Shrinking and Shrinking

To go from the green circle to the purple circle takes two operations:

• inscribe an equilateral triangle in the circle

• inscribe a circle in the equilateral triangle

Going from red triangle to yellow triangle takes also two operations:

• inscribe a circle in the equilateral triangle

• inscribe an equilateral triangle in the circle

They are the same two operations and the order in which they are executed does not change the amount of shrinkage accomplished.

So the answer is: YES, the areas are equal.

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Details

If the green circle has radius $R$ , the side of the orange triangle is $a=2\times R \times \cos 30^\circ=2R\dfrac{\sqrt3}{2}=R\sqrt3$

The radius of the purple circle is then $r=\dfrac a2\dfrac{1}{\cos30^\circ}=\dfrac{a}{2\sqrt3}=\dfrac{ R\sqrt3}{2\sqrt3}=\dfrac{R}{2}$

If the side of the red triangle is $A$ , then the radius of the blue circle is $r=\dfrac A2\dfrac{1}{\cos30^\circ}=\dfrac{A}{2\sqrt3}$

And the side of the yellow triangle $a$ is $a=2\times r \times \cos 30^\circ=2r\dfrac{\sqrt3}{2}=r\sqrt3=\dfrac{A}{2\sqrt3}\sqrt3=\dfrac A2$

Both dimensions are half of what they were, so they are in the same proportion and the areas remain equal to each other.

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Interestingly enough, the blue area is NOT equal the orange area.

To show this, we need the actual ratio between radius of the green circle $R$ and side of the red triangle $A$ . Their areas are equal, therefore:

$\pi R^2=\dfrac{\sqrt3}4 A^2$

$\dfrac AR=\dfrac{2\sqrt{\pi}}{^4\sqrt {3}}\approx2.69$

Orange triangle inscribed into the green circle has a side $a=R\sqrt3$

Blue circle inscribed into the red triangle has a radius $r=\dfrac A{2\sqrt3}$

So the ratio $\dfrac ar= R\sqrt3\times\dfrac {2\sqrt3}A= \dfrac {6R}A=\dfrac{3^{\frac54}}{\sqrt{\pi}}\approx2.23\not=2.69\approx\dfrac AR$

One purple area is equal to one yellow area, and one green area is equal to the red area. So, we get this. One blue area is equal to one orange area.