Hans Walser The Pascal Pyramid Published in: The College Mathematics Journal, Vol. 3, No. 5, November 2000, p. 383-392. Introduction The original purpose of this article was to describe how to build a model of the Pascal Pyramid (or Pascal Tetrahedron) displaying the arrangement in space of the trinomial n coefficients r s t = n!, r + s + t = n, n,r, s,t 0, that would be analogous to r!s!t! the Pascal Triangle (Figure ) displaying the arrangement in the plane of the binomial n coefficients r s = n!, r + s = n, n,r, s 0 (using the notation in [2]). r!s! 2 3 3 4 6 4 5 0 0 5 6 5 20 5 6 7 2 35 35 2 7 Figure : The Pascal triangle in a hexagonal lattice In Section we show that the rhombic dodecahedron is a space-filling polyhedron and that, when oriented properly, a collection of such polyhedra provides the sought-after Pascal Tetrahedron, with all the desired analogies intact. Section 4 gives very practical hints about how to construct a Pascal Pyramid. Some impatient readers may wish to begin with this section and construct the pyramid first. In fact, we wouldn t discourage this, since the mathematics of Section 3 might be easier to understand with the model in hand.
Hans Walser: The Pascal Pyramid 2 / 2 2. The trinomial coefficients n If you think of r s as the coefficient of ar b s in the expansion of a + b trinomial coefficient ( ) n, then the n r s t should be the coefficient of ar b s c t in the expansion of ( a + b + c) n. The term a r b s c t with r + s + t = n is easily seen to have the trinomial coefficient: n r s t = n! r!s!t! According to the Pascal Identity for the binomial coefficients n r s = n r s + n r s one might easily, and correctly, guess that the Pascal Identity for trinomial coefficients is: n r s t = n r s t + n r s t + n r s t Because we have now three parameters r, s, and t it is natural to try to display the trinomials in a three-dimensional lattice. For this purpose we need an analogy to the hexagons of Figure that tile the two-dimensional plane. We need a space filler, i.e. a polyhedron whose copies fill three-dimensional space without gaps or overlapping. A- bout space fillers see [3, p. 54-56]. It turns out that the rhombic dodecahedron works perfectly. 3. The rhombic dodecahedron To obtain the rhombic dodecahedron we build on every face of the cube (Figure 2(a)) a right pyramid with dihedral angles of 45 at the base (Figure 2(b)). Because of this special dihedral angle of 45 (between the shaded triangle ABD and the top face of the cube), the isosceles triangles of two adjacent pyramids with an edge of the original cube in common are coplanar and yield together a rhombus ( ABCD is an example, with BD an edge of the original cube). The surface of the compound of the cube and the six pyramids, is therefore bounded by twelve congruent rhombi; hence the name rhombic dodecahedron. The short diagonals of the rhombi are the edges of the inscribed cube. The long diagonals of the rhombi have as a consequence of the 45 dihedral angles the length of a diagonal of the square face of the inscribed cube (so that in Figure 2 the length of the line AC equals the length of the line BE ). Therefore the diagonals of the rhombi are in the ratio 2 :.
Hans Walser: The Pascal Pyramid 3 / 2 A E D B E D B C (a) (b) (c) Figure 2: The rhombic dodecahedron The rhombic dodecahedron has 4 vertices, but the vertices are of two different kinds. Six of them (the summits of the pyramids) are coincident with four rhombi, and the rhombi have acute angles at these vertices. The remaining 8 vertices (the vertices of the original cube) are coincident with only three rhombi each, and the rhombi have obtuse angles at these vertices. To prove that the rhombic dodecahedron is a space filler we first think of a cubical lattice. Then color the cubes in black and white such that when the face of one cube makes contact with another cube the two cubes always have different colors. Thus you get a three-dimensional chessboard. Now take the centers of the white cubes in each case and connect that center with the 8 vertices of the same white cube. This yields a dissection of every white cube into six congruent right pyramids with a common summit in the center of the white cube and bases on each of the six square faces of the white cube. The dihedral angle between the basic square and the lateral isosceles triangles is 45. Now think of gluing these pyramids to the adjacent black cubes. According to our construction of a rhombic dodecahedron we have now a gap-free and non-overlapping dissection of the three-dimensional space into rhombic dodecahedra. Figure 3 shows the analogous procedure for the two-dimensional case. In this case the new lattice is just a larger square lattice.
