This dissertation has been
microfilmed exactly as received
69-18,404
HOYLMAN, Douglas John, 1943THE DENSEST LATTICE PACKING OF
TETRAHEDRA.
University of Arizona, Ph.D., 1969
Mathematics
University Microfilms, Inc., Ann Arbor, Michigan
TI!L: J)I:NSE5 T L A T T I C I : PACKING
OF TF.TUAHKWIA
by
Douglas liovlr.ian
A
Dissertation Subnitttid to tho Fncultv of the
D] U' ARTM!::;T
O F MATIII-.MATICS
In Partial Fulfillnent oT tho ^.cnuircnonts
-For the Dcr-roe of
DOCTOR OF Piill.nsOPFY
Tn the Graduate College
TI:N UXIVI-T.SIT V
]969
OF APJ:'.O::A
THE UNIVERSITY OF ARIZONA
GRADUATE COLLEGE
I hereby recommend that this dissertation prepared under my
direction by
entitled
Pougla3 Iloylman
The Densest Lattice Packing of Tetrahedra
be accepted as fulfilling the dissertation requirement of the
degree of
Doctor of Philosophy
/
Dissertation
tion Director
A a,si
Date
-1,
I '1 \j L )
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^7
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STATIV II:;:? BY AUTHOR
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ACJ^.'O1 ,l,KPr!KMKNTS
I wish to ciiinresa nv sincere appreciation to Vrofcssor
lielr.iut Groeir.cr for his help niul ndvlcc durinr. the pronaration of
this; pnpcjr.
My thanUs also to !-'r. Charles ?!erchant for drnwinf,
the. Illustration.
i ii
taih.e o r contents
Pn^.e
LTS'f OF 1LLUSTRATI01iS
v
ABSTRACT
vi
CHAPTER 1:
INTRODUCTION'
1
CHAPTER 2:
STATEMENT OF RESULTS
3
CHAPTER 3:
LEWAS
7
CHAPTER 6:
COMPUTATIONS
CHAPTER 5:
RESULTS
' . "7
J'I
114
REFERENCES
117
iv
LIST 0!' ILLUSTUAT TO'iS
Figure
1.
The cubooctfiiioilron
Pa^o
K
6
v
AUSHIACT
The greatest possible densities for lattice packinrs of
tetrahedra and cubooctahedra are found by the methods outlined bv
Minkowski..
I'e use the fact that a densest lattice packing for a
^iven convex body is tjivon by a critical lattice for the correspond­
ing, difference body.
tices
(-1,1,1),
If
T
(1,-1,1),
is the regular tetrahedron with ver­
(1,1,-1),
difference body is the cubooctahedron
eous inequalities
£ 4.
K
(-1,-1,-1),
then its
described by the simultan­
|v| <_ 2, |
z j <_ 2,
and
|>;| + |v[ -(-|
zf
According to the results of Minkowski, an admissible lattice is
critical for
C,
|xj <_ 2,
and
K
and either
+ A
if it has as a basis
A - 15,
R - C
are on the boundary of
and
K.
(A, P., C)
C - A
or
such that
A + R,
R + C
A,
R,
and
C
Thus we consider all possible ways in
which cither set of six lattice points just listed can be distributed
amonp, the fourteen faces of
K,
with the restrictions Imposed by re­
quiring that all other lattice points except the origin not be in­
terior to
K,
and in each case find the smallest possible value for
the determinant of
than
Hence
196/27,
1.96/27
A.
lie show that in no case is the determinant less
and find all lattices for which this value is attained.
is the determinant of a critical lattice.
show that the densest lattice packing of
T
has density
From this we
18/49,
that the densest lattice packinp, of cubooctahedra has density
and
45/49.
vii
Since any two tetirahcilrn arc af finely equivalent, the result
is obtained for an arbitrary tetrahedron.
1S//I9
CHArn-n 1
INTRODUCTION
Minkowski (1904) set forth the fundamental techniques for
determining the densest possible lattice, packing of a p.ivcn convex
body, once its difference body is known.
In the same paper he ap­
plied these techniques to the determination of the densest lattice
packing of a regular tetrahedron, a problem which had also been su,°,Rested by Hilbert (1901, p. 319), and obtained a maximum density of
5 = 9/38 - .236... .
Unfortunately this result is not valid, since Minkowski made
the incorrect assumption that the difference body of a regular tet­
rahedron was a repular octahedron.
Pepper (see Hancock, 1939) demon­
strated that the actual difference body is a cubooctahedron (Figure
1).
From this it follows that
6 <_ .4,
and other estimates have
been given by Groemer (1961) and Ropers (1964).
the result
Groemer (1962) p.ave
6 > 18/49 = .367... .
In this paper it will be shown that in fact
6 = 18/49.
All
lattices which give such a packing are found; these consist of re­
flections, rotations, and combinations thereof, of a single lattice.
The results will be formally stated in Chanter 2; Chapter 3 includes
definitions, preliminary results, and division of the problem into
1
2
cases; Chnpter 4 contains the analysis of the individual eases; and
the proofs of the results are completed in Chapter 5.
CIIAPTKR 2
•STATEMENT OF RESULTS
All of the following definitions concern three-dimensional
Euclidean space with fixed Cartesian coordinates.
Let
lattice
A,
A
B,
C
be three linearly independent vectors.
generated by
mA + nfi + pC
where
A,
n,
n,
B,
p
C
is the collection of all points
are integers.
is said to form a bar,is for the lattice.
one basis.)
The parallelepiped having
The system
A(A)
of
A
{A, 11, C}
(A lattice has more than
OA,
Oil,
OC
its edges is called a fundamental panjlLelepiped of
terminant
The
as three of
A.
The d_e-
is the volume of a fundamental parallelepiped.
From elementary geometry it is known that this is equal to the abso­
lute value of the determinant of the coefficients of
depends only on the lattice
A,
not on the basis
A,
B,
C.
{A, IS, C}
A(A)
(see,
e.g., Cassels, 1959).
A bcidy is a compact, convex set with interior points.
J
be a body and
and
of
A
a lattice.
J + A = {J + q: q
J
J + p
if, whenever
and
6(A, J)
J + a
p
E
A}.
and
V.'e write
We say that
q
J + q - {>: + q: x
J + A
J + A
are distinct points of
is the ratio
3
E
J}
is a lattice packing
have no interior points in cornon.
of the packing
Let
A,
the bodies
The densi tv
where
Vol(wJ)
represents the volume of
J.
It can be shown (sec Cassels, 1959),
p. 142) that there always exists a lattice
'
'
is a packing and that, whenever
J H- A
A
o
such that
J + A
o
is a packing,
iS(Ao> J) <_ 6(A, J).
Then
J + A
for
o
(1)
is called an extremal packing, and we'write
J)*
I'10 difference body of
J
is the body
6
(J)
max
J - J
= {x-y: xeJ, y cJ}.
Henceforth we shall use
havinj* vertices
and
K
(-1,1,1),
T
to denote the regular tetrahedron
(1,-1,1),
(1,1,-1),
and
(-1,-1,-1),
to denote the cuhooctahedron described by the simultaneous
inequalities
lxl H
(sec Fip.ttre 1).
|y| < 2,
Than
K
\z\ <_ 2
and
jx| -I- jv| + \?.\ <4
(2)
may be shown to be the difference body of
(see Hancock, 1939, or Groemer, 1962).
T
Mow the fundamental results of
this paper may be stated:
Theorem 1.
If
T
is the tetrahedron j
just described,
» and
A0
is the lattice havinj* the basis
A = (2, -1/3, -1/3),
then
6
B = (-1/3, 2, -1/3),
(T) = S(T, A ) - 18/49,
max
o
6(T, A) = 18/49, then
A
and"if
-*
A
C « (-1/3, -1/3, 2),
(3)
is anv lattice with
is essentially equivalent to
Aq
(i.e.,
either lattice nay be obtained from the other by reflection in a co­
ordinate plane, rotation by a multiple of 90° about a coordinate axis,
or a combination of these).
C^roHarv.
If
T*
If
K
is any tetrahedron (rcp.ulnr or not), then
6 V(T*) = 18/49.
max
Theorem 2.
Aq
6
is the cubooctnhedron described by (2), and
is as in Theorem 1, and if
2;.1(
53
(-210 = S(vK, A ) = 45/49, and if
max
o
'
= 45/49, then
A
{(>:,y,z):(2x,2v,2z)
A
E
K},
is any
lattice with
J
is essentially equivalent to
Aq.
then
<S(-;K, A)
iio.T[>otjirjoaoi;n:') mj.f,
•[ runil p.}
(z-'.o'z)
(I'-'c'u)
(J-'o-'U)
(u'z'u
UVu)
(E'Z-'O)
CHAPTER 3
LKMMAS
Let
A lattico
A
R
A
be miy body with the origin
is said to bo admissib]e for
is j^i_tical for
R
0
R
as an interior point.
if
Ar\int R = {0}.
if it is admissible and if no other admissible
lattice has a smaller determinant.
As previously noted, every body
has a critical lattice.
Lcnina 3..
A + J
Let
R
be the difference body of the body
is a packing if and only if
A
is admissible for
J.
Then
R.
This follows from sections 3 and 4 of Minkowski (1904).
sequently, a critical lattice for
Lemma 2.
A,
15,
C,
Tf
{A, 11. C}
gives a densest packing for
is a basis of the lattice
are on the boundary of the difference body
none of the lattice points
2A ± li ± C,
R
A ± 213 + C,
admissible for
A ± II,
Con­
A i C,
A ± B ± 2C
ii ± C,
A,
R,
J.
if
and if
A ± il ± C,
is interior to
R,
then
A
is
R.
This follows from scction 14 of Minkowski's paper with the
observation that
A,
R,
first kind if and only if
C
determine n lattice octahedron of the
{A, 15, C}
is a basis.
(The implication
in one direction is an immediate consequence of section 11.
versely, if
{A, B, C}
Con­
is a basis, then the octahedron with vertices
7
8
±A,
±B,
±C
consists of Clio sot of a]l points
|
|+|
r
s[ + |t| ^3,
than
0
Lemma__3.
a basis
If
{A, B, C}
with
ancl this can contain no lattice points other
and the vertices.
the first kind since
rA + sB + tC
So it is a lattice octahedron, and of
3;-(A + D + C)
A
is not a lattice point.)
is a critical lattice for
such that
A,
B,
(I)
A - B, B - C, C - A
or
(II)
A + II, B + C, C + A
all lie on the boundary of
C
U,
then
A
has
and either
R.
This result is piven in section 21 of Minkowski's paper.
I
have combined his cases (II) and (ITT).
Hence if
A
is a critical lattice for
K,
it must meet the
conditions of Lemma 3 and be such that none of the lattice points
listed in Lenrna 2 are interior to
K.
V.'e consider all lattices meet­
ing these conditions and find one with the smallest possible deter­
minant.
This lattice will pive the densest packing.
V.'e proceed to consider the difference body
K
K
in more detail.
is a cubooctahedron, a setnircp.ulnr polyhedron (see Fejes Toth, 1953)
with six square faces and einht triangular faces.
It may be obtained
by "cutting off the corners" of a cube, or of a regular octahedron, as
far as the midpoints of the edpes; hence the nane.
Tt is symmetric
with respect to reflections in all coordinate planes, all coordinate
axes, and the oripin, and with respect to rotations of 00° or ISO0
about all coordinate axes,
9
l.'e use the following, notation for the faces of
be noted that every point of bdry
K
x « 2,
SJ:
x = -2,
S2:
y = 2,
S^:
y = -2,
S3:
z = 2,
S^:
z «= -2,
Tx:
x -I- y + z - 4,
T|:
-x - v - x. - A,
x > -2,
y > -2,
T2:
-x + y + z = 4,
x > -2,
y < 2,
T^:
x - y - z = 4,
x < 2,
y > -2,
z > -2
T^:
x - y + z = A,
x < 2,
y > -2,
z < 2
T^:
~x + y - z = 4,
T^!:
(It should
is assigned to exactly one race.)
Sx:
:
K.
|y| -i- |z[ < 2
J y| + M < 2
|xj + |z| £ 2,
|x| < 2
|x| + [|
ss < 2,
|x| + |y| < 2,
|x| < 2
|x| < 2,
|x| + |y| < 2,
x + y - z - A,
x < 2,
|x| < 2,
y < 2,
x> -2,
x < 2,
-x - y + z = A, x > -2,
|y| < 2
|y| < 2
z < 2
y < 2,
v < 2,
(4)
z > -2
z < 2
z > -2
z > -2
y > -2,
z < 2.
The problem will be broken up into cases according to the
distribution of the basis vectors
A,
B,
C
among the faces of
IC.
The following lemma will reduce the niu.ibcr of possible cases con­
siderably.
Lemma A.
If
A
is admissible for
cannot lie in the same triangular face of
K,
K.
two lattice points
10
Prjoof.
Suppose that
P2 =
^X2'
P.y symmetry, v;e p.ay restrict attention to the face
Vy ?2 r. Ani^, ?x i- ?2, P]L *= (x^
y2' 7"l} *
x^,
y^,
follows that
*1
y^,
'X1 " X2 ^
z^,
<
2'
=
X2 + y2 + Z2 =
'*'
z2'
>*2'
x ,
x^,
nre
y2,
'yl " y21
z2
<
!X1 ~ *J
+
x2
yl
—
I yl ~ y2^
+
~
y2
IZ1 ~ ?'21
1*1 ~ 7"l I
x^ - x9>
2.
Prom (5) it
<
2
C6
'
y^ - y2,
- z2
'^'1cn "sing (5) we have
X1
~
x2 + yl
" y2 'h IZ1 "
= k2 - Zj +|st1 - ?.21 - To
z2 ^
or
l2Cst2 ~ Vi
Taking (6) and (7) together, wc sec that
terior point of
the origin.
K.
Hut
P^ - P2
A
P^ - P2
C7)
is an in­
is a lattice point distinct from
This contradicts the assumption that
Note that, since both
and
K
A
was admissible.
are symmetric about the
origin, we have as a corollary that two lattice points cannot lie
on opposite triangular faces
(1\
and
T|)
unless one lattice point
is the negative of the other.
The following well-known result from analysis will be useful.
Lemma 5.
bet
>
Assume without loss of gen­
—
=
^
arc all positive, hence
2'
have the same sign, unless one is zero.
x^ ~
'''
*ess than
Now at least two of the three numbers
erality that
7^),
*'C 'inve
X1 + yl +
and
'1^•
f(x., ..., x } be a real-valued function of
1
n'
n
real variables which is differentiable at every point of a closed and
11
bounded subset;
S
of
n-dimensional F.nclidean space.
sumes a minimum value on
S, and every minimum of
either at a critical point or on the boundary of
point of
f
f
Then
on
S.
S
f
as­
occurs
(A critical
is a point at which all first partial derivatives of
f
are simultaneously zero.)
Wc now enumerate all possible essentially different locations
of
A,
B,
C
on the faces of
K.
Lemma 4 and the symmetry of
K
will be used frequently.
a)
All on tho same square face.
The most general case is
1.
Ac S^,
b)
Let
A
B e S^,
C e S^.
Two on one square face, one on another square face.
and
R
be on
S^.
Then the face containing
C
may be either
or one of tho four square faces touching (having a common vertex
with)
S^.
But the latter four cases are essentially the sane, so we
have the two possibilities .
2.
A e S^,
D e Sp
C e
3.
A e
B e S^,
C c S2.
c)
Let
A
and
On three different square faces.
be on
S^.
If
B
is on
Sj,
then taking
C
on any of the
four remaining square faces gives essentially the sane situation, so
we may take
4.
A e
If
B
,
B E S^,
C E S2>
is on one of the other four square faces, we may take that to
12
be
Then Caking
C
that of
4,
and
and
li E Sj,
C E Sy
5.
A £ S^,
d)
Let
on
Sj
S^
or
gives n situation similar to
give symmetric, cases, so v.'e have
Two on ouo square face, one on a triangular face.
A
and
11
be on
S^.
Then the triangular facc must adjoin either
or
S|,
giving two essentially different cases represented by
6.
A e S^,
B e
C c T^,
7.
A e S^,
11 e S^,
C c T^.
e)
Let
and
Two on opposite square faccs, one on a triangular face.
A e S^,
B e S|.
Then all eight triangular faces give essentially
the same situation, so wo rnay take
8.
A
E
S1,
11 c S'v
f)
Take
C c T .
Two on touching square faces, one on a triangular face.
A e S^,
B e S^.
neither, or just one.
9.
A e S^,
B e Sj,
The triangular facc may adjoin both of these,
So we have
C E T^,
10.
A c S^,
D c S^,
C e T|,
11.
A e
B E
C E ^2*
,
g)
,
and
One on a square facc, two on (necessarily different) tri­
angular faces.
Let
A e S^.
If one of the triangular faces adjoins
B c T^.
Then as possible locations for
T^
T^;
and
T^;
four, cases arc
T£?
C
S^,
we may take
we have the symmetric pair
and the symmetric pair
T^
and
T^.
So the
13
12.
A
E
13.
A
E
14.
A
C
15.
A
E
IJ c T
l
C c Ti. ^ J
B E T
c
E T1
r.
c
E T
c
E T'
v
V
V
V
V
sa'
E
B E
V
V
2'
2
,
and
-
3'
If, on the other hand, neither triangular face adjoins
take
15 e T2,
ant^
this leaves the symmetric pair
T|,
as locations for
C,
so we have
16.
A e S1,
B e T2,
C
Tj, and
17.
A t
Re T2,
C e T|.
h)
E
we may
and
T^,
or
All in triangular faces.
If two of the faces have a common vertex, we take
Then the four choices
T^,
T^,
T/(,
Tj!
A c Tj,
Be T^.
Rive symmetric situations,
so we have
18.
A
E
Tlf
B
E
T2,
C
E
T3.
If no two of the faces are touching, we take
A c
three faces
P.
T„,
3
T. ,
4
T'
1
are available for
ant'
and
C.
Then all
three pairs of faces Rive symmetric situations, so we have
19.
A e T2,
B e T3,
C e T^.
only the
CHAPTER 4
COMPUTATIONS
By combining each of the nineteen eases enumerated at the end
of the previous chapter with the (I) or (II) of Lemma 3, we obtain
thirty-eif*ht cases.
In this chapter each case will be considered in­
dividually, and (if the ease can, in fact; occur) the smallest value
for the determinant of a latticc meeting the necessary conditions for
a critical lattice will be found.
for "det(A, B, C)",
V.'e will use "det" as an abbreviation
except where otherwise noted.
Then in each case
it will be shown that
|det[ > — = 7.259... .
It should be noted that the cases in which a particular lattice
point lies in a square face (these faces include their edf>es, except
for some vertices) are always treated before those in which the point
lies in a triangular face (these do not include their edp,es).
Thus,
when a strict inequality is included in the list of constraints on the
variables, it v/ill always be found that the points satisfying the cor­
responding equation have already been considered.
Ler.ma 5 will be ap­
plied to the closure of the region in
2-
n-(usually
or
3-)dimen-
sional space which is described by the applicable inequalities, and
on the boundaries described by strict inequalities it v/ill have already
14
15
been shown that
again.
|dct| >_ 196/27,
so that they need not bo considered
For example, in 6(1) we will often have
constraints describing the region.
minimum for
|det]
e < 2
as one of the
Tt will not be necessary to find n
at points for which
c = 2,
since we have already
considered such configurations in 1(1).
The method of division into subcases should be explained.
The
notation 3(1), for example, is used for the combination of case 3 of
the previous chapter with condition (I) of Lemma 3.
This case is di­
vided into cases 3(I)-1 and 3(I)-2; case. 3(I)— 2 is subdivided into
3(I)-2-l, 3(I)-2-2, 3(1)—2—3, and 3(I)-2-4, and some of these arc sub­
divided still further, usinp an obvious extension of this notation.
1(1).
A, 11, C c
A - Ji,
D - C,
In this case we see, from the description of
C - A c bdry K.
in (4), that
A,
II,
C
must be of the form:
A = (2, a, b)
where
|a| + [b| < 2;
B = (2, c, d)
where
|c| + |d| < 2;
C = (2, e, f)
where
|e] + |f| <_ 2.
Then we obtain
and
and
II — C — (0, c - e, d - f),
and each of these must be in
Since the intersection of
bdry K
formed by the four lines
y - ±2,
a - c = ±2
(9)
(10)
A - 15 = (0, a - c, b - d),
C - A = (0, e - a, f - b),
either
(8)
with the plane
?. - ±2
(11a)
or
x = 0
bdry K.
is the square
in that plane, we must have:
b - d = ±2;
(lib)
t*
ei ther
c -"if = ±2
(12a)
or
d - f = s.2: :
(12b)
either
e - a = ±2
(13a)
or
f - b = tl.
