ON STABILITY OF POLYTOPES
BY Wendy Finbow
SUBSIITTED IX P-iFtïL-U. FCZFILLMEST OF THE REQLXRE-UIEXTS FOR THE DEGREE OF
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Canada Contents
List of Tables
List of Figures vii
Acknowledgements X
Abstract xi
Abbreviations and Symbols Used xii
1 Introduction 1
2 Preliminary Results 2.1 Minkowski's Theorem ...... 2.2 Further Results ......
3 Monostatic Simplexes in 8, 9, and 10 Dimensions 24 3.1 The Search Method ...... 25 3.2 Cornputer Algorithm ...... 26 3.3 Error .4nalysis...... 33 3.4 The Main Results ...... 37
4 Loaded Polytopes 44 4.1 Preliminary Results ...... 44 4.2 Nonloadable Polytopes ...... 47 4.3 Loadable Polytopes ...... 52 4.4 Related Results ...... 89
Bibliograp hy 91 List of Tables
3.1 Maximum contribution of each simplex with five facets ...... 30 3.2 Maximumcontributionofeachsimplexwithsixfacets ...... 31 3.3 Maximum contribution of each simplex with seven facets ...... 32 3.4 Mawimum contribution of each sirnplex with eight facets ...... 33 3.4 Maximum contribution of each simplex with eight facets, continued. . 34 3.5 Maximum contribution of trees with 5, 6. 7. 8. and 9 nodes ...... 38 3.6 Contributions of vectors ...... 39
4.1 Loadable/Nonloadable Polytopes ...... 48 4.1 Loadable/rYonloadable Polytopes. continued ...... 49 List of Figures
2.1 Illustration of the Projection Criterion (tetrahedron) ......
3.1 Example of a straight tree ...... 3.2 Example of a tree with six nodes ......
4.1 Overview of a truncated tetrahedron ...... 4.2 Overlap of opposite triangular facets (icosahedron) ...... 4.3 Overlapping facets of the icosahedron ...... 4.4 Overlap of opposite pentagonal facets (dodecahedron)...... 4.5 Overlapping facets of the dodecahedron ...... 4.6 Overlapping facets of the tmncated icosahedron ...... 4.7 Overlapping facets of the truncated dodecahedron ...... 4.8 Overlapping facets of the cuboctahedron ...... 4.9 Overlap of a pentagonal facet onto a triangular facet of the icosidodec- ahedron ...... 4.10 Overlap of a triangular facet onto a pentagonal facet of the icosidodec- ahedron ...... 1.11 Overlap of square facets of the rhornbicoçidodecahedron ...... 4.12 Overlap of a square facet onto a pentagonal facet of the rhombicosido- decahedron ...... 4.13 Overlap of square facets onto triangular facets of a rhombicosidodeca- hedron ......
vii 4.14 Overlap of triangular facets onto square facets of the rhornbicosidodec- ahedron ...... 4.13 Overlap of pentagonal facets onto square facets of the rhombicosido- decahedron ...... 4.16 Overlapping facets of the rhombicosidodecahedron ...... 4-17 Overlapping square facets of a snub cube ...... 4-18 Overlap of square facets onto triangular facets of the snub cube ... 4.19 Overlap of triangular facets of the snub cube and snub dodecahedron 4.20 Overlap of triangular facets onto square facets of the snub cube ... 4.21 Overlapping facets of the snub cube ...... 4.22 Overlapping pentagonal facets of the snub dodecahedron ...... 4.23 Overlap of the pentagonal facets of the snub dodecahedron ...... 4.24 Overlap of triangular facets onto pentagonal facets of the snub dodec- ahedron ...... 4.25 Overlapping facets of the snub dodecahedron ...... I dedzcate my thesis to my Dad. Acknowledgements
1 should like to thank my supervisor, Dr. Dawson, for his support and encourage- ment over the last two years, and Dr. Thornpson, for his endless supply of corrections.
1 should also like to thank Dr. M. Heukaeufer' a professor in the department of Modem Languages at Saint Mary's University, and Connie hk. for their help in translating the paper by Minkowski, (111, from Gerrnan to English.
Lastly, 1should like to thank my fnends and famil. for ail their love and support. Abstract
A polytope is said to be monostatic if its center of gravity is above only one facet. We show that no monostatic simplex exists in eight or fewer dimensions?and there is no monostatic simplex in ten dimensions that fdls sequentially onto dl its facets.
We cda polytope loadable if there is a point in the interior of the polytope that is above only one facet. We classify some convex polytopes as either loadable or nonloadable. Abbreviat ions and Symbols Used
e,= (0,. . . 0,1,0,. . . ,O)=, where the 1 is in the ith position of the vector.
0 u 5 V. where u = (ul,.. . , and v = (vlo... v,JT implies that ui 5 viVi =
1, ...,72
xii Chapter 1
Introduction
-4closed and bounded region in space is said to be convez if given aqv tn-O points p and q in the region, then the line joining them is aiso completely contained in the region. (In other w-or&. tp + (1- t )q is inside the region for every t E [O. 11). A conves set wïth nonempty interior is called a convez body. Let a E Rn. and let b E R. Then a hyperplane is the set of d points x = (xl. . . . .x,) satisfiing the equation a - x = b. The vector a = (alt... .a,) points in the direction normal to the hyperplane. and if llall = 1. b is the distance fiom the origin 0. to the point in the hyperplane ciosesr to it.
-4 polytope is definecl. see. for example. [4]to be a conves region in Rn bounded by a finite number of hwyperplanes. -1 h-yperplane that bounds a polytope is cailed a supportzng hyperplane, or a support plane of the polytope. If HI.. . . . Hm is a finite set of hwyperplanesthat bound a polytope P. Hi n P. where i = 1.. . . . m is said to be a facet of P provided that it is maximal in the sense that no Hj n P. j = 1. . . . . m. j # i properly contains it. If H, f~ P is a facet? then H* is called a facet hypeîplane. For any facet F* of a polytope, the facet vector vi is the vector with length equal to the (n - 1)-dimensional volume of the facet, directeci outward and normal to the facet. We may describe a polytope P mith m facets in the following -: The facet Fi lies in the hyperplane ai-x = bit V2 = 1,. . .,m, with 11aJ1 = 1. The facet vector vi that describes Fi is a multiple of the vector Vi = 1, . . . , m. Then if R-edefine Yi = ~~/ll~~ll,and let A = (Yl,.. . ,y,)=. b = (bl,.- . .bJT, P is descrîbed as the region defined by -4.x 5 b. ( For v = (vil .. .,un) E Rn.w = (wl,.. ., w,) E Rn-we Say v 5 w ifui 5 wi, Vz = 1,. . . ,n.)
Let a polytope P E Rn be given. Let density of P. p(x), be a non-oegative, real- valued function defined on the points x E P. If the volume of P is V. then define the mas of P to be m = J Jv p(x)dCi. The center of grauity of P is
If for ail x,y E P, p(x) = p(y), then we say the polytope has unzfom density.
We Say that a point is above a facet of a polytope if the line parallel to the facet vector v and through the point meets the interior of the facet. .A polytope is said to be stable on a facet if and only if its center of gravity is above that facet. Therefore the position of the center of gravity of a polytope determines whether or not the polytope is stable on a particular facet. A regular polytope (constructed) of uniform density will be stable on dl of its facets, since its center of gravity will be located at the center of the polytope, above every facet.
Imagine a pardelogram with two 45' angles, such that two sides are tnice as long as the other ones. Such a pardelogram will be stable on any of the long sides, but if set on a short side, it will fa11 over, since the center of gravity of the object is not above that face. A convex polytope of uniform density will fall from its ith facet to another facet if the center of gravity of the polytope is not above the ith facet. The dihedral angle between the ith facet and the facet onto which the polytope falls must be obtuse, and if the polytope is stable on its jth facet, the center of gravity di be above that facet. Because the angle between any two facets in a polytope such that it falls from one facet to the other must be obtuse, the angle between the facet vectors must be acute. Note that when a polytope falls boom one facet to another. the position of the center of gavity of the polytope is lowered. so the polytope cannot continue to fa11 kom one facet to another facet forever. In fact, a polytope will never fall onto a facet from which it has already fallen. This means a polytope must be stable on at least one of its facets, a resting jacet; a polytope which is stable on only one of its facets is said to be monostatic. Given a monostatic polytope P. we define a pendtirnate facet of P to be a facet from which the polytope fails onto its resting facet.
A simplex in Rn is defined to be the convex hull of n + 1 independant points. Equivalently, it can also be defined in Rnto be a convex body which is bounded by exactly n + 1 facet-planes. In a paper published in 1969 by Conway and Guy. [3]. it was proved that no tetrahedron or polygon is monostatic, and it was asked in how few dimensions a monostatic simplex can exîst, if any. Dawson. [5] showed in 1985. that no monostatic simplex exists in six or fewer dimensions, and he found one that does exist in RLo.In 1996, it was shown by Dawson, Finbow, and Mak, [7] that t here are no monostatic simplexes in seven dimensions or fewer, and a monostatic sirnplex exists in RH that fails through each of its facets in turn.
In Chapter 3, Theorem 3, it is shown that no such simplex exists in eight dimen- sions. In addition, we show that in ten dimensions, no monostatic simplex exists which falls sequentially through al1 of its facets. In Chapter 2, proofs are provided of previously known results which will be needed in Chapter 3. These include an English translation of Minkowski's Theorem on equilibrated facet vecton.
In Chapter 4 we look at polytopes of non-uniform density. -4 polytope is loadable if there is a point in the interior of the polytope that is above only one facet. If the polytope were weighted in such a way as to move the center of gravity to this point, then such a weighting would yield a monostatic polytope. We investigate the polytopes in several well known classes to determine which ones are loadable. These classes are: the regdar solids, the zonotopes, and the Archimedean solids.
Findy, in Chapter 4, we study the construction of density functions which may be useful for optimizing the position of the center of gravity. Let K be a bounded convex body in Rn, and for each density function p on K. let c(p) be the correspond- ing center of gavity of K. We consider the set D of density functions p whose range is {dl,d2}, where dl > 0, and d2 < W. For a given linear functional f on K. we constmct (Theorem 5) p* E D such that j(c(p*))= mau{/(c(p)) : p E D}.
