Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}^3$.

share|cite|improve this question
This is incomprehensible. Maybe a link to the original would help us to decipher it. – Gerry Myerson Jun 30 '13 at 12:31
Just post it and someone who knows Danish can translate it. – BU982T Jun 30 '13 at 13:57
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 – matok Jun 30 '13 at 14:01
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. – Andreas Blass Jun 30 '13 at 14:57
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. – tomasz Jul 1 '13 at 12:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.