11.1. ** Given an equilateral triangle $\triangle K L M$ with side length $\frac{1}{4}$ inside (or on the boundary of) an equilateral triangle $\triangle A B C$ with side length 1, find the position of $\triangle K L M$ when the sum of the distances from point $A$ to the lines $K L$, $L M$, and $M K$ is maximized.
First, move points $L$ and $K$ to $BC$ and $AB$ respectively, and denote $\angle B L K = \angle A K M = \alpha$. When $60^{\circ} \leqslant \alpha \leqslant 120^{\circ}$, point $A$ is located inside $\angle K L M$, and the sum of distances is $8\left(S_{\triangle A K L} + S_{\triangle A L M} + S_{\triangle A K M}\right) = 8\left(S_{\triangle K L M} + 2 S_{\triangle A K M}\right)$. Thus, it becomes necessary to find an appropriate $\alpha$ that maximizes $S_{\triangle A K M}$, which is equivalent to maximizing $A K \cdot \sin \alpha$.

It is easy to see that $B K = \frac{\sin \alpha}{2 \sqrt{3}}$, so $A K \sin \alpha = \left(1 - \frac{\sin \alpha}{2 \sqrt{3}}\right) \sin \alpha$. The maximum value occurs when $\alpha = 90^{\circ}$.

When $0^{\circ} \leqslant \alpha < 60^{\circ}$, construct the symmetric point $N$ of point $K$ with respect to $L M$ (not shown in the figure). In this case, $\triangle M N L \cong \triangle K M L$, and point $N$ is inside $\triangle A B C$. Then, adjust the configuration to reduce it to the previous scenario.