}

Let \(r_{1}\) be the base circle radius, \(I_{1}\) the surface area, and \(V_{1}\) the volume of the cone, and \(r_{2}, I_{2}\) and \(V_{2}\) the corresponding measures of the cylinder. The length of the slant height of the cone is \(2 r_{1}\) due to the shape of the axial section.

The surface area of the cone consists of a circular disk with radius \(r_{1}\) and a conical mantle with base circle radius \(r_{1}\) and slant height \(s=2 r_{1}\). Therefore, we have:

\[
I_{1}=\frac{\pi}{2}\left(2 r_{1}\right)^{2}+\pi r_{1}^{2}=3 \pi r_{1}^{2}
\]

As the area of an equilateral triangle, the axial section and thus the cone have a height of \(2 r_{1} \frac{1}{2} \sqrt{3}\), and therefore

\[
V_{1}=\pi r_{1}^{2} r_{1} \frac{1}{3} \sqrt{3}=\frac{\pi r_{1}^{3}}{\sqrt{3}}
\]

The surface area of the cylinder consists of two circular disks with radius \(r_{2}\) and a cylindrical mantle with radius \(r_{2}\) and height \(2 r_{2}\). Therefore, we have:

\[
I_{2}=4 \pi r_{2}^{2}+2 \pi r_{2}^{2}=6 \pi r_{2}^{2}
\]

and also

\[
V_{2}=\pi r_{2}^{2} \cdot 2 r_{2}=2 \pi r_{2}^{3}
\]

Since \(V_{1}=V_{2}\), it follows from (2) and (4) that

\[
\frac{r_{1}}{r_{2}}=\sqrt[3]{2 \sqrt{3}}=\sqrt[6]{12}
\]

and thus from (1), (3), and (5) we get

\[
\frac{I_{1}}{I_{2}}=\frac{1}{2}\left(\frac{r_{1}}{r_{2}}\right)^{2}=\frac{1}{2} \sqrt[3]{12}=\sqrt[3]{\frac{3}{2}}
\]
\section*{Problem 3 - 051013}

How does the measure of the surface area of a cone, whose axial section is an equilateral triangle, compare to the measure of the surface area of a cylinder with a square axial section, if the measure of the volumes of both bodies is equal?

In both figures, the same units of measurement are assumed.