$A B C$ is a triangle from whose medians we construct another triangle; from the medians of this new triangle we construct yet another triangle, and so on to infinity. What is the sum of the areas of these triangles if the area of the original triangle is $t$?
I. Solution. The midpoints of the sides of triangle $ABC$ are $A', B', C'$. Extend $C'B'$ to $D$ so that $B'D = B'C'$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9d517a5cb408e1aa1ecbg-1.jpg?height=290&width=485&top_left_y=267&top_left_x=804)

Since $B'D$ is equal and parallel to $BA'$, it follows that $A'D = BB'$. But $CD$ is equal and parallel to $AC'$, so $AD = CC'$. Therefore, $\triangle ADA'$ is the triangle constructed from the medians of the original triangle. Let's calculate the area $t'$ of this triangle:

$$
t' = \triangle AA'B' + \triangle AB'D + \triangle DB'A'
$$

but

$$
\triangle AA'B' = \triangle AB'C' = \frac{1}{4} \triangle ABC = \frac{1}{4} t
$$

and similarly,

$$
\triangle AB'D = \frac{1}{4} t \text{ and } \triangle DB'A' = \frac{1}{4} t
$$

thus

$$
t' = \frac{3}{4} t
$$

Similarly, we find that the area of the triangle constructed from the medians of the triangle with area $t'$ is $t'' = \frac{3}{4} t' = \left(\frac{3}{4}\right)^2 t$, and so on.

Thus, the areas of the triangles form an infinite geometric series with the first term $t$ and common ratio $\frac{3}{4}$. Therefore, the sum of the series is:

$$
S = \frac{t}{1 - \frac{3}{4}} = 4t
$$

(Károly Szabó.)

II. Solution.

Let the medians of the triangle with area $t$ be $k_1, k_2, k_3$ and $k_1 + k_2 + k_3 = k$; then (K.M.L.IV. year, p. 63)

$$
t = \frac{4}{3} \sqrt{k(k - k_1)(k - k_2)(k - k_3)}
$$

The area of the triangle constructed from the medians is:

$$
t' = \sqrt{k(k - k_1)(k - k_2)(k - k_3)}
$$

and thus

$$
t' = \frac{3}{4} t
$$

The problem was also solved by: Dénes A., Détshy K., Erdős A., Fekete J., Freibauer E., Goldziher K., Kertész L., Koós A., Kornis Ö., Krisztián Gy., Lukhaub Gy., Manheim I., Perl Gy., Raab L., Roth M., Sasvári G., Schiffer H., Szabó I., Weisz Á., Weisz J.