Harry Potter And The Magical Sphere

For similar figures (which all spheres are), multiplying corresponding lengths by a factor of $k$ multiplies corresponding areas by a factor of $k^2$ and corresponding volumes by a factor of $k^3$ . Therefore multiplying a surface area by a factor $s$ would multiply a length by a factor of $s^{1/2}$ and multiplying a volume by a factor $v$ would multiply a length by a factor of $v^{1/3}$ .

For any radius $r_0$ , increasing the volume of the magic sphere by 20% means that the volume factor is $v = 1.2$ or $v=\frac{6}{5}$ , and the radius multiplies by a factor of $(\frac{6}{5})^{1/3}$ . Then decreasing the surface area of the magic sphere by 30% means the surface area factor is $s = \frac{7}{10}$ and the radius would be multiplied by $(\frac{7}{10})^{1/2}$ .

This means the final radius $r_f = r_0 (\frac{6}{5})^{1/3}(\frac{7}{10})^{1/2}$ and thus, $\frac{r_f}{r_0} = (\frac{6}{5})^{1/3}(\frac{7}{10})^{1/2} \approx .889$

Let the initial radius be $r$ . Then the initial volume will be $\frac{4}{3}\pi r^3$ . Let the volume and the radius after increment be $V_i$ and $x$ . Hence, we have $V_i = \frac{4}{3}\pi r^3 + \frac{20}{100} \times \frac{4}{3}\pi r^3$ $\frac{4}{3}\pi x^3 = \frac{4}{3}\pi r^3 \Bigg( 1 + \frac{1}{5} \Bigg)$ $\frac{4}{3}\pi x^3 = \frac{4}{3}\pi r^3 \times \frac{6}{5}$ $\frac{4}{3}\pi x^3 = \frac{8}{5}\pi r^3$ $x^3 = \frac{6}{5} r^3$ $x = \sqrt[3]{\frac{6}{5}}r\cdots (1)$ Let the surface area after decrement be $A_d$ and the final radius be $R$ . Therefore, $A_d = 4\pi x^2 - \frac{30}{100} \times 4\pi x^2$ $4\pi R^2 = 4\pi x^2\Bigg( 1-\frac{3}{10}\Bigg)$ $R^2 = \Bigg( \sqrt[3]{\frac{6}{5}}r\Bigg) ^2 \times \frac{7}{10}$ $R^2 = \Bigg( \sqrt[3]{\frac{36}{25}}\Bigg) r^2 \times \frac{7}{10}$ $\frac{R^2}{r^2} = \frac{7}{10}\sqrt[3]{1.44}$ $\Bigg( \frac{R}{r} \Bigg) ^2 \approx \frac{7 \times 1.129243}{10}$ $\Bigg( \frac{R}{r} \Bigg) ^2 \approx 0.7904701$ $\boxed{\frac{R}{r} \approx 0.889}$