Question

The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

Answer: (B)

Let height and radius of cone is $h$ and $r$ respectively, $h, r \in I$
Given volume of cone $=$ Surface area of cone
$\frac{1}{3}\pi r^{2}h = \pi rl + \pi r^{2}$
$\Rightarrow \frac{1}{3} \pi r^{2}h = \pi r \sqrt{h^{2} +r^{2} } + \pi r^{2}$
$\Rightarrow \frac{1}{3} rh = \sqrt{h^{2} +r^{2}} +r \left[r\ne0\right] $
$\Rightarrow rh -3r = 3 \sqrt{h^{2} +r^{2}}$
$\Rightarrow r^{2}h^{2} + 9r^{2} - 6hr^{2} = 9h^{2} + 9r^{2}$
$\Rightarrow h^{2} \left(r^{2}-9\right) = 6hr^{2} $
$\Rightarrow h = \frac{6r^{2}}{r^{2} -9}$
$\Rightarrow h = 6\left(\frac{r^{2}}{r^{2} -9}\right) $
$ \Rightarrow h = 6+ \frac{54}{r^{2} - 9} $
$h$ and $r$ are integer.
$\therefore r^2 - 9$ is a factor of $54$.
$\therefore r^2 - 9 = 1, 2, 3, 6, 9, 18, 27, 54$
$r^2 = 10, 11, 12, 15, 18, 27, 36, 63$
$\therefore r = 6$ only possible value.
$\therefore h = 6 + \frac{54}{36-9}$
$= 6 + 2=8 $
$\therefore r = 6, h = 8$