Some Elementary Aspects of 4-dimensional Geometry

We indicate that Heron's formula (which relates the square of the area of a triangle to a quartic function of its edge lengths) can be interpreted as a scissors congruence in 4-dimensional space. In the process of demonstrating this, we examine a number of decompositions of hypercubes, hyper-parallelograms, and other elementary 4-dimensional solids.


Introduction 8
Heron's formula gives the explicit relationship among the area, A, of a triangle and its edge lengths. We assume, now and throughout, that a triangle with edge lengths, a, b, c where 0 < a ≤ b ≤ c is given. The semi-perimeter is the quantity: s = (a + b + c)/2. One representation of the formula is The formula was written in Heron's (Hero's) book Metrica circa 60AD. According to wikipedia [8], the formula may have been known to Archimedes more than two centuries earlier. But Heron's book was a compendium of a number of known results. Here, we square both sides of the equation and multiply by the denominator to get the following expression This can be considered an expression that relates the 4-dimensional volumes of a number of hyper-solids. 9 As the text develops, we will further manipulate this expression all the while maintaining a 4-dimensional 10 awareness of the algebraic manipulations that are performed.

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Our primary purpose here is to indicate a 4-dimensional scissors congruence among the hyper-solids 12 that are involved in Heron's formula. But in the process, we will also develop several secondary goals 13 along the way. These are our own interpretations of conversations with several people. Most importantly, 14 we hope to develop the reader's intuition and interest in elementary 4-dimensional geometry.

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In particular, many algebraic identities, such as the binomial and multi-nomial theorems, have elegant 16 and useful interpretations in terms of relating the higher dimensional volumes of various n-cubes. The 17 symmetries in the expressions are manifest in the symmetries among the pieces of these decompositions. 18 We remark here that the algebraic proof of Heron's formula given in this form goes as follows. Suppose that we arrange the triangle in the coordinate plane as indicated in Fig. 1. In this way, one vertex is at the origin, the longest side (the side of length c) lies along the x-axis with a vertex at the point (p, 0) (Here a different coordinate is chosen so that in the new variables are of the same flavor). The remaining vertex lies at the point (r, s). In this way, a 2 = r 2 + s 2 , and b 2 = s 2 + (p − r) 2 . The left-hand-side of Heron's formula can be shown to be 4p 2 s 2 . The right-hand-side reduces to We expand the terms on the right-hand-side as follows: 19 2a 2 b 2 = 2(r 2 + s 2 )((p − r) 2 + s 2 ) = 2r 2 (p − r) 2 + 2s 2 (p − r) 2 + 2r 2 s 2 + 2s 4 . 2a 2 c 2 = 2(r 2 + s 2 )p 2 = 2r 2 p 2 + 2s 2 p 2 .
As such the formula has been proven. However, we point out that even in our most quiet moments, the simplification is quite tedious. After some preliminary work on developing higher dimensional models and the reader's 24 intuition on 4-dimensional geometry, our first task will be to show how to use 4-dimensional geometry 25 to encode the simplification. The second task that we endeavor upon is to capitalize upon a scissor's 26 congruence proof of the Pythagorean Theorem, to expand the 4-dimensional solids of size a 4 , b 4 , and 27 a 2 b 2 . Each of a 2 and b 2 is the square of a hypotenuse of a right triangle, and can each be written as 28 a sum of squares. To do so, one decomposes the square figure into five pieces and reassembles them.

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In 4-dimensions, the quartics a 4 , b 4 , and a 2 b 2 are decomposed into twenty-five pieces and reassembled 30 as the sum of four hypercubes (or more precisely, hyper-rectangles). The grouping and factoring that 31 is indicated above can also be seen in terms of 4-dimensional scissor's congruences. Meanwhile, there 32 is a little work to rearrange the factorization 4s 2 p 2 from four copies of a 4-dimensional figure that is a 33 parallelogram times itself into a hyper-rectangle.

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In Section 2, we discuss the n-cube from several points of view including its representation as a configuration space. Then we decompose it into right n-dimensional simplicies. This decomposition gives Nicomachus's Theorem [6]. Our interpretation yields that the cube [0, a] 4 can be decomposed as the union of four congruent figures, each of the form a right isosceles triangle times itself. In Section 3, we discuss the multinomial theorem as another decomposition of the n-cube, and we use this idea to organize the computation of the right-hand-side of Heron's formula. In Section 4, we demonstrate that the proof of the Pythagorean Theorem can be adjusted to a proof of a 4-dimensional result: if z 2 = x 2 + y 2 , and u 2 + v 2 = w 2 , then the hyper-cube [0, z] × [0, z] × [0, w] × [0, w] can be cut into 25 pieces that can be reassembled into a figure that is the product of a pair of sums of squares. This is the geometric interpretation of the identity In Section 5, we tie up some loose ends in the algebraic proof above. Section 6 discusses the 35 left-hand-side of the equation, and Section 7 is a conclusion. These are digits since they are fingers. They are binary since they are either up or down. Assign 52 the number 2 0 = 1 to your thumb; assign 2 1 = 2 to your index finger; assign 2 2 = 4 to your middle    67 We call this cube the hypercube. We begin this subsection with a simple case. Consider the possible configurations of a pair of drawers.
78 Let x denote the distance that the first drawer is opened, and let y denote the distance that the second 79 drawer is opened. Of course 0 ≤ x, y ≤ 1 -that is each drawer can be opened any amount from 0 80 meters to 1 meter. This situation can be subdivided into two cases. Either the drawer on the left is no 81 more open than the drawer on the right is, or vice versa. Thus either 0 ≤ x ≤ y ≤ 1 or 0 ≤ y ≤ x ≤ 1.  Now consider three drawers. Let x, y, and z denote the amount that each is pulled out. As before 0 ≤ x, y, z ≤ 1, but there is no relation among the openness of the drawers in general. We decompose this situation into six configurations: as indicated in Fig. 4. Using the coordinates of the 3 dimensional cube that are indicated, the cube has 89 been decomposed into six congruent tetrahedra. Each tetrahedron has three edges with length 1, two  By induction, the remaining (n − 1) drawers are ordered in (n − 1)! configurations.

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A right n-dimensional simplex consists of the point set The n-dimensional cube can be decomposed into n! such simplices that are all congruent. So the 102 n-dimensional volume of such a simplex is 1 n! .

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There is a typographical trick called the L7-trick 2 , to obtain the set of vertices for each simplex. Let n be fixed. Consider a row of n 0s (separated, if you like, by commas): 00 . . . 0. Also construct a column of n 1s. Now the column of 1s can be appended below any one of the 0s. Thus there are n possible ways of doing so. For each such shape, we can insert the (n − 1)! shapes that exist by induction. The first step of the induction is to consider the array 0 1 . In the next step, we have the arrays The rows are the coordinates of the vertices of the two triangles 0 ≤ x ≤ y ≤ 1 and 0 ≤ y ≤ x ≤ 1 into which the unit square was decomposed. For the cube, we start from the three arrays: and fill them respectively with the two arrays above.
In the next step, each of these six matrices are fed into the four patterns The resulting rows of each matrix are the five vertices of each 4-simplex. The union of these fills  In the unit n-cube, [0, 1] n consider the set For any fixed value x n = C, the cross-sectional set 1)-dimensional cube [0, C] n−1 whose volume 3 , then is C n−1 .

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Considering the situation of drawers, we can imagine a "high priority drawer" which can be opened 114 any distance between 0 and 1 meters. If it is opened to C meters, then the remaining drawers can only be 115 3 We will use the term volume to mean n-dimensional volume in one of n-chairs. All the subjects are also seated, and no-one is to stand taller than the sovereign. As the 117 sovereign stands, all the inferiors attempt to stand. The set of configurations of partially standing people 118 corresponds to the configurations of partially opened drawers and of the points in the pyramidal set P n . 119 Here, note that the subscript n on the expression P n indicates that the last coordinate is the position of 120 the sovereign.

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More generally, let Observe, that the n-cube is decomposed into n congruent pyramids. One can then interpret this decomposition as a geometric manifestation of the fact that This interpretation was given in [3] and earlier in [1]. One usually proves this by induction. Since n 2 (n + 1) 2 /4 + (n + 1) 3 = (n + 1) 2 n 2 /4 + (n + 1) = (n + 1) 2 4 (n 2 + 4n + 4), the proof follows easily. That formula is often used to compute the integral  First, we consider the product of a right isosceles triangle with itself. This will be As a configuration of drawers, the first drawer is no more open than the second while the third drawer 133 is no more open than the fourth. Alternatively, we consider a pair of married couples in which neither 134 husband speaks more than his own wife, but either husband may speak more than the wife of the other.
See Figure 6 for an illustration.

