We provide a tight result for a fundamental problem arising from packing squares into a circular container: The critical density of packing squares into a disk is $\delta=\frac{8}{5\pi}\approx 0.509$. This implies that any set of (not necessarily equal) squares of total area $A \leq \frac{8}{5}$ can always be packed into a disk with radius 1; in contrast, for any $\varepsilon>0$ there are sets of squares of total area $\frac{8}{5}+\varepsilon$ that cannot be packed, even if squares may be rotated. This settles the last (and arguably, most elusive) case of packing circular or square objects into a circular or square container: The critical densities for squares in a square $\left(\frac{1}{2}\right)$, circles in a square $\left(\frac{\pi}{(3+2\sqrt{2})}\approx 0.539\right)$ and circles in a circle $\left(\frac{1}{2}\right)$ have already been established, making use of recursive subdivisions of a square container into pieces bounded by straight lines, or the ability to use recursive arguments based on similarity of objects and container; neither of these approaches can be applied when packing squares into a circular container. Our proof uses a careful manual analysis, complemented by a computer-assisted part that is based on interval arithmetic. Beyond the basic mathematical importance, our result is also useful as a blackbox lemma for the analysis of recursive packing algorithms. At the same time, our approach showcases the power of a general framework for computer-assisted proofs, based on interval arithmetic.

翻译：我们为将方块包装成圆形容器所产生的根本问题提供一个紧凑的结果:将方块包装到磁盘中的关键密度是$\delta ⁇ frac{8\\5\pie} ⁇ =approx0.509美元。 这意味着将圆形或平方对象包装到圆形或方容器中的任何一组(不一定相等)方块$A\leq\frac{8\ ⁇ {8\%5}5}美元可以总是用半径1打入盘中;相比之下,对于任何$\varepsil>0$的盘中,总面积为$\prec{8\ ⁇ 5 ⁇ vävareblesslon$的一组方块无法包装,即使正方块可以旋转方块旋转。 以平方块法法的平方块内值分析也可以用到平方块的平方块内值。