360 Problems for Mathematical Contests [Andreescu] 9739417124 (1)

p

. . •

==

==

==

==

=

53

2. 1. PROBLEMS 40. Let

Pk denote the k-thf primer number.r;:Provemthat p + p + . . . + p > n +l

for all positive integers m and n .

41 . Let n be a positive integer. Find the sum of all positive integers less than 2n and relatively prime with n . 42. Prove that any number between n distinct divisors of n!.

1 and n! can be written as a sum of at most

43. Find the largest value of n such that the complementary set of any subset contains at least two elements that are relatively ith n elements of w prime.

{ I , 2, . . . , 1984}

a, if a2 � n then aln. (Un ) n� l' (Vn )n�l defined by Ul 3 , Vl = 2 and 45 . 3Consider the sequences Un+l1 = un + 4vn , Vn+l[ V22uJn + 3vn , n �1 1 . Define Xn Un + Vn , Yn Un + 2vn , for all n � . n � . Prove that Yn = x n 2 1 46. Define X n = 2,, - 1 + for all positive integers. Prove that 2 (i) X n = XIX 2 . . . Xknl- l + , n E ( ) 1 , , E k -:f. l (ii) Xk ' xz 3 (iii) X n ends in 7 for all n � . ( ) 1 and 47. Define the sequence an n � l by al 2an + J3a� - 2 a +l n 1 for all integers n � . Prove that an is an integer for all n. ( ) (b ) 1, 48. Define the sequences an n �o and n n �o, by ao 2an-l an = 1 + 2an2 _ l and bn = 1 _ 21a2n- l ( ) for all positive integers n . Prove that all terms of the sequence an n �O are irreducible b ( ) fractions and all terms of the sequence n n �o are squares. 3, Yo 2 , ( ) ( ) 49. Define the sequences Xn n �o and Yn n �O by Xo Xn = 3Xn- l + 4Yn-l and Yn 2(Xn)- l + 3Yn- l 1 4 ; ; for all positive integers n . Prove that the sequence zn n �o, where Zn + x y , 44. Find all positive integers n such that for all odd integers

=

=

N,

=

=

=

N

=

=

=

=

=

=

=

contains no prime numbers.

2. NUMBER THEORY 50. Let p be a positive integer and let X l be a positive real number. Define the ( ) sequence Xn n �l by Xn+l Vp2 1xn pvx; 1 for all positive integers Prove that among the first terms of the sequence there are at least [�] irrational numbers. ( ) 5 1 . Define the sequence xn n � o by 2)1) XXnn+l 0 ifX[and] only( -ifl)nx[�0] and for all 2:: o. � Find X n in closed form. ( ) 2 , a3 6 and 52. Define the sequence an n �O by ao 0, al 1, a2 54

+

=

+

+

n.

=

m

n =

+

=

n

=

Prove that n divides

=

=

=

n

an for all > O. 53. Let Xl X 2 X3 1 and X n+3 X n X n+IXn+ 2 for all positive integers =

=

=

=

+

Prove that for any positive integer m there is an integer k > 0 such that

Xk ·

m

n.

divides

(an ) n�O be the sequence defined by ao 0, al 1 and an+l - 3a2n an- l ( _ l )n for all integers > O. Prove that an is a perfect square for all 2:: O. 55. Let al a2 97 and an+l anan- l V(a; - l)(a;_ 1 1), > 1. Prove that 2 2an is a perfect square. a) 2 + J2 2an is a perfect square. b) 2 56. Let k 2:: be an integer. Find in closed form for the general term an of the sequence defined by ao 0 and an - a [ �] 1 for all > O. 57. Let ao al 3 and an+l tan - an- l for 1. Prove that an - 2 is a perfect square for all 1. 2 58. Let a and (3 be nonnegative integers such that a 4(3 is not a perfect square. ) ( Define the sequence xn n �O by 54. Let

=

+

=

=

n

=

n

=

=

+

+

n

-

+

=

=

=

= n

=

2::

n

n

2::

+

2.1. PROBLEMS

55

for all integers n 2:: 0, where and are positive integers. Prove that there is no positive integer no such that

X l X2 X2no - Xno-IXno +l. _

59. Let n > 1 be an integer. Prove that there is no irrational number

Va + Ja2 - 1 + Va - Ja2 - 1 is rational.

+ and - the expression ± 1 ± 2 ± 3 ± . . . ± (4n + 1), yields all odd positive integers less than or equal to (2n + 1)(4n + 1). 60. Prove that for different choices of signs

a such that

SOLUTIONS 1 . The problem is equivalent to finding the number of functions

1 : { I , 2 , 3 , 4, 5 , 6 , 7} {O , 1 , 2 , . . . , 9 } such that 1(1 ) f. 0, 1(1) f. 1 and 1(7) f. l. 1(1 ) E {2, 3, . . . , 9} , there are 8 possibilities to define 1(1) . For 1(7) there Because 9 1(2) 1(3) 1(4) 1(5) 1(6) there are 10 . are possibilities and for 8 9, 0 , 72, 0 , 1 5 . 1 5 numbers with the desired property. To are . . in Andrithere (Dorconclude, 11 ( 1979) , pp. 421 , Problem ca, Gazeta Matematica (GM-B) , No. 17999) d d d 2 . Let be the greatest common divisor of m and n. Hence m m l and n n l for some integers ml and n l . The numbers are 1 · ml 2 · ml p · ml , nl , . . . , n l nl [: ] and, since are relatively prime, there are , integers among them. Because d(m, n)p ] . [ n n gC . mtegers. follows that there are nl d n 11 ( 1 979 ) 429, Problem (DoringcAd(nm,drica,n) ItGazeta Matematica (GM-B) , No. , pp. 0:89) 3 . Define k pn and note that k is odd. Then dk + (p - d) k p[dk- 1 dk-2 (p d) + . . . + (p d) k -l ] d 1 to d [�] implies that p divides 1k + Summing up the equalities from 2k + . . . + (p - l) k , claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 49, 3813) Problem 55n • Then 4. Define m 55n +1 + 55n + 1 m5 m + 1 (m2 + m + 1 )(m3 - m2 1) 57 -+

=

=

--

=

--

ffi" n,

=

=

=

=

_

_

_

=

=

as

=

=

+

=

+

58

2. NUMBER THEORY 1 and, ( since factors are greater than , the conclusion 2follows. dn reescu, Titu Aboth 001) Korean Mathematics Competition, N = (�) - 5 (�) + 52 (�) - + 5n- ' (�) . Then 5. Let 5N = 1 - 1 + 5 (�) _ 52 (�) + 53 (�) _ . . . + 5n (�) = 1 + ( - 1 + W . Hence N = � W + l) = � [C2n + 1 )2 - (2 � r] � � = �" ['2n - 2 + 1'] [2n +" 2' + 1] = ' = � [(2 ;- - 1r + 2n-5] [(2 ;- + 1r + 2n- ] . Because greater than or equal to , both the numerator are 5. One ofn isthem 5, call 5N1 ,factors N1 > 1of, the N2greater than is divisible by it other being N = N1 N2 , where N1 and N2 are both integers greater than 1 , and we are done.. Then ( Titu Andreescu, Korean Mathematics Competition, 2001 ) 4 3 4 4 4 and 6. The given number is of the form m + � n , where m = n=4 4 =2 2 ' "

�

�

•

The conclusion follows from the identity

and the inequalities

4) 1 2 i1 3 2 2 2 _ 1 ( 56 4 i 2 - > m - mn + n > n ( n - m) = 2 4 > 2 2 (2 562- 1 - 22 4 )4 > 2 2 . 251 2 (42 562-1 -5 12 - 1) > 21 0. 5�t1 25 1 2 > 21 0.5 . 210.50 > 103 .5 . 103.50 > 102002 > (Titu Andreescu, Korean Mathematics Competition, 2002) 111 1 3 37 111 7.) The positive divisors of 2 are , , , 7 . So we have the following cases: yl [ ] 1 111 1 1 111 n �4 . 2) [ yl111] == 3, oror 3n 111<
�

.

.

$

$

$

$

�

2. 2. SOLUTIONS

( Titu Andreescu)

8. Note that

59

2a(a2 + 3b3 2 ) = (a + b)3 + (a b)3 -

The Fermat equation for n =

x3 + y3 = z 3

has no solution in positive integers (see T. Andreescu, D. Andrica, "An Introduction to Diophantine Equations" , GIL Publishing House, pp. Hence there is no integer c such that

2002, 87-93) .

if

a > b.

On the other hand, if

b > a then there is no integer

c

such that

This concludes the proof. Revista Matematidi Timi§oara (RMT) , No. Problem

( Titu Andreescu, 1911)

1 (1974) , pp. 24,

4 = q4 for some positive integer q. Then p = q4 + 4 and q > 1 . We obtain ( 2 2q + 2)(q2 + 2q + 2) , p= q 1 a product integers greater than , contradicting dreescu, ( Titu Aofntwo ) fact that p is a prime. 2003the Math Path Qualifying Quiz, 9 . Assume that p

-

_

10. The inequalities

0 � x + y + 8, which is 3 3 2) x2 +2 y 3 (x + 2 2) 2 y2 + x (y + 2) 2 is true. Without 3y = 2 x + 1 . 2x3k 1x2 + 3y 2k(x +1x2)+2 y (x + x2 k+. 3y0 (x + 1 ) 2 , hence 3x = 4k2 + 13k + 4. 2 x y + � y + k> (2k + 3)2 < 4k2 + 13k + 4 < (2k + 4)2 3 k E {O, 1 , 2, 3, 4} , y2 + 3x so y2 + x cannot be a square. It is easy to check that for 0 4 3 k 22 is not a (square but for = , y2 + x = = . Therefore the only solution is ) 1 1 (x, y)( Ti= A, d. ) tu n reescu

cannot hold simultaneously because summing them up yields false. Hence at least one of or < < loss of generality assume that < From < < we derive = Then = + and = for some integer and so If 5, then

2. NUMBER THEORY 1 7 + 1 2v'2 = (v'2 + 1 ) 4 and 1 7 - 1 2v'2 = (v'2 _ 1 ) 4 , so 1 1 . Note that (17 1 2v'2) n - ( 1 7 - 1 2v'2) n ( v'2 1) 4n - (v'2 _ 1 ) 4n 4v'2 4v'2 (v'2 1) 2n (v'2 - 1 ) 2n (v'2 + 1 ) 2n _ (v'2 _ 1 ) 2n 2 . 2v'2 -

60

+

_

_

+

+

+

Define

A = (v'2 + 1) 2n +2 (v'2 - 1 ) 2n

and

B = (v'2 1) 2n2v'2 - (v'2 _ 1) 2n +

U sing the binomial expansion formula we obtain positive integers that

x

and

2n v2 1 + ( ) = X + y v2, (v2 - 1) 2n = X - y v2 Then (v'2 + 1 ) 2n + (v'2 - 1) 2n (v'2 1) 2n _ (v'2 _ 1) 2n 2 2v'2 and y = X= AB is as integer, as claimed. and so

y such

+

Observe that

A B

so and are relatively prime. not a perfect square. We have

It

is sufficient to prove that at least one of them is

A = (v'2 + 1) 2n +2 (v'2 _ 1) 2n = [ (v'2 l ) nv'2(v'2 _ 1) n ] 2 - 1 +

and

Since only one of the numbers

(1)

l ) n - (v2 _ l) n v'2 v'2 ( 1) and (2) we derive is an Ainteger - depending on the parity of n - from the relations that ( D isinotA a dri square. This completes the proof. or4n285n) ca, Revista Matematica Timi§oara (RMT) , No. 1 ( 1981 ) , pp. 48, Problem 7 8 9 6 13 < 16 < 12. The claim is true for n = and n = , because U 6 = < < U7 = 21 U8 = . (v2 +

l) n

+

(v2

_ l) n

+

(v2 +

If n

� 8, then

2. 2.

61

SOLUTIONS

Un - 1 > r,;-;--:-: . � -_ v'UnUn+1 --'-+1 +- UnVU;; >- 2v'Un +1 - 1 = .1Fi v'Un 1 > 1 � 2Y. Un� oJUn- 1 2y 3

y. Un+1 - y Un

-

--=

and so between Un and Un+ 1 there is a perfect square.

(Dorin Andrica)

13. If n = 1, 2, 3 or 4 then n! + 5 = 6, 7, 11 or 29, so it is not a square. If n

then n! + 5 desired.

� 5,

= 5(5k + 1) for some integer k and therefore is not a perfect square, as (Dorin Andrica, Gazeta Matematidl. (GM-B ) , No. 8(1977) , pp. 321, Problem

16781; Revista Matematica Timi§oara ( RMT ) , No. 1 (1978), pp. 61, Problem 3254)

14. Suppose by way of contradiction that n2 +3n+3 is a cube. Hence n(n 2 +3n+3)

is a cube. Note that n (n2 + 3n + 3) = n3 + 3n2 + 3n = (n + 1) 3 - 1 and since (n + 1) 3 - 1 is not a cube, we obtain a contradiction. Gazeta Matematica ( GM-B ) , No. 8(1977) , pp. 312, Problem E5965; Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 28, Problem 3253)

(Dorin Andrica,

1 5 . Consider the product of 2p + 1 consecutive numbers

P(n) = (n + 1) (n + 2) . . . (n + 2p + 1) Observe that P(n) > (n + 1) 2P+1 . On the other hand, (n + l) + (n + 2) + . . . + (n + 2P + l) 2P+ 1 = (n + p + 1) 2p+1 P(n ) < 2p + 1 from the AM-GM inequality. then E {n + 2, . . . , n + pl . Assume by way of contradiction If P(n) that there is k E {2, 3, . . . , p} such that P(n) = (n + k) 2p+1 . Then (n + 1) (n + 2) . . . (n + k - 1) (n + k + 1) . . . (n + 2p + 1) (n + k) 2P . (1)

]

[

= m2p+1 ,

m

=

We have two cases: I. k p 1) If n == (mod p) , then (n + k) 2p is divisible by p2p. The left-hand side of the equality (1) is clearly not divisible by p2p, hence we reach a contradiction. 2) If n == r (mod p) , then the left-hand side of the equality (1) is divisible by p2 , because of the factors n + p - r and n + 2p - r, while the right-hand side is not, since (n + k) 2p == This is a contradiction. II. k E {2, 3, . . . , p - l }

= 0

r f. 0 , r2 P .

62

2. NUMBER THEORY

1) If n - k (mod p) , then the right-hand side of ( 1) is divisible by p2p , but the left-hand 2) nside -qis not.(mod p) , q f. k and q E {O , 1 , . . . ,p - I} , then the left-hand side of ( 1 ) is divisible by p. On the other hand (n + k)2p (q - k) 2 (mod p) � 0 (mod p) , 0 I kl because < q - < p. Both Andend ir ca)up in contradictions, so the problem is solved. (Dorincases 16. It suffices to prove that there are no integers in the interval � y'P � vp (n - n ,n + n] for n � [ JI -a2avp] + 1 . k Assume by way of contradiction that there is integer such that ny'P - -an < k <- ny'P + -.an Hence 2 2 n2p + an - 2ay'P < k2 <- n2p + an + 2ay'P. 1 a2 + 2a..;p < 1 an 2 ..,fP 1 a Observe that 2 - a > - . If n � [ J1 2 vp ] + , then n -a so 2p - 1 < k2 < n2p + 1 n k2 pn2 or vp � , which is false, since p is prime. It follows that (Dorin Andrica, Gazeta Matematica n (GM-B) , No. 8 ( 1977) , pp. 324, Problem 16804) 1 the claim is clear, so let n � 3. 17. For n Define Sn (�) + (;) + . . . + ( n�l) . Then ==

If

==

==

2"

2"

2"

2"

=

=

=

=

S!,- 2n -1 - 1 .

Sn

Sn

= or contains an odd is odd it follows that the sum Because number of odd terms, as desired. Revista Matematica Timi§oara (RMT) , No. pp. Problem

( Titu Andreescu, 5346)

2(1984) , 71 ,

) . we can assume that m :O; [iT : ( ': ) ( 0 , then 1 1984m, false.n If n 1 , then m 1984. If n 18. Because =

=

=

=

=

2.2. SOLUTIONS

63

m(m - 1) = 1984, with no integral solutions . n If n :::: 3, then (:2) :::: (�), so 1984 m(m - � (m - 2) . Hence 11904 :::: m3 - 3m2 + 2m or (m - 30)(m2 + 27m + 812) � -12456 < 0, and so m < 30. This m (m - 1) . . . (m - n + 1) does not contam. the . that (m) 198 4 , because implIes n n., factor 31 of 1984. To the solutions are m = 1984, n 1 and m = 1984, n 1983. Andreescu, ( Tituconclude, Revista Matematica Timi§oara (RMT ) , No. 1 (1985) , pp. 80, If = 2, then

::::

-Ir

=

=

Problem 5)

1 9 . The equation is equivalent to

(x + 2y) (x + 4y) - ( + 2y) =

3x

or

(x + 2y) (x + 4y

{

3,

- 3) = 3.

We have ( I')

or

(" ) {

X + 2y

x+

=

3 4y - 3 = 1

, with solution ( x, y)

=

(2, -21 ) (

)

x + 2y = 1 ' ( x, y ) = - 4, -5 . , WI. th soIutlOn = 2 x+ Note that there are no solutions in integers, as claimed. Revista Matematica Timi§oara (RMT) , No. 1 (1971), pp. 20, Problem 312) 11

4y - 3 3

( Titu A ndreescu,

20. The equation is equivalent to

22

2 2

x y - 2xy + 1 + x + y - 2xy + 2(x - y) (l - xy) = 4, or

2

2

(xy - 1) + (x - y) + 2 (x - y) (1 - xy) = 4. Hence (1 - xy + x - y) 2 = 4 and, consequently, 1 (1 + x) ( l - y)1 = 2. We have two cases: I. (1 + x) ( l - y) = 2. Then a) 1 + x = 2, 1 - y = 1, so x = 1, y = b) 1 + x = -2, 1 - y = -1, so x = y = 2. c) 1 + x = 1, 1 - y = 2, so x = y = -1 . d) 1 + x = -1, 1 - y = -2, so x = -2, y = II. (1 + x) ( l - y) = -2. Then

0.

-3,

0,

3.

2 . NUMBER THEORY

64

a) 1 + b) 1 + c) 1 + d) 1 +

x = 2, 1 y = -1, so x = 1, y = 2. x = -2, 1 - y = 1, so x = -3, y = O. x = 1, 1 - y = -2, so x = 0, y = 3. x =d -1, 1 - y = 2, so x = -2, Y = - l. A Ti ( tu n reescu, Revista Matematica Timi§oara (RMT) , No. 4-5(1972) , pp. 43, -

Problem 1383)

2 1 . The equation is equivalent to

We have the cases: 1 + pq, pq(l + pq) . 1) pq p2 q2 , so pq 1, 2) - pq p, - pq pq2 , so p( l + q) , pq( l + q) . 2 3) q(l + p), pq( l + p) . pq q, - pq p q, so 4) - pq p2 , - pq q2 , so p(p + q) , q(p + q) . 2pq, 5) pq pq, - pq pq, so 2pq. The equation is symmetric, so we have also: 6) pq( l + pq), 1 + pq. 7) pq( l + q) , p( l + q) . 8) pq( l + p) , q( l + p). 9) q(l + q) , p(p + q) . ( Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 45, Problem 3486)

X= y= X= y= x= y= X= Y= x= Y=

x = y = x = y = x = y = x = Y = x = y = x= y= x= y= x= Y= Tixtu=Andreescu,Y = -

-

-

-

Xn+l = 2xn + Yn

for all � l . By induction we obtain that

n

and

Yn+l = Xn + 2Yn

3n = xn2 - yn2 ' n _> 1 Denote by Pk the k-th prime number. Then x = xPlo ' Y = YPlo ' U = 3, v = Pk is a solution of equation UV for any integer k > O. Alternat ive solutixon2 -. Lety2 p= and q be two arbitrary primes, p � 3. Then pq = 2k + 1, for some positive integer k. Because 2k+ 1 = (k+ 1) 2 - k2 , it follows that all quadruples (pq + 1 pq - 1 ) . the equatIOn. . ( x, y, u, v ) = -- ' -2- ' P, q satIsfy D( orin Andri2ca) 23. The equation is equivalent to

(

x - y) (x - z) (y - z) = 2.

65

2.2. SOLUTIONS

2 Observe that (x - y ) + (y - z) = x -z. On the other hand, can be written as a distinct2 integers in the following ways product2 of three 1 1 i) 2= ( - ) · 2( - ) · , 1 1 ii) 2 = . . , 1 1 2 iii) = ( - ) · · ( - ) .

{ xx -- zy == l2l y-z= 2 { xX -- yz == l--1 y-z=l { xx -- yz == -21

Since in the first case any two factors do not add up to the third, we only have three possibilities: a) b)

c)

x,y, z) = (k + 1 , k , k - 1) for some integer k ; k k 1k 1k so (x, y,z ) = ( - , + , ) for some integer ;

so (

so (

x, y, z) = (k , k - 1 , k + 1) for some integer k .

=- . ( Titu yA-ndzreescu, Revista Matematidl. Timi§oara (RMT) , No. 1-2(1989) , pp. 97, 2 Problem ) 24. We have

xyz - (xy + yz + zx) + x + y + z - 1 = 1 , and, consequently, 1 1 1 1 ( x - ) ( y - ) (z - ) = . Because x, y, z are integers, we obtain I = y - 1 = z - 1 = 1, x 2 so x (=Tiy =Az =d . tu n reescu, Revista Matematica Timi§oara (RMT), No. 3(1971), pp. 26, Problem 487)

25. Multiplying by 8 yields

8xy (x2 + y2 ) = 16z4 or 4 (x y ) 4 = ( 2 z ) 4 , (x + y ) and so 4 24 4 ( x - y ) + ( z ) = (x + y ) . This is Fermat's equation for the case n = 4 and it is known that this equation has 2 0 0 solutions only if x -y = or z = (see T. Andreescu, "An Introduction 2002D., Andrica, to Diophantine Equations" , GIL Publishing House, 85-87) . _

_

pp.

2. NUMBER THEORY ± 0 Case I. x - Y = , then x = Y and Z = x. The solutions are x = y = m, z = ±m for any integer m. 0 0 O ( )4 ( )4 Case II. z = , then x - y = x + y and so x = or y = . The solutions are x = O , y = m, z = o and x = m, y = O, z = o for any m. Andreescu, ( Tituinteger 1 ( 1978) , Problem Revista Matematica Timi§oara (RMT) , No. 2813; Gazeta Matematica (GM-B), No. 11 ( 1981) , pp. 424, Problem 0: 264) ( ) ( ) ( ) 26. Consider the sequences Xn n �l ' Yn n �l ' Zn n � l , defined by Xn+2 = 5592xn ' Xl = 1 , X2 = 14 Yn+2 = 59922Yn ' YI = 73, Y2 = 4239 Zn+2 = zn , Zl = , Z2 = 1 for all n � . � + y� + z; = 59n , for all integers n � 1 . It is easy to check that x (Dorin Andrica, Romanian, Mathematical Olympiad - second round, 1979 , Revista 1 2 ( 1980) , pp. 58, Problem 4075) Matematica Timi§oara (RMT) , No. 66

If

If

27. Observe that the equation

( )

has infinitely many solutions in distinct nonnegative integers, for example

1

k � 0. (Xkl ,Ykl , Zkl ) and (Xk ' Yk ' Zk ) be two solutions of equation (1 ) with kl f. k2 . 2 2 2

for any integer Let Then

and multiplying yields

(Xk1 Xk ) n + (Xkl Yk )n + (Ykl Xk )n + (Ykl Yk ) n = (Zkl Zk )n- 1 2 2 2 2 2

This means that

2.2. SOLUTIONS

67

is a solution to the equation

kl f. k2

Since are arbitrary positive integers, the conclusion follows. For the second equation, the proof is similar, based on the fact that the equation has infinitely many solutions in distinct nonnegative integers, for example Xk

k � O. (Dorin Andrica)

for any integer

= 1 + kn ,

Yk

= k ( l + kn) ,

Zk

( 2)

= 1 + kn ,

x = 0, = 0 is a solution to the equation n + yn = xn -1 + yn- l x f. 1 Let a - be a rational number such that y = ax. Hence 28. It is clear that

Y

so

1 1 x = 1++aann- 1 ' y = a 1++aann-1 U sing the symmetry of the equation, we also have the solution 1 + an - 1 1 + an -1 x = a 1 + an , y = 1 + an f. 1 with a - rational. 1 , then again x = y = O. This concludes the solution. drica,n >Revista 2 (1981 ) , pp. 62, (Daor=in-A1nand Matematica Timi§oara ( RMT ) , No. 4578 ) Problem 3 2 3v 8 3 29. Let x = u , x + = , x + = t so U + + t = y. Then 3v - 3u = 2 and 3t - 3u = 8. It follows that 3U (3V-U - 1 ) = 2 and 3U (3t-u - 1) 8. 0 1 2 1 23 8 Hence = and 3v - = 3 , t - 1 = , therefore = , t = . 1 The is x = , y = . Andreescu, 2 ( 1978) , pp. 47, ( Titusolution Revista Matematica Timi§oara ( RMT ) , No. 281 2; Gazeta Matematica (GM-B) , No. 1 2 (1980) , pp. 496, Problem 18541 ) Problem l 2 1 l 1 l 30. If x, y � then + x and + y are both odd and (x + y ) is even. Hence If

v

=

u

the equation has no solutions.

v

2. NUMBER THEORY 1 . The equation becomes Consider the case x 2 ( 1 + y l) (1 + y) l 2 . If y � 3, then 3 divides ( 1 +y) l but and 2it( is notl ) difficult to notice the solution y 1 + y and y 1 does not2 satisfy the equality. not 2 or x , 1 due to the symmetry of the equation. 1 Hence x , y ( Titu Andreescu, Revista Matematidl. Timi§oara (RMT) , No. 2 ( 1977) , pp. 60, 3028; Gazeta Matematica (GM-B), No. 2 ( 1980) , pp. 75, Problem 0 : 118) Problem 3 1 . Without loss of generality we may assume that x � y � z. The equation is equivalent to l [x (x - 1 ) . . (z + 1 ) + y (y 1 ) . . . (z + 1) + 1] 2vl . z 3 3 If z2 � , then the right-hand side is divisible by but the left-hand side is not, so z � . We have two cases. 1 . Then we have l ! v I. z x + y 2 l - 1, or ! [x (x - 1 ) . . . (y + 1) + 1] 2vl - 1. y y1� 2, then the right-hand side is an even number but the left-hand side is odd, so y . Then l 2 ( 2V I - 1 - 1 ) . x 4 2 (2VI -1 - 1) 0 (mod 8) , false. If x � , then 1 , x 2, and x 3. the cases x It remains to examine V 2 2 ) ( x 21 , then 21 2 ( 2VII --11 - 11) , impossible. ! 2 2. If x 3 , then 2v orI v 3 , so v 3 x , then l -1 - 1 or v , false. Hence the only solution in this case is 2, 1 , z 1 , v 2 x 2. Now we have II. z xl + y l 2 v l - 2 , or l [x (x - 1 ) . . . (y + 1) + 1] 2 (2VI - 1 - 1 ) . y 4 2 (2VI - 1 - 1) 0 (mod 8) , false. If y � , the 1 , y 2 , or y 3. It follows that y l 2v 1 l - 3. Since x � 2 implies 2vl - 3 0 (mod 2) false, then If y 2, then x x 1, v 2 . l 2vl 4 3 so xl 4(2VI -2 - 1) . If y 3 , then x4 ( 2 V We must have v � . I-2 - 1 ) 0 (mod 8) , false. If x � , then 68

=

=

=

=

=

Y =

=

=

.

