Phenomenal Three-Dimensional Objects
by
Brennan Wade
A thesis submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Master of Science
Auburn, Alabama
May 9, 2011
Keywords: monostatic, stable, mono-monostatic
Copyright 2011 by Brennan Wade
Approved by
Andras Bezdek, Chair, Professor of Mathematics
Wlodzimierz Kuperberg, Professor of Mathematics
Chris Rodger, Professor of Mathematics
Abstract
Original thesis project: By studying the literature, collect and write a survey paper on
special three-dimensional polyhedra and bodies. Read and understand the results, to which
these polyhedra and bodies are related.
Although most of the famous examples are constructed in a very clever way, it was
relatively easy to understand them. The challenge lied in the second half of the project,
namely at studying the theory behind the models. I started to learn things related to
Sch onhardt?s polyhedron, Cs asz ar?s polyhedron, and a Meissner body. During Fall 2010 I
was a frequent visitor of Dr. Bezdek?s 2nd studio class at the Industrial Design Department
where some of these models coincidently were fabricated. A change in the thesis project
came while reading a paper of R. Guy about a polyhedron which was stable only if placed
on one of its faces. It turned out that for sake of brevity many details were omitted in the
paper, and that gave room for a substantial amount of independent work. As a result, the
focus of the thesis project changed.
Modi ed thesis project: By studying the literature, write a survey paper on results con-
cerning stable polyhedra.
The following is the outcome of the thesis:
1. A collection of elementary facts/theorems/proofs concerning stable tetrahedra.
2. A description of a double tipping tetrahedron constructed by A. Heppes.
3. A description of a 19 faceted polyhedron of R. Guy, which has only one stable face.
4. Description of the G omb oc, a recently discovered mono-monostatic 3D body.
ii
Acknowledgments
First of all, I would like to express my sincere gratitude to my advisor, Dr. Andras
Bezdek, for your constant guidance during Graduate School. Your patience, enthusiasm,
and incredible knowledge of mathematics ensured that I was successful in writing this thesis.
I would also like to thank my committee members Wlodzimierz Kuperberg and Chris Rodger
for your time and input. Thank each of you for your dedication to your students and Auburn
University.
I would not be in school for mathematics if not for my high school math teachers, Mrs.
Hardcastle and Mrs. Bronars. I believe that because of you I was able to see how important
math is for students to understand and apply to their lives to be successful. I really learned
and understand mathematics because of the way you laid the foundations through your
teaching, and I am so appreciative. Mrs Nichols, you lead your life by example. Thank you
for teaching us how to serve others. Thank you for being an example both in and out of the
classroom for me to follow. I am so thankful to you for the time that you put in to make
me the person and student that I am today.
I owe any success in my life rst to my Lord, Jesus Christ, and then to my family. Dad,
Mom, Elle and Parker, thank you for your constant love and support. Your encouragement
and constant praise has been such a blessing in my life. You made me believe that I could
accomplish anything that I put my mind to. Your a rmation encouraged me to try my
hardest and nish everything I started. Last, but not least, I want to thank my future hus-
band, Jeremy. Thank you for keeping me grounded through everything we?ve been through
thus far. I am so con dent and excited to start our lives together. Thank you for your
encouragement, patience, love, support, and leadership. You lead your life by example in
leaning on Christ for your strength.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction and outline of the thesis . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Geometry of tetrahedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.1 Things to know about the mass center of tetrahedra . . . . . . . . . . . . . . 3
2.2 A double tipping tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 All tetrahedra have at least two stable faces . . . . . . . . . . . . . . . . . . . . 9
4 A polyhedron, which has only one stable face . . . . . . . . . . . . . . . . . . . 11
4.1 (1) Setting up the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 (2) Evaluating the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4.3 (3) Algebraic/ trigonometric manipulation . . . . . . . . . . . . . . . . . . . 17
4.4 (4) Sum from n = 1 to m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.5 (5) Evaluating the sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.6 Evaluating m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5 Bodies which have exactly one stable and one unstable position. . . . . . . . . . 38
5.1 All two-dimensional objects have at least two stable equilibrium. . . . . . . . 38
5.2 The G omb oc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6 Cs asz ar?s polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
7 Szilassi?s polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
8 Sch onhardt Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
9 A Meissner body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
iv
List of Figures
2.1 All tetrahedra have no more than three obtuse dihedral . . . . . . . . . . . . . . 4
2.2 Top view of tetrahedron with only two stable faces . . . . . . . . . . . . . . . . 5
2.3 Top view of the double tipping tetrahedron rst position . . . . . . . . . . . . . 7
2.4 Top view of the double tipping tetrahedron second position . . . . . . . . . . . 8
3.1 No monostatic tetrahedron exist . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.1 Truncated Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.2 View 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.3 View 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4.4 Guy?s solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4.5 The "m" le that is run in MATLAB to determine our value for m . . . . . . . 36
4.6 Side view of the solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.7 The "m" le to determine a value for b. . . . . . . . . . . . . . . . . . . . . . . 37
5.1 Two-Dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.2 G omb oc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.1 Cs asz ar?s Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.2 Flat Pattern of Cs asz ar?s Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . 44
6.3 Pattern for top of Cs asz ar?s polyhedron . . . . . . . . . . . . . . . . . . . . . . 45
6.4 Pattern for base of Cs asz ar?s polyhedron . . . . . . . . . . . . . . . . . . . . . . 45
6.5 Completed Base of Cs asz ar?s polyhedron . . . . . . . . . . . . . . . . . . . . . . 45
7.1 Transparent Szilassi?s Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . 46
v
7.2 Szilassi?s Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
8.1 Sch onhardt?s Polyhedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
vi
Chapter 1
Introduction and outline of the thesis
Since tetrahedra and other special polyhedra occur so often in physics, in chemistry and
in crystallography, it?s particularly useful to understand their geometry. In Chapters 2-5
of this thesis we will study the stability properties of such bodies. Intuitively speaking we
want to know how many stable positions these bodies can have and those speci c positions.
Recall that in planar geometry, the centroid, geometric center, or barycenter of a plane gure
refers to the intersection of all straight lines that divide the gure into two parts of equal
moment about the line (equal moment means balanced position). This de nition extends to
any object in 3-dimensional space. The centroid of an object is the intersection of all planes
that divide the body into two parts of equal moment.
In physics, the mass center or the center of gravity refers to the point where the object
is supported while maintaining a balanced position. Informally, the center of mass (and
center of gravity in a uniform gravitational eld) is the average of all points, weighted by
the local density or speci c weight. This gives rise for using integral calculus at computing
the coordinates of the mass center. For special shapes and special density distributions this
computation is simple, otherwise it can be quite involved.
If a physical object has uniform density, then its center of mass is the same as the
centroid of its shape. In the subsequent sections, if we do not say otherwise, polyhedra are
assumed to be solids with uniform mass density. We will call a face of given polyhedra stable
if the polyhedra will rest on that particular face in a stable position on top of a horizontal
table.
The following is an outline of the thesis:
1
In Chapter 2 we study the geometry of tetrahedra. We will list and prove some basic
properties concerning the mass center of tetrahedra and also describe some special tetrahedra
to give an insight on what makes a face stable.
Chapter 3 contains the proof of Conway?s theorem, which claims that every tetrahedron
has at least two stable faces.
In Chapter 4 we explain a construction of Guy [7], which shows that there are polyhedra
with exactly one stable face. This is our main section where most of our independent
work was done. It turned out that a crucial step of the proof in Guy?s paper was given
without explanation. This step was about nding a closed formula for a quite elaborate
integral. Pages 26-36 contain these computations. Moreover Guy?s paper ended by saying
that if certain distance is su ciently large then the constructed model has the right stability
properties. We computed the numerical value of this measurement and then 3D printed
Guy?s model. The printed model will lead other students to a better understanding of Guy?s
construction.
Chapter 5 contains a brief description of a recently discovered body called the G omb oc.
The G omb oc is a body with exactly one stable and exactly one unstable position. In this
section we include the proof of planar theorem on the existence of stable and unstable
equilibriums. This result shows the di erence both in di culty and in nature of the 2D and
the 3D versions of our stability questions.
Finally, the Appendix contains Chapter?s 6-9 which are further reading on phenomenal
three-dimensional objects. They are not contained in the main part of the thesis because
they are not closely related, but much time was spent on this before the discovery of the
monostatic polyhedron [7].
2
Chapter 2
Geometry of tetrahedra
2.1 Things to know about the mass center of tetrahedra
In this section we assume that the tetrahedra have uniform mass density. The mass
center of such a tetrahedron always coincides with its centroid.
1. Simple coordinate calculus shows that the centroid of a tetrahedron with vertices
(xi;yi;zi);i = 1;:::4 is the point
P4
1xi
4 ;
P4
1yi
4 ;
P4
1zi
4
.
2. It follows from (1) that the centroid of a solid tetrahedron is located on the line segment
that connects the apex to the centroid of the base, moreover the centroid is at 1/4 of
the distance from the base to the apex.
3. It follows from (1) that the centroid of a solid tetrahedron is also the midpoint of three
segments each connecting the midpoints of the opposite edges.
4. Assume that the base ABC of tetrahedron ABCD lies on a horizontal surface. (3)
implies that this tetrahedron will tip over the edge BC if and only if the perpendicular
projection D0 of the vertex D onto the plane of ABC lies further from the edge BC than
the vertex A. Thus, an obtuse dihedral angle (angle between two neighboring faces) is
necessary for a tip, moreover a tetrahedron can tip from a face to another face only
if the later has larger area. This simple observation will play a very important role in
most of the proofs later.
5. The following is a natural question: At most how many obtuse dihedral angles can
a tetrahedron have? It turns out that the maximum is three. Although none of the
3
proofs of our subsequent results will use this bound, we present a proof here for sake
of completeness. The solution is by M. Klamkin [9]. There is at least one vertex of
the tetrahedron such that its corresponding face angles are all acute. This is necessary
because otherwise the sum of all face angles would be greater than 4 (the sum of two
face angles at a vertex is greater than the third face angle). Since there are four faces,
the sum must be equal to 4 .
The following two lemmas are for the face angles a;b;c, and the opposite dihedral
angles of a trihedral angle.
Lemma 2.1. If 2 >a b c, then B and C are acute.
Lemma 2.1 follows from the sign analysis of the spherical law of cosine: sinbsincsinA =
cosa cosbcosc;etc.
Lemma 2.2. If A B 2 >C, then a, b 2 >c.
Lemma 2.2 follows from the sign analysis of the spherical law of cosine: sinBsinC cosa =
cosA+ cosBcosC;etc.
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Figure 2.1: All tetrahedra have no more than three obtuse dihedral
Choose P as the vertex with all acute face angles in tetrahedron PQRS. By Lemma
1, we can take PQ and PS as the edges of the two acute dihedral angles. Now assume
4
these are the only acute dihedral angles and obtain a contradiction. By the application
of Lemma 2 to trihedral angles at vertices Q and S, it follows that angles PQS and
PSQ are both non-acute. This is impossible, so there are always at least three acute
dihedral angles in any tetrahedron.
6. Every tetrahedron has at least one stable face. To see this notice that a face of a given
tetrahedron is stable i it contains the perpendicular projection of its mass center. A
simple indirect reasoning shows that the face nearest to the mass center has the later
property.
7. As in (6) it follows that every tetrahedron with all acute dihedral angles has four stable
faces.
8. Figure 2.2 shows a tetrahedron which has only two stable faces.
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Figure 2.2: Top view of tetrahedron with only two stable faces
9. Heppes [8] discovered a \two-tip tetrahedron" which Guy [1] realized with edges 41, 26,
24, 20, 17 and 4 (amature carpenters may like to construct this). Section 2.2 contains
Heppes?s construction.
5
10. Dawson [2] includes an argument of Conway which proves that every tetrahedron has
at least two stable faces. Section 3.1 contains Conway?s proof.
2.2 A double tipping tetrahedron
Alad ar Heppes posed the following problem in SIAM Review [8]. Design a homogeneous
tetrahedron which, when placed lying on one of its faces on the top of a horizontal table, will
tip over to another face and then tip over again, nally coming to rest on a third face. Heppes
presents a solution and called his tetrahedron the double tipping tetrahedron.