(13b)
16
Now it is not possible that all three equations in one column (with any
choices of signs) hold, since, e.g., adding (11a) and (12a) given
a - e = 4, 0
or
-4,
which contradicts (13a).
Thus it might be that:
two equations in one column hold, and one (possibly two) in-the. other.
Using the symmetry of the situation wc nay assume without loss of gen­
erality that
a - c = 2,
and
.
(11c)
e - a = ±2,
(13a)
d - f = ±2.
(12b)
We consider the two cases arising from the possible sirns in (13a).
1(I)-1.
e - a = 2.
Adding (11c) and (13c) gives
(9) and (10) requires that
e - c = 4.
e = 2,
(13c)
This with the inequalities
c = -2,
d - 0
and
f = 0.
But
this contradicts the assumption (12b).
1(I)-2.
e - a » -2.
(13d)
Then using (11c), (12b) and (13d) to eliminate
c,
e
and
f,
we
obtain
det -
1(H).
2
a
b
2
a - 2
d
2
a - a
d + 2
A, B, C e
= ± 8,
and
|det| >
A+B, B+C, C+Ae bdry k.
This situation is easily shown to be impossible.
as in 1(1).
Then
A + C = (4, a + c, fc + d),
because of tho condition
|x| <_ 2
196
27 *
in (2).
Let
A
and
Ti
which cannot be in
be
K
17
2(1).
A, B E S1(
This is also impossible.
Let
C - A = (—A, c - n, f - b),
dition
|x[ <_ 2
C c Sj,
A - B,
A = (2, n, b),
B - C,
C - A c bdry K.
C = (-2, c, f).
which cannot be in
K
Then
because of the con­
in (2).
2(11)
A, B e Sj,
C c Sj,
A + B,
B + C,
C + A e bdry K.
Thin may ba shown to be impossible exactly ns in 1(11).
3(1).
A, 11 e S^,
C
E
S2,
A - B,
B - C,
C - A e bdry K.
Apnin from (4), we obtain
A = (2, a, b),
where
|n| + |b| ^2;
15 = (2, c, d),
where
|c| + [dj
C = (e, 2, f),
where
|c| + |f| <_ 2
Then
2;
A - B « (0, a - c, b - d),
C - A = (e - 2, 2 - a, f - b).
(14)
and
and
(15)
|e|
2.
(16)
B - C = (2 - c, c - 2, d - f),
1
Since (2) requires that
and
[x| < 2
and
|y| < 2, 'wc must have
a > 0,
Nov/
A - 15 must be in
®3'
^2
in
A
an<^
and
^3'
B,
bdry K,
s^-ncG
*n
c > 0,
e > 0.
(17)
and the only possible faces are
no ot'icr
face can
x
be zero.
we need only consider the two cases
A - B e
By symnctry
nn^
A - B t S^.
3(1-3).
A - B E S2«
Then fron (A) v?e have
a - c = 2.
(18)
|
a j £ 2,
(19)
Now (14) yields
and the condition
B - C E K,
together with (4) and (IS), yields
18
J 2 - c[ =|
4 - a| _< 2.
(20)
But (19) and (20) can both be satisfied only if
a - 2.
(23)
Then (18) p.ives
c » 0,
-
(22)
and (14) Rives
b = 0.
Now
C - A = (e-2, 0, f)
e - 2 < 0,
must be in
(23)
bdry K,
the only possible faces for
In the latter two eases vc have
f = 2
C - A
and
either of which, together with (16), ?,ives
we obtain
e = 0
directly,
and sincc (16) p.ives
are
S|,
f = -2
e = 0;
respectively,
and if
this can be in
K
C - A.
C - A e-S^,
llencc v?e have
e = 0
for any location of
ant'
Then
(24)
H - C ~ (2, -2, d - f),
and by (2)
only if
d - f = 0.
(25)
Usinp, equations (21) through (25), we obtain
dot =
Nov?
A - I' - C = (0, 0, -2d)
terior to
K.
2
2
0
2
0
d
0
2
d
= -3d.
is a lattice point, thus cannot be in­
Hence from (2) wc must have
! d! >_ 1.
Then |
d o t f >_ 8
196
27 *
3(I)-2.
A - B e S3<
Then
b -d = 2,
(26)
19
and (14) and (15) give
hcnce
We must have
dition that
b < 2,
d^-2,
(27)
d
b ^ 0.
(28)
0,
B - C ~ (2 - e, c - 2, d - f)
C - A = (e - 2, 2 - a, f - b)
(2) and (26), Rives
in
bdry K.
be in
K,
-2<f-b = f- d-2,
Mow the con­
together with
or
d - f <_ 0.
(29)
lrrom (15) and (16) we also have
and
2 - e >_ 0
(30)
c - 2 <_ 0.
(31)
Then from (29), (30) and (31) it follows that
the four faces
S^,
S^,
and
Tj,.
IJ - C
must be in one of
Wo. consider the corresponding
four cases.
3(I)-2-l.
B - C c S^.
Then
e = 0,
(32)
and
det =
2
a
d + 2
2
c
d
0
2
f
Now consider the lattice points
and
= 8 - 2f(a - c).
(33)
A - B - C = (0, a - c - 2, 2 - f)
A-B+C=(0,a-c+2,2+f).
From (14), (15), (16) and (17)
we have
and
|a - c| <2
(34)
|
fI < 2.
(35)
20
Thus if
if
a - c
a - c
nnd
and
f
f
arc both positive, then
arc both negative, then
A - H - C e int It;
A - K + C c int K.
!5ut
neither of these is possible, since we have assumed chat the latticc
is admissible for
K.
Hence either
a - c
sif;n, or one is zero; and (33) yields
3(I)-2-2.
B - C e S^.
and
f
are opposite in
det ^ 8 > 196/27.
Then
c = 0,
e > 0,
(36)
and, usinp, (26),
2
a
d + 2
det = 2 0
e
d
2
C - A = (e - 2, 2-a, f - b)
= 8 + ade - 2af,
(37)
f
must be in bdry K,
and we have: from (16)
and (36),
-2 < e - 2 < 0:
(38)
0 < 2 - a < 2;
(39)
f < 2;
(40)
f - b < 2.
(41)
from (17) and (19),
from (16) and (36),
and from (28) and (40),
It follows from (38)-(41) that
S^,
anc*
^3*
is in one of the four faces
S2,
k'° consider the four resulting cases.
3(I)-2-2-l.
det = 8 > 196/27.
C - A
C - A g ^2*
Then
a = 0,
so from (37) wo have
21
3(1)— 2—2—2.
gives
f = d,
C - A E S^.
Then
and from (37) we have
f - b - -2,
which with (26)
dot = 8 4- ad(e - 2) >_ 8 > 196/27
by (17), (28) and (36).
3(1>-2-2-3,
C - A
c
T .
Then
2 - e + 2 - a + f - b = 4,
or
a + b + e = f.
Now
H - C - (2 - e, -2, d - f)
is in
1C,
(42)
so (2) implies that
2 - e + 2+F-d<_4..
(A3)
But (42), (43) and (26) together yield
a
which contradicts (17).
3(1)-2-2-4.
-2,
Hence this case cannot occur.
C - A e T^.
Then from (4) we have
2 - e + 2 - a + b - f = 4,
2 - e < 2,
a - 2 > -2,
and
(44)
or
b = a + e + f;
(45)
c > 0;
(46)
a > 0;
(47)
or
or
f - b > -2,
which with (45) gives
a + e < 2.
(48)
e - f < 2,
(49)
a + e + f > 0,
(50)
We also obtain, from (16)
from (27) and (45),
22
and from (14) and (45),
2a + e + f < 2.
(51)
The equations (26), (36) and (45) nake it possible to write
and
C
in terms of
a,
e
and
f.
A,
1'
If will now be shown that the
inequalities (46)-(51), with one additional condition, are sufficient
to ensure the conditions of Ler.ina 2.
M < 2,
|y|<2, |
z
|< 2 ,
A « (2, a,
and
a + e + f).
Recall that
|x| + |y| + |zj < 4}.
Then
x = 2
by inspection, and
|y| + jz| = |a| + |a + c + f| = a + ja + e + f|
a + a + e + f < 2
A c
by (49) and (46),
c bdry K .
]J = (2, 0, a + e + f - 2).
Then
x = 2,
=ja+e+f-2|;
a + e + f - 2 <_ -a < 0
a + e + f - 2 >_ -2
by (50).
C = (e, 2, f).
by (46):
(49).
by (47)
by (51) or
a - (a + e + f) = -e - f <_ 2 - 2e < 2
so
int K - {(x, y, z):
Then
Hence
Hence
|x| + |z| ^2,
(47) and (48) respectively.
and
Thus
2
by (46).
c bdry K.
and
e - f < 2
since
since
x - 4 > 2.
|z| = 2.
since
by
by (46),
C z Sj c hdry K.
A + C = (2 + e, 2+ a, a + e + 2f) { int K
>
II c
|x|s=|o|-e<a + e<2
A + II - (4, a, 2a + 2e + 2f - 2) i int K
A - B = (0, a, -2) I int K
and
|x| + ]z| = je] -t- |f| - e H- |f|
by (47) and (51),
and
|y| 4- \z\
by (51) and (47), and
|y| + jz| <_ 2,
y = 2,
e + f < 2a -f- e + f <_ 2
and
x = 2 + e
23
A - C = (2 - c, a - 2, a
= 12 — q j -I-|
a - 21 +|
n
e) t int K
since
|x| + |y| + jz|
|- 2 - e + 2 - a + a -I- e = 4
e
by (46),
(47) and (4S).
A - 15 + C = (e, n + 2, 2 + f) 4 int 1C
since
y -• a H- 2 > 2
by (47).
A - II - C = (-c, a - 2, 2 - f)
forces the additional condition.
-2 < ~e < 0,
|
y
| <2,
-2 < a - 2 < 0
y. > -2,
and
A - B - C will be in
(In the first cane,
is the lattice, point which
Since the six inequalities p.ive
and
2 - f > 0,
we have
|x| < 2,
|x| + |y| H- |z| «e + 2- a + 2 - f,
int K
unlef.fi
either
f <_ 0
z >_ 2 1
and in the second,
or
so that
e >_ a + f.
Hence we assume (52) as a seventh condition on
(52)
|x| + |y| +|
?.| ^4.)
A,
It
and
C.
2A + B + C - (6 + e, 2a + 2, 3a+3c + 4f - 2) i int K
x=6 + e>6>2
since
by (46).
2A + II - C = (6 - e, 2a - 2, 3a H- 3e + 2f - 2) i int K
x=6-c>6-(a + e)>4>2
by (47) and (48).
2A - R + C = (2 + e, 2a + 2, a + e + 2f + 2) i int K
x = 2 + e > 2
since
since
by (46).
2A - B ~ C = (2 - e, 2a - 2, a + e *1- 2) { int K
z = a + e-H2>2
since
by (46) and (47).
A + 213 + C = (6 -I* e, a + 2, 3a + 3e + 4f - 4) i int K
x = 6 H* e > 6 > 2
by (46).
since
24
A + 211 - C - (6 - c, a - 2, 3a + 3c + 2f - 4) i int K
x =-6 - c > 6 - (a + e) > 4 > 2
by (47) and (48).
2, 4 - a - e) i int K
A - 2B + C = (c - 2, a
y = a + 2 > 2
since
sincc
by (47).
A - 2B - C ~ (-2 - e, a - 2, 4 - a - e - 2f) i int K
|x| = 2 + e > 2
since
by (46).
A + B + 2C = (4 + 2e, a + 4, 2a + 2e + 4f - 2) i int K
x = 4 + 2c > 4 > 2
by (46).
A + B - 2C = (4 - 2c, a - 4, 2a + 2c - 2) { int K
|y| = 4 - a > 4 - (n *t- e) > 2
since
by (46) and (48).
A - B + 2C = (2e, a + 4, 2 4- 2f) t int K
> 4 > 2
since
since
y = a + 4
by (47).
A - B - 2C = (-2c, a - 4, 2 - 2f) i int K
> 4 - (a + e) > 2
since
[y| « 4 - a
by (46) and (48).
Hence any lattice with basis
A,
]},
C
as in (14)-(16) vrherc
the variables meet the conditions (26), (36), (45), (46)-(52)
missible.
To find the minimum value of det under these conditions, we
will use the theorem from analysis mentioned in Chapter 3.
det =
is ad­
2
a
a + o + f
2
0
a + c + f - 2
e
2
f
is a function of
a,
e,
and
Then
2
2
= 8 + a e + aef 4- ac - 2ae - 2af (53)
f,
to be minimized an the region deg
scribcd by (46)-(52).
only the solutions
Since the equation
a = 0
and
e -• 2,
det = ac - 2a = 0
has
there are no critical points
25
iri the Interior of this region.
The faces
e = 0,
a = 0,
Hence the minimum occurs on a boundary.
a + e = 2
which arc not included in the re­
gion 3end immediately to previous cases C3(I)— 2—1, 3(f.)—2—2—1» and
3(I)-2-2-2
respectively) and need not be considered again.
We now con­
sider the remaining faces, arising from (49)-(52).
3(I)-2-2-4-l.
e - f = 2.
Then, eliminating
f,
we obtain
fxom 1[53)
dot - 8 H- a^e -I* 2ae2 - 6ae + 4a,
(54)
from (50)
a + 2c > 2,
(55)
from (51) 2a + 2e <_ 4,
which is redundant in view of (45), and from
(52) either
a
e <_ 2
(47) and (48),
or
2,
both of which wo already have from (46),
lience we must now minimize the right side of (54) on
the region described by (46)-(48) and (55).
Testing for critical points,
we obtain the conditions
— det = 2ae + 2e2 - 6e + 4 = 0
o tl
(56)
and
~ det = a2 + 4ae - 6a = 0.
3e
(57)
lri view of (47), we may write (57) as
a + 4e = 6,
and using (58) to eliminate
a
(58)
fro:n (56) gives
3e2 - 3e - 2 = 0.
(59)
Because of (46) we need to consider only the positive solution of (59),
ij \ 4 —
3 + /33
which is e =
r
.
— O
,
/CQ\
Ihen from (58),
a
18 ~ 2/33"
,
J
. /c/\
and (54)
26
. „
__
22/33 n n, s 196
det - 22 - —--—> 7.95 > -^y-.
gives
The only edge which must be considered is
a + 2e - 2.
(60)
Substituting (CO) in (54) gives dot = 16 - lGe
>_ S,
so that
dot > 7.95
3(I)— 2—2—4—2.
(53), we obtain
on the edge
3(I)-2-2-4~3.
2
2
8 + 8(1 - c) *
e - f - 2.
a + c + f - 0.
det = 8 + 2a
2
8c '
Then, eliminating
f
from
^,8.
2a + e + f = 2.
Then (53) becomes
det = 8 - 4a + lit?" + 2ac - a^e;
(49) yields
yields
2a + 2e
a < 2,
4,
(61)
which is redundant in view of (48); (50)
which is already required by (46), (47) and (48); and
(52) becomes
either
2a + e
2
or
a + 2e
2.
(62)
Thus the region for which (61) must be minimized is that described by
(46), (47), (48) and (62).
Testing for critical points, we obtain
dc
which has the solutions
det = 2a - a2 = 0,
a = 0
and
a = 2.
(63)
Hence there are no in­
terior critical points, and the minimum occurs otv the boundary.
The
only edges which must be considered are those determined by (62);
2a + e = 2
and
a + 2e = 2.
3(I)-2-2-4-3~l.
2a + e = 2.
Substitution in (61) gives
det = 8 - 2a2 + 4a3,
(64)
a < 1.
(65)
and (46) yields
27
Now
d
2
det - - 4a + 12a « 0
da
has the solutions
so that the minimum of det on the interval
the three values
= 214/27,
and
0,
1/3,
Ko1.;
1.
a « 1 ^ del = 10,
0
a = 0
a
1
a = 0 «> dot = 8,
so the minimum is
ant]
a = 3/3,
occurs at one of
n = 1/3 -'dot
214/27,
which is
greater than 196/27.
3(I)-2-2-4-3-2.
a + 2e = 2.
Substitution in (61) pives
det « 8 - 2a + 2a2 + |a3,
and cither (46) or (48) yields
a < 2.
has the positive solution
352 - 56/7 >
_ 203
jjy"jj"
, t
,,
0
a = 2, det = 16,
(67)
a - ——and at this value (66) p,ives
196
~xj"
>
Nov/
_2 + l\a + |a2 = 0
l
-i.
da
det =
(66)
_.
®ince
„
.
a = 0,
at
, ,
0
det = 8,
and at
,
. .
, f
.
, , 352 - 56/7
we have a minimum determinant of
^
.
3(I)-2-2-4-4.
f - 0.
Then (53) pives
2
2
det = 8 + a e + ne - 2ae
(63)
and (51) becomes
2a + e
2,
(69)
(48) and the relations arising from (49) and (50) are redundant, and
the region of definition is p,iven by (46), (47) and (69).
Testing
for critical points, ve obtain
—• det = 2ae + e2 - 2e = 0
da
(70)
and
3e
det = a2 + 2ae - 2a = 0.
Subtracting (71) from (70) pives
e
2
2
- a
- 2e + 2a = 0,
(e - a)(e + a - 2) = 0.
(71)
or
(72)
28
Since (47) and (69) p.ive
e + a - 2 < 0,
(72) reduces to
e = a,
and
substitution in (70) p,ives
3c2 - 2e = 0.
(73)
liy (46), the only acceptable solution of (73) is
det = 208/27
at the critical point.
e = 2/3,
which p.ives
The edge arising from (69) was
considered in 3(IJ-2-2-4-3-1.
3(I)-2-2-4-5,
e = a + f.
Then (53) gives
dot = 8 + 2ae^ + 2a^ - 4ae,
(74)
and (51) yields
a + 2e
2;
(75)
this with (46) and (47) describes the region of admissibility.
Testing
for critical points, we have
~~ det = 2e^ + 4a - 4e - 0
3a
(76)
~ det = 4ae - 4a = 0.
(77)
and
Since
a ^0
(76) gives
by (47), (77)- ^ives
e = 1.
a = 1/2, and we obtain
Substituting this value in
det «* 7.5.
The cdfie arising from
(75) was considered in 3(l)-2-2-4-3-2.
3(I)-2-3.
B - C e S^.
Then, by the description of
in-"
(4), we have
d - f = 2,
and
(78)
e > 0
(79)
c > 0.
(80)
e > 2
(81)
We also have
29
by (16),
0< a< 2
<82)
by (14) and (.17), and
b >_ 0.
Than C - A «= (e - 2, a - 2, 0),
by (79) and (82), have
det =
a = 0.
(28b)
and for this to be in
K,
we must,
Hence we obtain
2
0
b
a
c
b - 2
e
2
b
= 8 + bc(2 - e) > 8 >
196
27
because of (80), (81) and (28b).
B - C e T^,
3(I)-2-/,.
- d - 4,
2-e+2-c + f
or
2 - e < 2,
c + d + e = f;
(83)
e > 0;
(84)
c > 0;
(85)
or
c - 2 > -2,
and
Then (A) p.ives:
or
d - f > -2,
which with (83) p,ives
c + e < 2. .
We also have
that for
a >_ 0 from (17).
(86)
Then from these inequalities we see
C - A => (e - 2, 2 - a, c + e - 2)
be in either
or
3(I)-2-4-l.
T^.
to be in
This plves two cases.
C - A e; Sj.
Then
a = 0,
and
bdrv K,
it must
30
2
0
d H- 2
dct = 2 0
g
2
d
- 8 + 2c
+ 2cd - cde.
(87)
c + d + e
The conditions (84), (85), (86),
d < 0
and
(2Sa)
c - d < 2,
(88)
the latter obtained from (15), are sufficient to ensure the hypotheses
of Lemma 2; this may bo verified by the straightforward but tedious
process employed in case 3(I)-2-2-4,
Hence we may consider the values
of det on the region described by these five inequalities.
Testing for
interior critical points, we see from
9
det = -cd = 0
3e
that there are none.
The edp.es
c = 0,
(89)
e = 0,
c + a - 2
cases previously considered, and v:e need only look at
lead to
d = 0
and
c - d
= 2.
3(I)-2-4-l-l.
d- = 0.
Then (87) becones
3(I)-2-4-1-2.
c - d - 2.
det = 8 - 4c + 4c
det = 8 + 2c
>8.
Then (87) becomes
2
2
+ 2ce - c e,
and the region to be considered is described by (84)-(S6).
(90)
There are
no interior critical points> since
8e
has only
c - 0
and
c = 2
det = 2c - c^ = 0
as solutions,
(91)
lience the minimum occurs
at a boundary point, and these have all been considered previously.
31
3(
- e = 4,
I
)
.
C - A c T ^ .
