Báo cáo khoa học: Một số phép biến đổi bảo toàn cạnh và góc của tam giác

f (x) [0; π] f (A)
f (B) f (C)
ABC
f (x) = x f (x) =
π
3
f (x) [0; π]
f (x) > 0, f (x) + f (y) + f (π − x − y) = π, ∀x, y ∈ (0; π) , x + y < π.
y → 0
+
f (x) + f (0) + f (π − x) = π, ∀x ∈ (0; π)
f (π −x) = π −f (0) −f (x) , ∀x ∈ (0; π) .
f (x) + f (y) + (π − f (0) − f (x + y)) = π, ∀x, y ∈ (0; π) , x + y ≤ π
f (x) + f (y) = f (x + y) + f (0) , ∀x, y ∈ [0; π] , x + y < π.
f (x) = f (0) + g (x) g (x) [0; π]
f (0) + g (x) + f (0) + g (y) = f (0) + g (x + y) + f (0) , ∀x, y ∈ [0; π] , x + y < π
⇔ g (x) + g (y) = g (x + y) , ∀x, y ∈ [0; π] , x + y < π.
g (x) [0; π]
g (x) = αx f (x) = f (0) + αx
f (0) = β f (x) = αx + β
α β f (x) > 0 ∀x ∈ (0; π) x+y < π f (A)+f (B)+f (C) = π

αx + β > 0, ∀x ∈ (0; π) ;
αA + β + αB + β + αC + β = π.
⇔

αx + β > 0, ∀x ∈ (0; π) ;
α (A + B + C) + 3β = π.
⇔

αx + β > 0, ∀x ∈ (0; π) ;

απ + 3β = π.
⇔



αx + β > 0, ∀x ∈ (0; π) ;
β =
(1 − α) π
3
.
f (x) = αx +
(1 − α) π
3
, ∀x ∈ (0; π) .
x → 0
+
(1 − α) π
3
≥ 0 ⇔ α ≤ 1.
x → π
−
απ +
(1 − α) π
3
≥ 0
α ≥ −
1
2
−
1

2
≤ α ≤ 1
−
1
2
< α < 1 f (x)
α = −
1
2
f (x) = −
1
2
x +
π
2
0 < x < π f (x) > f (π) = 0 f (x) > 0 ∀x ∈ (0; π)
α = 1 f (x) = x
f (x) = αx +
(1 − α) π
3
, −
1
2
≤ α ≤ 1.
f (x)
−
1
2
≤ α ≤ 1 A B C
A

1
B
1
C
1
A
1
= αA +
(1 − α) π
3
, B
1
= αB +
(1 − α) π
3
, C
1
= αC +
(1 − α) π
3
,
α < −
1
2
A B C
max {A, B, C} <
(α − 1) π
3α
A
1

B
1
C
1
A
1
= αA +
(1 − α) π
3
, B
1
= αB +
(1 − α) π
3
, C
1
= αC +
(1 − α) π
3
,
α < −
1
2
max {A, B, C} <
(α − 1) π
3α
⇒ A <
(α − 1) π
3α
⇒ 3αA + (1 − α) π > 0 ⇒ αA +

(1 − α) π
3
> 0 ⇒ A
1
> 0.
B
1
> 0 C
1
> 0 A
1
+ B
1
+ C
1
= π
α > 1 A B C
min {A, B, C} >
(α − 1) π
3α
A
1
B
1
C
1
A
1
= αA +
(1 − α) π

3
, B
1
= αB +
(1 − α) π
3
, C
1
= αC +
(1 − α) π
3
,
α > 1
min {A, B, C} >
(α − 1) π
3α
⇒ A >
(α − 1) π
3α
⇒ 3αA + (1 − α) π > 0 ⇒ αA +
(1 − α) π
3
> 0 ⇒ A
1
> 0.
B
1
> 0 C
1
> 0 A

1
+ B
1
+ C
1
= π
α = −
1
2
A B C A
1
B
1
C
1
A
1
=
π −A
2
, B
1
=
π −B
2
, C
1
=
π −C
2

A
1
=
B + C
2
, B
1
=
C + A
2
, C
1
=
A + B
2
α =
1
2
A B C A
1
B
1
C
1
A
1
=
π + 3A
6
, B

1
=
π + 3B
6
, C
1
=
π + 3C
6
A
1
=
4A + B + C
6
, B
1
=
4B + C + A
6
, C
1
=
4C + A + B
6
α = −
2
3
A B C max {A, B, C} <
5π
6

