On the number of tetrahedra with minimum,
unit, and distinct volumes in threespace^{†}^{†}thanks: A preliminary
version of this paper appeared in the Proceedings of the 18th
ACMSIAM Symposium on Discrete Algorithms (New Orleans, LA,
2007), ACM Press, pp. 11141123.
Abstract
We formulate and give partial answers to several combinatorial problems on volumes of simplices determined by points in 3space, and in general in dimensions. (i) The number of tetrahedra of minimum (nonzero) volume spanned by points in is at most , and there are point sets for which this number is . We also present an time algorithm for reporting all tetrahedra of minimum nonzero volume, and thereby extend an algorithm of Edelsbrunner, O’Rourke, and Seidel. In general, for every , , the maximum number of dimensional simplices of minimum (nonzero) volume spanned by points in is . (ii) The number of unitvolume tetrahedra determined by points in is , and there are point sets for which this number is . (iii) For every , the minimum number of distinct volumes of all fulldimensional simplices determined by points in , not all on a hyperplane, is .
1 Introduction
Typical Erdős type problems in extremal discrete mathematics ask for the minimum or maximum number of certain configurations over all inputs of a given size. They are easy to formulate but often extremely hard to answer. Their impact on mathematics and computer science has been enormous, not only because of specific algorithms based on combinatorial bounds but also because they have triggered the development of theoretical and practical methods that turned out to be applicable elsewhere.
Some of the most simply formulated yet notoriously hard Erdős type problems occur in combinatorial geometry. In 1946, Erdős [21] asked two questions on distances: (1) at most how many times can a given distance occur among points in the plane; (2) what is the minimum number of distinct distances determined by points in the plane? The difficulty of the, so called, unit distance and distinct distance problems is still to be measured. Erdős and Purdy [23, 24] generalized the unit and distinct distance problems to congruent (or repeated) simplices: What is the maximum number of congruent dimensional simplices among points in , for ? No asymptotically tight bound is known for this problem for , (and it is trivial for by Lenztype constructions). Pach and Sharir [31] derived bounds on the maximum number of occurrences of the same angle determined by a planar element point set; recently Apfelbaum and Sharir [3] studied the analogous problem in threespace. A recent book by Braß, Moser, and Pach [8] and a survey of Pach and Sharir [32] provide substantial details on these and other similar problems. In the sequel, we focus on problems about the extremal number of simplices with certain volume properties in a point set.
In 1967, A. Oppenheim (see [22]) asked what is the maximum number of unit area triangles determined by points in the plane. Erdős and Purdy gave an upper bound, and also showed that a suitable section of the integer lattice yields such triangles [23]. The currently best upper bound, , due to Dumitrescu, Sharir and Cs. Tóth [15], recently improved an older bound of Pach and Sharir [31]. Answering further questions of Erdős and Purdy [23], Braß, Rote, and Swanepoel [9] showed the following two results: (1) The maximum number of triangles of maximum area (or of maximum perimeter) determined by points in the plane is exactly . (2) The maximum number of triangles of minimum (nonzero) area determined by points in the plane is .
In 1982, Erdős, Purdy, and Straus [35, 25] considered the generalization of the problem of distinct triangle areas to higher dimensions and posed the following problem: Let be a set of points in not all in a hyperplane. What is the minimal number of distinct volumes of fulldimensional simplices with vertices in ? It is easy to see that by taking sets of about equally spaced points on parallel lines through the vertices of an simplex. Erdős, Purdy, and Straus conjectured that equality holds at least for sufficiently large (see also [13]). Very recently, Pinchasi [33] confirmed the conjecture in the plane after earlier work by Burton and Purdy [10] and Dumitrescu and Cs. Tóth [14]: The minimum number of distinct (nonzero) triangle areas determined by noncollinear points is . In this paper we give a first partial result for by proving for every , which is optimal up to a multiplicative constant. As expected, the three dimensional analogues of combinatorial problems in the plane are often much harder. Only the recent few years saw intensifying work on these problems in threespace [3, 4, 5, 26, 29, 34].
