Non-Euclidean Triangle Centers

# Non-Euclidean Triangle Centers

## Abstract

Non-Euclidean triangle centers can be described using homogeneous coordinates that are proportional to the generalized sines of the directed distances of a given center from the edges of the reference triangle. Identical homogeneous coordinates of a specific triangle center may be used for all spaces of uniform Gaussian curvature. We also define the median point for a set of points in non-Euclidean space and a planar center of rotation for a set of points in a non-Euclidean plane.

## 1 Introduction

Clark Kimberling’s on-line Encyclopedia of Triangle Centers [6] is a collection of thousands of Euclidean triangle centers. It provides descriptions and trilinear coordinates for each center, along with additional information. There does not appear to be a similar collection for non-Euclidean triangle centers, which can also be given similar coordinate ratios.

In Non-Euclidean Geometry [2], H. S. M. Coxeter describes the use of homogeneous coordinates for non-Euclidean spaces of uniform Gaussian curvature. Coxeter mentions the homogeneous coordinates of three triangle centers. These are the incenter (), the orthocenter (), and the intersection of the medians (). He notes that these are the same as the Euclidean trilinear coordinates.

## 2 Euclidean coincidence

In Euclidean geometry there is more coincidence for triangle centers than in non-Euclidean geometry. For example, in Euclidean geometry, the center of rotation of the vertices, the center of rotation of the triangle, the intersection of the medians, and the intersection of the area-bisecting cevians are the same point. In non-Euclidean geometry these are all distinct. Although their trilinear coordinates in Euclidean geometry are identical, in non-Euclidean geometry the homogeneous coordinates are not.

In non-Euclidean space, the homogeneous coordinates of the circumcenter, with , are (). In Euclidean space, must be , so that the trilinear coordinates for the circumcenter are also (), since . But in non-Euclidean geometry, the triangle center with homogeneous coordinates () is not the circumcenter of the triangle. It is a different center, which happens to coincide with the circumcenter in Euclidean space.

## 3 Generalized trigonometric functions

We shall employ generalized trigonometric functions, since they apply to all spaces of uniform Gaussian curvature. For a space of uniform Gaussian curvature , the generalized sine function is defined as

 sing(x) = ∞∑i=0(−K)ix2i+1(2i+1)!=sinx√K√K=sinhx√−K√−K = x−Kx3/3!+K2x5/5!−K3x7/7!+… = xif K=0 = sin(x)if K=1 = sinh(x)if K=−1.

We pronounce the same as “singe x.” It allows us to express the law of sines for any space of uniform Gaussian curvature as

 singasinA=singbsinB=singcsinC.

The generalized cosine function is defined as

 cosg(x) = ∑i≥0(−K)i(2i)!x2i=cosx√K=coshx√−K = 1−Kx2/2!+K2x4/4!−K3x6/6!+… = 1if K=0 = cos(x)if K=1 = cosh(x)if K=−1.

We pronounce as if it rhymed with “dosage x.” We can show that

 cosg2x+Ksing2x=1.

Finally, the generalized tangent function is defined as

 tang(x) = singxcosgx=tanx√K√K=tanhx√−K√−K = xif K=0 = tan(x)if K=1 = tanh(x)if K=−1.

We pronounce as if it rhymed with “flange x.”

## 4 Homogeneous coordinates

Coxeter describes the homogeneous coordinates of a point as a triple ratio . These coordinates are equivalent if multiplied by the same value, so that for any nonzero number , in the same way that we can multiply the numerator and denominator of a fraction by the same value to obtain an equivalent fraction. Given a reference triangle in a space of uniform Gaussian curvature , we can obtain the homogeneous coordinates of a point by first calculating the directed distances of the point from the edges of the reference triangle. The distances are positive (negative) if the point is on the same (opposite) side of the triangle edge as the remaining vertex. As shown in Fig. 1, the directed distances of the point from the edges , , and of the reference triangle are , , and respectively. The homogeneous coordinates of are then . Since in Euclidean geometry, these homogeneous coordinates are equivalent to trilinear coordinates when the Gaussian curvature .

As with trilinear coordinates, we can also specify the homogeneous coordinates of a line as a triple ratio . A point is on a line only if , so that the line through points and is as shown in in Coxeter. The homogeneous coordinates of the vertices of the reference triangle are , , and . The sides are , , and .

