Arrangement of Central Points on the Faces of a Tetrahedron
IInternational Journal of Computer Discovered Mathematics (IJCDM)ISSN 2367-7775 © IJCDMVolume 5, 2020, pp. 13–41Received 6 August 2020. Published on-line 30 September 2020web: © The Author(s) This article is published with open access . Arrangement of Central Pointson the Faces of a Tetrahedron
Stanley Rabinowitz
545 Elm St Unit 1, Milford, New Hampshire 03055, USAe-mail: [email protected]:
Abstract.
We systematically investigate properties of various triangle centers(such as orthocenter or incenter) located on the four faces of a tetrahedron. Foreach of six types of tetrahedra, we examine over 100 centers located on the fourfaces of the tetrahedron. Using a computer, we determine when any of 16 con-ditions occur (such as the four centers being coplanar). A typical result is: Thelines from each vertex of a circumscriptible tetrahedron to the Gergonne pointsof the opposite face are concurrent.
Keywords. triangle centers, tetrahedra, computer-discovered mathematics, Eu-clidean geometry.
Mathematics Subject Classification (2020).
Introduction
Over the centuries, many notable points have been found that are associated withan arbitrary triangle. Familiar examples include: the centroid, the circumcenter,the incenter, and the orthocenter. Of particular interest are those points thatClark Kimberling classifies as “triangle centers”. He notes over 100 such pointsin his seminal paper [10].Given an arbitrary tetrahedron and a choice of triangle center (for example, thecircumcenter), we may locate this triangle center in each face of the tetrahedron.We wind up with four points, one on each face. What can be said about thesepoints? For example, do the 4 points form a tetrahedron similar to the originalone? Could these 4 points ever lie in a plane? Might they form a regular tetrahe-dron? Consider the 4 lines from the vertices of the tetrahedron to the centers inthe opposite faces. Do these 4 lines concur? Might they have the same length? This article is distributed under the terms of the Creative Commons Attribution Licensewhich permits any use, distribution, and reproduction in any medium, provided the originalauthor(s) and the source are credited. a r X i v : . [ m a t h . HO ] J a n Arrangement of Central Points on the Faces of a Tetrahedron
In this paper, we investigate such questions for a large collection of triangle centersand for various types of tetrahedra.A typical result is: The lines from each vertex of a circumscriptible tetrahedronto the Gergonne points of the opposite face are concurrent.For information about what you need to know about triangle centers and centerfunctions, we give a short overview in Appendix A.We make extensive use of areal coordinates (also known as barycentric coordi-nates) when analyzing points associated with triangles, such as the faces of atetrahedron. For the reader not familiar with areal coordinates, we give the ba-sics in Appendix B.For points, lines, and planes in space, we make heavy use of tetrahedral coordi-nates. For the reader not familiar with tetrahedral coordinates, we present theneeded information in Appendix C.Throughout this paper, the notation [
XY Z ] will denote the area of triangle XYZ.2.
Coordinates for the Face Centers
When referring to an arbitrary tetrahedron (the reference tetrahedron), we willusually label the vertices A , A , A , and A . The lengths of the sides of thebase ( (cid:52) A A A ) will be a , a , and a , with edge a i opposite vertex A i . In thetetrahedron, the edge opposite the edge of length a i will have length b i . See Figure1a. Figure 1. edge labelingThus, we have A A = a , A A = a , A A = a , A A = b , A A = b , A A = b . If the tetrahedron has its opposite edges of equal length, then the tetrahedron iscalled an isosceles tetrahedron. See Figure 1b (not to scale). It is clear that in anisosceles tetrahedron, the four faces are congruent because they each have sides oflength a , a , a . In a sense, the isosceles tetrahedon “looks the same” from eachvertex. This four-fold symmetry makes the isosceles tetrahedron be the figurein space that corresponds to the equilateral triangle in the plane. An equilateraltriangle has 3 identical sides and an isosceles tetrahedron has 4 identical faces. Theface of the tetrahedron opposite vertex A i will be called face i of the tetrahedron.Its area will be denoted by F i . tanley Rabinowitz If ( x , x , x ) are the areal coordinates for a triangle center, then the tetrahedralcoordinates for the corresponding center on face 4 of our reference tetrahedron A A A A is ( x , x , x , P on face4 (triangle A A A ) of the tetrahedron. Then[ P A A A ] : [ P A A A ] : [ P A A A ] = [ P A A ] : [ P A A ] : [ P A A ]since these 4 tetrahedra have a common altitude from A .We will frequently have occasion to pick a point on each face of the tetrahedron.In such a case, the point on face i will be labelled P i . It is often necessary tolocate such a point based on its areal coordinates in face i . We must be carefulhow we set up the coordinate system on each face. Note that in an arbitrarytetrahedron, each face has the property that the labels associated with each edge( a i or b i ) contains one label with subscript 1, one with subscript 2, and one withsubscript 3. In order to maintain the 4-fold symmetry exhibited by an isoscelestetrahedron, under the mapping A → A → A → A we want the faces totransform as follows: (cid:52) A A A → (cid:52) A A A → (cid:52) A A A → (cid:52) A A A . Note that in each face, our labelling starts with the vertex opposite the edge whoselabel has subscript 1, then proceeds to the vertex opposite the edge whose labelhas subscript 2 and finally ends with the edge whose label has subscript 3. Thisinduces the following correspondence between the edges:( a , b , b , b , a , a ) → ( b , a , b , a , b , a ) → ( b , b , a , a , a , b ) → ( a , a , a , b , b , b ) . This mapping is shown in figure 2.
Figure 2. mappingLet us now consider the mapping which takes A into A in this 4-fold symmetry.We start, by finding the tetrahedral coordinates for the point, P , with arealcoordinates ( x , x , x ) in face 1 of the reference tetrahedron. The first coordinate“ x ” refers to the area formed by the point P and the side of the triangle witha “1” as subscript. In this case, face 1 has sides of length a , b , and b , sothe side we need is the side of length a . On face 1, this side is opposite vertex A of the reference tetrahedron and so the “ x ” coordinate will appear as the 4thtetrahedral coordinate. Proceeding in this manner, we find that P has tetrahedralcoordinates (0 , x , x , x ). This point wants to map to a point, P , with the sameareal coordinates in face 2. In face 1 ( (cid:52) A A A ), the coordinates correspond toareas associated with edges a , b , and b . In face 2 ( (cid:52) A A A ), the correspondingedges are b , a , and b . Point P has tetrahedral coordinates ( x , , x , x ) because Arrangement of Central Points on the Faces of a Tetrahedron on face 2, edge b is opposite vertex A (so x moves to the 3rd coordinate in thetetrahedral system), a is opposite vertex A (so x moves to the 4th coordinatein the tetrahedral system), and b is opposite vertex A (so x moves to the 1stcoordinate in the tetrahedral system).In other words, given a point in the plane with areal coordinates ( x , x , x ), thecorresponding points in the faces of our reference tetrahedron are:Face 1 : (0 , x , x , x )Face 2 : ( x , , x , x )Face 3 : ( x , x , , x )Face 4 : ( x , x , x , f ( a , a , a ) , f ( a , a , a ) , f ( a , a , a )) , then the corresponding points on the faces of the tetrahedron are:Face 1 : (0 , f ( b , b , a ) , f ( b , a , b ) , f ( a , b , b ))Face 2 : ( f ( b , b , a ) , , f ( b , a , b ) , f ( a , b , b ))Face 3 : ( f ( b , b , a ) , f ( b , a , b ) , , f ( a , b , b ))Face 4 : ( f ( a , a , a ) , f ( a , a , a ) , f ( a , a , a ) , . Kimberling [11] and [12] has collected the trilinear coordinates for over 40,000centers associated with a triangle. He lists the trilinear coordinates in terms ofthe sides a , b , c of the reference triangle and trigonometric functions of A , B , C ,the angles of the reference triangle. Only the first coordinate is given, for if thiscoordinate is f ( a, b, c, A, B, C ), then the other coordinates are f ( b, c, a, B, C, A )and f ( c, a, b, C, A, B ) respectively.We wish to study points associated with a tetrahedron based on the lengths ofthe 6 edges of the tetrahedron. The six edge lengths are independent quantities.Involving other quantities such as the face areas or trigonometric functions of theface or dihedral angles would yield expressions containing dependent variablesand would complicate the process of determining if such expressions are identi-cally 0 for all tetrahedra. We thus need to remove the presence of angles fromKimberling’s data. Since all the trigonometric functions present can be expressedin terms of sine’s and cosine’s of the angles of the reference triangle, the followingformulas suffice to remove all reference to these angles:sin A = 2 Kbc cos A = b + c − a bc where K denotes the area of the reference triangle. The first formula comes fromthe well-known area formula: K = bc sin A ; and the second formula is The Lawof Cosines. Similar expressions hold for angles B and C .Factors (such as K or a + b + c ) that would be common to all three coordinatesare then removed. tanley Rabinowitz The presence of a K in the denominator of any fraction involved is cumbersomeand was removed by replacing terms of the form x/ ( y + zK ) by x ( y − zK ) / ( y − z K ). This leaves all square roots in the numerators.The variable K is then replaced by its equivalent expression in terms of the sidesof the triangle (Heron’s Formula), namely K = 14 √ a b + 2 b c + 2 c a − a − b − c . Since tetrahedral coordinates are barycentric, if ( x, y, z, w ) are the coordinatesfor one of the above centers in our reference tetrahedron, then the tetrahedralcoordinates for the corresponding center in a tetrahedron with vertices P , P , P ,and P are xP + yP + zP + wP , where xP denotes the scalar product of x andthe vector P , etc. Algorithmically, the desired coordinates are the dot productof the vectors ( x, y, z, w ) and ( P , P , P , P ).3. Types of Tetrahedra
The types of tetrahedra investigated are listed in the following table.
Types of Tetrahedra Considered
Tetrahedron Type Geometric Definition Algebraic Condition i = 1 , , a i = b i Circumscriptible edges are tangent to a sphere a i + b i = constantIsodynamic symmedians are concurrent a i b i = constantOrthocentric opposite edges are perpendicular a i + b i = constantHarmonic n/a 1 /a i + 1 /b i = constant Only tetrahedra that have the requisite 4-fold symmetry were studied. Thus,for example, trirectangular tetrahedra are not included in this study. We wouldhave liked to have investigated isogonic tetrahedra (ones in which the ceviansto the points of tangency of the insphere are concurrent, ([2, p. 328]), but thecorresponding algebraic condition was too messy to be manageable. The conceptof a harmonic tetrahedron was invented for this study and has a few interestingproperties, but perhaps not enough to warrant future study. The other types oftetrahedra are well known and information about them can be found in [2].4.
Methodology
For this study, we considered the first 101 triangle centers listed in [11], X1 throughX101, as well as a few other centers, listed in the following table. Arrangement of Central Points on the Faces of a Tetrahedron
Triangle Centers Considered
Triangle Center TrilinearsX1–X101 see [11]Y1 1 / ( a ( b + c ) − abc )Y2 a/ (( b − a )( c − a ))Y3 a ( b + c )Y4 1 / ( a ( b + c ))Y5 a/ ( b + c )Y6 a ( b + c )Y7 1 / ( a ( b + c ))Y8 ( b + c ) /a Y9 a ( b + c − a )Y10 1 / ( a ( b + c − a ))Y11 a/ ( b + c − a )Y12 a ( b + c − a )Y13 1 / ( a ( b + c − a ))Y14 b + c − bc/a r-power point a r Z1 a r ( b + c )Z2 a r ( b + c )Z3 a r ( b + c − a )Z4 a r ( b + c − a )Z5 a r ( b + c − a )Z6 a r ( b + c )Z7 a r ( b + c + bc )Z8 2 a r + b r + c r Z9 ( b r + c r ) /a Z10 ( b r + c r − a r ) /a Z11 ( b r + c r + 2 a r ) /a power center a r g [ b, c ]arbitrary center f [ a, g [ b, c ]]areal center f [ a, b, c ] /a For each type of tetrahedron considered, and for each triangle center considered,we computed the tetrahedral coordinates of these centers on each face of thetetrahedron.Once we had located these four centers, we then used Mathematica to run a bar-rage of tests on these four points to see if they satisfied any special properties.Since these tests involved algebraic coordinates (i.e. we were not looking at spe-cific tetrahedra with numerical sides), any results found constitute a proof thatthe result is true and not merely a conjecture based on numerical evidence. Theseresults are stated in sections 5 through 10. The proofs are by coordinate geome-try, mechanically performed by the Mathematica program which was written tocompute all the necessary lengths and coordinates and then confirm the claimedresults symbolically.Most of these results are new, however, some of them may have previously ap-peared in the literature. We give references, when known. A few related resultsappeared as problem 3 in the 15th Summer Conference of the International Math-ematical Tournament of Towns, [15]. tanley Rabinowitz First a few definitions.
Definition 1.
The original tetrahedron is known as the reference tetrahedron . Definition 2.
The tetrahedron formed by the four centers is called the centraltetrahedron . Definition 3.
The line segment from a vertex of the reference tetrahedron to thecenter on the opposite face is called a cevian . Definition 4.
Four skew lines in space are said to form a hyperbolic group if thereis an infinite number of lines that meet all four of these lines.According to Altshiller-Court ([2, p. 10]), “Such a group is often the space analogof three concurrent lines in the plane.”
Definition 5.
The four skew lines are part of an infinite family of lines that forma ruled surface known as a hyperboloid of one sheet.
Definition 6.
By a space center of a tetrahedron, we mean one of: centroid,circumcenter, incenter, Monge point, or Euler point. These are described in thefollowing table.
Tetrahedron Centers Considered
Space Center Descriptioncentroid intersection point of medianscircumcenter center of circumscribed sphereincenter center of inscribed sphereMonge point symmetric of circumcenter with respect to centroidEuler point center of 12-point sphereMore background information about these centers is given in Appendix E.The properties that were checked for are listed in the table below.
Properties Considered
Property 1 The cevians to the four centers are concurrent.Property 2 The cevians to the four centers form a hyperbolic group.Property 3 The four centers are coplanar.Property 4 The four centers are collinear.Property 5 The normals to the faces at the centers concur.Property 6 The faces of the central tetrahedron are parallel to the facesof the reference tetrahedron.Property 7 The central tetrahedron is isosceles.Property 8 The central tetrahedron is regular.Property 9 The central tetrahedron is isodynamic.Property 10 The central tetrahedron is circumscriptible.Property 11 The central tetrahedron is orthocentric.Property 12 The central tetrahedron is similar to the reference tetrahedron.Property 13 The cevians to the four centers have the same length.Property 14 The central tetrahedron has a space center in commonwith some space center of the reference tetrahedron.Property 15 The central tetrahedron has a space center on the Euler lineof the reference tetrahedron.Property 16 The reference tetrahedron has a space center on the Euler lineof the central tetrahedron. Arrangement of Central Points on the Faces of a Tetrahedron
If the cevians concurred, we also checked to see if the point of concurrence wasa space center of the reference tetrahedron or if it lied on the Euler line of thereference tetrahedron. Also, if the four cevians formed a hyperbolic group, wecomputed the center of the hyperboloid for which these cevians were generatorsand checked this point to see if it was a space center of the reference tetrahedron(or on its Euler line).To find the center of the hyperboloid, we used the following result:
Proposition 4.1 ([13]) . If L , L , and L are three lines that determine a hyper-boloid of one sheet, then if one draws planes through each of these lines parallelto the two others, then we get a parallelepiped. The center of this parallelepiped isthe center of the hyperboloid. Thus, our test was as follows: Use formula 17 to find the plane through L andparallel to L . (See Appendix D for formulas using tetrahedral coordinates.)Let X be the point of intersection of L with this plane (found via formula 18).Similarly, find the plane through L parallel to L . Let Y be the intersection ofthis plane and L . Then the center of the hyperboloid is the midpoint of segment XY .To make some of the computation of properties 1-16 easier, we first checked theproperty for a specific tetrahedron with numerical sides. If the property was false(using exact arithmetic) for this numerical case, then we did not bother checkingto see if the property was algebraically true in general.5. Results found for Arbitrary Tetrahedra
The following results were discovered and proven by our computer program.
