TITLE: fitting rectangle inside another rectangle in diagonal
QUESTION [10 upvotes]: Having the following case, where we want to fit a rectangle (c,d) inside another rectangle (a,b). We know that c>b or d>a so it will not fit horizontally, so we need to check if it will fit diagonally.
Example:

What would be the condition to check if the rectangle (c,d) will fit (a,b) or not.
Thank you in advance,

REPLY [3 votes]: Let $2a\geq2b$ and $2c\geq2d$ be the sidelengths of the two rectangles, and put $f:=\sqrt{c^2+d^2}$.
When $R_{cd}\subset R_{ab}$ can be realized then it can be done with common center $M$.
A necessary condition for such an inclusion is $d\leq b$, because otherwise $R_{cd}$ contains a disk of radius $d>b$, which cannot be covered by $R_{ab}$. If $c\leq a$ as well then $R_{cd}$ can be embedded in $R_{ab}$ with aligned sides.
Therefore it remains to consider the case $c>a\geq b\geq d$ which is shown in the following figure.

The short sides of $R_{cd}\subset R_{ab}$ are tangent to the circle (shown in blue) with radius $c>a$ and center $M$. This circle intersects $R_{ab}$ in four arcs cutting off the vertices of $R_{ab}$. It follows that the two shorter sides of $R_{cd}$ lie in two opposite corner regions of $R_{ab}$. Consider such a side. Its endpoints lie on a similar arc of a circle  with radius $f$ and center $M$. Therefore the  length $2d$ of this side is at most equal to the distance $|PQ|$ in the figure. This means that we have
$$(2d)^2\leq \left(b-\sqrt{f^2-a^2}\right)^2+\left(a-\sqrt{f^2-b^2}\right)^2\ .$$
Expressing $f$ in terms of $c$ and $d$, and simplifying, you obtain the condition you were looking for.