Algorithmic problems on sets of points in Euclidean space have been often attacked using the duality transform and the machinery of constructing hyperplane arrangements. The first applications of this technique can be found in early works of Chazelle, Edelsbrunner, Guibas, Lee, O’Rourke and Seidel [11, 17, 18]: among others, the problem of computing a minimum area triangle in a given set of points. A time algorithm for such a task was given independently in [11] and [17]. If there are degeneracies in the set, the algorithm returns zero area and a triplet of collinear points. An extension of the algorithm for finding a minimum volume simplex among points in was given in [18]: it runs in time, but again reports zero volume if there are points on a hyperplane. Here we further extend the algorithms for two and three dimensions to report all simplices of minimum nonzero volume, within the same and running times, respectively.
Our contribution.
In this paper, we address classical problems on minimum, unit, and distinct volume simplices. We show that every set of points in threespace determines minimum (positive) volume tetrahedra, and there are points sets in that span minimum volume tetrahedra. Our techniques generalize to arbitrary dimensions: For every , , any set of points in determines minimum (positive) volume dimensional simplices (each having vertices), and there are points sets in that span minimum volume dimensional simplices. In threespace, we also give a time algorithm for reporting all tetrahedra of minimum nonzero volume, and thereby extend an early algorithm of Edelsbrunner, O’Rourke, and Seidel. We prove that every set of points in determines at most unitvolume tetrahedra, and there are point sets that span unitvolume tetrahedra. Finally, we show that for every , any set of points in , not all in a hyperplane, determines at least fulldimensional simplices of distinct volumes, and there are point sets for which this number is ; this gives a first answer to the question of Erdős, Purdy, and Straus.
Organization.
Section 2 presents a collection of tools we use from Euclidean geometry, previous results and extensions of previous results adapted to our goals. We prove our main theorems on the number of minimum, unit, and distinct volume tetrahedra in Sections 3, 4, and 5, respectively. To simplify notation, we assume in our proofs that is an integer.
2 Toolbox
We will frequently apply a classic result of Szemerédi and Trotter [36] on the number of pointlines incidences in the plane. (Since collinear points remain collinear under any affine transformation, the result holds in arbitrary dimensions.) The constant factor hidden in the asymptotic notation has been significantly improved by the cutting method and by the theory of crossing numbers [32]; the current best constant is due to Pach et al. [30]. The SzemerédiTrotter bound comes in two equivalent formulations; we also state two immediate corollaries that we use in our proofs. Given a point set in , for any integer , a line is called rich if it is incident to at least points of . We denote by the set of rich lines.
Theorem 1
(SzemerédiTrotter [36]). Given points in , , the number of rich lines, , is
Theorem 2
(SzemerédiTrotter [36]). The number of pointline incidences among points and lines in , , is
Corollary 1
Given points in , , the number of pointline incidences among the points and the rich lines, , is .
Note that in Theorem 2, the term is responsible for the incidences on lines containing a single point.
Corollary 2
Given points and lines in , , each line containing at least two points, the number of pointline incidences among them is .
Incidence bounds have been key components in many results in combinatorial geometry. Some SzemerédiTrottertype bounds were found for pointplane incidences in the space but they are either not known to be tight or they hold in severely restricted settings only (e.g., [7, 12, 20, 27]). Instead of these multidimensional bounds, we apply planar incidence bounds on the projections of a finite point set onto planes in certain directions determined by . The SzemerédiTrotter bound on pointline incidences is tight in the worst case, but does not hold for points with multiplicities. Since the planar projection of a threedimensional point set may have an irregular distribution of multiplicities, we partition the point sets (and the planar multiset) into subsets of roughly the same multiplicities. We demonstrate this technique in Section 4, where we aggregate SzemerédiTrotter bounds for various substructures and give an upper bound on the number of incidences of lines and points with multiplicities in the plane, and ultimately a bound on incidences in threespace.
