Library / /434 of 740

, 2023

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $ABCD$ and $WXYZ$ be two squares that share the same center such that $WX \parallel AB$ and $WX < AB$ . Lines $CX$ and $AB$ intersect at $P$ , and lines $CZ$ and $AD$ intersect at $Q$ . If points $P$ , $W$ , and $Q$ are collinear, compute the ratio $AB / WX$ .

Solution

Solution:

Without loss of generality, let $AB = 1$ . Let $x = WX$ . Then, since $BPWX$ is a parallelogram, we have $BP = x$ . Moreover, if $T = XY \cap AB$ , then we have $BT = \frac{1 - x}{2}$ , so $PT = x - \frac{1 - x}{2} = \frac{3x - 1}{2}$ . Then, from $\triangle PXT \sim \triangle PBC$ , we have
$\begin{aligned} \frac{PT}{XT} = \frac{PB}{BC} &\Longrightarrow \frac{\frac{3x - 1}{2}}{\frac{1 - x}{2}} = \frac{x}{1} \\ &\Longrightarrow 3x - 1 = x(1 - x) \\ &\Longrightarrow x = \pm \sqrt{2} - 1 \end{aligned}$
Selecting only the positive solution gives $x = \sqrt{2} - 1$ . Thus, the answer is $\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.