Let A, B, C, D denote the vertices of a tetrahedron T, and let G0 denote the orthogonal
projection of its centroid G, on the "horizontal" coordinate plane z=o. We compute the x
coordinate of G0 by averaging the x coordinates of the vertices of T. The z coordinate of G0
is obviously 0. We will be using the mass center?s projection to see if the solid will tip over,
and if it does tip over, the side with which it will rotate over. Heppes presents the following
coordinates for the tetrahedron. The initial position of the tetrahedron is described by the
following coordinates where 0 < < 1.
A ( -7, -8(1- ), 0),
B ( -1, 0, 0),
C ( 1, 0, 0),
D ( 7, 8, 8).
For simplicity we choose = 12
A ( -7, -4, 0),
B ( -1, 0, 0),
C ( 1, 0, 0),
D ( 7, 8, 8).
We have G0 = (0, 1, 0). Notice that G0 falls outside of the triangle ABC on the y-axis
therefore the tetrahedron will tip over side BC.
6
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Figure 2.3: Top view of the double tipping tetrahedron rst position
In order to determine coordinates of T after the tip, we need to evaluate each coordinate
individually. The tetrahedron is tipping over side BC, so the coordinates of B and C are not
going to change. Since the tetrahedron is ipping straight over the x-axis, all xth coordinates
of the vertices will remain the same. Both y and z coordinates of A and D will be determined
through the right triangles with legs x,y. In the case of both vertices these form 45 ;45 ;90
triangles. These right triangles are known to have a side ratio of 1:1:p2, with p2 being the
hypotenuse. The triangle associated with A has a hypotenuse of 4. With the given ratios,
we know that the new y and z coordinates will have a length of 4( 1p2)= 4
p2
2 = 2
p2. The
triangle associated with D has side lengths 8 and 8. With the given ratios, we know that
the new y coordinate will have a length of 8(p2)= 8p2. The tetrahedron is tipping onto
triangle BCD, so the z coordinate of D will become 0.
This gives us the following coordinates for the vertices of T after the rst tip.
A ( -7, -2p2, 2p2),
B ( -1, 0, 0),
C ( 1, 0, 0),
D ( 7, 8p2, 0).
7
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Figure 2.4: Top view of the double tipping tetrahedron second position
The projection for the mass center is G0 = (0, 3
p2
2 , 0). This is also outside of triangle
BCD, and our tetrahedron will tip over the side BD and come to rest on the base ABD.
Heppes found a tetrahedron with only two stable faces that tips over twice when placed
on a horizontal surface. Richard Guy [1] modi ed this example and found that with edges
41, 26, 24, 20, 17, and 4, and with opposite pairs summing to 45, 43, and 44, an amateur
carpenter could construct such a tetrahedron! In the next chapter we will discuss if there
are tetrahedrons with only one stable face.
8
Chapter 3
All tetrahedra have at least two stable faces
J. Conway and R. Guy presented the following problem in the SIAM Review[7]: Show
that any homogeneous tetrahedron will rest in stable position when lying on any one of at
least two of its faces.
Michael Goldberg [7] published an incorrect solution to prove that all tetrahedra have
at least two stable faces. Goldberg said that a tetrahedron is always stable when resting
on the face nearest to the center of gravity. He stated that if you project the apex and the
edges onto the stable base, then the projection of the center of gravity will fall onto one
of the projected triangles or on the projected edges. If it lies within a projected triangle,
then a perpendicular from the center of gravity to that corresponding face will meet within
the face, thus proving that it is another stable face. If it projects onto an edge, then both
corresponding faces are stable faces.
Apparently there are two problems concerning the above underlined sentence. A) Con-
way constructed a tetrahedron with non-uniform density which is stable only on one of the
faces. Goldberg?s solution makes no explicit use of the position of the mass center of a
tetrahedron, thus in view of Conway?s example, it cannot be complete. B) We constructed a
tetrahedron, which is stable on two faces, but not on those which Goldberg claims, thus his
proof cannot be completed. It was not until 1985 that R. J. Dawson [2] included Conway?s
proof in his paper. For the sake of completeness we also include Conways proof here.
De nition 3.1. Bodies with just one stable equilibrium are called monostatic.
Theorem 3.1. No tetrahedron is monostatic.
Proof (Conway): A tetrahedron will not tip about an edge unless the dihedral angle at
that edge is obtuse. Thus, an obtuse angle is necessary for a tip. In view of (2) in 2.1 and
9
the volume formula of the tetrahedron, the closer the center of mass is to a tetrahedron?s
face, the larger the area of the face. Thus a tetrahedron can only tip from a smaller face to
a larger one.
Suppose there exists some tetrahedron which is monostatic. Let A and B be faces with
the largest and second-largest areas. There are two ways for the tetrahedron to roll onto face
A. Either it rolls from B onto A and from C onto A, or it rolls from another face C, onto
B, and then from B onto A. Either way, one of the two largest area faces has two obtuse
dihedral angles (call these sides with dihedral angles e and f). Either e or f is a side shared
with the other of the two largest faces.
Since sides e and f are both obtuse dihedral angles, the vertex v must be outside the
intersection of the extended lines of e and f. The projection of D shows that the face is
larger than A or B, thus leading us to a contradiction.
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Figure 3.1: No monostatic tetrahedron exist
10
Chapter 4
A polyhedron, which has only one stable face
R. K. Guy [7] approached this problem in an attempt to de ne how to develop a mono-
static body with only 1 stable equilibrium. His goal was to nd the polyhedron which is
stable on only one face with the least number of faces. In view of chapter 3, the minimum
number of faces is between 4 and 19. The exact value is still unknown today.