T h e n v?c h a v e
2 - c + 2 - n + 2 - e
or
n + c + 2o = 2:
and
2 - a < 2,
(92)
which with (92) pives
c + 2e < 2.
From (14) we have
a + b <_ 2,
(93)
and substituting from (26) anc! (92) gives
c + 2e > 2 + d.
(94)
c - d < 2
(88)
We may obtain the inequality
as in the previous case.
Then (84), (85), (8S), (93) and (94)
are suf­
ficient to describe a region in uhich the lattice is admissible, as may
be shovm by checking the lattice, points listed In T.euma 2.
2
2 - c - 2e
d + 2
= 8 - 4c - 4d - 4e + 4c2 + 4c2 (95)
c
det -
We have
2
c + d + e
+ 4cd + 6ce + 6de - 2de
2
- 2cde.
Testing, for critical points in the interior of the repion, v.*e obtain
1 a- dot
2 9c
- -2 + 4c + 2d -(• 3e - de =.0,
1 3_
det = -2 + 2c + 3e - c
2 3d
and
- ce ~ 0,
v v~ det = -2 + 4c + 3c + 3d - 2de - cd = 0.
I oe
Solving (96) for
d
(96)
(97)
(98)
yields
4c .-f- 3e - 2
i " —-e"--2
•
(99)
and substitution of (99) into (9S) qives
2e2 + 8ce + 4c2 - 8c - 3e + 2 = 0.
(100)
32
Solving (97) for
c
gives the two solutions
e - 2,
which will not
yield an interior point of the region, and
e - 1 - c.
(101)
Substituting (101) into (100) yields
2c2 + c - 1 = 0.
(102)
The only solution of (102) satisfying (85) is
then obtain
e = 1/2,
and front (99),
c = 1/2.
d = -1,
From (101) vc
These arc then the co­
ordinates on the only interior critical point, and at this point
det - 7.5.
= 2 + d,
We must also consider the faces
c - d - 2
and
c + 2e
arising from (88) and (94) respectively.
3(I)-2~4-2-l.
c - d = 2,
Then elimination of
d
from (95)
pives
det = 16 - 16c - 16e + 8c2 + 8c2 + 16ce - 2c2e - 2ce2,
(103)
and the region to be considered is described by (84), (85) and (93).
Searching for critical points, we obtain the equations
and
~
det = -8 + 8c + 8c - e2 - 2ce - 0
l 3c
(104)
™
det = -8 + 8e + 8c - c2 - 2ce - 0.
2. 3 e
(105)
Subtracting (105) from (104) pives
c2 = e2,
006)
and in view of (84) and (85) we must have
c = e.
Then (104) pives
3c2 - 16c + 8 = 0.
The only solution of (107) which is less than
c ~ e =
and putting (10S) into (103) pives
(107)
2
is
T
8 - 2/ O
^
,
,ino^
(108)
/To
,
2224 - 640
det =
^
11
200
"Jf
>
196
~27~*
33
All boundary portions
of the region in question have been previously
considered.
3(I)-2-4-?.-2,
c + 2c = 2 + d.
.Substituting for
d
in (95)
p.ives
det « 16 - lGc - 24e + 8c2 + 20e2 + 24ce - 6ce2 - 2c2e - 4e3, (109)
and again the region of admissibility is described by (84), (85) and
(93).
Testing for critical points Rives the equations
IJi 9 c
and
det = -S + 8c + 12c - 3e2 - 2ce « 0
(110)
™
det = -12 + 20e + 12c - 6ce - c2 - 6e2 = 0.
/ 9c
Solvinp (110) for
c
in terns of
e
(Ill)
and substituting into (111) yields
the equation
F(e) = 3e* - 16e3 - 28e2 + 176c - 96 = 0.
(112)
By (S4), (85) and (93), we need only look for solutions of this equa­
tion between
0
and
1.
Since
F(0) < 0
and
either one or three solutions of (132) between
were three solutions, then
in that interval, and
F(1) > 0,
0
and
1.
there are
If there
F'(e) = 0 -would have at least two solutions
F,r(e) = 0,
at least one.
But
F"(e) « 36e2 - 96e - 56,
and the solutions of
outside the interval.
the interval.
FM(e) = 0
are
(113)
(A ± /30)/3,
both of which are
We conclude that (112) has only one solution in
This solution was found by Horner's method to be in the
range
0.6280 < e < 0.6281.
Fron (110) we then obtain
0.25S3 < c < 0.2592,
(114)
and (109) pives
34
det > 7.4100 >
]96
Ap.ain all boundary portions have been previously
considered.
3(11).
A e S1,
B c Sj,
C c S2,
A + 1J,
B + C,
C 4- A E bdry K.
This may be shown to be inpor.sible just nr. was 1(11).
4(1).
A e
E c q'
1,
,
C c «2,
This is impossible.
Let
= (4, a - c, b - d)
which cannot be in
|
x| _< 2
A - 15,
A = (2, a, b),
IJ - C,
C - A E bdry K.
R = (-2, c, d).
K
Then
A - Pi
because of the condition
in (2).
4(11).
A E S1,
11 E Sjj,
C E S2,
A + B,
B + C,
C + A
c
bdry U.
Then, from the description (4) of the faces, VJO have
A = (2, a, b),
B = (-2, c, d),
C = (e, 2, f),
Then
where
|a| + |b| <_ 2;
where
|c| H- |d|
where
2;
|c| -1- |fj < 2
A + B » (0, a + c, b + d),
C + A = (e + 2, a + 2, b + f).
(115)
and
and
(116)
|e] < 2.
(117)
B + C = (e - 2, c + 2, d + f),
For all of these to be in
K,
and
we
clearly must have
a <_ 0,
Then
A + IJ
r.:ust be in
c < 0
S^,
and
or
S^.
e = 0.
(118)
anc'
But
^3
P .ive
symmetric configurations, so that we need consider only the two canes
and
S^.
4(II)-l.
A + B E
Then
2
det =
-2
0
a
a + c = -2,
b
-a - 2 J
2
and
f
= -4(b + d + f).
(119)
35
Nov
A + D + C = (0, 0, b + d + f)
so we r.uiRt hnve
>
j b + d + f| >_ 2,
cannot; bo nn interior point of
K,
and from Q19) we have |
d c 11 >_ y
196
2 7"'
A(II)-2.
A + E E S3>
Then
b + d = 2,
(120)
and
det =
2
a
2 - d
-2
c
d
0
2
f
= -8 + 2f(a + c)
liy (115), (116) and (118), v.'c have
-2 <_ a + c + 2 <_ 2.
Since
(122)
A + K + C - (0, a+c+2, f+2) cannot be interior to
must have either equality on one side of (122), or
f >_ 0.
K,
we
Hence there
are three possible cases to consider.
4(II)-2-l.
a + c + 2 = -2.
and by (115) and (116) we nust have
Then we nust have
b = d = 0.
a = c = -2,
Fiut this contradicts
(120).
4(II)—2—2.
(121) skives
a + c
f >_ 0.
|det| >_ S >
50).
E bdry K.
Then by (lift)a = c = 0,
and
|det| = 8 >
4(II)-2-3.
have
2 = 2.
Then, since
a + c <_ 0
by (IIP,), we
fron (121).
A E S^,
In casje
_5
B e $2»
C
E
^3'
alone v.'c sjiall
A - 31,
n - C,
C - A
all souare _fac_es_ as closed.
Because of the c.ain in syr.inetry, this assumption will elir.iinate more
repetition than it causes.
Then we have
36
A = (2, a, b),
where
|n[ + |bj
B = (c, 2, d),
where
|c[ +|
d| £ 2,
C = (e, f, 2),
whore
|e| + |f| <2.
From these we obtain
2,
(123)
and
(12A)
(125)
A - B = (2 - c, a - 2, b - d),
• (c - e, 2 - f, d - 2),
three points to be in
and
K,
B - C
C - A = (e - 2, f - a, 2 - b).
For all
we must by (2) have
a, b, c, d, c, f >_ 0.
Then, since by (123) and (1.24) we have
A - B
must be in one of the faces
But
x
and
and
y),
A - li e S^.
-c
0,
S2,
or
f.~ 0.
a - 2 _<_ 0,
S^,
S^,
T^
S^,
Then
2 - f >_ 0,
A
or
and
T^.
15,
c = 0,
and
and
B - C = (-e, 2 - f,
d - 2 <_ 0,
we must have
T£.
B - C e R|.
5(I)-1-1.
have
and
so we have only five cases to consider.
Since
B - C c Sj,
2 - c >_ 0
pive svranctric situations (interchange
5(I)-1.
d - 2).
(126)
Then
e = 2,
and fron (125) we r>iust
Hence this is a special instance of the situation in
5(I)-l-2.
B - C e S2.
5(I)~1~2.
-a, 2 - b),
Since
C - A c Sj,
S^,
e - 2 <_ 0,
S3
5(I)-l-2-l.
or
Then
-a _< 0
f = 0,
and
and
2 - b > 0,
Tj.
C - A E S|.
dot =
Then
2
a
b
0
2
d
0
0
2
C - A - (e - 2,
e = 0,
= 8 > 196
27 *
and
we must liave
37
5(I)-l-2-2.
b = 0,
C - A c S|.
Then
a = 2,
so from (123)
and
det =
2
2
0
0
2
d
e
0
2
- 8 + 2de > 8 >
196
27
of (126).
5(I)-l-2~3.
C - A c S
det ~
2
a
0
0
2
d
e
0
2
= 8 + ade > S >
196
"2 7
by (126)
5(I)-l~2-/i.
2 - e < 2,
a < 2,
and
C - A e TjJ.
Then
2 - c + a + 2 - b = 4,
or
a = b + e;
(127)
e > 0;
(128)
which gives
which with (127) becomes
2 - b < 2,
b + e < 2;
(129)
b > 0.
(130)
or
From (124) and (126) we also have
0 < d
2.
(131)
Then
det =
2
b + c
b
0
2
d
e
0
2
- 8 + bde + de
- 2be,
(132)
38
and this must be minimized on the region described by (128)-(13])
Testing for critical points, v:c obtain
(133)
tv" det - 2de - 2e ~ 0,
oh
which has only the solutions
terior critical points.
e = 0
and
so there arc no in­
The only faccs of this repion which do not
lead immediately to previous canes are
comes
cl = 2,
det ~ 8 + 2e^ > 8 >
and
d = 2,
d = 0.
for which (132) be­
In the latter case (132)
flives
det = 8 - 2be.
Now the lattice point
in the interior of
(134)
A - B - C = (2 - e, b + e - 2, b - 2)
K,
cannot be
and in view of (128)-(130) wc must: then have
2 - e + 2 - b - e - t - 2 - b ^ 4 ,
which reduces to'
b + G < 1.
(135)
Then, from the well-known inequality between the arithnetic and peometric neans,
e <
we obtain
be
1/4,
5(l)-l-3.
b + e
(136)
which with (134) £ives
B - I
det =
s1
3"
Then
d = 0,
and
•
2
a
b
0
2
0
e
f
2
- b)
-jj- > '-pf1 *
det
= 8 - 2be.
must be in
r.ust have
(137)
bdry K,
C + A c S|,
S2»
and since
S2*
S3'
T2
30
or
,
anil there arc six ca:;cs to consider.
5(I)-l-3-l.
a
C - A e S£.
Then
e
0,
and from (137)
C - A e S^.
Then
F - a - 2.
c
196
dot = 8 > -£y.
5(I)-l-3~2.
ables arc between
• (125) .".ives
0
e. = 0,
and
f = 0,
we nust have
and (137) "ives
5(T.)-l-3-3.
have
2,
a - 2.
C - A e S^.
a = 0
det = fi >
Then
Then (123) rives
and
f - 2.
Then
1 06
f - a = -2,
b = 0,
Since all vari­
and
nnd we r.uiRt
106
dot = R > ™-
fron (137).
5(I)-l--3-'t.
C - A e S^.
Then
b = 0,
and (3 37) rives
C - A e T2.
Then
2 - e -I- f - a + 2 - b « A,
det = 8 > ~|y.
5(I)-l-3-5.
or
a + b + e = f:
(138)
e > 0;
(139)
c - 2 > -2,' or
f - a < 2,
and
Nov;
which with (133) Rives
2 - b < 2,
b + e < 2:
(]^(0)
b > 0.
(l'rt)
or
A + T. - C = (2-e, 2-b-e, b-2),
5(I)--l-2-4, obtaining
5(1)-1-3-6.
det >
1
and we r:av proceed as in
106
C - A e Tj.
Then
2-e + a- f + 2 - b M,
or
b + e + f - n.
(1A 2)
40
We obtain the inequalities (139)-(I'll) nr. before.
= (2 - e, b + e - 2, b - 2),
t, • •
. „. 15
obtainmc
det
5(l)-4-4.
Mow
A - II - C
and we may an.ain proceed as in 5(l)-l-2-4,
196
27"'
B - C c T^.
Then
e + 2 - f + 2 - d = 4,
e = d + f;
e < 2,
(143)
which with (143) gives
d + f < 2;
2 - f < 2,
and
or
'
or
2 - d < 2,
f > 0;
(145)
d > 0.
(146)
or
C - A = (d + f - 2, f - a, 2 - b )
(144)-(146) v.'e have
5(I)-l-4-l,
bdrv K.
anc'
^4*
C - A
C - A c
.a
0
d + f
f
arc
Then
S^,
b » 0,
0 <_ 2 - b
and
0
2d
With the chanpe of variables
+ ad
+ adf - 2df.
(147)
2
e ->- d,
d -> a,
to the situation in 5(I)-l-2-4 (except that
b •> f,
C - A e T2.
Then
this is identical
A + B - C,
must be considered), and we obtain
5(l)-l-4-2.
By (123) and
and
2
det =
must be in
-2<d + f- 2<0, -2 < f - a < 2
so the only possible faces for
A - 1) - C,
(144)
rather than
det ^
2-d-f+f-a+2-b«4
or
a + b + d = 0.
(148)
41
ISut (14S) contradicts (146) and (126).
5(I)-1-4-3.
- b = 4,
C - A t T '
4
So this case is impossible.
Then
2 - d - f + n - f + 2
or
a » b + d + 2 f ;
and
2 - b < 2,
(149)
or
b > 0.
(150)
We also have
d,f > 0
have
which by (149) nay be v.'rittcn
a < 2,
from (145) and (146).
From (123) and (150) v:e
h + d + 2f < 2.
051)
then
2
det =
b + d + 2f
0
2
d + f
f
b
= 8 + bdf + bd2 + d3 + 3d2f + 2df2
- 2bd - 2bf - 2df,
(152)
and tlie region to bo considered is described by (145), (146), (150) and
(151).
Since
4r det = df + d2 ~ 2d - 2f = 0
db
implies either
points.
d = 0
d + f = 0
d = 2,
there arc no interior critical
All faces lead to previous cases:
to 5(1)—1—3,
equivalent to
f = 0
a = 2,
Remark.
lent to the case
to 5(1)-1-2,
x ->• y,
y ->• z,
Ii - C e S^.
b = 0
and
which by (123) fives
This completes the case
change of variables
11 - C e
or
(153)
to 5(1)-1-4-1,
b + d + 2f = 2
b = 0
attain.
A - I> e
z -> x,
is
Ilv nakinf*. the
we sc-e that this is equiva­
Hence we may exclude the possibility
from further consideration in 5(1), since
A - P» e
has
42
already been shown to require
tation
x -• z,
lent to
y -> x,
A - 11 E Sj.
det > 15/2.
y,
7. ->
Similarly, with the permu­
we see that
Moreover, since
C - A e S
A - I) t
is also equiva­
is equivalent to
A - 1! c Sj
as previously noted, we may in similar fashion exclude
B - C e
and
C - A e S*.
5(I)-2.
A - I! e Sj,
knowns are between
0
and
(123) we then obtain
bdry U.
Since
eliminates
f = 2.
a = 0.
b - d = 2,
we must have
Now
and
This with (125) p,ives
d = 2.
bdry K.
2
0
2
c
2
0
0
2
2
(124) pives
Since
e = 0,
c = 0,
ing remark excludes
S2»
(125) p.ives
We obtain
f = 0,
det =
Remark.
and
a
0
0
2
2
2
0
2
must be in
C - A e
and
and
19 fi
by (126).
b - d = -2,
and
and we must have
B - C = (-e, 2 - f, 0)
2 - f >_ 0,
we must have
2
5(I)-2
Then
From
and since the previous remark
= 8 + 4c >_ 8 >
-c < 0
d = 0.
C - A - (e-2, f, 0)
f >_ 0,
A - B e S^.
and since all un-
b = 2,
S| from consideration, vo. must have
5(T.)-3,
be in
2,
e - 2 < 0
det =
b - 0,
Then
and since the preced­
B - C e S^.
= 8 + 4a > 8 >
196
27
must
Then
e = 2,
and
by (126).
and 5(I)-3 may now be used, by a change of
variables like that in the previous remark, to exclude the possibili­
ties
B - C t S^,
B - C £ Sj,
C - A £ Sj
and
C - A e SjJ
from
43
further consideration in 5(1).
5(I)-A.
and
2 - a < 2,
Ue have
A — IS e T^.
Tlicn
(154)
a > 0.
(155)
or
By Lerana 4,
B - C
there), and the remarks exclude
B - C.
This leaves
cannot be in
S^,
B - C e
c — e + 2 -'f + 2 - d = 4,
and
2 - f ^ 0,
(since
S|,
and
IJ - A
is
as locations
as the only possibility.
Then
or
c = d + e + f.
Using (154) and (156) we obtain
Since the third coordinate is
tivc, and
e-2^0.
and cannot be in
C - A c T^.
C-A
(156)
C - A = (e - 2, f - a, 2 - a - 2d
- e - f).
have
or
b = a + c + d;
B - C = (c - e, 2 - f, d - 2) e bdry K,
d - 2^0.
of
2-c + 2-a+b-d = 4,
2 - b,
cannot be in
S|,
or
'L'j!
it must he nonnep.a-
since
by the remarks.
C - 15
is there,
So we must
Then
2 - e + f - a + 2 - a - 2 d ~ e - f = 4 ,
or
a + d + e = 0.
But (157) contradicts (155) and (126).
(157)
Hence the conditions of this
case cannot be satisfied.
Remark.
and
As before, we nay henceforth exclude
B - C e T^
c - A e T^.
5(I)-5.
A - B
e
T^•
Then
2 - c + 2 - a + d - b = 4,
d = a + b -h c;
or
(158)
t\h
and
2 - c < 2,
or
c > 0.
We have
(159)
B - C = (c-c, 2-f, d - 2) e bdry K,
and
d - 2 < 0.
and
T^
Then, since the remarks exclude
as possible locations of
B - C e T^.
B - C,
W e tiave, usinp (158),
where
S^,
2 - f >_ 0
S|,
S^,
the only possibility is that
e - c + 2 - f + 2 - a - b - c = 4,
or
e = a + b + 2c + f.
Also
C - A = (e - 2, f - a, 2 - b)
c ~ 2 _< 0
and
and
2 - b ^ 0.
(160)
must be in
bdry K,
The remarks have excluded
from consideration.
Thus
C - A z T^.
where
Sj,
S^,
S2'
Then, usinp, (160),
2 - a - b - 2 c - f + a - f + 2 - b = 4 ,
or
b + c + f = 0.
(161)
But (161) is impossible in view of (159) and (126).
Ucnco this situa­
tion cannot occur.
5(11).
C + A e bdry K.
A c
B c S2,
C e S3,
A + B,
B 4- C,
We arc still rep.ardiny, all square faccs as closed.
Thus we have
A = (2, -a, -b)
where
|a| + |b| < 2,
062)
B = (-c, 2, -d)
where
|c| + |d| <_ 2,
C *= (-e, -f, 2)
where
|e| + j f-| <_ 2.
and
(163)
(164)
The minus sip.ns are included so that all literal symbols will represent
nonnegative quatities.
In fact, since
B + C = (-c -e, 2-f, 2-d),
and
A + P> = (2 - c, 2 - a, -b - d),
C + A = (2 - e, -a - f, 2 - b)
45
are nil in
1C, we must have
a, b, c, d, e, f >_ 0
in view of the inequalities (2) defining
R,
(165)
Mote also that because of
(162), (163) nnd (164), we have
a, b, c, d, o, fj<2.
Then the only possible faces of
S^,
and
T^.
Hut
K
in which
and ftive
terchanging the roles of
A
and
(166)
A + B
can lie are
S^,
symmetric situations (by in­
11,
x
and
v),
so there are only
three cases to consider.
5(1I)-1.
< 4,
A + B e S^.
Then
nnd
2 -1* 2 - a + b + cl
or
b + d
a.
