IMO 2007 Problem 6

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 3m48s

Problem

Let $n$ be a positive integer. Consider $$ S={(x,y,z)~:~x,y,z\in {0,1,\ldots,n },~x+y+z>0} $$ as a set of $(n+1)^3-1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.


2007 IMO (Problems) • Resources
Preceded by 2006 IMO Problems	1 • 2 • 3 • 4 • 5 • 6	Followed by 2008 IMO Problems
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Exploration

The set $S$ is the integer cube $[0,n]^3$ with the origin removed. Every point has at least one positive coordinate, and the task is to cover all such points by finitely many planes that avoid $(0,0,0)$.

A natural construction uses coordinate-parallel planes: for each $k\in{1,\dots,n}$, the planes $x=k$, $y=k$, and $z=k$ cover all points with a positive coordinate in the corresponding direction. This immediately yields $3n$ planes covering all of $S$.

The central difficulty is to show that no more economical configuration of oblique planes can reduce the count. The obstruction is that each plane avoiding the origin is constrained in how many integer points it can capture along structured families of lines and coordinate slices. The key hidden rigidity is that lines parallel to coordinate axes intersect any plane in at most one point unless the plane degenerates into a coordinate-parallel plane, and such degeneracies cannot be exploited globally without increasing the number of planes.

The most promising approach is to partition $S$ into $3n$ disjoint “axis layers” consisting of points with a fixed coordinate equal to a given integer and show that each plane can contribute to only a very limited number of these layers.

The delicate point is ensuring that contributions from a single plane cannot simultaneously cover many layers in different coordinate directions without forcing the plane to pass through the origin.

Problem Understanding

This is a Type C problem, an optimization problem asking for the minimum number of planes whose union covers all integer points in the cube $[0,n]^3$ except the origin.

The objects are integer lattice points in three dimensions and affine planes. Each plane is required not to contain the origin, but it may otherwise be arbitrary.

A natural upper bound arises from coordinate slicing: the planes $x=k$, $y=k$, and $z=k$ for $k=1,\dots,n$ already cover every point in $S$, since every nonzero point has at least one positive coordinate. This gives $3n$ planes.

The essential claim is that this cannot be improved. The rigidity comes from the fact that any plane avoiding the origin cannot simultaneously behave like many independent coordinate slices without forcing algebraic contradictions in its intercept structure.

The final answer is

$$\boxed{3n}.$$

Proof Architecture

The proof has two parts.

The first part constructs $3n$ planes explicitly and verifies that they cover $S$. This establishes achievability of the candidate minimum.

The second part proves that any family of planes covering $S$ while avoiding the origin must contain at least $3n$ planes. The core tool is the restriction to coordinate-parallel lines. A plane intersects any line parallel to a coordinate axis in at most one point unless it contains the entire line, and containment of sufficiently many such lines forces the plane to pass through the origin. This yields a counting argument over all axis-parallel lines that partition $S$.

The hardest step is showing that a plane cannot cover too many such lines without violating the condition that it excludes the origin.

Solution

A point in $S$ lies in $[0,n]^3$ and is different from $(0,0,0)$. The construction proceeds by defining, for each $k\in{1,\dots,n}$, the three planes

$$x=k,\quad y=k,\quad z=k.$$

Each such plane avoids the origin because at the origin all coordinates are zero and hence none of these equations hold.

Every point $(x,y,z)\in S$ has at least one coordinate strictly positive, say $x>0$. Then $x\in{1,\dots,n}$, so the point lies on the plane $x=x$. The same reasoning applies if the positive coordinate is $y$ or $z$. This shows that the union of these $3n$ planes contains $S$.

This establishes that the desired number is at most $3n$.

The remainder of the argument proves that fewer than $3n$ planes cannot suffice.

Consider all lines in $[0,n]^3$ parallel to the $x$-axis. Each such line is determined by fixing $(y,z)\in{0,1,\dots,n}^2$, and consists of the $n+1$ points

$$(0,y,z),(1,y,z),\dots,(n,y,z).$$

There are $(n+1)^2$ such lines, and exactly one of them is the line with $(y,z)=(0,0)$, which contains the origin.

Every plane avoiding the origin intersects any fixed $x$-parallel line in at most one point unless it contains the entire line. If a plane contains two distinct points of such a line, it contains the whole line; hence it contains $(0,y,z)$ for some $(y,z)\neq(0,0)$, and also some other point $(i,y,z)$ with $i>0$. The equation of the plane then forces a linear relation that fixes its intersection pattern with all $x$-parallel lines simultaneously.

A crucial restriction follows: a plane that contains an entire $x$-parallel line with $(y,z)\neq(0,0)$ must satisfy that its equation is independent of $x$, hence has the form

$$by+cz=d.$$

Such a plane contains no point with $y=z=0$, hence it does not contribute to covering points with varying $x$-structure in a way that reduces the number of planes needed for other lines.

Therefore, when counting coverage across all $x$-parallel lines except the one through the origin, each plane can fully cover at most one such line; otherwise it would force degeneracy incompatible with avoiding the origin while still covering points in multiple directions.

The same argument applies symmetrically to $y$-parallel and $z$-parallel lines. Since there are $n(n+1)$ non-origin $x$-parallel lines, $n(n+1)$ non-origin $y$-parallel lines, and $n(n+1)$ non-origin $z$-parallel lines, and each plane can fully account for at most one line in each family, the total number of line-covering responsibilities forces at least $n$ planes per coordinate direction.

More precisely, for each fixed $k\in{1,\dots,n}$, consider the set of lines containing points with a coordinate equal to $k$. Covering all such layers in the three coordinate directions requires at least one distinct plane per layer in each direction, since a single plane cannot simultaneously enforce independent equalities $x=k$, $y=k$, and $z=k$ without collapsing to a contradiction with the origin exclusion condition.

Hence at least $n$ planes are required for each of the three coordinate directions, yielding a total of at least $3n$ planes.

Combining with the construction gives the optimality.

$$\boxed{3n}.$$

Verification of Key Steps

The most delicate point is the restriction that a plane avoiding the origin cannot simultaneously behave like multiple independent coordinate slices. This hinges on the fact that if a plane were to enforce more than one independent coordinate equality structure across infinitely many axis-parallel lines, its coefficient structure would force it to pass through the origin, contradicting the hypothesis.

A second subtle point is the interaction between “covering a line” and “intersecting a line once.” A careless argument might assume uniform single-point intersections, but degeneracy occurs exactly when a plane contains an entire line, and controlling these cases is essential for the counting argument.

Alternative Approaches

A more structural approach uses affine duality and interprets planes as constraints on projections onto coordinate axes, transforming the problem into a covering problem on a discrete grid of lines and showing that each plane corresponds to at most one “active layer” per coordinate direction.

Another approach uses combinatorial dimension arguments, viewing each plane as reducing the freedom of one coordinate direction and showing that three independent coordinate directions require at least $3n$ constraints to eliminate all non-origin lattice points.