Constructing a Tetrahedron with Prescribed

Constructing a Tetrahedron with Prescribed Heights and Widths Lu Yang1,2 and Zhenbing Zeng1 1

2

Institute of Theoretical Computing, East China Normal University, Beijing 100083, China [email protected] Chengdu Institute of Computer Applications, Chinese Academy of Sciences, Chengdu 610041, China [email protected]

Abstract. Employing a method of distance geometry, we present a symbolic solution to the following problem: express the edge-lengths of a tetrahedron in terms of its heights and widths. Keywords: generalized Cayley-Menger algebra, widths of a tetrahedron, geometric constraint solving.

1

Introduction

A typical geometric constraint problem requires to find a configuration of points, lines and planes with prescribed pair-wise constraints between these geometric objects. A pair-wise constraint may be the distance or the angle between them. For example, in 5P1L problem, one considers a configuration of 5 points and 1 line in 3-space that is constrained as in Fig. 1. All constraints are distances. Now, let us discuss an nontypical geometric constraint problem. Given are a set of constraints on four points P1 , P2 , P3 , P4 in E3 : the distances from each point to the plane determined by the other three points, namely, d(P1 , P2 P3 P4 ), d(P2 , P3 P4 P1 ), d(P3 , P4 P1 P2 ), d(P4 , P1 P2 P3 ), and the distances from the line determined by each pair of points to the line determined by the other two, namely, d(P1 P2 , P3 P4 ), d(P1 P3 , P2 P4 ), d(P1 P4 , P2 P3 ). We want to find a realization of the four-point-configuration. In other words, how to reconstruct a tetrahedron from its heights and widths. Here P1 , P2 , P3 , P4 stand for the vertices of a tetrahedron, and h1 , h2 , h3 , h4 the heights respectively. By width we mean the distance between a pair of edges with no intersection. By τij denote the width between Pi Pj to its opposite, for 1 ≤ i < j ≤ 4. Clearly that τ12 = τ34 , τ13 = τ24 , τ14 = τ23 . So each tetrahedron has three

This work is supported in part by NKBRPC-2004CB318003 and NNSFC-10471044.

F. Botana and T. Recio (Eds.): ADG 2006, LNAI 4869, pp. 203–211, 2007. c Springer-Verlag Berlin Heidelberg 2007

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Pm Pm 3 4 PP @ P P PP PP @ @ PP @ Pm Lm 5 P PP @ PP PP @ P P @ P @ Pm Pm 1 2 Fig. 1. The 5P 1L problem: graph vertices represent 5 points and 1 line, while graph edges distances

widths, namely, τ12 , τ13 , τ14 . The tetrahedron reconstructing means to express the six edge-lengths dij = Pi Pj in terms of the heights h1 , h2 , h3 , h4 and widths τ12 , τ13 , τ14 . Since any tetrahedron has 6 freedoms up to isometries, it is possible to establish a formula connecting the seven quantities h1 , h2 , h3 , h4 , τ12 , τ13 , τ14 The following, which was proved repeatedly in literatures (see [2] and [10] for the history and proofs), gives a very simple equality. Lemma 1. The following relation 1 1 1 1 1 1 1 + 2+ 2+ 2 = 2 + 2 + 2 2 h1 h2 h3 h4 τ12 τ13 τ14 holds for any tetrahedron. So the above problem can also be formulated as: Given any six quantities out of the four heights and three widths of a tetrahedron, can we express the six edge-lengths in terms of the given quantities? This problem has been mentioned by some geometers in different occasions before but still remained unsolved. In this paper we shall present a solution to this problem by making use of the so-called metric equations and the Jacobi’s Theorem about the minors of a matrix. The key is the following formula connecting the dihedral angles, heights and widths of a tetrahedron. Theorem 1. Let hi and hj be the heights, τij the width of a tetrahedron. Draw a triangle ABC with AB =

1 , hi

AC =

1 , hj

BC =

1 , τij

then

A = θij ,

where θij stands for the dihedral angle opposite to the edge Pi Pj . In view of this theorem, we may at first construct a tetrahedron similar to that with given heights and widths, and then compute the similarity constant.

Constructing a Tetrahedron with Prescribed Heights and Widths

205

The paper is organized as follows: §2 gives a brief description to the generalized Cayley-Menger Algebra and uses it to derive heights and widths in terms of the edge-lengths (Lemma 2); §3 introduces a formula for dihedral angles (Lemma 3) and proves Theorem 1; §4 shows a method to determine the six edge-lengths of a tetrahedron from its dihedral angles. Lemmas 2 and 3 were established in earlier works of one of the authors of this paper with other collaborators; we shall give a sketch to their proofs and the prerequisite for convenience to readers.

