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.


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 ...


7

Here is one solution Thanks for the fun! Bonus shapes:


7

Unless I made a mistake somewhere, this solution is unique: To get there, I used a couple of rules of thumb: Apart from those, there were a couple of slightly mind-boggly deductions required, but all in all, everything seemed extremely well designed, and no guesswork was needed at any point. Progress, part 1: Progress, part 2: Progress, part 3: ...


7

I guess, I'm too late, but I propose


6

Rev 4 - added 1 x 9 incorrect but still trying, added 1 x 5 using Florian's clue Rev 3 - added 1 x 8 solution Rev 2 - added 2 x 6 solution Rev 1 - added 2 x 2 solution Thanks for the clarification, Somnium. I will put some more answers here. 1 x 5 solution using Florian's clue (area = 54) 1 x 8 solution (area = 560) 1 x 9 incorrect still trying ...


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