XXVI - III - Task 2

On the surface of a regular tetrahedron with edge length 1, a finite set of segments is chosen in such a way that any two vertices of the tetrahedron can be connected by a broken line composed of some of these segments. Can this set of segments be chosen so that their total length is less than $1 + \sqrt{3}$?
On the surface of a regular tetrahedron with edge length $1$, one can choose a finite set of segments such that any two vertices of the tetrahedron can be connected by a broken line composed of some of these segments, and the total length of the segments belonging to this set is less than $1 + \sqrt{3}$.
Consider, in particular, the rhombus $ABDC$ formed by unfolding two faces of the given tetrahedron onto a plane (Fig. 10). Then $AB = 1$ and $\measuredangle BAC = 60^\circ$. Let $P$ be the midpoint of segment $\overline{BC}$, and $Q$ be a point in triangle $ABP$ from which the sides $\overline{AP}$ and $\overline{BP}$ are seen at an angle of $120^\circ$. Then $\measuredangle AQB = 120^\circ$ as well. Let $x = AQ$, $y = BQ$, $z = PQ$. Since $AP = \frac{1}{2}\sqrt{3}$, $BP = \frac{1}{2}$, and $\measuredangle APB = 90^\circ$, comparing the area of triangle $ABP$ to the sum of the areas of triangles $AQB$, $BQP$, and $AQP$ gives $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} (xy + yz + zx) \cdot \sin 120^\circ$, which simplifies to

Since $\cos 120^\circ = -\frac{1}{2}$, applying the law of cosines to triangles $AQB$, $BQP$, and $AQP$ yields

Adding the equations (2), (3), (4), and equation (1) multiplied by $3$ gives $2(x + y + z)^2 = \frac{7}{2}$, and thus $x + y + z = \frac{\sqrt{7}}{2}$.
If $Q'$ is the point symmetric to $Q$ with respect to point $P$, then the set of segments $\overline{AQ}$, $\overline{BQ}$, $\overline{QP}$, $\overline{PQ'}$, $\overline{GQ}$, $\overline{DQ}$ has the property mentioned in the problem, and the sum of the lengths of these segments is $\sqrt{7}$. This number is less than $1 + \sqrt{3}$.