# Circumcenter of Mass and Generalized Euler Line

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## Abstract

We define and study a variant of the center of mass of a polygon and, more generally, of a simplicial polytope which we call the Circumcenter of Mass (CCM). The CCM is an affine combination of the circumcenters of the simplices in a triangulation of a polytope, weighted by their volumes. For an inscribed polytope, CCM coincides with the circumcenter. Our motivation comes from the study of completely integrable discrete dynamical systems, where the CCM is an invariant of the *discrete bicycle *(*Darboux*)* transformation* and of *recuttings* of polygons. We show that the CCM satisfies an analog of Archimedes’ Lemma, a familiar property of the center of mass. We define and study a *generalized Euler line* associated to any simplicial polytope, extending the previously studied Euler line associated to the quadrilateral. We show that the generalized Euler line for polygons consists of all centers satisfying natural continuity and homogeneity assumptions and Archimedes’ Lemma. Finally, we show that CCM can also be defined in the spherical and hyperbolic settings.

## Keywords

Triangulation Center of mass Circumcenter Archimedes Lemma Euler line## 1 Introduction

*Circumcenter of Mass*of the polygon; we shall show that it does not depend on the choice of the point \(O\).

We learned about this construction from [1] where this center is mentioned as a point invariant under the transformation of the polygon called *recutting*. In [15], we proved that the circumcenter of mass is invariant under another map, called the *discrete bicycle* (or discrete Darboux) transformation; see [9, 13]. Although our motivation comes from the study of these discrete dynamical systems, we believe that the circumcenter of mass is an interesting geometric construction and it deserves attention on its own right.

If one replaces the circumcenter of a triangle by its centroid then, instead of \({\text {CCM}}(P)\), one obtains the center of mass of a polygon \(P\) (more precisely, of the homogeneous lamina bounded by \(P\)), which we denote by CM\((P)\).

In the case of CM\((P)\), one may safely allow degenerate triangles, whose centroids are well defined and whose areas vanish, and which consequently do not contribute to the sum. However, not all degenerate triangles are allowed in the definition of the circumcenter of mass: although their areas vanish, their circumcenters may lie at infinity, in which case formula (1) fails to make sense. This discontinuity with respect to the location of point \(O\) is a subtle point of the definition of \({\text {CCM}}(P)\).

The contents of this paper are as follows. In Sect. 2, we prove that the definition of the circumcenter of mass is correct, that is, does not depend on the choice of the point \(O\).

- 1.
The circumcenter of mass satisfies the Archimedes Lemma: if a polygon is a union of two smaller polygons, then the circumcenter of mass of the compound polygon is a weighted sum of the circumcenters of mass of the two smaller polygons. As a consequence, one may use any triangulation of \(P\) to define \({\text {CCM}}(P)\).

- 2.
If \(P\) is an equilateral polygon then \({\text {CCM}}(P)={\text {CM}}(P)\).

- 3.
One can naturally extend the notion of the circumcenter of mass to smooth curves. However, this continuous limit of the circumcenter of mass coincides with the center of mass of the lamina bounded by the curve.

We investigate whether there are other natural assignments of “centers” to polygons, satisfying natural properties: the center depends analytically on the polygon, commutes with dilations, and satisfies the Archimedes Lemma. We show that all such centers are points of the generalized Euler line.

We also show that the Euler line is sensitive to symmetries, both Euclidean (reflections, rotations), and others, such as equilateralness.

In Sect. 5, we show that the construction of the circumcenter of mass extends to simplicial polyhedra in \({\mathbb {R}}^n\) in a rather straightforward manner. In Sect. 6, we extend our constructions to the spherical and hyperbolic geometries.

Whenever possible, we try to provide both geometric and algebraic proofs, so some statements in this paper are given multiple proofs.

## 2 Correctness of Definition. Degenerate Triangles

Let us prove that the circumcenter of mass is well defined.

