# Problema 1135

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• Publicado : 19 de agosto de 2010

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Problem 11325. Let P be a point within a triangle ABC. Let the lines AP, BP and CP intersect the sides BC, CA and AB at L, M and N , respectively. Show that |AP ||BP ||CP | ≥ 8|P L||P M ||P N |, withequality if and only if P is the centroid of ABC.

Let O = A be the origin for a system of vectors as shown in the picture above. The idea is to reduce the question to a two-variable calculusproblem. To this end, we introduce the parameters t, s ∈ (0, 1) that deﬁne the positions of L and N as l = tb + (1 − t)c and n = sb. The position vector of the point of intersection of the lines AL and CNis obtained by ﬁnding scalars α, β for which p = αl = βn + (1 − β)c. Using the fact that {b, c} is a linearly independent set, we ﬁnd s t α= , β= , t + s − st t + s − st which gives sl . p= t + s −st The position vector of the point M is obtained as the combination rp+(1−r)b for which the component in b vanishes. This occurs exactly when r = (t + s − st)/(t + s − 2st). With this |AP | α s = = |PL| 1−α t(1 − s) and , |CP | β t = = , |P N | 1−β s(1 − t)

|BP | 1 s(1 − t) + t(1 − s) = = . |P M | r−1 st

Therefore

|AP | |BP | |CP | ≥8 |P L| |P M | |P N | 1 1 is equivalent to proving thatthe function ϕ(s, t) = + attains at t = s = 1 its unique 2 t(1 − s) s(1 − t) minimum value 8 in the region (0, 1) × (0, 1). Note that ϕ(s, t) must have at least one minimum point in the interior ofthe square since it tends to inﬁnity at the boundary. The set of equations ϕs = ϕt = 0 corresponds to 1 1 1 1 − 2 =0 , − 2 = 0. 2 2 t(1 − s) s (1 − t) s(1 − t) t (1 − s) Hence s2 (1 − t) = t(1 − s)2 ,s(1 − t)2 = t2 (1 − s) , from where after division s/(1 − t) = t/(1 − s), that is, s(1 − s) = t(1 − t). The parabola y = x(1 − x) is symmetric with respect to the line x = 1 , thus either s = t or s =1 − t. If s = t then inserting in 2 the ﬁrst equation gives s = 1 − s, that is, s = 1 = t. If s = 1 − t then s3 = (1 − s)3 , whereby again 2 1 s = 1 − s and s = 2 = t. Thus the minimum value of ϕ...