=

-

=

=

=

If

=

=

==

If

=

=

=

=

If

=

=

=

=

=

=

=

=

=

=

Y =

=

=

=

=

==

=

=

=

=

=

==

=

=

=

=

=

==

2.2. SOLUTIONS 69 2 3 1 Hence x3 , x l or2vx 8 and all those3cases lead to a contradiction. , then5 x l - . Then5 v � and xl 23 (2VI -3 - 1) � 623 7. It If y follows x � and because x does not yield a solution and x � implies 23 (2v! -3that - 1 ) 0 (mod 16) , which is false, we do not obtain a solution here. In case II we have found only x = 1 , y = 1 , 2, v 2 , which does not satisfy the condition x � y � To conclude, we have the solution - from case I x 2, Y 1 , 1 , v = 2 and, due to the symmetry of the equation, we also have x 1 , y 2, 1, v 2 and 1 , y 1 , 2, v 2. x ( Titu Andreescu, Revista Matematidl. Timi§oara (RMT) , No. 2 ( 1981) , pp. 62, 4576) Problem =

=

=

=

=

=

•

=

==

z =

=

z.

=

=

z =

=

=

z =

=

=

=

z =

=

32. The equation is equivalent to

or

[X2 . . . Xn - (n - l) x2 . . . Xn-l - . . . - 2X2 - 1] 1 . X I 1 Hence X l = and 2 [X3 . " Xn - (n - 1) x3 ,Xn-l - - 3X3 - 2] 2. X 2 and f:. Since X2 Xl, it follows that X2 Xf:.3 [X4 Xn -f:.(n - l) x4 Xn- l - 3 - 4X4 - 3] 3. Because X 3 X2 and X3 Xl , we obtain X3 . Continuing with the same procedure we deduce Xk k for all k. Remark. Turning back to the equation we find thethatidentity 1 + 1 1 . 1 + 2! . 2 + . . . + (n - 1) ! (n - 1 ) nIno ( Titu Andreescu, Revista Matematidl. Timi§oara (RMT), No. 3 ( 1973) , pp. 23, 1 509) Problem 1 the claim is true. Using the identity 33. We proceed by induction on n . For n (X2 - y2 )(U2 v2 ) = (xu + YV) 2 - (xv + YU) 2 =

• •

•

. •

=

•

. •

=

=

.

.

. •

•

.

=

=

=

=

_

70

2. NUMBER THEORY

and the fact that the claim holds for we deduce that the property is valid for n + as desired. We have n n n = + = Pn =

n

Alternative solution. II (a� b� ) II (ak - bk ) II (ak bk ) k= l k= l k= l

1,

-

where An , Bn are integers. Hence as

claimed .

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 ( 1975) , pp. 45, 2239; Gazeta Matematica (GM-B ) , No. 7( 1975) , pp. 268, Problem 1 5212) Problem 34. Subtracting the equalities yields

(x + y - xy) + (v + y - vt) = (x + y - xy)(v + t - vt) , or [(x + y - xy) 1][(v + t - vt) - 1] 1 , so (1) (1 - x)(1 - y)(1 - v)(1 t) 1 . It follows that 1 1 - xl = 1 1 y l = 1 1 v i 1 1 - tl = 1 , ( 1) we obtain and using (x, y, t) (0, 0, 0, 0) , (0, 0, 2, 2) , (0 , 2, 0 , 2) , (0 , 2, 2, 0) , (2, 0 , 0 , 2) , ( 2, 0 , 2, 0) , (2, 2, 0 , 0) and (2, 2, 2, 2) . Turning back to the system we obtain (x, y, v, t) (0 , 0, 0 , 0 , 0) , (0 , 0 , - 4, 2, 2) , (0 , 2, 0 , 0, 2) , (0 , 2, 0 , 2, 0) , (2, 0 , 0 , 0, 2) , (2, 0 , 0 , 2, 0) , (2, 2, - 4, 0, 0) and (2, 2, 24, 2, 2) Ti A d (RMT) , No. 2 ( 1978) , 46, ( tu343n reescu, Revista Matematica Timi§oara 8 5 1 ; Gazeta Matematica (GM-B) , No. ( 19 1 ) , 216, Problem 18740) Problem 35. We start withd ab useful llemma. ively prime positive integers, then there are positive Lemma. I f a an are re at integers u and v such that au - bv 1 . =

-

-

-

z,

z,

-

=

=

=

=

pp .

=

pp.

2.2. SOLUTIONS

Proof. Consider the numbers

71

1 . 2, 2 · a, . . . , (b - 1) . a

(1) b Whenk divided by the remainders of these numbers are distinct. Indeed, otherwise b k f. 2 E {I, 2, . . . , - I} such that we have l kl a = Pl b + r, k2 a = P2 b + r for some integers PI , P2 ' Hence b Ikl - k2 1 0 (mod b) , which is Since a and Ikare relatively prime it follows that 1 l - k2 1 < b. false because b On the otherkhand, none of the numbers listed in (1) is divisible by . Indeed, if so, then there is E {I, 2, . . . - I} such that k . a = P . b for some integer p. k k = kl d, P = PI d, for d Let be the kgreatest common divisor of kand p. Hence b k d( ) 1 . Then l a = PI and since ged(a, b) = 1, we = p , with ge some kintegers l l I' PI b k b have I = , PI = a. This is false, because l < . follows that one of the b so Itthere b numbers from (1) hasb the remainder 1 when divided by is U E {I, 2, . . . , - I} such that au = v + 1 and the lemma is proved. We prove now that the system { axbx -- ytyz+-de == 0O b d d( b) with a, , e, nonnegative integers and ge a, = 1 has at least a solution in nonneg­ ative integers. d( b) Because ge 1 a, = 1 using the lemma, there are positive integers u and v such b that au - v = . Hence x = eu + dv, y = ad + be, z = v, t = u, is a solution the system. ( Titu Antodreescu, Revista Matematidl. Timi§oara (RMT) , No. 2(1977) , pp. 60, ==

::;

,n

Problem 3029)

36. Consider the determinant

� = 1l 12 Xpl - l p2 - l X

X

X

1

P

p = II ( pp-l

X

X

i , i=l i >i

Xi - Xj )

72

2. NUMBER THEORY

Summing up all columns to the first one and applying the hypothesis yields

�

==

0 p ) ) II (mod p , hence (Xi - Xj ) 0 (mod p . Because p is}a prime number, it follows that) there are distinct positive integers 2 {I k, E( D , i , .A. . ,p such that Xk - Xl 0 (mod p . d i ) or n n r ca ==

i,j=l i>j

==

1

37. Let

a = gcd(al' a2 , . . , an ) and b - gcd ( al +2 a2 ' a2 +2 a3 , . . . , an +2 al ) 12 Then ak = aka, for some integers ak, k = , , . . . , n. It follows that ak +2ak+l = ak +2ak+l a, (1) where an+l = al and an+l a l' Since ak are odd numbers, ak are also odd, so ak +2ak+l are mtegers. . ak +2ak+l for all so a divides b. From relation (1) it follows that a divides ak +2ak+ l = f3k b, for some mtegers . f3k. Then On the other hand, 2b) ak + 0 (mod ak+ l { I 2 } Summing up from k = 1 to k = n yields for all k E , , 2 (al + a2 + . . . + an) 0 (mod 2b) 2) Since al, a2 , . . . , an are all odd al + a2 + . . . + an � 0 (mod , hence (3) al + a2 + . . .2+ an 0 (mod b) . Summing up for k = 1, 3, . . . , n - implies al + a2 + . . . + an- l 0 (mod 2b) and furthermore al + a2 + . . . + an- l 0 b)(mod b) . (4) 2 ) ( Subtracting (4) from (3) implies an 0 (mod , then using relation we obtain ak ( Ti0 (modA db) for all k. Hence bl a and the proof is complete. tu28 1n4)reescu, Revista Matematica Timi§oara (RMT) , No. 1(1978) , pp. 47, Problem ·

=

==

. . .,n .

==

n,

==

==

==

==

==

38. It is known that

cp(kl) = cp(k)cp(l) for any relatively prime positive integers k and

l.

2.2. SOLUTIONS 73 On the other hand, it is easy to see that if P is a prime number, then (pl ) pI _ pl - I

n - 1, n � 1. (n + 1) Then 2 n PI + P2 + . . . + Pn > 2 - n n2 . The inequality al" + a;" : . . . + a;:' ;:: ( a1 + "" : . . . + an ) m holds for any positive real numbers aI, a2 , . . . , an . Hence m m+ 1 : l" ) . . + Pn r p . + ( : J n , >n P + pr + . . . + P::' ;:: n ( as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 45, 3483) Pro blem S L d and let

=

=

=

=

=

=

=

=

1'2

=

d <2n gcd(d, n ) = l

=

2. NUMBER THEORY

74

Note that

d(n, d) = 1 gcd(n, n - d) = 1 gcd(n, n + d) = 1 (1) gc d1, d2 , , dcp(n) be the numbers less than n and relatively prime with n. From ( 1 ) weLetdeduce that d1 + dcp(n) = n d2 + dcp(n) - l = n {:}

{:}

• • •

hence d
d = ncp2(n) .

On the other hand,

L d = L (n + d) = ncp(n) + L d = (n) 3ncp(n) ( ) ncp 2 - 2 . - ncp n + -Therefore (n) 3ncp(n) 2 ( ) ncp s = 2 + 2 = ncp n . 2 ( 1981 ) , pp. 61 , (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 4574) Problem 3 42 . We proceed by induction. For n = ! the claim is true. Assume that the 1 1 hypothesis holds for n - . Let < k < n and let kl , q be the quotient and the 0 � q < n and 0 � kl < � < = remainder when k is divided by n. Hence k k i n + q, n n-! = (n - I)! . n From the inductive hypothesis, there are integers di < d� < . . . < d� , s � n - 1di, suchd�that dil(n d�- I) ! , i = 1 ,02, . . . , s andd kl =d di + d� d+ . . . + d�d. Hencedi + . . . + n + q. If q = ,! then k = 1 + 2 + . . . + s , where i = n , ik== 1n, 2, .+. .n, s, are distinct divisors of dn . 0 d If q f. , thend kl = 1 + 1 2+ . . . + s+1 ,dwhereI ! di = ndL iI , 2, . . . , s, and ds+1 = q. ! i It dS+1is clear d thatd i n , =d , , . . . , s and d s+1 n , since diq < dn. On the other hand, < 1 < 2 < . . . < s , because S+1 = q < n ! � n = 1 . Therefore k can be written of at most n distinct divisors of n , claimed. 2 ( 9 83 ) dreescu, ( Titu Aansum 1 , pp. 88, Revista Matematica Timi§oara (RMT) , No. C4 1 0) Problem : d
n < d <2n gcd( d , n ) = l

_

d
_

rh

as

as

75

2.2. SOLUTIONS 43. If

n � 992, take the set with all 992 odd numbers from {I, 2, . . . , 1984}.

Its complementary set has only even numbers, any two of them not being relatively prime. Hence n � 991. Let c be the complementary set of a subset with 991 elements of the set {I, 2, . . . , 1984}. Define D = {c + 11 c E C } If C n D = 0, then C U D has 2 · 993 = 1986 elements but C u D C {I, 2, . . . , 1985}, which is false. Hence C n D f. 0, so there is an element a E C n D. It follows that a E C and a + 1 E C and since a and a + 1 are relatively prime we are done. Revista Matematica Timi§oara ( RMT) , No. 1(1984) , pp. 102, Problem C4:7) .

( Titu Andreescu,

n, hence n < (a + 2) 2 . If a � 7, then a - 4, a - 2, a are odd integers which divide n. Note that any two of these numbers are relatively prime, so (a - 4) (a - 2)a divides n. It follows that (a - 4) (a - 2)a < (a + 2) 2 Oso a3 - 6a2 + 8a < a2 + 4a + 4. Then a3 - 7a2 + 4a - 4 � 0 or a2 (a - 7) + 4(a - 1) � . This is false, because a � 7, hence a = 1, 3 or 5. If a 1, then 12 � n < 32 , so n E {I, 2, . . . , 8} . If a = 3, then 32 � n < 52 and 1 . 31n, so n E {9, 12, 15, 18, 21, 24}. If a 5, then 52 � n < 72 and 1 . 3 . 51n so n E {30, 45}. Therefore n E {2, 3,( D4, 5,i 6, A 7, 8, 9, 12, 15, 18, 21, 24, 30, 35}. or n ndrica and Adrian Ghioca, Romanian Winter Camp 1984; Revista 44. Let a be the greatest odd integer such that a2 <

=

=

P.

Matematica Timi§oara (RMT) , No. 1(1985) , pp. 78, Problem T.21) 45 . We prove by induction that

u� - 2v� = 1, For have

n = 1 the claim is true. Assuming that the equality is true for some n, we

U�+ l - 2V�+1 hence (1) is true for all We prove now that

=

n�

(3u + 4Vn ) 2 - 2(2un + 3V ) 2 = u� - 2v� = 1 1.

n

n

Y

2x� - � = 1,

Indeed,

n n

n

(2)

n�1

2x� - Y� = 2(u + V ) 2 - (u + 2v ) 2 as

(1)

n � 1.

n

=

u� - 2v� = 1,

claimed. It follows that

(Xn V2 - Yn) (Xn V2 + Yn) 1, n � 1. Notice that xn ..,fi + Yn > 1 so 0 < xn V2 - Yn < 1, n � 1. Hence Yn [Xn V2] , as claimed. =

=

2. NUMBER THEORY (Dorin Andrica, Gazeta Matematica (GM-B) , No. 11 (1979) , pp. 430, Problem 0 :97)

76

46. ( i) We have

hence

Xk = 22k -1 + 1 = 22k-2 .2 + 1 = (Xk-l - 1 ) 2 + 1 = X2k-l - 2Xk- 1 + 2, Xk(1 )- 2 = kXk- l2(Xk-kl - 2) .

Multiplying the relations

from = to = n yields

(1)

X l = 3, we obtain 2 (2) = X l X 2 . . . Xn - l + X n 2. for all n For a different proof use the identity x2n - 1 -1 I = nII-l (X2k- 1 + 1) . X - k=l d( d d(xn' X2 ) = . . . = ( ii ) From relation (2) we obtain gc x x gc n' gck d(xnl, Xn - l ) = 1 for all n, hence gcd(Xk ' xz ) = 1 for all distinct positive integers and . (2) it follows that Xn (iii ) Since X2 = 5 and X I X2 . . . X n - l is odd, using relation end with 7 for all integers n 2:: 3. (Doridigit n Andrica) and since

2::

47. We have

Then and

a�+2 + a�+l - 4an+2 an+1 = - 2, Subtracting these relations yields a�+2 - a� - 4an+l (an+2 - an ) = and (an+2 - an)(an+2 + an - 4an+d = 0, n 2:: 1 . ( ) Since the sequence an n 2:1 is increasing, it follows that °

77 2.2. SOLUTIONS Taking into account that al = 1, a2 = 3 and inducting on n we reach the conclu­ sion.( Ti A d tu n reescu, Gazeta Matematica (GM-B) , No. 11(1977), pp. 453, Problem 2 16947; Revista Matematica Timi§oara (RMT ) , No. 1(1978) , pp. 51, Problem 840) 2 2 2 2= 48. For n = we have a2 = 1+ 3" ' d( ) Suppose that an = Pn , where Pn , qn are integers and gc Pn' qn = 1. qn Then 2pnqn an+ 1 = qn2 + 2P2n ' d(2 � 2 �) and it suffices to prove that gc Pnqn' q + p = 1. d kk Assume by way of contradiction that there are integers and l' 2 such that 2Pnqn = kI d and q� + 2p� = k d. 2 2 2 2k � � d If = , then q = 2 - p , so qn is even. This is a contradiction since ql , q2 , ·d. . ,q2n are all� odd.(k k )d dl If > � then 2 �q =k d2 qn - lPn sodl qn' d( ) Since q + p = 2 it follows that Pn , which is false, because gc Pn' qn = 1. This proves our claim. We prove that � - 2p� = 1, n 2:: 2. (1) q 2 2 2 2 � � For n = we have q �- p = 32 - . 2 = 1. 2 � Suppose that q - p = 1. Because

is irreducible, we obtain Hence

qn2 + 1 - 2Pn2 +1 = (qn2 + 2P2n)2 - 8p2nqn2 = (qn2 - 2P2n) 2 = 1 , claimed. Hence �I bn = 21 1 q -2P2 - qn2 - I = 2 2 2 1 - an2 - I q P I 1 n n - - n- I 1 2 q b 2'rl- I and (soTi n isA adperfect square for all n 2:: . tu n reescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 63, as

_

--::-_.:.:-=-=_

_

Problem 3083; Gazeta Matematica ( GM-B ) , No. 1(1981), pp. 44, Problem C :88)

2. NUMBER THEORY

78

49. Inducting on n we obtain

xn2 - 2Yn2 - 1 , Hence Zn = 1 + 4X2n Yn2 = (X2n

_

2yn2 ) 2 + 4X2n yn2 = ( 2n + 2yn2 ) 2

= (x; + 2y� + 2xnYn ) (x; + 2y�

X

_

4X2n yn2 =

- 2xnYn) , for all n � 1.

Since both factors are greater than 1, it follows that all numbers Zn are not prime. Revista Matematica Timi§oara (RMT) , No. 2(1976), pp. 54, Problem 2571)

( Titu Andreescu,

50. We have

or Hence and

Xn Jp2 + 1Xn+ 1 ± PJX;+1 + 1. From hypothesis we have X n+l > Xn , hence Jp2 + 1Xn+ 1 -pJX�+ + 1. Xn = 1 =

(1)

But

(2) and so (3) by summing up the relations (1) and (2) . 2 + 1 is irrational. From (3) Because is a positive integer, it follows that we deduce that among any three consecutive terms of the sequence there is at least an irrational term. Hence there are at least irrational terms among the first m terms of the sequence. Revista Matematica Timi§oara (RMT) , No. 1-2(1980), pp. 68, Problem 4139; Gazeta Matematica ( GM-B ) , No. 6(1980), pp . 281, Problem C:48)

p

[;]

( Titu A ndreescu,

5 1 . Setting n = 0 and n = 1 yields From the given condition we obtain

Jp

Xl = x� and X2 = x� , hence Xl X2 X2n+ 1 = 2n+l + Xn2 and X2n Xn2 +l - X2n-l . X

=

=

=

1.

2.2. SOLUTIONS

79

Subtracting these relations implies hence We induct on

(1) (2)

X2n+1 = X2n + X2n- l , n � 1 .

n to prove that � 1. n , X2n = X2n-1 + X2n-2 (2)

X2 = Xl + Xo and assume that is true up to n . Then X2n+2 - X2n = Xn2 +2 - Xn2 - Xn2 + l + X�n2 - l (*)= (Xn+ l + Xn ) 2 - Xn2 - Xn2 +l + + (Xn+l - Xn ) 2 = X + l( 1+) X� = X2n+ l , and the induction hypothesis) . as claimed (the equality because of O. Because ( 1 ) and( )(2holds ) it follows � From relations that = for all n x X X + +2 + n n n l 0 Xo = and Xl = 1 the sequence (xn ) n 2:0 is the Fibonacci's sequence, hence Indeed,

*

( Titu Andreescu, Romanian Winter Camp 1984; Revista Matematidi Timi§oara 1 ( 1985) , pp. 73, Problem T.3) (RMT), No. 1 2 25 48 52. From the hypothesis it follows that a4 = , a5 = , a6 = . We have �l , a22 = 1 , a33 = 2 , a44 = 3 , a55 = 5 , a66 = 8 so an = Fn for all n = 1 , 2 , 3 , 4, 5 , 6 , where (Fn ) n 2: 1 is the Fibonacci's sequence. F n kFk We prove by induction that a for all n. Indeed assuming that a = = n k n n k 3 for � n + , we have 2 (n + l) Fn+1 - nFn = a2 n+4 = F2 (n + 3) Fn+32 +F(n + 2)F +2 n = (n + 3) n+32+( (n +3)F) n+2 -2F2 (n + l()Fn+1 -Fn(Fn+2 - Fn+l ) = = 3nF+ n+2F3 + n+( 2 - 2n +F 2) n+1 = 2 ( = n + ( ) n+4)3 +F n+2F- n + () ( n+4 3F- Fn+2 ) = = n + ( n+3 + n+2 ) = n + ) n+4 as desired. (Dorin Andrica, Revista Matematidl. Timi§oara (RMT) , No. 1 ( 1986) , pp. 106, C8 2) Problem : O 53. Observe that setting Xo = 0 the condition is safisfied for n = . k that Xk divides m. Let rt be the We prove that there is integer � m3 such 2 0 1 3 remainder of Xt when divided by m for t = , , . . . , m + . Consider the triples (ro , rl , r2 ) , (rl ' r2 , r3), ' . . , (rm3, rm3+ 1 , rm3 +2 ) ' Since rt can take m values, it follows -

-

-

-

-

-

'

2. NUMBER THEORY by the Pigeonhole Principle ( that at least) two triples are equal. Let p (be the smallest) number such that triple rp , rpO+l, rp+ 2 is equal to another triple rq, rq+l, rq+ 2 , p < q :::; m3 . We claim that p = . Assume by way of contradiction that p 2:: 1. Using the hypothesis we have rp+2 Tp- l + rprp+l (mod m) and rq+2 rq -l + rqrq+l (mod m) . rp d= rq,( rp+l = rq+l) and rp+2 = rq+2 , it follows that rp-l = rq - l so (rp- Since with the minimality of p. , which is a contradiction l ,rp ,rp+0 rq- l ,rq,rq+l 0 0 Hence = = ) m (mod Xq therefore and , r = rq so , ( Tiptu Andreescu ando Dorel Mihet, Revista Matematica. Timi§oara (RMT) , No. 106 , Problem C8: 1) 1(1986), pp. F6 , al = F12 , a2 = F? , 4 5 = so 25, = a 9, = 54. a 4, = a3 1, = a that Note ao 2 F F F F sequence. a3 = i , a4 = ; , a5 = i , where ( n )Fn2:o� is the Fibonacci Ff O We induct on n to prove that an = for all n 2: . Assume that ak = for all k :::; n . Hence (1) an - Fn2 ' an -l - Fn2- l ' an-2 = F�_2 ' 80

==

==

=

==

From the given relation we obtain

and

an - 3an- l + an-2 = 2( - l )n- r ,

n 2: 2.

