Math Problems

Mathproblems ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 4, Issue 1 (2014), Pages 231-262

Editors: Valmir Krasniqi, Jos´ e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Mohammed Aassila, Mih´aly Bencze, Valmir Bucaj, Emanuele Callegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mortici, Jozsef S´ andor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco Javier Garc´ıa Capit´an.

PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]

Solutions to the problems stated in this issue should arrive before June 15, 2014

Problems 88. Proposed by Hun Min Park, Korea Advanced Institute of Science and Technology, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 ≤ a, b, c, ≤ 1) satisfy the following equality;  X a − b a · =0 1 − ab 1 − a2 cyc Prove that X a = b = c. X (Note that means ’cyclic sum’ f (x, y, z) = f (x, y, z) + f (y, z, x) + f (z, x, y)) cyc

cyc

89. Proposed by Mohammed Aassila, Strasbourg, France. Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Prove that X1 < +∞. n n∈S

c

2010 Mathproblems, Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. 231

232

90. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. Let n be a positive integer, prove that n X √ √  b kc n , (−1) ≤ k=0

and determine the cases of equality. 91. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Calculate Z 1Z 1 ln(1 − x) ln(1 − xy)dxdy. 0

0

92. Proposed by D.M. B˘ atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇ au, Romania. Let {an }n≥0 be a sequence of positive integer numbers such that 5 does not divide an for all, n ∈ N,and let the sequence {bn }n≥0 be defined by bn = a2n L2n , for n ∈ N where {Ln }n≥0 is the sequence of Lucas numbers. Prove that bn is squarefree, (i.e. bn is not a perfect square), for every n ∈ N \ {1}. 93. Proposed by Anastasios Kotronis, Athens, Greece. For x ∈ (−1, 1), evaluate +∞ X

n+1

(−1)

n=1

  2n+1 x3 −1 n+1 x n tan x − x + − · · · + (−1) . 3 2n + 1

94. Proposed by Sava Grozdev and Deko Dekov (Jointly), Bulgaria. Find a point P in the plane of a given triangle ABC, such that the sum 2

2

2

|BP | |CP | |AP | + + b2 c2 a2 is minimal, where a = BC, b = CA and c = AB.

95. Proposed by Li Yin, Department of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603, China. An approximation formula of Wallis product. For all n ∈ N, then n  1 ln n Wn ∼ √ 1 − 2n π where Wn :=

(2n−1)!! (2n)!!

is Wallis product.

233

Solutions No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.

81. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. Find, in terms of a > 0, the minimum of a(x2 + y 2 + z 2 ) + 9xyz xy + yz + zx when x, y and z are nonnegative real numbers such that x + y + z = 1.. Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Answer: The minimum is 2a if a ≤ 1 and a + 1 if a ≥ 1. Proof Let x + y + z = 3u, xy + yz + zx = 3v 2 , xyz = w3 . Let m be the searched minimum. We homogenize by writing a(x2 + y 2 + z 2 )(x + y + z) + 9xyz (xy + yz + zx)(x + y + z) which in terms of the variables (u, v, w) becomes a(9u2 − 6v 2 )3u + 9w3 −m≥0 9uv 2 that is 9w3 + R(u, v) ≥ 0 This is a linear function in w3 and then it holds if and only if it holds for the minimum value of w3 . The standard theory says that, once fixed the valued of (u, v), the minimum value of w3 occurs when at least one among the variables is zero or when two of the are equals. In the first case we set z = 0 and get a(x2 + y 2 ) ≥m xy and it is evident that m = 2a. As for the second case we set x = y and get (x − z)2 (2ax − 2x + az) a(x2 + y 2 + z 2 )(x + y + z) + 9xyz −a−1= (xy + yz + zx)(x + y + z) x(x + 2z)(2x + z) which is positive for any x, z ∈ R if and only if a ≥ 1. The minimum is thus the maximum between 2a and a + 1 which is 2a if a ≤ 1 while a + 1 if a ≥ 1. Solution 2 by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia. a) Let a ≥ 1. Schur’s inequality is equivalent to 9xyz x2 + y 2 + z 2 + ≥ 2(xy + yz + zx). x+y+z

234

Using the above inequality and the given condition x + y + z = 1, we have x2 + y 2 + z 2 + 9xyz ≥ 2(xy + yz + zx)

(1)

Also clearly, we have x2 + y 2 + z 2 ≥ xy + yz + zx. From this we find that (a − 1)(x2 + y 2 + z 2 ) ≥ (a − 1)(xy + yz + zx).

