Centroid and Circumcenter in Isosceles Tetrahedra

Let $AB=CD=a,\,$ $AC=BD=b,\,$ $AD=BC=c.\,$ We have $\displaystyle \overrightarrow{AG}=\frac{\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}}{4},$

so that

$\displaystyle\begin{align} AG^2 &= \left(\frac{\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}}{4}\right)^2\\ &=\frac{AB^2+AC^2+AD^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}+2\overrightarrow{AC}\cdot\overrightarrow{AD}+2\overrightarrow{AB}\cdot\overrightarrow{AD}}{16}\\ &=\frac{a^2+b^2+c^2+a^2+b^2-c^2-a^2+b^2+c^2+a^2-b^2+c^2}{16}\\ &=\frac{a^2+b^2+c^2}{8}, \end{align}$

implying $\displaystyle AG=\sqrt{\frac{a^2+b^2+c^2}{8}}.$ Due to the symmetry of the expression, we'd find that $\displaystyle AG=\sqrt{\frac{a^2+b^2+c^2}{8}}=BG=CG-DG,$ making $G\,$ the circumcenter.

$\displaystyle\frac{\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}}{4}\,$ such that

$\displaystyle\begin{align} R^2 &= \frac{AB^2+AC^2+AD^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}+2\overrightarrow{AC}\cdot\overrightarrow{AD}+2\overrightarrow{AB}\cdot\overrightarrow{AD}}{16}\\ &=\frac{3(AB^2+AC^2+AD^2)-(BC^2+BD^2+CD^2)}{16}. \end{align}$

Similarly,

$\displaystyle\begin{align} R^2&=\frac{3(AB^2+BC^2+BD^2)-(AC^2+AD^2+CD^2)}{16}\\ R^2&=\frac{3(AC^2+BC^2+CD^2)-(AB^2+AD^2+BD^2)}{16}\\ R^2&=\frac{3(AD^2+BD^2+CD^2)-(AB^2+AC^2+BC^2)}{16}. \end{align}$

All of these give us $AB = CD,\,$ $AC = BD\,$ and $AD = BC.$

Indeed, let $AB^2=u,\,$ $AC^2=v,\,$ $AD^2=w,\,$ $CD^2=m,\,$ $BD^2=n,\;$ and $BC^2=p.\, From

$3(u+v+w)-(p+n+m)=3(u+p+n)-(v+w+m)$

obtain $3(v+w)-(n+p)=3(n+p)-(v+w),\,$ and, subsequently, $v+w=n+p.\,$ Similarly, we get $w+u=p+m\,$ and $u+v=m+n.\,$ The three add up tp $u+v+w=m+n+p,\,$ and by subtracting $u=m,\,$ $v=n,\,$ and $w=p.$

The statement and the above proof have been communicated to me by Leo Giugiuc (30 Nov, 2016).