Consider a given cube's 3 mutually skew edges, and all triangles whose vertices lie on these edges respectively. Which one has the smallest perimeter?
I. solution. Let the points on the edges $AE, BC$, and $GH$ of the unit cube $ABCDEFGH$ be denoted by $L, M, N$ respectively. Let $AL=x, CM=y$, and $HN=z$. Our goal is to determine the values of $x, y, z$ for which the perimeter of triangle $LMN$ is minimized.

![](https://cdn.mathpix.com/cropped/2024_05_02_1f08843d39024866cbf4g-1.jpg?height=487&width=509&top_left_y=304&top_left_x=797)

We express the lengths of the sides using the Pythagorean theorem. From triangle $ABM$, $AM=\sqrt{1+(1-y)^{2}}$, and from triangle $LAM$, $LM=\sqrt{x^{2}+(1-y)^{2}+1}$. Similarly, $MN=\sqrt{1+y^{2}+(1-z)^{2}}$ and $NL=\sqrt{(1-x)^{2}+1+z^{2}}$. Let the sum of these be denoted by $K$. After some rearrangement, we can estimate $K$ from below using the relationship between the arithmetic and quadratic means.

$$
\begin{gathered}
K=\sqrt{6}\left(\sqrt{\frac{x^{2}+(1-y)^{2}+4\left(\frac{1}{2}\right)^{2}}{6}}+\sqrt{\frac{4\left(\frac{1}{2}\right)^{2}+y^{2}+(1-z)^{2}}{6}}+\right. \\
\left.+\sqrt{\frac{(1-x)^{2}+4\left(\frac{1}{2}\right)^{2}+z^{2}}{6}}\right) \geq \\
\geq \sqrt{6}\left(\frac{x+(1-y)+4\left(\frac{1}{2}\right)}{6}+\frac{4\left(\frac{1}{2}\right)+y+(1-z)}{6}+\frac{(1-x)+4\left(\frac{1}{2}\right)+z}{6}\right)=\frac{3}{2} \sqrt{6} \\
\quad
\end{gathered}
$$

Thus, $K \geq \frac{3}{2} \sqrt{6}$, and equality holds only when $x=y=z=\frac{1}{2}=1-x=1-y=1-z$, i.e., when the vertices of triangle $LMN$ are the midpoints of the respective edges.

Ákos Seress (Budapest, Fazekas M. Gyak. Gimn.)

II. solution. Let the edges be denoted by $p, q, r$, and their midpoints by $P, Q, R$. We will show that the perimeter of triangle $PQR$ is minimal among the triangles we are examining. Let the plane of triangle $PQR$ be denoted by $S$, and reflect the cube across the plane through $r$ that is perpendicular to $S$. Since the center of the cube is in $S$, the plane we reflect across forms a $45^{\circ}$ angle with $p$ and $q$. Therefore, the reflection of $p$ remains in the plane through $p$ that is perpendicular to $r$ and is perpendicular to $p$. Denote this by $p_{1}$, and the reflections of $q, P, Q$ by $q_{1}, P_{1}, Q_{1}$. Since $S$ is perpendicular to the reflection plane, $P_{1}, Q_{1}$ will also be in $S$.

Continuing step by step, reflect across the planes through $q_{1}$, $p_{2}$, $r_{3}$, and $q_{4}$ that are perpendicular to $S$, denoting the reflections of $p_{i}, q_{i}, r_{i}$ after the $(i+1)$-th reflection by $p_{i+1}, q_{i+1}, r_{i+1}$ (where $i=1,2,3,4$, and $r_{1}$ is identical to $r$). During this process, the points $P, R, Q_{1}, P_{2}, R_{3}, Q_{4}, P_{5}$ will all lie on a single line. Perform these reflections on any points $P^{*}, Q^{*}, R^{*}$ on the lines $p, q, r$ as well. The perimeter of triangle $P^{*} Q^{*} R^{*}$, doubled, will split into the length of the broken line $P^{*} R^{*} Q_{1}^{*} P_{2}^{*} R_{3}^{*} Q_{4}^{*} P_{5}^{*}$. This length cannot be smaller than the length of segment $P^{*} P_{5}^{*}$, which is equal to the length of $P P_{5}$ and twice the perimeter of triangle $PQR$.

Remark. In our diagram, we have placed the 6 cubes into a 6-story building with a $4 \times 3$ base to better visualize the spatial relationships, which resulted in strong perspective distortions.

![](https://cdn.mathpix.com/cropped/2024_05_02_1f08843d39024866cbf4g-2.jpg?height=2250&width=1458&top_left_y=127&top_left_x=325)