Geometry problem (find QS)

Question

Could someone help with this one? Been trying it for days, no success until now.

Consider the picture above, in which $ABCD$ is a square, $\angle DCQ = 45^\circ$, $\overline{AS} = 12$ and $\overline{DS} = 6$. Calculate the length of $\overline{QS}$.

(a) $6+ 3\sqrt{6}$

(b) $2+3\sqrt{6}$

(c) $6+\sqrt{6}$

(d) $3+6\sqrt{6}$

(e) $3+3\sqrt{6}$

Answer 1

Hint.

Given $L$ the square side and $\theta$ the square rotation $$ L\cos(\theta)+L_0\cos(\theta+45^{\circ})=12\\ \left(L\sin(\theta)\right)^2+\left(L_0\cos(\theta+45^{\circ})\right)^2 = 6^2 $$

solving this you have the values for $L$ and $\theta$

so

$$ QS = L\sin(\theta)+L_0\sin(\theta+45^{\circ}) $$

NOTE

Here $L_0 = \frac{L}{\sqrt 2}$

To solve this we have

$$ \frac{3}{2} L \cos (\theta )-\frac{1}{2} L \sin (\theta )=12\\ \frac{1}{2} L^2 \sin ^2(\theta )-\frac{1}{2} L^2 \cos ^2(\theta )-\frac{1}{2} L^2 \sin (\theta ) \cos (\theta )+\frac{3 L^2}{4}=36 $$

or

$$ \frac{3 c}{2}-\frac{s}{2}=\frac{12}{L}\\ \frac{s^2}{2}-\frac{1}{2} c^2-\frac{1}{2} s c+\frac{3}{4}=\frac{36}{L^2} $$

solving for $s, c$ we have

$$ \left( \begin{array}{cc} c & s \\ \frac{12}{L}-\frac{\sqrt{\frac{3}{10}} \sqrt{-L^2 \left(L^2-144\right)}}{L^2} & -\frac{3 \left(\sqrt{30} \sqrt{-L^2 \left(L^2-144\right)}-40 L\right)}{10 L^2} \\ \frac{\sqrt{\frac{3}{10}} \sqrt{-L^2 \left(L^2-144\right)}}{L^2}+\frac{12}{L} & \frac{3 \left(40 L+\sqrt{30} \sqrt{-L^2 \left(L^2-144\right)}\right)}{10 L^2} \\ \end{array} \right) $$

now using $c^2+s^2=1$ we have

$$ \left\{ \begin{array}{rcl} -\frac{48 \sqrt{\frac{6}{5}} \sqrt{-L^2 \left(L^2-144\right)}}{L^3}+\frac{720}{L^2}-4&=&0 \\ \frac{48 \left(75 L+\sqrt{30} \sqrt{-L^2 \left(L^2-144\right)}\right)}{5 L^3}-4&=&0 \\ \end{array} \right. $$

giving

$$ L = \left\{ \begin{array}{c} \sqrt{\frac{1}{5} \left(13-2 \sqrt{6}\right)} \\ \sqrt{\frac{468}{5}+\frac{72 \sqrt{6}}{5}} \\ \end{array} \right. $$

finally substititing the values for $c,s,L$ in

$$ QS = \frac{c L}{2}+\frac{3s L}{2} $$

we obtain

$$ QS = 6+3\sqrt 6 $$

Answer 2

Take $S$ as the origin and $D,A$ as $6e^{i\theta},-12$. The center of the square lies at

$$ \frac{1}{\sqrt{2}}(D-A)e^{\pi i/4}+A $$ and the point $Q$ lies at $$ A+\frac{3+i}{2}(D-A). $$ $\text{Re}(Q)=0$ leads to $\theta=\arccos\frac{\sqrt{6}-6}{10}\approx 110.8^\circ$ and to $$\text{Im}(Q) = QS = 6+3\sqrt{6}.$$