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

## Comments

Sort by:

TopNewestFor 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 \(c_{k}=c(\frac{a}{c})^{2k}\)

Log in to reply

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

Log in to reply

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.

Log in to reply

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)

Log in to reply

The corollary or the theorem?

Log in to reply

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

Log in to reply