B. If positive numbers $x, y, z$ can be the lengths of the three sides of a triangle, then $(x, y, z)$ is called a "triangle number". If $(a, b, c)$ and $\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)$ are both triangle numbers, and $a \leqslant b \leqslant c$, then the range of $\frac{a}{c}$ is $\qquad$.
B. $\frac{3-\sqrt{5}}{2}c \\
\frac{1}{c}+\frac{1}{b}>\frac{1}{a}
\end{array}\right. \\
\Rightarrow \frac{1}{c}+\frac{1}{c-a}>\frac{1}{c}+\frac{1}{b}>\frac{1}{a} \\
\Rightarrow \frac{1}{c}+\frac{1}{c-a}>\frac{1}{a} \\
\Rightarrow\left(\frac{a}{c}\right)^{2}-3\left(\frac{a}{c}\right)+1<0 \\
\Rightarrow \frac{3-\sqrt{5}}{2}<\frac{a}{c}<\frac{3+\sqrt{5}}{2}
\end{array}
$$

Since $\frac{a}{c} \leqslant 1$, we have,
$$
\frac{3-\sqrt{5}}{2}<\frac{a}{c} \leqslant 1 .
$$