
C7 Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that
the sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rectangle. Determine the dimensions of all rectangles with this property.


## Solution

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

The sum of areas of all of this squares is equal to $3 k l$. Besides of the obvious conditions $k \sqrt{3}m$ 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 $R_{m, n}=\frac{3 k l}{m n}$ 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 $R_{m, n}=\frac{k \sqrt{3} \cdot l \sqrt{3}}{m n}>\frac{(m-\sqrt{3})(n-\sqrt{3})}{m n}=$ $\left(1-\frac{\sqrt{3}}{m}\right)\left(1-\frac{\sqrt{3}}{n}\right) \geq\left(1-\frac{\sqrt{3}}{6}\right)^{2}=\frac{1}{2}+\frac{7}{12}-\frac{\sqrt{3}}{3}>\frac{1}{2}+\frac{\sqrt{48}}{12}-\frac{\sqrt{3}}{3}=0,5$

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 $R_{m, 2}=\frac{3 k}{2 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, 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 $R_{m, 3}=\frac{3 k}{3 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{3 m}=$ $\frac{\sqrt{3}}{3} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, 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 $R_{m, 4}=\frac{6 k}{4 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, 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 $R_{m, 5}=\frac{6 k}{5 m}>\frac{2 \sqrt{3} \cdot(m-\sqrt{3})}{5 m}=$
$\frac{2 \sqrt{3}}{5} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, 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 desired property.
