From:bdsm@poincare.WPI.EDU (Billy Don McConnell)Newsgroups:sci.math.researchSubject:Solve the Perfect Tetrahedron Equation!Date:11 Feb 93 02:26:21 GMTOrganization:Worcester Polytechnic Institute

Hello, everyone... A short time ago, I posted "3 'Simple' Quartics" to sci.math, seeking some _nice_ Solutions to a triple of fourth-degree Polynomials. I have received a few responses stating interest in the problem (and one good reference), but alas no Solution. In this post I intend to explain, in some detail, the Theory behind these Equations, in the hopes that a complete understanding of the Problem will generate a good approach to the Solution. I welcome any and all comments or suggestions concerning this topic (please respond by e-mail). I realize that this article may be excessively long for a post, but I have found myself unable to produce an "abstract" that adequately expresses the intriguing qualities of the subject matter. It is my hope that the content of this article will justify its use of bandwidth. Thanks in advance for your time. Don Mc Connell bdsm@fermat.wpi.edu HEDRONOMETRY: The Search for Solutions to the Perfect Tetrahedron Equation SUMMARY. "Hedronometry" is an extension of Trigonometry to 3-Dimensions, focussing on the relationships between the Angles and the Faces of a Tetrahedron. Tetrahedra have more degrees of freedom than Triangles, and consequently are not uniquely determined by, say, the Areas of their Faces in the way that Triangles are uniquely determined by the Lengths of their Edges. Yet there do exist a number of Hedronometric analogues to Trigonometric formulae -- the "Neo-Pythagorean Theorem" and two "Laws of Cosines" to name a few -- that provide vaguely familiar relationships between the Faces and Angles. These formulae are discussed first. Restricting our focus to "Perfect Tetrahedra" -- Tetrahedra for opposite Edges are Orthogonal (as Vectors) -- we find ourselves with a class of figures that _are_ uniquely determined by their Face-Areas (up to permutation of the Faces, of course). The "Perfect Tetrahedon Equation" -- "P.T.E." -- (of which the "3 'Simple' Quartics" were versions) describes the closer ties between Face-Areas and Dihedral Angles in Perfect Tetrahedra, providing the key to determining the overall shape of a Tetra- hedron whose Face-Areas are given. However, even though the P.T.E. represents a _nice_ relationship, its Solutions appear to be anything BUT nice. I am convinced, nevertheless, that a inspired simplification can be performed on the results I have been able to obtain, yielding forms of the Solutions that reflect the beauty of the Problem. The challenge I set forth, then, is to find that simplification: FIND THE "NICE" FORMS OF SOLUTIONS TO THE PERFECT TETRAHEDRON EQUATION. Note. "Hedronometry" is by no means a new subject. I understand that some of the ideas presented here were developed over a century ago, as aspect of "Tetrahedrometry". (I forget the name of the Mathematician who published two volumes on the subject). The 2 results I mentioned (Pyth. Thm. & Law of Cosines) can be found in fairly recent Journal publications (e-mail for references). But I have no idea if the problem I pose or the answers I seek already exist in the literature. Alright, now to begin. NOTATION: Let the Vertices of a Tetrahedron be labelled Q, A, B, and C. The Faces (and their Areas) opposite these Vertices we denote R, X, Y, and Z, respectively. An Edge (thought of as a Vector) through points U and V we denote UV; thus the Edges are QA, QB, QC, AB, BC, and CA. We use "| |" for the Length of an Edge. Dihedral Angles we denote like the Edges, hopefully without confusion. Now although the Face-Areas X, Y, Z, and R do not define a Tetrahedron we do have the following: THE NEO-PYTHAGOREAN THEOREM: If Tetrahedron QABC has a "Right Corner" at Q (that is, if Edges QA, QB, and QC are mutually perpendicular), then... 2 2 2 2 X + Y + Z = R (the Sum of the Squares of the Areas of the Leg-Faces is equal to the Square of the Area of the Hypotenuse-Face). PROOF. Exercise. It's pretty easy, requiring nothing more sophisticated than Heron's Formula for Triangle Area. COMMENT 1: Sadly, the Converse of this Theorem is not true... Having the "Sum of Squares" relation does not guarantee the existence of a Right Corner. Thus, there are "Pythagorean Tetrahedra" that are not "Right-Corner Tetrahedra". COMMENT 2: Happily, this idea extends to 'n' Dimensions as a Theorem for an n-dimensional "Right-Simplex", where n mutually perpendicular Edges meet at a Vertex. But this is a story for another day. Just as the (ordinary) Pythagorean Theorem can be viewed as a special case of the Law of Cosines, so the Neo-Pythagorean Theorem is a special case of... THE FIRST LAW OF COSINES ("LOC.1"): 2 2 2 2 R = X + Y + Z - 2 Y Z cos QA - 2 Z X cos QB - 2 X Y cos QC PROOF. Exercise. This, too, is easy, though it involves a great deal of symbol manipulation and a working knowledge of the formulae from Spherical Trigonometry (in particular, a Law of Cosines!) in order to relate Dihedral Angles to "Face-Angles" like <AQB. COMMENT: When Q is a Right Corner, QA, QB, and QC are all Right Angles, so that LOC.1 reduces to the Neo-Pythagorean Theorem. By Symmetry, we can find similar expressions for X^2, Y^2, and Z^2 in terms of other Faces and "opposite" Dihedral Angles. The four expressions combine to give... THE SECOND LAW OF COSINES ("LOC.2"): 2 2 2 2 2 R + X - 2 X R cos BC = H = Y + Z - 2 Y Z cos QA 2 2 2 2 2 R + Y - 2 Y R cos AC = J = Z + X - 2 Z X cos QB 2 2 2 2 2 R + Z - 2 Z R cos AB = K = X + Y - 2 X Y cos OC COMMENT 1: Notice that Faces R & X "bound" Angle BC, and Faces Y & Z "bound" Angle QA; moreover, BC and QA are "opposite" one another in the Tetrahedron. Similar observations hold for the other Equations. COMMENT 2: There is a geometric interpretation of the numbers H, J, and K. They are the Areas of "Pseudo-Faces", and can be calculated by: <QA>x<BC> <QB>x<CA> <QC>x<AB> H = --------- J = --------- K = --------- 2 2 2 Where 'x' is the Cross Product. This fact isn't very important now, but it will aid discussion later. Also, using H^2, J^2, and K^2 as they are used in LOC.2 will help with simplification. COROLLARY to LOC.2: 2 2 2 2 2 2 2 X + Y + Z + R = H + J + K PROOF. Simply set the sum of the second column from the statment of LOC.2 equal to the sum of the third column, then simplify using the statement of LOC.1. Okay, so now we know that all information about a Tetrahedron can be gleaned from six Numbers: X, Y, Z, R, H, J, and K, with the Corollary making one of these Numbers (you choose) unnecessary. These Areas and "Pseudo-Areas" give us Dihedral Angles, by LOC.2; the Dihedral Angles give us Face-Angles, by Spherical Trigonometry; and Face Angles together with Face Areas give us lengths of Edges. Unfortunately, all 6 of these Area Numbers are virtually independent of one another (but subject to a 'Triangle Inequality' of sorts). However, with "Perfect Tetrahedra" (defined below), we get a situation in which X, Y, Z, and R suffice to determine all information about the Tetra- hedron. DEFINITION: A _Perfect_Tetrahedron_ QABC is a Tetrahedron whose opposing Edges are orthogonal (when thought of as Vectors). That is, if <UV> represents the Vector going through Points U and V and '.' represents the Euclidean Dot-Product, then <QA>.<BC> = <QB>.<CA> = <QC>.<AB> = 0 NOTE: In a Perfect Tetrahedron, the Pseudo-Faces can be computed as |QA||BC| |QB||CA| |QC||AB| H = -------- J = -------- K = -------- 2 2 2 The following Lemma and its Corollary will be useful. LEMMA: Let UQV be the Angle between Edges (Vectors) QU and QV. In a Perfect Tetrahedron QABC, cos BQC cos CQA cos AQB --------- = --------- = --------- |QA| |QB| |QC| PROOF: Vector methods handle this nicely, although High School Trig can do the job as well. COROLLARY to LEMMA: The Face Angles at a Corner of a Perfect Tetrahedron are either all Acute, all Right, or all Obtuse; further, a Perfect Tetra- hedron can have _at most_ one "Right" Corner or "Obtuse" Corner. By this Corollary, we may assume that Q is not a Right Corner. Therefore, we may use the Lemma to write: |QA| = k cos BQC, |QB| = k cos CQA, |QC| = k cos AQB, for some k. Using these facts, and the formulae for the Areas of X, Y, and Z, |QB| |QC| sin BQC |QC| |QA| sin CQA |QA| |QB| sin AQB X = -----------------, Y = -----------------, Z = -----------------, 2 2 2 we can get X cos CQA sin BQC cos QB + cos QC cos QA --- = --------------- = (by Spherical Trig) ---------------------- Y cos BQC sin CQA cos QA + cos QB cos QC Y cos QC + cos QA cos QB ( Z cos QA + cos QB cos QC ) --- = ---------------------- (and --- = ---------------------- ) Z cos QB + cos QC cos QA ( X cos QC + cos QA cos QB ) Now, using the formulae for X/Y and Y/Z, as well as the Formula in LOC.1, we can solve for cos QA, cos QB, and cos QC in terms of X, Y, Z, and R. Thus, the Face Areas completely determine the Tetrahedron. So, the only thing to do is solve this system of Equations... ** THIS IS WHERE THE "3 'SIMPLE' QUARTICS COME IN ** NOTATION: Let A = cos QA, B = cos QB, and C = cos QC. Then the System we are trying to solve looks like this... X B + C A Y C + A B ( Z A + B C ) --- = ------- --- = ------- ( --- = ------- ) Y A + B C Z B + C A ( X C + A B ) 2 2 2 2 R = X + Y + Z - 2 X Y C - 2 Z X B - 2 Y Z C Maple V (Release 2) assimilates the basic idea here quite rapidly, giving the following "Solutions" (though in a slightly different form): 2 2 2 2 ( Y - X C ) ( X + Y + Z - R - 2 X Y C ) A = ---------------------------------------, 2 2 2 Z ( X + Y - 2 X Y C ) 2 2 2 2 ( X - Y C ) ( X + Y + Z - R - 2 X Y C ) B = ----------------------------------------, 2 2 2 Z ( X + Y - 2 X Y C ) ---- 3 3 4 2 2 3 2 2 2 2 ---- | 12 X Y C - 8 X Y C ( 2 X + 2 Y - Z - R ) | | | | | | / 2 2 2 2 2 2 2 2 \ | | 2 | ( 7 X + 7 Y - Z - R ) ( X + Y - Z - R ) | | | + X Y C | | | | | 2 2 2 2 | | | \ -4 ( X Y - Z R ) / | C = RootOf| | | | | 2 2 / 2 2 2 2 2 2 2 2 2 \ | | - C ( X + Y ) | 4 ( X Y + Z R ) - ( X + Y + Z + R ) | | | \ / | | | | | | 2 2 2 2 2 2 2 2 | | - X Y ( X + Y + Z - R ) ( X + Y - Z + R ) = 0 | ---- ---- Of course, C is the hard part, and here our investigations come to a screeching halt. The Quartic Polynomial of which C is a Root is a form of the Perfect Tetrahedron