http://dx.doi.org/10.5831/HMJ.2013.35.1.17
VOLUME FORMULAS OF A EUCLIDEAN TETRAHEDRON
Joon Suk Park and Yunhi Cho∗
Abstract. We survey and study several volume formulas of a Eu- clidean tetrahedron.
1. Introduction
In two- and three-dimensional Euclidean geometry, triangles and tetra- hedra are elementary objects. Just as there are many well-known area formulas for triangles, there exist many volume formulas for tetrahedra (see [10], [9], [11], [2], [4] and [3]).
We survey several volume formulas of a Euclidean tetrahedron in Section 2, and derive the following two theorems for volume formulas in Section 3. For the definitions of notations at the following theorems, we explain them at the next section.
Theorem 1 For three edges a12, a23, a31 of the facial triangle A0 with adjacent (to A0) dihedral angles α12, α23, α31, the volume V of the tetra- hedron is given by
V =2
3A02 sin α12sin α23sin α31
a12cos α12sin α23sin α31+ a23sin α12cos α23sin α31+ a31sin α12sin α23cos α31
.
Theorem 2 For a tetrahedron with dihedral angles α01, . . . , α23 and inradius r, its volume V is expressed as
V (α01, . . . , α23, r) = 1 6
(√
N0+√
N1+√
N2+√ N3)3
√N0N1N2N3 r3,
Received October 16, 2012. Accepted December 31, 2012.
2010 Mathematics Subject Classification. 51M04, 52A38, 52B15.
Key words and phrases. Euclidean tetrahedron, volume formula.
The authors would like to thank referee’s helpful suggestions.
∗Corresponding author
where Nl= 1 − cos2αli− cos2αlj− cos2αlk− 2 cos αlicos αljcos αlk with {i, j, k, l} = {0, 1, 2, 3}.
2. Known volume formulas of a Euclidean tetrahedron In order to study the subject, we need the following notations. Let v0, v1, v2, v3 be independent points in three-dimensional Euclidean space E3. Then, V or V3denotes the volume of the tetrahedron T = [v0, v1, v2, v3]; Ai, a triangle face of T , where vi ∈ A/ ifor i = 0, 1, 2, 3 (for simplicity, Ai also denotes the area of the triangle face); aij, the edge length of [vi, vj]; bij, the edge length of Ai ∩ Aj; αij, the dihedral angle at a given edge [vi, vj]; βij, the dihedral angle at a given edge Ai∩ Aj; and θij, the facial angle corresponding to the vertex vi and face Aj, where j ∈ {0, 1, 2, 3}\{i}. Note that aij = aji, bij = bji, αij = αji, βij = βji; however, in general, θij 6= θji, and aij = bkl, αij = βkl, where {i, j, k, l} = {0, 1, 2, 3}.
The volume of an n-dimensional simplex [v0, v1, . . . , vn], denoted by Vn, can be expressed in terms of the edge length aij (see §9.7 in [1]).
Thus, the volume V3 is given by (1) V32= (−1)3+1
23(3!)2 Γ(v0, · · · , v3) = 1
288Γ(v0, · · · , v3), where
Γ(v0, · · · , v3) =
0 1 1 1 1
1 0 a012 a022 a032
1 a102 0 a122 a132 1 a202 a212 0 a232 1 a302 a312 a322 0
= 0.
There is another volume formula based on two facial areas Ai, Aj, the common edge length bij(= akl), and the dihedral angle βij(= αkl), where {i, j, k, l} = {0, 1, 2, 3}. Then, the volume of the tetrahedron is given by
(2) V = 2
3bijAiAjsin βij = 2
3aklAiAjsin αkl, since we know that V = 13A0h = 13A0h0sin α23 = 13A0(2Aa 1
23) sin α23, where h0is the height of A1for the base [v2, v3]. Note that (2) is similar to the triangle area formula, V2 = 12ab sin γ, and it is essentially represented by six terms comprising the edge lengths and angles of T .
From (2), we derive V = 2
3
(12a03a23sin θ31)(12a12a23sin θ20) a23
sin α23
= 1
6 a03a32a21sin θ31sin θ20sin α23. Thus, we obtain third volume formula
(3) V = 1
6 aijajkaklsin θjlsin θkisin αjk, and this formula is also similar to the triangle area formula.
