Is it possible for a right triangle (drawn on some specific topology) to have all sides with equal lenth?

Question

Consider a triangle ABC drawn on some surface (for example, a soccer ball or top hat). Is it possible for lengths AB, BC, and CA to be equal while the angle between two of the lines is 90 degrees?

I am not very skilled with geometry, but I want to learn more! Here is my best attempt so far:

I assume that:

1. There is a right angle between lines AB and CA: $\vec{AB} \cdot \vec{CA} = 0$

2. The triangle takes us on a closed path: $\vec{AB} + \vec{BC} + \vec{CA} = 0$

3. The sides are of equal length: $|\vec{AB}| + |\vec{BC}| + |\vec{CA}| = 0$

I think the next step is to assume some geometry. I think I can map the surface height from the Cartesian coordinates as $s(\vec{x})$ (since $s$ would just be the height of the surface it is just a scalar. I envision $\vec{x}$ as having two dimensions). Since the lines are drawn on the surface, they should also be functions of $\vec{x}$. It seems like I should then rewrite the dot products somehow.

This is where I'm stuck. I'm not sure if it is even fair to define dot products of vectors on a surface. If I walk in a straight line on the surface, I can trace my path to form one line of the triangle. It seems like an OK assumption that those "straight line paths" can be parameterized using some small number of inputs, which is what $AB$ would represent.

I've looked through the first fundamental theorem of differential geometry but I'm not sure if that is relevant to my problem. It seems like it contains important information for my problem, but I have not been able to exploit it.

What would be my next step in solving this problem? Is it possible to construct a geometry that allows for a right angled triangle with three sides possible?

Answer 1

Sure it is, as is clear in this figure:

Answer 2

On a sphere $S^2$ you can draw a triangle with all angles $\pi/2$ (the sides are parts of the equator and of two orthogonal meridians).