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Equal angles in tetrahedron


If the angles between the three pairs of opposite sides of a tetrahedron are equal, prove that these angles are right.

Solution We use the identity \vec{AB}\cdot \vec{CD}+\vec{AC}\cdot \vec{DB}+\vec{AD}\cdot \vec{BC}=0 . Since the angles between opposite sides is the same for all three pairs of opposite edges, let’s say \alpha, we can conclude from the inner product properties that
\cos \alpha (\pm AB\cdot CD \pm AC\cdot BD \pm AD\cdot BC )=0.
Now suppose the second factor is 0. Then we can’t have all signs to be + or all to be – and we will get a formula like AB\cdot CD - AC\cdot BD + AD\cdot BC =0 or something equivalent with this. So we have equality in Ptolemeu’s inequality, which means that A,B,C,D are in the same plane and on the same circle, which is a contradiction with the fact that ABCD is a tetrahedron.
By the arguments above, the second factor of our product can never be 0, meaning that \cos \alpha=0, which means that all the angles between opposite sides are right ( 90^\circ )

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