Khóa luận tốt nghiệp toán học: Một số bài toán nhận dạng tam giác

∆ABC ABC
A, B, C A, B, C
a, b, c
A, B, C
h
a
, h
b
, h
c
A, B, C
m
a
, m
b
, m
c
A, B, C
l
a
, l
b
, l
c
A, B, C
R
r
r
a
, r

b
, r
c
A, B, C
p =
a + b + c
2
S
•
•
•
a
sin A
=
b
sin B
=
c
sin C
= 2R
sin A =
a
2R
; sin B =
b
2R
; sin C =
c

2R
a
2
= b
2
+ c
2
− 2bc cos A = (b −c)
2
+ 4bc sin
2
A
2
b
2
= c
2
+ a
2
− 2ca cos B = (c − a)
2
+ 4ca sin
2
B
2
c
2
= a
2
+ b

2
− 2ab cos C = (a − b)
2
+ 4ab sin
2
C
2
a − b
a + b
=
tan
A−B
2
tan
A+B
2
;
b − c
b + c
=
tan
B−C
2
tan
B+C
2
;
c − a
c + a
=

tan
C−A
2
tan
C+A
2
a
2
= b
2
+ c
2
− 4S cot A; b
2
= c
2
+ a
2
− 4S cot B; c
2
= a
2
+ b
2
− 4S cot C
cot A =
b
2
+ c
2

− a
2
4S
; cot B =
c
2
+ a
2
− b
2
4S
; cot C =
a
2
+ b
2
− c
2
4S
a = b cos C + c cos B = r

cot
B
2
+ cot
C
2

b = c cos A + a cos C = r


cot
A
2
+ cot
C
2

c = a cos B + b cos A = r

cot
A
2
+ cot
B
2

m
2
a
=
b
2
+ c
2
2
−
a
2
4
; m

2
b
=
c
2
+ a
2
2
−
b
2
4
; m
2
c
=
a
2
+ b
2
2
−
c
2
4
h
a
= b sin C = c sin B =
2S
a

;
h
b
= c sin A = a sin C =
2S
b
;
h
c
= a sin B = b sin A =
2S
c
l
a
=
2bc
b + c
cos
A
2
=
2
√
bc
b + c

p(p − a)
l
b
=

2ca
c + a
cos
B
2
=
2
√
ca
c + a

p(p − b)
l
c
=
2ab
a + b
cos
C
2
=
2
√
ab
a + b

p(p − c)
S =



































1
2
a.h
a
=
1
2
b.h
b
=
1
2
c.h
c
1
2
bc sin A =
1
2
ca sin B =
1
2
ab sin C
abc
4R
2R
2
sin A sin B sin C
pr

(p − a)r
a
= (p − b)r
b
= (p − c)r
c

p(p − a)(p − b)(p −c)
R =











a
2 sin A
=
b
2 sin B
=
c
2 sin C
abc
4S

r =











S
p
= (p − a). tan
A
2
= (p − b). tan
B
2
= (p − c). tan
C
2
a sin
B
2
sin
C
2
cos

A
2
=
b sin
C
2
sin
A
2
cos
B
2
=
c sin
A
2
sin
B
2
cos
C
2
4R sin
A
2
. sin
B
2
. sin
C

2
r
a
=
S
p − a
=
a cos
B
2
cos
C
2
cos
A
2
= p. tan
A
2
r
b
=
S
p − b
=
b cos
C
2
cos
A

2
cos
B
2
= p. tan
B
2
r
c
=
S
p − c
=
c cos
A
2
cos
B
2
cos
C
2
= p. tan
C
2
ABC
sin(A + B) = sin C; cos(A + B) = −cos C
sin
A + B
2

= cos
C
2
; cos
A + B
2
= sin
C
2
tan(A + B) = −tan C; cot(A + B) = −cot C
tan
A + B
2
= cot
C
2
; cot
A + B
2
= tan
C
2
sin A + sin B + sin C = 4 cos
A
2
. cos
B
2
. cos
C

