A tetrahedron is called isosceles if each pair of edges which do not share a vertex have equal length.
Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.
Let \(OABC\) be an isosceles tetrahedron and let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\).
By considering the lengths of \(OA\) and \(BC\), show that
\[2\mathbf{b}.\mathbf{c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2.\]
Show that
\[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
Let \(G\) be the centroid of the tetrahedron, defined by \(\overrightarrow{OG} = \frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})\).
Show that \(G\) is equidistant from all four vertices of the tetrahedron.
By considering the length of the vector \(\mathbf{a}-\mathbf{b}-\mathbf{c}\), or otherwise, show that, in an isosceles tetrahedron, none of the angles between pairs of edges which share a vertex can be obtuse. Can any of them be right angles?