Stacking Squares in a Triangle

Let the side the largest square be $a$ . We note that base length of the largest triangle is 4, while the base length of the second largest triangle is $a$ . Since the pattern repeats, we have the base length of third triangle is $a \times \frac a4 = \frac {a^2}4$ ; fourth, $\frac {a^3}{4^2}$ ; fifth, $\frac {a^4}{4^3}$ ; and so on. This means that the side lengths of the square of decreasing size are $a, \frac {a^2}4, \frac {a^3}{4^2}, \frac {a^4}{4^3} \cdots$ . Then the sum of the areas of the squares:

$\begin{aligned} A & = a^2+ \frac {a^4}{4^2} + \frac {a^6}{4^4} + \frac {a^8}{4^6} + \cdots \\ & = a^2 \left(1 + \frac {a^2}{4^2} + \frac {a^4}{4^4} + \frac {a^6}{4^6} + \cdots \right) \\ & = \frac {a^2}{1-\frac {a^2}{16}} = \frac {16a^2}{16-a^2} = \frac {16a^2}{(4-a)(4+a)} & \small \blue{\text{As }\tan 60^\circ = \frac a{2-\frac a2} = \frac {2a}{4-a} = \sqrt 3} \\ & = \frac {8\sqrt 3 a}{4+a} = \frac {192-96\sqrt 3}{8\sqrt 3 - 8} = \frac {24-12\sqrt 3}{\sqrt 3 -1} & \small \blue{\text{and }a = \frac {4\sqrt 3}{2+\sqrt 3} = 8\sqrt 3-12} \\ & = 6(2-\sqrt3)(1+\sqrt 3) = \boxed{6\sqrt 3-6} \end{aligned}$

Consider one trapezoid layer, where the short leg of one of the side triangles is $x$ , which makes the side of the square $\sqrt{3}x$ .

Then the ratio of the area of the square to the area of the trapezoid is $\frac{(\sqrt{3}x)^2}{(\sqrt{3}x)^2 + \sqrt{3}x^2} = \frac{1}{2}(3 - \sqrt{3})$ .

Since this ratio is maintained for the infinite layers of trapezoids in the diagram, and since the area of the whole equilateral triangle is $\frac{\sqrt{3}}{4} \cdot 4^2 = 4 \sqrt{3}$ , the sum of the areas of the stacked squares is $\frac{1}{2}(3 - \sqrt{3}) \cdot 4 \sqrt{3} = \boxed{6\sqrt{3} - 6}$ .