Both powers? [Math]

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From: jm bergot on 14 Jun 2010 08:43

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.

From: Simplane Simple Plane Simulate Plain Simple on 14 Jun 2010 12:56

From: James Burns on 14 Jun 2010 17:00

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.)

Jim Burns

From: Gerry Myerson on 14 Jun 2010 19:58

In article <4C169887.8010205(a)osu.edu>, James Burns <burns.87(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 (gerry(a)maths.mq.edi.ai) (i -> u for email)

From: jm bergot on 15 Jun 2010 15:53

GIGAthanks to you two who gave your electrons some
exercise with this item. If I had the fingers to
rapidly run a simple calculator at petaflop speed,
it still would have taken my a bit of time to find
the solutions you found. Of course, I'd have to
press the correct buttons==my fatal weakness.

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