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And it also can be decomposed into the four pyramidal sets and and we decompose P 4 into six 4-simplices: the three of which listed to the left correspond to the three that are listed to the left above. So, half of  To begin this section, in which an n-dimensional cube with variable edge lengths is decomposed into 157 several smaller sub-cubes, we first consider the case n = 1.

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A line segment whose length is the sum x + y of two positive quantities x and y can be cut into two 159 segments with the segment of length x on the left and the segment of length y on the right (or vice versa). 160 Similarly, suppose that x < y and a segment of length y − x is given. It could be produced by means of 161 cutting a segment of length x from the longer segment. Of course, we can, in general, discuss directed 162 segments with the direction "right" indicating positive length quantities, and the direction "left" used to 163 indicate negative quantities. In this way, each of x + y, x − y, and y − x can be thought of as being 164 built by composing directed segments of lengths x and y. It is conceptually easier to think in this fashion 165 rather than explicitly removing segments and hypercubes whose total volumes in algebraic expressions 166 happen to be negative.

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In case a given rectangular area of dimensions (x+y)-by-z can be decomposed as a pair of rectangular 168 areas x-by-z and y-by-z. As above, we can indicate (x − y)z as the difference between areas xz and 169 yz. We further exploit the distributive law as a scissors congruence when considering the case (x + 170 y)-by-(z + w). This rectangular area is decomposed as the union of four rectangles of size xz,xw, yz 171 and yw. Some of these scissors congruences are indicated in Fig. 7. We leave the reader to imagine 172 others.

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For convenience within cubes, hypercubes, and beyond, we dualize the pictures by indicating a vertex 174 with a label that indicates its volume (and is drawn red if the net volume is negative) and connecting 175 these by an edge if they share a codimension-1 face. If they share a codimension 2 face, then the dual 176 vertices will bound a polygon (always a rectangle in our cases), and so forth. The language of shared 177 polygonal vertices, edges, faces, and so forth is initially quite awkward (thus the term "codimension" is 178 used), but in our cases the situation is a bit more straight-forward than one might presume.

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Our main interest here is to consider the n-dimensional cube that is decomposed as a product of edges in the form This expression RH is called the right-hand-side of Heron's formula. When we use the distributive law to expand the product as the union of 81 hypercubes (more precisely hyper-rectangles), we want to identify which terms will cancel. It is not terribly difficult to determine that the product reduces to "Not terribly difficult" is a relative phrase. An electronic computer can do so easily; a human has to 180 develop some technique to organize and regroup the 81 terms that are initially involved in the expansion.

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Our purpose here will be to find within the hyper-solid the location of each piece. We will observe that 182 the pieces that cancel can be found on specific affine 3-dimensional planes.

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To help determine the addresses of the vertices in the dual complex, we will examine the terms in the multinomial expansion (x 1 + x 2 + · · · + x k ) n .
This situation is more general in that we are considering an n-dimensional quantity, but it is too specific 184 for the case at hand since the terms in the expansion of RH involve differing edge lengths. Still, once 185 the addressing problem is solved for the multinomial expansion, it will be easy to apply here.

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The quantity (x 1 + x 2 + · · · + x k ) n corresponds to decomposing [0, k x k ] n into k n pieces. The volumes of each piece is a product of x i s where each x i is taken from one factor of the decomposition of the edges of the n-cube. Thus there is a specific n-cube 4 of volume x i 1 x i 2 · · · · · x in where each i ∈ {1, . . . , k} and this th factor x i was selected from the th factor in the product expansion: In order to keep track of these terms, we consider the coordinates of the points in the integral lattice x i 1 x i 2 · · · · · x in in the expansion above.

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The next step in understanding the multinomial theorem is to count all of the terms that have the x i 1 x i 2 · · · · · x in and x j 1 x j 2 · · · · · x jn have the same volumes if and only if the corresponding multisets are x 2 x 2 x 1 , x 2 x 1 x 2 and x 1 x 2 x 2 ).