2
sin A + sin B + sin C = 2 sin
A + B
2
. cos
A − B
2
+ 2 sin
C
2
. cos
C
2
= 2 cos
C
2

cos
A − B
2
+ cos
A + B
2

= 4 cos
A
2
. cos
B
2

. cos
C
2
sin 2A + sin 2B + sin 2C = 4 sin A. sin B. sin C
sin 2A + sin 2B + sin 2C = 2 sin(A + B). cos(A − B) + 2 sin C. cos C
= 2 sin C[cos(A − B) − cos(A + B)]
= 4 sin A. sin B. sin C
cos A + cos B + cos C = 1 + 4 sin
A
2
. sin
B
2
. sin
C
2
cos A + cos B + cos C = 2 cos
A + B
2
. cos
A − B
2
+ 1 − 2 sin
2
C
2
= 1 + 2 sin
C
2


cos
A − B
2
− cos
A + B
2

= 1 + 4 sin
A
2
. sin
B
2
. sin
C
2
cos 2A + cos 2B + cos 2C = −1 − 4 cos A. cos B. cos C
cos 2A + cos 2B + cos 2C = 2 cos(A + B). cos(A −B) + 2 cos
2
C − 1
= −1 − 2 cos C[cos(A − B) + cos(A + B)]
= −1 − 4 cos A. cos B. cos C
tan A + tan B + tan C = tan A. tan B. tan C
tan(A + B) = −tan C
tan A + tan B
1 − tan A. tan B
= −tan C.
tan A + tan B + tan C = tan A. tan B. tan C
tan
A

2
. tan
B
2
+ tan
B
2
. tan
C
2
+ tan
C
2
. tan
A
2
= 1
tan

A
2
+
B
2

=
1
tan
C
2

tan
A
2
+ tan
B
2
1 − tan
A
2
tan
B
2
=
1
tan
C
2
,
tan
A
2
tan
B
2
+ tan
B
2
tan
C
2

+ tan
C
2
tan
A
2
= 1
cot A. cot B + cot B. cot C + cot C. cot A = 1
tan A + tan B + tan C = tan A. tan B. tan C
1
cot A
+
1
cot B
+
1
cot C
=
1
cot A. cot B. cot C
,
cot A. cot B + cot B. cot C + cot C. cot A = 1
cot
A
2
+ cot
B
2
+ cot
C

2
= cot
A
2
. cot
B
2
. cot
C
2
tan
A
2
. tan
B
2
+ tan
B
2
. tan
C
2
+ tan
C
2
. tan
A
2
= 1,
1

cot
A
2
cot
B
2
+
1
cot
B
2
cot
C
2
+
1
cot
C
2
cot
A
2
= 1,
cot
A
2
+ cot
B
2
+ cot

C
2
= cot
A
2
cot
B
2
cot
C
2
sin
2
A + sin
2
B + sin
2
C = 2 + 2 cos A. cos B. cos C
sin
2
A + sin
2
B + sin
2
C =
3
2
−
1
2

(cos 2A + cos 2B + cos 2C)
= 2 + 2 cos A. cos B. cos C
cos
2
A + cos
2
B + cos
2
C = 1 − 2 cos A. cos B. cos C
cos
2
A + cos
2
B + cos
2
C = 1 +
1
2
(cos 2A + cos 2B) + cos
2
C
= 1 − [cos(A − B) + cos(A + B)] cos C
= 1 − 2 cos A. cos B. cos C
ABC
b = a sin B = a cos C = c tan B = c cot C
c = a sin C = a cos B = b tan C = b cot B
a.h = b.c; b
2
= ab
′

; c
2
= ac
′
1
h
2
=
1
b
2
+
1
c
2
a
2
= b
2
+ c
2
A
B C
H
h
a
b
c
c
′

b
′
n a
1
, a
2
, , a
n
(n ≥ 2)
a
1
+ a
2
+ + a
n
n
≥
n
√
a
1
a
2
a
n
a
1
= a
2
= = a

n
n (a
1
, a
2
, , a
n
) (b
1
, b
2
, , b
n
)(n ≥ 2)
(a
1
b
1
+ a
2
b
2
+ + a
n
b
n
)
2
≤ (a
2

