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# Tag Info

341

It is impossible. Proof: Let the $7\times 28$ area be painted with black and white squares in a checkerboard pattern. Every piece will cover $2$ black and $2$ white squares, except the T-piece, which covers $3$ of one color and $1$ of another. Since there are $7$ T-pieces, a tiling that uses every piece cannot cover the same number of black and white ...

55

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TLDR: I'll fill the board and prove that the solution is unique. First, let's start by: I'll paint those green: Let's repeat those steps a few more times, using orange, blue, red and purple, in precisely that order: Now, We can easily fill the topmost white squares by that reasoning. They can't be filled in any other way: Now, let's look at: And by ...

22

I think that this tiling is a valid Tetris stack:

20

The trick to this puzzle is to: (And here are those tilings: the center was already given, and the rest are obtainable from these by rotation.)

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The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this or else you get this The latter can be easily improved by replacing the ones at the edges to give this A different way to get the same ...

19

How's this? I got this by square-izing the icosahedron graph:

14

2x2 - area 108 - optimal 2x3 - area 72 - optimal 2x4 - area 108 - optimal 3x4 - area 84 - optimal 1x5 - area 54 - optimal 2x5 - area 304 - optimal 3x5 - area 576 - optimal 2x6 - area 240 - optimal 1x7 - area 1034 1x8 - area 432 - optimal 3x8 - area 5880 1x9 - area 585 - optimal Note that by subdividing the yellow rectangles: 2x3 indirectly solves ...

13

Got the dissection part done. No idea about the native African word, though.

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Challenge 1: Fit the four yellow birds into the tray, no overlapping. Rotation and reflection are allowed. Challenge 2: Fit the four yellow birds and the blue piece into the tray, no overlapping. Rotation and reflection are allowed. Challenge 3: Fit the four yellow birds and the red piece into the tray, no overlapping. Rotation and reflection are allowed. ...

13

I think the answer is using the following coloring: For other board sizes, Reasoning:

12

Here is a proof that $12$ is the smallest possible number of regions in any feasible solution. Consider an arbitrary division of an arbitrary rectangle into $n$ regions, such that every region has exactly five neighboring regions. We translate this picture into a so-called planar graph: each of the $n$ regions then translates into a vertex/point, and there ...

12

Finally arrived at this solution after playing around on https://www.scholastic.com/blueballiett/games/pentominoes_game.htm for way longer than I care to admit!!!! :) First pair (UI-TF) Second pair (WX-PY) Third pair (VZ-LN)

12

First, a generalizable solution for $1 \times n$, $n$ is even. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no W-pentominos can be shortened. This is a way to tile This is optimal for this $a \times b$ because And here is a way to tile I've found two more, one for 1x4: and a rather large one ...

11

I found these solutions while playing around on https://www.scholastic.com/blueballiett/games/pentominoes_game.htm.

11

The answer is Here's the proof, step by step (EDIT: the result was right, but the specific argument had a hole; I think I've patched that now) :

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I think the answer is this Explaining $16$ possible solutions Some notes on how the solution is obtained.

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People have given some good upper bounds, how about a lower bound. However we can improve this ... However we can improve this ...

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Here's another proof of the lower bound in Sriotchilism O'Zaic's answer.

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They are trying to prove that How are they doing it? Will they succeed? A special case of the result being considered may be found at Maths SE (spoilers, obviously). The L-ish proof here uses the same underlying idea, but with the difference that

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I can prove that the answer is exactly Several people, including Jaap Scherphuis, have shown that the square can be covered with this many pentominoes, so it only remains to show that at least this many pentominoes are needed. (A matching lower bound). Let us start with the magic board given by A. Rex: As a first lower bound, However, Therefore, we can ...

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9

I think I've found a solution for an 8x8 square. I do not know if it is the minimum solution or how to prove that: It was definitely fun to try and find this! Took me a while. Excellent puzzle. Some comments on how I got to the solution (Rather a chronology than a full deduction):

9

The 35 hexominos can be tiled like this: How did I find the tiling?

9

I brute-forced a solution that I think is the smallest possible. I stumbled upon it while trying to fill a bigger rectangle:

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Here's one solution. The second solution can be achieved by: .

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Since a single line doesn't do damage, it is possible to do To achieve this, there are a couple of requirements: To get these pieces one after the other For this to occur so that the final piece also clears the board, we get these constraints on the number of pieces $X$: Given these, the smallest $X$ that satisfies both requirements is This is a small ...

8

Let's start with the easy ones. 1x1 1x2 1x3 1x4 1x6 These ones took me a while. 1x5 2x3

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Added 1x8. Added 1x5 and 1x6. Replaced 1x3. 1 x 1 (Area = 9), 1 x 2 (Area = 9), 1 x 3 (Area = 42), 1 x 4 (Area = 56) 1 x 5 (Area = 165), 1 x 6 (Area = 156) 1 x 8 (Area = 432) 2 x 2 (Area = 36), 2 x 3 (Area = 104)

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