41st International Mathematical OlympiadIMO 2000 Problems and SolutionsTaejon, KoreaJuly 13-25, 2000 Problem 6A_{1} A_{2} A_{3} is an acute-angled triangle. The foot of the altitude from A_{i} is K_{i} and the incircle touches the side opposite A_{i} at L_{i}. The line K_{1} K_{2} is reflected in the line L_{1} L_{2}. Similarly, the line K_{2} K_{3} is reflected in L_{2} L_{3} and K_{3} K_{1} is reflected in L_{3} L_{1}. Show that the three new lines form a triangle with vertices on the incircle. Back to Competition Problems page SolutionLet O be the centre of the incircle. Let the line parallel to A_{1} A_{2} through L_{2} meet the line A_{2} O at X. We will show that X is the reflection of K_{2} in L_{2} L_{3}. Let A_{1} A_{3} meet the line A_{2} O at B_{2}. Now A_{2} K_{2} is perpendicular to K_{2} B_{2} and O L_{2} is perpendicular to L_{2} B_{2}, so A_{2} K_{2} B_{2} and O L_{2} B_{2} are similar. Hence K_{2} L_{2}/L_{2} B_{2} = A_{2} O/O B_{2}. But O A_{3} is the angle bisector in the triangle A_{2} A_{3} B_{2}, so A_{2} O/O B_{2} = A_{2} A_{3}/B_{2} A_{3}.Take B'_{2} on the line A_{2} O such that L_{2} B_{2} = L_{2} B'_{2} (B'_{2} is distinct from B_{2} unless L_{2} B_{2} is perpendicular to the line). Then angle L_{2} B'_{2} X = angle A_{3} B_{2} A_{2}. Also, since L_{2} X is parallel to A_{2} A_{1}, angle L_{2} X B'_{2} = angle A_{3} A_{2} B_{2}. So the triangles L_{2} X B'_{2} and A_{3} A_{2} B_{2} are similar. Hence A_{2} A_{3}/B_{2} A_{3} = X L_{2}/B'_{2} L_{2} = X L_{2}/B_{2} L_{2} (since B'_{2} L_{2} = B_{2} L_{2}). Thus we have shown that K_{2} L_{2}/L_{2} B_{2} = X L_{2}/B_{2} L_{2} and hence that K_{2} L_{2} = X L_{2}. L_{2} X is parallel to A_{2} A_{1} so angle A_{2} A_{1} A_{3} = angle A_{1} L_{2} X = angle L_{2} X K_{2} + angle L_{2} K_{2} X = 2 angle L_{2} X K_{2} (isosceles). So angle L_{2} X K_{2} = 1/2 angle A_{2} A_{1} A_{3} = angle A_{2} A_{1} O. L_{2} X and A_{2} A_{1} are parallel, so K_{2} X and O A_{1} are parallel. But O A_{1} is perpendicular to L_{2} L_{3}, so K_{2} X is also perpendicular to L_{2} L_{3} and hence X is the reflection of K_{2} in L_{2} L_{3}. Now the angle K_{3} K_{2} A_{1} = angle A_{1} A_{2} A_{3} because it is 90 - angle K_{3} K_{2} A_{2} = 90 - angle K_{3} A_{3} A_{2} (A_{2} A_{3} K_{2} K_{3} is cyclic with A_{2} A_{3} a diameter) = angle A_{1} A_{2} A_{3}. So the reflection of K_{2} K_{3} in L_{2} L_{3} is a line through X making an angle A_{1} A_{2} A_{3} with L_{2} X; in other words, it is the line through X parallel to A_{2} A_{3}. Let M_{i} be the reflection of L_{i} in A_{i} O. The angle M_{2} X L_{2} = 2 angle O X L_{2} = 2 angle A_{1} A_{2} O (since A_{1} A_{2} is parallel to L_{2} X) = angle A_{1} A_{2} A_{3}, which is the angle between L_{2} X and A_{2} A_{3}. So M_{2} X is parallel to A_{2} A_{3}; in other words, M_{2} lies on the reflection of K_{2} K_{3} in L_{2} L_{3}. If follows similarly that M_{3} lies on the reflection. Similarly,
the line M_{1} M_{3} is the reflection of K_{1}
K_{3} in L_{1} L_{3}, and the line
M_{1} M_{2} is the reflection of K_{1}
K_{2} in L_{1} L_{2} and hence the triangle
formed by the intersections of the three reflections is just M_{1}
M_{2} M_{3}.
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