Think about taking a long cylinder and diagonally cut o one end, then at the opposite
angle chop o the other end. This truncated cylinder can lie stable on the table on it?s
longest "side", but in no other position. It only has one stable equilibrium. Intuitively we
can think about how to prove that it only has one stable equilibrium. Think about a normal
cylinder; the mass center of the object goes through the center of the circle. After we cut the
edges diagonally, there is more mass below the center of the circle. Thus the mass center is
below the center of the circular center of the truncated cylinder. This shows that the bottom
of our cylinder is a stable equilibrium. If we place the cylinder down anywhere except the
bottom of the cylinder then it will roll because the perpendicular from the mass center to the
side will fall outside of the cylinder. Thus, it only has one stable equilibrium. We are going
to evaluate a gure that is similar to the truncated cylinder, however it is multi-faceted and
much more complicated.
Figure 4.1: Truncated Cylinder
11
Guy?s solid is a 17-sided prism. The rst view (see Figure 4.2) is only half of the side
view because they are symmetrical. Basically you are looking into the 17-sided prism from
the 2 ends. The second view (see Figure 4.3) is looking at the prism from the front, and it
is also symmetrical so we are only showing half of it once again.
In general, we are taking a (2m 1) sided prism. The section is made up of 2m similar
right triangles with angle = m at point O. This means that our m is 9, and we have 18
right triangles around the prism. The reasons we have 1 less side face is due to the two
triangles at the bottom that meet to form just 1 side face. Our angle from O is = 9 .
When on it?s stable equilibrium, the longest hypotenuse is r = ro going from O to the
top. This is where we start "counting," or naming o everything. We will name the rest
of the hypotenuses rn = rcosn , 0 < n < m. Our side length rm = s, so it is collinear
with r = r0. We can conclude from this that the measure of the height of this prism is
r +s = r0 +rm. (rm = rcosm respectively).
Figure 4.2: View 1
12
Figure 4.3: View 2
The vertices other than O (on the boundary) lie on two symmetrical equiangular spiral
S. So, we start with r, then make the next hypotenuse (r1) 9 from r. Then connect the
two, so that there is a right angle where the side and r1 meets. Continue repeating until the
prism is complete.
Our "view 2" (see Figure 4.3) of the cylinder is called the half length. It is represented
as y = fx+g where x is measured vertically downwards from O and where f and g denote
the following numbers
f = br +s
g = a+ brr +s
Notice that when x moves below O it?s positive, and above it is negative.
We want to choose b su ciently large compared to a because we want the mass center
below O and above P. For the de nition of P see Figure 4.3. P is the point where the
perpendicular line through Q to R intersects the vertical line containing O. P moves based
o of the length of b because as it increases or decreases the right angle must remain. Since
the angle PQR must remain a right angle, adjusting b moves P up or down.
We describe the half width (looking into the prism from the side) with depth x as well.
We are again measuring x below O. It is going to be represented in terms of z as z = px+q.
13
We will represent p and q in the following way as well:
p = zn zn 1x
n xn 1
q = zn 1xn znxn 1x
n xn 1
We will represent xn and zn in terms of r in the following way:
xn = rncosn = r(cos )ncosn
zn = rnsinn = r(cos )nsinn
So, looking at Figure 4.2, our x0 is all the way at the top with zo. Then x1 (since x
is measured downward from O) goes up to where r1 forms a right angle with the edge of the
prism. The distance from the point ( x1, 0) to the point where r1 forms a right angle is z1.
Continue this process until n = m 1, in which case we have xn and zn where xn = xm 1
and zn = zm 1. Then we reach xm and zm where rm meets the bottom of our prism and
rm = s.
It seems that the reason Guy constructed the prism this way is so that one could easily
compute the coordinates of the mass center. The reason that each of the gures only shows
half of the actual views is that they are symmetrical. To check to see where there is a stable
face, drop a perpendicular from the mass center. If it is on the face, then it stays there when
sitting on the face. If the dropped perpendicular falls outside of the face, then it tips over.
For example, with a normal cube sitting on one of its faces we can drop a perpendicular, and
it will fall inside the face (meaning the face will not tip over). In our other possible example
the cube is stretched and crooked to where if we drop a perpendicular from the mass center,
it would fall outside of the face. This means that after modi cation it will not stand on that
face, and it will tip over. This way of thinking was most likely how Guy began forming this
prism.
14
They knew the body would be multi-faceted, however, it needed to be somewhat simple
so that they could easily compute the mass center. They want the perpendicular from the
mass center to only fall on one face and none of the others. After all the computation is
carried out we know that in the z plane we are alright with the 18 faces because when we
drop a perpendicular from the mass center it falls outside of the faces. We also need to take
care of the other 2 faces which we evaluated in the y plane. We need the mass center to be
below O, but above P. P will take care of the other points when we drop the perpendicular
because all of the points above it will not make a perpendicular with the two side faces, thus
they are not stable. It has to be in between the two for it to work. That is when picking
the correct proportion of a and b comes into play.
4.1 (1) Setting up the integral
So we know that we have constructed the prism symmetrically so that the y and z
coordinates of the mass center are equal to 0. They are trivial (there is only one face where
we drop a perpendicular from the mass center and land on a face). We are only going to
have to set up the integral to nd the rst moment relative to the yz plane. Our goal is to
show that the rst moment is positive so that the mass center lies below 0. We are going to
slice the prism parallel to the yz coordinate plane. Since they are both symmetrical, we are
going to have 2 of each, giving us 4 on the outside of the integral. My slices are going to go
from the very top of the prism where xn is x1, all the way down to the bottom of the prism
where xn is xm. The product of xyz, as a function of x, depends on the subinterval [xi 1;xi],
thus our slices are going to be from xn 1 to xn. Since the integral only takes care of one of
the slices, we will need to sum from n = 1 to m. Thus giving us the following integral to
compute the rst moment (4.1) below O of the material solid:
4
mX
n=1
Z xn
xn 1
xyzdx (4.1)
15
The following excerpt (Figure 4.4) is from Guy?s paper [7]. It shows how he evaluates
1. (the labeling 1... 5 is by us). As we go through 2-5 it turns out that a substantial amount
of algebra had to be redone.
1
2
3
4
5
Figure 4.4: Guy?s solution
4.2 (2) Evaluating the integral
The following completes Part 1 of the solution in Figure 4.4. We substitute the equations
y = fx+g and z = px+q so that the integral is in terms of x.
