 ... can be folded onto the surface of an octahedron in a way that perfectly covers the entire octahedron with no gaps and no overlaps.

How can it be done?

• Another puzzle for @WeatherVane, perhaps? Feb 23 '20 at 8:02
• How do you come up with these? :-) Feb 23 '20 at 10:45
• @ÉbeIsaac sadly I missed it, but there is a good answer. Feb 23 '20 at 17:56
• @Randal'Thor At this point, my steps are: 1. Create a simple primitive (cube, octahedron, etc.) in a 3D modeling program (Wings3D, in this case), 2. Divide the faces of the primitive into a diagonal grid. 3. Experiment with different ways to cut along the grid lines until I find something aesthetically pleasing. 4. Use UV-unwrapping to get a distorted, imprecise version of the flattened shape. 5. Redraw a pixel perfect version of the shape in a bitmap editor (Krita, in this case). 6. Profit! Feb 23 '20 at 20:01

area 56 little triangles, and an octahedron has 8 sides. Therefore the side of the octahedron must be sqrt(7) little-triangle-sides. Since $$7=1^2+1\cdot2+2^2$$, a line of length $$\sqrt7$$ may be had by going two units in one direction and one in a direction $$120^{\circ}$$ away from it. We would like the $$\sqrt7$$-triangle grid we work with to have the centre of the shape at the centre of one triangle, and the little projecting nubs at the centre of another so that they can fit together nicely.