1
+ a
2
2
+ + a
2
n
)(b
2
1
+ b
2
2
+ + b
2
n
)
a
1
b
1
=
a
2
b
2
= =
a
n
b

n
(a
n
) (b
n
).
a
1
+ a
2
+ + a
n
n
.
b
1
+ b
2
+ + b
n
n
≤
a
1
b
1
+ a
2
b
2

+ + a
n
b
n
n

a
1
= a
2
= = a
n
b
1
= b
2
= = b
n
(a
n
) (b
n
)
a
1
+ a
2
+ + a
n
n

.
b
1
+ b
2
+ + b
n
n
≥
a
1
b
1
+ a
2
b
2
+ + a
n
b
n
n
ABC A, B, C a, b, c
BC, CA, AB
|a − b| < c < a + b; |b − c| < a < b + c; |c − a| < b < c + a
a ≥ b
A ≥ B
∆ABC
a) cos 2A +
√

3(cos 2B + cos 2C) +
5
2
= 0
b)
√
1 + 2 cos
2
A
sin B
+
√
1 + 2 cos
2
B
sin C
+
√
1 + 2 cos
2
C
sin A
= 3
√
2
2 cos
2
A − 1 + 2
√
3 [cos(B + C). cos(B − C)] +

5
2
= 0

2 cos A −
√
3 cos(B − C)

2
+ 3 sin
2
(B −C) = 0,



sin(B − C) = 0
cos A =
√
3
2
cos(B − C) = 0

A = 30
0
B = C = 75
0
ABC A = 30
0
; B = C = 75
0

(1; 1; 1)

1
√
2
;
1
√
2
;
√
2 cos A


1
√
2
+
1
√
2
+
√
2 cos A

2
≤ 3

1
2

+
1
2
+ 2 cos
2
A

,
2(1 + cos A)
2
≤ 3(1 + 2 cos
2
A).
√
1 + 2 cos
2
A ≥
2
√
2
√
3
cos
2
A
2
,
√
1 + 2 cos
2

B ≥
2
√
2
√
3
cos
2
B
2
,
√
1 + 2 cos
2
C ≥
2
√
2
√
3
cos
2
C
2
.
T =

(1 + 2 cos
2
A)(1 + 2 cos

2
B)(1 + 2 cos
2
C), T ≥ 0
T ≥

2
√
2
√
3

3
cos
2
A
2
cos
2
B
2
cos
2
C
2
.
√
1 + 2 cos
2
A

sin B
+
√
1 + 2 cos
2
B
sin C
+
√
1 + 2 cos
2
C
sin A
≥ 3
3

T
sin A sin B sin C
.
√
1 + 2 cos
2
A
sin B
+
√
1 + 2 cos
2
B
sin C

+
√
1 + 2 cos
2
C
sin A
≥
√
6
3

cot
A
2
cot
B
2
cot
C
2
,
cot
A
2
cot
B
2
cot
C
2

= cot
A
2
+ cot
B
2
+ cot
C
2
cot
A
2
+ cot
B
2
+ cot
C
2
≥ 3
3

cot
A
2
cot
B
2
cot
C
2

> 0
cot
A
2
cot
B
2
cot
C
2
≥ 3
√
3.
√
1 + 2 cos
2
A
sin B
+
√
1 + 2 cos
2
B
sin C
+
√
1 + 2 cos
2
C
sin A

≥ 3
√
2













2 cos A −1 = 0
2 cos B −1 = 0
2 cos C − 1 = 0
cot
A
2
= cot
B
2
= cot
C
2
A = B = C = 60
0

ABC A = B = C = 60
0
ABC
cos 2A + 2
√
2 cos B + 2
√
2 cos C = 3
ABC.
cos
2
A + 2
√
2 sin
A
2
. cos
B − C
2
− 2 = 0,
cos A(cos A −1) +

