On a vector equality Nguyen Tien Lam Abstract In this paper, we recall a vector equality and give some of its applications. Let us start by introducing the notations that will be used in this paper. By T we will denote the triangle A 1 A 2 A 3 . The inscribed circle of T has center I and is tanget to the side opposite to vertex A i at B i , i = 1, 2, 3. Denote by a i and h i , i = 1, 2, 3 the lengths of the side opposite to vertex A i and the length of the height from that same vertex, respectively. Let K, p, R, r be the area, semiperimeter, circumradius, and inradius of T. Denote by K 1 , p 1 , r 1 the area, semiperimeter, and inradius of B 1 B 2 B 3 . Theorem. I is the center of mass of the system B 1 , B 2 , B 3 with masses 1 h 1 , 1 h 2 , 1 h 3 or −−→ IB 1 h 1 + −−→ IB 2 h 2 + −−→ IB 3 h 3 = 0. (1) Proof. We have A 1 B 2 = A 1 B 3 = p − a 1 , A 2 B 3 = A 2 B 1 = p − a 2 , A 3 B 1 = A 3 B 2 = p − a 3 . It is not difficult to prove that for every point M on the side A i A j , i = j, i, j ∈ {1, 2, 3} we have −−→ IM = MA j A i A j · −−→ IA i + MA i A i A j · −−→ IA j . When M = B 1 we get −−→ IB 1 = B 1 A j A i A j · −−→ IA i + B 1 A i A i A j · −−→ IA j or equivalently a 1 −−→ IB 1 = (p −a 3 ) −−→ IA 2 + (p −a 2 ) −−→ IA 3 . Similarly, we obtain a 2 −−→ IB 2 = (p −a 1 ) −−→ IA 3 + (p −a 3 ) −−→ IA 1 and a 3 −−→ IB 3 = (p −a 2 ) −−→ IA 1 + (p −a 1 ) −−→ IA 2 . Adding the above equalities side by side, we obtain a 1 −−→ IB 1 + a 2 −−→ IB 2 + a 3 −−→ IB 3 = (2p −a 2 −a 3 ) −−→ IA 1 + (2p −a 3 −a 1 ) −−→ IA 2 + (2p −a 1 −a 2 ) −−→ IA 3 . Mathematical Reflections 6 (2008) 1 Note that 2p = a 1 + a 2 + a 3 and recall that a 1 −−→ IA 1 + a 2 −−→ IA 2 + a 3 −−→ IA 3 = 0, thus a 1 −−→ IB 1 + a 2 −−→ IB 2 + a 3 −−→ IB 3 = 0. Using the formula a i = 2K h i we conclude that −−→ IB 1 h 1 + −−→ IB 2 h 2 + −−→ IB 3 h 3 = 0. Alternative proof. For i = 1, 2, 3 we know that −−→ IB i IB i is a unit vector perpendicular to the side opposite to vertex A i , its direction being out of the triangle. By applying the “Porcupine Theorem” we obtain a 1 −−→ IB 1 IB 1 + a 2 −−→ IB 2 IB 2 + a 3 −−→ IB 3 IB 3 = 0. Note that IB 1 = IB 2 = IB 3 = r and by using the formula a i = 2K h i , i = 1 , 2, 3 we deduce that −−→ IB 1 h 1 + −−→ IB 2 h 2 + −−→ IB 3 h 3 = 0. The proof is thus c omplete and we are ready to show som e of the results due to this theorem. Problem 1. Prove that B 1 B 2 2 h 1 h 2 + B 2 B 3 2 h 2 h 3 + B 3 B 1 2 h 3 h 1 = 1. (2) Solution. By squaring both sides of the equation (1) and using the following identity 2 −−→ IB i −−→ IB j = IB i 2 + IB j 2 − B i B j 2 = 2r 2 − B i B j 2 we obtain 3 i=1 1 h 2 i + 2 · 3 i=1 1 h i h j r 2 − B 1 B 2 2 h 1 h 2 + B 2 B 3 2 h 2 h 3 + B 3 B 1 2 h 3 h 1 = 0. Using the fact that 1 h 1 + 1 h 2 + 1 h 3 = 1 r it follows that B 1 B 2 2 h 1 h 2 + B 2 B 3 2 h 2 h 3 + B 3 B 1 2 h 3 h 1 = 1. We will now present four corollaries following from Problem 1. Mathematical Reflections 6 (2008) 2 Corollary 1. The following inequality holds a 1 a 2 B 1 B 2 + a 2 a 3 B 2 B 3 + a 3 a 1 B 3 B 1 ≤ 4 √ 3 3 