# Tag Info

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If I'm not mistaken, I believe that it is Because

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Started by scaling the goal, I put one-by-one with some trial-and-error in the end to find this solution:

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No, you cannot:

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Apologies for MS Paint quality. Numeric feature:

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Having a Blokus Trigon set at hand was a big help! I arrived at a different solution (looks like there's more than one after all): My approach was rather straightforward, here's how I went around it: Hint 1: Hint 2:

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Here's one 5 piece solution: To make this into a square: To make it into an equilateral triangle: Here's a different 5 piece solution: Here's an answer with You will see that However, which is still a problem, it's possible to just solve it the original way without any flips. To solve this, you can simply take a symmetrical sliver out of the blue piece ...

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@Daniel_Mathias gave a very helpful link which has all the 12x5 solutions in a text file. So some simple code allows us to see that of the 1010 12x5 solutions, there are 264 with 1 straight cut. But, sadly, none with 2 or more cuts. A few examples of the former are: 12 FFPPP IIIIINN LFFPP ZZWNNNT LFXUU VZWWTTT LXXXU VZZWWYT LLXUU VVVYYYY 81 ...

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There are no solutions in three rectangles. For an index of all pentomino solutions: Pentominos home Index of 5x12 solutions A particular solution you may be interested in: #747

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Here's simple 2-D pattern that seems to tile quite efficiently: The area of the each tile (blue square) is $21\times21 = 441$ tiles, and it contains $4\times14=56$ generators tiles, for a ratio of $\frac{56}{441} \approx 12.7\%$ The trick here is that The final pattern looks like this: POST-TICK EDIT: managed to find an even better pattern with $\mathbf{... 2 I was going to post a different answer, only to realize it was the same as Magma's. So I had to find a new one: 2 Like @Omega Krypton I have solved one part, Task 2: 2 The beginnings of a solution, where warmer colors correspond to areas, and cooler colors to perimeters (Last updated 6-26): Reasoning: Cont. I believe that also 2 I decided to place the wind generators together pinwheel-fashion: The repeated section looks like this: To calculate the efficiency: There are$70+70+16+9 = 165$empty squares,$4\cdot14 = 56$filled squares, for an efficiency of$\frac{56}{221} = 25.339\%\$. Sadly this is not quite as efficient as Bass's solution, but ever so close. In my first attempt I ...

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As mentioned in the question, here is an example of an answer: This 20x7 tile setup can be validly tile-replicated, as the required open space of the left generator which is "out of bounds" correctly loops to the right to coincide with the open space of the right one: Since we've established validity, the ratio is: 28 generator tiles/140 total tiles=20%

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The answer is no, and not only for rectangular shapes, but any shape of room, with the assumption that there exists a tiling.And the proof is as follows:

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