The circumcenter O of the triangle ABC does not lie on any side or median. Let the midpoints of BC, CA, AB be L, M, N respectively. Take P, Q, R on the rays OL, OM, ON respectively so that ∠OPA = ∠OAL, ∠OQB = ∠OBM and ∠ORC = ∠OCN. Show that AP, BQ and CR meet at a point.
Solution
We show that the circumcircle ABC is the incircle of PQR. Then (AR/AQ) (CQ/CP) (BP/BR) = 1 since AR = BR, AQ = CQ, BP = CP (equal tangents) and hence PA, QB, RC are concurrent by Ceva's theorem.
So let the tangents to the circumcircle at B and C meet at P'. It is sufficient to show that ∠OP'A = ∠OAL, for then it follows that P' = P (since there is obviously a unique point on the ray OL at which the segment OA subtends the ∠OAL).
This is curiously difficult to prove. Let LP' meet the circle at K. Then ∠KCP' = ∠KBC (P'C tangent) = ∠KCB (KO perpendicular to BC, since L midpoint) = ∠KCL (same angle). So KC bisects ∠P'CL. Hence KP'/KL = CP'/CL. But obviously CP'/CL = BP'/BL. So K, C and B lie on the circle of Apollonius, the points X such that XP'/XL is constant. But A also lies on the circle of Apollonius. Hence KA bisects ∠P'AL. (See Canada 71/9 if you are not familiar with the circle of Apollonius.) But ∠OP'A + ∠KAP' = ∠OKA = ∠OAK (OA and OK radii) = ∠OAL + ∠KAL. So ∠OP'A = ∠OAL.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002