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The area of prime sided triangles

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  • cino hilliard
    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
    Message 1 of 4 , Aug 7, 2004
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      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
    • Adam
      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
      Message 2 of 4 , Aug 8, 2004
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        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
      • Adam
        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
        Message 3 of 4 , Aug 8, 2004
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          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,
          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
        • cino hilliard
          ... Hi Adam et al, 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
          Message 4 of 4 , Aug 8, 2004
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            >From: "Adam" <a_math_guy@...>
            >To: primenumbers@yahoogroups.com
            >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,
            >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
            Hi Adam et al,

            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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