29th IMO shortlist 1988/15 solution

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