$\mathbf{I 2 . 3}$ 在下圖中, 正方形 $A B C D$ 及 $X Y Z W$ 相等而且互相交睤使得頂點 $A$ 位在 $X Y Z W$ 的中心及綫段 $A B$ 將絨段 $Y Z$ 邊分為 $1: 2$ 。若 $X Y Z W$ 的面積與交疊部分的面積比率為 $c: 1$,求 $c$ 的值。
In the figure below, identical squares $A B C D$ and $X Y Z W$ overlap each other in such a way that the vertex $A$ is at the centre of $X Y Z W$ and the line segment $A B$ cuts line segment $Y Z$ into $1: 2$. If the ratio of the area of $X Y Z W$ to the overlapped region is $c: 1$, determine the value of $c$.
$A$ is the intersection of the diagonals $X Z$ and $Y W$. Suppose $A B$ cuts $Y Z$ at $P$ and $A D$ cuts $W Z$ at $Q$. By the property of squares, $\angle P A Q=\angle P Z Q=90^{\circ}$
$$
\begin{array}{l}
\therefore \angle P A Q+\angle P Z Q=90^{\circ}+90^{\circ}=180^{\circ} \\
\Rightarrow A, P, Z, Q \text { are concyclic (opp. } \angle \text { s supp.) } \\
\angle A P M=\angle A Q N \text { (ext. } \angle \text { s cyclic quad.) }
\end{array}
$$

Let $M, N$ be the mid-points of $Y Z$ and $W Z$ respectively.
It is easy to show that $A M=A N$ and $\angle A M P=\angle A N Q=90^{\circ}$
$$
\begin{array}{l}
\therefore \triangle A P M \cong \triangle A N Q \text { (A.A.S.) } \\
\text { Area of shaded region }=\text { area of } A P M+\text { area of } A M Z Q \\
=\text { area of } A N Q+\text { area of } A M Z Q \\
=\text { area of } A M Z N \\
=\frac{1}{4} \times \text { area of } X Y Z W \\
\therefore c=4 \\
\end{array}
$$