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# Geometry

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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 $\frac{r_4}{r_1}.$

Dec 17, 2023

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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).

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.

Dec 17, 2023