Convex Polygons

Date: 2/5/96 at 9:19:44
From: Sweet Home Middle School
Subject: Convex polygons et al.

Dr. Math:

* Given a fixed area, how can I maximize the perimeter of a 
convex polygon?

* How do you find the area of an oval?

*If a convex polygon had 1400 sq. yards, what would be the 
maximum perimeter?

*What convex polygon has the greatest perimeter but has an area 
of 1400 sq yds?

*What is the largest possible perimeter for an area of 1400 sq 
yds?

Thanks in advance for your responses.

Mrs. Reimer's Math class.
Sweet Home Middle School
Amherst, N.Y.  14226  716-837-3500, fax 716-837-3397
mislib@ns.moran.com

Date: 2/7/96 at 14:39:1
From: Doctor Elise
Subject: Re: Convex polygons et al.

Hi!

The last 3 questions are basically the same question, but 
without a little more restriction on the problem, it can't be 
solved.

Okay.  A circle encloses the most area for the least perimeter, 
and you can sort of think of a circle as a convex polygon with 
an infinite number of very, VERY short sides.  If you then try a 
regular polygon with fewer sides, like an octagon with 8 equal 
sides that encloses the same area, you'll have a slightly larger 
perimeter.  

If you try a regular pentagon next, it will have an even larger 
perimeter than the octagon if it encloses the same area.  And so 
on until you hit the polygon with the fewest number of sides - a 
triangle.

So a triangle is the convex polygon with the greatest ratio of 
perimeter to area.  But wait a minute.  I've been talking about 
REGULAR polygons, meaning polygons where all the sides are the 
same length.  Think about this for a minute.  Draw 2 triangles, 
one with all 3 sides the same length, and then one really tall, 
skinny one with 2 long equal sides and one really short side at 
the base.  The tall, skinny one has to be pretty tall before it 
even comes close to enclosing the same area as the regular one, 
doesn't it?  And the taller it gets, the longer those 2 long 
sides get, so the bigger the perimeter is.  

You get the maximum perimeter by making the triangle infinitely 
tall and skinny, so it's really almost a line.  But there isn't 
any boundary on how tall the triangle is, so there isn't any way 
to calculate the perimeter.

So, the way you maximize the perimeter of a convex polygon is to 
make it really long and flat.

To actually find a value for the perimeter, what you need on 
this problem is a requirement that the convex polygons are 
REGULAR polygons.  Then, the regular polygon with the maximum 
perimeter is an equilateral triangle.  If we call the length of 
a side 'a', then the perimeter is 3aand the area is 1/2 a * h, 
where h is the height of the triangle.  'h' is the perpendicular 
bisector of a side that also bisects one of the angles.

Using the pythagorean theorem, we know that h^2 + (1/2a)^2 = a^2 
so 

h^2 = 3/4 a^2
h = sqrt(3) * a/2

So the area of the triangle = 1/2 a * sqrt(3) * a/2  
which is a^2 * sqrt(3)/4

If you already know the area is 1400, then

a^2 * sqrt(3)/4 = 1400

a^2 = 4 * 1400/sqrt(3)

solve this for a, multiply it by 3 sides on a triangle, and 
that's your perimeter.

Good luck in the competition!

-Doctor Elise,  The Math Forum