, 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 $A \leq C$, 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.