Delving deeper Christopher M. Kribs-Zaleta
Painting the Pyramid
ne hallmark of a mathematician is the can measure or count about a cube: numbers of ver- instinct to extend and generalize. This article tices, edges, and faces, surface area, and volume. In Otakes a common high school problem relating particular, the unit cubes at the vertices of the large algebra, geometry, and patterns and extends it. The cube are those that have paint on 3 faces (for n > 1). extension seems simple enough at first glance but The remaining unit cubes along the edges of the large proves to have a number of interesting complications. cube are those that have paint on 2 faces, and the cor- Painting the Cube (NCTM 1989; Reys 1988) is a responding formula 12(n – 2) that emerges from the popular and mathematically rich problem used in mid- data is formed by measuring the length of one edge, dle and high school mathematics courses (as well as discounting the two corner cubes, and then multiply- some courses for preservice teachers). In that problem, ing by the number of edges. The remaining unit a large cube is assembled from small unit cubes, and cubes on the faces of the large cube are those with the exposed faces of the large cube are then painted one painted face, and they are counted by measuring (see fig. 1). The problem asks how many of the unit the area of a face, discounting the edges and corners, cubes will have paint on 0, 1, 2, . . . faces, as a func- and multiplying by the number of faces. Finally, the tion of the edge length n of the large cube (measured number of completely unpainted cubes inside the in small cubes). Painting the Cube blends algebra and large cube, as well as the total number of unit cubes geometry in a way that allows students at various needed to form the large cube, is given by invoking levels to use the concrete context to explore linear, the volume formula for a cube. quadratic, and cubic functions and relationships. High Now suppose we change the problem slightly, to school and college students typically approach this consider a “pyramid” made in the same way, namely, problem by building models for the first several cases, a tetrahedron of side length n formed with unit tet- gathering data in a table, generalizing the patterns into rahedra. The analogous question appears to be, If we formulae, and finally justifying the formulae. assemble a number of small unit tetrahedra to form The formulae that arise in this problem have a a large tetrahedron with edge length n, how many very nice correspondence to all the quantities one small tetrahedra will have paint on 0, 1, 2, 3, or all 4 faces? The alert reader, recalling that tetrahedra do This department focuses on mathematics content that appeals to secondary school not tessellate three-dimensional space, may suspect teachers. It provides a forum that allows classroom teachers to share their mathemat- that the question is not quite as simple as it sounds. ics from their work with students, their classroom investigations and projects, and their other experiences. We encourage submissions that pose and solve a novel or UNDERSTANDING THE QUESTION interesting mathematics problem, expand on connections among different math- If we follow the basic approach described above for ematical topics, present a general method for describing a mathematical notion Painting the Cube, the first thing we will do is con- or solving a class of problems, elaborate on new insights into familiar secondary struct models for small n, gather data, and look for school mathematics, or leave the reader with a mathematical idea to expand. Send submissions to “Delving Deeper” by accessing mt.msubmit.net. patterns. (Go on; get yourself a bunch of little tet- “Delving Deeper” can accept manuscripts in ASCII or Word formats only. rahedra and play along.) The case n = 1 obviously corresponds to a single unit tetrahedron with paint Edited by Al Cuoco, [email protected] on all 4 faces. When we attempt to assemble the n = Center for Mathematics Education, Education Development Center 2 case, however, we run into trouble—as the earlier Newton, MA 02458 remark about tessellating three-space should have prepared us to expect. As shown in figure 2, when E. Paul Goldenberg, [email protected] Center for Mathematics Education, Education Development Center we place four unit tetrahedra toe-to-toe to cover the Newton, MA 02458 vertices of the larger tetrahedron, we see that the space left in between them is not tetrahedral. If we
276 Mathematics Teacher | Vol. 100, No. 4 • November 2006 Copyright © 2006 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
Number of unit cubes with paint on… edge length 0 faces 1 face 2 faces 3 faces TOTAL 1 (1) 2 8 8 3 1 6 12 8 27 4 8 24 24 8 64 n (n – 2)3 6(n – 2)2 12(n – 2) 8 n3
Fig. 1 A 3 × 3 × 3 cube and data gathered for the Painting the Cube problem
Three- dimensional Headprint Footprint
Fig. 2 Four-unit tetrahedra stacked to make a 2 × 2 × 2 tetrahedron consider the vacant space in the middle as a solid, we see that four triangles touch each corner of this new solid: It is octahedral in shape. It is at this point that we realize we must rephrase the questions we are investigating. To get a complete understanding of the make-up of the large tetrahe- dron, we will want to know what kinds of objects are required to build the nth large tetrahedron, how many of each type are required, and how many of Fig. 3 Three views of each of the top four layers of a large tetrahedron. White figures them have paint on how many faces. In fact, as we are upright unit tetrahedra, shaded figures are octahedra, and black figures are inverted will see shortly, there is one more type of object to unit tetrahedra. count, which does not appear until the third layer. shape, denoted in black in figure 3, is also required EXPLAINING THE PATTERNS to build subsequent layers of the larger tetrahedra, Figure 3 shows the nth layer of a multilayered tet- so we add a column to our data tables (see table 1), rahedron (counting down from the top) for n = 1, 2, as one more type of object to count. This is the last 3, 4. If we count and sum each of these (e.g., adding type of object required to build the larger tetrahe- the first four layers for the n = 4 case), we will gen- dra, although it is not until n = 5 that an unpainted erate the data in tables 1 and 2. upright tetrahedron appears (all the inverted tet- Note that when adding a layer (moving from the rahedra are unpainted), visible in the center of the n = 3 to n = 4 case, for example), the counting must footprint of the fourth layer in figure 3, and it is not be done anew. Some tetrahedra may have paint until n = 6 that an unpainted octahedron appears. on a different number of faces in the larger case, There are many patterns to explore in these because a given tetrahedron may no longer be on data, even after they have all been identified and an outside edge or face. the formulae derived. Veterans of Painting the Note also that in the three-dimensional view of Cube will think to observe that unit polyhedra with the third layer, it is evident that when we place three paint on 3 faces are again corner pieces, polyhedra octahedra among the six upright unit tetrahedra, with paint on 2 faces occur along edges, polyhedra there is a space left in the center of the layer, between with paint on 1 face appear in the centers of faces the octahedra. The shape of the space is that of an of the large tetrahedron, and polyhedra with paint inverted (i.e., point-down) unit tetrahedron. This on no faces fill the interior. Indeed, the formulae
Vol. 100, No. 4• November 2006 | Mathematics Teacher 277 Table 1 Numbers of Unit Tetrahedra Used in Making Large Tetrahedra Number of upright tetrahedra with paint on . . . edge Inverted length 0 faces 1 face 2 faces 3 faces TOTAL (no paint) 1 (1) 2 4 4 ()n−4() n − 3 ()n − 2 3 6 4 10 1 6 4 4 12 4 20 4 5 1 12 18 4 ()n−435() n − 3 ()n − 2n() n +110() n + 2 6 6 6 4 24 24 4 56 20
