2007 IMO Problems/Problem 6
Problem
Let be a positive integer. Consider
as a set of
points in three-dimensional space.
Determine the smallest possible number of planes, the union of which contain
but does not include
.
Solution
We will prove the result using the following Lemma, which has an easy proof by induction.
Lemma Let ,
and
. If
is a polynomial in
that vanishes on all points of the grid
except at the origin, then
Proof. We will prove this by induction on . If
, then the result follows trivially. Say
.
WLOG, we can assume that
.
By polynomial division over
we can write
Since
is a monomial, the remainder
must be a constant in
, i.e.,
is a polynomial in two variables
,
.
Pick an element of
of the form
and substitute it in the equation. Since
vanishes on all such points, we get that
for all
.
Let
and
.
For every point
in
we have
where
.
Therefore, the polynomial
vanishes on all points of
except the origin. By induction hypothesis, we must have
. But,
and hence we have
.
Now, to solve the problem let be
planes that cover all points of
except the origin. Since these planes don't pass through origin, each
can be written as
. Define
to be the polynomial
. Then
vanishes at all points of
except at the origin, and hence
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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