Declaración del Eurobasket sobre Seguridad Vial

so

P

is the incenter of tetrahedron

AB CD.

135

Remark.

The inequality holds for convex polyhedra circumscribed about a sphere.

(Titu Andreescu,

Romanian IMO Selection Test,

1982;

Revista Matematica Timi§oara (RMT) , No.

1(1982),

pp.

82,

Problem

4910)

60.

All summations here range from

i = 1

to

i = 4.

Let 0 be the circumcenter and

R

be the circum radius of

A1A2A3A4.

By the Power-of-a-point Theorem,

GAi ·GA� =

R2 - OG2,

for

1 � i � 4.

Hence the desired inequalities are equivalent to and

(R2 - OG2) L

�

L GAi•

Now

(1)

follows immediately from

(1)

(3)

136 3. G EOMETRY

by the Arithmetic-Geometric-Mean Inequality. To prove (3) , let P denote the vector from 0 to the point P. Then

(4)

This is equivalent to (3), since the last term of

(4)

vanishes. By Cauchy-Schwarz Inequality,

so

4LGA� �

(LGAi)2

and

1

_{GAi GAi � 2 1 1}

₂)_{2 1}

16

GAi � GAi.

Hence (2) also follows from (3) .

(Titu Andreescu,

IMO

1995

Shortlist)

Chapter

4

1.

Prove that

2.

Prove that for all x E

JR.

PROBLEMS

3 X X

+

27r 3 X

+

47r 3

cos -

+ cos3

--

+

cos --= -cos x

3 3 3 4

3.

Evaluate the sum

n-1

Sn

=

L

sin kx cos(n - k)x.

k=l

4.

Evaluate the sums

Sl

= sin x cos 2y

+

sin 2x cos 3y

+ .. . +

sin(n - l)x cos ny,

S2

= cos x sin 2y

+

cos 2x sin 3y

+

.

.

.

+

cos(n - l)x sin ny.

5.

Evaluate the products

1) PI = (1 - tg1°)(1 - tg2°) . . . (1 - tg8g0)j

2) P2 = (1

+

tg1 °) (1

+

tg2°) . . . (1

+

tg44°).

6.

Prove that

(4 cos2 go - 3) (4 cos2 27° - 3) = tan go,

7.

Let x be a real number such that sec x - tan x = 2. Evaluate sec x

+

tan x.

8.

Evaluate the product

7r where Ixl

< 2n+2'

140

and

4. TRIGONOMETRY

9.

Let a, b,

e,

d, x be real numbers such that x

f:. k7r, k

E .z and sinx sin 2x sin 3x sin 4x

10.

Let a, b,

e,

d E

[O,7r]

such that

2 cos a

+ 6

cos b

+

7 cos

e +

9 cos d

=

°

2 sin a -

6

sin b

+

7 sin

e

- 9 sin d

= 0.

Prove that 3 cos(a

+

d)

=

7 cos(b

+ e

)

.

11.

Prove that if

arccos a

+

arccos b

+

arccos

e = 7r,

then

12.

Let a, b,

e

be positive real numbers such that

Prove that

ab

+

b

e + e

a

= 1.

1

1

1

arctg -_a

+

arctg -b

+

arctg -

_e = 7r.

13.

Let x and y be real numbers from the interval

(

0,

i)

such that cos2 (x - y)

=

sin 2x sin 2y

7r

Prove that x

+

y

=

2 '

14.

Consider the numbers a, (3" E

(0, i)

such that

�(1

₂ - tga) (l - tg(3)

(1

- tg,)

= 1

- (tga

+

tg(3

+

tg,).

7r

Prove that a

+

(3

+

,

= "4'

15.

Let a,

b E (0, i),

Prove that

if and only if a

=

b. a

)

2

+ (

cos2 a

)

2

= 1

sm b cos b

16.

Prove that

7r

for all

0

< a, b < 2" 4.1. PROBLEMS sin3 a cos3 a

--

_{sin b}

+ --

_{cos b -}> sec(a - b)

17.

Let a, (3 be real numbers with (3

� 1.

