, 2010

Let $a$ and $b$ be the side lengths of the plot. Assume that the cuts divided the side of length $a$ to parts of length $x$ and $a-x$ where $x$ being the greater part, and the side of length $b$ to parts of length $y$ and $b-y$ where $y$ being the greater part. Then the area of the largest piece was $xy$. The condition that this area equals the sum of the areas of the other three pieces can be written as follows:
$$xy = (a-x)y + x(b-y) + (a-x)(b-y).$$

Dividing both sides by $x(a-x)y(b-y)$, one obtains
$$\frac{1}{(a-x)(b-y)} = \frac{1}{x(b-y)} + \frac{1}{(a-x)y} + \frac{1}{xy}.$$
If the price that every musketeer paid for the plot was 1, then the l.h.s. of the last equality is precisely the price per area unit of the piece with area $(a-x) \cdot (b-y)$. Analogously, the r.h.s. equals the sum of the prices per area unit of the other three pieces. Hence multiplying the sides of this equality by the number of area units per acre, the claim of the problem follows.