(12) Among all the circumscribed circular cones of a sphere with radius $R$, the total surface area of the cone with the minimum total surface area is $\qquad$ .
(12) Let $x = A H$ be the radius of the base circle of the frustum of a cone, $C H$ be its height, and $S$ be its total surface area, then
$$
S = \pi x^{2} + \pi x A C = \pi x^{2} + \pi x(C D + x).
$$

From $\triangle C D O \backsim \triangle C H A$, we find $C D = \frac{2 R^{2} x}{x^{2} - R^{2}}$, then
$$
\begin{array}{c}
S = \pi x\left(x + x + \frac{2 R^{2} x}{x^{2} - R^{2}}\right) = \frac{2 \pi x^{4}}{x^{2} - R^{2}}, \\
\frac{1}{S} = \frac{x^{2} - R^{2}}{2 \pi x^{4}} = \frac{1}{2 \pi x^{2}}\left(1 - \frac{R^{2}}{x^{2}}\right), \\
\frac{2 \pi R^{2}}{S} = \frac{R^{2}}{x^{2}}\left(1 - \frac{R^{2}}{x^{2}}\right),
\end{array}
$$

When
$$
\frac{R^{2}}{x^{2}} = 1 - \frac{R^{2}}{x^{2}},
$$

i.e., $x = \sqrt{2} R$, $\frac{2 \pi R^{2}}{S}$ reaches its maximum value $\frac{1}{4}$, hence $S$ has a minimum value of $8 \pi R^{2}$.