MHB Prove Quadrilateral Divider: Equal Area & Perimeter

  • Thread starter Thread starter SatyaDas
  • Start date Start date
AI Thread Summary
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.
SatyaDas
Messages
22
Reaction score
0
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.
 
Mathematics news on Phys.org
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. :)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top