JBMO Shortlist

Solution:

Let $ABCD$ be a rectangle with $AB = m$ and $AD = n$ where $m, n$ are natural numbers such that $m \geq n \geq 2$. Suppose that inside of the rectangle $ABCD$ is placed a rectangular lattice consisting of some identical squares whose areas are equal to $3$, where $k$ of them are placed along the side $AB$ and $l$ of them along the side $AD$.

The sum of areas of all of these squares is equal to $3kl$. Besides the obvious conditions $k\sqrt{3} < m$ and $l\sqrt{3} \leq n$ $\textbf{(1)},$ by the assumption of the maximality of the lattice consisting of these squares, we must have $(k+1)\sqrt{3} > m$ and $(l+1)\sqrt{3} > n$ (2).

The proposed problem is to determine all pairs $(m, n) \in \mathbb{N} \times \mathbb{N}$ with $m \geq n \geq 2$, for which the ratio $R_{m, n} = \frac{3kl}{mn}$ is equal to $0.5$ where $k, l$ are natural numbers determined by the conditions (1) and (2).

Observe that for $n \geq 6$, using (2), we get
$$R_{m, n} = \frac{k\sqrt{3} \cdot l\sqrt{3}}{mn} > \frac{(m-\sqrt{3})(n-\sqrt{3})}{mn} = \left(1-\frac{\sqrt{3}}{m}\right)\left(1-\frac{\sqrt{3}}{n}\right) \geq \left(1-\frac{\sqrt{3}}{6}\right)^2 = \frac{1}{2} + \frac{7}{12} - \frac{\sqrt{3}}{3} > \frac{1}{2} + \frac{\sqrt{48}}{12} - \frac{\sqrt{3}}{3} = 0.5$$
So, the condition $R_{m, n} = 0.5$ yields $n \leq 5$ or $n \in \{2,3,4,5\}$. We have 4 possible cases:

Case 1: $n = 2$. Then $l = 1$ and thus as above we get
$$R_{m, 2} = \frac{3k}{2m} > \frac{\sqrt{3} \cdot (m-\sqrt{3})}{2m} = \frac{\sqrt{3}}{2} \left(1-\frac{\sqrt{3}}{m}\right)$$
which is greater than $0.5$ for each $m > \frac{\sqrt{27} + 3}{2} > \frac{5+3}{2} = 4$, hence $m \in \{2,3,4\}$. Direct calculations give $R_{2,2} = R_{2,4} = 0.75$ and $R_{2,3} = 0.5$.

Case 2: $n = 3$. Then $l = 1$ and thus as above we get
$$R_{m, 3} = \frac{3k}{3m} > \frac{\sqrt{3} \cdot (m-\sqrt{3})}{3m} = \frac{\sqrt{3}}{3} \left(1-\frac{\sqrt{3}}{m}\right)$$
which is greater than $0.5$ for each $m > 4\sqrt{3} + 6 > 12$, hence $m \in \{3,4, \ldots, 12\}$. Direct calculations give $R_{3,3} = 0.(3)$, $R_{3,5} = 0.4$, $R_{3,7} = 4/7$, $R_{3,9} = 5/9$, $R_{3,11} = 6/11$ and $R_{3,4} = R_{3,6} = R_{3,8} = R_{3,10} = R_{3,12} = 0.5$.

Case 3: $n = 4$. Then $l = 2$ and thus as above we get
$$R_{m, 4} = \frac{6k}{4m} > \frac{\sqrt{3} \cdot (m-\sqrt{3})}{2m} = \frac{\sqrt{3}}{2} \left(1-\frac{\sqrt{3}}{m}\right)$$
which is greater than $0.5$ for each $m > \frac{\sqrt{27} + 3}{2} > \frac{5+3}{2} = 4$. Hence $m = 4$ and a calculation gives $R_{4,4} = 0.75$.

Case 4: $n = 5$. Then $l = 2$ and thus as above we get
$$R_{m, 5} = \frac{6k}{5m} > \frac{2\sqrt{3} \cdot (m-\sqrt{3})}{5m} = \frac{2\sqrt{3}}{5} \left(1-\frac{\sqrt{3}}{m}\right)$$
which is greater than $0.5$ for each $m > \frac{12(4\sqrt{3} + 5)}{23} > \frac{12 \cdot 11}{23} > 6$, hence $m \in \{5,6\}$. Direct calculations give $R_{5,5} = 0.48$ and $R_{5,6} = 0.6$.

We conclude that: $R_{i, j} = 0.5$ for $(i, j) \in \{(2,3); (3,4); (3,6); (3,8); (3,10); (3,12)\}$.

These pairs are the dimensions of all rectangles with the desired property.