2020 USAMO Problems/Problem 2
Problem
An empty cube is given, and a grid of square unit cells is drawn on each of its six faces. A beam is a rectangular prism. Several beams are placed inside the cube subject to the following conditions:
- The two faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are possible positions for a beam.)
- No two beams have intersecting interiors.
- The interiors of each of the four 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 if its long edge is parallel to x-axis. Similarly for and .
Define a beam's location as (direction, ( face's location in 2D coordinate).
For example, (y, 2, 4) indicates the beam with vertex (1, 0, 3) and (2, 2020, 4)
Apparently beam needs the other or beams to touch its faces, or beams to touch its faces. Similarly for and beam.
If there are only 1-dir or 2-dirs beams, it is easy to approve that is the minimal number beams.
(for example, if only use and beams, all the beams's xz faces can only be touched by beams. In the other word, beams will be stacked to meet xz faces touch requirements in that xz layer)
Consider cases with all beams. WLOG there is a beam and it needs or beams to touch its faces (unless it touches the cube surface).
And this beam also needs a or to touch it's faces. And so on until one which touches cube surface. So from face perspective, it needs beams.
Similarly from and face perspective, it also needs and beams.
Consider one beam has four faces and it can be counted twice. So there should be at least beams.
Here is one solution with 3030 beams.
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