11. On the positive direction of the $x$-axis, take a sequence of points $\left\{A_{j}\right\}, j=1,2, \cdots$, and in the first quadrant on the parabola $y^{2}=\frac{3}{2} x$, take a sequence of points $\left\{B_{k}\right\}, k=1,2, \cdots$, such that $\triangle A_{k-1} B_{k} A_{k}(k=1,2, \cdots)$ are all equilateral triangles, where $A_{0}$ is the origin. Then the side length of the 2005th equilateral triangle is
11.2005.

Let the side length of the $n$-th equilateral triangle be $a_{n}$. Then the coordinates of the vertex $B_{n}$ of the $n$-th equilateral triangle on the parabola are
$$
\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}, \sqrt{\frac{3}{2}\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}\right)}\right) .
$$

From the $n$-th equilateral triangle, we can obtain the $y$-coordinate of $B_{n}$ as
$$
\sqrt{a_{n}^{2}-\left(\frac{1}{2} a_{n}\right)^{2}}=\frac{\sqrt{3}}{2} a_{n} \text {. }
$$

Thus, $\frac{\sqrt{3}}{2} a_{n}=\sqrt{\frac{3}{2}\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}\right)}$.
Therefore, $\frac{1}{2} a_{n}^{2}=a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}$.
From this, we get
$$
a_{1}+a_{2}+\cdots+a_{n}=\frac{a_{n}}{2}+\frac{1}{2} a_{n}^{2},
$$

and $a_{1}+a_{2}+\cdots+a_{n-1}=\frac{a_{n-1}}{2}+\frac{1}{2} a_{n-1}^{2}$.
(1) - (2) yields
$$
a_{n}=\frac{1}{2}\left(a_{n}-a_{n-1}\right)+\frac{1}{2}\left(a_{n}-a_{n-1}\right)\left(a_{n}+a_{n-1}\right) \text {. }
$$

Rearranging gives $\left(a_{n}-a_{n-1}-1\right)\left(a_{n}+a_{n-1}\right)=0$.
Since $a_{n}+a_{n-1} \neq 0$, we have $a_{n}-a_{n-1}=1$.
Taking $n=1$ in equation (1), we get $\frac{1}{2} a_{1}=\frac{1}{2} a_{1}^{2}$.
Since $a_{1} \neq 0$, it follows that $a_{1}=1$.
Therefore, the side length of the 2005th equilateral triangle is 2005.