SLICING THE CUBE
A thesis submitted to the
Kent State University Honors College
in partial fulfillment of the requirements
for University Honors
by
David Zach
May, 2011
Thesis written by
David Zach
Approved by
, Advisor
, Chair,
Department of Mathematical Sciences
Accepted by
, Dean, Honors College
ii
TABLE OF CONTENTS
LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . .
v
1 INTRODUCTION AND HISTORY . . . . . . . . . . . . . . . . . . . .
1
2 OPTIMAL INTERSECTIONS OF Qn WITH LINES . . . . . . . . . .
6
√
3 A SOLUTION FOR t >
n−1
2
. . . . . . . . . . . . . . . . . . . . . . .
21
REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
iii
LIST OF FIGURES
1
Q2 with a general, intersecting ` . . . . . . . . . . . . . . . . . .
7
2
Minimal lines of intersection with Q2 for various t . . . . . . . .
8
3
Q3 with St2 , v, and ` . . . . . . . . . . . . . . . . . . . . . . . . .
9
4
Projections of ` and St2 onto Q2 . . . . . . . . . . . . . . . . . . .
10
5
The maximal lines of intersection with Q2 for various t . . . . . .
12
6
` contained in a plane H sharing the same point of tangency to St2 12
7
A fixed point x with hypercone intersecting an opposite face . . .
15
8
The maximal ` for Case I . . . . . . . . . . . . . . . . . . . . . .
17
9
The maximal ` for Case II . . . . . . . . . . . . . . . . . . . . . .
19
10
The maximal ` is restricted to a face of the cube when t ≥
11
A possible intersecting hyperplane H with t ∈ (
1
2
. .
19
. . . .
22
12
ai , ai , and Fi near a corner of Q3 . . . . . . . . . . . . . . . . . .
23
13
The maximal hyperplane for t ∈ (
√
√
iv
√
n
n−1
,
2
2 ]
√
n
n−1
2 , 2 ]
. . . . . . . . . . . .
27
ACKNOWLEDGEMENTS
I would like to thank my advisor Dr. Artem Zvavitch for his guidance and
support throughout this process, along with my research partners James Moody,
New York University, and Corey Stone, University of Nebraska-Lincoln. The new
material presented in this thesis is the result of research done in the summer of
2010 at the Kent State University Mathematics Research Experience for Undergraduates, National Science Foundation Grant #0755318, with Moody and
Stone, under Dr. Zvavitch. I would also like to thank my defense committee
members, Drs. Feodor Dragan, Brett Ellman, and Jenya Soprunova, and my
Honors College alumni sponsors, Michael Sagert and Angel O’Neal. My deepest
gratitude and apologies go out to those friends, family, and especially roommates
who endured my erratic behavior during the many frantic and sleepless nights
spent writing and formatting this paper.
v
CHAPTER 1
INTRODUCTION AND HISTORY
To the common mathematical passer-by, there are few objects that seem
simpler than the cube. It is an object with which even children are familiar. A
high school geometry class explains it as a three-dimensional analog to a square:
All edges of a cube have equal length, yielding an easily visualizable object whose
volume and surface area are easily computable. Cubes are frequently seen and
used in our day-to-day lives, and have been for millennia, so one might naturally
assume that mathematicians have a firm grasp on all the properties possessed
by cubes.
This, however, is not the case. Several simple questions about cubes remain
unanswered. We refer to Chuanming Zong [Zo] for a review of many of the
properties of the unit cube as well as a related open problem. This paper will
focus on one such question in particular: What are the optimally sized cross
sections of the cube?
A typical high school geometry student is often expected to visualize the
various polygons that can be obtained from taking different cross sections of a
cube. Such an exercise leads very naturally to the question of which cross section
is biggest, and which is smallest. Yet, this question baffled mathematicians for
years.
Although this problem has now been solved for a cube in any dimension, the
1
2
proofs for both the maximal and minimal case have only been found within the
past forty years. Additionally, the proof is highly nontrivial and, surprisingly, it
relies heavily on probabilistic arguments rather than geometric ones.
Again, this discovery leads naturally to more questions. What if our cross
section must cut a specific distance away from the center of the cube? Progress
has been made toward solving this problem, but even in the visualizable, threedimensional case, a solution has not been found.
Let us begin by formalizing the questions posed. We will consider the unit,
n-dimensional cube, Qn := [− 12 , 12 ]n . This cube has edges of length 1, volume 1,
and is centered conveniently at the origin on our set of n coordinate axes. We
then define H to be an (n − 1)-dimensional hyperplane of Rn .