The following notation is used throughout this thesis: ei = (0, .. . .O, 1.0, . . . ,O)=. where the 1 is in the ith position of the vector. O = (0, . . . ,o)~. and 1 = (1, . . . .1)*. If v = (u~,.. .,un), w = (utlt.. .,wn)E Rn,, then V = v/llvll. and v 5 w means Chapter 2
Preliminary Result s
In this chapter, we develop definitions and lemmas that will be used in Chapter 3 to prove various results on monostatic simplexes.
2.1 Minkowski's Theorem
-4 set of vectors {m,xi,. . . ,x,,,) in Rnis equzlzbrated [B. 111 if:
ii) {q} spans Rn,and
iii) (Vz # f,i,j =O, .. . ,m,a 3 O) # axj .
Given an rn x n matrix A, each of whose rows are unit vectors. and a vector b E Rm such that b 2 0, we define the polytope Pa4(b)= {x E Rn : .-lx 5 b)' and V,(b) to be the n-dimensional volume of P4(b). When no ambiguity can arise. the subscript A wilI be dropped. If vi is the ith row of A, then the number rl is called the ith tangentid parameter of the polytope if rl 2 0, and the plane v,-x= rf is a supporting hyperplane of the polytope. If wf is the ith tangential parameter of P(w) then P(w) = P(w*),where w*= (w;,. .. ,w;)=. We cal1 the vector w* a tangential vector of P(w).
Let TA = {b E Rm : 3w E Rmsuch that b is the tangential vector of P4(w)). We note that Tq is a cmex cone in Rm. Indeed. if u, w E TA,then for each i there exist y, z E Rn,SUC~ that y E P(u),and z E P(w) SUC~that vi*y= ui and Vi.2 = Wi. Hence vi*(Ay+ (1 - X)Z) = Xvi*y+ (1 - X)vi.z = Aui + (1 - A)wi (2-1) and therefore Xy+(l-X)z E P(Xu+(l -X)w), and we dsosee that Xy+(l-X)z is contained in the hyperplane vi*x= Aui + (1 - A) wi, thus Au+ (1- A) w is a tangential vector. The set TAis also a cone, since P(Aw) = XP(w) for al1 A.
Theorem 1 is a generalization from 3 to n dimensions of a theorem discovered by Minkowski which originally appeared in 1897, [II], in German. For the proof we dl need two technical lemmas.
Lemma 1 (Minkowski, [Il]) Let {vi}zR=,be a set of equzlzbrated vectors in Rn. Then there ezists a bounded polytope with m fucets, whose ichfacet vector is a multiple of vi
Proof. Let Yi = vi/llvill The system .Ix 5 1, where d = (Yl,Y*, . . . . defines the polytope P(1). The m planes described by the equation =tu = 1 are tangent to the unit ball in Rn,thus P(1)contains the ball. Furthemore, the point of tangency of the i" plane with the unit ball is contained in no other hyperplane, and hence each hyperplane intersects P(1) in a facet. Therefore 1 is a tangential vector for P(1),P(l) has rn facets and its ichfacet vector is a multiple of Yi.
Findy we show that P(1)is bounded. Ifq E P(I),then vi*q= II~illYi*q5 II~ill, for i = 1, . . ., m. On the other hand, since {vi) is equilibrated, for ail q E P(1). But then P(l)is a subset of the bounded parallelepiped B obtained as follows: let S c {vi)ZLbe a basis for Rn and define B = nvEs{x: a, 5 v - x 5
Lemma 2 (Brunn, [2])Let A and B be conuez sets in Rn. and let Q4) and V(B)be the volumes of the sets A and B respectivefg. Then for t E (O. 1) we have dv(t4+ (1 - t)B) 2 t + (1 - t) uith equality if and only if -4 is obtained ftom B by dzlation and translation.
Theorem 1 (Minkowski, [Il])Every equilibrated set of uectors in Rn is the set of fucet vectors for a unique conuez polytope up to translation: and the set of facet vectors of any convez polytope is equilibrated.
Proof. Let P be any convex polytope, and let P have rn facets Fi. mhere the m facets have an arbitras. ordering. The following construction and result ing formulas for the derivative will be useful in both halves of the proof.
Consider any q in the interior of P and define the one parameter family of poly- topes P(t)= tP + (1- t)q. Let C(t)be the volume of P(t),and for i = 1.. . . . m. let C&) be the volume of the convex hd, E(t)of {q) U (tFi + (1 - t)q),d,(t) be the (n - 1)-dimensional volume of tF, + (1- t)q, and hi(t)be the length of the perpen- dicular àropped fiom q to the hyperplane containing tFi + (1- t)q. Then it is easy to see that C(t)= CgR=,Ci(t), k(t) = th(l),and &(t) = tn-'.li(l). NOW,since Pi(t) is a pyramid, Ci(t)= iAih(t) = :tn%(l)hi(l),and hence dCi(t) = tn-lAi(l)hi(l)= Ai(t)hi(l). If V denotes the total volume of P, we have both
V = &hi(l) and dV = didhi. i= 1 n i= 1 Now, let {vilbe the set of facet vectors of such a polytope P. we will show that
{v,) is an equilibrated set. Choose q in the interior of P. there exists E > O such that qk = q + cep is in the interior of P for k = 1: 2,. . . , n? and then b- 2.3 we have both
where qki is the length of the perpendicdar dropped from qk to the hyperplane containing Fi,V k = 1, . . ., n. But then
for k = 1,. . . , n and hence m
Therefore, {y) is equilibrated, for, since P has non-empty interior. {vi)spms Rn. and convexity assures that vi # avi, Vi # j, a 2 O.
Conversely, let {vi)g"=,ean equilibrated set of vectors in Rn. Let -4 be the matrix whose ith row is the vector Ci,where Yi = vi/j(vi11, let T = T4be the convex cone of tangential vectors, and for each x E T, let V(x)= I.A(x) be the n-dimensional volume of P(x) = PA(x).Suppose x = (X~~....X,)~,and set r = mâ~~<~~~{~~}. Xow P(l)is bounded, as shown in Lemma 1, and P(;x) C P(1).hence if we define p so that V(1) = pdn we obtain
This implies that P(x)is bounded.
Let A,(x) be the area of the ith facet of P(x). Xote that .-li(x)and V(x) are continuous functions of x; indeed V is differentiable and by Equation 2.3. 61- =
Setting f(x) = (r,x) x E Tl we have by Lemma 2.
for x, u E Tl and hence the set
T T W = {W = (wl1... ,w,+l) : (utl,.. ., wJT E T and O 5 w,+l 5 /(L.I,.. . . w,) ) is convex. We further note that it is a cone with apex at the origin. For x E T. x =
(xll...,x,)~. Set x = (21,.. .,x,+~)*,where X,+I = f(x) and let L: = {x:x E T). Observe t hat and hence we have
i= 1 is the tangent hyperplane to W at the point x E U. It follows that for dl q E 11,: q =
To prove existence of the polytope, consider ÇVl = {w E W : w,+~ = 1). let d = ( IIv~ll,. - - ,IIvm 11, O)=, and define the real-valued function (T on CVl by set ting O(W) = d-W.First, it is established that the minimum value of o is attained on LC;.
We do this by showing that the set Z = {w E 1.h : d-w 5 p XE1IIv~~~) is com- pact and nonempty. Indeed V(p1) = pnV(l) = 1 so r = (p.. . . .p. 1)= E 11'; and d-r = pEhl Ilvill: and therefore we see that Z # 0. On the other hand, Z is com- pact since d=(wl,.. . ,w,+~)~ 5 p llvi1l implies chat mi 5 fi zz"=,Ivi11. where M = m~~~i~n{llvill).
Note that the minimum value of o is positive: for if w E T is such that TV E 1.b;
~d r = ~u~<~<~{w~),- - then by 2.7, p = JI ~v(w)5 r and hence d-W 2 .Ur 2 Mp.
Let p, = (pl,. . ., P,+~)~ = (pl, . . . p,. l)= be the point where the minimum
OCCUIS, let p = (po,.. . ,Pm), and let the minimum va,lue of 0 be a. Then we have d-po = a. If x E W, = -E IVl, so ~-X/X,+~2 a, which implies d-x 2 ~m+l î/&F) ax,+l. This implies that XEl ll~~ll~~= axrn+i supports the cone FI; at p, and hence must be the same hyperplane as 2.10 with x = po. Comparing coefficients and not ing that V(pj = 1, we see that
Now P(p) is a polytope whose ith facet has n - 1dimensional volume nllvill/a. There- fore P = P (($ p) is a polytope with the correct face vecton.
To see that this po-ytope is unique up to translation, suppose the set of face vectors of both P(p) and P(q) is {vl,.. . ,v,). Since ll~ill> 0 W ~e sec that P and q are in T. Observe that since Ai(p) = Ai(q) = Ilvill, we have by 2-10 and 2-11 which implies V(p) 5 V!q). Symmetrically, V(q) 5 C'(p). so Y(p) = Ce(q).
Xote that this implies by equation 2.10 that the hyperplane that supports CV at p and the hyperplane that supports W at q are identical. and the line {w + (1- t)q: t E [O, 11) also lies on this plane. But then
so /(tp + (1 - t)q)= t f (p) + (1- t)f (q). But then by Lemma 2. rve have that P(p) can be obtained from P(q) \la translation and dilation. and since C'(p) = C'(q). the dilation factor is 1.
2.2 Further Results
In this section, we concentrate on results about simplexes. Given a simplex in Rn with facet vectors Q, . . . ,%, we dehe pi(xj) = ~=x,/ll~ll= IIx~~~ COSO~~~ where Qij is the angle between vectors and xj. In the following lemrna we determine what conditions must hold in order for a simplex to fd from a facet i to a facet j. The Projection Criterion, which originally appeared in [5], gives an inequality involving the area of the ith facet of the simplex, and the projection of the jth face vector onto the ith face vector. Lemma 3 (Projection Criterion, Dawson, [5]) A simplez wzth/acet vectors {v*} will tip from the ith facet to the jl facet if and only zf // < pi(xj)-In particular, the ith facet is smaller (that is, the (n- 1)-dimensional volume) then the jth facet.
Proof. Let F, be the facet described by xi, and vi be the vertex opposite Fi in the simplex. The center of gravity of a simplex, c = ,1 (1:z vi). Let L be the (n - 2) - dimensional hyperplane containing Fin F,, and without loss of generality, assume that the origin O is the foot of the perpendicular dropped from v, to L. See Figure 2.1.
