21*. $n$ wire triangles are arranged in space such that:

$1^{\circ}$. Any two of them share one common vertex.

$2^{\circ}$. At each vertex, the same number $k$ of triangles meet.

Find all values of $k$ and $n$ for which the specified arrangement is possible.
21. Let $k$ be the number of triangles converging at one vertex, and we will prove that $k \leqslant 3$. Indeed, suppose that $k>3$. We will number our triangles with Roman numerals, and their vertices with Arabic numerals. Consider one of the triangles, for example, triangle $I$; its vertices will be denoted by the numbers 1, 2, and 3. Since we assume that $k>3$, at least three more triangles are adjacent to vertex $1$ of triangle $I$, which we will number as $II$, $III$, and $IV$ (Fig. 64) (if $k>4$, then more triangles are adjacent to vertex $I$ besides $I$, $II$, and $IV$, but they will not interest us). Now let $V$ be some triangle adjacent to vertex 2 of triangle $I$. Triangle $V$ is different from triangles $I$, $II$, and $IV$ (otherwise, it would have two common vertices with triangle $I$, namely, 1 and 2). According to the problem's condition, each of the triangles $I$, $II$, and $III$ has one common vertex with triangle $V$ (as with any other of our triangles). This common vertex in none of these cases can coincide with vertex 2 (otherwise, one of the triangles $II$, $III$, and $IV$ would have two common vertices with triangle $I$, namely, vertices 1 and 2). Since, besides vertex 2, triangle $V$ has only two more vertices, at least two of the triangles $II$, $III$, and $IV$ touch triangle $V$ at the same vertex. But then these two triangles have two common vertices: vertex 1 and one of the vertices of triangle $V$. Since the presence of two common vertices in two triangles is prohibited by the problem's conditions, it follows that $k>3$ cannot be.

Thus, $k$ can only be 1, 2, or 3.

The case $k=1$ is completely uninteresting; in this case, there is only one triangle (otherwise, every pair of triangles would have to have a common vertex, and $k$ could not equal 1).

Nevertheless, we must count one triangle as a solution to the problem (condition $1^{\circ}$ can also be considered satisfied here, as no two triangles exist).

Now let's consider the more interesting cases when $k=2$ or $k=3$.

Let $I$ be, as before, one of our triangles, having vertices 1, 2, and 3. To each of these vertices, $(k-1)$ triangles are adjacent, and all these $3(k-1)$ triangles are obviously different (if two triangles adjacent to two vertices of triangle $I$ coincided, it would mean that triangle $I$ has two common vertices with one of our triangles, which is impossible). Furthermore, apart from these $3(k-1)$ triangles and triangle $I$, no other triangles can exist; otherwise, there would be triangles that do not have common vertices with triangle $I$. Thus, when $k=2$, the total number of triangles can only be: $n=3 \cdot 1+1=4$ (Fig. $65, a$), and when $k=3$ - only $n=3 \cdot 2+1=7$ (Fig. $65, b$).

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-077.jpg?height=617&width=497&top_left_y=512&top_left_x=361)

Fig. 64.

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-078.jpg?height=456&width=940&top_left_y=858&top_left_x=134)

Fig. 65.