3-cha 2. $k$ wire triangles are arranged in space such that:

1) any 2 of them have exactly one common vertex,
2) in each vertex, the same number $p$ of triangles meet.

Find all values of $k$ and $p$ for which the specified arrangement is possible.
Solution 2. Answer: $(k, p)=(1,1),(4,2)$ or $(7,3)$.

First, we prove that $p \leqslant 3$. Suppose that $p \geqslant 4$. Take triangle $\Delta_{1}$. To its vertex $A$ adjoins triangle $\Delta_{2}$. To the vertex $B \neq A$ of triangle $\Delta_{2}$ adjoin triangles $\Delta_{3}, \Delta_{4}, \Delta_{5}$; all of them are different from $\Delta_{1}$, since otherwise triangles $\Delta_{1}$ and $\Delta_{2}$ would have two common vertices. By the condition, each of the triangles $\Delta_{3}, \Delta_{4}, \Delta_{5}$ has a common vertex with triangle $\Delta_{1}$, and this vertex is different from $A$. But from this it follows that two of the triangles $\Delta_{3}, \Delta_{4}, \Delta_{5}$ have a common vertex different from $B$.

If $p=1$, then $k=1$ (if there were at least 2 triangles, then they would have a common vertex, and then $p \geqslant 2)$

Suppose that at each vertex $p \geqslant 2$ triangles meet. Fix one of the triangles. To each of its vertices adjoin $p-1$ triangles, i.e., a total of $3(p-1)$ triangles adjoin it. All these triangles are different, and there are no other triangles, since any two triangles must have a common vertex. Thus, we get a total of $3(p-1)+1=3 p-2$ triangles.[^4]

To construct a configuration for which $(k, p)=(4,2)$, one can take an octahedron and remove half of its faces, leaving only the triangles that do not share common sides. To construct a configuration for which $(k, p)=(7,3)$, one can take a tetrahedron, place a triangle $A B C$ inside it, and for each vertex of triangle $A B C$ take a triangle formed by this vertex and one of the two non-adjacent edges of the tetrahedron (for each vertex of triangle $A B C$, a different pair of non-adjacent edges of the tetrahedron is taken).