A Class of Orthohomological Triangles

Transcript

1 A Class of Orthohomologcal Trangles Prof. Claudu Coandă Natonal College Carol I Craova Romana. Prof. Florentn Smarandache Unversty of New Mexco Gallup USA Prof. Ion Pătraşcu Natonal College Fraţ Buzeşt Craova Romana. Abstract. In ths artcle we propose to determne the trangles class A orthohomologcal wth a gven trangle A nscrbed în the trangle A ( A B AC C AB ). We ll remnd here the fact that f the trangle A nscrbed n A s orthohomologc wth t then the perpendculars n A B respectvely n C on CA respectvely AB are concurrent n a pont P (the orthologcal center of the gven trangles) and the lnes AA BB CC are concurrent n pont (the homologcal center of the gven trangles). To fnd the trangles A t wll be suffcent to solve the followng problem. Problem. Let s consder a pont P n the plane of the trangle A and A ts pedal trangle. Determne the locus of pont P such that the trangles A and A to be homologcal. Soluton. Let s consder the trangle A A(100) B(010) C (001) and the pont P ( α βγ ) α β γ. The perpendcular vectors on the sdes are: CA ( ) ( ) ( ) U a a b c a b c U a b c b a b c UAB a b c a b c c The coordnates of the vector are (0 11) and the lne has the equaton x. The equaton of the perpendcular rased from pont P on s: x y z α β γ a a b c a b c We note A ( xyz ) because A we have: x and y z = 1. The coordnates y and z of A can be found by solvng the system of equatons 1

2 We have: x y z α β γ a a b c a b c y z α γ α β y = z a a b c a a b c ( ) ( ) α ( a b c ) γa y y = 1 α ( a b c ) βa ( ) ( ) α ( a b c ) βa y α a b c γa = z α a b c βa α a b c βa α a b c γa y = 1 ( α β γ) a y = 1 α a b c βa ( ) t results α y = ( a b c ) β a z 1 y α 1 ( ) ( ) a b c α = = β = α γ a a a b c. Therefore α α A 0 ( a b c ) β ( a b c ) γ a a. Smlarly we fnd: β β ( ) 0 ( B a b c α a b c ) γ b b γ γ C ( a b c ) α ( a b c ) β 0 c c. We have:

3 α ( a b c ) γ AB a αccos B γ a = = AC α cos ( a b c ) β αb C βa a β ( a b c ) α b βacosc αb = =. BA α ccos A b ( a b c ) γ β γ a γ ( a b c CA ) β c γbcos A βc = = CB γ acos B c ( a b c γ α ) α c (We took nto consderaton the cosne s theorem: a = b c bccosa). In conformty wth Ceva s theorem we have: AB CA =1. AC BA CB aγ αccos B bα βacos C cβ γbcos A = ( )( )( ) = ( aβ αbcosc)( bγ βccos A)( cα γacos B) ( )( cos cos cos ) ( )( cos cos cos ) aα b γ c β A B C bβ c α a γ B A C ( )( ) cγ a β b α cosc cos Acos B. Dvdng t by abc we obtan that the equaton n barycentrc coordnates of the locus L of the pont P s: α γ β β α γ ( cos A cos BcosC) ( cos B cos AcosC) a c b b a c γ β α ( cosc cos Acos B). c b a We note da db d C the dstances orented from the pont P to the sdes CA respectvely AB and we have: α da β db γ d = = = C. a s b s c s The locus L equaton can be wrtten as follows: cos cos cos d d d A B C d d d cos B cos A cos C Remarks. ( )( ) ( )( ) A C B B A C dc db da cosc cos Acos B 0 ( )( ) = 3

4 1. It s obvous that the trangle s A orthocenter belongs to locus L. The orthc trangle and the trangle A are orthohomologc; a orthologcal center s the orthocenter H whch s the center of homology.. The center of the nscrbed crcle n the trangle A belongs to the locus L because da= db= dc = r and thus locus equaton s quckly verfed. Theorem (Smarandache-Pătraşcu). If a pont P belongs to locus L then also ts sogonal 4 P belongs to locus L. Proof. Let P( α βγ ) a pont that verfes the locus L. equaton and P ( α β γ ) ts sogonal αα ββ γγ n the trangle A. It s known that = =. We ll prove that P L.e. a b c α γ β ( cos A cos BcosC) a c b α γ b β c ( cos A cos BcosC) a b c α ( γ b β c )( cos A cos BcosC) ab c α γ ββ c c γγ β ( cos A cos BcosC) ab c γ β αβγ βc γb ( cos A cos BcosC) ab c γ β αβγ β c γ b ( cos A cos BcosC) ab c βγ α α βγ 1 β γ bc ( cos A cos BcosC). a αβγ b c b c We obtan that: αβγ α γ β ( cos A cos BcosC) αβγ a c b ths s true because P L. Remark. We saw that the trangle s A orthocenter H belongs to the locus from the precedent theorem t results that also O the center of the crcumscrbed crcle to the trangle A (sogonable to H ) belongs to the locus. Open problem: What does t represent from the geometry s pont of vew the equaton of locus L?

5 In the partcular case of an equlateral trangle we can formulate the followng: Proposton: The locus of the pont P from the plane of the equlateral trangle A wth the property that the pedal trangle of P and the trangle A are homologcal s the unon of the trangle s heghts. Proof: P α βγ a pont that belongs to locus L. The equaton of the locus becomes: Let ( ) ( ) ( ) ( ) α γ β β α γ γ β α Because: ( ) ( ) ( ) α γ β β α γ γ β α = αγ αβ βα βγ γβ γα = = αβγ αγ αβ βα βγ γβ γα αβγ = = αβ γ β αγ γ β α γ β βγ γ β = ( ) ( ) ( ) ( ) ( γ β) α ( β α ) γ ( β α ) ( β α)( α γ)( γ β) = =. We obtan that α = β or β = γ or γ = α that shows that P belongs to the medans (heghts) of the trangle A. References: [1] C. Coandă Geometre analtcă în coordanate barcentrce Edtura Reprograph Craova 005. [] Multspace & Multstructure. Neutrosophc Trandscplnarty (100 Collected Papers of Scences) vol. IV North European Scentfc Publshers Hanko Fnland