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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 ...
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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 ...
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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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I believe this works as a short proof.
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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 ...
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One possible way is to use ...
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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 ...
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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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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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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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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 ...
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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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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 ...
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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 ...
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If I'm not mistaken, I believe that it is
Because
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It is only possible in
Those cells are
Because
An example of valid configuration is
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Can I help you build the tower?
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I see the answer is found already, still I have a simpler one to present.
A 'special triangle' contains at least two trapezoids of 'diagonal' direction (green here). Here all diagonal triangles are far apart, so you don't have to look for errors.
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Answer to part 1:
Divide an infinite checkerboard into $2\times 2$ blocks, then color each block alternately white and black, as shown:
$$
\begin{array}{ccccc|ccccc}
&&\vdots&&\vdots&&\vdots&&\vdots&&\\
\cdots & B & B & W & W & B & B & W & W &\cdots\\
\cdots & B & B & W &...
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I have a satisfying arrangement:
I used this tool. I just let it run a few times until it had only used the desired pieces. The shortcoming is that one cannot set how many instances of each piece are allowed.
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I'll one-up you guys and prove this stronger statement: If $m\geq n$ are the dimensions of a rectangle that admits a nonsplittable domino tiling using more than one domino, then $m \geq 6$ and $n \geq 5$.
It's stronger because it also proves the nonexistence of very long yet thin nonsplittable domino tilings, such as $60\times3$ or $999 \times 4$.
Proof:
...
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Started by scaling the goal, I put one-by-one with some trial-and-error in the end to find this solution:
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This puzzle is known as the mutilated chessboard problem. The other answer correctly explains that such a covering is impossible because it would require an equal number of black and white squares (since each domino must cover one black and one white square), which the corner-cut board does not have.
The link above contains a slightly more general result as ...
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An equilateral triangle with sidelength $L$ can be divided into $L^2$ equilateral triangles with sidelength $1$. A trapezoid with sidelengths $2,1,1,1$ covers three of these triangles, so $L^2$ must be divisible by $3$. $L$ has to be an integer (because if it is tiled by the trapezoids, its sidelength must be the sum of sidelengths of the trapezoids), so $L$ ...
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