Two triangles, $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$, are positioned such that the sides $A_{1} B_{1}$ and $A_{2} B_{2}$, $B_{1} C_{1}$ and $B_{2} C_{2}$, as well as $A_{1} C_{1}$ and $A_{2} C_{2}$ are parallel. Connect vertex $A_{1}$ to vertices $B_{2}$ and $C_{2}$, vertex $B_{1}$ to vertices $C_{2}$ and $A_{2}$, and vertex $C_{1}$ to vertices $A_{2}$ and $B_{2}$. What can be the area of the hexagon formed by the midpoints of the connecting segments, if the areas of the original triangles are $T_{1}$ and $T_{2}$?
Solution. The two triangles are clearly similar. Let $A_{1} B_{1}=c_{1}, B_{1} C_{1}=a_{1}, A_{2} B_{2}=c_{2}$, and name the other sides similarly. We can assume that $T_{1} \leq T_{2}$, i.e., $a_{1} \leq a_{2}$, and it is known that

$$
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}
$$

Let the midpoint of $A_{1} B_{2}$ be $F_{A B}$, the midpoint of $B_{1} A_{2}$ be $F_{B A}$, the midpoint of $A_{1} C_{2}$ be $F_{A C}$, and name the others similarly. Let the area of the hexagon be $T$.

First case: $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ are not collinear (Figure 1)

![](https://cdn.mathpix.com/cropped/2024_05_02_545ccfa69635081d9fc2g-1.jpg?height=395&width=596&top_left_y=499&top_left_x=754)

Figure 1

From the theorem of parallel intercepts, it follows that $F_{A B} F_{A C}=\frac{a_{2}}{2}, F_{B A} F_{C A}=\frac{a_{1}}{2}, F_{B C} F_{C B}=\frac{a_{1}+a_{2}}{2}$, all parallel to $B_{1} C_{1}$, and a similar statement holds for the other sides. By considering the direction of the segments, we find that the sides of the resulting hexagon are $\frac{a_{1}}{2}, \frac{b_{2}}{2}, \frac{c_{1}}{2}, \frac{a_{2}}{2}, \frac{b_{1}}{2}, \frac{c_{2}}{2}$.

Due to the corresponding angle equalities, by extending the sides of lengths $\frac{a_{2}}{2}, \frac{b_{2}}{2}$, and $\frac{c_{2}}{2}$, we obtain a triangle with side lengths $\frac{2 a_{1}+a_{2}}{2}, \frac{2 b_{1}+b_{2}}{2}$, $\frac{2 c_{1}+c_{2}}{2}$, whose area is $\frac{3}{4} T_{1}$ larger than the hexagon, let this be $T^{\prime}$. The triangle is similar to $A_{1} B_{1} C_{1}$, with the ratio of the sides being

$$
\frac{2 a_{1}+a_{2}}{2 a_{1}}=1+\frac{a_{2}}{2 a_{1}}=1+\frac{1}{2} \sqrt{\frac{T_{2}}{T_{1}}}
$$

and the ratio of the areas is:

$$
\frac{T^{\prime}}{T_{1}}=\left(1+\frac{1}{2} \sqrt{\frac{T_{2}}{T_{1}}}\right)^{2}=1+\sqrt{\frac{T_{2}}{T_{1}}}+\frac{T_{2}}{4 T_{1}}
$$

from which $T^{\prime}=T_{1}+\sqrt{T_{1} T_{2}}+\frac{T_{2}}{4}$. Therefore, $T=\frac{T_{1}+T_{2}}{4}+\sqrt{T_{1} T_{2}}$.

Second case: $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ are collinear (Figure 2).

In this case, similarly to what was observed in the first case, it can be established that $F_{A B} F_{A C}=\frac{a_{2}}{2}, F_{B A} F_{C A}=\frac{a_{1}}{2}, F_{B C} F_{C B}=$ $\frac{a_{2}-a_{1}}{2}$, all parallel to $B_{1} C_{1}$, and similarly for the others. By considering the directions, we find that the sides of the hexagon are $\frac{a_{1}}{2}, \frac{b_{2}-b_{1}}{2}, \frac{c_{1}}{2}, \frac{a_{2}-a_{1}}{2}, \frac{b_{1}}{2}, \frac{c_{2}-c_{1}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_545ccfa69635081d9fc2g-1.jpg?height=398&width=572&top_left_y=2013&top_left_x=771)

Due to the corresponding angle equalities, by extending the sides of lengths $\frac{a_{2}-a_{1}}{2}, \frac{b_{2}-b_{1}}{2}$, and $\frac{c_{2}-c_{1}}{2}$, we obtain a triangle with area $T^{\prime \prime}$, whose sides are $\frac{a_{1}+a_{2}}{2}, \frac{b_{1}+b_{2}}{2}, \frac{c_{1}+c_{2}}{2}$, and $T^{\prime \prime}=T+\frac{3}{4} T_{1}$, and it is similar to $A_{1} B_{1} C_{1}$. The ratio of the sides of $T^{\prime \prime}$ and $T_{1}$ is:

$$
\frac{a_{1}+a_{2}}{2 a_{1}}=\frac{1}{2}+\frac{1}{2} \sqrt{\frac{T_{2}}{T_{1}}}
$$

From this,

$$
\frac{T^{\prime \prime}}{T_{1}}=\left(\frac{1}{2}+\frac{1}{2} \sqrt{\frac{T_{2}}{T_{1}}}\right)^{2}=\frac{1}{4}+\frac{1}{2} \sqrt{\frac{T_{2}}{T_{1}}}+\frac{T_{2}}{4 T_{1}}
$$

from which

$$
T^{\prime \prime}=\frac{T_{1}}{4}+\frac{1}{2} \sqrt{T_{1} T_{2}}+\frac{T_{2}}{4}, \quad \text { so } \quad T=\frac{T_{2}-2 T_{1}}{4}+\frac{1}{2} \sqrt{T_{1} T_{2}}
$$

This expression also gives the area when $a_{1}=a_{2}$, and the hexagon degenerates into a triangle.