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Sep 24, 2017 at 7:01 answer added user44143 timeline score: 4
Sep 2, 2017 at 20:28 comment added Anthony Quas Hmmm... Maybe I meant $d=d+2r$? If I'm right that there's a reformulation like this, it will clearly make your problem much easier.
Sep 2, 2017 at 18:08 comment added Tom Solberg @AnthonyQuas, if that were the case, then it would also be the same as taking $r=d+r$ and $d=0$. If we take $d+r=1$ and have $p=(0,1)$, then your set is empty, whereas the new set I just described is not.
Sep 2, 2017 at 9:51 comment added Anthony Quas Why isn't this the same question as taking $r=0$ and $d=d+r$?
Sep 1, 2017 at 19:36 history edited Tom Solberg CC BY-SA 3.0
Joseph O'Rourke's notation suggestion
Sep 1, 2017 at 19:05 answer added Joseph O'Rourke timeline score: 2
Sep 1, 2017 at 19:04 comment added Joseph O'Rourke You seem to use $\ell$ in two different senses...
Sep 1, 2017 at 17:52 history asked Tom Solberg CC BY-SA 3.0