Subfield: Combinatorics


Question: Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively.

![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=163&width=312&top_left_y=455&top_left_x=632)

S-tetrominoes

![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=163&width=323&top_left_y=455&top_left_x=1112)

Z-tetrominoes

Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes.

Solution:
We may assume that polygon $P$ is the union of some squares of an infinite chessboard. Colour the squares of the chessboard with two colours as the figure below illustrates.

![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=665&width=874&top_left_y=978&top_left_x=591)

Observe that no matter how we tile $P$, any S-tetromino covers an even number of black squares, whereas any Z-tetromino covers an odd number of them. As $P$ can be tiled exclusively by S-tetrominoes, it contains an even number of black squares. But if some S-tetrominoes and some Z-tetrominoes cover an even number of black squares, then the number of Z-tetrominoes must be even.
Let us assign coordinates to the squares of the infinite chessboard in such a way that the squares of $P$ have nonnegative coordinates only, and that the first coordinate increases as one moves to the right, while the second coordinate increases as one moves upwards. Write the integer $3^{i} \cdot(-3)^{j}$ into the square with coordinates $(i, j)$, as in the following figure:

| $\vdots$ |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 81 | $\vdots$ |  |  |  |  |
| -27 | -81 | $\vdots$ |  |  |  |
| 9 | 27 | 81 | $\cdots$ |  |  |
| -3 | -9 | -27 | -81 | $\ldots$ |  |
| 1 | 3 | 9 | 27 | 81 | $\ldots$ |

The sum of the numbers written into four squares that can be covered by an $S$-tetromino is either of the form

$$
3^{i} \cdot(-3)^{j} \cdot\left(1+3+3 \cdot(-3)+3^{2} \cdot(-3)\right)=-32 \cdot 3^{i} \cdot(-3)^{j}
$$

(for the first type of $S$-tetrominoes), or of the form

$$
3^{i} \cdot(-3)^{j} \cdot\left(3+3 \cdot(-3)+(-3)+(-3)^{2}\right)=0
$$

and thus divisible by 32. For this reason, the sum of the numbers written into the squares of $P$, and thus also the sum of the numbers covered by $Z$-tetrominoes in the second covering, is likewise divisible by 32 . Now the sum of the entries of a $Z$-tetromino is either of the form

$$
3^{i} \cdot(-3)^{j} \cdot\left(3+3^{2}+(-3)+3 \cdot(-3)\right)=0
$$

(for the first type of $Z$-tetrominoes), or of the form

$$
3^{i} \cdot(-3)^{j} \cdot\left(1+(-3)+3 \cdot(-3)+3 \cdot(-3)^{2}\right)=16 \cdot 3^{i} \cdot(-3)^{j}
$$

i.e., 16 times an odd number. Thus in order to obtain a total that is divisible by 32, an even number of the latter kind of $Z$-tetrominoes needs to be used. Rotating everything by $90^{\circ}$, we find that the number of $Z$-tetrominoes of the first kind is even as well. So we have even proven slightly more than necessary.

Final answer: 证明题，略