Z xn
xn 1
xyzdx =
Z xn
xn 1
x(fx+g)(px+q)dx
16
=
Z xn
xn 1
x(fpx2 +fqx+gpx+gq)dx =
Z xn
xn 1
(fpx3 +fqx2 +gpx2 +gpx)dx (4.2)
When we evaluate the integral we get:
fpx
4
4 +fq
x3
3 +gp
x3
3 +gq
x2
2
xn
xn 1
= 14fp(x4n x4n 1) + 13fq(x3n x3n 1) + 13gp(x3n x3n 1) + 12gq(x2n x2n 1) (4.3)
So the integral in (4.2) has value:
= 14fp(x4n x4n 1) + 13(fq +gp)(x3n x3n 1) + 12gq(x2n x2n 1) (4.4)
The above equation is given in Part 2 of the solution in Figure 4.4.
4.3 (3) Algebraic/ trigonometric manipulation
Then when we substitute p = zn zn 1xn xn 1 and q = zn 1xn znxn 1xn xn 1 into our equation we get
the following (The equation is broken up into three parts so that we can see how each form
is manipulated):
= 14f
zn zn 1
xn xn 1
(x4n x4n 1) (4.5)
+13
f
zn 1xn znxn 1
xn xn 1
+g
zn zn 1
xn xn 1
(x3n x3n 1) (4.6)
+12g
zn 1xn znxn 1
xn xn 1
(x2n x2n 1) (4.7)
We will start by evaluating 4.5:
1
4f
zn zn 1
xn xn 1
(x2n x2n 1)(x2n +x2n 1)
17
= 14f
zn zn 1
xn xn 1
(xn xn 1)(xn +xn 1)(x2n +x2n 1)
= 14f(zn zn 1)(xn +xn 1)(x2n +x2n 1)
= 14f(zn zn 1)(x3n +xnx2n 1 +x2nxn 1 +x3n 1)
= 14f (x3nzn)+(xnx2n 1zn)+(x2nxn 1zn)+(x3n 1zn) (x3nzn 1) (xnx2n 1zn 1) (x2nxn 1zn 1) (x3n 1zn 1)
For future manipulation of the equation, we will break 4.5 up into:
= 14f (x3nzn) (x3n 1zn 1)
+14f (xnx2n 1zn) + (x2nxn 1zn) + (x3n 1zn) (x3nzn 1) (xnx2n 1zn 1) (x2nxn 1zn 1)
Now we will evalute 4.6:
1
3
h
f
xnzn 1 xn 1zn
xn xn 1
+g
zn zn 1
xn xn 1
i
(xn xn 1)(x2n +xnxn 1 +x2n 1)
= 13
hf(xnzn 1) f(xn 1zn) +g(zn zn 1)
(xn xn 1)
i
(xn xn 1)(x2n +xnxn 1 +x2n 1)
= 13[f(xnzn 1) f(xn 1zn) +g(zn) g(zn 1)](x2n +xnxn 1 +x2n 1)
= 13
h
f (x3nzn 1) + (x2nxn 1zn 1) + (xnx2n 1zn 1) (x2nxn 1zn) (xnx2n 1zn) (x3n 1zn)
+g (x2nzn) + (xnxn 1zn) + (x2n 1zn) (x2nzn 1) (xnxn 1zn 1) (x2n 1zn 1)
i
For future manipulation of the equation, we will break 4.6 up into:
1
3g
(x2
nzn) (x
2
n 1zn 1)
+13
h
f (x3nzn 1) + (x2nxn 1zn 1) + (xnx2n 1zn 1) (x2nxn 1zn) (xnx2n 1zn) (x3n 1zn)
+g (xnxn 1zn) + (x2n 1zn) (x2nzn 1) (xnxn 1zn 1)
i
18
Now we will evaluate 4.7:
1
2g
(xnzn 1) (xn 1zn)
(xn xn 1)
(xn xn 1)(xn +xn 1)
= 12g (xnzn 1) (xn 1zn) (xn +xn 1)
= 12g (x2nzn 1) + (xnxn 1zn 1) (xnxn 1zn) (x2n 1zn)
Combining 4.5, 4.6, and 4.7 to simplify further we have:
1
4f
(x3
nzn) (x
3
n 1zn 1)
+13g (x2nzn) (x2n 1zn 1)
+14f (xnx2n 1zn) + (x2nxn 1zn) + (x3n 1zn) (x3nzn 1) (xnx2n 1zn 1) (x2nxn 1zn 1)
+13f (x3nzn 1) + (x2nxn 1zn 1) + (xnx2n 1zn 1) (x2nxn 1zn) (xnx2n 1zn) (x3n 1zn)
+13g (xnxn 1zn) + (x2n 1zn) (x2nzn 1) (xnxn 1zn 1)
+12g (x2nzn 1) + (xnxn 1zn 1) (xnxn 1zn) (x2n 1zn)
When we add the third and fourth rows
1
4f
(x
nx2n 1zn) + (x2nxn 1zn) + (x3n 1zn) (x3nzn 1) (xnx2n 1zn 1) (x2nxn 1zn 1)
+13f (x3nzn 1) + (x2nxn 1zn 1) + (xnx2n 1zn 1) (x2nxn 1zn) (xnx2n 1zn) (x3n 1zn) ;
they are combined to get:
1
12f
(x
nx2n 1zn) (x2nxn 1zn) (x3n 1zn) + (x3nzn 1) + (xnx2n 1zn 1) + (x2nxn 1zn 1)
19
When we add the fth and sixth rows
1
3g
(x
nxn 1zn) + (x2n 1zn) (x2nzn 1) (xnxn 1zn 1)
+12g((x2nzn 1) + (xnxn 1zn 1) (xnxn 1zn) (x2n 1zn));
they are combined to get:
1
6g
(x2
nzn 1) + (xnxn 1zn 1) (xnxn 1zn) (x
2
n 1zn)
Combining 4.5, 4.6, 4.7, and the simpli cations we get:
1
4f
(x3
nzn) (x
3
n 1zn 1)
(4.8)
+13g (x2nzn) (x2n 1zn 1)
+ 112f (xnx2n 1zn) (x2nxn 1zn) (x3n 1zn) + (x3nzn 1) + (xnx2n 1zn 1) + (x2nxn 1zn 1)
+16g (x2nzn 1) + (xnxn 1zn 1) (xnxn 1zn) (x2n 1zn)
The above equation is given in Part 3 of the solution in Figure 4.4.