Theorem 5.1.
Consider the centroids on each face of an arbitrary tetrahedron.Then(a) The faces of the central terahedron are parallel to the corresponding faces ofthe reference tetrahedron.(b) The cevians to the centroids concur at the centroid of the reference tetrahedron.(c) The central tetrahedron is similar to the reference tetrahedron.(d) The central centroid coincides with the reference centroid.(e) The central circumcenter coincides with the reference Euler point.(f ) The central Monge point lies on the reference Euler line (at 2/3).(g) The central Euler point lies on the reference Euler line (at 8/9).(h) The reference circumcenter lies on the central Euler line (at 4).(i) The reference Monge point lies on the central Euler line (at -2).
Theorem 5.2.
For an arbitrary tetrahedron, the normals at the circumcenters ofeach face concur at the circumcenter of the reference tetrahedron.
Theorem 5.3.
For an arbitrary tetrahedron, the lines to the r -power points forma hyperbolic group. These include the incenters, the centroids, and the symmedianpoints. Note.
Results found for specific tetrahedra that are immediate consequences ofresults in this section for arbitrary tetrahedra will not necessarily be listed againbelow. tanley Rabinowitz Theorem 5.4.
For fixed r , the a r + b r + c r points of an arbitrary tetrahedronform a central tetrahedron that has the same centroid as the reference tetrahedron. Results found for Isosceles Tetrahedra
The following results were discovered and proven by our computer program.
Theorem 6.1.
Consider an arbitrary center on each face of an isosceles tetrahe-dron. (The same type of center is considered on each face.) Then(a) The cevians to these centers have the same length.(b) The central tetrahedron is isosceles.(c) The central tetrahedron has the same centroid as the reference tetrahedron.(d) The cevians form a hyperbolic group. Results found for Circumscriptible Tetrahedra
The following results were discovered and proven by our computer program.
Theorem 7.1.
For a circumscriptible tetrahedron (in which a i + b i = t , i = 1 , , ),(a) The cevians to the Gergonne points concur.The 4th coordinate of the intersection point is ( a + a − a )( a + a − a )( a + a − a ) .(b) The cevians to the Nagel points concur.The 4th coordinate of the intersection point is a + a + a − t .This equals S − S where S i is the sum of the edges at A i and S = (cid:80) S i .(c) The Feuerbach points are coplanar.(d) The normals at the incenters concur.(e) The normals at the X40 points concur.Note that X40 is collinear with the incenter and circumcenter. Theorem 7.2.
For a circumscriptible tetrahedron, the lines to the following tri-angle centers form a hyperbolic group:(a) Gergonne points (and their inverses)(b) Nagel points (and their inverses)(c) Mittenpunkts (and their inverses)(d) X41 points (and their inverses)(e) Feuerbach points (and their inverses). Results found for Isodynamic Tetrahedra
The following results were discovered and proven by our computer program.
Theorem 8.1.
For an isodynamic tetrahedron,(a) The cevians to any power point concur.The 4th coordinate of the intersection point is a a r +12 a .(b) The Feuerbach points are coplanar.(c) The X44 points are coplanar.(d) The Lemoine axes are coplanar.(e) The circumcenter of the X76 points (3rd power point inverses) coincides withthe reference centroid. Arrangement of Central Points on the Faces of a Tetrahedron
Theorem 8.2.
For an isodynamic tetrahedron, the lines to the following trianglecenters form a hyperbolic group:(a) Spieker centers (and their inverses)(b) X37 points (and their inverses)(c) X38 points (and their inverses)(d) Brocard midpoint (and their inverses)(e) X42 points (and their inverses)(f ) X106 points(g) X107 points(h) X108 points(i) X109 points(j) X110 points(k) X111 points Results found for Orthocentric Tetrahedra
The following results were discovered and proven by our computer program.
Theorem 9.1.
For an orthocentric tetrahedron (in which a i + b i = t , i = 1 , , ),(a) The cevians to the orthocenters concur.The 4th coordinate of the intersection point is ( a + a − a )( a + a − a )( a + a − a ) .(b) The cevians to the isotomic conjugates of the orthocenters concur.The 4th coordinate of the intersection point is a + a + a − t . This equals T − T where T i is the sum of the squares of the edges at A i and T = (cid:80) T i .(c) The centroid of the 9-point centers coincides with the reference centroid.(d) The centroid of the orthocenters coincides with the reference Monge point.(e) The circumcenter of the orthocenters lies on the reference Euler line.(f ) The Monge point of the orthocenters lies on the reference Euler line.(g) The centroid of the X53 points coincide with the reference Monge point. Theorem 9.2.
For an orthocentric tetrahedron,(a) The normals at the circumcenters concur.(b) The normals at the centroids concur.(c) The normals at the orthocenters concur.(d) The normals at the nine point centers concur.(e) The normals at the De Longchamps points concur.
Note that these five centers lie on the Euler line and have constant ratio distancesapart.
Theorem 9.3.
For an orthocentric tetrahedron, the lines to the following trianglecenters form a hyperbolic group:(a) circumcenters(b) Crucial points (and their inverses)(c) X25 points(d) X48 points (and their inverses)
Results found for Harmonic Tetrahedra
The following results were discovered and proven by our computer program. tanley Rabinowitz Theorem 10.1.
For a harmonic tetrahedron,(a) The Feuerbach points are coplanar.(b) The cevians to the X117 points (and their isotomic conjugates, the X102points) concur.
Theorem 10.2.
For a harmonic tetrahedron, the lines to the following trianglecenters form a hyperbolic group:(a) X43 points (and their inverses)(b) X102 points(c) X117 points
Note that X
43 = 1 /b + 1 /c − /a , X
102 = a (1 /b + 1 /c − /a ), and X
117 =(1 /b + 1 /c − /a ) /a .11. General Results about Concurrent Cevians
The data collected by our program suggested (but did not prove) the followingresults. Thus, independent proofs are needed.
Lemma 11.1.
Let P = (0 , y , z , w ) and P = ( x , , z , w ) be two points onfaces 1 and 2 of the reference tetrahedron. Then the condition that the lines A i P i , i = 1 , intersect (or be parallel) is z w = z w . Proof.
From formulas 7 and 11, the condition is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y z w x z w (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . which reduces to the formula claimed. (cid:3) Geometric interpretation.
Let P and P be points on the faces oppositevertices A and A , respectively, of tetrahedron A A A A . Then lines A P and A P intersect if and only if[ P A A ][ P A A ] = [ P A A ][ P A A ] . Here “intersect” also includes being parallel.Figure 3 shows a top view of tetrahedron A A A A with a point taken on faces1 and 2. The condition is that the product of the areas of the yellow trianglesequals the product of the areas of the green triangles. Lemma 11.2. If A i P i , i = 1 , , , meet in pairs, then all three lines meet at apoint.Proof. Let the lines meet in pair at points Q , Q , Q . Then the plane through Q , Q , and Q contains the three lines and hence the three vertices A , A , A . Thus A , A , A would lie on a plane other than the base plane, a contradiction. (cid:3) Arrangement of Central Points on the Faces of a Tetrahedron
Figure 3. A P intersects A P if product of yellow areas equalsproduct of green areas Corollary 11.3.
Let P = (0 , y , z , w ) , P = ( x , , z , w ) , P = ( x , y , , w ) be three points on faces 1, 2, and 3 of the reference tetrahedron. Then the conditionthat the lines A i P i , i = 1 , , concur (or be parallel) is z w = z w , y w = y w , x w = x w . Corollary 11.4 ( The Concurrence Condition ) . The condition for the con-currence of cevians to two centers (each with center function F ( a, b, c ) ) on faces1 and 2 of the reference tetrahedron is F ( b , a , b ) F ( a , b , b ) = F ( a , b , b ) F ( b , a , b ) . Theorem 11.5. If P and P are points on faces 1 and 2 of our reference tetra-hedron such that the cevians to P and P meet, then the cevians to the isotomicconjugates of P and P meet.Proof. In areal coordinates, the isotomic conjugate of ( x, y, z ) is (1 /x, /y, /z ).The concurrence condition therefore becomes1 F ( b , a , b ) · F ( a , b , b ) = 1 F ( a , b , b ) · F ( b , a , b )which is equivalent to the original condition. (cid:3) Corollary 11.6 ([2, p. 139]) . If the four cevians to corresponding face centersconcur, then the four cevians to the isotomic conjugates of these centers alsoconcur.
Corollary 11.7.
If cevians to the triangle centers with center function F ( a, b, c ) concur, then so do cevians to the centers with center function F ( a, b, c ) r for any r . Theorem 11.8.
The centroid is the only triangle center with the property that inany isosceles tetrahedron, the cevians to these face centers concur.Proof.
Suppose F ( a, b, c ) = af ( a, b, c ) is such a center function. The algebraiccondition for this to be true is obtained by substituting b i = a i in the concurrence tanley Rabinowitz condition to get F ( a , a , a ) = F ( a , a , a ) . This implies that(1) F ( a , a , a ) = − F ( a , a , a )or(2) F ( a , a , a ) = F ( a , a , a )for all a , a , and a .If condition (1) holds, then we would have F ( a , a , a ) = − F ( a , a , a ) F ( a , a , a ) = − F ( a , a , a ) F ( a , a , a ) = − F ( a , a , a )since the equality must be true for all values of its arguments. Since F ( a, b, c ) = F ( a, c, b ), multiplying these three equations together yields 1 = −
1, a contradic-tion.If condition (2) holds, then we would have F ( a , a , a ) F ( a , a , a ) = 1or F ( a , a , a ) : F ( a , a , a ) : F ( a , a , a ) = 1 : 1 : 1so that F represents the centroid. (cid:3) Corollary 11.9.
The centroid is the only triangle center with the property thatin any tetrahedron, the cevians to these face centers concur.Proof.
Since the cevians concur for any tetrahedron, they must surely concur forany isosceles tetrahedron. But the previous theorem rules this possibility out. (cid:3)
The following lemma is well known:
Lemma 11.10 (Power Lemma) . If f ( x ) is a nonzero function satisfying f ( xy ) = f ( x ) f ( y ) for all x and y , then f ( x ) = x r for some constant r . Theorem 11.11.
The power points are the only triangle centers with the propertythat in any isodynamic tetrahedron, the cevians to these face centers concur.Proof.
The concurrence condition becomes F ( ta , a , ta ) F ( a , ta , ta ) = F ( a , ta , ta ) F ( ta , a , ta )for all a , a , a , and t . We can write this as F ( ta , a , ta ) F ( a , ta , ta ) = F ( ta , a , ta ) F ( a , ta , ta ) . Since this is true for all a , it will be true if we replace a by t/a to get F ( a , a , ta ) F ( a , a , ta ) = F ( ta , ta , ta ) F ( ta , ta , ta ) . Arrangement of Central Points on the Faces of a Tetrahedron
Since F is homogeneous, we have F ( a , a , ta ) F ( a , a , ta ) = F ( a , a , a ) F ( a , a , a ) . Let t = a x to get F ( a , a , x ) F ( a , a , x ) = F ( a , a , a ) F ( a , a , a ) . The right-hand side is independent of x , and so we can define(3) G ( a , a ) = F ( a , a , x ) F ( a , a , x ) . Thus F ( a, b, c ) = G ( a, b ) F ( b, a, c ) = G ( a, b ) G ( b, a ) F ( a, b, c )and so(4) G ( a, b ) G ( b, a ) = 1for all a and b . Similarly, F ( a, b, c ) = G ( a, b ) F ( b, a, c ) = G ( a, b ) F ( b, c, a )= G ( a, b ) G ( b, c ) F ( c, b, a ) = G ( a, b ) G ( b, c ) F ( c, a, b )= G ( a, b ) G ( b, c ) G ( c, a ) F ( a, c, b ) = G ( a, b ) G ( b, c ) G ( c, a ) F ( a, b, c )and so(5) G ( a, b ) G ( b, c ) G ( c, a ) = 1for all a , b and c . Using the homogeneity of F in equation (3), we can divide allarguments by a to get G ( a /a ,
1) = F (1 , a /a , x/a ) F ( a /a , , x/a ) = F ( a , a , x ) F ( a , a , x ) = G ( a , a ) . Thus, if we define g ( x ) = G ( x, , then we get analogs of equations (4) and (5): g ( ab ) g ( ba ) = 1and g ( ab ) g ( bc ) g ( ca ) = 1 . Now g ( ab ) g ( bc ) = 1 /g ( ca ) = g ( ac ) . Let x = a/b and y = b/c to get g ( x ) g ( y ) = g ( xy ) tanley Rabinowitz for all x and y . By the Power Lemma, we must have g ( x ) = x r for some r . Hence F ( a, b, c ) : F ( b, c, a ) : F ( c, a, b ) = F ( a, b, c ) F ( a, b, c ) : F ( b, c, a ) F ( a, b, c ) : F ( c, a, b ) F ( a, b, c )= 1 : G ( b, a ) : G ( c, a )= 1 : g ( ba ) : g ( ca )= 1 : (cid:16) ba (cid:17) r : (cid:16) ca (cid:17) r = a r : b r : c r and thus the center is a power point. (cid:3) Theorem 11.12.
Suppose the edges of a tetrahedron satisfy the condition h ( a i ) + h ( b i ) = t , i = 1 , , , for some function h ( x ) and some constant, t . Then thecevians to the face centers with (areal) center function [ h ( b ) + h ( c ) − h ( a )] r concurfor any r .Proof. The concurrence condition is[ h ( a ) + h ( b ) − h ( b )] r [ h ( b ) + h ( b ) − h ( a )] r = [ h ( b ) + h ( b ) − h ( a )] r [ h ( a ) + h ( b ) − h ( b )] r . This is equivalent to[ h ( a ) + h ( b ) + h ( a ) − t ] r [ h ( b ) + h ( b ) + h ( b ) − t ] r = [ h ( b ) + h ( b ) + h ( b ) − t ] r [ h ( a ) + h ( b ) + h ( a ) − t ] r which is easily seen to be an identity. (cid:3) Corollary 11.13.
In a circumscriptible tetrahedron, the cevians to the face cen-ters with areal center function ( b + c − a ) r concur. This includes the Gergonnepoint and its isotomic conjugate, the Nagel point. Corollary 11.14.
In an orthocentric tetrahedron, the cevians to the face centerswith areal center function ( b + c − a ) r concur. This includes the orthocenterand its isotomic conjugate. Corollary 11.15.
In a harmonic tetrahedron, the cevians to the face centers withareal center function (1 /b + 1 /c − /a ) r concur. This includes the X117 point andits isotomic conjugate, the X102 point. Corollary 11.16.
In an isodynamic tetrahedron, the cevians to the power pointsconcur. This includes the incenter, centroid, symmedian point and their isogonaland isotomic conjugates.Proof.
Take h ( x ) = log x in the previous theorem. (cid:3) Conjecture 11.17.
Suppose the edges of a tetrahedron satisfy the condition h ( a i )+ h ( b i ) = t , i = 1 , , , for some power function h ( x ) = x n and some constant, t . Ifcevians to four corresponding face centers are concurrent, then the center functionfor these face centers must be of the form [ h ( b ) + h ( c ) − h ( a )] r for some r . Theorem 11.18.
If cevians to the points h ( b ) + h ( c ) − h ( a ) concur, then the edgesof the tetrahedron satisfy h ( a i ) + h ( b i ) = t , for i = 1 , , . Arrangement of Central Points on the Faces of a Tetrahedron
Proof.