Let us prove that these homogeneous coordinates are those defined by Coxeter. In equation 12.14, Coxeter states that the distance from a point to a line is

 sin−1|{xY}|√(xx)√[YY]for K=1sinh−1|{xY}|√(xx)√−[YY]for K=−1

Then the distances from a point to the three edges of the reference triangle are

 Missing dimension or its units for \hskip

Taking the generalized sine of these distances gives us the homogeneous coordinates, since the denominators are equivalent. Thus, the generalized sines of the directed distances of a point from the edges of the reference triangle are identical to the homogeneous coordinates described by Coxeter.

These homogeneous coordinates for points and lines allow us to do linear operations in spaces of constant curvature. As described above, we can tell if a point is on a line and determine a line passing through two points. Three points (, , and ) are collinear if

 ∣∣ ∣∣x0x1x2y0y1y2z0z1z2∣∣ ∣∣=0.

The intersection point of two lines and is

 (X1Y2−X2Y1:X2Y0−X0Y2:X0Y1−X1Y0).

## 5 Finding the homogeneous coordinates of a triangle center

Let us find the homogeneous coordinates of the intersection of the medians of a non-Euclidean triangle. First we want to find the homogeneous coordinates of the midpoint of an edge of the triangle. Let be the midpoint of edge . Let be on edge such that is orthogonal to edge , and let be on edge such that is orthogonal to edge . Then the homogeneous coordinates of are . Using the generalized law of sines on right triangle , we have

 singMaFbsinC=singa21.

Similarly,

 singMaFcsinB=singa21.

Substituting, we have the homogeneous coordinates

 Ma=(0:singMaFb:singMaFc)=(0:sinCsinga2:sinBsinga2)=(0:sinC:sinB).

Similarly, and . Now we can determine the homogeneous coordinates of the medians , , and to be

 AMa=[0:−sinB:sinC]BMb=[sinA:0:−sinC]CMc=[−sinA:sinB:0]

The point at the intersection of any two of these medians is

 M=(sinBsinC:sinAsinC:sinAsinB)=(cscA:cscB:cscC).

The use of the linear algebra of homogeneous coordinates makes this exercise easy.

## 6 Coordinate conversion

The stereoscopic projection of a space of uniform curvature projects points on an embedded curved space onto a Euclidean subspace, which we shall make tangent to the embedded curved space. We place the origin at the tangent point of the curved space, and we know the Gaussian curvature of the curved space. We shall want to convert coordinates of points on the projection to and from points on the embedded curved space. We shall also want to convert homogeneous coordinates to and from these coordinates as well.

Consider a space of uniform curvature of dimension . We can assign a point on its stereographic projection ordinary Cartesian coordinates such as or polar coordinates such as , where is the distance of the point from the origin and is a sequence of direction cosines such that and . For points on the embedded curved space, we require an additional coordinate , which will have an imaginary value in the case where .

In Fig. 3, we see cross sections of the embedded space and the image space on which it is projected, a horizontal line. The coordinates of , the stereographic projection, are . We can convert these to the coordinates of , which are . The equations are

 r2√1/K=r′2√1/K+x0r′2+(√1/K+x0)2=1/K

Solving, we have

 x0=−2r2(4+Kr2)√1/Kr′=4r4+Kr2r=2√−x0Kx0+2√Kr=2−2√1−Kr′2Kr′

Note that is less than when and greater than when . Also the first coordinate is negative when and positive imaginary when .

Conversion of the homogeneous coordinates is more complicated. We first need the coordinates of the vertices of the triangle of reference in the embedded space. We then find the equations of the three planes that pass through the edges of the reference triangle and the center of the sphere. Next we translate each plane in the direction of its normal (positive is in the direction of the remaining vertex of the reference triangle) by the amount of the corresponding homogeneous coordinate. These planes will intersect at a point, which we can project onto the sphere from its center.

Why does this method work? In Fig. 4, we have drawn a cross section of an embedded sphere with radius . Point is the foot of a perpendicular drawn from a triangle center to an edge of the triangle. Letting the homogeneous coordinates of be , we draw a plane parallel to the plane through the edge at and the center such that the translation is equal to . Point is the intersection of the plane with line . The length of the arc is , and we can see that

 sinh0OM=EMOM,

which means that by definition. From similar triangles, we can see that

 EMDM′=singh0x0=OMOM′.

