In triangle $P Q R$, the midpoint of side $P Q$ is $A$, the midpoint of $Q R$ is $B$, and the midpoint of $R P$ is $C$. We enlarge triangle $A B C$ from its centroid $S$ by a factor of $k$, where $1<k<4$. The sides of the enlarged triangle intersect the sides of triangle $P Q R$ at points $D_{1} D_{2}, E_{1} E_{2}$, and $F_{1} F_{2}$, respectively. For what value of $k$ will the area of the hexagon $D_{1} D_{2} \ldots F_{1} F_{2}$ be twice the area of triangle $A B C$?
Let the vertices of the triangle after enlargement be denoted by $K, L, M$, where $K$ is the vertex opposite to $P$. We know that the triangle $ABC$ was created by a $\frac{1}{2}$-scale reduction from the centroid $S$ of the triangle $RQP$, meaning the centroids of these two triangles coincide. Let the intersection of the median $PB$ and the side $AC$ be $P_1$, then $PP_1 = \frac{1}{2} PB, PS = \frac{2}{3} PB$, from which $PS = 4P_1S$. The segment $LM$ intersects the segment $PP_1$, and thus also the sides $PA$ and $PC$ of the triangle $PQR$. This means that the $D_1D_2E_1E_2F_1F_2$ is a real (convex) hexagon. If, however, $k \leq 1$, then the image triangle $KLM$ is smaller than the original triangle $ABC$; if $k \geq 4$, then the image triangle contains the triangle $PQR$, the corresponding sides do not intersect, and no hexagon is formed.

$$
\text{1985-01-019-1.eps}
$$

1. Figure

Let $1 < k < 4$, and let the intersection of the segments $PP_1$ and $D_1F_2$ be $P_2$. Due to the similarity of the triangles $ABC$ and $MKL$,

$$
\frac{P_2S}{P_1S} = k
$$

and thus

$$
\frac{PP_2}{P_1S} = \frac{PS - P_2S}{P_1S} = \frac{PS}{P_1S} - \frac{P_2S}{P_1S} = 4 - k
$$

The triangles $PD_1F_2$, $F_1E_2R$, and $D_2QE_1$ are each similar to the triangle $PQR$, and the similarity ratio, based on $PB = 3 \cdot BS = 3 \cdot (2 \cdot P_1S)$, is

$$
\lambda = \frac{PP_2}{PB} = \frac{PP_2}{6P_1S} = \frac{4 - k}{6}
$$

in all three cases. Therefore, the small triangles are congruent, and their area is $\lambda^2$ times the area of the triangle $PQR$.

Let the area of the triangle $PQR$ be $T$. The area of the hexagon can be obtained by subtracting the sum of the areas of the three small triangles from $T$. This must be twice the area of the triangle $ABC$. The area of the triangle $ABC$ is $T / 4$, so the following equation must hold: $T - 3\lambda^2T = T / 2$, which means $\lambda^2 = 1 / 6$. Substituting $(4 - k) / 6$ for $\lambda$ according to (1), we get $(4 - k)^2 = 6$, which, under the condition $1 < k < 4$, is only satisfied by $k = 4 - \sqrt{6}$.

Thus, the value of $k$ can only be $4 - \sqrt{6}$. If we choose $k$ to be this value, then according to (1), the congruent triangles $PD_1F_2$, $F_1E_2R$, and $D_2QE_1$ are created by a $\lambda = \sqrt{6} / 6$-scale reduction from the triangle $PQR$, so the area of the hexagon is

$$
T - 3\lambda^2T = T - \frac{1}{2}T = \frac{1}{2}T
$$

which is indeed twice the area of the triangle $ABC$.

Remarks. 1. The equation $(4 - k)^2 = 6$ obtained as a condition is also satisfied by $k = 4 + \sqrt{6}$. What could be the geometric meaning of this second solution? When the triangle $ABC$ is enlarged by a factor of $(4 + \sqrt{6})$, the sides of the resulting triangle $MKL$ do not intersect the sides of the triangle $PQR$, but only their extensions. The resulting $D_1D_2E_1E_2F_1F_2$ figure is a "looped" hexagon. If we define the "area" of this hexagon by subtracting the area of the "loops" from the area of the triangle $PQR$, then for this hexagon as well, its "area" is twice the area of the triangle $ABC$ (2. Figure).

$$
\begin{gathered}
\text{1985-01-019-2.eps} \\
\text{2. Figure}
\end{gathered}
$$

2. The maximum score was awarded to those who also justified why $k = 4 - \sqrt{6}$, which is a solution to the equation $(4 - k)^2 = 6$, is also a solution to the problem (and not just to this equation), and why the other solution, $k = 4 + \sqrt{6}$, is not a solution.