Stacked cylinders inscribed in a cone!

Let the base radius and height of the cylinder be $r$ and $h$ and those of the inscribed cylinder be $x$ and $y$ . Then we note that $\dfrac y{r-x} = \dfrac hr \implies y = h - \dfrac hr x$ and the volumn of the cylinder is $V = \pi x^2y = \pi x^2\left(h - \dfrac hrx\right)$ . To find $V_{\max}$ , $\dfrac {dV}{dx} = \pi h \left(2x - \frac {3x^2}r\right)$ . Equating $\dfrac {dV}{dx} = 0$ , $\implies x = \dfrac 23 r$ , $y = \dfrac 13h$ , and $V_{\max} = \pi \left(\dfrac 23r\right)^2 \cdot \dfrac h3 = \dfrac 49 V_{\text{cone}}$ , since $V_{\text{cone}} = \dfrac 13 \pi r^2 h$ .

Let $V_{\text{cone}} = V_1$ and the subsequence volumes of smaller cones be $V_2, V_3, V_4, \cdots$ . Since $y = \dfrac 13 h$ , then the height of the next smaller similar cone is $\dfrac 23 h$ . Therefore $\dfrac {V_2}{V_1} = \left(\dfrac 23\right)^3 = \dfrac 8{27}$ . As the patent repeats, we have $\dfrac {V_n}{V_{n+1}} = \dfrac 8{27}$ .

Then the sum of volumes of all cylinder is:

$\begin{aligned} S & = \dfrac 49 V_1 + \dfrac 49 V_2 + \dfrac 49 V_3 + \dfrac 49 V_4 + \cdots \\ & = \dfrac 49 (V_1 + V_2 + V_3 + V_4 + \cdots) \\ & = \dfrac 49 V_{\text{cone}} \left(1 + \frac 8{27} + \left(\frac 8{27} \right)^2 + \left(\frac 8{27} \right)^3 + \cdots \right) \\ \implies \frac S{V_{\text{cone}}} & = \dfrac 49 \cdot \frac 1{1-\frac 8{27}} = \frac {12}{19} \end{aligned}$

Therefore $a+b = 12+19 = \boxed{31}$ .

Let $R_{1}$ and $r_{1}$ be the radius of the cone and the initial cylinder and $H_{1}$ and $h_{1}$ be the height of the cone and the initial cylinder.

$V_{cone} = \dfrac{\pi}{3}R_{1}^2H_{1}$ and $V_{1} = \pi r_{1}^2h_{1}$ the volume of the initial cylinder.

From the above diagram we have the proportion:

$\dfrac{R_{1}}{R_{1} - r_{1}} = \dfrac{H_{1}}{h_{1}} \implies h_{1} = \dfrac{H_{1}}{R_{1}}(R_{1} - r_{1}) \implies V_{1} = \dfrac{\pi H_{1}}{R_{1}}(R_{1}r_{1}^2 - r_{1}^3) \implies$

$\dfrac{dV_{1}}{dr_{1}} = \dfrac{\pi H_{1}}{R_{1}}(2R_{1} - 3r_{1}) = 0$ and $r_{1} \neq 0 \implies \boxed{r_{1} = \dfrac{2}{3}R_{1}} \implies \boxed{h_{1} = \dfrac{1}{3}H_{1}}$

$H_{2} = H_{1} - h_{1} = \dfrac{2}{3}H_{1}$ and $\dfrac{r_{1}}{r_{1} - r_{2}} = \dfrac{H_{2}}{h_{2}} \implies h_{2} = \dfrac{H_{2}}{r_{1}}(r_{1} - r_{2})$

$\implies V_{2} = \dfrac{\pi H_{2}}{r_{1}}(r_{1}r_{2}^2 - r_{2}^3) \implies$ $\dfrac{dV_{2}}{dr_{2}} = \dfrac{\pi H_{2}}{r_{1}} r_{2}(2r_{1} - 3r_{2})$ and $r_{2} \neq 0 \implies$

