\section*{Problem 3 - 301033}

Let \(F\) be the surface of a regular tetrahedron \(A B C D\). The midpoints of the segments \(A B\) and \(C D\) are \(M\) and \(N\), respectively.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1580.jpg?height=249&width=620&top_left_y=595&top_left_x=387)

Figure a)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1580.jpg?height=249&width=605&top_left_y=595&top_left_x=1065)

Figure b)

Figures a and b illustrate the process of "cutting a surface \(F\) along a curve \(k = X Y\)".

This curve \(k\), which runs inside the surface \(F\), turns into a curve \(k_{1}\) running from \(X\) to \(Y\) and another curve \(k_{2}\) running from \(Y\) to \(X\) after cutting.

Together, the curves \(k_{1}\) and \(k_{2}\) form a new closed (i.e., self-closing) boundary curve of the cut surface \(F\).

a) If the tetrahedral surface \(F\) is cut in this way twice, namely along the segment \(A B\) and also along the segment \(C D\), the cut surface \(F\) can be bent such that the boundary curves formed from \(A B\) and \(C D\) become two circular lines lying in parallel planes.

The surface \(F\) thus becomes the lateral surface of a right cylinder.

b) If the tetrahedral surface is cut both along the curve consisting of the segments \(C M\) and \(M D\), and along the curve consisting of the segments \(A N\) and \(N B\), the surface \(F\) can also be bent such that the boundary curves become circular lines and \(F\) becomes the lateral surface of a right cylinder. Investigate which of the two cylinders mentioned in a) and b) has the larger volume!
}

Each of the two resulting cylindrical surfaces has as the circumference of its base and top circle twice the length of one of the curves along which the tetrahedron face \(F\) is cut.

The height of the relevant cylinder is the length of the perpendicular dropped from a point on one of the cutting curves to a - in the original position of \(F\) straight - segment of the other cutting curve, provided that this perpendicular lies entirely within the tetrahedron face. This results in (see figures):
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1580.jpg?height=352&width=894&top_left_y=1868&top_left_x=588)

If \(a\) is the edge length of the tetrahedron \(A B C D\), then in case a) the cylinder has the base circumference \(u_{1}=2 A B=2 a\) and the height \(h_{1}=\frac{a}{2} \sqrt{3}\).

In case b) the cylinder has the base circumference \(u_{2}=2(C M+M D)=2 a \sqrt{3}\) and the height \(h_{2}=\frac{a}{2}\). For the heights, we have

\[
\frac{h_{1}}{h_{2}}=\sqrt{3}
\]

For the base radii \(r_{1}\) and \(r_{2}\), we have

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

Thus, for the volumes \(V_{i}=\pi r_{i}^{2} h_{i}(i=1,2)\) of the two cylinders, we have

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

and therefore \(V_{2}>V_{1}\), i.e., the cylinder mentioned in b) has the larger volume.

Adapted from \([5]\)