Let pLL(x) denote the vector projection of a point x into LI. and define A(& n 4)to be the (n - 2)- dimensional volume of F, n Fj. Since Ilxi(l is the area of
= l*llpLl(vj)ll.-I(Fin-L n F,), and similarily 11~~11= t herefore follows t hat
Let M be the hyperplane containing LL fi Fi,and define pLtI(x)to be the scalar projection of the point x into M. Then
Given that the simplex falls £rom its ith facet onto its jth facet, its center of gravity c must not be above the ith facet, therefore pM(c)< O. Then holds if and onlv if
and the statement of the lemma foliows.
Figure 2.1: Illustration of the Projection Criterion (enahedron) The following lemma domus to project the facet vectors of a simplex into the plane spanned b- 3 and x,,, thereby simplif'ing any calculations to do with the projection of a facet of the simplex onto W.
Lemma 4 (Dawson, Finbow, Mak, [?]) Let Bi, be the angle between cectors x, and xj in Rn. For any Hven value of Bo, E (0. a). cos Bol cos 012- - - COS em-lm is m&màzed for Bi i+l E (O. ~/2)~helien Bol = B12 = = Bm-lm = Bh/m-
Proot We need to show three things in order to prove this lemma. one. the prod- uct cos OO1 cos BI2 COS 0,- 1 m is ma.-edden the vectors {x, } Lie in a plane. two. the above product is ma.cimized when the vector x, lies in the cone generated b-x,-~ ad and three, it is mashized when OO1 = OI2 = . . . = 0rn-l m.
Take any three vectors {q?x, Y xr } in Rn. If al1 t hree do not lie in a plane. then we can project the jfi vector into the plane spanned by x, and xk. and ob- tain a new vector $ to replace xj. If tj is the angle betn-een x, and x,. then COS B;, COS 5, > cos 19, cos Ojk. This is because q, < Bij. and cosine is a decreasing POS- itive function on (O. ;i/2)' w COS Bij < COS q,,. similady n-e see that COS qk> COS ojk. so COSO,~COSO~~~ahere O*jj'Ojk € (0. ~12).is maximized when the three vectors lie in a plane. Since this is the case for any set of three vectors. it is t hen rrue for al1 vectors.
Sel? we must show that the product cos Ool cos e12- cos Om-l, is maximized when x, is a vector lyhg in the cone generated by q-l and the plane. for if this were SO, we can say that the product is maximized when Bi-1 i-1 = i i Bi i-1. This can be proved by contradictiony so suppose xi does not lie betn-een x,-1 and x+ll Say! q+l lies between q-1 and 4, instead. The angle Bi-l > Bi-l i,l. so cos Bi i-l < COS Oi-, i,, . Therefore interchanging xi and X+I ni11 increase the product of the cosines of those angles.
The argument above tells us that the product of the cosines is maximized n-hen ali the vectors lie in a plane' and the sum of all the angles between the vectors is equal to the angle between the first vector m, and the last vector. h.If a?0 E (O, r), and a + ,O is constant, then 1 COS CE cos ,û = - [cos(cr + 8) + cos(a - O)] (2.18) 2 is maximized when a - /3 = O, that is, when O = B. Hence cos eOlcos - COS for Qi-l i E (0, '2) Vz is maximized when al1 the angles between each of the wctors is the same. 1
In light of Lemma 3, for vectors G,xj, we write x, \A xj (read xi falls onto Xj) if and only if llqll < pi(xj) For i = O, 1,. . ., let
Given a simplex in Rn, with facets Fo, . . . , F, which falls from Fito Fo t hrough other facets, the number ai represents a bound on how small the projection of an individual facet vector x, onto xo can be. The following lemma. which first appeared in [5], determines the bound on how smd ai cm be for a given 2.
Lemma 5 (Dawson, [5]) If x, \ q-i \ - - \ q,then and by the Projection Criterion (Lemma 3), kJ.LIlx,ll < COS Oij,
Multiplying both sides of Equation 2.21 by cos O&, we see that
a- L 11411,me, > min ~,co~e~IIcos~~,+~ llxoll j=o
The product in 2.22 is maximized when the angles Oj j+i lie in a plane by Lemma 4, so c~~~B~~+~2 19,. Adding a to both sides and solving, we find (T - Bot) + 6, j+i 2 r. The product of the cosines is maximized when the angles are equal, that is, when a - Bo, = Oo1 = . . . = 1 ~/i+ 1, and we find that
Since cos(n - O~)= - cos go, ,
Finally, since - (cos (&))i+L < O, we have, by 2.22 and 2.23,
PO(%)= (rn" cos 0 ) llqll
Lemrna 5 implies that PO(%) ai = min - llxall It is also clear that a0 = 1. If xl \ qlthen Bol must be acute, so al 2 0. and by the projection criterion, a-1 2 1.
For a simplex in Rn with facet vectors qt...,%, where llxoll 2 3 ll~~ll, optimizing the projection of each individual vector onto one particular vector will give bounds that enable us to show that no monostatic simplex exists in five or fewer dimensions as follows: Let a set of six vectors in R5 be given and suppose, for a contradiction, that they are the facet vectors of a monostatic simplex. There are two cases to consider, if the simplex has 1 penultimate facet, and if the simplex has more than 1 penultimate facet. For the case where the sirnplex has 1 penul- timate facet xl, assume without loss of generality that llxl 11 = 1- The projection of a onto xl is bounded below by a-1, and the projection of a vector onto it- self is a*. .4.s well, the projection of the ilh vector, qtonto XI is bounded belon. by ai-1. The sum of the projections of the vectors xo, . . . , xj onto xl is bounded below by a-l + a. + al + a:! + a3 + a4 zz 1.278443219 > 0: and therefore the vec- tors cannot be equilibrated. In the other case, where there are two penultimate facets, and we project each vector onto q,so without loss of generality, assume 11% 11 = 1. Then the mm of the projection of the vectors must be bounded below b- a0 + 2a1 + a2 + as + a4 x 0.27843219 > 0, and we see that no set of six vectors in R5 cmbe equilibrated, and therefore no monostatic simplex exists in R5.
Optimizing the projection of two vectors simultaneously onto a will give us better bounds, and allow us to show that no monostatic simplexes exist in sLu and seven dimensions.
The parameter bi is defined to be the minimum of the combined projections of xl and onto the vector W. For any bi we have:
Hence if the system tips from xl to Q, we get bi > COS*^^^ - COS^^^. Setting x = cos OO1, we get bi 1 x2 - x. This is minimized when z = 8,therefore,
\Ire now have bounds on vector projections that are tight enough to show that no monostatic simplex exists in R6. (see [SI). Let a set of seven vectors in R6, 3,. . . ,~g, where 11% 11 2 - 3 11% 11 be given, and suppose, for a contradiction, that they are the facet vectors of a monostatic simplex. Again there are two cases to consider, depending on whether the simplex has 1 or more than 1 penultimate facet. If there is only one penultimate facet, the sum of the projections of al1 the vectors onto the vector describing the penultimate facet, xl, is bounded below by a-1 + a. + a* + a3 + a* + b5 = 1.02843219 > O, contradicting the assumption that the vectors are equilibrated. If there is more than one penultimate facet, the sum of the projections of al1 the vectors onto the vector describing the largest facet, xo, is bounded below by a0 +ai + a2 + as + a4 + bs = 0.28432189 > O, and no set of 7 vectors in R6 which obey the various inequalities required by the projection criterion for a monostatic simplex is equilibrated, therefore there is no monostatic simplex in Rb For sorne calculations, we will need to improve the bound on bi. Lemma 6 (Dawson, Finbow, Mak, [7])Suppose x, \ xi-1 \
bi 2 cos2(ir ) + COS(Z~)cosi (fif), where y = (a - Boi)/i
Proof. -4s was seen, it is straightforward to determine a lower bound on bi = min{po(xL+ q)/llm(l : (3 x.,-*, . . . .XL E Rn)xi\ xi-1 \ . \ q). However. -I4 is a somewhat rough estimate, and when the number of vectors lying in the cone generated by Q and q,and hence the angle Ba is small. then bi will be larger than
-I.4 This is because the angle between any two facets in a simplex such that the simplex falls from one facet to the other, must be obtuse. Hence the angle between the facet vectors must be acute.
First, note that
by the Projection Criterion, and similarily,
Note that COS Bli = - COS(* - Oli)- By Lemma 4, the product in 2.30 is maximized when the vectors lie consecutively in a plane, so cor + + Bi-i < Bo*, and the angles Bi-i are dl equal to &,Ji. This implies that the product is maximized when T - Ooi = Oi-1i = Bol, thus Bo, = Since 7 = r - BO1/i,Bol = 7r - i7 Substituting these into Equation 2.31. Lemma 7 follows.
In particular, 1
To see this, suppose that x2 \ xl \ a.Since b2 = minpo(xl + x2)/llxali, Lemma 7
Differentiating the right-hand side of Equation 2.33 with respect to 7 and minimizing, we see that
O = -4 cos(27) sin(2y) - 2 sin(2y) cos2y - 2 cos(27) sin cos
= - sin y cos y[8(2 cos2 7 - 1) + 4 cos2 nf + 2(2 cos2 - l)] = (12~0s~~-5)(-2suiycosy).
Solving for y, nie find the smaller of the two solutions is
Substituting y = arccos(,/5/12) into Equatioo 2.33, we see that 62 2 2 .
In solving for each remaining bi, i = 3,d, 5, we obtain the following numerical solutions:
Lemma 7 (Dawson, Finbow, Mak, [7]) If O 5 i 5 j, then ai 2 aj, and b, 1 bj. Lemma 7 relies on the fact that the projection of a vector v onto another vector u is bigger than the projection of a third vector w onto u- if the angle between u and w is greater than the angle between u and v, and llvll 2 Ilwll To show bj 2 bk, where j is less then or equal to k, it is sufficient to show that bi 5 biVl for al1 i > 0.
A key step to showing that no monostatic simplex elusts in Riand R8 requires that given two penultimate facets Fiand F', such that the simplex never falls onto F., we find a bound for the minimum of pl(x. + Q), where x. and q are the facet vectors corresponding to facets F. and Fo respectively.
Lemma 8 (Dawson, Finbow, Mak, [7])Suppose xt \ % and x. \ Q. Then
To prove Lemma 8, we consider two cases: one when the angle between facet vectors xl and x. is les than or equal to g, and the other case, when BI. > 5. Let a monostatic simplex with two penultimate facets described by the vectors xl and x. which both fall onto a final vector be given. The goal is to find an upper bound on how small pl (x. + q)can be.
