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Example

If $AB$ is the door width, when its possible for a regular polygon to pass through the door?

Intuitively, the minimum polygon "width" must be lesser than $AB$, so I think the answer to this question is when $2 \times a < AB$, where $a$ is the measure of the apothem.

Am I correct? If so, how can I prove that $2 \times a$ is the minimum "width" of the polygon?

MrBr
  • 367
  • Turn it sideways: it is infinitely thin. – Michael Aug 27 '19 at 23:38
  • Yes, you're right. What about through an 1-dimensional door? – MrBr Aug 27 '19 at 23:43
  • Is the shape constrained on the plane, or can you rotate it above and below? – Gabe Aug 27 '19 at 23:46
  • The shape is constrained on the plane. – MrBr Aug 27 '19 at 23:47
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    For regular $n$-gon with even $n$, the condition is indeed $2a \le AB$. If you can move the $n$-gon pass the door, then at certain time, the centroid will cross $AB$. At that moment, the intersection of the regular $n$-gon with $AB$ will be a line segment of length $s$ such that $2a \le s \le AB$. This shows the min-width $\ge 2a$. When $AB = 2a$, it is clear how to push the $n$-gon through the door. So the min-width $= 2a$. No idea what is the right condition for odd $n$. – achille hui Aug 28 '19 at 04:26
  • The problem is considerably more intricate if the polygon is not regular, and not convex. – Joseph O'Rourke Aug 28 '19 at 23:19
  • Yes if the diameter of circumferential circle of regular polygon is less than the width of the door. For hexagon this diameter is twice as the measure of side. so what you claim is true only for hexagon. – sirous Aug 30 '19 at 12:30

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