Triangle Bordered With Squares C˘at˘alin Barbu Abstract The triangle determined by the centers of the squares built outside on the sides of an acute triangle is called the exterior Vecten triangle. Similarly we define the interior Vecten triangle if the squares are built to the interior of the triangle. In this article we give properties for squares with sides equal to half of triangles’s sides that are built with respect to the midpoints of the sides of the triangle. Denote by O a , O b , O c the centers of the squares A b A b A c A c , B c B c B a B a , C a C a C b C b , A b , A c ∈ [BC], B c , B a ∈ [CA], C a , C b ∈ [AB] constructed externally on the sides BC, CA, AB; by O ” a , O ” b , O ” c the centers of the squares A b A ” b A ” c A c , B c B ” c B ” a B a , C a C ” a C ” b C b constructed internally; by M a , M b , M c , A a , B b , C c the midpoints of the segments BC, CA, AB, B a C a , C b A b , A c B c ; by G a , G b , G c , G ” a , G ” b , G ” c the centroids of triangles O b M a O c , O c M b O a , O a M c O b , O ” b M a O ” c , O ” c M b O ” a , O ” a M c O ” b ; by G 1 , G 2 , G 3 , G ” 1 , G ” 2 , G ” 3 the centroids of triangles O a O ” b O ” c , O b O ” c O ” a , O c O ” a O ” b , O ” a O b O c , O ” b O c O a , O ” c O a O b , and {A } = B c B a ∩ C a C b , {B } = C a C b ∩ A b A c , {C } = A b A c ∩ B c B a . Let O be the circumcenter of triangle ABC. Consider a complex plane with origin at O. Denote by the corresponding lowercase letter the coordinate of a point denoted by an uppercase letter. Then m a = b + c 2 , a b = 3b + c 4 , a c = 3c + b 4 . Point O a is obtained from point A b by a rotation of center M a and angle 90 ◦ . Hence o a = m a + i(a b − m a ) = b + c 2 + i · b − c 4 . Likewise, o b = c + a 2 + i · c − a 4 , o c = a + b 2 i · a − b 4 , (1) o ” a = b + c 2 − i · b − c 4 , o ” b = c + a 2 − i · c − a 4 , o ” c = a + b 2 − i · a − b 4 , (2) a b = 3b + c 4 + i · b − c 2 , a c = 3c + b 4 + i · b − c 2 b c = 3c + a 4 + i · c − a 2 , b a = 3a + c 4 + i · c − a 2 , c a = 3a + b 4 + i · a − b 2 , c b = 3b + a 4 + i · a − b 2 (3) a ” b = 3b + c 4 − i · b − c 2 , a ” c = 3c + b 4 − i · b − c 2 , b ” c = 3c + a 4 − i · c − a 2 , b ” a = 3a + c 4 − i · c − a 2 , c ” a = 3a + b 4 − i · a − b 2 , c ” b = 3b + a 4 − i · a − b 2 (4) Mathematical Reflections 4 (2010) 1 The coordinates of the midpoints A a , B b , C c of the segments B a C a , C b A b , A c B c are a a = 6a + b + c 8 + i · c − b 4 , b b = 6b + c + a 8 + i · a − c 4 , c c = 6c + a + b 8 + i · b − a 4 . (5) Theorem 1. Lines AO a , BO b and CO c are concurrent. Proof. Because triangles BO a C, CO b A and AO c B are similar isosceles triangles, the concurrence follows from Kiepert’s Theorem [1]. Theorem 2. Lines AO ” a , BO ” b and CO ” c are concurrent. Proof. The proof is similar to the previous one. Theorem 3. Lines AA , BB and CC are concurrent. Proof. The proof follows from the fact that triangles ABC and A B C have parallel sides. Remark 1. If the squares have the sides equal with the sides of the triangle, then lines AO a , BO b and CO c are concurrent in the external Vecten point, lines AO ” a , BO ” b and CO ” c are concurrent in the internal Vecten point, AA , BB and CC are concurrent in the point of Lemoine, the A B C triangle beeing Grebe’s triangle of triangle ABC [1]. Theorem 4. Triangles ABC, O a O b O c , O ” a O ” b O ” c , A b B c C a , A c B a C b , A ” b B ” c C ” a , A ” c B ” a C ” b have the same centroid G. Mathematical Reflections 4 (2010) 2 Proof. Triangles ABC, O a O b O c , O ” a O ” b O ” c , A b B c C a , A c B a C b , A ” b B ” c C ” a , A ” c B ” a C ” b have the same cen- troid because o a + o b + o c 3 = o ” a + o ” b + o ” c 3 = a b + b c + c a 3 = a c + b a + c b 3 = a ” b + b ” c + c ” a 3 = a ” c + b ” a + c ” b 3 = a + b + c 3 = g. Theorem 5. Lines A a O a , B b O b and C c O c are concurrent. Proof. The equations of lines A a O a , B b O b and C c O c are: z o a − a a − z o a − a a + o a a a − a a o a = 0, z o b − b b − z o b − b b + o b b b − b b o b = 0, and z o c − c c − z o c − c c + o c c c − c c o c = 0. By adding up the previous equations we obtain o a a a − a a o a + o b b b − b b o b + o c c c − c c o c = 0. The last relation is equivalent to [a(b + c) − a(b + c)] + [b(c + a) − b(c + a)] + [c(b + a) − c(b + a)] = 0, which is clear. Theorem 6. Lines A a O ” a , B b O ” b and C c O ” c are concurrent. Proof. The proof is similar with Theorem’s 6 proof. Theorem 7. Lines A a M a , B b M b and C c M c are concurrent. Proof. The proof is similar with Theorem’s 6 proof. Theorem 8. Lines A O a , B O b and C O c are concurrent. Proof. The proof is similar with Theorem’s 6 proof. Theorem 9. Lines A O ” a , B O ” b and C O ” c are concurrent. Proof. The proof is similar with Theorem’s 6 proof. Theorem 10. The following relations are true: AO a ⊥B a C a , BO b ⊥A b C b , CO c ⊥A c B c and AO a = B a C a , BO b = A b C b , CO c = A c B c . Mathematical Reflections 4 (2010) 3 Proof. Because b a − c a = i b+c−2a 2 + i b−c 4 = i(o a − a), we have AO a ⊥B a C a and b a − c a = i(o a − a) = (o a − a) , hence AO a = B a C a . In the same way we demonstrate the other relations. Theorem 11. If by A [XY Z] we denote the area of triangle XY Z, then A [O a O b O c ] = 7 16 A [ABC] + 1 16 3R 2 + BC 2 + CA 2 + AB 2 2 . Proof. Because A [ABC] = i 4 a a 1 b b 1 c b 1 = i 4 [bc + ca + ab − ac − ba − cb] and BC 2 + CA 2 + AB 2 = |c − b| 2 + |a − c| 2 + |b − a| 2 = (c − b)(c − b) + (a − c)(a − c) + (b − a)(b − a) = 2[3R 2 − (bc + ca + ab + ac + ba + cb)], we have bc + ca + ab + ac + ba + cb = 3R 2 − BC 2 + CA 2 + AB 2 2 (where |a| = |b| = |c| = R is circumradius of triangle ABC). We obtain A [O a O b O c ] = i 4 o a o a 1 o b o b 1 o c o c 1 = i 4 · 1 16 · [7(bc + ca + ab − ac − ba − cb) + 4i(bc + ca + ab + ac + ba + cb) − 24iR 2 ], hence A [O a O b O c ] = 1 16 7A [ABC] − 3R 2 − BC 2 + CA 2 + AB 2 2 + 6R 2 = 1 16 7A [ABC] + 3R 2 + BC 2 + CA 2 + AB 2 2 . Theorem 12. If by A [XY Z] we denote XY Z triangle’s area, then A [O ” a O ” b O ” c ] = 7 16 A [ABC] − 1 16 3R 2 + BC 2 + CA 2 + AB 2 2 . Proof. The proof is similar to the previous one. Theorem 13. If {P 1 } = O a O c ∩ BC, {P 1 } = O a O b ∩ BC, {Q 1 } = O b O a ∩ CA, {Q 1 } = O b O c ∩ CA, {R 1 } = O c O b ∩ AB, {R 1 } = O c O a ∩ AB, then points P 1 , P 1 , Q 1 , Q 1 , R 1 and R 1 are on the same conic. Mathematical Reflections 4 (2010) 4 Proof. Because ABC and O a O b O c are homological triangles (See Theorem 1), then by using Salmon’s theorem [2] the conclusion follows. Theorem 14. If P 2 } = O ” a O ” c ∩ BC, {P 2 } = O ” a O ” b ∩ BC, {Q 2 } = O ” b O ” a ∩ CA, {Q 2 } = O ” b O ” c ∩ CA, {R 2 } = O ” c O ” b ∩ AB, {R 2 } = O ” c O ” a ∩ AB, then points P 1 , P 1 , Q 1 , Q 1 , R 1 and R 2 are on the same conic. Proof. The proof is similar to the previous one. Theorem 15. The following relations are true: G a G⊥BC, G b G⊥CA, G c G⊥AB and BC = 12G a G, CA = 12G b G, AB = 12G c G. Proof. We have g a = m a +o b +o c 3 = g + i · c−b 12 , hence g a −g c−b = i · 1 12 . Since G a G⊥BC and g a −g c−b = i · 1 12 = 1 12 , we obtain 12G a G = BC. Theorem 16. The following relations are true: G ” a G⊥BC, G ” b G⊥CA, G ” c G⊥AB and BC = 12G ” a G, CA = 12G ” b G, AB = 12G ” c G. Proof. We have g ” a = m a +o ” b +o ” c 3 = g − i · c−b 12 , then G ” a G⊥BC and 12G ” a G = BC. Corollary 1. The centroid of the triangle ABC is the midpoint of the segments G a G ” a , G b G ” b and G c G ” c . Theorem 17. Triangles G a G b G c , G ” a G ” b G ” c and ABC have the same centroid G Proof. We have g a +g b +g c 3 = g ” a +g ” b +g ” c 3 = a+b+c 3 , as desired. Theorem 18. The following relations are true: G a G b ⊥AM a , G c G a ⊥BM b , G a G b ⊥CM c and AM a = 6G b G c , BM b = 6G c G a , CM c = 6G a G b . Proof. We have g b − g c = i 6 (a − m a ), then G b G c ⊥AM a and AM a = 6 · G b G c . Theorem 19. The following relations are true: G ” b G ” c ⊥AM a , G ” c G ” a ⊥BM b , G ” a G ” b ⊥CM c and AM a = 6G ” b G ” c , BM b = 6G ” c G ” a , CM c = 6G ” a G ” b . Proof. We have g ” b − g ” c = − i 6 (a − m a ), then G ” b G ” c ⊥AM a and AM a = 6 · G ” b G ” c . Remark 2. The sides of triangles G a G b G c and G ” a G ” b G ” c have lenghts equal to one-sixth of the lengths if the medians of triangle ABC.The existence of triangles G a G b G c and G ” a G ” b G ” c implies, also, the existence of a medians triangle(a triangle that has sides of equal lenght with medians the lengths of the triangle ABC) [1]. Corollary 2. Quadrilaterals G b G c G ” b G ” c , G c G a G ” c G ” a , G a G b G ” a G ” b are parallelograms. Proof. By Theorems 20 and 21 we get G b G c G ” b G ” c , G c G a G ” c G ” a , G a G b G ” a G ” b and G b G c = G ” b G ” c , G c G a = G ” c G ” a , G a G b = G ” a G ” b . Mathematical Reflections 4 (2010) 5 Corollary 3. Triangles G a G b G c and G ” a G ” b G ” c are congruent. Theorem 20. The pairs of triangles G a G b G c and ABC, G ” a G ” b G ” c and ABC are bilogic. Proof. The solutions results from Theorems 16-21, the common center of orthology being centroid of the triangle ABC. Theorem 21. If by A [XY Z] we denote XY Z triangle’s area, then A [G a G b G