VMO

a. Consider the rectangle $ABCD$ where $AB=CD=120$ and $AD=BC=100$. Divide the rectangle into $10$ sub-rectangles of size $30\times 40$ as shown below. Consider $9$ centers of the flower pots. By the pigeonhole principle, it is clear that there exists a sub-rectangle that does not contain any center.

Suppose that rectangle is $XYZT$ where $XY=ZT=40$, $XT=YZ=30$. Consider one more rectangle $X^{\prime }{Y}^{\prime }{Z}^{\prime }{T}^{\prime }$ lying inside $XYZT$ such that the sides of two rectangles are pairwise parallel and the gaps are equal to $2.5$.



It is easy to check that $X^{\prime }{Y}^{\prime }{Z}^{\prime }{T}^{\prime }$ does not share any point with the pots so we can build a house on this plot.

b. Consider a rectangle $ABCD$ where $AB=CD=120$ and $AD=BC=100$. Let $L$ be the perimeter of the lake. According to the problem, there exist points $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ in $L$ such that $$A{A}^{\prime },B{B}^{\prime },C{C}^{\prime },D{D}^{\prime }\le 5.$$ Since the lake is a convex polygon, then $A^{\prime }{B}^{\prime }$, $B^{\prime }{C}^{\prime }$, $C^{\prime }{D}^{\prime }$ and $D^{\prime }{A}^{\prime }$ do not overlap. Hence, $$|L|\ge A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{A}^{\prime }.$$ Denote $A_1$ as the projection of $A^{\prime }$ to $AD$ and $A_2$ as the projection of $A^{\prime }$ to $AB$. Similarly, we can define $B_1$, $B_2$, $C_1$, $C_2$, $D_1$, $D_2$ and we have $$A_1{A}^{\prime }+A^{\prime }{B}^{\prime }+B^{\prime }{B}_1\ge A_1{B}_1\ge AB=120.$$ Similarly, we also have $$B_2{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{C}_2\ge 100,$$ $$C_1{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{D}_1\ge 120,$$ $$D_2{D}^{\prime }+D^{\prime }{A}^{\prime }+A^{\prime }{A}_2\ge 100.$$ Hence, $$A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{D}^{\prime }+D^{\prime }{A}^{\prime }+(A^{\prime }{A}_1+A^{\prime }{A}_2+B^{\prime }{B}_1+B^{\prime }{B}_2+C^{\prime }{C}_1+C^{\prime }{C}_2+D^{\prime }{D}_1+D^{\prime }{D}_2)\ge 440.$$ Finally, applying the Cauchy-Schwarz 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 that the length of $L$ does not exceed $440-20\sqrt{2}$. $\square$