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According to this note, three medians of a hyperbolic triangle in the Poincare Disk $\Bbb{D}=\{z\in\Bbb{C} : |z|\lt 1\},\,ds=\dfrac{|dz|}{1-|z|^2}$ are concurrence. But how can I prove this?

Can we write this concurrent point in terms of vertices?

Bumblebee
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    http://www.maths.gla.ac.uk/wws/cabripages/hyperbolic/ceva.html – Moishe Kohan Mar 03 '17 at 18:16
  • @MoisheCohen: Thank you for your interesting link. It helped me lot. I edited my question to clarify the position of this hyperbolic triangle centroid. In Euclidean geometry we can easily find this point by taking the average of vertices. So I think in hyperbolic case also there should be a way to find this point. Do you have any idea about this? – Bumblebee Mar 05 '17 at 05:31
  • Sure: Use the hyperboloid model and repeat the averaging. – Moishe Kohan Mar 05 '17 at 15:14
  • Compare my answer here: http://math.stackexchange.com/questions/2101606/formula-for-midpoint-in-hyperbolic-3-space/2138748#2138748 – Moishe Kohan Mar 05 '17 at 15:22
  • @MoisheCohen: I am not familiar with hyperbolic geometry and I have never use hyperboloid model :). Thank you for you help. I post an answer following your idea. Cold you please look at it? – Bumblebee Mar 05 '17 at 18:40
  • I did not check your formulae but they look reasonable. I suggest you learn the hyperboloid model: Many (but not all) things are so much more transparent when you use it. My formula for the centroid in this model would be: $(A+B+C)/\sqrt{<A+B+C, A+B+C>}$, where $A, B, C$ are points on the upper sheet of the hyperboloid and $<u, v>$ is the Lorentzian inner product. – Moishe Kohan Mar 05 '17 at 19:59

2 Answers2

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Here I follow the @Moishe Cohen's guide lines and these conversion formulas.

Let $a, b, c$ be three points on the disk $\Bbb{D}.$

Now the corresponding points on the upper sheet of the hyperboloid of the hyperboloid model are given by $A=\left(\dfrac{1+|a|^2}{1-|a|^2},\dfrac{2a}{1-|a|^2}\right),$ $B=\left(\dfrac{1+|b|^2}{1-|b|^2},\dfrac{2b}{1-|b|^2}\right)$ and $C=\left(\dfrac{1+|c|^2}{1-|c|^2},\dfrac{2c}{1-|c|^2}\right).$

Centroid of the triangle formed by A, B and C is given by the formula described in Moishe Cohen's answer.

Bumblebee
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I suggest you learn the hyperboloid model and avoid conversion to the unit disk model until you absolutely have to. Many (but not all) things are so much more transparent when you use the hyperboloid model. My formula for the centroid in the hyperboloid model is: $$(A+B+C)/\sqrt{<A+B+C, A+B+C>},$$ where $A, B, C$ are points on the upper sheet of the hyperboloid and $<u, v>$ is the Lorentzian inner product, where the 2-sheeted hyperboloid is given by the equation $<u,u>=1$. As for the proof, you first check the midpoint formula: $$ (P+Q)/\sqrt{<P+Q, P+Q>} $$ and then use it to show that the centroid of a hyperbolic triangle (in the hyperboloid model) is the same as the normalization of the affine centroid of the corresponding affine triangle with the same set of vertices.

Moishe Kohan
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