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 aa and bb be the foots of a right triangle with hypotenuse cc, we have

c2a2=n=0(bc)2n \dfrac {c^{2}}{a^{2}} = \sum_{n=0}^\infty (\dfrac{b}{c})^{2n} And c2b2=n=0(ac)2n\dfrac{c^{2}}{b^{2}}=\sum_{n=0}^\infty (\dfrac{a}{c})^{2n}

Note by Hjalmar Orellana Soto
4 years, 1 month ago

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

Deeparaj Bhat - 3 years, 6 months ago

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For the triangle in the above photo we know by euclidean theorem that a2=pca^{2}=pc, and h=abch=\frac{ab}{c}. So it is true that - a1=c1hba_{1} = \frac{c_{1}h}{b} - a1=b2cabcb=a(bc)2a_{1}=\frac{\frac{b^{2}}{c}\frac{ab}{c}}{b}=a(\frac{b}{c})^{2} It's easy to see that c1=c(bc)2c_{1}=c(\frac{b}{c})^{2} by euclidean theorem. And the same with b1b_{1} to get b1=b(bc)2b_{1}=b(\frac{b}{c})^{2}. If you take it with the next triangles at that side you'll find that an=a(bc)2na_{n}=a(\frac{b}{c})^{2n}, bn=b(bc)2nb_{n}=b(\frac{b}{c})^{2n} and cn=c(bc)2nc_{n}=c(\frac{b}{c})^{2n} And with the triangles at the other side you'll find that ak=a(ac)2ka_{k}=a(\frac{a}{c})^{2k}, bk=b(ac)2kb_{k}=b(\frac{a}{c})^{2k} and ck=c(ac)2kc_{k}=c(\frac{a}{c})^{2k}

Hjalmar Orellana Soto - 4 years, 1 month 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 - 4 years, 1 month 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)

John Doe - 4 years, 1 month ago

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The corollary or the theorem?

Hjalmar Orellana Soto - 4 years, 1 month ago

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

John Doe - 4 years, 1 month ago

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