2.1 Degenerate and nondegenerate planes
We extensively apply the concepts of degenerate and nondegenerate planes in threespace. They were introduced recently by Elekes and Tóth [20]—here, we use the same concepts with a different terminology. Consider a point set . For a constant , , we say that a plane is degenerate if at most points of are collinear. In this paper, we fix and say that a plane is nondegenerate if it is degenerate; otherwise is degenerate. A nondegenerate plane is always spanned by , while a degenerate plane may contain only one point, or collinear points. If lies in a plane , we refer to as degenerate or nondegenerate according to the above definition.
We apply several consequences of a well known combinatorial geometric result due to Beck [6].
Lemma 1
(Beck [6]). Consider a set of points in the plane. If at most points of are collinear, then determines at least distinct lines.
Corollary 3
Consider a set of noncollinear points in the plane. If at most points of are collinear, then determines at least nondegenerate triangles.
Proof. If every line contains fewer than points, then by Lemma 1, determines line segments. If a line contains points, then there are segments along this line. In either case, each of these line segments is the base for at least distinct nondegenerate triangles. Since we count every triangle at most three times, we obtain distinct triangles.
For a set of points in the plane, let denote the number of unit area triangles determined by . Let be a constant such that for every set of points in the plane. It is conjectured that can be arbitrarily small; we can assume due to the bound of Pach and Sharir [31] (the current best bound gives , cf. [15]). We use the general parameter when we further chisel the bound on unit area triangles in the special case of degenerate planes. We apply these results in Section 4 in a charging scheme, where we distribute unit area triangles spanned by a planar point set among line segments in that plane.
Lemma 2
Let be a set of points in the plane, and be a line incident to exactly points of . Then determines at most unit area triangles.
Proof. Put , and let us denote the lines parallel to containing at least one point of by , where is the number of such lines. For every , let . We partition the unit area triangles into three subsets:

counts triangles with all three vertices in ;

counts triangles with two vertices on and one in ;

counts triangles with one vertex on and two in ;
(i) By the assumption, .
(ii) For every line , , there are at most unit area triangles , with and . Therefore, .
(iii) determines less than line segments . For segments nonparallel to , there are at most two vertices such that has unit area (because has to lie on one of two lines parallel to ), so we have less than unit area triangles of this kind. For segments on a line parallel to , there are at most unit area triangles , with and . Summing over all lines , there are at most triangles of unit area (since ) of this second kind. Consequently . Altogether we have at most unit area triangles in .
Corollary 4
There is a constant with the following property. If a set of points in the plane determines nondegenerate triangles, then there is a subset of points and for each there is a set of pairwise nonoverlapping line segments spanned by such that (i) all segments in have a common endpoint and (ii) the sets of segments , , are pairwise disjoint.
Proof. First assume that is nondegenerate. By Corollary 3, we have and spans distinct lines. For any , , at most lines are incident to more than points by Theorem 1. There is a constant such that the set of lines incident to at least 2 and at most points still contains lines. For each line , choose an arbitrary point among the at most points in and place into the set . The resulting sets are disjoint and each contain at most lines. For at least points , the set contains at least lines.
Next assume that is degenerate, that is, there is a line incident to points. In this case, every point off forms triangles with point pairs from , and so . Each of the points off the line is incident to lines spanned by . For any constant , , at most lines are incident to more than points in by Theorem 1. There is a constant such that the set of lines incident to at least 2 and at most points contains elements. For each line in , choose an arbitrary incident point and assign the line to . For at least points , the resulting set contains at least lines.
Corollary 5
If a set of points in the plane determines nondegenerate triangles, then at most of them have unit area.
Proof. First, suppose that is nondegenerate. By Lemma 1, , hence .
2.2 Beck’s lemma in higher dimensions
Lemma 3
(Beck [6]). For any there exist constants such that, for any set of points in , at least one of the following holds.

a hyperplane contains more than points of ; or

the tuples of span at least distinct hyperplanes.