Equation. Fortunately, the P.T.E. is only Quartic, so that it is solvable by Radicals. _Un_fortunately, the Solutions look absolutely hideous, even when small values are substituted in for X, Y, Z, and R. In an attempt to simplify the Solutions, one might try to simplify the Equation: One obtains a slightly less complicated version of the P.T.E. by subsituting 2 2 2 X + Y - K C = ------------ ( from the "Definition of K" in LOC.2 ) to get 2 X Y 8 6 2 2 2 2 3 K - 4 K ( X + Y + Z + R ) 4 / 2 2 2 2 2 2 2 2 2 2 2 2 2 \ - K | ( X + Y + Z + R ) - 2 ( X - R ) ( Y - Z ) + 2 ( Y - R ) ( Z - X ) | \ / 2 2 2 2 2 2 - ( Z - R ) ( X - Y ) = 0 Observe that this is a _Quartic_ in K^2 (and it has no "linear" term...is this good?) This is, essentially, one of the "3 'Simple' Quartics" from my previous post (although in that post I had replaced X^2 with X, Y^2 with Y, ... , and I also made some other variable assignments in an attempt to simplify things). By cycling Variables K -> J -> H -> K , Z -> Y -> X -> Z , R -> R we can transform the K^2-Quartic to a J^2-Quartic and the H^2-Quartic, completing the list of "3 'Simple' Quartics". I mention these other two Quartics because, while the Solutions to any _one_ of them look hopelessly complex, we can observe that, since the Solutions must satisfy the simple relation shown in the LOC.2 Corollary, there must be some method in the madness. To plead this case a little more convincingly, let me remind the Reader that any Quartic Equation 4 3 2 a q + b q + c q + d q + e = 0, has Roots of the form b q = - --- + (**STUFF**) 4a Consequently, the Solutions to the three P.T.E.'s have the form 2 2 2 2 2 X + Y + Z + R H = -------------- + (**STUFF.1**) 3 2 2 2 2 2 X + Y + Z + R J = -------------- + (**STUFF.2**) 3 2 2 2 2 2 X + Y + Z + R K = -------------- + (**STUFF.3**) 3 This gives that 2 2 2 2 2 2 2 H + J + K = X + Y + Z + R + **STUFF.1** + **STUFF.2** + **STUFF.3** In accordance with the Corollary to LOC.2, we can conclude that **STUFF.1** + **STUFF.2** + **STUFF.3** = 0 Now, the **STUFF**'s are pretty complicated expressions, involving complex numbers and nested Radicals. But as complicated as these thing are, they MUST cancel each other out in the end, and this is encouraging. Somehow, some way, these **STUFF**'s can be artfully manipulated into expressions that "obviously" cancel. I have searched for over 5 years for this method of artful manipulation, but to no avail. Maple hasn't been much help in the general case: I have waited for over an hour at a sitting for Maple to churn out even the _ugly_ Solutions to the P.T.E., but the task seems to require a great deal more time than I can spend waiting. Let's take a specific example to show the extent of the problem for even simple numerical values: Assign: X = 1 , Y = 2 , Z = 5 , R = 9. COMMENT: The values are chosen to avoid some special circumstances, and in the hopes that they -as prime powers- might be 'recognizable' inside a complicated expression. These give the following P.T.E. system: 8 6 4 3 H - 444 H + 16353 H - 2822400 = 0 8 6 4 3 J - 444 J + 15345 J - 3415104 = 0 8 6 4 3 K - 444 K + 5265 K - 28224 = 0 Maple solves these Equations readily. In evaluated form, we get... 2 -8 H = 86.30112783 + .2120961320 10 i 2 -8 J = 22.08279673 - .1249277220 10 i 2 -6 K = 2.616075639 - .2103278201 10 i (Of course, there are really four choices for H^2, four for J^2, and four for K^2, since these quantities are Roots of a Quartic. However, one can easily pick out a particular set H^2, J^2, K^2 that satisfies the Corollary to LOC.2; in fact the twelve total Solutions can be partitioned into four such sets.) In complicated mess form (brace yourselves)... 3/4 ( 1/2 2/3 1/3 )1/2 2 %2 + ( 3700 %3 - %2 ( %1 - 3684 %1 + 2047089 ) ) H = 37 + --------------------------------------------------------------- 1/6 1/4 2 %1 %2 3/4 ( 1/2 2/3 1/3 )1/2 2 - %5 + ( -587708 %6 - %5 ( %4 - 4132 %4 + 1389201 ) ) J = 37 + ---------------------------------------------------------------- 1/6 1/4 2 %4 %5 3/4 ( 1/2 2/3 1/3 )1/2 2 - %8 + ( -550708 %9 - %8 ( %7 - 8612 %7 + 329681 ) ) K = 37 + --------------------------------------------------------------- 1/6 1/4 2 %7 %8 where ______ 2/3 1/3 %1 = 2532898313 + 2279200 i \/416363, %2 = %1 + 1842 %1 + 2047089, ____ %3 = 64 \/ %1 , ______ 2/3 1/3 %4 = 252741049 + 2507120 i \/416363, %5 = %4 + 2066 %4 + 1389201, ____ %6 = 64 \/ %4 , ______ 2/3 1/3 %7 = 119179929 + 227920 i \/416363, %8 = %7 + 4306 %7 + 329681, ____ %9 = 64 \/ %7 Notice that this could be simplified a little bit (although Maple doesn't seem to realize this fact); more importantly, observe that the **STUFF**'s are the fractions following the 37's in each line...and these **STUFF**'s have to cancel; but getting them into "obvious cancelling form" requires some bit of insight that I just haven't acquired. In this numerical case, we can perhaps see better that the problem becomes one of "denesting" the nested Radicals. "Denesting" is a method of algebraic manipulation with its core in the theory of Field Extensions. As an example, ____________ / ___ can be "denested" to ___ ___ \/ 5 + 2 \/ 6 \/ 2 + \/ 3 . Susan Landau -- in "Simplification of Nested Radicals", SIAM J. Comput., 21 (Feb.1992) 85-110 -- explains Denesting and provides an algorithm for denesting arbitrarily "deep" Radical Expressions. (I freely admit that I do not fully comprehend the steps involved, but I have only begun to look into this avenue of investigation.) I am not sure if the algorithm will readily transfer to the general X-Y-Z-R case, since the steps require knowledge of Irreducible Polynomials and such things, which I should think would be extremely difficult to determine when dealing with variables instead of _numbers_. At this point, I'd be happy to see the denested forms of the three Expressions above; examination of these might provide some valuable insight to the general case, fostering a good guess or two that could save a lot of time. Having mentioned time, I find that perhaps I should not waste much more of the Reader's. I'll close up this post with a listing of some Special Case Solutions to the P.T.E.'s....Perhaps someone can discern the Grand Pattern from these scattered examples. The Solutions for H^2, J^2, and K^2 are ordered such that repective Solutions for these three quantities satisfy the Corollary to LOC.2. Some of the Roots are extraneous (see discussion below), and '**' represent Roots that are too complicated to write out. CASE ROOTS OF P.T.E.s 2 2 "EQUI-HEDRAL" 2 2 2 4 X 4 X X = Y = Z = R H = J = K = ---, ---, 0, 0 3 3 +-+-+-+-+-+ 2 2 2 2 "TRISO-HEDRAL" 2 2 2 3 X + R 3 X + R X = Y = Z R H = J = K = --------, --------, 0, 0 3 3 +-+-+-+-+-+ "DOUBLY 2 2 2 ___ 2 2 ___ 2 2 BISO-HEDRAL" H = X + R + 2 / S , X + R - 2 / S , (X + R), (X - R) X = Y, Z = R -------------- -------------- 3 3 2 2 2 ___ 2 2 ___ 2 2 J = X + R + 2 / S , X + R - 2 / S , (X - R), (X + R) -------------- ---------------- 3 3 2 2 2 ___ 2 2 ___ K = 4 ( X + R - / S ), 4 ( X + R + / S ), 0, 0 ------------------ ------------------ 3 3 4 2 2 4 S = X - X R + R +-+-+-+-+-+ "PYTHAGOREAN" 2 2 2 2 2 2 2 H = Y + Z, **, **, ** R = X + Y + Z 2 2 2 J = X + Z, **, **, ** 2 2 2 K = X + Y, **, **, ** +-+-+-+-+-+ "FLAT" 2 2 R = X + Y + Z H = ( Y + Z ), **, **, ** 2 2 J = ( X + Z ), **, **, ** 2 2 K = ( X + Y ), **, **, ** NOTE: I should mention something about extraneous Roots. In the Root List for each of H^2, J^2, and K^2 in the cases above, the last three Roots can be considered extraneous. Let me explain. Notice that if H = 0, then (by the formula given for calculating H from the Cross Product) either |QA| = 0 or |BC| = 0; so either Y = Z = 0 or X = R = 0. Similarly, if J = 0, then either X = Z = 0 or Y = R = 0; and if K = 0, then either X = Y = 0 or Z = R = 0. Now the Triso-hedral/Equi-hedral case, the '0' Roots for the triple H, J, K imply that X = Y = Z = R = 0. But in this case the formula in the "first" entries of the Root List give us the answer. Thus, there is no need for the '0' entries. Moreover, the "second" entry is unnecessary since it simply copies the "first". In the Doubly Biso-hedral case, H^2 can equal (X - R)^2 only if we have X = Y = Z = R = 0, which means that H's value can be computed from its "first" Root formula, and so can the corresponding values of J and K; similarly, for when J^2 = (X - R)^2. So the last two entries in these Root Lists are extraneous. Moreover, one can show that X^2 + R^2 - 2sqrt(X^4 - X^2*R^2 + R^2) is non-positive, and reaches zero only when X^2 = R^2; but if these are identical, then the "second" Root formulae for H^2 and J^2 yield 0, so that we get X = Y = Z = R = 0, and find that, again, the "first" Root formulae suffice. In the Pythagorean case, one can use arguments about Edges to show that the Tetrahedron _must_ be a Right-Corner Tetrahedron; that is, the Neo-Pythagorean is an "if and only if" Proposition in the realm of Perfect Tetrahedra. (Another nice thing about Perfect Tetrahedra!) Thus, A = B = C = 0, and the "first" Roots give the Solutions, so we don't even need the '**'s. One can show that in the Flat Case, the Tetrahedron is in fact flat: all Faces lie in the same Plane, and the Vertex Q lies at the Ortho- centre of Face R. Here, H must be Y + Z, J must be X + Z, and K must be X + Y (by Edge arguments). The '**' Solutions in this case are extraneous as well. Is it true in general that 3 of the Solutions for each of H^2, J^2, and K^2 are extraneous? I don't know. Numerical experimentation has always lead to twelve values (all together) which could be grouped into H-J-K triplets satisfying LOC.2. Some of the numbers have been negative or even complex which _does_ make them extraneous as Area- Values. (In the numerical example above, it may be that the Imaginary components are _supposed_ to be zero.) My GUESS is that there is only one valid triplet (except in special cases where we have repeated Roots), and that "3/4"s of each P.T.E. contains useless information, but I don't know for sure yet if that is the case. Well, anyway, that's about all I care to inflict upon you for now. Anyone interested in other aspects of Hedronometry (such as its analogue in Hyperbolic Space among other things) should feel free to inquire. I have a little notebook full of bits and pieces (but nothing especially *deep*). Let me close now by saying that, although it is *trivial* Mathematics, Hedronometry has been a pet project of mine since my High School Days. I've been struggling with various forms of the P.T.E. off and on for nearly 10 years, and I'd very much like to get its Solution pinned down. So I appreciate all the help I can get in this matter. Any questions, comments, suggestions, or what-have-you's will be graciously accepted; again, please respond by e-mail, so as not to use up even more bandwidth on this subject. I thank you all for your time and tolerance. Regards, Don Mc Connell bdsm@fermat.wpi.edu

