Let us consider $n$-dimensional simplex $K$ in $n$-dimensional Euclidean space. Let $r_0$ be the radius of the inscribed sphere of $K$, and let be $r_1, r_2, \cdots, r_{n+1}$ be each radius of the exsphere of $K$.

Then, here is my question.

Question : How can we represent $r_0$ by $r_1, r_2, \cdots, r_{n+1}$?

Remark : This question has been asked previously on math.SE without receiving any answers.

Motivation : I've known the followings :

In the $n=2$ case, $$r_0=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}}.$$ In the $n=3$ case, $$r_0=\frac{2}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}}.$$

Then, I reached the following conjecture (of course, this is just a conjecture) : $$r_0=\frac{n-1}{\sum_{i=1}^{n+1}\frac{1}{r_i}}.$$

I don't have any good idea for $n$ in general. Can anyone help?

  • $\begingroup$ What is an escribed sphere? You might want to define this in the question. $\endgroup$ – Igor Rivin Nov 24 '13 at 15:50
  • $\begingroup$ @IgorRivin: Thank you for pointing it out. I hope this is better. $\endgroup$ – mathlove Nov 24 '13 at 16:09

Let $S_1,\dots,S_{n+1}$ be the $(n-1)$-dimensional volumes of the corresponding faces, and $V$ be the $n$-dimensional volume of the simplex. Then $\displaystyle V=\frac{r_0S}n=\frac{r_i(S-2S_i)}n$, where $S=\sum_i S_i$. Thus $$ \frac {n-1}{r_0}=\frac {(n-1)S}{nV}=\frac{\sum_i (S-2S_i)}{nV} =\sum_i\frac{1}{r_i}, $$ as required.

  • $\begingroup$ Sorry, I can't understand why $\frac{r_0S}n=\frac{r_i(S-2S_i)}n$ ? $\endgroup$ – lanse7pty Apr 16 '18 at 15:39
  • $\begingroup$ The relation $V=\frac{r_i(S-2S_i)}n$ is obtained in the same way as the area of a triangle is expressed via its exradius. $\endgroup$ – Ilya Bogdanov Apr 22 '18 at 5:48

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