a)1) Assume that there are at least three red segments with endpoint $A_1$, the others endpoints being say (at least ) $A_2,A_3,A_4$. Then, if one of the segments $A_2A_3,A_3A_4,A_4A_2$ is red, we have a red triangle, with $A_1$. Otherwise, $A_2A_3A_4$ form a blue triangle. In any case, we have a contradiction. Thus, there are at most two red segment from any point. Similarly, we have at most two blue segment from any point. Since there are exactly 4 segments from any point, we are done. 2) Now, consider the graph whose vertices are the five points and the edges are the red segments (that is, we delete the blue ones). From a), we know that each vertex has degree 2, which is well-known to be an even number. Thus, the graph has an eulerian closed path (which is a closed path which uses each edge only once). b) Am I missing something or it is more trivial than the preceeding questions? Pierre.