B + C = (-e, 2 - f, 2 - d) c bdry K,
can only have
B + C
5(II)-1-1.
have
c = 0,
f = 0,
so because of (165) and (3 66) we
or
^3
B + C e Sj.
(167)
^2'
Then
e = 2,
so from (164) we must
and this is a special cane of 5(II)-l-2.
5(II)-l-2.
B + C e S2.
Then
f = 0,
and
e + 2 + 2 - d <_ 4,
or
' e < cl.
C + A - (2 - e, -a, 2 - b) c bdry K,
the.only possible faccs for
5(II)-l-2-l.
C + A
and because of (165) and (166)
are
C + A c S^.
dot =
(168)
Then
2
-a
-b
0
2
-d
0
0
2
Sp,
e - 0,
-9-6-.
- S
b > -12?
and
and
T^.
46
5(II)-l-2-2.
have
b = 0,
C *F A c S.},
a » 2,
so from (162) we
and this ir. a special ease of 5(II.)-1-2-3 below.
5(II)-l-2-3.
C + A E S^.
det -
Now
Then
Then
2
-a
0
0
2
-d
-c
0
2
b = 0,
and
= 8 - ade.
(169)
A+l) + C= (2-e, 2-a, 2-d) cannot be in the interior of
and because of (165) and (166) wc r:ust have either that one of
and
c
is zero, in which ease (169) j^ives (Jot =8,
2 - e + 2 - a + 2 - dj^/i>
a,
K,
d
or that
which way be written
a + d + e <_ 2.
(170)
From thiii and tlie well-known inequality between the aritlinctic and
p.corcctric means of nonncnativc quantities,
3
a + d + c
/adc <
we have in the latter case
ade
8/27, and (169) Rives
dct >_ 208/27
> 196/27.
5(H)-l-2-4.
2 - e < 2,
a < 2,
and
C + A E T3.
Then
2-c + a + 2-b = 4,
or
a - b + c:
(171)
e > 0:
(172)
b + e < 2;
(173)
b > 0.
(174)
or
which with (171) pives
2 - b < 2,
or
47
Combining (171) with the inequalities (167) and (168) pivos
d = e.
(175)
Making the subs Li tuti.oris indicated by (17.1.) and (175), we have
2
-b - e
-b
0
2
-e
-e
0
2
det =
Now
K.
2
3
= S - 2bc - be - c .
(176)
A + 15 + C = (2-e, 2 - b - e, 2-b-e) cannot be interior to
By (172)-(174), the only possibility is that
+ 2 - b - e > 4,
2 - e + 2 - b - e
or
2b + 3e < 2.
(177)
We then consider the problep of minimizing the expression in (176) on
the region described by (172), (3 74) and (177).
Since
3
2
vv; det = -2e - e =0
3b
lias only the solutions
critical points.
e = 0
and
e = -2,
(178)
there are no interior
The only edpe which does not lead immediately to a
previous case is
#
2b + 3e = 2.
(179)
On this edfte we have, from (176),
*>
(180)
det = 8 - 2e + 2e^ +
where
0 < e < 2/3
(the rir.ht-hand inequality is from (179) and (174)).
The end-points of this interval ftive det = 8
and 208/27
respectively.
Testing for a minimum,
A
-j~
de
det =
-2
has as its only positive solution
10
o
-4 +^—.
2/7
(181)
+ 4e +|e2 «•2
e =
At this critical
48
point v:e have
det
5(II)-l-3.
< 4,
352 - 56>/7
203
19 f>
>
>
27
2 7"
~2 7 _ "
B + C c S
Then d = 0,
3*
and
c + 2 - f + 2
or
e < f.
(182)
We have
d e t =
Since
2
-a
-b
0
2
0
-c
-f
2
C + A = (2 - e, -a -f, 2 - b)
view of (165) and (166), have
5(II)-1-3-1.
i
have
= 8 - 2bc.
must be in
C + A E S^,
(183)
bdry K,
S^,
or
C + A e S^.
Then
e = 0,
so
C + A
Then
a + f = 2,
v.'e must, in
T^.
from (183) we
.
o
196
det = 8 > "27"*
5(II)-l-3-2.
= (2 - e, 0, 2 - b).
S^.
e
and
This cannot be nn interior point of
(165) we must have either
e - 0
or
b = 0.
A + 15 -1- C
K,
so by
In cither case (183) Rives
det = 8 > 196/27.
5(II)-l-3-3.
C + A e S^.
Then
b = 0,
hence (183) frives
C + A e T^.
Then
2-c + a+ £ + 2- b°4,
det » 8 > 196/27.
5(T.I)-l-3-4.
or
2 - e < 2,
a + f < 2,
a + f = b + e:
(184)
e > 0;
(185)
or
which with. (184) cives
'(9
and
2 - b < 2,
Now we have
(2 85)
b > 0.
(187)
or
A + K + C = (2 - e, 2 - 1> - e, 2 - b),
cannot be in the interior of
and obtain
b + e < 2;
K,
and since this
we may procccd as in 5(I)-l-2-4
det >_ 7.5 > 196/27.
5(Il)-l-4.
B + C c T2.
Then
e + 2 - f + 2 - d = 4,
e = d + f;
e < 2,
Nov;
(188)
which with (188) p,ivcs
2 - f < 2,
and
d + f < 2;
(189)
f > 0;
(190)
d > 0.
(191)
or
2 - d < 2,
or
C + A = (2 - d - f, -a - f, 2 - b) E bdry K,
(166), (189), (190) and (191) we must have
and
C + A
and from (165),
in one of
S^,
•
5(II)-l-4-l.
C + A E Sj.
Then
A + ]> + C = (2 - d - f, 0, 2 - b - d),
(166), (189), (190) and (191) show that
-2<2-b-d<2.
K,
or
Hence
A + P. + C
a + f = 2.
Ue obtain
and the inequalities (165),
0<2-d-f<2
and
must be an interior point of
which is impossible for an admissible lattice.
5(II)-l-4-2.
+ a + f + 2 < 4 ,
or
C + A e S3>
Then
b = 0,
and
2 - d - f
50
a < d.
liut we have
d < a
(192)
from (107), so
a = d.
Mow
(193)
A + li + C = (2-d-f, 2-d-f, 2-d),
2
de t =
-4
0
0
= 8 - 2dC - d2f - d3,
2 - d
-d - f
-f
(194)
2
and vc may proceed as in 5(II)-1-2-4.
5(IT.)—1—4—3.
- 4,
C + A c T3.
Then
2-d-f + n + f + 2- b
or
a = b + d;
*a + f < 2,
and
(195)
which with (195) pives
2 - b < 2,
b + d -I- f < 2;
(196)
b > 0.
(1.97)
or
From (190) and (191) we have that
d
and
f
are also positive.
Nov;
A + 15 + C = (2-d-f, 2-b-d-f, 2-b-d) cannot be interior
to
K.
The inequalities (190), (191), (196) and (197) show that all
three coordinates of
have
A + B +0
are between
0
and
2 - d - f + 2 - b - d - f + 2 - b - d > _ 4 ,
2,
so we must
or
b + 2d + f <_ 1.
(198)
Then we must find tlie minimum value for
2
det =
0
-d - f
-b - d
2
-f
-
-b
d
2
= 8 - bd2 - bdf - d3 - d2f - 2bd - 2bf
- 2df
(199)
51
on the region described by (190), (191), (197) and (198).
Since the
i
equation
JL del = -l»d - d2 - 2b - 2d = 0
dt
implies that either
points.
The faces
b + d - 0
b = 0,
or
d = -2,
d=0, f=0
(200)
there are no critical
of the region lead to pre­
viously considered cases, (respectively, 5(IT)-l-4-2, 5(lI)-l-3, and
5(lI)-l-2), and we need only consider the face
b + 2d + £ ° 1,
Using (201) to eliminate
b
(201)
from (199), v:e obtain
det = 8 - 2d - 2f *!• 3d2 + 2f2 + 3df + 2d2f + df2 + d3.
(202)
The region under consideration is described by (190), (191) and
2d + f < 1,
(203)
the latter condition arising from (201) and (197).
region lead to previous case (those: noted above).
All edp,es of this
Testing for critical
points, we obtain the equations
"~j det = -2 + 6d + 3f + 4df + f2 + 3d2 = 0
del
(204)
-|f- dot » -2 + 4f + 3d + 2d2 + 2df = 0.
dr
(205)
and
Solving (205) for
f
and substituting into (20/0 yields the equation
5dA + 32d3 + 79d2 + 60d - 4 = 0.
(206)
Equation (206) has exactly one positive solution, which lies between
.061
and
.062.
Then (205) yields
.438 < f < .440,
and (202) cives
det > 7.486 > 196/27.
Renark.
Similarly as in the rerinrk following 5(X)—1—4—3, we
52
may exclude the possibilities
and
C + A c
A + B r. S^.
= (2 - c - e, 2-a-f, 0).
K,
Then
Since
c = e = 0;
b -!• d « 2,
11 + C
and
A + II + C
cannot be an interior
or
a = f = 2.
respectively, nust1 be in
wo have case 5(JI)-1 a<7,ain.
dot =
c = e - 2;
H + C = (-c - e, 2.- f, 2 - d)
C + A = (2 - e, -a - f, 2 - b),
Finally, if
2
0
-c
2
b - 2
-e
0
2
K.
a - f = 0,
Tlie.
and
If
then
-b
C + A = (2 - e, 0, 2 - b):
since
must, because of (165), have cither
(20V) pives
A
a = f = 0;
latter two are inpoKsible, since
and
C + A c
we nust, in view of (165) and (160), have one of four
circumstances:
c ~ 0,
B + C e
throughout Lite remainder of case 5(11).
5(II)-2.
point of
H + C e
=8- 2be
C H- A
e = 0
or
(207)
mist be in
b = 0.
bdry K,
wo
In either case,
det = 8 > 196/27.
Remark.
Henceforth we r.iay, in the sar.:e nanner as in the re­
mark following 5(I)-l-4-3, exclude the possibilities
1> + C e S^,
C + A c S^.
5(II)-3.
2 - c < 2,
2 - a < 2,
A +.B e Tz.
Then
2 - c + 2 - a + b + d = h,
or
a + c = b + d;
(203)
c > 0;
(20 f J)
a > 0;
(210)
or
or
53
and
b + d < 2.
V.'e have fi + C = (-c - e, 2 - f, 2 - d),
and
2 - d > 0
*
by (165) and (166).
(211)
whore
Since
-c - c £ 0,
S',
J,
excluded by the remarks as possible locations for
]j
C c Tg •
Then
c •!- e + 2 - f -h 2 - d = A(
and
tr,
S
2 - f j> 0,
have been
J
I> + C,
wc must have
or
c + q = d -b f;
c + e < 2,
2 - f < 2,
and
which with (212) p.ives
d + f < 2;
(213)
f > 0;
(214)
d > 0.
"(215)
or
2 - d < 2,
or
In a similar manner it may be shown that
must b a in
(212)
T^,
s o that w c have
C + A = (2 - e, -n - f, 2 - b)
2 - e + a + f + 2 - b = 4,
- a + f = b -I- e;
2 - e < 2,
and
o r
(216)
or
2 - b < 2,
e > 0;
(217)
a + f < 2;
(218)
b > 0.
(219)
or
The three equations (20S), (212), (216) are not independent, so only
two variables can be eliminated from the expression for the determin­
ant.
Ug choose to eliminate
c
and'^e,
obtaining
5/)
2
dct =
Since
to
K,
-a
-b
a - b - d
2
-d
b - a - f
- f
- 8 + abd •!• nbf + 2a2 + 2b2 - adf - bdf
- a2d - b2f - 2df - 2bC - 2nd - 4nb.(220)
2
A -I- 11 + C = (2 - d - f, 2 - a - f, 2 - b - d)
cannot be interior
and since, by the inequalities (210), (211), (213), (214)»
(215), (218) and (21(J), all three coordinates of this point are be­
tween
0
and
2,
we mast have
2 - d - f + 2 - a - f + 2 - b - d>_'t,
or
a + b + 2d + 2f <_ 2.
(221)
Uencc we may consider the problem of ninir.iizinp. the expression in
(220) on the region described (210), (214), (215), (219) and (221).
In
testing for critical points, wc obtain the equation
s
2
T-r dot «= ab - af - bf - a - 2f - 2a = 0.
3d
Solving for
f
in (222) piver.
c
£
Since
b
a > 0
(222)
and
b < 2,
a -2)
M ^(b
- t+t+ t'
we have
arc both positive, so is
(223)
b - a - 2 < 0,
a + b + 2,
right side of (223) is negative, but
f
and since
a
and
Hence the expression on the
must be positive.
It follows
that there are no critical points in the region under consideration.
The faces
a = 0,
b = 0,
d = 0,
and
f = 0
of this region lead to
situations which have been excluded from consideration as beinp, equiva­
lent to case 5(II)-1 (for example,
a = 0
t>ives
A + B e ^2^ •
There­
fore the only portion of the boundary to consider is the face
a + b + 2d + 2f = 2.
(224)
55
Using (224) to eliminate
a
from (220), v.'e obtain
det - 16 + 24bd + 24df + lGbf + Si.2 + 8f2 4- 20d2 - 16b - 16f - 24(1
- 2b2d - 2df2 - 2b2f - 2bf2 - 6bd2 - Gd2f - Sbdf - 4d3
with the conditions
b,
d,
f > 0
(225)
and, from (224) and (210),
b + 2d + 2f <. 2.
(226)
As just noted, no boundary points of this region need be considered.
Tcstinp, for critical points, v:e obtain
--- det = 24d + 16f + 16b - 16 - 4bd - 4bf - 2f2 - 6d2 - 8df - 0, (227)
det = 24b + 24f + 40d - 24 - 2b2 - 2f2 - 12bd - 12df - 8bf
- 12d2 = 0,
(228)
and
yf- det = 24d + 16b + 16f - 16 - 4df - 2b2 - 4bf - 6d2 - 8bd = 0. (229)
Subtracting (229) from (227), v?e obtain
(b - f)(b -I- 2d
2b2 - 2f2 + 4brl - 4df - 0,
f) = 0.
or
(230)
Since all variables represent positive quantities, (230) inplien that
b « f.
Substituting
b
for
f
in (227) and (228) yields
—• G.24d + 32b - a 6 - 12bd - 6b2 - 6d?" = 0
(231)
and
48b + 40d - 24 - 12b2 - 12d2 - 24bd = 0.
Dividing the left side of (232) by
4d + Sb - 4 = 0,
2
(232)
and subtracting from (231) p.ives
or
d + 2b = 1.
Usinf, (233) to eliminate
d
from (231), and dividing, by (-2),
3b2 + 2b - 1 = 0.
(233)
yields
(234)
*56
The only positive solution of (234) is
b = 1/3.
(230), (224), (20S) and (21.2), we obtain
2
det =
Then, froin (233),
d = f = a = c = e = 1/3,
-1/3
-1/3
-1/3
2
-1/3
-1/3
-1/3
2
and
» 196/27.
Here for tlie -first tine the minimum possible deterninant has been
obtained.
6(1).
e bdry K.
A c S1>
n c S3 ,
C e
A - 11,
B - C,
C - A
Let
A = (2, a, b),
where |
a|
B - (2, c, d),
where
]b
| < 2,
|c| + |d| <_
C = (e, f, t \ - e - f),
where
'
2,
o < 2,
and
f < 2,
Note that the inequalities in (237) inp.lv that
A - B = (0, a - c, b-d),
we must have cither
A,
B
and in
y,
z,
(236)
and
e + f > 2.
e > 0,
f > 0.
(237)
Then
3»-C = (2-e, c - f, cl + e -I- f - 4),
C - A - (e-2, f-a, 4 - b - e - f).
K,
a - c = i2
Since.
or
A - 1)
b - d = ±2.
must he in
|d|
2
(238)
from (236), we have
|c| < _ 2 - |d| <^2 + d,
(239)
also from (236), we have
j c[ < b.
Usinp (238) p,ives
0 < 2 - e < 2:
bdry
we nay assume without loss of peneralitv that
b ^ 0,
and since
and
Bv syndic try in
b - d = 2.
Then since
(23f>)
(240)
B - C = (2-c, c-f, b + e + f-6).
from (236) and (237),
c-f <2;
From (237),
and fron (235)
57
and (237),
b + e + f - 6 < 0.
sible faces for
11 - C
arc.
These relations show that the only pos­
S^,
1^
and
T^.
So we have four
cases to consider,
6(X)-1.
15 - C e Sg.
Then
c - f = -2,
or
c = f - 2.
We have
C - A = (e - 2, f - a, 4 - b - c - f).
- 2 < 0;
froin (235) and (237),
f - a > -2;
and (239),
-2 < 4 - b - e - f < 2.
for
are
C - A
there.)
(241)
S„.
1
T„,
2
T'
5
and
-2 < e
and fron (235), (237)
Hence the only possible faces
T'.
4
(TJ
1
is excluded since
-C
is
Thus there are four possibilities.
6(i)-1-1.
C - A c Sj.
Then
f - a = 2,
dot =
The point
2
f - 2
b
2
f - 2
b - 2
e
f
4 - o - f
and
(242)
~ 4e 4 4f - 2ef.
A - IS - C = (~e, -f, e + f - 2)
0 < f < 2,
e + f
and
0<o+f-2<2
e + f - 2 > 'i,
(243)
cannot be interior to
since the lattice is assured to be admissible.
that
From (237),
Since
K,
0 < e < 2,
by (237), the only possibility is
or
c + f 1 3.
(244)
Thus the problem is now to minimize the determinant in (243) under the
conditions (237) and (244).
3c
has only the solution
Since
dct = 4 - 2f = 0
f = 2,
there are no critical points.
(245)
The
5S
edftes
c = 2 and
f - 2
of the boundary lead to eases 1(1) and 3(1)
respectively, am! these have already been considered.
e + f = 3,
we obtain from (243)
det = 12 = 6e 4- 2c:2 = 15/2 + 2(e - 3/2)2
6(1)-1-2.
= 4,
On the edee
C - A c T
2 >
Then
15/2 > 196/27.
2 - a + f - a + 4 - b - e - f
or
a = 2 - b - 2e;
and
f - a < 2,
which with (246) ^ives
b
Ue also have
(246)
b
0,
f < 2
+ 2e + f < 4.
and
e + f > 2
(The inclusion of (247) makes the relation
(247)
from (239) and (237).
e < 2
also follows from these four inequalities that
b < 2.)
e > 0,
f > 0,
and
Then
2
det =
redundant, and it
2 - b - 2e
b
f - 2
b - 2
f
4 - e - f
= -32 + Sb + 20e + 20f + 4be - 2bf
- 6ef - 2f2 - b2e - 2boZ - bef,
(248)
Testing for critical points, we obtain
~ r det = 8 + 4e - 2f - 2be - 2o2 - ef = 0
(249)
3b
and
Solving (250) for
f
+ be2 - 8 + 4b - 0,
e = 2
(250)
det = 20 - 2b - 6e - 4f - be = 0.
3b
and substituting into (249) pives
or
(b - 2)2(e - 2) - 0,
as its only solutions.
which has
Se - 4be - 2e
b = 2
and
Thus there are no interior critical
2
59
points.
All hut one of the faces of the rer.ion described by (237),
(239) and (247) fall into cases v.'hich have already been considered.
The exception is the face
b = 0,
for which (248) i»ivc.s
det = -32 -I- 20c + 20f - 6cf - 2f2,
(251)
and the rep,ion to be considered is described by (237) and
2e + f < 4,
(252)
g
the. latter a consequence of (247).
- 6f - 0,
which has
f = 10/3
From (251) we have
as solution.
dC
det = 20
Hence there are. no criti­
cal points in the repion, and no boundary points need be considered.
6(I)-l-3.
- 4 = 4,
C - A c T^.
Then
2-e+F-a + b + e+ f
or
b - a = 6 - 2f.
(253)
b - a < 2
(254)
6 - 2f > 2
(255)
But
by (235),
by (237).
and
Hence this case cannot occur.
6(I)-l-4.
- f - 4,
C - A
e
Tj.
Then
2 - c + a - f + 4 --- b - e
or
a - b = 2e •!• 2f - 2.
(256)
I'ut
a - b < 2
(257)
2e + 2f - 2 > 2
(253)
by (235), and
60
by (237).
Hence this cane is also impossible.
6(I)-2.
Ii - C c Sy
Then
d + e + f - A - -2,
v/hich with
(238) pives
b + e + f = A.
Nov;
C - A = (e - 2, A - a - b - e, 0).
(259)
Since
A - a - b - e > - 2
(260)
0 < 2 - c < 2
(261)
by (235) and (237), and
by (237), the only possible fact; for
- e = 2 ,
C - A
is
S^.
Then
A - a - b
or
a + b + e - 2;
(262)
and
2
2 - b - e
b
c
b - ">
A - b - e
b
det = 2
e
= 16 - Sb - Se + 2b
2
+ 2e
2
- bcc - 2b e - be .
