PSEUDOFACES OF TETRAHEDRA
BDS MCCONNELL
The Law of Cosines for tetrahedra states that
W 2 = X 2 + Y 2 + Z 2 − 2Y Z cos ∠OA − 2ZX cos ∠OB − 2XY cos ∠OC
where W := area4ABC, X := area4OBC, Y := area4OCA, and Z := area4OAB,
and where ∠OA, ∠OB, and ∠OC are the dihedral angles along corresponding edges. The
resemblance to the Law of Cosines for triangle is already striking, but can be made moreso.
Combine the above equation with its counterparts
X 2 = W 2 + Y 2 + Z 2 − 2Y Z cos ∠OA − 2ZW cos ∠AB − 2W Y cos ∠CA
Y 2 = X 2 + W 2 + Z 2 − 2W Z cos ∠AB − 2ZX cos ∠OB − 2XW cos ∠BC
Z 2 = X 2 + Y 2 + W 2 − 2Y W cos ∠CA − 2W X cos ∠BC − 2XY cos ∠OC
to get a family of relations
Y 2 + Z 2 − 2Y Z cos ∠OA = W 2 + X 2 − 2W X cos ∠BC
Z 2 + X 2 − 2ZX cos ∠OB = W 2 + Y 2 − 2W Y cos ∠CA
X 2 + Y 2 − 2XY cos ∠OC = W 2 + Z 2 − 2W Z cos ∠AB
that teeter on the very brink of exactly matching the triangular Law of Cosines. We push
them over that brink by formally defining P , Q, and R such that
Y 2 + Z 2 − 2Y Z cos ∠OA = P 2 = W 2 + X 2 − 2W X cos ∠BC
Z 2 + X 2 − 2ZX cos ∠OB = Q2 = W 2 + Y 2 − 2W Y cos ∠CA
X 2 + Y 2 − 2XY cos ∠OC = R2 = W 2 + Z 2 − 2W Z cos ∠AB
We say that P , Q, and R are the areas of “pseudofaces” of the tetrahedron . . . whatever
“pseudofaces” might be. Note that, although pseudo, these faces together are as good as
the “real” faces (at least when squared):
P 2 + Q2 + R2 = W 2 + X 2 + Y 2 + Z 2
With judicious substitution and rearrangement, one can use pseudofaces to convert a
“hedronometric” volume formula that favors one vertex
!
1
−
2
cos(∠OA)
cos(∠OB)
cos(∠OC)
4
V 4 = X 2Y 2Z 2
81
− cos2 (∠OA) − cos2 (∠OB) − cos2 (∠OC)
Date: 6 March, 2005.
1
2
BDS MCCONNELL
into something more symmetrical
1
V4 =
81
2W 2 X 2 Y 2 + 2W 2 X 2 Z 2 + 2W 2 Y 2 X 2 + 2X 2 Y 2 Z 2 + P 2 Q2 R2
−P 2 W 2 X 2 + Y 2 Z 2 − Q2 W 2 Y 2 + Z 2 X 2 − R2 W 2 Z 2 + X 2 Y 2
!
So, here we see that the collection of faces and pseudofaces determine a tetrahedron
(and its volume) uniquely. This type of result isn’t true for just faces alone.
Pseudofaces revealed
Although initially defined merely to make a few equations look nice, pseudofaces in fact
have a geometric interpretation: viewing a tetrahedron in a direction perpendicular to the
plane determined by opposite edges (for instance OA and BC, as in Figure (1a) below)
reveals a quadrilateral. That quadrilateral is a pseudoface, insofar as its area is computed
via a pseudoface formula.
A
O
a
C
hB
A
OA
hC
B
O
w
B
w
C
Figure 1. Views of a Tetrahedron. (a) Perpendicular to OA and BC; (b) parallel to OA.
