Disclaimer: I don't know how to prove it formally but the intuitive proof(key) is enough.
Hint:
Still no answerAsif Hossain wrote: ↑Thu May 20, 2021 9:50 pmProve that any closed loop is inscribed in a square.(It is not the unsolved topology problem rather a brain teaser)
Disclaimer: I don't know how to prove it formally but the intuitive proof(key) is enough.
Hint:
After rotating 90 degrees or 180 degrees? (If we label one side as blue and one side as brown after rotating 180 degrees the |brown side| becomes |blue| and |blue side| becomes |brown| )Anindya Biswas wrote: ↑Mon May 31, 2021 3:24 pmLet's imagine a big rectangle $ABCD$ (named in clockwise direction) inside of which a closed loop is lying.
Now start moving the side $AB$ towards $CD$ until it touches the closed loop. Do this for every sides, pull them towards the opposite side until they touches the loop.
Following this algorithm, we will find a rectangle in which the loop is inscribed.
Now if it is a square, we are done. If $AB>BC$, start rotating the rectangle counterclockwise while keeping it's sides tangent to that loop (this may change it's side lengths.)
After rotating $90$ degrees, It becomes the same rectangle we started with, but now point $A$ goes to the place where point $B$ initially was, and other point changed their positions accordingly.
Therefore, now we have $BC>CD\Rightarrow BC>AB$
So, we started with $AB>BC$ and ended up with $AB<BC$ after a continuous change.
So, somewhere in between we must get a rectangle where $AB=BC$. And that will be a square.
After rotating $180$ degrees, the side lengths stays the same with what they started with. This shouldn't guarantee that at some angle the smaller side and the bigger side actually becomes same. In the other hand, rotating $90$ degrees leaves the smaller side bigger and the bigger side smaller which guarantees that at some point they were same.Asif Hossain wrote: ↑Wed Jun 02, 2021 9:59 pmAfter rotating 90 degrees or 180 degrees? (If we label one side as blue and one side as brown after rotating 180 degrees the |brown side| becomes |blue| and |blue side| becomes |brown| )Anindya Biswas wrote: ↑Mon May 31, 2021 3:24 pmLet's imagine a big rectangle $ABCD$ (named in clockwise direction) inside of which a closed loop is lying.
Now start moving the side $AB$ towards $CD$ until it touches the closed loop. Do this for every sides, pull them towards the opposite side until they touches the loop.
Following this algorithm, we will find a rectangle in which the loop is inscribed.
Now if it is a square, we are done. If $AB>BC$, start rotating the rectangle counterclockwise while keeping it's sides tangent to that loop (this may change it's side lengths.)
After rotating $90$ degrees, It becomes the same rectangle we started with, but now point $A$ goes to the place where point $B$ initially was, and other point changed their positions accordingly.
Therefore, now we have $BC>CD\Rightarrow BC>AB$
So, we started with $AB>BC$ and ended up with $AB<BC$ after a continuous change.
So, somewhere in between we must get a rectangle where $AB=BC$. And that will be a square.
.Anindya Biswas wrote: ↑Mon May 31, 2021 4:00 pmDoes this matches your reasoning? I highly doubt if "rotating the rectangle and keeping it's side tangent to the loop" is a continuous action or not.