现有一个90度的直角过道，过道宽度为一米，然后我现在需要把一个宽0.5米的矩形沙发推过这个过道，仅考...

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现有一个90度的直角过道，过道宽度为一米，然后我现在需要把一个宽0.5米的矩形沙发推过这个过道，仅考虑二维平面，请问这个家具的最大长度是多少？

Réflexion

Exploring rotations and dimensions of squares and rectangles

I’m trying to figure out the movement of a 1x1 square in a unit square. It can’t quite rotate, as the bounding box would be too big. It can slide around, though, and a 1xL rectangle with L>1 would need rotation to fit, which isn’t possible in this situation. So, for a square, length equals width.

Thinking through square and rectangle movement in corridors

I’m considering a 1x1 square in a 1-wide horizontal corridor. It can move along the corridor without rotating, even through corners, and then shift upward as long as it stays aligned. The formula says the width can't be 1, but it’s more about the square's orientation and passage without rotation. For rectangles, at least one must rotate if L > W. For a square (L = W = 1), no rotation is needed.

Thought for 2m 12s

Réponse

最大长度约为：

\boxed{1.828\text{ 米}}

建模如下：过道宽 $1\text{ m}$ ，沙发宽 $d=0.5\text{ m}$ 。当沙发以角度 $\theta$ 通过拐角时，能通过的最大长度为：

L(\theta)=\csc\theta+\sec\theta-\frac{d}{\sin\theta\cos\theta}

由于两个过道等宽，最“卡”的位置出现在对称角度：

\theta=45^\circ

代入 $d=0.5$ ：

L_{\max} = \sqrt2+\sqrt2-\frac{0.5}{(1/\sqrt2)(1/\sqrt2)} =2\sqrt2-1

L_{\max}\approx 2.828-1=1.828\text{ m}

所以，在理想二维、刚性矩形、无额外安全间隙的条件下，这个宽 $0.5\text{ m}$ 的矩形沙发最大长度约为 1.83 米。

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