Solution. Let's decompose the triangle - as shown in the diagram - into trapezoids by drawing parallels to the $AC$ side (where the bases of the triangles are located) through the vertices of the equilateral triangles on the $BC$ leg. The first such trapezoid is $BCDL$. The altitude $FM$ of the $\triangle FCD$ is the midline of this trapezoid, and the base $CD$ is the height of the trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_02_10533cb1b6e29c02144eg-1.jpg?height=501&width=464&top_left_y=343&top_left_x=817)

Since the area of the trapezoid is

$$
\frac{BC + LD}{2} \cdot CD = FM \cdot CD
$$

the area of the $\triangle FCD$ is exactly half the area of the trapezoid $BCDL$. By similarity, the same situation applies to the other trapezoids and the corresponding equilateral triangles. Since we have divided the (original) right triangle into these infinitely many trapezoids, the sum of their areas is the area of the right triangle, $\frac{36 \cdot 36}{2} = 648$. Thus, the sum of the areas of the infinitely many equilateral triangles is half of this, 324 square units.
In an isosceles right triangle, one of the legs is 36 units. Starting from the right-angle vertex on one of the legs, we draw an infinite sequence of equilateral triangles, each touching the next, such that the third vertex of each inscribed triangle always lies on the hypotenuse, and the sides opposite these vertices fill the leg. Determine the sum of the areas of the equilateral triangles.