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        riddles >> medium >> Triangles and quadrilaterals with given perimeter
        
(Message started by: ecoist on Feb 6th, 2008, 10:53am)
Title: Triangles and quadrilaterals with given perimeter
Post by ecoist on Feb 6th, 2008, 10:53am
I found simple proofs of:

Among all triangles with a given perimeter, the equilateral triangle has the largest area.

Among all quadrilaterals in the plane, the square has the largest area.

But I was told that this is a "classic" resut.  Does anyone know this result?  Does it hold for planar n-gons with n>4?
Title: Re: Triangles and quadrilaterals with given perime
Post by JohanC on Feb 6th, 2008, 1:01pm
I don't have an answer to your questions, but you might be interested in Thomas Hales' 2001 paper about the Honeycomb Conjecture (http://arxiv.org/abs/math.MG/9906042/). It gives the impression that it's not too obvious.
Title: Re: Triangles and quadrilaterals with given perime
Post by Sir Col on Feb 6th, 2008, 2:26pm
It is a "classic" result and is fairly easy to prove that a fixed perimeter n-gon has its area maximised when it is regular.

"Dr. Rob" provides an overview of a non-calculus approach at this link:
http://mathforum.org/library/drmath/view/53668.html
Title: Re: Triangles and quadrilaterals with given perime
Post by tiber13 on Feb 7th, 2008, 4:08am
well, since the circle has largest area of all shapes, the  equilateral triangle and normal square are closest to the shape
Title: Re: Triangles and quadrilaterals with given perime
Post by ecoist on Feb 7th, 2008, 9:29am
Thanks for the link, Sir Col.  Simple proofs, indeed, for triangles and quadrilaterals.  However, none of the proofs there show that a pentagon with given perimeter and maximum area must be regular.  Both proofs show that all sides have the same length but assume that the vertices lie on a circle.
Title: Re: Triangles and quadrilaterals with given perime
Post by pex on Feb 7th, 2008, 10:13am

on 02/07/08 at 09:29:58, ecoist wrote:
Thanks for the link, Sir Col.  Simple proofs, indeed, for triangles and quadrilaterals.  However, none of the proofs there show that a pentagon with given perimeter and maximum area must be regular.  Both proofs show that all sides have the same length but assume that the vertices lie on a circle.

"Dr. Rob" only requires knowledge of the fact that for given (not necessarily equal) side lengths, the quadrilateral of largest area is cyclic. This is not hard to show.

See the attached picture. The are of the quadrilateral is (ab sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + cd sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/2. Further, using the cosine rule to express the length of the diagonal in two ways, a2+b2-2ab coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif = c2+d2-2cd coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif.

Using the standard Lagrangian technique for constrained optimization with respect to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, maximal area is obtained in a situation satisfying
(1/2)ab coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif ab sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif = 0;
(1/2)cd coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif cd sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = 0.
Eliminating the Lagrangian multiplier http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif, coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif/sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif = - coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, or tanhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif = - tanhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif. The only way this can be satisfied with 0 < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif < 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif is if http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif; that is, the quadrilateral is cyclic.
Title: Re: Triangles and quadrilaterals with given perime
Post by ecoist on Feb 7th, 2008, 10:58am
Gosh, pex!  My proof for the quadrilateral does not require that "the quadrilateral of largest area is cyclic"!

Let ABCD be a quadrilateral in the plane labelled clockwise, with maximum area among quadrilaterals with the same perimeter.  Select diagonal AC.  Then B lies on an ellipse with foci A and C.  Hence, if AB=/=BC, then we can move B along the ellipse, preserving AB+BC, until AB=BC, where the area of triangle ABC is larger.  Thus, all sides of ABCD must equal.  Hence ABCD is a rhombus.  Since the area is maximum, this rhombus must be a square.
Title: Re: Triangles and quadrilaterals with given perime
Post by rmsgrey on Feb 7th, 2008, 11:27am

on 02/07/08 at 10:58:58, ecoist wrote:
Since the area is maximum, this rhombus must be a square.

That's not obvious unless you allow the result that you claim not to require, or show it some other way...
Title: Re: Triangles and quadrilaterals with given perime
Post by Sir Col on Feb 7th, 2008, 11:37am

on 02/07/08 at 09:29:58, ecoist wrote:
Thanks for the link, Sir Col.  Simple proofs, indeed, for triangles and quadrilaterals.  However, none of the proofs there show that a pentagon with given perimeter and maximum area must be regular.  Both proofs show that all sides have the same length but assume that the vertices lie on a circle.

Not quite, "Dr. Rob" shows that an n-gon of fixed perimeter has maximum area when it is a regular. He does this in three steps:
(1) Show that the polygon must be convex.
(2) Show that the polygon must equilateral.
(3) Show that the polygon must be equiangular.
Title: Re: Triangles and quadrilaterals with given perime
Post by ecoist on Feb 7th, 2008, 11:40am
Why isn't it obvious, rmsgrey?  The area of a rhombus is the length of a side times the distance between opposite sides.  If the rhombus is not a square, then the distance between opposite sides is less than a side length.
Title: Re: Triangles and quadrilaterals with given perime
Post by ecoist on Feb 7th, 2008, 11:54am
Sorry, Sir Col, I couldn't find any argument there even remotely related to (3).
Title: Re: Triangles and quadrilaterals with given perime
Post by rmsgrey on Feb 7th, 2008, 12:25pm

on 02/07/08 at 11:40:58, ecoist wrote:
Why isn't it obvious, rmsgrey?  The area of a rhombus is the length of a side times the distance between opposite sides.  If the rhombus is not a square, then the distance between opposite sides is less than a side length.

It's not so obvious when you're more used to calculating the area of a rhombus by treating it as a kite rather than a parallelogram...
Title: Re: Triangles and quadrilaterals with given perime
Post by Sir Col on Feb 7th, 2008, 1:02pm

on 02/07/08 at 11:54:45, ecoist wrote:
Sorry, Sir Col, I couldn't find any argument there even remotely related to (3).


Here is "Dr. Rob's" non-calculus approach...

Step 1: Show that a polygon with a fixed perimeter and largest area must be convex. Do this by considering any vertex with an interior angle of more than 180 degrees, and constructing the diagonal connecting the two vertices adjacent. The diagonal and the two edges will form a triangle. Reflect that triangle in the line that is the diagonal extended, and you have a polygon with a larger area and the same perimeter.

Step 2: Show that a convex polygon with a fixed perimeter and largest area must have all sides of the same length. Do this by looking at two unequal adjacent sides, and the triangle formed by them and the diagonal connecting their two outside ends. Show that the isosceles triangle with the same base and perimeter has a larger area, so replacing the constructed triangle with the isosceles one will give a polygon with the same perimeter and a larger area.

Step 3: Show that a convex polygon with equal sides, fixed perimeter, and largest area, must have all its interior angles of the same measure. Do this by looking at two adjacent angles of unequal measure, and the three sides forming them. Connect a diagonal from the outside ends of the sides to form a quadrilateral. Show that the isosceles trapezoid formed by the diagonal and three sides of the same length as the sides of the polygon has the same perimeter and larger area than the quadrilateral constructed, so replacing the three given sides and two included angles with three sides of the same lengths and two equal included angles forms a polygon with the same perimeter and larger area. (The fact that, given the lengths of the sides, the quadrilateral with the largest area is cyclic, that is, all its vertices lie on a single circle, should help here.)

These three steps show that among all polygons with n sides and a fixed perimeter, the regular n-gon has the largest area.


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