1 − 2 sin
2
A
2

+ 2
√
2 sin

A
2
. cos
B − C
2
− 2 = 0,
cos A(cos A − 1) −

√
2 sin
A
2
− cos
B − C
2

2
− sin
2
B − C
2
= 0.
∆ABC cos A ≥ 0 cos A −1 < 0.
cos A(cos A −1) −

√
2 sin
A
2
− cos

B − C
2

2
−sin
2
B − C
2
≤ 0,









cos A = 0
√
2 sin
A
2
= cos
B − C
2
sin
B − C
2
= 0


A = 90
0
B = C = 45
0
ABC A = 90
0
; B = C = 45
0
ABC
sin A + sin B + sin C =
3 +
√
3
2
A < B < C.
A, B, C A + C = 2B.
ABC A + B + C = π B =
π
3
.
sin A + sin
π
3
+ sin C =
3 +
√
3
2
sin A + sin C =

3
2
,
cos
C − A
2
=
√
3
2
= cos
π
6
.
C > A ABC









C − A
2
=
π
6
C + A =

2π
3
B =
π
3







A =
π
6
B =
π
3
C =
π
2
ABC A =
π
6
; B =
π
3
; C =
π
2

ABC
sin
2
A + sin
2
B =
2n+1

sin
2
C
A B C.
C
C > 90
0
cos C ∈ (−1; 0) cos C =
a
2
+ b
2
− c
2
2ab
a
2
+ b
2
< c
2
.

a
sin A
=
b
sin B
=
c
sin C
= 2R,
sin
2
A + sin
2
B < sin
2
C
sin C ∈ (0; 1) sin
2
C ∈ (0; 1). sin
2
C <
2n+1
√
sin
2
C.
sin
2
A + sin
2

B <
2n+1
√
sin
2
C.
C < 90
0
2n+1
√
sin
2
C < 1.
sin
2
A + sin
2
B = 1 + cos C. cos(A −B).
C < 90
0
A, B cos C > 0 cos(A − B) > 0.
2n+1

sin
2
C = sin
2
A + sin
2
B = 1 + cos C. cos(A − B) > 1

C = 90
0
C = 90
0
.
∆ABC 60
0
sin A + sin B + sin C
cos A + cos B + cos C
=
√
3
sin A + sin B + sin C =
√
3(cos A + cos B + cos C),
sin

A −
π
3

+ sin

B −
π
3

+ sin

C −

π
3

= 0,
2 sin

C
2
−
π
6

−cos
A − B
2
+ cos

C
2
−
π
6

= 0,




sin


C
2
−
π
6

= 0
cos
A − B
2
= cos

C
2
−
π
6

= cos

π
3
−
A + B
2



A =
π

3
B =
π
3
C =
π
3
ABC 60
0
.
ABC
sin A + sin B + sin C =
r + 4R sin
2
C
2
2R sin
C
2
C = 120
0
ABC r = 4R sin
A
2
sin
B
2
sin
C
2

sin A + sin B + sin C = 4 cos
A
2
. cos
B
2
. cos
C
2
4 cos
A
2
. cos
B
2
. cos
C
2
= 2

sin
A
2
sin
B
2
+ sin
C
2


,
4 cos
A
2
. cos
B
2
. cos
C
2
= 2

sin
A
2
sin
B
2
+ cos
A
2
cos
B
2
− sin
A
2
sin
B
2


.
cos
C
2
=
1
2
C = 120
0
.
ABC C = 120
0
.
ABC M = cos
2
A + cos
2
B + cos
2
C − 1.
M = 0 ABC
M < 0 ABC
M > 0 ABC
M = cos
2
A + cos
2
B + cos
2

C − 1
= −2 cos A. cos B. cos C
M = 0 cos A. cos B. cos C = 0. ∆ABC
M < 0 cos A. cos B. co s C > 0. cos A, cos B, cos C
∆ABC
cos A > 0, cos B > 0 cos C > 0. ABC
M > 0 cos A. cos B. cos C < 0.
ABC
ABC
a)

b
2
+ c
2
≤ a
2
sin A + sin B + sin C = 1 +
√
2
b)

cos A + cos B + cos C =
√
2
cos
2
A + cos
2
B + cos

2
C ≥ 1
cos A =
b
2
+ c
2
− a
2
2bc
b
2
+ c
2
≤ a
2
cos A ≤ 0.
π
2
≤ A < π,
cos
A
2
≤ cos
π
4
=
√
2
2

.
sin A + sin B + sin C = sin A + 2 cos
A
2
. cos
B − C
2
≤ 1 + 2.

