13.386 On the sides $AB, BC, AC$ of an equilateral triangle $ABC$, points $A_1, B_1, C_1$ are located such that $AA_1 = BB_1 = CC_1 = x$. The side of the triangle is $a$. Find such $x$ for which the ratio of the areas of triangles $A_1B_1C_1$ and $ABC$ is equal to $m$. Within what limits can the value of $m$ vary?
Solution. By the cosine theorem for $\Delta A_{1} B B_{1}$, we have:

$$
\begin{aligned}
& A_{1} B_{1}^{2}=x^{2}+(a-x)^{2}-2 x(a-x) \cos 60^{\circ} \Rightarrow m=\frac{A_{1} B_{1}^{2}}{A B^{2}}=\frac{x^{2}+(a-x)^{2}-x(a-x)}{a^{2}} \Rightarrow \\
& \Rightarrow 3 x^{2}-3 a x+a^{2}-m a^{2}=0 \Rightarrow x=\frac{a}{6}(3 \pm \sqrt{12 m-3})
\end{aligned}
$$

The problem makes sense if $3-\sqrt{12 m-3}>0 \Leftrightarrow\left\{\begin{array}{l}9>12 m-3, \\ 12 m-3 \geq 0,\end{array} \Leftrightarrow \frac{1}{4} \leq m<1\right.$.

Answer: $\frac{a}{6}(3 \pm \sqrt{12 m-3})$ for $\frac{1}{4} \leq m<1$.