IMO 2000 SL G8

Origin: RUS | Category: Geometry

Problem

A1A2A3 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 K1K2 is reflected in the line L1L2. Similarly, the line K2K3 is reflected in L2L3, and K3K1 is reflected in L3L1. Show that the three new lines form a triangle with vertices on the incircle.

Solution

Denote by \alpha1, \alpha2, \alpha3 the angles of \triangleA1A2A3 at vertices A1, A2, A3 respec- tively. Let T1, T2, T3 be the points symmetric to L1, L2, L3 with respect to A1I, A2I, and A3I respectively. We claim that T1T2T3 is the desired triangle.

Denote by S1 and R1 the points symmetric to K1 and K3 with re- spect to L1L3. It is enough to show that T1 and T3 lie on the line R1S1. To prove this, we shall prove that \angleK1S1T1 = \angleK′K1S1 for a point K′ on the line K1K3 such that K3 and K′ lie on different sides of K1. We show first that S1 \inA1I. Let X be the point of intersection of lines A1I and L1L3. We see from the triangle A1L3X that \angleL1XI = \alpha3/2 = \angleL1A3I, which implies that L1XA3I is cyclic. A1 A2 A3 L1 L2 L3 T1 T2 T3 I X K1 K3 S1 K′ We now have \angleA1XA3 = 90◦= \angleA1K1A3; hence A1K1XA3 is also cyclic. It follows that \angleK1XI = \angleK1A3A1 = \alpha3 = 2\angleL1XI; hence X1L1 bisects the angle K1X1I. Hence S1 \inXI as claimed. Now we have \angleK1S1T1 = \angleK1S1L1 + 2\angleL1S1X = \angleS1K1L1 + 2\angleL1K1X. It remains to prove that K1X bisects \angleA3K1K′. From the cyclic quadrilat- eral A1K1XA3 we see that \angleXK1A3 = \alpha1/2. Since A1K3K1A3 is cyclic, we also have \angleK′K1A3 = \alpha1 = 2\angleXK1A3, which proves the claim.