Get Outta My Way! Topic: Algebra/Geometry/Sets. Level: AIME/Olympiad.

Problem: (Math Olympiad Treasures - 2.1.16) Seven reals numbers are given in the interval $ (1,13) $. Prove that at least three of them are the lengths of a triangle's sides.

Solution: Let $ x_1, x_2, \ldots, x_7 $ with $ x_1 \le x_2 \le \cdots \le x_7 $ be the reals. Suppose no three of them are the lengths of a triangle's sides. This means $ x_{i+2} \ge x_{i+1}+x_i $ for $ i = 1,2,\ldots,5 $.

We have $ x_2 \ge x_1 > 1 $. Then

$ x_3 \ge x_2+x_1 > 2 $.

Continuing this, we have

$ x_4 \ge x_3+x_2 > 3 $

$ x_5 \ge x_4+x_3 > 5 $

$ x_6 \ge x_5+x_4 > 8 $

$ x_7 \ge x_6+x_5 > 13 $,

a contradiction. Hence there must be three that are the length's of a triangle's sides. QED.

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Comment: From the proof we can easily see the generalization of $ n $ real numbers in the interval $ (1,F_n) $. This is a good example of a problem that doesn't really relate to geometry at all but the triangle condition is simply used as a way to restrict selection of the reals.

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Practice Problem: (Math Olympiad Treasures - 2.1.11) Consider $ n $ red and $ n $ blue points in the plane, no three of them being collinear. Prove that one can connect each red point to a blue point with a segment such that no two segments intersect.