[數學] 邊長為a的正方形, 其中三角(逆時針, 從左上開始)至正方形一內點的距離分別為1, 2和3, 求邊長a和對應長度為1和2的線段夾角

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。