Problem (Combinatorics):
Determine all integers $m$ for which the $m \times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1,2,3, \ldots, 10$ in some order.

Solution:
The solution naturally divides into three different parts: we first obtain some bounds on $m$. We then describe the structure of possible dissections, and finally, we deal with the few remaining cases.

In the first part of the solution, we get rid of the cases with $m \leqslant 10$ or $m \geqslant 14$. Let $\ell_{1}, \ldots, \ell_{5}$ and $w_{1}, \ldots, w_{5}$ be the lengths and widths of the five rectangles. Then the rearrangement inequality yields the lower bound

$$
\begin{aligned}
\ell_{1} w_{1} & +\ell_{2} w_{2}+\ell_{3} w_{3}+\ell_{4} w_{4}+\ell_{5} w_{5} \\
& =\frac{1}{2}\left(\ell_{1} w_{1}+\ell_{2} w_{2}+\ell_{3} w_{3}+\ell_{4} w_{4}+\ell_{5} w_{5}+w_{1} \ell_{1}+w_{2} \ell_{2}+w_{3} \ell_{3}+w_{3} \ell_{4}+w_{5} \ell_{5}\right) \\
& \geqslant \frac{1}{2}(1 \cdot 10+2 \cdot 9+3 \cdot 8+\cdots+8 \cdot 3+9 \cdot 2+10 \cdot 1)=110
\end{aligned}
$$

and the upper bound

$$
\begin{aligned}
\ell_{1} w_{1} & +\ell_{2} w_{2}+\ell_{3} w_{3}+\ell_{4} w_{4}+\ell_{5} w_{5} \\
& =\frac{1}{2}\left(\ell_{1} w_{1}+\ell_{2} w_{2}+\ell_{3} w_{3}+\ell_{4} w_{4}+\ell_{5} w_{5}+w_{1} \ell_{1}+w_{2} \ell_{2}+w_{3} \ell_{3}+w_{3} \ell_{4}+w_{5} \ell_{5}\right) \\
& \leqslant \frac{1}{2}(1 \cdot 1+2 \cdot 2+3 \cdot 3+\cdots+8 \cdot 8+9 \cdot 9+10 \cdot 10)=192.5
\end{aligned}
$$

As the area of the square is sandwiched between 110 and 192.5 , the only possible candidates for $m$ are 11, 12, and 13 .

In the second part of the solution, we show that a dissection of the square into five rectangles must consist of a single inner rectangle and four outer rectangles that each cover one of the four corners of the square. Indeed, if one of the sides the square had three rectangles adjacent to it, removing these three rectangles would leave a polygon with eight vertices, which is clearly not the union of two rectangles. Moreover, since $m>10$, each side of the square has at least two adjacent rectangles. Hence each side of the square has precisely two adjacent rectangles, and thus the only way of partitionning the square into five rectangles is to have a single inner rectangle and four outer rectangles each covering of the four corners of the square, as claimed.

Let us now show that a square of size $12 \times 12$ cannot be dissected in the desired way. Let $R_{1}, R_{2}, R_{3}$ and $R_{4}$ be the outer rectangles (in clockwise orientation along the



boundary of the square). If an outer rectangle has a side of length $s$, then some adjacent outer rectangle must have a side of length $12-s$. Therefore, neither of $s=1$ or $s=6$ can be sidelengths of an outer rectangle, so the inner rectangle must have dimensions $1 \times 6$. One of the outer rectangles (say $R_{1}$ ) must have dimensions $10 \times x$, and an adjacent rectangle (say $R_{2}$ ) must thus have dimensions $2 \times y$. Rectangle $R_{3}$ then has dimensions $(12-y) \times z$, and rectangle $R_{4}$ has dimensions $(12-z) \times(12-x)$. Note that exactly one of the three numbers $x, y, z$ is even (and equals 4 or 8 ), while the other two numbers are odd. Now, the total area of all five rectangles is

$$
144=6+10 x+2 y+(12-y) z+(12-z)(12-x)
$$

which simplifies to $(y-x)(z-2)=6$. As exactly one of the three numbers $x, y, z$ is even, the factors $y-x$ and $z-2$ are either both even or both odd, so their product cannot equal 6 , and thus there is no solution with $m=12$.

Finally, we handle the cases $m=11$ and $m=13$, which indeed are solutions. The corresponding rectangle sets are $10 \times 5,1 \times 9,8 \times 2,7 \times 4$ and $3 \times 6$ for $m=11$, and $10 \times 5,9 \times 8,4 \times 6,3 \times 7$ and $1 \times 2$ for $m=13$. These sets can be found by trial and error. The corresponding partitions are shown in the figure below.


Final answer: 11,13