Identities and inequalities in a quadrilateral

Vol. 17, No.2, October 2009, pp 754-763 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

754

Identities and inequalities in a quadrilateral

Ovidiu T. Pop³²

ABSTRACT. In this paper we will prove some identities and inequalities in cyclic and tangential quadrilaterals.

1. INTRODUCTION

LetABCDbe a convex quadrilateral and we note AB =a,BC =b,CD=c, DA=d,BD=e, AC=f, s= a+b+c+d

2 ,AC∩BD={M}, the

measure of the angles AM B is ϕ and ∆ is the area of quadrilateralABCD.

If ABCD is a cyclic and tangential quadrilateral, letR,r be the radii of the circumscribed circle, respectively inscribed circle of quadrilateral ABCD.

It is well-known that the sides a,b, c,d are the solutions of the equation (see [4])

x⁴−2sx³+(s²+2r²+2rp

4R²+r²)x²−2rs(p

4R²+r²+r)x+r²s²= 0 (1.1) and the following inequalities are true (see [3])

2 q

2r(p

4R²+r²−r)≤s, (1.2)

with equality if and only if ABCDis a square (when R=r√

2) and isosceles trapezoid (whenR6=√

2),

s≤p

4R²+r²+r, (1.3)

with equality if and only if ABCDis a orthodiagonal quadrilateral and

32Received: 12.09.2009

2000Mathematics Subject Classification. 51M16.

Key words and phrases. Cyclic and tangential quadrilateral.

2 q

2r(p

4R²+r²−r)≤s≤p

4R²+r²+r, (1.4) when the equalities hold simultaneous if and only ifABCD is a square (whenR=r√

2), and ifR6=√

2 at least one inequality from (1.4) is strict.

On the other hand, the L. Fejes T´oth Inequality R≥√

2 (1.5)

holds, with equality if and only if ABCDis a square and the following identities

ef = 2r(p

4R²+r²+r) (1.6)

and

∆ =sr (1.7)

hold.

If quadrilateralABCD is cyclic, then

e²= (ac+bd)(ab+cd)

ad+bc , (1.8)

ef =ac+bd, (1.91)

e

f = ab+cd

ad+bc (1.10)

and

16R²∆² = (ab+cd)(ac+bd)(ad+bc), (1.11) called Girard’s relation.

2. IDENTITIES AND INEQUALITIES WITHR₁, R₂, R₃, R₄ LetABCD be a convex quadrilateral and we note withR₁, R₂, R₃, R₄ the radii of the circumscribed circles to the trianglesAM B,BM C,CM D and DM A.

Lemma 2.1. The following identity

R₁+R₂+R₃+R₄ = sef

2∆ (2.1)

and

a= 2sR₁

R₁+R₂+R₃+R₄, b= 2sR₂

R₁+R₂+R₃+R₄, c= 2sR₃

R₁+R₂+R₃+R₄, d= 2sR₄

R₁+R₂+R₃+R₄ (2.2) hold.

Proof. From the sine theorem we deduce that R₁ = a

2 sinϕ,R₂= b 2 sinϕ, R₃ = c

2 sinϕ,R₄ = d

2 sinϕ, from where R₁+R₂+R₃+R₄ = a+b+c+d

2 sinϕ = s

sinϕ. But ∆ = efsinϕ

2 and then (2.1) follows. On the other hand, we have that R₁

a = R₂ b = R₃

c = R₄

d = 1 2 sinϕ, from whereb= R₂

R₁a,c= R₃

R₁a andd= R₄

R₁a. From the relations above, by summing, we obtain that

2s=a+b+c+d=a+R₂

R₁a+R₃

R₁a+ R₄

R₁a= a

R₁(R₁+R₂+R₃+R₄) and then relations from (2.2) also result.

Lemma 2.2. If quadrilateralABCDis a cyclic and tangential quadrilateral, then

a= 2sR₁

√4R²+r²+r. (2.3)

Proof. From (2.2), by taking (2.1), (1.6) and (1.7) into account, relation (2.3) is obtained.

Theorem 2.1. IfABCD is a cyclic and tangential quadrilateral, then R₁, R₂, R₃, R₄ are the solutions of the equation

16s²x⁴−16s²(p

4R²+r²+r)x³+4(s²+2r²+2rp

4R²+r²)(p

4R²+r²+r)²x²

−4r(p

4R²+r²+r)⁴x+r²(p

4R²+r²+r)⁴ = 0. (2.4) Proof. By taking (2.3) into account, we have a solution for equation (1.1) by replacing one from (2.3) in (1.1), after calculus, we obtain thatR₁ verifies equation (2.4).