Prove that

(1 +

2 sin2 a)t3

+ (1 +

2 cos2 a)t3 � 213+1

for all a E

JR.

18.

Let x be a real number, x E

[-1,1].

Prove that

< x2n +

(1 -

x2)n <

1

2n-1 -

-

for all positive integers n.

19.

Prove that

sec2n x

+

cosec2nx

�

2n+1 for all integers n

�

° and for all x E

(0, �) .

20.

Prove that

(1 +

sin x) (l

+

cos x) :::;

� + V2

for all real numbers x.

21.

Find the maximal value of the expression

E =

sin Xl COS X2

+

sin x2 cos x3

+ ... +

sin xn cos Xl , when Xl , X2 , . . . , Xn are real numbers.

22.

Find the extreme values of the function f

: JR

-t 1R,

f (x)

=

a cos 2x

+

b cos x

+ e,

where a, b,

e

are real numbers and a, b >

0.

23.

Let ao, aI , . . . , an be numbers from the interval

(0,

7r /2) such that

tan

(

ao -

�) +

tan

(

a1

-�) + . . . +

tan

(

an -

i)

� n -

1.

Prove that

tan ao tan a1 . . . tan an

�

n n+ 1 .

24.

Find the period of the function if p, q are positive integers.

f(x) = cospx

+

cos qx, x E 1R

142

4.

TRIGONOMETRY

a

2

- 5

25.

Let aD = � + V3 +

v'6

and let

an+1

=

_{2 an} ( n

₊

₂

) for n >

-O.

Prove that

(2n

-37r

)

an = cot

- 2

for all n.

26.

Let n be an odd positive integer. Solve the equation cos nx =

2n-1

cos x.

27.

Solve the equation

A sin2 x + B sin 2x + C = 0, where A, B, C are real parameters.

28.

Solve the equation

. . . 3

sm x cos y + sm y cos z + sm z cos x = '2

29.

Prove that the equation

. . 2

. 3 . 4 3

sm x sm x sm x sm x = 4

has no real solutions.

30.

Solve the system of equations

{

2 sin x + 3 cos y = 3 3 sin y +

2

cos x = 4.

31.

Solve the system of equations

{ X

sin y + -x2 cos Y = �

7r

2

x + y = 4'

32.

Solve the system of equations

1

37r

x + y + Z = 4 tgx + tgy + tgz = 5 tgx . tgy . tgz = 1.

33.

Prove that in any triangle

abc a cos A + b cos B + ccos C =

--2 _2R

and

and

and

4.1.

PROBLEMS

34.

Prove that in any triangle

A B . C A B C . A

cos3 - sin - sm - = cos - cos - cos

2 2 2

2 2

- sm2 -

2 '

35.

Let n be a positive integer. Prove that in any triangle

L

sin nA sin nB sin nC =

(_I)n+1

+ cos nA cos nB cos nC

L

cos nA cos nB sin nC = sin nA sin nB sin nCo

36.

Consider a triangle ABC such that

sin A sin B + sin B sin C + sin C sin A = A (1 + sin A) (1 + sin B) (1 + sin C) = 2(A + 1) Prove that triangle AB C has a right angle.

37.

Let A > 1 be a real number and let ABC be a triangle such that

aA

cos B + bA cos A =

cA

a2A-1

cos B +

b2A-1

cos A =

C2A-1•

Prove that the triangle is isosceles.

38.

Prove that the triangle ABC is equilateral if and only if

A B C 1

tg"2 + tg2" + tg2" = 4S

(a2

+ b2 +

c2).

39.

Let ABC be a triangle such that

sin2 B + sin2 C =

1

+ 2 sin B sin C cos A. Prove that triangle ABC has a right angle.

40.

Let ABC be a triangle such that

(

cot

�r

+

(

2 cot

�r

+

(

3 cot

�r

=

(�;r,

143

where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle ABC is similar to a triangle

T

whose side lengths are all positive integers with no common divisor and determine these integers.

144

4.

TRIGONOMETRY

41.

Prove that in any triangle

. A B C

_{< -.} 9 2 2 2 - 8

42.

