Fallbrook Epsilon Academy - Monthly Problem 11398

11398. Suppose acute triangle ABC has its middle-sized angle at A. Suppose further that the incenter I is equidistant from the circumcenter O and the orthocenter H. Show that angle A has measure 60 degrees and that the circumradius of IBC is the same as that of ABC.

Solution by the Fallbrook Epsilon Academy Problem Group, Fallbrook, CA.


Some definitions:

incenter
the point where the 3 angle bisectors of the triangle meet
circumcenter
the point where the 3 perpendicular bisectors of the triangle meet; also, the center of the circle (the circumcircle or circumscribed circle) containing the 3 vertices of the triangle
orthocenter
the point where the 3 altitudes of the triangle meet
circumradius
the radius of the circumcircle; it is equal to abc/sqrt((a+b+c)(b+c-a)(c+a-b)(a+b-c)) where the lengths of the 3 sides are a,b,c