The concurrency condition (for centers on faces 1 and 2) becomes[ h ( a ) + h ( b ) − h ( b )][ h ( b ) + h ( b ) − h ( a )]= [ h ( b ) + h ( b ) − h ( a )][ h ( a ) + h ( b ) − h ( b )] . Simple algebra transforms this into the equation h ( b )[ h ( a ) + h ( b )] = h ( b )[ h ( a ) + h ( b )]from which we conclude that h ( a ) + h ( b ) = h ( a ) + h ( b ) . By symmetry, analogous results are true for any two faces of the reference tetra-hedron, so h ( a i ) + h ( b i ) is constant for all i . (cid:3) Corollary 11.19 ([2, p. 299]) . If cevians to the Nagel points concur, then thetetrahedron is isodynamic.
Corollary 11.20 ([2, p. 299]) . If cevians to the Gergonne points concur, then thetetrahedron is isodynamic.
Corollary 11.21.
If cevians to the orthocenters concur, then the tetrahedron isorthocentric.
Corollary 11.22.
If cevians to the (1 /b + 1 /c − /a ) centers concur, then thetetrahedron is harmonic. Corollary 11.23.
For a fixed r (cid:54) = 0 , if cevians to the r -power points concur, thenthe tetrahedron is isodynamic. This generalizes proposition 841 of [2] (which was for the Symmedian point only).
Proof.
The a r centers are the same as the a r b r c r /a r centers, so the result followsby taking h ( x ) = log x . (cid:3) General Results about Hyperbolic Lines
Theorem 12.1.
Let P = (0 , y , z , w ) , P = ( x , , z , w ) , and P = ( x , y , , w ) be three points on faces 1, 2, and 3 of the reference tetrahedron. Then the condi-tion that there is a line through vertex A that meets all three of the lines A i P i , i = 1 , , is z x y = y z x . This line is called a spear line. The spear line meets face A A A at the point ( x y , y y , y z , . (This point is known as the spear trace.)Proof. This result was found by computer but could easily be carried out byhand. Formula 16 gives us the equation of the plane, E , through A and A P .Any spear line must clearly lie in this plane. Formula 18 determines the point, Q , where line A P meets plane E . Then A Q must be the desired spear line.Similarly, we can find the point Q , where line A P meets plane E . Then A Q must also be the desired spear line. Thus the condition is that points A , Q , and Q colline. Formula 3 gives us this condition. Upon simplifying the result, thecomputer came up with z x y = y z x as the algebraic condition. We can thenfind the intersection of the common line A Q Q and the plane A A A to getthe coordinates of the spear trace. (cid:3) tanley Rabinowitz Geometric interpretation.
Let P , P , and P be points on the faces opposite vertices A , A , and A ,respectively, of tetrahedron A A A A . Then there is a line through A thatmeets lines A P , A P and A P if and only if[ P A A ][ P A A ][ P A A ] = [ P A A ][ P A A ][ P A A ]where [ XY Z ] denotes the area of triangle
XY Z .Figure 4a shows a top view of tetrahedron A A A A . Figure 4b then shows apoint taken on faces 1, 2, and 3. The condition is that the product of the areasof the yellow triangles equals the product of the areas of the green triangles. Figure 4.
There is a line through A that meets A i P i , i = 1 , , F ( a, b, c ) = af ( a, b, c ) is a hyperbolic center functionif the cevians from the vertices of a tetrahedron to these centers on the oppositefaces form a hyperbolic group. Proposition 12.2 ([2, p. 11]) . If four given mutually skew lines passing throughthe four vertices of a tetrahedron are such that through each vertex it is possibleto draw a spear line meeting the three lines passing through the remaining threevertices, then the four given lines form a hyperbolic group.
Corollary 12.3 (Hyperbolic Condition) . The condition that F ( a, b, c ) = af ( a, b, c ) be a hyperbolic center function is: F ( b , a , b ) F ( b , b , a ) F ( b , a , b ) = F ( b , b , a ) F ( b , a , b ) F ( b , b , a ) . This follows from the coordinates for the corresponding center on each face foundin section 2 and the preceding Proposition. Note that symmetry conditions implythat we need only find the condition that one spear line exists (instead of all 4).
Corollary 12.4.
In an isosceles tetrahedron, all center functions are hyperbolic.Proof.
When b i = a i , the hyperbolic condition becomes F ( a , a , a ) F ( a , a , a ) F ( a , a , a ) = F ( a , a , a ) F ( a , a , a ) F ( a , a , a ) Arrangement of Central Points on the Faces of a Tetrahedron which is clearly an identity since a center function F ( a, b, c ) is symmetric in b and c . (cid:3) Theorem 12.5. If F ( a, b, c ) is a hyperbolic center function, then so is a r F ( a, b, c ) q .Proof. The hyperbolic condition becomes b r F ( b , a , b ) q b r F ( b , b , a ) q b r F ( b , a , b ) q = b r F ( b , b , a ) q b r F ( b , a , b ) q b r F ( b , b , a ) q which is an immediate consequence of the original condition. (cid:3) Theorem 12.6.
If cevians to corresponding centers on each face of a tetrahedronform a hyperbolic group, then so do cevians to the isotomic conjugates of thosecenters.Proof.
Since in areal coordinates, the isotomic conjugate of a center ( x, y, z ) is(1 /x, /y, /z ), the hyperbolic condition becomes1 F ( b , a , b ) 1 F ( b , b , a ) 1 F ( b , a , b ) = 1 F ( b , b , a ) 1 F ( b , a , b ) 1 F ( b , b , a )which is equivalent to the original condition. (cid:3) Corollary 12.7 ([2, p. 332]) . If cevians to corresponding centers on each face of atetrahedron form a hyperbolic group, then so do cevians to the isogonal conjugatesof those centers.Proof.
This is because in areal coordinates, the isogonal conjugate of a center( x, y, z ) is ( a − /x, b − /y, c − /z ). Thus the result follows from the previous twotheorems. (cid:3) Theorem 12.8.
Suppose the edges of a tetrahedron satisfy the condition h ( a i ) + h ( b i ) = t , i = 1 , , , for some function h ( x ) and some constant, t . Then the ce-vians to the face centers with center function [ h ( b )+ h ( c ) − h ( a )] q form a hyperbolicgroup.Proof. The hyperbolic condition becomes[ h ( a ) + h ( b ) − h ( b )] q [ h ( b ) + h ( a ) − h ( b )][ h ( a ) + h ( b ) − h ( b )] q =[ h ( b ) + h ( a ) − h ( b )] q [ h ( a ) + h ( b ) − h ( b )][ h ( b ) + h ( a ) − h ( b )] q which immediately follows from[ h ( a ) + h ( b ) + h ( a ) − t ] q [ h ( b ) + h ( a ) + h ( a ) − t ] q [ h ( a ) + h ( b ) + h ( a ) − t ] q =[ h ( b ) + h ( a ) + h ( a ) − t ] q [ h ( a ) + h ( b ) + h ( a ) − t ] q [ h ( b ) + h ( a ) + h ( a ) − t ] q which is identically true. (cid:3) Corollary 12.9.
In a circumscriptible tetrahedron, the cevians to the face centerswith center function a r ( b + c − a ) form a hyperbolic group. This includes the Ger-gonne point, the Nagel point, the Mittenpunkt, the X41 point, and their isogonaland isotomic conjugates. Corollary 12.10.
In an orthocentric tetrahedron, the cevians to the face centerswith center function a r ( b + c − a ) form a hyperbolic group. This includes theorthocenter, circumcenter, the crucial point, the X25 point, the X48 point, andtheir isogonal and isotomic conjugates. tanley Rabinowitz Corollary 12.11.
In a harmonic tetrahedron, the cevians to the face centers withcenter function a r (1 /b + 1 /c − /a ) form a hyperbolic group. This includes the X43point, the X102 point, the X117 point, and their isogonal and isotomic conjugates. Theorem 12.12.
The power points are the only triangle centers with the propertythat in any tetrahedron, the cevians to these face centers form a hyperbolic group.Proof.
Let F ( a, b, c ) = af ( a, b, c ) be a hyperbolic center function. Then F mustsatisfy the functional equation F ( b , a , b ) F ( b , b , a ) F ( b , a , b ) = F ( b , b , a ) F ( b , a , b ) F ( b , b , a ) . This equation must be true for all values of a , a , a , b , b , and b . Rewritingthis as(6) F ( b , a , b ) F ( b , b , a ) = F ( b , a , b ) F ( b , b , a ) · F ( b , b , a ) F ( b , a , b )shows that F ( b , a , b ) /F ( b , b , a ) is independent of a , so we may define(7) G ( b , b ) = F ( b , a , b ) F ( b , b , a ) . Substituting this in equation (6) yields G ( b , b ) = G ( b , b ) G ( b , b )for all b , b , and b . If we then let H ( z ) = G ( b , z )we find that G ( x, y ) = H ( y ) H ( x )for all x and y . From the definition of G , (equation 7), we get H ( y ) H ( x ) = F ( x, z, y ) F ( y, x, z )for all x , y , and z . Since F is homogeneous, this implies that H ( ty ) H ( tx ) = H ( y ) H ( x )for all x , y , and t . Letting K ( y ) = H ( y ) /H ( x ) gives K ( ty ) K ( y ) = H ( ty ) H ( y ) = H ( tx ) H ( x ) = K ( tx ) . Letting x = 1 shows that K ( ty ) = K ( t ) K ( y )for all t and y , so by the Power Lemma we have K ( x ) = x r for some constant r .Denoting H (1) by c , we have H ( x ) = cK ( x ) = cx r . Arrangement of Central Points on the Faces of a Tetrahedron
The coordinates for the center are F ( x, y, z ) : F ( y, z, x ) : F ( z, x, y ) = F ( x, y, z ) F ( x, y, z ) : F ( y, z, x ) F ( x, y, z ) : F ( z, x, y ) F ( x, y, z )= 1 : H ( x ) H ( y ) : H ( x ) H ( z )= 1 H ( x ) : 1 H ( y ) : 1 H ( z )= x r : y r : z r and thus the center is a power point. (cid:3) Theorem 12.13.
In an isodynamic tetrahedron, the center function a r g ( b, c ) ishyperbolic for any symmetric homogeneous function g .Proof. The hyperbolic condition becomes b r g ( a , b ) b r g ( b , a ) b r g ( a , b ) = b r g ( b , a ) b r g ( a , b ) b r g ( b , a ) . Since the tetrahedron is isodynamic, this is equivalent to b r g ( a , t/a ) b r g ( t/a , a ) b r g ( a , t/a ) = b r g ( t/a , a ) b r g ( a , t/a ) b r g ( t/a , a ) . Using the fact that g is homogeneous, we see that this is equivalent to b r g ( a a , t ) b r g ( t, a a ) b r g ( a a , t ) = b r g ( t, a a ) b r g ( a a , t ) b r g ( t, a a )which is easily seen to be an identity since g is symmetric. (cid:3) Corollary 12.14.
In an isodynamic tetrahedron, the cevians to the face centerswith center function a r ( b q + c q ) ± form a hyperbolic group. This includes theSpieker center, the Brocard midpoint, the X37, X38, and X42 points, and theirisogonal and isotomic conjugates. Conjecture 12.15.
The center functions a r g ( b, c ) , where g is a symmetric ho-mogeneous function, and r is arbitrary, are the only hyperbolic center functionsfor an isodynamic tetrahedron. General Planarity Results
Theorem 13.1.
No triangle center has the property that in any isosceles tetra-hedron, the corresponding face centers are coplanar.Proof.
The algebraic condition for planarity (found by computer) is one of(8) F ( a , a , a ) + F ( a , a , a ) = F ( a , a , a )(9) F ( a , a , a ) + F ( a , a , a ) = F ( a , a , a )(10) F ( a , a , a ) + F ( a , a , a ) = F ( a , a , a )(11) F ( a , a , a ) + F ( a , a , a ) + F ( a , a , a ) = 0 . If condition (11) holds, then the center ( α, β, γ ) in areal coordinates would satisfy α + β + γ = 0, a contradiction.If condition (8) holds, then since it must be valid for all variables a , a , a , wesee that conditions (9) and (10) would also hold. Adding these three equationsyields condition (11) which we have already seen yields a contradiction. tanley Rabinowitz The same contradiction is reached if we assume (9) or (10) holds. (cid:3)
Corollary 13.2.
No triangle center has the property that in any tetrahedron, thecorresponding face centers are coplanar.
The following condition was found by our computer program.
Theorem 13.3.
The Feuerbach points on the faces of a tetrahedron are coplanarif the edges of the tetrahedron satisfy the following condition: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a + b a + b a + b a b a b a b (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . In particular, the Feuerbach points are coplanar for circumscriptible, isodynamic,and harmonic tetrahedra. The Feuerbach points would also be coplanar for tetra-hedra that satisfy other simple relationships, for example, ones in which a i b i + a i + b i were constant.14. General Results about Concurrent Normals
Theorem 14.1.
Let P and P be any two points on faces 1 and 2 of tetrahedron A A A A , respectively. If normals to the faces at P and P concur, then ( P A ) + ( P A ) = ( P A ) + ( P A ) . Proof.
From right triangles
P P A , P P A , P P A , and P P A , we have( A P ) + ( P P ) = ( A P ) = ( A P ) + ( P P ) and ( A P ) + ( P P ) = ( A P ) = ( A P ) + ( P P ) . Subtracting one equation from the other gives us the desired result. (cid:3)
The converse is also true.
Theorem 14.2 (Tabov [14]) . Let P and P be any two points on faces 1 and 2of tetrahedron A A A A , respectively. If ( P A ) + ( P A ) = ( P A ) + ( P A ) , then normals to the faces at P and P concur. Miscellaneous Conjectures
The following conjectures are backed up by the data, but I don’t have formalproofs.
Conjecture 15.1.
If the central tetrahedron is isosceles, then the reference tetra-hedron is isosceles.
Conjecture 15.2.
If the central tetrahedron is regular, then the reference tetra-hedron is regular.
Conjecture 15.3.
If the central tetrahedron is similar to the reference tetrahe-dron, then the center must be the centroid.
Conjecture 15.4.
