ON AN ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED

Transcription

1 ON AN ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED CEZAR LUPU, ŞTEFAN SPĂTARU Abstract. In this note we give a refinement of an inequality obtained by Torrejon [10] between the area of a triangle and that of an inscribed triangle. Our approach is based on using complex numbers and some elementary facts on geometric inequalities. 1. Introduction and Main result Let us consider a triangle ABC on each of the sides BC, CA and AB and fix arbitrary points A 1, B 1, C 1. As pointed out in [10], [7] a question with a long history is the following Erdos-Debrunner inequality: (1) min{area(ac 1 B 1 ); area(c 1 BA 1 ); area(b 1 AC)} area(a 1 B 1 C 1 ). Later, Janous [11] generalized inequality (1) by proving () M 1 {area(ac 1 B 1 ); area(c 1 BA 1 ); area(b 1 AC)} area(a 1 B 1 C 1 ), where M 1 denotes the harmonic mean of the areas of triangles mentioned in the above inequality. Moreover, Janous formulated a more general question which is extended and solve by Mascioni [7], [8]. Using a different method, Frenzen, Ionaşcu and Stănică [5] proved Janous conjecture independently of Mascioni. The purpose of this note is to extend the result obtained by Torrejon [10] regarding the areas of triangles A 1 B 1 C 1 and ABC when the points A 1, B 1, C 1 satisfy a certain metric property. In fact our main result is given by the following Theorem 1.1. Let ABC be a triangle and let A 1, B 1, C 1 be on BC = a, CA = b and AB = c respectively with none of A 1, B 1, C 1 coinciding with a vertex of ABC. If AB + BA 1 AC + CA 1 = BC + CB 1 AB + AB 1 = AC + AC 1 BC + BC 1 =, Key words and phrases. Erdos-Debrunner inequality, Schur s inequality, area of an inscribed triangle. 010 Mathematics Subject Classification. 6D15, 51 Fxx, 97G30, 97G70. 1

2 CEZAR LUPU, ŞTEFAN SPĂTARU area(a 1 B 1 C 1 ) ( ( ) ) 1 area(abc) + s 4 area(abc) 1, 4(a + b + c)(a + b + c ) + 1 where s is the semi-perimeter of triangle ABC. For we obtain Corollary 1.. ([6]) Let ABC be a triangle and let A 1, B 1, C 1 be on BC = a, CA = b and AB = c respectively with none of A 1, B 1, C 1 coinciding with a vertex of ABC. If AB + BA 1 = AC + CA 1, BC + CB 1 = AB + AB 1, AC + AC 1 = BC + BC 1, area(a 1 B 1 C 1 ) area(abc) 4(a + b + c)(a + b + c ). Clearly, by the Arihmetic-Geometric mean inequality, we have and by Theorem 1.1 we obtain (a + b + c)(a + b + c ), Theorem 1.3. ([10]) Let ABC be a triangle and let A 1, B 1, C 1 be on BC = a, CA = b and AB = c respectively with none of A 1, B 1, C 1 coinciding with a vertex of ABC. If AB + BA 1 AC + CA 1 = BC + CB 1 AB + AB 1 = AC + AC 1 BC + BC 1 =, ( ) 1 4 area(a 1 B 1 C 1 ) area(abc) + s 4 area(abc) 1, + 1 where s is the semi-perimeter of triangle ABC. For we derive Aassila s inequality

3 ON AN ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED 3 Corollary 1.4. ([1]) Let ABC be a triangle, and let A 1, B 1, C 1 be on BC, CA, AB, respetively, with none of A 1, B 1, C 1 coinciding with a vertex of ABC. If AB + BA 1 = AC + CA 1, BC + CB 1 = AB + AB 1, AC + AC 1 = BC + BC 1, 4 area(a 1 B 1 C 1 ) area(abc). Our approach in computing the area of the triangle A 1 B 1 C 1 will be different from the one given by Torrejon in [10] and it is based on the geometric image of complex numbers in the plane combined with a geometric inequality.. Proof of Theorem 1.1 First of all, we prove the following equality: area(a 1 B 1 C 1 ) = S (s a)(s b)(s c) + s ( 1 +1 ) (a + b + c). abc For simplicity denote by a, b, c the sidelenghts of triangle ABC, s its semiperimeter, S its area, z A, z B, z C the affixes of the points A, B, C and by z A1, z B1, z C1 the affixes of the points A 1, B 1, C 1. Firstly, = a + b + c = (AB + AB 1 ) + (BC + CB 1 ) = ( + 1)(c + AB 1 ) and consequently AB 1 = + 1 c and CB 1 = CB AB 1 = b Analogously, we have c = a + 1 = + 1 a. Denote z A1, z B1, z C1 BC 1 = CA 1 = + 1 a, + 1 b, BA 1 = + 1 c, AC 1 = + 1 b, the affixes of A 1, B 1, C 1, and they are given by z A1 = ( +1 b)z B + ( +1 c)z C a

4 4 CEZAR LUPU, ŞTEFAN SPĂTARU z B1 = ( +1 c)z C + ( +1 a)z A b z C1 = ( a)z +1 A + ( b)z +1 B c Now the formula for the area of triangle A 1 B 1 C 1 is = Im ( cyc area(a 1 B 1 C 1 ) = Im( cyc ( +1 b)z B + ( +1 c)z C a ( abc Im z B z C [c( + 1 b)( + 1 cyc + 1 abc Im( cyc z B z C [c( + 1 b)( + 1 ( ) s(1 ) z B z C [b + s b (. z A1 z B1 ) = ( c)z +1 C + ( a)z ) +1 A b ] ) c) + b( c)( ) z C z B a( + 1 b)( + 1 c)) c)+b( c)( ) a( s(1 ) +s c)+c( s( 1) s(1 ) a( + s b)( + s c)]) + 1 b)( + 1 c)]) s( 1) +s b)( s( 1) +s c) z B z C [(s b)(s c)(b+c a)+s (b(s b+s c) c(s b+s c)+a(s b s+c))]) +s ( ) (b + c + a)] z B z c [(s a)(s b)(s c)+s (ab ac+ac ab)+s ( ) (a+b+c)]) abc Im( cyc z B z c [(s a)(s b)(s c) + s ( ) (a + b + c)]) = (s a)(s b)(s c) + s ( 1 +1 ) (a + b + c) Im( abc cyc = S (s a)(s b)(s c) + s ( 1 +1 ) (a + b + c) abc which is equivalent to z B z c )

5 ON AN ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED 5 area(a 1 B 1 C 1 ) = S (s a)(s b)(s c) + s ( 1 +1 ) (a + b + c), abc where we used the fact that z A z A and z B z C + z C z B are real numbers hence have the imaginary part 0. Therefore, we derived abc s area(a 1 B 1 C 1 ) = S s(s a)(s b)(s c) + s 4 ( ) S Now using Heron s formula we finally obtain abc s area(a 1 B 1 C 1 ) = S 3 + s 4 ( ) S In this moment, we are only left to prove the following inequality sabc (3) 4S (a + b + c)(a + b + c ). Using abc = 4RS, we get the following equivalent inequality a b c 16S (a + b + c)(a + b + c ), 9 which is succesively equivalent to a b c 16S a + b + c, 9 9R a + b + c, which is evident since the distance between the circumcenter O and barycenter G is given by Leibniz identity OG = 9R (a + b + c ) and the proof of inequality (3) ends. Now, we can conclude ( ) 1 S 3 + s 4 S 4S (a + b + c)(a + b + c ) area(a 1 B 1 C 1 ), + 1 which is finally equivalent to area(a 1 B 1 C 1 ) ( S + s 4 4(a + b + c)(a + b + c ) ( ) ) 1 S and thus our theorem is proved. Remark. In fact, inequality (3) can be rewritten in the following form: (4) (b + c a)(c + a b)(a + b c) 9a b c (a + b + c)(a + b + c ).

6 6 CEZAR LUPU, ŞTEFAN SPĂTARU By Schur s inequality, we have (xy + yz + zx) (x + y + z ) If we put x = a, y = b and z = c, we have which is equivalent to (4) 9xyz x + y + z. (a b + b c + c a ) (a 4 + b 4 + c 4 ) 9a b c a + b + c (a + b + c)(b + c a)(c + a b)(a + b c) 9a b c a + b + c. References [1] M. Aassila, Problem 1717, Math. Mag. 78 (005), 158. [] A. Bager, Solution to Problem 60, Elem. Math. 1 (1957), 47. [3] M. Bataille, Solution to Problem 1857, Math. Mag. 84 (011), [4] H. Debrunner, Problem 60, Elem. Math. 11 (1956), 0. [5] C. L. Frenzen, E. J. Ionascu, P. Stanica, A proof of two conjectures related to Erdos- Debrunner inequality, J. Ineq. Pure Appl. Math. 8 (007), art. 68, 13 pp. [6] C. Lupu, T. Lupu, Problem 1857, Math. Mag. 83 (010), 391. [7] V. Mascioni, On the Erdos-Debrunner inequality, J. Ineq. Pure Appl. Math. 8 (007), art. 3, 5 pp. [8] V. Mascioni, An extension of the Erdos-Debrunner inequality to general power means, J. Ineq. Pure Appl. Math. 9(008), art. 67, 11pp. [9] D. S. Mitrinovic, Analytic Inequalities, Springer Verlag, [10] R. Torrejon, On a Erdos inscribed triangle inequality, Forum Geom. 5 (005), [11] W. Janous, A short note on the Erdos-Debrunner inequality, Elem. Math. 61 (006), University of Pittsburgh, Department of Mathematics, Pittsburgh, 301 Thackeray Hall, PA-1560, USA E mail address: cel47@pitt.edu, lupucezar@gmail.com Harvard University, Cambridge, MA-0138, USA E mail address: spataru stefan96@yahoo.com