Newsgroups:sci.math.researchFrom:"Greg Kuperberg" <gk00@midway.uchicago.edu>Subject:Re: Solve the Perfect Tetrahedron Equation!Date:Fri, 12 Feb 1993 17:08:03 GMTOrganization:University of Chicago

For those who have not been following this discussion, Don describes a tetrahedron by the area of its four faces and the length of the cross product of every pair of opposite edges, and he wants an expression for the dihedral angles. He has also restricted to the case where every pair of opposite edges is perpendicular and wants the dihedral angles in terms of the four areas. The straightforward approach with vector calculus in R^3 yields algebraic equations for the cosines of the angles that seem unnecessarily complicated. In article <1lcdgd$shl@bigboote.WPI.EDU> you write: >THE FIRST LAW OF COSINES ("LOC.1"): > > 2 2 2 2 > R = X + Y + Z - 2 Y Z cos QA - 2 Z X cos QB - 2 X Y cos QC > > PROOF. Exercise. This, too, is easy, though it involves a great deal > of symbol manipulation and a working knowledge of the formulae > from Spherical Trigonometry... Here is another solution. X,Y,Z, and R are your four areas, and let cAB be the cosine of the angle between faces with area A and B. The key to this equation is the shadow principle: At high noon, a plate with area A at an angle alpha from the ground casts a shadow of area cos(alpha) A. By resting the tetrahedron on the ground on each face and applying the shadow principle, it follows that: R = cRX X + cRY Y + cRZ Z X = cRX R + cXY Y + cXZ Z Y = cYX X + cRY R + cYZ Z Z = cZX X + cZY Y + cRZ R Now multiply each equation by the left-hand side and subtract the last three equations from the first. The real lesson here is that you can describe a tetrahedron by a 4 x 4 symmetric matrix of numbers: ( R^2, -cRX R X, -cRY R Y, -cRZ R Z) (-cRX R X, X^2, -cXY X Y, -cXZ X Z) (-cRY R Y, -cXY X Y, Y^2, -cYZ Y Z) (-cRZ R Z, -cXZ X Z, -cYZ Y Z, Z^2) As long as all four rows add to zero and the matrix is positive semidefinite (in other words v^T M v>0 always), it corresponds to an actual Euclidean tetrahedron. Your three pseudo-areas (cross-products of opposite edges) should be somewhere in this matrix also, but I have to think about exactly where. Finding them will hopefully solve or at least clarify your question.

Newsgroups:sci.math.researchFrom:geoff@math.ucla.edu (Geoffrey Mess)Subject:Re: Solve the Perfect Tetrahedron Equation!Date:Sun, 14 Feb 1993 02:37:35 GMTOrganization:UCLA, Mathematics Department

In article <9302121708.AA09604@midway.uchicago.edu> "Greg Kuperberg" <gk00@midway.uchicago.edu> writes: > In article <1lcdgd$shl@bigboote.WPI.EDU> you write: > >THE FIRST LAW OF COSINES ("LOC.1"): > > > > 2 2 2 2 > > R = X + Y + Z - 2 Y Z cos QA - 2 Z X cos QB - 2 X Y cos QC > > > > PROOF. Exercise. This, too, is easy, though it involves a great The area vector of a plane region on the boundary of a solid in 3-space has magnitude the area of the region and direction the outwards normal. The area vectors of the faces of a solid in 3-space add up to zero, so the sum of the area vectors of two of the faces has the same length of the sum of the area vectors of the other two, and the sum of the area vectors of three of the faces has the same length as the area vector of the other. The second and first law of cosines follow. The argument applies to polyhedra with any number of faces and in any dimension. and in particular, as Billy Don McConnell points out, to simplices in n-space. I suppose this result is to be found in Grassmann's Ausdehnunglehre, though I've never looked. The statement that the area vectors add up to zero can be thought of as a limiting case of the statement that the integral of the normal vector of a smooth closed surface with respect to surface area is zero, which can be deduced from Stokes's theorem. You can get the polyhedral case directly from a suitably formulated version of Stokes's theorem. Essentially it means that if fluid is flowing at a velocity constant in space and time, as much fluid flows out of the tetrahedron (or other solid) as flows into it. The vector area of the pseudoface tells you how much fluid crosses a surface which spans the skew quadrilateral which divides the tetrahedron into two triangles, if the fluid is flowing with unit speed and the best direction is chosen for its velocity vector. I hope Billy Don McConnell will post something about tetrahedra in hyperbolic space, as many people are interested in them. -- Geoffrey Mess Department of Mathematics, UCLA Los Angeles, CA. geoff@math.ucla.edu NeXTmail welcome.

