3A. Given a triangle $ABC$ with area $P$. From its medians, another triangle is constructed, then from the medians of the second triangle, a third triangle is constructed, and so on. Generally, the $(n+1)$-th triangle is constructed from the medians of the $n$-th triangle. Calculate the sum of the areas of all triangles in the resulting sequence.
Solution. Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the medians in $\triangle A B C$. Let $A X A_{1}$ be a triangle such that $A_{1} X\left\|C C_{1}, B B_{1}\right\| A X$. The sides of triangle $A X A_{1}$ have the same lengths as the medians of triangle $A B C$. From the way triangle $A X A_{1}$ is constructed, it is clear that every triangle constructed from the medians of $\triangle A B C$ has the same area as $\triangle A X A_{1}$. The quadrilateral

![](https://cdn.mathpix.com/cropped/2024_06_05_e95f2dbd17df8adc2c7cg-09.jpg?height=466&width=450&top_left_y=262&top_left_x=1015)
$X B A_{1} C_{1}$ is a parallelogram ($A X B B_{1}$ is a parallelogram, because, from the construction of $\triangle A X A_{1}$, it follows that $\overline{A X}=\overline{B B_{1}}, \quad A X \| B B_{1}$; hence $B X \| A_{1} C_{1}, \overline{A_{1} C_{1}}=\frac{1}{2} \overline{A C}$ and $\overline{B X}=\frac{1}{2} \overline{A C}$), so it follows that point $Y$ is the midpoint of segments $A_{1} X$ and $C_{1} B$ (intersection of the diagonals) and $\overline{A Y}=\frac{3}{4} \overline{A B}$. Therefore, triangles $A Y A_{1}$ and $A X Y$ have the same area. If $h$ is the height of triangle $A B C$ dropped from vertex $C$, then it is clear that the height of triangle $A Y A_{1}$ dropped from vertex $A_{1}$ is $\frac{1}{2} h$. Let $P_{1}$ be the area of triangle $A X A_{1}$. Then

$$
P_{1}=2 \frac{1}{2} \cdot \frac{3}{4} \overline{A B} \cdot \frac{1}{2} h=\frac{3}{4} P
$$

If $P_{2}, P_{3}, \ldots, P_{k}, \ldots$ are the areas of the second, third, $\ldots, k$-th triangle, etc., it is clear that

$$
\begin{gathered}
P_{2}=\left(\frac{3}{4}\right)^{2} P, P_{3}=\left(\frac{3}{4}\right)^{3} P, \ldots, P_{k}=\left(\frac{3}{4}\right)^{k} P, \ldots \\
P+P_{1}+P_{2}+\ldots+P_{k}+\ldots=P\left(1+\frac{3}{4}+\frac{3^{2}}{4^{2}}+\ldots+\frac{3^{k}}{4^{k}}+\ldots\right)=P \frac{1}{1-\frac{3}{4}}=4 P
\end{gathered}
$$