If the cevians to the corresponding face centers have the samelength, then the tetrahedron is isosceles. Arrangement of Central Points on the Faces of a Tetrahedron
Appendix A. Center Functions and Trilinear Coordinates
In this appendix, we give the basic information about center functions and trilinearcoordinates that the reader needs to know.Before giving the definition of a triangle center, let us review the concept oftrilinear coordinates. If
ABC is a fixed reference triangle in the plane (with sidesof lengths a , b , and c ), and if P is an arbitrary point in the plane, then the trilinearcoordinates of P are ( α, β, γ ) where α , β , γ are the signed distances from P tothe sides BC , CA , AB , respectively. The three coordinates satisfy the condition(12) aα + bβ + cγ = 2 K where K is the area of (cid:52) ABC . If α , β , and γ are any three real numbers (with aα + bβ + cγ (cid:54) = 0), then there is a unique point P in the plane whose trilinearcoordinates are proportional to α : β : γ . Thus ( α, β, γ ) may be considered to bethe trilinear coordinates of P even if condition (12) is not satisfied. If condition(1) is satisfied, then we refer to the coordinates as exact trilinear coordinates.A center function is a nonzero function f ( a, b, c ) that is homogeneous in a , b , and c and symmetric in b and c . In other words, a center function must satisfy thefollowing two conditions for all a , b , c , t , and some integer r :(C1) f ( ta, tb, tc ) = t r f ( a, b, c )(C2) f ( a, c, b ) = f ( a, b, c )A center is an ordered triple α : β : γ given by α = f ( a, b, c ) , β = f ( b, c, a ) , γ = f ( c, a, b )for some center function f ( a, b, c ). See [10] for more details. If f is a polynomial,then the center is called a polynomial center.If P = ( α, β, γ ), then the point ( α − , β − , γ − ) is denoted by P − and is calledthe isogonal conjugate of P . For the geometric meaning of isogonal conjugates,consult [1].More information about center functions and trilinear coordinates can be foundin [10], [11], and [12]. Appendix B. Areal Coordinates
In this appendix, we give the basic information about areal coordinates that thereader needs to know.Let
ABC be a fixed reference triangle in the plane. If P is an arbitrary point inthe plane of (cid:52) ABC , then the areal coordinates of P are given by ( x, y, z ) where x = [ P BC ] / [ ABC ] ,y = [ P CA ] / [ ABC ] ,z = [ P AB ] / [ ABC ] . The three areal coordinates satisfy the condition(13) x + y + z = 1 . Areal coordinates are also known as barycentric coordinates. tanley Rabinowitz If x , y , and z are any three real numbers (with x + y + z (cid:54) = 0), then there is aunique point P in the plane whose areal coordinates are proportional to x : y : z .Thus ( x, y, z ) may be considered to be the areal coordinates of P even if condition(13) is not satisfied. The three coordinates are proportional to the areas formedby P and the sides of the reference triangle ABC . If condition (13) is satisfied,then we refer to the coordinates as exact areal coordinates.Areal coordinates, ( x, y, z ), can be transformed to trilinear coordinates, ( α, β, γ ),and vice versa, by the following formulas: α = x/a, β = y/b, γ = z/c where a , b , and c are the lengths of the sides of the reference triangle.If P = ( x, y, z ), then the point ( x − , y − , z − ) is denoted by P T and is calledthe isotomic conjugate of P . For the geometric meaning of isotomic conjugates,consult [1]. In terms of trilinear coordinates, the isotomic conjugate of ( α, β, γ ) is( α − /a , β − /b , γ − /c ).More information about areal coordinates can be found in [3] and [9]. Appendix C. Tetrahedral Coordinates
In this appendix, we give the basic information about tetrahedral coordinates thatthe reader needs to know.The 3-dimensional analog of areal coordinates are tetrahedral coordinates. Let[ T ] denote the volume of a tetrahedron T . Let A A A A be a fixed referencetetrahedron in space with edges of lengths A A = a , A A = a , A A = a , A A = b , A A = b , and A A = b , so that the edges of lengths a i and b i areopposite each other, i = 1 , , P is an arbitrary point in space, then the tetrahedral coordinates of P are givenby ( x , x , x , x ) where x = [ P A A A ] /V,x = [ P A A A ] /V,x = [ P A A A ] /V,x = [ P A A A ] /V, where V = [ A A A A ] is the volume of the reference tetrahedron. The fourtetrahedral coordinates satisfy the condition(14) x + x + x + x = 1 . If x i , i = 1 , , , P in space whose tetrahedral coordinates are proportional to x : x : x : x . Thus ( x , x , x , x ) may be considered to be the tetrahedralcoordinates of P even if condition (14) is not satisfied. The four coordinates areproportional to the volumes formed by P and the faces of the reference tetrahedron ABCD . If condition (14) is satisfied, then we refer to the coordinates as exacttetrahedral coordinates.For more information about tetrahedral coordinates, see [4] and [7]. Arrangement of Central Points on the Faces of a Tetrahedron
Appendix D. Formulas
In this appendix, we collect together formulas about tetrahedral coordinates thatare needed in this paper. All points are given using exact tetrahedral coordinates.
POINTS
Formula 1: Coordinates of a point ([7, p. 65]):( x, y, z, w )where x + y + z + w = 1. The coordinates are proportional to the volumesof the tetrahedra formed by the point and the faces of the reference tetrahe-dron. If a more symmetric notation is needed, we will use the alternate form( X , X , X , X ).Formula 2: Condition for 4 points ( x i , y i , z i , w i ), i = 1 , , , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z w x y z w x y z w x y z w (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0or equivalently, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − x y − y z − z x − x y − y z − z x − x y − y z − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Formula 3: Condition for 3 points ( x i , y i , z i , w i ), i = 1 , , x − x x − x = y − y y − y = z − z z − z = w − w w − w . Formula 4: Square of distance between points ( x , y , z , w ) and ( x , y , z , w )([7, p. 66]):( y − y )( z − z ) a + ( z − z )( x − x ) a + ( x − x )( y − y ) a +( x − x )( w − w ) b + ( y − y )( w − w ) b + ( z − z )( w − w ) b . This can be written in the symmetrical form: − (cid:88) i,j d i,j ( X i − X (cid:48) i )( X j − X (cid:48) j )representing the square of the distance from ( X , X , X , X ) to ( X (cid:48) , X (cid:48) , X (cid:48) , X (cid:48) ).In this symmetrical notation, the summation symbol (cid:88) i,j means (cid:88) i,j =1 i Formula 5: General equation of a straight line ([7, p. 67]): x − x K = y − y L = z − z M = w − w N where K + L + M + N = 0. The line passes through the point ( x , y , z , w ).The quadruple, ( K, L, M, N ), represents the direction of the line and is calledthe direction vector. Two lines are parallel if and only if they have the samedirection vector (or a multiple thereof). Note that some of K, L, M, N may be0 because the condition ( x − x ) /K = ( y − y ) /L is really an abbreviation for( x − x ) L = ( y − y ) K which does not involve any possible divisions by 0.Formula 6: Parametric equation of a straight line:( x + Kt, y + Lt, z + M t, w + N t ) . This formula yields all the points on the line through ( x , y , z , w ) with directionvector ( K, L, M, N ) as t varies through the real numbers.Formula 7: Equation of line through ( x , y , z , w ) and ( x , y , z , w ): x − x x − x = y − y y − y = z − z z − z = w − w w − w . Formula 8: Coordinates of point that divides the line joining points ( x , y , z , w )and ( x , y , z , w ) in the ratio µ : λ ([7, p. 65]):( λx + µx λ + µ , λy + µy λ + µ , λz + µz λ + µ , λw + µw λ + µ ) . Formula 9: Condition for two lines ( x − x i ) /K i = ( y − y i ) /L i = ( z − z i ) /M i =( w − w i ) /N i , i = 1 , K : L : M : N = K : L : M : N or equivalently K K = L L = M M = N N . Formula 10: Condition for two lines ( x − x i ) /K i = ( y − y i ) /L i = ( z − z i ) /M i =( w − w i ) /N i , i = 1 , a ( L M + L M ) + a ( K M + K M ) + a ( K L + K L )+ b ( K N + K N ) + b ( L N + L N ) + b ( M N + M N ) = 0 . Formula 11: Condition for two lines ( x − x i ) /K i = ( y − y i ) /L i = ( z − z i ) /M i =( w − w i ) /N i , i = 1 , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z w K L M N x y z w K L M N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0or equivalently (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − x y − y z − z K L M K L M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Arrangement of Central Points on the Faces of a Tetrahedron Formula 12: Direction cosines of the line ( x − x ) /K = ( y − y ) /L = ( z − z ) /M =( w − w ) /N : KF V σ , LF V σ , M F V σ , N F V σ , where V is the volume of the reference tetrahedron, F i is the area of face i (theface opposite vertex A i ), and σ is determined from a LM + a M K + a KL + b KN + b LN + b M N = − σ . The direction cosines are the cosines of the angles that the line makes with thenormals