From:hillman@math.washington.edu (Christopher Hillman)Date:14 Jan 1997 12:57:19 GMTNewsgroups:sci.mathSubject:Re: finding volume of tetrahedron with one vertex at origin and others

In article <32DAA78E.7078@qlink.queensu.ca>, Paul Turkstra <6pt@qlink.queensu.ca> writes: |> How do I find the volume of a tetrahedron with vertices at the origin, |> (1,2,0), (0,1,3) and (1,1,1) using determinants and the like? From a submission which apparently never made it into the FAQ for this newsgroup: THE VOLUME OF SIMPLICES Simplices (plural of ``simplex'') are convex bodies which generalize the notions of triangles and tetrahedra to n-dimensional euclidean space, E^n. A k-dimensional simplex has k+1 (affinely independent) vertices, (k+1)k/2! edges, (k+1)k(k-1)/3! two dimensional faces, and so forth up to k+1 (k-1)-dimensional faces. Just as triangles can exist in E^3, k-simplices can exist in E^n for any n >= k. Any convex polytope (the generalization of polygons and polyhedra to E^n) can be partitioned into simplices (just as any convex polygon can be partitioned into triangles), so in principle if you can compute the volume of arbitrary simplices, you can compute the volume of any convex polytope. There are several pretty and easily implemented formulae which give the volume of a k-dimensional simplex sitting in euclidean n-space, S, in terms of different kinds of given information, such as the coordinates of the vertices or the lengths of the edges. These formulae can be used by anyone who knows how to compute a determinant, although software such as Mathematica will probably be needed to evaluate really large determinants. I will write Vol_k for ``k-dimensional volume'' (really k-dimensional Lebesgue measure, for those who know what this means); thus Vol_1 means length, Vol_2 means area, and so forth. Unless stated otherwise, S will denote a k-dimensional simplex in E^n, where n \geq k. Vol_k(S) will mean the k-dimensional volume of S. A hyperplane in E^n is an (n-1)-dimensional plane. 1. COMPUTING THE VOLUME OF A SIMPLEX WITH KNOWN VERTICES. Let S have vertices given by the n-dimensional row vectors v_0, v_1, ... v_k. Let w_1 = v_1 - v_0, w_2 = v_2 - v_0, ... w_k = v_k - v_0 and let W be the k by n matrix whose rows are the row vectors w_j, 1 \leq j \leq k. Then the GRAM DETERMINANT FORMULA says | det W W^t |^{1/2} Vol_k (S) = ------------------ k! where W^t is the transpose of W. Note that W W^t is a k by k matrix, so this formula makes sense. 2. COMPUTING THE VOLUME OF A SIMPLEX WITH KNOWN EDGE LENGTHS. Let v_0, v_1, ... v_k be the (unknown) vertices of S and suppose that || v_i - v_j || = d_{ij} is known. Let D be the k+2 by k+2 matrix | 0 1 1 .... 1 | | 1 0 d_{01}^2 .... d_{0k}^2 | D = | ... .... .... .... | | 1 d_{k0}^2 d_{k1}^2 .... 0 | Then the CAYLEY-MENGER DETERMINANT FORMULA says | det D |^{1/2} Vol_k(S) = --------------- 2^{k/2} k! This generalizes the HERON FORMULA for the area of a triangle in terms of its edge lengths. 3. COMPUTING THE VOLUME OF A SIMPLEX WITH KNOWN HYPERFACES. Let S be the n-simplex in E^n whose faces are the n+1 hyperplanes whose Cartesian equations are given by a_{i0} + a_{i1} x_1 + a_{i2} x_2 + ... a_{in} x_n = 0 where 0 <= i <= n. Note that the x_j are variables standing for real numbers and the a_{ij} are real constants. Let A be the n+1 by n+1 matrix with elements a_{ij} and let A_{i0} be the cofactor matrix of A with respect to a_{i0}. Then the KLEBANER-SUDBURY-WATTERSON DETERMINANT FORMULA says | det A |^n Vol_n (S) = --------------------------------------- n! det A_{00} det A_{10} ... det A_{n0} Note this is valid only for an n-simplex in E^n. REFERENCE: See the paper by Peter Gritzmann and Victor Klee, ``On the complexity of some basic problems in computational convexity II: Volume and mixed volumes''. This is a state of the art survey paper discussing the general problem of the efficient computation of volume of convex bodies in E^n, and discusses such fundamental topics as the Brunn-Minkowski theory of mixed volume (a sort of average of the volume of several bodies) the volume of Minkowski sums of polytopes, and efficient dissection into simplices, among other things. The Gram, Cayley-Menger, and KSW formulae are discussed in sections 3.6.1, 3.6.2, and 3.6.3 respectively of this paper. This paper is available via anonymous FTP from "dimacs@rutgers.edu" (login as anonymous and use your email address as the password; cd to the directory "pub/dimacs/TechnicalReports/TechReports"). It will also appear in physical form in the proceedings of a NATO conference to be published in 1994 under the title Polytopes: Abstract, Convex,i and Computational. Chris Hillman

From:eppstein@ics.uci.edu (David Eppstein)Date:28 Jan 1997 11:00:52 -0800Newsgroups:sci.mathSubject:Re: [HELP]Solid angle

Xavier SERPAGGI wrote: >> Does anybody know a simple way to calculate the solid angle of four >> points, or even three. Ok, suppose you are trying to calculate the solid angle of a tetrahedron vertex. Translate the vertex to the origin (i.e. subtract its coordinates from the coordinates of the other three vertices), and normalize the other three points (call them a,b,c) so they are all at unit distance from the origin (e.g. if a has coordinates a1,a2,a3, divide each coordinate by sqrt(a1^2+a2^2+a3^2)). Then the solid angle E can be found from the formula E = 2 tan^{-1} (a . (b x c))/(1 + b.c + c.a + a.b) where x is a cross product and the periods are dot products. Reference: F. Eriksson, "On the measure of solid angles", Math. Mag. 63, no. 3, 1990, pp. 184--187. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:mcconnel@vvm_See.Sig_.com (Don McConnell -- Read .sig before replying)Date:Wed, 30 Jul 1997 16:34:48 -0600Newsgroups:sci.mathSubject:Re: What is Tetrahedrometry ?

