A triangle has an area of 2. The lengths of its medians equal the lengths of the sides of a second triangle. The lengths of the medians of the second triangle equal the lengths of the sides of a third triangle. In general, the medians of the zth triangle have the same lengths as the sides of the (z+1) triangle. Find the limit of the sum of the areas of all the triangles thus formed in this infinite sequence.

I tried to solve it by creating a right triangle with lengths 2 and 2 but finding the medians became very difficult so I discarded that method. Is there a cleaner method for solving this?


The area of the triangle given by the medians is just $\frac{3}{4}$ of the area of the original triangle:

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hence the wanted sum is:

$$ 2\sum_{n=0}^{+\infty}\left(\frac{3}{4}\right)^n = \color{red}{8}.$$


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