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Re: FW: Pythagorean theorem (fwd)
Posted:
Feb 17, 2000 5:07 AM
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On Wed, 16 Feb 2000, Lambrou Michael wrote:
>
>
> On Wed, 16 Feb 2000, Michael de Villiers wrote:
>
> >
> >
>
> > Dear Jason
> > Some 20 years ago, while I was teaching at a high school, an 11th Grade
> > student of mine asked a similar question: "do exponents exist if the
> > triangle is not right-angled?". Intrigued by this, we worked together and
> > eventually came up with the following theorems:
> > THEOREM 1
> > For any triangle with sides c > b >= a (and angle C opposite c) there
> > exists a number p > 1, so that c^p = a^p + b^p with:
> > (i) p < 2 for angle C > 90 deg
> > (e.g. for triangle with angle C = 133 deg, angle B = 30 deg, and
> > angle A = 17 deg; c= 5, b = 3.4175, and a = 2 gives 5^1.1398 = 2^1.1398 +
> > 3.4175^1.1398 to 4 decimals)
> > (ii) p = 2 for angle C = 90 deg
> > (iii) p > 2 for angle C < 90 deg
> > (e.g. for triangle with angle C = 75.5 deg, angle B = 46.5 deg, and
> > angle A = 58 deg; c = 8, a = 6, and b = 7 gives 8^3.4579 = 7^3.4579 +
> > 6^3.4579 to 4 decimals)
> > THEOREM 2
> > For any triangle with sides c < b <= a there exists a negative number p, so
> > that c^p = a^p + b^p.
> > {e.g. for triangle with c = 1.0353, b = 2, and a = 2 gives 1.0353^(-1.0526)
> > = 2^(-1.0526) + 2^(-1.526) to 4 decimals}
> >
> > Strict proofs require Rolle's Theorem, and is therefore at undergraduate
> > level.
>
> The whole question is easy, and you can do with less than Rolle's theorem:
> For instance, from the cosine rule it is easy to see that for obtuse
> angled triangles we have a^2 > b^2 + c^2. So the continuous function
> f(x)= a^x - b^x - c^x is positive for x=2 and clearly negative if x=0. By
> continuity it is zero somewhere in between.
Yes, it is quite intuitive; I should've said Rolle or the Intermediate
Value Theorem.
> The case for acute angled triangles is no harder (use that for large x
> the power a^x goes faster to infinity that the other two together).
>
> Regards, Michael Lambrou
>
>
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