TITLE: Insert squares into square
QUESTION [5 upvotes]: Let $ABCD$ be a square, $AB=2a$
Is it possible to insert two disjoint squares, both of side $a$ into $ABCD$?

REPLY [2 votes]: 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.