A Big Doubt (Own Theorem)

A few years ago I discovered this property and it supposed to be my own Geometry Theorem. I want to know if is it real. Have you seen a Theorem like this?

As a corollary we can say:

If we let \(a\) and \(b\) be the foots of a right triangle with hypotenuse \(c\), we have

\( \dfrac {c^{2}}{a^{2}} = \sum_{n=0}^\infty (\dfrac{b}{c})^{2n} \) And \(\dfrac{c^{2}}{b^{2}}=\sum_{n=0}^\infty (\dfrac{a}{c})^{2n}\)

Note by Hjalmar Orellana Soto
3 years, 2 months ago

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For the triangle in the above photo we know by euclidean theorem that \(a^{2}=pc\), and \(h=\frac{ab}{c}\). So it is true that - \(a_{1} = \frac{c_{1}h}{b}\) - \(a_{1}=\frac{\frac{b^{2}}{c}\frac{ab}{c}}{b}=a(\frac{b}{c})^{2}\) It's easy to see that \(c_{1}=c(\frac{b}{c})^{2}\) by euclidean theorem. And the same with \(b_{1}\) to get \(b_{1}=b(\frac{b}{c})^{2}\). If you take it with the next triangles at that side you'll find that \(a_{n}=a(\frac{b}{c})^{2n}\), \(b_{n}=b(\frac{b}{c})^{2n}\) and \(c_{n}=c(\frac{b}{c})^{2n}\) And with the triangles at the other side you'll find that \(a_{k}=a(\frac{a}{c})^{2k}\), \(b_{k}=b(\frac{a}{c})^{2k}\) and \(c_{k}=c(\frac{a}{c})^{2k}\)

Hjalmar Orellana Soto - 3 years, 2 months ago

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You would need to explain the set up with some words, in order to explicitly convey what you intend.

Calvin Lin Staff - 3 years, 2 months ago

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I've seen to problem of this sort. The closed form of \(a_{k}\) wasn't given, but we had to sum a very similar series.

Deeparaj Bhat - 2 years, 7 months ago

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My calculations in Maple show me that it's true for triangles :

(a=4, b = 8.01, c = 8.95)

(a=3, b = 4, c = 5)

(a=4.9875077, b=12.429690, c=13.392999)

Fatum Altum - 3 years, 2 months ago

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The corollary or the theorem?

Hjalmar Orellana Soto - 3 years, 2 months ago

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I think it's theorem, I have not seen it before.

Fatum Altum - 3 years, 2 months ago

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