There is no triangle with rational area and all odd sides.

Proof: Let the three sides be a=2k+1,b=2m+1,c=2n+1. Use Heron's

formula area=sqrt[s*(s-a)*(s-b)*(s-c)]=sqrt[(a+b+c)(-a+b+c)(a-b+c)

(a+b-c)/16] where s=(a+b+c)/2 is the semiperimeter. The numerator is

a mess, but it is also 3 mod 8 (once you distribute all the k's to

l's to n's, etc.) and so not a perfect square. #

Adam

--- In

primenumbers@yahoogroups.com, "cino hilliard"

<hillcino368@h...> wrote:

> Hi,

> I was playing with prime sided triangles and area and deduced the

following.

>

> Theorem 1: The area of a prime sided triangle is irrational.

>

> The area of a triangle of sides a < b < c is easily derived as

>

> A = 1/4*sqrt((-a+b+c)*(a-c+b)*(a+c-b)*(a+c+b))

>

> for all real a,b, c <=a+b

>

> Is it sufficient to prove Theorem 1 by proving if a+b+c is square

then a+c-b

> cannot be square

> and if a+c-b is square then a+c+b cannot be square?

>

> If so, maybe someone can fill in the details.

>

> I would guess

>

> Theorem 2: The area of an odd sided triangle is irrational.

>

> Theorem 3: The area of a square sided triangle is irrational

>

> Theorem 4: The area of a non-pythagorean sided triangle is

irrational

>

> Fractions are allowed, ie., (3/2)^2 + (4/2)^2 = (5/2)^2

>

> are also true.

>

> CLH