Summing up these equalities yields

(2) Using the relations (1) and (2) we obtain

an+l = 2F� + 2F�_1 - F�_2 (Fn + Fn_ 1 ) 2 + (Fn - Fn_ 1 ) 2 - F�_2 = =

as desired.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1986), pp. 108,

Problem C8:8)

a2 - 1 and 2 + 2a ask for the substitution a � (b2 + b� ) . 2 The equality � ( 2 + � ) = 97 implies ( + �) = 196, hence + � = 14. Setting 55. The expressions b

b

=

b

b

81 2.2. SOLUTIONS 2 2 = c yields (c + �) = 16, thus c + � = 4. Let c = 2 + va. We will prove by induction that F a = � ( C4 . + 4�. ) , n 2: 1, n C F where n is the nth Fibonacci number. Indeed, this is true for n = 1, n = 2 and, assuming that 4 41_ ) ' k <_ n a k = �2 (e Fk + _ e Fk b

implies

and

2 + J2 + 2an = 2 + e2Fn + 21 = (eFn + -1 ) 2 e Fn eFn Let for all positive integers m. +� dn reescu,thatUSAemMathematical ( TituusAmention em is an integer Olympiad Shortlist, 1997) 56. We induct on m to prove that km n < km+1 , then an = 1 + m. (1) if k For m = 0 the claim is true, since if 1 n < , then [ �] = 0 hence an = l + ao = 1 + 0. k k k k Assume that (1) holds for m. Then if m+1 n < m+2 , we have m � < m+ 1 , so km [�] < km+1 . --

�

�

�

�

�

Using the inductive hypothesis, we deduce that a [�] = 1 + m so

an = 1 + a [ �] = 1 + (m + 1)

as claimed. Therefore for any positive integer n with m � logk n < m + 1, we have = 1 + m thus 1 + [log kn]. Revista Matematica Timi§oara (RMT) , No. 2(1982) , pp. Problem 4997)

an

(aTintu= Andreescu,

an - 2 = b � ,

57. We prove that where bo all n 2:: 1. This is clear for n = 0 and n = 1.

66,

= -1, b1 = 1 and bn+1 = 3bn - bn- 1 for

2. NUMBER THEORY 82 b� . Subtracting the relation an = k Assume that for any = � n we have ak -2 7an- l - an-2 from an+l = tan - an l yields an+l = 8an - 8an l + an 2 or - 8( - 8 ( ( an+l -2 = an -2) - an-l - 2) + an-2 -2) Hence an+l -2 = 8b� - 8b�b_ 1 + b�b�_2 = 8b�3b- 8b�b _1 + (3bn-b� 1 - bn )2 = 9b� - 6bn n-1 + l = ( n - n_ 1 ) 2 = +l = Alternative solution. The general term- of the sequence is given by ( v'5 ) 4n - 2 ( 1 v'5 ) 4n-2 1 _ + an - 2 + 2 ' n 2:: 0 _

Hence

( 1 + V5) 2n-l + ( 1 - V5) 2n-l 2 2

and an inductive argument proves that

is an integer for all n.

( Titu Andreescu)

58. Note that all terms of the sequence are positive integers. Assume by way of contradiction that there is a positive integer no such that

X�o = Xno - lXno +l ' Then Xno +l _- xno - t xno Xno - l where t is rational. (3 Because Xno +l o:xno + Xno- l, it follows that Xno+l 0: + (3 � xno Xno - l or t2 - o:t - (3 = O. --

--

,

=

=

� 0:2 4(3 is not

The last equation has no rational roots since the discriminant = + a perfect square. This is a contradiction and hence the problem is solved. From the condition of the problem we obtain

Alternative solution.

(1)

83

2.2. SOLUTIONS

Assume by way of contradiction that there is a positive integer no such that 2 _ X l X no+l Xno - no '

() (a2 + 4,8)x�o = (Xno+l + ,8xno - d2 4,8 is not a square. This completes the proof. 2+ is false, since a which A (Dorin ndrica) From the relations 1 and (2) we deduce that

(2 )

59. Assume by way of contradiction that there is an irrational number a such

that is rational. Define a = \ja + a2 - 1 and observe that \ja - a2 - 1 = � .

J

J

Hence A = a + � is rational.

a

a

We prove that an + �n is rational. a Indeed, 1 2

(

is rational and a3 +

U sing the identity

:3 = (a + � r

(

1

= ak - 1 + _ ak + � ak a k -1

)

1 2

a + _2 = a + a a

-

-2

3 (a + �)

is rational.

a+

ak - 2 + _ ak- 2

) ( �) ( _

a

1

)

k,

it follows by induction that ak + -\- is rational for all positive integers hence a 1 . . 1 n l a +IS rat Ona . an Thus a + a2 - 1 + a - a2 - 1 = 2a is rational, which is false. The solution is complete. Romanian IMO Selection Test 1977; Revista Matematidi Timi§oara RMT , No. 1 1978 , pp. 78, Problem

J J ( Titu Andreescu, ( ) ( )

3344) 3 4 60. We induct on n. For n = 1, from ±1 ± 2 ± ± ± 5 we obtain all odd positive )( integers less than or equal to (2 + 1 4 + 1) = 15: + 1 22 + 33 + 44 55 31 1+ + + 1+2+3 4+5 5 -

-

-

-

-

=

-

=

=

84

2. NUMBER THEORY

- 11 + 22 - 33 + 44 + 55 = 97 -1 - 2 +3 +4 +5 = 11 +1 - 2 + 3 + 4 + 5 = 13 - 1 + 2 + 3 + 4 + 5 = 15 + + ± +(4 + 1 )+ = ± ± 2 ± 1 . . . n + with suitable Assume that from ) . + and (2n +choices 1)(4n of+ 1signs we obtain all odd positive integers less than or equal to (4n+ 2) + (4n+ 3) + (4n+ 4) - (4n+ 5) = O . Hence from ± 1 ±2± Observe that . . . ± (4n + 5) for suitable + and - we obtain all odd positive numbers choices of signs 2 )(4 ( ) 1 1 less than or equal to n + n+ . It suffices to obtain all odd integers m such that (1) (2n + 1 )(4n + 1 ) < m :s; (2n + 3)(4n + 5) (2n + 3)(4n + 5) - (2n + 1 )(4n + 1) 8 7 There are 2 = n+

such odd integers m. We have

(2n + 3)(4n + 5) = + 1 + 2 + . . . + (4n + 5) (2n + 3)(4n + 5) - 2k = + 1 + 2 + . . . + (k - 1 ) - k+ + (k + 1 ) + . . . + (4n + 4) + (4n + 5) , k = 1 , 2, . . . , 4n + 5 and (2n + 1 )(4n + 5) - 2l = + 1 + 2 + . . . + (l - 1) - l + + (l + 1) + . . . +( 1 )(4n + 4) + (4n + 5) , l = 1 , 2, . . . , 4n + l. Hence from are obtained, desired. Andrica,mGazeta 2 ( 1986) , pp. 63, Problem C570) ( Dallorinnumbers Matematica ( GM-B ) , No. as

Chapter 3 GE OMETRY

PROBLEMS 1. Triangle

ABC has side lengths equal to a, b, c. Find a necessary and sufficient

condition for the angles of the triangle such that a2 , b2 , e can be the side lengths of a triangle. 2. Prove that the triangle whose side lengths are equal to the length of the medians

ABC has area equal to 3/4 of the area of triangle [ABC] . ABC the median AM meets the internal bisector BN at P. Let Q 3. In triangle C AB . Prove that triangle BNQ is isosceles. be the point of intersection of lines P and ABC the midline parallel to AB meets the altitudes from A and B 4. In triangle D E AC intersects the altitudes from A and C at points F and G . The midline parallel to at points andDC. II BF I GE Prove that . ABC. Lines AM, BM, CM intersect LetCA M be a point in the interior of triangle 5.BC ' ' ' A sides , , AB at points 'B ,AB' C, C , respectively. ' A Denote 'A by 81 , C82'B, 83 , 84, 85, 86 ' A B C B C the areas of triangles M , M , M , M , M and M , respectively. of a triangle

Prove that if

81 82

+ ABC+ 83 84

85 - 3, 86 -

then M is the centroid of the triangle

. ABC and let M, P, Q be three 6. Let M be a point in the interior of triangle AB, BC and line CA. colinear points on sides Prove that if

AC+

SMAN 8MBN

then MP is skew-parallel with

5MBP = 8M C P

.

2J

SMAQ , 8McQ

ABC. Prove c and 8 be the sides lengths and the area of a triangle P ABC such that that if is a point interior to triangle aPA + bPB + cPC = 48 7. Let a, b ,

87

3. GEOMETRY

88

then P is the orthocenter of the triangle. 8. Let 11 , 12 be the incenters of triangles A l Bl Cl and A2 B2 C2 . Prove that if II and 12 divide the internal bisectors Al A� , A2 A� in the same ratios they divide the internal bisectors Bl BL B2 B� , then triangles A l Bl Cl and A2 B2 C2 are similar. 9. On side BC of a triangle ABC consider points M and N such that ifAjJ

CAN.

Prove that

MB MC

==

NB 2: 2 AB + NC AC

1 0 . Let M be a point on the hypotenuse BC of a right triangle ABC and let points N, P be the feet of the perpendiculars from M to AB and AC. Find the position of M such that the length NP is minimal. 1 1 . Let ABC be an equilateral triangle and let P be a point in its interior. Let the

lines AP, BP, CP meet the sides BC, CA, AB at the points AI , Bl , Cl respectively. Prove that

1 2 . Let S be the set of all triangles ABC for which

5( 1 + 1 + 1 )3 6 min{P, BQ, CR} = r ' AP BQ CR

where r i s the imradius and P, Q, R are the points of tangency of the incircle with sides AB, BC, C A, respectively. Prove that all triangles in S are isosceles and similar to one another . 1 3. Let ABC be a triangle inscribed in a circle of radius R, and let P be a point in the interior of ABC. Prove that PB PA PC BC2 CA2 AB2 R '

+

+

2: 1

1 4 . Let I b e the incenter of an acute triangle ABC. Prove that

AI · BC

+ BI · CA + CI · AB = SABe

if and only if triangle ABC is equilateral. 1 5 . Let ABC be a triangle such that

(8AB - 7BC - 3CA) 2 = 6(AB2 - BC2 - CA2 ).

Prove that A = 60°.

3 . 1. PROBLEMS

89

P be a point in the plane of triangle ABC such that the segments PA, 16. Let PB and PC are the sides of an obtuse triangle. Assume that in this triangle the PA. Prove that BAlJ is acute. obtuse angle opposes the side congruent to ABC has the following property: there is an interior point P such 1 7. Triangle ° 0 20° , PcA = 30° , and PAC = 40° . Prove that the triangle 1 that PAJj = , PEA = ABC is isosceles. ABC be a triangle such that max{A, B } = C + 30° . Prove that ABC is 18. Let right angled if and only if !!. = J3 + 1 . r ABC there is a point P such that 1 9 . Prove that in the interior of any triangle PAB, PBC, PCA are equal. the circum radii of triangle 20.CThe ' of the triangle ABC touches the sides AB , BC, CA at the ' A' incircle B Prove that the perpendiculars from the midpoints of 'B' B' C, ' C, 'A,' respectively. Apoints , , to AB , BC , CA, respectively, are concurrent. I. Let Ci , i = 1 , 2 , 3 , AAA 2 1 . Let I 2 3 be a non-isosceles triangle with the incentre Ai Ai+1 and Ai Ai+2 (the addition of indices be the smaller circle through I tangent to CHI and Ci+2 . 3) B i 1 2 3 being mod . Let i , = , , , be the secondApoint of intersection of BI ABI ABI Prove that the circumcenters of the triangles I I , 2 2 , 3 3 are collinear. AB and AC of a triangle ABC consider points B' and C' such that 22. On sides AB' ' AC' ' k B B + C C = is constant. BC' and CB' . Find the locus of the intersection point of lines ABC be an equilateral triangle of side length 1 . Find the locus of points 23. Let P such that PA · PB · PC PA PC PB { } 2 max , , = PA . PB + PB . PC + PC · PA - 1 24. Prove that in any acute triangle

Ja2 b2 - 482 + Ja2 c2 - 482 = a2 .

25. Prove that a triangle in which r,;.-

y. Ta

is equilateral.

r,;::

�

+ y rb + y rc = Jrarbrc -r .

.

-'--

3. GEOMETRY

90

26. Prove that a triangle is equilateral if and only if

+ + --+-+-

1 1 1 1 1 1

-

m2a

-

m2b

-

rb2

m2c - ra2

. rc2

1 a,bb, c has circum radius equal to . Prove that a + + c ;::: abc.

27. A triangle with side lengths

28. Prove that in any triangle

a)(2 ) � 6R b � (p - p - c -;:-. 29. Prove that in any triangle 1 1 1 1

+ + h2b

h2a

30. Prove that in any triangle

Jr arb 31 . Prove that in any triangle

+

Jrb rc

h2c

+

;:::

3r2

Jrc ra ;::: gr.

32. Prove that

for any triangle .

� ib if and only if (p - b)(p - c) � 4e '

33. Let ABC be a triangle. Prove that A

34. Three equal circles of radii r are given such that each one passes through the

centers of the other two. Find the area of the common region. 35. Let ABCD be a nonisosceles trapezoid with bases AB and CD. Prove that

+

AC 2 - BD 2 AB CD AD 2 _ BC2 - AB - CD ' 36. Let ABC D be a trapezoid with bases AB and CD. Prove that if

(AB

+

CD) 2 = BD 2 + AC2

91

3.1. PROBLEMS

then the diagonals of the trapezoid are perpendicular. 37. Prove that if in a trapezoid with perpendicular diagonals the altitude is equal to the midline, then the trapezoid is isosceles. 38. Let ABCD be a trapezoid with bases AB and CD. Prove that

AB 2 - BC2 + AC2 CD 2 - AD 2 + AC2

AB AB2 - AD2 + BD2 = CD 2 _ BC 2 + BD2 = CD '

39. Prove that a trapezoid whose difference of the diagonal lengths is equal to the difference of nonparallel side lengths is isosceles. 40. Let ABCD be a cyclic quadrilateral. Prove that

lAB - CD I + l AD - BC I 2: 21AC - BD l · 41 . Prove that a right trapezoid whose altitude length is equal to the geometrical

mean of the lengths of its bases has perpendicular diagonals.

E and F be the projections of the vertices A and B of a trapezoid ABCD E and F onto BD and AC, respectively of on line CD . Let M and N be the projections E and let P and Q be projections of and F onto BC and AD, respectively. 42. Let

Prove that the quadrilateral MN PQ is cyclic.

43. Let ABC D be a convex quadrilateral that is not a parallelogram and let M

and N be the midpoints of the diagonals AC and BD . Prove that numbers AB + CD,

BC + AD,

AC + BD,

2MN

can be the side lengths of a cyclic quadrilateral.

E. E Point F is the reflection of C across . Prove that lines AF and B D are perpendicular 44. Let ABCD be a cyclic quadrilateral and let lines AB and CD meet at point

if and only if lines AB and CD are perpendicular.

45. Find all cyclic quadrilaterals having the side lengths odd integers and the

area a prime number. 46. A cyclic quadrilateral has the side lengths a,

and the semiperimeter p . Prove that

b, c, d, the diagonal lengths

e,

f

92

3. GEOMETRY 47. A cyclic quadrilateral has the side lengths a,

I e and .

Ib

{

:::; 1 , then II :::; V2. Ie -

Prove that if max l a - c l , - d l }

b, c, d and the diagonal lengths

48. A cyclic quadrilateral has area S and semiperimeter p . Prove that if S =

then the quadrilateral is a square.

(�) 2 ,

49. Let ABCD be a convex quadrilateral and let P be the intersection point of its diagonals. Prove that SPAB + SPCD = SPBC + SPDA if and only if P is the midpoint of AC or BD. 50. Let M be a point on the circumcircle of the cyclic quadrilateral ABCD and

let points A', B', C', D' be the projections of M onto AB, BC, CD, DA, respectively. Prove that (i) lines A' B' , C'D' and AC are concurrent (ii) lines B'C' , D' A' and BD are concurrent. 5 1 . A convex quadrilateral ABCD with area 2002 contains a point

P in its interior

such that PA = 24, PB = 32, PC = 28, and PD = 45. Find the perimeter of ABCD . 52. Find the locus of points P in the plane of a square ABC such that

max(PA, PC) =

1

V2

(PB + PD).

. LetA P be the set of all quadrialterals with the same diagonal lengths and let (P)5 3and 2 (P) be the lengths of the segments determined by the midpoints of two opposite sides of a quadrilateral p E P.A�(P) A� (P) Prove that for all p E P the sum + is constant and find the value of >'1

this constant .

54. Let ABCDEFGHIJKL be a regular dodecagon and let

dius. Prove that

4

R be the circumra­

AB AF = and AB2 + AC 2 + AD 2 + AE2 + AF2 = 0 2 . + AF AB

1R

93

3 .1. PROBLEMS

55. Prove that if inside a convex poligon there is a point such that the sum of the

squares of its distances to the vertices of the poligon is twice the area of the poligon, then the poligon is a square. 56. Let AI A2 . . . An be a cyclic polygon and let P be a point on its circumcircle.

Let PI , P2 , . . . , Pn be the proj ections of P onto the sides of the polygon. Prove that the product n

II PA��

i= l P��

is constant.

57. Let AI A2 . . . A2n be a cyclic polygon and let M be a point on circumcircle. Points KI , K2 ' . . . , K2 n are the projections of M onto sides its AI A2 , A2 A3 , . . . , A2 n AI and points HI , H2 , . . . , Hn are the projections of M onto diagonals AI An+ I , A2 An+2 , . . . , An A2n · Prove that

MKI · MK3 . . . MK2 n-l = MK2 · MK4 . · . MK2 n = MHI · MH2 . . . MHn · 58. Find the circumradius of a cyclic polygon with 2n sides if n sides have the length a and n sides have the length

b.

59. Let P be a point in the interior of a tetrahedron ABCD such that its projec­

tions AI , BI , CI , DI onto the planes (BCD) , (CDA) , (DAB) , (ABC), respectively, are all situated in the interior of the faces. If S is the total area and r the inradius of the tetrahedron, prove that SBeD PA l

+

Se DA PBI

When does the equality hold?

+

SDAB PCI

+

SABe PDI

> -

� r

60. Let AIA2 A3 A4 be a tetrahedron, G its centroid, and A� , A� , A�, A� the points where the circumsphere of AI A2 A3A4 intersects GAl , GA2 , GA3 , GA4 respectively. Prove that

1+1+1+1 1+1+1+1 GA l · GA2 · GA3 · GA4 � GA� . GA� · GA� . GA�

and

GA�

GA�

GA�

GA4

�

GAl

GA2

GA3

GA4 ·

SOLUTIONS 1 . The positive real numbers

only if Because

a2 , b2 , c2 can be the side lengths of a triangle if and (1)

2 + b2b 2 + a2 - b2 c a cos B = ' cos C = 2a ' 2 ca O 0 0 relation (1) is equivalent to cos A > , cos B > , cos C > . Hence the necessary and sufficient is that triangle AB C is acute. ( Doricondition n 07Andrica, Revista Matematidl Timi§oara (RMT) , No. 2 ( 1978) , pp. 48, - c2

Problem 35 )

A', B', C' be the midpoints of sides B C, CA, AB respectively. Construct point M such that BC M C' is a parallelogram . Note that AC' C M and B B' M A' are also parallelograms hence AM = CC' and A'M = BB'. Hence the triangle determined by the medians is AA' M. M A 2 . Let

B

A'

C

Let N be the intersection point of lines B' C' and AA'. Because AC' A' B' is a parallelogram, we have

C'N = B'N = !B'M 95

2 '

3. GEOMETRY

96 SO

BI is the centroid of triangle AAI M . Hence

Alternative solution: Consider the vectors = BC, b = AC , c = AB , ma = AAI , mb = BBI and observe that a

We obtain

1-

(-b - 2c) = 43 (c -b),

ma mb = 4 (b + c) 3 I l henceTima A m b = 4) 1c bl and the conclusion follows. ( tu ndreescu AC at P, then by Ceva's theorem we obtain B 3. If P meets MB NC 1 QBA MC Q ' ' NA = . MB = MC, we have Because QBA NAC Q N' QN I BC and then NBC = CiiiB. Therefore BN angle --xiiC, it follows that NBC = QiiN. On the other hand, because Bbisects NQ is isosceles, desired. Hence CiiiB = QiiN and triangle A d Ti ( tu n reescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1978) , pp. 66, 3286) Problem ABC and let AI be the midpoint of side H be the 4. Let orthocenter of triangle I I BC . Because FA I AC, it follows that FA 1- BH. Moreover BAI 1- HF, so AI is the BHF. Hence orthocenter of triangle HAI 1- BF. ( 1) I D I AB AI D HC HD AI D A D On the other hand, , so 11it follows that . Since l HA C thus is the orthocenter of triangle DC 1- HAI . ( 2) x

x

x

x

x

as

3.2. SOLUTIONS

97

A

B �------�-*�--� F

Note that GH 1- AB and EH 1- AG . Since ABI I A'E and AG I IGA', we have GH 1- A' E and EH 1- GA' , so H is the orthocenter of triangle A' EG. Therefore

(3)

A'H 1- GE . 2 ( ) and (3) we obtain From relations ( 1 ) , DG II BFI IGE,

as desired.

( Titu Andreescu)

5 . Using Ceva's theorem, we obtain

G'A A'B B'G - · _ · - = 1' G' B A' G B ' A

hence

S1 S3 S5 . . = S2 S4 S6

l.

The given condition now reads S1 S2

so This implies

+ + S3 S4

.3 / S1 . S3 . S5 S5 = 3V S2 S4 S6 S6

G'A A'B B'G = l. = = G' B A' G B ' A

It follows that A' , B' , G' are the midpoints of the triangle's sides and M is the centroid, as claimed. Revista Matematica Timi§oara (RMT) , No. 1 ( 1974) , pp . Problem 1904; Gazeta Matematica ( GM-B ) , No. 2 ( 1979 ) , pp. 63, Problem 0:16)

( Titu Andreescu,

23,

98

3. GEOMETRY 6. Using Menelaus' theorem yields

NA AB . QC . NB BC QA

Hence

= 1.

SMAN . 5MBA . SMCQ 5MBN SMC A SMAQ -

or

SMAN 5MBP . 5MBN SMCP A

1,

= SMQA ' SMCQ

B

From the condition in the hypothesis it follows that 5MBP + 5MBP = 2 V. j SMAN , . 5MBN SM CP SMCP

SMAN SBMN

and so

SMAN 5MBP ' 5MBN SMCP

Thus

NA NB

PB PC

hence N P and AC are skew-parallel . Revista Matematica Timi§oara (RMT) , No. Problem Gazeta Matematica (GM-B), No. Problem pp.

( Titu Andreescu, 3287;

1 ( 1978) , pp. 66, 17607) 2 ( 1979) , 56,

7. Let AI , Bl , Cl be the feet of triangle ' s altitudes and let A', B', C' be the pro­

jections of P onto its sides. If P is the orthocenter of the triangle, then the equality holds. Assume by way of contradiction that P is not the orthocenter of the triangle and aPA

+ bPN + cPC = 4S.

Then at least two of the inequalities PA + PA � AAl ,

PB

+ PB' � BBl ,

PC + PC' � CCI

are strict.

A

3.2. SOLUTIONS

It follows that

99

l ) b (BB PBI) (CC PC/ ) AA PA PA PC bPB ( a + +c > a l - + l - +c l or AA bBBl + cCCl - (aPAI + bPBI + cPC/ ) . 4S > a l + 6S - 2S = 4S , which is false. Hence ( 1978) , pp. 74, ( Titu A4Snd>reescu, Revista Matematica Timi§oara (RMT ) , No. 2 3 89 980 6 ) ( Problem ; Gazeta Matematica (GM-B) , No. 2 1 , pp. 64, Problem 18 122) 8. Using the angle bisector theorem and Van Aubel's theorem, it follows that IA BIA CIA b + c k IAI = BI C + CIB = -a- = l IB AlB CI B c + a k and IBI = AI C + CI A = -b- = 2 IC _ a + b k Let ICI - C - 3 and note that kl + 1 = -;;:2p k2 + 1 = b' 2p k3 + 1 = -2p , ' c where p is the semiperimeter. Hence kl 1+ 1 + -k2 1+ 1 + -k3 1+ 1 = 1 -_

so

Furthermore, since

k2 + 1 k3 + 1 (k2 + 1) (k3 + 1 ) 1 A [ ( kl + 1 ) 2 ' cos = !2 k 1 + k 1 _ 3+ 2+

1 00

3. GEOMETRY

we observe that cos A depends only on kl and k3 • Hence cos A = cos Al and, analo­ gously, cos B = cos BI , thus triangles ABC and AIB I CI are similar. Romanian Regional Mathematical Contest " Grigore Moisil" , 1999; Revista Matematica Timi§oara ( RMT ) , 1999, pp.

(Dorin Andrica,

87)

9. Using Steiner's theorem, we obtain

MB · NB AB2 = MC · NC AC2 Applying the AM-GM inequality yields MB MC

NB . / MB · NB AB + NC =2 � 2V MC . NC AC '

as desired. Note that the equality case occurs only if AM and AN coincide with the internal bisector of angle A.

( Titu Andreescu)

10. The quadrilateral APMN is a rectangle, therefore NP = AM. Hence NP is

minimal if AM ..1 BC, so M is the foot of the altitude from A.

( Titu Andreescu)

1 1 . Applying the law of cosines to triangle A l BI C we obtain

and using the inequality x2 get

+

y 2 - xy � xy, which holds for all real numbers x, y, we

AI B� � AI C , BI C. A

101

3.2. SOLUTIONS

�

Similarly, we obtain Bl Cr Bl A · CIA and CI A� three inequalities together, we obtain

� Cl B . Al B . MUltiplying the

Now the lines AAI , BBI , CCI concur, so and after substituting and taking square roots, we have Al Bl . Bl CI . CI AI

� Al B · BI C · CIA,

the desired inequality. Equality holds if and only if CAl CBl , ABI ACI and BCI BAI , which in turn holds if and only if P is the center of triangle ABC . IMO 1996 Shortlist)

=

( =Titu Andreescu,

=

1 2 . We start with the following lemma. Lemma. A, B, C

Let

tan

A

-

2

be the angle of a triangle ABC. Then

+ tan -2 + tan -2 tan -2 + tan -2 tan -2 B

B

C

A

C

= 1.

P f. We present two arguments. h. Since Firoorst approac

+ ,8) [1 - tan a tan ,8] = tan a + tan ,8, tan(90 ° - a) = cot a = l/ tan a, and A/2 + B/2 + C/2 = 90° , the desired identity follows from ( tan - tan - + tan - tan - = tan - tan - tan - ) = 2 2 2 2 2 2 2 = tan 2 tan (�2 + 2 ) [l - tan �2 tan C2 J = = tan � tan (90° - � ) [1 - tan � tan �J = tan(a

A

B

B

B

C

B

A

+

C

C

A C = 1 - tan '2 tan '2 ' Second approach. Let a, b, c, r, s denote the side lengths, inradius and semiperime­ ter of triangle ABC, respectively. Then SABe = rs, AP = s - a, and tan(A/2) = r/(s a) . Hence A ( e tan - = SAB 2 s s - a) -

.

Likewise,

B

tan 2

(s -eb) = -:sSAB

--

and tan

C

2

( s ec) . = sSAB -

1 02

3. GEOMETRY

Hence

A B tan - tan -

by Heron's formula.

2 2 + tan -B2 tan C2 + tan -C2 tan A2 = b) ) + + ( a c ( 8 ( 8 ( 8 ) ) S �BC b) = = 82 ( 8 - a ) ( 8 - ( 8 - c) 1 S�BCb ) = , 8 ( 8 - a )( 8 - ( 8 - c) -

0

-

A

Q

B

C

Without loss of generality assume that AP = min{AP, BQ, CR}. Let = tan( A;2 ), = tan( ii/2) , and = tan( C;2 ) . Then AP = BQ = and CR = The condition given in the problem statement becomes

/r z. y

z

and the equation in the lemma is

r/x,

2x + 5y + 5z = 6 ,

1 xy + yz + zx = . (1) and (2) yields Eliminating x from 5y2 + 5z2 + 8yz - 6y - 6z + 2 =

Completing squares, we obtain

r/y,

x

( 1) ( 2)

O.

(3 y _ 1 ) 2 + (3 z _ 1 ) 2 = 4 ( y - Z ) 2 . 3 1 3 ( 1)/3, z = (v + 1 )/3) gives Setting y - 1 = u, z - = v ( Le. , y = u + 5u2 + 8uv + 5v2 = 82 4 25 < 0, the only Because the discriminant of this quadratic equation is real solution to the equation is u = v = Hence there is 4only / 3) set of /3, yone= zpossible = 1 . Thus values for the tangents of half-angles of ABC (namely x = all triangles in S are isosceles and 4similar / 3 4/ 2 /3 andtoy =onez another. Indeed, we have x = r/AP = = r/BQ = r/CQ =2 1 = 1 , so 3 4 1 we can set r =5 , AP = AR = , and BP = BQ = CQ = CR = . This leads to 24 1 BC = . By scaling, all triangles in S are similar to the triangle AB = AC = and 558 with side lengths , , . O.

O.

x

103

3.2. SOLUTIONS

We can also use half-angle formulas to calculate C

. C= . B = sm sm

2 tan -

2

1 + tan2 C "2 3 : 4 : 5 and BC : AC : BC = 5 : 5 : 8. From this it follows that AQ : QB : BA = Alternative solution. By introducing the variables p = y + z and q = yz - 1 , ( 1) and (2) become 2x + 5p = 6 and xp + q = 0, respectively. Eliminating x relations

yields

p(6 - 5p) Note that

(3 )

+ 2q = O .

y and z are the roots of the equation (4 ) t2 - pt + (q + 1) = O . (3 ) ( 4) Expressing q in terms of p in , and substituting in , we obtain the following quadratic equation in t: 2t pt + 5p2 - 26p + 2 - 0 . O 2 $ . Hence the equation has real This equation has discriminant - (3p 2) 1/3. solutions if p = 2/3, and y = z = Note.only We can also let x = AP, = BQ, z = CR and use the fact that r (x + y + z) = SABC = Jxyz (x + y + z) _

Y

to obtain a quadratic equation in three variables. Without loss of generality, we may set = Then the solution proceeds as above. ( USA Mathematical Olympiad,

x Ti 1 . A d tu n reescu, 13. Let a,

2000)

b, c, A, B, C be the side lengths and angles of triangle ABC. Let X, Y, Z

be the feet of the perpendiculars from P to lines BC, CA, AB, respectively. Recall the inequality (the key ingredient in the proof of the Erdos-Mordell inequality): PA sin A � PY sin C

+ PZ sin B.

( 1)

104

3. GEOMETRY

BC PY PZ, BC. = = 90° , AZPYY Z PA A AP M . Z Y PX=. 90° . = = PZBX ZM = P Z B . Y = PY C . = Y Z � Y + M Z. 2R PA � cPY + bPZ. a bPB � aPZ + cPX and cPC � bPX + aPY. Using these Likewise, we have inequalities, we obtain PA PB PC PX ( b c ) PY ( c a ) PZ ( a b ) a2 + b2 + 7 � c3 + b3 + a3 + + b3 + a3 � 2PX 2PY 2PZ � be + co: + � (AM-GM inequality) 4SABC 1 abc Y=ZR ' BC and so on. This Equality in the first step requires that be parallel to ABC. Equality in the second step P occurs if and onlyb if is the circumcenter of ABC is equilateral and P requires that a = = c. Thus equality holds if and only if is its( Ti center. tu Andreescu, USA IMO Team Selection Test, 2000) H is the orthocenter of a triangle ABC then 14. It is known that if AH · BC + BH · CA + CH · AB = 4SABC . YZ

This says that the length of is greater than equal to its projection onto the latter being equal to the sum of the lengths of the projections of and onto as a diameter In fact, since AYP Azp is cyclic with of its circumcircle. By the Extended Law of Sines, sin Let and N be the feet of perpendiculars from and to the line Since 1fZp BxP , sin Similarly, N is cyclic. Hence x:iPZ 13 and sin Thus (1) is equivalent to N Multiplying by and using the Extended Law of Sines, (1) becomes

c3

A

3. 2. SOLUTIONS

1 05

I H. Al, Bl, C1 I H ' ' ' A, B , C A , B , C I BC, CA, AB , IA + lA' � AA1 , IB + IB' � BB1, IC + IC' � CC1 are strict. It follows that ' ' ' ·a IA + b · IB + c · IC > a(AA1 - IA ) + b(BBl - IB ) + c(CCl - IC ) ,

We prove that == Assume by way of contradiction that points and are distinct. Let be the feet of the altitudes from and let be the projection of onto the sides respectively. Hence at least two of the inequalities

or

which is false. Therefore 1 =

H and ABC is an equilateral triangle, as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 ( 1981) , pp. 67, 46 16) Problem 1 5 . Using standard notations, we have

64c2 + 49a2 + 9b2 - 112ac - 48bc + 42ab = 6c2 - 6a2 - 6b2 .

This is equivalent to

15b2 + 2b(21a - 24c) + 55ab2 - 1 12ac + 58c2�= O.0 Viewing this as a quadratic equation in , the condition � is satisfied. That is 441a2 - 1008ac + 576c2 - 825a2 + 1680ac - 870c2 � O. The last relation is equivalent to 6(64a2 - 112ac + 49c2 ) � 0, 8 7c. Substituting back into the given condition (8 7 ) 2 or a 3- C 7b� O. follows that a yields a = . We obtain It

=

106

3. GEOMETRY

hence triangle ABC is similar to triangle A' B' C' having sides 7, 3, 8. In this triangle 3 2 82 - 72 cos AI = . 3 . 8 = 2' It follows that A = A7 = Korean Mathematics Competition,

1

2+

60 ° .

2002) ( Titu Andreescu, Fi l i 16. rst so ut on. By the Cauchy-Schwarz Inequality, ..jPB2 + PC2 ..jAC2 + AB 2 2:: PB · AC + PC · AB.

Applying the (Generalized) Ptolemy's Inequality to quadrilateral ABPC yields

+ PC . AB 2:: PA . BC. Because PA is the longest side of an obtuse triangle with side lengths PA, PB , PC, we have PA y'PB2 + PC2 , and hence P A . BC 2:: ..jPB2 + PC2 . BC. P B . AC >

Combining these three inequalities yields

..jAB2 AC2

+

2:: BC,

implying that angle BAC is acute. With some careful argument, it can be proved that quadrilateral ABPC is indeed convex. We leave it as an exercise for the reader. Let D and Q be the feet of the perpendiculars from B and P to line AC, respectively. Then DQ � BP. Furthermore, the given conditions imply that Ap2 > Bp2 PC2 , which can be written as Ap2 - PC2 > BP2 . Hence, AQ 2 2:: AQ 2 QC2 = (Ap2 PQ 2 ) (Cp2 PQ 2 ) =

Note. Second solution. +

_

_

_

_

AP2 - PC2 > BP2 2:: DQ 2 . Let I be the ray AC minus the point A. Note that, since PA > PC, Q lies on ray If D did not lie on then AQ would be less than or equal to DQ, a contradiction. Thus, D lies on and angle BAC is acute. B =

l.

l,

l,

o

P o

o�----�o�----�o

A

D

Q

C

1 07

3 . 2. SOLUTIONS

Third solution. Set up a coordinate system on the plane with A (0, 0) , B (a, O) , C (b, c) , and P (x, y) . Without loss of generality, we may assume that ab > O0 and that c > O . Proving that angle BAC is acute is equivalent to proving that > . Since PA2 > PB2 + PC( 2 , ) x2 + y2 > x a 2 + y2 + (x _ b)2 + (y _ C)2 . Hence 2b ( ) 2b b ( o > x - a 2 - x + 2 + y - C) 2 � - x. O. follows that b > 0, as desired. Since P A > P B, we have x > � > Fourth solution. We first prove the following Lemma. F f i W X Y Z in the plane, Lemma. or any our po nts , , , =

=

=

=

_

It

Proof. Pick an arbitrary origin 0 and let w, x, y, z denote the vectors from 0 to W, X, Y, Z, respectively. Then I w - x l2 + I x - Yl2 + I y - Zl 2 + Iz - w l2 - Iw - Y l 2 - I x - Zl 2 w · w + x · x + y . y + z · z - 2I (w , x + x · y + y . z + z · w - w · y - x · z) w + y - x - Z 12 , which is always Equality holds if and only if w + y x + z, which is true Wnonnegative. XYZ is a (possibly if and only if degenerate) parallelogram. Applying the Lemma to points A, B, C, P gives 0 :::; AB2 + Bp2 + PC2 + CA2 _ AP2 - BC2 (PB2 + PC2 PA2 ) + (AB2 + AC2 - BC2 ) < 0 < + (AB2 + AC2 - BC2 ) AB2 + AC2 - BC2 . Therefore AC is acute. Fifth solutangle ion.° InBthis 0° 90° . Note - 1 takes on values between and solution, sin 90 , since PB < PA. Applying the Law of Sines to triangle PAB yields that -p;fjj < =

=

=

=

=

=

0

=

=

_

=

-

sin PAB =

It

follows that

Since PA2

PB - PB sin ABP :::; PA PA '

-

> PB 2 +

_ PB PAB :::; sm 1 PA ' PC2 , we have similarly •

- < ' - 1 PC . _ 1 VPA2 _ PB 2 . PAC _ sm < sm PA PA

1 08

3. GEOMETRY

Thus

-

-

--

1 PB 1 BAC � BAP + PAC < sin- PA + sin then If = sin - 1

()

Hence

��,

()) sin (90° -

=

() cos

=

-

JPA2 - PB2 PA

JPA2 - PB2 () . V'l - sin2 PA J . 1 PA2 _ PB2 =

. - 1 PB + smBAC < sm PA and angle B AC is acute. C

=

PA

90° ,

P

P A B

B

As we mentioned at the end of the first solution, the conditions in the problem imply that quadrilateral ABPC is indeed convex. Thus, the diagram on the right-hand side is not possible, but this solution does not depend on this fact.

Sixth solution.

/

I

/'

I

(

\

I

\

\

"-

--

...--

I

\

'-...

_

,/

'\.

I

/

_

""

B

;'

P

-....

--

C

,/

...--

./

I

Note that PA2 > PB 2 + PC2 . Regard P as fixed and A, B, C as free to rotate on circles of radii PA, PB, PC about P, respectively. As A, B, C vary, will be maximized when B and C are on opposite sides of line PA and and are right angles, i.e., lines AB and AC are tangent to the circles passing through B and C. Without loss of generality, we assume that PA > PB � PC. In this case, ABPC is cyclic and AB 2 = PA2 - PB 2 > PC2 , and similarly AC2 > PB 2 . Hence on

IfA1J Jfijp ;[(j'p

109

3.2. SOLUTIONS

the circumcircle of ABPC, arcs AB and AC are bigger than arcs PC and PB, respectively. Thus, IiPC > BAiJ. Because these two angles are supplementary, angle BAC is acute. B -

,.,

A

P /'

( Titu Andreescu, USA Mathematical Olympiad, 2001) 17. All angles will be in degrees. Let x

=

F0B. Then PiiC

=

80 - x. By the

Law of Sines (or the trigonometric form of Ceva' s Theorem) ,

PB PC sin � sin !!.!!. sin � = PA PB PC PA sin PAB sin PBC sin PCA x sin 40 cos 10 20 sin x sin 40 = sin sin10 sin(80 � - x) sin 30 4 sinsin(80 - x)

1

=

=

------

---------

B

A

C

b The identity 2 sin a cos

formula) now yields

1= so

=

b) b) sin(a - + sin(a + (a consequence of the addition

2 sin x(sin 30 sin 50) sin(80 - x)

+

=

sin x(1 2 cos 40) , sin(80 - x)

+

2 sin x cos 40 = sin(80 - x) - sin x = 2 sin(40 - x) cos 40. This gives x = 40 - x and thus x triangle ABC is isosceles.

=

20. It follows that AcE = 50

=

BAiJ, so

3. GEOMETRY

110

Alternative solution. Let D be the reflection of A across the line BP. Then triangle APD is isosceles with vertex angle 2 (180 - EPA) = 2 (PAii + AiiP) = 2 (10 + 20) = 60, APi5 = 2JiBA = 40. Since 1fAC = 50, we have DB .l AC. and so is equilateral. Also, ISiiA = B

C D Let E be the intersection of DB with CPo Then

= 180180CEi5 = 180 - (90 - AcE) = 90 + 30 = 120 and so PEi5 + I5A.J5 = 60. We deduce that the quadrilateral APED is cyclic, and therefore DEA = DPA = . 60 Finally, we note that I5EA = = 15EC. Since AC DE, we deduce that A across the line DE, which implies99that BA = BC, as desired. and (CTiare Asymmetric d 6) 1 tu n reescu, USA Mathematical Olympiad, 90° , B = 30° 18. Let max{ A, B} = A. If triangle ABC is right-angled, then A = 60° . In order to find !!:. , we may assume that ABC is the triangle with sides and C = r 1 2b 1 a = , = , c = y'3. We have R = and S y'3 2 2 r - - + 12+ J3 - 3 +y'3J3 ' 3 + J3 v'3 1 R so = -- = +. r y'3 R J3 1 Conversely, assume that = r + . From the identity r = 4R sm· 2A sm. "2B sm. "C2 PEi5

-

-

-

.l

_

ABC

--

8

-

-

_

_

---

3.2. SOLUTIONS

it follows that or

111

r = 4(V3 + l)r sin � sin � sin �

Then

0 - 1 ( A - C A + C) . B = cos -2- - cos -2- sm 2" ° , we obtain A C 30 = and, since B B 1 V2 + J6 o4- = ( sin 2 ) sin 2 ' B · Lettmg sm . 2" = x Yl.elds 0 1 + V2 + 4- _ 0 J6 x2 -, x B 5° B 5° V2 V2 J6 . whose solutlOns are x = and x = 2 ' AIt follows that B2" =301 ° Aor 2"90=° . B The is not because � . Hence = , = and ABCacceptable, C = second 60° . Thussolution is right angled. triangle ( Titu Andreescu, Korean Mathematics Competition, 2002) ABC three circles equal to the circum­ 1 9. Construct in the exterior of triangle ABC circle that pass through two vertices P of the triangle. ( By the five-coin theorem the circles will have a common point , as desired see Dorin Andrica, Csaba Varga, Daniel 2002 , pp.Vacare�u, 5 1-56) . " Selected Topics and Problems in Geometry" , PLUS, Bucharest, Alternative solution. Let H be the orthocenter of triangle ABC. The reflections ABC. H of across the sides of the triangle are points of the circumcircle of triangle HAB, HBC, HCA are equal to the circumcircle of Therefore circles of ABC and forthe Pcircum H the claim holds. 2 ( 978) , pp. 74) (Dorin Andr=ica, Revista Matematica Timi§oara (RMT), No. 1 A'B' , B' C' , C' A' by Co, Ao, Bo, respectively, and 20. Denote the midpoints of the ' C' perpendiculars in question by le, lA , lB. Consider the centroid of triangle A'Bthree . -4-

4

_

4

_

4

4

112

3 . GEOMETRY C

1 2,

Since AoG : GA' = BoG : GB' = CoG : GC' = : the dilatation h with center G and coefficient takes Ao, Bo, Co to A' , B' , C', respectively. Since dilatations carry straight lines into parallel lines, h transforms le into the line through C' perpendicular to AB. But C' is the point of tangency of the incircle and AB, so this line passes through the incenter of triangle ABC. The same applies to the images of l A and lB under h. Since the images of l A , lB, le under h are concurrent, so are lA , IB , le themselves. ( Titu Andreescu, Romanian IMO Selection Test,

-2

1986)

2 1 . Because triangle AI A2 A3 is not isosceles, it is not difficult to see that the

circumcenter of the triangles AI BI I, A2 B2 I, A3 B3 1 are defined. We start with a simpIe lemma. Lemma. Let ABC be a triangle with the incenter I. Let T be the circumcenter of the triangle BIC. Then T lies on the internal bisector of the angle A. Proof. Let us draw the external bisectors of the angles B and C as shown in the figure below. B

A

C

They intersect at the excenter E, which lies on the internal bisector of the angle A. Since BE ..1 BI and CE ..1 CI, the quadrilateral BECI is cyclic with the center of the circumscribed circle on IE. This center will be also the circumcenter of BIC. The lemma is proved.

3.2. SOLUTIONS

113

i

Let us prove the main statement. For = 1, 2, 3 we denote by the center of the Clearly, lies on the circle and by the circumcenter of the triangle By the lemma, also lies on the same bisector. Thus internal bisector of the angle are perspective from the point By Desargues ' and the triangles theorem these triangles are perspective from a line. This is to say that if we denote = 1, 2, 3, to be the point of intersection of the lines and then are collinear. But since is the perpendicular bisector the points these points are exactly is the perpendicular bisector of and of respectively. the circumcenter of the triangles A student not familiar with Desargues' theorem may proceed from the point as follows. Applying Menelaus ' theorem to the triangles respectively, one and to the triples of points can, observing usual agreement about the signs, write:

Q Ai+1 IAi+2 . i Oi Ai' Ti I. T1 T2T3 0i+1 0i+2 Ti+1 Ti+2 ' Qi, i Ti+1 Ti+2 QQ Q Bi li Ail 01 i,+120, i+32 A1 B1 I, A2B2I, A3 B3 I, Remark. 101 02 , 102 03, 103 0 1 (T1' T2 , Q3 ) , (T2 ' T3, Q 1 ) , (T3 ' Tt, Q2 ) , 01 T1 IT2 02 Q3 IT1 . 02 T2 . 01 Q3 - 1 , IT3 02 T2 Q3 Q 1 03 T3 . IT2 . 02 Q 1 - , IT1 03T3 0 1 Q2 01 T1 . IT3 . 03 Q2 - . MUltiplying them all one gets 0 2 Q 3 03 Q 1 01 Q2 0 1 Q 3 . 0 2 Q 1 . 03 Q 2 = QQ Q meansi that lthei points 1 , 2 , 3 are collinear. which Alternat ve so ut on. This proof will be based on inversion. We take the incenter I Ci

Ti 0 1 02 03

_

--

--

--

_

_

1

1

1,

to be the center of the inversion and the power of the inversion is arbitrary. Using primes to denote images of points under the inversion we have the following " dual" figure shown below.

11 4

B'I

3. GEOMETRY

B2' Gi is a straight line B�+I B�+2 ' with these lines Indeed, the imageBof the circle � B�B� . The line AiAi+1 will be transformed into the circle forming the triangle ri+2 , with the side AiAi+1 becoming the arc Ai Ai+1 which does not contain I. Note AAA I that all these circles have equal radii since the distances from to the sides of I 2 3 were equal. E I, E2 , E3 are three circles passing through the common point Let us note that if I and no two of them touch, then their centres are collinear if and only if there is i- I through which all these three circles pass. another common point Ei being the circumcircle of Ai Bi I. Since the We will useEthis observation for AiB� , the desired result is to show that the lines A� BL to the line Ainversion B� are iconcurrent. A� A� A� � B� , A�takes For this, it suffices to show that the triangles B� B� B� are homothetic, which is the same to say that their corresponding sides and r r r PPP are parallel. Since the radii of the circles I , 2 , 3 are equal, the triangle I 2 3 formed their centre has its sides parallel to the corresponding sides ofAthe triangle B� B�B�by. The A A� into I 2 � � / 1 centred at takes thePtriangle homothety of ratio P2 P3. Therefore the the triangle whose vertices are the midpoints of the triangle I A A A� � � and PI P2 P3 are also parallel and the corresponding sides of the triangles result( Tifollows. tu Andreescu, IMO 1997 Shortlist) I be the intersection point of lines GB' and G'B and let A' be the 22. Let AI and BG . intersection point of lines We have AB' AG' AI B'B + G' G = lA" I from Van Aubel 's theorem, therefore ::, is constant. Hence the locus of point is a BG line (segment dreescu) to . Titu Anparallel J

3 . 2 . SOLUTIONS

115

23. Without loss of generality assume that

PC = max{PA, PB, PC } .

The condition in the hypothesis is

PB . PC + PA . PC = PA . PB + 1 PC PA · PB + 1 · 1 or 1 - PA . 1 + PC . 1 . PACB is a From the converse of the second theorem of Ptolemy it follows that AB C P cyclic quadrilateral. Note that cannot be , or P otherwise the denominator of side equals OA. Hence the ABCright-hand B, C. the locus of point is the circumcircle of triangle without the vertices , ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1985) , Problem C7: 3) 24. We have Ja2 b2 4S2 + Ja2 c2 - 4S2 = Ja2 b2 - a2 b2 sin2 C+ + J-a2 c2 - a2 c2 sin2 B = ab cos C + ac cos B = = ab a2 +2ab2b c2 + ac a2 +2acc2 - b2 = a2 as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 8 ( 1971) , pp. 25, 1006) Problem _

25. The relation is equivalent to

+..j y'rb + Fc = -,1 r rarbre 1 1 1 1 or + + = -. Fay'rb FcFa Fay'rb r On the other hand, 1 1 1 1 -ra + -rb + -re = -r , so 1 1 1 1 1 1 + + = - + - + -. --Fa,y'rb y'rbFc FcFa, ra rb re Then ( 1 1 ) 2 + ( 1 1 ) 2 + ( 1 1 ) 2 = 0, - ..fib y'rb - Fc Fc - Fa ' It follows so ra( Ti that the triangle is equilateral, as desired. ( 974) = rb A= rdeFa, tu1903n )reescu, Revista Matematica Timi§oara (RMT) , No. 1 1 , pp. 2 1 , Problem Fa,

---

116

3 . GEOMETRY 26. If the triangle is equilateral the conclusion is true,

To prove the converse, we assume by way of contradiction that the triangle is not equilateral and say that Then

b f:. c, 2 a2 (b + C) 2 - a2 2 +� c b -2 =p(p - a) m= -4> 4 b � ) � and likewise m 2: p (p - , m 2: p(p - c) , It follows that 1 1 1 <1( 1 1b 1 ) m� p- p--b ab + p-m� + m� + - + p--c = -p2 + a + c + ca S2 On the other hand 1 1 1 1 _p2 + a2 + b2 + c2 �-r + -r� + -r� = -S2 [(p _ a) 2 + (p b)2 + (p C) 2] = ---=S2 b b b f:. 2 2 Since c then a + 1c + ca 1< a +1 b2 + 1c hence 1 1 < m2e ra + rb + re m2b + m2a + which d 2 ( 977) 66 ( Tiistua Acontradiction, 306n3)reescu, Revista Matematidi Timi§oara (RMT) , No, 1 , pp, , _

_

2

2"

_____

2'

Problem

27. We know that

,sm -A2 sm, -B2 sm, -C2 -< -81 ' A B C 'A'B' C so 2 cos -2 cos -2 -> sm sm sm , cos On the other hand, 'sm A + sm' B + sm' C = 4 cos "2A cos 2"B cos 2"C ' (1) gives so inequality A B C A B C sin + sin + sin 2: 4 sin sin sin , 1 Since the circumradius is , we have b = 2 sin B, c = 2 sin C, = a 2 sin A, (2) yields and relation a + b + c 2: abc,

as

claimed,

(1) (2)

3. 2 .

