問題：
解答：
Let $A:(0,0)$, $B:(0,a)$, $C:(a,a)$, $D:(a,0)$ be the corners of the square.
The equations of the circles centered at $A$, $B$ and $D$ with radii $2$, $1$ and $3$ respectively are
$x^2+y^2=4$ ------- $(1)$
$x^2+y^2-2ay+a^2=1$ ------- $(2)$
$x^2+y^2-2ax+a^2=9$ ------- $(3)$
The point $K:(m,n)$ in the square satisfies these equations simultaneously, where $m>0$ and $n>0$
From (1) and (2), (2) and (3), we have
$2an=a^2+3>0$
$2am=a^2-5>0$
Sum of their squares is
$(2ay)^2+(2ax)^2=4a^2(x^2+y^2)=16a^2$
Thus,
$(a^2+3)^2+(a^2-5)^2=16a^2$
$a^4-10a^2+17=0$
Using quadratic equation to solve $a^2$, which is the area of the square.
$Area=a^2$$\\=\frac{10 \pm \sqrt{10^2-4(17)}}{2}
\\=5 \pm 2 \sqrt{2}
\\=5 + 2 \sqrt{2}$
(Since $a^2>5$, reject $a^2=5-2 \sqrt{2}$)
Therefore $a= \sqrt{5 + 2 \sqrt{2}}$
Consider $\triangle ABK$, by cosine law, we have
$\begin{align*}
a^2&=1^2+2^2-2(1)(2) \cos \theta
\\\cos \theta &=-\frac{1}{\sqrt{2}}
\\\theta &=135 ^\circ
\end{align*}$
P.S. 這個順便測試在Mathb.in寫的東西直接複製過來怎麼，完全沒問題。在Mathb.in編輯簡單多了，寫了的東西可立即顯示，這邊預覽跑半天呢。