3 Minimum volume tetrahedra
We show that points determine at most minimum volume tetrahedra in three space. The upper bound is based on a new charging scheme which assigns each tetrahedron of minimum volume to one of its four faces. We then extend our charging scheme and show that for any fixed , , the number of dimensional simplices of minimum (nonzero) volume in is , where the constant of proportionality depends only on and . This bound is best possible apart from the constant factor.
Theorem 3
The number of tetrahedra of minimum (nonzero) volume spanned by points in is at most , and there are point sets for which this number is . Given points in , all tetrahedra of minimum nonzero volume can be reported in time and working space.
Lower bound.
The lower bound construction is simple. Form a rhombus with unit sides in the plane from two equilateral triangles and with a common side . Extend it to a prism in 3space, and make a unit volume tetrahedron with vertices , , , and , on the four vertical lines. Replace each of , , , and , with equally spaced points with interpoint distances along these lines, for a sufficiently small (assume is divisible by 4). Observe that each tetrahedron with one vertex on each of the four lines has volume very close to 1, but the minimum volume is , given by tetrahedra with two consecutive vertices on a line, and two vertices on any two of the other three lines. The number of such tetrahedra is . All other tetrahedra have zero volume.
Upper bound.
Let be a set of points in . Denote by the tetrahedron determined by four noncoplanar points and by its volume. Similarly, let denote the tetrahedron determined by the endpoints of two line segments and . The key to our upper bound is the following charging scheme: assign every (nondegenerate) tetrahedron determined by to one of its four faces as follows. Assign to a triangle face of maximum area among the faces adjacent to a diameter of . We show that at most a constant number of minimum volume tetrahedra are charged to every triangle, and this yields the desired upper bound.
Consider a (nondegenerate) triangle with . Let be a diameter of . Choose a coordinate system such that is the origin, lies on the axis, and lies in the plane. Refer to Fig. 1.
Let , and denote by the height from in . Assume that the minimum volume of a tetrahedra spanned by is . All points with must lie in two horizontal planes at distance from the plane. Let us consider the plane for now. By our assignment, all points must lie in the interior of an axisaligned rectangle in the plane (otherwise would not be the diameter or would have larger area than ; note that point cannot lie on the boundary of ). Partition the open rectangle into two rectangles and
Claim 1
For , there is at most one point such that is minimum.
Proof. Assume there are two points for some . We will pick a side so that is not parallel with . Then is a nondegenerate tetrahedron. We show that .
Assume first that is not parallel to , and pick . The segment lies in rectangle , ; observe that at most one element of may lie on the boundary of . Denote by and the two lines containing the two sides of parallel to the axis. The volume of strictly increases if we move to position and to position along the line such that and . The volume of is , hence (e.g., the base triangle of lies in the plane, and its height is ). A contradiction.
Next assume that is parallel to , and pick . In this case, and are strictly in the interior of . If we replace and by points and on the line such that has the same extent as the rectangle , then . Draw two lines parallel to through and . Let these two lines intersect the plane at points and . (Here we use the fact that if the vertices of a tetrahedron are on three parallel lines, one can shift the single points along the corresponding lines, and the volume remains the same.) Since we moved and along lines parallel to , the volume of is the same as that of , and this volume is (e.g., the base triangle of lies in the plane, and its height is ). A contradiction.
Symmetrically, the plane also contains at most 2 points with , hence the number of minimum volume tetrahedra is at most .
Reporting minimum nonzero volume tetrahedra.
Next, we devise an algorithm for counting and reporting all minimum (nonzero) volume tetrahedra spanned by a given set of points in threespace.
A plane and a point form a slab, which is defined as the open region bounded by and the plane , which is parallel to and incident to . We say that the slab is empty, if it is disjoint from . Braß, Rote, and Swanepoel [9] observed that if is a minimum area triangle in the plane, then the relative interior of the line segment and the slab formed by the line through and the point is empty of . This observation readily generalizes to threespace: If is a minimum volume tetrahedron, then the triangle is a minimum area triangle in the plane spanned by , and the open slab formed by and point is empty.
Let denote the set of planes spanned by . We compute for every empty slab formed by a plane and point , the number of tetrahedra whose base is a minimum area triangle in and whose 4th vertex lies in a plane parallel to and incident to ; we also record the volume of these tetrahedra (which might not be the global minimum). We then sum up these quantities for those empty slabs where this volume is the minimum. We can compute the number of minimum volume tetrahedra, since each of them is counted exactly four times, once for each face. If we record all minimum area triangles in every plane and every point in plane , then we can also report all minimum volume tetrahedra (we have already shown that there are only of them).
Our algorithm works on the dual arrangement . Consider the (wellknown) duality transform which maps a point to the (nonvertical) plane with equation . Conversely, a nonvertical plane with equation is mapped to the point . Duality satisfies , for any point , and , for any nonvertical plane . The duality preserves pointplane incidences and it reverses the abovebelow relationship, which is understood with respect the the axis. Furthermore, it preserves vertical distances for pointplane pairs. More precisely, point is above plane , if and only if point is above plane , and the vertical distance between the point and the plane in each pair is the same. Every plane spanned by the point set corresponds to a vertex of the arrangement. An empty slab determined by and a point corresponds to a vertical line segment between and plane that does not pierce any plane of the arrangement .
First we solve a planar problem: count the minimum (nonzero) area triangles in a plane efficiently. For a plane , let , let be the number of lines spanned by , and let denote the number of minimum (nonzero) area triangles determined by . Let denote the set of lines spanned by .
Proposition 1
Consider a plane . Given the list of points in , and for every line the length and the number of shortest (that is, minimum length) segments in , we can compute in time.
Proof. For every line , , compute the number of minimum area triangles in the empty slabs formed by (that is, triangles such that one side is a shortest segment and a third vertex is a closest point to ). Since the number of shortest segments along is given, it suffices to check the distance of all points of to the line . We can enumerate these triangles in time. Summing this over all lines, can be computed in time.
We use the SzemerédiTrotter Theorem (Theorem 1) to bound the time to compute for all planes .
Proposition 2
With the notation defined above, we have .
Proof. Denote by the set of lines incident to at least but fewer than points of , for . By Corollary 2, we have
We can now count the minimum (nonzero) volume tetrahedra spanned by . Assume that the entire dual arrangement of and its incidence graph is available; it can be computed in time, and stored in space [1, 16]. The arrangement is a cell complex with faces of dimension 0, 1, 2, and 3. For every vertex, we store the incident planes, and the incident ridges (lines formed by the intersection of two planes); for every ridge, we store the incident vertices (sorted along the ridge) and planes of the arrangement (sorted around the ridge).
We perform the following three tasks: (1). Preprocess the lines spanned by . For every line , compute the number of shortest segments in and record their (common) length. The sorted list of points along each line can be extracted from the arrangement. Since there are at most lines, each containing no more than points, this can be done in time. (2) For every vertex , compute the number of minimum (nonzero) area triangles lying in the plane . By Proposition 1, this can be done in time for a plane . By Proposition 2, the total time over all planes in is . The list of points in each plane is also available from the arrangement. (3) Sweep the dual arrangement with a horizontal plane. For every vertex , drop a vertical line through and find the face of the cell hit by this line, and the intersection point of the vertical line with : the face can be of dimension 0, 1, or 2. Again, the number of planes in incident to can be extracted from the arrangement. The empty slab formed by and determines tetrahedra with a minimum area base triangle in and a 4th point in plane . For every vertex of the arrangement, we record and the volume of the corresponding tetrahedra. If this volume turns out to be the minimum volume, all these tetrahedra can also be reported in a second pass over the data. Since the total number of minimum volume tetrahedra is , all reporting takes time.
If the entire dual arrangement and its incidence graph is available, then our algorithm requires time and space in the real RAM model of computation (where algebraic operations over reals have unit cost). The space requirement can be reduced to if we use a plane sweep algorithm to perform tasks (2) and (3). It sweeps the arrangement with a horizontal plane and keeps in memory the cells intersecting the sweep plane, whose total complexity is by the zone theorem [19]. When the sweep plane arrives at a vertex , it can provide the set of planes and ridges incident to and so we can perform task (2). The plane sweep algorithm takes time (since it maintains an event queue, which is updated in time for each vertex of the arrangement). The time can be reduced to while maintaining an working space by the topological sweep method of Edelsbrunner and Guibas [17] (originally designed for line arrangements in the plane, and adapted to threedimensions by Anagnostou et al. [2]).
