2025 SSMO Team Round Problems/Problem 3
Problem
A rectangle is divided as shown into nine smaller rectangles. The areas of the five smallest rectangles are 
and 
. What is the largest possible area of the original rectangle?
![[asy] unitsize(1cm); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); draw((1,0)--(1,3)); draw((1.7,0)--(1.7,3)); draw((0,0.8)--(4,0.8)); draw((0,2)--(4,2)); [/asy]](http://latex.artofproblemsolving.com/4/4/5/44512addc6f4a2c445e557c9dfbbbc7b739825f2.png)
Solution
Without loss of generality, assume that the sub-rectangle with area 
is in the bottom-left corner. We label the areas and the side lengths of the sub-rectangles as shown.
![[asy] unitsize(1cm); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); draw((1,0)--(1,3)); draw((1.7,0)--(1.7,3)); draw((0,0.8)--(4,0.8)); draw((0,2)--(4,2)); label("$1$",(0.5,0.4)); label("$P$",(0.5,1.4)); label("$Q$",(0.5,2.5)); label("$R$",(1.35,0.4)); label("$S$",(2.85,0.4)); label("$W$",(1.35,1.4)); label("$X$",(1.35,2.5)); label("$Y$",(2.85,1.4)); label("$Z$",(2.85,2.5)); label("$a$",(-0.2,0.4)); label("$b$",(-0.2,1.4)); label("$c$",(-0.2,2.5)); label("$d$",(0.5,-0.3)); label("$e$",(1.35,-0.35)); label("$f$",(2.85,-0.3)); [/asy]](http://latex.artofproblemsolving.com/1/a/3/1a30c967ab7c7332ac7e33629f5093e699a7eb89.png)
Then, we have
\begin{align*}
W &= abde = PR\\
X &= acde = QR\\
Y &= abdf = PS\\
Z &= acdf = QS.
\end{align*}
Because 
and 
, we must have 
. By analogous reasoning, we can conclude that 
as well. Therefore, 
.
We know that have 
. Thus, in order to maximize the area of the original rectangle, it suffices to maximize the value of 
. We have
\begin{align*}W+X+Y+Z &= PR+QR+PS+QS \\
&= (P+Q)(R+S).
\end{align*}
We know that 
, so the expression 
is maximized when 
, which is achievable; for example, consider 
. This yields 
as the largest possible value of . Hence, the largest possible area of the original rectangle is 
.
~Sedro
