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The following rather squashed and bullet-riddled lowercase lambda:

a squashed and bullet-riddled lowercase lambda

...can be wrapped onto the surface of a cube in a way that perfectly covers the entire cube, with no gaps and no overlaps.

How can this be done?

plasticinsect
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    I could be wrong but the sum of all the black squares does not come with an exact cube, i.e. (all black squares / 6) should be x ^ 3, where x is the size of the side of the cube. – rhsquared Aug 03 '20 at 09:20
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    @rhsquared I think it should be x ^ 2, because the size of the cube is just a aquare. – athin Aug 03 '20 at 09:34
  • @athin, sorry, you are right. That's what I meant. All squares / 6 should give a number which is exact square. – rhsquared Aug 03 '20 at 09:36
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    @rhsquared That's only if you assume that the side of the square needs to be an integer multiple of the reference squares which does not necessarily need to be true (see previous questions by OP) – hexomino Aug 03 '20 at 09:44
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    @rhsquared If the area seems impossible, it's because you have not had a certain important insight needed for this puzzle yet. This is all part of the challenge. – plasticinsect Aug 03 '20 at 16:24
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    darn it all, I was just about to make a post with this exact title. ;) – Greg Martin Aug 03 '20 at 18:51

1 Answers1

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Here is the cube I made:

There are $246$ squares in the cube surface.
That gives $41$ squares per face.
So the side length is $\sqrt{41}$.
This is the hypotenuse of the right triangle $4 \times 5$
and shows how the paper must be slanted.
The cube was made 'on the fly' folding as needed.

enter image description here

These cubes are getting harder to solve and construct.
There are some small 'ears' that I can't easily fold and join by hand.

Update:

This shows how the paper should be folded. You can see that the folds slightly miss some of the grid points, leaving several tiny ears that need to be folded into a corresponding hollow.

My first fold was found by noticing that the two blocks (top and bottom) each has a $4 \times 5$ right triangle with the hypotenuse joining an internal angle and an external angle, so there were my first two edges. The rest followed from there.

enter image description here

I also showed graphically how the area of each face is $41$ with four blue $4 \times 5$ right triangles. Each has an area of $10$ and there is a $1$ unit hole in the middle.

Weather Vane
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    +1 for SUPERB explanation! – Jeremy Dover Aug 03 '20 at 11:29
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    Excellent work! Yes, the fiddly little corners may be difficult fold down perfectly, but I don't think it matters. They are so small, it's obvious that they fit. I'm honestly just really impressed that anyone can make these out of physical paper at all. – plasticinsect Aug 03 '20 at 16:49
  • The finished cube was only about 3cm size and needed tweezers to apply some of the sticky tabs inside it. Afterwards I realised I was working with an A4 size print that I had reduced a bit so that the grid pitch was an exact match for some quadrille ruled paper I have (to try different angles and alignments). – Weather Vane Aug 03 '20 at 17:23
  • Would it make things easier if you printed the shape split across multiple sheets? (taping them together into one shape after they've been cut out) Then you could make it as large as you want. – plasticinsect Aug 03 '20 at 17:36
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    I guess so, and if the next puzzle is "hairy tapeworm" I will have to. – Weather Vane Aug 03 '20 at 17:48
  • Oooh! Good idea! :) – plasticinsect Aug 03 '20 at 17:56