4.4 (4) Sum from n = 1 to m
Applying the sum to the rst and second row, 14f (x3nzn) (x3n 1zn 1) + 13g (x2nzn)
(x2n 1zn 1) , means summing from n = 1 to m. These two terms yield 14f (x3mzm) (x0z0)
and 13g (x2mzm x20z0) . These both equal zero because z0 = zm = 0. In order to sum the
third and fourth lines, we will recall the following formulas: r = r0, rn = rcosn ;0 0.
To determine the value for b based o of given values for a, , r, and s I wrote a
MATLAB program (See Figure 4.7). This made it easier to nd values of b based o of
a given a. Remember that r is equal to length from the origin to the top of our solid,
s = rcos9 = rk9, and = =9.
35
Figure 4.5: The "m" le that is run in MATLAB to determine our value for m
Figure 4.6: Side view of the solid
For the results when a = 0:25 and r = 1, b must be greater than 7.9762. For the given
r value, the ratio between a and b is 0.0313.
In order to ensure that P is below the mass center, we want b long enough compared
to a so that the right angle Q will cause PQ to intersect the center of the solid below the
bottom of the solid.
Since a = 0:25, and the height of the solid is r +s = 1 + 1 cos9 1:57, we know that
the diagonal from Q to the center and bottom of the gure is 1.59. Using similar triangles,
we nd that any b 9:8624 and a = 0:25 ensures that P is below the the mass center.
36
Table 4.1: Results for m
m value equation result
1 0
2 -1
3 -0.903076
4 -0.667969
5 -0.452051
6 -0.279457
7 -0.146033
8 -0.043859
9 0.034274
10 0.094100
Figure 4.7: The "m" le to determine a value for b.
37
Chapter 5
Bodies which have exactly one stable and one unstable position.
Guy?s construction of a polyhedron with just one stable equilibrium is fascinating and
important because of the fact that it is a polyhedron with the least known number of
faces and just one stable equilibrium (it is still an open question). Guy?s construction
had 1 stable equilibrium, and 3 unstable equilibrium. There are bodies with less than 4
equilibrium. These bodies are called mono-monostatic bodies. G abor Domokos? interest was
to nd a three-dimensional body with just one stable and one unstable equilibrium. The
following provides a proof that no monostatic bodies exist in two-dimensions, and then the
construction of a three-dimensional mono-monostatic body.
5.1 All two-dimensional objects have at least two stable equilibrium.
G abor Domokos was a Civil Engineering teacher in Hungary in the 1980?s. He enjoys
the mathematical side of things, and often times would discuss interesting problems with
Andy Ruina. Ruina had a friend named Jim Papadopoulous and both were working on
a simple conjecture. Papadopoulous started working with two-dimensional (2D) problems
with plywood to see if he could nd shapes with only one stable equilibrium.
We are considering these bodies as resting on a horizontal surface in the presence of
uniform gravity. Monostatic bodies are possible to construct; for example, the popular
children?s toy called a "Comeback Kid" is a monostatic body. Looking at homogeneous,
convex monostatic bodies in the 2D case, one can prove that:
Theorem 5.1. Among planar (slab-like) objects rolling along their circumference no mono-
static bodies exist.
38
In 2D Papadopoulous found that a square, for example, had four positions for which
it was stable. He looked at an ellipse, which is stable when horizontal on one of the two
atter parts. It is unstable when balanced on either end. Imagine balancing an upright egg.
Papadopoulous? conjecture was that no matter what convex shape you draw and cut out, it
has at least two orientations where it is stable.
The following ideas are published by G. Domokos and P. V arkonyi [4]. Consider a
convex, homogeneous planar "body" B and a polar coordinate system with origin at the
center of gravity of B. Let the continuous function R(?) denote the boundary of B. It is
easy to see that non-degenerate stable/unstable equilibria of the body correspond to local
minima/ maxima of R(?).
Assume that R(?) has only one local maximum and one local minimum. Meaning that
there is only one point on the boundary that is closest to the origin, and one point of the
boundary that is furthest away from the origin. Finally a simple continuity argument shows
that there exists exactly one value ? = ?0 for which R(?0) = R(?0 + ). Rotate a directed
chord which passes through O clockwise around O. Assume initially the chord points toward
the closest boundary point. Notice that initially O is closest to the head of the directed chord,
while at the end O is closest to the tail of the directed chord.
Figure 5.1: Two-Dimensional case
39
From this we know R(?) > R(?0) if > ? ?0 > 0, and R(?) < R(?0) if <
? ?0 < 0. The straight line discussed above ? = ?0 is the same as ? = ?0 + . It
is passing through the origin O and cuts B into a "thin" (R(?) < R(?0)), and a "thick"
(R(?) >R(?0)) part. This implies that O can not be the center of gravity, which contradicts
our initial assumption.
De nition 5.1. Mono-monostatic bodies are convex, homogeneous bodies with fewer than
four equilibria.
5.2 The G omb oc
In 1995 Domokos attended an International Congress on Industrial and Applied Math-
ematics in Hamburg. There were many lectures, but the one that everyone attended was
Vladimir Igorevich Arnold?s lecture. A returning theme of that particular talk was the recur-
rence of the integer 4 in many seemingly unrelated results. His lecture reminded Domokos
of the 2D objects which always have at least four equilibrium (two unstable and two stable).
He proposed a counter example to Arnold, saying that he has found a three-dimensional
object with just one stable equilibrium. Arnold immediately explained that it was not a
counter example because the gure has one stable equilibrium, but three unstable. Arnold
did, however, conjecture that "convex homogeneous bodies with fewer than four equilibria
(mono-monostatic bodies) may exist."
Domokos moved on to construct a three-dimensional solid which has less than four
equilibrium. About ten years later, Domokos constructed the G omb oc to prove that Arnolds
conjecture is indeed correct. The problem was solved in 2006 [4]. It is called the G omb oc,
and is a convex three-dimensional homogeneous body which, when resting on a at surface,
has just one stable and one unstable point of equilibrium.
