Example 12 In the plane, two equilateral triangles $A B C$ and $K L M$ have side lengths of 1 and $\frac{1}{4}$, respectively, with $\triangle K L M$ inside $\triangle A B C$. Let $\Sigma$ denote the sum of the distances from $A$ to the lines $K L$, $L M$, and $M K$. Try to find the position of $\triangle K L M$ when $\Sigma$ is maximized.
Solution: First, translate $\triangle K L M$ so that $K$ is on side $A B$ and $L$ is on side $B C$ (Figure 13-12(1)), in this way, the new value of $\Sigma$ is $\geqslant$ the original value of $\Sigma$. Note that
$$
\angle B L K=120^{\circ}-\angle B K L=\angle A K M \text {, }
$$
denote it as $a$.
Consider the following two cases:
(1) $60^{\circ} \leqslant \alpha \leqslant 120^{\circ}$.
In this case, $A$ is inside $\angle K L M$ (Figure 13-12(2)), and
$$
\begin{aligned}
\Sigma & =8\left(S_{A K L}+S_{A L M}+S_{A K M}\right) \\
& =8\left(S_{K L M}+2 S_{A K M}\right) .
\end{aligned}
$$

Since the value of $S_{X L M}$ is fixed, $\Sigma$ reaches its maximum when and only when $S_{A K M}$ reaches its maximum, and $S_{A K M}$ and $A K \sin a$ reach their maximum values simultaneously.
By the Law of Sines, we get
$$
\begin{aligned}
B K & =\frac{2}{\sqrt{3}} K L \cdot \sin \alpha \\
& =\frac{\sin \alpha}{2 \sqrt{3}} .
\end{aligned}
$$

Therefore, $A K=1-B K$
$$
=1-\frac{\sin \alpha}{2 \sqrt{3}} \text {. }
$$

The function $f(\alpha)=(2 \sqrt{3}-\sin \alpha) \sin \alpha\left(60^{\circ} \leqslant \alpha \leqslant 120^{\circ}\right)$ reaches its maximum value when $\sin \alpha=1$, i.e., $\alpha=90^{\circ}$.
(2) $0^{\circ} \leqslant \alpha<60^{\circ}$.
Let $N$ be the reflection of $K$ about the line $L M$ (Figure 13-12(3)). Since $\alpha<60^{\circ}, B L<K L \sin 60^{\circ}<$ $\frac{1}{2} \leqslant L C$, $N$ is inside $\triangle A B C$, and the $\Sigma$ values of the two triangles $K L M$ and $N M L$ are the same (the distance from $A$ to $K L$ equals the distance from $A$ to $M N$ plus the height of $\triangle K M L$; the distance from $A$ to $K M$ equals the distance from $A$ to $L N$ minus the height of $\triangle K M L$), and $\angle M L B \geqslant 60^{\circ}$, which reduces to case (1). Case (1) has already provided the required maximum value.