I want to construct two quadrilaterals on the unit sphere with same interior angles $\alpha_1,…,\alpha_4$ and the same perimeter, but which are not congruent to each other. Is that possible? How can one construct a counterexample or how can one proof that this is impossible, respectively?
For triangles this is impossible. But I have no idea how to examine this situation for quadri
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The answer is that you can find quadrilaterals on a sphere that have the same angles in the same order, and have the same perimeter, but are not congruent. This is a lot like finding parallelograms in a plane that have the same angles and the same area/perimeter-squared ratio, and the same area, but are not congruent.
In the plane, that problem is easy: The $A/P^2$ ratio is a function of the height/width ratio, which attains a maximum (for a given corner angle $\theta$) at some ratio $r_{\max}$, and falls off smoothly near that optimal ratio. Plot $A/P^2$ versus $r$, draw a line horizontal cutting the curve near the maximum, and read off two values of $r$ that can share the same area and perimeter but will not be congruent parallelograms. Although the algebra is fairly easy here, I have intentionally not provided it because in the spherical case, it won't be easy.
Let's call two such non-congruent parallelograms with equal areas, perimeters, and angles "fraternal twin parallelograms."
On the unit sphere, consider a family $\{q_i\}$ of quadrilaterals which may have different areas $A_i$ (and therefore different interior angle sums $S_i = A_i +2\pi$ since the angle sum depends on the area) but which all share the same ratios of angles to that sum: $$ \begin{array}{cc} \theta^A_i = \alpha S_i & \theta^B_i = \beta S_i \\ \theta^D_i = \delta S_i & \theta^C_i = \gamma S_i \\ (\alpha + \beta + \gamma + \delta = 1) \end{array} $$ If we find two non-congruent $q_i, q_j$ in that family sharing the same $S_i = S_j$ we will have found two quadrilaterals satisfying your proposition.
Let's specify the family's $\alpha, \beta, \gamma,\delta$ as fixed values such that $\alpha = \gamma$, $\beta = \delta$, and $\alpha \neq \beta$. Now choose among that family of quadrilaterals two very small ones which are close to a pair of planar fraternal twin parallelograms, having a corner angle of $2\pi\alpha$. . These can in general be quite different in height/width ratio, and they will have very nearly equal areas and perimeters on the sphere. But since these are of non-zero size, in general the areas or perimeters may disagree on the sphere.
But we have two parameters to adjust: the height-width ratio of the starting parallelogram we had projected onto the sphere, which will push the area to perimeter-squared ratio in one direction or the other; and the overall size chosen in that projection, which will push the area and the square of the perimeter in the same direction. By small adjustments to both of these for one of the parallelograms, we can achieve a match in both area and perimeter, while remaining in the same family of quadrilaterals.
Therefore, there exist fraternal twin pairs of spherical parallelograms.
Unlike the case for planar quadrilaterals, you can't find non-congruent pairs of equi-perimetric quadrilaterals on the sphere both of which have all four angles equal.
Here are a pair of non-congruent spherical quadrilaterals having the same angles. $$ \begin{array}{c|c} Q_1 & Q_2 \\ (\theta = 85.7867979^\circ, \phi = -149.7665879^\circ) & (\theta = 85.4337583^\circ, \phi = -131.4321518^\circ) \\ (\theta = 87.0198508^\circ, \phi = -45.3478986^\circ) & (\theta = 86.5371734^\circ, \phi = -81.18809126^\circ) \\ (\theta = 85.7867979^\circ, \phi = +30.23341212^\circ) & (\theta = 85.4337583^\circ, \phi = 48.56784815^\circ) \\ (\theta = 87.0198508^\circ, \phi = +134.6521011^\circ) & (\theta = 86.5371734^\circ, \phi = +98.81190874^\circ) \\ \mbox{Q1 Angles} & \mbox{Q2 Angles} \\ 70.06877926 & 70.06877926 \\ 110.1435593 & 110.1435593 \\ 70.06877926 & 70.06877926 \\ 110.1435593 & 110.1435593 \\ \mbox{Q1 Sides (radians)} &\mbox{ Q2 Sides (radians)} \\ 0.10005 & 0.06196 \\ 0.07875 & 0.12712 \\ 0.10005 & 0.06196 \\ 0.07875 & 0.12712 \end{array} $$
The way I got this, was to start with a rhombus cetered on the origin of a plane, with side 1 in the X direction and a 70 degree angle at the lower left; this implies a height of .9397. The rhombus has maximal area to perimeter-squared ratio for all 70-degree parallelograms. Now I form parallelograms P1 by reducing the height by 0.2, and P2 by starting from an increased height by 0.2, and then adjusting the height such that P1 and P2 have the same $A/P^2$ ratio; the height of P2 comes out to 1.19376.
Now I parallel project both of these onto a sphere of radius 10, which gives two quadrilaterals with slightly different angular excess (and hence forced to have different angles). I now adjust the $70^\circ$ lower-left angle in P2 by the discrepancy in the corresponding angle in the quadrilaterals (and projecting again); since there is still an angular difference discrepancy, the other angle is still different. Now I deal with the angular discrepancy, by decreasing the the area of P2 keeping the height and angles constant. This gives two quadrilaterals with equal angular excess, but spoils the exact agreement on the "$70^\circ$" angles.
Two further iterations of this using an approximate Newton's algorithm gives the spherical quadrilaterals above, which as you can see have angles that match to 8 decimal points and have quite different side lengths.