Now, let us derive another volume formula.
From the formula (2), we have V = 2
3a02A1A3sin α02
= 2
3a02
(1
2a01a02sin θ03)(1
2a02a03sin θ01) sin α02
= a01a02a03
6 sin θ03sin θ01sin α02.
Then, by a small sphere truncation around the vertex v0 and spherical cosine law, we know
cos θ03=cos α01cos α02+ cos α03
sin α01sin α02
and cos θ01= cos α02cos α03+ cos α01
sin α02sin α03
.
So sin θ01 and sin θ03 are obtained. In particular, sin θ03=
√1 − cos2α01− cos2α02− cos2α03− 2 cos2α01cos2α02cos2α03
sin α01sin α02 .
Hence we have the following volume formula:
(4)
V = a01a02a03 6
1 − cos2α01− cos2α02− cos2α03− 2 cos α01cos α02cos α03 sin α01sin α02sin α03
.
In (4), if we replace the dihedral angles by facial angles around the vertex v0, we can derive useful volume formulas (see [3] and [6] for the first one, and [8] and [7] for the second one in (5)) similar to a triangle
area formula and Heron’s formula:
(5) V = 1
3 a01a02a03
psin θ sin(θ − θ01) sin(θ − θ02) sin(θ − θ03),
where θ = θ01+ θ02+ θ03 2
= 1
6 a01a02a03p
1 − cos2θ01− cos2θ02− cos2θ03+ 2 cos θ01cos θ02cos θ03.
From now, we consider volume formulas with a special condition.
In the case of a rectangular triangle, the rectangular condition for a tetrahedron, i.e., α01= α02= α03= π2, gives a simple volume formula:
(6) V (a01, a02, a03) = a01a02a03
6 .
If a tetrahedron has the three consecutive rectangular dihedral angles condition, i.e., α20 = α03 = α31 = π2, then we call it an orthoscheme or a birectangular tetrahedron T . Let x = a03 and y = a02; then, from the orthoscheme condition, we get 4 rectangular facial triangles of T . Therefore, we have a012+ y2 = a122 and x2+ a232 = y2; hence, we get x = √
a122− a012− a232. In addition, we know that V = 13a01A1 =
1
3a01(12xa23).
Therefore, we can simply represent its volume as (7) V (a01, a12, a23) = a01a23
6
pa122− a012− a232.
There is another volume formula [2] with the condition that T can be embedded as a face of a rectangular 4-simplex. This formula is expressed as
(8) V (a01, . . . , a23) = 1 6
p4u · p(p − a12)(p − a23)(p − a31) − q2,
where u = a122+ a302 = a132+ a202 = a102+ a232, p =a12+ a23+ a31
2 , q = a12a23a31.
We know that the three angles of any Euclidean triangle do not de- termine the triangle, since we have the α + β + γ = π angle condition.
Similarly, any Euclidean tetrahedron has a dihedral angle condition, i.e.,
the determinant of the Gram matrix is 0:
1 − cos β01 − cos β02 − cos β03
− cos β10 1 − cos β12 − cos β13
− cos β20 − cos β21 1 − cos β23
− cos β30 − cos β31 − cos β32 1
=
1 − cos α23 − cos α13 − cos α12
− cos α23 1 − cos α03 − cos α02
− cos α13 − cos α03 1 − cos α01
− cos α12 − cos α02 − cos α01 1
=0.
Thus, one dihedral angle is determined by the other 5 dihedral angles.
Therefore, if we want to derive a volume formula with all dihedral angles, we have to add one variable.
We know an area formula (V2 = 2R2sin α sin β sin γ) of a triangle with three angles and circumradius R, and a volume formula with all dihedral angles and circumradius R (see [4]), that is,
(9) V (α01, . . . , α23, R) = 32 3
N0N1N2N3 M32
R3, where
(10) Nl = 1 − cos2αli− cos2αlj − cos2αlk− 2 cos αlicos αljcos αlk
with {i, j, k, l} = {0, 1, 2, 3}
and
M = −
0 sin2α01 sin2α02 sin2α03
sin2α10 0 sin2α12 sin2α13 sin2α20 sin2α21 0 sin2α23 sin2α30 sin2α31 sin2α32 0
.