Edelsbrunner and Guibas [17] and Chazelle et al. [11] gave quadratic time sweepline algorithms for the problem of finding a triangle of minimum (possibly zero) area. It is easy to rewrite our 3D algorithm for the planar problem and report all triangles of minimum nonzero area determined by a set of points in time and space. When the sweep line arrives at vertex of the dual arrangement, pairs of lines incident to with consecutive slopes correspond to minimum length segments determined by the points of whose duals intersect at . By drawing vertical segments at each vertex of the arrangement, the algorithm examines all triangles whose base is a minimum length segment on some line of , and the third vertex is closest to the line through the base.
Generalization to simplices in space
Theorem 3 can be generalized to arbitrary dimensions.
Theorem 4
For every (fixed) with , the number of dimensional simplices of minimum (nonzero) volume spanned by points in is , and there are point sets for which this number is .
Proof. Lower bound. Consider parallel lines such that any two are at unit distance apart from each other; place equally spaced points on each line (assume is a multiple of ). Every simplex with positive volume has two vertices on one line and one vertex on each of the other lines. If the simplex has minimum volume, the vertices along the same line must be consecutive. Conversely, every simplex with two consecutive vertices on one line and one arbitrary point on each of the other two lines has the same (minimum) volume. The number of such simplices is .
Upper bound.
Let be a set of points in . Assign every (nondegenerate) simplex determined by to one of its faces as follows. We choose the vertices of a face of in steps. First choose an arbitrary vertex of . In step , choose a vertex of which is at maximum (Euclidean) distance from the dimensional affine subspace spanned by . We show that at most a constant number of minimum volume simplices are assigned to every simplex spanned by . This yields an upper bound.
Assume that the minimum volume of a simplex is 1. Consider a (nondegenerate) simplex such that , , is a furthest point in from the affine subspace . Choose an orthogonal coordinate system such that is the origin and , , is the dimensional subspace spanned by the first coordinate axes. Denote the extents of the axisaligned bounding box of by ; and choose so that . So is the volume of the bounding box of any unit volume simplex with a dimensional face in the coordinate system described. By the choice of the points , we have . If our charging scheme assigns a minimum volume simplex to , then , since we chose such that it is at least as far from as .
Every point such that is a simplex of a given volume lies in a hypersurface , which is the Minkowski sum of the dimensional Euclidean space spanned by (in the first coordinates) and a sphere of radius (in the remaining coordinates). Every point has a unique decomposition into the vector sum with and .
If the simplex , , is assigned to , then must satisfy constraints: is at most as far from the subspace as for . These constraints are satisfied if the th coordinate of is at most , . That is, lies in an axisaligned rectangle of extents , for .
Partition into congruent axisaligned rectangles of extents , each for . Partition into convex regions of diameter each. These two partitions give a partition of the Minkowski sum into regions. Next we show that each region contains at most one point of . This confirms that at most minimum simplices are assigned to , completing our proof.
Suppose, by contradiction, that are assigned to and both lie in the same region of . Since , both and are simplices of minimum volume. We show that spans a simplex of a strictly smaller positive volume, which is a contradiction. Consider the maximum index for which the set is a nondegenerate simplex. (Such an index exits, since is a noncollinear triple.) We show that the simplex has smaller volume than . Note that it is enough to compare the volumes of the simplices and . To compare the volumes of and , it is enough to compare the distances of and to the affine subspace spanned by , that is, and . If we shift the line segment in an affine subspace parallel to , then the volumes of and do not change, so we may assume that segment is orthogonal to . Then, we have . The first coordinates of the vector are parallel to , and so they can be ignored when computing . Each of the coordinates of is at most a fraction of the corresponding coordinate of ; the th coordinate of is , and each of the last coordinates of is at most . Hence
This confirms that is strictly closer to than , hence .