Prove that in any triangle

� � �

(

. 2 A . 2 B . 2 C)

- + - + -

_{be ea ab -}> 4

sm - + sm - + sm -_{2 2 2}

.

43.

Prove that in any triangle

cos A _{+ cos B + cos C}

_{> �.}

a3

b3

e3 - 16p3

44.

Prove that in any triangle

2 A

2 B

2 C

sec - sec - sec - 9

__ 2 +

> _.

be

ea

ab

-p2

45.

Prove that in any triangle

!!.

> 3V3.

r -

46.

Let

ABC

be a triangle. Prove that

3A

3B

3C

A - B

B - C

C

-

A

sin 2 + sin 2 + sin 2 :s cos

+

cos

+

cos

47.

Find the number of ordered pairs

(a, b)

such that

(a+bi)2002

=

a-bi, a, b E

R

48.

Find

Imz5 min -­

zEC\1R Im5

Z

and the values of z for which the minimum is reached.

49.

Let

ZI, Z2, .. . , Z2n

be complex numbers such that

IZll

=

IZ21

= . :

.

=

IZ2nl

and arg

ZI

:s arg

Z2

:s . . . :s arg

Z2n

:s 7r. Prove that

50.

For all positive integers k define

Ak

= {z

E

<C

I Zk

=

I}.

Prove that for any integers m and n with 0 < m < n we have

Al U A2 _{U · · · U} Am

C

An-m+1

U

An-m+2

U · · · U

An.

4.1.

PROBLEMS

145

51.

Let

ZI, Z2, Z3

be complex numbers, not all real, such that

IZ11

=

IZ21

=

I

za

l

=

1

and 2(ZI +

Z2

+

Z3) - 3Z1Z2Z3 E JR .

Prove that

52.

Let n be an even positive integer such that

�

is odd and

co, C1, · .. ,cn-1

the

complex roots of unity of order n. Prove that

n-l

II (a

+

bc%)

=

(a�

+

b�)2

k=O

for any complex numbers

a

and

b.

53.

Let n be an odd positive integer and

co, C1, .. . , cn-l

the complex roots of

unity of order n. Prove that

n-1

II (a

+

bc%)

=

an

+

bn

k=O

for all complex numbers

a

and

b.

54.

Let

ZI, Z2, Z3

be distinct complex numbers such that

IZ11

=

IZ21

=

IZ31

= r.

Prove that

55.

Let

ZI,Z2, Z3

be distinct complex numbers such that

IZ1 1

=

IZ21

=

IZ31

= r

and

Z2 f:. Z3·

Prove that

56.

If

Z

is a complex number satisfying Iz

3 + z-31

:s 2, the inequality show that Iz +

z-11

:s 2.

57.

The pair

(Zb Z2)

of nonzero complex numbers has the following property: there is a real number

a E

[-2, 2] such that z? -

aZI Z2

+ z� =

O.

Prove that all pairs

(zf , z�) , n = 2, 3, . . . , have the same property.

58.

Let

A1A2 .. . An

be a regular polygon with the circumradius equal to 1. Find

_n

the maximum value of max

II

P

Aj

when P describes the circumcircle.

146 _4. TRIGONOMETRY

59.

Let

n

be an odd positive integer and let aI , a2 , . . . , an be numbers from the interval [0, 71'].

Prove that

60.

Let n be a positive integer. Find the real numbers

ao

and

akl,

k, l

=

1, n, k >

l,

such that

sin2

nx = ao

₊

_{L akl}

_{cos 2(k -}

_l

_)x sm x _19<k5n

for all real numbers x f:. m7l', m E Z.

SOLUTIONS

1.

We prove that

2 k7l' 21 -_ n (2n - 1) 3 '

k=1

n+

for all integers n >

O.

Consider the equation with roots

sin(2n + l)x

=

0, 71' 271' n7l' 2n + 1 ' 2n + 1 ' . . . , 2n + 1 .

Expressing sin(2n + l)x in terms of sin x and cos x, we obtain

sin(2

n

+ l)x

= (

2n

:

1

)

cos2n x sin x _

e

n

:

1

)

COs2n-2 sin3 X + ...