(2)

Add (1) and (2) we get a(x2 + y 2 + z 2 ) + 9xyz ≥ (a + 1)(xy + yz + zx) or equivalently a(x2 + y 2 + z 2 ) + 9xyz ≥ a + 1. xy + yz + zx This inequality becomes equality when x = y = z = 31 . Hence   a(x2 + y 2 + z 2 ) + 9xyz min = a + 1. xy + yz + zx b) Let 0 < a < 1. From (1) we have, a(x2 + y 2 + z 2 ) + 9axyz ≥ 2a(xy + yz + zx).

(3)

On the other hand we have a(x2 + y 2 + z 2 ) + 9xyz ≥ a(x2 + y 2 + z 2 ) + 9axyz

(4)

From (3) and (4) we have a(x2 + y 2 + z 2 ) + 9xyz ≥ 2a(xy + yz + zx) or equivalently a(x2 + y 2 + z 2 ) + 9xyz ≥ 2a. xy + yz + zx with equality when x = y = 21 , and z = 0. Hence   a(x2 + y 2 + z 2 ) + 9xyz = 2a. min xy + yz + zx Also solved by Arkady Alt, San Jose, California, USA; Moti Levy, Rehovot, Israel; and the proposer. 82.Proposed P by Anastasios Kotronis, Athens, Greece. 1 Let F (n) := k≥1 (kn+1)k! . Determine the sequence {cm }m≥0 such that m−1  X ck  lim nm F (n) − = cm n→+∞ nk k=0

where, for m = 0, the sum is considered to be 0. Solution 1 by Moti Levy, Rehovot, RIsrael. Actually, the problem deals with
1 xn finding asymptotic expansion of F (n) = 0 e − 1 dx. ∞

j−1

X (−1) 1 1 1 = = 1 kn + 1 kn 1 + kn kj j=1

1 nj

235 ∞ X k=1

∞

∞

j−1

X X (−1) 1 = (kn + 1)k! k j k! j=1 =

k=1 ∞ X j=1

1 nj

∞ j−1 1 X (−1) . nj k j k! k=1

Let ( {cm }m≥0 =

(−1)

m−1

∞ m−1 X (−1)

k m k!

k=1

then

∞ X k=1

) m≥0

1 c3 c1 c2 = + 2 + 3 + ··· . (kn + 1)k! n n n

The sequence {cm } can be expressed by the generalized hypergeometric function, cm = (−1)

m−1

m+1 Fm+1

(1, 1, . . . , 1; 2, 2, . . . , 2; 1) .

For m = 1, in particular, c1 = 2 F2 (1, 1; 2, 2; 1) = −γ + Ei (1) ∼ = 1.3179 Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. . X F (n) = k≥1

∞

X 1 X (−1)j 1 = (kn + 1)k! k!kn j=0 (kn)j k≥1

The absolute convergence of the series allows us to take the limit n → +∞ under the sum so for m = 0 we have lim F (n) = 0,

(c0 = 0)

n→+∞

Let m = 1. n

m

F (n) −

m−1 X k=0

ck nk

! = nF (n) =

k≥1

whose limit n → +∞ is evidently c1 = cm =

∞ X (−1)m−1 k=1

nr+1

k!k m

∞ X 1 X (−1)j k!k j=0 (kn)j

∞ X k=1

1 . By induction we suppose that k!k

for any 0 ≤ m ≤ r. We have

  ! r ∞ r j X X X X cm (−1) 1 c m = nr+1  − F (n) − m j+1 j+1 m n n k!k n m=0 m=1 j=0 k≥1   r ∞ r p−1 X p−1 X X X X 1 (−1) 1 c (−1) m = nr+1  + − p p p p m n k!k n k!k n p=1 p=r+1 m=1 k≥1

and the limit as n → ∞ yields (−1)r

k≥1

∞ X k=1

1 . k!k r+1

236

Solution 3 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. Let Ω = {z ∈ C : −1}. For z ∈ Ω we consider G(z) defined by the formula G(z) =

∞ X

1 p! (p + z) p=1

Clearly, this is a series of analytic functions on Ω (namely: z 7→ 1/(k! (z + k)),) that converges to G uniformly on every compact subset of Ω. This proves that G, itself, is analytic in Ω and that, for every m ≥ 0 and every z ∈ Ω, we have G

(m)

(z) =

∞ X

(−1)m m! p! (p + z)m+1 p=1

In particular, for |z| < 1 we have G(z) =

∞ X G(m) (0) m z m! m=0

Thus for z in the neighborhood of 0, and for every m ≥ 0 (with the same convention as in the statement of the problem,) we have zG(z) =

m−1 X

ck z k + O(z m )

k=0

with c0 = 0 and ck = G(k−1) (0)/(k − 1)! when k is a positive integer. But F (n) = 1 1 n G( n ), so for large n, and for every nonnegative integer m, we have   m−1 X ck 1 + O . F (n) = nk nm k=0

