I have been asked to provide a solution for this problem: "A regular tetrahedron with side 1 m is divided into two pieces with a flat section so that the cutting surface is a quadrangle. What is the smallest possible perimeter of such a quadrilateral?"
I know that the smallest possible perimeter for a quadrilateral occurs when the "length" and the "height" are equal to each other. The problem I am facing is that the height is not vertical relative to the length.
So how would the solution look like?
Imagine a plane parallel to an edge and parallel to the opposite edge of this edge, such that the distance of the plane to the line is $h$. Then the perimeter will be after some calculations (using sine, cosine and tangent with $\frac{\pi}{6}$ and $\frac{\pi}{3}$ as angles) equal to $\frac{2}{3}\sqrt{3}h+2$. Now $h$ can be be $0 < h < \frac{\sqrt{3}}{2}$, hence the perimeter $p$ is $2 < p < 3$ and be chosen to be arbitrarily close to $2$.