Four Fools, One Pool

Applying the Law of Cosines , in $\triangle ADB$ , $\begin{aligned} AB^2&=AD^2+BD^2-2\cdot AD\cdot BD\cdot \cos\angle ADB \\ &=(2x)^2+(y-x)^2-2\cdot (2x)\cdot (y-x)\cdot \cos 120^{\circ} \\ &=4x^2+y^2+x^2-2xy+2xy-2x^2 \\ &=3x^2+y^2 \end{aligned}$ Therefore, $\begin{aligned} {\color{#D61F06} \text{Red}}&=3\cdot \frac{x^2\sqrt{3}}{4}+\frac{y^2\sqrt{3}}{4} \\ &=(3x^2+y^2)\cdot\frac{\sqrt{3}}{{4}} \\ {\color{#69047E}\text{Purple}}&=\frac{AB^2\sqrt{3}}{4} \\ &=(3x^2+y^2)\cdot\frac{\sqrt{3}}{{4}} \\ \\ \therefore\; &\boxed{{\color{#D61F06} \text{Red}}={\color{#69047E}\text{Purple}}} \end{aligned}$

The straight line between the apex / summit of the leftmost little red equilateral triangle and the shared point of two red equilateral triangle of different sizes are perpendicular to the left slope side of big red equilateral triangle, so the two lines would be two legs of a right triangle and the purple's side would be the hypothenuse, no matter the size of the two biggest triangles' side length. Since the problem is asking about area, the smallest three can be rearranged into a bigger equilateral triangle by halving them and reconstructed with the new side being twice the height of the original.

Let the sides be p, q and r with q < r < p, and after reconstruction, 3 equilateral triangles combined into a bigger one with sides Q.

Q = √3 × q

The right triangle / Pythagorean equation would be
p² = r² + Q²

Multiply each term with 0.5(sin 60°)

(1/2)p²(sin 60°) = (1/2)r²(sin 60°) + (1/2)Q²(sin 60°)

(1/2)p²(sin 60°) = (1/2)r²(sin 60°) + (1/2)(3q²)(sin 60°)
= (1/2)r²(sin 60°) + (3) × (1/2)(q²)(sin 60°)

Area of big purple = Area of big red + Area of 3 small reds

Purple = Red

The equality hold for r ≤ q too as long as the overlapping area is accounted for both red and purple or none at all.