In this paper We present the Smarandache's Orthic Theorem in the geometry of
the triangle

Patrascu, I.

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Scientia MagnaVol. 6 (2010), No. 2, 26-28Smarandache’s Orthic TheoremEdited by Ion PatrascuFratii Buzesti College, Craiova, RomaniaAbstract In this paper We present the Smarandache’s Orthic Theorem in the geometry ofthe triangle.Keywords Smarandache’s Orthic Theorem, triangle.§1. The main resultSmarandache’s Orthic TheoremGiven a triangle ABC whose angles are all acute (acute triangle), we consider A′B′C ′, thetriangle formed by the legs of its altitudes.In which conditions the expression:‖A′B′‖ · ‖B′C ′‖+ ‖B′C ′‖ · ‖C ′A′‖+ ‖C ′A′‖ · ‖A′B′‖is maximum?AB CC′ B′ya− xzxc− zb− yA′Proof. We have4ABC ∼ 4A′B′C ′4AB′C ∼ 4A′BC ′. (1)We note‖BA′‖ = x, ‖CB′‖ = y, ‖AC ′‖ = z.It results that‖A′C‖ = a− x, ‖B′A‖ = b− y, ‖C ′B‖ = c− z.B̂AC = B̂′A′C = B̂A′C ′; ÂBC = ÂB′C ′ = Â′B′C ′; B̂CA = B̂C ′A′ = B̂′C ′A.From these equalities it results the relation (1)Vol. 6 Smarandache’s Orthic Theorem 274A′BC ′ ∼ 4A′B′C ⇒ A′C ′a− x =x‖A′B′‖ , (2)4A′B′C ∼ 4AB′C ′ ⇒ A′C ′z=c− z‖B′c′‖ , (3)4AB′C ∼ 4A′B′C ⇒ B′C ′y=b− y‖A′B′‖ . (4)From (2), (3) and (4) we observe that the sum of the products from the problem is equalto:x(a− x) + y(b− y) + z(c− z) = 14(a2 + b2 + c2)− (x− a2)2 − (y − b2)2 − (z − c2)2,which will reach its maximum as long as x = a2 , y =b2 , z =c2 , that is when the altitudes’ legsare in the middle of the sides, therefore when the 4ABC is equilateral. The maximum of theexpression is 14 (a2 + b2 + c2).§2. Conclusion (Smarandache’s Orthic Theorem)If we note the lengths of the sides of the triangle 4ABC by ‖AB‖ = c, ‖BC‖ = a, ‖CA‖ =b, and the lengths of the sides of its orthic triangle 4A∗B∗C∗ by ‖A∗B∗‖ = c∗, ‖B∗C∗‖ =a∗, ‖C∗A∗‖ = b∗, then we proved that:4(a∗b∗ + b∗c∗ + c∗a∗) ≤ a2 + b2 + c2.§3. Open problems related to Smarandache’s Orthic The-orem1. Generalize this problem to polygons. Let A1A2 · · ·Am be a polygon and P a point insideit. From P we draw perpendiculars on each side AiAi+1 of the polygon and we note by Ai′ theintersection between the perpendicular and the side AiAi+1. A pedal polygon A1′A2′ · · ·Am′is formed. What properties does this pedal polygon have?2. Generalize this problem to polyhedrons. Let A1A2 · · ·An be a poliyhedron and P apoint inside it. From P we draw perpendiculars on each polyhedron face Fi and we note by Ai′the intersection between the perpendicular and the side Fi. A pedal polyhedron A1′A′2 · · ·Ap′is formed, where p is the number of polyhedron’s faces. What properties does this pedalpolyhedron have?References[1] Caˇtaˇlin Barbu, Teorema lui Smarandache, Teoreme fundamentale din geometria tri-unghiului, Chapter II, Teoreme fundamentale din geometria triunghiului, Section II. 57, EdituraUnique, Bacaˇu, 2008, p. 337.28 Edited by Ion Patrascu No. 2[2] Jo´zsef Sa´ndor, On Smarandache’s Pedal Theorem, Geometric Theorems, DiophantineEquations, and Arithmetic Functions, AR Press, Rehoboth, 2002, 9-10.[3] Ion Paˇtras¸cu, Smarandache’s Orthic Theorem, http://www.scribd.com/doc/28311593/Smarandache-s-Orthic-Theorem.[4] F. Smarandache, Eight Solved and Eight Open Problems in Elementary Geometry, inarXiv.org, Cornell University, NY, USA.[5] F. Smarandache, Proble`mes avec et san. . . proble`mes!, Somipress, Fe´s, Morocco, 1983,Problem 5. 41, p. 59.

Eight Solved and Eight Open Problems in Elementary Geometry, in arXiv.org,

On Smarandache’s Pedal Theorem, Geometric Theorems, Diophantine Equations, and Arithmetic Functions,

Proble`mes avec et

Smarandache’s Orthic Theorem,

Teorema lui Smarandache, Teoreme fundamentale din geometria triunghiului, Chapter II, Teoreme fundamentale din geometria triunghiului, Section II. 57, Editura Unique, Bacaˇu,

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Smarandache's Orthic Theorem

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