## Task 3/64

At each corner of a rectangle $F=ab$, a segment $x$ is cut off, and specifically,

a) from each corner in the counterclockwise direction,

b) from two opposite corners in both directions

The resulting points are the vertices of a parallelogram. For which segment $x$ is the area of the parallelogram extreme?
a) $\quad F_{1}=a b-(a-x) x-(b-x) x=a b-(a+b) x+2 x^{2} ; \quad F_{1}^{\prime}=-(a+b)+4 x ; \quad F_{1}^{\prime \prime}=4>0$
b) $\quad F_{2}=a b-x^{2}-(a-x)(b-x)=(a+b) x-2 x^{2} ; \quad F_{2}^{\prime}=(a+b)-4 x \quad F_{2}^{\prime \prime}=-4<0$

From $F_{1}^{\prime}=0$ it follows that $x_{E}=\frac{1}{4}(a+b)$. From $F_{2}^{\prime}=0$ it follows that $x_{E}=\frac{1}{4}(a+b)$.

In both cases, one must set $x=\frac{1}{4}(a+b)$ to obtain an extremum. One finds a minimum in case a) and a maximum in case b). (Figure)

$$
F_{1 \min }=a b-\frac{(a+b)^{2}}{8} \quad ; \quad F_{2 \max }=\frac{(a+b)^{2}}{8}
$$

It is always the case, even in the extreme cases, that $F_{1}+F_{2}=F$. The sum of the remaining triangles is also equal to $F$. From them, one can compose $F$ in various ways, including forming a trapezoid with perpendicular diagonals.

Determination, special cases:

a) If the smaller side is denoted by $b$, then $x \leq b$ must hold to obtain a parallelogram lying within the rectangle. In the extreme case, $\frac{a+b}{4} \leq b$, which means $b \geq \frac{a}{3}$. In the limiting case $b=\frac{a}{3}$, two vertices of the extreme area coincide with opposite vertices of the rectangle, $F_{1 \min }=b^{2}, F_{2 \max }=2 b^{2}$.

b) If one starts with a square with $F=a^{2}$, the extreme figures coincide with the square of the midpoints. This appears as the minimum of all inscribed squares and the maximum of all inscribed rectangles. The sides are parallel to the diagonals of the square.