4 Unit volume tetrahedra
Lower bound.
As mentioned in the introduction, Erdős and Purdy showed that a suitable section of the integer lattice has unitarea triangles [23]. This immediately gives a lower bound of unit volume tetrahedra (e.g., by placing two such lattice sections, with points each, in two parallel planes).
Upper bound.
Pach and Sharir [31] showed that every point of a set of points in the plane is incident to at most unit area triangles spanned by . This bound follows easily from the SzemerédiTrotter bound on the number pointline incidences: For every point , there is a unique line parallel to such that every point above gives a unit area triangle ; there are point line incidences between and the lines . This bound is tight, since the SzemerédiTrotter bound is tight: there are point sets with unit triangles having a common vertex: Place a set of points and a set of lines with incidences in the plane; determines a point set with , and so spans unit area triangles incident to .
We follow a similar strategy in three dimensions. Instead of bounding directly the number of unitvolume tetrahedra with a common vertex or a common edge, we design a charging scheme. We assign every unit volume tetrahedron to a line segment lying in the plane containing one of its four faces. Ideally, a tetrahedron is assigned to one of its edges, but this does not always hold in our charging scheme. We show that at most tetrahedra are assigned to every segment; which immediately gives an bound on the number of unit volume tetrahedra.
Theorem 5
The number of unitvolume tetrahedra determined by points in is .
Proof. Let be a set of points in , and denote the set of all planes spanned by . We follow the convention that in a tetrahedron , vertex lies above the plane spanned by . For every triangle , every vertex for which has unit volume lies in a plane parallel to plane . We decompose the set of unit volume tetrahedra into two subsets:

contains every where is nondegenerate;

contains every where is degenerate.
Consider . For two distinct points , let denote the set of planes spanned by containing the segment .
Assigning triangles to segments.
We design a charging scheme where we assign every triangle to a segment in the plane . This induces a charging scheme for unit volume tetrahedra: If is assigned to segment , then we assign every unit volume tetrahedron to segment , too.
The assignment is done in each plane independently. Consider a plane . We proceed in two stages: in the first stage we compute, for every triangle , a collection of segments in ; in the second stage, we assign to one of the segments in . The assignment is fairly elaborate; intuitively, we pursue two goals: (1) every segment should be assigned to “few” triangles of the same area, and (2) a segment should not be assigned to a triangle if the plane contains a “rich” line parallel to . We continue with the details.
Denote by the number of points in , and by the number of triangles spanned by . Corollary 4 tells us that there is a set of points, for some absolute constant , such that each is incident to a set of line segments determined by , and the line segments in , , are all distinct. Denote the collection of all these segments by . We have . For every triangle , let denote the set of segments in parallel to any one of the richest lines in the plane . Let be the set of segments to which may be assigned. Since for every , we have .
For two points , let denote the number of points in that do not lie on the line through . If , then one endpoint of is in , and by Corollary 4, we have . Hence, we have . By Corollary 5, determines at most triangles of any given area, with (say) . We are now ready to assign triangles to segments: Assign every triangle determined by to a segment such that (i) every segment is assigned to triangles and (ii) every segment is assigned to triangles of any given area.
Multiplicities.
Let denote the set of triangles for which is nondegenerate and is assigned to segment . We say that the multiplicity of a triangle is the number of triangles in with the same area as that of . For every , let denote the set of triangles for which . Since is assigned to at most triangles of any given area, the multiplicity of no triangle can exceed , for a sufficiently large constant ; that is, for .
Aggregate the triangles assigned to the same segment in various planes by letting
Projection to a plane.
Fix a line segment . Project along lines parallel to onto a plane orthogonal to . The image of under the projection is a multiset , where the multiplicity of each point is the number of points of on a line through parallel to . Similarly, the projection of a subset is a multiset . The projection of the line through is a single point , and every plane parallel to is projected into a line in . An illustration is provided in Fig. 2.