2007 USAMO Problems/Problem 4

Problem

An animal with $n$ cells is a connected figure consisting of $n$ equal-sized cells. ${}^1$ The figure below shows an 8-cell animal.

(insert picture here)

A dinosaur is an animal with at least 2007 cells. It is said to be primitive it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.

${}^1$ Animals are also called polyominoes. They can be defined inductively. Two cells are adjacent if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.

Solution

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Note that for the dinosaur to have $\ge$ 4014 cells then the answer is $4(2007-1)+1=8025$ .

Note that for $8025$ , we can use one cell with four arms of length 2006 sticking out from it.

Let an n-dino denote an animal with $n$ or more cells.

We show by induction that an n-dino with 4n-2 or more animal cells is not primitive. (Note: if it had more, we could just take off enough until it had 4n-2, which would have a partition, and then add the cells back on)

Base Case: If n=1, we have two boxes, which are clearly not primitive.

Inductive Step: Assume any 4n-2 cell animal can be partitioned into two or more n-dinos.

For a given 4n+2-dino, take off any four cells (call them w,x,y,z) to get an animal with 4n-2 cells.

This can be partitioned into two or more n-dinos, let's call them A and B. This means that A:B, where : denotes a connection.

Well if both A and B are n+1-dinos or if w,x,y,z don't all attach to one of them, then we're done.

So assume A has n cells and thus B has at least 3n-2 cells, and that w,x,y,z are added to B. So B has 3n+2 cells total.

Let C denote the cell of B attached to A. There are 3n+1 cells on B besides C. Thus, of the three sides of C not attached to A, one of them must have n+1 cells by the pigeonhole principle. It then follows that we can add A, C, and the other two sides together to get an n+1 dino, and the side of C that has n+1 cells is also an n-dino, so we can partition the animal with 4n+2 cells into two n+1-dinos and we're done.

Solution by AoPS user Calc Rulz.

2007 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
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2007 USAMO Problems/Problem 4

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