()n−4() n − 3 ()n − 2 n() n +1()n + 2 ()n−2() n − 1 n n 2(n – 3)(n – 2) 6(n – 2) 4 6 6 6
Table 2 n() n +1()n + 2 ()n−2() n − 1 n (n − 5)(n− 4)( n − 3) Numbers of Unit6 Octahedra Used in Making Large Tetrahedra 6 ()n−4() n6 − 3 ()n − 2 6 Number of octahedra with paint on . . . edge ()n−2() n − 1 n (n − 5)(n− 4)( n − 3()n) −1 n()n + 1 length 0 faces()n− 46 () n − 3 ()n − 2 1 face 2 faces 3 faces 6 n() nTOTAL+16()n + 2 1 6 6 (n − 5)(n− 4)( n − 3) ()n−1 n()n + 1 2 (1) n() n +16()n + 2 6 ()n−2() n − 1 n 3 4 4 6 6 4 ()n−1 n()n + 1 6 4 10 5 ()n−26 () n − 1 n 4 12 4 (n − 520)(n − 4)( n − 3) 6 6 6 1 12 18 4 35
(n − 5)(n− 4)( n − 3) ()n−1 n()n + 1 n 2(n – 4)(n – 3) 6(n – 3) 4 6 6
for numbers of()n upright−1 n()n +tetrahedra 1 parallel exactly edge, subtracting 2 to account for the corners, and the formulae for unit6 cubes in Painting the Cube. multiplying by the number of edges, hence, 6(n – 2). Formulae for numbers of octahedra and inverted The same kind of reasoning gives 6(n – 3) octahedra tetrahedra have the same structure, but with the sandwiched between them. Counting unit polyhedra edge-length n reduced by 1 and 2, respectively, in each face of the large tetrahedron is equivalent to because of the way octahedra are always sand- measuring the area of each such face and multiply- wiched between upright tetrahedra, and inverted ing by the number of faces (4), after accounting for tetrahedra are always sandwiched between octahe- the corners and edges—but here the area is split dra. In any row of k upright tetrahedra, there are between unit tetrahedra and unit octahedra. Finally, always k – 1 octahedra between them. counting unit polyhedra in the interior of the large The reason for the similarity of structure of the tetrahedron is equivalent to measuring its volume, formulae is the similarity of the basis for counting after accounting for those on the outer surfaces (or each type of object: measurement of lengths, areas, without a correction factor to give the totals for each and volumes. The large tetrahedron has four verti- type)—but here the volume is divided among upright ces, so there will always be four unit tetrahedra with tetrahedra, octahedra, and inverted tetrahedra. 3 faces painted, and likewise four unit octahedra Thus, we are prepared for the easily recognizable adjacent to them. (In each case the exception is the constant and linear patterns that emerge in the table first entry in the table, where all four of them are the data. The quadratic pattern that emerges in two of same object, as is true for the cube.) Counting unit the columns and the cubic pattern that emerges in tetrahedra (that are not “corner tetrahedra”) along five more may be more difficult to identify by inspec- the edges is equivalent to measuring the length of an tion, because dividing the areas and volumes among
278 Mathematics Teacher | Vol. 100, No. 4 • November 2006 different types of polyhedra forces us to count in triangular patterns rather than the square ones to which we are accustomed.
TRIANGULAR GEOMETRY AND TRIANGULAR NUMBERS The familiar formulae for area and volume, which appear in the solutions to Painting the Cube, are based on objects made from squares (including cubes), down to the very units in which we measure them. That is,
A = l × w Fig. 4 Removing one row of triangles from an equilateral is an appropriate formula for the area of a rectangle triangle of side length n produces a triangle of side length in terms of its length and width only if the area is n – 1; removing one layer of tetrahedra and octahedra from given in square units, and likewise our formulae a tetrahedron of side length n produces a tetrahedron of for volume hold only if the resulting measurements side length n – 1. are given in cubic units. “What other kind of units would we use?” you may ask, especially if you note that multiplying units properly requires that
