A GEOMETRIC INEQUALITY INVOLVING A
MOBILE POINT IN THE PLACE OF THE TRIANGLE
A Geometric Inequality
XIAO-GUANG CHU
YU-DONG WU
Suzhou Hengtian Trading Co. Ltd.,
Suzhou 215128, Jiangsu
People’s Republic of China
EMail: srr345@163.com
Department of Mathematics
Zhejiang Xinchang High School
Shaoxing 312500, Zhejiang, China
EMail: yudong.wu@yahoo.com.cn
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
Received:
04 March, 2008
JJ
II
Accepted:
27 August, 2009
J
I
Communicated by:
S.S. Dragomir
Page 1 of 16
2000 AMS Sub. Class.:
Primary 51M16; Secondary 51M25, 52A40.
Key words:
Bottema’s inequality, Euler’s inequality, Fermat’s sum, triangle, mobile point.
Abstract:
By using Bottema’s inequality and several identities in triangles, we prove a
weighted inequality concerning the distances between a mobile point P and three
vertexes A, B, C of 4ABC. As an application, a conjecture with regard to
Fermat’s sum P A + P B + P C is proved.
Acknowledgements:
The authors would like to thank Dr. Zhi-Gang Wang and Zhi-Hua Zhang for their
enthusiastic help.
Dedicatory:
Dedicated to Professor Lu Yang on the occasion of his 73rd birthday.
Go Back
Full Screen
Close
Contents
1
Introduction and Main Results
3
2
Preliminary Results
5
3
The Proof of Theorem 1.2
13
A Geometric Inequality
4
Remarks
15
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 2 of 16
Go Back
Full Screen
Close
1.
Introduction and Main Results
For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, ∆ the area, p
the semi-perimeter, R the circumradius and r the inradius, respectively. In addition,
supposing that P is a mobile point in the plane containing 4ABC, let P A, P B, P C
denote the distances between
PP and A, B, C, respectively. We
P will customarily use
the cyclic symbol,Qthat is:
f (a) = f (a) + f (b) + f (c),
f (a, b) = f (a, b) +
f (b, c) + f (c, a), f (a) = f (a)f (b)f (c), etc.
The following inequality can be easily proved by making use of Bottema’s inequality:
(1.1)
(P B + P C) cos
B
C
A
+ (P C + P A) cos + (P A + P B) cos
2
2
2
p2 + 2Rr + r2
≥p·
.
4R2
JJ
II
J
I
Page 3 of 16
A
B
C
P A cos + P B cos + P C cos ≥ p,
2
2
2
Go Back
Full Screen
2
q1 = cos2
2
2
B−C
C −A
A−B
p + 4R + 2Rr + r
+ cos2
+ cos2
=
,
2
2
2
4R2
we easily get
(1.4)
vol. 10, iss. 3, art. 79, 2009
Contents
and
(1.3)
Xiao-Guang Chu and Yu-Dong Wu
Title Page
Here we choose to omit the details. From inequality (1.1) and the following known
inequality (1.2) and identity (1.3) (see [3, 4, 6]):
(1.2)
A Geometric Inequality
PA + PB + PC ≥ p ·
cos2
B−C
2
+ cos2 C−A
+ cos2
2
cos A2 + cos B2 + cos C2
A−B
2
.
Close
Considering the refinement of inequality (1.4), Chu [2] posed a conjecture as
follows.
Corollary 1.1. For any 4ABC,
(1.5)
cos B−C
+ cos C−A
+ cos A−B
2
2
2
PA + PB + PC ≥ p ·
.
cos A2 + cos B2 + cos C2
A Geometric Inequality
The main object of this paper is to prove Conjecture 1.1, which is easily seen to
follow from the following stronger result.
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Theorem 1.2. In 4ABC, we have
A
B
C
(1.6) (P B + P C) cos + (P C + P A) cos + (P A + P B) cos
2
2
2
B−C
C −A
A−B
≥ p · cos
+ cos
+ cos
−1 .
2
2
2
Title Page
Contents
JJ
II
J
I
Page 4 of 16
Go Back
Full Screen
Close
2.
Preliminary Results
In order to prove our main result, we shall require the following four lemmas.
Lemma 2.1. In 4ABC, we have that
(2.1)
q2 = cos
B−C
C −A
A−B
p2 + 2Rr + r2
· cos
· cos
=
,
2
2
2
8R2
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
A
B
C
p
cos · cos · cos =
,
2
2
2
4R
(2.2)
(2.3)
(2.4)
vol. 10, iss. 3, art. 79, 2009
C
B−C 2
B
q3 =
cos cos cos
b + c 2 − a2
2
2
2
4
2
p + 2Rrp − r (2R + r) (4R + r)2
=
,
4R2
q4 =
1
2
X
cos2
B
C
+ cos2
2
2
b 2 + c 2 − a2
Contents
JJ
II
J
I
Page 5 of 16
Go Back
(2R + 3r) p2 − r (4R + r)2
=
,
2R
Full Screen
and
(2.5)
Title Page
X
Close
X
B−C
C −A
A−B
− cos
cos
2
2
2
Q
2
4
p∆
(b − c)
= q1 − 3q2 − 2 2 2 Q
,
abc
(X + x)
Q=
cos
where
p
X = a bc (p − b) (p − c),
p
Z = c ab (p − a) (p − b),
(2.6)
Y =b
p
ca (p − c) (p − a),
and
x = (b + c) (p − b) (p − c) ,
z = (a + b) (p − a) (p − b) .
(2.7)
y = (c + a) (p − c) (p − a) ,
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Proof. The proofs of identities (2.1) and (2.2) were given in [6]. Now, we present
the proofs of identities (2.3) – (2.5). By utilizing the formulas
r
r
p (p − a)
A
(p − b) (p − c)
A
,
sin =
,
cos =
bc
bc
2
r2
b + c (p − b) (p − c)
B−C
,
cos
=
bc
2
a
and (see [5, pp.52])
Y
a = 4Rrp,
=
Contents
JJ
II
J
I
Page 6 of 16
X
a = 2p
X
and
bc = p2 + 4Rr + r2 ,
Go Back
Full Screen
we get that
q3 = p
Title Page
X (b + c) (p − b) (p − c)
p
4a2 b2 c2
X
Close
2
2
2
b +c −a
a2 bc
bc (b + c) (c + a − b) (a + b − c) b2 + c2 − a2
=
p
[6(ab + bc + ca)2 (a + b + c)3 − 8(ab + bc + ca)3 (a + b + c)
2
2
2
4a b c
− (ab + bc + ca)(a + b + c)5 − 2abc(ab + bc + ca)(a + b + c)2
+ 8abc(ab + bc + ca)2 − abc(a + b + c)4 − 4(a + b + c)a2 b2 c2 ]
=
p4 + 2Rrp2 − r (2R + r) (4R + r)2
4R2
A Geometric Inequality
and
1X
2 B
2 C
cos
+ cos
b 2 + c 2 − a2
2
2
2
X
A
=
a2 cos2
2
p X 3 X 4
=
p
a −
a
abc p 5
=
(ab + bc + ca)(a + b + c)2 − 2(ab + bc + ca)2
abc 2
1
5
4
− (a + b + c) − (a + b + c)abc
2
2
(2R + 3r) p2 − r (4R + r)2
.
=
2R
Thus, identities (2.3) and (2.4) hold true.
With (1.3), (2.1) and the formulas of half-angles, we obtain that
−p2 + 8R2 − 2Rr − r2
8R2
X
1
= 2 2 2
x [bc (b + c) − (c + a) (a + b) (s − a)],
abc
q1 − 3q2 =
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 7 of 16
Go Back
Full Screen
Close
and
X
B−C
C −A
A−B
Q=
cos
− cos
cos
2
2
2
P
X [bc (b + c) − (c + a) (a + b) (s − a)]
=
.
a2 b 2 c 2
It is easy to see that
∆2 (b − c)2
X −x=
,
X +x
A Geometric Inequality
∆2 (c − a)2
Y −y =
,
Y +y
∆2 (a − b)2
and Z − z =
.
Z +z
Then
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
a2 b2 c2 [Q − (q1 − 3q2 )]
X
=
[bc(b + c) − (c + a)(a + b)(p − a)](X − x)
X
∆2 (b − c)2
=
[bc(b + c) − (c + a)(a + b)(p − a)]
X +x
2
X
(a − b)(a − c)(b − c)
=
p∆2
.
(X + x)
Therefore,
Contents
JJ
II
J
I
Page 8 of 16
Go Back
Full Screen
(a − b)(a − c)(b − c)2
a2 b2 c2 (X + x)
p∆2 (a − b)(a − c)(b − c) X b − c
=
,
a2 b 2 c 2
X +x
Q − (q1 − 3q2 ) =
X
p∆2
Close
where
b−c
c−a
a−b
+
+
X +x Y +y Z +z
(b − c) (Y − X + y − x) (a − b) (Y − Z + y − z)
=
+
(X + x) (Y + y)
(Z + z) (Y + y)
p (p − c) (b − c) (b − a) pabc (p − c) (b − c) (b − a)
=
+
(X + x) (Y + y)
(X + x) (Y + y) (X + Y )
p (p − a) (a − b) (b − c) pabc (p − a) (a − b) (b − c)
+
+
(Z + z) (Y + y)
(Z + z) (Y + y) (Z + Y )
p (b − c) (a − b)
= Q
[(p − a) (X + x) − (p − c) (Z + z)]
(X + x)
pabc (b − c) (a − b)
Q
+
(X + Y ) (Y + Z) (X + x)
× [(p − a) (X + x) (X + Y ) − (p − c) (Z + z) (Y + Z)]
2
−∆ (b − c) (a − b) (a − c)
Q
=
(X + x)
i
Y
p (b − c) (a − b) (a − c) h
Q
+
abc
(p − a) − ca (p − b) Y
(Z + X) (X + x)
Y
Y
pabc (b − c) (a − b) (a − c) h
Q
+
abc
(p − a) − Y
(p − a)
(X + Y ) (Z + Y ) (X + x)
Q
abc (Y − abc) (p − a)
+
Z +X
Q
Q
abc (pb + ca) (p − b) (p − a) − ca (p − b) Y (p − a)
+
Z +X
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 9 of 16
Go Back
Full Screen
Close
−∆2 (b − c)(a − b)(a − c)
pabc(b − c)(a − b)(a − c)
Q
Q
=
+
(X + x)
(X + Y )(Z + Y ) (X + x)
nhY
i
p
·
(p − a) − (p − b) ca(p − c)(p − a) (X + Y )(Y + Z)
h
i
Y
p
+ abc
(p − a) Y − abc + pb(p − b) + ca(p − b) − (p − b) ca(p − c)(p − a)
o
Y
+(Z + X)(abc − Y )
(p − a)
=
Xiao-Guang Chu and Yu-Dong Wu
−∆2 (b − c)(a − b)(a − c)
Q
,
(X + x)
vol. 10, iss. 3, art. 79, 2009
which implies the assertion (2.5).
Title Page
Lemma 2.2. For any 4ABC,
s
A
B
C
A
B
C
p
cos cos cos
cos + cos + cos
≥
.
(2.8)
2
2
2
2
2
2
2R
Proof. From Euler’s inequality R ≥ 2r, abc = 4Rrp, a + b + c = 2p and the law
of sines, we obtain that
(2.9)
2R2 p ≥ 4Rrp ⇐⇒ R2 (a + b + c) ≥ abc
⇐⇒ sin A + sin B + sin C ≥ 4 sin A sin B sin C.
Taking
Contents
JJ
II
J
I
Page 10 of 16
Go Back
Full Screen
Close
π−A
A→
,
2
we easily get
(2.10)
A Geometric Inequality
cos
π−B
B→
,
2
and
π−C
C→
,
2
A
B
C
A
B
C
+ cos + cos ≥ 4 cos cos cos .
2
2
2
2
2
2
Inequality (2.8) follows immediately in view of (2.10) and (2.2).
Lemma 2.3. In 4ABC, we have
(2.11)
X
cos
p4 + 2Rrp2 − r (2R + r) (4R + r)2
B
C 2
cos
b + c 2 − a2 ≥
.
2
2
4R2
Proof. By employing (2.3) and the formulas of half-angles, inequality (2.11) is equivalent to
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
(2.12)
X
B
C 2
cos cos
b + c 2 − a2
2
2
X
B
C
B−C 2
≥
cos cos cos
b + c2 − a2 ,
2
2
2
vol. 10, iss. 3, art. 79, 2009
Title Page
or
(2.13)
Contents
i
X (b2 + c2 − a2 ) h p
a
bc
(s
−
b)
(s
−
c)
−
(b
+
c)
(s
−
b)
(s
−
c)
≥ 0,
a2 bc
that is
(2.14)
X ∆2 (b2 + c2 − a2 )
·
(b − c)2 ≥ 0,
abc
a (X + x)
where X, Y, Z and x, y, z are given, just as in the proof of Lemma 2.1, by (2.6) and
(2.7), respectively.
Without loss of generality, we can assume that a ≥ b ≥ c to obtain
a (X + x) ≥ b (Y + y) ≥ c (Z + z) ,
and
(a − c)2 ≥ (b − c)2 ,
JJ
II
J
I
Page 11 of 16
Go Back
Full Screen
Close
and thus
(b − c)2
(c − a)2
≤
.
a (X + x)
b (Y + y)
Hence, in order to prove inequality (2.14), we only need to prove that
∆2 (b2 + c2 − a2 )
(c2 + a2 − b2 )
2
2
(2.15)
(b − c) +
(c − a) ≥ 0.
abc
a(X + x)
b(Y + y)
We readily arrive at the following result for a ≥ b ≥ c,
(c2 + a2 − b2 )
∆2 (b2 + c2 − a2 )
2
2
(b − c) +
(c − a)
abc
a(X + x)
b(Y + y)
(c − a)2
∆2 2
≥
b + c 2 − a2 + c 2 + a2 − b 2
abc
b (Y + y)
2
2
∆
(c − a)
= 2c2 ·
·
≥ 0.
abc b (Y + y)
This shows that the inequality (2.15) or (2.11) holds true. The proof of Lemma 2.3
is thus complete.
Lemma 2.4 (Bottema’s inequality, see [1, pp. 118, Theorem 12.56]). Let ∆0 denote the area of 4A0 B 0 C 0 , and a0 , b0 , c0 the side-lengths of 4A0 B 0 C 0 , respectively.
Then
(2.16)
2
(a0 P A + b0 P B + c0 P C)
1
≥ [a02 (b2 + c2 − a2 ) + b02 (c2 + a2 − b2 ) + c02 (a2 + b2 − c2 )] + 8∆∆0 .
2
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 12 of 16
Go Back
Full Screen
Close
3.
The Proof of Theorem 1.2
Proof. It is easy to show that
B
C
+ cos ,
2
2
A
B
c0 = cos + cos
2
2
a0 = cos
b0 = cos
C
A
+ cos ,
2
2
and
are three side-lengths of a certain triangle. By using Bottema’s inequality (2.16), in
order to prove inequality (1.6), we only need to prove that
r
2
Y
AX
A 1X
B
C
8∆
cos
cos +
cos + cos
b 2 + c 2 − a2
2
2
2
2
2
X
2
B−C
2
≥p
cos
−1
2
or
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 13 of 16
r
(3.1) 8∆
A Geometric Inequality
Y
AX
A
cos
cos + q4
2
2
X
C 2
B
b + c2 − a2 + 2p2 Q ≥ p2 (q1 + 1) .
