Given a circle C, and two points A, B inside it, construct a right-angled triangle PQR with vertices on C and hypoteneuse QR such that A lies on the side PQ and B lies on the side PR. For which A, B is this not possible?
Solution
Since ∠APB = 90o, P must lie on the circle C and on the circle diameter AB. So a necessary condition is that these circles should intersect. Conversely, if the circle do intersect, then it is clear that we can take a point of intersection as P. Then extend PA to meet the circle C again at Q and extend PB to meet the circle C again at R.
© John Scholes
jscholes@kalva.demon.co.uk
11 Oct 2003
Last corrected/updated 11 Oct 03