Equilateral/Equiangular Quadrilaterals on a Sphere

Question

I have recently been taking a Geometry class and I am a little bit confused about spherical geometry. I know there can be equiangular spherical quadrilaterals but does this also imply that the quadrilaterals are equilateral?

Answer 1

Here is an example of an equilateral quadrilateral with unequal angles:

Let the angles in two side-sharing equilateral triangles $ABC$ and $DBC$ be $2\pi/5$; then by the law of cosines, the sides will all be $\pi/5$. Let the shared side be $BC$.

Angle $G = \angle ABD$ and the long diagonal $g = AD$ are determined from triangle $ABD$ which is triangle $AGD$, with sides $AG = DG = a = \pi/5$ and angles $DAG = ADG = \frac12 \angle BAC= \frac{A}2 = \pi/5$, as follows:

By the law of sines, $$ \frac{\sin A/2}{\sin a} = \frac{\sin G}{g} \implies \frac{\sin G}{g}=\frac{\sin \pi/5}{\sin \pi/5} = 1 $$ so either $g=G$ or $g = \pi - G$. The $g=G$ solution in the context of the quadrilateral gives a degenerate quadrilateral where $A$ and $D$ coincide (two overlapping triangles) so we take $g = \pi - G$.

By the law of cosines, $$ \cos g = \cos a \cos a + \sin a \sin a \cos G \implies -\cos G = \cos^2 a + sin^2 a \cos G\\ \cos G = -\frac{\cos^2 a}{1+\sin^2 a} = -\frac{\cos^2 (\pi/5)}{1+\sin^2 (\pi/5)} $$ which simplifies to $$ G = \cos^{-1}\left(- \frac{11+4\sqrt{5}}{41} \right) \approx 119.1^\circ $$

Thus we have (in degrees) a quadrilateral $ABDC$ whose angles are respectively $72,119.1,72,119.1$, with all four sides = $\pi/5$.