Isosceles Tetrahedra Characterization via Inequalities


Isosceles tetrahedron (i.e., the tetrahedron whose opposite edges are pairwise equal) has unexpectedly rich personality:

The list could be continued (the last two items may make you expect another one). Below I list three additional characterization collected by Leo Giugiuc. The first one is due to Gojko Klajdzic, the second to Marcin Mazur, the third is due to Kadir Altintas (all proofs are by Leo Giugiuc).

The theorems below specify inequalities between elements of the tetrahedron that become equalities only if the latter is isosceles. Where mentioned, $G\,$ is the centroid, $O\,$ the circumcenter, $d=OG,\,$ $R\,$ the circumradius of the tetrahedron.

$AG + BG + CG + DG \le 4\sqrt{R^2 - d^2}$

The proof can be found on a separate page.

$8R^2 \ge AB\cdot CD + AC\cdot BD + AD\cdot BC$

We have: $2AB\cdot CD \le AB^2 + CD^2,\,$ $2AC\cdot BD \le AC^2 + BD^2\,$ and $2AD\cdot BC \le AD^2 + BC^2,\,$ so that

$\displaystyle 2(AB\cdot CD + AC\cdot BD + AD\cdot BC) \le \sum_{all}AB^2 =16(R^2-d^2) \le 16R^2.$

Equality holds iff $AB = CD,\,$ $AC = BD\,$ and $AD = BC.$

$\displaystyle\sum_{all}AB\le 4R\sqrt{6}$

We have

$\displaystyle\begin{align} \left(\sum_{all}AB\right)^2 &\le 3(AB+CD)^2+3AC+BD)^2+3(AD+BC)^2\\ &= 3\left(\sum_{all}AB^2\right)+6(AB\cdot CD+AC\cdot BD+AD\cdot BC)\\ &\le 48R^2+48R^2, \end{align}$

implying $\displaystyle\sum_{all}AB\le 4R\sqrt{6}$.

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Copyright © 1996-2018 Alexander Bogomolny

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