I had a typical A4, 80 gsm sheet of paper and I had determined that its dimensions were: 210 x 297 x 0.11 mm. The last dimension was obtained using an old micrometer I possess and is close to the average of four measurements taken in various places on that piece of paper. Thus the volume of this sheet of paper is approximately 6861 mm cubed and its surface area slightly more than 124740 mm squared. This gives a surface area to volume ratio of about 18.2 to 1. The smallest surface to volume ratio value would be to form a sphere with the material that makes up the sheet of paper. This is trivial to calculate: the radius is close to 11.8 mm, the surface area is 1746 mm squared and so the surface to volume ratio is a little over 0.25 to 1. To the best of my knowledge, it is not possible to fold a sheet of paper into sphere in the conventional meaning of 'fold'. The challenge is this: What is the smallest surface area to volume ratio a sheet of A4 80 gsm can be folded into such that it can be readily unfolded back into its original shape (not including creases) and critically, when in its folded shape, no objects like hefty books are used to confine it to its largest three dimensions. What do I mean by the largest three dimensions? Well, basically the measurement of its length by width by breadth so that nothing sticks out. I don't see why a clever designer couldn't use the paper's own weight to help compress itself when sitting on a desk as this aspect is, to some extent, inescapable in practice. But I'm thinking a folded sheet of paper should be able to maintain it's folded shape or rather, maximum surface area as defined by its largest three dimensions, for, say, five minutes. No tape. Obviously I'm not including surface area inside the folds. That would defeat the purpose of the challenge as the surface area of a folded sheet of paper does not, in fact, change. As you will have already deduced, a criteria where the surface area is determined by the largest three dimensions, would mean that if a sphere with a radius of 11.8 mm could actually be constructed by folding, it would be then be determined to have a surface area of about 3342 mm squared and a volume of about 13144 mm cubed. Nearly twice the actual surface area and volume! But therefore, a surface area to volume ratio not dissimilar to that of the sphere of a little over 0.25. I just can't think of a more convenient way of determining a surface area of a folded object. Most of the more low ratio structures I've constructed have been cuboid anyway. So using the 'volume as defined by the largest three dimensions' rule, the lowest ratio achievable would be 2166:6861 or about 0.32. My best efforts to date is getting close to a ratio of 2 to 1. I have, in some attempts, used a weighty book to hold my folding in place, but made my measurements of its three largest dimensions after removing the book and then after five minutes to observe if my compressed structure was inclined to expand. This challenge offers no prizes except the gratification in the knowledge that you've broken the 2 to 1 ratio barrier that has eluded me. I'd be interested in knowing the sequence and types of folds used to accomplish it. If anyone's interested, I'll post the folds that got me close to the 2 to 1 ratio. And yes, I do realise there are better things to do with one's life than fold paper. But you never know where these sort of things can lead. Perhaps people used to say similar things about the quest for determining large prime numbers.