2

Generalized Cayley-Menger Algebra

A very natural attempt to represent the edge-lengths of a tetrahedron with its heights and widths is firstly to establish certain formulas for expressing heights and widths in terms of the edge-lengths, and then solve these equations by taking the edge-lengths as unknowns. It is very easy to compute the heights from the edge-lengths since a height equals three times the quotient of the volume V by the area of a facet, that is, V =

1 · A(P2 P3 P4 ) · h1 , 3

where A(P2 P3 P4 ) is the area of triangle P2 P3 P4 , and both the volume and area can be expressed in terms of edge-lengths by using Cayley-Menger determinants, V2 =

1 · D(P1 , P2 , P3 , P4 ), 23 · (3!)2

A(P2 P3 P4 )2 = −

1 · D(P2 , P3 , P4 ), 22 · (2!)2

where the Cayley-Menger determinants associated to point sets {P1 , P2 , P3 , P4 } and {P2 , P3 , P4 } are defined by 0 d212 d213 d214 1 2 d12 0 d223 d224 1 2 2 D(P1 , P2 , P3 , P4 ) = d13 d23 0 d234 1 , d214 d224 d234 0 1 1 1 1 1 0 and

0 2 d D(P2 , P3 , P4 ) = 23 2 d24 1

d223 0 d234 1

2 g24 d234 0 1

1 1 ; 1 0

and the corresponding matrices are called Cayley-Menger matrices associated to {P1 , P2 , P3 , P4 } and {P2 , P3 , P4 }, respectively (see [1,6]). The formulas for heights can be written in short form as follows. For any n × n matrix M and integers j, k satisfying 1 ≤ j, k ≤ n, by Mj,k denote the j, k

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minor of M . Let M and D be the Cayley-Menger matrix and Cayley-Menger determinant associated to the vertices of a tetrahedron P1 P2 P3 P4 respectively, then the heights of the tetrahedron can be expressed in following way. h21 = −

D , 2M1,1

h22 = −

D , 2M2,2

h23 = −

D , 2M3,3

h24 = −

D . 2M4,4

The Cayley-Menger Algebra was generalized in 1980s-1990s([12,13,14,15,9,11]) to configurations consisting of points, oriented hyperplanes and hyperspheres (shown as in Fig.2) in En . normal vector

'$

r o &% (O, r) with r > 0

normal vector

'$

r o &% (O, r) with r < 0

Fig. 2. oriented hyperspheres in the plane

In the following we quote the generalized Cayley-Menger Algebra for pointplane configurations. Given an n-tuple of points and oriented hyperplanes in an Euclidean space, P = (p1 , p2 , · · · , pn ), we define a mapping g : P × P → R by letting – g(pi , pj ) be the square of the distance between pi and pj if both are points, – g(pi , pj ) be the signed distance from pi to pj , if one is a point and the other an oriented hyperplane, 1 – g(pi , pj ) be − cos(pi , pj ), if both are oriented hyperplanes. 2 By gij denote g(pi , pj ) and G denote the matrix (gij )n×n , and let δ = (δ1 , δ2 , · · · , δn )

where δi =

1, if Pi is a point, 0, if Pi is a hyperplane

for i = 1, . . . , n. Then, let M (p1 , p2 , · · · , pn ) =

G δT δ 0

which is called the Cayley-Menger matrix of P, and let G δT D(p1 , p2 , · · · , pn ) = δ 0 which is called the Cayley-Menger determinant of P. The following theorem is an extension to the classical Cayley-Menger determinant.


207

Theorem 2. Let D(p1 , p2 , · · · , pn ) be the Cayley-Menger determinant of an ntuple of points and oriented hyperplanes in d-dimensional space. If n ≥ d + 2, then (1) D(p1 , p2 , · · · , pn ) = 0. By appropriately selecting the sets of the geometric elements, formula (1) creates polynomial equations connecting widths and edge-lengths of a tetrahedron. For instance, construct a plane Π12 that passes through points P3 , P4 and parallels to line P1 P2 , and consider the configuration formed by the following five elements: p1 = P1 , p2 = P2 , p3 = P3 , p4 = P4 , , p5 = Π12 , as in Figure 3.

sP1 EQQ E Q Q E Q c Q c E τ Q 12c E Q cE Qs P4 P2 s c @ @ E c I @ @ E c @ @ c E Π12 @ E @ E @ Es P@ 3 Fig. 3. The plane Π12 passes points P3 , P4 and parallels to line P1 P2

Following the notations in the last page, for 1 ≤ i, j ≤ 4, we have that gij := g(pi , pj ) = Pi Pj2

and

1 g(p1 , p5 ) = g(p2 , p5 ) = τ12 , g(p3 , p5 ) = g(p4 , p5 ) = 0, g(p5 , p5 ) = − . 2 Substituting to the metric equation we get 0 g12 g13 g14 τ12 1 g12 0 g23 g24 τ12 1 g13 g23 0 g34 0 1 = 0. g14 g24 g34 0 0 1 τ12 τ12 0 0 −1/2 0 1 1 1 1 0 0 This immediately implies the next lemma.

