Scientia Magna Vol. 6 (2010), No. 1, 7-8
Smarandache’s ratio theorem Edited by M. Khoshnevisan Neuro Intelligence Center, Australia E-mail: mkhoshnevisan@neurointelligence-center.org Abstract In this paper we present the Smarandache’s ratio theorem in the geometry of the triangle. Keywords Smarandache’s ratio theorem, triangle.
§1. The main result Smarandache’s Ratio Theorem If the points A1 , B1 , C1 divide the sides k BC k= a, k CA k= b, respectively k AB k= c of a triangle 4ABC in the same ratio k > 0, then k AA1 k2 + k BB1 k2 + k CC1 k2 ≥
3 2 (a + b2 + c2 ). 4
Proof. Suppose k > 0 because we work with distances. k BA1 k= k k BC k, k CB1 k= k k CA k, k AC1 k= k k AB k . We’ll apply tree times Stewart’s theorem in the triangle 4ABC, with the segments AA1 , BB1 , respectively CC1 : k AB k2 · k BC k (1 − k)+ k AC k2 · k BC k k− k AA1 k2 · k BC k=k BC k3 (1 − k)k, where k AA1 k2 = (1 − k) k AB k2 +k k AC k2 −(1 − k)k k BC k2 . Similarly, k BB1 k2 = (1 − k) k BC k2 +k k BA k2 −(1 − k)k k AC k2 . k CC1 k2 = (1 − k) k CA k2 +k k CB k2 −(1 − k)k k AB k2 . By adding these three equalities we obtain: k AA1 k2 + k BB1 k2 + k CC1 k2 = (k 2 − k + 1)(k AB k2 + k BC k2 + k CA k2 ), 1 which takes the minimum value when k = , which is the case when the three lines from the 2 enouncement are the medians of the triangle. 3 The minimum is (k AB k2 + k BC k2 + k CA k2 ). 4