$r_{2} = \dfrac{2}{3}r_{1} = (\dfrac{2}{3})^2 R_{1} \implies h_{2} = \dfrac{H_{2}}{r_{1}}(\dfrac{2}{3} - \dfrac{4}{9})R_{1} = \dfrac{H_{2}}{r_{1}}(\dfrac{2}{9})R_{1} = \dfrac{2H_{1}}{3} * \dfrac{3}{2R_{1}}(\dfrac{2}{9})R_{1} =$ $\dfrac{2}{9}H_{1} = \dfrac{1}{3}(\dfrac{2}{3}) H_{1}$

$\therefore \boxed{r_{2} = (\dfrac{2}{3})^2 R_{1}}$ and $\boxed{h_{2} = \dfrac{1}{3}(\dfrac{2}{3}) H_{1} }$

$H_{3} = H_{2} - h_{2} = \dfrac{4}{9}H_{1}$ and $\dfrac{r_{2}}{r_{2} - r_{3}} = \dfrac{H_{3}}{h_{3}} \implies h_{3} = \dfrac{H_{3}}{r_{2}}(r_{2} - r_{3}) \implies$

$V_{3} = \dfrac{\pi H_{3}}{r_{2}}(r_{2}r_{3}^2 - r_{3}^3) \implies$ $\dfrac{dV_{3}}{r_{3}} = \dfrac{\pi H_{3}}{r_{2}} r_{3}(2r_{2} - 3r_{3}) = 0$ and $r_{3} \neq 0 \implies$

$r_{3} = \dfrac{2}{3}r_{2} = (\dfrac{2}{3})^3 R_{1} \implies h_{3} = \dfrac{H_{3}}{r_{2}}(\dfrac{4}{9} - \dfrac{8}{27})R_{1} =\dfrac{H_{3}}{r_{2}}(\dfrac{4}{27})R_{1} =$

$\dfrac{4H_{1}}{9} * \dfrac{9}{4R_{1}} * \dfrac{4}{27} R_{1} = \dfrac{4}{27}H_{1} = \dfrac{1}{3}(\dfrac{2}{3})^2 H_{1}$

$\therefore \boxed{r_{3} = (\dfrac{2}{3})^3 R_{1}}$ and $h_{3} = \boxed{\dfrac{1}{3}(\dfrac{2}{3})^2 H_{1}}$

In General:

$\boxed{r_{n} = (\dfrac{2}{3})^n R_{1}}$ and $\boxed{h_{n} = \dfrac{1}{3}(\dfrac{2}{3})^{n - 1} H_{1}}$ $\implies V_{n} = \pi r_{n}^2h_{n} = \dfrac{\pi}{3}(\dfrac{2}{3})^{3n - 1} R_{1}^2H_{1} =$

$\dfrac{\pi}{3}(\dfrac{2}{3})^{3n} (\dfrac{3}{2}) R_{1}^2H_{1} = \dfrac{\pi}{2}(\dfrac{8}{27})^n R_{1}^2H_{1}$

$\implies S = \sum_{n = 1}^{\infty} V_{n} = \dfrac{\pi R_{1}^2H_{1}}{2}(\dfrac{8}{27})(\dfrac{27}{19}) =$ $\dfrac{4}{19}\pi R_{1}^2 H_{1} = \dfrac{4}{19}(\dfrac{1}{3}\pi R_{1}^2H_{1})3 =$

$\dfrac{12}{19}V_{cone} \implies \boxed{\dfrac{S}{V_{cone}} = \dfrac{12}{19}} = \dfrac{a}{b} \implies a + b = \boxed{31}$ .

Note: We could have wrote: $V_{n} = \dfrac{3}{2}(\dfrac{2}{3})^{3n} V_{cone} \implies$ $S = V_{cone}\dfrac{3}{2}\sum_{n = 1}^{\infty} (\dfrac{8}{27})^n =$

$\dfrac{3}{2}\dfrac{8}{27}\dfrac{27}{19} V_{cone} = \dfrac{12}{19}V_{cone}$