Note that there was a mistake in the sign of the last term of the definition of Nl in [4].
3. two volume formulas of a Euclidean tetrahedron
In Section 2, we surveyed nine volume formulas (1), (2), . . . , (9).
Now we introduce two new volume formulas of a Euclidean tetrahedron.
Let a, b, c denote the sides of a Euclidean triangle, and let α, β, γ denote the corresponding angles for three sides, respectively. Then, the
area (V2) of the triangle is given by V2 = 12ab sin γ. In addition, we have V2= 1
2c2 tan α tan β tan α + tan β = 1
2c2 sin α sin β sin α cos β + cos α sin β by using V2 =
1
2bc sin α·12ca sin β
1
2ab sin γ = 12c2 sin α sin β
sin γ = 12c2 sin α sin β sin(α+β).
There is a similar type volume formula for a tetrahedron correspond- ing to the above area formula.
Theorem 1. For three edges a12, a23, a31 of the facial triangle A0
with adjacent (to A0) dihedral angles α12, α23, α31, the volume V of the tetrahedron is given by
(11)
V =2
3A02 tan α12tan α23tan α31
a12tan α23tan α31+ a23tan α12tan α31+ a31tan α12tan α23
=2
3A02 sin α12sin α23sin α31
a12cos α12sin α23sin α31+ a23sin α12cos α23sin α31+ a31sin α12sin α23cos α31.
Proof. First, suppose that α12, α23, and α31 are acute angles. Let us calculate the length v0H, where v0H is a height with base [v1, v2, v3].
If we denote the feet of perpendiculars from H to [v1, v2], [v2, v3], and [v3, v1] as E, F , and G, respectively, then ∠v0EH, ∠v0F H, and ∠v0GH are the same to α12, α23, and α31, respectively. Let EH = x, F H = y, and GH = z. Then, 4v0EH, 4v0F H, and 4v0GH are right triangles.
Thus, we have
v0H = x tan α12= y tan α23= z tan α31, y = tan α12
tan α23
x, z = tan α12 tan α31
x.
In the plane [v1, v2, v3], its area A0 is given by 2A0 = a12x + a23y + a31z.
Here, we can obtain the value of x:
x = 2A0
a12+ a23tan α12
tan α23 + a31tan α12
tan α31
, and by substitution, we get
v0H = x tan α12
= 2A0tan α12tan α23tan α31
a12tan α23tan α31+ a23tan α12tan α31+ a31tan α12tan α23.
Therefore, we obtain V = 1
3A0· v0H
= 2
3A02 tan α12tan α23tan α31
a12tan α23tan α31+ a23tan α12tan α31+ a31tan α12tan α23. We considered only the case wherein α12, α23, and α31 are acute angles.
Furthermore, we can easily derive and verify the formulas for other cases.
In the expressions for V in Theorem 1, the first tangent expression requires all non-rectangular dihedral angles, i.e., α, β, γ 6= π2; however, the second expression dose not need to require the condition.
From now, we introduce another volume formula similar to the for- mula (9) with the inradius r instead of the circumradius R. We can find an area formula of a triangle with three angles and inradius r, and a volume formula with all dihedral angles and inradius r.
For a triangle, we know that V2 = Rr(sin α + sin β + sin γ) by V2 =
a+b+c
2 r and the sine law, and we have already obtained V2 = 2R2sin α sin β sin γ.
Hence, we obtain
V2 = 1 2
(sin α + sin β + sin γ)2 sin α sin β sin γ r2.
Now, let us derive a volume formula with dihedral angles and inradius r.
Theorem 2. For a tetrahedron with dihedral angles α01, . . . , α23and inradius r, its volume V is expressed as
(12) V (α01, . . . , α23, r) = 1 6
(√
N0+√
N1+√
N2+√ N3)3
√N0N1N2N3 r3. Proof. From the formula (3), we define
(13) a(i, j, k, l) := aijajkakl= 6V
sin θjlsin θkisin αjk
, also we obtain another formula from the formula (13):
(14) a(k, i, j, l) a(i, k, l, j) = aij
akl = sin θkjsin θlisin αkl sin θilsin θjksin αij.
