# Square into square

### Square into square

Let ABCD be a square, |AB|=2. Let EFGH be a unit square included in ABCD (every point of EFGH is inside ABCD). If O is the center of ABCD, is it possible for O to stay outside EFGH?
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### Re: Square into square

Suppose it is possible to tilt and position the smaller square in such a way that it doesn't contain the centre of the larger square (which is orientated orthogonally). We can always move the smaller square vertically so that it still doesn't contain the centre and either it touches the top or bottom edge of the large square. Similarly we can then move it horizontally so that it touches the left or right edge of the larger square without containing the centre. So now we have a solution where the smaller square has been moved into one of the corners (it touches the two sides adjacent to that corner). We can assume that this corner is the lower left one (if it isn't just rotate our entire diagram by a multiple of 90 degrees to make that the case).

Rather than try to work out if the centre lies in the tilted smaller square, it is easier to tilt the entire diagram so that the smaller square is orthogonal and the larger square is tilted. Furthermore if we say that $$(0,0)$$ is the lower left corner of the smaller square all we have to do is work out the coordinates $$(x,y)$$ of the centre of the larger square and show that $$0\leq x\leq 1$$ and $$0\leq y\leq 1$$ always occur no matter the angle of the tilt of the larger square.

If you rotate the larger square anticlockwise by an angle of $$\theta$$ (where $$0\leq \theta\leq 90^\circ$$) and place the smaller square orthogonally in what was the lower left corner of the larger square, you can work out with a little trigonometry that the coordinates of the centre are
$$(1-s+c(1-c),c+s(1-c))$$
where $$c=\cos\theta$$ and $$s=\sin\theta$$

It is easy to show that both coordinates are at least $$0$$.

If $$1-s+c(1-c)>1$$ then $$c>s+c^2\geq s^2+c^2=1$$ which is a contradiction so the $$x$$-coordinate lies between $$0$$ and $$1$$.

If $$c+s(1-c)>1$$ then $$0>1-c-s+cs=(1-c)(1-s)$$ which is a contradiction so the $$y$$-coordinate lies between $$0$$ and $$1$$. Consequently the centre always lies in the smaller square no matter how we tilt (or position) things.

Hope this helped,

R. Baber.
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