A Class of Orthohomological Triangles Prof. Claudiu Coandă, National College “Carol I”, Craiova, Romania. Prof. Florentin Smarandache, ENSIETA, Brest, France. Prof. Ion Pătraşcu, National College “Fraţii Buzeşti”, Craiova, Romania.
Abstract. In this article we propose to determine the triangles’ class Ai Bi Ci orthohomological with a given triangle ABC , inscribed în the triangle ABC ( Ai ∈ BC , Bi ∈ AC , Ci ∈ AB ). We’ll remind, here, the fact that if the triangle Ai Bi Ci inscribed in ABC is orthohomologic with it, then the perpendiculars in Ai , Bi , respectively in Ci on BC , CA , respectively AB are concurrent in a point Pi (the orthological center of the given triangles), and the lines AAi , BBi , CCi are concurrent in point (the homological center of the given triangles). To find the triangles Ai Bi Ci , it will be sufficient to solve the following problem. Problem. Let’s consider a point Pi in the plane of the triangle ABC and Ai Bi Ci its pedal triangle. Determine the locus of point Pi such that the triangles ABC and Ai Bi Ci to be homological.
Solution. Let’s consider the triangle ABC , A(1, 0, 0), B(0,1, 0), C (0, 0,1) , Pi (α , β , γ ), α + β + γ = 0 . The perpendicular vectors on the sides are: ⊥ U BC ( 2a 2 , − a 2 − b 2 + c 2 , − a 2 + b 2 − c 2 )
and
⊥ U CA ( −a 2 − b2 + c 2 , 2b2 , a 2 − b2 − c 2 )
the
point
⊥ U AB ( − a 2 + b 2 − c 2 , a 2 − b 2 − c 2 , 2c 2 ) uuur The coordinates of the vector BC are (0, −1,1) , and the line BC has the equation x = 0 . The equation of the perpendicular raised from point Pi on BC is:
x
y
α 2a
z
β 2
γ
−a −b +c 2
2
2
=0
−a +b −c 2
2
2
We note Ai ( x, y, z ) , because Ai ∈ BC we have: x = 0 and y + z = 1 . The coordinates y and z of Ai can be found by solving the system of equations
1
x y β α 2 2 2 2 2a − a − b + c y + z = 0
We have:
y⋅
α
γ
2a 2 − a 2 + b 2 − c 2
z
γ
=0
−a +b −c 2
= z⋅
2
2
α
β
2a 2 − a 2 − b 2 + c 2
,
y α ( −a 2 + b 2 − c 2 ) − 2γ a 2 = z α ( − a 2 − b 2 + c 2 ) − 2 β a 2 , y+ y⋅
y⋅
α ( − a 2 + b 2 − c 2 ) − 2γ a 2
α ( −a 2 − b 2 + c 2 ) − 2β a 2
= 1,
α ( −a 2 − b 2 + c 2 ) − 2β a 2 + α ( − a 2 + b 2 − c 2 ) − 2γ a 2 α ( −a 2 − b2 + c 2 ) − 2β a 2
y⋅
−2a 2 (α + β + γ )
α ( −a 2 − b 2 + c 2 ) − 2 β a 2
it results
y= z = 1− y = 1− β −
α 2a
α
2a
2
2
(a
(a
2
2
= 1,
= 1,
+ b2 − c 2 ) + β
+ b2 − c2 ) = α + γ −
Therefore,
α 2a 2
(a
2
+ b2 − c2 ) .
α α Ai 0, a 2 + b2 − c2 ) + β , a 2 − b2 + c2 ) + γ . 2 ( 2 ( 2 a 2 a Similarly we find: β β Bi 2 ( a 2 + b 2 − c 2 ) + α , 0, −a 2 + b2 + c 2 ) + γ , 2 ( 2b 2b γ γ Ci 2 ( a 2 − b 2 + c 2 ) + α , −a 2 + b2 + c 2 ) + β , 0 . 2 ( 2c 2c We have:
2
α uuur a 2 − b2 + c2 ) + γ 2 ( Ai B α c cos B + γ a uuur = − 2a =− α α b cos C + β a AC i a 2 + b2 − c2 ) + β 2 ( 2a
β uuur a 2 + b2 − c2 ) + α 2 ( Bi C β a cos C + α b uuur = − 2b =− . α β c cos A + γ b 2 2 2 Bi A ( −a + b + c ) + γ 2a 2
γ uuur −a 2 + b 2 + c 2 ) + β 2 ( Ci A γ b cos A + β c uuur = − 2c =− γ γ a cos B + α c Ci B a 2 − b2 + c2 ) + α 2 ( 2c (We took into consideration the cosine’s theorem: a 2 = b 2 + c 2 − 2bc cos A ). In conformity with Ceva’s theorem, we have: uuur uuur uuur Ai B Bi C Ci A uuur ⋅ uuur ⋅ uuur = −1 . AC Bi A Ci B i
( aγ + α c cos B )( bα + β a cos C )( cβ + γ b cos A) = = ( aβ + α b cos C )( bγ + β c cos A )( cα + γ a cos B ) aα ( b2γ 2 − c 2 β 2 ) ( cos A − cos B cos C ) + bβ ( c 2α 2 − a 2γ 2 ) ( cos B − cos A cos C ) + +cγ ( a 2 β 2 − b 2α 2 ) ( cos C − cos A cos B ) = 0 . Dividing it by a 2b 2 c 2 , we obtain that the equation in barycentric coordinates of the locus L of the point Pi is:
α γ2 a c2
−
β2
( cos A − cos B cos C ) +
b2
β α2 b a2
−
γ2
( cos B − cos A cos C ) + c2
γ β2 α2 + 2 − 2 ( cos C − cos A cos B ) = 0 . cb a We note d A , d B , d C the distances oriented from the point Pi to the sides BC , CA respectively AB , and we have: α d A β d B γ dC = , = , = . a 2s b 2s c 2s The locus’ L equation can be written as follows: d A ( dC2 − d B2 ) ( cos A − cos B cos C ) + d B ( d A2 − dC2 ) ( cos B − cos A cos C ) + + dC ( d B2 − d A2 ) ( cos C − cos A cos B ) = 0
Remarks.