But this will be true for the other edges as well, so that . This means that the intersection of the planes at can be centrally projected to obtain the center . Let’s do examples with positive and negative Gaussian curvature.

Converting homogeneous coordinates when . Let and the projected vertices of the triangle of reference be , , and . Then the polar coordinates are , , , , , and . Then we can calculate the coordinates on the embedded sphere using our formulas. We have

 A′=(−25,1625,1225)B′=(−15,−1225,925)C′=(−15,−925,−1225).

Since the center of the embedded sphere is at , the three planes in which the triangle edges are embedded are

 OB′C′: 15x0+28x1+4x2+15=0 OC′A′: 12x0−60x1+65x2+12=0 OA′B′: 288x0+105x1−500x2+288=0

The normals to these planes allow us to compute the cosines of the angles of the triangle of reference, which are

 cosA=3534429√3259321cosB=−105229√16769cosC=248√326729

From these we can compute the sines of the angles.

 sinA=3862529√3259321sinB=360529√16769sinC=515√326729

We want to translate the planes above by the homogeneous coordinates, for which we shall use the intersection of the medians, (). The new planes are

 15x0+28x1+4x2 = √152+282+42cscA−15 12x0−60x1+65x2 = √122+602+652cscB−12 288x0+105x1−500x2 = √2882+1052+5002cscC−288

Their intersection is

 (319√1336321611113945−1,−29√1336321611113945,87√1336321611856575).

We want a central projection of this point on the sphere centered at , which is

 (55√3131−1,−5√3131,9√3131).

Finally we want to relocate this point to its stereographic projection, which is

 (−553(√3131−55),953(√3131−55)).

This point is shown at the left in Fig. 5.

Converting homogeneous coordinates when . As before, the projected vertices of the triangle of reference are , , and , and the polar coordinates are , , , , , and . Then we can calculate the coordinates on the embedded sphere using our formulas. We note that the first coordinate is imaginary when the sphere has negative curvature. We have

 A′=(23i,1615,45)B′=(14i,−35,920)C′=(14i,−920,−35).

Since the center of the embedded sphere is at , the three planes in which the triangle edges are embedded are

 OB′C′: 3ix0−7x1−x2=3 OC′A′: 84ix0+600x1−625x2=84 OA′B′: 288ix0−75x1+700x2=288

The normals to these planes allow us to compute the cosines of the angles of the triangle of reference, which are

 cosA=506692√306856798489cosB=1039√16919921cosC=3827√30486329

From these we can compute the sines of the angles.

 sinA=223875√306856798489sinB=3980√16919921sinC=3980√30486329

We translate the planes above by the homogeneous coordinates, which we shall take as those for the intersection of the medians, (). The new planes are

 3ix0−7x1−x2 = √−32+72+12cscA+3 84ix0+600x1−625x2 = √−842+6002+6252cscB+84 288ix0−75x1+700x2 = √−2882+752+7002cscC+288

Their intersection is

 ((√12581128738049712818−1)i,√12581128738049178204500,13√1258112873804959401500).

The central projection of this point on the sphere centered at is

 ((250√60978−1)i,1√60978,117√60978).

Finally we want to relocate this point to its stereographic projection, which is

 (1761(250−√60978),39761(250−√60978)).

This point is shown at the right in Fig. 5.

## 7 The median point

In this section we discuss an important non-Euclidean center point that we call the median point. In our list of triangle centers below, we provide homogeneous coordinates for the median points of the vertices, edges, and interior of a triangle. To define the median point, we first assign coordinates of the form to points in a space of constant curvature . We select an origin in the space and let represent the distance from the origin of a point in the space. We select a set of orthogonal rays from the origin and let represent the set of direction cosines determined by these rays. As shown in Fig. 6, let the origin be and the rays be and . (There would be more rays if the space had more than two dimensions.) If we draw a line to any point in the space, the coordinates will be , where is the length of , and is the cosines of the angles between and each of the rays. As shown in Fig. 6(a), the coordinates of are .