Theorem 2.2. If quadrilateralABCD is a cyclic and tangential quadrilateral, then the following identities

XR₁ =p

4R²+r²+r, (2.5)

XR₁R₂ = (s²+ 2r²+ 2r√

4R²+r²)(√

4R²+r²+r)²

4s² , (2.6)

XR₁R₂R₂= r(√

4R²+r²+r)⁴

4s² , (2.7)

R₁R₂R₂R₄= r²(√

4R²+r²+r)⁴

16s² , (2.8)

XR²₁= (√

4R²+r²+r)²(s²−2r²−2r√

4R²+r²)

2s² , (2.9)

X 1 R₁ = 4

r, (2.10)

X 1

R₁R₂ = 4(s²+ 2r²+ 2r√

4R²+r²) r²(√

4R²+r²+r)² , (2.11) and

X 1

R₁R₂R₃ = 16s² r²(√

4R²+r²+r)³ (2.12) hold.

Proof. It results from Theorem 2.1.

Theorem 2.3. If quadrilateralABCD is a cyclic and tangential quadrilateral, then the following inequalities

4r ≤X

R₁≤2R√

2, (2.13)

6r² ≤R²+r²+rp

4R²+r²≤X

R₁R₂ ≤ (5√

4R²+r²−3r)(√

4R²+r²+r)³ 64R²

≤ (5√

4R²+r²−3r)R√ 2

4 ≤ (10R√

2−3r)R√ 2

4 , (2.14)

4r³≤r(√

4R²+r²+r)²

4 ≤X

R₁R₂R₃≤(√

4R²+r²+r)⁵

128R² ≤R³√

2, (2.15)

r⁴≤r²(√

4R²+r²+r)²

16 ≤R₁R₂R₂R₄≤r(√

4R²+r²+r)⁵ 512R² ≤R⁴

4 , (2.16) 8r⁴

R² ≤ (√

4R²+r²+r)³3(√

4R²+r²−5r)

32R² ≤X

R²₁ ≤2R², (2.17) 4√

2

R ≤X 1

R₁, (2.18)

12

R² ≤ 8(5√

4R²+r²−3r) r(√

4R²+r²+r)² ≤X 1 R₁R₂ ≤

≤ 16(R²+r²+r√

4R²+r²) r²(√

4R²+r²+r)² ≤ 3R²

r⁴ (2.19)

and 8

rR²≤ 512R² r(√

4R²+r²+r)⁴≤X 1

R₁R₂R₃≤ 16 r²(√

4R²+r²+r)≤ 4

r³ (2.20) hold.

Proof. It results from Theorem 2.2 and from inequalities (1.2), (1.3) and (1.5).

IDENTITIES AND INEQUALITIES WITHr₁, r₂, r₃, r₄ In this section, we consider that ABCD is a cyclic and tangential

quadrilateral and we note withr₁, r₂, r₃, r₄ the radii of the inscribed circles to the triangles AM B,BM C,CM D andDM A.

Lemma 3.1. The following identities AM = eda

ab+cd, (3.1)

BM = eab

ab+cd, (3.2)

CM = ebc

ab+cd (3.3)

and

DM = ecd

ab+cd (3.4)

hold.

Proof. From the similarity of trianglesABM andDCM, respectivelyADM and BCM, we have that AM

DM = BM

CM = AB DC = a

c and AM

BM = DM

CM = AD BC= d

b. From the equalities above, it results that BM = b

dAM,CM = c

aBM = bc

adAM and DM = c

aAM. If we chooseBM and DM in the equalityBM +DM =e, then we obtain b

dAM+ c

aAM =e, from where relation (3.1) is obtained and after that, by replacing relations, relations (3.2)-(3.4) follow.

We note with s₁,T₁ the semiperimeter, respectively the area of the triangle AM B andα = sr²(√

4R²+r²+r) 2R(2R+r+√

4R²+r²).

Lemma 3.2. We have that

r₁ = α

c. (3.5)

Proof. By takings=a+c=b+d, (3.1),(3.2) and (1.6)-(1.11) into account, we calculate

s₁ = 1

2(AB+M B+M A) = 1 2

Ã

a+ eab

ab+cd+ eda ab+cd

!

= a 2

Ã

1 +e(b+d) ab+cd

!

= a 2

Ã

1 + s ab+cd

!

=



 s

(ac+bd)(ab+cd) ad+bc





= a 2

Ã

1 +s ef

p(ab+cd)(ac+bd)(ad+bc)

!

= a 2

Ã

1 +s ef 4R∆

!

= a 2

Ã

1 + ef 4Rr

!

= a 2

Ã

1 +2r(√

4R²+r²+r) 4Rr

! ,

from where

s₁ =a2R+r+√

4R²+r²

4R . (3.6)

We have thatT₁ = M A·M B·sinϕ

2 and by taking ∆ = efsinϕ

2 and (3.1), (3.2), (1.10), (1.11), (1.6) into account, it results that

T₁= 1 2

e²a²bd (ab+cd)²

2∆

ef = e f

a²bd∆

(ab+cd)² = a²bd∆

(ab+cd)(ad+bc)

= a²bd∆ef

(ab+cd)(ac+bd)(ad+bc)= a²bd∆ef 16R²∆² , from where

T₁= a²bd(√

4R²+r²+r)

8sR² . (3.7)

Because abcd= ∆² =s²r², from (3.6) and (3.7) it results that r₁ = T₁

s₁ = abcd(√

4R²+r²+r) 8sR²

4R c(2R+r+√

4R²+r²), and than (3.5) follows.

Remark 3.1. Similarly, we obtain that r₂ = ^α_d,r₃=^α_aandr₄ =^α_b. Remark 3.2. Because R₁ = ^{a·M B·M A}_4T

1 , with the help of relations (3.1), (3.2) and (3.7), we can calculate R₁, respectivelyR₂, R₃ and R₄.

Theorem 3.1. IfABCD is a cyclic and tangential quadrilateral, then r₁, r₂, r₃, r₄ are the solutions of the equation

16R⁴(2R+r+p

4R²+r²)⁴x⁴−16rR³(p

4R²+r²+r)²(2R+r+p

4R²+r²)³x³+

+4r²R²(s²+ 2r²+ 2rp

4R²+r²)(p

4R²+r²+r)²(2R+r+p

4R²+r²)²x²−

−4s²r⁴R(p

4R²+r²+r)³(2R+r+p

4R²+r²)x+

+s²r⁶(p

4R²+r²+r)⁴= 0. (3.8) Proof. By takingcfrom (3.5) as a solution of the equation (1.1), by replacing c from (3.5) in (1.1), after calculus, we obtain thatr₁ verifies equation (3.8).

Theorem 3.2. If quadrilateralABCD is a cyclic and tangential quadrilateral, then

Xr₁= r(√

4R²+r²+r)² R(2R+r+√

4R²+r²), (3.9) Xr₁r₂ = r²(s²+ 2r²+ 2r√

4R²+r²)(√

4R²+r²+r)² 4R²(2R+r+√

4R²+r²)² , (3.10) Xr₁r₂r₂ = s²r⁴(√

4R²+r²+r)³ 4R³(2R+r+√

4R²+r²)³, (3.11) r₁r₂r₂r₄= s²r⁶(√

4R²+r²+r)⁴ 16R⁴(2R+r+√

4R²+r²)⁴, (3.12) Xr²₁ = r²(√

4R²+r²+r)²(8R²+ 2r²+ 2r√

4R²+r²−s²) 2R²(2R+r+√

4R²+r²)² , (3.13)

X 1

r₁ = 4R(2R+r+√

4R²+r²) r²(√

4R²+r²+r) , (3.14) X 1

r₁r₂ = 4R²(s²+ 2r²+ 2r√

4R²+r²)(2R+r+√

4R²+r²)² s²r⁴(√

4R²+r²+r)² (3.15)

and

X 1

r₁r₂r₃ = 16R³(2R+r+√

4R²+r²)³ s²r⁵(√

4R²+r²+r)² . (3.16) Proof. It results from Theorem 3.1.

Theorem 3.3. If quadrilateralABCD is a cyclic and tangential quadrilateral, then the following inequalities

8r³(√ 2−1) R² ≤X

r₁≤2√ 2(√

2−1)R, (3.17)

24r⁶(√ 2−1)²

R⁴ ≤X

r₁r₂ ≤ 3R⁴(√ 2−1)²

2r² , (3.18)

32r⁹(√ 2−1)³

R⁶ ≤X

r₁r₂r₃≤ R⁶(√ 2−1)³

2r³ , (3.19)

64r¹²(√ 2−1)⁴

R⁸ ≤r₁r₂r₂r₄ ≤ R⁴(√ 2−1)⁴

4 (3.20)

and

4√ 2(√

2 + 1)

R ≤X 1

r₁ ≤ 2R²(√ 2 + 1)

r³ (3.21)

hold.

Proof. It results from Theorem 3.2 and from inequalities (1.2), (1.3) and (1.5).

REFERENCES

[1] Minculete, N.,Characterization of a tangential quadrilateral,Forum Geometricorum 9 (2009), 113-118

[2] Mitrinovi´c, D. S., Peˇcari´c, J. E. and Volonec, V.,Recent Advances in Geometric Inequalities, Kluver Academic Publishers, Dordrecht, 1989

[3] Pop, O., Inegalit˘at¸i ˆın patrulater, Gazeta matematic˘a B, 5-6 (1988), 203-206 (Romanian)

[4] Pop, O.,Identit˘at¸i ¸si inegalit˘at¸i ˆıntr-un patrulater, Gazeta matematic˘a B, 8 (1989), 279-280 (Romanian)

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