Let $f(uv)$ denote the edge weight of an edge $uv$. In every triangle, there is one large edge and two small edges. Say a vertex $v$ is weak if, for all triangles containing $v$, the large edge of that triangle is incident to $v$. Let $AB$ be the edge with maximum weight $f(AB)=M$. We claim that $A$ is weak: indeed, suppose for the sake of contradiction that there exists some triangle $ACD$ where $f(AC) + f(AD) = f(CD)$. By maximality of $M$, the edge weights of triangle $BCD$ are $\{M-f(AC), M-f(AD), f(AC)+f(AD)\}$, but it is easy to check that we cannot add two of these to make the third: either we get an edge with weight $0$ or we get $f(CD)=M$, contradicting distinctness. Now we construct out vertex-labeling: set $w(A)=0$ and for each other vertex $X$ set $w(X) = f(AX)$. By construction, this vertex labeling works on all edges incident to $A$, so consider an arbitrary edge $XY$ not containing $A$. Since $A$ is weak, if we WLOG assume that $w(X)<w(Y)$, then $AY$ is the large edge of triangle $AXY$ and we obtain $f(XY) = f(AY) - f(AX) = w(Y)-w(X)$. This means that our vertex labeling works on $XY$, so it works on all edges and we are done.

nice solution i have 2 solutions for this one is induction if we use an special induction we really just don't need to do anything but to check n=3 and n=4 for my second solution pay attention that if we say that weights are distance then the graph would be a line with some points on it take this as a hint it will solve the problem easily

Mr.C wrote: nice solution i have 2 solutions for this one is induction if we use an special induction we really just don't need to do anything but to check n=3 and n=4 for my second solution pay attention that if we say that weights are distance then the graph would be a line with some points on it take this as a hint it will solve the problem easily Do you mean to say like assigning a vector coordinate at each vertex?

yayitsme wrote: Mr.C wrote: nice solution i have 2 solutions for this one is induction if we use an special induction we really just don't need to do anything but to check n=3 and n=4 for my second solution pay attention that if we say that weights are distance then the graph would be a line with some points on it take this as a hint it will solve the problem easily Do you mean to say like assigning a vector coordinate at each vertex? yeah sure , if $E(u,v)$ is the weight of the edge between $u,v$ there exists a represintation of the ghraph in the "plane" such that $D(u,v)=E(u,v)$ you need to prove this in the first part, next you can easily check that the geaph will form a line , and the rest is easy ...

Mr.C wrote: one is induction if we use an special induction we really just don't need to do anything but to check n=3 and n=4 how induction works? (check n=3 and n = 4 is easy how is the rest?)