Let f be a real-valued function on the plane
such that for every square ABCD in the plane,
f(A) + f(B) + f(C) + f(D) = 0.
Prove or disprove that f(P) = 0
for every point P in the plane?
It is so! Proof:
Draw a square ABCD. Place bisectors on each side:
E bisects AB, F bisects AC, G bisects BD, and H bisects CD. Place P at the center of the square. We have formed 5 new squares: AEFP,BEGP, CFHP, DGHP, and EFGH.
0 = (f(a) + f(e) + f(f) +f(p))
+ (f(b) + f(e) + f(g) +f(p))
+ (f(c) + f(f) + f(h) +f(p))
+ (f(d) + f(g) + f(h) +f(p))
= (f(a) + f(b) + fc) +f(d))
+2 (f(e) + f(f) + f(g) +f(h))
+4*f(p))
= 0 + 0 + 4*f(p)
= 4*f(p)
Therefore, f(p) = 0.
Since every point is the center of a some square, f(p) = 0 for all points in the plane.
Edited on November 16, 2010, 4:07 pm
Squaring the square (spoiler)
