$13.1 .22 \star$ Three vertices of a rhombus are located successively on the sides $A B$, $B C$, and $A D$ of a square with side length 1. Find the area of the figure filled by the fourth vertex of the rhombus.
Let $K$, $L$, $N$ be the vertices of the rhombus on the sides $AB$, $BC$, $AD$ of the square (as shown in Figure a). Note that the length of $KB$ is equal to the distance from point $M$ to the line $AD$. Therefore, if point $K$ is fixed, the possible positions of point $M$ fill a line segment $M_1 M_2$ parallel to the side $AD$. The lower position of point $M$, $M_1$, corresponds to the case where $N_1 = A$, and the higher position of point $M$, $M_2$, corresponds to the case where $L_2 = C$.

To determine the positions of points $M_1$ and $M_2$ related to the position of point $K$, we establish a coordinate system as shown in Figure b. It is easy to calculate that if the x-coordinate of point $K$ is $x > 0$, then $M_1 = (-x, \sqrt{2x})$, $M_2 = (-x, \sqrt{1-2x} + 1)$. Using the symmetry of the set of points $M$ about the y-axis, we find that the desired set is the shaded area in Figure c, with the curved boundaries being parabolas, and the areas of the regions labeled 1 and 2 are equal. Therefore, the area filled by the fourth vertex of the rhombus is equal to the area of the square $ABCD$, which is 1.

Fig a
Fig b
Fig c