11 7

SOLUTIONS

Alternative solution. Let Zl , Z2 , Z3 be the afixes of points A, B , C such that I zd = 1 IZ2 1 = I Z3 1 = . We have BC = IZ - z3 1 , b = AC = IZ3 - zI i , c = AB = IZI - z 1 · a= 2

2

U sing the identity

Z; ( Z2 - Z3 ) Z� ( Z3 - zd Z� ( ZI - Z2 ) = ( Zl - Z2 ) (Z2 - Z3 ) (Z3 - Zl )

+

and triangle inequality it follows

+

abc = IZ I - z2 1 1z2 - z3 1 1z3 - zd ::; IZl l 2 1z2 - z3 1 IZ2 1 2 11z3 - zl l + IZ3 1 2 1z1 - z2 1 = = IZ2 - z3 1 IZ3 - zl l IZI - z2 1 = a b c.

+

+ + + + 2 (1978) 49 ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. , pp. , 35 0258 98 3 ( ) 1 1 1 11 Problem Gazeta Matematidi , Problem ; No. 2 (1988) , pp. 78; , Problem 21353) (GM-B), No. 28. For any positive real numbers x, y, Z we have 8xyz ::; (x + y)(y + z) (z + x) Setting = a - b + c, Z = a + b - c, x = -a + b + c, gives ( -a + b + c) (a - b + c) (a + b - c) ::; abc, Y

so

It follows that then Hence

3 <- 2 (p - a) (pabc- b) (p - c) = �2 . pabc S2 = abc E 6R = 6 4S . S = ' r desired. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT), No. 2 (1974) , pp. 51 , 2028) Problem a2 L...J (p - b) (p - c) "

as

29. Using the inequality

3 (a2 + b2 + c2 ) � (a + b + C)2 ,

3. GEOMETRY a2 +4Sb22+ C2 � 3Sp22 ' 1 1 1 1 h� + h� + h� � 3r2'

118

we obtain Hence as

desired.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , 66, 3062) Problem pp.

30. From the inequality

1 1 1 1 1 Jrcra + -Jrbrc + -Jrar1 b ' -ra + -rb + -rc >- -1 1 1 1 J J > J -r - -rcra + --. rarb rbrc + --

we obtain Then hence as

Jrarb + Jrbrc + Jrcra � 9r,

claimed.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978) , 64, 3277) Pro blem pp.

3 1 . We have

and likewise

b2 + C2 a2 J (b + C)2 _ a2 "'; ( ) J 4 = pp-a ma = -2- - 4 �

It follows that

as

ma mbmc � p"';p(p - a)(p - b)(P - c) = pS = ra rbrc, desired. ( Titu A ndreescu, Revista Matematica Timi§oara (RMT), No. 1 ( 1978) , 64, 3276) pp.

Problem

32 . By the AM-GM inequality,

so

8p3 � 27abc. Hence

(a + b + C) 3 � 27abc, b c S 2 7R 2p2 > 27 a 4S - · -p = r,

3.2. SOLUTIONS

11 9 desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1973) , pp. 43, 1 585) Problem as

33. We have

It follows that hence as

desired.

A 7r . A 1 _2 <- _6 sm 2 "2 ' V (P - bb)(P - C) < �2 ' e (p - b)(P - e) 4'be {:}

S;

S;

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1984) , pp. 67, 5221 ) Problem 0 O0 34. Let 1 , 2 , 3 be the centers of the three circles and S the area of the common

region.

0 1 , O2 , 03 which subtend the arcs three sectors with centers 0 0 02 0The 0 0 3, 1 3, 2 1 , respectively, cover the 2 surface of area S and twice more 0 0 0 r ) the surface of triangle 1 2 3 (which is � . On the other hand, the area of 2 these three circular sectors equals the area of a semicircle, which is � 7rr . Hence 1 2-7rr2 = S + 2 . r24J3 ' therefore 2 V3) . S = �2 r (7r (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 (1978) , pp. 50 , 3522) Problem --

1 20

3. GEOMETRY 35. The parallel BD through C meets AB at point E. By Stewart's formula, we

obtain

AC2 · BE + CE2 . AB - CB 2 · AE = AB · BE · AE Because CE = BD and BE = CD, we deduce AC2 . CD + BD2 . AB - BC2 . (AB + CD) = AB . CD . (AB + CD) C

D

B

A

E

Drawing the parallel to AC through D and using similar computations yields BD 2 . CD + AC2 · AB - AD 2 . (AB + CD) = AB · CD · (AB + CD)

( 2)

(1)

( 1)

Subtracting the relation from gives (AC2 - BD 2 ) (AB - CD) = (AD2 - BC2 ) (AB + CD),

(2)

as desired.

(Dorin Andrica, Gazeta Matematidi (GM-B), No. 9( 1977) , Problem 6852; Revista 1 2 (1980) , pp. 64, Problem 4119) Matematidi Timi§oara (RMT), No. 36. Let I be the intersection point of the diagonals AC and BD. Since

AB · BD AB · AC and IB = AB + CD AB + CD the condition in the statement becomes IA =

90° , as desired. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2 (1978) , pp. 59, 3524) Problem Hence JfilJ =

37. Let ABCD be the trapezoid. Point I is the intersection of diagonals and

M, N are the midpoints of AB and DC. In a right triangle the length of the median corresponding to the hypothenuse is half of lenght of the the hypothenuse. Hence AB CD IM = and IN = .

2

2

3. 2. SOLUTIONS D F N C

12 1

B M E Then IM + IN = AB +2 CD = EF, which of the midline and hencelAB the lengthICD of the altitude.lAB It followsICD that IM andis theIN length therefore are also altitudes in triangles and and ABCD are isosceles. Thus is isosceles, as claimed. ( Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 1 (1978) , pp. 48, 28 1 7) Problem A

38. From the Law of Cosines we deduce that

Note that

2AB . AC cos 1fAC = AB2 - BC2 + AC2 2DC · AC cos DcA = CD2 - AD2 + AC2 2AB . DB cos DiiA = AB2 - AD2 + DB2 2DC · DB cos Cfiiij = CD2 - BC2 + DB2

(1) (2) (3 ) ( 4)

1fAC = DcA, DiiA = Cfiiij , (2) (3) and (4) yields ) ( 1 so dividing relations and , AB2 - BC2 + AC2 AB2 - AD2 + DB2 AB ' CD2 - AD2 + AC2 = CD2 - BC2 + DB2 = DC as desired. D i A di ) ( or n n r ca b the lengths of the bases, c, d be the lengths of the nonparallel sides 39. Let a, dd and 1 , 2 be the lengths of the diagonals. From Euler's theorem for quadrilaterals, it

follows that Hence

(d1 - d2 )2 + 2d1 d2 = (c - d) 2 + 2 (ab + cd) ,

1 22

3. GEOMETRY

ab +

and d1 - d2 = C - d implies d1 d2 = cd. From Ptolemy's Theorem, we deduce that the trapezoid is cyclic and so isosceles, as claimed. Revista Matematica Timi§oara (RMT) , No. pp. 48, Problem

( Titu Andreescu, 1 ( 1978) , 281 7) First solution. Assume the opposite. Then lAC - BDI > lAB - CDI or 40.

l AC - BDI > l AD - BCI. Without loss of generality, l AC - BDI > lAB - CDI , otherwise switch B and D. We have (1) AC2 - 2AC . BD BD2 > AB2 - 2AB . CD CD 2 and, from Euler's relation,

+

+

(2) where M and N are the midpoints of AC and BD, respectively. From and (2) , AD 2 + BC2 - 2AC · BD > 4MN2 - 2AB · CD.

(1)

(3)

Let P be the midpoint of AB. Then NP = AD /2, MP = BC /2 and since MN � INP - MPI , it follows that 4MN2 � (AD - BC) 2 . (4)

(3 )

From and (4) , -2AC · BD > -2AB · CD - 2AD · BC, in contradiction with Ptolemy's Theorem. We are done. The cyclicity is essential. The inequality fails if ABCD is a parallelogram. Let E be the intersection of AC and BD. Then the triangles ABE and DCE are similar, so if we let x = AE, = BE, = AB, then there exists k such that kx = l)E, = CE, = CD. Now

Note. Second solution.

y z li l AB - CDI = I k - z and ) (k )1 1 1 l l AC - BDI = I (kx + y - y + z = Ik - . Ix y · Since I x - y l ::; z by the triangle inequality, we conclude that l AB - CDI � lAC - BD I , and similarly IBC - DAI � lAC - BD l. These two inequalities imply the desired Thirresult. d solution. Let 20', 2,8, 2,, 28 be the measures of the arcs subtended by AB, BC, CD, DA, respectively, and take the radius of the circumcircle of ABCD 1 to be . Assume without loss of generality that ,8 � 8. Then 0' + ,8 + , + 8 = 7r, and ky

kz

-

(by the Extended Law of Sines)

lAB - CDI = 21 sin - sin

a

1'1 = I Sin a; l' cos a � l' I

1 23

3.2. SOLUTIONS

and

lAC - BD I = 2 1 sin(o: + t3l - sin(t3 + "Ill = I Sin 0:; "I cos (0:; + t3) I . 0 ( + ) / 2 ( + ) /2 + (3 7r/ 2 (by the assumption (3 6) and the Since [0 , / 2] , we conclude that l AB cosine function is nonnegative and decreasing on A C A C BD A D BC CD I � l - I , and similarly l - I � l - BD l . ( Titu Andreescu, USA Mathematical Olympiad, 1999) E be the intersection point of the diagonals. Consider AD < BC the 41 . Let AB the altitude. Since basis of the trapezoid and AB2 = AD . BC, then AB BC AD = AB ' ABC and ABD are similar. so the right triangles "I

::;

a

,

::;

a

,

::;

7r

::;

On the other hand we have

BCA + CAB = 90°

-

hence

-

AEB = 90° so the diagonals are perpendicular, as claimed. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 (1972) , pp. 28, 1164)

It follows that

-

Problem

I the intersection point of the diagonals AF and BE of the rectangle ABF42.E. LetNoticebethat NI is the median of the right triangle ANF with hypothenuse AF, so IN = AF2 = IA Likewise, BE IB IM = BE2 = IE ' IQ = AF2 = IF, fP = 2 = . fA = IE = IF = IB, it follows that 1M = IN = IP = IQ . Hence MNPQ Since

is cyclic, as desired.

124

A

3. GEOMETRY B

Q D E ( Titu Andreescu)

F

C

< al :::; a2 :::; a3 :::; a4 < al + a2 + a3, then there is a cyclic quadrilateral having side lengths aI, a2AB , a3, aBC4. CD DA AC b d j Denote by a, , c, , e, , m the lengths of the segments BD , MN, respectively. Without loss of generality assume that ,b + d, � a +, c. , , P the midpoint of the side BC. The segments MP and NP are midlines Let beCAB BDC, so in triangles and MP = "21 a and NP = "21 c. Then 2m = 2MN < 2MP + 2NP = a + c so 2m < a + c < b + d. 0 On the other hand, if is the intersection point of the diagonals, we have b + d = BC + DA < BO + OC + DO + OA = AC + BD = e + j, b d j 2 hence m < a + c < + < e + j . 2 b d It suffices to prove that e + < m + a + c + + . Note that e < c + d, j < b + c, e < a + b, j < a + d. Summing up these inequalities yields e + j < a+ b + c+ d < a + b + c+ d + 2m and (the complete. Andisreescu, Tituproof 66 Revista Matematidi Timi§oara (RMT) , No. 1 ( 1978 ) , pp. , 43. By Sturm's theorem, we know that if 0

Problem 3288; Gazeta Matematidi ( GM-B ) , No. 10 ( 1981 ) , pp . 402, Problem C148 )

125

3.2. SOLUTIONS

44. Let I be the intersection point of lines BD and AF. The parallel to BD through C meets line AF at point T. First we consider AF .l BD and prove that AB .l CD. F

F

A

�------�

D

C

i) Assume that D lies on the segment CEo Then .ArC = we obtain On the other hand CTIIBD so

( 1 ) (2) ° 90 ,

90° . Since 1fAC : Ifi5C, (1) (2 )

Relations and imply ifAE : EaT , therefore EATC is cyclic. hence AB .l CD, as desired. It follows that BEG = ii ) Assume that C lies on the segment DE. In the right triangle CTF, TE is the median, so Because CTIIBD, we have

(3) ( 4) (5)

Ere : EaT.

EaT : CnB . Also,

(3)°, (4) , (5) , we obtain Ere : 1fAC. Hence ATEC is cyclic, then AEC = 90 , and AB CD, desired. JfiiC =

so from

.l

as

Conversely, consider that AB .l CD. i) If D is on the segment CE, then ME == AcE On the other hand AcE == Aci5 == AiiJ5 , so ME == AiiJ5 , and FBIE is cyclic. It follows that BiF = 1iEF = hence B D .l AF, as desired. ii ) If C is on the segment DE, then .

90° ,

(6)

3. GEOMETRY

126

( 7)

AcE := AiiJ.

( 7)

From (6) and we obtain ME := AiiJ, hence FEB! is cyclic. Note that lfEF = 90°, so ifiF = 90° and BD ..L AF, as desired. Revista Matematica Timi§oara (RMT) , No. 1 (1986) , pp. 106, Problem C6:4)

( Titu Andreescu,

45. Let

a, b, c, d be the side lengths of the quadrilateral and let S be its area.

Because the quadrilateral is cyclic, we have

S2 (p - a)(p - b)(P - c)(P - d) . =

N umbers

a, b, c, d are odd, hence p=

is an integer . If p is odd, then p p which is false. Hence p is even.

(1)

d a + b +c+ 2 ---

--

- a, - b, - c, - d are even and so 82 is divisible by 16, p

p

12 7

3.2. SOLUTIONS 1

1

b c d. Since S is a prime number, a (1) p - d =p - c = l4md2p - a2 = 2p - b = £ b 1 + 1 . Hence a = = + Summing up these equalities yields p = S so p = S p d and c = = S . The quadrilaterals are either rectangles or kites. Andreescu, 2 (1977) , pp. 66, ( Titurequired Revista Matematidi Timi§oara (RMT ) , No. 7 30 ) 6 Problem

Without loss of generality assume that � � � from relation we obtain

46. From the AM-GM inequality it follows that

ba cd < ( a + b +4 c + d ) 4 - -p41 4 . Hence 16abcd � p4 , or 8 (ac + bd)2 -p4 � 8 (a2 c2 + b2d2 ) . The desired inequality is now obtained from Ptolemy's Theorem: bd = eJ. ac+ ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 3 (1973) , pp. 36 , 1811 ; Gazeta Matematidi (GM-B) , No. 8 (1980) , pp. 364, Problem 18370) Problem _

47. Let

m

be the length of the segment determined by the midpoints of the

diagonals. From Euler 's Theorem for quadrilaterals we have

( 1)

128

3. GEOMETRY

and from Ptolemy's Theorem, From relations

ac + bd = e f.

(2 )

( 1) and (2) we obtain

- c/ , / b - d/ } � 1 , then 2 = 1 + 1 � (a - C)2 + (b - d) 2 = (e - f) 2 + 4m2 � (e - f) 2 . Hence l e - fl � V2, desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 ( 1978) , pp. 51 , 3527) Problem Since max{/a

as

48. The quadrilateral is cyclic so

S=

J(p - a)(p - b)(P - c)(P - d)

2 (�) , we have {j(p - a)(p - b)(P - c)(P - d) = P.2 = (p - a) + (p - b) + (p - c) + (p - d) 4 b Note that this is the equality case in the AM-G M inequality, hence p - a = p - = b d d p - c( Ti= p -A .dIt follows that a = = c = , so the quadrilateral is a 1square. tu213n6)reescu, Revista Matematica Timi§oara (RMT) , No. ( 1977) , pp. 24, Problem Since S =

�

SPAB ' SPCD = SPBC ' SPDA , since both are equal to PA . PB · PC · PD · sin P. The numbers SPAB, SPCD and SPBC, SPDA have the same sum and the same product, thus SPAB = SPBC and SPCD = SPDA or SPAB = SPDA and SPBC = SPCD, i.e. P is the midpoint of AC or BD, as desired. 4�. Observe that

( Titu Andreescu, Korean Mathematics Competitions, 2001 )

50. (i) Let MI and Mil be the projections of point M onto diagonals AC and

BD.

1 29

3.2. SOLUTIONS D

We recall the Simpson 's theorem: the projections of a point of the circumcircle of a triangle onto the sides of the triangle are collinear. Applying this result to triangles M', are collinear and yields that and are collinear. Hence the lines and meet at M' , as claimed. and we deduce that Mil (ii) From Simpson's Theorem for triangles the conclusion follows. and Since Mil is a point of is on the lines Revista Matematidi Timi§oara (RMT), No. pp. Problem Romanian Regional Mathematical Contest " Grigore MoisH" ,

C' , M' , D ' ABD BDC AC, 1 ( 1979) , 995)54, 1 1 AC BD 5 1 . We have 2 . , SABCD AC BD with equality if and only if . Since 1 AC BD 2002 1-= SABCD 2 . 52 · 77 2002 PD ) 21 (AP +ACPC) . (BP + = 2- = , BD P . Thus, it follows that the diagonals and are perpendicular and intersect at AB V24 32 40 BC = V282 + 322 = 4Vl13, CD = V282 + 452 = 53, and ABCD is therefore DA == V4522 ++ 2422 == 51 . , The perimeter of 144 + 4V113 = 4(36 + V113) . ( Titu A ndreescu, American Mathematics Contest 12 (AMC 12 - Contest B) , 2002, 24) Problem ABCD . Assume without loss of gener­ 52. Let a bePA thePC side length of the square P A ) = . We have ality that m ( , y'2PA = PB + PD. A' B' 'A B' C'D,' AC , B' C' D'A' . (Dorin Andrica, 3855 ;

ABC DAC

S;

S;

S;

ax

S;

-

3. GEOMETRY

130

.j2PA = aPB + aPD , hence Then a BD · PA = AD · PB + AB · PD . PDAB is a cyclic From the converse of the Ptolemy's Theorem it followsABCD that P quadrilateral, therefore lies on the circumcircle of square . Conversely, using theDPtolemy's Theorem we deduce that any point of the cir­ ABC cumcircle of square has the given P property.

P

ABCD .

It follows that the locus of point is the circumcircle of square , be a point with the given property and assume Let > Point are considered like in the diagram. > Note that so

Alternative solution. P (x y) y O . APC , B , C, DPA , V2PC PB PD = +

Squaring both sides yields

2X2 + 2 (y + b)2 = (x - b) 2 + (x + b)2 + 2y2 + 2y'(x - b)2 + y2 + 2y'(x + b)2 + y2 , then 2by = y'(x - b)2 + y2 + y'(x + b)2 + y2 , hence (x2 + y2 + b2 _ 2bx)(x2 + y2 + b2 + 2bx) = 4b2y2 . It follows that

so Thus

2 + y2 = b2 x P BD that contains A. Likewise, for and 0so point lies on Pthe semicircle of diameter BD that contains C and y :::; we deduce that lies on the semicircleABCD of diameter finally we obtain the circumcircle of square .

3.2. SOLUTIONS y A(O, a)

131

P(x, y) 0) D (a,

x

C(O, -a)

( Titu Andreescu, Romanian IMO Selection Test, 1981; Revista Matematica

Timi§oara (RMT) , No. 2(1981), pp. 87, Problem 4751) 53. Let ABCD be a quadrilateral from the set

P and let

midpoints of sides AB, CD, BC, AD, respectively. The Euler ' s relation for the parallelogram

MNPQ is MP2 + PN2 + NQ2 + Q M2 = MN2 + PQ2 ,

or

C

M

A On the other hand, we have

MQ = DB , MP = AC 2 2

so Hence

B

M, N, P, Q be the

13 2

which is, clearly, a constant.

3. GEOMETRY

(Dorin Andrica)

54. From the law of cosines we derive

AB = 2R sin ;

or

AF = 2R sin �;. 2 The first equality is equivalent to .sm 512rr sm. 512rr ---rr + -rr- = 4, . sm sin 12 5rr 12. rr ' 1 = 4 sm sm ' 12 12 Furthermore, rr . rr or 1 = 2 sm' '6rr 1 = 4 cos sm

12 which is clear. For the second equality, we have

and

12

AC2 - 4R2 sm. 2 2rr AD 2 - 4R2 sm. 2 3rr AE2 - 4R2 sm. 2 412rr ' AF2 -- 4R2 sm. 2 512rr ' 12 ' 12 ' It reduces to 2rr . 2 3rr . 2 4rr . 2 5rr 5 2 2 rr . = sm - + sm - + sm - + sm - + sm 12 12 12 12 12 2 ' which dreescu, Revista Matematica Timi§oara (RMT) , No. 6(1971), pp. 27, ( TiistualsoA nclear. Problem 821) A A2 . . . An be the given polygon and let S be the area of AI A2 . . . An. 55. Let 1 M There is a point inside the polygon such that Ln MA% = 2S. k=1 We write n A A L = SA kM A k+l ' n+l = I . S k=1 •

Hence

-

-

3 .2. SOLUTIONS 133 In order to have equality everywhere we must have A MAk = MAk+l , k = 1 , 2, . . . , n. 1 sin k7i'i.A'k+ 1 = and M the circumcenter of a cyclic polygon and all sides subtent It follows ° Thatthatis theispolygon 90 is a square, desired. arcs (of Titu A. ndreescu, Revista Matematidi 2 ( 1978) , pp. 50, Timi§oara (RMT) , No. 3528) Problem ABC with the altitude AA' and the circumradius R the following 56. In a triangle equality holds: AB · AC = 2R · AA' . (1) A A A let PI , P2 , . . . , Pn be R Let be the circum radius of the polygon I 2 . . . n and P AA AA AA the projections of a point on the circumcircle onto the sides l 2 , 2 3, . . . , n l, respectively. ( ) 1 for triangles PAiAi+I , where An+l = AI , yields Applying PAi · PAi+1 = 2R · PPi , i = 1 , 2, . . . , n. as

Hence

which is a constant,

as

claimed. 2 ( 1980) , pp. 65, (Dorin Andrica, Revista 1 Matematidi Timi§oara (RMT), No. 41 23) Problem ( 1 ) from the previous problem to triangles MAI A2 , 57. Applying relation A M A MA3 A4 , MA5 6 , • • • , 2n I A2n , we obtain MA2n 1 . MA2n = 2R · MK2n l. Multiplying these equalities yields MA2n MKI . MK3 · · MK2n 1 -_ MAI · MA (2) 2 . . . 2 n Rn · For the triangles

relation

( 1 ) yields MA2 · MA3 = 2R · MK2 , MA4 • MA5 = 2R · MK4 , . . , .

3. GEOMETRY

134

Multiplying these equalities gives

MK2 . MK4 · · · MK2n = MAl . M2AR2 . . . MA2n (3) nn Similarly, we obtain MA2n MH1 . MH2 · · · MHn _- MAI · MA 2 · 2n Rn (4) A A M A A M A M A I n+ I ' 2 n+ 2 , . . . , n 2 n . by applying relation (1) to triangles 3 ) ( From , and (4) we draw the conclusion. Andrica,(2)Revista (Dorinequalities Matematica Timi§oara (RMT) , No. 2(1981), pp. 68,

Problem 4622)

58. Let x and y be the measures of the arcs subtended by the sides a and b,

respectively. We have nx + ny = 27r,

or

27r

x+y= -

.

n Let R be the circumradius of the polygon. Then x

. sm 2=

27r Now y = - x so

n

(

a ' y= b and sm 2 2R ' 2R

)

b X 7r 7r X 7r . x y SI. n 2 = sm ;; - 2 = sm ;; cos 2 - cos ;; sm 2= .

hence

.

