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Let $T$ be a $2$-dimensional simplex in $\mathbb{R}^2$. A circle $C(x,y,r) \subset \mathbb{R}^2$ is given by its center $(x,y) \in \mathbb{R}^2$ and radius $r\ge 0$. Show that the set of circles in $T$ can be considered a $3$-dimensional simplex in $\mathbb{R}^3$.

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  • $\begingroup$ This is incomprehensible. Maybe a link to the original would help us to decipher it. $\endgroup$ – Gerry Myerson Jun 30 '13 at 12:31
  • $\begingroup$ Just post it and someone who knows Danish can translate it. $\endgroup$ – user70962 Jun 30 '13 at 13:57
  • $\begingroup$ Danish text: Lad nu T betegne et 2-dimensionalt simplex (en trekant) i R2. En cirkel C = C(x; y; r) R2 er givet ved dens centrum (x; y) 2 R2 og radius r 0. Vis at mngden af cirkler indeholdt i T kan opfattes som et 3-dimensionalt simplex i R3. SIDE 2 $\endgroup$ – matok Jun 30 '13 at 14:01
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    $\begingroup$ I think if you just change "amount" to "set" in the question, it becomes sensible. To be really accurate, consider the set of triples $(x,y,r)$ such that the circle with center $(x,y)$ and radius $r$ lies insider the given triangle $T$. The problem is to show that this set of triples is a $3$-simplex. $\endgroup$ – Andreas Blass Jun 30 '13 at 14:57
  • $\begingroup$ I think Andreas is on to something here. I will not vote to reopen the question just yet (until it is edited), but I will not vote to leave it closed, either. $\endgroup$ – tomasz Jul 1 '13 at 12:16

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