A
1
B
1
C
1
A
1
=
5π −6A
9
, B
1
=
5π −6B
9
, C
1
=
5π −6C
9
A
1
=
5B + 5C − A
9
, B
1
=
5C + 5A − B

9
, C
1
=
5A + 5B − C
9
α = −
4
5
A B C max {A, B, C} <
3π
4
A
1
B
1
C
1
A
1
=
3π −4A
5
, B
1
=
3π −4B
5
, C
1

=
3π −4C
5
A
1
=
3B + 3C − A
5
, B
1
=
3C + 3A − B
5
, C
1
=
3A + 3B − C
5
α = −1
A B C max {A, B, C} <
2π
3
A
1
B
1
C
1
A
1

=
2π
3
− A, B
1
=
2π
3
− B, C
1
=
2π
3
− C
A
1
=
2B + 2C − A
3
, B
1
=
2C + 2A − B
3
, C
1
=
2A + 2B − C
3
α = −2

A B C max {A, B, C} <
π
2
ABC A
1
B
1
C
1
A
1
= π −2A, B
1
= π −2B, C
1
= π −2C
A
1
= B + C −A, B
1
= C + A −B, C
1
= A + B − C
α = 2
A B C min {A, B, C} >
π
6
A
1
B

1
C
1
A
1
= 2A −
π
3
, B
1
= 2B −
π
3
, C
1
= 2C −
π
3
A
1
=
5A − B − C
3
, B
1
=
5B − C − A
3
, C
1

=
5C − A − B
3
α = 4
A B C min {A, B, C} >
π
4
A
1
B
1
C
1
A
1
= 4A − π, B
1
= 4B − π, C
1
= 4C −π
A
1
= 3A − B − C, B
1
= 3B − C −A, C
1
= 3C −A −B
−
1
2

≤ α ≤ 1
A
1
= αA +
(1 − α) π
3
= αA +
(1 − α) (A + B + C)
3
=
(1 + 2α)
3
A +
(1 − α)
3
B +
(1 − α)
3
C.
B
1
C
1
α
1
=
1 + 2α
3
β
1

= γ
1
=
1 − α
3
α
1
β
1
γ
1
≥ 0 α
1
+ β
1
+ γ
1
= 1
A
1
= α
1
A + β
1
B + γ
1
C, B
1
= α
1

B + β
1
C + γ
1
A, C
1
= α
1
C + β
1
A + γ
1
B
A B C A
1
B
1
C
1







A
1
= αA + βB + γC;
B

1
= αB + βC + γA;
C
1
= αC + βA + γB,
α β γ ≥ 0 α + β + γ = 1
A
1
B
1
C
1
> 0 A
1
+ B
1
+ C
1
=
(α + β + γ) (A + B + C) = 1.π = π.
A
1
B
1
C
1
A B C
A =
(α
2

− βγ) A
1
+ (γ
2
− αβ) B
1
+ (β
2
− γα) C
1
α
3
+ β
3
+ γ
3
− 3αβγ
;
B =
(α
2
− βγ) B
1
+ (γ
2
− αβ) C
1
+ (β
2
− γα) A

1
α
3
+ β
3
+ γ
3
− 3αβγ
;
C =
(α
2
− βγ) C
1
+ (γ
2
− αβ) A
1
+ (β
2
− γα) B
1
α
3
+ β
3
+ γ
3
− 3αβγ
,

α β γ
1
3
α β γ ≥ 0
1
3
α + β + γ = 1 A B C A
1
B
1
C
1







A
1
= α
1
A + β
1
B + γ
1
C;
B
1

= α
1
B + β
1
C + γ
1
A;
C
1
= α
1
C + β
1
A + γ
1
B,
α
1
=
α
2
− βγ
α
3
+ β
3
+ γ
3
− 3αβγ
;

β
1
=
γ
2
− αβ
α
3
+ β
3
+ γ
3
− 3αβγ
;
γ
1
=
β
2
− γα
α
3
+ β
3
+ γ
3
− 3αβγ
,
α β γ
α = sin

2
ϕ, β = cos
2
ϕ, γ = 0.
A B C A
1
B
1
C
1
A
1
= α
1
A + β
1
B + γ
1
C;
B
1
= α
1
B + β
1
C + γ
1
A;
C
1