Figure (1a) surrounds a pseudoface with a rectangle of dimensions a := |OA| and w, so
that the area of the pseudoface satisfies
1
P 2 = a2 w2
4
From Figure (1b), we compute the distance w from the triangular Law of Cosines, using
the dihedral angle ∠OA and the lengths hB and hC (which are altitudes to OA of 4OAB
and 4OAC, respectively):
w2 = h2B + h2C − 2hB hC cos∠OA
PSEUDOFACES OF TETRAHEDRA
3
Thus,
1
P 2 = a2 h2B + h2C − 2hB hC cos ∠OA
4
1
1
1
1
= a2 h2B + a2 h2C − 2 · ahB · ahC · cos ∠OA
4
4
2
2
= Y 2 + Z 2 − 2Y Z cos ∠OA
A slightly different analysis, involving sighting the tetrahedron along BC, would yield
P 2 = W 2 + X 2 − 2W X cos ∠BC
Projecting the tetrahedron into planes determined by the other two pairs of opposite
edges gives pseudofaces Q and R.
Assorted Pseudoface Formulas
A Volume Formula. If p is the distance between edges OA and BC —the “pseudoaltitude” corresponding to the pseudoface P — then
1
V = pP
3
Proof. Project points B and C to points B 0 and C 0 in the plane through OA parallel to
P , and project points O and A to points O0 and A0 in the plane through BC parallel to
P . Then OB 0 AC 0 and O0 BA0 C are copies of the pseudoface P , and form bases of a right
prism with height p and volume pP . We can carve the tetrahedron out of the prism by
removing the height-p tetrahedron pairs OO0 BC and AA0 BC (whose bases together form
O0 BA0 C), and BB 0 OA and CC 0 OA (whose bases together form OB 0 AC 0 ). Each of these
tetrahedron pairs has one-third the volume of the prism, so that the original tetrahedron
makes up the last third of that volume.
Cosine Formulas. We naturally define “the” angle between two (real) faces of a tetrahedron as the dihedral angle containing the interior of the tetrahedron (or as the supplement
of the angle between two inward-pointing —or two outward-pointing— normal vectors).
Pseudofaces have no inherent position relative to the interior of the tetrahedron, and so we
cannot define the angle between pseudofaces without ambiguity or artifice. We will accept
the ambiguity and let vectors provide the artifice.
Recall that, for vectors u and v subtending angle θ,
u · v = kukkvk cos θ
and
ku × vk = kukkvk sin θ
and that u × v is perpendicular to both u and v. Thus, we have formulas such as these:
1
1
1
1
Y 2 = kc × ak2
Z 2 = ka × bk2
W 2 = k(a − b) × (c − b)k2
X 2 = kb × ck2
4
4
4
4
1
1
Y Z cos ∠OA = − (c × a) · (a × b)
ZX cos ∠OB = − (a × b) · (b × c)
4
4
1
XY cos ∠OC = − (b × c) · (c × a)
4
4
BDS MCCONNELL
−→
−−→
−−→
where a := OA, b := OB, and c := OC.
(Note that, although there are more direct approaches, one can recover the Law of
Cosines for Tetrahedra by expanding the vector formula for W 2 in terms of the other
expressions.)
With pseudofaces, we have these unambiguous area formulas
1
P 2 = ka × (c − b)k2
4
1
Q2 = kb × (a − c)k2
4
1
R2 = kc × (b − a)k2
4
For angles, we make a subjective selection of normal vector (which depends upon our
preference for vertex O), and declare the angle between two pseudofaces is the angle (not
the supplement of the angle) between the two corresponding normal vectors. In particular,
writing θP for the preferred angle between pseudofaces Q and R, we have
QR cos θP
1
:= − (b × (a − c)) · (c × (b − a))
4
= ...
