- Aug 8, 2004Whoops, 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,

Adam

--- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:

> 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 - << Previous post in topic Next post in topic >>