### **Lemma 2.1**

### *Proof*

Let \(C_{A}\), \(C_{B}\), \(C_{C}\) and \(C_{D}\) be the circumcenters of \(\triangle BCD\), \(\triangle ACD\), etc., see Fig. 2. It suffices to see that \(C_{A}C_{C}\cap C_{B}C_{D}\) divides the diagonals of \(C_{A}C_{B}C_{C}C_{D}\) at the ratio corresponding to the areas of the triangles. We show this for one pair, and by symmetry, it will follow for the other.

An easy angle count shows that \(\angle C_{B} C_{D} C_{D}=\angle BAC\), \(\angle C_{A} C_{D} C_{B}=\angle A C B\), \(\angle C_{C} C_{B} C_{D} = \angle C A D\), \(\angle C_{A} C_{C} C_{D} =\angle A B D\), \(\angle C_{C} C_{A} C_{D} = \angle C B D\). Indeed, the angles \(C_{A} C_{D} C_{B}\) and \(A C B\) have pairwise perpendicular sides, see Fig. 2, and likewise with the other pairs of angles.

We use Lemma 2.1 to prove that the circumcenter of mass is well defined. We wish to show that \({\text {CCM}}(\oplus _{i=0}^{n-1}V_{i}V_{i+1}O)\) is independent of \(O\). We proceed by induction on \(n\), the number of vertices of the polygon \(P\).

Next, we deduce coordinate expressions for the circumcenter of mass. Let \(O\) be the origin and let \(V_i=(x_i,y_i)\).

### **Proposition 2.2**

### *Proof*

### *Remark 2.3*

We now use Proposition 2.2 to give a different proof of the independence of \({\text {CCM}}(P)\) of the choice of the origin \(O\). We wish to show that parallel translation of the point \(O\) through some vector results in translating \({\text {CCM}}(P)\) by the same vector.

### *Remark 2.4*

Lemma 2.1 follows from the independence of \({\text {CCM}}(P)\) on the choice of the point \(O\). Indeed, choosing the point \(O\) close to vertex \(A\) on the bisector of the angle \(DAB\), see Fig. 4, one obtains, in the limit, two safe degenerate triangles \(OAD\) and \(OAB\). With this choice of \(O\), one has \({\text {CCM}}(ABCD)={\text {CCM}}(ABC\oplus CDA)\), and likewise for the the other pair of triangles.

## 3 Properties of the Circumcenter of Mass. Archimedes Lemma

The center of mass satisfies the Archimedes Lemma which states the following: if an object is divided into two smaller objects, the center of mass of the compound object lies on the line segment joining the centers of mass of the two smaller objects; see, e.g., [3]. It turns out, the circumcenter of mass satisfies the Archimedes Lemma as well.

### **Theorem 1**

### *Proof*

### *Remark 3.1*

One subtelty of Archimedes Lemma has to do with degenerate triangulations. At first sight, a counterexample to Theorem 1 is a right isosceles triangle \(P\), which is bisected into two smaller right isosceles triangles \(Q\) and \(R\) along its axis of symmetry. In this case CCM(*P*) is the midpoint of the base of *P*, which is not on the line joining CCM(*Q*) and CCM(*R*), the midpoints of its other two sides.

The issue is that there is a hidden degenerate triangle. Label the isosceles triangle *ABC* with *B* having a right angle and let *D* be the foot of the perpendicular from B to side AC. If we perturb *D* away from line *AC*, we obtain an additional triangle *ACD*. This triangle has minute area but its circumcenter is far away. In this situation everything works out as expected. The problem occurs when we disregard this triangle because of its zero area. In other words, the CCM of *ABC* is the weighted average of the CCMs of *ABD*, *BCD* and *ACD*.

As a consequence of Theorem 1, one can use any triangulation of \(P\) to define \({\text {CCM}}(P)\), not only a triangulation obtained by connecting point \(O\) to the vertices. In particular, we have the following, expected, corollary.

### **Corollary 3.2**

The circumcenter of mass of an inscribed polygon is the circumcenter.

### *Proof*

Let \(P\) be inscribed, and let \(O\) be the circumcenter. Consider a triangulation by diagonals. Then \(O\) is the circumcenter of each triangle involved, and hence the circumcenter of mass is \(O\).

