The following tri-bladed boomerang shape:
... 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?
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Sign up to join this communityThe following tri-bladed boomerang shape:
... 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?
The shape has
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.
This leads us to the following picture:
having drawn which, the easiest thing is to cut it out and Just Do It. I hope you will believe (since it's true) that actually solving it to make the untidy thing shown below was less effort than faking it would have been :-).