Given: the situation described in the figure below, with $\alpha=45^\circ$, $\overline{AS}=12$, $\overline{DS}=6$.
Find: $\overline{QS}$.
Question from a math contest. I've tried some ideas but developments are not going through. Hints and answers are welcomed.
On
After the excellent answer given by Aretino, I came up with another solution, based on midsegment theorems and Pitagoras.
By adding some construction lines, define points O, K, T, P and G. Our goal is to find $QS=m+n$. First, notice that
$$\triangle BEA\cong \triangle AFD\cong \triangle KDC\ \ \ \text{and}\ \ \ \triangle DGQ\cong \triangle ATO.$$
Let $m=BA=DF=KD$, $x=BE=AF=KC$ and $n=QG=OT$.
Consider the trapezoid $BFDE$ and notice that $OT$ is a midsegment, given that $O$ is a midpoint of $ED$. Therefore, by the Trapezoid midsegment theorem: $$n=\dfrac{m+x}{2}\ \ \ (1).$$ On $\triangle APC$ notice that $O$ is a midpoint for $AC$ and by the triangle midsegment theorem, $T$ is a midpoint for $AP$. Therefore, $AT=TP=FS=(x-12)$, and as $PF=m$, it is true that $$3(12-x)+m=12\ \ \Leftrightarrow\ \ m=3x-24\ \ \ (2).$$ Consider the right triangle $\triangle FSD$ and by Pitagoras, using the given fact that $DS=6$, $$(12-x)^2+m^2=6^2\ \ \ (3).$$ Now there are 3 equations, (1), (2) and (3), that can be easily solved for the unknowns $x$, $m$ and $n$. Substituting $m$ from (2) in (3) leads to $$x^2-84x-342=0\ \ \Leftrightarrow\ \ x=\dfrac{42\pm 3\sqrt{6}}{5}.$$ Finally, by replacing this last result in (2) and then in (1) to define $m$ and $n$, it follows $$m+n=6\pm 3\sqrt{6}.$$ But only the solution $6+3\sqrt{6}$ is acceptable as $6-3\sqrt{6}<0$. Therefore $$QS=6+3\sqrt{6}.$$
Draw $DG$ parallel to $AS$, with $G$ on $QS$ and set $\angle FAD=\theta$, so that $\angle GDQ=45°+\theta$. As $DG=FS$ you have the equation $$ 12-L\cos\theta={L\over\sqrt2}\cos(45°+\theta), $$ where $L=AD$. This can be solved for L: $$ L={24\over3\cos\theta-\sin\theta}. $$ On the other hand $FS=\sqrt{6^2-L^2\sin^2\theta}=12-L\cos\theta$. Plugging here the above result for $L$ and squaring one can solve for $\theta$: $$ \tan\theta={1+4\sqrt6\over19}. $$ Finally, express $QS$ in terms of $L$ and $\theta$ to obtain the final result: $QS=6+3\sqrt6$.