Alternatively, one can use formula (2). For a circumscribed polygon, all \(|V_i|\) are equal, hence (2) yields zero, that is, the origin \(O\). Rotating back \(90^{\circ }\) is \(O\) as well.\(\square \)

### *Remark 3.3*

It follows from Theorem 1 that the circumcenter of mass of a quadrilateral is the intersection point of the perpendicular bisectors of its diagonals. These perpendicular bisectors may be parallel; this happens exactly when the area of the quadrilateral is equal to zero, see Fig. 6.

Next, we consider the case when \(P\) is equilateral, that is, all the sides have equal lengths.

### **Theorem 2**

If \(P\) is an equilateral polygon then \(\mathrm{{CCM}}(P)=\mathrm{{CM}}(P)\).

### *Proof*

We present a computational proof based on Proposition 2.2 and Remark 2.3. It would be interesting to find a geometric argument as well.

### **Theorem 3**

One has \({\mathrm{{CCM}}}(\gamma )={\mathrm{{CM}}}(\gamma )\).

### *Proof*

Thus one may view the circumcenter of mass of a polygon as a “different” discretization of the center of mass of a lamina bounded by a continuous curve.

## 4 Generalized Euler Line. Other Centers

The line through the center of mass and the circumcenter of a triangle is called the Euler line. The points of this line are affine combinations \(C_t:= t {\text {CM}} + (1-t) {\text {CCM}}\); thus \(C_0={\text {CCM}}\) and \(C_1={\text {CM}}\). For example, the orthocenter of a triangle lies on the Euler line and is given by \(3{\text {CM}}-2{\text {CCM}}\), that is, the orthocenter is the point \(C_{3}\).

For a fixed \(t\), one can repeat the construction of the circumcenter of mass of a polygon to obtain the center \(C_t(P)\): triangulate \(P\) and take the sum of the centers \(C_t\) of the triangles, weighted by their areas. The resulting center is again independent of the triangulation, and all these centers lie on a line that we call the *generalized Euler line* of the polygon \(P\). Since the ratios of distances on the Euler line of a triangle are transferred over to those on the generalized Euler line, we conclude that \(C_t(P) = t {\text {CM}}(P) + (1-t) {\text {CCM}}(P)\) for every polygon \(P\).

We note that, for the case of quadrilaterals, the generalized Euler line was discussed in [12]. One remark we may add to the discussion is the next result, which follows easily from our approach.

### **Proposition 4.1**

Next, we shall show that the above described centers \(C_t(P)\) are the only ones satisfying natural assumptions. Namely, assume that one assigns a “center” to every polygon so that the center depends analytically on the polygon, commutes with dilations, and satisfies the Archimedes Lemma.

### **Theorem 4**

A center satisfying the above assumptions is \(C_t(P)\) for some \(t\).

### *Proof*

Consider an isosceles triangle with base of length \(2\) and base angles \(\alpha \). By symmetry, the center is on the axis of symmetry. Let it be at height \(f(\alpha )\) above the base. For example, if the center is CM then \(f(\alpha )={(\tan \alpha )}/{3}\), and if the center is \({\text {CCM}}\) then \(f(\alpha )=-\cot 2\alpha \). From the symmetry of the equilateral triangle, \(f({\pi }/{3})={1}/{\sqrt{3}}.\)

Any triangle can be triangulated into three isosceles triangles, hence the function \(f(\alpha )\) determines the center uniquely. We shall prove that \(f\) satisfies a certain functional equation, solve it, and deduce the desired result.

We now show that the generalized Euler line is highly sensitive to symmetries. As such, one might consider utilizing the Euler line to detect symmetries, or as a certificate for their non-existence. The symmetries involved are of different type, some being Euclidean and some of another sort, such as equilateralness.

As we saw in the previous section from Theorem 2, if \(P\) is an equilateral polygon, then the Euler line of \(P\) degenerates to a point.

### **Theorem 5**

- 1.
If \(P\) has a line of reflection symmetry \(L\), then \(E\) is either \(L\) or a point on \(L\).

- 2.
If \(P\) has a center \(C\) of rotational symmetry and none of the extensions of sides of \(P\) pass through the center, then \(E=C\).

- 3.
Assume that the sides \(S_1,S_2 \ldots , S_n\) of \(P\) satisfy \(|S_1|\!=\!|S_2|\!=\!\ldots =|S_{n-1}|\). Then \(E\) is orthogonal to side \(S_n\).

### *Proof*

- (1)
First, we note that \(P\) cannot have sides which are subsets of \(L\) because then at least one such side would have three edges incident to a vertex. Thus we may place \(O\) on \(L\) and triangulate. Since every triangle on one side of \(L\) has a mirror image counterpart on the other side of \(L\), the circumcenter of mass lies on \(L\) and so does the center of mass.

- (2)
We let \(O=C\) and triangulate from \(O\). To each triangle in the triangulation, there is a class of triangles obtained from it via rotation. The vector sum of the centers of mass of the triangles of this class is equal to \(O\).

- (3)
Let \(V_1,V_2,\ldots ,V_n\) be the vertices of \(P\). We reflect \(P\) in side \(S_n\). Denote the image of a vertex under reflection by a prime. Consider the polygon \(Q=V_1 V_2 \cdots V_n V_{n-1}' \cdots V_{2}'\). This polygon has twice the area of \(P\) and since it is equilateral, its Euler line is a point (the center of mass). Since \(Q\) has \(S_n\) as an axis of symmetry, its center of mass lies on \(S_n\) (or its extension). By Archimedes’ Lemma (Theorem 1), the Euler line of \(Q\) is the weighted sum of the Euler lines of \(P\) and its reflection. Since the area of the reflection is the same as that of \(P\), the weights are the same. Let \(e\) and \(e'\) denote the two Euler lines. These are symmetric about \(S_n\), so the Euler line of \(Q\) must be the midpoint of every pair of mirror points of \(e\) and \(e'\). This is only possible if \(e=e'\) and are orthogonal to \(S_n\).\(\square \)

### *Remark 4.2*

The generalized Euler line that we have defined is as a property of the polygon, rather than the set of its vertices. For example, if a different choice of edges is given to a quadrilateral, e.g., rather than considering \(ABCD\), we consider \(ABDC\), its center of mass and circumcenter of mass will be different, and consequently its Euler line also.

In the case of a quadrilateral, a definition of the Euler line which is a property of the set of vertices is given in [10], following the work [14]. The authors of [14] construct a point which is a “replacement” to the circumcenter of a quadrilateral when the quadrilateral is no longer cyclic. This point, called the “isoptic point”, is a center of the quadrilateral which depends only on the set of vertices, in the sense that it is independent of the choice of edges for the four vertices of the quadrilateral, e.g., it is the same for \(ABCD\) and \(ABDC\). The Euler line is then defined as the line connecting the isoptic point and the centroid of the vertices of the quadrilateral. In [16], the second author extends the definition of the isoptic point to \((n+2)\)-polytopes in \(\mathbb {R}^n\), thus constructing an Euler line which depends only on the set of vertices of the \((n+2)\)-tope.

## 5 In Higher Dimensions

Similarly to the 2-dimensional case, we have the following result.

### **Theorem 6**

The circumcenter of mass is well defined, that is, does not depend on the choice of point \(O\).

As a preparation to the proof, we state two lemmas. The first one is well known and we do not prove it. Let \(S=(V_0,V_1,\ldots ,V_n)\) be a simplex in \({\mathbb {R}}^n\).

### **Lemma 5.1**

The second lemma is lesser known so, for completeness, we provide a proof. Let \(S=(V_0,V_1,\ldots ,V_n)\) be a simplex and \(C(S)\) its circumcenter.

### **Lemma 5.2**

### *Proof*

### *Proof of Theorem 6*

Consider the quadratic map \(Q:{\mathbb {R}}^n \rightarrow {\mathbb {R}}\) given by the formula \(Q(V)=|V|^2\). The graph of \(Q\) is a paraboloid. For \(V\in {\mathbb {R}}^n\), let \(\widetilde{V} \in {\mathbb {R}}^{n+1}\) be the vector \((V,Q(V))\). If \(S\) is a simplex in \({\mathbb {R}}^n\), let \(\widetilde{S}\) be its lift in \({\mathbb {R}}^{n+1}\).

