\section*{Problem 19}

A polygonal line connects two opposite vertices of a cube with side 2 . Each segment of the line has length 3 and each vertex lies on the faces (or edges) of the cube. What is the smallest number of segments the line can have?

\section*{Answer \(\quad 6\)}


\section*{Solution}

![](https://cdn.mathpix.com/cropped/2024_05_09_2c0a57c089c74a6cc642g-117.jpg?height=251&width=345&top_left_y=931&top_left_x=250)

Suppose one endpoint of a segment length 3 is at A. Evidently the other end could be at the edge midpoints B, C, D. It could also be on the circular arc connecting B and C (with center O and radius \(\sqrt{5}\) ). Similarly, it could be on arcs connecting C and D, or B and D. We claim that if \(\mathrm{X}\) is a point of one of these arcs other than its endpoints, then the only possible segment length 3 with an endpoint at \(X\) (and the other endpoint on the surface of the cube) is AX. wlog we can consider \(X\) to be on the arc BC. Take axes with origin O, so that A is \((0,0,2)\). Suppose \(X\) is \((a, b, 0)\) and that the other endpoint of the segment is \(Y(x, y, z)\). Then \(X Y^{2}=(x-a)^{2}+(y-b)^{2}\) \(+z^{2}=a^{2}+b^{2}+z^{2}-x(2 a-x)-y(2 b-y)=5+z^{2}-x(2 a-x)-y(2 b-y)\). But \(a, b>1\) since \(X\) is not an endpoint of the arc, so (2a-x) and (2b-y) are both positive. Hence \(-x(2 a-x)-y(2 b-y) \leq 0\) with equality iff \(x=y=0\). Similarly, \(z^{2} \leq 4\) with equality iff \(z=2\). Hence \(X Y^{2} \leq 9\) with equality iff \(\mathrm{Y}=\mathrm{A}\), which proves the claim.

Thus if the next link of the polygonal line goes from A to anywhere except B, C, D, then it has to go back to A. So a minimal line must go to B, C, or D.

Now from \(\mathrm{D}\) the line can only go to \(\mathrm{A}\) or \(\mathrm{O}\). For if it goes to \(\mathrm{Z}(\mathrm{x}, \mathrm{y}, \mathrm{Z})\), then we have \(\mathrm{DZ}^{2}=(\mathrm{x}-\) \(2)^{2}+(y-2)^{2}+(z-1)^{2} \leq 2^{2}+2^{2}+1^{2}=3^{2}\) with equality iff \(x=0, y=0\) and \(z=0\) or 2 .

So let us take A as the starting point of the polygonal line. wlog the first segment is AD. Then the second segment must be DO (for a minimal line). Thus the best we can do with 2 segments is to move along an edge. It takes three such moves to get to the opposite corner, and hence at least 6 segments. But it is obvious that it can be done with 6 segments.

