### 29th IMO 1988 shortlist

**Problem 15**
ABC is an acute-angled triangle. H is the foot of the perpendicular from A to BC. M and N are the feet of the perpendiculars from H to AB and AC. L_{A} is the line through A perpendicular to MN. L_{B} and L_{C} are defined similarly. Show that L_{A}, L_{B} and L_{C} are concurrent.

**Solution**

Let O be the circumcenter. It is sufficient to show that O lies on L_{A}.

Angle AOB = 2 angle ACB. Hence angle BAO = 90^{o} - C.

Angles AMH and ANH are both 90^{o}, so AMHN is cyclic. Hence angle AMN = angle AHN = 90^{o} - angle HAN = C. Hence the angle between L_{A} and AB is 90^{o} - C. Hence O lies on L_{A}.

29th IMO shortlist 1988

© John Scholes

jscholes@kalva.demon.co.uk

30 Dec 2002

Last corrected/updated 30 Dec 02