Let $ABCD$ be a rectangle with side lengths $AB = CD = 5$ and $BC = AD = 10$. $W, X, Y, Z$ are points on $AB, BC, CD$ and $DA$ respectively chosen in such a way that $WXYZ$ is a kite, where $\angle ZWX$ is a right angle. Given that $WX = WZ = \sqrt{13}$ and $XY = ZY$, determine the length of $XY$.
Problem
Source: SAMO 2017 Q2
Tags: geometry
Tintarn
08.08.2017 19:14
Let $AW=a$. Then by similarity, $BX=AW=a, AZ=BW=5-a$ and hence $a^2+(5-a)^2=13$ i.e. $a=2$ or $a=3$. By symmetry, w.l.o.g. $a=2$.
If $CY=x$, then $x^2+8^2=XY^2=YZ^2=(5-x)^2+7^2$ and hence $10x=5^2+7^2-8^2=10$ i.e. $x=1$ and hence $XY=\sqrt{x^2+64}=\sqrt{65}$.
Fortified
08.11.2017 10:06
DylanN wrote: Let $ABCD$ be a rectangle with side lengths $AB = CD = 5$ and $BC = AD = 10$. $W, X, Y, Z$ are points on $AB, BC, CD$ and $DA$ respectively chosen in such a way that $WXYZ$ is a kite, where $\angle ZWX$ is a right angle. Given that $WX = WZ = \sqrt{13}$ and $XY = ZY$, determine the length of $XY$. Were you the winner of the SAMO in 2012 and 2011???