Once Again, You Forgot The Radius!

Let's find the ratio of areas between the two circles and the larger circle as shown in the left column given in the problem statement.

Let $r_2$ denote the radii of the two equal circles enclosed inside a circle with minimal radius $R_2$ .

The sum of diameters of the two small circles is equal to the diamater of the encompassing circle: $2(d_2) = R_2 \Rightarrow 2(2 r_2) = 2(2R_2) \Rightarrow 2r_2 = R_2$ .

The area of the black region is equal to the sum of areas of the two small circles, $\pi r_2^2 + \pi r_2^2 =2 \pi r_2^2$ .

The area of the grey region is equal to area of the encompassing circle minus the area of the small circles, $\pi R_2^2 -2 \pi r_2^2 = 4\pi r_2 -2 \pi r_2^2 = 2 \pi r_2^2$ .

The ratio of areas between these two regions is $\dfrac{\pi r_2^2}{2\pi r_2^2 } =\dfrac12 \qquad \qquad (1)$ .

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Let's find the ratio of areas between the three circles and the larger circle as shown in the left column given in the problem statement.

Since we're finding the ratio of areas, we can assume that the radii of all three circles is $r_3 = 1$ .

Let $O_1,O_2$ and $O_3$ denote the center of the three smaller circles, then $O_1 O_2 O_3$ is an equilateral triangle. Now let $O_4$ denote the center of this triangle, and let $s = O_4O1 = O_4 O_2 = O_4 O_3$ , then $\measuredangle O_1 O_4 O_2 = \measuredangle O_1 O_4 O_3 = \measuredangle O_2 O_4 O_3 = \dfrac{360^\circ}3 = 120^\circ$ .

By cosine rule , $2s^2 - 2s^2 \cos(120^\circ) = 2^2 \Rightarrow s= \sqrt{\dfrac43}$ . And so the radius of the encompassing circle, $R_3$ is equal to $R_3 = s+r_3 = \sqrt{\dfrac43} + 1$ .

Like above, taking the ratio of the two regions gives us $\dfrac{3 \pi}{\pi \left( \sqrt{\dfrac43} + 1 \right)^2 - 3\pi} \approx 1.8826$ .

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Let's find the ratio of areas between the four circles and the larger circle as shown in the left column given in the problem statement.

Since we're finding the ratio of areas, we can assume that the radii of all three circles is $r_4 = 1$ .

By joining the center of these four smaller circles, we get a square of side length 2 and diagonal $2\sqrt2$ (By Pythagorean theorem ), thus the diameter of encompassing circle is $2r_4 + 2\sqrt2 = 2\sqrt2 + 2$ .

Likewise, we can work out the areas to be $\dfrac{ 4\pi}{\pi (\sqrt2 + 1)^2 - 4\pi} \approx 2.18$ .

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Comparing these values show that the answer is 4.