# 2020 USAMO Problems/Problem 2

## Problem

An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:

• The two $1 \times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are $3 \cdot 2020^2$ possible positions for a beam.)
• No two beams have intersecting interiors.
• The interiors of each of the four $1 \times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.

What is the smallest positive number of beams that can be placed to satisfy these conditions?

## Solution

Take one vertex of the cube as origin and establish 3D coordinates along the cube's edges.

Define a beam as $x-dir$ if its long edge is parallel to x-axis. Similarly for $y-dir$ and $z-dir$.

Define a beam's location as (direction, ($1 \times 1$ face's location in 2D coordinate).

For example, (y, 2, 4) indicates the beam with vertex (1, 0, 3) and (2, 2020, 4)

Apparently $x$ beam needs the other $x$ or $y$ beams to touch its $xy$ faces, $x$ or $z$ beams to touch its $xz$ faces. Similarly for $y$ and $z$ beam.

If there are only 1-dir or 2-dirs beams, it is easy to approve that $2020 \times 2020$ is the minimal number beams.

(for example, if only use $x-dir$ and $y-dir$ beams, all the $x-dir$ beams's xz faces can only be touched by $x-dir$ beams. In the other word, $2020 x-dir$ beams will be stacked to meet xz faces touch requirements in that xz layer)

Consider cases with all $3-dirs$ beams. WLOG there is a $x-dir$ beam and it needs $x-dir$ or $y-dir$ beams to touch its $xy$ faces (unless it touches the cube surface).

And this $y-dir$ beam also needs a $x-dir$ or $y-dir$ to touch it's $xy$ faces. And so on until one which touches cube surface. So from $xy$ face perspective, it needs $2020$ beams.

Similarly from $xz$ and $yz$ face perspective, it also needs $2020$ and $2020$ beams.

Consider one beam has four $1 \times 2020$ faces and it can be counted twice. So there should be at least $2020 \times 3 \div 2=3030$ beams.

Here is one solution with 3030 beams.

$(x, 1, 1),\ (y, 1, 2),\ (z, 2, 2),$

$\cdots ,$

$(x, (2n+1), (2n+1)),\ (y, (2n+1), (2n+2)),\ (z, (2n+2), (2n+2)),$

$\cdots ,$

$(x, (2019, 2019)),\ (y, 2019, 2020),\ (z, 2020, 2020)$

 2020 USAMO (Problems • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions