Difference between revisions of "2021 IMO Problems/Problem 1"
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==Video Solutions== | ==Video Solutions== | ||
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | ||
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https://youtu.be/0Vd4ZBEr3o4 | https://youtu.be/0Vd4ZBEr3o4 | ||
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https://www.youtube.com/watch?v=a3L9O7b1WYg [disclaimer: only a sketch of the solution] | https://www.youtube.com/watch?v=a3L9O7b1WYg [disclaimer: only a sketch of the solution] | ||
Revision as of 11:26, 24 July 2021
Problem
Let be an integer. Ivan writes the numbers
each on different cards. He then shuffles these
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
Video Solutions
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]
https://www.youtube.com/watch?v=a3L9O7b1WYg [disclaimer: only a sketch of the solution]
Solution
If we can guarantee that there exist cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains
cards that sum to a perfect square. Assume the perfect squares
,
, and
satisfy the following system of equations:
where
,
, and
are numbers on three of the cards. Solving for
,
, and
in terms of
,
, and
tells us that
,
, and
. We can then substitute
,
, and
to cancel out the
s in the denominatior, and simplifying gives
,
, and
. Now, we have to prove that there exists three numbers in these forms between
and
when
. Notice that
will always be the least of the three and
will always be the greatest of the three. So it is sufficient to prove that there exists numbers in the form
and
between
and
.
For two numbers in the form of and
to be between
and
, the inequalities
must be satisfied. We can then expand and simplify to get that
Then, we can complete the square on the left sides of both inequalities and isolate
to get that
Notice that
must be an integer, so there must be an integer between
and
. If
and
differ by at least
, then we can guarantee that there is an integer between them (and those integers are the possible values of
). Setting up the inequality
and solving for
tells us that
always works. Testing the remaining
numbers (
to
) manually tells us that there is an integer between
and
when
. Therefore, there exists a triplet of integers
with
when
such that every pair of the numbers sum to a perfect square. By the pigeonhole principle, we know that
of the numbers must be on cards in the same pile, and hence, when
, there will always be a pile with
numbers that sum to a perfect square.
~Mathdreams