37th Iranian Mathematical Olympiad

Let $d_{X,Y}$ be the weight assigned to the edge $XY$ of the graph. Clearly, in each triangle the maximum weight of an edge is the sum of the two others. Let $d_{A,B}$ be the maximum weight among all weights assigned to the edges. Now assign $0$ to vertex $A$ and $d_{A,X}$ to any other vertex $X$. Obviously, this labeling works for all edges including $A$. We claim that it works for all other edges.

Assume to the contrary that the labeling doesn't work for an edge $XY$ of the graph. So $d_{X,Y}$ is not the difference between $d_{A,X}$ and $d_{A,Y}$, which means $d_{X,Y}=d_{A,X}+d_{A,Y}$. Note that $B$ cannot be any of $X,Y$. Since $d_{A,B}$ is the maximum edge number, we have $d_{B,X}=d_{A,B}-d_{A,X}$ and $d_{B,Y}=d_{A,B}-d_{A,Y}$. Now consider triangle $BXY$. If $d_{X,Y}$ is the maximum weight of an edge in this triangle, we have $$d_{X,Y}=d_{B,Y}+d_{B,X}⟹\begin{array}{rl}[t]d_{X,Y}& =d_{A,B}-d_{A,Y}+d_{A,B}-d_{A,X}\\ & =2d_{X,Y}=2d_{A,B}.\end{array}$$ Which is impossible since the weights are distinct. So the maximum edge weight in triangle $BXY$ is not $d_{X,Y}$. Without loss of generality, assume that $d_{B,Y}$ is the maximum one. So we have $$d_{B,Y}=d_{B,X}+d_{X,Y}⟹\begin{array}{rl}[t]d_{A,B}-d_{A,Y}& =d_{A,B}-d_{A,X}+d_{A,X}+d_{A,Y}\\ & =d_{A,Y}=0.\end{array}$$ Which is again a contradiction. This completes the proof. ■