MHB Prove Quadrilateral Divider: Equal Area & Perimeter

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For any quadrilateral, there exists at least one straight line that can divide it into two equal parts in both area and perimeter. The discussion highlights the need for a proof of this statement, with participants sharing their attempts and solutions. Clarity in the wording of the problem is emphasized, as some find it challenging to grasp initially. The thread has been moved to a more appropriate forum for mathematical challenges. This topic invites further exploration of geometric properties and proofs related to quadrilaterals.
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Prove that for any quadrilateral there exists at least one straight line that divides the given quadrilateral into 2 equal parts in area and perimeter.
 
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I realized I asked in the wrong forum. Don't know how to delete it.
 
Moved to Challenge Questions and Puzzles, assuming that was the intended destination.
 
Satya said:
Prove that for any quadrilateral there exists at least one straight line that divides the given quadrilateral into 2 equal parts in area and perimeter.
My attempt.
Every line through the centroid of the quadrilateral divides the area into 2 equal halves.
Consider the function that maps the angle of a line through the centroid to the perimeter on one side minus half the total perimeter.
Over a full period, this function is either the zero function, or it has positive maxima with corresponding negative minima.
If it is the zero function, we are done, since any line through the centroid divides the perimeter into 2 equal halves.
So assume that at least 1 positive maximum exists, which means that the perimeter on one side is greater than the perimeter on the other side.
Then there must be a corresponding negative minimum at a distance of half a period from that maximum.
So according to the intermediate value theorem, the function must take the value 0 somewhere between that maximum and minimum.
QED.
 
Klaas van Aarsen said:
My attempt.
Every line through the centroid of the quadrilateral divides the area into 2 equal halves.
Consider the function that maps the angle of a line through the centroid to the perimeter on one side minus half the total perimeter.
Over a full period, this function is either the zero function, or it has positive maxima with corresponding negative minima.
If it is the zero function, we are done, since any line through the centroid divides the perimeter into 2 equal halves.
So assume that at least 1 positive maximum exists, which means that the perimeter on one side is greater than the perimeter on the other side.
Then there must be a corresponding negative minimum at a distance of half a period from that maximum.
So according to the intermediate value theorem, the function must take the value 0 somewhere between that maximum and minimum.
QED.

I think this is right solution. Only thing is that I needed to read it multiple times to understand the wordings. :)
 
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