Function on plane

Let f be a function from the set of points in the plane to the set of real numbers ,with the property that for any square ABCD,

f(A)+f(B)+f(C)+f(D) = 0

Prove that f(P) = 0 , for any point P in the plane.

Not a doubt.

2 Answers

[2] This is a q from putnam 2009.

Take any point A in plane. we will prove f(A) = 0

We have from the construction in figure

[f(A) + f(D) + f(C') + f(B') ]+ [f(A) + f(B') + f(C'') + f(D')] + [f(A) + f(D') + f(C''') + f(B)]+[f(A) + f(D) + f(C) + f(B)] = 0

=> 4 f(A) + [f(C) + f(C') + f(C'') + f(C''')] + 2[f(B) + f(D) + f(B') + f(D')] = 0

But from figure [f(C) + f(C') + f(C'') + f(C''')] = 0, [f(B) + f(D) + f(B') + f(D')]= 0 ( they r also squares)

this gives f(A) = 0 [1]

Bless you for this nice solution

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω