Square with same Perimeter and Area as a Triangle

Date: 10/16/96 at 1:2:14
From: Todd Edwards - a.k.a mathman
Subject: Square with same perimeter and area as a triangle

Dr. Math:

I am perplexed!  I've been hunting for a square/triangle combo with 
the same perimeters and areas.  Is this possible?  I believe this is 
an application of semi-perimeters (is that what it is called?).

I've been fiddling around with this for 3 or 4 hours now on CABRI with
no success.

A solution in CABRI (as opposed to GSP) would be unbelievably
appreciated by both myself (and my students)!

Thank you so much,
Todd Edwards
Upper Arlington High School, Columbus, OH

Date: 10/17/96 at 13:53:24
From: Doctor Yiu
Subject: Re: Square with same perimeter and area as a triangle

Dear Todd,

This is a very interesting question. It is indeed  IMPOSSIBLE to
find a square and a triangle with the same perimeter and the same 
area.


Here is the reasoning. It relies on  two basic facts.

(1)  A formula for  the area of a triangle.
      
There are many ways of finding the area of a triangle. One of these 
is the formula that gives the area in terms of the lengths of the 
sides. This is the Heron formula: if  the lengths of the sides of a   
triangle are a, b, and c, and if the SEMIPERIMETER is denoted by s,  
namely,

s = (a+b+c)/2, 

then the area A is given by

A^2 = s(s-a)(s-b)(s-c).             (HERON)


(2)  An inequality involving the arithmetic mean and geometric mean 
     of three positive numbers.

If you have three positive numbers, say X, Y, Z, there are many 
different ways of taking "averages". Two of the most common averages 
are

(i)    the ARITHMETIC  MEAN:   (X+Y+Z)/3, and

(ii)   the GEOMETRIC MEAN:  the  CUBE root of   XYZ.

Now, it is true that  for all positive numbers X, Y, Z, the arithmetic 
mean is always NO LESS THAN the geometric mean. In other words,

Cube root of  XYZ   < or =  (X+Y+Z)/3,

or

3(Cube root of XYZ) < or =  X+Y+Z.

If you take the cube (third power) of both sides, then

27XYZ  < or = (X+Y+Z)^3.                     (AGI)

Since we shall make use of this inequality below, let's label it  
(AGI)


Now,  consider a triangle with sides  a, b, c, and semiperimeter s. 
Its area A is given by the formula (HERON) above.

Suppose there is a square with the same perimeter and same area.

Then, the perimeter of the square being  2s, each of its sides has
length  (2s)/4 = s/2.

The area of the square is then  (s^2)/4.

This means that the area of the triangle is also   A =  (s^2)/4.

Plugging into the formula  (HERON), we now have

(s^4)/16  =  s(s-a)(s-b)(s-c).

Cancelling a common factor  s, and multiplying by 16,  we have

16(s-a)(s-b)(s-c) =  s^3.            (******)

We shall use   (AGI) to see that this is indeed impossible.

The numbers s-a, s-b, and s-c are POSITIVE. (In fact, since the sum of
the lengths of any two sides of a triangle must exceed the third, say,
a + b > c, we have  a+b+c > 2c, or  2s > 2c, or  s > c. This means s-c
is positive. For the same reason,  s-a and s-b are also positive). 

Note that their sum is  exactly s, because

(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s.

Now, using  (AGI) with X = s-a, Y = s-b and Z = s-c, 

from (AGI):    27XYZ < or = (X+Y+Z)^3,         

we have

27(s-a)(s-b)(s-c) < or =  s^3.

and the equation  (*****) above is impossible.

This completes the demonstration that there is no triangle/square pair
equal in perimeter and area.

Sincerely,

-Doctor Yiu,  The Math Forum

Date: 10/17/96 at 14:19:19
From: Doctor Ken
Subject: Re: Square with same perimeter and area as a triangle

Hi Todd-

Here's a little intuition as to why this can't work.  Let's say we 
choose our perimeter first.  Call it P.  Once you've chosen P, then 
you've completely determined the square.  It has side length P/4, and 
area P^2/16.  

The only freedom you have at this point is how the length P is 
distributed among the three sides of the triangle you're making.  So 
the question is whether there's a triangle with perimeter P and area 
P^2/16.

That's where Yiu's proof comes in.

-Doctor Ken,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: Thu, 17 Oct 1996 21:00:15 -0400 (EDT)
From: Todd Edwards
Subject: Re: Square with same perimeter and area as a triangle

Thank you so much for your thorough proof regarding my square/triangle
problem.  I can't tell you enough how happy I was to hear from you.  I 
was really excited by this problem, but quite frankly haven't had a 
lot of time to try and solve it.  It's exciting to know that a 
resource such as yours exists.  I told all of my students about Dr. 
Math and have encouraged them to visit your Web Site.

I've used a good number of lessons from your site and have found them
really great!  Keep up the great work.  You help me feel "connected" 
to a larger mathematics community - which is important for high school 
teachers!

Thanks again,
Todd Edwards
Upper Arlington High School - Columbus, Ohio