Equable Cone

A cone with base radius $r$ and height $h$ , has volume, surface area and slant height $V=\frac13 \pi r^2 h;\;\;\;S=\pi r^2+\pi r\sqrt{r^2+h^2};\;\;\;s=\sqrt{r^2+h^2}$ respectively. Equating $V$ and $S$ , $\begin{aligned}\frac13 \pi r^2 h &= \pi r^2+\pi r\sqrt{r^2+h^2} \\ rh &= 3r+3\sqrt{r^2+h^2} \\ rh-3r &=3\sqrt{r^2+h^2} \\ r^2 h^2-6r^2 h+9r^2 &=9r^2+9h^2 \\ r^2 h^2-6r^2 h &=9h^2 \\ r^2 h-6r^2 &=9h \end{aligned}$

We need to solve this in positive integers. Using the usual factorising trick, $\begin{aligned} r^2 h-6r^2 &=9h \\ r^2 h-6r^2-9h+54&=54 \\ \left(r^2-9\right)(h-6)&=54 \end{aligned}$ so that $r^2=9+\frac{54}{h-6}$

We can now just check factors of $54$ ; the only solution comes when $h=8$ , $r=6$ and we find the slant height $s=\boxed{10}$ .

Since the volume and surface area have the same numerical value, $\frac{1}{3}\pi r^2 h = \pi r^2 + \pi r l$ , which rearranges to $3l = rh - 3r$ , so $3l$ must also be an integer.

Substituting $h = \sqrt{l^2 - r^2}$ into the above equation, this rearranges to $l = r + \cfrac{9}{r + 3} - \cfrac{9}{r - 3}$ .

This can be rearranged to $3l(r + 3) = 3r(r + 3) + 54 + \cfrac{162}{r - 3}$ and $3l(r - 3) = 3r(r - 3) + 54 - \cfrac{162}{r + 3}$ , which tells us that $r - 3$ and $r + 3$ are factors of $162$ .

The difference between these two factors is $(r + 3) - (r - 3) = 6$ , and the only positive factors of $162$ (which include $1, 2, 3, 6, 9, 18, 27, 54, 81, 162$ ) that are $6$ apart are $3$ and $9$ , which means $r - 3 = 3$ or $r = 6$ , and $l = r + \cfrac{9}{r + 3} - \cfrac{9}{r - 3} = 6 + \cfrac{9}{6 + 3} - \cfrac{9}{6 - 3} = \boxed{10}.$

Volume $V=\frac13\pi r^2h$ , surface area $S=\pi r^2+\pi r\sqrt{r^2+h^2}$ .

Through algebraic manipulation, $V=S\iff r=3\sqrt{\frac h{h-6}}$ .

$r,h\in\mathbb N\implies\sqrt{\frac h{h-6}}\in\mathbb N\implies \frac h{h-6}=n^2$ for $n\in\mathbb N\implies7\leq h$ (as $h-6>0$ ), and $h\leq12$ (as $\frac h{h-6}\geq2$ ( $2$ not $1$ since never does $h=h-6$ ))

$7\leq h\leq12\implies7\geq n^2\geq2$

$4$ is the only square in this range.

$\implies n^2=4\implies h=8\implies r=6\implies\sqrt{r^2+h^2}=10$