2R '

b 7r X 7r SIn X . sm cos = - + cos -. 2 2R n 2 Squaring boths sides yields b X 7r . X 7r 2 X = b2 7r . 2+ 2- cos - sm - + cos2 - sm2 sm cos 2R n 2 2 2 4R2 or b b2 X 7r 7r . X 7r . X . 2. 21 - sm = + 2 - cos - sm - + cos2 - sm sm -. 2 2R 2 2 2 4R x a . ' Smce sm 2 = we 0b tam sin2 � (4R2 - a2 ) = b2 + 2ab cos � + a2 cos2 � . .

-

n

·

( n 2R '

Therefore

)

n

--

n

n

n

n

R=

(Dorin Andrica)

.

--

n

� Ja2 + 2ab cos :':n + b2 • 2 sin n

n

2

3.2. SOLUTIONS

59. We have from the Cauchy-Schwarz Inequality that

for any positive numbers Setting n = 4,

135

ai , bi , i = 1 , 2, . . . , n.

yields

3V " SPA BCD S2 - , V where is the volume of the tetrahedron. Then " SPBCDA 3VS2 = SS2 = � ' r r L....J

L....J

>

>

as desired. The equality occurs if and only if

a Jai bi i 1234 b i 1234 for = , , , hence i = a - I , = , , , . Then PAl = PBI = PCI = PDI , ABCD . P so is the kincenter of tetrahedron RTiemarA. The inequality holds for convex polyhedra circumscribed about a sphere. 982 ( tu ndreescu, 1 Romanian Selection49Test, ( 1 (1982) , pp.IMO82, Problem 10) ; Revista Matematidi Timi§oara RMT ) , No. i 1 i 4 0 and 60. All summations here range from = to = . Let be the circumcenter A A A A GA GA �= R be the 4 2 I · 3 circum radius of By the Power-of-a-point Theorem, . i 4 OG i 1 R2 - 2 , for � � . Hence the desired inequalities are equivalent to ( 1) and (R2 - OG2 ) L G� � L GAi • ( 1 ) follows immediately from i Now (3) �= V b;

3. GEOMETRY

136

) P denote the vector (4)

by the Arithmetic-Geometric-Mean Inequality. To prove (3 , let from to the point Then

0

P.

(4) vanishes. By Cauchy-Schwarz L GAi L GA1 2:: 16, and i GA1 i > "GA ·. GA1 i > -161 (" (GA · ) 2 ) 2 " 4-1 "GA2. " ). 3 ( from follows also (2) Hence ( Titu Andreescu, IMO 1995 Shortlist) )

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

6

� 6

-

6

Z

6

-

6

�

C hapter 4 TRIGON OMETRY

PROBLEMS 1 . Prove that

27r 37r 7r ctg2 7 + ctg2 7 + ctg2 7

=

5.

2. Prove that

3 X + 47r X X + 2 7r cos3 - + cos3 -- + cos3 = - cos x 4 3 3 3 --

for all x E JR. 3. Evaluate the sum

Sn =

n- 1

L sin kx cos(n - k)x. k=1

4. Evaluate the sums

S1

=

sin x cos 2y + sin 2x cos 3y + . . . + sin(n

S2

=

cos x sin 2y + cos 2x sin 3y + . . . + cos(n - l )x sin ny.

-

l)x cos ny,

5. Evaluate the products

1) PI = (1 - tg1°) (1 - tg2 ° ) . . . (1 - tg8g 0 ) ; 2 ) P2 = ( 1 + tg1 ° ) (1 + tg2° ) . . . ( 1 + tg44°). 6. Prove that

(4 cos2 go - 3) (4 cos2 27° - 3)

=

7. Let x be a real number such that sec x - tan x 8. Evaluate the product

where Ixl

< 2n7r+2 • 1 39

tan go.

=

2. Evaluate sec x + tan x .

4. TRIGONOMETRY

140

9. Let a, b, c, d, x be real numbers such that x f:. k7r, k E Z and sin x sin 2x sin 3x sin 4x

=

a

--b

=

-

c

d

-'

10. Let a, b, c, d E [0, 7r] such that

2 cos a + 6 cos b + 7 cos c + 9 cos d = °

and

2 sin a - 6 sin b + 7 sin c - 9 sin d = 0 .

Prove that 3 cos(a + d) = 7 cos(b 1 1 . Prove that if

+ c) ,

arccos a + arccos b + arccos c = 7r,

then

1 2 . Let a, b, c be positive real numbers such that

ab + bc + ca = Prove that

-1

1

1.

-1

arctg + arctg - + arctg = 7r, b a c

13. Let x and y be real numbers from the interval

Prove that x + y = 2 ' 7r

cos2 (x - y) = sin 2x sin 2y

(o , �) such that

( i) such that 21 ( 1 - tga) (l - tg(3) ( 1 - tg,) = 1 - (tga + tg(3 + tg,) ,

14. Consider the numbers a , (3 "

E 0,

7r

Prove that a + (3 + , = 4"

( �), Prove that ( Si�2 a ) 2 + ( cos2 a ) 2 = 1

15. Let a, b E 0,

if and only if a = b.

sm b

cos b

141

4. 1. PROBLEMS 16. Prove that

sin3 a cos3 a + sin cos

b -> sec(a - b)

b

--

7r

b

for all < a, < 2" '

0

--

1 7. Let a, (3 be real numbers with (3 2:: 1. Prove that

(1 + 2 sin2 a) (3 + (1 + 2 cos2 a) (3

2::

2(3+ 1

for all a E JR. 1 8 . Let x be a real number, x E [- 1, 1] . Prove that

for all positive integers n .

_1_ < x2n + (1 _ x2 ) n < 1 2n- 1 -

19. Prove that

sec2 n x + cosec2 n x 2:: 2 n+ 1 for all integers n 2:: and for all x E

0

(0, %) .

20. Prove that

(1 + sin x) (1 + cos x) for all real numbers x.

:s

� + V2

2 1 . Find the maximal value of the expression

E = sin Xl COS X2 + sin x2 cos x3 + . . + sin x n cos X l , .

when X l , X 2 , . . . , Xn are real numbers. 22. Find the extreme values of the function f : JR

b f (x) = a cos 2x + cos x + b O b where a, , are real numbers and a, > .

-t

1R,

c,

c

23. Let ao, a I , . . . , an be numbers from the interval

tan ao -

(

Prove that

(0,

7r

/2) such that

�) + tan (a1 - � ) + . . . + tan (an - �) tan ao tan a1 . . . tan an

2::

n

n+ 1 .

24. Find the period of the function

f(x) = cospx + cos qx, if p, q are positive integers.

x E JR

2::

n

- 1.

12

4. TRIGONOMETRY

4

25. Let

aD = + + �

V3

for all n .

an+1 = 2(aa2nn-+25 for 2:: O. Prove that an = cot ( -2n3--37r) -2

v'6 and let

)

n

26. Let n be an odd positive integer. Solve the equation

cosnx = 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

. cosy + s.m y cos z+sm. z cos x = 3 Prove that the equation . . 2xsm. 3xsm. 4x = 3 smxsm

'2

Slll X

29.

4

has no real solutions. 30. Solve the system of equations

=3 { 32sisinn xy ++ 23cosy cos x = 4.

{ xsix+yn y=+:1 - x cosy = V;

31. Solve the system of equations

4'

2

32 . Solve the system of equations

1

x+y+ Z = 437r tgx + tgy +tgz = 5 1 tgx . tgy . tgz = .

33. Prove that in any triangle

acosA + b cos B + c cos C = 2R2abc

4. 1 . PROBLEMS 34. Prove that in any triangle A B C A B C � cos3 "2 sin "2 sin "2 cos "2 cos "2 cos "2 � sm. 2 "2A ' �

1 43

�

=

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

L sin nA sin nB sin nC (_l) n+l + cos nA cos nB cos nC and L cos nA cos nB sin nC sin nA sin nB sin nCo ABC such that 36. Consider a triangle A B B C C A A sin sin + sin sin + sin sin =

=

=

and

( 1 + sin A) (l + sin B) (l + sin C) 2 (A + 1) AB C has a right angle. Prove that triangle ABC be a triangle such that 1 37. Let A > be a real number and let a>' cos B + b>' cos A c>' and a2>.- 1 cos B + b2>.- 1 cos A C2>'- 1 . Prove that the triangle is isosceles. ABC is equilateral if and only if 38. Prove that the triangle A B C 1 ( 2 b2 2 ) tg "2 + tg "2 + tg "2 a + +c . ABC be a triangle such that 39. Let in2 B + sin2 C 1 + 2 sin B sin C cos A. s ABC =

=

=

= 4S

=

Prove that triangle 40. Let

has a right angle.

ABC be a triangle such that

(cot � r (2 cot �r + (3 cot �r (�;)" , +

=

where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine these integers.

ABC

T

4. TRIGONOMETRY

144

41. Prove that in any triangle

A B C 9 . � sm - cos - cos - < - . 2 2 2 -8 L..J

42 . Prove that in any triangle

a-b2 + -b2 + -e2b - (sm. 2 -A + sm. 2 -B + sm. 2 -C ) . e ea a 2 2 2 >

4

43. Prove that in any triangle

� e3 - 16p3 . 44. Prove that in any triangle A 2B 2C 2 sec - sec - sec be 2 + ea _2 + ab 2_ p29 45. Prove that in any triangle 3V3. !!. ABC be a triangle. Prove that 46. Let 3A . 3B . 3C A - B B - C C A .sm < cos -- + cos -- + cos -- . + sm - + sm - 2 2 2 2 2 2 ( b) ( bi) 2002 = a- bi , a, b E 47. Find the number of ordered pairs a, such that a + cos A

a3

+

cos B

b3

+

__

__

cos C

>

__

> _. -

> r -

-

R

48. Find

Imz 5 z EC\1R Im5 z and the values of z for which the minimum is reached . min

-­

ZI, Z2 , . . . , Z2n be complex numbers such that IZl l = IZ2 1 = . . . = IZ2n l and arg ZI arg Z2 . . . :::; arg Z2n Prove that 49. Let :::;

:::;

:::;

7r.

50. For all positive integers k define

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

Al A2 U

U···U

Am An m+1 U An m+2 U · · · U An . C

4. 1. PROBLEMS

145

Z1 ,Z2 ,Z3) be3 complex numbers, not all real, such that I Z1 1 = IZ2 1 = I Z3 1 = 1 Z2 + Z3 - Z1 Z2Z3 E JR.

5 1 . Let and 2 ( Z1 +

Prove that

� is odd and co, C1 , . . . , cn- 1 the complex roots of unity of order n. Prove that nII- l ( b % ) ( � b � ) 2 k=O ba + c = a + for any complex numbers a and . 53. Let n be an odd positive integer and co, C 1 , . . . , cn - l the complex roots of unity of order n. Prove that nII-1 ( b % ) n bn kb=O a + c = a + for all complex numbers a and . I 1 I 1 I 1 54. Let Zl,Z2 ,Z3 be distinct complex numbers such that Z1 = Z2 = Z3 = Prove that 1 1 1 1 I Z1 - z2 11z1 - z3 1 -- + I Z2 - zd lz2 - z3 1 + I Z3 - zl llz3 - z2 1 >- -. ---r2 I 1 I 1 I 1 55. Let Zl ,Z2 , Z3 be distinct complex numbers such that Z1 = Z2 = Z3 = r and Z2 f:. Z3· 52. Let n be an even positive integer such that

r.

Prove that

I 1 If Z is a complex number satisfying z3 + z - 3 :::; 2, the inequality show that I z + 56. 1 z -1 2. ( ) following property: 57. The pair ZI, Z2 of nonzero complex numbers has the O 2 there is a real2number 3 a E [-2, 2] such that z? - aZl Z + z� = . Prove that all pairs (zf, z�), n = , , . . , have the same property. AA A 1 58. Let 1 2 . . . n be a regular polygon with the circumradius equal to . Find IIn PAj when P describes the circumcircle. the maximum value of max j=l :::;

.

4.

146

TRIGONOMETRY

59. Let n be an odd positive integer and let interval [0, 71'] . Prove that

aI, a2, ... , an be numbers from the

L cos(ai - aj ) 2:: 1 --n · 2 ijn

l� < �

n be a positive integer. Find the real numbers ao and akl , k, 1 , n, l k > , such that L akl cos 2 (k - l) x sin2 -sm. -2nxx ao + 9
60. Let

=

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

1

SOLUTIONS 1 . We prove that

k7r n (2n - 1) ' 3 n+

(1)

� cot2 21 � k= l _

for all integers > O. Consider the equation

n

(2n + l)x 0, 27r with roots 7r 2n + 1 ' 2n + 1 ' . . . , 2nn7r+ 1 . (2 Expressing sin n + l)x in terms of sin x and cos x, we obtain 2n : 1 2 ( ( ) cos2n x sin x en : 1) COs2n-2 sin3 X + . . . sin n + l}x n 1 n 1 Sin2 n+ 1 X ( e : ) cot2n - e : ) cot2 n -2 x + . . . ) k7r 2 2 · SIn · 2n+l x 0 , we have , k - 1 , , . . . , n. Slnce Set x n + 12 ( n : 1) cat2n en: 1) cot2n-2 x + . . sin

=

_

=

X

=

-

=

--

-I-r

-

X

Substituting y = cot2 x yields l n y

_

.

= O.

en : ) en : l) yn- l + .

. . = 0,

-

with roots cot2 cot2 �, ficients and roots, we obtain

2n7r+_,1 2n + 1 . . . , cot2 2nn7r+ 1 . Using the relation between coef_ 2n + 1 ( ) n ( 2n - 1 ) k7r n 3 2 � cot2 n + 1 en : 1 ) 3 desired conclusion follows. 2 ( 979) , 51 , Dorin An ndr3,ica,theRevista (Setting Matematidi Timi§oara (RMT), No. 1- 1 �

=

=

pp.

Problem 3831)

147

148

4. TRIGONOMETRY

Applying the identity cos t = 4 cos3 3"t - 3 cos 3"t ' t E 1R for t = x, t = x + 271", t = x + 471" and summing up the three relations, we obtain X +-271" + cos3 X +-471" ) 3 cos x = 4 (cos3 3'X + cos3 3 3 X + 471" ) . -3 (cos -X3 + cos X +3 271" + cos -3 On the other hand, X +-471" = 2 cos 6471" cos 2x + 471" + cos x +271" + cos X + 271" cos 3"x + cos 6 3 3- = 3 2 x � 2" = 0 = (2 cos ; + 1) cos and (the follows. ir ca, Revista Dordesired in Andidentity Matematica Timi§oara (RMT) , No. 2(1975), pp. 44, Problem 2124) 3. We have 2.

--

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, n-1 = -- sinnx. 8 n (Dorin Andrica, Gazeta Matematica2 (GM-B), No. 8(1977), pp. 324, Problem .

+

+

so

16803; Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 30, Problem 3055) 4. Note that 81 + 82 = sin(x + 2y) + sin(2x + 3y) + . . . + sin[(n - l)x + ny] and 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 ) = .

.

149

4.2. SOLUTIONS h n l sm. [y + (n - I) 2hI ] .sm T .sm -h I S2 - SI = sin(y + h2 ) + sin(y + 2h22 ) + . . . + sin(y + (n - I ) h ) = 2

h l . [ ( h I ] . n h2 . [ ( h 2 ] n . SI = 1 sm T sm y +h n - 1) 2 I sm T sm y +h n - 1) 2 2 2 s. 22 sm. 21 and h; ] hl [ ( hI ] [ n n 2 � ( S 1 sin 2 sin Y +h n - 1) 2 + 1 sin sin Y +h n - 1) . 2-2 2 sm. 22 sm. 2I (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), 65,

Hence

m

_

Problem 3056 ) 5 . We have PI = 0 because of the factor 1 - tan 45 ° = O. On the other hand we have 44° + sin 44°) = P2 = (cos 1 ° + sincos1 0)1 °. .. .. .(cos cos 44° � cos 10 + � sin 10 � cos 440 + � sm 440

(

(.;2)

) ( • • •

pp.

)

= �--------------� 44 --�----------------�

cos 1 ° . . . cos 44° sin 46° . . . sin 89° (v'2) 44 = 222 . = ( Titu Andreescu) cos 1 ° . . . cos 440 2

= 4cos3 x - 3cosx, so 4cos2 X - 3 = -- for all x f. (2k +6. 1)We. 90°,havek Ecos3x cos x Z. Thus 27° . cos 8 1° cos 81° sin 9 ° (4cos2 9 ° 3)(4 cos2 270 3) = cos cos go cos 270 = cos 9° = cos 9° = tan 9° , desired. ( Titu Andreescu) _

as

_

cos3x

1 50

4. TRIGONOMETRY

From the identity 1 + tan2 x = sec2 x it follows that 1 = sec2 x - tan2 x = (sec x - tanx)(secx + tan x) = 2(secx + tanx), = 0.5. so secx Andxreescu, ( Titu+ tan American High School Mathematics Examination, 1999, Prob­ lem 15) 8 . Since 1 + tan2 2kx cos2 2kx (1 - tan2 2k x)2 - cos2 2k+l X for all Ixl < 2:+2 ' it follows that cos2 2k xl = cos2 2x Pn = IIn cos2 k= l 2k+ X cos2 2n+1 x ' 7.

(Dorin Andrica) 9.

Let

4x - A sin3x - sin2x - sinsin-x - d b a Then sin2 4x = 2sin2 2x(l - sin2 2x). Because sin2 4x = 2 sin2 2x(1 - sin2 2x), we obtain dA2 = 2b2 (1 A A2 b2 ). On the other hand, sin 3x = C, sin x = a, and since sin 3x = sin x (3 - 4 sin2 x), we have = (a(31 ) - 4A2 a2 ). A Eliminating from the relations and (2) yields 2a3 (2b3 - � ) = b4 (3a ) desired. (D i A d i ) _

_

_

_

c

.

_

c

-c ,

as

or n n r ca

as

Rewrite the two equalities 2 sin a - 9 sin d = 6 sin b - 7 sin 2 cos a + 9 cos d = - 6 cos b - 7 cos By squaring the two relations and adding them up we obtain 1 85 + 8 cos(a + d) = 85 + 42 cos(b + ) and the conclusion follows. 10.

c

c.

c ,

(1)

(2)

4.2. SOLUTIONS

( Titu Andreescu, Korean Mathematics Competition, 2002)

15 1

From the hypothesis it follows that b sinearccos a + arccos + arccos c) = o. Using the identity L cosacos,Bsin'Y = sin(a + ,B + 'Y) + sin a sin,Bsin'Y and the formulas sin(arccosx) = � and sin(arcsinx) = x, x E [-1, 1], we obtainsb a � + be� + eaJ1=b2 = )(1 - a2 )(I - b2 )(I - c2 ) , as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 64, Problem 3054) 1 2 . The identity arctgx + arctgy + arctgz = arctg 1 x-+(xyy ++zyz- +xyzzx) + k7r implies b + be + ea - 1 1 1 1 a b + e) + k7r. b = arctg arctg + arctg + arctg e ( a ea + b a Because ab + be + ea = 1, we obtain arctg -a1 + arctg -1b + arctg -1e = k7r, where k is integer. Note that 0 < arctgx < "27r for all real x > 0, hence 1 1 37r 1 o < arctg � + arctg z; + arctg c < 2· Therefore k = 1 and arctg -a1 + arctg -1b + arctg -e1 = 7r, as claimed. ( 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 (cos x cos y - sin x sin y) 2 = 0 11.

4. TRIGONOMETRY

152 Hence

cos2 (x + y) = 0, and (since (0, %), we obtain x + y = %, as desired. Titu x,AnydEreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1977) , pp. 42, Problem 2826) 14. Expanding yields 1 (1 - tana - tan,B - tan')' + tan a tan,B + tan,Btan,),+ "2 + tan,), tan a - tan a tan ,B tan,),) = = 1 - (tan a + tan,B + tan,),), or tan a + tan,B + tan')' - tan atan,B tan,), = = 1 - tanatan,B - tan,Btan,), - tan')'tana. (1) Since a,,B, ')' E (0, i), we have ° < a + ,B + ')' < hence tan a + tan,B + tan')' f= tan a tan ,B tan,),. From relation (1) we derive 1 - tan a tan,B - tan,B tan')' - tan,), tan a tan a + tan,B + tan,), - tan a tan ,B tan ')' = 1, therefore cot(a + ,B + ')') = 1. Hence( Tia +A,B +d ')' = i, desired. tu n reescu, Revista Matematica Timi§oara (RMT) , No. 1(1973) , pp. 42, Problem 158 2) 1 5 . The relation in the statement is equivalent to 4 a cos4 a ) = 1, (sm. 2 b + cos2 b) ( -sin cos b sm. -2-b + � or sin2 b sm4 a + cos4 a + -cos .sm-22-bb sm. 4 a + � cos b cos4 a = 1. It follows that sin2 bb cos4 a = 1 , . 2 acos2 a + -cos.-2-b sm. 4 a + � 1 - 2sm cos sm2 b hence COS b . sin b cos2 a) 2 = 0. ( --:-2 a sm sm b cos b Furthermore, cos b sm. 2 a = -sin b cos b cos2 a, sm b 7r ,

as

•

--

--:--

153

4.2. SOLUTIONS

or tan2 a = tan2 b. Because a, b E (0, -i), we obtain a = b. The converse and we are done. Alterna it ve soislutclear ion. From the given relation we deduce that there is a number e E (0, -i) such that cos2 a sin2 a -sm. -b- = sine and -cos--b = cose Hence sin2 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) 2 Since a, b, e E (0, i), we have b e = ° and a = b + e, hence a = b, ( Titu A ndreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1977) , desired.41 , Problem 2825 ; Gazeta Matematica (GM-B), No. 11 ( 1977) , 452, Problem 16934) 16. Multiplying the inequality by sinasinb + cosacosb = cos(a - b), we obtain the equivalent form 3 a cos3 -a ) (sin a sin b + cos a cos b) 2:: 1 . ( Sin -sm. -b- + -cos b But this follows from Cauchy-Schwarz Inequality, because, according 1to this in­ equality, d ) side is greater than or equal to (sin2 a + cos2 a)2 = . ( Ti theA left-hand as

pp .

pp.

tu n reescu

17.

Using the inequality

for m 2:: 1 we obtain (1 + 2 sin2 a)p + (1 + 2 cos2 a)p :::: 2 ( 2 + 2 sin2 � + 2cos2 a ) = 2,8+1 , desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1974) , 30, Problem 1942) [ 1 1] 18. Because x E - , , there is a real number y such that x = siny. It suffices to prove that 1 >- sin2n y + cos2n y > � - 2n 1 P

as

pp.

-

4. TRIGONOMETRY

154

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 + cos2 = 1, desired. For the right-hand side we use the inequality x? + x� - ( Xl + X2 ) n 2 2 Hence sin2n y + cos2n y - ( sin2 + cos2 y ) n 2 2 claimed. Alternative Solution. By setting u = x2 and v = 1 - x2 the inequality becomes [0, 1] and u + v = 1, we have un � U, vn � v, 1 n n E u,v Because 1. u v + � � 2n- l un + vn � U + v = 1. Also, by the power mean inequality, implying 1_ un + vn - 2 ( u +2 V ) n = 2 ( !2 ) n = _ -l . 2n (D i A d i ) y

y

as

>

y

>

as

>

or n n r ca

We have sec2 x = tan2 x + 1 2:: 2 tan x and cosec2 x = cot2 x + 1 2:: 2 cot x by the AM-GM inequality. It follows that sec2n x + cosec2n x 2:: 2n (tann x + cotn x) . Since tann x + cotn x 2:: 2, we obtain sec2n x + cosec2nx 2:: 2n+l , desired. Alternative Solution. Using the AM-GM inequality we obtain sec2n x + cosec2n x - 2Jsec2n xcosec2nx = 2 sinn x1cosn x 12 = 2 n+ l n _ x - 2n+lNo. 3(1975), 104, Problem sin (Dorin Andrica, Gazeta Matematidi (GM-B), 19.

as

>

_ _

>

pp.

14900)

20.

We have (1

.

