The shape Using this, we can make a guess for how the cube might be folded: Once that fold is done, the shape looks more like this: A drawing of the finished product: And an animation of the whole process:


This seems to work: Below, I printed out the shape, and cut off the excess. The white parts are for glueing; if everything works out as planned, all of them will be covered by the coloured bits around the black squares. Joy, it all worked! Here's the final cube, with some white "intentionally" showing through between the pieces, highlighting the borders: ...


The shape can be folded like this


I can do it in folds, by


I managed to make Like so:


My Answer: Red lines are where you would go, yellow lines indicate moving over the fold.


Why not open with a short riddle to start with—perhaps a couplet? For example, on the ears, you could write in suitably mysterious type: Here you are at number one To start, undo what has been done


You could have a small part of the QR code visible on the folded origami. Just small enough that it's not immediately obvious what it is at first.


Well, you beat me by 30 minutes, but I worked hard on this so I'm posting it anyway :)


This can be done To figure out how to do it, Here are some images of the covering:


The cube can be covered by


I can do it in just three folds and a single unfold. Proof that this is a rhombus:


I have managed to come up with Following are the images of the paper with each fold.


If folded as shown below, it can be closed along the matching letters, to form a 12 sided polyhedron with six quadrilateral faces and two sets of three isosceles triangular faces. The quads are arranged in two triplets that join to form two right corners (as found in a cube), and the triangles 'join' the two groups of quads together. The shape has 120 degree ...


If you had some sort of super paper that defied all laws of physics, and was $0.05 \text{mm}$ thick, then folding it a hundred times would give a thickness of: $0.05 \times 2^{100} \text{mm}$ $\approx 6.34\times10^{28} \text{mm}$ Note that the observable universe has a width of around: $8.8\times10^{29} \text{mm}$ Using the paper folding formula $W = \pi t \...


First, we need to figure what is the minimum amount of superconductor that we'll lose when making the cut. Optimally, we can cut the large cube net into two smaller cube nets without any waste at all. Let's start by trying to find such a cube net: The blue and yellow parts are pretty simple cube wraps, so after the cut, we can easily make the required two ...


Here is a 6-step way:


I can do it with 11 folds also.


I can do it in just: Initial configuration: First: We have: Now: We get the mark: Finally: You get: And the required distance is: Why this works:


This is the solution I intended, though I've accepted Penguino's Florian F's since it has the smallest number of folds. Dashed lines are mountain folds, and dotted lines are valley folds. Perhaps I'll make a 3D representation later, but the special thing about the resulting shape is that it can be made by cutting a cube into two halves as below, rotating ...


Here's a solution with 2 folds. Fold the 11" side in half to get a 5.5" X 8.5" rectangle. Fold the 4.5" side on the 8.5" side as shown. [see image for better understanding]




Game 1 A bit too late, but constructed differently.


Consider what happens by doing $2$ folds, one in each direction. So after $2k$ folds For this to be a cube we need: So the number of folds we need is In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which ...



This is a classic. Any zig-zag pattern will do e.g. (sorry for the bad drawing): To get the cut, first we have to concertina fold like this, leaving a bit on each end which we make sure not to cut under any circumstances (this can be achieved by folding the bits it out of the way of any cuts that will harm it: Then we need to fold the thin rectangle like ...


Using a long and thin rectangular strip, you can get up to Here's how:


There is no largest area, but the area can be made as close as wanted to 75. Let's imagine our rectangle. It has diagonal $AC$, centre $O$; let's say $AB$ is the longest side and $H$ is the middle of $AB$. The rectangle will be folded along a line $p$, which is perpendicular to $AC$ at point $O$. $p$ also crosses $AB$ at some point $M$. It is easy to note ...


Game 2

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