A Proof of Smarandache-Pătraşcu’s Theorem using Barycentric Coordinates

A Proof of Smarandache-Pătraşcu’s Theorem using Barycentric Coordinates Prof Claudiu Coandă The National College “Carol I” Craiova, Romania In this article we prove the Smarandache–Pătraşcu’s Theorem in relation to the inscribed orthohomological triangles using the barycentric coordinates.

Will remind the following: Definition Two triangles ABC and A1 B1C1 , 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 P of the perpendiculars on the triangle ABC ’s sides from above definition is the orthological center of triangles ABC and A1 B1C1 .

Smarandache–Pătraşcu Theorem If the triangles ABC and A1 B1C1 are orthohomological, then the pedal triangle A1' B1'C1' of the second center of orthology of triangles ABC and A1 B1C1 , 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 B1C1 (See figure).

A

B’1

C1

B1

C1’

P P’ B

B

A1’

A1

The perpendicular vectors on the sides are: U BC⊥ ( 2a 2 , − a 2 − b 2 + c 2 , − a 2 + b 2 − c 2 )

U CA⊥ = ( − a 2 − b 2 + c 2 , 2b 2 , a 2 − b 2 − c 2 )

U AB⊥ = ( − a 2 + b 2 − c 2 , a 2 − b 2 − c 2 , 2c 2 )

We know that:

JJJG JJJG JJJG A1 B B1C C1 A JJJG ⋅ JJJG ⋅ JJJG = −1 (1) A1C B1 A C1 B and we want to prove that: JJJG JJJG JJJG A1' B B1'C C1' A JJJG (2) ⋅ JJJG ⋅ JJJG = −1 A1'C B1' A C1' B We will show that: JJJG JJJG JJJG JJJG JJJG JJJG A1 B B1C C1 A A1' B B1'C C1' A JJJG ⋅ JJJG ⋅ JJJG ⋅ JJJG ⋅ JJJG ⋅ JJJG = 1 A1C B1 A C1 B A1'C B1' A C1' B implies the relation (2) The equation of the line BC is x = 0 , and the equation of the line PA1 is 0

y

z

α

β

γ

=0

2a 2 − a 2 − b 2 + c 2 − a 2 + b 2 − c 2

It results that:

y⋅

α

γ

= z⋅

α

β

.

2a − a + b + c 2a − a − b 2 + c 2 Because y + z = 1 , we find: α α ⎛ ⎞ A1 ⎜ 0, a 2 + b2 − c2 ) + β , a 2 − b2 + c2 ) + γ ⎟ .

2 ( 2 ( 2a 2a ⎝ ⎠ Similarly: −β 2 ⎛ −β ⎞ B1 ⎜ 2 ( − a 2 − b 2 + c 2 ) + α , 0, a − b2 − c2 ) + γ ⎟ , 2 ( 2b ⎝ 2b ⎠ −γ 2 ⎛ −γ ⎞ C1 ⎜ 2 ( − a 2 + b 2 − c 2 ) + α , a − b2 − c2 ) + β , 0 ⎟ 2 ( 2c ⎝ 2c ⎠ We’ll make the following notations: − a 2 + b 2 − c 2 = i; − a 2 − b 2 + c 2 = j , a 2 − b 2 − c 2 = k , And we have: 2

2

2

2

2

2

−α −α ⎛ ⎞ A1 ⎜ 0, j + β, i +γ ⎟ 2 2 2a 2a ⎝ ⎠ −β ⎛ −β ⎞ B1 ⎜ 2 j + α , 0, k +γ ⎟ 2 2b ⎝ 2b ⎠ −γ ⎛ −γ ⎞ C1 ⎜ 2 i + α , k + β, 0⎟ 2 2c ⎝ 2c ⎠ −α JJJG i +γ 2 A1 B JJJG = − 2a ; −α A1C j+β 2a 2 −β JJJG j +α 2 B1C JJJG = − 2b ; −β B1 A k +γ 2b 2 −γ JJJG k+β 2 C1 A c 2 JJJG = − −γ C1 B i +α 2c 2 If P ' (α ', β ', γ ') is the second center of orthology of the triangles ABC and A1 B1C1 , and A1' , B1' , C1' are the projections of P ' on BC , AC , AB respectively, similarly, we’ll find:

−α ' JJJG i +γ ' ' 2 A1 B 2 a JJJG =− ; −α ' A1'C β + ' i 2a 2 −β ' JJJG j +α ' ' 2 B1C 2 b JJJG = − ; −β ' B1' A k +γ ' 2b 2 −γ ' JJJG k+β' 2 C1' A JJJG = − 2c −γ ' C1' B i +α ' 2c 2 It is known the theorem [2] 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: αα ' ββ ' γγ ' = 2 = 2 . a2 b c On the other side:

⎛ −α ⎞ ⎛ −α ' ⎞ αα ' 2 αγ ' α ' γ JJJG JJJG i + γ ⎟ ⎜ 2 i + γ '⎟ i − 2 i − 2 i + γγ ' ' ⎜ 2 4 U A1 B A1 B 2a 2a ⎝ ⎠ ⎝ ⎠ 4 a 2a 2a JJJG JJJG ⋅ = = = 1; ' αα αβ α β ' ' ' V1 2 A1C A1C ⎛ −α j + β ⎞ ⎛ −α ' j + β ' ⎞ ⎜ 2 ⎟⎜ 2 ⎟ 4a 4 j − 2a 2 j − 2a 2 j + ββ ' ⎝ 2a ⎠ ⎝ 2a ⎠ ββ ' 2 βα ' αβ ' JJJG JJJG j − 2 j − 2 j + αα ' B1C B1'C 4b 4 U 2b 2b JJJG ⋅ JJJG = = 2, B1 A B1' A ββ ' k 2 − βγ ' k − β ' γ k + γγ ' V2 4b 4 2b 2 2b 2 JJJG JJJG γγ ' k 2 − γβ ' k − γ ' β k + ββ ' U C1 A C1' A 4c 4 2c 2 2c 2 JJJG ⋅ JJJG = = 3 γγ ' 2 γα ' γ 'α V3 C1 B C1' B i − 2 i − 2 i + αα ' 4c 4 2c 2c The only thing left to be proved is that: a2 b2 c2 U U U3 1 2 2 2 2 U1 U 2 U 3 ⋅a ⋅b = 1. ⋅ ⋅ = 1 if and only if c V1 V2 V3 V1 V2 V3 We show that b2 c2 a2 U = V , U = V , U1 = V3 2 1 3 2 a2 b2 c2 b2 ββ ' 2 βα ' αβ ' b2 1 αα ' βα ' αβ ' U j j j = − − + αα ' = 2 2 j 2 − 2 j − 2 j + ββ ' = V1 2 2 2 2 2 2 2 a 4a b 2a 2a a 4a a 2a 2a 2 2 c γγ ' γβ ' γ 'β c ββ ' γβ ' γ 'β U 3 = 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 2 2 a αα ' αγ ' α ' γ a γγ ' αγ ' α ' γ U1 = 2 2 i 2 − 2 i − 2 i + 2 γγ ' = 4 i 2 − 2 i − 2 i + αα ' = V3 2 c 4a c 2c 2c c 4c 2c 2c

References

[1] [2] [3]

Ion Pătrașcu and Florentin Smarandache, A Theorem about Simultaneous Orthological and Homological Triangles, in arXiv:1004.0347v1, Cornell University, NY, USA, 2010.

Claudiu Coandă, Geometrie Analitică în coordonate baricentrice, Editura Reprograph, Craiova, 2005.

Mihai Dicu, The Smarandache-Pătrașcu Theorem of Orthohomological Triangles, http://www.scribd.com/doc/28311880/Smarandache-PatrascuTheorem-of-Orthohomological-Triangles, 2010.