Smarandache’s Cevians Theorem (II) Edited by Dr. M. Khoshnevisan Neuro Intelligence Center, Australia E-mail: mkhoshnevisan@neurointelligence-center.org
Abstract. In this paper we present the Smarandache’s Cevians Theorem (II) in the geometry of the triangle. Smarandache’s Cevians Theorem (II) In a triangle Δ ABC we draw the Cevians AA1 , BB1 , CC1 that intersect in P . Then: PA PB PC AB ⋅ BC ⋅ CA ⋅ ⋅ = PA1 PB1 PC1 A1 B ⋅ B1C ⋅ C1 A Solution 6. In the triangle Δ ABC we apply the Ceva’s theorem: AC1 ⋅ BA1 ⋅ CB1 = − AB1 ⋅ CA1 ⋅ BC1 (1) In the triangle Δ AA1 B , cut by the transversal CC1 , we’ll apply the Menelaus’ theorem: AC1 ⋅ BC ⋅ A1 P = AP ⋅ A1C ⋅ BC1 (2) In the triangle Δ BB1C , cut by the transversal AA1 , we apply again the Menelaus’ theorem: A B1
C1 P
C B
A1
BA1 ⋅ CA ⋅ B1 P = BP ⋅ B1 A ⋅ CA1 (3) We apply one more time the Menelaus’ theorem in the triangle Δ CC1 A cut by the transversal BB1 : AB ⋅ C1 P ⋅ CB1 = AB1 ⋅ CP ⋅ C1 B (4) We divide each relation (2), (3), and (4) by relation (1), and we obtain: PA BC B1 A (5) = ⋅ PA1 BA1 B1C
PB CA C1 B = ⋅ PB1 CB1 C1 A PC AB A1C = ⋅ PC1 AC1 A1 B
(6) (7)
Multiplying (5) by (6) and by (7), we have: PA PB PC AB ⋅ BC ⋅ CA AB1 ⋅ BC1 ⋅ CA1 ⋅ ⋅ = ⋅ PA1 PB1 PC1 A1 B ⋅ B1C ⋅ C1 A A1 B ⋅ B1C ⋅ C1 A
but the last fraction is equal to 1 in conformity to Ceva’s theorem. Unsolved Problem related to the Smarandache’s Cevians Theorem (II). Is it possible to generalize this problem for a polygon?
References: [1] F. Smarandache, Problèmes avec et sans… problèmes!, Problem 5.40, p. 58, Somipress, Fés, Morocco, 1983. [2] F. Smarandache, Eight Solved and Eight Open Problems in Elementary Geometry, in arXiv.org, Cornell University, NY, USA. [3] M. Khoshnevisan, Smarandache’s Cevian Theorem I, NeuroIntelligence Center, Australia, http://www.scribd.com/doc/28317760/Smarandache-s-Cevians-Theorem-II