Hans Walser: The Pascal Pyramid 4 / 2 Figure 3: Chessboard with dissection of the white squares We have already observed that the rhombic dodecahedron has two different types of vertices. This means that, with a set of congruent rhombic dodecahedra, we can build two different kinds of pyramids. Figure 4 shows the layer-by-layer construction of a pyramid built on a square 4 4 base. In this pyramid every rhombic dodecahedron has a summit vertex with 4 acute rhombus angles.
Hans Walser: The Pascal Pyramid 5 / 2 Figure 4: Pyramid on square base The other possibility, shown layer-by-layer in Figure 5, is a triangular based pyramid. In this version every rhombic dodecahedron has a summit vertex with three obtuse rhombus angles.
Hans Walser: The Pascal Pyramid 6 / 2 Figure 5: Pyramid on triangular base This pyramid is very close to the regular tetrahedron. It is this pyramid we will use to get the three-dimensional analogue of the Pascal Triangle of Figure. 4. The Pascal Pyramid as a space lattice Figure 6(a) shows the compact Pascal Pyramid and Figure 6(b) shows the trinomial entries of each layer of it individually (on its faces, since we cannot see the interior!).
Hans Walser: The Pascal Pyramid 7 / 2 2 2 2 2 3 3 3 6 3 3 3 4 3 3 6 4 4 4 6 2 4 2 4 6 2 6 4 4 (a) (b) Figure 6: The Pascal trinomial Pyramid The Pascal Identity for trinomials (see Section ) means that every trinomial coefficient is the sum of the trinomial coefficients in the three adjacent rhombic dodecahedra in the layer above. These are the rhombic dodecahedra sitting on the roof of the rhombic dodecahedron in question. The rhombic dodecahedra on the oblique edges of the pyramid have only one adjacent neighbor above and therefore we have on the edges an infinite repetition of the number. Similarly the dodecahedra on the faces of the pyramid have only two adjacent neighbors above and therefore we have on each face a copy of the entries in the two-dimensional Pascal Triangle. 5. Technical hints for constructing braided rhombic dodecahedra There are many ways to construct rhombic dodecahedra. We will describe a method using paper strips which have to be braided together, which is fairly quick and easy to do. For more about braiding other polyhedra see [], [4], [5]. We begin by observing that if you take two opposite vertices of the rhombic dodecahedron which have three rhombi incident at their vertices and think of one as the North Pole N and the other as the South Pole S, as shown in Figure 7, then you will have about the equator a zigzag strip consisting of six rhombi.
Hans Walser: The Pascal Pyramid 8 / 2 N S Figure 7: Two poles and the corresponding equatorial strip Since we have four such pairs of opposite poles, we also have four equatorial zigzag strips going around the rhombic dodecahedron. It turns out that we can braid a model of the rhombic dodecahedron with four of these zigzag strips. For technical reasons we add two additional rhombi to every strip (shaded in Figure 8). These two rhombi will overlap the two first rhombi on the completed model and thereby add stability to it. Figure 8: One of the four equatorial zigzag strips The zigzag strips can be made in a very efficient way. Since the diagonals of the rhombi are in the ratio 2 :, we get for the acute angle of the rhombi: ( ) = 2arctan 70.53 2 Now take an 8 2 " by " sheet of paper and fold it three times in parallel folds as shown in Figure 9. Figure 9: Folding three times Then you will have a strip with 8 layers. Draw on this strip rhombi with the acute angle 70.53 (Figure 0(a)) the 8 2 " by " paper should yield five rhombi as shown in Figure 0(b).
