8
JOSEPH ZAKS
Case#
x 0,l X U x l, 2
X U x
2,2 x2,3
Impossible by
1 0 0 0 0 9 0 Baston [2], Theorem 10, p. 196.
2 0 0 1 0 7 1 Chapter 5.
3 0 0
2
0 5
2 Chapter 6.
4 0 1 0 1 5 2 Theorem 3.
5 2 0 0 0 6 2 Theorem 1.
6 0
0 3 0 3 3 Chapter 7.
7 0 1
1 1
3 3 Theorem 3.
8 1 0 0 2 3 3 Baston's Theorem 8; Theorem 2.
9 1 2 0 0 4 3 Chapter 7.
10
2 0 1 0 4 3
i t
11 0
0 0 4 0 4 Theorem 2.
12 0 0
4
0
1 4 Chapter 8.
13 0 1 2
1 1 4
Theorem 3.
14 0 2 0 2 1 4 Theorem 3.
15 1 0 1 2 1 4 Theorem 2.
16 0 4 0 0 2 4 Chapter 8.
17 1 2 1 0 2 4
t i
18 2 0
2
0
2 4 u
19 2 1 0 1 2
4
Theorem 4.
20 0 4 1 0 0 5 Chapter 9.
21 1 2 2 0 0 5
M
22 2 0 3 0 0 5
I t
23
1
3 0
1
0
5
H
24 2 1 1 1 0 5 Theorem 4.
Table 1. The 24 solution of the equations (3, 4, 6, 7, 8).
Baston proved that case #1 (of our table 1 ) is impossible (Theorem 10, p. 196 of [2]). He
observed that all the facets of members of F cannot lie on just eight planes (Theorem 8, p. 186 of
[2]), thus case #11 is also impossible. We shall give a simpler proof that case #11 is impossible .
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