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Is the circle externally tangent to the three excircles of an irregular non-Euclidean triangle internally tangent to the incircle of the triangle, the tangent point being a generalized Feuerbach point? In Euclidean plane geometry, the circle externally tangent to the excircles of an irregular triangle is internally tangent to the incircle at the Feuerbach point. My calculations indicate that this should similarly obtain for a non-Euclidean triangle, but I have found no proof. (It is possible that my calculations could, with difficulty, be turned into an analytic proof.) A circle is internally (externally) tangent to another circle if the distance between their centers is the difference (sum) of their radii. A triangle is irregular if it is not equilateral.

In response to a helpful comment, my calculations indicate that the additional tangency does not obtain in higher dimensions, so I have revised the question to apply to only a two-dimensional space.

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Is this true for Euclidean geometry? The sphere defined by Feuerbach himself (through the four barycentres of the faces of the tetrahedron) does not touch the insphere of the tetrahedron. If it is not true, is there any reason to believe that things are nicer for non-Euclidean geometry? –  user11235 Apr 8 '11 at 14:22
I've found synthetic proof for it to the two dimensional non-Euclidean case. But article is in Russian. mccme.ru/free-books/matpros/mpd.pdf#page=155 –  akopyan Apr 12 '11 at 6:31
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Akopyan pointed out to me that Hart proved this. Hart's 1861 article can be found at http://books.google.com/books?id=y9xEAAAAcAAJ&pg=PA260#v=onepage&q&f=false. Akopyan's article in translation appears at http://arxiv.org/pdf/1105.2153.pdf. My dynamic illustration appears at http://demonstrations.wolfram.com/NonEuclideanTriangleContinuum.

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