The feet of the altitudes from A, B, C are P, Q, R respectively, and P, Q, R are distinct points. The altitudes meet at O. Show that if ABC is acute, then O is the center of the circle inscribed in the triangle PQR, and that A, B, C are the centers of the other three circles that touch all three sides of PQR (extended if necessary). What happens if ABC is not acute?
Solution
Suppose first that ABC is acute-angled. ∠OPC = ∠OQC = 90o, so OPCQ is cyclic. Hence ∠OPQ = ∠OCQ = 90o - ∠A. Similarly, ∠OPR = ∠OBR = 90o - ∠A. Hence OP bisects ∠QPR. Similarly for the other angles. So O is the incenter of PQR.
Now suppose ∠A is obtuse. Then A is the orthocenter of OBC, which is acute-angled, so A is the incenter of PQR. O the the intersection of the bisector of ∠QPR and the perpendicular to the bisector of ∠PRQ, so it is the excenter of PQR opposite to P.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03