The numbers $S_1=2^2, S_2=2^4,\ldots, S_n=2^{2n}$ are given. A rectangle $OABC$ is constructed on the Cartesian plane according to these numbers. For this, starting from the point $O$ the points $A_1,A_2,\ldots,A_n$ are consistently marked along the axis $Ox$, and the points $C_1,C_2,\ldots,C_n$ are consistently marked along the axis $Oy$ in such a way that for all $k$ from $1$ to $n$ the lengths of the segments $A_{k-1}A_k=x_k$ and $C_{k-1}C_k=y_k$ are positive integers (let $A_0=C_0=O$, $A_n=A$, and $C_n=C$) and $x_k\cdot y_k=S_k$. a) Find the maximal possible value of the area of the rctangle $OABC$ and all pairs of sets $(x_1,x_2,\ldots,x_k)$ and $(y_1,y_2,\ldots,y_k)$ at which this maximal area is achieved. b) Find the minimal possible value of the area of the rctangle $OABC$ and all pairs of sets $(x_1,x_2,\ldots,x_k)$ and $(y_1,y_2,\ldots,y_k)$ at which this minimal area is achieved. (E. Manzhulina, B. Rublyov)
Problem
Source: 2019 Belarusian National Olympiad 10.7
Tags: number theory, algebra, geometry, rectangle, analytic geometry
03.09.2019 10:03
Cool problem. Note that we're trying to find extremal values of $OA\cdot OC=(x_1+x_2+\dots+x_n)(y_1+y_2+\dots+y_n)$. a) The answer is $(2^{2n}+n-1)(1+2^2+2^4+\dots+2^{2n-2})$. Claim: We have the maximal case when one of $y_n,x_n$ is equal to $2^{2n}$. Proof: Assume not, let $x_n=2^a,y_n=2^b$ and $x_{n-1}=2^c, y_{n-1}=2^d$. But then it is easy to check that $$(2^{2n}+1+1+\dots 1)(1+2^2+\dots+2^{2n-2})\geq (2^a+2^{c}+2^{2n-4}\dots+2^2)(2^b+2^{d}+2^{2n-4}+\dots+2^2)$$which leads to a contradiction. $\blacksquare$. W.l.o.g let $x_n=2^{2n}$. Caim: Maximum is achieved when $y_{n-1}=2^{2n-2}$. Proof:Assume not and let $x_{n-1}=2^c$ and $y_{n-1}=2^d$, we have that \begin{align*} (2^{2n}+1+x_1+\dots+x_{n-2})(1+2^{n-2}+y_1+\dots+y_{n-2}) & \geq (2^{2n}+2^c+x_1+\dots+x_{n-2})(1+2^d+y_1+\dots+y_{n-2}) \\ \Longleftrightarrow 2^{4n-2}+1+\sum_{i=1}^{n-2}y_i+(1+2^{2n-2}) \sum_{i=1}^{n-2}x_i & \geq 2^{2n+d}+2^c+2^c\sum_{i=1}^{n-2}y_i+(2^d+1) \sum_{i=1}^{n-2}x_i \\ \implies 2^{4n-2}+1+\sum_{i=1}^{n-2}y &\geq 2^{2n+d}+2^c+2^c \sum_{i=1}^{n-2}y \end{align*}which is true since $2^{4n-2}$ dominates everything else. $\blacksquare$ Now, with a simple induction and with an identical inequality as in the above claim, we can show that the maximum is indeed achieved when $y_i=2^{2i}$ for all $i\leq n-1$ which finishes the proof. b) The answer is $2^2(2^n-1)^2$. By Cauchy Schwarz, we have \begin{align*} (x_1+x_2+\dots+x_n)(y_1+y_2+\dots+y_n) &\geq (\sqrt{x_1y_1}+\sqrt{x_2y_2}+\dots+\sqrt{x_ny_n})^2 \\ &=(2+2^2+2^3+\dots+2^n)^2 \\ &=(2^{n+1}-2)^2. \end{align*}Equality holds at $\frac{x_i}{y_i}$ are equal for $1\leq i \leq n$ wich is easy to find that $x_i=\sqrt{\frac{S_i}{S_j}}\cdot x_j$ for all $i\neq j$, so $x_1=\frac{x_n}{2^{n-1}}$ and since $x_1 \in \{1,2,4\}$ we have that $x_n \in \{2^{n-1},2^n,2^{n+1}\}$. So, all equality cases are $x_i=\sqrt{\frac{S_i}{S_n}}\cdot x_n$ where $x_n$ is in the given set and $x_n$ is the same for all $x_i$'s.