It remains to notice that the total signed volume of \(p_i \pi _1(\widetilde{\partial P})\) is equal to zero since \(\partial P\) is a cycle. Therefore the coefficients of \(|O|^2,o_2,o_3,\ldots ,o_n\) in (9) vanish, and this sum does not depend on \(O\), as needed. \(\square \)

Similarly to Proposition 2.2, one has an explicit formula for \({\text {CCM}}(P)\) that follows from Lemmas 5.1 and 5.2 and the proof of Theorem 6.

Let \(F=(V_1,\ldots ,V_n)\) be a face of a simplicial polyhedron \(P\subset {\mathbb {R}}^n\). Let \(A(F)\) be the matrix whose columns are the vectors \(V_1,\ldots ,V_n\), and let \(A_i(F)\) be obtained from the matrix \(A(F)\) by replacing its \(i\)th row by the row \((|V_1|^2,\ldots ,|V_n|^2)\). Choosing \(O\) as the origin yields the following formula for the \(i\)th component of the circumcenter of mass.

### **Proposition 5.3**

This rational function of the coordinates of the vertices of the polyhedron \(P\) has homogeneous degree one: it is a ratio of polynomials of degrees \(n+1\) and \(n\).

As an application of this formula, let us prove an analog of Corollary 3.2.

### **Corollary 5.4**

The circumcenter of mass of an inscribed polyhedron is its circumcenter.

### *Proof*

Choose the origin \(O\) at the circumcenter and, without loss of generality, assume that the circumscribing sphere is unit. We wish to prove that \({\text {CCM}}(P)=O\); to be concrete, consider the \(n\)th component of \({\text {CCM}}(P)\), as given in Proposition 5.3.

The matrices \(A_n(F)\) have \(n\)th row consisting of \(1\)s, that is, these are the matrices from Lemma 5.1. It follows that, up to a factor, \({\text {CCM}}(P)_n\) is the signed volume of the projection of \(\partial P\) to \({\mathbb {R}}^{n-1}\) along the last coordinate direction. This volume vanishes, as in the proof of Theorem 6, and we are done.\(\square \)

A multi-dimensional version of Theorem 1 holds as well, and for the same reason: the contribution of the simplices whose bases belong to the cut cancel out. Thus, with the self-explanatory notation, one has the Archimedes Lemma.

### **Theorem 7**

If a simplicial polyhedron \(P\) is decomposed into simplicial polyhedra \(Q\) and \(R\) then \(\mathrm{{CCM}}(P)=\mathrm{{CCM}}(Q\oplus R).\)

It follows that one can find the circumcenter of mass using any triangulation of \(P\).

Similarly to the 2-dimensional case, one can take an affine combination of the centroid and the circumcenter of a simplex to define the center \(C_t(P)\); as \(t\) varies, these centers lie on a line that we call the Euler line of the polyhedron.

The Euler line of a simplex is well studied: this is the line through the centroid and the circumcenter, see [5, 7] and the references in the latter article.

The Euler line of a simplex contains its *Monge point*. This is the intersection point of the hyperplanes through the centroids of the \(n-2\)-dimensional faces of a simplex in \({\mathbb {R}}^n\), perpendicular to the opposite 1-dimensional edges (in dimension two, the Monge point is the orthocenter). In our notation, the Monge point is \(C_{(n+1)/(n-1)}\). Thus we obtain a definition of the Monge point of a simplicial polyhedron in \({\mathbb {R}}^n\).

## 6 On the Sphere

In this section, we consider a version of the theory on the sphere (and, to some extent, in the hyperbolic space). Let us start with \(S^2\).

A circle on the unit sphere in \({\mathbb {R}}^3\) is its section by an affine plane, that is, a plane not necessarily through the origin which is the center of the sphere. To an oriented circle we assign its center; the choice of the two antipodal centers is made using the right-hand rule. Given a spherical triangle \(ABC\), its circumcircle is oriented by the cyclic order of the vertices, and this determines the circumcenter of the oriented triangle.

With these preparations, we can define the circumcenter of mass of a spherical polygon and study its properties, similarly to the Euclidean case.

Let \(P=(V_1,\ldots , V_n)\) be a spherical polygon and let \(W\in S^2\) be a point, distinct from the vertices of \(P\). Consider the triangles \(W V_i V_{i+1}\) where the index \(i\) is cyclic. Let \(O_i\) be the circumcenter of \(i\)th triangle, taken with the mass \(m_i\), equal to the area of the plane triangle \(W V_i V_{i+1}\). Define \({\text {CCM}}(P)\) to be the center of mass of the collections of points \((O_i,m_i)\).

### **Theorem 8**

The point \({{\text {CCM}}}(P)\) does not depend on the choice of \(W\).

### *Proof*

The Archimedes Lemma holds, the same way as in the Euclidean case. We also have an analog of Theorem 2.

### **Theorem 9**

If \(P\) is an equilateral spherical polygon then its circumcenter of mass coincides with the center of mass of the lamina bounded by \(P\), that is, \({{\text {CCM}}}(P)=\mathrm{{CM}}(P)\).

### *Proof*

### **Theorem 10**

One has \(\mathrm{{CCM}}(\gamma )=\mathrm{{CM}}(\gamma )\).

### *Proof*

Finally, we make a brief comment on the hyperbolic case. In this case, the Euclidean space is replaced with the Minkowski space, the pseudo-Euclidean space of signature \((1,n)\), and the sphere \(S^n\) by the pseudo-sphere, the hyperboloid model of the hyperbolic space \(H^n\). One can still define cross-product, and one can use formula (13) for the circumcenter of mass. The center of mass is also defined in the hyperbolic case, see [6].

However, the geometrical interpretation is not as straightforward as in the spherical case. To fix ideas, consider the case \(n=2\), the hyperbolic plane.

One has three possibilities for the vector (13): it may be space-like, time-like, or null. In the first case, the line spanned by this vector intersects the upper sheet of the hyperboloid \(z^2-x^2-y^2=1\), and we obtain a point of the hyperbolic plane. In the last case, we obtain a point on the circle at infinity, and in the second case, a point on the hyperboloid of one sheet \(z^2-x^2-y^2=-1\) (outside of our model of \(H^2\)).

This issue is already present in the case of a triangle. The intersection of the upper sheet of the hyperboloid \(z^2-x^2-y^2=1\) with an affine plane may be a closed curve, disjoint from the null cone \(z^2=x^2+y^2\); it may be a closed curve tangent to the null cone; or it may be an open curve intersecting the null cone. A triangle is circumscribed by a curve of constant curvature, the intersection of the hyperboloid \(z^2-x^2-y^2=1\) with the plane of the triangle. If the curvature is greater than one, this curve is a circle; if the curvature equals one, it is a horocycle; and if the curvature is less than one, the curve is an equidistant curve. These three cases correspond to the three relative positions of the plane and the null cone.

All three types of curves are represented by circles in the Poincaré disc model. In the first case, the circle lies inside the Poincaré disc, in the second it is tangent to the boundary, and in the third it intersects the boundary. Only in the first case the center of the circumcenter of a triangle is a point of the hyperbolic plane.

### *Remark 6.1*

The results on the circumcenter of mass in the Euclidean case can be obtained in the limit \(R\rightarrow \infty \) from the spherical case, where \(R\) is the radius of the sphere.

### *Remark 6.2*

A proof that the Bicycle (Darboux) Transformation [15] preserves the CCM of triangles and quadrilaterals in spherical and hyperbolic geometry is entirely analogous to the one in Euclidean space. This fact and conservation of area under the Transformation allows to fully describe the dynamics of these polygons under the Bicycle Transformation.

## Notes

### Acknowledgments

It is a pleasure to acknowledge interesting discussions with V. Adler, A. Akopyan, I. Alevi, Yu. Baryshnikov, D. Hatch, I. Rivin, O. Radko, A. Sossinsky, A. Veselov. This project originated during the program Summer@ICERM 2012; we are grateful to ICERM for support and hospitality. S. T. was partially supported by the NSF Grant DMS-1105442.

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