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Re: FW: Pythagorean theorem (fwd)
Posted:
Feb 17, 2000 5:07 AM


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 rightangled?". 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 > >