*1- 2bc + 6be
(263)
\!c have, the conditions
0 .< e < 2
(26A)
|c| < b < 2
(265)
from (237) and
from (2A0), (259) and (237).
Testing for critical points, we see that
there are none in the interior of the region described by (26A) and
(265) since
3c
has the solutions
b = 0
det = 2b - be = 0
and
e = 2.
(266)
Three of the faces of the rep.ion
61
need not he considered here:
e = 2
yields case 1(1);
C e S^>
a situation equivalent to that of
to keep
C e K,
require that
There reinain the faces
l> = 2,
b = c
and
3(1);
and
b = 2
f*ivcs
e - 0
would,
the situation just discussed.
b = -c,
which v.*ill now be con­
sidered.
6(l)-2-l.
b = c.
Then fro-i (263),
det = 16 - 8b -• 3e + Ab2 + 2e2 + 6bc - 2b2e - be2,
(267)
under the conditions (264) and
0 £ b < 2.
(263)
Testing for critical points, we obtain the equations
det = -3 + Sb + 6e - Abe - o2 = 0
(269)
and
v~ det = -8 + Ac + 6b - 2b2 - 2be = 0.
<3e
Factoring (269) j»ives
(e - 2)(Ab + e - A) - 0,
(270)
and since
e < 2
by
(26A) we have
e = A ~ Ab.
(271)
3b2 - 9b + A = 0,
(272)
Usinp, (271) in (270) p/ives
and the only solution of (272) which satisfies (265) is
b =
9 - /33
g
,
from uhich (271) and (267) yield det - 22 -
> 7.95 > 196/27.
As in 6(I)-2, all ed<;cs of the refion other t!;an
b = 0
regarded.
If
may be dis­
b - 0, (267) becones
det = 16 - So + 2e2 = 8 + 2(e - 2)2 > 8 > 196/27.
6(I)-2-2.
b = -c.
Then (263) fiives
(273)
62
det = 16 - 8b - 8e + 2e2 + Gbe - be2,
where the conditions (264) and (268) n^ain apply-.
(274)
Since
~~ del » -8 + 6e - e2 = 0
0 l>
has the. solutions
points.
e = 2
and
e = 4,
(275)
tliere are no interior critical
As in 6(I)-2-1, the only edpe to consider is
b = 0,
ancl
there we obtain (273) apain.
6(I)-3.
= 4,
R - C E T^.
Then
2-c + c- f + 6 - b - c - f
or
c - b » 2e + 2f - 4.
nut
c - b
0
by (240), and
2e + 2f - 4 > 0
(276)
by (237).
Hence this
case cannot occur.
6(I)-4.
= 4,
R - C e
.
Then
2-e + £- c+ 6- b- c- f
or
b + c + 2e - 4;
f - c < 2,
and
(277)
which with (277) t»ives
b + 2c + T < 6:
(278)
b + c + f > 4.
(279)
6-b-c-f<2, or
Now C — A — (e - 2, f-a, 4-b-e-f) e bdry K.
The conditions on
thejvariables involved are the same as they were in 6(I)-1, where it
was shown that
C - A
had to be in
S„.
2
1' ,
2
T'
3
or
T!.
4
Hut the
latter two faces nay be eliminated as they were in 6(1)—1—3 and 4, and
T2
is excluded by Lenna 4, since
C - A c S^,
so that
C - 11
is there.
Thus we must have
63
f - a = 2.
(280)
We also obtain the conditions
0 < b ^ f < 2;
the first is a
(281)
consoquonco. of (279) and (237), the second of (280) and
(235), and the last war. stated in (237).
Then the; conditions (273),
(279) and (281) spccifv the repion on which the determinant is to be
minimised (note that the relations
these).
are implicit in
V.'e have
2
det =
0<e<2<e+f
2
f - 2
b
4 -• I* - 2e
e
f
b - 2
4 - e - f
« 48 - 8b - 24e - If.f + 4c2 + 2F2
2
- Abe + 2bf + 4ef + b e + 2bo
2
+ bef.
(282)
Testing for critical points, we obtain the equations
—•* dot = -16 + 4f + 2b + 4e + be ~ 0
3e
(283)
and
~ dot = -S - 4e + 2f + 2be + 2e2 + of == 0.
3b
Solving (283) for
f
and substituting into (284), we obtain
(e - 2)(4c - b(c - 2)) = 0.
This yields
e = 2
(284)
and
b =
ac
->
neither of which is satisfied in
the rep,ion (the latter bccausc the ripht side nust bo negative).
there are no critical points.
(285)
Hence
The only face which docs not lead to a
previous case is
b = f.
Usin? (286) in (282) pives
(286)
64
det = 48 - 24e - 24f + 4e2 + 4f2 + 2e2f + 2ef2.
(287)
Testing for critical points, wc obtain
9e
det = -24 + So + 4ef + 2f2 - 0
(288)
yf- dut = -24 + Sf + 2e2 + /ief - 0.
(289)
Subtracting (289) from (28S) pives
2(e - f)(4 - o - f) = 0.
Since
o < 2
and
f < 2,
(290)
ue wust have
g — f.
(291)
3o2 + 4e - 12 = 0,
(292)
Usinjy, this in (288) pives
and the only positive solution of this is
.
(287) pives
. ,
2224 - 640/lb'
clet =
6(11).
E bdry K.
4 - e - f)
A c Sx,
where
7(11).
A + B,
11 + C,
C + A
I! c
C E T?>
e > 0,
K
then
since
A e S ,
A - 15,
B - C,
A = (2, a, b),
C - A
B
C
(-c, f,
C - A - (-e -2, f - a, 4-b-e-f),
-c - 2 < -2.
r.
E $v
C E 1'2,
A + II,
B + C,
C + A
A E
s
,
B E
,
C e 1'2,
A - B,
B - C,
C - A
This situation can be shown to be impossible as v/as 4(1).
8(11).
e bdry K.
C E Tx,
This situation can bo shown to be impossible as v/as l(TI).
8(1).
c bdry K.
196
'2T'
This is impossible, bccausa if
which cannot be in
c bdry K.
B e S ,
>
This situation can be shown to be impossible as v/as 1(T1).
7(1).
E bdry K.
A e 5^
200
c = —•-•, which with
A e Sj,
Be Sj,
C E 1'2,
This is impossible, because if
A + I),
U + C,
B = (-2, c, d,)
C + A
and
65
C = (-e, f, A - c - f) where
of
-2 - e < -2,
9(1)
E bdry K.
e > 0,
then
hencc cannot be in
Ac
C
1'. c S2,
15 + C
has nn
x-coordinate
i!
C
K.
E Ti»
A
"
" c»
~
A
Then we have
A = (2, a, b), where
]a| + |b| <_ 2,
(293)
(294)
B = (c, 2, d),
where
|c| + |d| £ 2, |c] < 2,
C = (e, f, 4 - e - f),
where
e < 2, f < 2,
and
Then, since
and
A - B = (2 - c, a - 2, b - d) nust be in
e -I- f < 2. (295)
K,
we have
a >_ 0, c > 0.
We consider first of all the special case
because of (293), and
(296)
a = 2.
In this case
C - A = (e - 2, f - 2, A - e - f).
conditions (295) this must be an interior point of
dicts the assumption that it is a boundary point.
K,
Jlut by the
which contra­
Thus we. must have
a < 2,
and from (294) we know
terchanging
A
with
Ii
c <. 2.
and
x
that, as possible locations of
following pairs are equivalent:
T^.
(297)
Thus the problem is unchanged by in­
with
v.
A - U,
S,
JL
and
Usinp this symmetry uq see
the two faces in each of the
Si,
..A-
S_
J
and
Si,
J
Hence only the first of each pair wi31 be considered.
that, since
2 - c > 0
b - 0
by (294) and
faces are the only ones possible.
a - 2 < 0
T„
j
and
Note also
by (297), these six
Hence there are three cases to con­
sider.
9(I)-1.
A - B e Sy
Then
c = 0,
.. (298)
66
and
B - C = (-e, 2 - f, d + e + f - 'i)
0 < 2 - f < 2
and
d + c + f - A < 2
possibilities are that
Since
-2 <
-C
or
Tiicn
T^.
d+e+f-4=-2, or
d « 2 » c - f,
and
(299)
C - A = (e - 2, f - a, 4 - b - e - f) c bdry K.
- 2 < 0
by (295),
4-b-e-f>-2
- 2 < f -- a < 2
C - A
9(I)-l-l-l.
are
C - A c
Since
-2 < e
by (295), (296) and (297), and
by (293) and (295), and since
possible locations for
< 0,
by (295) and (294), the only
li - C c
» - C c S^.
9(I)-l-l.
bdry K.
E
T^,
Then
-C c T^,
and
the only
T^.
A-b-e-f-2, or
b = 2 - c - f,
(300)
and
det =
2
a
2 - e - f
0
2
2 - c - f
e
f
4 - c - f
= 16 - 8c - 81: + 2e~ + 2f2 + 2ac + 4ef - acf,
(301)
which wil] be Minimized on the rcp.ion described by (295), (296) and
(297).
We obtain
--- det = 2c - e2 - ef - 0,
d cl
and
e + f = 2,
(302)
which lias the solutions
e - 0
so there are no in­
terior critical points.
The only face of the rcjiion which must be
considered is
a = 0.
(303)
det « 16 - 3c - Sf + 2c2 + 2f2 + 4ef,
(304)
On this face, (301) becomes
67
and testing for critical points vc obtniu
3 - dqt = -8 + ac +
—-
= 0,
DC
which contradicts (295).
(305)
The region now under consideration is that
described by the relations (295), no that no boundary points need be
considered.
C - A E T .
9(I)-l-l-2.
- f = 4,
Then
2-c+£-a + 4-l>-c
or
a + b + 2c - 2,
(306)
and
dct =
Now
2
a 2 - a - 2e
0
2
2 - e - f
e
f
4 - e - f
= 16 - 8c - 8f + Ae^ + 2f^ + 4ac + 2ef - ae^
- aef.
A+B-C= (2-e, 2+a-f, -a- 2e)
cannot be interior to
and from the relations (295), (296) and (297) we have that
<2,
2 + a - f > 0,
2 + a - f >_ 2,
and
-a - 2e < 0.
Uence vc must have cither
which f.ivas
-a - 2e _< -2,
(308)
which f.ives
a + 2e >_ 2;
or
2 - e
K,
0 < 2 - d
a > f;
or
(307)
2 -I- a - f + a + 2c > A,
2a + e
(309)
which p.ives
f.
Uut if (303) holds, then by (295) and (296) so must (310).
(310)
Thus either
(309) or (310) must hold, alonn with (295), (296) and (297).
™- dot = 4e 3a
- of = 0
How
(311)
68
implies that cither
e = 0
terior critical, points.
a + 2e = 2,
and
or
e + f - 4,
so thnt there are no in­
V.'e have three faces Co consider:
a - 0,
2a -1- o - f.
9 CT)—1—1—2—1.
n = 0.
Then (309) and (310) bccoi-ci either
e > 1
(332)
e > f,
(313)
det = 16 - Se - 8f -I- Ac2 + 2f2 + 2ef,
(314)
or
and (307) gives
where the ration is described by (295) and (312)—(3D.3).
Testing for
critical points, v:e have
3
. dot = -8 + 3e +2f = 0
de
(315)
and
d i"
det « -8 + /if + 2e « 0.
Solving (315) and (316) simultaneously gives
docs not satisfy either (312) or (313).
(316)
e = 4/7,
f
n
12/7,
which
Hence there are no critical
points in the region.
It is necessary to consider the edp.es arising
from (312) and (313).
If
c -• 1,
then (31'i) gives
+ 2f2 « 15/2 + 2(f - 3/2)2 > 15/2 > 196/27,
(314) Rives
and
if
det » 12 - 6f
e = f,
then
dot = 16 - 16f + Sf2 = 8 + 8(f - I)2 > B > 196/27.
9(I)-1-1-2-2.
a'+ 2e = 2.
Since under this assumption
(309) is satisfied, we may disregard (310) and consider only the in­
equalities (295) and
e < 1,
the latter resulting from (296).
vields
Substituting
(317)
a = 2e - 2
into (307)
69
dot = 16 - Sf - 6g2 + 2f2 + 2g3 + 2c2f.
(31B)
Testing for critical points, wo have
3e
dct = -12e + 6c2 -h 4ef ~ 0
(319)
dot « -3 + 4f + 2e2 = 0.
(320)
and
3c
From (319) we have
3c + 2f = 6,
since by (295)
Then substitution into (320) pj.ves
solutions
c = 1
and
e = 2.
e
2
e
cannot be?, zero.
- 3e H- 2 = 0,
which has the
Thus there arc no critical points in
the interior of the region, and the only edp.e which need be considered
is
e = 1,
for which as in the previous case,
2
dot » 12 - 6f + 2f *
> 15/2 > 196/27.
9(I)-l-l-2-3.
f,
2a + e » f.
Then wo eliminate the variable
and (30/) becomes
det = 16 - 16a - 16c + 8a2 + Sc2 + 16ae - 2nc2 - 2a2e.
The conditions of
f < 2
and
e + f > 2
(321)
fron (295) become
2a + e < 2
(322)
a + e > 1
(323)
and
respectively.
The relations
e < 2
(295) and
a >_ 0
(296) remain in
forcc; (297) is now redundant, and (309) does not apply.
Nov: the de­
terminant (321) is like that in equation (103), case 3(I)-2-4-2-l, ex­
cept for the substitution of
a
for
c.
Thus it has the same criti­
cal point (note that in this case also both .variab]cs are nonnenntivc)
,
, .
•
, P
and a determinant of
222/. - 640/10 ^ 200
>
196
.
xs obtained there.
70
The edp,e
n = 0
in the. present ense v?ns considered in 9(T)-1-1~2-1,
and the other eddies as usual belong to earlier cnr.cs.
9(l)-l-l-3.
-4 = 4,
C - A c T^.
Then
2-e+f-a+b+e+f
or
b - a = 6 - 2 f .
Cut
b - a <_ 2
by (293), and
6 - 2f > 2
by (295).
Hence this ease
is impossible.
9(I)—1—1—A.
- f = 4,
C - A E Tj|.
Then
2- C + a- f + 4 - b- e
or
a - b = 2 e + f - 2 .
But
a - b <_ 2
by (293), and
2c + 2f - 2 > 2
by (295).
Hence this
case cannot occur.
9 CI)—1—2.
or
d + 2e = 6.
B - C e T2<
But
d _< 2
Then
e+2-f+d+e+f-4=4,
by (294), and
a < 2
by (295).
Hence
this case is impossible.
9(I)-l-3.
U - C e Ty
Then
e+2-f+4-d-e-f=4,
or
d = 2 - 2f;
and
(324)
4-d-e-f<2, which with (324) c.ives
e > f.
(325)
Mow
C-A= (e-2, f - a, 4-b-e-f),
< 0
from (295):
-2 < f - a < 2
4-b-e-f>-2
e
and since
-2 < e - 2
fron (295), (296) and (297); and
by (293) and (295).
T^
and we have
Since
-C £ Tj
and
IS - C
may be elininnted as it was in 9(I)-l-l-4, ve see
that the only possible locations for
C - A
are
and
T^.
71
C - A e S^.
9(I)-X-3-l.
Then
4-b-e-f-2, or
(320)
b » 2 - e - f,
and wo have
det -
2
a
0
2
e
f
Nov; sincc
point of
2 - c - f
= 16 - So. - 8f + 2.o
2 - 2f
'if + 2ac -I- 2cf - 2aef.
(327)
A - c - f
A H- II - C = (2 - e, 2 + a - f, -2f)
K,
and since
(295) and (296), and
0 < 2 - e < 2
-2f < 0
cannot be. an interior
by (295),
2 + a - f > 0
by
by (295), at least one of Che following
inequalities r.iust hold:
2 + a - f >_ 2,
-2f <_ -2,
or
or
a >_ f;
(328)
f >_ 1;
2 - e + 2 + a - f + 2f >_ 4,
or
(329)
a + f _> e.
(330)
The problem now is to nininiae (327) on the repion described by (295),
(296), (297) and (325), which way be summarized in
0
a < 2,
0<f<e<2<e+f, and by (328) or (329) o_r (330).
If a critical
point exists in this region, it must satisfy
3
det = 2e - 2ef ~ 0.
3a
Since
point,
e > 0,
(331) is satisfied only when
det = 12 - 6e + 2e
2
>_ 15/2 > 196/27.
which must be considered are
9(l)-l-3-3-l.
a = 0,
a = 0.
a = f,
(331)
f = 1.
But for any such
The faces of this region
f = 1,
and
a + f = e.
Under this assumption we see by (295)
that (32S) cannot hold, and by (325) that (330) cannot hold.
(329) must hold.
From (327) we have
Hence
72
det = 16 - Se - Sf + 2e 2 + 4f2 + 2cf
(332)
to mininizci on the region described by (295), (325) and (329).
then ~ det - -8 + Sf + 2e = 0
ar
together yield
in this region.
cannot hold, sincc (325) and (329)
Sf + 2e > lOf >_ 10.
On the edf.e
Hut
f = 1
Hence there are no critical points
\te have
det » 12 - 6e + 2e
15/2,
as shown above.
9(I)-1-3-1-2.
a = f.
Then (327) p,ivcs
det « 1G - 8e - 8f + 2e2 + 4f2 + Aef - ef2,
(333)
and tlie region on which this is to be considered is described by (295)
and (325).
Testing for critical points, we have
-- det = -8 + 4e +
- f2 = 0
(334)
dG
and
det = -8 + 8f + 4e - 2ef = 0.
(335)
Subtracting (335) fron (334) .skives
f(f - 2e + 4) = 0.
But ve have
f > 0
and
f~- 2e + 4 > 0
critical points in tlie region.
(336)
by (295), so there arc no
lio boundary points need to be con­
sidered.
9(I)-l-3-l-3.
f ~ 1.
Then, as noted in 9(T)-l-3-l,
9(I)-l-3-l-4.
a + f = c.
dot
> 15/2.
Then, eliminating
a
from (327),
we have
det = 16 - Se - Sf + 4e2 + 4f2 - 2e2f + 2ef2,
(337)
which must be minimized on tlie region described by (295) and (325).
Testing for critical ooints we have
73
V- det « -S + So - 4ef + 2f2 = 0
3c
(338)
~ dot = -S -1- Sf - 2o2 + 4cf - 0.
(339)
and
Factoring (330) gives
(f - 2)(f + 2 - 2e) = 0,
and since
f < 2
by
(295) we must have
f + 2 - 2e » 0.
(340)
Using (340) in (339) gives
3e2 + 4e - 12 = 0,
which has
c =
as its only positive solution.
„.M
. ,__,N
,
(340) and (337) uc have
, .
2224 - 640/10
200
det =
^
>
196
Then from
,
,
VT
No boundary
points need be considered.
9(I)-l-3-2.
- f = 4,
C - A e Tg.
Tlien
2-e + f-a + 4- b- e
or
a + b + 2e = 2;
and
(341)
4-b-e-f<2, which with (341) gives
a + e < f.
But (342) contradicts (325) and (290).
(342)
Hence this situation cannot
occur.
9(I)— 2.
A - li e S3.
Then
b - d - 2,
and
]} - C = (c ~ e, 2-£, b+e+f-6).
(294) and (295),
0 < 2 - f < 2
(343)
Since
by (295), and
b + c + f - 6 < 0
(293) and (295), the only possible locations for
and
T|j.
-2 < c - e < 2
B - C
are
S^,
by
by
T^
74
9(l)-2-l.
B - C c S^.
Then
b + c + f - 6 « -2,
or
b = 4 - e - f,
and from (344) we have
<
0
from (295) and
(344)
C - A - (c - 2, f - a, 0).
-2 < f - a < 2
is ail interior point of
K.
Since
-2 < e - 2
from (295), (296) and (297),
C - A
This contradicts the assumption that the
lattice was admissible.
9(I)-2-2.
li - C e T^.
Then
c - e + 2 - f + G - b - e - f = 4
(345)
6 - b_- e - f < 2.
(346)
and
Combining (345) and (346) frives
e + f > 2
by (295) and
e + f < c,
c < 2
by (29-4).
15 - C E ' 1 ' y
9(I)-2-3.
which is impossible since
Then
b+e+f-G>-2,
or
b + e + f > 4.
Ke have
< 0
C - A = (e - 2, f - a, 4 - b - e - f),
by (295),
-2 < f - a< 2
-b-e-f<0
T^
and
and since
by (293), (295) and (347), all facos of
T|. irtny be excluded as possible locations of
we would have
b - a
-C
is there.
If
by (293), and
not be in any face of
9(I)-3.
