1×H矩形内三个正方形的总边长至多为H
Three Squares in a Rectangle
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中文总结 AI 辅助
该数学研究证明1×H矩形内三个内部不交的任意旋转正方形总边长至多为H，解答了MathOverflow问题，推导了x≥3/2时的最大总边长精确值，还涉及Erdos问题#106的相关结论。
中文摘要 AI 辅助
设H≥2，三个内部两两不交、可任意旋转的正方形位于矩形[0,1]×[0,H]内。我们证明它们的边长之和至多为H，且H≥2的条件不能弱化；H=2的情况解答了2021年MathOverflow上提出的一个问题。该证明用包围三角形的矩形半周长的两个初等下界替代了早期部分论证中用到的内接正方形估计，因此无需对分隔线的斜率施加限制。我们推导了当x≥3/2时，1×x矩形内三个正方形的最大总边长的精确值，并证明不存在轴对齐的 Guillotine 切割能验证Erdos问题#106中f(5)>2。
英文摘要
For $x\ge1$, let $G_3(x)$ be the maximum sum of the side lengths of three pairwise interior-disjoint, arbitrarily rotated squares contained in a $1\times x$ rectangle. We determine this function exactly: $G_3(x)=x+\tfrac12$ for $1\le x\le\tfrac32$, $G_3(x)=2$ for $\tfrac32\le x\le2$, $G_3(x)=x$ for $2\le x\le3$, and $G_3(x)=3$ for $x\ge3$. This completes the $n=3$ case of the rectangular square-packing question posed by Richard Stanley in a 2021 MathOverflow comment. The values for $\tfrac32\le x\le3$ follow from a strip theorem stating that three squares in $[0,1]\times[0,H]$, $H\ge2$, have total side length at most $H$. For $x\ge3$, the formula is immediate because each square has side length at most $1$. The range $1\le x\le\tfrac32$ is handled by combining the two-square theorem with an additional semi-perimeter estimate for a triangle whose two nonhorizontal sides have opposite slopes. In particular, the special case $x=2$ answers the question asked in the MathOverflow post. In connection with Erdős problem #106, we also prove that every five-square packing in the unit square with an axis-parallel guillotine cut has total side length at most $2$.
