6. At the vertices of a square with a side length of 4, there are four cities. These cities need to be connected by roads such that it is possible to travel from any city to any other. Propose at least one variant of such roads with a total length of less than 11.

Hint. The following statement may be useful in solving the problem (which can be used without proof). Let the internal angles of triangle $ABC$ be less than 120 ${ }^{\circ}$. The sum of the distances $AT + BT + CT$ from point $T$ to the vertices of the triangle is minimal if the sides of the triangle are seen from point $T$ at an angle of $120^{\circ}$ ( $T$ - the Torricelli point of the triangle). If one of the angles of the triangle is greater than or equal to $120^{\circ}$, then the point of minimum sum of distances will be the vertex of this angle.
Solution. Suppose our road system has no intersections, that is, there is one road connecting the vertices of the square $E, F, G$, and $H$ sequentially. Then its length will be no less than three sides of the square (the letter "П" in Fig. (a)), that is, no less than 12 (a road connecting adjacent vertices of the square is no less than its side).

Thus, the desired (and ideally the shortest) road system must have intersections (for example, as in Fig. (d)). To understand how the total length of the roads can be effectively reduced, it is useful to first note some properties that the shortest road system must possess:

i. The shortest road system consists of segments connecting intersections and the vertices of the square. This is obvious, as the shortest path from one point to another is a straight line segment. Without proof, although this is almost obvious, for any system of cities, the shortest road system connecting them is a connected graph without closed paths, whose edges are segments.

ii. Each intersection must be connected to at least three vertices of the graph (vertices of the square or intersections). (If it is connected to only two, then it makes little sense: it can be removed, and these two vertices can be connected directly; it will be shorter.)

iii. The angle between any two roads coming out of one intersection (or from one vertex of the square) cannot be less than $120^{\circ}$. Indeed, let roads $AB$ and $AC$ come out of point $A$, and angle $BAC$ is less than $120^{\circ}$. Then roads $AB$ and $AC$ can be replaced by roads with a smaller total length. If in triangle $ABC$ all internal angles are less than $120^{\circ}$, then roads $AB$ and $AC$ can be replaced by roads $TA, TB$, and $TC$, where $T$ is the Torricelli point of triangle $ABC$ (Fig. (b)). If, for example, angle $B$ is greater than $120^{\circ}$, then $AB$ and $AC$ can be replaced by $AB$ and $BC$ (Fig. (c)).

iv. Exactly three roads come out of one intersection at angles of $120^{\circ}$ (otherwise, the length of the roads can be reduced). This immediately follows from properties (ii) and (iii).

Suppose the road system has one intersection. If it is connected to all vertices (Fig. (g)), then the length of the roads is no less than the sum of the diagonals, which is $8\sqrt{2} > 11$. If it is connected to only three vertices (Fig. (d)), then $T$ is the Torricelli point of triangle $EFH$, and vertex $G$ is connected to $F$. In this case, the length of the roads is approximately 11.7. Moreover, property (iii) is violated, as $\angle THG < 120^{\circ}$, which means the length of the roads can be reduced by adding another intersection (as in the case in Fig. (b)).

Suppose there are two intersections. By symmetry, place them on an axis of symmetry parallel to two sides of the square (Fig. (e)) such that roads come out of each intersection at angles of $120^{\circ}$ (property (iv)). In this case, the total length of the roads is $4(\sqrt{3}+1) < 10.92$.

Answer: For example, a road system with two intersections on an axis of symmetry parallel to two sides of the square (Fig. (e)). Roads come out of each intersection at angles of $120^{\circ}$. Their total length is $4(\sqrt{3}+1) < 11$.

Comment. The road system in Fig. (e) is called the Steiner network of the given four points, the vertices of the square $E, F, G$, and $H$. Without proof, note that this system has the minimum length among all possible ones.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-3.jpg?height=431&width=146&top_left_y=2165&top_left_x=1360)

(r)

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-3.jpg?height=228&width=174&top_left_y=2118&top_left_x=1546)

(b)

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-3.jpg?height=200&width=152&top_left_y=2392&top_left_x=1545)

(d)

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-3.jpg?height=189&width=215&top_left_y=2158&top_left_x=1725)

(B)

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-3.jpg?height=194&width=159&top_left_y=2392&top_left_x=1736)

(e)

## Interregional School Olympiad Based on Departmental Educational Organizations

in Mathematics

