357/MR00. problem / Math - Solved Math Problems

MR-357. problem

We write a cone on the sphere. The center of the sphere divides the height of the cone into two sections such that the longer section equals the geometric mean of the shorter section and the total height. Determine the volume ratio of the sphere to the cone.

Source: MathReference: Generic Math Textbook - 357.)

Algebra → Exponential equations and inequalities → Exponential equations

a

R = \sqrt{x \cdot (x + R)}

R^{2} = x (x + R)

(1) R^{2} = x^{2} + x R

(2) R^{2} = x^{2} + r^{2}

(1) \land (2) \Rightarrow R^{2} = x^{2} + x R = x^{2} + r^{2}

(3) x R = r^{2}

(2) \land (3) \Rightarrow R^{2} = x^{2} + x R

x^{2} + R x - R^{2} = 0

x_{1, 2} = \frac{- R \pm \sqrt{R^{2} - 4 \cdot (- R^{2})}}{2}

x_{1, 2} = \frac{- R \pm \sqrt{5 R^{2}}}{2}

x_{1, 2} = \frac{- R \pm R \sqrt{5}}{2} = \frac{R (- 1 \pm \sqrt{5})}{2}

x_{2} < 0 \Rightarrow x = x_{1}

x = \frac{\sqrt{5} - 1}{2} R

V_{1} = \frac{4 R^{3} π}{3}

V_{2} = \frac{r^{2} πH}{3} = \frac{r^{2} π (x + R)}{3}

\frac{V_{1}}{V_{2}} = \frac{\frac{4 R^{3} π}{3}}{\frac{r^{2} π (x + R)}{3}}

\frac{V_{1}}{V_{2}} = \frac{4 R^{3} π}{r^{2} π (x + R)}

= \frac{4 R^{3} π}{x R π (x + R)}

= \frac{4 R^{2}}{x (x + R)}

= \frac{4 R^{2}}{(\frac{\sqrt{5} - 1}{2}) \cdot R \cdot (R + \frac{\sqrt{5} - 1}{2} \cdot R)}

= \frac{4 R^{2}}{R^{2} \cdot (\frac{\sqrt{5} - 1}{2}) \cdot (1 + \frac{\sqrt{5} - 1}{2})}

= \frac{4}{(\frac{\sqrt{5} - 1}{2}) \cdot (1 + \frac{\sqrt{5} - 1}{2})}

= \frac{4}{(\frac{\sqrt{5} - 1}{2}) \cdot (\frac{2 + \sqrt{5} - 1}{2})}

= \frac{4}{(\frac{\sqrt{5} - 1}{2}) \cdot (\frac{\sqrt{5} + 1}{2})}

= \frac{4}{\frac{{(\sqrt{5})}^{2} - 1^{2}}{2^{2}}}

= \frac{4}{\frac{5 - 1}{2}}

= \frac{4}{\frac{4}{2}}

= \frac{4}{2}

\frac{V_{1}}{V_{2}} = 2

\frac{V_{1}}{V_{2}} = 2