6. What is the maximum area that a convex quadrilateral can have, in which both segments connecting the midpoints of opposite sides are equal and have a given length d?
SOLUTION. Let $ABCD$ be the considered convex quadrilateral and $K, L, M$ and $N$ be the midpoints of its sides $AB, BC, CD$ and $DA$, respectively. The quadrilateral $KLMN$ is always a parallelogram (sometimes called the Varignon parallelogram of quadrilateral $ABCD$), because its sides $KL, LM, MN, NK$ are the midlines of triangles $ABC, BCD, CDA, DAB$, respectively. Therefore, $KL \| AC \| MN$ and $LM \| BD \| NK$. If both diagonals $KM$ and $LN$ in parallelogram $KLMN$ have the same length $d$, then $KLMN$ is a rectangle, in which $KL \perp LM$, and thus $AC \perp BD$. We have proven that the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are perpendicular to each other.

The area of parallelogram $KLMN$ is equal to half the area of quadrilateral $ABCD$ (see auxiliary problem 1), so quadrilateral $ABCD$ has the maximum area when the corresponding rectangle $KLMN$ has the maximum area. Among all rectangles with a given diagonal $d$, the square has the largest area, which is $\frac{1}{2} d^{2}$. The maximum possible area of quadrilateral $ABCD$ is therefore $d^{2}$. Any quadrilateral with equal and perpendicular diagonals of length $d \sqrt{2}$ (such as the one in Fig. 6.1) has this maximum area.

![](https://cdn.mathpix.com/cropped/2024_04_17_b5d4019ab45eeeca280ag-7.jpg?height=432&width=440&top_left_y=1374&top_left_x=516)

Fig. 6.1

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