2007 USAMO Problems/Problem 4
Problem
An animal with cells is a connected figure consisting of
equal-sized square cells.
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.
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
cells, one with
cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
Solution
Solution 1
Let a -dino denote an animal with
or more cells.
We show by induction that an -dino with
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 , we have two boxes, which are clearly not primitive.
Inductive Step: Assume any cell animal can be partitioned into two or more
-dinos.
For a given -dino, take off any four cells (call them
) to get an animal with
cells.
This can be partitioned into two or more -dinos, let's call them
and
. This means that
, where : denotes a connection.
If both and
are
-dinos or if
don't all attach to one of them, then we're done.
So assume has
cells and thus
has at least
cells, and that
are added to
. So
has
cells total.
Let denote the cell of
attached to
. There are
cells on
besides
. Thus, of the three sides of
not attached to
, one of them must have
cells by the pigeonhole principle. It then follows that we can add
,
, and the other two sides together to get an
dino, and the side of
that has
cells is also an n-dino, so we can partition the animal with
cells into two
-dinos and we're done.
Thus, our answer is cells.
Solution 2
For simplicity, let and let
be the number of squares. Let the centers of the squares be vertices, 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:
where
,
,
,
are the numbers of verticies on each branch, WLOG
. Note
.
Claim: If , then we must be able to divide the animal into two dinosaurs.
Chose a vertex,
, for which
is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element). We have that
, so
. 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 . Now move to the first point on the branch at
. We have a new quadruple
) where
.
Now consider the maximal element of that quadruple. We already have . WLOG
, then
so
, so
is the maximal element of that quadruple.
Also , so
. But that is a contradiction to the minimality of
. Therefore, we must have that
, so we have a partition of two dinosaurs.
Maximum: .
Consider a cross with each branch having
verticies. Clearly if we take partition
verticies, we remove the center, and we are not connected.
So :
.
See also
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |