Scientia Magna Vol. 6 (2010), No. 1, 36-39
A proof of Smarandache-Patrascu’s theorem using barycentric coordinates Claudiu Coanda The National College “Carol I” Craiova, Romania Abstract In this article we prove the Smarandache-Patrascu’s theorem in relation to the inscribed orthohomological triangles using the barycentric coordinates. Keywords Smarandache-Patrascu’s theorem, barycentric coordinates.
Definition. Two triangles and ABC and A1 B1 C1 , where A1 ∈ BC, B1 ∈ AC, C1 ∈ AB, are called inscribed ortho homological triangles if the perpendiculars in A1 , B1 , C1 on BC, AC, AB respectively are concurrent. Observation. The concurrency point of the perpendiculars on the triangle ABC’s sides from above definition is the orthological center of triangles ABC and A1 B1 C1 . Smarandache-Patrascu Theorem. If the triangles ABC and A1 B1 C1 are orthohomo0 0 0 logical, then the pedal triangle A1 B1 C1 of the second center of orthology of triangles ABC and A1 B1 C1 , and the triangle ABC are orthohomological triangles. Proof. Let P (α, β, γ), α + β + γ = 1, be the first orthologic center of triangles ABC and A1 B1 C1 (See Figure 1).
Fig. 1 The perpendicular vectors on the sides are: ¡ ¢ ⊥ UBC = 2a2 , −a2 − b2 + c2 , −a2 + b2 − c2 , ¢ ¡ ⊥ UCA = −a2 − b2 + c2 , 2b2 , a2 − b2 − c2 , ¡ ¢ ⊥ UAB = −a2 + b2 − c2 , a2 − b2 − c2 , 2c2 . We know that: −−→ A1 B −−→ A1 C
−−→ B1 C · −−→ B1 A
−−→ C1 A · −−→ = −1, C1 B
(1)
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A proof of Smarandache-Patrascu’s theorem using barycentric coordinates
37
and we want to prove that: −− → 0 A1 B −− → 0 A1 C
−−0→ B1 C · −− → 0 B1 A
−−0→ C1 A · −− → = −1. 0 C1 B
−−→ B1 C · −−→ B1 A
−−→ C1 A · −−→ C1 B
−− → 0 A1 B · −−→ 0 A1 C
(2)
We will show that: −−→ A1 B −−→ A1 C
−−0→ B1 C · −− → 0 B1 A
−−0→ C1 A · −− → = 1. 0 C1 B
implies the relation (2) The equation of the line BC is x=0, and the equation of the line P A1 is ¯ ¯ ¯ 0 ¯ y z ¯ ¯ ¯ ¯ ¯ α ¯ = 0. β γ ¯ ¯ ¯ ¯ ¯2a2 −a2 − b2 + c2 −a2 + b2 − c2 ¯ It results that: ¯ ¯ ¯ α y· ¯¯ ¯2a2
¯ ¯ ¯ ¯ ¯ ¯ α ¯ = z· ¯ ¯ 2 2 2 2¯ ¯2a −a + b + c ¯ γ
¯ ¯ ¯ ¯ = 0. 2 2 2¯ −a − b + c ¯ β
Because y+z=1, we find: ´ ³ α α A1 0, 2 (a2 + b2 − c2 ) + β, 2 (a2 − b2 − c2 ) + γ . 2a 2a Similarly:
¶ −β 2 −β 2 2 2 2 2 (−a − b + c ) + α, 0, (a − b − c ) + γ , 2b2 2b2 ¶ µ γ 2 −γ 2 2 2 2 2 C1 (−a + b − c ) + α, 2 (a − b − c ) + β, 0 . 2c2 2c µ
B1
We will make the following notations: −a2 + b2 − c2 = i, −a2 − b2 + c2 = j, a2 − b2 − c2 = k, And we have:
µ
