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.


The combined area of the X's is The solution:


Here's a diagram showing both the parts of the shape that make up each face of the cube in its own colour, and trying to give some indication of how they meet up when folded.


Ernie's jigsaw puzzle isn't as straightforward as it seems, as it's actually: One way of assembling the pieces legally is: How will you know when you have succeeded? PS Ernie definitely has a sense of humour about him. After all, when you texted him with "Bored", he replied...


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


Here is the cube I made: Update:


The cube can be covered by


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


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


I made folds like this: These puzzles are getting harder and harder. In this case it was impossible for me to eyeball where any face would be, and it only started to make sense once I started cutting and folding.


As I don't have a camera handy, I have had to unfold my (pink) cube before I could show it to you. Its sides are Sqrt(52) = Sqrt(4^2+6^2) units long.


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


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 \...


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 ...


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:


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 ...


It could be done with some slight folding of the paper, be careful though.


Here is a 6-step way:


I can do it with 11 folds also.


The shape has 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 :-).


The golden ratio has been described as something popping up all over nature, science and arts, so we shouldn't be too surprised at bumping into it here. Implementation in terms of actual folding. With the benefit of hindsight I notice that this is very similar to textbook construction of golden ratio: Pictures Alternative proof that $h^2 = 1/3$. I'll ...


If I'm not mistaken, it can be done as follows:


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

Only top voted, non community-wiki answers of a minimum length are eligible