Let S be the set of all points in the plane. Find all functions f:S→R such that for all nondegenerate triangles ABC with orthocenter H, if f(A)≤f(B)≤f(C), then f(A)+f(C)=f(B)+f(H).
We will show that the only solutions are the constant functions, which obviously satisfy the condition.
Assume there exist two points P and Q such that f(P)≠f(Q). WLOG f(P)>f(Q), and f(P)=a+b,f(Q)=a−b, where b>0. Now consider points R and S such that PRQS is a square. Observe that R is the orthocenter of △PRQ. We consider the possible values of f(R):
If f(R)≥f(P)>f(Q), then f(R)+f(Q)=f(P)+f(R)⟹f(Q)=f(P), a contradiction.
If f(P)>f(R)≥f(Q), then f(P)+f(Q)=f(R)+f(R)⟹f(R) is average of f(P) and f(Q).
If f(P)>f(Q)>f(R), then f(R)+f(P)=f(Q)+f(R)⟹f(Q)=f(P), a contradiction.
Therefore, we conclude that f(R)=a. But now consider the point O the center of PRQS. From △POR, we get f(O)=(f(P)+f(R))/2=(2a+b)/2, while from △QOR, we get f(O)=(f(Q)+f(R))/2=(2a−b)/2, which forces b=0. A contradiction.