= α
1
C + β
1
A + γ
1
B,
α
1
=
sin
4
ϕ
sin
6
ϕ + cos
6
ϕ
;
β
1
=
−sin
2
ϕ.cos
2
ϕ
sin
6
ϕ + cos

6
ϕ
;
γ
1
=
cos
4
ϕ
sin
6
ϕ + cos
6
ϕ
,
A
1
B
1
C
1
A B C A
1
B
1
C
1
α
A
2

= αA, B
2
= αB, C
2
= αC + (1 −α) π,
A
2
+ B
2
+ C
2
= π A
2
> 0 B
2
> 0 C
2
> 0 A
2
B
2
C
2
0 < α ≤ 1 A B C A
2
B
2
C
2
A

2
= αA, B
2
= αB, C
2
= αC + (1 −α) π,
A
2
> 0 B
2
> 0 C
2
> 0
α =
1
2
A B C A
2
B
2
C
2
A
2
=
A
2
, B
2
=

B
2
, C
2
=
π + C
2
C
2
0 < α ≤ 2 A B C
C A
2
B
2
C
2
A
2
= αA, B
2
= αB, C
2
= αC + (1 −α) π,
A
2
> 0 B
2
> 0
C >
π

2
⇒ C
2
= αC + (1 −α) π > α
π
2
+ (1 − α) π =
(2 − α) π
2
≥ 0 ⇒ C
2
> 0.
α = 2
A B C C
A
2
B
2
C
2
A
2
= 2A, B
2
= 2B, C
2
= 2π −C
−
1
2

≤ α < 0 A B C
A
3
B
3
C
3
A
3
= αA + mπ, B
3
= αB + nπ, C
3
= αC + pπ,
m ≥ −α, n ≥ −α, p ≥ −α, m + n + p = 1 −α,
A
3
+ B
3
+ C
3
= α (A + B + C) + (m + n + p) π = απ + (1 −α) π = π.
A < π ⇒ −αA < −απ ⇒ −α >
−αA
π
.
m ≥ −α
m >
−αA
π

⇒ αA + mπ > 0 ⇒ A
3
> 0.
B
3
> 0 C
3
> 0
m ≥ −α n ≥ −α p ≥ −α 1 − α = m + n + p ≥ −3α
α ≥ −
1
2
1 − α ≥ −3α
α = −
1
4
m =
1
4
n =
1
4
p =
3
4
A B C A
3
B
3
C

3
A
3
= −
A
4
+
π
4
, B
3
= −
B
4
+
π
4
, C
3
= −
C
4
+
3π
4
,
ABC C A
3
B
3

C
3
A
3
=
π
2
− A, B
3
=
π
2
− B, C
3
= π −C,
C
3
ABC C A
3
B
3
C
3
A
3
=
π
2
− A, B
3

=
π
2
− B, C
3
= π −C,
C
3
sin A + sin B + sin C ≤
3
√
3
2
,
cos A cos B cos C ≤
1
8
0 < sin A sin B sin C ≤
3
√
3
8
,
sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
sin

π −A
2

+ sin


π −B
2

+ sin

π −C
2

≤
3
√
3
2
.
cos
A
2
+ cos
B
2
+ cos
C
2
≤
3
√
3
2
.

cos

π −A
2

cos

π −B
2

cos

π −C
2

≤
1
8
.
sin
A
2
sin
B
2
sin
C
2
≤
1

8
sin 2

π −A
2

+ sin 2

π −B
2

+ sin 2

π −C
2

= 4 sin
π −A
2
sin
π −B
2
sin
π −C
2
.
sin (π −A) + sin (π −B) + sin (π − C) = 4 sin
π −A
2
sin

π −B
2
sin
π −C
2
.
sin A + sin B + sin C = 4cos
A
2
cos
B
2
cos
C
2
8 sin
A
2
sin
B
2
sin
C
2
≤ 1 ⇔ 32 sin
A
2
sin
B
2

sin
C
2
.cos
A
2
cos
B
2
cos
C
2
≤ 4cos
A
2
cos
B
2
cos
C
2
⇔ 4

2 sin
A
2
cos
A
2


2 sin
B
2
cos
B
2

2 sin
C
2
cos
C
2

≤ 4cos
A
2
cos
B
2
cos
C
2
⇔ 4 sin A sin B sin C ≤ 4cos
A
2
cos
B
2
cos

C
2
.
sin A sin B sin C ≤ cos
A
2
cos
B
2
cos
C
2
sin 2A + sin 2B + sin 2C ≤ sin A + sin B + sin C
ABC
sin 2 (π −2A) + sin 2 (π −2B) + sin 2 (π − 2C)
≤ sin (π −2A) + sin (π − 2B) + sin (π − 2C)
⇔ −sin 4A − sin 4B − sin 4C ≤ sin 2A + sin 2B + sin 2C.
sin 2A + sin 2B + sin 2C + sin 4A + sin 4B + sin 4C ≥ 0
sin

