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From: achille on 16 Jun 2010 04:55
On Jun 15, 7:58 am, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>
wrote:
> In article <4C169887.8010...(a)osu.edu>, James Burns <burns...(a)osu.edu>
> wrote:
>
>
>
> > jm bergot wrote:
> > > Can the area and the perimeter of a triangle
> > > both be powers?  For the triangle with sides
> > > 40,74,102 the perimeter is 216=6^3 and the area
> > > is 1224=35^2 - 1, almost a winner.
>
> > You almost surely can find triangles like that.
> > You have three parameters under your control,
> > the sides a, b and c, and you have two constraints
> > on those sides, the values of the perimeter
> > and the area.
>
> > One way of getting at least some example solutions
> > might be to hunt around for a triangle that
> > completes one requirement (perimeter a power or
> > area a power), and then scale the triangle so
> > that the first requirement is still met, plus
> > the second requirement is met as well.
>
> > For example, through a bit of trial-and-error,
> > I find that a triangle with sides 5, 5, 6
> > has a perimeter of 16 and an area of 12.
> >http://en.wikipedia.org/wiki/Heron%27s_formula
>
> > If we multiply each side by 12^2, (720,720,864),
> > the new perimeter will be 16*12^2 = 48^2 and the
> > new area will be 12*(12^2)^2 = 12^5.
>
> > (Another way to do that sort of thing is, if
> > the area is already a square, b^2, then scale
> > the sides by the size of the perimeter, p,
> > so the new perimeter is p*p = p^2 and the new
> > area is b^2*p^2= = (b*p)^2. There are probably
> > other ways.)
>
> Indeed, one could apply this technique to OP's triangle.
> Multiply the sides by 1224^3 to get a triangle whose perimeter
> is a cube and whose area is a 7th power.
>
> Finding a triangle whose sides have no common factor
> might be more of a challenge.
>
> --
> Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email)

Here is another way to generate a solution:

For (x,y,z,t) any solution of x^4+y^4+z^4 = t^4,

Let a = 2(t^4 - x^4), b = 2(t^4 - y^4), c = 2(t^4 - z^4),
be the sides of a triangle, then

the perimeter P
= a + b + c = 4 t^4 = (2 t^2)^2
AND the area A
= sqrt( ( a+b+c)(a+b-c)(a+c-b)(b+c-a) / 16 )
= sqrt( (4 t^4)(4 z^4)(4 y^4)(4 x^4) / 16 )
= sqrt( 16 ( x y z t )^4 )
= ( 2 x y z t )^2
are both squares.

For example, using Frye's solution listed on Wiki
(URL: http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
)

95800^4 + 217519^4 + 414560^4 = 422481^4.

one find a solution

a: 63549041528076691840642
b: 59240172952323281920000
c: 4645784879631809920642

with P = 356980390722^2
A = 7299401976791799744000^2
and gcd(a,b) = g(a,c) = g(b,c) = 2.

Please visit TPiezas's site (URL: http://sites.google.com/site/tpiezas/016
)
for more information how to find solution for x^4+y^4+z^4 = t^4.






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