Determine all integers

for which there exist

points

in the plane, no three collinear, and real numbers

such that for
, the area of

is
.
_______
I claim that for

the given statement is impossible. Consider 5 points in the plane with corresponding numbers

and

We have 3 possible arrangements of these 5 points.
Arrangement 1: They form a convex pentagon. By triangulating the pentagon, we see that the area is

for

Thus

is constant. It follows that

meaning that all 5 numbers are equal. But this is impossible since it would imply that

are collinear.
Now note that if we have three points with a fourth inside the triangle, the fourth point has a value equal to the additive inverse of the average of the three points. Additionally, given convex quadrilateral

we have
Arrangement 2: Four points form a convex quadrilateral with the fifth point inside. The four outer points form 4 triangles. The fifth point must lie within two of these triangles. Then by the above note, two adjacent vertices on the quadrilateral must have equal values. This makes the quadrilateral a trapezoid, so the two other points must be equal as well. However, the fifth point forms a convex quadrilateral with three of the four points of the original quadrilateral. From the above note, this implies that the fifth point must lie on one of the parallel sides of the trapezoid.
Arrangement 3: Two points lie inside a triangle formed by the other three points. By the above note, the two points inside must have equal values. Hence, the segments they form must be parallel to all three sides of the triangle, which is obviously impossible.
When

we can arrange the four points in a square and assign equal values to all four.
This post has been edited 1 time. Last edited by KingSmasher3, Oct 10, 2013, 3:09 AM