The shape is not unique, has countless varieties, and most of which are very close to
a sphere with very strict shape tolerance. These bodies are hard to visualize, describe, or
identify. They are neither " at" nor "thin." The solution has curved edges and resembles a
40
Figure 5.2: G omb oc
sphere with a squashed top. It rests in its stable equilibtirum, and its unstable equilibrium
position is obtained by rotating the gure 180 about the horizontal axis. Theoretically it
will rest on this point, but the smallest perturbation will bring it back to the stable point.
There is not a speci c description of how to create the G omb oc available at this time.
It has been proven that there exists a body that is convex with its center of gravity at
the origin. Unfortunately, from the numerical proof it suggests that d must be very small
(d < 5 10 5) in order to satisfy convexity together with other restrictions. The created
object with these properties is extremely similar to a sphere and nearly impossible to visually
see the di erence in the construction.
41
Appendix: For general appreciation and also to help me remember what was done during
the project, the description of Sch onhardt?s polyhedron, Cs asz ar?s polyhedron, and a
Meissner body are included here.
42
Chapter 6
Cs asz ar?s polyhedron
Theorem 6.1. The tetrahedron and Cs asz ar?s polyhedron are the only two known polyhedra
without any diagonals.
De nition 6.1. In a polyhedron, a line between two nonadjacent vertices is called a diagonal.
The Cs asz ar polyhedron is named after Akos Cs asz ar, who discovered it in 1949. Akos
Cs asz ar is a Hungarian mathematician specializing in general topology and real analysis.
Figure 6.1: Cs asz ar?s Polyhedron
It is very di cult to visualize the Cs asz ar Polyhedron because of its complexity. In an
attempt, to help you understand what it looks like, there is a at pattern of the polyhedron.
It would look like this if it was completely unfolded (See Figure 6.2).
In geometry, the Cs asz ar polyhedron is a nonconvex polyhedron, topologically a torus,
with 14 triangular faces. When every pair of vertices is connected by an edge, the polyhedron
has thus no diagnols. Therefore, the Cs asz ar polyhedron has no diagonals because every
43
Figure 6.2: Flat Pattern of Cs asz ar?s Polyhedron
pair of vertices is connected by an edge. The graph is isomorphic with the skeleton of a
six-dimensional simplex, the 6-space analogue of the tetrahedron.
It may be easier to visualize the polyhedron if you understand how to construct it. It is
not di cult to make a paper model of the Cs asz ar polyhedron. Copy the two patterns below
and cut the copies out. Color the seven shaded triangles on both sides. Crease the paper to
make "mountain folds" along each M line, and the "valley folds" along each v line. [6]
1. With the pattern for the base, fold the two largest triangles to the center and tape
the A edges to each other. Turn the paper over. Fold the two smaller triangles to the
center and tape the B edges together to obtain a completed base.
2. The six-faced conical top is formed by taping the C edges together. Place it on the
base as shown in the drawing of the completed model. It will t in two ways. Choose
the t that joins white to shaded triangles, then tape each of its six edges to the
corresponding six edges of the base.
The polyhedron has 7 polyhedron vertices, 14 faces, and 21 polyhedron edges.
The Cs asz ar polyhedron and tetrahedron are the only two known polyhedra without
any diagonals for which there are no interior diagonals.
44
C
C
M
M
M
V
V
V
V
Figure 6.3: Pattern for top of Cs asz ar?s polyhedron
A
A
B
B
M
M
M
M
V
V
V
Figure 6.4: Pattern for base of Cs asz ar?s polyhedron
Figure 6.5: Completed Base of Cs asz ar?s polyhedron
45
Chapter 7
Szilassi?s polyhedron
Theorem 7.1. The tetrahedron and Szilassi polyhedron are the only two known polyhedra
in three-dimensions in which each face shares an edge with another face.
The Szilassi polyhedron, named after Hungarian mathematics Lajos Szilassi, who dis-
covered it in 1977, is the dual to the Cs asz ar polyhedron with 7 faces, 14 vertices, and
21 edges. While the Cs asz ar polyhedron has the fewest possible vertices of any toroidal
polyhedron, the Szilassi polyhedron has the fewest possible faces of any toroidal polyhedron.
Figure 7.1: Transparent Szilassi?s Polyhedron
De nition 7.1. In geometry, polyhedra are associated into pairs called duals, where the
edges correspond to the faces of the other. The dual of the dual is the original polyhedron.
The dual of a polyhedron with equivalent vertices is one with equivalent faces, and of one
with equivalent edges is another with equivalent edges.
Duality means that every face of the Szilassi heptahedron has an edge in common with
each of the other 6 faces.
46
Figure 7.2: Szilassi?s Polyhedron
Theorem 7.2. The four color theorem states that given any separation of a plane into
contiguous (adjacent) regions, no more than four colors are required to color the regions of
the map so that no two adjacent regions have the same color.
De nition 7.2. Two regions are called adjacent if they share a border segment, not just a
point.
The Four Color Problem dates back to 1852 when Francis Guthrie, while trying to
color the map of the counties of England, notices that four colors su ced. He asked his
brother Frederick if it was true that any map can be colored using four colors in such a
way that adjacent regions receive di erent colors. Frederick Guthrie then communicated
the conjecture to DeMorgan. A year later the rst proof by Kempe appeared, and its
incorrectness was pointed out 11 year later by Heawood. Another failed proof was presented
by Tait in 1880. Although they both failed, Kempe discovered what became known as
Kempe chains, and Tait found an equivalent formulation of the Four Color Theorem in
terms of 3-edge-coloring.
47
Chapter 8
Sch onhardt Polyhedron
De nition 8.1. A tetrahedralization is a partition of the input domain, point set or poly-
hedron, into a collection of tetrahedra, that meet only at shared faces (vertices, edges, and
triangles).
The analogue of triangulation in three dimensions is called tetrahedralization.
If additional vertices (Steiner points) are allowed when constructing the tetrahedraliza-
tion, then all polyhedra are tetrahedralizable.
Tetrahedralization turns out to be signi cantly more complicated than triangulation.
Theorem 8.1. Any polyhedra can be triangulated with O(n2) Steiner points and P(n2) tetra-
hedra.
Proof: Extend a vertical "wall" from each edge of the polyhedron boundary, up and
down from that edge until is reaches some other part of the boundary. These walls divide the
polygon into generalized cylinders. Triangulating the top and bottom faces of the cylinders
partitions the polygon into O(n2) triangular prisms. Each vertical prism side is crossed at
most once by a polyhedron edge, so the prisms are polyhedra with at most twelve vertices.
Triangulate the faces of these polyhedra, making sure that tetrahedra from di erent prisms
will meet face to face, and then triangulate each prism with at most 20 tetrahedra incident
to the single interior Steiner point (Du [5]).
In geometry, the Sch onhardt Polyhedron is the smallest non-convex polyhedron that
cannot be triangulated into tetrahedra without adding new vertices.
It can be formed by two congruent equilateral triangles in two parallel planes so that
the line through the centers of the triangles is perpendicular to the planes. Consider an
48
equilateral triangular prism, where the base triangle ABC lies directly below the upper
triangle A0B0C0.
Suppose that the edges of the prisms are constructed from wire and that the edges AB0,
BC0, and CA0 have been added with extra wire. If we slowly rotate the upper triangle in
one direction, then the rectangular faces of the prism bend outwards along the extra edges
AB0, BC0, and CA0. If we rotate in the other direction, however, the rectangular faces of
the prism bend inward, yielding a polyhedron which is not convex.
When we have rotated a full 60 , then the three edges AB0, BC0, and CA0 intersect at
the center of the gure. Sch onhardt?s Polyhedron is obtained when the rotation is by some
intermediate value, for example 30 . In this instance the line segments AB0, BC0, and CA0
lie outside the polyhedron([3]).
Figure 8.1: Sch onhardt?s Polyhedron
We take the three vertices on the bottom triangle, and we need to connect it to a fourth
additional vertex to form a tetrahedron. If we connect it to any of the above three vertices,
then the edge is outside of the polyhedron because every diagonal that is not a boundary
edge lies completely in the exterior. It implies that there can be no triangulation of it without
new vertices because there is simply no interior tetrahedron (all possible tetrahedra spanned
by four of its six vertices would introduce new edges).
49
Sch onhardt proved that this is the smallest example of an untetrahedralizable polyhe-
dron. Bagemihl extended this example to construct a polyhedron of n vertices with the same
properties for every n 6.
50
Chapter 9
A Meissner body
De nition 9.1. A three-dimensional body is of constant width if it has the following property:
Whichever way it is clamped between two parallel plates, both plates are always at exactly the
same distance from each other.
Take a regular tetrahedron, which has four points in space that are arranged in such
a way that they are equidistant from each other. Place spheres of equal size at each of
the four tetrahedron vertices. The spheres expand at the same rate until they touch and
have to penetrate each other. The surface of each sphere intersects the three tetrahedron
vertices opposite the center of the sphere. This forms a Reuleaux tetrahedron, which is
the intersection of four spheres at the vertices of a regular tetrahedron. The result is a
body looking like a bulging tetrahedron. The Reuleaux tetrahedron has the same face
structure as a regular tetrahedron, but with curved faces: four vertices, and four curved
faces, connected by six circular-arc edges. Basically if a at plate is placed on three equal
Reuleaux tetrahedrons and pushed back and forth, it will move almost parallel to the surface
of the table. In other words, it will move as though it is placed on three spherical balls with
equal size, without wobbling. As noted above, the Reuleaux tetrahedron is almost constant
width.
Since a Reuleaux tetrahedron is only of almost constant width, mostly the body touches
two plates between which it is clamped with one of its vertices and one point on the opposite
surface of the solid. In cases like this, the two plates are at the same distance from each
other, but the two possible points of contact may be on two opposite edges of the solid. At
this place on the solid, the width is greater. For two edge midpoints, it is maximally p3 -
p2
2 1.0249 times larger than the smallest width.
51
It is possible, to make the Reauleaux tetrahedron into a body of constant width by
rounding three of its edges. First, pick a face of the original tetrahedron. Extend the chosen
face in all three directions so that it intersects the Reuleaux tetrahedron. Then, extend the
adjacent triangles so that they intersect with the original triangle. Remove the part of the
surface of this body which is located between the extensions of two adjacent lateral surfaces
of the tetrahedron. In the new space, replace it with a sphere so that there are three smooth
and three original "edges." The resulting gure is called a Meissner body (Weber [10]).
De nition 9.2. When three edges of a Reuleaux tetrahedron that meet at a common vertex
are rounded as described, the resultant solid is a body of constant width, called a Meissner
body.
52
Bibliography
[1] H. T. Croft, K. J. Falconer and R. Guy, Unsolved problems in geometry, Problem
B12, Springer, (1991), 61.
[2] R. Dawson, Monostatic simplexes, The American Mathematical Monthly, 92, (8),
(1985), 541-546.
[3] N. Do, Art gallery theorems, Mathellaneous, 288-294. http://www.austms.org.au/
Gazette/2004/Nov04/mathellaneous.pdf
[4] G. Domokos and P. L. V arkonyi, Mono-monostatic bodies, The Mathematical Intelli-
gencer, 28, (2006), 34-38.
[5] D. Du, Computing in Euclidean geometry, World Scienti c Publishing Co., (1992), 68-
71.
[6] M. Gardner, The Cs asz ar polyhedron, Time Travel and Other Mathematical Bewilder-
ments, W. H. Freeman and Company, (1988), 140-152.
[7] M. Goldberg and R. Guy, Problems and solutions, SIAM Review, 11, (1), (1969), 78-82.
[8] A. Heppes, A double tipping tetrahdreon, SIAM Review, 9, (1967), 599-600.
[9] M. Klamkin, Acute dihedral angles, Mathematics Magazine, 61, (5), (1988), 320.
[10] C. Weber, What does this solid have to do with a ball?, (2009). http://www.swiss
educ.ch/mathematik/material/gleichdick/docs/meissner en.pdf
53