The original question now reads: What are the optimal values of |H ∩ Qn |,
where | · | denotes the standard Euclidean volume. We take careful note that
this yields an (n − 1)-dimensional volume, and not an n-dimensional volume.
One might notice that the minimal case has a rather trivial answer: If we
consider H to be the hyperplane x1 + x2 + · · · + xn =
n
2,
then for any n ≥ 2.
we have that H intersects Qn only at the point ( 12 , 12 , . . . , 12 ), giving us that
|H ∩ Qn | = 0. Thus, we invoke one further stipulation; that H must contain the
origin. In other words, H is not just a hyperplane, but is actually an (n − 1)dimensional subspace of Rn .
For Q2 and Q3 , a well-informed conjecture is still not difficult to develop.
One can visualize that a subspace parallel to a face of the cube is again the unit
cube in a dimension one lower, and thus, its volume is 1. If one were to tilt this
3
subspace in any one direction, then it would acquire more volume. There is little
reason to believe that this situation would be any different in n dimensions
This claim is, in fact, correct, but the proof of this intuitive result is quite
far from trivial. The first such proof was published by Douglas Hensley [He1] in
1979, via rather unexpected methods. Hensley, himself, was acutely aware of this
peculiarity, stating in his introduction, “A curious feature of the solution to this
straightforward problem of geometry is that it is largely based on probabilistic
considerations”.
Hensley also provided the calculations to determine an upper bound of 5
for the maximal case, but conjectured that the actual maximal volume was
√
2,
which would later prove to be correct.
Hensley’s motivation came from a more general statement he claimed was
conjectured by Dr. Anton Good. The claim stated that |Pk ∩ Qn | ≥ 1, where
Pk is any k-dimensional subspace of Rn , for k < n. Hensley’s proof addressed
the case when k = n − 1, but later the same year, Jeffrey Vaaler gave a full
proof of Good’s conjecture [Va]. His method was similar to Hensley’s, employing
probability theory and inequalities via comparison to the normal distribution,
peppered with a healthy dose of measure theory, linear algebra, and Minkowski
addition. Zong [Zo] gives an abridged version of this proof.
A separate generalization came seven years later from Keith Ball. In his
1986 paper, Ball simplified Hensley’s result and also verified Hensley’s maximal
conjecture [Ba1] (extended further in [Ba2]). Furthermore, he showed that the
maximal result applied not just to (n − 1)-dimensional subspaces of Rn , but to
4
any hyperplane of Rn .
Ball then defined our volume of intersection as a continuous probability density function that is dependent on the normal vector to the hyperplane and the
distance from the hyperplane to the origin. He used this density along with the
Brünn-Minkowski inequality [Ga], [K], to show that the maximal volume occurs
when our hyperplane is, in fact, an (n − 1)-dimensional subspace of Rn .
Yet another alternate proof to Hensley’s result for minimal intersection was
also given by Ball. His probabilistic approach also yielded that the same minimal
value of |H ∩ Qn | = 1 is achieved by any hyperplane H that is parallel to a face
of Qn .
Ball’s proof of the the maximal intersection was more complicated. He again
relied on probability theory, but was also forced to use harmonic analysis and an
integral inequality. He successfully confirmed Hensley’s conjecture that |H ∩ Qn |
achieves a maximum value of
√
2, and showed that it does so when H contains
an (n − 2)-dimensional face of Qn .
The majority of Ball’s paper focused on proving a particular integral inequality that served as an essential lemma in his proof for the maximal volume
of intersection. His proof of this inequality, however, is less than elegant, employing a significant amount of estimation and “...a lot of unpalatable numerical
computations some of which [are] almost impossible to verify without a calculator,” according to Nazarov and Podkorytov [NP]. They provided a much cleaner
proof of Ball’s integral inequality in 2000. The proof of this inequality utilized
probability and measure theory (see [K] for more details).
5
Further investigations of this problem have been made. Oleszkiewicz and
Pelcyński [OP] proved an analog to Ball’s theorem in the complex space Cn .
Barthe and Koldobsky [BK] have published work dealing with optimal volumes
of symmetric slabs of Qn . In Koldobsky’s book [K], he discussed the cube as
an n-dimensional unit ball under the L∞ -norm. Here he cites Nazarov and
Podkorytov’s proof for maximal sections, and also gives an integral formula for
the volume of intersection that was given by Pólya [P].
In the next two sections, we will present new results concerning this problem
based on research done in the summer of 2010. The research was conducted at
the Kent State University Mathematics Research Experience for Undergraduates
with the help of Jams Moody, New York University, and Corey Stone, University
of Nebraska-Lincoln. Unlike the work that has come before, our work is almost
entirely geometric in nature, to which the complexity of the proofs, in part, can
be attributed. In section 2, we give the optimal lengths of intersection between
Qn and a line ` that is a specified distance t away from the origin. In section 3,
we provide a partial solution to the initial hyperplane problem, namely for when
√
t>
n−1
2 .
Although the methods of solution are rather cumbersome, the data
could be useful to help predict the results of other related problems as well.
CHAPTER 2
OPTIMAL INTERSECTIONS OF Qn WITH LINES
In an attempt to gain insight as to how hyperplanes interact with Qn , we
begin with a simpler question: How do smaller-dimensional analogs interact with
Qn ? We consider a line ` that is a specified distance t ≥ 0 from the origin, and
work toward finding the optimal values of |` ∩ Qn | via an entirely geometric approach.
Minimal Case:
Theorem 1. For a line ` that is distance t away from the origin, we have that






1,





min |` ∩ Q2 | = √2 − 2t,

`∈R2








0,
√
t ∈ [0, 21 ( 2 − 1)]
√
t ∈ [ 21 ( 2 − 1), 12 ]
t ∈ ( 12 , ∞)
Proof: We begin by conducting a simple investigation for the minimal values in
two dimensions. Consider first that ` must be a fixed distance t away from the
origin. Since distance is defined by the perpendicular to the line, then this is
equivalent to specifying that it must be tangent to the circle of radius t that is
centered at the origin.
We then take the standard polar parametrization and note that ` must pass
6
7
through the point (t cos θ, t sin θ), where θ is the angle from the positive x-axis.
Since we know that it must be tangent to x2 + y 2 = t2 , we find that the equation
of the line ` must be
x cos θ + y sin θ = t.
Figure 1 depicts Q2 with this circle and our line `. Highlighted in red is
` ∩ Q2 .
Figure 1: Q2 with a general, intersecting `
Due to the symmetries of the cube, we need only consider θ ∈ [0, π4 ]. We
note that for t > 12 , we have the trivial result |` ∩ Q2 | = 0 when we take θ = 0
(See Figure 2c), so we consider t ≤ 12 .
Under our domain, ` must intersect Q2 on its upper side, y =
1
2 , |x|
≤
1
2.
Using some basic trigonometry, we can see that ` will intersect the bottom side
8
of Q2 when 0 ≤ θ ≤
√
π
− arcsin t 2, yielding
4
|` ∩ Q2 | = sec θ.
On the other hand, ` will intersect the right side of Q2 , when
θ≤
√
π
− arcsin t 2 ≤
4
π
, yielding
4
|` ∩ Q2 | =
cos θ + sin θ − 2t
2 cos θ sin θ.
This break in the domain of θ occurs where the line ` passes through the corner
1
1
2, −2
of Q2 . Note that for a fixed t, this |` ∩ Q2 | is a continuous function of
π
θ on [0, ]. From here, we find the critical values and determine our absolute
4
minimums on this interval (Figure 2).
√
For t ∈ [0, 12 ( 2 − 1)], the minimum value of |` ∩ Q2 | = 1 occurs when θ = 0.
√
√
For t ∈ [ 12 ( 2 − 1), 12 ], the minimum of |` ∩ Q2 | = 2 − 2t occurs when θ = π4 .
For t ∈ ( 12 , ∞), the minimum of ` ∩ Q2 | = 0 occurs when θ = 0.
Remark: We should take careful note that min |` ∩ Q2 | is a non-increasing function of t. This is a direct result of the Brünn-Minkowski inequality.
√
(a) 0 ≤ t ≤ 12 ( 2−1)
(b)
1
2
1
(
2
√
2 − 1) ≤ t ≤
(c) t >
Figure 2: Minimal lines of intersection with Q2 for various t
1
2
9
Claim. Our minimal values for |` ∩ Qn | are exactly the same as for |Q2 ∩ `|.
Similarly to our two-dimensional case, if we know that ` is a fixed distance t
away from the origin, then it must be tangent to the (n−1)-sphere of radius t that
is centered at the origin, Stn−1 := {x ∈ Rn : |x| = t}. Let v = (v1 , v2 , . . . , vn )
be the vector from the origin to this point of tangency on Stn−1 . Thus, we know
that |v| = t and that ` runs perpendicular to v. Figure 3 depicts an example for
n = 3.
Figure 3: Q3 with St2 , v, and `
Consider that ` must pass through two distinct faces of our cube Qn . By
symmetry, we can assume that one face is given by {( 21 , x2 , x3 , . . . , xn )||xi | ≤ 12 }.
Then the other face is either opposite to the first, i.e.
