VII OM - I - Task 3

In a square $ABCD$ with area $S$, vertex $A$ is connected to the midpoint of side $BG$, vertex $B$ is connected to the midpoint of side $CD$, vertex $C$ is connected to the midpoint of side $DA$, and vertex $D$ is connected to the midpoint of side $AB$. Calculate the area of the part of the square that contains its center.
The solution to the problem is illustrated in Fig. 1, where $ M $, $ N $, $ P $, $ Q $ denote the midpoints of the sides, and $ O $ the center of the square; through the vertices $ A $, $ B $, $ C $, $ D $ lines parallel to the lines $ DQ $, $ AM $, $ BN $, $ CP $ have been drawn. If the figure is rotated by $ 90^\circ $ around the point $ O $, point $ A $ will fall on point $ B $, point $ Q $ on point $ M $, etc., and the figure will overlap with itself. From this, it is easy to conclude that the figure consists of a grid made up of 9 equal squares. According to the Pythagorean theorem, the area of square $ ABCD $ equals the sum of the areas of the squares constructed on the legs $ AK $ and $ BK $ of the right triangle $ AKB $. Denoting the sought area by the letter $ x $, we therefore have