MHB Squaring and Cubing Equations: Explained!

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The discussion clarifies the relationship between the volumes and areas of spheres in terms of their radii. It explains that the volume of a sphere is proportional to the cube of its radius, while the surface area is proportional to the square of its radius. The equation v2/v1 = a2/a1 demonstrates that when these ratios are squared or cubed, they yield consistent results, specifically showing that (V2/V1)^2 equals (r2/r1)^6. Similarly, it is shown that (A2/A1)^3 also equals (r2/r1)^6, regardless of the constants involved. This highlights the mathematical consistency in the relationships between volume and area as they relate to radius.
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Can you explain the part of the solution where v2/v1 = a2/a1were being squared and cubed repectively?

Thanks!
 
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$$(r^3)^2 = (r^2)^3$$
 
To provide a little more detail: It is known that the volume of a sphere of radius $r$ is $cr^3$ for some $c$ (in fact, $c=\frac43\pi$) and its area is $dr^2$ ($d=4\pi$). Therefore,
\[
\left(\frac{V_2}{V_1}\right)^2=\left(\frac{cr_2^3}{cr_1^2}\right)^2=\left(\frac{r_2}{r_1}\right)^6
\]
One can similarly show that $(A_2/A_1)^3=(r_2/r_1)^6$ regardless of $d$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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