√
2
2

.1 = 1 +
√
2,
sin A + sin B + sin C = 1 +
√
2







sin A = 1
cos
A
2

=
√
2
2
cos
B−C
2
= 1



A =
π
2
B = C =
π
4
ABC A =
π
2
; B = C =
π
4
cos
2
A + cos
2
B + cos
2
C = 1 −2 cos A. cos B. cos C

cos
2
A+cos
2
B+cos
2
C ≥ 1 cos A. cos B. co s C ≤ 0.
ABC A ≥ 90
0
B, C < 90
0
.
1 −
√
2 = 2 sin
2
A
2
− 2 sin
A
2
. cos
B − C
2
≥ 2 sin
2
A
2
− 2 sin
A

2
,
sin
2
A
2
− sin
A
2
≤
1 −
√
2
2
.
x = sin
A
2
x ∈

√
2
2
; 1

π
2
≤ A < π)
f(x) ≤
1 −

√
2
2
f(x) = x
2
−x f
′
(x) = 2x−1 > 0 x ∈

√
2
2
; 1

,
f(x) x ∈

√
2
2
; 1

.
f

√
2
2

=

1 −
√
2
2
f(x) ≥
1 −
√
2
2
f(x) =
1 −
√
2
2
.





sin
A
2
=
√
2
2
cos
B − C
2

= 1

A = 90
0
B = C = 45
0
ABC A = 90
0
; B = C = 45
0
ABC
a) p tan
B
2
tan
C
2
= p − c
b)
r
R
= 2 sin
2
C
2
+
1
4
cos
2

C
2
=
1
2

1 +
a
2
+ b
2
− c
2
2ab

=
p(p − c)
ab
,
cos
C
2
=

p(p − c)
ab
sin
C
2
=


1 − cos
2
C
2
=

(p − a)(p − b)
ab
.
tan
C
2
=

(p − a)(p − b)
p(p − c)
,
tan
B
2
=

(p − a)(p − c)
p(p − b)
.
p

(p − a)(p − b)
p(p − c)


(p − a)(p − c)
p(p − b)
= p − c,
a = b.
ABC B.
r
R
= 4 sin
A
2
. sin
B
2
. sin
C
2
cos A + cos B + cos C = 1 + 4 sin
A
2
. sin
B
2
. sin
C
2
.
1 +
r
R

= cos A + co s B + cos C,
r
R
= −2 sin
2
C
2
+ 2 sin
C
2
. cos
B − C
2
.
4 sin
2
C
2
− 2 sin
C
2
. cos
B − C
2
+
1
4
= 0,
1
4

sin
2
B − C
2
+

2 sin
C
2
−
1
2
cos
B − C
2

2
= 0,





sin
B − C
2
= 0
2 sin
C
2

−
1
2
cos
B − C
2
= 0



B = C
sin
C
2
=
1
4
ABC C.
ABC
a) h
a
=
√
bc. cos
A
2
b) m
a
=
√

bc. cos
A
2
ABC
l
a
=
2bc
b + c
cos
A
2
h
a
≤ l
a
,
h
a
≤
2bc
b + c
cos
A
2
b + c ≥ 2
√
bc
2bc
b + c

≤
√
bc.
h
a
≤
√
bc. cos
A
2
,
b = c.
ABC A.
m
2
a
=
1
4
(b
2
+ c
2
+ 2bc cos A).
1
4
(b
2
+ c
2

