IMO 2002 SL G7

Origin: BUL | Category: Geometry

Problem

The incircle Ωof the acute-angled triangle ABC is tangent to BC at K. Let AD be an altitude of triangle ABC and let M be the midpoint of AD. If N is the other common point of Ωand KM, prove that Ωand the circumcircle of triangle BCN are tangent at N.

Solution

Let k be the circle through B, C that is tangent to the circle Ωat point N ′. We must prove that K, M, N ′ are collinear. Since the statement is trivial for AB = AC, we may assume that AC > AB. As usual, R, r, \alpha, \beta, \gamma denote the circumradius and the inradius and the angles of \triangleABC, re- spectively. We have tan \angleBKM = DM/DK. Straightforward calculation gives DM = 1 2AD = R sin \beta sin \gamma and DK = DC−DB −KC−KB = R sin(\beta − \gamma) −R(sin \beta −sin \gamma) = 4R sin \beta−\gamma sin \beta 2 sin \gamma 2, so we obtain tan \angleBKM = sin \beta sin \gamma 4 sin \beta−\gamma sin \beta 2 sin \gamma = cos \beta 2 cos \gamma sin \beta−\gamma . To calculate the angle BKN ′, we apply the inversion \psi with center at K and power BK \cdotCK. For each object X, we denote by < X its image under \psi. The incircle Ωmaps to a K B C A k Ω N −\to \psi <B <C <U < N <Ω K <k

line <Ωparallel to <B <C, at distance BK\cdotCK 2r from <B <C. Thus the point @ N ′ is the projection of the midpoint <U of <B <C onto <Ω. Hence tan \angleBKN ′ = tan \angle<BK @ N ′ = <U @ N ′ <UK

BK \cdot CK r(CK −BK). Again, one easily checks that KB \cdot KC = bc sin2 \alpha 2 and r = 4R sin \alpha 2 \cdot sin \beta 2 \cdot sin \gamma 2 , which implies tan \angleBKN ′ = bc sin2 \alpha r(b −c)

4R2 sin \beta sin \gamma sin2 \alpha 4R sin \alpha 2 sin \beta 2 sin \gamma 2 \cdot 2R(sin \beta −sin \gamma) = cos \beta 2 cos \gamma sin \beta−\gamma . Hence \angleBKM = \angleBKN ′, which implies that K, M, N ′ are indeed collinear; thus N ′ \equivN.