K,
6 - 2f > 2
K
by (295).
-2 < 4
other than
C - A.
C - A
2-e+f-a+b+e+f-4=4, or
2
-2 < e - 2
by (295), (296) and (297), and
also excluded by Lemma 4, since
But
(347)
T| is
were in
b-a=6-2f.
Thus
C - A can­
and this situation cannot occur.
A - 11 e T3".
Then
2-c+2-a+b-d=4, or
b - a + c H- d;
(34S)
75
2 - c < 2,
2 - a < 2,
and
or
c > 0;
(349)
a > 0;
(350)
or
b - d < 2,
which with (348) p,ivcs
a + c < 2.
Ue have
< 2
(351)
II - C - (c - c, 2 - f, d + e + f - 4),
by (294), (295) and (349),
+ f - 4 < 2
0 < 2 - f < 2
by (294) and (295), nnd sincc
occupicd, the only available faccs for
T^
15 - C
and sincc
-2 < c - e
by (295), and
and
arc
d + c
arc already
S^,
' and
T
Then there are three subcases.
9(I)-3-l.
B - C e S^.
Then
d + e + f - 4 « -2,
d - 2 - e - f.
Using (352) and (343) to eliminate
f - a, 2 - a - c).
Since
by (295), (296) and (297),
(351), we must have
= 4,
and
(352)
and
-2 < c - 2 < 0
C - A c T^
9(I)-3-l-l.
b
cl,
we have
bv (295),
0 < 2 - a - c < 2
or
C - A e T2<
or
C - A = (c - 2,
-2 < f - a < 2
by (349),(350) and
l'J .
Then
2-e + F- a + 2- a- c
or
c = f - 2a - e,
(353)
and we have, using (353), (352) and (348),
det ~
2
a
2 - a - 2e
f - 2a - e
2
2 - c - f
= 1G - 8c - 8f + 8a2 + 4e2 + 4f2 + 3ae
f
4 - c - f
+ 2e2f - 2cf2 - 2a2e - 2ac2 + Aacf.
e
(354)
76
The region in quor,tion is described by (295), (349), (350), (351) and
(353).
e = 2
All of the faces of this region fall- into earlier car.es:
and
f = 2
lent of 5(1),
9(I)-1,
and
into equivalents of 3(1),
c = 0
into 9(T.)-1,
a + c = 2
a - 0
into 9(I)-2.
e + f - 2
into an equiva­
into an equivalent of
Testing for critical points,
ve obtain the following equations from (354):
-r— det = -8f + 4ef + 16a - 4ae -b 8e - 2e2 = 0,
(355)
<3a
det = -S - 2f2 + 4af + 4ef - 2a2 + 8a - 'iae + Se = 0,
(356)
and
~ det = -8 + 8f - 4of - 8u + 4nc + 2c2 =0. '
3f
Addition of (355) and (357) yields
8a + 8e - 8 = 0,
or
a = 1 - e.
Using this to eliminate
a
cannot have
since
e = 1 - f,
(358)
from (356)» wc obtain
e +"f > 2
e
2
2
= (f - 1) .
3e
2
- Se + l\ - 0.
and
and
f
e = 2/3.
from (355) ftives
e = 2,
which is impos­
The latter solution Rives
f = 5/3,
det = 196/27.
9(l)-3-l-2.
= 4 ,
(359)
The solutions of this are
sible by (295), and
a = 1/3,
a
VJe
by (295), so we must have
e = f - 1.
Using (358) and (359) to eliminate
(357)
C - A
E
TJ.
Then
2-e + a- f + 2- a- c
or .
c h- e + f = 0.
But
c > 0
impossible.
by (3A9), and
e + f > 2
by (295).
Hence this case is
77
9(T)-3-2.
- f = 4,
B - C c T^.
Then
c-e + 2- f + A- d- e
or
c - d = 2c + ,2f - 2,
But
c - d^2
by (294), and
2e + 2f - 2 > 2
by (295).
So this
situation cannot occur.
9(I)-3-3.
- A = A,
B - C E 1' 2 .
Then
e-c+2-f + d + e + f
or
d - c - 6 - 2g.
But
d - c < 2
by (294), and
6 - 2e > 2
by (295).
Hence this
situation is impossible.
9(11).
e bdry K.
A e S1#
B E S2>
This is impossible.
C E Tlt
B + C,
A
and
C
C + A = (e+2, f + a, 4 + b - c - f),
and
e + 2 > 2
that this cannot be in
10(1).
e bdry K.
A c S1,
where
cannot be in
K
10(11).
K.
e > 0.
since
C + A
be as in 9(1).
Then
by (295), so
K.
B e S?,
This is impossible.
e + f - d)
e bdry
Let
A + IJ,
C c Tj,
Let
Then
A - B,
A = (2, a, b),
U - C,
C - A
C = (-e, -f,
C - A = (-e - 2, ~E -a, e + f - b - A)
-e - 2 < -2.
A c Sr
B
E
S2,
C c Tj,
A + P.,
B + C,
C + A
Let
A = (2, -a, b),
B = (-c, 2, d),
where
where
|a| + ]b|
[c| + [d| <2
and
2,
(360)
|c| < 2,
(361)
and
C ~ (-e, -f, c + f - 4),
where
e < 2,
f < 2
and
e + f > 2.(362)
78
Sincc, under these conditions,
v/e will use
det(A, B, C)
det in this case only to represent the positive quantity
-det(A, B, C).
Then
det - det(A, IS, -C) -
Since
will always be negative,
2
-a
b
-c
2
d
e
f
A - e - f
C + A= (2-e, -a - E, b + c + f - 'i)
have
-a - f
-2,
must bo in
K,
and since it follows from (362) that
we must
f > 0,
we
must have
a < 2.
A -h B = (2 - c, 2 - a, b + d)
and
2 - a
2,
must also be in
2 - c > 0
since
-C e T^,
S^,
and
K,
so that
and
2 - a > 0
and
a >_ 0.
(364)
by (361) and (363) respectively, and
the only possible locations for
T^.
But the locations
and
S£
A H* B
with
v.
are
,
A
with
li
and
lience there are only four cases to consider.
10(ii)-1.
A + B e
Then
2 - c ~ 2,
or
c = 0,
and we have
by (362),
S£,
are equivalent,
since tho problen regains unchanged by interchanging
x
2 - c < 2,
or
c > 0
Since
(363)
(365)
B H* C = (-e, 2 - f, d + e + f - A).
0 < 2 - f < 2
by (362),
and
and (362), the possible locations for
10(I1)-1-1.
II + C e S^.
-2 < -e < 0
d + c + f-.4<2
B + C
Then
Sincc
are
by (361)
and
d + e + f - A = -2,
Ty
or
79
and
e + 2 - f <_ 2,
We have
< 2
d = 2 - e - f,
(366)
c < f.
(367)
or
C + A = (2 - c, -a -f, b + e + f - 4),
by (362),
-a - f < 0
by (362)and (364), and
by (360) and (362), the possible faces for
and
and since
C + A
0 < 2 - c
b + e + f - 4 < 2
are
S^,
,
lo(n)-i-i-i.
C + A e
Then
-n - f = -2,
or
a = 2 - f,
and
2~e + 4~b-o-f<_2,
(368)
or
4 < b + 2e + f.
(369)
Now, using (368), (366) and (365), v:c have
- f).
Since
A + B e S^,
we must have
A + IJ - (2, f, 2 + b - e
f-t-e+f-2-b<_2,
or
e + 2f < b + 4.
Conbininp. (369), (367) and (370), v;e have
+ 2f <_ 2b + 4,
(370)
4 <_ b + 2c + f <_ b + e
hence
b >_ 0.
Now
A + 11 + C - (2 - e, 0, b - 2),
(362) and
-2 < b - 2 <_ 0
and we have
from (371) and (360).
cannot be on interior point of
K,
(371)
0 < 2 - e < 2
Since
it oust be that
from
A + 15 + C
b - 2 = -2,
b = 0.
or
(372)
Then, using (372), (363), (366) and (365),
2
det =
0
f - 2
0
2 - e - f
4 - e - f
= 16 - So - 3f + 2e2 + 2f2 + 6eC - e2f
- cf2,
(373)
80
which must be minimized under the conditions (362) and (367).
Testing
for critical points, we have
9
dot = -8 + 4e + Cif - 2ef - f
3e
= 0
(374)
and
3
yjf dot = -8 +
+ Go. - c
2
- 2cf - 0,
Subtraction flives (e - f)(e + f - 2) =0,
by (362), wc must have
e = F.
which has the solutions
e = 2
the latter case (373) fiives
Then (374) becomes
det = 208/27.
2
3
det - 16 - lGe + lOe - 2e ,
e = A/3.
2
= 0,
In
The only edp.e of the region
e = f.
On this edjje,
and a minimum is obtained at the critical
e = 4/3,
)-l-l-2.
C + A e S^.
10(Il
Then
b = 2 - e - f;
and
e + f - 2 > 0
-8 + 10e - 3c
(excluded by (370)) and
which does not lead to a previous case is
point noted above,
and since
(375)
2-e + a+ f_<2,
b + e -I- f - 4 = -2,
or
(376)
or
a + £ <_ e.
(377)
But in combination with (364) and (367), (377) shows that we nust have
a = 0
and
det =
e = f.
Then
2
0
2 - 2e
0
2
2 - 2e
e
e
4 - 2e
10(II)-l-l-3.
-4 - 4,
= 16 - 16c + Se2 = 8 + 8(e - I)2 _> 8 > 196/27.
C + A e T^.
Then
or
a + b = 6 — 2f.
2—e+a+f+b+c+f
81
But
a + b <_ 2
by (300), and
6 - 2f > 2
by (362).
Hence this case
is impossible.
C + A t Tj,
10(II)-l-l-4.
- f = A,
Then
2-e + a + f + 4- b- e
or
b = a + 2 - 2e;
(378)
a + f < 2;
and
(379)
4-b-e-f<2, which with (378) r.ivcs
c < a + f.
l.'e also have
more,
K,
e + f > 2
(380)
from (362), and
a >_ 0
A + B + C » (2-e, 2 - a - f, a- 2e)
and since
0 < 2 - e < 2
(364) and (379), and
by (362),
a - 2e < 2
from (364).
Further­
cannot be interior to
0<2-a-f<2
by (362),
by (363) and (362), there are at most
three possibilities:
a ~ 2e <_ -2;
2 - e + 2 - a - f + .a - 2 e > 4 ,
ing (362); or
-a
3e + f £ 0 ,
Hence (381) must hold.
2 - e
3
2
v dot = 4e - e - ef = 0
9a
+ aef.
has the solutions
e = 0
(382)
and
both of which contradict (362), there arc no critical noints
in the region.
Two portions of the boundary must be considered: those
determined by (364) and by (381).
and
Then we have
= 16 - 8e - 8f + 4e^ + 2f^ + 2ef - 4ae •(- ae^*
4 - 4 - f
e + f = 4,
2a -!• f <_ e,
2 + a - 2e
det =
Since
contradict­
2-e + 2- a- f + 2e-a^4, which sives
contradicting (3S0) and (364).
2
which r.ives
(381)
a = 2e - 2.
Hence we consider1 the cases
a = 0
82
10(II)-l-l-4-l.
a = 0.
Then (381) reduces to
c > 1,
(333)
and (382) becones
dct » 16 - 8e - 8f + 4e2 -I- 2f2 + 2ef.
(384)
Since the equation
--- det = 8 - 8e - 2f = 0
3e
controdictr.
the conditions
e
1
are no interior critical points.
(383) and
f > 0,
from (362), there
On the edp.e
e = 1,
v?e have
det
= 12 - 6f + 2f2 > 15/2 > 196/27.
10(II)-l-l-4-2.
•
a + 2 « 2e.
Then (364) p.ives
e > 1,
(385)
det = 16 - 8f - 6o2 + 2f2 + 2c2f + 2e3.
(336)
arid (391) becomes
Testing for criticnl points in the rcpion described by (362) and (385),
v/e obtain the equations
V- det = -12e + 4ef + 6e2 = 0
3e
(387)
and
—• det - -3 +
Since
e
+ 2c2 «• 0.
(388)
cannot be zero by (385), v;e may write (387) as
and usinp, this to eliminate
This has the solutions
critical points.
+ 2f2 > 15/2.
1
f
and
Alonp, the cdf»e
from (388) uc obtain
2,
e
2
2f = 6 - 3e,
- 3e + 2 = 0.
so that there are no interior
e = 3,
we apain have
det - 12 - 6f
83
10(II)-l-2.
= h,
i! + C e T2.
Then
e + 2- f + d + e + f- 4
or
d + 2c - 6.
But
d <_ 2
from (361), and
c < 2
from (362), so that this situation
cannot occur.
B + C c T 'y Tho.ii
10(II)-l-3.
= 't,
e + 2- f-l-4-d-c-f
or
d = 2 - 2f;
and
(3C9)
d+c+f-4>-2, which with (339) pives
e > f.
We apain have
(390)
C + A - (2 - e, -2 - f, b + e + f - 4), and the situa­
tion is like that of 10(IT)-1-1, except that
T^
is now occupied lw
-B - C.
or
Tjj.
Thus
C -t* A
is in either
10(II)-l-3-l.
S^,
C + A e S^.
S^,
Then
a + f = 2,
(391)
which with (360) and (304) fives
|b|< f.
(392)
Then we must minimize
2
det =
f - 2
b
2
2 - 2f
f
F 16 - Se - 8f + 4f
(393)
ii - e - f
on the rcpion described by (362) and (392).
points in the rep.ion, since
(362).
+ Gcf - 2be - 2ef'
There are no critical
g
—• det = -2e = 0
cannot hold in view of
The only portions of the boundary which do not correspond
to
8/.
earlier cases arc those pivcu by (392).
tions
b - f
and
l.'e must consider the two situa­
b - -f.
10(II)-l-3-l-l.
b = f.
Then (393) becon».er»
dct = 16 - 8e - 8f + 4f2 + 4ef - 2ef2,
9
2
det = *-S + 4f - 2f =0
3e
and there aro no critical points since
no real roots.
has
Ko boundary points need be considered.
10(IT.)-l-3-l-2.
b = -f.
Then (393) becomes
dct « 16 - 8e - 8f + 4f2 + Sef - 2ef2,
8
2
— det = -8 + Sf - 2 f = 0
and there are no critical points since
only
f = 2
ae
as a solution, and this cannot happen in the region under
consideration.
10(11
No boundary points need he considered.
)-l-3-2.
C + A e Sy
Then
b + e + f - 4 = -2,
or
b = 2 - e - f.
Usinp (394), (389) and (365) we have
-2f).
(394)
A -)• IS •+ C = (2 - e, 2 - a - f,
Since this cannot be an interior point of
0 < 2 - c < 2
by (362),
fact that
C + A e K,
-2f £ -2,
pivinp
0< 2-a- f < 2
and
-2f < 0
K,
and since
by (362), (364) and the
by (362), we rr.ust have: either
f >_ 1;
or
has
2 - c + 2 - a - f + 2f>_4,
which
(395)
f t i v e s a + e <_ f .
I3ut t h e
latter inequality contradicts (390) and (364), so that (395) r.uist
hold.
Combining (395), (390), (364), (363) and (362), v:e have
< e < 2,
0<^a<2.
Then we obtain
1 <. f
85
det =
2
-a
0
2
e
f
2 - e - f
= 16 - 8e - 8f + 2c2 + 4f2 + 2ef - 2ae + 2aef.
2 - 2f
4 - c - f
There are no critical points in the interior of the region, since
£
--- del = 2e - 2e£ = 0
3a
imnlies that either
e = 0
or
only faces v:hich do not lead to a previous case are
dct « 12 - 6e + 2e
2
>_ 15/2,
that in 10(TI)-l~l-4-l,
lO(II)-l-3-3.
- f = 4,
and
with
e
f = 1.
f *= 1,
The
for which
a = 0,
which pives a situation like
and
interchanged.
f
C + A c
Then
2 - e -I- a -I- f -t- 4 - b - e
or
s
b
a + 2 - 2e;
(396)
and
a + f < 2.
Then by (3%), (389) and (365) v/e have
- 2f).
< 2,
Since this is in
S^,
(397)
A + 15 = (2, 2 - a, a + 4 - 2e
we must have
2 - a + 2e + 2f - a - 4
or
e + f <_ a + 2.
(39S)
On the other hand,
A -)• IS + C = (2 - e, 2 - a - f, a - e - f)
be interior to
and since
< 2
K,
0 < 2 - e < 2
by (397), (362) and (364), and
by (362),
a - e - £ < 0
cannot
0 < 2 - a - f
by (363) and (362),
we must have either
a - e - f < -2
or
2 - e + 2 ~ a - f + e + f - a ^ 4 ,
(399)
which reduces to
a = e + f - 2,
(400
86
and we obtain
dec =
2
2 - c - f
0
2
f - c
= 16 - S£ + 4f2 - 8cf + 2c2f + 2ef2,
2 - 2f
f
(402)
4 - c - f
to be minimized on the region described by (362).
The ed^.es lead to
previous cases, and in testing for critical points ue obtain
•£~ dot = -8f + Acf + 2f2 = 0,
3e
(403)
and
~ det = -8 + 8f - 8e + 2e2 + 4ef - 0.
Sf
Since
f
(403).
- 0.
cannot be zero by (362), we nay write
Elimination of
f
f = 4 - 2e
in (404) ftives the equation
The only positive solution of this quadratic is
,
,„ *
and we obtain
(404)
1 ,
2224 - 640/10
det =
"27
51
199
~27~
>
196
~2T
But if (400) should hold, then we r.iust have
3c
e =
2
from
+ 4e ~ 12
2 .|_ 2 /]q
„
.
critical point.
a = 0
by virtue of (373).
3
In this case
det =
2
0
2 - 2e
0
2
2 - 2C
e
f
= 16 - 8e - 8f + 4e2 + 4f2 = 8 + 4(e - l)2
+ 4(f - 1)
4 - e - f
10(II)-2.
-
A + n e S3<
> 8.
Then
b + d = 2,
(405)
and it follows from (369) that
c <_ 2 - d = b.
(406)
a + b < 2,
(407)
From (360) we have
87
and since
that
A + R = (2 - c, 2 - n, 2)
2 - c + 2 - a £ 2,
is in
K,
it must be the case
or
«*i + c >_ 2.
(40S)
But if (406), (407) and (408) arc to hole! siniultanoously, equality
must occur in all three.
Thus
b = c = 2 - a.
Usinf, (409) and (403), we obtain
and
(409)
B + C =• (-b -e, 2-f", e + f- b-2)
C + A = (2-e, b-f-2, b+e+f-4).
Since
B + C e K,
we
have
b -f - e <_ 2
and
b + e + 2- f + b + 2- e- f<_4,
(410)
or
b .> f.
Since
C + A e IC,
ue have
(411)
b - f - 2 >_ -2,
or
b >_ f.
(412)
Combining (411) and (412) wc see that
b = f,
e + f £ 2.
e + f > 2
Hut we have the condition
which with (410) cives
in (362).
Hence
this situation cannot occur.
10(II)-3.
and
A + D
Then
E
b + d = -2,
or
b = -d - 2;
(413)
a + c > ?..
(414)
2-c+2-a£2, or
Combininp (413) and (361) we have
C £ d + 2 = -b,
(415)
and frot.i (360) we have
a - b < 2.
(416)
88
The inequalities (414), (415) nnd (416) together imply that
c = -b = 2 - a.
Using (417) anc! (413) wc obtain
Since this must be in
v/hich reduces to
K,
f > 2,
(417)
15 + C - (-c -e, 2-f, c + e+ f-6).
v;e liave
c + e + 2 - f + 6 - c - e - f<_4,
Rut (362) pives
f < 2.
llcnce this case docs
not occur.
10(II)-4.
A + B e T/(.
—-
2 - c < 2,
and
2 - c + 2 - a - b - d M, or
b = -a - c - d;
(418)
or
2 - a < 2,
We have
Then
c > 0;
(419
a > 0.
(420)
or
B + C = (-c -e, 2- f, d+e+f -4).
(419) and (362),
0<2-£<2
by (362), and
T^.
-c - e < 0
d + c + f - 4 < 2
(361) and (362), the only possible locations for
and
Since
B + C
are
by
by
S^,
S^,
Thus v;e have-four subcases to consider.
10(II)-4-l.
B + C e Sj.
Then
-c - e = -2,
c = 2 - e.
or
(421)
UsinR (418), we have
C + A = (2 - e, -a - f, 2e + f - a - d - 6).
Since
by (362),
0<2-e<2
-a - f < 0
2e+f-a-d-6=b+e+f-4<2
possible faces for
C + A
10(II)-4-1-1.
are
S^,
by (360) and (362), the only
S^,
C + A c S£.
by (362) and (420), and
T^
and
T^.