Moreover, according to (1) we have ck = (−1)k−1

∞ X

1 . p! pk p=1

This yields the desired conclusion. Note that the ck ’s do not seem to have a closed form. but they can, alternatively, be expressed as integrals: Z Z ∞  (−1)k−1 X 1 ∞ k−1 −pt (−1)k−1 ∞ k−1  e−t ck = t e dt = t e − 1 dt. (k − 1)! p=1 p! 0 (k − 1)! 0 Also solved by Moubinool Omarjee, Paris, France, and the proposer. Editor’s Comment: The proposer of this problem indicated that it is a generalization of problem U278 of Mathematical Reflections. 83.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technology,Daejeon, South Korea. Let {an } strictly increasing sequence of positive integers such that gcd(ai , aj ) = 1 for any i, j(i < j). Let bn = an+1 − an . Prove that the sequence {bn } is unbounded(has no upper bound).

237

Solution 1 by Shmuel Isaac and Moti Levy, Rehovot, Israel (Jointly). Let us define the sequence {pn }, where pn is the largest prime number that divides an . Since ai and aj are relatively prime (for i 6= j) then pi 6= pj . Let A (x) be the number of all {an } sequence terms, which are less or equal to x, i.e., aA(x) ≤ x,

and aA(x)+1 > x.

Let P (x) be the number of {pn } sequence terms, which are less or equal to x. By the definitions above, A (x) ≤ P (x) ≤ π (x) , (1) where π (x) is the number of prime numbers less or equal to x. Let X be a positive integer, X ≥ 2a1 . By definition of A (x) ,we have aA(X)+1 > X, hence X X = . aA(X)+1 − a1 ≥ X − 2 2 Clearly, A(X) X bn = aA(X)+1 − a1 . n=1

Suppose, on the contrary, that the sequence {bn } is bounded by the positive constant M > 0. Then aA(X)+1 − a1 ≤ A (X) M, X ≤ aA(X)+1 − a1 ≤ A (X) M. 2 By equations ((1)) and (2),

(2)

π (X) A (X) 1 ≥ ≥ X X 2M Multiplying both sides by ln X > 0, 1 π (X) ≥ ln X. X/ ln X 2M π(X) On one hand, the Prime Number Theorem states that limX→∞ X/ ln X = 1 but on 1 the other hand limX→∞ 2M ln X = ∞; This is a contradiction, which leads us to the conclusion that the sequence {bn } is unbounded.

Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let pn be the smallest prime number that divides an . Since gcd(ai , aj ) = 1 for every distinct i and j we conclude that p1 , p2 , . . . , pn are distinct primes from the interval [1, an ] and consequently, π(an ) ≥ n. where π(x) represents the number of primes p that are smaller or equal to x. Now, according to the weak form of the prime number theorem, there exists an absolute positive constant A such that π(x) ≤ A lnxx hence an an n ≤ π(an ) ≤ A ≤A ln an ln n where with a0 = 0, we have an = Pn we used the trivial Pn inequality n ≤ an (since, n ln n (a − a ) ≥ 1 = n.) Thus, a ≥ , and in particular k k−1 n k=1 k=1 A an lim = +∞. (3) n→∞ n

238

Now, suppose that the sequence {bn } is bounded by some constant M then it follows immediately that n−1 an − a1 1X = bk ≤ M n n k=1

and consequently lim supn→∞ ann ≤ M which is a absurd according to (3). This contradiction proves that {bn } cannot be bounded. Remark. Refining upon the proof presented above we see that we have lim inf n→∞

an ≥ 1, n ln n

and

lim sup n→∞

bn > 1. ln n

Solution 3 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. The sequence {aj } with the smallest entries aj for any j is the sequence of the prime numbers. Any other sequence {a0j } is such that a0j ≥ aj and then asymptotically a0j ≥ c0 j ln j for any j large enough. We argue by contradiction by assuming that there exists a constant c1 such that bn ≤ c1 . This implies that n X aj+1 − aj = an+1 − a1 ≤ nc1 k=1