<

( 1 + sinx) 2 + (1 + COSX) 2

+ mnx) (1 + coo x ) 2 - 2 + 2(sinx + cos x)2 + (sin2 x + cos2 x) _

_

=

In . v

as

1 55

4. 2. SOLUTIONS 7r ) 3 In 3 3 . = 2 + (sm x + cos x) = 2 + 2 sm (x + '4 � 2 + 2, v

desired. Note for x = � + 2k7r, where k is integer. Solutionholds Alternathattiveequality . Expanding the left-hand side, we see that 1 + (sin x + cos x) + � sin 2x = 1 + ../2 sin (x + �) + � sin 2x �

- + ../2 + �2 = �2 + ../2. A d Ti 2 ( 1978) , pp. 47, ( tu n reescu, Revista Matematidl Timi§oara (RMT) , No. Problem 3500) 2 1 . We have E = sin X l COSX2 + sinx2 COSX 3 + . . . + sinxn cos Xl � 2 + cos2 n 2 2 2 2 -< sin Xl +2 cos X2 + sin X2 +2 cos X3 + . . . + sin Xn 2 Xl = 2 Therefore the maximal value of E is n2 and it is reached, for example, when 7r X l = X 2 = . .A. = X n = '4 ' D( orin ndrica, Revista Matematidi Timi§oara (RMT), No. 2( 1977) pp. 65, Problem 3058) b f b 22. Because a, > 0, it follows that the maximal value of is a + + Setting y = cos X yields f (x) = a(2y2 - 1) + by + c = 2ay2 + by + c - a. If :a E [- 1 , 0) , then the minimal value of f is � b2 + 8a(c - a) 8a 8a b ) If - 4a E ( - 1 , then the minimal value of f is f ( - 1 ) = 2a - b + c - a = a - b + c. D i A di 1 ( 1981) , 52, ( or n n r ca, Revista Matematidi Timi§oara (RMT) , No. Problem 4315) Fi S l i b 1 23. rst o ut on. Let k = btan( ak - 7r / 4) , k = 0, , . . . , n . It follows from the 1 1 hypothesis that for each k, - k and <

1

,

c.

-

-

- 00 ,

pp.

<

<

1 + b k � L ( 1 - bl ) . 09¥k5;n

( )

1

4. TRIGONOMETRY 156 thek Arithmetic-Geometric-Mean Inequality to the positive numbers 1 bz, I Applying k 1 1 1 = 0, , . . . , - , + , . . . ,n, we obtain ( 2) L ( 1 - bl ) � n ( II ( 1 bl )) l in 09::fk �n O�l::fk �n 2 ) ) ( ( 1 From and it follows that ( n ) lin n !! ( 1 + bk ) � nn+ 1 Do ( 1 - bz ) n , _

and hence that Because

1 + bbk 1 + tan (ak - �) -1 - k -- 1 - tan (ak - �) = tan ( (ak - �) + �) = tan ak , the conclusion Second Solutfollows. ion. We first prove a short lemma: Let W,X,Y,z be real numbers with x + y = w + z and I x - y l < I w - z l . Then wz
and let

::;

Then - 1 < bk < 1 and

(1) tit. = G � :� ) G � :: ) = 1 + b1 + bbi�' _ 1 j+ +k b b b b 1 1 1 < < . . . + n � nb - b, it follows First note that because k and o + + 1 b b for all 0 j, k n with j f:. k . Next note that if j + k > 0 and that bj f:. bjk,+thenk >it0follows ( 1) that the value of tj tk can be fromb the lemma applied to closer together and with the made smaller by replacing b j and bk by two numbers b b < same sum. In particular, if j 0, then replacing j and k by their average reduces the problem to the case where bi > 0 for allb i . We . may now successively replace the i 's by their arithmetic mean. As long b the i are not equal, one is greater than the mean and another one is less than the mean. We can replace one of this pair by the arithmetic mean of all the bi 'S, and the ::;

::;

as

4.2.

1 57

SOLUTIONS

other by a positive number chosen so that the sum of the pair does not change. Each such change decreases the product of the ti 'S. It follows that for a given sum of the bi 'S, the minimum 1 producti is attained when all of the bi 'S are equal. In this case we n have bi - -n + 1' for each , so >

tO t1 . . . tn ;:::

( : : �) n+1

n+ 1

=

C;) n+ l = nn+ 1 .

This proof. Solution. the Thirdcompletes We present a solution based on calculus. We set a = bo + b1 + . . . + bn , where - 1 < bi < 1 , and assume that a ;::: n - 1 . We then show that the product lIn 1 + bk k=O 1 - bk attains its minimum when all of the bk 's are equal, that is, their common value is l( a n + 1). The desired inequality will follow immediately. We proceed by induction. The2 case n = 1 was established in the discussion of (1 ) in the previous solution. For n ;::: , set ' -1 nL bk = a = a - bn n - 2. k=O The last inequality follows from a ;::: n - 1 and bn < 1 . Set b = bn and c = a' ln , so b + nc = a. By the induction hypothesis, 1 + bk ) 1 + bn ( 1 + c ) n 11 + b . 1 k=O - bk 1 - bn - 1 - c - b Thus we need to prove that (1) ( 11 +- cc ) n (�) ( n + 1 + a ) n+ 1 , 1 - b n + 1 - a l 1) where the right-hand side is obtained by substituting a (n + for each bk , k = 0, 1, . . . ,n, in the product. Next, recall that a is fixed, and that b + nc = a. Thus we can eliminate b from ( 1 ) to obtain the equivalent inequality >

(IT

>

> _

n+ 1 ( G � � r G � : � :�) � :: � � : )

(2)

Now bring all terms in (2) to the left-hand side of the inequality, clear denomi­ nators, and replace c by x. Let the expression on the left define a function f with

158

4. TRIGONOMETRY

0 � x < 1 , I(x) attains its To establish (2) it isJ( sufficient to show that for minimum value at x = a n + 1) . Towards this end we differentiate to obtain ) t (x) = n (a - (n + l )x)((l + x)n- 1 (n + 1 - a) n+1 - ( 1 - x)n- 1 (n + 1 + a)n+1 (a - (n + l ) x) g (x) , = n where g (x) = (01 + x) n- 1 (n + 1 - a) n+1 ( 1 - x) n- 1 (n + 1 + a) n+1 . It is clear that a f' ( _ n +_1 ) , so we check the second derivative. We find 0 ( ll !" ( n : l ) = -n n + g ( n : l ) > , ' so I has a local minimum at x = (aJ() n + 1) . But I (x) could have another zero, t, obtained the equation g x = O. Because ' ( ) by solving )( ( 1 g x = n 1 + x) n-2 (n + 1 - a)n+1 + (n - 1 )(1 - x)n-2 (n + 1 + a)n+1 is( obviously one1))solution to (the equation positive for all x E [0 , 1 ) , there is at (most 0 ) ) J( l g x = in this interval. ItJ(is easy1) to check1 that g a( ) n + < 0 and g > O. Thus there is a real number t, a "n + < t < , with g' t = O. For this t we have I (t) = n (a - (n + l )t)g (t) < O. I , and no other extrema exist on the interval t Thus, is a local maximum for (0 , 1 ) . 1(1) � l (aJ(n + 1 )) . Note that the case x 1 The only thing left is to check that is also an extreme case with bo = b1 = . . . = bn- 1 = 1 . This case does not arise) in I( 1 our problem, but we must check to be sure that on the interval 0 � x < , x has a minimum at x = aJ(n + 1 ) . We have 1( 1) = 2n ( 1 + a - n)(n + 1 - a) n+1 � 0, since n - 1 � a � n + 1 , andJ(l (aJ(1n) + 1 )) = 0 (by design). Thus I(x) indeed attains a unique x=a n+ . Andreescu,at USA ( Tituminimum Mathematical Olympiad, 1998 , Problem 3) T d 24. Let be the greatest common divisor of p and q. We prove that = � is the lowest positive period of the function f . It is clear that I(x + T1 ) = I(x) I. for all real x, thereforeTT is 0a period of function T AT1 and l(x+ T1 ) = A 0 > > Suppose there are and an integer such that 1 T 0 I(x) for all real x. Then I( 1 ) = 1( ) = 2 , so cos pT1 + cos qT1 2, =

-

=

-

=

=

=

4.2. SOLUTIONS 1 59 therefore COSpT1 = COSqT1 = 1 . It follows that T1 = 2k 7r = 2k27r P q O k k 2 ' for some integers 1 T> . 27rA T AT Smce = 1 and 1 = d ' k1 k2 1 k (Ad) , q = k2 ( Ad) . = q = Ad and so p = 1 P d d(p, q) , so A = 1 , hence T = T1 desired. On the other hand, = gc (Dorin Andrica, Revista Matematidt Timi§oara (RMT) , No. 2 ( 1978) , 75, Problem 3695) l

•

as

25.

pp.

We have

7r 2 cos2 7r 1 + cos 17r2 cos 7r cot 24 = -sm.--247r- = 2sm. -7r cos24 -7r --=-==. -17r2 = sm 24 24 24 1 + cos (i - �) 1 + V; + 4 v: + VB + v'2 VB4 v'24 VB - v'2 sin (i - �) v'2) + ( VB + v'2)2 4(VB + v'2) + 8 + 4V3 4 ( VB + 6-2 4 = 2 + v'2 + V3 + VB = ao + 2. 2n-3 ) - 2 is true for n = O. Hence an = cot ( T 2n-3 ) , where b = an + 2, n � 1 . The It suffices to prove that bn = cot ( T n recursive relation becomes )2 5 ( bn - 2 bn+1 2 - 2b ' n b 2 -1 b 3 7r ' yields 2k -3or n+1 = ;b . Assuming, inductively, that bk = cotck, where Ck = n 1 bk+1 = cot2 2coCkt Ck- = cot(2Ck ) = cot Ck+1 , and (weTi areAdone. tu ndreescu, Korean Mathematics Competition, 2002) 1 26. If n = , then all real numbers x are solutions to the equation . ---���� = --����

=

_

_

_

_

_

_

1 60

4. TRIGONOMETRY

Let n > 1 and note that cosnx = (�) cosn X - (�) cosn-2 xsin2 x + . . . + 1 ) ( n 1) cos x sinn- 1 x. +(n We have two 1cases: � a) x f:. (2k + ) 2 for any integer k . Then I cosnxl = I cosxl l (�) cosn - 1 X - (�) cosn -3 sin2 X + . . . + ' ; ) n 1 (n � 1) sinn - 1 x l ::; +( ::; I cos x l ((�) I cosn-1 x l + (�) i cosn -3 X sin2 xl + . . . + (n � 1) I sinn -1 x l ) < n-l

-2-

_

hence there are no solutions in this case. b) x = (2k + l) i for some integer k. Then cos x = 0 and cosnx = 0, sinceAln is odd,i soS {l (2ki + 1 ) i l k integer } is the set of solutions. ternat ve o ut on. With the substitution x = i - y the equation becomes ( 1) cos (n � - ny) = 2n - 1 siny. Because n is odd, ( 1) is equivalent to ± sinny = 2n - 1 siny. Taking modules gives ( 2) I sinnyl = 2n -1 1 sinyl · But I sin nYI � n l sinyl for all y in lR, hence n l sinyl � 2n - 1 I sinyl · If y f:. k7r , k E Il , then n � 2n - 1, which implies n E { 1 , 3 } . The case n = 1 original equation reduces to cos3x = 4cosx, that is 4iscosclear3 X -and3 cosforx =n 4=cos3 x.theTaking 0 into account that cos x f:. , this yields cos2 x = �, which is not possible. It( Tifollows Il . that y = k7r , which gives the solutions x = (2k + l) i , k E304 A d 7 978 ( ) 1 tu n reescu, Gazeta Matematidi (GM-B),1 No2 ( 1. 980) , pp. , Problem 1 7297; Revista Matematidl Timi§oara (RMT), No. , pp. 63, Problem 4107)

4.2. SOLUTIONS

The equation(Ais equivalent to G ) + sin2 x + 2B sinxcosx + G cos2 X = 0 We have the following cases: i) A + G = 0 and G 1= O . Then 2B ' cos x = 0 or cot x = - 0 hence 1 )1r I k Z (2k 2 k 1 k E Z} . ; x E E { U { arcctg ( - g ) + 1r } ii) A + G 1= 0 and G = O . Then sm. x = 0 or tan x = - A2B+ G ' hence {k1r 1 k E Z } U { arctg (-}:c ) + k1r1 k E Z } . x E iii)) AA = GB = 0G = O . Then any real number is a solution. iv = = and B 1= O . Then sin 2x = 0 Zand so k� 1 k E } . x E { equation is equivalent to0 v) A + G 1= 0 and G (1=A O . The G G ) + tan2 x + 2Btanx + = , hence B ± JB2 AG + G2 A+G tan x = for B2 + G2 � AG . It{ follows thatk l k Z } { k 1 k E Z} arctgY + 7r E x E U arctgY1 + 7r 1 if B2(D+ Gi2 �AAGd .i Otherwise there are no solution. ( ) 1 ( 1978) , or n n r ca, Revista Matematica Timi§oara RMT , No. Problem 3429) 28. The equation is equivalent to 2 sin x cos Y + 2 sin Y cos z + 2 sin z cos x = 3, or (sin x - cosy)2 + (siny - COS Z)2 + (sinz - COS X)2 = O. It follows that sin x = cosy, siny = cosz, sinz = cosx. Hence

16 1

27.

-

_

pp.

89,

162

4. TRIGONOMETRY Z+X=

(4k3 + 1) '27r

for some integers k1 ' k2 ' k3 and therefore [4(k1 - k2 + k3 ) + 1] 7r ' y = [4(k1 + k2 - k3 ) + 1] 4"7r ' X= 4" and [4( k k k ) 7r k k k Z = - 1 + 2 + 3 + 1] 4" ' 1 , 2 , 3 E Z. ( Titu Andreescu, Gazeta Matematica (GM-B ) , No. 11 ( 1977) , 451, Problem 16931; Revista Matematica Timi§oara (RMT) , No. 1-2 ( 1979) , 52, Problem 3835) 29. Note that sin x sin 2x sin 3x sin 4x = � (cos 3x - cos 5x) (cos x - cos 5x) = 1 = 4 ( cos2 5x - cos 3x cos 5x - cos 5x cos x + cos x cos 3x ) = 1 6 3 = 8" ( 2cos2 5x - cos2x + cos8x - cos4x + cos6x ) < 8" = 4' hence( Tithe equation has no solution. A d tu n reescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1977) , pp. 41, Problem 2923) 30. Squaring both equations and summing up yields 4(sin2 x + cos2 x) + 9 (cos2 + sin2 y) + 12 (sinxcosy + sinycosx) = 25, or 13 + 12 sin (x + y) = 25. Hence sin (x + y) = 1, and so (4k + 1 ) '27r X+y= for some integer k. It follows that sin x = cosy and siny = cosx. Turning back to the system we obtain 4 .smx = cosy = 53 and smy . = cos x = 5 ' 4 hence 3 tanx = 4' tany = 3 · Note that sinx, cosx, siny, cosy are all positive, therefore x = arctan 43 + 2k7r pp.

Y

pp.

4.2. SOLUTIONS

163

4 217r and y = arctan 3' + , l k for some . Andreescu,andRevista 2 ( 1978) , 74, ( Tituintegers Matematica Timi§oara ( RMT ) , No. Problem 3694) [ 1 1] 31 . Observe that x E - , and x = sin (arcsin x) , � = cos(arcsin x) . From the first equation we obtain . .j2 cos(y - arcsmx ) = 2' then y - arcsin x = ± � + 2k1l". . x + y = "4 ' we get Usmg x + arcsin x = � ± � + 2k1l", k. for some Caseinteger 2k1l". Because x E [- 1 , 1] and arcsin x E [-i, iJ , we 1 . x + arcsin x = i have k = 0, hence x + arcsmx = 2' Therefore arcsmx = 2' - x, or x = cosx. For this equation there is only one solution Xo E (0, � ). The system has the solution Case 2 . x + arcsin x = 2k1l".x = Xo, = 4 - Xo Using similar arguments, k = 0, so arcsin x = -x. This equation has the unique solution x = ° so the system has the solution pp.

11"

11"

•

•

11"

Y

11"

-

x = o, ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1 977) , 41 , 2824 ) Problem pp.

32 .

Using the formula + tany + tanz - tanxtanytanz tan (x + y + z) = 1tanx - tanxtany - tanytanz - tanztanx

---=�---------=�--

----

4. TRIGONOMETRY

164

we have

- 1 = tan -37r4 = 1 - tan x tan y - tan5 -y1tan z - tan z tan x

-------

Hence

tan x tan y + tan y tan z + tan z tan x = 5. The equation

t3 - 5t2 + 5t - 1 = 0

has roots tan x, tan y, tan z, from the relations between the roots and the coefficients. On the other hand, the equation has the roots 1, 2 + V3, 2 - V3, hence {x, y, z} == 7r 57r 7r . 4' + k7r, + h7r, + p7r , for some mtegers 12 12 Revista Matematica Timi§oara (RMT), No. 8(1971), pp. 27, Problem 1018)

{ Ti A d ( tu n reescu,

}

k, l ,p.

33. Using the Extended Law of Sines we obtain a cos A +

b cos B + c cos C = 2R( sin A cos A + sin B cos B + sin C cos C)

==

= R(sin 2A + sin 2B + sin 2C) = R(2 sin(A + B) cos(A - B) + sin 2C) = A-B-C A-B + C = cos = 2R sin C(cos(A - B ) + cos C ) = 4R sin C cos 2 2 = -4R sin C cos as desired.

( i - B) cos ( i - A)

==

4R sin A sin B sin C ==

;!�,

( Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1977),

Problem 3060)

34. Because A + B + C = 7r, we have

B A B C . C . "2 cos "2 = sm cos "2 + sm "2 cos "2 '

Hence

B C A . A C . cos - + sm cos - = sm - cos - ' 2 2 2 2 2 A B . B C A . cos - = sm cos - + sm - cos - ' 2 2 2 2 2 A . B B C . C - cos - sm cos - sm - cos 2 2 2 2 2 B . C A . A C cos - sm - cos - sm - cos 2 2 2 2 2

.A

C B . B A cos - sm - cos - sm - cos 2 2 2 2 2

= 0,

pp.

65,

4.2. SOLUTIONS

165

because the first column is the sum of the other two. Computing the determinant, we obtain A . B . C A A B C ' 2- sm - = cos - cos - cos sm cos3 - sm 2 2 2 2 2 2 2' as desired.

L (Dorin Andrica)

L

35. Denote

E1 = L sin nA sin nB cos nC

Observe that

and

&

=

L cos nA cos nB sin nCo

(cos A + i sin A) (cos B + i sin B) (cos C + i sin C) = = cos(A + B + C) + i sin(A + B + C) = = cos 7f + i sin 7f =

By de Moivre's formula,

- 1.

(cos nA + i sin nA) (cos nB + i sin nB) (cos nC + i sin nC) = ( _ l) n . Expanding the brackets yields

- E1 + iE2 + cos nA cos nB cos nC - i sin nA sin nB sin nC = ( _ l)n . Hence E1 = ( _ 1) n+1 + cos nA cos nB cos nC and = sin nA sin nB sin nC. (Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 1(1978) , pp. 65, Problem 3278) 2 36. Subtracting from the second equality the first multiplied by yields O 1) (sin A - (sin B - l) (sin C - 1) = . 1 , so AB C is a right triangle. Hence sin A, sin B or sin C is A d Ti ( tu n reescu, Revista Matematidi Timi§oara (RMT), No. 2(1978), pp. 49, 1 &

Problem 35 4)

37. Note that

a cos B + b cos A The system of linear equations

{ aa2A

b

- c = O.

cos B + cos A - = 0 =0 cos B + cos A cos B + cos A

a A- 1

bA

c A b2 A- 1- c - C2 A- 1 = 0

4. TRIGONOMETRY

166

has the solution (cos B, cos A, - 1) and is homogeneous. Therefore the determinant Ll = is zero. On the other hand,

a a2A-A 1 a

b b b2AA- 1

c c2AA- 1 C

1 1 1 b b 1 A-1 A A Ll = a c ( a ) ( b ) ( c - 1) = A-1 2 A- 1 2 cA-1 2 a b ( b )( ) )(b = a c aA- 1 _ A- 1 aA - 1 _ cA - 1 A - 1 _ cA - 1 . b b or c = a, hence the triangle is isosceles. Therefore (Dorina A=nd,rica,= cRevista Matematidi Timi§oara (RMT) , No. 2(1977), pp. 89 , Problem 3199)

38. The relation is equivalent to

J(P - b)(p - c) J(P - c)(p - a) J(P - a)(p - b) 1 2 2 2 Jp(P - a) + Jp(p - b) + Jp(p - c) = 4S (a + b + c ) or 1 L( P - a)(p - b) = 4S1 (a2 + b2 + c2 ) . S Expanding the brackets yields _p2 + ab + bc + ca = 41 (a2 + b2 + c2 ) , then 4(ab + bc + ca) = a2 + b2 + c2 + (a + b + C)2 . It follows that (a - b)2 + (b - C)2 + (c - a)2 = 0, b hence( Tia = A =dc. tu n reescu, Revista Matematica Timi§oara (RMT) , No. 2 (1972) , pp. 28, ....--;= .:..-..: 7='====;�

-

-

Problem 1160)

39. By the Extended Law of Sines,

On the other hand, so

a = 2R sin A, b = 2R sin B, c = 2R sin C. sin2 B + sin2 C = sin2 A + 2 sin B sin C cos A.

4. 2. SOLUTIONS

167

From the hypothesis we have sin2 B + sin2 C = + 2 sin B sin C cos A,

1

1 . It follows that A = �, hence the triangle ABC is right, as (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2( 1977) , 52,

therefore sin2 A = desired.

pp.

Problem 3838)

2 7

40. Because 6 2 + 32 + 2 = 2 and

s

A B C A B C + cot + cot = cot - cot - cot 2 2 2 2 2 the given relation is equivalent to 2 2 2 = (62 + 32 + 22 ) cot + 2 cot + 3 cot - = cot

r

-

-

2'

-

(1)

[ ( �) ( !) ( � ) ]

(

A

B

= 6 cot "2 + 6 cot 2"

+ 6 cot 2C ) 2

This means that we have equality in the Cauchy-Schwarz inequality. It follows that

A cot ­ 2 6

__

B 2 cot 2

C

3 cot 2

2 3 7 ) ( 1 Plugging back into gives cot � = , cot � = �, and cot � = �. Hence by 28 130 Thus the side the Double angle formulas, sin A = !.- , sin B = , and sin C = �. 45 25 65 lengths 26, 40, and . ( Tituof ATnaredreescu, USA Mathematical Olympiad, 2002, Problem 2) 41 . Summing up the formulas

ra = 4R sm. 2A cos 2B cos 2C ' rb = 4R sm. 2B cos 2C cos 2'A rc = 4R sm. 2C cos 2B cos 2A yields B C ra +4rb + rc '" Slll . A R " 6 2" cos 2 cos 2 = 4 R On the other hand ra + rb + rc = + r, hence 4R + r 1 r C '" sm . A cos B cos 2 = ----;w:- = + 4R " 2 6 2"

4. TRIGONOMETRY

168

Because � �

r,

it follows that A B C 1 9 . < 1 + '" sm cos - cos - = L...J 2 8 8' 2 2 -

as

desired.

-

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 49,

Problem 3510)

42. We have

or

Because � + �

a2 b e 2 cos A + - = - + - . be e b � 2, we obtain A a2 � 2(1 - cos A) = 4 sin2 "2 be

and likewise

b2 ' B -> 4 sm2 - ' 2 ae Summing up these inequalities yields

(

)

A . 2B . 2C � � 2 . 2> 4 sm + sm - + sm - ' -+-+-be ea ab 2 2 2 as

desired.

( Dorin Andrica)

43. We have

so Likewise, and

2be cos A b2 +1=2 2 a a 2ae cos B b2

+ ae2

2'

+ 1 - ab22 + eb22 _

2ab cos C 1 a2 b2 + - 2 + 2' e2 e e Summing up these equalities implies 2be cos A 2ae cos B 2ab cos C b2 3+ + + b2 a2 a2 e2 _

_

( + b2 ) + a2

4. 2. SOLUTIONS hence

169

be cos A

ea cos B ab cos C a2 + b2 + e2 -> �2 ' 3 cos A cos B cos C > -+ + -. 3 3 3 a b 2abe e

and moreover

--

By the AM-G M inequality,

'

--

( + ) = 16p813 '

3 3 3 3 � 2abe 2 a b + e

therefore

cos A

-as

cos B cos C > 81 + ---,;s + � - 16pS '

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1975), pp. 46,

Problem 2134)

44. We have

= (p

2=

B 1 be ea A , sec2 , sec2 2" = --A p(p - b) cos2 - P - a) 2 so it suffices to prove that 1 1 1 9 > -. p-a p- b p-e - p Setting = p - a, P - b, = P - e in the inequality

Y=

x

yields

sec2

2 = p(p C

ab

- e)

,

-- + -- + -Z (x + Y + Z) ( .!. + � + � ) � 9 x Y Z 1 1 1 ) > 9, P ( -- + -- + -- p-a p-b p-e

and the solution is complete.

