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Comment Complicity (Score 1) 69

The phone companies are largely to blame. Itâ(TM)s not that difficult to set up a policy forbidding domestic calls from presenting a callback number that doesnâ(TM)t belong to the same company that owns the line from which a call is placed. Attempting to violate it should kill your lines instantly. Not that it made a lick of sense for clients to be allowed to specify an arbitrary caller id in the first place! Foreign networks should be required to have the same basic guarantees as part of peerage, just like any other technical protocols that have to be followed. Violate the protocol, you lose peerage. Iâ(TM)d guess this would be doable globally within a couple of years.

Comment Proof sketch (Score 1) 2

Let's see, quick sketch: If X, Y, and Z are distinct points, and A is equidistant from each of them, then they specify a unique circle C centered on A. If X, Y, and Z are also points on a circle C', then there must be at least three intersections between the circles C and C'. But the only intersection cases of two circles is none, one (tangent), two, or infinity (same circle). Therefore C and C' are the same circle, and A is the center of that circle (so XA, YA, and ZA are radii). That's a nice theorem to have handy, and very intuitive to prove *and* to use. Congratulations.

Comment The Land of Ire (Score 1) 1376

Don't forget -- almost everybody that posted here, *even the religious zealots*, would be subject to raids, enormous fines, and probably imprisonment if they lived in The Land of Ire and they decided to start enforcing that anti-speech BS. I certainly have no intention of ever visiting such a place (and yes, I'm aware of worse situations in seemingly harmless countries).

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