¶ −α −α A1 0, 2 j + β, 2 i + γ , 2a 2a ¶ µ −β −β j + α, 0, k + γ , B1 2b2 2b2 ¶ µ −γ −γ i + α, 2 k + β, 0 , C1 2c2 2c −−→ −α A1 B 2a2 i + γ = − ; −−→ −α A1 C 2a2 j + β −−→ −β B1 C 2b2 j + α ; −−→ = − −β B1 A 2b2 k + γ
38
Claudiu Coanda
If P 0
0
0
³
0
0
α ,β ,γ
0
No. 1
−−→ −γ C1 A 2c2 k + β = − . −−→ −γ C1 B 2c2 i + α
´
is the second center of orthology of the triangles ABC and A1 B1 C1 , and
0
0
A1 , B1 , C1 are the projections of P on BC, AC, AB respectively, similarly, we will find: 0 −− → 0 0 −α A1 B 2a2 i + γ ; = − −− → 0 −α0 0 A1 C 2a2 j + β 0 −−0→ 0 −β B1 C 2b2 j + α = − ; −−0→ 0 −β 0 B1 A 2b2 k + γ 0 −−0→ 0 −γ C1 A 2c2 k + β . −−0→ = − −γ 0 0 C1 B 2c2 i + α
It is known the theorem [2]. 0 0 0 0 Theorem. Given two isogonal conjugated points P (α, β, γ) and P (α , β , γ ) with respect to the triangle ABC (BC=a, CA=b, AB=c), then: 0
0
0
ββ γγ αα = 2 = 2 . a2 b c On the other side: −−→ A1 B −−→ A1 C
´ ¢ ³ −α0 ¡ −α 0 −− → 0 2a2 i + γ 2a2 i + γ A1 B ´= · −−→ = ¡ ¢ ³ −α0 0 0 −α A1 C 2a2 j + β 2a2 j + β −−→ B1 C −−→ B1 A
−−0→ B1 C · −− → = 0 B1 A
ββ 2 4b4 j − ββ 0 2 4b4 k −
−−→ C1 A −−→ C1 B
−−0→ C1 A · −− →= 0 C1 B
γγ 2 4c4 k γγ 0 2 4c4 i
0
0
− −
0
0
αγ αα 2 4a4 i − 2a2 i αβ 0 αα0 2 4a4 j − 2a2 j
0
βα 2b2 j − βγ 0 2b2 k − 0
γβ 2c2 k − γα0 2c2 i −
− −
0
0 α γ 2a2 i + γγ 0 0 α β 2a2 j + ββ
=
U1 ; V1
0
0 αβ 2b2 j + αα 0 β0 γ 2b2 k + γγ
=
U2 ; V2
=
U3 . V3
0
0 γ β 2c2 k + ββ 0 γ0 α 2c2 i + αα
The only thing left to be proved is that: U1 U2 U3 · · =1 V 1 V 2 V3 if and only if a2 c2 U1
V1 We show that
·
b2 a2 U2
V2
·
c2 b2 U3
V3
= 1.
b2 c2 a2 U = V , U = V , U1 = V3 ; 2 1 3 2 a2 b2 c2 0
0
0
0
0
0
0 0 ββ 2 βα αβ b2 αα βα βα b2 U = j − j − j + αα = 4 j 2 − 2 j − 2 j + ββ = V1 ; 2 2 2 2 2 2 2 a 4a b 2a 2a a 4a 2a 2a
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A proof of Smarandache-Patrascu’s theorem using barycentric coordinates
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0
0
0
0
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0
0 c2 γγ γβ γβ c2 0 ββ γβ γβ U3 = 2 2 k 2 − 2 k − 2 k + 2 ββ = 4 k 2 − 2 k − − 2 k + γγ = V2 ; 2 b 4c b 2b 2b b 4b 2b 2b
0
0
0
0
0
0
0 a2 αα αγ αγ a2 0 γγ αγ αγ U1 = 2 2 i2 − 2 i − 2 i + 2 γγ = 4 i2 − 2 i − − 2 i + αα = V3 . 2 c 4a c 2c 2c c 4c 2c 2c
References [1] Ion Patrscu and Florentin Smarandache, A Theorem about Simultaneous Orthological and Homological Triangles, in arXiv.org, Cornell University, Ny, USA, 2010. [2] Mihai Dicu, The Smarandache-Patrascu Theorem of Orthohomological Triangles, http:// www.scribd.com/doc/28311880/Smarandache-Patrascu-theorem of Orthohomological-Triangles. [3] Claudiu Coanda, Geometrie Analitica in coordonate baricentrie, Editura Reprographh, Craiova, 2005.