3
1. It is obvious that the triangle’s ABC orthocenter belongs to locus L. The orthic triangle and the triangle ABC are orthohomologic; a orthological center is the orthocenter H , which is the center of homology. 2. The center of the inscribed circle in the triangle ABC belongs to the locus L, because d A =d B = d C = r and thus locus’ equation is quickly verified. Theorem (Smarandache-Pătraşcu). If a point P belongs to locus L, then also its isogonal P ' belongs to locus L. Proof. Let P (α , β , γ ) a point that verifies the locus’ L. equation, and P ' α ' , β ' , γ ' its isogonal
(
in the triangle ABC . It is known that
α γ '
'2
αα ' a
2
=
ββ ' b
2
=
γγ ' c
2
)
. We’ll prove that P ' ∈ L, i.e.
β '2
( cos A − cos B cos C ) = 0 b2 α ' γ '2b 2 − β '2 c 2 ∑ a b2c 2 ( cos A − cos B cos C ) = 0
∑ac
2
−
α'
∑ ab c (γ 2 2
⇔∑
b − β '2 c 2 ) ( cos A − cos B cos C ) = 0 ⇔
'2 2
α ' γ ' ββ 'c 2 c 2γγ ' β ' − ( cos A − cos B cos C ) = 0 ⇔ ab 2c 2 γ β
α ' β 'γ ' β c 2 γ b 2 − ( cos A − cos B cos C ) = 0 ⇔ ab 2 c 2 γ β α ' β 'γ ' β 2 c 2 − γ 2b 2 ⇔∑ 2 2 ( cos A − cos B cos C ) = 0 ⇔ ab c βγ ⇔∑
⇔∑
2 α α ' β 'γ ' 1 γ2 2 2 β 2 2 ⋅ b c 2 − 2 ( cos A − cos B cos C ) = 0 . a αβγ b c c b
We obtain that:
α ' β 'γ ' α γ 2 β 2 ∑ a c2 − b2 ( cos A − cos B cos C ) = 0 , αβγ this is true because P ∈ L. Remark. We saw that the triangle ‘s ABC orthocenter H belongs to the locus, from the precedent theorem it results that also O , the center of the circumscribed circle to the triangle ABC (isogonable to H ), belongs to the locus. Open problem: What does it represent from the geometry’s point of view the equation of locus L? 4
In the particular case of an equilateral triangle we can formulate the following: Proposition: The locus of the point P from the plane of the equilateral triangle ABC with the property that the pedal triangle of P and the triangle ABC are homological, is the union of the triangle’s heights. Proof: Let P (α , β , γ ) a point that belongs to locus L. The equation of the locus becomes:
α ( γ 2 − β 2 ) + β (α 2 − γ 2 ) + γ ( β 2 − α 2 ) = 0
Because:
α ( γ 2 − β 2 ) + β (α 2 − γ 2 ) + γ ( β 2 − α 2 ) = αγ 2 − αβ 2 + βα 2 − βγ 2 + γβ 2 − γα 2 = = αβγ + αγ 2 − αβ 2 + βα 2 − βγ 2 + γβ 2 − γα 2 − αβγ =
= αβ ( γ − β ) + αγ ( γ − β ) − α 2 ( γ − β ) − βγ ( γ − β ) = = ( γ − β ) α ( β − α ) − γ ( β − α ) = ( β − α )(α − γ )( γ − β ) . We obtain that α = β or β = γ or γ = α , that shows that P belongs to the medians (heights) of the triangle ABC .
References:
[1] [2]
C. Coandă , Geometrie analitică în coordanate baricentrice, Editura Reprograph, Craiova, 2005. Multispace & Multistructure. Neutrosophic Trandisciplinarity (100 Collected Papers of Sciences), vol. IV, North European Scientific Publishers, Hanko, Finland, 2010.
5