From: bdsm@poincare.WPI.EDU (Billy Don McConnell)
Newsgroups: sci.math.research
Subject: Solve the Perfect Tetrahedron Equation!
Date: 11 Feb 93 02:26:21 GMT
Organization: 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.research
From: "Greg Kuperberg" <gk00@midway.uchicago.edu>
Subject: Re: Solve the Perfect Tetrahedron Equation!
Date: Fri, 12 Feb 1993 17:08:03 GMT
Organization: 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$shl@bigboote.WPI.EDU> you write:
>THE FIRST LAW OF COSINES ("LOC.1"):
>
> 2 2 2 2
> R = X + Y + Z - 2 Y Z cos QA - 2 Z X cos QB - 2 X Y cos QC
>
> PROOF. Exercise. This, too, is easy, though it involves a great deal
> of symbol manipulation and a working knowledge of the formulae
> from Spherical Trigonometry...
Here is another solution. X,Y,Z, and R are your four areas, and let
cAB be the cosine of the angle between faces with area A and B. The
key to this equation is the shadow principle: At high noon, a plate
with area A at an angle alpha from the ground casts a shadow of area
cos(alpha) A. By resting the tetrahedron on the ground on each
face and applying the shadow principle, it follows that:
R = cRX X + cRY Y + cRZ Z
X = cRX R + cXY Y + cXZ Z
Y = cYX X + cRY R + cYZ Z
Z = cZX X + cZY Y + cRZ R
Now multiply each equation by the left-hand side and subtract the last
three equations from the first.
The real lesson here is that you can describe a tetrahedron by
a 4 x 4 symmetric matrix of numbers:
( R^2, -cRX R X, -cRY R Y, -cRZ R Z)
(-cRX R X, X^2, -cXY X Y, -cXZ X Z)
(-cRY R Y, -cXY X Y, Y^2, -cYZ Y Z)
(-cRZ R Z, -cXZ X Z, -cYZ Y Z, Z^2)
As long as all four rows add to zero and the matrix is positive
semidefinite (in other words v^T M v > 0 always), it corresponds to an
actual Euclidean tetrahedron. Your three pseudo-areas (cross-products
of opposite edges) should be somewhere in this matrix also, but I have
to think about exactly where. Finding them will hopefully solve or
at least clarify your question.
Newsgroups: sci.math.research
From: geoff@math.ucla.edu (Geoffrey Mess)
Subject: Re: Solve the Perfect Tetrahedron Equation!
Date: Sun, 14 Feb 1993 02:37:35 GMT
Organization: UCLA, Mathematics Department
In article <9302121708.AA09604@midway.uchicago.edu> "Greg Kuperberg"
<gk00@midway.uchicago.edu> writes:
> In article <1lcdgd$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 GMT
Newsgroups: sci.math
Subject: 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 -0800
Newsgroups: sci.math
Subject: 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 -0600
Newsgroups: sci.math
Subject: 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)
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------+------ * HAVE A FUNCTIONAL DAY! * ------+------
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From: ksbrown@seanet.com (Kevin Brown)
Date: Sat, 02 Aug 1997 23:08:47 GMT
Newsgroups: sci.math
Subject: 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.
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