## 15213RE: [PrimeNumbers] Re: The area of prime sided triangles

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• Aug 8, 2004
>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,
>
>
> > 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. #
> >

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 =
1/2hc,
A = 1/4*sqrt(4*a^2*c^2 - (a^2-b^2+c^2)^2). This reduces to
(c+b-a)(a-c+b)(a+c-b)(a+b+c)
in the discriminant.

Now Create the table of possible modulo 4 numbers a,b,c can assume and the
corresponding
remainders after the additions in each of the factors followed by the
remainders of the products of
these factors mod 4.

Rem of
Sides mod 4
mod 4
a b c (c+b-a) (a-c+b) (a+c-b) (a+b+c)
(c+b-a)*(a-c+b)*(a+c-b)*(a+b+c)
1 1 1 1 1 1 3
3
1 1 3 3 3 3 1
3
1 3 1 3 3 3 1
3
1 3 3 1 1 1 3
3
3 1 1 3 3 3 1
3
3 3 1 1 1 1 3
3
3 3 3 3 3 3 1
3
3 1 3 1 1 1 3
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.

Cino
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