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Answer: Reasoning:


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


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


Partial answer, Tasks 2-4 Finished Task 2 Finished Task 3 (what a doozy) Finished Task 4 (easier than 3)


Note: Answer:


No, you cannot:



Apologies for MS Paint quality. Numeric feature:


Partial Answer Finished Task 1


Partial Answer - finished Task 1


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:


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 ...


Sooo much NSFW! But thanks anyway :) Task 1 Task 2 Task 3 Task 4 - not solved yet.


@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 ...


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


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{...


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:


Like @Omega Krypton I have solved one part, Task 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


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 ...


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%


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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