# Tag Info

336

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 ...

48

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.)

18

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

13

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

13

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

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

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

10

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) :

10

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

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:

9

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

9

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

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):

8

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

8

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)

8

I believe I have found multiple solutions to the puzzle. It is possible that there is a nuance to the rules that I missed that disqualifies one or both of them. My understanding of the rules did not allow for a full logical deduction of the solution, so these were both found by brute force. The sudoku is on the left, while the tetrominoes are on the right: ...

8

Here are a couple of $1 \times n$ solutions, which I believe to be optimal. They seem to follow some pattern(s), some of which are generalizable (spoiler ahead): The solutions for $1 \times 10$, $1 \times 18$ and $1 \times 20$ (all below) also seem to form some kind of generalizable family. As @JaapScherphuis notes in the comment, there's a (probably non-...

8

Here is my (now improved) solution for $2\times3$ rectangles plus N-pentomino(s): And here is my solution for $2\times5$ rectangles plus N-pentomino(s): Here are some more solutions, though for these I used computer assistance. $1\times6$ rectangles plus N-pentomino(s): The solutions given by others for $1\times3$ and $1\times5$ generalise to give ...

8

If I understand the game correctly,

7

Golden Dragon has solved it, but for the record, this is as intended:

7

A few additions... firstly for the 1x5 a different answer, smaller area but more rectangles than Len's so just for interest 1 x 5 (Area = 160) 3 x 4 (Area = 400) This one's the same area as Len's but fewer rectangles 1 x 8 (Area = 432)

7

Finally, a more interesting heptomino :) (in the sense that previous ones all had generalizable solutions who looked very much like this hexomino) Here's the minimal solution for $1 \times 2$: and for $2 \times 2$: For $3 \times 5$: My program found another one for $2 \times 7$: a very narrow one for $1 \times 10$: another one for $1 \times 11$: and ...

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