{(− 12 , x2 , x3 , . . . , xn )||xi | ≤ 12 }, or by symmetry again, we can reorient so that it
is {(x1 , 21 , x3 , x4 , . . . , xn )||xi | ≤ 12 }.
We then project our cube and our line onto the x1 x2 -coordinate plane (Figure
4). The cube will simply project down to Q2 , and our line ` will project to a
10
line `0 such that
|`0 ∩ Qn | = cos α|` ∩ Qn | ≤ |` ∩ Qn |,
where α is the angle of inclination between ` and `0 . Notice that `0 passes either
through opposite sides of the cube (the first case above), or adjacent sides of Q2
(the second case).
Figure 4: Projections of ` and St2 onto Q2
Note also that our projected line `0 must be a distance t0 ≤ t away from
the origin. Since we’ve already established that min |` ∩ Q2 | is a non-increasing
function of t, then we can see that |`0 ∩ Q2 | ≥ |` ∩ Q2 | for any `. We also can see
that `0 is the same line as ` if and only if ` is contained in the x1 x2 -coordinate
plane. Clearly, then, the minimum values (dependent on t) are the same as those
in our two-dimensional case.
11
Maximal Case: For the maximal case, we will describe our results in geometric
terms, providing numerical answers at the end.
Theorem 2. For a line ` ∈ Rn that is distance t ∈ [0, 12 ] away from the origin,
the maximal length |` ∩ Qn | occurs when ` travels through a corner of Qn and
passes through an edge containing the opposite corner. For t ≥ 21 , the maximal
` is confined to a face of Qn .
Proof: We, again, begin with the two-dimensional case. Using the data from
our minimal case, we get that |` ∩ Q2 | is maximized at θ =
√
π
− arcsin t 2 for
4
t ∈ [0, 21 ]. That is, the maximal ` goes through a corner of the square to an
opposite side (Figure 5a), and the maximum value is
2
p
.
2 − (2t)2 + 2t
Furthermore, for t ∈ ( 12 ,
θ=
√
2
2 ],
the maximal length of intersection occurs when
π
(Figure 5b), and the length, as mentioned before, is
4
√
2 − 2t.
√
For t ≥
2
2 ,
the maximal length of intersection is trivially 0.
12
(a) 0 ≤ t ≤
1
2
(b)
1
2
√
<t≤
2
2
Figure 5: The maximal lines of intersection with Q2 for various t
Consider a general argument in n dimensions. We again use the notion that
a line ` that is distance t > 0 from the origin is actually tangent to the sphere
Stn−1 at a specific point. Furthermore, ` is contained in a two-dimensional plane
H, which is tangent to Stn−1 at the same point (Figure 6). Thus, ` ∩ Qn is
contained within H ∩ Qn , a convex polygon (both H and Qn are convex, so their
intersection is as well).
Figure 6: ` contained in a plane H sharing the same point of tangency to St2
13
Lemma 1. Let x be a point contained in a convex polygon P and let ` be a line
passing through x. Then,
max |` ∩ P |
`
is obtained when ` passes through a vertex of P .
Proof: Assume that ` passes through x and does not pass through a vertex of
P . Then it must pass through two distinct edges of P , E1 and E2 . Let a1 , a2 be
the points of intersection of ` with E1 and E2 , respectively, and let d1 , d2 be the
distance from x to E1 and E2 , respectively.
Note then that |` ∩ P | = dist(a1 , a2 ) = dist(a1 , x) + dist(x, a2 ). Let α be the
angle between ` and the perpendicular from x to E1 , and consider dist(a1 , x) as
a function of α on [− π2 , π2 ]. Then,
dist(a1 , x)(α) =
d1
,
cos α
and on our domain,
dist(a1 , x)00 (α) = d1
1 + sin2 α
≥ 0.
cos3 α
Thus, dist(a1 , x) is a convex function. Similarly, for some angle β, we will
have that
dist(x, a2 )(α) =
d2
,
cos (α + β)
which is also convex. The sum of convex functions is convex, so therefore, our
14
intersection |` ∩ P | = dist(a1 , x) + dist(x, a2 ) is also a convex function of α. A
convex function achieves its maximum on the boundary, which would imply that
our line ` is parallel to E1 . However, in a polygonal setting, ` would not be able
to extend along E1 infinitely, and would be restricted first by the intersection of
E1 with another edge, i.e. a vertex of P .
Remark: This result is not true when we consider a line through a point in a
polytope K in Rn . If K is centrally symmetric, then it does hold, as we need only
to maximize the distance from our point to the boundary, which must happen at
an extremal point (i.e., a vertex). However, for a nonsymmetric counterexample,
consider the simplex in R3 given by points (0, 1, 1), (0, −1, 1), (1, 0, −1), (−1, 0, −1).
The maximal line through the origin extends from (0, 0, 1) to (0, 0, −1), and does
not intersect a vertex of our polytope.
As we stated before, ` is contained in a two-dimensional plane H that is
tangent to Stn−1 at the same point as `. Thus, ` ∩ Qn is contained in H ∩ Qn .
But H ∩Qn is a convex polygon. By Lemma 1, the maximal ` must pass through
a vertex of H ∩ Qn , which corresponds with an edge of our cube Qn .
Now, we can assume that ` passes through a fixed point x := (x0 , 21 , . . . , 12 ), |x0 | ≤
1
2
on an edge of Qn and that ` is distance t away from the origin. We then
construct a hypercone with x as the apex and which is tangent to our sphere
Stn−1 . The height of this cone will travel through the origin and the point
−x = (−x0 , − 21 , . . . , − 12 ). Figure 7 shows this situation in three dimensions,
with x on a different (but symmetrically equivalent) edge.
15
Figure 7: A fixed point x with hypercone intersecting an opposite face
If t ≤ 12 , then for a hyperplane containing a face of Qn such that −x is at most
a distance
1
2
away (i.e. x is at least a distance
1
2
away, so the face is “opposite”
to x), then the intersection of the hyperplane containing our face of Qn with our
cone is an (n − 1)-dimensional ellipsoid. Furthermore, any one face of Qn cannot
contain an entire such ellipsoid, because the center of the ellipsoid falls either
directly on the boundary of the face (the center is −x when we intersect the
cone with the face containing −x), or projects to be completely outside of the
face (the line containing x and −x will not intersect the hyperplanes containing
other opposite faces on Qn , itself).
We project our vertex x onto the same face as our ellipsoid, to a point x0 .
Then, for a point y on the boundary of the ellipsoid, we have that
dist2 (x, y) = dist2 (x, x0 ) + dist2 (x0 , y)
by the Pythagorean theorem. Since dist2 (x, x0 ) is a constant, we need only to
16
maximize dist2 (x0 , y). Since x0 is guaranteed to lie outside of our ellipsoid, and
since an ellipsoid is a convex body, then we have that dist2 (x0 , y) is a convex
function. Thus, dist2 (x, y) and also dist(x, y) are convex functions. Therefore,
its maximum values occurring on the boundary, i.e. where the ellipsoid intersects
with an edge of Qn , and so we can now conclude that a line of maximal length
must pass through two distinct edges on Qn .
We now have two cases, which, upon further inspection, yield the same result.
Case I: The two edges are parallel.
The only such edge that does not share a face with x itself is the edge that
contains −x. So we consider the intersection of the two-dimensional plane that
is uniquely defined by these lines with Qn . This plane also contains the origin,
since these edges are opposite from each other on the cube, and thus intersects
Stn−1 in a great circle, which also has radius t.
The result is that our line ` passes through the two opposite of edges of side
length 1 in a 1 by
√
n − 1 rectangle, and is tangent to a circle of radius t that has
its center at the center of the rectangle. We can again set this rectangle on the
coordinate axes so that the centers of the rectangle and circle are at the origin,
and the rectangle has vertices (± 12 , ±
√
this rectangle at the points (x0 ,
√
n−1
2 )
n−1
2 ).
` thus intersects the boundary of
√
and (y0 , −
n−1
2 ).
As with our two-dimensional minimum problem, we can use basic calculus
to solve for the optimal length. Again, let θ be the angle between the positive
x-axis and the line perpendicular to `. By symmetry, we need only consider
h
i
θ ∈ 0, arcsin √1n − arcsin √2tn . Our length of intersection is a function of θ,
17
|` ∩ Qn | =
√
n − 1 sec θ. We find that the maximum of |` ∩ Qn | occurs at θ =
arcsin √1n − arcsin √2tn and the maximum distance is
√
n n−1
|` ∩ Qn | = p
.
(n − 1)(n − (2t)2 ) + 2t
This value for θ seems a bit complicated, but geometrically, it makes perfect
sense. As we can see in Figure 8, it forces ` to intersect the bottom side of the
rectangle exactly at the corner, which is also a corner of the cube Qn . since x
is a corner of the cube, then −x is the opposite corner of the cube, and thus
the other endpoint of this maximal ` could be located on any of the n edges
extending from −x.
Figure 8: The maximal ` for Case I
Case II: The two edges are skew.
In this case, we consider the three-dimensional space defined by these lines
and its intersection with Qn . The result is a 1 by 1 by
√
n − 2 rectangular prism.