inch × inch = square inch. Top Second Third Fourth However, it is possible to define alternative units; layer layer layer layer and in special cases, such as the problem we are currently engaged in solving, new definitions are Fig. 5 The triangular pattern in which each type of polyhe- sometimes helpful. For example, we might make dron in the large tetrahedron is spaced definitions such as the following: There is, however, a difference: that is, “taking triangular inch—a unit of area equal to that of one away on each end” manifests itself differently an equilateral triangle with side length one inch with triangles. When working with squares and tetrahedral inch—a unit of volume equal to that cubes, removing all the outer rows or layers results occupied by a tetrahedron with side length one inch in a smaller square or cube whose edge-length is two less than the original; hence the prevalence of (n – 2) More informally, in this problem we need to know, in the Painting the Cube formulae. With a triangle or not how many unit squares fit into the area occupied tetrahedron, however, removing just one outer row or by one face of the large tetrahedron, but rather how layer (of both a unit triangle and a unit tetrahedron) many unit equilateral triangles fit into it. Likewise, results in a smaller triangle or tetrahedron whose edge- we need to know how many unit tetrahedra fit inside length is one unit less than the original, as illustrated the volume occupied by the large tetrahedron. in figure 4. Therefore, removing all three outer rows The answers to these two questions are surpris- of a face of the large tetrahedron results in a triangle of ingly easy. If we imagine taking a unit equilateral side length n – 3, while removing all four outer layers triangle and magnifying first its length (or base) of the large tetrahedron results in a tetrahedron of side and then its height by a factor of n, we can see that length n – 4. Thus, the number of unit polyhedra with its area should increase by a factor of n each time, paint on one side is 4(n – 3)2, as each of the four sides with the resulting magnification being of factor n2. has (n – 3)2 triangles not adjacent to an edge. Likewise, Therefore, the familiar formula the unpainted tetrahedron inside the outermost layer has a volume of (n – 4)3 unit tetrahedra. A = s2 Dividing the counts among the different types of polyhedra gives rise to two types of triangular also holds here: The area of an equilateral triangle patterns. If we observe the “footprints” shown in of side length s inches is s2 triangular inches; that is, figure 3 (or better yet a physical model in front of s2 unit triangles fit inside each face of the large tet- us), we can see that the alternation of upright tet- rahedron. Likewise, we find (by making three simi- rahedra and octahedra makes each of the two types lar magnifications of a unit tetrahedron) that the of polyhedra occur on each face of the large tet- volume of the large tetrahedron is s3, or n3, times rahedron in the triangular pattern represented in that of the unit tetrahedron. figure 5. The numbers 1, 3, 6, 10, and so on have
Vol. 100, No. 4• November 2006 | Mathematics Teacher 279 ()n−3() n − 2 2
()n−4() n − 3 . 2
1 2 ()n−3() n − 2 n− 3− n − 3 + n − 3 = , 2 ( ) ( ) ( ) 2
1 2 (n − 3)(n − 4) n− 3− n − 3 = . been known as triangular numbers (Conway 2which( can) be( obtained) by several2 different combi- and Guy 1996) since at least the time of Pythago- natorial approaches, is ras (Boyer 1968), and we can easily see that the n() n +1()n + 2 nth triangular number is the sum of the numbers . from 1 to n: namely, the well-known formula 6 n(n + 1)/2. (For consistency, we define the nth From this expression we obtain the formulae for