Consider a set of points , where . The coordinates of are . If , we shall say that the coordinate origin point is the median point of that set of points. If there is just one point, it must be its own median point. If there are two points, they must be equidistant from in opposite directions, so that the median point of two points is the midpoint of the line joining them. When there are more points, we have a simple process for determining their median point.

As shown in Fig. 6(b), if we embed our curved space in a Euclidean space, there is a simple way of transforming our coordinates for our space of curvature into Euclidean coordinates of the space in which the sphere is embedded. The cross section of the sphere includes the center of the sphere , the coordinate origin point , and a particular point . We see that

 MPOP=sinr√K,

and thus

 MP=sinr√K√K=singr.

Similarly, .

We let be the origin for our Euclidean coordinates with the direction from to being the axis and the other axes having the same orientation at as they do on the sphere. Then the Euclidean coordinates for on the sphere become . For example, if and , our Euclidean coordinates would be

 (cosgr√K,singr(cosπ6,cosπ3))=(√22,√64,√24).

Now we consider a set of points such that their median point is , which by definition means that

 0=∑ising(ri)θi.

The Euclidean coordinates for these points become . Let us find the Euclidean centroid of these points. It is simply

 \nolimits is allowed only on operators

In other words, all but the first Euclidean coordinate is zero, so the Euclidean centroid must lie on the line . Thus, if we have a set of points in a curved space that we embed in a Euclidean space, the median point of those points is found by finding the Euclidean centroid of the points (a point within the sphere), and then projecting it onto the sphere from the center of the sphere. Of course this won’t work if the Euclidean centroid is in fact the center of the sphere, but in that case we say that the median point is not defined.

This use of the Euclidean centroid gives us a method for finding a median point. We know that the centroid of a set of points must lie on a line joining the centroids of two nonempty subsets of these points when each point lies in one of the subsets. Thus the centroid of the vertices of a Euclidean triangle must lie on a line joining one vertex to the midpoint of the edge joining the other two vertices. But if the median point of a set of points lies on a central projection of the centroid, the same relationship holds for the median points of the subsets, since a central projection preserves straight lines. We call this central projection of the line joining the median points of the two subsets a median. Thus, the median point of a set of points in a space of constant curvature must lie on a median joining the median points of two complementary nonempty subsets of those points. In other words, we can locate the median point of a non-Euclidean triangle at the intersection of its medians. For four points, we take a median joining the midpoints of two pairs and repeat that with a different set of pairs as shown in Fig. 7. This will also work to find the median point of the vertices of a non-Euclidean tetrahedron.

## 8 A planar center of rotation

If we consider a rigid set of points in a non-Euclidean plane of constant Gaussian curvature , we can determine if a point in that plane is a center of rotation for these points as they rotate in the plane. Using the same polar coordinate system that we used to define the median point, we say that the coordinate origin is a planar center of rotation of a rigid set of points if

 0=∑ising(2ri)θi.

Why is this point a center of rotation? Lamphere [8] has shown that the centrifugal force of a particle rotating in non-Euclidean space of constant Gaussian curvature is equivalent to

 mv2tangr.

In non-Euclidean space the circumference of a circle of radius is , so that the velocity of objects rotating about a fixed point at a fixed number of revolutions per unit time is proportional to . Assuming that our points have equivalent masses, the centrifugal force of a point is proportional to

 sing2rtangr=singrcosgr=sing(2r)2.

So the centrifugal force is proportional to , and in order for these forces to cancel each other out at the coordinate origin , the value of must be zero.

If is negative, this point is unique. When is positive, however, this is not the case. Consider a single point. One center of rotation is the point itself. Another center of rotation is any point on the polar of the point. For two points a fixed distance apart, one center of rotation is the midpoint of the line connecting them. Other centers of rotation include the poles of that line and the intersections of that line and the polar of the midpoint.

Non-Euclidean rotations in more than two dimensions are complex. In Euclidean space all axes of rotation of a rigid object pass through a common point. In non-Euclidean space, this apparently is not true. See Gunn [4].

In Fig. 8, we show the median point and the interior center of rotation for triangles in spaces of positive and negative curvature. As shown, if we extend the ray from the center of rotation to each point to double its length, the median point of the set of new points at the end of the extended rays will coincide with that interior center of rotation.