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Lemma 2. Let P1 P2 P3 P4 be a tetrahedron, M and D the Cayley-Menger matrix and Cayley-Menger determinant associated respectively, then 1 1 2M1,1 + 2M2,2 − 4M1,2 2M1,1 + 2M3,3 + 4M1,3 , 2 = , 2 = τ12 D τ13 D 1 2M1,1 + 2M4,4 − 4M1,4 . 2 = τ14 D This result can be found in [10]. Combining this with the height formula it is very easy to prove Lemma 1 by the following computation: (M1,1 + M2,2 − 2M1,2 ) + (M1,1 + M3,3 + 2M1,3 ) + (M1,1 + M4,4 − 2M1,4 ) = −M1,1 − M2,2 − M3,3 − M4,4 . If some six out of the seven quantities, four heights and three widths, are given as constant numbers, we can get the seventh by Lemma 1 and obtain the edge-lengths from height and width formulas by solving a set of equations. But if heights and widths are given in symbols, it is still too complicated to do elimination. Our strategy for this problem is to find certain geometric invariants as intermediate variables that way maybe makes the matter easier. Therefore, the dihedral angles come to mind naturally.

3

Dihedral Angles as Intermediate Variables

In what follows we recall a known result for computing the dihedral angles of a tetrahedron from its edge-lengths. Lemma 3 (Law of Cosine of Dihedral). Let P1 1P2 P3 P4 be a tetrahedron, M its Cayley-Menger matrix, and θij the dihedral angle opposite to edge Pi Pj . Then −Mi,j cos(θij ) = Mi,i · Mj,j for 1 ≤ i < j ≤ 4. A proof of this lemma can be found in [14]. It involves the application of Theorem 2 combining with the Jacobi’s theorem about the minors of matrices (see [7,4,3,8]). Theorem 3 (Jacobi’s Theorem). For any n×n matrix M and 1 ≤ j < k ≤ n, the equality 2 Mj,j · Mk,k − Mj,k = Mj,j ;k,k · det(M ) holds, where Mj,j ;k,k stands for the determinant of the submatrix of M by deleting j, k-th rows and j, k-th columns . We just show the sketch of a proof to the Lemma 3. Consider the geometric configuration formed by four vertices of a tetrahedron and two facets P2 P3 P4 and P1 P3 P4 . Let p1 = P2 P3 P4 , p2 = P1 P3 P4 , p3 = P1 , p4 = P2 , p5 = P3 , p6 = P4 .


and

⎡

−1/2 − cos(θ12 )/2 h1 ⎢ − cos(θ12 )/2 −1/2 0 ⎢ ⎢ h 0 0 1 ⎢ g 0 h M =⎢ 2 12 ⎢ ⎢ 0 0 g 13 ⎢ ⎣ 0 0 g14 0 0 1

0 h2 g12 0 g23 g24 1

0 0 g13 g23 0 g34 1

0 0 g14 g24 g34 0 1

209

⎤ 0 0⎥ ⎥ 1⎥ ⎥ 1⎥ ⎥, 1⎥ ⎥ 1⎦ 0

the corresponding Cayley-Menger matrix, where θ12 is the dihedral angle opposite to edge P1 P2 . Then we have det(M ) = 0,

M1,1 = 0,

M2,2 = 0

according to Theorem 2, hence M1,2 = 0 according to Theorem 3. Expanding the determinant M1,2 , and substituting the height formula for h1 , h2 into it, we get the law of cosine for the tetrahedron as in Lemma 3. Now we are ready to prove Theorem 1. Proof to Theorem 1. Let ABC be a triangle formed by AB = Then we have

cos(A) =

1 , hi

AC =

1 , hj

1 1 1 + 2− 2 h2i hj τij

BC =

2·

1 . τij

1 1 · hi hj

.

Substitute the formulas for heights and widths into it, we get −Mi,j , cos A = Mi,i · Mj,j and therefore, A = θ12 according to Theorem 3.

4

Determine Edge-Lengths by Dihedral Angles

By means of Theorem 1, we can determine the shape (that is, all the dihedral angles) of a tetrahedron if given the heights and widths. In what follows we show a way to compute edge-lengths via the dihedral angles. The following result ([5]) is the key for this task. Lemma 4. Let P1 P2 P3 P4 be a tetrahedron, Ai the area of the facet opposite to Pi for i = 1, 2, 3, 4, dij the edge-length of Pi Pj , θij the dihedral angle opposite to Pi Pj , 1 ≤ i < j ≤ 4, and V the volume. Then V =

2 Ai Aj sin(θij ). 3dij

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Combining this with Lemma 2, 1 2M1,1 + 2M2,2 − 4M1,2 , = τ12 2 D and the height formula A1 =

3V , h1

A2 =

3V h2

we get the following procedure for computing edge-lengths: 36 sin2 (θij ), 1 ≤ i < j ≤ 4, h2i h2j ⎡ ⎤ 0 0 0 0 g12 g13 g14 1 0 0 0 ⎢ g12 1⎥ ⎢ 0 00 g23 g24 ⎥ 0 ⎢ m = ⎢ g13 g23 0 g34 1 ⎥ ⎥, 0 0 0 ⎣ g14 g24 g34 0 1⎦ 1 1 1 1 0 2m1,1 + 2m2,2 − 4m1,2 V = · τ12 , det(m) 0 = gij

dij =

6V sin(θij ), hi hj

1 ≤ i < j ≤ 4.

Now, the tetrahedron has been reconstructed as we asked for.

Acknowledgement The authors would like to extend their thanks to the anonymous referee for valuable comments and suggestions.

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