Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Iran

Morteza placed 88 points on the plane. Then he writes down all the areas of the triangles constructed by these points (a total of 5656 numbers). Prove that he can always assign positive or negative signs to these numbers such that the sum of these numbers becomes zero.

Solution

First of all consider 44 points AA, BB, CC and DD.

Figure 1
Figure 2

i) If convex hull of these points is a triangle (without loss of generality let DD be the inner point)
S△ABC−S△ABD−S△ACD−S△BCD=0 S_{\triangle ABC} - S_{\triangle ABD} - S_{\triangle ACD} - S_{\triangle BCD} = 0

ii) If convex hull of these points is a quadrilateral
S△ABC−S△ABD+S△ACD−S△BCD=0 S_{\triangle ABC} - S_{\triangle ABD} + S_{\triangle ACD} - S_{\triangle BCD} = 0

Let A1A_1, A2A_2, A3A_3, …\dots, A8A_8 be 88 problem's 88 points. Now consider these sets of 44 points, it is easy to see that each triangle of 5656 triangle is exactly in one of these sets, and from above it is possible to put −- and ++ between areas in every set such that the result becomes zero, hence we are done.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.