VN IMO Booklet

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 $X^{\prime }{Y}^{\prime }{Z}^{\prime }{T}^{\prime }$ lying inside $XYZT$ such that the sides of the two rectangles are pairwise parallel and the gap equals $2.5$, then $X^{\prime }{Y}^{\prime }{Z}^{\prime }{T}^{\prime }$ has the size $25\times 35$. It is clear that the rectangle $X^{\prime }{Y}^{\prime }{Z}^{\prime }{T}^{\prime }$ 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 $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ lying on $L$ such that $$A{A}^{\prime },B{B}^{\prime },C{C}^{\prime },D{D}^{\prime }\le 5.$$ Since the lake is a convex polygon, then $A^{\prime }{B}^{\prime },B^{\prime }{C}^{\prime },C^{\prime }{D}^{\prime },D^{\prime }{A}^{\prime }$ do not overlap. Hence, the length of $L$ is at least $$A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{A}^{\prime }.$$ Denote $A_1$ as the projection of $A^{\prime }$ to $AD$, $A_2$ as the projection of $A^{\prime }$ to $AB$, and similarly define the points $B_1,B_2,C_1,C_2,D_1,D_2$. We have $$A_1{A}^{\prime }+A^{\prime }{B}^{\prime }+B^{\prime }{B}_1\ge A_1{B}_1\ge AB=120.$$ By the same way, we get $$\begin{array}{rl}{B}_2{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{C}_2& \ge 100,\\ C_1{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{D}_1& \ge 120,\\ D_2{D}^{\prime }+D^{\prime }{A}^{\prime }+A^{\prime }{A}_2& \ge 100.\end{array}$$ These imply that $$A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{A}^{\prime }+(A^{\prime }{A}_1+A^{\prime }{A}_2+B^{\prime }{B}_1+B^{\prime }{B}_2+C^{\prime }{C}_1+C^{\prime }{C}_2+D^{\prime }{D}_1+D^{\prime }{D}_2)\ge 440.$$ Finally, by applying Cauchy-Schwarz's inequality, we have $$A^{\prime }{A}_1+A^{\prime }{A}_2\le \sqrt{2(A^{\prime }{A}_1^2+A^{\prime }{A}_2^2)}=\sqrt{2A^{\prime }{A}_2^2}\le 5\sqrt{2}.$$ Similarly, we also have $B^{\prime }{B}_1+B^{\prime }{B}_2\le 5\sqrt{2}$, $C^{\prime }{C}_1+C^{\prime }{C}_2\le 5\sqrt{2}$, $D^{\prime }{D}_1+D^{\prime }{D}_2\le 5\sqrt{2}$.

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