+
cos cos
2
2
With identities (1.3), (2.4), (2.5), together with Lemma 2.2 and Lemma 2.3, in order
to prove inequality (3.1), we only need to prove that
p
(2R + 3r) p2 − r (4R + r)2 p4 + 2Rrp2 − r (2R + r) (4R + r)2
8∆ ·
+
+
2R
2R
4R2
Go Back
Full Screen
Close
"
+ 2p2
#
Q
−p2 + 8R2 − 2Rr − r2
p∆4 (b − c)2
− 2 2 2Q
8R2
abc
(X + x)
2
p + 4R2 + 2Rr + r2
2
≥p
+1
4R2
or
(3.2)
Q
−p4 + (4R2 + 20Rr − 2r2 ) p2 − r (4R + r)3
2p3 ∆4 (b − c)2
≥ 2 2 2Q
.
4R2
abc
(X + x)
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
From the known identities (see [5])
∆ = rp and
Title Page
(b − c)2 (c − a)2 (a − b)2 = 4r2 [−p4 + (4R2 + 20Rr − 2r2 )p2 − r(4R + r)3 ],
Contents
inequality (3.2) is equivalent to
Y
(3.3)
(X + x) ≥ 2r4 p5 .
For X ≥ x, and with the following two known identities (see [5, pp.53])
Y
Y
(b + c) = 2p(p2 + 2Rr + r2 ),
(p − a) = r2 p,
we obtain
Y
JJ
II
J
I
Page 14 of 16
Go Back
Full Screen
Close
(X + x) ≥ 8
Y
x=8
Y
(b + c)
Y
2
(p − a)
= 16r4 p3 (p2 + 2Rr + r2 ) > 16r4 p5 > 2r4 p5 .
Therefore, inequality (3.3) holds. This completes the proof of Theorem 1.2.
4.
Remarks
Remark 1. From inequalities (1.2) and (1.6), it is easy to see that inequality (1.5)
holds.
Remark 2. In view of
X
cos
X
B−C
p2 + 4R2 + 2Rr + r2
B−C
≥
cos2
=
2
2
4R2
X
B−C
p2 + 2Rr + r2
⇐⇒
cos
−1≥
,
2
4R2
it follows that inequality (1.6) is a refinement of inequality (1.1).
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 15 of 16
Go Back
Full Screen
Close
References
[1] O. BOTTEMA, R.Ž. DJORDEVIĆ, R.R. JANIĆ, D.S. MITRINOVIĆ AND
P.M.VASIĆ, Geometric Inequalities, Wolters-Noordhoff Publishing, Groningen,
The Netherlands, (1969).
[2] X.-G. CHU, A conjecture of the geometric inequality, Website of Chinese
Inequality Research Group http://www.irgoc.org/bbs/dispbbs.
asp?boardid=5&Id=2507&page=17. (in Chinese)
[3] X.-G. CHU, A class of inequality involving a moving point inside triangle, J.
Yibin Univ., 6 (2006), 20–22. (in Chinese)
[4] M.S. KLAMKIN, Geometric inequalities via the polar moment of Inertia, Math.
Mag., 48 (1975), 44–46.
[5] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND V. VOLENEC, Recent Advances in
Geometric Inequalities, Acad. Publ., Dordrecht, Boston, London, (1989).
[6] X.-Z. YANG AND H.-Y. YIN, The comprehensive investigations of trigonometric inequalities for half-angles of triangle in China, Studies of Inequalities, Tibet
People’s Press, Lhasa, (2000), 123–174. (in Chinese)
A Geometric Inequality
Xiao-Guang Chu and Yu-Dong Wu
vol. 10, iss. 3, art. 79, 2009
Title Page
Contents
JJ
II
J
I
Page 16 of 16
Go Back
Full Screen
Close