177. Tetrahedron - through a straw. Take a flexible thin-walled cylinder of diameter $d$, for example, a cocktail straw. What is the edge $e$ of the maximum regular tetrahedron that can be pushed through this straw?
177. If we form a parallelogram from four equilateral triangles of side $e$ placed adjacent to each other, then in this parallelogram, the length of any segment drawn parallel to its longer side will be equal to $2 e$. From this, it follows that if we form a regular tetrahedron from this parallelogram and then draw any plane perpendicular to the segment connecting the midpoints of opposite edges of the tetrahedron, the bimedian, the perimeter of the resulting section will be equal to $2 e$.

Therefore, if we align the bimedian with the axis of the cylinder, the tetrahedron can be pushed through a flexible thin-walled cylinder, the perimeter of whose cross-section is equal to $\pi d=2 e$. Thus,
![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-133.jpg?height=430&width=1378&top_left_y=518&top_left_x=341)

the edge of the desired tetrahedron is $e=\frac{\pi d}{2}$. In practice, it is useful to slightly bend the edges of one end of the cylinder outward to make it easier to push the tetrahedron inside the cylinder.

If the tetrahedron is positioned differently relative to the axis of the cylinder, then some plane perpendicular to this axis will pass through a vertex of the tetrahedron and intersect two edges not emanating from this vertex. As can be easily seen from our net, the perimeter of such a section $A E F A^{\prime}$ will be greater than $2 e$, and we will not be able to push the tetrahedron through the cylinder in this position. From this, it follows that the edge of the largest tetrahedron that can be pushed through a cylinder of diameter $d$ is $\frac{\pi d}{2}$.

$[M . M ., \mathbf{3 9}, 133$ (March 1966).]