Let us determine the homogeneous coordinates of the interior center of rotation of the vertices of an isosceles triangle with apex as shown in Fig. 9. Note that when the Gaussian curvature is positive, there are exterior centers of rotation at the poles of . The homogeneous coordinates will be . First we determine . We know that the three vectors must cancel out at point , which requires that

 sing2RC=2cosθsing2RA,

where . Note that is half of so that

 2cosg2FA=1+cosgAB.

In right triangle ,

 cosθ=tangRFtangRA

and

 cosgRA=cosgRFcosgFA

so that

 sing2RC = 4tangRFtangRAsingRAcosgRA = 4tangRFcosg2RA = 4tangRFcosg2RFcosg2FA = sing2RF(1+cosgAB).

From right triangle , we have

 cosgCF=cosgACcosgFA,

so that

 cosg2CF=2cosg2CF−1=4cosg2AC1+cosgAB−1.

Also

 cosg2CF = cosg2RCcosg2RF−Ksing2RFsing2RC 4cosg2AC1+cosgAB−1 = cosg2RCcosg2RF−Ksing22RF(1+cosgAB) = cosg2RCcosg2RF−(1−cosg22RF)(1+cosgAB) = √1−Ksing22RCcosg2RF−(1−cosg22RF)(1+cosgAB) = √1−Ksing22RF(1+cosgAB)2cosg2RF −(1−cosg22RF)(1+cosgAB) = √1−(1−cosg22RF)(1+cosgAB)2cosg2RF −(1−cosg22RF)(1+cosgAB).

The only unknown is . Solving, we obtain (the positive root)

 cosg2RF=cosgAB+cosg2AB+4cosg2AC(1+cosgAB)√cosg2AB+8cosg2AC.

This allows us to calculate either or from the formula

 cosg2RF=1−2Ksing2RF=2cosg2RF−1.

We now turn our attention to . By the law of sines

 singRE=singAB2singRCsingAC=√(1−cosgAB)/(2K)singRCsingAC.

Also

 singRC=sing(CF−RF)=cosgRFsingCF−singRFcosgCF.

In right triangle

 cosgCF=cosgACcosgFA=√2cosgAC√1+cosgAB.

This allows us to calculate and therefore as functions of and (and ), which we leave to the reader.

We are now in a position to calculate the homogeneous coordinates of by evaluating . We have

 singREsingRF = √(1−cosgAB)/(2K)(cosgRFsingCF−singRFcosgCF)singACsingRF = √(1−cosgAB)/(2K)(singCF/tangRF−cosgCF)singAC = \nolimits is allowed only on operators = singAB(√(cosgAB−cosg2AC)/(2K)/tangRF−cosgAC)singAC(1+cosgAB).

Now let’s work on . We have

 1tangRF = cosgRFsingRF=√K1+cosg2RF1−cosg2RF =    ⎷K(1+cosgAB)(√8cosg2AC+cosg2AB−cosgAB)+4cosg2AC(1+cosgAB)(√8cosg2AC+cosg2AB−cosgAB)−4cosg2AC = √8KcosgAC√cosgAB−cosg2AC(1+cosgAB)(√8cosg2AC+cosg2AB−cosgAB)−4cosg2AC.

Plugging this into our earlier equation for and simplifying, we get

 singREsingRF=singABsingACcosgAC(√8cosg2AC+cosg2AB−cosgAB−2)4cosg2AC−(1+cosgAB)(√8cosg2AC+cosg2AB−cosgAB).

Finally we have the homogeneous coodinates of . Substituting for and for , the value of is

 (cscA(cosgb(√8cosg2b+cosg2c−cosgc−2))
 :cscB(cosgb(√8cosg2b+cosg2c−cosgc−2))
 :cscC(4cosg2b−(1+cosgc)(√8cosg2b+cosg2c−cosgc))).

We can convert this to functions of only the angles of the triangle with a law of cosines:

 Missing dimension or its units for \hskip

Having found the homogeneous coordinates of , the interior planar center of rotation for the vertices of an isosceles triangle, we are still far from obtaining the homogeneous coordinates for the same center of a scalene triangle, which we pose as (presumably open) problem 4 below.

## 9 A short list of non-Euclidean triangle centers

Here and . Note that many of the homogeneous coordinates are equivalent to the trilinear coordinates in Euclidean space.

1. The incenter, the center of the incircle.

2. The vertex median point, the intersection of the medians.

3. The circumcenter, the center of the circumcircle.

4. The orthocenter, the intersection of the altitudes.

5. The Euler circle center, the center of the circle that is externally tangent to the three excircles of the triangle.

6. The symmedian point, the intersection of the symmedians.

7. The Gergonne point, the intersection of the cevians through the touch points of the incircle.

 (cscAsing(s−a):cscBsing(s−b):cscCsing(s−c))

8. The Nagel point, the intersection of the cevians through the touch points of the excircles.

 (sing(s−a)sinA:sing(s−b)sinB:sing(s−c)sinC)

9. The mittenpunkt, the intersection of the lines through the midpoint of each edge and the center of the excircle on the other side of that edge from its opposite vertex.

 (−sinA+sinB+sinC:sinA−sinB+sinC:sinA+sinB−sinC)

10. The edge median point.

 \nolimits is allowed only on operators

11. The Feuerbach point, the common point of the incircle and the Euler circle (the circle that is externally tangent to the three excircles).

 (1−cos(B−C):1−cos(C−A):1−cos(A−B))

12. The incenter of the medial triangle, with edges , , and . We can use the generalized law of cosines () to evaluate these edges, replacing, e.g., , , and with , , and .

 ⎛⎝cosgb2singcm+cosgc2singbmsinga2:cosgc2singam+cosga2singcmsingb2:cosga2singbm+cosgb2singamsingc2⎞⎠

13. The triangle median point, the median point of the points in the interior of the triangle. We need to take the limit, , when .

 (a−bcosC−ccosB:b−ccosA−acosC,c−acosB−bcosA)

14. The polar median point, the median point of the points in the interior of the polar triangle, or the limit of this point when .

 (π−AsinA:π−BsinB:π−CsinC)

## 10 Problems

Here are some problems regarding the homogeneous coordinates of non-Euclidean triangle centers. Perhaps only the first two have known solutions.

1. What is the polar median point (triangle center 14 above) called in Euclidean geometry?

2. Show that the limit when of the generalized law of cosines (triangle center 12 above) is equivalent to .

3. What is the geometric description of the non-Euclidean triangle center with homogeneous coordinates ?

4. What are the homogeneous coordinates of the interior planar center of rotation of the vertices of a non-Euclidean triangle?

5. What are the homogeneous coordinates of the interior planar center of rotation of the edges of a non-Euclidean triangle?

6. What are the homogeneous coordinates of the interior planar center of rotation of the interior of a non-Euclidean triangle?

## 11 Barycentric coordinates for non-Euclidean triangles

While Coxeter’s homogeneous coordinates discussed here are trilinear coordinates in the Euclidean case, other authors use homogeneous coordinates that are barycentric coordinates in the Euclidean case. G.Horvàth [3] uses what is equivalent to the generalized polar sines [7] of the triangles , , and as the coordinates for point in triangle . These coordinates can be obtained from Coxeter’s by multiplying by the generalized sines of the edges to get .

### References

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3. Àkos G.Horvàth, On the hyperbolic triangle centers, Studies of the Univ. of Žilina 27/1 (2015) 11-34.
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8. Robert L. Lamphere, Solution of the direct problem of uniform circular motion in non-Euclidean geometry, Amer. Math. Monthly 109 (2002) 650-655.
9. George M. Minchin, A Treatise on Statistics. Clarendon Press, second ed., 1880, p. 271, https://books.google.com/books?id=O65AAAAAIAAJ&pg=PA271.
10. Robert A. Russell, Non-Euclidean Triangle Continuum, Wolfram, 2011, http://demonstrations.wolfram.com/NonEuclideanTriangleContinuum/.
11. George Salmon, On the circle which touches the four circles which touch the sides of a given spherical triangle, Quarterly J. of Math. 6 (1864) 67-73, https://books.google.com/books?id=lhULAAAAYAAJ&pg=PA67.
12. Abraham A. Ungar, Hyperbolic Triangle Centers: The Special Relativistic Approach, Springer, 2010.
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