12. $A B C D$ and $E F G H$ are squares of side length 1, and $A B / / E F$. The overlapped region of the two squares has area $\frac{1}{16}$. Find the minimum distance between the centres of the two squares.
(2 marks)
$A B C D$ 和 $E F G H$ 皆是邊長為 1 的正方形, 且 $A B / / E F$ 。兩個正方形的重疊部分的面積為 $\frac{1}{16}$ 。求兩個正方形的中心的距離的最小可能值。
(2 分)
12. $\frac{\sqrt{14}}{4}$
12. Let $I$ and $J$ be the centres of $A B C D$ and $E F G H$ respectively, with the orientation as shown. Let $A B$ meet $E H$ at $N, B C$ meet $G H$ at $M$ with $H M=x$ and $H N=y$. Let also $K$ be a point such that $I K / / A B$ and $J K / / E H$. Note that $I K=\frac{1}{2}+\frac{1}{2}-x=1-x$, and similarly $J K=1-y$. Since $x y=\frac{1}{16}$, we have
$$
\begin{aligned}
I J^{2} & =I K^{2}+J K^{2} \\
& =(1-x)^{2}+(1-y)^{2} \\
& =x^{2}+y^{2}-2(x+y)+2 \\
& =x^{2}+y^{2}+2 x y-2 x y-2(x+y)+2 \\
= & (x+y)^{2}-2\left(\frac{1}{16}\right)-2(x+y)+2 \\
= & (x+y)^{2}-2(x+y)+\frac{15}{8} \\
= & (x+y-1)^{2}+\frac{7}{8}
\end{aligned}
$$

Hence the minimum value of $I J$ is $\sqrt{\frac{7}{8}}=\frac{\sqrt{14}}{4}$, which occurs when $x+y=1$ and $x y=\frac{1}{16}$, i.e.
$$
\{x, y\}=\left\{\frac{2+\sqrt{3}}{4}, \frac{2-\sqrt{3}}{4}\right\}
$$