So we can calculate akl by
(15) akl3 =
(alkakiaij)2
akl
aij
2
(aikaklalj)
aik
ajl
= 6V sin θilsin θjksin2αij sin2θkjsin2θlisin αklsin αiksin αjl. Also we can express sin θij in terms of dihedral angles αij by applying the spherical cosine law at the spherical triangle cut by a small sphere with center at a vertex of the tetrahedron.
sin2θij = 1 − cos2θij
= 1 − cos αikcos αil+ cos αij
sin αiksin αil
2
= 1 − cos2αij− cos2αik− cos2αil− 2 cos αijcos αikcos αil
sin2αiksin2αil . Hence we get (see (10))
(16) sin θij =
√Ni
sin αiksin αil.
By substitution of (15) and (16) into an area formula, we derive that Ai= 1
2aklajlsin θli
=
9V2sin2θilsin αijsin αik
2 sin θjksin θlisin θkjsin2αjlsin2αkl
13
= 9V2Ni
2pNjNkNl
!1
3
. Therefore we have
V = r
3(A0+ A1+ A2+ A3)
= r 3
9V2 2
13 N0
√N1N2N3
13 +
N1
√N0N2N3
13
+
√ N2
N0N1N3
1
3
+
√ N3
N0N1N2
1
3
!
and finally we get
V = r3 6
N0
√N1N2N3
13 +
N1
√N0N2N3
13 +
N2
√N0N1N3
13 +
N3
√N0N1N2
13!3
.
The condition of a regular tetrahedron implies that cos α = 13; hence, the volume of a regular tetrahedron with inradius r is 8√
3r3 by the above theorem.
A Coxeter diagram is a convenient means of describing tetrahedra.
From [5], there are only three Euclidean Coxeter diagrams of dimension 3.
These three Euclidean tetrahedra T (α01, α02, α03, α12, α13, α23), are exactly T (π2,π3, π3, π3, π3, π2), T (π4,π2,π3,π2,π2,π4) and T (π2,π2,π2,π4,π3,π3).
Then, the volumes of the three Euclidean Coxeter tetrahedra are calcu- lated by the above theorem.
Example 3 The volumes of Coxeter tetrahedra with inradius r are expressed as
V π 2,π
3,π 3,π
3,π 3,π
2, r
= 32√ 2 3 r3, V
π 4,π
2,π 3,π
2,π 2,π
4, r
= 28 + 20√ 2 3 r3, V π
2,π 2,π
2,π 4,π
3,π 3, r
= 25 + 22√ 2 3 r3.
To the best of our knowledge, (11) and (12) have not been derived in other studies; however, there is a possibility that they are known results.
Now, we list and summarize the formulas with variables:
(1) = V (a01, . . . , a23)
(2) = V (a01, . . . ,acij, . . . , a23, βij) (3) = V (aij, ajk, akl, θjl, θki, αjk) (4) = V (a01, a02, a03, α01, α02, α03) (5) = V (a01, a02, a03, θ01, θ02, θ03) (11) = V (a12, a23, a31, α12, α23, α31)
(9) = V (α01, . . . , α23, R) (12) = V (α01, . . . , α23, r).
The uniqueness (up to isometry) of a tetrahedron with the fixed variables condition is trivial, and hence, we need not show it here.
There is a possibility that we don’t know the existence of other nice and symmetric volume formulas of a Euclidean tetrahedron with some geometric variables.
So we end this section with the following problem.
Problem 4 Find nice and symmetric volume formulas of a Euclidean tetrahedron with the given geometric variables, and, if necessary, show the uniqueness (up to isometry) of the tetrahedron with the fixed vari- ables condition.
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Joon Suk Park
Department of Physics, Yonsei University, Seoul 120-749, Korea.
E-mail: [email protected]
Yunhi Cho
Department of Mathematics, University of Seoul, Seoul 130-743, Korea.
E-mail: [email protected]