18
Again, it will contain the origin, because the edge opposite x in the prism is the
one that contains −x, and thus, its intersection with our sphere Stn−1 is St2 , the
three-dimensional sphere of radius t. If we recenter it on the origin, we get a
rectangular prism with vertices (± 12 , ± 12 , ±
the boundary of the prism at (x0 , 12 ,
√
√
n−2
2 )
n−2
2 ),
(Figure 9). Our line ` intersects
and at one of (± 12 , y0 , −
symmetry they are equivalent, so we will consider ( 12 , y0 , −
Let D(x, y) be the distance between (x, 12 ,
√
n−2
2 )
√
√
n−2
2 ).
By
n−2
2 ).
and ( 12 , y, −
√
n−2
2 ).
We use
the method of Lagrange multipliers to optimize the function
1
1
D2 = (x − )2 + (y − )2 + n − 2
2
2
with the constraint that our line is t away from the origin, i.e.
|(x, 12 ,
√
n−2
2 )
× ( 12 , y, −
D2
√
n−2 2
2 )|
= t2 .
The only valid solutions that result are if either x = − 21 or y = − 12 . In either
case, our line fits into a two-dimensional plane that intersects our rectangular
prism in a rectangle exactly as above. Thus, our two cases are actually the
same case, and we validate that ` travels from a corner of the cube to an edge
containing the opposite corner, with length of intersection as listed above.
Now we may consider when
1
2
√
≤ t ≤
n
2 .
This time, intersecting our hy-
percone with those faces closest to x will result in paraboloids. By the same
arguments of convexity, ` must pass through two distinct edges of the cube,
19
Figure 9: The maximal ` for Case II
but this time, they must be on adjacent faces. A similar Lagrange multipliers
argument forces our point x to a corner of the cube, but this implies that x is
contained in any face to which is it adjacent, so our argument is reduced to an
(n − 1)-dimensional case, as shown in Figure 10.
Figure 10: The maximal ` is restricted to a face of the cube when t ≥
1
2
20
By the Pythagorean theorem, the effective radius in this (n − 1)-dimensional
q
case becomes t2 − 14 . By applying these values recursively, we can get an
explicit formula for the maximum length of intersection:
√
√ m−1 m
, the maximum value of
For 1 ≤ m ≤ n − 1, and for all t ∈
,
2
2
|` ∩ Qn | is
√
[n − (m − 1)] n − m
q
p
(n − m)[n − (2t)2 ] + 2 t2 −
√
and for t ∈
,
m−1
4
√ n−1 n
,
, the maximum value is
2
2
√
r
2−2
t2 −
n−2
.
4
Remark: Not surprisingly, the maximum of |` ∩ Qn | strictly decreases as t
increases. However, it is interesting to note that when viewed as a function of
√
t, the maximum of |` ∩ Qn | is continuous everywhere except at
n−1
2 ,
where
there is a jump discontinuity. It seems reasonable that this might happen with
other-dimensional intersections as well. Maybe intersecting a two-dimensional
√
plane with Qn will always yield a jump discontinuity in its maximum at
n−2
2 ,
or
maybe a maximal hyperplane will always experience a jump discontinuity at 21 .
Of course, there is also no guarantee that these would be the only discontinuities
as n increases, and as the dimension of the intersecting space increases.
CHAPTER 3
√
A SOLUTION FOR t >
n−1
2
We will now introduce a partial solution to our original proposed problem:
What are the optimal volumes of intersection of the cube, Qn and a hyperplane
H with unit normal vector ξ that is distance t > 0 from the origin?
√
For t >
n
2 ,
our answers are trivial. The points on Qn that are furthest away
√
from the origin are the corners, which are precisely at a distance of
n
2
away.
√
n
2
Thus, a hyperplane that is a distance greater than
away from the origin will
not intersect Qn at all, leaving both the maximum and minimum volumes of
intersection to be 0.
Theorem 3. Let H := ξ ⊥ + tξ be a hyperplane in Rn , where ξ ∈ Rn is a unit
vector and ξ ⊥ is the (n − 1)-dimensional subspace of Rn with ξ as its normal
√
vector and let
n−1
2
√
n
2 .
<t≤
Then,
nn
|H ∩ Qn | ≤
(n − 1)!
1 − 2t
2
n−1
,
with equality occurring when ξ = ± √1n , . . . , ± √1n .
√
Proof: Let
n−1
2
√
<t≤
n
2 .
It is important to understand the geometric im√
plications of this restriction. Again, we know that
√
origin to any corner of Qn . Similarly,
n−1
2
n
2
is the distance from the
is the distance from the origin to
any two-dimensional face (i.e. an edge) of Qn . Thus, if our hyperplane is further
21
22
away, it cannot contain any two-dimensional edge, nor can it separate two corners from the rest of the cube, as then it would also separate the edge connecting
them (Figure 11).
