1. On the side $AB$ of the equilateral triangle $ABC$, a point $M$ is chosen. Through the point $M$, lines are drawn parallel to the sides $BC$ and $AC$, resulting in points $P \in BC$ and $Q \in AC$. Determine the position of point $M$ such that the segment $PQ$ has the smallest length.
Solution. By drawing the parallel lines, triangles $\triangle M B P, \triangle A M Q$ are obtained, which are similar to $\triangle A B C$, and thus are equilateral. Let $P^{\prime}, Q^{\prime}$ be the feet of the perpendiculars dropped from points $P, Q$ to side $A B$ respectively. Let $\overline{A M}=x$. Then $\overline{M B}=a-x$. From the right triangle $\triangle Q R P, \overline{Q P}^{2}=\overline{Q R}^{2}+\overline{P R}^{2}$. From the equilateral triangles, we have

$$
\overline{Q Q^{\prime}}=\frac{x \sqrt{3}}{2}, \overline{P P^{\prime}}=\frac{(a-x) \sqrt{3}}{2}
$$

On the other hand, $\overline{P R}=\overline{P P^{\prime}}-\overline{Q Q^{\prime}}$, so by substitution we get $\overline{P R}=\frac{a \sqrt{3}}{2}-x \sqrt{3}$. Similarly,

$$
\overline{Q R}=\overline{Q^{\prime} P^{\prime}}=\overline{Q^{\prime} M}+\overline{M P^{\prime}}=\frac{x}{2}+\frac{a-x}{2}=\frac{a}{2} \text {. }
$$

Finally,

$$
\overline{Q P}^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{a \sqrt{3}}{2}-x \sqrt{3}\right)^{2}=3 x^{2}-3 a x+a^{2}
$$

The segment $P Q$ has the minimum length if $\overline{Q P}^{2}$ has the minimum value. Since $\overline{Q P}^{2}$ is a quadratic function of the variable $x$, its graph is a parabola which has a minimum, and this occurs at the vertex $x=-\frac{-3 a}{6}=\frac{a}{2}$. Therefore, the desired position of point $M$ is at a distance $\frac{a}{2}$ from point $A$.