A. 416. In a right triangle ABC, point F is the perpendicular foot of C on hypotenuse AB. Let point Q on AB be the perpendicular foot of a certain point P on leg AC. The line through B which is perpendicular to BP intersects line CF at point R. Show that the intersection of circles with diameters AC and FQ, different from F, lies in the interior of segment PR.
Deadline expired on 15 February 2007.