Firstly, we present an obvious lemma.
Lemma 1. If for two triangles $ABC$ and $A'B'C'$, we have $AB \le A'B'$, $AC \le A'C'$ and $\angle BAC \le B'A'C'$, then $ABC$ can be placed into $A'B'C'$.

Let $XYZ$ be an equilateral triangle with center $O$. We denote the radius of circumcircle of this triangle and the length of its altitudes by $R$ and $h$, respectively (Clearly, $h = \frac{\sqrt{3}}{2}R$). Let $P$ be a point close to $O$ such that $OP < \frac{h-R}{2}$ and angles $\angle OXP$, $\angle OYP$ and $\angle OZP$ are all in the interval $(0, 30^\circ)$. We claim that if $A$ puts the points $X, Y, Z, O$ and $P$; he would be able to construct the desired triangles.

Now we go to some cases, according to the place of sixth point, say $Q$.

* If $Q$ lies inside the triangle, then $OQP$ is located completely inside $XYZ$.

* If $Q$ lies inside the circumcircle of $XYZ$ but outside of the triangle, we have
$$QP < QO + OP \le R + OP < h < XY, \quad QO \le R < h < XZ$$
On the other hand, since $Q$ is outside of $XYZ$, $OQ \ge h - R > 2OP$ and $QP > OQ - OP > OP$. So $OP$ is the shortest side of triangle $OPQ$. Therefore, $\angle OQP \le 60^\circ$. Now since $QP < XY$, $QO < XZ$ and $\angle OQP \le 60^\circ = \angle YXZ$, according to the lemma, triangle $OQP$ can be placed inside $XYZ$.

* If $Q$ lies outside the circumcircle of $XYZ$, triangles $QXY$, $QYZ$ and $QZX$ cover the triangle $XYZ$. So one of them, for example $QYZ$ contains $O$. Therefore, triangle $OYZ$ is located inside $QYZ$. On the other hand, since $OPX$ can be placed into $OYZ$ ($P$ is in one of congruent triangles $OXY$, $OYZ$ or $OZX$), we obtain that $OPX$ can be placed into $QYZ$.