# «1. INTRODUCTION. It is well known that the three altitudes of a triangle are concurrent at the so-called orthocenter of the triangle. So one might ...»

Altitudes of a Tetrahedron

and Traceless Quadratic Forms

Hans Havlicek Gunter Weiß

1. INTRODUCTION. It is well known that the three altitudes of a triangle are

concurrent at the so-called orthocenter of the triangle. So one might expect that

the altitudes of a tetrahedron also meet at a point. However, it was already pointed

out in 1827 by the Swiss geometer Jakob Steiner (1796–1863) that the altitudes of

a general tetrahedron are mutually skew, for they are generators of an equilateral hyperboloid. This is a hyperboloid with the following rather peculiar property: each nontangential plane that is perpendicular to a generator meets the hyperboloid along an equilateral hyperbola, i.e., a hyperbola with orthogonal asymptotes.

There are many papers, especially from the nineteenth century, that deal with the altitudes of tetrahedra and related topics. An excellent survey article with many historical footnotes is the paper of N.A. Court [7]. Another major source is the article of M. Zacharias in the Encyclopedia of Mathematical Sciences [26] (completed in 1913). Also, the book of H. Schr¨ter [24] contains in section 28 some interesting o remarks on older papers that are not cited elsewhere.

Some of the many special points and lines associated with triangles extend to tetrahedra, others do not. More precisely, there are often a number of possibilities for generalizing a concept (like the orthocenter) from the plane to 3-space (or even n-space), since separate notions in higher-dimensional geometry may coincide in the plane. We illustrate this in section 3, where we discuss the Monge point of a tetrahe- dron. If the deﬁnition of the Monge point is applied directly to the planar case, then the orthocenter of a triangle is obtained. Table 1 starts with some well-known “note- worthy points” of a triangle and lists their existence in higher dimensions, together with short remarks or references.

Dimension n=2 n=3 n≥4 Object Triangle Tetrahedron n-Simplex Centroid G Yes Yes Yes (See [20], [22].) Circumcenter C Yes Yes Yes (See [20], [22].) Incenter I Yes Yes Yes (See [20], [22].) Orthocenter H Yes Sometimes Sometimes (sec. 2) (See [17], [20].) Monge point M Yes: M = H Yes (sec. 3) Yes (See [17], [20].) (sec. 3) Euler line e Yes Yes (sec. 4) Yes (See [17], [20].) (G, C, M ∈ e) (G, C, H ∈ e) Table 1.

Clearly, this list is far from being complete. For example, Feuerbach’sfamous 9-point circle for the triangle is generalized in [10] to the n-dimensional case as a 1 Feuerbach hypersphere, and in [7] two 12-point spheres for an orthocentric tetrahedron are described. However, in higher dimensions many questions are still open and have not been treated systematically. Further references are given in section 7.

There are two major reasons for revisiting the subject “altitudes of a tetrahedron.” On the one hand, we want to visualize the results, since there are hardly any pictures in the cited papers. The ﬁgures in this article have been prepared with the computer algebra system Maple. Clearly, a ﬁgure must not replace a formal proof, but we are convinced that ﬁgures can assist the reader in better understanding spatial relationships. Also, in certain instances the idea behind a proof is rather immediate from an adequate illustration.

On the other hand, we aim at a modern coordinate-free presentation in terms of analytic geometry based on a Euclidean vector space, whereas a lot of papers on the subject are written in terms of synthetic projective geometry. Thus the prerequisite for reading this article is knowledge of basic linear algebra.

We shall see that the quadric surface carrying the altitudes of a general tetrahedron appears as a level set of a traceless (trace = 0) quadratic form Q∗. In fact, there is a natural link between tetrahedra and certain traceless quadratic forms that will lead us to an explicit expression for Q∗.

It follows from (4) and (2) that any two of these equations are linearly independent, whereas (3) implies that all three equations are linearly dependent. Thus (7) describes three diﬀerent planes that meet at the altitude hl, any two of which are already suﬃcient to determine hl. Figure 1 shows those planes through the altitude h3 that are perpendicular to A0 A1 and A0 A2, respectively.

