2007 USAMO Problems/Problem 4

Problem

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

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

Solution 1

Let a $n$ -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 $\displaystyle n=1$ , 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.

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.

Thus, our answer is $\displaystyle 4(2006) + 1 = 8025$ cells.

Solution 2

For simplicity, let $k=2007$ and let $n$ be the number of squares. Let the centers of the squares be verticies, and connect any centers of adjacent squares with edges. Suppose we have some loops. Just remove an edge in the loop. We are still connected since you can go around the other way in the loop. Now we have no loops. Each vertex can have at most 4 edges coming out of it. For each point, assign it the quadruple: $(a,b,c,d)$ where $a$ , $b$ , $c$ , $d$ are the numbers of verticies on each branch, wlog $a\ge b\ge c\ge d$ . Note $a+b+c+d=n-1$ .

Claim: If $n=4k-2$ , then we must be able to divide the animal into two dinosaurs. Chose a vertex, $v$ , for which $a$ is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element). We have that $4a\ge a+b+c+d=4k-3$ , so $a\ge k$ . Hence we can just cut off that branch, that forms a dinosaur.

But suppose the remaining verticies do not make a dinosaur. Then we have $b+c+d+1\le k-1 \iff n-a\le k-1\iff a\ge 3k-1$ . Now move to the first point on the branch at $a$ . We have a new quadruple p,q,r,b+c+d+1) where $p+q+r=a-1\ge 3k-2$ .

Now consider the maximal element of that quadruple. We already have $b+c+d+1\le k-1$ . WLOG $p\ge q\ge r\ge 0$ , then $3p\ge p+q+r=a-1\ge 3k-2\implies p\ge k$ so $p>k-1=b+c+d+1$ , so $p$ is the maximal element of that quadruple.

Also $a-1=p+q+r\ge p+0+0$ , so $p<a$ . But that is a contradiction to the minimality of $a$ . Therefore, we must have that $b+c+d+1\ge k$ , so we have a partition of two dinosaurs.

Maximum: $n=4k-3$ . Consider a cross with each branch having $k-1$ verticies. Clearly if we take partition $k$ verticies, we remove the center, and we are not connected.

So $k=2007$ : $4*2007-3=8025$ .

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

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