√
Figure 11: A possible intersecting hyperplane H with t ∈ (
√
n
n−1
,
2
2 ]
Each corner is the intersection of precisely n edges, so H cannot intersect
more than n edges. Furthermore, since H cannot be contained in an (n − 1)dimensional face, then it must intersect exactly n edges (or none at all) and
separate one corner from the rest of Qn (provided that it intersects Qn ).
Let a1 , a2 , . . . , an ∈ Rn be the points of intersection of H with each respective
edge extending from a particular corner (we need only consider one, due to
symmetry). For simplicity, we use the corner ( 21 , . . . , 12 ).
We note that each (( 12 , . . . , 12 ) − ai ) is orthogonal to each (( 12 , . . . , 12 ) − aj ) for
i 6= j at our corner ( 21 , . . . , 21 ). Thus, if we consider the convex hull of our points
( 12 , . . . , 21 ), a1 , . . . , an , we have an n-dimensional right simplex, where our corner
of Qn is the orthogonal vertex and H ∩ Qn is precisely the face opposite this
23
vertex. Note that we can consider this simplex to be an n-dimensional pyramid.
Let ai = |( 12 , . . . , 12 ) − ai | for 1 ≤ i ≤ n. That is, ai is the distance from the
orthogonal corner to the points of intersection of the edges of Qn and H. Let
Fi , 1 ≤ i ≤ n be the face of our n-simplex that is opposite from the vertex ai , as
shown in Figure 12.
Figure 12: ai , ai , and Fi near a corner of Q3
It is important to note that Fi is still an (n − 1)-dimensional right simplex,
with our same corner of Qn as the orthogonal corner. As a result it, is easy to
find the volume of our simplex.
We first recall that the n-dimensional volume of an n-dimensional pyramid
is given by |Pn | =
1
n |B|h,
where B is the base of the pyramid and h is the
perpendicular height to that base. If we consider one of our Fi to be the base,
then the perpendicular height of our simplex is exactly ai . Thus, the volume of
our simplex is
1
|Fi |ai .
n
24
However, since Fi is an (n − 1)-dimensional right simplex, we can again remove
a vertex aj and consider the base to be the face opposite aj , so the volume is
now
1
|Fi |ai aj .
n(n − 1) j
We proceed by induction and determine that the volume of our simplex is actually
n
1 Y
ai ,
n!
i=1
and that
|Fi | =
n
Y
1
aj .
(n − 1)!
j=1
j6=i
From [A], we have an n-dimensional analog of the Pythagorean theorem:
Lemma 2. For a right n-simplex with opposite face F and other faces F1 , . . . , Fn ,
|F |2 = |F1 |2 + · · · + |Fn |2 =
n
X
|Fi |2 .
i=1
In our case the opposite face is our intersection H ∩ Qn , so we have
2

|H ∩ Qn |2 =
n
X
i=1
|Fi |2 =
2 X
n
n
n Y
n
Y
X


1
1


a
=
a2j .
j
 (n − 1)!
(n − 1)!
j=1
j6=i
i=1
Thus, we seek to maximize our
n Y
n
X
i=1 j=1
j6=i
a2j .
i=1 j=1
j6=i
We can now solve for the distance to the origin in terms of our lengths
25
a1 , . . . , an . Consider that each coordinate of ai is 12 , except for the i-th coordinate, which is
1
2
− ai . We attempt to find our unit normal vector ξ.
Since ξ is normal to our hyperplane H, then for any vector u in our hyperplane, we will have that u · ξ = 0. Consider that ai is a vector in H for 1 ≤ i ≤ n.
Then ai − aj is also in H for 1 ≤ i, j ≤ n, i 6= j.
Thus, we know that (ai − aj ) · ξ = 0 for all i, j, so ai · ξ = aj · ξ. Since
most terms of ai , aj are
be ( 12 − ai ) −
1
2
components are
1
2,
we have that for ai = aj , the i-th component will
= −ai , the j-th component is
1
2
−
1
2
1
2
− ( 12 − aj ) = aj , and all other
= 0. Thus, we get
(ai − aj ) · ξ = 0
−ai ξi + aj ξj = 0
a j ξj = a i ξi
for all i 6= j. Thus, each ξi must be some constant multiple of
1
ai .
