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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asked Feb 23, 2020 at 6:58
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$\begingroup$ Another puzzle for @WeatherVane, perhaps? $\endgroup$– Ébe IsaacFeb 23, 2020 at 8:02
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$\begingroup$ How do you come up with these? :-) $\endgroup$– Rand al'ThorFeb 23, 2020 at 10:45
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$\begingroup$ @ÉbeIsaac sadly I missed it, but there is a good answer. $\endgroup$– Weather VaneFeb 23, 2020 at 17:56
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1$\begingroup$ @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! $\endgroup$– plasticinsectFeb 23, 2020 at 20:01
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1 Answer
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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 :-).
answered Feb 23, 2020 at 15:26
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$\begingroup$ I keep making these harder and harder, and still someone manages to nail the solution within a few hours. Excellent work! $\endgroup$ Feb 23, 2020 at 19:47
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3$\begingroup$ Now I'm scared about what the next one might be like. Some Banach-Tarski monstrosity, perhaps :-). $\endgroup$– Gareth McCaughan ♦Feb 23, 2020 at 20:46
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$\begingroup$ @GarethMcCaughan I look forward seeing origami-version of it :):) $\endgroup$ Feb 23, 2020 at 22:25