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 A1 , B1 , C1 . As pointed out in [10], [7] a question with a long history
is the following Erdos-Debrunner inequality:
(1)
min{area(AC1 B1 ); area(C1 BA1 ); area(B1 AC)} ≤ area(A1 B1 C1 ).
Later, Janous [11] generalized inequality (1) by proving
(2)
M−1 {area(AC1 B1 ); area(C1 BA1 ); area(B1 AC)} ≤ area(A1 B1 C1 ),
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 A1 B1 C1 and ABC when the points A1 , B1 , C1 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 A1 , B1 , C1 be on BC = a, CA = b
and AB = c respectively with none of A1 , B1 , C1 coinciding with a vertex of ABC.
If
AB + BA1
BC + CB1
AC + AC1
=
=
= α,
AC + CA1
AB + AB1
BC + BC1
then
9abc
area(A1 B1 C1 ) ≤
4(a + b + c)(a2 + b2 + c2 )
area(ABC) + s4
α−1
α+1
2
!
area(ABC)−1 ,
Key words and phrases. Erdos-Debrunner inequality, Schur’s inequality, area of an inscribed
triangle.
2010 Mathematics Subject Classification. 26D15, 51 Fxx, 97G30, 97G70.
1
2
CEZAR LUPU, ŞTEFAN SPĂTARU
where s is the semi-perimeter of triangle ABC.
When α = 1 we obtain
Corollary 1.2. ([6]) Let ABC be a triangle and let A1 , B1 , C1 be on BC = a,
CA = b and AB = c respectively with none of A1 , B1 , C1 coinciding with a vertex of
ABC. If
BC + CB1
AC + AC1
AB + BA1
=
=
= α,
AC + CA1
AB + AB1
BC + BC1
then
area(A1 B1 C1 )
9abc
≤
.
area(ABC)
4(a + b + c)(a2 + b2 + c2 )
Clearly, by the Arihmetic-Geometric mean inequality, we have
(a + b + c)(a2 + b2 + c2 ) ≥ 9abc,
and by Theorem 1.1 we obtain
Theorem 1.3. ([10]) Let ABC be a triangle and let A1 , B1 , C1 be on BC = a,
CA = b and AB = c respectively with none of A1 , B1 , C1 coinciding with a vertex of
ABC. If
AB + BA1
BC + CB1
AC + AC1
=
=
= α,
AC + CA1
AB + AB1
BC + BC1
then
α−1
4 area(A1 B1 C1 ) ≤ area(ABC) + s
α+1
where s is the semi-perimeter of triangle ABC.
4
2
area(ABC)−1 ,
When α = 1 we derive Aassila’s inequality
Corollary 1.4. ([1]) Let ABC be a triangle, and let A1 , B1 , C1 be on BC, CA, AB,
respetively, with none of A1 , B1 , C1 coinciding with a vertex of ABC. If
AB + BA1 = AC + CA1 ,
BC + CB1 = AB + AB1 ,
AC + AC1 = BC + BC1 ,
then
4 area(A1 B1 C1 ) ≤ area(ABC).
Our approach in computing the area of the triangle A1 B1 C1 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.
ON A ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED
3
2. Proof of Theorem 1.1
First of all, we prove the following equality:
2(s − a)(s − b)(s − c) + s2 · ( α−1
)2 (a + b + c)
α+1
area(A1 B1 C1 ) = S ·
abc
.
We use complex numbers. For simplicity denote by a, b, c the sidelenghts of triangle ABC, s its semiperimeter, S its area, zA , zB , zC the afixes of the points A, B, C
and by zA1 , zB1 , zC1 the afixes of the points A1 , B1 , C1 .
First of all, 2s = a + b + c = (AB + AB1 ) + (BC + CB1 ) = (α + 1)(c + AB1 ) and,
2s
− c and
consequently AB1 =
α+1
2s
2sα
2s
+ c = 2s − a −
=
−a
CB1 = CB − AB1 = b −
α+1
α+1
α+1
.
Analogously we have
2s
− a,
α+1
2s
CA1 =
− b,
α+1
2sα
BA1 =
− c,
α+1
2sα
AC1 =
− b,
α+1
the affixes of A1 , B1 , C1 , and they are given by
BC1 =
Denote zA1 , zB1 , zC1
zA1
2s
2sα
( α+1
− b)zB + ( α+1
− c)zC
=
a
zB1 =
2sα
2s
( α+1
− c)zC + ( α+1
− a)zA
b
2s
2sα
( α+1
− a)zA + ( α+1
− b)zB
zC1 =
.
c
Now the formula for the area of triangle A1 B1 C1 is
X
2 area(A1 B1 C1 ) = Im(
zA1 zB1 ) =
cyc
2sα
2s
2sα
2s
X ( α+1
− b)zB + ( α+1
− c)zC ( α+1
− c)zC + ( α+1
− a)zA
= Im
·
a
b
cyc
=
1
· Im
abc
X
cyc
!
!
2s
2s
2sα
2sα
zB zC c(
− b)(
− c) + b(
− c)(
)
α+1
α+1
α+1
α+1
4
CEZAR LUPU, ŞTEFAN SPĂTARU
+
=
X
2sα
2s
1
· Im(
zC zB a(
− b)(
− c))
abc
α+1
α+1
cyc
X
1
2s
2s
2sα
2sα
2sα
2s
·Im(
zB zC [c(
−b)(
−c)+b(
−c)(
)−a(
−b)(
−c)])
abc
α
+
1
α
+
1
α
+
1
α
+
1
α
+
1
α
+
1
cyc
X
s(1 − α)
s(α − 1)
s(α − 1)
1
s(1 − α)
·Im(
+s−b (
+s−c)+c(
+s−b)(
+s−c)
=
zB zC [b
abc
α
α
α
α
cyc
−a(
=
s(α − 1)
s(1 − α)
+ s − b)(
+ s − c)])
α
α
X
1
1−α
·Im(
·(b(s−b+s−c)−c(s−b+s−c)+a(s−b−s+c))])
zB zC [(s−b)(s−c)(b+c−a)+s·
abc
1
+
α
cyc
+s2 · (
=
1−α 2
) (b + c + a)]
1+α
X
1
1−α
α−1 2
·Im(
zB zc [2(s−a)(s−b)(s−c)+s·
(ab−ac+ac−ab)+s2 ·(
) (a+b+c)])
abc
1+α
α+1
cyc
=
X
1
α−1 2
· Im(
zB zc [2(s − a)(s − b)(s − c) + s2 · (
) (a + b + c)])
abc
α
+
1
cyc
X
)2 (a + b + c)
2(s − a)(s − b)(s − c) + s2 · ( α−1
α+1
· Im(
zB zc )
=
abc
cyc
)2 (a + b + c)
2(s − a)(s − b)(s − c) + s2 · ( α−1
α+1
= 2S ·
abc
which is equivalent to
2(s − a)(s − b)(s − c) + s2 · ( α−1
)2 (a + b + c)
α+1
,
abc
where we used the fact that zA zA and zB zC + zC zB are real numbers hence have
the imaginary part 0.
Now we have that
area(A1 B1 C1 ) = S ·
abc · s
α−1 2
· area(A1 B1 C1 ) = S · s(s − a)(s − b)(s − c) + s4 · (
) ·S
2
α+1
Now using Heron’s formula we finally obtain
abc · s
α−1 2
· area(A1 B1 C1 ) = S 3 + s4 · (
) ·S
2
α+1
In this moment, we are left to prove the following inequality
(3)
sabc
(a + b + c)(a2 + b2 + c2 )
≥ 4S 2 ·
.
2
9abc
ON A ERDOS INSCRIBED TRIANGLE INEQUALITY REVISITED
5
Using abc = 4RS, we get the following equivalent inequality
2sa2 b2 c2 ≥ 16S 2
(a + b + c)(a2 + b2 + c2 )
,
9
which is succesively equivalent to
a2 b2 c2 ≥ 16S 2 ·
a2 + b 2 + c 2
,
9
9R2 ≥ a2 + b2 + c2 ,
which is evident since the distance between the circumcenter O and barycenter G
is given by Leibniz identity OG2 = 9R2 − (a2 + b2 + c2 ) and the proof of inequality
(3) ends.
Now, we can conclude
3
S +s
4
α−1
α+1
2
S ≥ 4S 2 ·
(a + b + c)(a2 + b2 + c2 )
area(A1 B1 C1 ),
9abc
which is finally equivalent to
9abc
area(A1 B1 C1 ) ≤
4(a + b + c)(a2 + b2 + c2 )
4
S+s ·
α−1
α+1
!
2
·S
−1
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) ≤
9a2 b2 c2
.
(a + b + c)(a2 + b2 + c2 )
By Schur’s inequality,
2(xy + yz + zx) − (x2 + y 2 + z 2 ) ≤
9xyz
,
x+y+z
by putting x = a2 , y = b2 and z = c2 , we have
2(a2 b2 + b2 c2 + c2 a2 ) − (a4 + b4 + c4 ) ≤
9a2 b2 c2
a2 + b 2 + c 2
which is equivalent to
(a + b + c)(b + c − a)(c + a − b)(a + b − c) ≤
which gives inequality (4).
9a2 b2 c2
,
a2 + b 2 + c 2
6
CEZAR LUPU, ŞTEFAN SPĂTARU
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
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C. L. Frenzen, E. J. Ionascu, P. Stanica, A proof of two conjectures related to ErdosDebrunner inequality, J. Ineq. Pure Appl. Math. 8(2007), art. 68, 13 pp.
C. Lupu, T. Lupu, Problem 1857, Math. Mag. 83 (2010), 391.
V. Mascioni, On the Erdos-Debrunner inequality, J. Ineq. Pure Appl. Math. 8(2007), art. 32,
5 pp.
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Ineq. Pure Appl. Math. 9(2008), art. 67, 11pp.
D. S. Mitrinovic, Analytic Inequalities, Springer Verlag, 1970.
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University of Pittsburgh, Department of Mathematics, Pittsburgh, 301 Thackeray Hall, PA 15260, USA and University of Craiova, Department of Mathematics, A. I. Cuza 13, RO–200585, Craiova, Romania
E–mail address: [email protected], [email protected]
International Computer High School of Bucharest, 428 Mihai Bravu St., Sector
3, Bucharest, Romania
E–mail address: spataru [email protected]