Cutting a spherical surface into mutually non-congruent pieces of equal area Question: For what values of integer $n$ can the surface of a sphere be partitioned into $n$ convex and mutually non-congruent pieces of same area? (convexity could be viewed as geodesic convexity). If such a partition exists for some $n$'s, one can ask if one can achieve a 'fair' convex partition - pieces with same area and same perimeter - for some of those n's.
Note: one can replace sphere by general ellipsoids or tori or polyhedrons.
Speculation: Even a negative result like: "the spherical surface (or maybe even any ellipsoid or say, toroid) cannot be cut into any number of mutually non-congruent convex pieces of same area (or for that matter, 'same area and same perimeter')" could be interesting.
 A: There is no solution for $n=4$, by the following result.
Proposition. If $ABC$ is a spherical triangle of area $\pi$, then there is a partition of the sphere into four copies of $ABC$, and that is the unique partition into four triangles of area $\pi$ with $ABC$ as one of the triangles.
Proof of existence. By the relation of area and excess, the angles within $ABC$ sum to $2\pi$. So consider a point $P$ with $\angle ABP= \angle BCA$ and $BP=AC$. Then \begin{align}
\angle PBC
&= 2\pi - \angle ABC - \angle ABP \text{, since the angles at }B\text{ sum to }2\pi\\
&= 2\pi - \angle ABC - \angle BCA \text{, by construction}\\
&= \angle CAB,\phantom{2\pi - \angle ABC - }\text{ since the angles in }ABC\text{ sum to }2\pi.
\end{align}
From this equality we can prove others to show that the triangles $ABC$, $APB$, $BPC$, $CPA$ are congruent, and therefore all of area $\pi$. The final result has the following diagram —- where, because this is a sphere, the area outside the triangle is also a triangle.

Proof of uniqueness. Suppose $ABC$, $AOB$, $BOC$, $COA$ are all of area $\pi$. If $AOB \subset APB$ or $BOC \subset BPC$ or $COA \subset CPA$ then the triangle with $O$ would have area of less than $\pi$. Since this is impossible, $O$ must avoid all those inclusions, and therefore $O=P$. $\square$
