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 .