We now consider those uniquely determined planes through Ak, Ai, and Aj that are parallel to the ﬁrst, the second, and the third plane of (7), respectively. This yields the linear system

Proof. We read oﬀ from the second equation in (7) (after interchanging i with l) and the second equation in (8) that hi and nl are both contained in the plane with

In Figure 2 the line n3 meets three out of the four altitudes of T. Also, under orthogonal projection onto the plane A0 A1 A2 the line n3 is mapped to the orthocenter of the triangle with vertices A0, A1, A2, and the altitudes h0, h1, and h2 project to the altitudes of that triangle. We can see that the foot of h3 is on no altitude of the opposite triangle. Hence h3 does not meet any other altitude of the tetrahedron. So we can seek a criterion for deciding whether or not two altitudes meet at a point.

It will be convenient to say that two lines (with or without a common point) are orthogonal if their direction vectors are orthogonal.

Theorem 2. An altitude hi meets an altitude hj precisely when

Now (10) implies that bkl · ai = bkl · aj. So two of these four equations are the same.

Hence hi ∩hj = ∅, for by (4) the three planes given by those equations are not parallel to a line.

Conversely, let P be a common point of hi and hj. Put p = OP. We infer from

that (10) is satisﬁed.

By symmetry, condition (10) is also necessary and suﬃcient for the altitudes hk and hl to have a point in common. Hence, if two altitudes intersect, then so do the other two. Moreover, the reader will easily verify that (10), ni ∩ nj = ∅, and nk ∩ nl = ∅ are mutually equivalent.

**At a ﬁrst glance the following result may be somewhat surprising:**

** Theorem 3. If one altitude meets two other altitudes then all altitudes are concurrent.**

Proof. Suppose that hi meets hj and hk. Then (10) implies bkl · bij = bjl · bik = 0 and (5) shows that bil · bjk = 0, i.e., hi meets also hl. By Theorem 2, any two altitudes intersect. Clearly, the four altitudes are not coplanar. So they have a common point.

4 A tetrahedron with exactly one pair of opposite orthogonal edges is called semiorthocentric. Figure 3 shows such a tetrahedron (A0 A3 ⊥ A1 A2 ). Observe that the foot of h3 is on exactly one altitude of the opposite triangle. In Figure 4, however, all four altitudes are concurrent at a point. Such a tetrahedron is said to be orthocentric, the orthocenter being the point where the altitudes meet. In an orthocentric tetrahedron every edge is orthogonal to its opposite edge. Furthermore, each altitude hi coincides with the orthocentric perpendicular ni.

We refer to [1, p. 371], [6], and [9] for other proofs of Theorems 1, 2, and 3.

3. THE MONGE POINT OF A TETRAHEDRON. For an orthocentric tetrahedron the intersection of all planes given by (6) is its orthocenter. However, we focus on an arbitrary tetrahedron T. Then among the planes (6) there are two that are perpendicular to Ai Aj and pass through Ak and Al, respectively. These two planes are either identical or disjoint. (In Figure 1 two such pairs of parallel planes can be seen.) In either case their midplane is the plane with equation

** bij · (ak + al − 2x) = 0. (11)**

This midplane is orthogonal to the edge Ai Aj, and it passes through the midpoint of the opposite edge Ak Al rather than the midpoint of the edge Ai Aj, as the perpendicular bisector of Ai Aj does. In general, the midplane (11) is therfore not the perpendicular bisector of the edge Ai Aj. The equation of that plane is given in equation (13). The tetrahedron T has six midplanes.

Now for an arbitrary tetrahedron T there will be no orthocenter, but T will have a point discovered by Gaspard Monge (1746–1818) that is now called its Monge point.

**The construction of this point goes as follows:**

** Theorem 4. All six midplanes of a tetrahedron are concurrent at a point.**

Proof. From (4) we know that the vectors b01, b02, and b03 are linearly independent.

Hence there exists a unique common point, say M, of the (mutually nonparallel) planes whose equations are

where we have used equations (2), (3), and (9). So M lies in the midplane that is perpendicular to the edge A1 A2. A similar calculation shows that M belongs to the remaining midplanes as well.

Figure 5 displays a tetrahedron and its Monge point M. It lies on that line of the plane spanned by the parallel lines hi and ni which is equidistant from both; one such line (dotted) is illustrated for i = 3. If we deform T by “sliding” the vertex A3 down along the line h3, holding all other vertices ﬁxed, then the center of gravity of T remains an inner point of T, while the circumcenter of T approaches “negative inﬁnity,” since the radius of the circumscribed sphere of T tends to inﬁnity. We shall see in section 4 that the center of gravity of T is the midpoint of M and the circumcenter of T. Hence the Monge point M “moves up” along the dotted line.

