Ratio of volumes of n-simplexes

Question

I have a simplex $S \subset \mathbb{R}^n$ created by unit vectors $v_1, v_2,\ldots,v_{n+1}$ such that $v_i\in \mathbb{R}^n$. Let us fix $v_{n+1}$ and consider the other vectors. Clearly the volume of simplex is $\frac{|\text{det}(L)|}{!n}$ where $L=[v_1-v_{n+1}, v_2-v_{n+1},\ldots,v_n-v_{n+1}]$. Now choose $v_i,v_j$ such that $\|(v_i-v_j)\|_2$ is maximum and $v_i,v_j \neq v_{n+1}$. Take $u=0.5(v_i+v_j)$. Let us consider two new simplexes $S_1$ and $S_2$ which are created from vectors of $S$ such that $S_1$ contains $u$ instead of $v_i$ and $S_2$ contains $u$ instead of $v_j$. Can we say something about vol$(S_1)$/vol($S$) ? Any possible upperbound on this ratio ?

Answer 1

You may assume $v_{n+1}=0$. Write $\Lambda$ for the volume form ("det" in your question). Then $$\Lambda\left({v_1+v_2\over2},v_2,\dots,v_n\right)={1\over2}\Lambda(v_1,v_2,\ldots,v_n)+{1\over2}\Lambda(v_2,v_2,\ldots,v_n)\ .$$ Here the second term on the right hand side is $=0$ since it contains two equal entries. It follows that the ratio you are interested in is always $={1\over2}$.