# tangram cannot get a square with a little square hole in the center

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

Why?

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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:

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