cac chuyen de boi duong hoc sinh gioi toan 2013 tai nam dinh

Mục lục
Lời nói đầu
Nguyen Van Mau
On the solutions of some classes of functional equations with transformed
argument
Dam Van Nhi
A new inequality and identity (M,N)
Nguyen Dang Phat
Some National Olympiad Problems in plane geometry
Le Anh Vinh
Elementary Counting Problems
Đặng Huy Ruận
Hai phương pháp giải bài toán trò chơi bốc vật
Trần Nam Dũng
Một số phương pháp xây dựng bài toán về dãy số
Tạ Duy Phượng
Sơ lược giới thiệu di sản sách toán trong thư tịch Hán Nôm
Nguyễn Minh Tuấn
Lời giải cho một lớp các bất đẳng thức đồng bậc
Nguyễn Bá Đang
Ứng dụng tính chất tam giác đồng dạng
Trần Xuân Đáng
Sử dụng đạo hàm để chứng minh bất đẳng thức
Hoàng Minh Quân
Phương trình bậc bốn và hệ thức hình học trong tứ giác hai tâm
Nguyễn Thùy Trang
Hàm số mũ và phương trình hàm liên quan
Nguyễn Đình Thức
Một số ứng dụng lượng giác trong bài toán dãy số
Nguyễn Hoàng Cương
Phép quay véctơ và một số ứng dụng

Nguyễn Thị Giang
Tính chia hết trong các bài toán về dãy số
và phương trình hàm trên tập số nguyên
Nguyễn Hữu Thiêm
Ánh xạ và một số bài toán liên quan

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57
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156
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201

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Phạm Văn Nho
Vài nét về lịch sử phát triển lí thuyết số
Trần Quang Vinh
Một số phương pháp giải phương trình và hệ phương trình đại số

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ON THE SOLUTIONS OF SOME CLASSES OF FUNCTIONAL
EQUATIONS WITH TRANSFORMED ARGUMENT
NGUYEN VAN MAU
Abstract
We deal with some functional equations with transformed arguments
in real plane. By an algbraic approach we solve some kinds of functional
equations with the reflection arguments.
f (x, y)±f (2p−x, y)±f (x, 2q−y)+f (2p−x, 2q−y) = h(x, y), (x, y) ∈ Ω,
(0.1)
where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is
given.
In applications, we formulate the necessary and sufficient condition
for solvability of the following functional equations
f (xy)±f ((1−x)y±f (x(1−y))+f ((1−x)(1−y)) = h(xy), ∀x, y ∈ (0, 1).
and
f (x + y) ± f (−x + y) ± f (x − y) + f (−x − y)) = h(x + y), ∀x, y ∈ (−1, 1)
and describe the formulae of the general solution f (xy) and f (x + y), respectively.
1

3


September 25, 2013

1

Representations of some classes of two-variable
functions with reflection argument

In this section we will describe some classes of two-variable functions with
transfromed argument. Namely, we deal with two-variable functions being
skew symmetric about a given point (p, q).
Definition 1.1. Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)
is the center (centeral point) of Ω. Function f (x, y) defined in Ω is said to be
even-even (or even in both variables or skew symmetric) about the point (p, q)
iff
f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω.
Definition 1.2. Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)
is the center (centeral point) of Ω. Function f (x, y) defined in Ω is said to be
even-odd about the point (p, q) iff
f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω.
Remark 1. Similarly, we have the definitions of odd-even and odd-odd funtions.
In a special case, we have
Definition 1.3. Function f (x, y) defined in R × R is said to be even-even iff
f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀x, y ∈ R.
The following natural questions arise:
∗

2000 Mathematics Subject Classification. Primary 39B99, 39B62, 39B22, 39B32, 39B52.

4


Problem 1.1. How to describe the two-variable function f (x, y) in the cases

f (x, y) is even-even in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω.

(1.1)

Solution. Note that
f (2p − x, 2q − y) = f (x, 2q − y) = f (x, y), ∀x, y ∈ Ω,
so we can write
1
f (x, y) = [f (x, y) + f (x, 2q − y) + f (2p − x, y) + f (2p − x, 2q − y)]. (1.2)
4
Now we prove that the function f (x, y) is even in both variables (x, y) if and
only if there exists a function g(x, y) defined in R × R such that
1
f (x, y) = [g(x, y) + g(x, 2q − y) + g(2p − x, y) + g(2p − x, 2q − y)].
4

(1.3)

Indeed, if f (x, y) is of the form (1.3) then it is easy to check the conditions
(1.1) are saistified and if f (x, y) is even then it has the form (1.2) and then
the form (1.3) with g = f.
Corollary 1.1. The two-variable function f (x, y) is even in both variables
(x, y), i.e.
f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀(x, y) ∈ R.

(1.4)

iff it is of the form
1

f (x, y) = [g(x, y) + g(x, −y) + g(−x, y) + g(−x, −y)].
4

(1.5)

where g(x, y) is an arbitrary function defined in R × R.
Similarly, we have
Problem 1.2. How to describe the two-variable function f (x, y) in the cases
f (x, y) is even-odd in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω.
Solution. Note that
f (2p − x, 2q − y) = f (x, 2q − y) = −f (x, y), ∀x, y ∈ Ω,
5

(1.6)


so we can write
1
f (x, y) = [f (x, y) + f (2p − x, y) − f (x, 2q − y) − f (2p − x, 2q − y)]. (1.7)
4
Now we prove that the function f (x, y) is even-odd in both variables (x, y) if
and only if there exists a function g(x, y) defined in R × R such that
1
f (x, y) = [g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.8)
4
Indeed, if f (x, y) is of the form (1.8) then it is easy to check the conditions
(1.6) are saistified and if f (x, y) is even-odd then it has the form (1.7) and
then the form (1.8) with g = f.
Similarly, we can formulate the following representations

Theorem 1.1. The two-variable function f (x, y) in the cases f (x, y) is eveneven in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω.

(1.9)

is of the form
1
f (x, y) = [g(x, y) + g(2p − x, y) + g(x, 2q − y) + g(2p − x, 2q − y)]. (1.10)
4
Theorem 1.2. The two-variable function f (x, y) in the cases f (x, y) is evenodd in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω.

(1.11)

is of the form
1
f (x, y) = [g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.12)
4
Theorem 1.3. The two-variable function f (x, y) in the cases f (x, y) is oddeven in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = −f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω.

(1.13)

is of the form
1
f (x, y) = [g(x, y) − g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)]. (1.14)
4

6

Theorem 1.4. The two-variable function f (x, y) in the cases f (x, y) is oddood in both variables (x, y) about the point (p, q), i.e.
f (2p − x, y) = −f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω.

(1.15)

is of the form
1
f (x, y) = [g(x, y) − g(2p − x, y) − g(x, 2q − y) + g(2p − x, 2q − y)]. (1.16)
4
Now we consider the special case of two-variable function f (x, y) defined in
the set Ω0 = (0, 1) × (0, 1) in the cases f (x, y) is even - even in both variables
1
(x, y) about the point 0, . So theorem 2.1 can be formulate in the following
2
form.
Corollary 1.2. The two-variable function f (x, y) in the case when f (x, y) is
1 1
even-even in both variables (x, y) about the point
, , i.e.
2 2
f (1 − x, y) = f (x, y) and f (x, 1 − y) = f (x, y), ∀x, y ∈ (0, 1)

(1.17)

is of the form
1
f (x, y) = [g(x, y) + g(1 − x, y) + g(x, 1 − y) + g(1 − x, 1 − y)].
4

(1.18)

Corollary 1.3. The two-variable function f (x, y) in the case when f (x, y) is
1 1
odd-odd in both variables (x, y) about the point
, , i.e.
2 2
f (1 − x, y) = −f (x, y) and f (x, 1 − y) = −f (x, y), ∀x, y ∈ (0, 1)

(1.19)

is of the form
1
f (x, y) = [g(x, y) − g(1 − x, y) − g(x, 1 − y) + g(1 − x, 1 − y)].
4

2

(1.20)

Functional equations for two-variable funtions induced by involutions

In this section we will solve the following functional equations
f (x, y) + f (2p − x, y) + f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω
(2.1)

7


and

f (x, y) − f (2p − x, y) − f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω,
(2.2)
where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is given.
Denote by X the set of all functions defined on X and X = L0 (X, X),
where L0 (X, X) denotes the linear space of all linear operators A : X → X
with dom A = X. It is easy to check that X is an algebra (linear ring) over
field R.
Consider the following linear elements (operators) V and W in X as follows
(V f )(x, y) = f (2p − x, y),

(W f )(x, y) = f (x, 2q − y), f ∈ X.

(2.3)

It is easy to see that V and W are involution elements, i.e. V 2 = I and
W 2 = I, where I is an identity element of X . Moreover, they are commutative,
i.e. V W = W V.
Now we rewrite (2.1) in the form
Kf := (I + V + W + V W )f = h.

(2.4)

Lemma 2.1. Operator K defined by (2.4) is an algebraic element with charateristic polynomial
PK (t) = t2 − 4t.
(2.5)
Proof. Note that the following identities hold
V K = K, W K = K, V W K = K.
So K 2 = 4K and K is not a scalar operator, which toghether imply PK (t) =
t2 − 4t, which was to be proved.
Theorem 2.1. The general solution of the homogeneous equation Kf = 0 is

of the form
1
f (x, y) = [3g(x, y)−g(2p−x, y)−g(x, 2q−y)−g(2p−x, 2q−y)], g ∈ X. (2.6)
4
Proof. By Lemma 2.1, from equality (K − K)f = 0, ∀f ∈ X, we find

1
4

K2 −

4K f = 0 ⇔ (K 2 − 4K)f = 0 ⇔ K(K − 4I)f = 0. Hence (K − 4I)X ⊂
ker K. On the other hand, if ϕ ∈ ker K then Kϕ = 0 and K(K − 4I)ϕ =
(K − 4I)Kϕ = 0. It follows ϕ ∈ (4I − K), which was to be proved.

8


Theorem 2.2. The non-homogeneous equation (2.1) (Kf = h) is solvable if
and only if the following condition
Kh = 4h.

(2.7)

If it is the case, then the general solution of (2.1) is of the form
f (x, y) = 3g(x, y) − g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y) (2.8)
1
+ [3h(x, y) − h(2p − x, y) − h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.
4
Proof. Suppose that the equation (2.1) is solvable and f0 is a solution. Then

By Lemma 2.1 from equality Kf0 = h it follows K 2 f0 = Kh ⇔ 4Kf0 = Kh ⇔
4h = Kh.
Suppose that the condition (2.7) is satisfied. Write the non-homogeneous
equation (2.1) in the form Kf = 41 Kh or in the equivalent form
1
K f − h = 0.
4

(2.9)

Theorem 2.1 gives the general solution of (2.9) in the form
1
1
f = h + (4I − K)ψ ⇔ f = h + (4I − K)ψ, ψ ∈ X,
4
4
i.e. it has the form (2.8).
Now we consider (2.2). Write it in the form
Lf := (I − V − W + V W )f = h.

(2.10)

Lemma 2.2. Operator L defined by (2.4) is an algebraic element with charateristic polynomial
PL (t) = t2 − 4t.
(2.11)
Proof. The proof follows from the following identities
−V L = L, −W L = L, V W L = L.
So L2 = 4L and L is not a scalar operator, which toghether imply PL (t) =
t2 − 4t, which was to be proved.
Theorem 2.3. The general solution of the homogeneous equation Kf = 0 is

of the form
1
f (x, y) = [3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)], g ∈ X.
4
(2.12)
9


Proof. By the same method as for theorem2.1.
Theorem 2.4. The non-homogeneous equation (2.1) (Lf = h) is solvable if
and only if the following condition
Lh = 4h.

(2.13)

If it is the case, then the general solution of (2.1) is of the form
f (x, y) = 3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y) (2.14)
1
+ [3h(x, y) + h(2p − x, y) + h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.
4
Proof. Suppose that the equation (2.2) is solvable and f0 is a solution. Then
By Lemma 2.2 from equality Lf0 = h it follows L2 f0 = Lh ⇔ 4Lf0 = Lh ⇔
4h = Lh.
Suppose that the condition (2.13) is satisfied. Write the non-homogeneous
equation (2.2) in the form Lf = 14 Lh or in the equivalent form
1
L f − h = 0.
4

(2.15)

Theorem 2.1 gives the general solution of (2.15) in the form
1
1
f = h + (4I − L)ψ ⇔ f = h + (4I − L)ψ, ψ ∈ X,
4
4
i.e. it has the form (2.14).

3

Special cases

1
Now we consider some special cases of equation when q = p = . In that case,
2
1 1
the center point of Ω is
, . and (2.1) is of the form
2 2
f (x, y)+f (1−x, y)+f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1). (3.1)
and
f (x, y)−f (1−x, y)−f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1). (3.2)
In this case, the role of x and y in the left side of (3.1) are the same.
Now return to the function f (t), we can formulate the following
10


Theorem 3.1. The function f (t) satisfying the conditions
f ((1 − x)y) = f (xy), ∀x, y ∈ (0, 1)

(3.3)

if and only if there exists a function g(t) such that
1
f (xy) = [g(xy) + g((1 − x)y) + g(x(1 − y)) + g((1 − x)(1 − y))].
4

(3.4)

Proof. From (3.7) we find f (x(1 − y)) = f (xy) and
f ((1 − x)(1 − y)) = f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1).
So we can write f (t) in the form
1
f (xy) = [f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))].
4
Last equality gives us the proof of the theorem.
Theorem 3.2. The functional equation
f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy)

(3.5)

is solvable if and only if there exists h(t) satisfying the following condition
h(x(1 − y)) = h(xy), ∀x, y ∈ (0, 1).

(3.6)

So now we shall examine the equation (3.6).
1
1

, 1 into (3.6), we get
Putting x = t, y = , where t ∈
2t
2
1
1
1
h t−
=h
, ∀t ∈
,1 .
2
2
2
or

1
1
, ∀x ∈ 0, .
2
2
1−y
Similarly, putting xy = t, then x(1 − y) = t
and
y
h(x) = h

h(t) = h

1−y

t , ∀y ∈ (0, 1).
y

(3.7)

(3.8)

(3.7) and (3.8) together inply h(t) ≡ const .
So, the necessary condition for solvability of (3.1) if
h(x, y) = h(y, x), ∀x, y ∈ (0, 1).
Now we formulate the similar results as in the previous section.
11

(3.9)


Theorem 3.3. The general solution of the homogeneous equation
f (x, y) + f (1 − x, y) + f (x, 1 − y) + f (1 − x, 1 − y) = 0, ∀x, y ∈ (0, 1) (3.10)
is of the form
1
f (x, y) = [3g(x, y) − g(1 − x, y) − g(x, 1 − y) − g(1 − x, 1 − y)], g ∈ X. (3.11)
4
Theorem 3.4. The non-homogeneous equation
f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)
(3.12)
is solvable iff h(t) ≡ const in (0, 1). If it is the case, then the general solution
of (3.14) is of the form
1
f (xy) = [c+3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], c ∈ R, g ∈ X.
4

(3.13)
Proof.
Theorem 3.5. The function f (t) is a general solution of the homogeneous
equation
f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = 0, ∀x, y ∈ (0, 1) (3.14)
if and only if there exists a function g(t) such that
1
f (xy) = [3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.15)
4
Remark 2. The homogeneous equation (3.14) was posed by Sahoo and Sander
in 1990 (see [3]-[4]). The continuous solutions were found by Z. Daroczy and
A. Jarai in [5].
Now we deal with the equation induced by addition of arguments. We
consider the special cases of equation when q = p = 0 In that case, the center
point of Ω is (−1, 1) and (2.1) is of the form
f (x, y) + f (−x, y) + f (x, −y) + f (−x, −y) = h(x, y), ∀x, y ∈ (−1, 1). (3.16)
Theorem 3.6. The function f (t) is a general solution of the homogeneous
equation
f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.17)
if and only if f (x) is an odd function in (−2, 2).
12


Proof. If f is odd in (−2, 2), then f (x − y) = −f (−x + y) and f (−x − y) =
−f (x + y). Hence,
f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1).
Conversely, suppose f is a solution of (3.17). Putting x = 0, y = 0 into (3.17)
we find f (0) = 0. Similarly, putting y = x, into (3.15) and ussing the equality
f (0) = 0, we find f (−2x) = −f (2x), i.e. f is odd in (−2, 2).
Theorem 3.7. The non-homogeneous equation

f (x+y)+f (−x+y)+f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.18)
is solvable iff h(t) ≡ const in (−2, 2). If it is the case, then the general solution
of (3.14) is of the form
1
f (t) = h(0) + g(t),
(3.19)
4
where g is an arbitrary odd function in (−2, 2).
Proof. Suppose equation (3.18) is solvable and f is its solution. Putting y = x
and y = −x into (3.18), we find h(2x) = h(0), x ∈ (−1, 1), i.e. h(t) ≡ h(0) in
(−2, 2).
If h(t) ≡ h(0) in (−2, 2) then we can reduce equation (3.18) to the equation
ϕ(x + y) + ϕ(−x + y) + ϕ(x − y) + ϕ(−x − y)) = 0, ∀x, y ∈ (−1, 1), (3.20)
1
where ϕ(t) = f (x)− . Hence, the solution (3.19) follows from theorem 3.6.
4
Now we return to the equation
f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = 0.

(3.21)

By the same way as previous equations, we have
Theorem 3.8. The function f (t) satisfying the conditions
f ((1 − x)y) = −f (xy), ∀x, y ∈ (0, 1)

(3.22)

if and only if there exists a function g(t) such that
1
f (xy) = [g(xy) − g((1 − x)y) − g(x(1 − y)) + g((1 − x)(1 − y))].

4

13

(3.23)


Proof. From (3.23) we find f (x(1 − y)) = −f (xy) and
f ((1 − x)(1 − y)) = −f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1).
So we can write f (t) in the form
1
f (xy) = [f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))].
4
Last equality gives us the proof of the theorem.
Theorem 3.9. The functional equation
f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy)

(3.24)

is solvable if and only if there exists h(t) ≡ 0.
Theorem 3.10. The non-homogeneous equation
f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)
(3.25)
is solvable iff h(t) ≡ 0 in (0, 1). If it is the case, then the general solution of
(3.14) is of the form
1
f (xy) = [3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.26)
4
Theorem 3.11. The function f (t) is a general solution of the homogeneous
equation

f (xy)−f ((1−x)y)−f (x(1−y))+f ((1−x)(1−y)) = 0, ∀x, y ∈ (0, 1) (3.27)
if and only if there exists a function g(t) such that
1
f (xy) = [3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.28)
4
Remark 3. The equation (3.27) was posed firstly by K. Lajko in [8] for X = R
and then by Sahoo and Sander in 1990 (see [3]-[4]). The differentiable solutions
were found by C.J. Eliezer in [6].
Theorem 3.12. The non-homogeneous equation
f (x+y)−f (−x+y)−f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.29)
is solvable iff h(t) ≡ const in (−2, 2). If it is the case, then the general solution
of (3.27) is of the form
1
f (t) = c + (g(t) − g(−t)), t ∈ (−2, 2),
2
where g is an arbitrary function in (−2, 2), c = f (0).
14

(3.30)


Proof. Suppose equation (3.31) is solvable and f is its solution. Putting y = x
and y = −x into (3.31), we find h(2x) = −h(0), x ∈ (−1, 1), i.e. h(t) ≡ −h(0)
in (−2, 2). Put x = 0 = y into (3.31) we find h(0) = 0.
So (3.30) is of the form
f (x + y) − f (−x + y) − f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.31)
and it has solution of the form (3.30).

References
[1] T. Acze’l, Lectures on functional equations and their applications, Academic Press, New York/San Francisco/London, m1966.

[2] M. Kuczma, B. Choczewski, R. Ger, Interative Functional Equations, Cambridge University Press, Cambridge/New York/Port
Chester/Melbourne/Sydney, 1990.
[3] P.K. Sahoo, T. Riedel, Mean Value Theorems and Functional Equations,
World Scientific, Singapore/New Jersey/London/HongKong, 1998.
[4] B.R. Ebanks, P.K. Sahoo and W. Sander, Determination of measurable
sum from infomation measures satisfying (2, 2)−additivity of degree (α, β),
Radovi Matematicki, vol.6, 77-96, 1990.
[5] Z. Daroczy and A. Jarai, On the measurable solution of a functional equation of the information theory, Acta Math. Acad Sci. Hungaricae, vol.34,
105-116, 1979.
[6] C.J. Eliezer, A solution to f (1−x)(1−y))+f (xy) = f (x(1−y))+f (1−x)y,
Aequationes Math., vol.46, 301, 1993.
[7] Gy. Maksa, Problem, Aequationes Math., vol.46, 301, 1993.
[8] K. Lajko, What is the general solution to the equation f (1 − x)(1 − y)) +
f (xy) = f (x(1 − y)) + f (1 − x)yAequationes Math., vol.10, 313, 1974.
[9] D. Przeworska - Rolewicz, Algebraic analysis, PWN - Polish Scientific
Publishers and D. Reided Publishing Company, Warszawa - Dordrecht,
1988.
[10] D. Przeworska - Rolewicz and S. Rolewicz, Equations in linear space,
Monografie Matematyezne 47, PWN - Polish Scientific Publishers,
Warszawa, 1968.
[11] Ng. V. Mau, Boundary value problems and controllability of linear systems
with right invertible operators, Dissertationes Math., CCCXVI, Warszawa,
1992.
15


INEQUALITY AND IDENTITY (M, N )
Dam Van Nhi
Pedagogical University Ha Noi
136 Xuan Thuy Road, Hanoi, Vietnam.

Email:
2010 Mathematics subject classification: 26D05,26D15,51M16
Abstract. In this paper we introduce the inequalty and identity (M, N ), of which Hayashi
is a spcial case: Inequality, Identity (M, N ). And after then we will present some interesting applications.
Keywords. Hayashi’s inequality, point, triangle, polygon.

1

Introduction

In Euclidian geometry, the Hayashi’s Inequality in R2 states: Suppose given a triangle
ABC of the lengths of sides a, b, c. Then, with any point M, we have an inequality
aM B.M C + bM C.M A + cM A.M B

abc

(see [3, pp. 297, 311]). In this paper we propose to give a new inequality and it’s
identity, which is a generalization of the above inequality, and after then we want to
give some interesting applications about triangle.

2

Inequality and Identity (M, N )

Now we prove an inequality, of which Hayashi is a special case. Use this result we can
give some new interesting inequalities. We give the following:
Theorem 2.1. Let A1 A2 . . . An be a polygon, s be an integer, s < n, and arbitrary
points N1 , N2 , . . . , Ns , M in euclidean plane R2 we have the following inequality
s

s

M Nj

j=1

M Ai
i=1

Ak Nj

n

j=1
n
k=1

Ak Ai .M Ak
i=k

16

, (M, N ).


(i) If s = 0, we have Hayashi inequality.
(ii) If n = 3, s = 1, and A, B, C, N belong to the circle with the centere M we have the
inequality aAN + bBN + cCN 4SABC .
Proof. Suppose that Ak have affixe ak , M has affixe z and Nh affixe zh . Using the
s

s

(ak − zj )

n

j=1

(z − zj ) =

Lagrange interpolation formula, we have

j=1

(ak − ai )

k=1

(z − ai ) and
i=k

i=k
s

s

|z − zj |
deducing

|z − ai |

|ak − zj |

n

j=1

j=1

|ak − ai ||z − ak |

k=1

i=1

. From this, we deduce geometric inequal-

i=k

ity
s

s

M Nj

Ak Nj

n

j=1
n

j=1

M Ai

k=1

i=1

(i) When s = 0 we have

Ak Ai .M Ak

.

i=k

M Nj = 1 =

Hayashi inequality for the polygon

Ak Nj and the inequality (M, N ) becomes the
n
1
.
A A .M Ak
M Ai k=1 i=k k i

1

n
i=1

(ii) When n = 3, s = 1 and A, B, C, N belong to the circle with the centere M we have
abc
the inequality
aAN + bBN + cCN or aAN + bBN + cCN 4SABC . .
R
Remark 2.2. Denote N as the center of circumcircle. From the iequality aAN +bBN +
cCN
4SABC by the inequality (M,N) (ii) we deduce R(a + b + c) 2r(a + b + c) or
R 2r [Euler].
Corollary 2.3. Suppose that O, I and G are respectively the centre of circumcirle and
incircle of ABC. Denote the radii of circumcircles of the triangles GBC, GCA, GAB
by R1 , R2 , R3 , respectivly. Let ra , rb , rc be the the radius of circumcircle of the triangles IBC, ICA, IAB, and let R1 , R2 , R3 be the radii of circumcircles of the triangles
OBC, OCA, OAB, respectivly. We will have
(i) R2

abc
.
a+b+c

(ii) R1 + R2 + R3

3R (see [1]).

(iii)

ra
rb
rc
+
+
ha hb hc

R
with ha , hb , hc are the lengths of altitudes of ∆ABC.
r

(iv)

R1 x R2 y R3 z
+
+
R when ∆ABC is not obtituse and x, y, z are the distances
ha
hb
hc
from O to the 3 sides.
17


Proof. (i) Applying the inequality (M,N) (ii) we obtain aOB.OC+bOC.OA+cOA.OB
abc
abc or R2
.
a+b+c
(ii) Applying the inequality (M, N ) we obtain aGB.GC + bGC.GA + cGA.GB

abc.
SABC
abc
abc
abc
Since aGB.GC = 4R1 SGBC = 4R1
= 4R1
= R1
, bGC.GA = R2
and
3
3.4R
3R
3R
abc
abc
abc
abc
cGA.GB. = R2
therefore R1
+ R2
+ R3
abc or R1 + R2 + R3 3R.
3R
3R
3R
3R
(iii) Applying the inequality (M, N ) we have aIB.IC + bIC.IA + cIA.IB abc. Since
ra rabc
ra rabc

rb rabc
aIB.IC = 4ra SIBC = 2ra ra = 4
=
, bIC.IA =
and cIA.IB =
ha 4R
ha R
hb R
rc rabc
ra rabc ra rabc ra rabc
ra
rb
rc
R
there is
+
+
abc or
+
+
.
hc R
ha R
ha R
ha R
ha hb hc
r
(iv) Applying the inequality (M, N ) we have aOB.OC +bOC.OA+cOA.OB abc. Since
x abc
x abc

y abc
aOB.OC = 4R1 SOBC = 2R1 xa = 4R1
= R1
, bOC.OA = R2
and
ha 4R
ha R
hb R
z abc
R x abc R2 y abc R3 z abc
Rx Ry Rz
cOA.OB = R3
we have 1
+
+
abc or 1 + 2 + 3
hc R
ha R
hb R
hc R
ha
hb
hc
R.
Proposition 2.4. Suppose given a triangle ABC with the lengths of sides a, b, c respectively and R is the radius of circumcircle of ABC. Let’s I, Ja , Jb , Jc are the centres of
incircle and escribed circles of ∆ABC, respectively. Then, with any point M, we have
abcM I
M A.M B.M C
√
MI a + b + c

(ii)
M A.M B.M C
(i)

(iii)
(iv)

aAI
bBI
cCI
+
+
.
MA MB MC
√
√
√
b+c−a
c+a−b
a+b−c
√
+ √
+ √
.
caM B
bcM A
abM C

M Ja + M Jb + M Jc
M A.M B.M C

AJa + AJb + AJc BJa + BJb + BJc CJa + CJb + CJc
+
+
.
bcM A
caM B
abM C

M Ja .M Jb + M Jb .M Jc + M Jc .M Ja
AJa .AJb + AJb .AJc + AJc .AJa
M A.M B.M C
bcM A
BJa .BJb + BJb .BJc + BJc .BJa CJa .CJb + CJb .CJc + CJc .CJa
+
+
.
caM B
abM C

Proof. (i) Applying the inequality (M, N ) we have
MI
M A.M B.M C

AI
BI
CI
+
+
.

bcM A caM B abM C

bc(b + c − a)
ca(c + a − b)
ab(a + b − c)
(ii) Since IA2 =
, IB 2 =
, IC 2 =
therefore
a +√b + c
a+b+c
√
√a + b + c √
MI a + b + c
b+c−a
c+a−b
a+b−c
√
+ √
+ √
.
M A.M B.M C
caM B
bcM A
abM C

18


(iii) Applying the inequality (M, N ) to n = 3, s = 1, we have the three inequalities

M Ja
M A.M B.M C
M Ja
M A.M B.M C
M Jc
M A.M B.M C

AJa
BJa
CJa
+
+
bcM A caM B abM C
AJb
BJb
CJb
+
+
bcM A caM B abM C
AJc
BJc
CJc
+
+
.
bcM A caM B abM C
M Ja + M Jb + M Jc
On adding the three inequalities, we find the inequality
M A.M B.M C
AJa + AJb + AJc BJa + BJb + BJc CJa + CJb + CJc

+
+
.
bcM A
caM B
abM C
(iii) Applying the inequality (M, N ) to n = 3, s = 1, we have the three inequalities
M Ja .M Jb
M A.M B.M C
M Jb .M Jc
M A.M B.M C
M Jc .M Ja
M A.M B.M C

AJa .AJb BJa .BJb CJa .CJc
+
+
bcM A
caM B
abM C
AJb .AJc BJb .BJc CJb .CJc
+
+
bcM A
caM B
abM C
AJc .AJa BJc .BJa CJc .CJa
+
+
.

bcM A
caM B
abM C
M Ja .M Jb + M Jb .M Jc + M Jc .M Ja
On adding the three inequalities, we find the inequality
M A.M B.M C
AJa .AJb + AJb .AJc + AJc .AJa BJa .BJb + BJb .BJc + BJc .BJa
+
+
bcM A
caM B
CJa .CJb + CJb .CJc + CJc .CJa
.
abM C
Corollary 2.5. Let be the triangle ABC with the lengths of sides a, b, c and R is the
radius of circumcircle of ABC. Denote O, H the center of circumcircle and the orthocenter of ABC. Then, with any point M, we have the inequality:
aAH bBH c CH
abcM O.M H
+
+
.
RM A.M B.M C
MA
MB
MC
if M belongs to the circle with the centre O and the radius R, we obtain the inequality
√
√
√
abcM H

a 4R2 − a2 b 4R2 − b2 c 4R2 − c2
+
+
.
M A.M B.M C
MA
MB
MC
Proof. Applying the inequality (M, N ) to the case n = 3, s = 2 we have the inequality:
M O.M H
M A.M B.M C

AO.AH BO.BH CO.CH
+
+
.
bcM A
caM B
abM C
abcM O.M H
aAH bBH c CH
+
+
. Since AH =
Thus, we obtain the inequality
RM A.M B.M C
MA
MB
MC
√

√
√
abcM H
4R2 − a2 , BH = 4R2 − b2 and CH = 4R2 − c2 we obtain
M A.M B.M C
√
√
√
a 4R2 − a2 b 4R2 − b2 c 4R2 − c2
+
+
.
MA
MB
MC
19


Corollary 2.6. Suppose given the triangle ABC with the lengths of sides a, b, c, respectively. I, G, H the center of inscribed circle and the barycenter and the orthocenter of
∆ABC. Then, with any point M, we always have the inequality
(i)

abcM I 2
M A.M B.M C

aAI 2 bBI 2 c CI 2
+
+
.
MA

MB
MC

(ii)

abcM G2
M A.M B.M C

aAG2 bBG2 c CG2
+
+
.
MA
MB
MC

(iii)

abcM H 2
M A.M B.M C

a(4R2 − a2 ) b(4R2 − b2 ) c(4R2 − c2 )
+
+
.
MA
MB
MC

Proof. Applying the inequality (M,N) to the case n = 3, s = 2, N1 ≡ N2 ≡ N, we have

MN2
M A.M B.M C

AN 2
BN 2
CN 2
+
+
.
bcM A caM B abM C

aAI 2 bBI 2 c CI 2
abcM G2
+
+
and
MA MB MC
M A.M B.M C
aAG2 bBG2 c CG2
abcM H 2
+
+
. Then we have (i) and (ii). If N ≡ H we have (iii):
MA
MB
MC
M A.M B.M C
a(4R2 − a2 ) b(4R2 − b2 ) c(4R2 − c2 )
+
+

.
MA
MB
MC

Therefore, we obtain the inequality

abcM I 2
M A.M B.M C

Proposition 2.7. Suppose the polygon A1 A2 . . . An is inscribed in the circle with the
center O and radius R. Then, with any s < n points N1 , . . . , Ns in the plane A1 A2 . . . An ,
s

s

Ak Ni

n

ONi

i=1
n

we always have the inequality

i=1

k=1

Ak Ai

Rn−1

. When R = 1 we obtain

i=1,i=k
s
n

Ak Ni
i=1
n

k=1

s

ONi .
Ak Ai

i=1

i=1,i=k

When n = 3, s = 1 and a1 = A2 A3 , a2 = A3 A1 , a3 = A1 A2 we obtain the inequality
ON
a1 A1 N + a2 A2 N + a3 A3 N 4SA1 A2 A3
.

R
s
n

Proof. The inequality

s

Ak Ni

ONi

i=1
n

k=1

i=1

Rn−1

Ak Ai

follows from the inequality (M, N )

i=1,i=k
s
n

when M ≡ O. With R = 1 we have

Ak Mi
i=1
n

k=1

OMi .
Ak Ai

i=1,i=k

20

s
i=1


Example. Giving the triangle ABC with the lengths of sides a, b, c and R is the radius of circumscribed circle; r1 , r2 , r3 . are the radius of escribed circles. Let’s da , db , dc
the distances from the center of circumscribed circle to the center of escribed circles.
Then, with any point D belong to the circumscribed circle of ∆ABC we always have the
inequality:
√
(i)

(ii)
+

da db dc
a+b+c

R3

√
√
bc
ca
ab
DJa .DJb .DJc
√
√
.
+ √
+ √
+
x b + c − a y c + a − b z a + b − c xyz a + b + c
√
√
√
(R + 2r1 )(R + 2r2 )(R + 2r3 )
bc
ca
ab
√
+√
+√
3
R (a + b + c)
b+c−a
c+a−b
a+b−c

√

DJa .DJb .DJc
√
.
xyz a + b + c

Proof. (i) 
We consider M ≡ O.
bc(a + b + c)
ca(a + b − c)
ab(a − b + c)


Ja A2 =
, Ja B 2 =
, Ja C 2 =



b+c−a
b+c−a
b+c−a

ca(a + b + c)
ab(b + c − a)
bc(b + a − c)
2
2
2

Since Jb A =
there, Jb B =
, Jb C =

c
+
a
−
b
c
+
a
−
b
c
+
a
−
b



bc(c + a − b)
ca(c + b − a)
ab(a + b + c)

Jc A2 =
, Jc B 2 =
, Jc C 2 =
a+b−c

a+b−c
a+b−c
da db dc
√
√
√
√
bc
ca
ab
DJa .DJb .DJc
a+b+c
√
√
.
fore we obtain
+ √
+ √
+
3
R
x b + c − a y c + a − b z a + b − c xyz a + b + c
(ii) Since d2a = R2 + 2Rr1 , d2b = R2 + 2Rr2 , d2c = R2 + 2Rr3 therefore
√
√
√
(R + 2r1 )(R + 2r2 )(R + 2r3 )
bc
ca
ab

√
√
√
+
+
+
R3 (a + b + c)
b+c−a
c+a−b
a+b−c
DJa .DJb .DJc
√
.
xyz a + b + c
Now, we illustrate the advantage of this identity .... by addressing several important
problems of elementary Geometry. Firstly, we use the functions sin and cosin to create
the identity under the form of trigonometry.
Without generality, we can assume that the radius R of the circle C equal to 1. Suppose
that every point Ak has affixe ak = cos αk + i sin αk , and M has affixe z = cos u +
i sin u and every Nh has affixe zh = cos uh + i sin uh . Following the interpolation formula

21


s

s

(z − zj )
Lagrange, we have

j=1
n

(ak − zj )

n

j=1

=

(z − at )

(z − ak )

k=1

u − uj
2i sin
e
2
j=1
s

(ak − at )
t=k

t=1

or

u − αt
2i sin
e
2
t=1
n

i(u + uj )
2
i(u + αt )
2

i(αk + uj )
αk − uj
2
2i sin
e
2
j=1
.
i(u + αk )
i(αk + αt )
u − αk
αk − αt
2
2
2i sin
e

2i sin
e
2
2
t=k
s

n

=
k=1

iuj
iαt
We reduce all the factors 2i, e 2 and e 2 . And we receive the relationship
u − uj
2
j=1
=
n
u − αt
sin
2
t=1
s

sin

αk − uj
i(s + 1 − n)(αk − u)

2
j=1
2
.
u − αk
αk − αt e
sin
sin
2 t=k
2
s

n

k=1

sin

From this relationship, we deduce 2 identities below:
u − uj
2
j=1
=
n
u − αt
sin
2
t=1
s

sin

αk − uj
2
(s + 1 − n)(αk − u)
j=1
u − αk
αk − αt cos
2
sin
sin
2 t=k
2
s

n

k=1

sin

αk − uj
2
(s + 1 − n)(αk − u)
j=1
and
sin
= 0. With this result, we have
u
−

α
α
−
α
k
k
t
2
k=1 sin
sin
2 t=k
2
just built the identities under the form of trigonometry and geometry for the inequality
(M, N ) as below:
s

n

sin

Proposition 2.8. Assume that the polygon A1 A2 . . . An is inscribed in the circle with
the center O and radius R. Taking s + 1 points N1 , . . . , Ns and M also belonging to this
circle C. Assuming that the coordinate Ak (cos αk ; sin αk ), k = 1, 2, . . . , n; the coordinate
Nj (cos uj ; sin uj ), j = 1, 2, . . . , s and the coordinate M (cos u; sin u). Then, we will have
these identities
s
s
u − uj
αk − uj
sin

sin
n
2
2
(s + 1 − n)(αk − u)
j=1
j=1
(i) n
=
cos
u
−
α
α
−
α
u − αt
k
k
t
2
k=1 sin
sin
sin
2
2
2
t=k
t=1
22

αk − uj
n
2
(s + 1 − n)(αk − u)
j=1
(ii)
sin
= 0.
α
−
α
u
−
α
k
t
k
2
k=1 sin
sin
2 t=k
2
s

sin

αk − u1
2

(iii)
α
− αt = 0 when n = 3, s = 1.
k
k=1
sin
2
t=k
3

sin

n−1
u − uj
αk − uj
sin
n
2
2
j=1
j=1
(iv) n
=
u − αk
αk − αt when s = n − 1.
u − αt
k=1 sin
sin
sin
2 t=k

2
2
t=1
n−1

sin

n−2
n−2
uj
αk − uj
αk − uj
sin
sin
n j=1
n j=1
2
2
2
αk
j=1
=
cot
(v) n
and
αk − αt
αk − αt = 0 when s = n − 2
αt
2
k=1

k=1
sin
sin
sin
2
2
2
t=k
t=k
t=1
and u = 0.
n−2

sin

Remark 2.9. If the quadrilateral ABCD is inscribed in the circle we have
DB
aDA2 cDC 2
bDB 2
by (iii) or
+
=
. Using the relationship
ca
DA
DC
DB

DA DC
+

=
bc
ab

DA2 .DB.DC.a + DC 2 .DA.DB.c = DB 2 .DC.DA.b.
Hence DA2 SDBC + DC 2 SDAB = DB 2 SDCA [Feuerbach].
Proposition 2.10. Suppose the polygon A1 A2 . . . An is inscribed in the circle with
the radius R = 1. Taking s + 1 points N1 , . . . , Ns and M also belonging to this circle C. Assuming that the coordinate Ak (cos αk ; sin αk ), k = 1, 2, . . . , n; the coordinate
Nj (cos uj ; sin uj ), j = 1, 2, . . . , s and the coordinate M (cos u; sin u). Then, with the proper
choices of + or − we will have the identities
s

s

M Nj
(i)

j=1
n

±

n

=
k=1

M At

Ak Nj

j=1

M Ak

Ak At

cos

(s + 1 − n)(αk − u)
, (M, N ).
2

t=k

t=1
s
n

(ii)
k=1

±

Ak Nj
j=1

M Ak

Ak At

sin

(s + 1 − n)(αk − u)
= 0.
2

t=k

23


n−2

n−2

M Nj
(iii)

j=1
n

n

±
j=1

=
M At

t=1

n−2

Ak Nj
Ak At

k=1

αk
cot
and
2

n

±

Ak Aj
j=1

Ak At

k=1

t=k

= 0.

t=k

Corollary 2.11. Asuming that the points A1 , A2 , . . . , An , M in order belong to the circle
C with the center O. We convent n + 1 := 1. Then, we have the identities
n

(i)

(−1)r

r=1

cos(n − 1)∠M Ar+1 Ar
=
M Ar
Ar Ak
k=r

n

(ii)

(−1)r

r=1

1
n

.

M Ak

k=1

sin(n − 1)∠M Ar+1 Ar
= 0.
M Ar
Ar Ak
k=r

Proof. These identities follow from the identity (M, N ) with s = 0
Corollary 2.12. Let the quadrilateral ABCD be inscribed in the circle C with the center
O. Let a = BC, b = CA, c = AB. Then, we have 2 identities:
(i)

a cos(OD, OA) b cos(OD, OB) c cos(OD, OC)
abc
−
+
=−
.
DA
DB
DC
DA.DB.DC

a sin(OD, OA) c sin(OD, OC)
b sin(OD, OB)
+
=
.
DA

DC
DB
Proof. These identities follow from the identity (M, N ) with n = 3, s = 0.
(ii)

3

Conjecture

Despite of not having been proven yet, these following results are still hoped to be true:
Open Problem 3.1. Suppose given a triangle ABC with the lengths of sides a, b, c
respectively and R is the radius of circumcircle of ABC. Let’s I, Ja , Jb , Jc are the
centres of incircle and escribed circles of ∆ABC, respectively. Then, with any point M,
we have
M Ja .M Jb .M Jc
AJa .AJb .AJc BJa .BJb .BJc CJa .CJb .CJc
(i)
+
+
.
M A.M B.M C
bcM A
caM B
abM C
√
√
√
M Ja .M Jb .M Jc
bc
ca

ab
√
√
√
√
b+c−a
c+a−b
a+b−c
a+b+c
(ii)
+
+
.
M A.M B.M C
MA
MB
MC
Open Problem 3.2. Giving the triangle ABC with the lengths of sides a, b, c and R
is the radius of circumscribed circle; r1 , r2 , r3 . are the radius of escribed circles. Let’s
da , db , dc the distances from the center of circumscribed circle to the center of escribed
circles. Then, with any point D belong to the circumscribed circle of ∆ABC we always
have the inequality:
24


√
(i)

(ii)

da db dc
a+b+c
R3

√

√
√
bc
ca
ab
√
+ √
+ √
.
x b+c−a y c+a−b z a+b−c
√
√
√
(R + 2r1 )(R + 2r2 )(R + 2r3 )
bc
ca
ab
√
+√
+√
.
R(a + b + c)
b+c−a
c+a−b

a+b−c

References
[1] T. Andreescu and D. Andrica, Proving some geometric inequalities by using complex
numbers, Eduatia Mathematica Vol 1. Nr. 2 (2005),19-26.
[2] T. Hayashi, Two theorems on complex numbers, Thoku Math. J., 4(1913/14), pp.
68-70.
[3] D.S. Mitrinovic, J.E. Pecaric and V. Volenec, Recent Advances in Geometric Inequalities, Acad. Publ., Dordrecht, Boston, London 1989.

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