Stacking It Right

Let the leg of the right isosceles triangle have length $A$ . Then $\begin{cases} x=A+A\sin { { 45 }^{ ° } } \\ y=A-A\sin { { 45 }^{ ° } } \end{cases}$ We can then find the ratio by factoring out $A$ from the numerator and denominator: $\dfrac{x}{y}=\dfrac{1+\dfrac{\sqrt2}{2}}{1-\dfrac{\sqrt2}{2}}=\dfrac{(2+\sqrt2)^2}{(2-\sqrt2)(2+\sqrt2)}=3+2\sqrt2$ $=\large \color{#20A900}{\boxed{\sqrt8+\sqrt9}}$

Relevant wiki: Tangent - Sum and Difference Formulas

The ratio $\dfrac xy$ can expressed as $\tan(A + B )$ .

Let $PQ = QR = RT = k$ . Then by Pythagorean theorem , $PR^2 = PQ^2 + QR^2 = k^2 + k^2 \Rightarrow PR = k\sqrt2$ .

So $\tan B = \dfrac{QR}{PQ} = \dfrac kk = 1$ and $\tan A = \dfrac{RT}{PR} = \dfrac k {k\sqrt2} = \dfrac1{\sqrt2}$ .

Hence, by compound angle formula ,

$\begin{aligned} \dfrac xy = \tan (A+B) &=& \dfrac{\tan A + \tan B}{1 - \tan A \tan B} \\ &=& \dfrac{1 + \frac1{\sqrt2}}{1 - 1 \cdot \frac1{\sqrt2}} \\ &=& \dfrac{\sqrt2 + 1}{\sqrt2 - 1} \\ &=& \dfrac{(\sqrt2 + 1)^2}{(\sqrt2 - 1)(\sqrt2 + 1)} \\ &=& \dfrac{2 + 1 + 2\sqrt2}{2 - 1} \\ &=& 3 + 2\sqrt2 = \boxed{\sqrt8 + \sqrt9} \; . \end{aligned}$