ON A ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED

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1 ON A 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 conjectures 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 11 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 area(a 1 B 1 C 1 ) AB + BA 1 AC + CA 1 = BC + CB 1 AB + AB 1 = AC + AC 1 BC + BC 1 =, ( ( ) ) 1 area(abc) + s 4 area(abc) 1, 4(a + b + c)(a + b + c ) + 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 where s is the semi-perimeter of triangle ABC When 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 11 we obtain (a + b + c)(a + b + c ), Theorem 13 ([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 When we derive Aassila s inequality Corollary 14 ([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 Torrejon s one [10] and it is based on the geometry of complex numbers and combining with a straightforward geometric inequality we obtain the conclusion of the main result

3 ON A ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED 3 Proof of Theorem 11 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 We use complex numbers 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 afixes of the points A, B, C and by z A1, z B1, z C1 the afixes of the points A 1, B 1, C 1 First of all, = 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 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 z A1 z B1 ) = ( c)z +1 C + ( a)z ) +1 A b ] ) c) + b( c)( )

4 4 CEZAR LUPU, ŞTEFAN SPĂTARU + 1 abc Im( cyc z B z C [c( + 1 b)( + 1 ( ) s(1 ) z B z C [b + s b ( 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 ) 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 Now we have that 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 left to prove the following inequality (3) sabc 4S (a + b + c)(a + b + c )

5 ON A ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED 5 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) By Schur s inequality, (xy + yz + zx) (x + y + z ) by putting x = a, y = b and z = c, we have which is equivalent to 9a b c (a + b + c)(a + b + c ) 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, which gives inequality (4)

6 6 CEZAR LUPU, ŞTEFAN SPĂTARU 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, 1970 [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), 3-35 University of Pittsburgh, Department of Mathematics, Pittsburgh, 301 Thackeray Hall, PA 1560, USA and University of Craiova, Department of Mathematics, A I Cuza 13, RO 00585, Craiova, Romania E mail address: cel47@pittedu, lupucezar@gmailcom International Computer High School of Bucharest, 48 Mihai Bravu St, Sector 3, Bucharest, Romania E mail address: spataru stefan96@yahoocom