IMO 1998 SL 3

Origin: UKR

Problem

Let I be the incenter of triangle ABC. Let K, L, and M be the points of tangency of the incircle of ABC with AB, BC, and CA, respectively. The line t passes through B and is parallel to KL. The lines MK and ML intersect t at the points R and S. Prove that \angleRIS is acute.

Solution

Lemma. If U, W, V are three points on a line l in this order, and X a point in the plane with XW \perpUV , then \angleUXV < 90◦if and only if XW 2 > UW \cdot V W. Proof. Let XW 2 > UW \cdotV W, and let X0 be a point on the segment XW such that X0W 2 \geqUW \cdot V W. Then X0W/UW = V W/X0W, so that triangles X0WU and V WX0 are similar. Thus \angleUX0V = \angleUX0W + \angleWUX0 = 90◦, which immediately implies that \angleUXV < 90◦. Similarly, if XW 2 \leqUW \cdot V W, then \angleUXV \geq90◦. Since BI \perpRS, it will be enough by the lemma to show that BI2 > BR\cdotBS. Note that \triangleBKR ∼\triangleBSL: in fact, we have \angleKBR = \angleSBL = 90◦−\beta/2 and \angleBKR = \angleAKM = \angleKLM = \angleBSL = 90◦−\alpha/2. In particular, we obtain BR/BK = BL/BS = BK/BS, so that BR \cdot BS = BK2 < BI2. Second solution. Let E, F be the midpoints of KM and LM respectively. The quadrilaterals RBIE and SBIF are inscribed in the circles with diameters IR and IS. Now we have \angleRIS = \angleRMS +\angleIRM +\angleISM = 90◦−\beta/2 + \angleIBE + \angleIBF = 90◦−\beta/2 + \angleEBF. On the other hand, BE and BF are medians in \triangleBKM and \triangleBLM in which BM > BK and BM > BL. We conclude that \angleMBE < 1 2\angleMBK and \angleMBF < 1 2\angleMBL. Adding these two inequalities gives \angleEBF < \beta/2. Therefore \angleRIS < 90◦. Remark. It can be shown (using vectors) that the statement remains true for an arbitrary line t passing through B.