HMMT-Nov 2023 guts, Problem 11

Let $A B C D$ and $W X Y Z$ be two squares that share the same center such that $W X \| A B$ and $W X<A B$. Lines $C X$ and $A B$ intersect at $P$, and lines $C Z$ and $A D$ intersect at $Q$. If points $P, W$, and $Q$ are collinear, compute the ratio $A B / W X$.

Solution: Solution: Without loss of generality, let $A B=1$. Let $x=W X$. Then, since $B P W X$ is a parallelogram, we have $B P=x$. Moreover, if $T=X Y \cap A B$, then we have $B T=\frac{1-x}{2}$, so $P T=x-\frac{1-x}{2}=\frac{3 x-1}{2}$. Then, from $\triangle P X T \sim \triangle P B C$, we have \[ \begin{aligned} \frac{P T}{X T}=\frac{P B}{B C} & \Longrightarrow \frac{\frac{3 x-1}{2}}{\frac{1-x}{2}}=\frac{x}{1} \\ & \Longrightarrow 3 x-1=x(1-x) \\ & \Longrightarrow x= \pm \sqrt{2}-1 \end{aligned} \] Selecting only positive solution gives $x=\sqrt{2}-1$. Thus, the answer is $\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$. $\fbox{\sqrt{2}+1}$.

Answer: \sqrt{2}+1