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Elana
Posts:
22
Registered:
12/6/04


Re: Graph Theory Question #1
Posted:
May 2, 2005 6:02 PM


Is this on the right track:
Given vertices A, B, and C, there are 3 segments between them: AB,BC and CA. Now, consider triangle ABC formed by these vertices. Assume, without loss of generality, that this triangle is not equilateral. That is, distances between vertices A, B, and C are not equal. That means, that at least one angle > 60 and at least one is < 60. So, assume, that AB was the first segment picked. Assume that BC was the second picked. That means that segment AC (the third, unpicked choice) is the largest of the three, and thus it must be opposite to the largest angle (the angle between AB and BC). And since we have estableshed that in a nonequilateral triangle at least one angle is > 60, the largest angle (the one that's accross from AC) must be larger than 60. Therefore, whenever 2 edges meet at a vertex, it's equivalent to chosing two of the shortest edges in a triangle formed by the three vertices, and the angle between them is the largest angle (opposite the largest, nonpicked, side), and thus it must be > 60.
Does this make sense??
Thanks, Elana