c ] = A [G ” a G ” b G ” c ] = 1 48 A [ABC] . Proof. Triangle G a G b G c is similar with the medians triangle M 1 M 2 M 3 , so m( G a G b G c ) = m( M 1 M 2 M 3 ) = 180 ◦ − m( BGC). We have A [G a G b G c ] = G a G b · G a G c · sin G a G b G c 2 = CM c 6 · BM b 6 · sin( BGC) 2 = 1 16 A [BGC] = 1 16 · 1 3 A [ABC] = 1 48 A [ABC] . Triangles G a G b G c and G ” a G ” b G ” c are congruent, so A [G a G b G c ] = A [G ” a G ” b G ” c ] = 1 48 A [ABC] . Mathematical Reflections 4 (2010) 6 Theorem 22. The following relations are true: G 1 G⊥BC, G 2 G⊥CA, G 3 G⊥AB and BC = 6G 1 G, CA = 6G 2 G, AB = 6G 3 G. Proof. We have g 1 = o a +o ” b +o ” c 3 = g + i · b−c 6 , so g 1 −g b−c = i 6 . Because G 1 G⊥BC and g 1 −g b−c = i 6 , we obtain BC = 6G 1 G. Theorem 23. The following relations are true: G ” 1 G⊥BC, G ” 2 G⊥CA, G ” 3 G⊥AB and BC = 6G ” 1 G, CA = 6G ” 2 G, AB = 6G ” 3 G. Proof. The proof is similar to the one of the previous theorem. Theorem 24. Triangles G 1 G 2 G 3 , G ” 1 G ” 2 G ” 3 and ABC have the same centroid. Proof. We have g 1 +g 2 +g 3 3 = g ” 1 +g ” 2 +g ” 3 3 = a+b+c 3 = g, as desired Corollary 4. The centroid of the triangle ABC is the midpoint of the segments G a G ” a , G b G ” b , G c G ” c , G 1 G ” 1 , G 2 G ” 2 and G 3 G ” 3 . Corollary 5. Points G ” 1 , G a , G, G ” a , G 1 are collinear and G ” 1 G a = G a G = GG ” a = G ” a G 1 . Theorem 25. Quadrilaterals G 1 G ” 3 G 2 G, G ” 1 G 3 GG 2 and G 3 G ” 2 G 1 G are parallelograms. Proof. Because g + g ” 3 = g 1 + g 2 = 2g c , quadrilateral G 1 G ” 3 G 2 G is a parallelogram. Similarly, quadrilaterals G ” 1 G 3 GG 2 and G 3 G ” 2 G 1 G are parallelograms. Corollary 6. Hexagons G ” a G c G ” b G a G ” c G b and G 1 G ” 3 G 2 G ” 1 G 3 G ” 2 are homotetic, the center of ho- motethy being the centroid of triangle ABC. Corollary 7. The points G a , G b , G c , G ” a , G ” b and G ” c are same conic Proof. We have G a G ” c G ” a G c , G b G ” a G ” b G a , G c G ” b G ” b G c . As two parallel lines have the same infinity it follows that the lines are intersected in the same point of the infinity. Then, the inter- section points between the pair of parallel lines enumerated above belong to the infinite line and from Pascal’s reverse theorem the conclusion follows. Corollary 8. Points G 1 , G 2 , G 3 , G ” 1 , G ” 2 and G ” 3 are on the same conic. References [1] C. Barbu, Teoreme fundamentale din geometria triunghiului, Ed. Unique, Bac˘au, 2008. [2] G. Salmon, Trait´e de g´eom´etrie analytique, Ed. Gauthier –Villars, Paris, 1903. [3] T. Andreescu, D. Andrica, Complex Numbers from A to Z, Birkh¨auser, Boston, 2006. [4] D. Andrica, K. Nguyen, A note on the Nagel and Gergonne points, Creative Math. & Inf., 17 (2008), 127-136. [5] R. Musselman, The triangle bordered with squares, American Mathematical Monthly, 43 (1936), 539-548. [6] J. Neuberg, Bibliographie du Triangle et du T´etraˇcdre, Mathesis, 37 (1923), 289-293. C˘at˘alin Barbu “Vasile Alecsandri” College, Bac˘au, Romania E-mail address: kafka mate@yahoo.com Mathematical Reflections 4 (2010) 7