Insert squares into square Let $ABCD$ be a square, $AB=2a$
Is it possible to insert two disjoint squares, both of side $a$ into $ABCD$?

 A: Let $\mathcal C$ be the circle with radius $a$ inscribed in the square $\mathcal A$ of side $2a$ centered, say, at point $O= (1,1)$ and consider the distance $d = d(\mathcal C,\mathcal S)$ where $\mathcal S$ is any square inserted in $\mathcal A$; if $\mathcal S \cap \mathcal C$ is not empty  then $d = 0$, which is obvious. On the other hand the biggest square $\mathcal S$ contained in $\mathcal A$ and out of $\mathcal C$ --and having a point in common with $\mathcal C$, so $d = 0$ again-- has its diagonal equal to $(\sqrt2-1)a$ (there are four, each placed in one corner of $\mathcal A$) then its length $l=\frac{\sqrt2-1}{2}a<a$ . Consequently, if $\mathcal S$ and $\mathcal C$ are disjoint forcibly $l<a$. Therefore the answer to the problem is negative. 
A: Suppose the big square with side length $2$ is $ABCD$ (starting from $A$ in the lower left corner and going clockwise.) And the small square with side length $1$ is $EFGH$. I'll prove that $EFGH$ must contain $I$ (the center of $ABCD$) Somebody else came up with the idea, but I don't remember his name so I can't give credit.
I'll assume that $E\in[AD]$ and that $F\in[AB]$, i.e. two corners of $EFGH$ are in contact with $ABCD$. We can use this to prove it for the general case. But I don't feel motivated enough to think about it. I believe we can do it just by translation of $EFGH$ until it is in contact with two sides.
We'll use Cartesian coordinates for our squares. So we have:
$A(0,0),\ B(0,2),\ C(2,2),\ D(2,0)$ and $E(x,0),\ F(0,y)$
$s.t.\ 0\leq x,y\leq 1 $ and $ x^2+y^2=1$
We need to determine the coordinates for $G$ and $H$.
You can get $G$ by rotating $E$ around $F$ by $\displaystyle\frac\pi2$.
$$(x_G-0,y_G-y)=rot_{\pi/2}(x,-y)=(y,x)$$
$$x_G=y\ ;\ y_G=y+x $$
Now similarly for $H$, you get:
$$x_H= y+x\ ;\ y_h=x$$
Now we need to show that $I(1,1)$ is in our rectangle.
We'll show that $I$ is: a) under $(FG)$, b) above $(EH)$, c) to the left of $(GH)$ and d) to the right of $(EF)$. I'll only prove the first two, since you can use the same method for the others.
a)
The equation for $(FG)$ is: $Y=y+\frac{x}{y}X$
$$1\leq y+\frac xy\cdot1 = \frac{x+y^2}{y}=\frac{x+1-x^2}{y}=1+\frac{1-y}{y}+\frac{x(1-x)}{y}$$
So $I$ is under $(FG)$:
b) The equation for $(EH)$ is: $Y=\frac{x}{y}(X-x)$
$$1 \geq \frac{x}{y}(1-x)=\frac{x}{y(1+x)}(1-x^2)=\frac{xy}{1+x}$$
So $I$ is above $(EH)$
c) and d) are true because of symmetry. So $I$ is inside $EFGH$
If two squares of side length $1$ are inside $ABCD$ they both contain $I$, so they can't be disjoint.