Hello ... Peter wrote: Will someone please tell us something about Tetrahedrometry ? And David Eppstein responded: You might try my web page http://www.ics.uci.edu/~eppstein/junkyard/hedronometry.html I wrote the "Hedronometry: The Search For Solutions to the Perfect Tetrahedron Equations" post. Thanks to David for archiving it as part of his impressive collection. I thought I would post two informational updates here: First: I am no longer at Worcester Polytechnic Institute or the University of Washington, by the way; my current e-mail address is mcconnel @ vvm . com (without the spaces ... I'm just a bit paranoid about address-harvesters). Incidentally, I promised references to anyone who e-mailed me about the topic ... in the years since the post, I've lost track of those references. Sorry. Second: I essentially "solved" the Perfect Tetrahedron Equations (PTEs) recently. At least, I solved something equivalent to the problem posed by them: I found a Heron-like formula that gives the volume of a perfect tetrahedron in terms of the areas of its faces. (A "perfect" tetrahedron is one whose pairs of opposite edges determine pairs of orthogonal vectors.) With face areas and volume known, various substitutions into other formulae can yield things like dihedral and face angles, and areas of "pseudo- faces" (the primary target of the previously posted PTEs); thus, all aspects of a Perfect Tetrahedron are derivable from the areas of its faces. My solution isn't explicit, but is encoded in a single quartic, instead of the four I'd given before; the one quartic is far more complicated to express than the other four, but I haven't done a great deal of analysis to see what I can fix up. Here 'tis: Let X, Y, Z, and R be the areas of faces of a perfect tetrahedron, and let W = 81 * volume^4 . Then W is the root of the following "Heron Quartic": 0 = 27 W^4 - W^3 * ( << 6, 0, 0, 0 >> - 33 << 4, 2, 0, 0 >> ) ( + 114 << 2, 2, 2, 0 >> ) - W^2 * ( <<10, 2, 0, 0 >> - 4 << 8, 4, 0, 0 >> ) ( - 43 << 8, 2, 2, 0 >> + 6 << 6, 6, 0, 0 >> ) ( + 42 << 6, 4, 2, 0 >> + 408 << 6, 2, 2, 2 >> ) ( - 204 << 4, 4, 4, 0 >> - 218 << 4, 4, 2, 2 >> ) - W * ( <<14, 2, 2, 0 >> - 6 <<12, 4, 2, 0 >> ) ( - 57 <<12, 2, 2, 2 >> + 15 <<10, 6, 2, 0 >> ) ( + 18 <<10, 4, 4, 0 >> + 165 <<10, 4, 2, 2 >> ) ( - 20 << 8, 8, 2, 0 >> - 12 << 8, 6, 4, 0 >> ) ( + 109 << 8, 6, 2, 2 >> - 398 << 8, 4, 4, 2 >> ) ( - 6 << 6, 6, 6, 0 >> + 478 << 6, 6, 4, 2 >> ) ( - 324 << 6, 4, 4, 4 >> ) - X^2 Y^2 Z^2 R^2 (X+Y+Z-R)^2 (X+Y-Z+R)^2 (X-Y+Z+R)^2 (-X+Y+Z+R)^2 * (X+Y-Z-R)^2 (X-Y+Z-R)^2 (X-Y-Z+R)^2 ( X+Y+Z+R)^2 (I hope I've caught all of the typos.) Here, <<a, b, c, d>> := SUM A^a B^b C^c D^d , with the sum taken over all permutations ABCD of XYZR. Although the cubic, quadratic, and linear coefficients are fairly unenlightening when expanded in this way, I'll point out that the constant term expands to something far worse; I'm hoping that the other coefficients can be "reasonably" expressed in a more compact form. (Again, I haven't given it much thought.) I suspect that at most one of the four roots are real and non-negative for non-degenerate tetrahedra. This is certainly the case in the special cases, which I'll list below: ======================================================================== Equi-Hedral: X=Y=Z=R = P Heron Quartic: W^3 ( -64 P^6 + 27 W ) = 0 Volume: V = 0 (triple root), V^4 = 2^6 P^6 / 3^7 Triso-Hedral: X=Y=Z = P, R = Q Heron Quartic: ( 27 W - Q^2 (9 P^2-Q^2 ) ) * ( W + P^2 ( P^2-Q^2 )^2 )^3 = 0 Volume: V^4 = - P^2 (P^2-Q^2)^2/3^4 (triple root, imaginary V) V^4 = Q^2 (9 P^2-Q^2)/3^7 Doubly Biso-Hedral: X=Y = P, Z=R = Q Heron Quartic: W^2 ( 27 W^2 + 32 W (P^2+Q^2)(2 P^2-Q^2)(P^2-2 Q^2) - 256 P^4 Q^4 (P^2-Q^2)^2 ) = 0 Volume: V = 0 (double root) V^4 = 2^4/3^7 ( -2 P^6 + 3 P^4 Q^2 + 3 P^2 Q^4 - 2 Q^6 +- 2 ( P^4 - P^2 Q^2 + Q^4 )^(3/2) ) (V imaginary when +- is - ; V real when +- is +) Right: R^2 = X^2 + Y^2 + Z^2 Heron Quartic: / 27 W^3 \ | + 4 W^2 ( 16 <<6,0,0>> + 12 <<4,2,0>> ) | | ( - 39 <<2,2,2>> ) | (W-4 X^2 Y^2 Z^2) | + 16 W ( 8 <<8,4,0>> - 16 <<8,2,2>> ) |=0 | (+ 15 <<6,6,0>> + <<6,4,2>> ) | | (+ 30 <<4,4,4>> ) | | + 64 (X^2+Y^2+Z^2) (XY+YZ+XZ)^2 | \ * (XY+YZ-XZ)^2 (XY-YZ+XZ)^2 (-XY+YZ+XZ)^2 / Volume: V^4 = 4 X^2 Y^2 Z^2 / 81 V^4 = not yet explicitly determined. ===================================================================== At any rate, I'd welcome any insights about how to simplify the Heron Quartic. Pointers to discussions about expressions I've denoted with <<a,b,c,d>> notation would be helpful. Right now, though, I'm reasonably satisfied that the formula is easy to apply to numerical examples. Regards, Don |y=f(x) |y=f(-x) ,+-. /::::B.D.S."Don"McConnell::::\ ,-+. ,' | `-' ::: mcconnel @ vvm . com ::: `-' | `. `-' | :::::::::::::::::::::::::::: | `-' ------+------ * HAVE A FUNCTIONAL DAY! * ------+------ | | ( Remove "_See.Sig_" from my address to respond via e-mail. ) ( -------N---O------J---U---N---K------M---A---I---L------- ) ( This article is not to be construed as a solicitation for ) ( commercial e-mail promoting goods or services of any kind ) ( ---DO--NOT--ADD--THIS--ADDRESS--TO--ANY--MAILING--LIST--- )

From:ksbrown@seanet.com (Kevin Brown)Date:Sat, 02 Aug 1997 23:08:47 GMTNewsgroups:sci.mathSubject:Re: What is Tetrahedrometry ?

B.D.S."Don"McConnell wrote: > ...I found a Heron-like formula that gives the volume of a > perfect tetrahedron in terms of the areas of its faces. (A > "perfect" tetrahedron is one whose pairs of opposite edges > determine pairs of orthogonal vectors.)... My solution isn't > explicit, but is encoded in a single quartic, instead of the > four I'd given before; the one quartic is far more complicated > to express than the other four, but I haven't done a great deal > of analysis to see what I can fix up... At any rate, I'd welcome > any insights about how to simplify the Heron Quartic. Pointers > to discussions about expressions I've denoted with <<a,b,c,d>> > notation would be helpful... This may not be of any use to you, but here's an old post on this subject. It's interesting that this approach seems to invariably lead to a quartic equation (although yours is the first I've seen where the quartic is in terms of the volume itself, rather than one of the edge lengths). ================================================================ In general there is no complete generalization of Heron's formula giving the volume of a general tetrahedron in terms of the areas of its faces, because the face areas don't uniquely determine the volume (in contrast to the case of triangles, where the three edge lengths determine the area). However, it IS possible to derive a "Heron's formula" for tetrahedrons if we restrict ourselves to just those that would fit as the "hypotenuse face" of a right 4D solid. (Notice that EVERY triangle is the face of a right tetrahedron, which explains why Heron's formula is complete for triangles). To review, remember that Heron's formula for triangles is essentially equivalent to Pythagoras' Theorem for right tetrahedrons. If A_xyo, A_xoz, and A_oyz denote the areas of the three orthogonal faces of a right tetrahedron, and A_xyz denotes the area of the "hypotenuse face", then it's easy to show that (A_xyz)^2 = (A_xyo)^2 + (A_xoz)^2 + (A_oyz)^2 (1) This is not only a generalization of Pythagoras's Theorem, it's also essentially Heron's formula for triangles. Note that if a,b,c denote the three orthogonal edge lengths of the tetrahedron, then the areas A_xyo, A_xoz, and A_oyz are simply ab/2, ac/2, and bc/2. Furthermore, the edges of the hypotenuse face d,e,f are directly related to a,b,c according to the 2D Pythagorean theorem a^2 + b^2 = d^2 a^2 + c^2 = e^2 (2) b^2 + c^2 = f^2 so we can express the squares of the areas A_xyo, A_xoz, A_oyz in equation (1) in terms of d^2, e^2, and f^2 to give the ordinary Heron's formula explicitly. Incidentally, it might be (and has been) argued that this derivation does not apply to obtuse triangles, because the "hypotenuse face" of a right tetrahedron is necessarily acute (i.e., each of its angles must be less than 90 degrees). However, ANY triangle can be the hypotenuse face of a right tetrahedron, provided the orthogonal edge lengths and areas are allowed to be imaginary. (It's interesting to speculate on whether the ancient Greeks might have been aware of this aspect of the problem.) Anyway, we can do the same thing for tetrahedrons based on the generalized Pythagorean theorem for volumes of *right* 4D solids (V_wxyz)^2 = (V_wxyo)^2 + (V_wxoz)^2 + (V_woyz)^2 + (V_oxyz)^2 (3) If we let a,b,c,d denote the orthogonal edge lengths of the 4D solid, then the volumes of the four orthogonal "faces" are simply abc/6, abd/6, acd/6, and bcd/6, so equation (3) can be rewritten as (V_wxyz)^2 = (abc/6)^2 + (abd/6)^2 + (acd/6)^2 + (bcd/6)^2 (3') Also, the areas of the four faces of our "hyoptenuse" solid can be expressed in terms of a,b,c,d by means of the 3D Pythagorean theorem (ab)^2 + (ac)^2 + (bc)^2 = 4(A_1)^2 (ab)^2 + (ad)^2 + (bd)^2 = 4(A_2)^2 (4) (ac)^2 + (ad)^2 + (cd)^2 = 4(A_3)^2 (bc)^2 + (bd)^2 + (cd)^2 = 4(A_4)^2 Thus, given the four face areas A_1, A_2, A_3, A_4, we have four equations in the four unknowns a,b,c,d, so we can solve for these values and then compute the area using (3'). This is the point at which people usually turn away from this approach, for two reasons: [1] everything we're doing is restricted to the special tetrahedrons that can serve as the hypotenuse of a "right" 4D simplex, so we're certainly not going to end up with a general formula applicable to every tetrahedron (and such formulas are already available anyway via the usual determinants), and [2] it turns out to be somewhat messy to solve the set of equations (4). Pressing on anyway, we can reduce (4) to a single quartic in the square of any of the four variables a,b,c or d. Arbitrarily selecting c, and letting A,B,C,D denote 4 times the squares of the face areas (i.e., the right hand sides of equations (4)), we can express the quartic in x=c^2 with coefficients that are functions of B and the elementary symmetric functions of A,C,D s1 = A+C+D s2 = AC+AD+CD s3 = ACD In these terms the quartic for x=c^2 is [12B] x^4 + [3s1^2 - 12s2 + 14Bs1 - B^2] x^3 + [2B(s1^2 + 6s2) - B^2 s1 - s1^3 + 4s1s2 - 36s3] x^2 + [2B(s1s2 + 6s3) - B^2 s2 + 4s2^2 - s1^2 s2 - 12s1s3] x - [s3 (s1-B)^2] = 0 Of course, the analagous quartics can be given for a^2, b^2, and d^2, but once we have any one of them we can more easily compute the others. For example, given c we can compute b from the relation ___________________ / (c^2 + A)(c^2 + D) b = -c +- / ------------------- \/ (c^2 + C) and the values of a and d follow easily, allowing us to compute the volume using equation (3'). It would be nice if we could express the volume as an explicit function of the face areas, but I don't know if such a formula exists. _______________________________________________________________ | MathPages /*\ http://www.seanet.com/~ksbrown/ | | / \ | |___________/"I know thee not, old man. Fall to thy prayers."__|