(Dorin Andrica)

45. By the AM-GM inequality,

p Then

= (p - a) + (p - b) + (p - e) � 3 \!(P -

a ) (p - b) (P - e),

p3 � 27(p - a) (p - b) (p - e)

so

=

p4 � 27p(p - a) (p - b) (p - e) 2782 • It follows that p2 � 3V38, and since 8 pr, !!. � 3va, as desired. Revista Matematica Timi§oara (RMT), No. 2(1982), pp. 66, Problem 4993)

( Titu Andreescu,

= r

4. TRIGONOMETRY

170

0° < a, (3" < 90° and a + (3 + , = 90° . We have ' B-C . ( a + 2, ) = . . 3A - cos sm: - = sm 3 a - cos ((3 - , ) = sm 3a - sm "'2 2 A

B

C

46. Let a = 2 ' (3 = 2 ' , = 2 ' Then

and

= 2 cos(2a + ,) sin(a - ,) = -2 sin(a - (3 ) sin(a - ,) . In exactly the same way, we can show that C-A . ( (3 - , ) . 3B - cos - = - 2 sm ' ((3 - a ) sm sm "'2 2

A-B . 3C - cos . (, - a) Slll . (, - (3 ) . - = - 2 sm sm T 2 Hence it suffices to prove that

sin(a - (3) sin(a - ,) + sin((3 - a) sin((3 - ,) + sin(, - a) sin(, - (3 ) � O.

Note that this inequality is symmetric with respect to a, (3 " , we can assume without loss of generality that < a < (3 < , < Then regrouping the terms on the left-hand-side gives

0°

90° .

a)] , ° ° 0 0 9 which is positive = sin x is increasing for < x < . Alternat ive Sasolutfunction ion. We ykeep the notation of the first solution. We have sin(a - (3 ) sin(a - ,) + sin(, - (3 ) [sin(, - a) - sin((3 -

sin 3a = sin a sin 2a + sin 2a cos a;

cos((3 - a) = sin(2a + ,) = sin 2a cos , + sin , cos 2a; cos((3 - ,) = sin(2, + a) = sin 2, cos a + sin a cos 2,; sin 3, = sin , cos 2, + sin 2, cos , . It follows that sin 3a + sin 3, - cos((3 - a) - cos((3 - ,) = = (sin a - sin ,) (cos 2a - cos 2,) + (cos a - cos ,) (sin 2a - sin 2,) = = (sin a - sin ,) (cos 2a - cos 2,) + 2(cos a - cos ,) cos(a + ,) sin(a - ,). Note that sin x is increasing and cos x is decreasing for < x < Since o < a" , a + , < each of the two products in the last addition is less than or equal to O. Hence

90° ,

0

sin 3a + sin 3, - cos((3 - a) - cos((3 - ,) � O. In exactly the same way, we can show that sin 3(3 + sin 3a - cos(, - (3) - cos(, - a) � and

0

sin 3, + sin 3(3 - cos(a - ,) - cos(a - (3) � O.

90° .

4.2. SOLUTIONS

Adding the last three inequalities gives the desired result. USA IMO Team Selection Test, 2002, Problem 1)

Z

171

( Titu Andreescu, bi bi I l J 2 b2 47. Let z = a + , = a - , and z = a + • The given relation becomes 2002 =

z.

Note that

Z

from which it follows that

I z l (l z l 200l - 1) = O. Il ( b) I z i = 1 . In the case I z l = 1, we have Z2002 = Hence z = 0, and a, = (0, 0) , or I l 200 2 1 200 1 which is equivalent to Z 3 = z = z = . Since the equation Z 3 = has 2003 1 distinct solutions, there are altogether + 2003 = 2004 ordered pairs that meet the required Andreescu, American Mathematics Contest 12A, 2002, Problem 24) ( Tituconditions. a, b be real numbers such that z = a + bi, b i- O. Then 1m z5 = 5a4b 01 a248.b2 +Letb5 and � ) 4 10 � ) 2 Imz5 = ( b - ( b + 1. Im5 z 2 Setting x = (�) yields 5 2 ( 1)2 - 4. 1m Z 5 = 5x - lOx + 1 = 5 x 1m z ( 1 ± i ) , a i- O. 1 The minimum value is -4 and is obtained for x = Le. for z = a ( Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 1(1984) , pp. 67, Problem 5221) Ml , M2 , . . . , M2n be the points with the complex coordinates 49. Let AA A MM Mn Mletn+l. I , 2 , . . . , n be the midpoints of segments l 2n , MZl 2, ZM2 ,2n l,, Z2n , and , Z

z·

5

. • .

.

. •

172

4. TRIGONOMETRY

M i 12 The points i , = , n lie on the upper semicircle Mn Mandn+lwitarhe M2n l ,in. . .origin Ml M2n , Mcentered 2 , radius 1. Moreover, the lengths of the chords OAl , OA2 , . . . , OAn are increasing.- Thus in a decreasing order, hence Z1 , � Z2n 1 1 Z2 + ;2n-' 1 :::; ... :::; 1 Zn +2Zn+ 1 1 and Al the conclusion follows. ternative S1olut2 ion. Consider Zk = r(cos tk +i sin tk ) , k = 1, 2, . . . , 2n and observe that for any j = , , . . . , n, we have i ) 2] d2 2 [( = r costj +Cost2 n _j+ + (sin tj + s nt2 n _j+l = )] 22 = r [ + 2(cos tj cos t2 n-j+l + sin tj sin t2 n - j+l = 2 2 [1 ( )] 4 2 2 t2n-j+l2 - tj . = r + COS t2 n - j+l - tj = r cos l 2 t2 j - tj and the inequalities I Therefore Zj + Z2 n-j+l = r cos n - �1 I ZI + z2n l :::; I Z2 + Z2n-l l :::; . . . :::; IZn + zn+l l :::; :::; :::; are equivalent to t2n - t l � t2 n- l - t2 � . . . � tn+l - tn · Because 0 tl t2 . . . t2n :::;(D7r, the last inequalities are obviously satisfied. n Andrica, Revista Matematica Timi§oara (RMT), No. 1(1984), pp. 67, ori5222 ) Problem A 12 1 50. Let p = , , . . . ,m and 2let Z E p e Then zP = . l ,n-m+ , . . . , n are m consecutive integers, and, since p :::; m, Note that n-m+ 2 } k. k 1 there is an integer p divides E { n - m + , n - m)+' , . . . ,n suchA that A U k = kip. It follows that zk = (zp k = 1 , so Z E k n-m+l An-m+2 U Let A . . . R n , ask claimed. can be obtained by using the fact that emar . An alternative( solution )( 1 an(-k a1n)(- lk -1 1) . . . (an(- k+l1-) 1) a - a - 1 - 1) . . . akis an( Dinteger integers a > and n > . in Aforndrallica,positive Romanian Mathematical Regional Contest " Grigore Moisil" , or 1997) tk + i sin tk,)k E3{ I , 2, 3 } . 1R 5 1 . Let Zk = cos 2 ( The condition Zl + Z2 + Z3 - ZlZ2 Z3 E implies ) 3 ( ) (1) 2(sin t l + sin t2 + sin t3 = sin tl + t2 + t3 . :::;

:::;

C

U

4.2. SOLUTIONS 173 ( ) Assume by way of contradiction that max tl' t2 , t3 < i, hence tl , t2 , t3 < i' Let . functl. On IS . concave on [0 ' '67r ) ' so t = tl + t32 + t3 E (0 , '67r ) ' The sme ( 2) 1( . '3 sm tl + sm(.2) t2 + sm. t3 ) � sm. tl + t32 + t3 . From the relations (1) and we obtain ) ( 2 sin t l + t + t3 < . t l + t32 + t3 ' - sm then 3 2 sin t � sin t. It follows that 0 4 sin3 t - sin t � , 0 i.e. sin2 t � �. Hence sin t � �, then t � i, which contradicts that t E ( , i), (tl , t2 , t3 ) i, as desired. Therefore ( Titu Andmax 2)reescu, Revista Matematidi Timi§oara (RMT), No. 1(1986), pp. 91, Problem 586 2 (28 + 1) and b f:. 0, otherwise the claim is obvious. Consider a 52. Let n = 2 � and the polynomial complex number a such that a f = xn - 1 = (X - eo )(X - el ) . . . (X - en- I ) ' 2

;:::

=

We have

and hence

f

ioo ) a ion d , 1 + a (-' n (� r =

Therefore

(

•

•

•

( +

nII( - l + b � ) bn nII-l � bn nII- l ( 2 � ) a e = (� + e ) = a + e = k=O k=O k=O b b b ] [( ) = n f (T) f (-T) n a2 2s+1 + 1 2 = n [(� t+l + If 2s+1b + b2S+1 ) 2 ( � b �)2 ( b 2 2 a ( ) s+1 2s+1 = a + . 2000) (Dorin Andrica, =Romanian Mathematical Olympiad - second round, 0 b b 53. If a = 0, then claim is obvious, so consider the case when a f:. and f:. O. =

We start with a useful lemma.

174

4. TRIGONOMETRY h l

If co, C , , cn-l are t e comp ex roots of unity of order n, where n is an odd integer, then 1 n-l A B A B II ( + Ck ) = n + n , for all complex numbers A ank=Od B . Proof. Using the identity n- l nx - 1 = II (x - ck ) k=O A . for x = - B Ylelds An 1) n- l ( A ) ( B - n + = - !! B + Ck , and the conclusion follows. b 2 Consider the equation X + a = 0 with roots Xl and X2 . Since Lemma.

• • •

0

we have

nII-l ( b � ) bn nII- l ( }( 2 ) bn nII- l ( ) nII- l 2 ) a + c = k=O ck - Xt ck - X = k=O ck - Xl k=O (€k - X k=O A B B A Using the lemma for = -Xl, = 1 and then for = -X2 , = 1 gives nII- l ( } ( ) n 1 c k - xt = -X l + = 1 X� , k=O nII- l ( 2 ) ( 2 )n 1 1 ck - X = -X + = - x�. -

Hence

k=O

nII- l ( b T:) n a + c = b ( l - x�)(l - x�) = k=O = bn [l + (XlX2 ) n - (x� + x�)] = bn [1 + (�) n] = an + bn , }n = o. since( DXlXi2 =A �dand x? + x� = x? + (_xt or n n rica, Romanian Mathematical Olympiad - second round, 2000) 54. Consider the triangle with vertices of complex coordinates Zl, Z2 , Z3 and the R circumcenter origin of the complex plane. Then the circumradius equals IZI I = I Z2 1 = I Zin3 1 the = r and the side lengths are b I a = IZ2 - z3 1 , = Zl - Z3 1 , c = IZI - Z2 1 .

4.2. SOLUTIONS 175 The desired inequality is equivalent to 1 1 R12 -ab + -be + e-1a >- ­ bc 48 4pr i. e. b a � a+ + e R2 = }f = ]f R � 2r,i which or ( D is Euler's inequality for a triangle. A d i or n n r ca, Revista Matematidi Timi§oara (RMT) , No. 2(1985), pp. 82, Problem 5720) AAA A 55. Let I, 2 , 3 and be the pointsAof complex coordinates ZI, Z2 , Z3 and let A A ) 2 2 3 ZA =A aZA + (1 - a z3, a E Hence point lies on the line and the triangle l 2 3 hasB its circumcenter in the origin ofA the complex planeA .A2 A 3 Al APoint � Al B ,issothe foot of the altitude from l in the triangle l . It follows that R

We have

I Z2 - z3 1 h I ZI - z2 11zl - z3 1 i A I ZI - z2 11zl - z3 1 · IZ2 2r- z3 1 8A 1 A 2 A 3 2 2 2 therefore h = IZI - z2 11zl - z3 1 ' 2r as desired. y _

_

s

n

l

_

(Dorin Andrica, Romanian Mathematical Olympiad - final round, 1984) I 1/ 1 56. Denote z + z by r. From the hypothesis,

4.

176

TRIGONOMETRY

3 ( )( 1 )2 :::; O . This implies Hence r3 :::; 2 + r, which by factorization gives r - 2 r + r 2,( Ti desired. 1987; Revista tu Andreescu, Romanian Mathematical Olympiad first round, 1 ( 1987) , pp. 75, Problem 6 191 ) Matematidl Timi§oara ( RMT) , No. � ? Zl 57. Denote t = 2 , t E
as

==

If

i

�

�

n

59. We will use the following auxiliary result:

177 4.2. SOLUTIONS 1) Let C be a semicircle of unit radius and (IMO 1973, Problem P P1 , PLemma. 2 , . . . , n points on C, where � 1 is an odd integer. Then I OP1 + OP2 + . . . + OPn l � 1 , 0 where is the center of C. idea is to show that the orthogonal projection of the vector sum f. The key OP OP OP1 Proo 2 +...+ + n onto some lme has length not less than 1 (see S . Savchev, T. Andreescu, " Mathematical Miniatures" , The Mathematical Association of America,l 1 2003, pp. 75) . Let 2k -0 P�. From the considerations of symmetry, the line �

--=t

�

--=-=-t

�

-::::-=+

•

n =

containing the middle vector the fact that n is odd!) .

A

n

is a natural candidate for such a line (here we use

o-------��--�

B

It is technically convenient to consider l as an axis with positive direction deter­ As is well known, the projection of the sum of several vectors is equal mined by to the sum of their projections. Hence it suffices to prove that the sum of the signed onto is lengths of the projections of and the orthogonal greater than or equal to Denote the diameter of C by and = and also projections of and onto by We have

OP� . OP1 , OP2 , . . . , OP2k -1 OP� , OP� , . . . , OP2k ! l AB 1. A B l A1 B1. OPk 1 OP1 + OP2 + . . . + OPk 1 � (k - 1) OA1' OPk+1 + OPk+2 + . . . + OP2k 1 � (k - 1) OB1. - 1 OP OB OA OP 1 This is because OA j � OB1 for j , . . . , k - and j � 1 for j 1 1 k + , . . . , 2k - . Since 1 + 1 0, the proof is complete. Consider the complex numbers 1 , 2, . . Zk cos O:: k + sin O::k , k PP P and the points 1, 2 , . . . , n with complex coordinates Zl , Z2 , . . . , Zn' =

=

i

=

0

=

.

,n

4. TRIGONOMETRY

178

Using the above Lemma we have . . . + l � 1 , or

Ikt=1 COSO:k itk=l O:k I

zn

+

It follows that as

I OP� + OP� + . . . + OP� I � 1, hence IZI + Z2 sin

desired .

� 1.

(Dorin Andrica, Revista Matematidi Timi§oara

Problem C:58)

+

(RMT) , No. 2(1983) ,

pp.

90,

60. Using the identities

sin nx cos ( n + l )-X · =--.--'------'S =� L..J cos 2 JX sm x

1 j= l

and

sin nx sin ( n + l ) x · =S2 = � --.---. 2JX L..J sm sm x

we obtain On the other hand ,

j= l S12 + S22

=

( sinsinnxx ) 2

S� + si = ( cos 2x + cos 4x + . . . + cos 2nx) 2 + + ( sin 2x + sin 4x + . . . + sin 2nx) 2 = =

n+

L:

(cos 2kx cos 2lx + sin 2kx sin 2lx) =

=x+2

hence

( sinsinnxx ) 2 = n + --

Set

l L: cos 2 ( k )x, 1 9< k�n -

l

cos 2(k - )x .

l �l
(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1995)

Chapter 5 M ATHEMATICAL ANALYSIS

PROBLEMS

1

1 . Let :::; a < f3 be real numbers . Prove that there are integers m, n that a < ifiii < f3.

> 1 such

(an ) n�O and (bn ) n�o be the sequences of integers defined by (1 + . 1o3) 2n+1 = an + bn 3, n E ( ) (b ) Find a recursive relation for each of the sequences an n � O and n n � o , ( ) 3. Study the convergence of the sequence xn n � O satisfying the following prop­ erties:) 1 Xn > 1 , n = 0,1 1, 2, . . . 1 ) Xn + 1 ( 2) - Xn+1 - -- = -2 Xn+1 Xn - 1 , n = 0, 1 , 2, . . . ( ) 4. Study the convergence of the sequence Xn n � l defined by Xl E (0, 2) and � 1 = + V2xn - x X n+1 1 for n � . ( ) 5 . Consider the sequence of real numbers Xn n� l such that i + x� + . . . + X2n x = O. lim n�oo n Prove that Xl + X2 + . . + Xn = 0. 1m n�oo n Is the converse true? b ( ) (b ) 6 . Let an n � l and n n � l be sequences of positive numbers such that an > n n 1 for all n > . (an) n� l is increasing and (bn)n� l is unbounded, then the sequence Prove that if (Cn ) n� l ' given by Cn = an+1 - an , is also unbounded. ( ) 7. Let 0 < a < a be real numbers and let Xn n � l be defined by X l = a and 1) (a + Xn- 1 +10') 2 Xn = Xn- 1 + (a + , n � 2. 18 1 2. Let

. 10

v

1·

V

.

N.

182

5. MATHEMATICAL ANALYSIS

Prove that the sequence is convergent and find its limit. 8. Find a sequence

(an ) n2:l of positive real numbers such that ) ( lim an+l - an n-+oo (Jan+l - �) O . lim n-+oo = 00

and

=

(Xn )n2: l be an increasing sequence of positive real numbers such that X 1·1m n 0 n-+oo n h ( Prove that there is a sequence nk 2: l of positive integers such that 9. Let

2 =

.

(3 be real numbers and let (Xn ) n2: l , (Yn )n2: l , (Zn ) n2: l be real sequences such that � (3 } � max{ x + O'Yn , y + xn :::; Zn for all n ;::: l . 1 ( 2 (32 ) for all n ;::: 1 a) Prove that Zn ;::: - 8 0' + . (3 ) ( ) ( ) ( 2 2 1 b ) If n-+oo lim Zn - -8 0' + , prove that the sequences Xn n 2:b Yn n 2: l are convergent and find their limits. ( ) ( ) 1 1 . The sequences x n n 2:l and Yn n 2: l are defined by Xl 2, Yl = 1 and Xn+l x� + 1, Yn+l /= XnYn for all n ;::: 1. a) Prove that Xn Yn < V7 for ( )all n ;::: 1. / b ) Prove that the sequence Zn n ->l, Zn xn Yn , is convergent and lim Zn < V7. n-+oo ) ( ;::: 0 and a f= 0 be real numbers and let Xn n2:l be an increasing 1 2 . Let sequence of real numbers such that C¥ ( ) a. lim n x + x l n n n-+oo 10. Let

a,

=

=

=

a

=

Prove that the sequence is bounded if and only if a > 1 . 1 3. Evaluate

14. Evaluate

1n-+oom kl:n k((nk -+ k)!l) ! (+n (-k k)!+ 1) =O 1.

Ln ( -k2 ) � +l n-+oo k= l n lim

183

5.1. PROBLEMS

L -1 ' q > 1; ii (4 1l) +l ' q > 1 ( ) � n + qn . n=l nq( n ) 16. Let Xn n � l be an increasing sequence of positive integers such that X n+2 2 Xn > Xn+l for all n � 1. Prove that the number L... ­ 101Xn n=l 15. Evaluate

(i)

00

00

+

00

(J - "'

is irrational.

An L (k!1) k=l n

17. Prove that

is irrational for all 18. Let

k,

s

=

n � 1.

00

be positive integers and let numbers such that

aI, a2 , . . . , ak , bl , b2 , . . . , bs be positive real

n � 2. k 21)) al a . . . ak bl b bs. 2 2 ) ( 19. Let X n n � l be a sequence with X l 1 and let X be a real number such that

for infinitely many integers Prove that =

s;

=

. • .

=

Prove that

20. Let

such that

IT (1 - �) Xn+l n=l

= e -x •

A f:. ±1 be a real number. Find all functions f : R R and ( f ln x + A ln y ) 9 (JX) + 9 (..fij) for all x, y E (0, 00). �

9

:

(0, 00)

�

R

=

21. Let f be a continuous real-valued function on the interval

[a, b] and let ml , m2 O be real numbers such that m l m 2 > . Prove that the equation m f (x) � a b-) x b -2x has at least a solution in the interval (a, . =

+

184

5. MATHEMATICAL ANALYSIS

b (0 , 1 /2) , and let 9 be a continuous 22. Let a and be real numbers in the interval ( ( )) ( ) b () real-valued function such that g g x = ag x + x for all real x. Prove that g x ex for some constant e. f [0,00) such that 23. Find all continuous functions f2 (X + y) - f2 (X - y) = 4f(x) f (y) for all real numbers x, y. ( -00, 0] and g : f Prove that if the continuous functions 24. (i) [0 , 00) have a fixed point, then f + 9 has a fixed point. [0 , 1] and 'ljJ [1 ,00) have (ii) Prove that if'ljJthe continuous functions rp a fixed point, then rp has a fixed point. 25. Let rp be a differentiable function at the origin and satisfying rp(O) = 0 . Evaluate () ;� � [rp x + rp (�) + ' " + rp (;)] , where n is a positive integer. =

:

�

�

:

:

:

�

�

�

�

�

�

�

:

�

�

�

�

26. Let a be a positive real number. Prove that there is a unique positive real number f-£ such that p? > for all >

f [a, b') 27. Let' b f( ) and f (a) = j (b) . :

�

�

XIL - aIL-

x

x 0.

be a twice differentiable function on

[a, b) such that f(a)

=

,\ the equation f" (x) - ,\(f' (X)) 2 = b) has at least a solution in the interval (a, . f [0, 2] (0 , 1] that are differentiable at the origin and 28. Find all functions satisfies f(2x) = 2f2 (X) - 1 , x E [0, 1] Prove that for any real number

°

:

�

29. Let ,\ be a positive integer. Prove that there is a unique positive real number

f) such that

x 0.

for all real number >

f

: � � � be a function continuous at the origin and let distinct positive real numbers.

30. Let

'\, f-£

be two

185

5. 1. PROBLEMS

Prove that the limit

f( >..x) - f(J-tx) 1�O -"----'-- X --'--"a: f exists and is finite if and only if is differentiable at the origin. ( ) 31 . The sequence Xn n � 1 is defined by 1. m

n�oo(0 nX] n = - 2. Vnxn . ) O 1 32. Let Xo E , and Xn+ l Xn - arcsin(sin 3 x n , n � . Evaluate lim n�oo f JR be a twice differentiable function with the second derivative 33. Let : JR nonnegati ve. Prove that f(x + I' (x)) � f(x) , x E b 34. Let a < be positive real numbers. Prove that the equation Y a( ; T+ aX bY ( b) has at least a solution in the interval a, . ) 35 (. Find with proof if there are differentiable functions cp : JR JR such that cp( x ) and cp' x are integers only if x is integer. f [ b] JR be a differentiable function. 36. Let : a, O O Prove that for any positive integer n there are numbers < 2 < . . . < n in the b ) ( interval a, such that f(b) - f(a) f' (OI ) + f' (02 ) + . . . + f' (On ) b-a n f 37. Let , 9 : JR JR be differentiable functions with continuous derivatives such that f ( x) + 9 ( x ) l ' ( x ) - g' ( x ) for all x E JR. f(x) - g (x) Prove that if Xl, Xf( real solutions of the equation 2 are) two( consecutive ) ( ) 0, then the equation x + g x = 0 has at least a solution in the interval X l , X2 . f' f 38. Let : [-�, �] (-1, 1) be a differentiable function whose derivative is continuous and nonnegative. Prove that there exists Xo in [-�, �] such that (f(XO ))2 + (1' (XO ))2 � 1. Prove that lim

=

�

R

=

�

�

fh

�

=

=

�

186

5.

MATHEMATICAL ANALYSIS

39. Prove that there are no positive real numbers

x and y such that Y 2 X x2 y - = x + y . (n+1) ( n 40. a) Prove that if x � y � n : 1 ) for some integer n � 2, then yX n+¢Y � ifY + n +V'x. b) Prove that 2n + 1 , n > 1 > n n +l� _ n+ 4 1 . Let Xl , X2 , . . . , Xn be positive real numbers such that Xl + X2 + . . . + Xn +

+

+ n'n V 'H

3.

Prove that

42. Let

f

:

=

1.

�

f(c) = 0 for someJR c E JRJR. be a function with a noninjective antiderivative. Prove that h f JR be continuous functions. Prove that 43. Let h , , . . . , n JR ( ) h( ) f ( ))d max( h x , x , . . . , n x x �

:

is a derivative and evaluate

44. Evaluate

p'

45 . Let be a polynomial of odd degree such that has no multiple zero and let : JR � JR be a function such that is a derivative.

f

p

f op f Prove that is a derivative. 1 f I I be a function with an antiderivative F that 46. Let = (0, 00) and let satisfies the condition F (x)f G) = x, 1 -+ 1R, g (x) = F (x)F (�) is a constant function and I for all x in . Prove that f then find . :

g

:

�

187

5. 1. PROBLEMS 47. Let n >

1 be an integer and let

that

f

:

[0, 1]

�

JR be a continuous function such

f(x) dx = 1 + -1 + . . . + -1 . 1 n 2 1 a

Prove that there is a real number

Xo E (0, 1), such that f(xo ) 1 - xg 1 - Xo f [a, b] 48. Consider the continuous functions , =

:

9

Prove that the equation

�

R

!(x) /." g(t)dt = g (x) f !(t) dt ( b) has at least a solution in the interval a, . [ b] JR be a continuous function such that 49. Let f a, { !(X) dx of O. b Prove that there are numbers a < a < f3 < such that { !(x) dx (b - a)!(p) . [ b] [c, dJ be a bijective 50. Let a, c be nonnegative real numbers and let f a, increasing function. ( b) Prove that there is a unique real number E a, such that { !(t)dt = (IJ - a)e + (b - lJ) d. 5 1 . Let