Since ξ is a
unit vector, we normalize, and we can see that
( a11 , a12 , . . . , a1n )
ξ = qP
n
1
i=1 a2i
Consider now that ai − tξ is also contained in H, and thus, (ai − tξ) · ξ = 0 for
all i. We simplify:
26
(ai − tξ) · ξ = 0
ai · ξ = tξ · ξ
Thus,
( a11 , . . . , a1n )
1
2
(1, . . . , 1, 1 − 2ai , 1, . . . , 1) · q
Pn 1 = t|ξ|
2
2
i=1 ai
1
a1
+ ··· +
1
ai−1
+
1
ai (1
2
− 2ai ) +
qP
n
1
1
ai+1
+ ··· +
1
an
=t
i=1 a2i
This finally yields that
P
n
1
i=1 ai
qP
n
2
i=1
−2
1
a2i
= t.
From here, we apply the method of Lagrange multipliers again, attempting
to optimize the function
f (a1 , . . . , an ) =
n Y
n
X
a2j ,
i=1 j=1
j6=i
with the constraint that our plane is still a distance t away from the origin, i.e.
2
P
( ni=1 a1i ) − 2
= t2 .
P
4 ni=1 a12
i
We find that the only solution to this problem is when a1 = a2 = · · · = an . This
yields that ξ = ( √1n , . . . , √1n ). That is, geometrically, our normal vector points
27
directly to our corner ( 12 , . . . , 12 ) (Figure 13).
√
Figure 13: The maximal hyperplane for t ∈ (
√
n
n−1
2 , 2 ]
By plugging in a := a1 = · · · = an into our constraint function for t above,
we find that
√
n
(1 − 2t),
2
a=
√
which yields that for t ∈ (
√
n
n−1
,
2
2 ],
v
v
u n
uX
n Y
n
uX
u
1
1
2
t
u
sup |H ∩ Qn | =
a
=
a2(n−1)
j
(n − 1)! t
(n − 1)!
ξ
i=1 j=1
j6=i
i=1
√
p
n
1
2(n−1)
=
na
=
a(n−1)
(n − 1)!
(n − 1)!
√
n−1
√
n
n
(1 − 2t)
=
(n − 1)!
2
nn
1 − 2t n−1
=
.
(n − 1)!
2
REFERENCES
[Al] S. A. Alvarez, Note on an n-dimensional Pythagorean theorem, unpublished note, www.cs.bc.edu/ alvarez/NDPyt.pdf (1997).
[Ba1] K. Ball, Cube slicing in Rn , Proc. Amer. Math. Soc. 97 (1986), 465-473.
[Ba2] K. Ball, Volumes of sections of cubes and related problems. Geometric
aspects of functional analysis (1987-88), 251-260.
[BGMN] F. Barthe, O. Guédon, S. Mendelson and A.Naor, A probabilistic approach to the geometry of the lpn -ball, Ann. Probab. 33 (2005), no.
2, 480-513.
[BK] F. Barthe, A. Koldobsky, Extremal slabs in the cube and the Laplace
transform. Adv. Math. 174 (2003), no. 1, 89-114.
[Ga] R.J. Gardner, Geometric tomography, Encyclopedia of Mathematics and
its Applications, Second Edition 58. Cambridge University Press, Cambridge, 2006.
[Ha] H. Hadwiger, Gitterperiodische Punktmengen und Isoperimetrie.
Monatsh. Math. 76, (1972), 410-418.
[He1] D. Hensley, Slicing the cube in Rn and probability. Proc. Amer. Math.
Soc. 73 (1979), no. 1, 95-100.
[He2] D. Hensley, A geometric inequality with applications to linear forms.
Pacific J. Math. 83 (1979), no. 2, 543-553.
[K] A. Koldobsky, Fourier analysis in convex geometry, Mathematical Surveys and Monographs, 116. American Mathematical Society, Providence,
RI, 2005.
[NP] F. Nazarov, A. Podkorytov, Ball, Haagerup, and distribution functions. Complex analysis, operators, and related topics, 247-267, Oper. Theory Adv. Appl., 113, Birkhauser, Basel, 2000.
[OP] K. Oleszkiewicz, A. Pelczyński, Polydisc slicing in Cn . Studia Math.
142 (2000), no. 3, 281-294.
[P] G. Pólya, Berechnung eines bestmimten Integrals, Math Ann. 74 (1913),
204-212.
[Va] J. D. Vaaler, A geometric inequality with applications to linear forms,
Pacific J. Math. 83 (1979), 543-553.
28
[W] R. Webster, Convexity, Oxford Science Publications. The Clarendon
Press, Oxford University Press, New York, 1994.
[Zo] Chuanming Zong, What is known about unit cubes, Bull. Amer. Math.
Soc. (N.S.) 42 (2005), no. 2, 181-211.
29