A4. In the figure, $A Q P B$ and $A S R C$ are squares, and $A Q S$ is an equilateral triangle. If $Q S=4$ and $B C=x$, what is the value of $x$ ?
Solution: The answer is $4 \sqrt{3}$.

Solution 1: Since $\triangle A Q S$ is equilateral, $A Q=Q S=A S$. Since $Q S=4, A Q=A S=4$. Since $A Q P B$ and $A S R C$ are squares, $A B=A Q=4$ and $A C=A S=4$. Since $\triangle A Q S$ is equilateral, $\angle Q A S=60^{\circ}$. Therefore, $\angle B A C=360^{\circ}-90^{\circ}-90^{\circ}-\angle Q A S=120^{\circ}$.

Drop the perpendicular from $A$ to side $B C$ and let this perpendicular intersect $B C$ at $M$. Then by symmetry, $M$ is the midpoint of $B C$ and $\angle B A M=\angle C A M=\angle B A C / 2=120 / 2=$ $60^{\circ}$. Therefore, $\triangle A B M$ is a $30-60-90$ triangle. Consequently,
$$
\frac{B M}{B A}=\frac{\sqrt{3}}{2} \text {. }
$$

Hence, we have $B M=4 \sqrt{3} / 2=2 \sqrt{3}$. Similarly, $C M=2 \sqrt{3}$. Therefore, $B C=B M+C M=$ $4 \sqrt{3}$.

Solution 2: By Solution 1, $A B=A C=4$ and $\angle B A C=120^{\circ}$. By the Cosine Law, we have
$$
\begin{aligned}
B C & =\sqrt{A B^{2}+A C^{2}-2 \cdot A B \cdot A C \cdot \cos \angle B A C} \\
& =\sqrt{4^{2}+4^{2}-2 \cdot 4 \cdot 4 \cdot \cos 120^{\circ}} \\
& =\sqrt{32-32 \cdot(-1 / 2)}=\sqrt{32+16}=\sqrt{48}=4 \sqrt{3} .
\end{aligned}
$$

Therefore, $x=4 \sqrt{3}$.
Let $M, N$ be the midpoints of $B C$ and $Q S$, respectively. By symmetry, $M, A, N$ are collinear and the line $M N$ is perpendicular to lines $Q S$ and $B C$. By Solution $1, \angle Q A S=60^{\circ}$ and $\angle B A C=120^{\circ}$. Therefore, by symmetry, $\angle Q A N=30^{\circ}$ and $\angle B A M=60^{\circ}$. Since $\triangle A Q S$ is equilateral, $\angle A Q N=60^{\circ}$ and $\angle A B M=180^{\circ}-\angle B A M-\angle A M B=180^{\circ}-60^{\circ}-90^{\circ}=$ $30^{\circ}$. Since $A B=A Q, \triangle A N Q$ is congruent to $\triangle B M A$. Therefore, $B M=A N$. By the Pythagorean Theorem,
$$
B M=A N=\sqrt{A Q^{2}-Q N^{2}}=\sqrt{4^{2}-2^{2}}=\sqrt{12}=2 \sqrt{3} .
$$

Hence, $x=B C=2 \cdot B M=2 \sqrt{3}$.
Solution 3: