4. There are 51 cities distributed in a square area with a side length of 1000 kilometers. It is planned to lay down 11000 kilometers of road network in the area. Can all the cities be connected?
(Continued from page 49) Therefore, the total length of all these roads does not exceed $1000 \times 6 + 100 \times 50 = 11000$ kilometers. This road network can connect all 51 cities in the region.
5. In the tetrahedron $\mathrm{ABCD}$, take points $\mathrm{A}^{\prime}$, $\mathrm{B}^{\prime}$, $\mathrm{C}^{\prime}$, and $\mathrm{D}^{\prime}$ on the faces $\mathrm{BCD}$, $\mathrm{ACD}$, $\mathrm{ABD}$, and $\mathrm{ABC}$ respectively, such that each pair of these points lies in the same plane with two vertices of the tetrahedron. Prove that the lines $\mathrm{AA}^{\prime}$, $\mathrm{BB}^{\prime}$, $\mathrm{CC}^{\prime}$, and $\mathrm{DD}^{\prime}$ intersect at one point.

Proof: We know that: if three lines intersect each other pairwise and are not coplanar, then they have only one common point. In fact, if these three lines intersect at three different points, then these three points determine a plane, and each line has two points in this plane, so these three lines will be coplanar, which contradicts the given condition of not being coplanar.

Now consider points $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ (Figure 7). They lie in the same plane with two vertices of the tetrahedron, which can only be $\mathrm{A}$ and $\mathrm{B}$. Otherwise, this plane would be one of the faces of the tetrahedron, and thus $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ would be on the same face, which is a contradiction. Suppose $\mathrm{A}$, $\mathrm{B}$, $\mathrm{A}^{\prime}$, and $\mathrm{B}^{\prime}$ lie in the plane $\mathrm{ABM}$, where $\mathrm{M}$ is on the edge $\mathrm{CD}$ (Figure 7). Since $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ lie on $\mathrm{BM}$ and $\mathrm{AM}$ respectively, $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$ intersect. Similarly, we can prove that $\mathrm{AA}^{\prime}$, $\mathrm{BB}^{\prime}$, $\mathrm{CC}^{\prime}$, and $\mathrm{DD}^{\prime}$ intersect pairwise, and any three of them are not coplanar. By the previous proof, these four lines have only one common point.

Solution: It can be done, and we provide a scheme for laying out the road network.

First, lay a road parallel to the boundary of the square and from one side to the other through one of the cities in the region, with a length of 1000 kilometers (Figure 6, cities are represented by small circles). Then, take two points 100 kilometers from each end of this road; between these two points, take a point every 200 kilometers, making a total of three points, plus the two initial points, making five points in total. Lay five roads perpendicular to the first road through these five points, with their ends also reaching the boundary. Now, from each city, lay a road along the shortest route to connect with one of these five roads. The length of each of these additional roads does not exceed 100 kilometers, and there are no more than 50 such roads. Thus,