triangular number to be zero if n < 1.) total numbers1T of1() nupright+ 2T2() nunit + 3 tetrahedra,T3( n))(+1O1 unit n)( + 2octaO2 - n)(+ 3O3 n) Now each face of the large tetrahedron has (n – 3)2 hedra, and inverted unit tetrahedra given in tables =1 2()n − 3() n − 2 + 2 6(()n − 2 + 34 unit triangles in its interior. The number of these 1 and 2. belonging to unit tetrahedra is the (n – 3)rd triangular The remaining two+ 1formulae, 2()n − 4 ()for n − numbers 3 +2 6 ()ofn − 3 + 3 4 number, unpainted upright unit= 4n tetrahedra2 . and octahedra, can be obtained very simply, by subtracting the ()n−3() n − 2 numbers of painted polyhedra of each type from the VV=8 − 4VV= 4 2 totals derived above.OT The resultTT is that the number (again, see the “footprints”()n−3() nin − figure 2 3), while the of unpainted upright unit tetrahedra is the (n – 4)th number of triangles ()belongingn−4() n − to 3 unit octahedra is tetrahedral number,n() n and+1 ()then + number 2 () ofn− unpainted2() n − 1 n ()n−1 n()n + 1 2 . V + V + V the (n – 4)th triangular number,2 octahedra is the (n – 5)th6 tetrahedralT number;6 these T 6 O are, respectively, the numbersn() n +1(n of + upright 2))( tetrahen−2)( n- − 1)(n n−1)( n n + 1) ()n4() n 3 − − = VT + VT + 4VT 1 2 ()n−3() n − 2. dra and octahedra in the aforementioned6 interior6 6 n− 3− n − 3 + n − 3 = 2 , ( ) ( ) ( ) tetrahedron of side length n – 4. Note, however, 2 2 VT We can easily see that these two quantities sum to that the total number= ofn() ncompletely + 1 (()n+2 unpainted + () n − 2 ()n unit − 1 n + 4()n − 1 n()n + 1 6 1 2(n – 3)2. Thus, the total()n −numbers3() n − 2 of unit tetrahedra polyhedra does not sum to (n – 4)3, for two reasons. n− 3− n − 32 + n − 3 = , V ( )1 ( ) ( ) (n − 3)(n − 4) 3 T 3 2 and n octahedra− 3− n −with 3 = paint on2 1 face. are four times One is that each octahedron= (6n )), occupies= n V more volume ( ) ( ) T 2the above quantities, i.e., 2(2n – 3)(n – 2) and than a tetrahedron; the other6 is that each of the ()n−3() n − 2 2(n – 4)(n – 3), respectively. four exterior layers removed to produce the interior 1 2 (n − 3)(n − 4())n−3() n − 2 2 (nIt− is3 )also−( possible n − 3) =n to() n derive+1()n +these 2 . formulae by tetrahedron contains some inverted unit tetrahedra. 2 2 . 2 observing that the tetrahedra6 () andn−3 octahedra() n − 2 occur ()n−4() n − 3 in matched pairs except for the row2 of n – 3 tet- CONSERVATION OF. SURFACE n() n +1()n + 2()n−4() n − 3 rahedra along1T the() n bottom,+ 2T() n so + 3thatT( n .the))(+1 numberO n)( + 2.ofO n)(+AR3OE nA) AND V2OLUME tetrahedra in each1 group2 of6 (n –3 3)2 triangles21 is 2 We3 can also use the principles of conservation =1 2()n − 3()n() n−−4 2() n −+ 32 6(()n − 2 + 34 1 .2 of surface () narea−3 and() n − volume 2 to verify our vari- 2 n− 3− n − 3 + n − 3 = , 1 12T() n + 2T() n + 3T( n()))(n+(1O3 () n))( + 2 2(O n)()+ ous3(O formulae. n)) Each face of a large tetrahedron 1 +12 2()n − 43 () n2− 3−1+2 − 6()n −2 3 + 33 4 2 (n− 3) −( n − 3) + (n − 3) = , 2 2 =1 22()n − 3() n − 2 + 2 6(()n − 2 + 3(of4 edge length n) contains n unit equilateral 2 = 4n . ()n−3() n − 2 2 1 ()n−3() n − 2 triangles, so the total surface area should be 4n (n− 3) −( n − 3) + (n − 3) = 1, 2 (n − 3)(n − 4) 2 while2 the number of octrahedra+1 2()n − 4 is() n −2 3 +2 6()nn− −33 −+small3 n − 4 3 triangles. = If we add. the numbers of painted 1 2 (n − 3)(n −( 4) ) ( ) VVn−=38−2 − n 4 −VV 3 ==4 2 . 2 ( OT= )4n .( TT) faces, say Ti(n) for the number of unit tetrahedra 2 2 ()n4() n 3 1 2 (n − 3)(n − 4) with i painted faces in− the large − tetrahedron of n− 3− n − 3 = . . ( ) ( ) edge lengthn() n n,+1 and()n +O 2(n) 2for the numbers of unit 2 VVn()= n +81() −n 4 +VV 2 = 4 ()2n−2() n − 1 n ()n−1 n()n + 1 i . OT TTV n+() n +1()n + 2 V + V6 6 T 6 . T octrahedra,6 we Ofind Inspection (and figure 3) also reveals6 that 2 n() n +1(n() + n 2+))(1()n +n 2 −21)( n − 1)(n n−1)( n n + 1) ()n−3() n − 2 n() n 1()n 2 ()n2() n. 1(nn− 3) ()−n( n 1 − 3n())n+ ( 1n − 3) =4 , assembly of the individual= + layers + (faces)VT +− we con −- VT1T+−() n + 2 +T() n + 3TV(T n))(+1O n)( + 2O n)(+ 3O n) 6 V + 6 2 6 V + 1 62 V 3 2 1 2 3 sidered in the previous 6paragraph1T() n +T leads 2T() n to + a63 Tsecond( n))(+ 1TO n)( + 2O6 n)(+ 3OO n) 1 2 3 1 2=1 2()nV −3 3() n − 2 + 2 6(()n − 2 + 34 triangular pattern, in= nwhichn() n() n+ +11 the(n(()n + triangular+ 22))( + () n −n 2patterns−()2n)( − n1 −n 1)(+n 4()n − 1n n−()1n)(+ n 1n + 1T) = 1T() n 2T=()1 nV 2()+n3T − (3 n()))( n −1O 2 n+)(12V 2 6O(()+n − n)(22 3+O 3 n4) 4V(n − 3)(n − 4) of figure 5 are stacked atop1 one+ another2 T + to3 form+ a1 + Tn2− 3+−+1 n3 −2() 3n6 −=4()T n − 3 +2 6.