A triangle is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.
For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.
Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.
To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with is odd and thus actually !
We enumerate all small triangles in the picture as , and denote by the number of edges with endpoints and in each triangle . Thus, if say the vertices of are labeled by , then , and so on …
Observe now that obviously we have
where is the number of triangles whose vertices are labeled , while is the number of those triangles labeled by or . (Actually it is easily seen that for such triangles, while if the vertices of are and otherwise.) All we have to show is that is odd.
Let denote the number of -edges lying inside the original triangle and let be the number of -edges lying on the boundary of . Every interior -edge lies in two triangles and thus it is counted twice in the sum , while every boundary -edge is counted only once. In conclusion we get
Hence is odd if and only if is odd. It is therefore enough to show that is odd.
According to the hypothesis of the problem, edges labeled or can occur only on the -edge of the large triangle . We start walking along the edge of the triangle starting at the vertex toward the vertex . Now, only when we first pass an edge labeled will we arrive at the first vertex labeled . A number of vertices labeled may now follow, and only after we have passed a segment do we reach a label , and so on. Thus after an odd number of segments or we arrive at vertices labeled , and after an even number of such segments we arrive at vertices labeled . Since the last vertex we will reach is the vertex of the big triangle , it follows that the total number of segments or lying on the side of the big triangle must be odd! The same reasoning applies for each of the other edges of the big triangle , so we deduce that , the total number of or -edges lying on the boundary of , must be odd. Proved
It is obvious. As a result of this numbering we get following diagram: