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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 $\sqrt{2/3} \approx 0.816$.
     

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And here is Wlodzimierz's much better partition. Each diagonal has length $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]

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For $k=8$, the $4 \times 2$ partition has diagonal $\sqrt{5}/4 \approx 0.559$. Here are two more partitions. Left is rational; right is better but is irrational, $x=\frac{2}{3}-\frac{\sqrt{\frac{7}{3}}}{6} \approx 0.412$:
   ![Sq8Rects][3]

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 $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]

---

For $k=8$, the $4 \times 2$ partition has diagonal $\sqrt{5}/4 \approx 0.559$. Here is a better, irrational partition, $x=\frac{2}{3}-\frac{\sqrt{\frac{7}{3}}}{6} \approx 0.412$:
   ![Sq8Rects62][3]

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 $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]

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 $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]

---

For $k=8$, the $4 \times 2$ partition has diagonal $\sqrt{5}/4 \approx 0.559$. Here are two more partitions. Left is rational; right is better but is irrational, $x=\frac{2}{3}-\frac{\sqrt{\frac{7}{3}}}{6} \approx 0.412$:
   ![Sq8Rects][3]

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 $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]

---

Here is a candidate for $k=8$ that satisfies Wlodzimierz's pattern (but I don't know that it is an optimal partition):
     ![Sq8Rects][3]

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 $\sqrt{2257}/72 \approx 0.660$:
     ![WK5Rects][2]