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From a tangram or seven-piece puzzle (first picture), we cannot get a square with a little hole in the center (the second cartoon), the hole is also a square


enter image description here enter image description here

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The picture is not consistent with your link. – WimC Dec 22 '13 at 20:00
@WimC sorry. It's ok – ziang chen Dec 22 '13 at 20:24

With the original piece set (as shown below), the problem isn't impossible:

enter image description here enter image description here

For the revised version of the problem, impossibilty can be shown in steps. First we argue that the shape of the figure must be a $1\times1$ square inside a $3\times 3$ square: take one of the shortest sides as length $1$; then the side length of the original square is $2\sqrt{2}$, and the area is $8$. Assume that the square-in-square diagram has its sides of the form $m$ or $n\sqrt{2}$; since the squares are aligned then the inner square and outer square must either both be rational or must both be a rational multiple of $\sqrt{2}$, so we must have either $a^2-b^2=8$ or $2c^2-2d^2=8$. (It's a bit complicated to show that this alignment is 'forced', but the short version of the argument comes down to the fact that $\sqrt{2}$ is irrational and so pieces must wind up with their vertices at integral locations, possibly after a scaling by $\sqrt{2}$)

If $a^2-b^2=8$, then $(a-b)(a+b)=8$; since $a-b$ and $a+b$ have the same parity, then it must be that $a-b=2$, $a+b=4$, and so $a=3$, $b=1$.

If $2c^2-2d^2=8$, then $c^2-d^2=4$, and a similar argument as the above goes through to show that this is impossible.

This implies that the setup must be a $1\times1$ square inside a $3\times3$ square. But now consider the two $(2,2,2\sqrt{2})$ triangles. These must be placed in opposite corners of the outer square, leaving two identical area-$2$ notch-shaped regions to be filled, by the remaining three distinct pieces of area 1 and two area-$\frac12$ triangles. But by inspection this area-2 'notch' region can't be filled by any two of the area-1 pieces (using the square leaves two disconnected triangles, as does the area-1 triangle, and using the parallelogram leaves another parallelogram-shaped region), so there aren't enough small triangles to go around. Note that dissecting any of these area-1 regions into two area-$\frac12$ triangles lets us solve the problem.

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(I should note that this started as a proof of impossibility: 'By area arguments, the inner square must be 1 unit and the outer square three units, where a unit is the shortest tangram side; then the red piece has to go with its long edge along an edge of the outer square, and then... oh, wait.') – Steven Stadnicki Dec 22 '13 at 19:27
oh. really sorry, I uploaded a wrong picture. the problem is ok, now – ziang chen Dec 22 '13 at 20:21
@ziangchen I've added a solution for the updated problem! – Steven Stadnicki Dec 22 '13 at 21:07
Thank you very much – ziang chen Dec 22 '13 at 21:08
@ziangchen I'll try to flesh out the argument for 1) a bit more when I have time; as I said, the short version is that all vertices of tangram pieces must lie on integer coordinates. 2), on the other hand, is easy; literally the only place they can be placed in the square-ring is in a corner (they overlap the center hole if they're placed anywhere else), and if placed in adjacent corners they overlap each other along their shared side. – Steven Stadnicki Dec 25 '13 at 6:31

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