Hans Walser: The Pascal Pyramid 9 / 2 α (a) α (b) Figure 0: Cutting the rhombi Now cut along the oblique lines. To avoid having the different layers slip while you are cutting, cut in the direction against the main fold as shown by arrows in Figure 0(a). Unfold a cut section and you will have a zigzag strip with eight rhombi. This strip contains valley folds and mountain folds. Since the strip has to go in a circular way around a convex model, you must change some of the folds so that you have all folds in the same direction along the entire strip. To construct the braided the model you need four strips. It will be easier to braid the model if you have four strips of different colors. The author has had students build these models in class, where every student began with a sheet of the same size, but different students had different colors. Once the strips were made, the students did a kind of stock exchange Who will exchange a red strip for a blue one?. The basic idea of braiding (or weaving) is that on the completed model every strip goes alternately over and under the strips it crosses as it goes around the model. This is analogous to the most elementary weave in the plane shown in Figure (a). At the vertices of our rhombic dodecahedron you have either four strips meeting (these are the vertices incident with four rhombi at their acute angles), or only three strips meeting (at the vertices with three rhombi at their obtuse angles). The easiest way to braid the model is to begin with three strips arranged in the situation shown in Figure (b) where the vertex is in the center of each strip. Think of this vertex being the South Pole. Then fix the strips in the proper position using a small bit of clear tape on what will be the inside of the finished model. You can see what the finished model will look like by overlapping the 2 end-rhombi of each strip. Do this so that you can see what you want to have when you finish. Then take the fourth strip and weave it about the equator and finally close the model at the North Pole by tucking the extra rhombi. Every strip will follow the path of a great circle. Remembering that each strip goes over and under the strips it meets, will enable you to complete the construction, if you are patient.
Hans Walser: The Pascal Pyramid 0 / 2 (a) (b) Figure : Braiding Don t worry if you are not quite successful at the first attempt no one is born a master. (The author once had some students compete in building one-colored rhombic dodecahedra. It took the champion two and a half minutes to complete the model beginning with a flat sheet of paper, a pencil, the rhombus pattern, and a pair of scissors!) The Figure 2 displays the braided model. Figure 2: Braided rhombic dodecahedron This braiding procedure allows a class to mass produce congruent rhombic dodecahedra. This way you will have enough polyhedra to show the space filler property and to build the pyramids of Figures 4 and 5. It is useful to have a base on which to build, and display, your pyramids. You can build these bases with the same zigzag strips just braid them in a flat way. With 6 strips of Figure 8 you can braid the buckled 4 4 base of Figure 3 appropriate to the pyramid of Figure 4. Of course this base is not flat in a geometrical sense, it looks more like a corrugated egg holder, where the eggs that really fit are rhombic dodecahedra. This is basically a two way weave meaning that strips run in 2 directions with each strip going alternately over and under each strip it meets.
Hans Walser: The Pascal Pyramid / 2 Figure 3: Square base The Figure 4 displays a base with a triangular lattice, appropriate to the pyramid of Figure 5 and the Pascal Pyramid of Figure 6. For this base you need strips in three different colors. In each color you need five strips with 2,4,6,8,0 rhombi. Figure 5(a) displays the base seen from above and Figure 5(b) the braiding structure, which is a three way weave. Figure 4: Triangular base (a) (b) Figure 5: The braiding structure The author wishes to thank Peter Hilton and Jean Pedersen for challenging him to build this pyramid and for their help in preparing the manuscript.
Hans Walser: The Pascal Pyramid 2 / 2 References. Peter Hilton / Jean Pedersen: Build Your Own Polyhedra. Menlo Park: Addison- Wesley 994. ISBN 0-20-49096-X 2. Peter Hilton / Jean Pedersen: Relating Geometry and Algebra in the Pascal Triangle, Hexagon, Tetrahedron, and Cuboctahedron (to appear) 3. Alan Holden: Shapes, Space, and Symmetry. New York: Columbia University Press 97. ISBN 0-23-03549-7 4. Hans Walser: Geometrie zum Anfassen. Mathematik Lehren, Heft 65, August 994, p. 56-59 5. Hans Walser: Der Goldene Schnitt. Second edition. Stuttgart-Leipzig: B.G. Teubner Verlagsgesellschaft 996. ISBN 3-854-25-5 Hans Walser Department of Mathematics Santa Clara University Santa Clara, California 95053, USA and Mathematikdepartement ETH Zürich CH 8092 Zürich, Switzerland hwalser@bluewin.ch