**The cube**

**Yes, the pentominoes cover a closed surface!**

**A cube, which has the edge equal to square root of 10, has an area of:**

**SQR(10) * SQR(10) * 6 = 60**

**The twelve pentominoes also. Now we try a solution at least. And this is a solution:**

**You could check the surface in the following picture. The pentominoes has been inclined of arctan(1/ 3) and refolded on the edges. I don't know how many solutions have the problem. If anyone knows it or knows how to calculate it, please send me an e-mail.**

**I greatly enjoy seeing the pentominoes cover a closed surface! Imagine, now, to blow up the cube like a balloon, a sphere. This also will be covered with pentominoes. But we must tame the definition, not 'Squares', but 'Quadrangular Segments of Spherical Bowl'. (like a field on the Earth). Personally I'll paint a soccer ball like the picture below.**

**The next picture tells the same things, but it's a grapho. It symbolizes the edges of a polyhedron with 60 quadrangular faces. This polyhedron has 62 vertices, 54 with 4 edges and 8 with 3 edges. The edges are obviously 120. The 60th quadrangle has the same perimeter of the grapho and it's closed to infinity. You could recognize the deformed twelve pentominoes in the picture.**

**With regard to infinity, I ask: could the twelve pentominoes tile the plane, without constitute rectangles or separation straight lines? The answer is yes. And each pentomino has his symmetric.**

**In how many ways? I don't know.**

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