364
It is impossible.
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Answer:
Reasoning:
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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 ...
48
The answer is:
Suppose we colour the floor of the room under the tiles like so:
extending up to the edge of the grid.
Then:
So:
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This paper by Joel Haddley and Stephen Worsley answers a slightly different question - finding monohedral disc dissections where not all pieces touch the centre - but the results generally apply to this problem too.
My favourite answer is this one:
Note that this one has interior pieces that don't even have a single point on the boundary.
There is an ...
34
This is a minor upgrade on @sybog64's answer:
One way of thinking about it is to start with this
configuration and then taking groups of 2 slices and rotating each group by 120°.
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I believe this works as a short proof.
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I haven't found a perfect solution, my pieces are symmetrical but not identical
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The best you can do is one with an area of 30 (5 x 6):
Disproving smaller cases
2 x 2 and 2 x 3
2 x anything
3 x 3
3 x 4
3 x anything
4 x 4
4 x 5
4 x 6
5 x 5
So that's definitely not an elegant mathematical proof, but I think it'll hold up. And since the image at the top of this answer shows that 5x6 can be done, that is the smallest possible ...
28
One possible way is to use ...
24
Here is a proof that 5x6 is the smallest possible rectangle.
A rectangle of size $x$ by $y$ has $\frac{xy}{2}$ dominoes and $x+y-2$ potential lines. All of these lines must be blocked by at least one domino which has one square on each side. However, if the line divides the rectangle into two even areas, then one domino blocking it would leave an odd area ...
22
It is not possible. The area of a $10 \times 10$ checkerboard is $100$, so it takes $25$ T pieces to have the same area. The checkerboard has the same number of red and black squares, but each piece covers three of one color and one of the other. $25$ pieces cannot cover $50$ squares of each color, the most even they can get is $51-49$
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I have a 14x14 (28 case) rectangle. It is completely symmetrical. Someone beat that.
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I think that this tiling is a valid Tetris stack:
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UPDATE 2: To put OP out of their misery find now at the very bottom of this post an answer to what they probably mean.
UPDATE: OP has changed the rules, so this is no longer valid, but see bottom of this post for an answer to the modified question which I assume is still not what OP has in mind.
I say it is
"Proof":
With the new rules the answer ...
21
COMPLETED GRID
The first step:
Next:
An important side note:
Moving on:
The top shaded region:
Hopefully, finishing up:
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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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Solution!
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FINITE PORTION OF ANSWER
This can be extended infinitely in all directions - see my route to solving below for how.
First a detour to explain how I made a tool (which competing answers could also use, perhaps improving on the finite number of differently-coloured tiles) by which ideas can be quickly tried using an Excel spreadsheet:
Select all, and set ...
17
Here is yet another solution with 9 pieces. This one is nice and symmetrical.
I have been trying to think of a way to show that 8 will not work by arguing in terms of the number of edge squares that need to be covered. However, I have not got very far with this. Here is a diagram I have been pondering.
Update
I haven't really given this a lot more thought....
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Here is a general solution for n>6.
Explanation:
Here is a proof of why there is no solution for n=6.
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I think I have the answer:
The first step is to determine the color of the square numbers.
Next, let us try the
Again, the flag could be in
Knowing this, we get
A few more
When you now consider
Below the Finnish flag there must be
It remains to place
Whew! That was a long explanation. But I hope it makes sense.
An excellent puzzle, as usual!
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I think I've found a really easy proof.
Every tile with vertical sides needs to have two other tiles with vertical sides adjacent to it, or the vertical boundary of the hexagon. For a given tile with vertical sides, following these adjacent tiles yields a specific path to both vertical sides of the hexagon.
This means that every tile with vertical sides ...
16
I want to post an answer that is more intuitive than mathematical.
This picture perfectly represents it:
White, grey and black are used to highlight the diamonds with the same orientation.
The right picture shows a weird solid, I guess everyone can see it.
Well, it's intuitive to see that, for any configuration, the black area is equivalent (white and grey ...
16
Colour in all cells in the top horizontal row in Red, the second in Blue, then Green, Red, Blue and so on.
R R R R R R
B B B B B B B
G G G G G G G G
R R R R R R R R R
B B B B B B B B B B
G G G G G G G G G G G
R R R R R R R R R R
B B B B B B B B B
G G G G G G G G
R R R R R R R
B B B B B B
Thanks to Mike Earnest for the ...
15
The Burr Tools freeware tool tells us there are exactly
How to use Burr Tools to solve this problem yourself:
The first tab is the Entities tab, where you define your shapes - both the puzzle pieces and the target shape. For each shape, click the New button, then paint it in the rectangular grid below. The slider next to the grid allows you to paint ...
15
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If I'm not mistaken, I believe that it is
Because
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Our mutilated cube
Let's first consider
What about
Finally we must consider
So the final answer is:
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