=

=

sin2n+, X

(e

n

:

1

)

cat2n X _

e

n

:

1

)

cot2n-2 x + . . .

)

S - k7l' _k

-1 2 S" 2n+l 0 h

et x - --, - , , . . . , n. Ince SIn x , we ave

2n + 1

(

2n

:

1

)

_cat2n X _

en:

1

)

cot2n-2 x + . . .

= O.

Substituting y

=

cot2 x yields

en:

l

)

yn -

en:

l

)

yn-l + . . .

=

0,

(1)

with roots cot2 -2 71' , cot2

�,

. . . , cot2 2 n7l' l ' Using the relation between coef-

n

+ 1 2n + 1 n +

ficients and roots, we obtain

n

(

2n + 1

)

2 k7l' 3

cot _{2n + 1}

_{= en:}

₁

)

Setting n

=

3, the desired conclusion follows.

n(2n - 1) 3

(Dorin Andrica,

Revista Matematidi Timi§oara (RMT), No. 1-2(1979), pp. 51, Problem 3831)

148 4. TRIGONOMETRY

2.

Applying the identity

t

t

cos

t =

4 cos3 3" - 3 cos 3"'

t

E lR

for

t =

x,

t =

x

+

271",

t

= x + 471" and summing up the three relations, we obtain

( X X

+ 271"

X

+ 471"

)

3 cos x

=

4 cos3 3' + cos3 + cos3 -

-3 cos - + cos --

( X X

₃ + 271" ₃

+

cos

--X

+ 471"

)

.

3 On the other hand,

x x

+

271"

X

+ 471" 471" 2x + 471" X

+

271"

cos - + cos -- + cos _{3 3}

--

₃

=

2 cos - cos ₆

---

₆

+

cos

--

₃

=

= (

2 cos

�

+ 1

)

cos x 21T

=

0 and the desired identity follows.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 44, Problem 2124)

3.

We have

28n

=

8n + 8n

=

=

sin x cos(n - l)x + sin 2x cos(n - 2)x

+

. .

.

+ sin(n - l)x cos x+

+

sin(n - l)x cos x + sin(n - 2)x cos 2x + . . . + cos(n - l)x sin x

=

=

sin nx + sin nx + . .

. +

sin nx

=

(n - 1) sin nx, so

n - 1

8n

=

(Dorin Andrica,

Gazeta Matematica (GM-B), No. 8(1977) , pp. 324, Problem 16803; Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 30, Problem 3055)

and

4.

Note that

81 + 82

=

sin(x + 2y)

+

sin(2x + 3y)

+ .. . +

sin[(n - l)x + nyJ 82 - 81

=

sin(2y - x) + sin(3y - 2x) + .

. . +

sin[ny - (n - l)x] .

Setting

x + y

=

hI and y - x

=

h2 yields

81 + 82

=

sin(y + hI ) + sin(y + 2hI ) + . . . + sin(y

+

(n - l)hI)

=

Hence and 4.2. SOLUTIONS . nhl

. [

( 1) hI

]

sm T sm y + n - 2 . hI sm 2

82 - 81

=

sin(y

+

h2) + sin(y

+

2h2) + . . . + sin(y

+

(n - 1)h2)

=

. nh2

.

[

h2

]

sm T sm y + (n - 1) 2 . h2 sm 2 . nhl

.

[

( 1) hI

]

1 sm T sm y + n - 2 81

=

₂

.

_h 1 sm 2

1 sin sin

[Y

+ (n

- ]

1 sin

�

sin

[Y +

(n - l)

�]

82

=

_{2 h + 2 . h2 .}

sin

-f

sm 2

149

(Dorin Andrica,

Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 65, Problem 3056)

5.

We have PI

=

0 because of the factor 1 - tan 45°

= O.

On the other hand we have

( cos 1 ° + sin 1 0) .

.

. (cos 44° + sin 44°) _

P2

=

_{cos 1 ° . . . cos 44 °}

-

(../2 ../2 . ) (../2

440

y'2 .

