## Task 29/74

We are looking for all cones where the measurements of the radius, height, and slant height are natural numbers and where the measurements of the surface area and volume are equal.
If the measures of radius, height, and slant height are denoted by $r, h$, and $s$ respectively, where $r, h, s \in \mathbb{N}$ and $r, h, s > 0$, the problem requires the validity of the equation

$$
\frac{1}{3} \pi r^{2} h = \pi r(r + s)
$$

with the side condition $r^{2} + h^{2} = s^{2}$. This leads to

$$
\frac{r h}{3} = r + s \rightarrow r h - 3 r = 3 s \rightarrow r^{2} h^{2} - 6 r^{2} h - 9 h^{2} = 0
$$

and since $h \neq 0$,

$$
r^{2} h - 6 r^{2} - 9 h = 0 \rightarrow r^{2} = \frac{9 h}{h - 6} = 9 + \frac{54}{h - 6}
$$

For $r^{2}$ to be a natural number, $k = h - 6$ must be a divisor of 54. Thus, the possible values for $k$ are $1, 2, 3, 6, 9, 18, 27, 54$, corresponding to $h = 7, 8, 9, 12, 15, 24, 33, 60$.

However, only $k = 2$ with $h = 8$ yields a perfect square $r^{2}$, ensuring that $r$ is also a natural number. Therefore, there is exactly one such cone with the measures $r = 6, h = 8, s = 10$. For this cone, $V = O = 96 \pi$.