-
Dear Clark and friends,
Happy New Year!
Consider the circle tangent internally to each of the excircles of
triangle ABC. This circle can be constructed as the inversive image of the
nine-point circle in the radical circle of the excircles. It has long been
known that the radius of this circle is (r^2+s^2)/(4r). See, for example,
Problem 1864 and solution, Crux Math. 20 (1994) 174--175. What is interesting
is that the center is the point X(970) in ETC. As such, this center lies on
the Brocard axis. Is there a simple explanation of this fact?
Best regards
Sincerely
Paul Yiu
Σάββατο 19 Οκτωβρίου 2019
HYACINTHOS 4619
Εγγραφή σε:
Σχόλια ανάρτησης (Atom)
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου