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Old 06-26-2018, 09:15 PM
Pallen Pallen is offline
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Join Date: Jun 2018
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Default Problem 2.2

I've been having trouble proving that the break point is 5 for problem 2.2 I don't know what special properties of pentagons might come into play.

For one thing, I know that in the case where at least one point is in the convex hull of the other four points, then the dichotomy with -1 for that one point and +1 for all others won't work. Therefore, we must only consider the case where no point is in the convex hull of the four other points (so that, I think, we have a convex pentagon). But from here I'm completely stuck.
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