Home > Combinatorics, IMO, Olympiad > Romanian TST II Problem 3

Romanian TST II Problem 3

Given a positive integer n, determine the maximum number of edges a simple graph of n edges vertices may have, in order that it won’t contain any cycles of even length.

Romanian TST 2011

Categories: Combinatorics, IMO, Olympiad Tags: ,
  1. maxim bogdan
    July 1, 2011 at 10:09 am

    It is easy to check that any two cycles in this graph cannot intersect after (an) edge(s) – without forming a even length cycle – . So, practically we have some disjoint cycles and we have to articulate them, or to adjoin an edge (or more) between two vertices (one from the first cycle involved and one from the second). In order to have a maximum number of edges it is better to replace all the superior order cycles (5,7,9…) by triangles. For instance, instead of having a pentagon we can replace it by two triangles fixed in one vertex. For the pentagon we would have 5 vertices and 5 edges. With triangles we still have 5 vertices, but now we have 6 edges. Furthermore, we can see that it is better to fix triangles in a point, than linking them by a suplementary edge. In the former case we have 5 vertices and 6 edges. In the latter we would have 6 vertices and 7 edges which is not convenient for us because: \frac{7}{6}<\frac{6}{5}. Hence if n=2k+1 we are to link k triangles however we want (but without creating an extra-cycle). We've used 3k edges. Now, if n=2k we are to bring together k-1 triangles, using therefore 3(k-1) edges. Here we notice that one vertex remains isolated. All we can do is to join it to any vertex we want (but just one!). Thereby combining the last two facts we conclude that the aswer is: n-1+\lfloor\frac{n-1}{2}\rfloor.

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