An acute-angled triangle has circumradius R. Show that any interior point of the triangle other than the circumcenter is a distance > R from at least one vertex and a distance < R from at least one vertex.
Solution
Since the triangle is acute-angled, the circumcenter O must lie inside the triangle. So it must lie in (or on) one of the triangles PAB, PBC, PCA. Suppose it lies in PAB. Then PA + PB > OA + OB = 2R, so either PA or PB >R. Similarly, P must lie inside one of the triangles OAB, OBC, OCA and so one of the distances <R.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03