Problem 1800

a. Consider the rectangle $ABCD$ with $AB = CD = 120$ and $AD = BC = 100$. Divide it into 10 subrectangles of size $30 \times 40$ as follows.

Consider 9 centers of the given gardens. By the pigeonhole principle, there is some subrectangle that does not contain any point among them. Suppose it is the rectangle $XYZT$ with $XY = ZT = 40$, $XT = YZ = 30$. Consider one more rectangle $X'Y'Z'T'$ lying inside $XYZT$ such that the sides of the two rectangles are pairwise parallel and the gap equals $2.5$, then $X'Y'Z'T'$ has the size $25 \times 35$. It is clear that the rectangle $X'Y'Z'T'$ does not share any point with any garden, so the desired result will follow.

b. Consider the rectangle $ABCD$ with $AB = CD = 120$ and $AD = BC = 100$. Let $L$ be the boundary of the lake. From the given hypothesis, there are four points $A', B', C', D'$ lying on $L$ such that
$$AA',\ BB',\ CC',\ DD' \le 5.$$
Since the lake is a convex polygon, then $A'B',\ B'C',\ C'D',\ D'A'$ do not overlap. Hence, the length of $L$ is at least
$$A'B' + B'C' + C'D' + D'A'.$$
Denote $A_1$ as the projection of $A'$ to $AD$, $A_2$ as the projection of $A'$ to $AB$, and similarly define the points $B_1, B_2, C_1, C_2, D_1, D_2$. We have
$$A_1A' + A'B' + B'B_1 \geq A_1B_1 \geq AB = 120.$$
By the same way, we get
$$\begin{aligned} B_2B' + B'C' + C'C_2 &\geq 100, \\ C_1C' + C'D' + D'D_1 &\geq 120, \\ D_2D' + D'A' + A'A_2 &\geq 100. \end{aligned}$$
These imply that
$$A'B' + B'C' + C'D' + D'A' + (A'A_1 + A'A_2 + B'B_1 + B'B_2 + C'C_1 + C'C_2 + D'D_1 + D'D_2) \geq 440.$$
Finally, by applying Cauchy-Schwarz's inequality, we have
$$A'A_1 + A'A_2 \leq \sqrt{2(A'A_1^2 + A'A_2^2)} = \sqrt{2A'A_2^2} \leq 5\sqrt{2}.$$
Similarly, we also have $B'B_1 + B'B_2 \leq 5\sqrt{2}$, $C'C_1 + C'C_2 \leq 5\sqrt{2}$, $D'D_1 + D'D_2 \leq 5\sqrt{2}$.

From these inequalities, it is clear to conclude that the length of $L$ does not exceed $440 - 20\sqrt{2}$.