Austria 2010

There are two distinct possibilities for dissections that we must consider. The rectangles can either be in the form of three "strips" (i.e. all with one side of length $2010$) or there can be one such strip, with the other two rectangles resulting from a cut at right angles to the first cut.

We first consider the case of the three strips. Since the areas of the strips are all equal to $2010$ times their width, they must be such that the middle one has a width that is the arithmetic mean of the widths of the other two. Since $2010:3=670$, the width of the middle strip is certainly $670$, and the width of the smallest strip can be any integer less than or equal to $670$. There are therefore $670$ possible dissections in this case.

We now consider the other option. Naming the areas of the three strips $A$, $B$ and $C$ with $B=\frac{A+C}{2}$ and $A\le C$, we again have two possible cases. The rectangle with area $B$ can either be the strip or one of the other rectangles.

Let us assume that the strip has area $B$. Since its area is one third of the area of the square, this rectangle has the dimensions $670\times 2010$. The other two together therefore have the dimensions $1340\times 2010$, and the common edge must be of length $1340$. Since $A<C$ the shorter edge of the rectangle with area $A$ can be any integer from $1$ to $2010:2=1005$, and there are therefore $1005$ possible dissections in this case.

Finally, we assume that the strip, which has the dimensions $a\times 2010$, either has area $A$ or $C$. There must then exist an integer $b<2010$, so that we can write $B=(2010-a)\cdot b=670\cdot 2010$. Since $67^2|670\cdot 2010$ and $67^2>2010$, both $a$ and $b$ must be divisible by $67$, and we can write $a=67c$ and $b=67d$. Substituting, we therefore obtain $(30-c)\cdot d=300$ with $d<30$. Since $30-c<30$ also holds, we also have $d>10$. $d$ must therefore be a divisor of $300$ between $10$ and $30$, which yields possible values $12$, $15$, $20$ and $25$ for $d$. This yields possible values of $5$, $10$, $15$ and $18$ for $c$, which in turns yields possible values of $335$, $670$, $1005$ and $1206$ for $a$. The value $a=670$ was already counted in the previous case, however, since this is the case for which $A=B=C$ holds. We therefore have possible dissections that have not already been counted.

In total, we obtain $670+1005+3=1678$ possible dissections which the required properties.