If 01, 5 a, then cos&. -> O and so pl(x.) 2 O. This implies that pl(m + x*) = 1 pi (m) +pl (4)2 pi (m) Since $ cos Bol < i, we actually have cos Ool > $ cos Bol - 5. ho,çince = cos solthen Noting that BI. = Bol +Bo., and taking the derivative with respect to 80. to minimize the right hand side of Equation 2.36, we find that
O = - sin 4. cos(80i + 60.) - cos 80. sin(Ooi + Bo.) = - sin(Ool + 2Bo.). (2.37)
Since xl \ ~0 and x. \ a,801 and Bo. must be acute, so Bol + 2eo. = T!and solving for Oo,, we find that
00. = (T - eol)/z. (2.38)
Substituting this into equation 2.36, and using half angle formulas to simpli@. we find that
7r - 001 n + 001 = coseoi +cos ( * ) cos ( )
For the next reçult we need the following definition, which modifies the deletion of ai. We cm now prove that in R7,no monostatic simplex exists. (See [il). Let a set of eight vectors in R~,Q,. .. ,XT, where ll~lj2 - - > Ilx711 be @en, and suppose. (for a contradiction) that they are the facet vectors of a monostatic simplex. The proof has six different cases, depending on how many penultimate Eacets there are. and on how many facets fa11 onto them. Five cases are straightforward. and can be verified in the same way as we showed that no monostatic simplex exists in fewer than 7 dimensions above. The sixth case, in which there is a penultimate facet onto which the simplex never falls is more difficult, and requires the use of Lemma 8. There is a penultimate facet onto which the simplex never falls. Without loss of generalit- label the corresponding facet vector x~.If the other penultimate facet is described by xa, then the sum of the projections of these vectors onto xo is greater than. by Lemma 3,
Note that if the set {q)is equilibrated, the left hand side is equal to zero. and Bo, > arccos 113 = 70.52878". Projecting onto m, by Lemrna 3 we find
PO(^') > - cos Bo, cos' llvoll (" -i"~) . 1 L and a2(Ooa) > -119, a3(OOa)> -0.173211.. . , a4(aoa)> 9 + 6;.i '-: -0.207336 - . .. a5(Boa)> -0.229257. . . , and as(Ooa) > -0.24606. . .. Therefore the surn of the projection of dl the vectors ont0 the largest vector, vo is bounded below by a0 + 2al + ~~(0,~)+ a3(e0,) + a4(OOa)+ a5(Boo)+ a6(Ooa) = 0.03478 > 0 adthe set of vectors is not equilibrated. Hence no monostatic simplex exists in R'. Xote that the sum of the projections of eight vectors is baxely positive. Csing the lemmas and theorems established in this chapter, we are unable to show that monostatic simplexes do not exist in R8. Chapter 3
Monostatic Simplexes in 8, 9, and 10 Dimensions
In Chapter 2 ae shoaed how. optimizing the projection of one vector individually. or rao vectors added together onto a single vector. a-e could prove no monostatic simplex exists in 7 dimensions or fewer. The techniques del-eloped in the previous chapter are not sufficient on their onto show that no monostatic simples esists in R8. and a natural extension of them wodd be to try to optimize the projection of the sum of three or more vecsors onto a single vector. This is veq- dficult analgicallc so in this chapter ae use a cornputer to optimize the projecrion of the sum of three or more vectors onto a single vector.
The aigorithm for the program that fin& the overd maximum of the projection of the sum of 3 or more wcton onto a single vector is discussed in Section 3.1 and appears in Section 3.2. In 3.3. ae provide a suitable upper bound on the error that may occur as a rdtof using a cornputer search ro calculase this maximum. and in Section 34we use our resdts to show no monostatic simples e-dts in 8 dimensions. and various other resdts about simpleves in 9 and 10 dimensions. 3.1 The Search Method
Let a sirnplex in Rn which falls through each of its facets in turn be given. Let vo, . . . , v, be its facet vectors. Then without los of generality. llvo11 2 IIvl II 2 2 Ilv, 11. Suppose vo points dong the negative 2-&S. Then if llvn11 = 1, the projection of v, onto vo, po (vn)= cos a,where a E (O. ?r). We want to find a formula for the projection of each of the facet vectors onto one single facet vector. Define fo(a)= cos (Y. Chooçe rn = 1, .. . : n - 1, and ,/3 E (O. r). We place the rn + l vectors v,-,, . . . ,v, so that vn-, makes an angle of 4 with the x-ais, and has length 1: and the other m vectors are at angles les than 4. We define fm(3)= max C:.,-,+, po(vi).Recall the Projection Critenon. which states that in order for a simplex to fall from its ichfacet to its jch facet, llxill < ll~,llcos0~~.Since the simplex we are looking at falls sequentially through al1 its facets. then IIviii 11 < llvill If v,-,+l has inclination LY and length 1. then by the Projection Criterion, IIv~-~+~II< COS(Q-p). Then f,+,(a) = cosa+rnaiq,g<,cos(a-i3) fm(,3). Therefore, for a simplex in Rn which falls through each of its facets sequentially, the projection of the sum of its facet vectors onto the largest facet vector. vo, ivill be
where fo(a)= cos a.
Given a directed tree with nodes x and y, x is said to be the parent of y if there exists a directed edge from x to y. Similady, y is the child of x if there exists a directed edge from x to y. We define the falling pattern of a monostatic sirnplex to be the directed tree with one root node such that each node in the tree represents a facet of the simplex, and an edge joining two nodes indicates that the simplex falls from the child node onto the parent node. Note that the root node dlrepresent the largest facet of the simplex, since by the projection criterion, a simplex will only fa11 from a smaller facet onto a larger one. Given a simplex with falling pattern r, the root node of T describes the largest facet of the simplex. If the vector representing the facet corresponding to the root of r has length 1 and lies dong the negative x-axis,we dehe f,(~)to be the maximum projection of the facet vectors of the simplex onto the laxgest facet. Then. as in 3.1 and 3.2, we see that
f&) = COS ?T + max cos(x - a) f, (a), O We call f,(r) the contribution of the facet vectors of a simplez with fallzng pattern r. If s - û is the angle of inclination of the vector representing the root of T, we call f,(B) the contribution of the facet vectors comesponding to the nodes of ri onto the facet vector corresponding to the root node of Ti. When it is clear which is meant. we will simply use the term contribution. The program developed to compute f,(n) for a simplex in Rn maximizes COS(T - a) f, (a)by computing it for values of a increasing from O to T by constant steps of typically .O1 radians. The size of each step is small enough that the program corne sufficiently close to the actual maximum of the function. LVe develop in Section 3.3 an upper bound on the error that may occur as a result of not checking every possible value of ai. 3.2 Computer Algorithm In the program, the tree corresponding to a simplex's fdling pattern has been reg resented as an array, with its length equal to the number of nodes in the tree. The facets of the simplex are labeled from O to n; the tree is rooted at node O, which represents the facet the simplex is hypothetically stable on. Position O in the array represents node O (or facet O), position 1 represents node 1, and so forth. The entry in each position of the array represents that vertexes parent node. For example. the array * O 1 2 3 4 represents the tree in figure 3.1, and the anay * O O 1 1 3 represents the tree in figure 3.2. Note that in the second tree, the root node. node 0, has two children, vertexes I and 2. node 1 also has two children, nodes 3 and 1, and node 3 has a child, node 4. Figure 3.1 : A straight tree wifh 5 nodes Figure 3.2: The tree thd is represented by * O O 1 1 3 The algorithm to maximize the contribution of the facet vectors uses d-mamie programming. A table, lookup[i][a] is used to store the value of the maximum con- tribution for each node (facet vector) i7 where the angle between the node i and its parent node is a, so values used repeatedly only have to be calculated once. The number of rows in the table is equal to the number of values of a n-e are testing, and the number of columns is equal to the number of nodes in the tree. Note that fT,(i(a) = COS a+lookup[i - l][a]. Let maiNode - 1 = the number of nodes in the tree for n = rnaxNode - 1 downto O begin for cr = 0.001 to T /* a < a */ /* - create table */ begin if parent[n] = O /* the contribution of the oth vector is -1 */ Q!=r maxValue = -10 /* set maxValue to a large negative number */ for p = initialAngle to a /* initialAngle is a defined constant */ begin total = cos 0 for i = n + 1 to maxNode begi n if parent[i] = n begin if n > maxNode -0 /* ignore uninitialized values*/ total += lookup[i]p] end end /* end for i */ /* - pick P that gives the maximum value */ tempMax = total * cos(a - 4) if tempMax > maxValue rnaxvalue = tempMax end /* for 0 */ if parent[n ] = O finaltotal + = rnaxVaIue else lookup[n][a] = rnaxvalue end /* for a */ end /* for n */ finaltotal = finaltotal - 1 Running the program for each tree with n nodes, where n = 5, 6. 7, 8. and 9- we get the following results: Tree Max contribution *O123 -0.646452 *O013 -0.836238 *O012 -0.916686 *O122 -0.668376 *O022 -0.875011 *O001 -0.958343 *O112 -0.728672 *O000 -1 *O111 -0.775066 Table 3.1: Maximum contribution of each simplex with five facets Tree Max Contribution Tree hlax Contribution *O0134 -0.646452 *O1234 *O0123 -0.794581 *O1233 *O0133 -0.668376 *O0112 *O0014 -0.836238 *O0134 *O1123 -0.559205 *O0012 *O0114 -0.728672 *O1224 *O1222 -0.474729 *O0222 *O0014 -0.836238 *O1112 *00000 -1 *O1122 *O0011 -0.875011 *Ollll Table 3.2: Maximum contribution of each simplex with sk facets. Tree Max contribution 11 Tree Max Contribution *O12345 -0.109524 -0.401123 *O01235 -0.672477 *O12344 -0.626719 *O01125 -0.646452 *O11345 -0.413313 *O12334 -0.626719 *O00142 -0.340109 *O02245 -0.181203 *O12234 -0.559205 *O00123 -0-151234 *O02333 -0.7.33409 *O00012 -0.875029 *O12223 -0.733409 *O11134 -0.916686 *O12222 -0.666698 *000001 -0.498526 *000000 -0.553644 *O1 1344 -0.833354 *O01122 -0.323578 *O11335 -0.833354 *O00133 -0.36597 *O01133 -0.1948 *O11144 -0.875011 *OOOlll -0.35183 Table 3.3: Maximum contribution of each simplex with seven facets. Tree Mau Contribution Tree Mau Contribution *O123456 0.221 121 *O023456 -0.109524 *O013456 -0.109524 *O013256 -0.482691 *O123455 0.217241 *O023455 -0.116323 *O012445 -0.405491 *O013255 -0.504614 *O003456 -0.401123 *O1 13456 0.0152136 *O123446 0.206399 *O023446 -0.135403 *O012446 -0.405491 *O013446 -0.135403 *O001256 -0.604795 *O1 12356 -0.0781741 *O022356 -0.276'7O2 *O122456 0.12325.5 *O123345 O.l69?'77 *O023345 -0.19912 *O012245 -0.517548 *O001236 -0.752924 *O022346 -0.340109 *O122346 0.0879254 *O112436 -0.105599 *O001426 -0.672477 *O013345 -0.19912 *O023356 -0.181205 *O123356 O. 180043 *O133444 O. 19'72% *O023444 -0.151234 *O012444 -0.433072 *O013222 -0.61 1305 *O000156 -0.646452 *O111256 -0.152127 *O123336 O. 149558 *O000126 -0.794581 *O 111236 -0.216512 *O011156 -0.405603 *O012226 -0.573607 *O023336 -0.233457 *O122256 0.0465525 *O111234 -0.253588 *O000123 -0.875029 *O011145 -0.439499 *O000016 -0.836238 *O000155 -0.668376 *O111255 -0.164613 *O112226 -0.108613 *O001116 -0.615264 *O0001 12 -0.833334 *O1 11223 -0.242939 *O011133 -0.429861 *O122233 0.031726 *O122333 0.0788608 *O011333 -0.35183 *O0011 12 -0.733409 *O112333 -0.l5839LL Table 3.4: Maximum contribution of each simplex vith eight facets Tree Mau Contribution 1 Tr ee hIau Contribution *O111156 -0.283507 1 *O111143 -0.318402 *O000045 -0.498526 *O122226 -1 *O111111 0.0270083 *O122222 -0.553644 *0000001 -0.379735 *O112455 -0.413313 *O01 1253 -0.0105184 *O001446 -0.603683 *O001258 0.00841475 *O011355 0.116076 *O112245 -0.559205 *O001233 -0.36597 *O122344 -0.128149 *O122446 -0.323578 *O001226 -0.056221 1 *O001156 -0.711249 *O112236 -0 -50849 *O012244 O. lï'2O79 *O112222 -0.666698 *0000011 -0.308488 *O112444 -0.474729 -kOOO0116 -0.l996i'ï *O112255 -0.550147 *O001 122 -0.16239 *O111222 -0.775066 *O123333 -0.265314 *O012222 -0.650077 Table 3.4: Maximum contribution of each simplex with eight facets, continued. Note that each value t, corresponds to a simplex which falls sequentially through al1 its facets. On this bais, 1 conjecture that t, will be mâuimized over a straight tree with n nodes, Vn > 1. The talues computed thus far for tn do not include any estimates of the error that may arise because we do not test every value in the domain for the maximum. 3.3 Error Analysis We see in Section 3.2, that given a simplex in Rn where n = 6.. . . -8. the upper bound on the projection of al1 the facet vectors onto the largest facet vector of the simplex is less than the projection of the n + l facet vectors that correspond to a simplex which falls through each of its facets in turn onto the largest facet vector. Since gn(9) = cos(a - B)fn- 1(P) is continuous on [O. n]and differentiable on (O. a). by the Mean Value Theorem we see that the maximum. g, (30) of gn (3)can be es- timated as follows: Given that a cornputer algorithm searches the domain (O. *) sequentially by a given step size ACY= AB and hds a mavimum niue of gn(.3):then gn(P0) < gn (P) + b,(C).na/2,where C E (0,0 + 3~/2). We show that gn(P)is bounded. Firstly, note that the expression g, (,3)= cos(a - P)fn-i(P) is manmized when g#) = O. Hence if 3 = ,do is the point where the maximum occurs, then we have Note that fn(a)is bounded. Since f&) = cos a 5 1: and fl(O) = cos cr+coi2(cr/2) 5 2 it is easily shown by induction that fn(a)5 n + 1. Then fA( Since every value for cx in (O, r) cannot be tested in a computer search for the maximum of fn(a),we cornpute the maximum of fn(a)on a sequence of equally spaced values, and then add in the maximum error. Since the stepsize of the program is na, the maximum will occur within A42of some duefor a. Therefore the error in computing the maximum of gi(a) is less then iiafl?for i = 1.. . . n. The the algorithm cm be altered as follows to cornpute a suitable upper bound for t,. for n = maxNode - 1 downto O begi n for a = 0.001 to * /* a < r */ /* - create table */ begin if parentln1 = O /* the contribution of the 0" vector is -1 */ a=7r maxValue = -10 /* set maxValue to a large negative number */ for ,û = initialAnsle to a /* initialAngle is a defined constant */ begin total = cos p for i = n + 1 to maxNode begin if parent[i] = n begin if n > maxNode -,i3 /* ignore uninitialized values*/ total += lookup[i][/3] end end /* end for i */ /* - pick P that gives the maximum value */ tempMax = total * cos(a - 8) if tempMax > maxValue maxValue = tempMax end /* for 0 */ if parent[n ] = O finaltotal + = maxValue else lookup[n][a] = maxValue + error /* error =(n + l)*stepSizem */ end /* for a */ end /* for n */ finaltotal = finaltotal - 1 If we mn the program over al1 trees with 5 nodes (n = 3), Ive find that the maximum contribution , t5 = -0.637610. Doing this for n = 5? 6. 7' and 8. we see in table 3.3 the resulting values of t,. 3.4 The Main Results The foIlowing theorern says no monostatic simplex esists in eight dimensions. n Tree where max occurs t, = Max contribution 5 *O123 -0.63'7610 6 *O1234 - 0.382380 7 *O12345 -0.076420 8 *O123456 0.273371 9 *O1234567 0.661784 Table 3.5: Maximum contribution of trees with 5, 6, 7, 8, and 9 nodes Recall that if x, is the facet vector of the ith facet of a sirnplex, then for Xi. xjTwe write \ xj (read falh onto xj) if md only if ll~~ll< Pi(xj) We also previously dehed, for n = 0,1,. . ., The following table contains the values of some of the aforementioned constants. Note that in the previous chapter: where the dues of ai, a- 1. and bi were determined, we were minimizing the contribution of one facet vector, in the case of the ai'st and two, in the case of the bils, onto the largest facet. Since we are now maximizing the contribution of these vectors, the values of these constants switch their signs. (They are multiplied by - 1). Variable Value Variable Value Table 3.6: Contributions of vectors Theorem 2 No simplex in R8 is monostatic Proof. Label the facets in decreasing order of size (ie.. with Fo the largest facet. and with F8 the smallest), and let vi denote the facet vector corresponding to the facet Fi of the simplex. We consider four cases, depending on whether the simplex falls onto fi from one facet, two facets, three facets! or more than three facets. Note that and As well, and Case 1. There is one pendtirnate facet. The simplex falls onto Fo from FI only: and, fiom any other facet, eventually falls onto FI. contradict ing the assumption that the vectors are equilibrated. Case 2. There are two pendthnate facets, Fa and FJ. Fie consider three subcases. depending on how many facets there are from which the simplex falls (directly or indirectly) onto each penultirnate facet. Case 2A. There is a penultirnate facet (without loss of generality Fj)onto which the simplex never fa&. Let the facets boom which the simples falls onto Fa be (in decreasing order of size) Fal, Fa*. . . . Fa5: then the simplex fails from Fa,onto Fa in at most i falls. If the set {vi) is equilibrated, the left hand side is equal to zero. and Boa > arccos 113 x 70.52878". Therefore and the set of vectors is not equilibrated. Case 2B. There is exactly one facet, Fol, from which the simplex falls onto Fo. +poba + vgi) llvo Il and the set of vectors is not equilibrated. Case 2C. There are two facets, FBland Fo2, from which the sirnplex falls, directly or indirectly, onto Fg;and three facets, FaI,Fa*, and Fa3from which it falls onto Fa. .Assume, as before, that the facets Ffi are numbered in decreasing order of size. as are the F,. Then PO (vi C - - PO(VO)-+ ~~(vai)+ PO(V~~) + PO(VQI) + PO(V~+va31 i=o llvoll llv~ll llvoll llvo II llvoll llvo Il and the set of vectors is not equilibrated. Case 3. There are three penultimate facets. \Ve consider two subcases, depending on whether the simplex fa& onto one, or more than one penultimate face. Case 3A. There are two penultimate facets onto which the simplex never f&. Label the three penultimate facets Fa,Fp, and F,, and suppose the simplex falls through Fa only. and the set of vectors is not equilibrated. Case 3B. The simplex falls onto more than one penultimate facet. If the three penultimate facets were Iabeled Fal,Fol, and FTIrespectively. then the simplex falls directly from some facet Fp2 to a penultimate facet. Fal and it also falls directly fiom Fm to a penultimate facet FBl. Thus, the simplex falls from F8 to a penultimate face in at most four falls. From the three remaining facets. in decreasing order of size. the simplex reaches Fo in at most 2,3, and 4 falls. and the set of vectors is not equilibrated. Case 4. There are more than three penultimate facets. The simplex falls €rom F8 to a penultimate facet? Fa,in at most four falls. By assumption there are at least two more facets from which it fa& directly to Fo. From the three remaining facets' in decreasing order of size, the sirnplex reaches Fo in at most 2.3, and 4 fails. and the set of vectors is not equilibrated. As no set of 9 vectors in R~which obey the various inequalities required by the projection criterion for a monostatic simplex can be equilibrated, we conclude that there is no monostatic simplex in Re We now show that no monostatic simplex exists in 10 dimensions which falls sequentially through each of its facets, thereby veri&ing a conjecture made in Dawson. Finbow, and Mak, [7]. Theorem 3 No monostatic sàrnplex in RIo falls sequentially onto each of its facets. Proof. Note that -4s the set of 11 vectors in Rio which obey the Mnous inequalities required by the projection criterion for a monostatic simplex? and which obey the hypothesis of the theorem can be equilibrated, we conclude that no monostatic simplex in R~O falls sequentially onto each of its facets. A similar arguement can be used to show that no monostatic simplex exists in 9 dimensions or fewer mhich falls sequentially through al1 of its facets. In [il it was suggested that the lowest dimension a monostatic simplex which falls through each of its facets exists is 11. In [7]we see that a simplex has been found in 11 dimensions which falls through each of its facets sequentially. Therefore, eleven is the lowest dimension in which there exists a simplex that falls through each of its facets in turn. The genetic algorithm which was used to find this simplex was also used to search for a monostatic simplex in 9 dimensions, however, it was unsuccessful. Note that a genetic algorithm is not a systematic search, and it can get stuck on local maximums. and miss the global maximum. The conjecture made in [7] that there are no monos- tatic simplexes in Rg still seems plausible. Chapter 4 Loaded Polytopes We define the inteBor of a facet to be the points in the facet t hat do not belong to any edge ((n - ?)-dimensional facet) of the facet. By above a facet ive mean the foot of the perpendicular from the point to the facet is in the interior of the facet. In this chapter, we look at polytopes of non-uniform density, and Ive consider adding weight to a polytope the same way one might go about loading a die. We Say a polytope is loadable if there is a point in the interior of the polytope that is above only one facet. If the polytope is loadable, we can move its center of gravity to a point over only one facet. Then we cal1 the polytope loaded. Yote that the loaded polytope w41 be mono- static. In a paper by Dawson and Finbow. [6], we determine which of the following polytopes are loadable: the measure polytopes, the simpleses. the cross-polytopes. the zonotopes, and the remaining regular solids. In this chapter we classify these. along with the Archimedean solids, as eit her loadable or nonloadable. 4.1 Preliminary Results There are several conditions a polytope must satisfy in order to be classified as non- loadable. To begin with, a polytope will fall fiom a facet F to a facet G if its center of gravity is not above F, and the dihedral angle between F and G is obtuse. Lemma 9 (Dawson,Finbow) [6] Given a polytope in Rn. 2f one of its facets. or a set of its facets, is separated from the rernainzng facets by a boundary of (n- 2)- dimensional facets with nonobtuse angles, the polytope is nonloadable. Lemma 10 (Dawson, Finbow) /6] If o polytope and al1 its facets have central syrn- rnetry, it is nonloadable. Proof. If a given polytope and its facets are centrally symmetric. then each of its facets has an opposite facet which it is directly above. Hence, if a point is above one facet, it must also be above the opposite facet. so the polytope can not be loadable. Detemining which polytopes are loadable is equivalent to asking which polytopes have intenor points which are above only one facet. Lemma 11 The region above a facet of a convex polytope is convex. Proof. The region above a facet of a convex pol'ope is the intersection of the polytope with a prism, one of whose "caps" is that facet. Since prisrns are convex. and the intersection of any number of convex bodies is a conver body. the region above a facet of a convex body is convex. 1 Lemma 12 (Dawson, Finbow ) [6] [Boundury Princàple olytope which has a point O that is aboue every facet is loadable if und only if there is a point on some facet that is aboue no other facet. Proof. By hypothesis there is a point (O) which is above all the facets. and note that the set of points above any facet is convex. Suppose there is a point B which is above only one facet E. Then the point B' at which the ray met the boundary would be on the facet G. Suppose B' is above some facet Er'. Observe that B is between the points O and B', and O is above the facet F.By convexity of the region above a facet B must be above Et', and so Eu = E. In particular. since Br is above G, it follows that Et = E. It thereby foilows that Br must also be above (or, rather, on) that facet, and Br is a point on E that is above no other facet. - Conversely, suppose B' is a point above only E. Then OBr is above E because the set of points above any facet is convex. Hence. there must be interior points nearby on the segment OB1 that are above only E. A point p on a facet of a polytope that has a point above every facet (satisfving the hypothesis of the Boundary Principle) is called a loading point if it is above no other facet. If a polytope has a loading point, then by the Boundary Principle, it must be loadable. We are now ready to examine some polytopes. 4.2 Nonloadable Polytopes As has been shown by Conway, [3], no regular polygon is loadable. A11 Zn-gons are centrally symmetric, and therefore by Lemma 10, they are nonloadable. A11 regular 272 + 1-gons have a center, 0, which is above every face. However. it is easily shown that there are no points on one face that are above no other face, so by the contra- positive of Lemma 12, the regular 272 + 1-gons are nonloadable. Only two of the five platonic solids are nonloadable. The cube is one of thern, since each of its faces is centrally symmetric. By Lemma 10, it is nonloadable. The cube is a zonotope, a polytope whose every face in every dimension is centrally symmetric. Since zonotopes are (described to be) centrally symmetric, and have centrally sym- metric faces, they are al1 nonloadable, by Lemma 10. Some examples of zonotopes are the Zn-gons in R2, and a few of the -4rchimedea.n solids in R3. the truncated octahedron, tmncated hexahedron, rhombitruncated cuboctahedron and the rhombi- tmcated icosidodecahedron. The regular tetrahedron is also nonloada ble. In fact . every regular simplex in every dimension is nonloadable, by Lemma 9. In Rn,given a set of n mutually independant vectors q,each of which have the same magnitude. we define a measure polytope to be the polytope with facet vectors {fq}r=L=,. The measure polytopes are nonloadable by Lernma 9, as well. The 24cell, which has twenty-four octahedra for facets, and the 120-cell, which has 120 dodecahedra for facets in R4 are centrally symmetric, as are their facets? and they are nonloadable. Note that although the dodecahedron, for example, is centrally symmetric, its faces are not, so it does not follow by Lemma 10 that the dodecahedron is nonloadable. In fact, it has been shown in a paper by Dawson and Finbow, [6]that the dodecahedron is loadable. The truncated tetrahedron has four triangular faces and four hexagonal faces. It is nonloadable on a triangular face, since the projection of the hexagonal face directly Polyt ope Schlafli Loadablel Reason Syrnbol 'lonloadable 2n-gons In) nonloadable Lernma 10 2n + 1-gons {2n + 1) nonloadable Lemma 12 t etrahedron 1 43,31 aonloadable Lemma 9 f X'I 1 1 cube 1 14-31 1 nonloadable 1 Lemma 10 L'I dodecahedron (57 31 Ioadable Prop 2 icosahedron (3,s) loadable Prop 1 octahedron (3941 Ioadable Theorem 4 24-ce11 {3,4,3) nonloadable Lemma 10 120-ce11 {3,3,3) nonloadable Lemma 10 60û-ce11 {3,3,5) loadable Prop 3 3 t mncat ed nonloadable see figure 1.1 6 t et rahedron 1 1 1 f t runcated loadable Prop 4 icosahedron 3 truncated loadable Prop 5 { 10 73) oct ahedron I I I Table 4.1: Loadable/Nonloadable Polytopes Symbol Nodoadable nonloadabiiity) 1 3 t runcat ed 8 nonloadable Lemma 10 3 snub cube loadable 4 y5} snub dodecahedron 1 {:,5} 1 loadable Prop 11 1 L J t rhombicuboctahedron loadable Prop 8 rhombitmncated nonlaadable 3) 1 icosidodecahedron 1 1 1 3 icosidodecahedron - ,4 loadabIe Prop 7 1 3 rhombit runcated 1 3 / oonloadable 1 cuboctahedron ( 1 regular simplexes {3"-') nonloadable Lemma 9 I cross-polytopes {3n-2,4) loadable Theorern 4 in Rd,d > 2 1 measure {dl 3n-2) nodoadable polytopes zonotopes - nonloadable Table 4.1: Loadable/Nonloadable Polytopes, continued opposite a triangular face onto the trïangular face is the entire triangular face: hence there are no points in the truncated tetrahedron that are above one triangular face. and dso above no other face. It is nodoadable on a hexagonal face. since the pro- jection of the tnangular face directly above it and the three other hexagonal faces. cover the face completely, see figure 4.1. Figure 4.1 : OveMew of truncded tetrahedron 4.3 Loadable Polytopes To determine if a polytope is loadabie, we examine the projection of each facet of the polytope onto its opposite facet(s). If the union of the projection of al1 the facets of the polytope except one onto that one facet does not equal the facet. but only part of it, then the polytope is loadabie. Proposition 1 The icosahedron in R3 is loadable First note that the icosahedron has twenty facets. al1 of which are equilateral triangles. Two faces of the icosahedron which axe opposite each other are also oriented at 180" from each other. Because of this. the projection of any face F onto its opposite face, G, is a regular hexagon contained in G. see Figure 4.2. The vertices of F do not project onto G, but instead onto the three faces adjacent to G. If the side of a face has length one, then the length of each side of the hexagon contained in G is i, and the corners of F that project onto the neighboring faces of G are equilateral triangles with an edge length of f , and hence have an altitude of $. Since the angle between two faces of an icosahedron is approximately 138.183". the projection of each of these smd triangles into the faces adjacent to G is an isosceles triangle of height 6 sec(138.183') z 0.3936. The triangle is short enough that it will be completely contained in the hexagon on the adjacent face, the hexagon being the projection of that face's opposite face onto itself, see Figure 4.3. The projection of al1 the faces onto G will lie in such a hexagonal region contained in G' and the points in G not in the hexagon are d the points on G that axe not above any other face. Since the hexagon does not cover G entirely, see Figure 4.2, the icosahedron has a loading point. Figure 4.2: Overlap of opposite triangular facets [icosahedron). Figure 4.3: Overlapping facets of the icosahedron Proposition 2 (Dawson, Finbow) [6] The dodecahedron in R3 iS foadable. The dodecahedron is a regular polyhedra with twelve pentagonal faces. To show it is loadable, we must show that there is a point on any one face that is above no other face. Opposite faces of the dodecahedron are oriented at 180" from each ot her. so the projection of the points of the opposite one onto a given face forms a regular decagon inscribed in the face; see Figure 4.4. Note that the projection of the opposite face onto the face also projects five isosceles triangles, one onto each of the five adjacent faces. If the edge length of the dodecahedron is one. then the length of each side of the inscribed decagon is 9.The vertices of the pentagon that do not project into the decagon are isosceles triangles with a base equal to the length of the side of the decagon, and altitude 6.25 1M Since the dihedral angle of the dodecahedron is approximately 116.655", the projection of such a triangle wiU have the same base. with height = 0.3632., and so each triangle is completeiy contained in the decagon inscribed in the adjacent face. Recall that the decagon is the projection of that face's opposite face onto itself; see Figure 4.3. Thus the union of the projection of al1 the faces of the decagon, excluding one, onto the excluded face, is a regular decagon inscribed in the face. The inscribed decagon does not cover the entire face. so the dodecahedron has a loading point. Figure 4.4: Overlap of opposite pentagonal facets (dodecahedron). Figure 4.5: Overîapping facets of the dodecahedron Proposition 3 (Dawson, Finbow) [6] The 600-cell in R4 is loadable. The 600-ce11 is a regular polytope in four dimensions. nith 600 regular tetrahedra for facets. Oppoçite facets are oppositely oriented: one is rotated 180" from the ot her. so the projection of a facet onto its opposite facet is the intersection of the ta-Ofacets. This is a regular octahedron inscnbed in the tetrahedron. As n-as the case for the faces of the icosahedron and dodecahedron discussed above. n-e must also consider what happens to the projection of the tips of the tetrahedra. four smaller regular tetrahedra, into the facets adjacent to the facet .ne are projecting into. Each of these smdtetrahedra will have a base equal to one side of the octahedron that is inscribed in the tetrahedron facet, and its height nill just be the same as the height of any regular tetrahedron, aith the side of the octahedron as its base. Since the dihedral angle of the 600-ceil is appro-uimately 164.06", the height of the projection of the small tetrahedron is the secant of the dihedral angle of the 600-cell. multiplied by the height of the srnail tetrahedron. This is about 1.01 times the height of the small tetrahedron, so the projection of the tip of the tetrahedron is not contained entirel- inside the octahedron inscnbed in the facet. Even though this is the case. together the octahedron and the four tetrahedra do not fil1 the facet. so the 600-ce1 bas a loading point. Proposition 4 The tninuzted twsahedron is loadable. Proof. The tmncated icosahedron bas two different types of facets. twenty hexag- onal faces, and twelve pentagonal faces. Each hexagonal face is surrounded by three pentagonal faces and three hexagonal faces. and each pentagonal face is surrounded by five hexagonal faces. The hexagonal faces occur opposite themselves. and the projection of one of them onto its opposite face is the opposite hexagonal face. The pentagonal faces occur opposite themselves. hon-ever. opposite pentagonal faces are oppoûitly oriented. so the projection of one onto another is a regular decagon in- scribed in the other face; see Figure 4.4. The tips of the pentagonal face. isosceles triangles, project into the hexagonal faces adjacent to the opposite face. If the length of each edge of a pentagonal face is one. then the base of the isoceles triangles is 9. and their altitude is G.25 IO& The projection one of these triangles into a neigh- boring hexagonal face is an isosceles triangle with a base of y.and a height of 25-10fi 10 sec(l42.62') = 0.2044. The isosceles triangle n-ill be complet ely cont ained in the hexagon, see Figure 4.6. Since each pentagonal face is completely surrounded by hexagonal faces! the union of the projection of al1 the faces excluding one pentag- onal face, onto that face is a decagon inscribed in that face. therefore the tmncated icosahedron has a loading point. Figure 4.6: Ovetlapping facets of the tnincated icosahedron. Proposition 5 The truncated dodecahedron is loadable. Proof. The truncated dodecahedron has twenty triangular faces and twelve decago- nal faces. Each triangular face is surrounded by three decagonal faces. The decagonal faces are opposite other decagonal faces, and the projection of one onto its opposite one is the decagonal face. The triangular faces are opposite other trianglar faces. ro- tated 180". The projection of a triangular face onto the opposite one is a hexagon con- tained in the face; see Figure 4.2. The tips of the triangular face are isoceles triangles that have base 5 , and height 6. Since the angle between a triangular and a decagonal face is approximately 142.62O, such triangle projected into a decagonal faces adjacent to the opposite face has a base of 5, and a height of 6sec(142.62.. .O) = 0.3633 so it is contained entirely in the decagonal face, see Figure 4.7. The projection of a decagonal face into its opposite face is the opposite face, with nothing projecting onto any neighboring triangles, therefore the union of the projection of al1 the faces of a tnuicated dodecahedron except any one triangular face, onto that triangular face. is a hexagon, completely contained in the face. Hence. the truncated dodecahedron is loadable. Figure 4.7: Overiapping facets of the truncated dodecahedron. Proposition 6 The cuboctahedron is Zoadable. Proof. The cuboctahedron has six square faces and eight triangular faces. Each square face is directly above another square face, so the projection of a square face onto its opposite one is the whole face. Each triangular face is above another triangular face, rotated 180" to it. The projection of one triangular face onto its opposite one is a hexagon contained in the face, see Figure 4.2. Since each triangular face is surrounded by three square faces, the tips of the tnangular face will project isoceles triangles into the squares adjacent to the opposite tnangular face: see Figure 1.8. Since each of these small triangles has a base of f, and a height of and the angle between each of the faces is $, the projection of one such triangle into the square will be an isoceles triangle with a base of 5, and a height of 6 sec($) = 0.5. Therefore the projection of the triangle is completely contained in the neighboring square. The union of the projection of all the faces but one triangular face, into that face is just the hexagon contained in the face, and hence the cuboctahedron has a loading point. Figure 4.8: Overlapping facets of the cuboctahedron Proposition 7 The icosidodecahedron is luadable. Proof. The icosidodecahedron has twenty triangular faces and twelve pentagonal faces. Each pentagonal face is directly above and oppositely oriented to another pen- tagonal face, and it is surrounded by five triangular faces. Similarily, each triangular face is directly above and oppositely oriented to another triangular face, and it is surrounded by three pentagonal faces. The projection of a pentagonal face onto its opposite face is a regular decagon inscribed in the opposite face. see Figure 4.4. The tips of the pentagon project isosceles triangles into the triangular faces adjacent to the opposite face, see Figure 4.9. The tips of the pentagons will have a base of y? 25- 10 6 and a height of , ,, so their projections dlbe isoceles triangles which have the 25- 10J5 same base, and a height of sec (3)= 0.3249, since the angle between the faces is 9. The projection of the tips of the pentagon ni11 be completely contained in triangles adjacent to the opposite pentagon. since each of those hasi an altitude of & x 0.5774. The projection of a trkngular face into its opposite triangular face is a regular hexagon inscribed in the face, see Figure 4.2. The tips of the triangular face. which happen to be equilateral triangles, will project into the pentagonal faces adjacent to the opposite triangular face, see Figure 4.10. Since the base of one of these triangles is 3' its height will be 9,and its projection into the pentagonal face ni11 be an isoceles triangle with a base of f, and a height of 6sec(?) rr 0.5. Note that this isosceles triangle will be completely contained in the decagon inscnbed in the pentagon. This is because itç base of (3) is smaller then a side of the decagon, (y).Therefore. the union of the projection of al1 the faces of the icosidodecahedron, excluding a pentagonal face, onto that pentagonal face, is a regular decagon inscnbed in that face, and the icosidodecahedron has a loading point. Figure 4.9: Overlap of a pentagonal facet onto a triangular facet of the icosidodecahedron Figure 4.10: Overiap of a tnangular facet onto a pentagonal facet of the icosidodecahedron Proposition 8 The rhombicvboctahedrvn is loadable. Proof. The rhombicuboctahedron has eighteen square faces. and eight triangular faces. Each trianglar face is surrounded by three square faces, and is directly above another triangular face rotated 180" from it. Square faces are opposite square faces, so the projection of a square face onto its opposite one is the square face. The projection of a triangular face onto its opposite face is a regular hexagon inscribed in the face, see Figure 4.2. The tips of the triangular face are equilateral triangles with base equal $, and height equal 9. They each project a triangle into a square adjacent to the opposite triangle, see Figure 4.8. The projected triangle has a base of 4 that it çhares with the hexagon, and a height of $ sec(?) = 00.082 . since the dihedral angle of the rhombicuboctahedron is F. The projection of the tip of the triangle is completely contained in the square face. The union of the projection of al1 the faces of the rhombicuboctahedron excluding one triangular face. onto that face. is a regular hexagon contained in the face, therefore the rhombicuboctahedron has a loading point. Proposition 9 The rhombicosidodecahed~onis loadable. Proof. The rhombicosidodecahedron has twelve pentagonal faces, twenty triangu- lar faces, and thirty square faces, for a total of sixty-two faces. Each square face is surrounded by two of the triangular faces and two of the pentagonal faces. arranged alternately around its edges. A square face is directly opposite another square face. and it is oriented at 45' to it. Therefore, the projection of a square face onto another square face is a regular octagon inscribed in the face, see Figure 4.11. The corners of the square face that are not above the opposite face have a base offi - 1 and a height of q.Two of the corners project isosceles triangles with a base offi - 1 and a height of qsec(159.04...O) 1: 0.2218 onto the two triangular faces adjacent to the square face, see Figure 4.13. The other two corners of the square face project isoceles triangles with base fi - 1, and height 9sec(l48.278 ...O) = 0.2434 onto the pentagonal faces adjacent to the square face, see Figure 4.12 The projections of isoceles triangles in the faces adjacent to the square are contained in each of those faces respect ively. Each triangular face is directly above another triangular face, and is oriented at 180" from it. The projection of a triangular face onto another triangular face directly above it is a regular hexagon contained in the face. see Figure 1.2. The corners of the face project into the faces adjacent to the opposite face, which happen to be square faces. Each of the triangles that project into a square face have a base of f. and a height of 6.Since the angle between a square face and a triangular face is 159.04 . . .O, the projection of the corner of a triangular face onto a square face is an isosceles triangle with height 3, and base 6 sec(159.04 . . .O) zz 0.3091, see Figure 4.14. that is contained in the octagon inscribed in the square face. Each pentagonal face is directly above another pentagonal face, and is oriented at 180" from it. The projection of a pentagonal face into another one directly above it is a regular decagon inscribed in the face, see Figure 1.4. The corners of the face project into the faces adjacent to the opposite pentagonal face. which happen to be square faces. Each of these corners has a base of and a beight of c25;,'06. and since the angle between a square face and a pentagonal face of a rhombicosidodecahedron is 148.278.. .O, its projection into a square face would be an isosceles triangle with a base of 5,and a height of Bsec(148.278..,, .O) z 0.1910. see Figure 4.15. Then each square face has loading points, see Figure 4.16. Therefore the rhombicosi- dodecahedron is loadable. Figure 4.1 1 : Overlap of square facets of the rhombicosidodecahedron. Figure 4.13: Overlap of square facets onto pentagonal facets of the rhombicosidodecahedron Figure 4.12: ûveriap of square facets ont0 triangular facets of the rhom bicosidodecahedron. Figure 4.1 4: Overiop of tnangular facets onto square facets of the rhombicosidodecahedron Figure 4.15: Overlap of pentagonal facets onto square facets of the rhom bicosidodecahedron Figure 4.1 6: Overlapping facets of the rhombicosidodecohedron Proposition 10 The snub cube M loadable. Proof. The snub cube has thirty-two triangular faces and six square faces. Each square is directly above another square face which has been rotated 22.5" to it. and it's surrounded by four tliangular faces. The triangular faces are directly above other triangular faces, which are at 22.5' from each other. The projection of a square face onto its opposite face is an inegular octahedron inscribed in the face. see Figure 4.17. The corners of the square project into the neighboring triangular faces. These triangles have a base of 0.4005.. . and a height of 0.1532.. .. so their projections into the triangular faecs will have a base of 0.4005.. . and a height of 0.1532. . .sec(125.231. . .O) z 0.2657, and will therefore be contained entirely in the adjacent triangular faces; see Figure 4.18. The projection of a tnangular face that lies adjacent to a square face, onto its opposite face (which is also adjacent to a square face), is an irregular hexagon inscribed in the face' see Figure 4.19. The corners of the triangular face will project a triangle into the neighboring square facet. The triangle being projected will have a base of 0.3863.. . and a height of 0.1693.. . , so its projection into the square face will have the same base, and a height of 0.1693sec(125.231.. .O) = 0.2936, see Figure 4.10. The base of the projected tri- angle is not contained in a side of the octagon inscribed in the square face, however it does not equal the side of the square face, see Figure 1.21. Therefore, since the union of the projection of al1 the faces of a snub cube except one square face onto that square face does not entirely cover the face, the snub cube has a loading point on that face. Figure 4.1 7: Overlapping square faces of a snub cube Figure 4.18: Overlap of square face onto Mangulor facet of a snub cube Figure 4.19: Overlapping triangular facets of the snub cube and snub dodecahedron. Figure 4.20: Overlap of a tnangular facet onto a square facet of the snub cube. Figure 4.21 : Ovedapping facets ofa snub cube Proposition 11 The snub dodectzhedron is loadable. Proof. The snub dodecahedron has twelve pentagonal faces and eighty triangular faces. The angle between each pentagonal face is the same as in the dodecahedron. 116.655.. .O. Since there is a triangular face between each pentagonal face. the an- gle between a pentagonal face and a triangular face must be greater than 116.655 . . .' . Each pentagonal face is directly above another pentagonal face and it is rotated 22.5' from it. Therefore the projection of a pentagonal face ooto its opposite face is an irregular decagon inscribed in the face, see Figure 1.22. Since each pentagonal face is surrounded by triangular faces, the projections of the corners of the pentagonal face, which have a base of 0.4541.. ., and a height of 0.1382.. ., Ml1 have the same base, and a height of les than 0.1382 . . .sec(l16.635.. .O) = 0.3097: and therefore will be completly contained in the triangufar faces. see Figure 4.23. Thuty-six of the eighty triangular faces belonging to the snub dodecahedron are adjacent to a pentagonal face. One of these is directly opposite another triangular face (which is itself adjacent to a pentagonal face) and it is rotated 22.5" to it. Therefore the projection of a triangular face adjacent CO a pentagonal one. is an irregular hexagon contained in the face, see Figure 4.19. The corners of the triangles have a base of 0.3865.. . , which is shared with a side of the decagon inscribed in the pentagonal face, and a height of 0.1636 . . ., so its projection into the neighboring pentagonal face has height strictly less than 0.1636.. .sec(116.655. . .O) 0.3755. see Figure 4.24. Therefore the union of the projection of al1 the faces of the snub dodecahedron except one pentagonal face is an irregular decagon inscribed in the face, see Figure 4.25. Therefore it has a loading point, and the snub dodecahedron is loadable. Figure 4.22: Overlapping pentagonal facets of a snub dodecahedron Figure 4.23: Overlap of pentagonal facets ont0 a triangular facet of the snub dodecahedron Figure 4.24: Overiap of a triangular facet onto a pentagonal facet of the snub dodecahedron Figure 4.25: Overlapping facets of the snub dodecahedron. Theorem 4 (Dawson, Finbow) [6] Any cross-polytope in n dimensions, where n > Proof. A cross-polytope in Rn has 2n vertices. Given an orthonormal basis of Rn. Say {q)y='=,, the corresponding set of vertices of an n dimensional cross-polytope will be {fq):=,. Each facet of the cross-polytope is an (n - 1)-dimensional simplex, so each has a set of n of the 272 vertices. Given a particular facet of the cross-polytope, it will have a set of vertices {s,~}~,,,where si is either -1 or 1- s = x:=, Yi& is a vector normal to the facet in this basis. This caii be shown as follows: Let a = (al'. . . .a,) be a point in the facet. Then a = Er=, Xisixi, and thus since {xi):=, is an orthonormal set. Thus the facet lies in the hyperplane s - a = 1. Let F be the facet of the cross-polytope which has vertices XI, x2,. . . . x,, and let F be any other facet of the cross-polytope. Without loss of generality: F haç vertices XI, . . . ,h, .. . , -xn) The vector u = C:,, x* is normal to facet F, and the vector u' = Cg1 - CL+, is normal to Fr.In general, a point q = C:=l cix, is in F provided q - u = 1. The point q is above F' provided q > O and q = q' + Au'. where q' E F' and so qf-u'= 1 (as we saw above). Let p = (al,.. . ,a,) be a point in F. Then p u = 1, in other words, Observe that p is also above F if there exists k E R such that p + ku' E F'? and hence m and al+k,..., am+k>O>a,+l -kl ....a,-k. Subtracting 4.1 from 4.2 and solving for k yields k = C:=miiai . But substituting 2 this on the right of 4.3 gives, for m + 1 5 j 5 n,aj - ;x:=,+, ai < O and 1a, > n-2 -ajn where the summation is over {m + 1.. . .:j - 1. j + 1.. . . . n}. If n > 2. this cannot be satisfied when aj is sufficiently greater than the other coordinates. Thus. for any n > 2, there exists e > O such that every point mhose coordinates are a permutation of (1 - e, e - e2,. . . ,en-' - en, en) is in F but not over any FI: and thus the crosspolytope is loadable. On the other hand, when n = 2 the cross-polytope is a square. which is not loadable. 4.4 Related Results Let K be a bounded convex body in Rn-and for each density function p on K' let c(p) be the corresponding center of gravity of K. IVe consider the set D of density Functions p whose range is {dl,d2), where dl > 0, and d2 < x. For a given linear functiond f on K, we wish to constmct p* E D such that f (c(p*)) = mau{f(c(p)) : p E D). Clearly, if dl is arbitrarily srnail, f (c(p))WU come arbitrarily close to mabe~f (x) . and similady, if d2 is allowed to become arbitrady large. then f(c(p)) nrill come arbitrarily close to minxEnf (x). Therefore, suppose dl> 0. and d2 < x. Theorem 5 Given a convez body K in Rn, and a set of density funetions D on K that have mnge (dl,d2). Given a linear functionul f on K, then there ezists a number to E R such that a density funetion p* which rnuxhizes the center of gruuzty O/ K in the direction off às defined by Proof. Let a convex body: K E Rn be given. Let p E D. with range {di,d2}. and without loss of generality, assume dl < d2. It suffices to show that there is a p* of the desired type which yields a better center of gravity than p. Let m = min{f (x) : x E K),and let il1 = maw{f(x) : x E K). For evey t E R, set =l(t) = {X E K : f(x) 5 t) and B(t) = {x E K : f(x) > t). Let E4t) = A(t)n ~-~(d*), and Eg(t) = B(t) n p-l(d1). If the volume of p-'(di) is O, choose p' = dj, where j = (i + i)mod2, and we are done. Othenvise, I'(E.4(m)) = O < V(EB(m))and V(Ea(M))> O = V(EB(M)). Therefore there exists a number to E R such that V(E.&-,)) = V(EB(to)). Shen set Shen if CA is the center of gravity of EA, and c~ is the center of gravi@ of Es, we have f (CA) 5 to < f (CB)so f (~.4)d1+ f (cg)& I f (~.4)d2+ f (cB)~Then f (~(p*))> f (c(p)),and Theorern 5 follows. Bibliography [l] O. Bottema et al. Geometric Inequulzties, Wolters-'ioordhoff. Groningen. 1969 [2] H. Brunn, Referat über eine Arbeit: Exacte Grundlagen für eine Theorie der Ovale, Sitzungsber. Geometrie der Zahlen d.math.-physik. Classe der bayer. Akad. d. Wks. 1894, Bd. LW,S. 102 [3] J. H. Conway and R. Guy, Stability of Polyhedra. SIAM Reu. 11 (1969). 75-82 [4] H. S. 31. Coxeter, Regular Polytopes, Yded. SlacMillan. 1963 [5] R. Dawson, Monostatic Simplexes, Amencan Mathematical Monthly 92 (1985). 111-128 [6] R. Dawson, W. Finbow, What Shape is a Loaded Die'?. submitted to ilfathematical Intelligencer [7] R. Dawson, W. Finbow, P. b1&, Monostatic Simplexes II. to appear in Geometn'ae Dedicuta [8] B. Grünbaum, Convex Polytopes, Pure and Applzed Mathematics. Vol. 16. New York, 1967, 339-340 [9] R. K. Guy, Twenty odd questions in combinatorics, in Combinatorial iVuthematics and ikr Applications, R. C. Bose et al, eds., Chape1 Hiil? XI1970, 209-232 [IO] 4. Heppes, A Double-Tipping Tetrahedron, SIAM Rev. 9 (1967), 599-600 [Il] H. Minkowski, Ailgemeine LehrGtze über dir konvexe Polyeder. Nachr. Ges. Wiss. G6ttingen (1897), 198-219 [12] M. Wennuiger, Polyhedmn Models, Cambridge Gniversity Press, 1963 IMAGE NALUATION TEST TARGET (QA-3) APPLIED il IMGE. lnc -.- 1653 East Main Süwt --- - Rochester. NY 14609 USA ------Phone: 71â1482-0300 - Fa71 6/208-5989 O '993. Wied Image. ini. W Riehtr Reserved