MHB Prove Quadrilateral Divider: Equal Area & Perimeter

  • Thread starter Thread starter SatyaDas
  • Start date Start date
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. :)
 
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...
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...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top