imo.wolfram.com
header_a.gif Wolfram Research  
mainscorescontestantscompetition problemsaboutphotoIMO Facts


41st International Mathematical Olympiad


IMO 2000 Problems and Solutions

Taejon, Korea
July 13-25, 2000



Problem 6

A1 A2 A3 is an acute-angled triangle. The foot of the altitude from Ai is Ki and the incircle touches the side opposite Ai at Li. The line K1 K2 is reflected in the line L1 L2. Similarly, the line K2 K3 is reflected in L2 L3 and K3 K1 is reflected in L3 L1. Show that the three new lines form a triangle with vertices on the incircle.

Back to Competition Problems page


Solution
Let O be the centre of the incircle. Let the line parallel to A1 A2 through L2 meet the line A2 O at X. We will show that X is the reflection of K2 in L2 L3. Let A1 A3 meet the line A2 O at B2. Now A2 K2 is perpendicular to K2 B2 and O L2 is perpendicular to L2 B2, so A2 K2 B2 and O L2 B2 are similar. Hence K2 L2/L2 B2 = A2 O/O B2. But O A3 is the angle bisector in the triangle A2 A3 B2, so A2 O/O B2 = A2 A3/B2 A3.

Take B'2 on the line A2 O such that L2 B2 = L2 B'2 (B'2 is distinct from B2 unless L2 B2 is perpendicular to the line). Then angle L2 B'2 X = angle A3 B2 A2. Also, since L2 X is parallel to A2 A1, angle L2 X B'2 = angle A3 A2 B2. So the triangles L2 X B'2 and A3 A2 B2 are similar. Hence A2 A3/B2 A3 = X L2/B'2 L2 = X L2/B2 L2 (since B'2 L2 = B2 L2).

Thus we have shown that K2 L2/L2 B2 = X L2/B2 L2 and hence that K2 L2 = X L2. L2 X is parallel to A2 A1 so angle A2 A1 A3 = angle A1 L2 X = angle L2 X K2 + angle L2 K2 X = 2 angle L2 X K2 (isosceles). So angle L2 X K2 = 1/2 angle A2 A1 A3 = angle A2 A1 O. L2 X and A2 A1 are parallel, so K2 X and O A1 are parallel. But O A1 is perpendicular to L2 L3, so K2 X is also perpendicular to L2 L3 and hence X is the reflection of K2 in L2 L3.

Now the angle K3 K2 A1 = angle A1 A2 A3 because it is 90 - angle K3 K2 A2 = 90 - angle K3 A3 A2 (A2 A3 K2 K3 is cyclic with A2 A3 a diameter) = angle A1 A2 A3.

So the reflection of K2 K3 in L2 L3 is a line through X making an angle A1 A2 A3 with L2 X; in other words, it is the line through X parallel to A2 A3.

Let Mi be the reflection of Li in Ai O. The angle M2 X L2 = 2 angle O X L2 = 2 angle A1 A2 O (since A1 A2 is parallel to L2 X) = angle A1 A2 A3, which is the angle between L2 X and A2 A3. So M2 X is parallel to A2 A3; in other words, M2 lies on the reflection of K2 K3 in L2 L3.

If follows similarly that M3 lies on the reflection. Similarly, the line M1 M3 is the reflection of K1 K3 in L1 L3, and the line M1 M2 is the reflection of K1 K2 in L1 L2 and hence the triangle formed by the intersections of the three reflections is just M1 M2 M3.


Back to Competition Problems page