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Let IaIbIc be an ideal triangle in Poincare disk.

Denote:

a, b, c = the hyperbolic lines whose ends are Ib and Ic, Ic and Ia, Ia and Ib respectively.

r = the incircle of a, b and c..

A, B, C = the contact points of r and a, b, c respectively.

O = the hyperbolic circumcenter of ABC

r1 = the incircle of r, a and b.

A1, B1, C1 = the contact points of r1 and r, a, b respectively.

O1 = the hyperbolic circumcenter of A1B1C1.

For any ideal triangle, the radius of r = 0.5493061443... = log[3]/2.

The radius of r1 = 0.2554128119...

Questions:

(1)In any ideal triangle, is the radius of r1 the same?

(2)If so, replace the numerical solution of the radius of r1 by the exact one.

(3)We can construct the contact circles ri+1 of ri, a, b for i = 1 to infinity consecutively, compute their radii.

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  • It's not clear to me what you are asking. The answers to Questions (1) and (2) are right there printed in your question and in the diagram. – Lee Mosher Jun 20 '19 at 16:08

0 Answers0