Four circles are drawn. Let $A_1,$ $A_2,$ $A_3,$ $A_4$ be the areas of the regions, so $A_1$ is the area inside the smallest circle, $A_2$ is the area outside the smallest circle and inside the second-smallest circle, and so on. The areas satisfy

\[A_1 = \frac{A_2}{2} = \frac{A_3}{6}\]

Let $r_1$ denote the radius of the smallest circle, and let $r_4$ denote the radius of the largest circle. Find $\frac{r_4}{r_1}.$

kittykat Dec 17, 2023

#1**0 **

From the given relationships, we can analyze the scaling of the areas and radii:

Scaling of Areas:

A1=2A2 implies that the area of the second smallest region (A_2) is twice that of the smallest region (A_1). This indicates a doubling in radius.

Similarly, A2=4A3 implies a doubling in radius again for the third region (A_3), and A3=8A4 implies another doubling for the fourth region (A_4).

Scaling of Radii:

If the radius of the smallest circle is r1, the radius of the second smallest circle would be 2r1 (twice the radius).

Following this pattern, the radii of the third and fourth circles would be 4r1 and 8r1, respectively.

Therefore, the ratio of the largest circle's radius (r_4) to the smallest circle's radius (r_1) is:

r_4 / r_1 = 8

So, the largest circle has a radius 8 times greater than the smallest circle.

BuiIderBoi Dec 17, 2023