pK. (3) Solution. Knowing that a i = 2K h i , we can rewrite the inequality in the form 4K 2 B 1 B 2 h 1 h 2 + B 2 B 3 h 2 h 3 + B 3 B 1 h 3 h 1 ≤ 4 √ 3 3 pK. Because K = pr, the above inequality becomes B 1 B 2 h 1 h 2 + B 2 B 3 h 2 h 3 + B 3 B 1 h 3 h 1 ≤ √ 3 3r . We will now prove that the last inequality is true. Indeed, by applying the Cauchy- Schwarz inequality we obtain B 1 B 2 h 1 h 2 + B 2 B 3 h 2 h 3 + B 3 B 1 h 3 h 1 2 ≤ 1 h 1 h 2 + 1 h 2 h 3 + 1 h 3 h 1 B 1 B 2 2 h 1 h 2 + B 2 B 3 2 h 2 h 3 + B 3 B 1 2 h 3 h 1 . By using the well-known inequality ab + bc + ca ≤ 1 3 (a + b + c) 2 for all real number a, b, c and the equality 1 h 1 + 1 h 2 + 1 h 3 = 1 r it follows, from Problem 1, that B 1 B 2 h 1 h 2 + B 2 B 3 h 2 h 3 + B 3 B 1 h 3 h 1 2 ≤ 1 3r 2 or equivalently B 1 B 2 h 1 h 2 + B 2 B 3 h 2 h 3 + B 3 B 1 h 3 h 1 2 ≤ √ 3 3r . The equality takes places if and only if T is an equilateral triangle. Corollary 2. If T is an acute triangl, then max{a 1 , a 2 , a 3 } ≥ √ 3R (4) with equality if and only if T is an equilateral triangle. Solution. Draw three lines tanget to the circle circumscribed around T from its vertices. Suppose that these three lines intersect at three points C 1 , C 2 , and C 3 where C i is the vertex opposite to the side passing through A i , i = 1, 2, 3. The circle circumscribed to T is the incercle of C 1 C 2 C 3 . Thus R is the length of the inradius of C 1 C 2 C 3 . Denote by l 1 , l 2 , and l 3 the lengths of the altitudes in the triange C 1 C 2 C 3 . Applying the result of Problem 1 we obtain A 1 A 2 2 l 1 l 2 + A 2 A 3 2 l 2 l 3 + A 3 A 1 2 l 3 l 1 = 1 Mathematical Reflections 6 (2008) 3 or a 3 2 l 1 l 2 + a 1 2 l 2 l 3 + a 2 2 l 3 l 1 = 1. Hence 1 ≤ max{a 2 1 , a 2 2 , a 2 3 } · 1 l 1 l 2 + 1 l 2 l 3 + 1 l 3 l 1 ≤ 1 3 max{a 2 1 , a 2 2 , a 2 3 } 1 l 1 + 1 l 2 + 1 l 3 2 . Note that 1 l 1 + 1 l 2 + 1 l 3 = 1 R , thus it follows that 1 3R 2 max{a 2 1 , a 2 2 , a 2 3 } ≥ 1 and we conclude that max{a 1 , a 2 , a 3 } ≥ √ 3R. Equality occurs if and only if l 1 = l 2 = l 3 and thus C 1 C 2 = C 2 C 3 = C 3 C 1 . This imples that C 1 C 2 C 3 is equilateral which in turn means that A 1 A 2 A 3 is equilateral, and we are done. Corollary 3. p 2 1 ≤ pK 2R . (5) When does equality occur ? Solution. We introduce the following inequality a 2 x + b 2 y + c 2 z ≥ (a + b + c) 2 x + y + z for all pos itive reals a, b, and c to prove the above inequality with equality if and only if a x = b y = c z . Using the above inequality and the result of Problem 1 we obtain 1 = B 1 B 2 2 h 1 h 2 + B 2 B 3 2 h 2 h 3 + B 3 B 1 2 h 3 h 1 ≥ (B 1 B 2 + B 2 B 3 + B 3 B 1 ) 2 h 1 h 2 + h 2 h 3 + h 3 h 1 . Observe that B 1 B 2 + B 2 B 3 + B 3 B 1 = 2p 1 and h 1 h 2 + h 2 h 3 + h 3 h 1 = 4K 2 1 a 1 a 2 + 1 a 2 a 3 + 1 a 3 a 1 = 8K 2 p a 1 a 2 a 3 . By using the formulae K = a 1 a 2 a 3 4R and h 1 h 2 + h 2 h 3 + h 3 h 1 = 2Kp R it follows that p 2 1 ≤ pK 2R . Equality occurs if and only if B 1 B 2 h 1 h 2 = B 2 B 3 h 2 h 3 = B 3 B 1 h 3 h 1 or a 1 a 2 B 1 B 2 = a 2 a 3 B 2 B 3 = a 3 a 1 B 3 B 1 . Mathematical Reflections 6 (2008) 4 Because a 1 a 2 B 1 B 2 = a 2 a 3 B 2 B 3 then a 2 1 B 1 B 2 2 = a 2 B 2 B 3 2 . By the Law of Cosines we obtain B 1 B 2 2 = 2(p −a 3 ) 2 (1 −cos A 3 ) = 2(p −a 3 ) 2 1 − a 2 1 + a 2 2 − a 2 3 2a 1 a 2 . Therefore B 1 B 2 2 = 4(p −a 3 ) 2 (p −a 1 )(p −a 2 ) a 1 a 2 . Similarly B 2 B 3 2 = 4(p −a 1 ) 2 (p −a 2 )(p −a 3 ) a 2 a 3 . Thus, the inequality a 2 1 B 1 B 2 2 = a 2 B 2 B 3 2 is equivalent to a 1 (p − a 3 ) = a 3 (p − a 1 ) and so a 1 = a 3 . Analogously we deduce that a 2 = a 3 . Consequantly a 1 = a 2 = a 3 which proves that T is equilateral and we are done. Corollary 4. Let O be the circumcenter of T. Then OB 1 h 1 + OB 2 h 2 + OB 3 h 3 ≤ R r − 1. (6) Solution. By using (1) and the equality 1 h 1 + 1 h 2 + 1 h 3 = 1 r we obtain −→ OI = r −−→ OB 1 h 1 + −−→ OB 2 h 2 + −−→ OB 3 h 3 . On squaring both sides and using the identity 2 −−→ OB i −−→ OB j = OB i 2 +OB j 2 −B i B j 2 , i, j = 1, 2, 3 we get OI 2 = r 2 3 i=1 OB i 2 h 2 i + 1≤i<j≤3 OB i 2 + OB j 2 h i h j + 1≤i<j≤3 B i B j 2 h i h j . By using the result from Problem 1 and Euler’s theorem OI 2 = R 2 −2Rr it follows that OB 1 2 h 2 1 + OB 2 2 h 2 2 + OB 3 2 h 2 3 + OB 1 + OB 2 2 h 1 h 2 + OB 2 + OB 3 2 h 2 h 3 + OB 3 + OB 1 2 h 3 h 1 = R r − 1 2 . Because OB i 2 + OB j 2 ≥ 2OB i OB j , i, j = 1, 2, 3 and the identity (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca we obtain OB 1 h 1 + OB 2 h 2 + OB 3 h 3 ≤ R r − 1 2 . Mathematical Reflections 6 (2008) 5 Since R ≥ 2r > r we obtain the desired result. Equality occurs if and only if T is an equilateral triangle. Problem 2. Prove that MB 1 h 1 + MB 2 h 2 + MB 3 h 3 ≥ 1 (7) for all points M. Solution. We will use the follwing two facts | −→ u | · | −→ v | ≥ −→ u · −→ v and −−−→ MB i = −−→ MI + −−→ IB i , i = 1, 2, 3 to obtain 3 i=1 MB i h i ≥ 1 r 3 i=1 −−−→ MB i −−−→ MB i h i = 1 r 3 i=1 IB i h i −−→ MI + r 3 i=1 1 h i . The desired inequality follows if we keep in mind that 1 h 1 + 1 h 2 + 1 h 3 = 1 r . Equality occurs if and only if M = I. Problem 3. Prove that MB 1 2 h 1 + MB 2 2 h 2 + MB 3 2 h 3 = MI 2 r + r (8) for all points M in the plane. Solution. We have 3 i=1 MB i 2 h i = 3 i=1 −−→ MI + −−→ IB i h i = MI 2 3 i=1 1 h i + 2 3 i=1 −−→ IB i h i −−→ MI + 3 i=1 IB i 2 h i . Thus, 3 i=1 MB i 2 h i = MI 2 r + r and we are done References [1] Viktor Prasolov, 2006, Problems in plane and solid geometry ( translated and edited by Dimitry Leites.) [2] Kiran S.Kedlaya, 2006, Geometry Unbound. [3] Nguyen Minh Ha, 2005, Toan nang cao hinh hoc 10, Education Publishing House, Hanoi, Vietnam. [4] Euler Triangle Formula, http://mathworld.wolfram.com/EulerTriangleFormula.html. Mathematical Reflections 6 (2008) 6 . On a vector equality Nguyen Tien Lam Abstract In this paper, we recall a vector equality and give some of its applications. Let us start. that −−→ IB 1 h 1 + −−→ IB 2 h 2 + −−→ IB 3 h 3 = 0. Alternative proof. For i = 1, 2, 3 we know that −−→ IB i IB i is a unit vector perpendicular to the side opposite to vertex A i , its direction being out of the triangle.