## 15212Re: The area of prime sided triangles

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• Aug 8, 2004
Whoops, sorry, forgot the rest of the argument:
I need to also prove that there is no triangle with a side of 2.

First off, there is no isosceles triangle with prime sides and
ratioanl area. Suppose (p,p,q) were the sides of such a triangle.
Then inside the square root of Heron's formula is (2p+q)*(p+q)*(p+q)*
(2p). This is a perfect as (2p+q)*(2p) is a perfect square. If p=2,
then 2p+q=4+q is a perfect square, say 4+q=x^2, so q=x^2-4=(x+2)(x-2)
is prime only when x-2=1, x=3, q=5. The possible edges (2,2,5) do
not form a triangle. We need an extra factor of 2 somewhere, so
suppose q=2. Then (2p+q)*(2p)=4*(p+1)(p) is not a square.
Therefore, no isosceles triangle with prime sides and ratioanl area.

Secondly, suppose (2,p,q) was a triangle with prime sides and rationa
area, with 2<=p<=q. Then the sum of two sides (2+p) is larger than
the thrid, q, so 2+p>q and q=p or q=p+1. If q=p, then the triangle
has sides (2,p,q)=(2,p,p), is isosceles. Can't happen. If q=p+1,
the only primes that differ by 1 are 2 and 3, so teh triangle is
(2,2,3), no isoscles triangles.

Therefore, no triangle with prime sides and rational area has a side
of measure two. Combined with the earlier "no triangle with odd
sides has rational area," we see that no triangle with prime sides
has rational area.

Sorry, I forgot the two part before,

> 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. #
>
>
>
>
> --- 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
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