+ 2bc cos A) ≥
1
4
(2bc + 2bc cos A) = bc cos
2
A
2
,
m
2
a
≥ bc cos
2
A
2
m
a
≥
√
bc cos
A
2
b = c.
ABC A.
ABC
a)
2(1 + cos C)
sin C
= tan A + tan B
b)

sin A − sin B + sin C
sin A + sin B + sin C
= tan
B
2
. tan
C
2
ABC
4 cos
2
C
2
2 sin
C
2
. cos
C
2
=
sin(A + B)
cos A. cos B
2 cos A. cos B = 2 sin
2
C
2
cos(A+B)+cos(A−B) = 1−cos C, cos(A−B) = 1,
A = B.
ABC C.
2 sin

A + C
2
. cos
A − C
2
− 2 sin
A + C
2
. cos
A + C
2
2 sin
A + C
2
. cos
A − C
2
+ 2 sin
A + C
2
. cos
A + C
2
= tan
B
2
. tan
C
2
sin

A
2
. sin
C
2
cos
A
2
. cos
C
2
= tan
B
2
. tan
C
2
,
tan
A
2
= tan
B
2
A = B.
ABC C.
ABC A, B, C <
π
2
.

ABC
tan
2
A + tan
2
B = 2 tan
2
A + B
2
tan A + tan B =
4 sin
A + B
2
. cos
A + B
2
cos(A + B) + cos(A − B)
A, B, C < 90
0
cos(A + B) + cos(A − B) ≤ 1 + cos(A + B).
cos(A − B) = 1, A = B.
tan A + tan B ≥
4 sin
A + B
2
. cos
A + B
2
1 + cos(A + B)
= 2 tan

A + B
2
,
(tan A + tan B)
2
≥ 4 tan
2
A + B
2
,
(tan A − tan B)
2
≤ 0.
tan A = tan B A = B
ABC C.
ABC
a) 2 cos B. sin A. sin C
+
√
3

sin B + 2 sin
2
B
2
+ 4 sin
A
2
. sin
B

2
. sin
C
2

=
17
4
b) cos A +
1
3
(cos B + cos C) =
19
18
cos B =
a
2
+ c
2
− b
2
2ac
=
sin
2
A + sin
2
C − sin
2
B

2 sin A. sin C
,
2 cos B. sin A. sin C = sin
2
A + sin
2
C − sin
2
B,
cos A + cos B + cos C − 1 = 4 sin
A
2
. sin
B
2
. sin
C
2
,
2 sin
2
B
2
= 1 − cos B.
sin
2
A + sin
2
C − sin
2

B +
√
3 sin B +
√
3 cos A +
√
3 cos C =
17
4
,

cos A −
√
3
2

2
+

cos C −
√
3
2

2
+

sin B −
√
3

2

2
= 0,















cos A =
√
3
2
cos C =
√
3
2
sin B =
√
3

2









A =
π
6
C =
π
6
B =
2π
3
ABC B.
2

1 − 2 sin
2
A
2

+
4
3

cos
B + C
2
. cos
B − C
2
=
19
9
,
4 sin
2
A
2
−
4
3
sin
A
2
. cos
B − C
2
+
1
9
= 0,

2 sin
A

2
−
1
3
cos
B − C
2

2
+
1
9
sin
2
B − C
2
= 0,





sin
B − C
2
= 0
2 sin
A
2
−

1
3
cos
B − C
2
= 0



B = C
sin
A
2
=
1
6
ABC A.
ABC
a)
1 + cos B
sin B
=
2 sin A + sin C

4 sin
2
A − sin
2
C
b)


sin B = (
√
2 − cos C) sin A
sin C = (
√
2 − cos B) sin A
ABC
1 + cos B
sin B
=
2a + c
√
4a
2
− c
2
1 + cos B
1 − cos B
=
2a + c
2a − c
1
1 − cos B
=
2a
2a − c
c = 2a cos B.
a = b. ABC
C.

sin B −sin C = sin A(cos B −cos C),
sin
B − C
2

cos
B + C
2
+ sin A. sin
B + C
2

= 0.
cos
B + C
2
+ sin A. sin
B + C
2
> 0
sin
B − C
2
= 0,
B = C.
ABC A.
ABC
a) sin
A
2

. cos
3
B
2
= sin
B
2
. cos
3
A
2
b)
2
sin
2
A + sin
2
B
= 1 +
1
2
(cot
2
A + cot
2
B)