PROBLEM OF THE MONTH
NOVEMBER 2001
Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.
There are several possible solutions of this problem. Here is one possible approach: 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
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
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 -edges lying on the boundary of ,
must be odd. This finishes the proof of our claim.