440

)

T cos 1° + T sm 1° . . . T COS + T sm

(

../2

)

44

� cos

1

° . . . cos 44°

=

sin 46° . . . sin 89° _{cos 1 ° . . . cos 44°}

(v'2)

44

=

222 .

(Titu A ndreescu)

cos 3x

6.

We have cos 3x

=

4 cos3 x - 3 cos x, so 4 cos2

X

- 3

= --

_{cos x} for all x

f.

(2k + 1) . 90°, k E Z. Thus

cos 27° cos 81 ° cos 81 ° sin 9° _° (4 cos2 9° - 3) (4 cos2 27° - 3)

= -- . -- = -- =

_{cos 9° cos 27° cos 9° cos °}

=

tan 9 ,

as desired.

150 4. TRIGONOMETRY

7.

From the identity 1 + tan2 x

=

sec2 x it follows that

1

=

sec2 x - tan2 x

=

(sec x - tan x) (sec x + tan x)

=

2(sec x

+

tanx),

so sec x + tan x

=

0.5.

(Titu Andreescu,

American High School Mathematics Examination, 1999, Prob­ lem 15)

8.

Since

1

+

tan2 2kx cos2 2kx (1 - tan2 2kx)2

=

cos2 2k+1X

for all Ixl < it follows that

(Dorin Andrica)

9.

Let Then

sin x _ sin 2x _ sin 3x _ sin 4x _ ;\

- - _{- -} .

sin2 4x = 2 sin2 2x(l - sin2 2x).

Because sin2 4x = 2 sin2 2x(1 - sin2 2x), we obtain

d2

=

2b2 (1 _ ;\2b2).

On the other hand, sin 3x

=

AC, sin x

=

Aa, and since sin 3x

=

sin x (3 - 4 sin 2 x), we have

c

=

a(3 - 4A2a2). Eliminating A from the relations (1) and (2) yields

2a3 (2b3 - �) = b4 (3a - c),

as desired.

(Dorin Andrica)

10.

Rewrite the two equalities as

2 sin a - 9 sin d

=

6 sin b - 7 sin c

2 cos a

+

9 cos d = -6 cos b - 7 cos c.

By squaring the two relations and adding them up we obtain 85 + 18 cos(a + d) = 85

+

42 cos(b + c) ,

and the conclusion follows.

(1)

(2)

4.2. SOLUTIONS

(Titu Andreescu,

Korean Mathematics Competition, 2002)

11.

From the hypothesis it follows that

sin( arccos a + arccos b + arccos c)

= O.

Using the identity

L

cos a cos ,B sin'Y = sin(a + ,B

+

'Y)

+

sin a sin ,B sin 'Y

and the formulas

we obtains

sin(arccos x) =

�

and sin(arcsin x)

=

x, x E [-1, 1] ,

ab

�

+ bc

�

+ ca

J1=b2 =

a2) (1 - b2) (1 -c2), as desired.

151

(Titu Andreescu,

Revista Matematidi Timi§oara (RMT) , No. 2(1977), pp. 64, Problem 3054)

12.

The identity

x

+ y + z

- xyz

arctgx + arctgy + arctgz

=

arctg 1 _{- xy}

(

₊

_{yz zx}

_{+ ) +}

_k7r implies

1 1 1 ab + bc

+

ca - 1

arctg-

+

arctg- + arctg-= arctg b ( b

+

)

+

k7r.

a b c a c - a + c

Because ab + bc + ca = 1, we obtain

1 1 1

arctg - + arctg -+ arctg -= k7r,

a b c

where k is integer.

7r

Note that 0 < arctgx < - for all real x > 0, hence

2

1 1 1 37r

o < arctg� + arctgz; + arctgc < 2'

Therefore k

=

1 and

as claimed.

1 1 1

arctg- + arctg- + arctg-

=

7r,

a b c

(Titu Andreescu,

Revista Matematidi Timi§oara (RMT) , No. 1 (1977), pp. 42, Problem 2827)

13.

The given relation is equivalent to

( cos x cos y + sin x sin y) 2

=

4 sin x sin y cos x cos y,

or

152 _4. TRIGONOMETRY

Hence

cos2 (x +

y) =

0, and since

x, y

E

(o, �),

we obtain

x + y = �,

as desired.

(

Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 1(1977) , pp. 42, Problem 2826)

or

14.

Expanding yields 1

"2 (1 - tan a - tan ,B - tan 1 + tan a tan ,B + tan ,B tan 1+

+ tan 1 tan a - tan a tan ,B tan 1)

=

=

1 - (tan a + tan ,B

+

tan 1) ,

tan a

+

tan ,B + tan1 - tan atan ,B tan 1

=

=

1 - tan atan ,B - tan ,B tan 1 - tan 1 tan a.

Since a, ,B, 1 E

(

0,

i),

we have ° < a + ,B + 1 < 7r , hence

tan a + tan ,B + tan 1 1= tan a tan ,B tan 1. From relation (1) we derive

1 - tan a tan ,B - tan ,B tan 1 - tan 1 tan a

=

1 tan a

+

tan ,B + tan 1 - tan a tan ,B tan 1 '

therefore

cot(a + ,B + 1)

=

1. Hence a + ,B + 1

= i,

as desired.

(1)

(

Titu Andreescu,

Revista Matematica Timi§oara (RMT), No. 1(1973) , pp. 42, Problem 1582)

15.

The relation in the statement is equivalent to

or It follows that .

(

sin4 a cos4 a

)

(sm2 b + cos2 b) _{sm b cos b} +

=

1, • 4 4 cos2 b . sin2 b sm a

+

cos a + sm4 a

+

cos4 a

=

1. sin b cos b . 2 2 cos2 b . sin 2 b

1 - 2 sm a cos a + _{sm b} sm4 a + _cos cos4 a

=

1,

b

hence

Furthermore,

(

COS b . 2 _{a - -- cos a} sin b 2

)

2

_{= 0.}

sm b cos b

cos b . 2 sin b 2 sin b sm a

=

cos b cos a,

4.2. SOLUTIONS

or tan2 a

=

tan2 b.

Because a, b E

(

0,

-i),

we obtain a

=

b. The converse is clear and we are done.

153

Alternative solution.

From the given relation we deduce that there is a number e E

(

0,

i)

such that Hence sin2 a . sin b

=

sm e and cos2 a cos b

=

cos e

sin 2 a

=

sin b sin e and cos2 a

=

cos b cos e.

It follows that

1

=

cos(b - e) and cos 2a

=

cos(b + e)

Since a, b, e E

(

0,

i),

we have b - e

=

° and 2a

=

b + e, hence a

= b,

as desired.

(

Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 41, Problem 2825; Gazeta Matematica (GM-B), No. 11(1977) , pp. 452, Problem 16934)

16.

Multiplying the inequality by sin a sin b + cos a cos b

=

cos(a - b), we obtain the equivalent form

(

Sin3 a cos3 a

₊

)

_{(sm a sm b + cos a cos b) 2: 1.} . . sm cos

But this follows from Cauchy-Schwarz Inequality, because, according to this in­ equality, the left-hand side is greater than or equal to (sin2 a + cos2 a)2

=

1.

(

Titu Andreescu

)

17.

Using the inequality

for m 2: 1 we obtain

x? + xr

>

(

Xl + X2

) m

2 - 2

(1

+

2 8in2 alP + (1 + 2 C082 aJII � 2 + 2 8in2 + 2 cos2 a

)

p

=

2P+1 ,

as desired.

(

Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 1(1974), pp. 30, Problem 1942)

18.

Because X E [- 1, 1 J, there is a real number

y

such that

x =

sin

y.

It suffices

to prove that

154 4. TRIGONOMETRY

For the left-hand side note that

I

sin yl

� 1

and

I

cos yl

�

1, hence sin2n y

+

cos2n y

�

sin2n-2 y + COS2n-2 y

�

. . .

�

sin2

y +

cos2 Y

=

1, as desired.

For the right-hand side we use the inequality Hence

as claimed.

x?

+

x� >

(

Xl

+

X2

)

n

2 - 2

In document La mejora de la seguridad vial (página 51-55)