One, (20 points) In $\triangle A B C$, it is known that $\angle A>\angle B>$ $\angle C$, and $\angle A=2 \angle B$. If the three sides of the triangle are integers, and the area is also an integer, find the minimum value of the area of $\triangle A B C$.
Let $B C=a, C A=b, A B=c$.
As shown in Figure 6, construct the angle bisector $A D$ of $\angle B A C$, then
$$
\begin{array}{l}
\angle B A D=\angle D A C \\
= \angle B, \\
\angle A D C=\angle B+\angle B A D \\
= 2 \angle B .
\end{array}
$$

Figure 6

Therefore, $\triangle A C D \backsim \triangle B C A$. Thus,
$$
\frac{b}{a}=\frac{C D}{b} \text {. }
$$

By the Angle Bisector Theorem, we have $\frac{b}{c}=\frac{C D}{B D}$. Hence,
$$
\frac{b}{b+c}=\frac{C D}{C D+B D}=\frac{C D}{a} \text {. }
$$

From equations (1) and (2), we get
$$
\frac{b+c}{a}=\frac{a}{b} \text {. }
$$

Thus, $a^{2}=b(b+c)$.
If $(b, c)=d$, then from equation (1) we know $d \mid a$, so we can assume $(b, c)=1$. Therefore, we can let
$$
b=m^{2}, b+c=n^{2} \text {. }
$$

Then $a=m n, c=n^{2}-m^{2}$.
Since $\angle A>\angle B>\angle C$, we have $a>b>c$, i.e.,
$$
m n>m^{2}>n^{2}-m^{2} \text {. }
$$

Thus, $m1$, i.e.,
$$
m>\sqrt{2}+1 \text {. }
$$

Let the area of $\triangle A B C$ be $S$. By Heron's formula, we have
$$
\begin{aligned}
S & =\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)} \\
= & \frac{1}{4} \sqrt{\left(m n+n^{2}\right)\left(n^{2}-m n\right)\left(n^{2}+m n-2 m^{2}\right)\left(2 m^{2}+m n-n^{2}\right)} \\
= & \frac{1}{4} n(n+m)(n-m) . \\
& \sqrt{(2 m+n)(2 m-n)} .
\end{aligned}
$$

From equation (4), we know $m \geqslant 3$. It is easy to verify from equation (3):
When $3 \leqslant m \leqslant 7$, only when $m=5$, $n=6$, $\sqrt{(2 m+n)(2 m-n)}=8$ (a rational number), at this time,
$$
S=\frac{1}{4} \times 6 \times 11 \times 1 \times 8=132 \text {. }
$$

Next, we prove that when $m \geqslant 8, n \geqslant 9$, $S>162$.
From equations (3) and (4), we know
$$
\begin{array}{l}
(2 m+n)(2 m-n) \\
>3 m(2 m-\sqrt{2} m)=(6-3 \sqrt{2}) m^{2} \\
>(6-4 \sqrt{2}) m^{2}=(2-\sqrt{2})^{2} m^{2}, \\
n(n+m)(n-m) \\
>n\left(1+\frac{\sqrt{2}}{2} n\right) \times 1=\frac{1}{2}(2+\sqrt{2}) n^{2} .
\end{array}
$$

From equation (5), we have
$S>\frac{1}{4} \times \frac{1}{2}(2+\sqrt{2}) n^{2}(2-\sqrt{2}) m=\frac{1}{4} n^{2} m$.
Thus, when $m \geqslant 8, n \geqslant 9$, we have $S>162$.
Therefore, the minimum value of $S$ is 132, at this time, $m=5, n=6$.
So, when $a=30, b=25, c=11$, the area of $\triangle A B C$ is minimized, and the minimum value is 132.