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Please be obtuse!


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Pick three points on a plane at random (assume uniform probability density across the plane). What is the probability that the triangle so formed is obtuse angled?   Source: general internet, with s

Please pay attention - I prove clearly that any obtuse triangle corresponds to one of the acute-angled, which shows clearly that we have  50% chance to choose one or the type no matter in which way yo

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Ohhhhhh, this one is weird.

I didn't read the thread for this one, so forgive me if I am redundant, but here are my thoughts...

After choosing the first two points, there's three cases that would result in an obtuse triangle. One, the third point is not "between" the first two points. Two, the third point is within the circle whose diameter is the line segment between the first two points. Sooooooo, technically this area is infinite, but so is the remainder of the plane... However, the "obtuse" territory expands infinitely along two axes while the "acute" territory expands infinitely along one axis, so I would be tempted to say that the probability of an obtuse triangle is 1. That's my amateur take on it anyways.

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I looked through the first few replies, and I now realize that this problem is even weirder than I initially thought.

But here's another thought: plasmid's method of assuming that the first two points are the closest/farthest doesn't work on a finite plane, so maybe it's wrong in the infinite case. But at the same time, maybe not, because stating that the first two points are the closest/farthest clearly warps the joint probability distribution of their positions on a finite plane (because obviously they are more likely to be closer together if they are the closest vertices of the triangle), but not on an infinite plane. Probability and infinity are pretty fragile things it seems. :(

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