to the four faces of the reference tetrahedron. They are proportional tothe direction vector. PLANES Formula 13: General equation of a plane ([7, p. 69]): Ax + By + Cz + Dw = 0where not all coefficients are 0. The coefficients A , B , C , D , are proportional tothe directed distances from the plane to the vertices of the reference tetrahedron.Formula 14: Equation of the plane through 3 points, ( x i , y i , z i , w i ), i = 1 , , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z wx y z w x y z w x y z w (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Formula 15: Condition for planes A x + B y + C z + D w = 0 and A x + B y + C z + D w = 0 to be parallel ([7, p. 70]): A − D A − D = B − D B − D = C − D C − D . Formula 16: Equation of the plane through the point ( x , y , z , w ) and the line( x − x ) /K = ( y − y ) /L = ( z − z ) /M = ( w − w ) /N : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z wx y z w x y z w K L M N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Formula 17: Equation of plane through the line ( x − x ) /K = ( y − y ) /L = ( z − z ) /M = ( w − w ) /N and parallel to the line with direction vector ( K , L , M , N ): (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z wx y z w K L M N K L M N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Formula 18: Point of intersection of the line ( x − x ) /K = ( y − y ) /L = ( z − z ) /M = ( w − w ) /N and the plane Ax + By + Cz + Dw = 0:( x − rK, y − rL, z − rM, w − rN )where r = Ax + By + Cz + Dw AK + BL + CM + DN . tanley Rabinowitz Formula 19: Condition for the line ( x − x ) /K = ( y − y ) /L = ( z − z ) /M =( w − w ) /N to be parallel to the plane Ax + By + Cz + Dw = 0: AK + BL + CM + DN = 0 . Appendix E. Tetrahedron Centers In this appendix, we collect together information about various “centers” asso-ciated with a tetrahedron. We give the tetrahedral coordinates for the morewell-known such centers and explain why some of these centers were not includedin our study. CENTROID The centroid, G , of a tetrahedron ([2, p. 54]) is the center of gravity of unit massesplaced on the vertices. Thus it has barycentric coordinates of G = (1 , , , . The exact tetrahedral coordinates are (1 / , / , / , / INCENTER The incenter, I , of a tetrahedron ([2, p. 76]) is the center of the sphere inscribed inthe tetrahedron (touching each of the faces internally). If we let r be the inradiusof the tetrahedron, then the volume of IA A A is rF . Similarly for the otherthree volumes formed by I and the faces of the tetrahedron. These four volumessum to V , the volume of the tetrahedron. Thus r = 3 V / ( F + F + F + F ). Theincenter is equidistant from each face of the tetrahedron. Thus, the tetrahedralcoordinates are I = ( F , F , F , F ) . To convert to exact tetrahedral coordinates, each coordinate should be divided by F , the surface area of the tetrahedron. CIRCUMCENTER The circumcenter, O , of a tetrahedron ([2, p. 56]) is the center of the circum-scribed sphere. If ( O x , O y , O z , O w ) are the coordinates for the circumcenter of ourreference tetrahedron, and if R denotes the circumradius, then we can set up 4equations in O x , O y , O z , O w , and R : d ( O, A ) = R d ( O, A ) = R d ( O, A ) = R d ( O, A ) = R where d ( P , P ) denotes the distance between points P and P . This distanceformula is given by formula 4. Upon subtracting equation 2 from equation 1,equation 3 from equation 2, and equation 4 from equation 3, we get 3 linearequations in O x , O y , O z , and O w . Combining these with O x + O y + O z + O w = 1gives us 4 linear equations in 4 unknowns. These are straightforward to solve.The value of O x found is proportional to O x = a b ( b + b − a ) + a b ( b + a − b ) + a b ( a + b − b ) − a b b . Arrangement of Central Points on the Faces of a Tetrahedron The values of O y , O z , and O w are similar and can be obtained from O x by themappings given in display (5). Specifically, O y = a b ( a + b − b ) + a b ( b + b − a ) + a b ( b + a − b ) − b a b ,O z = a b ( b + a − b ) + a b ( a + b − b ) + a b ( b + b − a ) − b b a ,O w = a b ( a + a − a ) + a b ( a + a − a ) + a b ( a + a − a ) − a a a . MONGE POINT The Monge point, M , of a tetrahedron ([2, p. 76]) is the common intersectionpoint of the six planes through the midpoints of the edges of the tetrahedron andperpendicular to the opposite edges. The Monge point is the symmetric of thecircumcenter with respect to the centroid ([2, p. 77]) and thus its coordinates canbe found from them: M = 2 G − O. Definition 7. The three points G , O , and M lie on a straight line called the Eulerline of the tetrahedron ([2, p. 77]). EULER POINT The Euler point, E , corresponds to the nine-point center in the plane. It isfrequently called the twelve point center ([2, p. 289]) because it is the center of asphere that passes through 12 notable points in the tetrahedron. The Euler pointlies on the Euler line of the tetrahedron and divides the segment M O in the ratio1 : 2 and it divides the segment GM in the ratio 1 : 2. Thus its coordinates canbe found from the coordinates of those points: E = (2 G + M ) / . ORTHOCENTER The altitudes of a tetrahedron do not normally intersect. They intersect if andonly if the tetrahedron is orthocentric and in that case, the intersection point (theorthocenter) coincides with the Monge point of the tetrahedron ([2, p. 71]). Wedo not include the orthocenter of a tetrahedron as a distinguished point in ourstudy since it is not present in all tetrahedra. SYMMEDIAN POINT (Lemoine Point) In a tetrahedron, the cevians to the symmedian points on the opposite faces donot normally intersect. They intersect if and only if the tetrahedron is isody-namic ([2, p. 315]). The point of intersection is called the symmedian point ofthe tetrahedron. We do not include the symmedian point of a tetrahedron as adistinguished point in our study since it is not present in all tetrahedra. GERGONNE POINT In a tetrahedron, the cevians to the Gergonne points on the opposite faces donot normally intersect. They intersect if and only if the tetrahedron is circum-scriptible ([2, p. 299]). The point of intersection is called the Gergonne point ofthe tetrahedron. We do not include the Gergonne point of a tetrahedron as adistinguished point in our study since it is not present in all tetrahedra. NAGEL POINT In a tetrahedron, the cevians to the Nagel points on the opposite faces do notnormally intersect. They intersect if and only if the tetrahedron is circumscriptible tanley Rabinowitz ([2, p. 299]). The point of intersection is called the Nagel point of the tetrahedron.We do not include the Nagel point of a tetrahedron as a distinguished point inour study since it is not present in all tetrahedra. FERMAT POINT In a tetrahedron, the cevians to the points of tangency of the opposite faceswith the insphere do not normally intersect. If they intersect, the tetrahedron iscalled an isogonic tetrahedron ([2, p. 328]). The point of intersection is called theFermat point of the tetrahedron. In this case, the insphere touches the faces at theFermat point of each face. We do not include the Fermat point of a tetrahedronas a distinguished point in our study since it is not present in all tetrahedra. References [1] Nathan Altshiller-Court, College Geometry , 2nd edition. Barnes & Noble, Inc. New York:1952.[2] Nathan Altshiller-Court, Modern Pure Solid Geometry , 2nd edition. Chelsea PublishingCompany. The Bronx, NY: 1964.[3] Francisco Javier Garc`ıa Capit`an, Barycentric Coordinates , International Journal ofComputer Discovered Mathematics, 0(2015)32–48. [4] George S. Carr, Formulas and Theorems in Pure Mathematics , 2nd edition. Chelsea Pub-lishing Company. The Bronx, NY: 1970.[5] Julian Lowell Coolidge, A Treatise on the Circle and the Sphere . Chelsea Publishing Com-pany. The Bronx, NY: 1971.[6] Heinrich D¨orrie, 100 Great Problems of Elementary Mathematics: Their History and Solu-tion . Dover Publications, Inc. New York: 1965.[7] Percival Frost, Solid Geometry , vol. 1. Macmillan and Co. London: 1875.[8] Peter Gilbert, personal correspondence.[9] Sava Grozdev and Deko Dekov, Barycentric Coordinates: Formula Sheet , InternationalJournal of Computer Discovered Mathematics, 1(2016)75–82. [10] Clark Kimberling, Central Points and Central Lines in the Plane of a Triangle , Mathemat-ics Magazine, 67(1994)163–187.[11] Clark Kimberling, Triangle Centers and Central Triangles , Congressus Numerantium,129(1998)1–295.[12] Clark Kimberling, Encyclopedia of Triangle Centers , 2020, http://faculty.evansville.edu/ck6/encyclopedia/ETC.html [13] T. R. Lee, Solution to Problem 18523 (Proposed by A. M. Nesbitt), Mathematical Questionsand Solutions from the Educational Times (series 3), 5(1918)46.[14] Jordan Tabov, personal correspondence.[15] A. Zaslavsky, Problem 3 , 15th Summer Conference of the International Mathematical Tour-nament of Towns. 2003., 15th Summer Conference of the International Mathematical Tour-nament of Towns. 2003.