A rectangular plate is measured in two ways by equal segments on its sides. In the first case, starting from each corner and moving towards the next corner (in the direction of one perimeter of the rectangle), and in the second case, starting from the endpoints of one diagonal towards the two adjacent corners. Consider the area of the quadrilateral defined by the endpoints of the segments in both cases. For what segment measurement does the area reach an extreme value in each case? For what segment measurement will the areas of the quadrilaterals obtained in the two ways be equal?
I. Let the sides of the rectangle be $a$ and $b$, where $a \geq b (>0)$, and the length of the segment measured on the sides be $x$. Thus, the area of the resulting quadrilaterals as a function of $x$ is

$$
03 b$ in the case where the $x_0$ abscissa of the common vertex of the parabola representing $t_1$ (and the parabola representing $t_2$) is to the right of the interval (1), because

$$
x_{0}=\frac{a+b}{4}>\frac{3 b+b}{4}=b
$$

Therefore, $t_1$ in (1) is represented by an arc of the descending branch of the parabola, and thus $t_1$ takes its smallest value at the right endpoint of (1):

$$
x=b \quad \text{in which case} \quad t_{1 \text{ min }}^{\prime}=b^{2}, \quad \text{and} \quad t_{2 \max }^{\prime}=a b-b^{2}
$$

Furthermore, $t_1$ does not take its maximum value, and $t_2$ does not take its minimum value, as in the previous case.

II. The requirement $t_{1}=t_{2}$, according to the above, is satisfied if

$$
4 x^{2}-2(a+b) x+a b=0, \quad \text{or if} \quad x_{1}=a / 2, \quad x_{2}=b / 2
$$

The latter value can always be measured on the plate, while the former only if $a / 2 \leq b$, i.e., $b \leq a \leq 2 b$. In both cases, the measurement in both ways not only results in quadrilaterals of equal area but also in congruent quadrilaterals.

László Márki (Budapest, Fazekas M. High School III. grade)