## Saturday, April 8, 2006

### 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.

#### 1 comment:

1. very easy and obvious problem ;)