15213RE: [PrimeNumbers] Re: The area of prime sided triangles
- Aug 8, 2004
>From: "Adam" <a_math_guy@...>Hi Adam et al,
>Subject: [PrimeNumbers] Re: The area of prime sided triangles
>Date: Sun, 08 Aug 2004 23:01:44 -0000
>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,
>--- In email@example.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
> > 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
I argue as follows. Given the triangle a,b, base c, altitude h, where c = x
+ c-x and solving the two
right triangles formed by h,a,x and h,b,c-x for h and x we get with A =
A = 1/4*sqrt(4*a^2*c^2 - (a^2-b^2+c^2)^2). This reduces to
in the discriminant.
Now Create the table of possible modulo 4 numbers a,b,c can assume and the
remainders after the additions in each of the factors followed by the
remainders of the products of
these factors mod 4.
Sides mod 4
a b c (c+b-a) (a-c+b) (a+c-b) (a+b+c)
1 1 1 1 1 1 3
1 1 3 3 3 3 1
1 3 1 3 3 3 1
1 3 3 1 1 1 3
3 1 1 3 3 3 1
3 3 1 1 1 1 3
3 3 3 3 3 3 1
3 1 3 1 1 1 3
All possible odd sides compute to a remainder of 3 mod 4 which implies the
product cannot be
square and thus the area is irrational. I would think there is an argument
that would imply
(c+b-a), (a-c+b), (a+c-b) are equivelant,at least it is so empirically
above, therefore substantiating
my origional argument if (a+c-b) is square then (a+b+c) is not square and
vice versa and
(a+b-c)*(a+b+c) == 3 mod 4 suffices to prove the assertion.
- << Previous post in topic