2.
A
2

+ sin

2.
B
2


+ sin

2.
π + C
2

≤ sin
A
2
+ sin
B
2
+ sin
π + C
2
.
sin A + sin B − sin C ≤ sin
A
2
+ sin
B
2
+ cos
C
2
ABC C
cos
2A
2
+ cos

2B
2
+ cos
2C − π
2
≤
3
√
3
2
.
cos A + cos B + sin C ≤
3
√
3
2

C >
π
2

ABC C
0 < sin

π
2
− A

sin


π
2
− B

sin (π −C) ≤
3
√
3
8
0 < cos A cos B sin C ≤
3
√
3
8

C ≤
π
2

ABC C
0 < sin

π
2
− A

+ sin

π
2

− B

+ sin (π −C) ≤
3
√
3
2
0 < cos A + cos B + sin C ≤
3
√
3
2

C ≥
π
2

α β f (x) = αx + β
f (a) f (b) f (c)
ABC
f (a) f (b) f (c)
f (a) > 0, f (b) > 0, f (c) > 0, ∀∆ABC.
αa + β > 0, αb + β > 0, αc + β > 0, ∀∆ABC.
α ≥ 0 α < 0 β ABC a
αa + β < 0.
α = 0 β = 0 f (x) ≡ 0
α ≥ 0 β ≥ 0 α + β > 0 f (a) f (b) f (c)
a b c ABC
f (a) + f (b) > f (c) , f (b) + f (c) > f (a) , f (c) + f (a) > f (b) ,








αa + β + αb + β > αc + β
αb + β + αc + β > αa + β
αc + β + αa + β > αb + β







α (a + b) + β > αc
α (b + c) + β > αa
α (c + a) + β > αb.
α ≥ 0 β ≥ 0 α + β > 0.
α ≥ 0 β ≥ 0 α + β > 0 f (x) = αx + β f (a)
f (b) f (c)
ABC
α β f (x) =
1
αx + β
f (a) f (b) f (c)
ABC
a ≥ b ≥ c
g (x) =

1
x
g (a) g (b)
g (c) ABC
ABC a = b = 2 c = 1 g (a) =
1
a
=
1
2
g (b) =
1
b
=
1
2
g (c) =
1
c
= 1
g (a) + g (b) =
1
2
+
1
2
= 1 = g (c) .
f (a) f (b) f (c)
f (a) > 0, f (b) > 0, f (c) > 0, ∀∆ABC.
1

αa + β
> 0,
1
αb + β
> 0,
1
αc + β
> 0, ∀∆ABC
αa + β > 0, αb + β > 0, αc + β > 0, ∀∆ABC.
α ≥ 0 α < 0 β
ABC a
αa + β < 0
β ≥ 0 β < 0
ABC a
αa + β < 0
α = 0 β = 0 f (x)
α = 0 β > 0 f (x) =
1
β
f (a) = f (b) =
f (c) > 0 f (a) f (b) f (c)
α > 0 β = 0 f (x) =
1
αx
α > 0 β > 0 a ≥ b ≥ c
αa + β ≥ αb + β ≥ αc + β > 0.
1
αa + β
≤
1

αb + β
≤
1
αc + β
f (a) ≤ f (b) ≤ f (c) .
α > 0 β > 0 f (a) + f (b) > f (c)
ABC a ≥ b ≥ c
1
αa + β
+
1
αb + β
>
1
αc + β
, ∀∆ABC : a ≥ b ≥ c.
ABC a = b =
3d c = d d > 0
1
3dα + β
+
1
3dα + β
>
1
dα + β
, ∀d > 0.
2
3dα + β
>

1
dα + β
, ∀d > 0
2dα + 2β > 3dα + β, ∀d > 0,
β > dα ∀d > 0 d
α = 0 β > 0
f (x) =
1
αx + β
,
f (a) f (b) f (c)
ABC
f (a) f (b) f (c) ABC
a b c
1
a + b
,
1
b + c
,
1
c + a
1
a + b
>
1
2 (b + c).
b + c > a ⇒ b + 2c > a ⇒ 2b + 2c > a + b ⇒
1
a + b