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In the illustration Fig. 8, the 27 terms in the expansion of (x 1 + x 2 + x 3 ) 3 are grouped according to 202 the intersection with the plane x + y + z = K as also indicated in the table below.

245
We apply these decompositions to the factors a 4 , b 4 , and a 2 b 2 in the expansion of Here, a 2 = r 2 + s 2 , and b 2 = s 2 + (p − r) 2 . In addition, the terms of the form a 2 c 2 and b 2 c 2 can be In our next two illustrations, we indicate that products of the form (x 2 − y 2 )(x 2 − y 2 ) and products of the form (x 2 + y 2 )(z 2 + w 2 ), when thought of as hyper-volumes of 4-dimensional solids, can be reassembled into hyper-rectangles whose volumes are the expected terms: and (x 2 + y 2 )(z 2 + w 2 ) = x 2 z 2 + x 2 w 2 + y 2 z 2 + y 2 w 2 .
Such multiplications and factorizations were used in the last stages to illustrate that  x (xx-yy)(xx-yy)=(x(x-y)+y(x-y))(x(x-y)+y(x-y)) = x(x-y)x(x-y) +y(x-y)x(x-y) +x(x-y)y(x-y) +y(x-y)y(x-y) =(x+y)(x-y)(x+y)(x-y) (x+y)(x-y)(x+y)(x-y) Consider now the proof that the area of an arbitrary triangle as indicated in Fig. 1 is p · r/2. The 252 triangle is duplicated and reflected along a horizontal line and these two copies of the original triangle 253 are joined to create a parallelogram. Thus twice the area of the triangle is the area of the parallelogram.

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Meanwhile, the parallelogram is scissors congruent to a rectangle.

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If we consider the 4-dimensional figure that is a parallelogram-times-parallelogram, then it can be 256 decomposed into four copies of a triangle times a triangle. The decomposition is analogous to that 257 which is depicted in Fig. 6   The left-hand side of Heron's formula is the quantity 4p 2 r 2 . Sixteen copies of the original triangle times itself are reassembled into four copies of a rectangle-times-rectangle where the dimensions of the rectangle are p × r. Thus the sequence of equalities are all realized as scissors congruences on 4-dimensional hyper-solids.

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Our 4-dimensional geometric proof is complete.

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Even after more than two and a half centuries of study, many people -including many outstanding 269 mathematicians -have not developed an intuitive grasp of elements of 4-dimensional geometry. Of 270 course, many of us have a more esoteric understanding of higher geometry that is facilitated by our deep 271 understanding of linear algebra and differential forms. Our purpose here has been to demonstrate with 272 2-dimensional projections and with analytic descriptions, that many aspects of 4-dimensional geometry 273 can be understood. In addition, we have indicated that a number of combinatorial structures and identities 274 are more deeply understood in terms of aspects of higher dimensional geometry.

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So while an algebraic proof of Heron's formula is tedious, yet conceptually simple, a more geometric 276 proof that depends on the geometry of 4-spacial dimensions removes some aspects of the tedium. One 277 can envision the pieces fitting tightly together.

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Still, there is a cautionary statement to be made. We only perceive 2-dimensions. Those of us who 279 are sighted see the world projected onto our retinas. As I type these lines, my hands feel the surface of a desk that I presume to be solid. A 4-dimensional hyper-solid has a boundary that is 3-dimensional. I 281 can only imagine the surfaces of these pieces. As the 4-dimensional hyper-solid is projected to 3-space, 282 its boundary overlaps in the same way that a ball or orange projects to my eye and appears to be a disk. 283 The front and the back of the ball veil the inner structure.

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When we use a temporal metaphor to describe the space-time of the room in which we sit, the authors 285 move within the confines of this 10×11×10 cubic foot room for a period of, say, 1 hour. The walls, floor 286 and ceiling of the room form a substantial portion of the boundary as they evolve in time. The"us-then" 287 and the "us-now" -that is the interior of the room before the paragraph was written and the interior of   The initial work for this project were conducted when the second author was a UCUR (University 296 Committee for Undergraduate Scholarship) student at the University of South Alabama. It appeared in 297 [4], but we felt that the pictures and our techniques deserved a wider audience. We would like to thank