Two unequal circles intersect at A and B. The two common tangents touch one circle at P, Q and the other at R, S. Show that the orthocenters of APQ, BPQ, ARS, BRS form a rectangle.
Solution
Let L be the line of centers. Let the orthocenters of APQ, BPQ, ARS, BRS be H, K, H' K' respectively. Then APQ is the reflection of BPQ in L, and ARS is the reflection of BRS in L. So H is the reflection of K in L and H' is the reflection of K' in L. Hence HK and H'K' are both perpendicular to L. The line through A parallel to L is perpendicular to PQ and to RS, so it contains H and H'. So HH' is parallel to L. Similarly, KK' is parallel to L. Hence HKK'H' is a rectangle.
© John Scholes
jscholes@kalva.demon.co.uk
16 June 2002