A square is inscribed in an equilateral triangle, with each vertex of the square on the perimeter of the triangle. One side of the square cuts off a smaller equilateral triangle; into this, we again inscribe a square in the same manner, and so on. What fraction of the triangle is covered by the infinite sequence of squares?
I. solution. Let the three vertices of the large equilateral triangle be $A, B$, and $C$. The first inscribed square has two vertices on the same side of $A B C$. Let these two vertices be $D$ and $E$, and the other two vertices of the square be $F$ and $G$ (see the figure). Denote the side length of the square by $x$. Since $A D G$ and $B F E$ are right triangles with angles of $60^{\circ}$ at $A$ and $B$, respectively, we have $A D = B E = \frac{x \cdot \sqrt{3}}{3}$, and thus

$$
A B = A D + D E + E B = x\left(1 + \frac{2 \sqrt{3}}{3}\right)
$$

The area of the trapezoid $A B F G$ is therefore

$$
T_{A B F G} = \frac{1}{2} \cdot \left[x + x\left(1 + \frac{2 \sqrt{3}}{3}\right)\right] \cdot x = x^{2} \cdot \frac{3 + \sqrt{3}}{3}
$$

This is $\frac{3 + \sqrt{3}}{3}$ times the area of the square $E F G D$, which is $x^{2}$. We obtain the same relationship between the square inscribed in the triangle $G F C$ and the area of the corresponding trapezoid, and similarly for the areas of the subsequent squares and trapezoids. The infinite sequence of trapezoids covers the entire triangle $A B C$, and each trapezoid covers $\frac{3}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{2} \approx 0.634$ of the area of the corresponding square. Therefore, the infinite sequence of squares covers $\frac{3 - \sqrt{3}}{2}$ of the area of the triangle $A B C$.

II. solution. Using the notation from the first solution, the area of the triangle $A B C$ is

$$
T_{A B C} = A B^{2} \cdot \frac{\sqrt{3}}{4} = \left(1 + \frac{2 \sqrt{3}}{3}\right)^{2} \cdot \frac{\sqrt{3}}{4} \cdot x^{2}
$$

The triangle $C G F$ is similar to the triangle $C A B$, and the ratio of their similarity is the same as the ratio of the side lengths of the squares inscribed in them. Therefore, the side length of the square inscribed in the second triangle is

$$
x \cdot \frac{G F}{A B} = x \cdot \frac{1}{1 + \frac{2 \sqrt{3}}{3}} = x \cdot (2 \sqrt{3} - 3)
$$

By similar reasoning, the side length of the square inscribed in the $i$-th triangle ($i = 1, 2, 3, 4, \ldots$) is $x \cdot (2 \sqrt{3} - 3)^{i-1}$. Therefore, the area covered by the infinite sequence of squares is

$$
T = x^{2} \cdot \left(1 + (2 \sqrt{3} - 3)^{2} + (2 \sqrt{3} - 3)^{4} + \ldots\right)
$$

The sum of this infinite geometric series is, according to the known formula:

$$
T = x^{2} \cdot \frac{1}{1 - (2 \sqrt{3} - 3)^{2}} = x^{2} \cdot \frac{3 \sqrt{3} + 5}{8}
$$

The ratio of the covered area to the area of the triangle $A B C$ is therefore

$$
\frac{T}{T_{A B C}} = \frac{x^{2} \cdot \frac{3 \sqrt{3} + 5}{8}}{x^{2} \cdot \left(1 + \frac{2 \sqrt{3}}{3}\right)^{2} \cdot \frac{\sqrt{3}}{4}} = \frac{3 - \sqrt{3}}{2} \approx 0.634
$$

Based on the solutions of Márton Hablicsek (Fazekas M. Főv. Gyak. Gimn., 9th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_668a417e6f6513d08d09g-1.jpg?height=602&width=572&top_left_y=2084&top_left_x=771)