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18 Nov

It Never Stops with Pythagoras

In the previous blog I described a discovery of Hirotaka Ebisui and an observation by Thanos Kalogerakis, both concerning what's known as Vecten's configuration. Vecten's configuration is a generalization of the famous Bride's Chair that underlies Euclid I.47, generally identified as the proof of the Pythagorean Theorem, although by now there are hundreds of them. In response to the previous post, Long Huynh Huu has expanded the two results, with the tools from linear algebra. To cut that introduction short, earlier today I was informed by Thanos Kalogerakis of a post by Carlos Hugo Olivera Días at the Peru Geometrico facebook group that adds another (and most fundamental at that) link to the just described chain of discoveries.

To find $M,$ and along the way determine that it does indeed exist, I'll employ the Law of Cosine, three times in the $\Delta ABC$ and then in each of the flank triangles. I'll use the well-known property of cosine: $=\cos (\pi-\alpha)=-\cos\alpha:$

$a^2=b^2+c^2-2\cos C$
$b^2=c^2+a^2-2\cos B$
$c^2=a^2+b^2-2\cos A$
$x^2=c^2+a^2+2\cos B$
$y^2=b^2+c^2+2\cos C$
$z^2=a^2+b^2+2\cos A.$

Adding all six up gives $(a^2+b^2+c^2)+(x^2+y^2+z^2)=4(a^2+b^2+c^2),$ i.e.,

$x^2+y^2+z^2=3(a^2+b^2+c^2),$

which not only finds that $$\displaystyle M=\frac{1}{3}$$ but actually proves that the expression $$\displaystyle M=\frac{a^2+b^2+c^2}{x^2+y^2+z^2}$$ is independent of the underlying triangle.

Thanos Kalogerakis made an additional observation. As we know, for the right triangle $x^2+y^2=5z^2,$ i.e., $x^2+y^2+z^2=6z^2,$ while, by the Pythagorean theorem, $a^2+b^2=c^2,$ i.e., $a^2+b^2+c^2=2c^2.$ Combining that with the latest revelation $6z^2=6c^2,$ i.e., $z^2=c^2$ which could be discerned from Vecten's diagram for a right triangle. That is exactly the case when a flank triangle is congruent to the base triangle.