# Cover 3*2.6 rectangle with the least number convex polygons cut from sheets of given size 1.2*2.4

I want to cover a rectangle of given dimensions ($3\times2.6$) with a minimal number of convex polygons cut from rectangular sheets of material of given size ($1.2\times2.4$). These polygons don't have to be similar. How can I find the minimal number of sheets and tiling?

The smallest I've got to is with 5 rectangles cut out from 4 sheets of OSB:

• What's with the VTC? This is pretty well defined. Anyways, I would try to see if you can cover the rectangle with an overlap of the 4 smaller ones, perhaps trying diagonal orientations, and if you can't, you've proven that 5 pieces is the minimum. – greenturtle3141 May 6 '17 at 2:52
• What is VTC? Well defined is what? – Gleb May 6 '17 at 22:47
• Someone Voted To Close your question because they thought your question was unclear. – greenturtle3141 May 6 '17 at 23:12

I do not think it is possible, but it I'm finding it rather hard to prove rigorously. Here is my attempt. I have scaled everything by a factor of 10 to make it all integers.

We want to cut 4 pieces from the four 12x24 rectangles such that they form a 26x30 rectangle.

We can assume that each of the four pieces is cut from a different rectangle. If it is possible for those pieces to form the goal rectangle, then we may as well not cut them out and use the four rectangles themselves to cover the goal as long as we allow them to overlap.

Consider the 8 points marked in red in this picture:

It is straightforward but tedious to prove that you can cover no more than two of them by a 12x24 rectangle. The diagonal of 12x24 is $12\sqrt{3} \approx 26.83$, so can just about cover the short edge of the goal rectangle, but there is not enough width left to include a third red point. Therefore each rectangle must cover exactly two of the points.

Now look at the 5 points marked in blue in this picture:

No rectangle can cover three blue points (and two reds). Obviously we need at least one rectangle to cover two of the blue points. There are essentially only three ways to do this.

It seems that none of these allow the remaining area to be covered by the other three rectangles. (This assertion still needs proof.)

The best I managed to get is the following:

There are two very small uncovered triangles at the arrows.

• You can cover 3 red points with single rectangle. You need to put one point in the corner and then you have the question of whether you can put 2 points on the opposite side inside. Long side of that is 26.02, so you have about 0.8 left and if you put it symmetric you have just enough (note that diagonal is going through the middle of 2 points). I believe this is quite useless for covering the thing though. Your final covering idea seems to be good, maybe you could try rotating the top and bottom rectangle a little bit and moving these 2 tilted ones up/down slightly? – Zizy Archer May 8 '17 at 11:08
• @ZizyArcher I don't think you can cover 3 red points. The diagonal is $12\sqrt{5}-26 \approx 0.832$ longer than the side of $26$, and the two red points are $2.5$ apart. The grid lines in my pictures are 2x2, which may have misled you. It may indeed be possible to rotate the two horizontal rectangles anti-clockwise slightly to cover more of the top/bottom edges, but there is not much room in the middle for it. This may allow the tilted rectangles to be tilted a bit more and maybe cover a bit more of the currently uncovered bits. I'll have to do some calculations to see if that is enough. – Jaap Scherphuis May 8 '17 at 12:13
• Yeah, you are right - I considered them to be 1x1. If the spacing is 2.5, diagonal of these 2 points is 26.12, which means you have 0.7 left, which split by the middle gives you only about 1 unit worth of coverage instead of needed 2.5. – Zizy Archer May 8 '17 at 12:20