In the center of a square park with side $a$, there is a circular flower bed with diameter $d$. Outside the park and around the edge of the flower bed, a path leads. A straight path also leads from the vertices and midpoints of the sides of the square to the center of the square, towards the flower bed. What is the shortest path from one corner of the park to the opposite corner?
Let's follow the markings on the diagram. We want to get from vertex $A$ to vertex $B$ by the shortest possible route. Based on the symmetry with respect to the diagonal $AB$, we can assume that we will not cross this diagonal. If our route touches point $H$, then the path taken is $A-H-B$, and its length is $2a$. Otherwise, we must pass through point $G$, which means we must definitely take the path $E-G-J$. This is exactly a quarter of the circumference of the circle, so its length is $\frac{d \pi}{4}$. In addition to this, we must pass through either $A-F-E$ and $A-D-E$, or $J-I-B$ and $J-L-B$. This represents four possibilities. However, the length of the path that passes through points $F$ and $L$, or $D$ and $J$, is certainly between the lengths of the other two paths, so it is sufficient to consider the paths $A-F-E-J-I-B$ and $A-D-L-B$. To calculate the lengths of these paths, we need the lengths of segments $|AF|=|IB|$, $|FE|=|JI|$, and $|AD|=|LB|$. These are, respectively, $\frac{a}{2}$, $\frac{a-d}{2}$, and $\frac{a \sqrt{2}-d}{2}$. Accordingly, the lengths of the three possible paths are:

(1) We travel along the side of the square. The length of the path is

$$
A H+H B=2 a
$$

(2) First, we travel along the side of the square to the midpoint, then towards the center of the circle, then along the circumference to point $J$, and finally back outside. The path taken is:

$$
A F+F E+E J+J I+I B=2\left(\frac{a}{2}+\frac{a-d}{2}\right)+\frac{d \pi}{4}=2 a-d+\frac{d \pi}{4}
$$

(3) We travel along the diagonal to the circle, then travel half the circumference, and finally back along the diagonal to the vertex. The path taken is:

$$
A D+D L+L B=2\left(\frac{a \sqrt{2}-d}{2}\right)+\frac{d \pi}{2}=a \sqrt{2}-d+\frac{d \pi}{2}
$$

Since $\pi > 3$, we have $\frac{d \pi}{2} > \frac{d \pi}{4}$. This leads to the condition

$$
\frac{d}{a}<\frac{4(2-\sqrt{2})}{\pi} \approx 0.746
$$

Thus, if the diameter of the flower bed is longer than about three-quarters of the side of the park, it is advisable to take the "outer" path; otherwise, it is shorter to go into the flower bed. (But going into the flower bed and trampling the flowers is in poor taste.)