1
= − ((c × b) · (c × b) + (b × a) · (a × c)
4
+ (c × b) · (b × a) + (a × c) · (c × b))
=
−X 2 + Y Z cos ∠OA + ZX cos ∠OB + XY cos ∠OC
so that
2QR cos θP
= −2X 2 + (2Y Z cos ∠OA + 2ZX cos ∠OB + 2XY cos ∠OC)
= −2X 2 + X 2 + Y 2 + Z 2 − W 2
= Y 2 + Z2 − W 2 + X2
= P 2 + 2Y Z cos OA − P 2 + 2W X cos BC
= 2 (Y Z cos OA − W X cos BC)
Altogether, we have
Y 2 + Z 2 − W 2 + X 2 = 2QR cos θP = 2 (Y Z cos OA − W X cos BC)
Z 2 + X 2 − W 2 + Y 2 = 2RP cos θQ = 2 (ZX cos OB − W Y cos CA)
X 2 + Y 2 − W 2 + Z 2 = 2P Q cos θR = 2 (XY cos OC − W Z cos AB)
With other choices in how we define the angles θP , θQ , and θR , the above formulas could
be off by a sign (replacing an angle by its supplement changes the sign of the cosine);
the formulas, therefore, are only “unambiguously correct” in absolute value. Note that
we do have this unambiguous corollary: The pseudofaces of an equihedral tetrahedron —
W = X = Y = Z— are mutually perpendicular.
PSEUDOFACES OF TETRAHEDRA
5
Finally, what good are cosines without a corresponding Law of Cosines?
Q2 + R2 − 2QR cos θP = Z 2 + X 2 − 2ZX cos OB + X 2 + Y 2 − XY cos OC
+ Y 2 + Z2 − W 2 + X2
= Y 2 + Z2 − W 2
+ X 2 + Y 2 + Z 2 − 2ZX cos OB − 2XY cos OC
= Y 2 + Z 2 − W 2 + W 2 + 2Y Z cos OA
R2 + P 2 − 2RP cos θQ
= Y 2 + Z 2 + 2Y Z cos OA
= Z 2 + X 2 + 2ZX cos OB
P 2 + Q2 − 2P Q cos θR = X 2 + Y 2 + 2XY cos OC
Note that
Q2 + R2 − 2QR cos(supplement of θP ) = W 2 + X 2 + 2W X cos BC
R2 + P 2 − 2RP cos(supplement of θQ ) = W 2 + Y 2 + 2W Y cos CA
P 2 + Q2 − 2P Q cos(supplement of θR ) = W 2 + Z 2 + 2W Z cos AB
We leave the reader to investigate “second-pseudofaces” whose areas satisfy, say,
H 2 := Q2 + R2 − 2QR cos θP
although we will point out the geometric interpretation of such elements. They are pseudofaces of “exterior” tetrahedra to the given tetrahedron: Reflect point C in segment OA to
get C 0 and a new tetrahedron OABC 0 that shares face OAB with the original tetrahedron.
The faces OAC 0 (which is congruent to OAC) and OAB enclose the supplement of angle
bounded by OAC and OAB, hence the pseudoface corresponding to OA and BC 0 has area
given by
Y 2 + Z 2 − 2Y Z cos(supp. ∠OA) = Y 2 + Z 2 + 2Y Z cos OA = H 2
Whether these exterior tetrahedra attach to edges surrounding a vertex (O) or to edges
surrounding a face (4ABC) depends upon the preferred definition of the pseudoface angles.
Another Volume Formula. Making appropriate substitutions into the hedronometric
volume formula from the first section of this article, we can achieve this volume formula
!
1
+
2
cos
θ
cos
θ
cos
θ
1
P
Q
R
V 4 = P 2 Q2 R2
81
− cos2 θP − cos2 θQ − cos2 θR
(The 2 cos θP cos θQ cos θR term is subject to sign change under alternate definition of the
angles involved.)
An equihedral tetrahedron’s volume, therefore, (unambiguously) satisfies
1
V 2 = P QR
9
6
BDS MCCONNELL
For example, a regular
edge-length s has pseudofaces in the
p form of
√ tetrahedron with
2 /2. Thus, the tetrahedron’s volume is
2/2
and
area
s
s6 /72 =
squares
of
edge
length
s
√
s3 2/12.
B.D.S. McConnell
[email protected]