Then
a + f = 2.
(422)
89
Nov
A + 15 + C - (0, 0, 2c + 2f - 8)
and since
2e + 2f - 8 < 0
cannot be in the interior of
by (362), v:e must have
2e + 2f - 8
K,
-2,
or
e + f < 3.
Also, since
have
(423)
]J+C=(-2, 2-f, d + e + f-4) is in
2-f + 4-d-e-fi<2,
wc must
or
d >_ 4 - e - 2f.
Since
C + A = (2 - e, -2, 2e + 2f - d - 8)
2 - c -I- d + 8 - 2e - 2F < 2,
(424)
is in
S^,
we must have
or
. d <_ 3c + 2f - 8.
Together (424) and (425) p,ive
(425)
4 - c - 2f <_ 3e + .2f - 8,
or
e + f >_ 3.
Then from (423) and (426) v;c have
det =
e -I- f = 3,
2
1 - e
e - 2
2
d
e
3 - e
1
10(II)-4-1-2.
(426)
and
-d - 1
= 12 - 6e + 2e
C + A c S^.
Then
> 15/2.
2e + f - a - d - 6 = -2,
or
d = 2c + f - a - 4;
and
2 - c + a + f < 2,
(427)
or
a + f < e.
Also, since
have
15 + C = (-2, 2 - f, 3e -f- 2f - a - 8)
2-f + a + 8-3e-2f<_2,
(428)
is in
S^,
we must
or
a + 8 < 3e + 3f.
(429)
90
Mote that ue nay combine (428), (429) and (420) to obtain:
< e;
8 < a + 8 <_ 3e + 3f < 6e,
f < a + f
from which uc have
e > 4/3.
Since
to
(430)
A + Ji -h C = (0, 2 - a - f, 2e + f - a - 6)
K,
and since
0<2-a-f<2
2e + f - a - 6 < 0
- 6 £ -2,
cannot be interior
by (428), (420) and (362) and
by (420) and (362), wo must have that
2e + f - a
or
2e + f £ a + 4.
Combining (431) and (429), we have
(431)
2e H* f + 4 ^ 3c + 3f,
or
4 < e + 2f;
and combining (431) and (428), we have
(432)
2e + 2f <_ a + f + 4 <_ c + 4,
or
e + 2f < 4.
Combining (432) and (433) v.*e have
e 4- 2f = 4.
(429) and (431) p.ives
and
that
a = 4 - 3f.
The conditions
Elimination of
4 - 3f < a
3f - 4
e
in
respectively, so
Lxpressinp. a]l variables in terns of
2
det =
a < 4 - 3f
(433)
f,
we have
f - 2
2 - 2f
2
0
4 - 2f
f
f
a > 0 (420) and
e < 2
= 16 - 8f - 4f2 + 4f3.
(362) £ive
1 < f; < 4/3.
Since
r—
~ dct = -3 ~ 8f + 12f^ = 0 has the positive solution f = —v
»
df
'
3
2
2
and since d (det)/df
is positive there, this is the point at which
• •
i
r i
the minimum value of djp_t occurs.
vi .
i
-r
That value is
352 - 56/7
N
>
196
'27"*
91
10(IT)-4-l-3.
- d - 6 = 4,
C + A e T3.
Then
2-c+a+f+2e+f -a
or
e + 2 f = d + 8 .
Ijut
e + 2f < 6
by (362), nnd
d + 8 > 6
by (361).
Hence this can­
not occur.
10(II)-4-1-4.
- 2e - £ = 4,
and
C + A e T£.
Then
2-c + a + f + a + d + 6
or
d = 3e - 2a - 4;
(434)
a + f < 2;
(435)
a -I- d + 6 - 2e - f < 2,
which with (434) rives
e < a -(• f.
Since
(436)
II + C = (-2, 2 - f, 4e -I- f - 2a - 8)
2 - f + 2a + 8 - 4c - f < 2,
is in
S|,
we raust have
or
a + 4 < 2e + f.
On Lhe otlier hand, since
cannot be interior to
(420) and (362), and
only have
K,
(437)
A + U + C = (0, 2 - a - f, 2e + C - a - 6)
and since
0<2-a-f<2
2e+f-a-6<0
2e + f - a - 6 < -2,
by (435),
by (420) and (362), we can
or
2e + f _< a + 4.
(438)
2c + f = 2 + 4 .
(439)
Then (438) and (437) p.ive
Usinp, (439) to eliminate
det -
a,
we have
2
4 - 2e - f
f - 2
e - 2
2
4 - e - 2f
e
f
4 - e - f
= 4S - 24e - 24f + 4e2 + 4f2
+ 2ef
2
2
+ 2e f.
(440)
92
The region is described by (362) and the relations
> 4
e + f < 3,
which result from (435) and (A36) respectively.
need be considered.
e+2f
No boundary points
Testing for critical points, we have
det = -24 + Re + 2f2 + 4ef = 0
^',1^
De
and
4r det = -24 + Sf + 4ef + 2c2 = 0.
3f
Subtracting, (442) from (441) Rives
since
e + f - 4 < 0
becomes
e =
by (362), tee can only have
-24 + Se + 6e
-2 + 2/ld
y
•
2(e - f)(e + f - 4) = 0,
2
= 0,
Then (441)
which has as its only positive solution
,//M .
This with (440) rives
10(II)-4-2.
e = f.
and
B + C c S^.
. „
2224 - 640/10
199
det =
£7
>
Then
d + c. + f - 4 = -2,
196
or
d = 2 - e - f;
(443)
c + e < 2.
(444)
and
Since
C + A - (2 - e, -a - f, 2e + 2f - a - c - 6)
(418), and since
and (420), and
0 < 2 - e < 2
2e+2f-a-c-6<2
the only possible locations for
10(II)—4 — 2—1.
the points
A
the situation
C + A e S^.
and
B
A c S^,
C + A
C + A E S^.
-a - f < 0
fron (362), (419) and (420),
are
S^,
T^
and
T^.
Bv interchanging the roles of
and of the coordinates
U z S^,
f.rom (362)
C e TJ ,
x
and
y,
A + R c T^,
wo achieve
B + C t Sj,
But this case was treated in 10(TI)-4-l-2.
10(II)-4-2-2.
or
from (362),
froip (443) and
C + A e S^.
Then
2e + 2f - a - c - 6 = -2,
93
c - 2o -I- 2f - a - 4.
Using (445), (443) nncl (418) we obLain
4 - 2e - 2f),
• (445)
A + B = (0 + a - 2e - 2f, 2 - a,
11 + C = (4 + a - 3e - 2f, 2 - f, -2)
- c, -a - f, -2).
Since all of these are in
spectively, that:
6 -I- a - 2e - f < 2,
K,
and
C + A = (2
we must have, re­
or
a + 4 < 2e + 2f;
3 e H - 2 f - 4 - a + 2 - f _ < 2 ,
(446)
or
3e + f <_ a + 4;
and
2 - c + a + f _< 2,
or
a + f
But combining (446) and (447) pives
comb inin f, this with (448) pives
dicts (420).
(447)
e.
i'
3e + f <_ 2e + 2f,
a + f <_ f,
or
a <_ 0.
(448)
or
e <_ f;
and
This contra­
Hence this situation cannot occur.
10(lI)-4-2-3.
- a - c - 6 = 4 ,
C + A e T .
Then . 2 - e + a + f + 2c + 2f
or
e + 3 f = c + 8 .
Hut
e + 3f < 8
by (362) and
c -I- 8 > S
by (419).
l!ence this case
cannot occur.
10(II)—4— 2—.
+ a + c + 6 - 4 ,
C + A c TJ.
Then
2 - e + a + f - 2e - 2f
or
c = 3 e + f - 2 a - 4 ;
(449)
a + f < 2.
(450)
and
Using (449) to eliminate
-2).
Since this is in
c,
S^,
we have
P. + C = (2a + 4 - 4c - f, 2 - f,
v;e nust have
4e + f - 2a - 4 + 2 - f < 2,
94
or
2e _< a + 2.
On the other hnnd,
A + K + C - (2-c-e, 2 - a - f", a- 2c)
be iu the interior of
and (362),
(451)
K.
Since
0<2-a-f<2
= e + f- a- c- 4<0
0<2-c-e<2
cannot
bv (444), (419)
by (450), (420) and (362),
and
a - 2e
by (449), (419), (420) and (362), v?e nur.t have
either
a - 2e <_ -2
or, usin p. (449) attain,
which reduces to
(452)
2a + 6 - 4e - f + 2 - a - f + 2e - a > 4,
e + f
2.
But the latter inequality contradicts
(362), so that (452) must hold.
Takinp, (452) and (451) together, v;e
have
a = 2e -
(453)
Me nay then use (453), (445), (443) and (41C) to eliriinnte all variables
other than
e
is in
we must have
and
T,,
4
and
4 - 2e < 2,
f.
Since
A +. 1> = (2 + e - f, 4 - 2c, 2 - e - f)
2 + e - f < 2,
or
c < f,
(454)
e > 3.
(455)
or
The conditions (454), (455) and (362) are the only restrictions neces­
sary on
e
2
dot -
e - f
and
f.
2 - 2e
Then
0
2 - e - f
4 - e - f
= 16 - So + 4e2 - 8ef + 2e2f + 2ef2.
(456)
95
In testing For critical points V/G obtain
--- det = -8 + 3c: - 8f + 4eE + 2f2 « 0
og
(457)
7%- dct = -8e + 2g^ + 4ef - 0.
3£
(458)
and
Since
e r 0
eliminate
e
by (455), we have
frou; (457), we obtain
,
.
tive solution is
199 . 196
> ~~~ > "tl-s-.
~27~
27
,
f
-2 + 2/10
-^y
,
from (458).
2
3f * + 4f - 12 = 0.
. . .
which yields
Usin?, this to
The only posi-
2224 - 640/l0
det
No boundary 1points need be considered.
'
10(II)-4-3.
- 4 = 4,
e = 4 - 2f
B + C c T?.
Then
c + e + 2- f + il + e+f
or
c + d = 6 - 2e.
But
c + d _< 2
by (361), and
6 - 2e > 2
by (362).
Hencc this cose
cannot occur.
10(II)-4-4.
- f = 4,
Pi + C e T^.
Then
c + e + 2- f + 4- d- e
or
d = c + 2 - 2f;
and
(459)
•
c + e < 2.
(460)
We have
C + A = (2 - e, -a - f, c + 3f - a - 2c - 6).
- e < 2
by (362),
- 2c - 6 < 2
-a - £ < 0
by (420) and (362), and
by (419), (420) and (362), and since
only possible locations Cor
10(II)-4-4-l.
C + A
are
C + A c S^.
S^,
Then
a » 2 - f;
Since
0 < 2
e -I- 3f - a
—15 — C c T^,
and
the
T^.
-a - f = -2,
or
(461)
96
and
2-e + a + 2c + 6- e-3f<_2, which with (461) pives
c + 4 £ e + 2f.
Now
K.
<A62)
A + 15 + C = (2 - c - e, 0, e + 2f - c - 6)
Since
0 < 2 - c - e < 2
- c - 6 < 0
cannot he interior to
by (460), (419) and (362), and
by (362) and (419), we must have
e + 2f
e + 2f - c - 6 < -2,
e + 2f <_ c + 4.
or
(463)
Then combining (462) and (463) we must have
c + e + 2f - 4.
Usinp this to eliminate
2e + f - 6).
c,
we have
Since this is in
T^,
(464)
H + C - (4 - 2e - 2f, 2 - f,
we must have
4 - 2c - 2f > -2,
or.
(465)
e 4- f < 3;
and
2e + f - 6 > --2,
or
2e + f > 4.
Uc also have the conditions of (362).
det =
2
f - 2
4 - 2c - f
4 - e - 2f
2
e-2
e
f
4 - e - f
(466)
Then
= 48 - 24e - 24f + 4e2 H- 4f2
2
2
+ 2e f + 2ef .
This is the same expression as that obtained in 10(II)-4-1-4, equation
(440), and the region is the sane if the roles of
terchanged.
10(II)-4-4-2.
A
and
(Note that the determinant is symmetric in
Thus here as in the earlier cose we obtain
roles of
e
and
E
C + A e S^.
and of
x
and
in case 10(II)-4-2-3 is obtained.
199
det > ~<pj~
>
f
e
are in­
and
f.)
196
~2jT'
If, as before, we reverse the
v,
then the situation considered
97
10(II)-4-4-3.
+ 6 - e - 3f = 4,
C + A
c
Tj.
Then
2 - e + a + f + a + 2c
or
a = e + f- c- 2.
('<67)
c + 2 < e + f.
(463)
With (467), (420) fives
The region under consideration is that described by (362), (419), (460)
and (463).
2
det =
Then
2 + c - c - f
f - c - c
-c
2
c + 2 - 2f
c
f
4 - c - f
- 16 + 8c - 8f + 4c2 + 4f2 - 8cf
- 8cf + 2e2f + 2ef2.
(469)
Testing for critical points, we have
dot = 8 + 3c - 8f = 0,
(470)
det - -8f + 4ef + 2f2 = 0,
(471)
oC
dC
and
J- jet = _s + Sf - 8c - 8e + 2e2 + 4ef = 0.
Adding (470) and (472), we have
e
nor
f
-Se + 2cfc + 4ef = 0.
(472)
Since neither
can be zero by (362), we have
e + 2f = 4
(473)
2e + f » 4.
(474)
and, from (471)
Solving (473) and (474) simultaneously gives
(470) V7C obtain
= 196/27.
e - f = 4/3.
Then from
c = 1/3, and from (469) we obtain, once again,
No boundary points need be considered.
det
98
11(1).
c bdry K.
- c - f)
A E
This is irapossihle.
where
4 - b - e - f),
e > 0
C
K.
15 e
Then
K
,
This also is impossible.
cannot be in
K
Then
since
12(T).
C e T2>
A - 15,
B - C,
A = (2, a, b),
f > 0.
A e Sx,
be as in 11(1).
e
C. bdrv
„ K.
and
Let
which camiot be in
11(11).
e bclry
]; c S2,
C - A
C = (-c, f, 4
C - A = (-c - 2, f - a,
since
C c 1'2>
Let
-e - 2 < -2.
A + B,
li + C,
B = (c, 2, d),
C + A
and let
r. + C = (c - c, 2 + f, 4 + d - e r f),
which
2 + f > 2.
Ac S1,
1!
Tr
c
C e T3,
A - B,
B - C,
C - A
Let
,
A = (2, a, b),
B = (c, d, 4 - c - d),
where
where
|a| + |h| <_ 2;
c < 2,
d < 2,
and
('(75)
c + d >' 2;(476)
and
C = (e, -f, 4 - 4 - f),
Then
e < 2,
f < 2,
A - B - (2 - c, 2 - d, b + c + d - 4).
by (476);
a - d < 2
b + c + d-4<2
points nay be in
A - B
where
are
12(1)-1.
because
a < 2
and
e + f > 2.
We have
by (475) and
0 < 2 - c < 2
d > 0
by (476);
by (475) and (476)*, and by Lemma 4 no more lattice
T^
S3>
or
T^.
Thus the only possible locations for
"T/,
i1luI
I'v
A - B
c
S^.
Then
a - d = -2.
We have
< 2,
(47S)
B - C = (c-e, d -I- f, c+f-c-d).
(477), we obtain
(477)
-2 < c - e < 2,
0 < d *f f
and
and since no r.ore lattice points may be in
point must be in cither
T0
or
.
Since, usin?. (476) and
- 2 < e + f - c - d
T^
or
T^,
the
99
12(I)-1-1.
C* = B.
Then
IJ - C e S0.
A" c S,,,
- A* e bdry K,
and
lent to
A* c S^,
|det(A*, B", C")|
x
and
B* E S^,
A- = B » A,
C* c
B* = B - C,
A* - H*,
II-• - C*,
C)
|= [dot(A, B, C) j.
y
C*
By inter­
we sec that this: situation is equiva­
C" E T^.
But the "problem of nininizinr,
under these conditions was solved in case 6(1).
12(I)-l-2.
- d = 4,
B* c S2,
[dot(A*,
changing the roles of
Let
Ii - C e T^.
Then
e-c + d + f + e-hf-c
or
c = e + f - 2;
(479)
d + f < 2.
(480)
and
Usinf* (478), wo have
since
-2 < e - 2 < 0
and (477),
and
T^
C - A = (e - 2, 2 - d - f, 4 - b - e - f),
by (477),
0 < 2 - d - f < 2
and 4 - b - e - f > -2
13* = B - A,
A * - B",
B" - C * ,
(A*, B, C)
|
,
C* = C - A.' Then
C * - A " e bdry K ,
By reflection in the
out changing the other conditions.
by (480), (476)
by (475) and (477), and since
are not available, we must have
A* = -A,
and
C - A
in
A* e Sj,
S^.
B* c S2>
|det (A 1 '-',
and
yz-plane
Now wc let
ve nay take
C* c Sj,
, C'' ;|
) = Jdet
A" E
with­
Then the resulting situation is
that v;hich was treated in 5(1).
12(I)-l-3.
- f = 4,
B,- C E T^.
Then
c-e+d+f+c+d-e
or
c = e + 2 - d;
and
(481)
100
d + f < 2.
The analysis of possible locations for
12(I)-l-2, except that
Tj
(482)
C - A
proceeds as in case
cannot now be immediately excluded, so
that we have two locations to consider.
If
C - A e S^,
we may pro­
ceed from there as in 12(I)-l-2 and reduce the problem to case 5(1).
O n the other hand, if
+ 4- b- e- f = 4,
C - A £
,
then w e have
2 - e + 2 - d - f
or
2e + 2f < 4 - b - d.
But from (475) and (478) we have
(483)
2-d-b = -a-b<_2,
or
b + d > 0.
From (483) and (484) v?e obtain
(484)
2e + 2f <_ 4,
which contradicts (477).
llence this situation cannot occur.
12(I)-2.
A - B e S£.
Then
b + c + d - 4 = - 2 .
(485)
Just as in 12(1)-1, we have three possible locations for
T2
and
T^^
12(l)-2-l.
C* = 15.
Then
B - C E S0.
A" e S2,
K" c S^,
mutation of the coordinates
E
Slt
B - C:
B* e S2
- A* c bdry K
and
and
het
and
(y -> x,
C* e T^
A" = » - C,
C* c T^,
x •* z,
l.'e also have
z
B* = B - A,
and by a cyclic per­
y)
A* - li'-,
wo may take
B* - C*,
|det(A", BA, C*)
| = |det (A, B, C)|,
A"
C*
so that this
reduces to the situation of case 9(1).
12(I)-2-2.
B - C
e
T,,.
Then
e-c + d + f + e+ f-'c
101
c = e + f - 2;
(48f,)
d + f < 2.
(487)
and
Using (486) and (485), wo have
< e - 2 < 0
by (477),
C - A = (c - 2, -n - f, d).
-a - f < 2
by (475) and (477), and
by (477), and since tliere is already a lattice point in
possible locations of
C - A
12(I)-2-2-l.
= A - I>.
the
Then
xy-planc
,
B* e S2,
we may take
and
and
C - A c S£.
A" e
- A* c bdry K,
arc
C* E S^.
0 < d < 2
1^*
the only
T^.
Lot
and
Since -2
A« = A,
C',f t S^,
Also
B* = A ~ C,
C*
and by symmetry in
A:V - B*,
det(A, B, C) = det(A*, B*, C>,f) •
B* - C>':,
C*
Thus v/e have a
situation corresponding to that of 5(1).
12(I)-2-2-2.
C - A e T^.
Then
2 - e + a + f + d = 4 ;
and
a + f < 2,
(48S)
which with (488) r,ives
e < d.
Coinbininp. (489) and (487) Rives
(489)
e+f< d + f < 2,
which contradicts
(477).
12(I)-2-3.
- f = 4,
B - C E T^.
Then
c-c + d+ f + c + d- e
or
c + d = e + 2.
Usinp, (490) and (485), we have
cannot be in
K
since
C - A - (e - 2, -a - f, 4 - f),
4 - f > 2
by (477).
(490)
which
102
12(I)-3.
= 4,
A - B e T..
4
Then
2-c + a- d + 4- b- c- d
or
a — b = 2c + 2d - 2.
But
a - b
2
by (475), and
2c + 2d - 2 > 2
by (477), so that this
case is impossible.
12(l)-4.
- c - d = 4,
A - B c T^.
Then we have
2-c + d- a + 4- b
or
b = 2 - a - 2c;
(491)
d - a < 2;
and
(492)
4-b-c-d<2, which with (491) p.ives
a + c < d.
(493)
The consideration of possible locations for
12(I)-1, except that
face.
Hence
B - C
12(I)-4-l.
T0
B - C
is nov/ excluded sincc
is in either
B - C e S.,.
S„
I
or
proceeds as in
B - A
is in that
T..
4
Then
d + f = 2.
Usinp, (494) to eliminate
a + 2 c + d - e ).
> -2
Mow
by (493) and (476), and
-B.
•
we have
- 2 < e -* 2 < 0
- e = 4 - b - e - f > -2
bv
F,
C - A = (e-2, d-a-2,
b y ( 4 7 7 );
d - a - 2 < 0
B* = C - A,
C* = C.
d - a - 2 > c - 2
by (492);
by (475) and (477); and
Thus the only nossible faces for
12(I)-4~1-1.
(494)
C - A c S^.
Then
A* e S^,
with the change of variables
x -•
C. - A
Tj
are
In this case let
B* c Sj,
y ->• x,
and
z -• v,
a + 2c + d
is occupied
S„
3
and
T!.
4
A* - C - B,
C* c Tj,
v:e have
and
A« z S^,
103
11* c S2.
and
C* E T .
KG also have
/.« -
])••• - C*,
|det(As'!, B", C'-'O
|= |det(A, !), C)
|,
C* - A* e bdry K
so that this problem
be­
comes that of case 0(1).
12(J.)--A-l-2,
+ 2 c + d - e - A ,
C - A c Tj.
Tlien
2 - e + a + 2 - d +' a
or
(ACJ5)
a H* c = o.;
and
.
a + 2c + d ~ e < 2,
which with (495) f^ives
c + d < 2.
Rut this
contradicts (A77).
12(I)-A-2.
- f = A,
II - C e T/(.
d + f < 2,
c < 2
c-e + d + f + c + d- e
or
c
and
Then
d = e + 2;
which with (A96) pives
from (A76) and
2 < e + f
(A96)
c + f < c.
from (A77).
r>ut we have
lience this case cannot
occur.
12(11).
e bdry K.
then
A E Sx,
n e Tx,
This case is ir,'possible.
C E T3,
If
A
A + B,
and
A + B = (2 + c, a + d, A + b - c - d)
R
B + C,
C + A
are as in 12(T),
cannot be in
K.
since
B - C,
C - A
2 + c > 2.
13(1).
e bdry K.
A e Sj,
B E T1(
C E T^,
A - R,
Let
A = (2, a, b),
B = (c, d, A - c - d),
where
where
]a| -I- jbj £ 2;
c < 2,
d < 2,
and
(A97)
c + d > 2;
(A9S)
and
C = (e, -f, e + f - A),
where
e < 2,
f < 2,
and
e + f > 2. (A99)
104
Then
A - II = (2 - c, a - d, b + c + d - 4).
(49S),
a - d < 2
+ d - 4 < 2
by (497) and (498) (because
by (497) and (493), and since
by Lemna 4, the only possible locations of
and
T^.
0<2-c<2
and
and
arc excluded
A - B
8-c-d-e-f>0
angular faces with both
Sincc
arc
and
S^.
-2 < c - e < 2,
B - C
d + f > 0
by (498) and (499), and since the only tri­
y > 0
and
z > 0
are
and
which
B - C
are
Now we consider the four cases arising from the possible
locations of
A - B.
13(I)-1.
°r
b + c
S^,
are excluded by Lemma 4, the only faces which can contain
S£
by
d > 0),
Before considering these cases separately, we look at
= (c - e( d + f, 8-c-d-c- f).
and
Since
A - B c S^.
As just noted,
B - C
may be in
S2
s3.
13(I)-1-1.
C" = B.
Then
15* _ c*,
B - C c S2.
A"» z
,
Let
B* E S^,
C* - A* e bdry K
and
A-« = B - A,
C" E T^.
B* = 15 - C,
V'e also have
A* - B*f,
jdet(A», B», C*)
| = |dct(A, B, C)
|
.
But the minimum of this deterninant was found in case 6(1).
13(I)-1-2.
B - C e S3.
13(I)-1-1.
Then
A* E S2,
variables
y *+ x,
z -> y,
E T^,
B* e S^,
v. •> 2,
and
A»,
B*
and
C*! E TJ .
we obtain
C*
be as in
By the change of
A* E
B* e S2,
C"
and the situation corresponds to. that of 9(1).
13(l)-2.
13(I)-2-l.
= B.
Let
Then
13(I)-l-2.
Aft E S2,
A - B E S^.
Apain we have
B - C E S2.
B" e S^,
Let
C* E T^,
!i - C E
A* = B - C,
or
S^.
B* = B - A,
and we nay proceed as in
C*
105
13(I)-2-2.
C" = B.
Then
11 - C g S3<
A" e 5^,
Let
B» e S^,
C* e T^,
13(1)—1—J, this time interchanging
13(I)-3.
= 4,
A - B
c
1'3.
x
and
Then
B* = B - A,
and wc nay proceed as in
z.
2-c + d- a + b + c + d-/i
or
b
But
A* = 13 - C,
b - a <: 2
a = 6 - 2d.
from (497), and
6 - 2d > 2
front (498).
Hence this
situation cannot occur.
13(I)-4.
= 4,
A — II e T,.
Then
2-c + a- d + 4- b- c-d
or
a - b = 2c H- 2d - 2.
But
a - b <_ 2 by (497) and
2c + 2d ~ 2 > 2
by (498).
Hence tliis
case is also impossible.
13(IT.).
E bdry K.
A e S ,
B E T1,
C e Tj,
A + II,
B + C,
C + A
The sane argument as that used in 12(11) shows that tliis
is impossible.
14(1).
E bdrv K.
A e S1,
BE 1^,
C E
A - P>,
B - C,
C - A
This may be shown impossible by the same argument as that
used in 11(1).
14(]T).
E bdry K.
A E Slf
B E Ty
C
c
T?,
A + B,
B + C,
C H- A
Af,ain the reasoning used in 12(IT) shows that this is
impossible.
15(1).
£ bdrv K.
A E S1?
R c Tj ,
This is impossible.
If
C E
A - 15,
A = (2, a, b)
and
B - C,
C - A
C = (-e, f,
106
c + f - 4) where
e > 0,
and this cannot be in
15(11).
e bdry K.
then
K
since
A e
C - A = (-e -2, f-a, e+f-b-4),
-e - 2 < -2.
I*. c T1,
C c T^,
A + B,
E + C,
C + A
This ease also may be shown to be impossible just as was
12(11).
16(1).
c bdry K,
A c Sy
7'. e T2,
C c Tj,
A - B,
B - C,
If, in the argument used in 11(1), we replace
C
C - A
by
B,
we see that the present case is impossible.
16(11).
E bdry K.
A e Sv
r. c T2,
C. e TJ(,
A + B,
D + C,
C + A
Let
IJ = (-c, d, 4 - c - d),
where
d < 2,
(500)
and
C = (-e, -f, h - c - f),
Then
K,
where
B + C = (--c - e, d - f, 8 - c - d - e - f).
f < 2.
(501)
For this to be in
we must have
v,
-c - e >_ 2
(502)
and
8 -
c
- d - e - f < _ 2.
Conbininr, (502) and (503) p.ives
and (501).
But this contradicts (500)
lience this situation is impossible.
17(1).
e bdry K.
d + f > 'i.
(503)
A e
,
V, c T2»
c
e
A - B,
B - C,
C - A
The renarh made in case. 16(1) also applies here, and this
case is irpnossible.
107
17(H).
c bdry K.
A c Sr
B E T2,
C E T|,
A + B,
11 + C,
C + A
I.ot
A = (2, a, b),
I'» = (*~c, d, A - c - d),
where
where
|a| + |b| < 2:
c < 2,
d < 2,
and
(504)
c + d > 2:
(505)
and
C = (-e, -f, e + f - 4),
Then
where
e < 2,
f < 2,
A + B - (2-c, a + d, 4+b-c-d).
from (505):
a + d > - 2
by (504)
17(II)-l.
Since
and (505): and
by (504) and (505); and since
and
the only possible locations for
and
0 < 2 - c < 2
4 + b - c - d > - 2
arc excluded by I.cnnn 4,
A + B
A + BC S2.
e + f > 2. (506)
are
Sg,
^3*
^3
anc*
.
Then
a + cl = 2,
(507)
4 + b - c - d_<2.
(508)
and
V.'e have
B + C = (-c -• e, d - f, e + f ~ c - d).
From (505) and (506)
v
we obtain
-c - c < 0,
Since Lcrana 4 excludes
B + C
are
Sj, T^
17(II)—1—1.
and
-2 < d - f < 2,
T| and
and
-2<e+f-c-d<2.
the only possible locations for
T£.
B + C E Sj.
Then
c + e = 2.
Usinf. (507) and (509) we obtain
- d).
A + B + C = (0, 2-f, b+e+f-c
This cannot be interior to
(506) and
K,
and we have
b + e + f- c- d;<e+f-2< 2
the only possibility i s that
(509)
0 < 2 ~ f < 2
by (508) and (506).
by
Thus
ins
b + e + f - c - d < 2 .
But
C +" A"= (2-c, 2-d-f, b+c+f-4),
< c H- d - 6 < -2
that
C + A
by (510) and (505).
v;as in
17(XI)— X—2.
- f = 4,
K.
(510)
and
b + e + f - A
This contradicts the nssunntion
Hence this case cannot occur.
B + C e T^.
Tlien
c + e + d- f + c + d- c
or
c = f + 2 - d:
(511)
c + e < 2.
(512)
and
Combining (511) nnd (512), V.'G obtain
> 2
from (506) and
1 7( I I ) - l - 3 .
- d = 4,
ci < 2
e + f < d.
from (505).
R + C c Tj!.
Rut we have
e + f
Hence this case cannot occur.
Then
c + e + f - d + e + f - c
or
e + f - d + 2:
(513)
c + e < 2.
(514)
and
From (505), (5]3) and (514) v/e have 'i<c + d+ 2sc + e+ f<2
+ f,
which contradicts the relation
f < 2
in (505).
Hence this
case is impossible.
17(II)-2.
A + B c Sy
Then
4+b-c-d=2, or
b = c + d - 2:
(515)
a + d < 2.
(516)
and
109
Exactly as in 17(1I)-1, ue obtain
locations of
S|,
T^,
and
Tj!
as possible
B + C.
1 7 ( I I ) —2—1.
JJ + C c S j .
'i'hen
c + c = 2.
Usinp, (515) and (517), wo have
Since
a + d - f < 2
(506),
A *1- B + C
(53 7)
A + B + C = (0, a + d - f, e + f - 2),
b y (516) a n d (50f>) a n d
will be interior to
K
0 < c + f - 2 < 2
by
unless
a + d - f £ -2.
Cut if (51S) holds, then
be in
K,
sincc
(518)
C + A - (2 - e, a - f, b + e + f - 4)
2 ~ f _< -d - 2 < -2
by (505).
cannot
lleiicc this case
cannot occur.
17(II)-2-2.
B + C c T^.
This may be shown to be impossible
P. + C c T^.
This is impossible, as shown in
just as in 17(II)-l-2.
17(IT)-2-3.
case' 17(JI)-l-3.
17(II)—3.
- d = 4,
A + D E T3.
Then
2-c-a-d + 4 + b - c
or
b - a = 2c + 2d - 2.
But
b - a <_ 2
by (504), and
2c + 2d - 2 > 2
bv (505).
Hence this
cannot occur.
17(II)-4.
-• b = A,
A + B e T^.
Then
2 - c + a -h d + c H- d - 4
or
a - b = 6 - 2d.
But
a - b < 2
cannot occur.
by (504), and
6 - 2d > 2
by (505).
Hence this case
110
A E Tlt
18(1).
£ bdry K.
n e T ,
C
T ,
E
A - B,
B - C,
C - A
Let
A = (a, 1), t\ - a - b),
where
B - (-c, d, 4 - c - d),
a < 2,
b < 2,
and
a + b > 2:
(519)
where
c < 2,
d < 2,
and
c + d > 2:
(520)
where
c < 2,
f < 2,
and
e -1- f > 2.
(521)
and
C = (e, -f, h - e - f),
Then
K,
B - C = (-c - e, d + f, e+f-c-d).
Since this must be in
we have
c + e52
(522)
d + f <_ 2.
(523)
and
Combining (522) and (523) r.ives
+ e + f > h
from (520) and (521).
18(11).
e bdry K.
c + d + e+f<_4.
bet
A e 1'v
A,
B
and
b+d, 8-a-b-c-d).
C e 1'3,
A + B.
be. as dr. 18(1).
Since this is in
Then
K,
P, + C,
8-a-b-c-d <^2,
we nust have
19(1).
A e T,,
(52A)
whicli combined with (524) cives
But this contradicts (519) and (520).
D e T3,
C E T,,
A - B,
An arpur-ent sinilar to that in 18(1),
B
replacing
B
19(11).
c bdry K,
Let
C,
a + c >_ A.
Hence this case is irpossib]c.
e bdry K.
and
C + A
A + Pi = (a - c,
b + d < 2 ,
and
c + d
Hence this case cannot occur.
P. e T2,
C
I'ut we have
with
B - C,
A
C - A
replacing
shows that this case is impossible.
A e T,,
B e T3,
C e
A + P.,
B + C,
C + A
Ill
A = (-n, b, 4 - a - b),
where
a < 2,
b < 2,
and
n + b > 2:
(525)
ft = (c, -d, A - c - d),
where
c < 2,
d < 2,
and
c + d > 2:
(526)
and
C = (e, f, c + f - 4),
Then we have
where
c < 2,
f < 2,
and
e -I- f > 2.
A + ]'> = (c - a, b - d, 8 - a - b - c - d).
from (525) and (526), that
-2 < c - a < 2,
-a-b-c-d, and since
only- possible locations of
19(ll)-l.
T^
A + Pi
A + 1? e S3.
and
8-a~b-c-d-2, or
(528)
B + C = (c + e, f - d, e + f - c - d) must be in
0 < c + e,
0 < 8
T^.
a + b + c + d- G.
(527) we have
and
have been excluded, the
are
Then
Since we have,
-2 < b - d < 2
and
(527)
-2 < f ~ d < 2
and
K.
Rv (526) and
- 2 < c + f - c - d < 2 .
Since Lemma 4 excludes most triangular faces, the only nossible loca­
tions for
B - C
arc
19(II)*-1-1.
Sj
and
T^.
D - C e Sj.
Then
c + e = 2.
Nov;
C + A= (e-a, b + f, e+f - a-b).
have
-2 < e - a < 2,
0 < b + f,
.
(529)
By (525) and (527) we
-2<e+f-a-b<2, so that
with the use of Lemma 4, we see that the only possible locations for
C + A
are
S£
and
T^.
19(11)-1-1-1.
C + A e Rr
Then
b + f = 2.
Substitution of (530) and (529) into (528) t^ives
a + d = e + f + 2.
(530)
112
Puit
a -t- d < 4
by (525) and (526)> and
e + f + 2 > 4
by (527).
llence this case cannot occur.
19(II)—1—1—2.
- b = 4,
C + A t Tj.
Then
e-a + b + f + c + f- a
or
a = c + f - 2,
(531)
Usinr» (531) and (530) in (528) pives
b + d + f - 6,
which contradicts (525), (52G) and (527).
19(IT)~l-2.
- d = 4,
B + C e
ltencc this cannot occur.
Then
c + e+ f- d + e+ f- c
or
d
» e + f - 2.
(532)
As in 19(II)-1-*1, VJC may eliminate all possible locations other than
S£ 'and
B + C
T^
for
is there.
C + A,
Then
but here ve may also eliminate
C + A e S2.
,
since
and
b + f = 2.
Usinc; (532) and (533) in (52fi) pives
(525), (526) and (527).
IS(II)-2. - d = '1,
(533)
a + c + e = 6,
which contradicts
l'ence this case cannot occur.
A + B e Tj.
Then
c-a + b-d + 8- a- h- c
or
a + d = 2.
As in 19(T.I)~1, we obtain
is excluded, so we have
B + C e
or
R + C e S^.
Then
(534)
T^.
But in this case
c + e = 2.
Similarly, "oinp bach to 19(TI)-1-1, we nust have
T^
(535)
C + A e S£, so that
113
b + f = 2.
CombininR (534), (535) and (536) <*ives
a + b + c + d + G + f = 6.
But it follov's from (525), (526) and (527) that
> 6.
llencc this cast! cannot occur.
(536)
a + b + c + il + e + f
CHAT'TKR 5
RESULTS
The t!>nin theorems of this paper, which were stated at the end
of Chapter 2, nav now be proved.
ZJEHPX Pf, JJa.c0.r.G.m.
remarked in Chapter 3, the lattices
which pive packings for
T
are exactly those which are admissible
for the difference body
K
defined by (2)', and the lattice having
the smallest determinant and meeting the conditions of Lcrnas 2 and
3 will then p,ive the densest possible packing,
Chapter 4 we obtained
|det[ _> 196/27,
Since throughout
and since
det
196/27
was
in fact obtained in 5(Tl)-3, it follows that the lattice of 5(II)-3,
which is that piven by (3), yields an extrenal packinp.
Vol(T) « 8/3,
- 18/49,
we then have
6
(T) =
max
Since
(T, A ) = (8/3)/(196/27)
o
by the renark following Def. 3.
The division into cases of Chapter 4 combines essentially
equivalent lattices by utilizing the symmetries of
K,
sentially different lattices arc treated separately.
but es­
Hence if there
v/ere several essentially different critical lattices, they all would
h a v e b e e n found b y the computations.
- 196/27
V'e did i n fact obtain
three tines, with the following three lattices:
]14
f de11
115
(
A} = (2, -1/3, -1/3)
T)^ = (-1/3, 2, -1/3)
obtained in
Cx - (-1/3, -1/3, 2)
TA2 = (2, 1/3, 1/3)
A2
is the lattice with basis * ^2 ~
9(I)-3-l-l; and
~l/3)
obtained in
^C2 = (2/3, 5/3, 5/3)
{
A3 = (2, -1/3, -1/3)
B3 = (-1/3, 2, -1/3)
obtained in
C3 = (-A/3, -A/3, -A/3)
Let
A*, fi*,
in the
A1
"
A*,
h
=
C*
be the points obtained by reflecting
yz-plane
A2'
B*,
V
B1
(i.e., chanr.inn the sipn of
- B*,
C£,
and
and
ci
and
A2
- c| - r,*,
so that
is a reflection of
= -A3 - IS, - C 3 >
x).
Ar
s o that
A
t\^,
11^»
Then we have
A^
is Renerated by
«•Jc also have
l"
^2
A3>
A^ = A3>
T h u s a l l critical
lattices are essentially similar.
Proof of Corollary:
ping each vertex of
T*
There exists an affine transformation map­
onto a vertex of
T.
Since an affine trans­
formation naps lines onto lines and planes onto planes, it is a one-toone mapping of
T1*
T*
onto
T;
onto a lattice packinp of
moreover, it maps a lattice packing of
T.
If
D
is the determinant of this
transformation, then all volumes are multiplied by
the density of a packing is a ratio
by the transformation.
ID|,
and since
of volumes, it will be unchanged
Hence the maximum densitv i.s the same.
116
Proof of Theorem 2:
Since the body
-;K
is symmetric about the
origin, it follows easily that its difference body is
K.
Then, ,iust
as in Theorem 1, the lattice which pives the densest packing of
has
|det| = 196/27.
Since
following TjCnmn 3, that
= 45/49.
5
Vol(-;;i\) = 20/3,
max
-.'K
we obtain, by the remark
&>'K) = 6(-?K, A ) = (20/3)/(196/27)
o
REFERENCKS
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Berlin: Springcr-Vcrlng.
Fejes Totli, L. (1953) Lap.erunf.cn in der_Ehene_, auf der Kupel und
im Raun. llerlin: Sprinper-Vcrlatt.
Groemer, H. (1961) Eine Kernerkunw iiber Lagerun^en konvexer Kepcl.
Arch, der Math. 12^, 78-80.
. (1962) Ucber die dichteste pitterformif^c Laperunp
kongruentor Tetraeder. Monatsh. Math. 6_6, 12-15.
Hancock, II. (1939) Development of the. Minkowski Geometry of Numbers
New York: Macniilan.
Hilbert, D. (1901) Mathercatische Probler.ie, Archiv f. Math. u. Phys
3. Reihe, JL, 44-63, 213-237. (= Gesnrcnelte. Abhandlunp.en, TIT
290-329, New York: Chelsea, 1965).
Minkowski, H. (1904) Dichteste nitterformife Laperunri kon^ruenter
Korper, Nachr. K. Ges. V.'iss. Gottinnen 19_04, 311-355.
(= Gesannelte Abhandlunpen II, 3-42, Ferlin: Teubner, 1911).
Ropers, C. A. (1964) Packing and Coverinr.
University Press.
117
Cambridge: Cambridge