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

Question

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?

Answer 1

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.