing the regular compass with a rusty one, i.e., a compass that can only be used to draw unit circles. It should be noted though that the authors of [7] allow for the choice of arbitrary points while performing constructions, a process is usually not allowed in pure geometric constructions (see [8] for more details). Finally, we should also mention that similar versions of the theorem have been proven in elliptic and hyperbolic geometry settings (see [9,10]).

In this note we consider the theorem above in a new context. Specifically, we introduce a new tool called n-compass and investigate the validity of the theorem above when the classical compass is replaced by an n-compass.

The main purpose for the introduction of the n-compass is to have a tool allowing for the construction of a regular n-gon. So, just like a compass can be used to draw the circle determined by two points (the center of the circle and a point on the circle), an n-compass can be used to draw the regular n-gon determined by its center and one vertex.

As a physical tool, an n-compass would consist of a central arm and n outside arms, all connected at a single point. The outside arms have equal lengths and make equal angles with the central one. As the angle between the central arm and one of the outside arms is modified, the other angles change by the same amount. To obtain the actual drawing of a regular n-gon, the tips of the outside arms are assumed to be connected with a flexible “chalk line”. Moreover, to avoid physical constraints, the length of the central arm can be modified so as to make it possible for the construction of regular n-gons of different sizes.

By noting that an n-compass has n outside arms whose tips are radially 2π/n apart, it follows that a 1-compass is just a divider, while a 2-compass is a segment-doubling tool. In some sense, the classical compass can be thought of as an ∞-compass, although this association is rather loose. In addition, it should be noted that, unlike a compass, an n-compass cannot be swung; a regular n-gon is drawn by simply placing the tool on paper and considering the trace of the “chalk line”.

In defining what it means for a point to be constructible (with any tools), one usually thinks of a process consisting of several steps at the end of which the point in question is obtained as an intersection of objects constructed in one of the previous steps. To avoid any ambiguities in the definition, one must impose that no arbitrary points be used at any step. Hence, we must have a starter set in order to be able to get off the ground and this set should consist of at least two points. This is actually the starter set one has in mind in the statement of the MohrMascheroni theorem. However, it should be noted that the aforementioned theorem remains valid for any starter set consisting of at least two points.

We claim that the following version of the MohrMascheroni theorem, stated in terms of starter sets, holds in the context of n-compass constructions.

Theorem 2. Given a set S consisting of at least two points, any point that can be constructed with ruler and n-compass (n > 2) and starter set S can also be constructed with just an n-compass and the same starter set.

Proof: Clearly, the theorem follows as long as we could find a way to replicate a ruler. Since the sides of an n-gon consist of straight line segments, we will use these segments to create our ad-hoc ruler. In particular, given any two points X and Y, as a first step, we need to be able to construct a regular n-gon such that the line is determined by two of its vertices contains X and Y. However, this is not enough since one actually has to be able to extend the segment XY in both directions, as needed. So our approach consists of two steps represented by the following constructions:

Construction 1. Given two points X, Y, by using only an n-compass, it is possible to construct a regular n-gon such that the extension of one of the sides of this polygon contains both X and Y.

Construction 2. Given two consecutive vertices P₁ and P₂ of a regular n-gon (given with together with its center), it is possible to extend segment P₁P₂ indefinitely in both directions by using only an n-compass.

For Construction 1, let us first construct the regular n-gon centered at Y with “radius” YX and denote consecutive vertices of this n-gon (in clockwise orientation) by X = X₁, X₂, …, X_n. Similarly, the vertices of the regular n-gon centered at X with “radius” XY will be denoted by Y = Y₁, Y₂, …, Y_n, but this time the notation is in counterclockwise orientation (see Figure 1). With center at X₂, we now draw the regular n-gon with “radius” X₂X and label its vertices in a counterclockwise manner by X = Z₁, Z₂, …, Z_n. We claim that Z₂ is on line XY and hence X and Y are on line Z₁Z₂. Indeed, if n is equal to 6, then X₂ coincides with Y₂, Z₂ coincides with Y, and we are done. If n is different from 6, then X₂ and Y₂ are different and we can consider the parallel to line Y₁Y₂ through X₂ and denote its intersection with X₁Y₁ by W. Note that we are only referring to this line in order to prove that Construction 1 works, we do not need to con-

struct it. In fact, we will show that W coincides with Z₂, which will finish the proof. To this end, note that lines X₂Y₂ and X₁Y₁ are parallel due to symmetry. Hence, the quadrilateral Y₁Y₂X₂W is in fact a parallelogram, implying that the segments Y₁Y₂ and X₂W are congruent. But X₁X₂ and Y₁Y₂ are also congruent as equal sides of regular n-gons with the same “radius”. So we have that the segments X₂X₁, X₂Z₁, and X₂W are congruent and

Now clearly W and Z₂ must coincide since the relation above shows that W is the rotation of Z₁ in X₂ through 2π/n. It should be noted that the regular n-gon centered at X is not needed in the actual construction of Z₂. We only used side Y₁Y₂ of this n-gon to prove that X, Y, and Z₂ are collinear.

For Construction 2, let Q₁ be the center of the regular n-gon having P₁P₂ as one of its sides. Throughout this proof we will assume that the indexing order among the vertices of any regular n-gon corresponds to the counterclockwise orientation, considering now the regular n-gon centered at P₁ with “radius” P₁Q₁. Finally, let us also construct the regular n-gon centered at Q₂ with “radius” Q₂P₁. If we denote the vertices of this n-gon by R₁, R₂, …, R_n, by relabeling P₁ as R₂, we claim that R₁ is on line P₁P₂ (see Figure 2). Thus, the segment P₁R₁ extends the segment P₁P₂ on the P₁-side of P₁P₂. Clearly, this construction can be continued (on both sides of P₁P₂) to obtain an arbitrarily long extension of P₁P₂.

To prove the claim, notice that

and that

Since the sum of the measures of the three angles above is π, it follows that R₁, P₁, and P₂ are collinear points, as claimed.