()n − 3 + 3 4 6 6 ( ) (6 ) V +1 2()n − 4() n − 3 2+2 6()n − 3 + 3 4 2 tetrahedral pattern. The 3numbersT=13 2 ()thatn − 3result() n − 2are +1,2 6(()n −2 + 342 = (6n )),= n V = 4n. VT 4, 10, 20, . . . , each =then ()sum n +61 of(() nthe+2 firstT +2() nn −triangular 2 ()n − 1 n + 4()n − 1 n()n + 1 +1= 24()nn. − 4() n − 3 +2 6()n − 3 + 3 4 6 numbers. If we call these numbers the tetrahedral If we return to then() nn+ =1 2() casen + 2 illustrated in fig- V . 3 T 32 VV=8 − 4VV= 4 numbers (Conway and= (6 Guyn )), 1996),==4nn V .then we see that ures 2OT and 3, we canTT easily6 relate the volumes VV=8T − 4VV= 4 the total number of upright6 unitOT tetrahedra TTin a of unit tetrahedra and octahedra. Since the large large tetrahedron of side length n is the nth tetrahe- tetrahedron has all dimensions twice as large as a VV=8 − 4VV= 4 n() n +11T()n1() +n 2+ 2T2() n()n +−32T3()( n n −))(+ 11nO1 n)( +()n 2O−21 n n)(()n+ +3O 13 n) OT TT V + 3V + V dral number. As observed in then() nprevious+1()n + section, 2 () n−2unit() n −tetrahedron, 1 n ()n− its1 nvolume()Tn + 1 must be 2 , orT 8, times O 6 =1 2()n − 3() n6 − 2 + 2 6(()n − 26+ 34 VT + VT + VO the numbers of unit octahedra and inverted6 unit the6 volumen() of n +a 1unit(n +6tetrahedron 2))(n− (which2)( n − 1 we)(n can n−1)( n n + 1) n() n +1()n + 2 ()n−2() n − 1 n = ()n−1 n()n ++1 2V()n+ − 4() n − 3 +2V 6()n+ − 3 + 3 4 4V tetrahedra lag in this pattern by onen() and n +1 twoV(n+ +indi 2))(- ncall−2 V)(VT n). −+The 1)(n volumen −of1 )(aV n unitTn + 1octahedron,) V OT, can T T V T V 6 O 4V 6 6 ces, respectively, so that the number=6 of octahedra isT +6 therefore be foundT +6 by subtracting2 theT volumes of 6 6 = 4n .6 V the (n– 1)st tetrahedral number,n() and n +1 the(n number + 2))( ofn −2the)( n four − 1)(n unit tetrahedran−1)( n n used + 1) from the total: T V =Vn() n + 1 (()n+2 + () n4V−V 2 ()n − 1 n + 4()n − 1 n()n + 1 = T + T + TT 6 inverted unit tetrahedra is the (n= – 2)ndn()6 n + tetrahedral1 (()n+2 + () n − 26 ()n − 1 n + 4()n − 1 n6()n + 1 VVV=8 − 4VV6= 4 number. The value of the nth tetrahedral number, 3 TOT3 V TT = (6n )),= n V T =n() n + 1 (()Vn+2 + () n − 2 ()n − 1 n + 4()n − 1 n()n + 1 T = (6n3 )),T = n3 V 6 6 6 T 280 Mathematics Teacher | Vol. 100, No. 4 • November 2006 V n() n +1()n + 2 ()n−2() n − 1 n ()n−1 n()n + 1 3 T 3 V V V = (6n )),= n V T + T + O 6 T 6 6 6 n() n +1(n + 2))(n−2)( n − 1)(n n−1)( n n + 1) = V + V + 4V 6 T 6 T 6 T V =n() n + 1 (()n+2 + () n − 2 ()n − 1 n + 4()n − 1 n()n + 1 T 6 V = (6n3 )),T = n3 V 6 T ()n−3() n − 2 2
()n−4() n − 3 . 2
1 2 ()n−3() n − 2 n− 3− n − 3 + n − 3 = , 2 ( ) ( ) ( ) 2
1 2 (n − 3)(n − 4) n− 3− n − 3 = . 2 ( ) ( ) 2
n() n +1()n + 2 . 6
1T1() n + 2T2() n + 3T3( n))(+1O1 n)( + 2O2 n)(+ 3O3 n) =1 2()n − 3() n − 2 + 2 6(()n − 2 + 34 With+1 this 2()n fact − 4() we n − 3can + verify2 6()n − conservation3 + 3 4 of volume To make complete sense of those coefficients—in in= the4n 2formulae. given in tables 1 and 2. particular, to interpret the 1—it helps to think Adding the total volumes of upright tetrahedra, about the dimensionality of the parts that are being inverted tetrahedra, and octahedra yields counted. The cube has 8 zero-dimensional vertices, VVOT=8 − 4VVTT= 4 12 one-dimensional edges, and so on. The author’s n() n +1()n + 2 ()n−2() n − 1 n ()n−1 n()n + 1 analysis raised the idea of dimensionality in other V + V + V 6 T 6 T 6 O places as well, leading to a different sort of gener- n() n +1(n + 2))(n−2)( n − 1)(n n−1)( n n + 1) alization idea. Where the author moved from one = V + V + 4V 6 T 