HMMT February, 2023

Solution:
We split into cases on the integer $k=\left\lfloor x^{2}+y^{2}\right\rfloor$. Note that $x+y=k$ but $x^{2}+y^{2} \geq \frac{1}{2}(x+y)^{2}=\frac{1}{2} k^{2}$ and $x^{2}+y^{2}<k+1$, which forces $k \leq 2$.

If $k=0$, the region defined by $0 \leq x^{2}+y^{2}<1$ and $x+y=0$ is the diameter from $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ to $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, which has length $2$.

If $k=1$, the region $1 \leq x^{2}+y^{2}<2$ and $x+y=1$ consists of two segments, which is the chord on $x^{2}+y^{2}=2$ minus the chord on $x^{2}+y^{2}=1$. The former has length $2 \sqrt{(\sqrt{2})^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{6}$, and the latter has length $2 \sqrt{1^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{2}$. So the total length here is $\sqrt{6}-\sqrt{2}$.

If $k=2$, the region $2 \leq x^{2}+y^{2}<3$ and $x+y=1$ is the chord on $x^{2}+y^{2}=3$, which has length $2 \sqrt{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=2$.

Our final answer is $2+(\sqrt{6}-\sqrt{2})+2=4+\sqrt{6}-\sqrt{2}$.