Tiling the square with rectangles of small diagonals For a given integer $k\ge3$, tile the unit square with $k$ rectangles so that the longest of the rectangles' diagonals be as short as possible. Call such a tiling optimal. The solutions are obvious in the easy cases when $k$ is the square of an integer and for a few small values of $k$ only (unpublished). In each of the solved cases, the sides of all rectangles turn out to be rational and their diagonals are equal.
Question. In an optimal tiling, must the sides of all rectangles be rational and their diagonals be equal?
The analogous question for tiling the $n$-dimensional cube with rectangular boxes can be asked for every $n\ge3$ as well.
 A: In a new related question I give a conjecture  for the unit square which agrees with the $k=5$ and $k=8$ solutions here.
If $s^2 \lt k \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically:
If $k=s^2+t$ with $0 \lt t \le s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $b \times \frac1s $ and $t$ rows of $s+1$ rectangles $a \times \frac1{s+1}$ where $b^2+\frac{1}{s^2}=a^2+\frac1{(s+1)^2}$ and $(s-t)b+ta=1$
But if $k=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$  rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles  $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1$ 
Note that in case $k=s^2+s,$ either description gives all rectangles $\frac1s \times \frac1{s+1}.$
A: For $k=5$, is this the optimal partition? 
Rectangle sides 
$x=\frac{1}{6} \left(3-\sqrt{3}\right) \approx 0.21$ and $1-x$, 
and (now corrected) all diagonals of length$^2$ of $\frac{2}{3}$, and so length $\sqrt{2/3} \approx 0.816$.
     

And here is Wlodzimierz's much better partition. Each diagonal has length
$\sqrt{2257}/72 \approx 0.660$:
     

For $k=8$, the $4 \times 2$ partition has diagonal $\sqrt{5}/4 \approx 0.559$.
Here is a better, irrational partition,
$x=\frac{2}{3}-\frac{\sqrt{\frac{7}{3}}}{6} \approx 0.412$:
   