6 T 6 T three-dimensional building block (the cube) to V another (the tetrahedron), and redesigned units of =n() n + 1 (()n+2 + () n − 2 ()n − 1 n + 4()n − 1 n()n + 1 T 6 area and volume to make the investigation conve- V nient, one might also move (with either a triangular = (6n3 )),T = n3 V 6 T or a square figure) from dimension to dimension, investigating either problem (cube or tetrahedron) which is the volume of the large tetrahedron. in two dimensions or four. We can draw the two- dimensional case, which makes it easier to experi- CONCLUSION ment and gather data, but that, and some thinking, Painting the Pyramid, like Painting the Cube, may suggest ways of analyzing the four-dimensional brings together ideas from many different parts of case, which is harder to visualize. Does the idea of a mathematics. The geometric context provides the polynomial in (n – 2) generalize for n-dimensional opportunities for modeling and data gathering to square-like objects (square, cube, . . . )? If so, what identify patterns and the measurement notions by is the significance of the coefficients in each case? which to interpret them. The values of triangular If we count the same features of n-dimensional tri- and tetrahedral numbers are classic basic combi- angle-like objects (triangle, tetrahedron, . . . ), what natorial problems, and indeed appear as diagonals set of numbers (corresponding to the coefficients of Pascal’s triangle. The patterns in the formulae, of (n – 2)k in the case of the cube) do we get, and is especially the cubic ones, provide a nice illustration there a correspondingly “clean” polynomial pattern of the notion of translation, as we can see the shift that uses these numbers as coefficients? of the expression for the nth tetrahedral number by 1, 4, and 5, both in the data (and models) and in the REFERENCES arguments of the expression. Finally, the problem Boyer, Carl B. A History of Mathematics. 2nd ed. New provides access to a number of interrelated func- York: Wiley, 1968. tions, with no prerequisites other than determina- Conway, John H., and Richard K. Guy. The Book of tion and a little counting. Numbers. New York: Springer-Verlag, 1996. Two problems related by context are Eight Kahan, Steven. “Eight Blocks to Madness—A Logi- Blocks to Madness (Kahan 1974; Sobczyk 1974) cal Solution.” Mathematics Magazine 47 (1974): and Painting the Cube with n Colors. In the latter, 57–65. one must paint all six sides of each unit cube with National Council of Teachers of Mathematics one of n colors (e.g., colors 1 through n) so that the (NCTM). Curriculum and Evaluation Standards for n × n × n cube can be assembled n different ways: School Mathematics. Reston, VA: NCTM, 1989. In the first way, all exposed faces show color 1, in Reys, Robert E. “Discovery with Cubes.” Mathematics the second way, all exposed faces show color 2, and Teacher 81 (1988): 377–81. so on. (A very elegant solution exists to this prob- Sobczyk, Andrew. “More Progress to Madness Via lem.) One final question to consider: Is it possible ‘Eight Blocks.’” Mathematics Magazine 47 (1974): to form an octahedron of side length n using unit 115–24. ∞ octahedra and tetrahedra?
Editors’ notes: The bottom row of the table in fig- ure 1, rewritten slightly and shown here as figure 6, CHRISTOPHER KRIBS-ZALETA, suggests a polynomial in (n – 2), and the author has [email protected], teaches at the Uni- already commented that the coefficients (1, 6, 12, 8) of versity of Texas at Arlington, in this polynomial count significant features of the cube. Arlington, TX 76019. His interests include the development of computational strat- 0 faces 1 face 2 faces 3 faces egies for operations on rational numbers, as well as mathematical epidemiology and ecology. 1(n – 2)3 6(n – 2)2 12(n – 2)1 8(n – 2)0 Photograph by James Epperson; all rights reserved. Fig. 6 Bottom row of table in figure 1, rewritten slightly
Vol. 100, No. 4• November 2006 | Mathematics Teacher 281