1982 USAMO Problems/Problem 1
Problem
In a party with people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?
Solution
We induct on to prove that in a party with
people, there must be at least
people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people
, who do not know each other.)
Base case: is obvious.
Inductive step: Suppose in a party with people (with
), at least
people know everyone else. Consider a party with
people. Take
of the people (leaving another person,
, out) and apply the inductive step to conclude that at least
people know everyone else in the
-person group,
.
Now suppose that everyone in the group knows each other. Then take
of these people and
to deduce that
knows a person
, which means
knows everyone else. Then apply the inductive step on the remaining
people (excluding
) to find
people out of them that know everyone else (including
, of course). Then these
people and
, which enumerate
people, know everyone else.
Suppose that there exist two people who do not know each other. Because
, there exist at least one person in
, person
, who knows everyone else in
. Now, take
and observe that because
do not know each other, either
or
knows everyone else of
(by the problem condition), so in particular
and
know each other. Then apply the inductive step on the remaining
people (excluding
) to find
people out of them that know everyone else (including
, of course). Then these
people and
, which enumerate
people, know everyone else.
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least people know everyone else.
See Also
1982 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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