One. (20 points) Given an isosceles triangle with a vertex angle less than $60^{\circ}$, the lengths of its three sides are all positive integers. Construct a square outward on each side, such that the sum of the areas of the three squares is 2009. Find the perimeter of this isosceles triangle.
Let the length of the legs of the isosceles triangle be $a$, and the length of the base be $b$. According to the problem, we have
$$
2 a^{2}+b^{2}=2009 \text{. }
$$

Therefore, $a^{2}=\frac{2009-b^{2}}{2}\frac{2009}{3}>669$.
Thus, $669<a^{2}<1005 \Rightarrow 25<a<32$.
Since $a$ and $b$ are positive integers, it is verified that:
When $a=28$,
$$
b^{2}=2009-2 \times 28^{2}=441=21^{2} \Rightarrow b=21 \text{; }
$$

When $a=26$, $b^{2}=657$ is not a perfect square,
so, $b$ has no integer solution.
Similarly, when $a=27, 29, 30, 31$, $b$ has no integer solution.

In summary, the length of the legs of the required isosceles triangle is 28, the length of the base is 21, and the perimeter is $28+28+21=77$.