>
1
2 (b + c)
.
1
c + a
>
1
2 (b + c)
.
1
a + b
+
1
c + a
>
1
b + c
.
1
b + c
+
1
a + b
>
1
c + a
,
1
c + a

+
1
b + c
>
1
a + b
.
h
a
h
b
h
c
a
b c
1
h
a
1
h
b
1
h
c
2S =
a

1
h
a


=
a

1
h
b

=
a

1
h
c

.
1
h
a
1
h
b
1
h
c
a b c k =
1
2S
m
a

m
b
m
c
ABC AA
1
BB
1
CC
1
G C
1
AA
1
BB
1
BC
E Q C BB
1
C
1
Q P
EG =
1
2
BG =
1
2
.
2

3
m
b
=
1
3
m
b
;
EG
P C
=
C
1
G
C
1
C
=
1
3
⇔ EG =
1
3
P C.
P C = m
b
C
1
E

C
1
P
=
C
1
G
C
1
C
=
1
3
⇒ C
1
E =
1
3
C
1
P ;
C
1
E =
1
2
AG =
1
2
.

2
3
m
a
=
1
3
m
a
.
C
1
P = m
a
CC
1
= m
c
CC
1
P m
a
m
b
m
c
a b c A, B, C
min (A, B, C) ≥
π
12

√
ab
√
bc
√
ca
√
ab
√
bc
√
ca
√
ab +
√
bc ≤
√
ca

b
c
+

b
a
≤ 1.
min


b

c
,

b
a

≤
1
2
√
b
√
c
≤
1
2
b
c
≤
1
4
sin B
sin C
≤
1
4
sin B ≤
1
4
.

sin
π
12
=




1 − cos
π
6
2
=

2 −
√
3
2
>
1
4
.
α sin α =
1
4
0 < α <
π
6
sin B ≤ sin α
B ≤ α <

π
12
B ≥ min (A, B, C) ≥
π
12
.
B ≥ π−α A C
π
12
. min (A, B, C) ≥
π
12
.
1
a + 1
,
1
b + 1
,
1
c + 1
,
a b c
1
h
a
1
h
b
1

h
c
1
h
b
+
1
h
c
>
1
h
a
h
a
h
b
h
c
1
h
b
+
1
h
c
≤
1
h
a

1
h
a
1
h
b
1
h
c
h
a
= 1, h
b
=
√
5, h
c
= 1 +
√
5,
1
h
b
+
1
h
c
=
√
5

5
+
√
5 − 1
4
=
9
√
5 − 5
20
< 1 =
1
h
a
.
h
a
= 1, h
b
=
√
5, h
c
= 1 +
√
5.
a b c
a
2
+ b

2
+ c
2
< 2 (ab + bc + ca) .
a < b + c ⇒ a
2
< ab + ca.
b
2
< bc + ab, c
2
< ca + bc.
1
h
a
+
1
h
b
+
1
h
c
=
1
r
,
r
S = rp = r


a + b + c
2

.
1
r
=
a + b + c
2S
=
a
2S
+
b
2S
+
c
2S
=
1
h
a
+
1
h
b
+
1
h
c

.
1
h
a
1
h
b
1
h
c
1
h
2
a
+
1
h
2
b
+
1
h
2
c
< 2

1
h
a
h

b
+
1
h
b
h
c
+
1
h
c
h
a

.
1
r
2
=

1
h
a
+
1
h
b
+
1
h

c

2
< 4

1
h
a
h
b
+
1
h
b
h
c
+
1
h
c
h
a

.
1
h
a
h
b
+

1
h
b
h
c
+
1
h
c
h
a
>
1
4r
2
m
2
a
+ m
2
b
+ m
2
c
< 2 (m
a
m
b
+ m
b

m
c
+ m
c
m
a
)
1
h
a
1
h
b
1
h
c
1
1
h
a
+
1
h
b
,
1
1
h
b
+

1
h
c
,
1
1
h
c
+
1
h
a
h
a
h
b
h
a
+ h
b
,
h
b
h
c
h
b
+ h
c
,

h
c
h
a
h
c
+ h
a
h
a
h
b
h
a
+ h
b
+
h
b
h
c
h
b
+ h
c
>
h
c
h
a

h
c
+ h
a