This gives Figure 6, in which the Monge point is exterior to the tetrahedron. For the proper choice of A3 we could also produce a tetrahedron whose the Monge point is incident with a face.

A3

Let m := OM be the position vector of the Monge point M described in Theorem

4. Then (11) implies that bij · (ak + al − 2m) = bkl · (ai + aj − 2m) and multiplying this out leads to the relation

This pretty equation illustrates in another way the special role of the point M.

Clearly, if all altitudes of T are concurrent, then the Monge point M is their point of intersection. This is one rationale for the assertion that the Monge point serves as a substitute for the “missing orthocenter” of a general tetrahedron.

We sketch the deﬁnition of the Monge point in a more general context in order to illustrate that the orthocenter of a triangle can be considered as its Monge point.

An n-simplex S in n-dimensional Euclidean space has n + 1 vertices. For each edge there is a unique hyperplane that is perpendicular to that edge and that contains the 6 center of gravity of the n − 1 “opposite vertices,” i.e., the vertices not on the given edge. There are n+1 such hyperplanes, and they have a point in common—the 2 Monge point of the n-simplex (see [17] or [20]). For n = 3 (tetrahedron) this is in accordance with Theorem 4, since each edge has two opposite vertices whose center of gravity is just their midpoint. For n = 2 (triangle) each edge has a single opposite vertex that is its own center of gravity. In this situation, the three altitudes of the triangle replace the hyperplanes from the general case, and their common point (the orthocenter) coincides with the Monge point.

4. THE EULER LINE IN SPACE. We consider the circumcenter C, the center of gravity G, and the Monge point M of the tetrahedron T with vertex set {A0, A1, A2, A3 }. An equation

describes the perpendicular bisector of the edge Ai Aj, so that OC is the only solution of the linear system of all six equations (13).

If we ﬁx indices i and j, then the midplane of the planes represented by (11) and (13) has an equation of the form

This plane contains the midpoint of the segment CM. From (4) we know that there are three linearly independent vectors among the bij, whence the only solution of the linear system comprising all equations (14) is 1 OG = (ai + aj + ak + al ).

4 This means that the center of gravity is the midpoint of the segment CM. In other words, whenever two of the points G, C, and M are diﬀerent, their join can be considered as an analog of the Euler line in 3-space. (We remind the reader that the Euler line of a triangle contains the center of gravity, the circumcenter, and orthocenter of the given triangle.)

**We add in passing that in the n-dimensional setting there is also an Euler line:**

the center of gravity divides the segment formed by the circumcenter and the Monge point internally in the ratio 2 : (n − 1) (see [17] or [20]). This explains why in the plane (orthocenter = Monge point) the ratio on the Euler line is 2 : 1, whereas in three dimensions it is 2 : 2.

If we change to another orthonormal basis, then Σ changes to a congruent matrix Σ = ΩT ΣΩ, where Ω is an orthogonal matrix and ΩT denotes the transpose of Ω.

As Ω is orthogonal, we have ΩT = Ω−1. The matrices Σ and Σ are thus similar. It is well known that similar matrices have the same trace. So it makes sense to speak of the trace tr Q of a quadratic form Q, as long as we restrict ourselves to orthonormal bases.

Consider, for example, arbitrary vectors c and d of E with Cartesian coordinates (γ1, γ2, γ3 ) and (δ1, δ2, δ3 ), respectively, and the quadratic form

and there are several cases to consider. These depend on the rank of Q, meaning the rank of any associated matrix.

Case 1: rank Q ≤ 1. Then at most one coeﬃcient σrr is nonzero and (18) shows that σ11 = σ22 = σ33 = 0, i.e., Q is the zero-form.

Case 2: rank Q = 2. Then the basis can be chosen in such a way that σ33 = 0.

This means that σ22 = −σ11 = 0 and Q(x) = σ11 (ξ1 − ξ2 )(ξ1 + ξ2 ). Consequently, Q(x) = 0 describes a pair of orthogonal planes with equations ξ1 = ξ2 and ξ1 = −ξ2, respectively.

Case 3: rank Q = 3. Then Q is indeﬁnite by (18) and Q(x) = 0 is the equation of a quadratic cone (with vertex at the origin) that is called an equilateral cone. Its generators (i.e., the lines entirely contained in the cone) have the following remarkable