Problem: [Silverman]
Strictly speaking, two men are brothers-in-law if one is married
to a full sister of the other. What is the largest number of men
possible, every pair of whom are connected by the brother-in-law
relationship?
Solution: [Silverman]
The only way for three man to mutual brothers-in-law is for A to
be the husband of B's sister, B of C's sister, and C of A's sister.
Thus, if D is to be a mutual brother-in-law of A, B, and C, then
in particular, A, B, and D form a mutual brother-in-law triple,
implying D is married to the sister of either A or B, say A. Then
C and D are each married to a sister of A and cannot be brother-in-law
unless one of them is A's brother, implying a violation of the law
against consanguinity. Hence, the maximum possible number of mutual
brother-in-law is three.
Solution: [Sillke]
Define the relation A->B as A is married to a sister of B.
If B<-A and A->C with B!=C then A, B, C cannot be mutual related.
Proof: B and C must be brothers. But to marry a sister in forbidden. qed.
We conclude that at most one arrow points out of every men.
For four men, we can draw at most four arrows
but we must have six brothers-in-law relations.
References:
- David L. Silverman;
A Problem of Relations (Solution to problem 129)
Journal of Recreational Puzzles 4 (Apr. 1971) 147
- Heinrich Hemme;
Das Hexeneinmaleins,
Vandenhoeck & Ruprecht, G\"ottingen, 2000, ISBN 3-525-40738-6
Problem 10: Schw"ager.
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