but this contradicts a0j ≥ c0 j ln j if n is large enough. Also solved by Arthur Handle, and the proposer. 84.Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, China. Let 1 < p < ∞, we can generalize the inverse of arcsin as follows: Z x 1 dt, 0≤x≤1 arcsinp (x) = (1 − tp )1/p 0 and Z 1 πp 1 dt. = arcsinp (1) = p 1/p 2 0 (1 − t )  πp  The inverse of arcsinp (x) on 0, 2 is called the generalized sine function and d denoted by sinp . The generalized cosine function is defined as cosp (x) = dx sinp (x). Similarly, the generalized inverse hyperbolic sine function Z x 1   dt , x ∈ [0, ∞), (1 + tp )1/p arcsinhp (x) = 0   −arcsinh (−x) , x ∈ (−∞, 0) p generalizes the classical inverse hyperbolic sine function. The inverse of arcsinhp is called the generalized hyperbolic sine function and denoted by sinhp . The generalized hyperbolic cosine function is defined as coshp (x) = sinh0p (x). For p > 2 and π x ∈ (0, 2p ), prove that ln

x sinp x − x cosp x < sinp x p sinp x

and

ln

sinhp x x coshp x − sinhp x > . x p sinhp x

One incomplete solution was received, so the problem remains open.

239

85.Proposed by D.M. B˘ atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇ au, Romania. Let n be a positive integer. Prove that n n X Fk2   X Fk3  (Fn+2 − 1)5 · ≥ Lk L2k (Ln+2 − 3)2 k=1

k=1

where Fn , respectively Ln represents the nth Fibonacci number respectively the nth Lucas number. Solution 1 by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia. Using the following well known indentities L1 + L2 + · · · + Ln = Ln+2 − 3 and F1 + F2 + · · · + Fn = Fn+2 − 1 we have !3

n X

n X F2

·

Lk

!

n X F3

k

k=1

k=1

k L2k k=1

Lk

! ≥

n X

!5 Fk

.

(1)

k=1

It is enough to prove the above inequality. Applying the H¨older’s inequality, we get n X

!3/5

n X F2 k

·

Lk

k=1

=

k=1 n  X

3/5 Lk

 53

n X k=1

·

n X F3

!1/5

k

k=1

L2k

    2 1/5 !5 1/5  3 1/5 !5 1/5 n n X X F F k k  ·  · Lk L2k k=1 k=1 ! n 3/5 X F Fk . · k2/5 = Lk k=1

!3/5 

2/5

3/5

Lk

·

Lk

k=1

≥

!1/5

Fk

1/5

Lk

Hence (1) is proved. Solution 2 by Moti Levy, Rehovot, Israel. A more general version of this problem appeared in ”The Fibonacci Quarterly”, Volume 51, Number 4, November 2013. ! n ! n m+p+2 X X F p+1 Fkm+1 (Fn+2 − 1) k ≥ m > 0, p > 0. (2) m+p , Lm Lpk (Ln+2 − 3) k k=1

k=1

To prove (2), we use the inequality (3) due to J. Radon, n X xp+1 k

k=1

ykp

Pn p+1 ( k=1 xk ) ≥ Pn p , p > 0, and xk ≥ 0, yk > 0, for 1 ≤ k ≤ n. ( k=1 yk )

(3)

By Radon’s inequality, n X F m+1 k=1

!

k

n X F p+1

Lm k

k=1

k

Lpk

!

Pn p+m+2 ( k=1 Fk ) ≥ Pn p+m ( k=1 Lk )

(4)

240

The following sums are well known, n X k=1 n X

Fk = Fn+2 − 1

(5)

Lk = Ln+2 − 3

(6)

k=1

The required result is obtained by substituting (5) and (6) in (3). Solution 3 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We will use the following lemma: Lemma. Let p, λ, µ be real numbers with p > 1, and 0 < λ, µ < 1. Then for every positive numbers a1 , . . . , an and b1 , . . . , bn we have: n n 2(1−λ)p X X ak a2λp (a1 + · · · + an )2p k ≤ · . 2µ(p−1) 2(1−µ)(p−1) (b1 + · · · + bn )2p−2 k=1 bk k=1 bk

Proof. Let x1 , . . . , xn and y1 , . . . , yn be positive numbers such that n n X X xk = yk = 1 k=1 1 p

Let r > 1 be defined by 1=

n X

xk =

n X xk 1/r

k=1

k=1

+

yk

k=1

1 r

= 1, then by H¨older’s inequality we have ! p1 ! r1 ! p1 n n n X X X xpk xpk 1/r · yk ≤ yk = p/r y p−1 k=1 k=1 k k=1 yk

That is, by the Cauchy-Schwarz inequality: 1≤

n X xpk

y p−1 k=1 k

=

n X

xλp k µ(p−1) y k=1 k

·

(1−λ)p xk (1−µ)(p−1) yk

v u n uX ≤t k=1

x2λp k 2µ(p−1)

·

yk

n X

2(1−λ)p

xk

2(1−µ)(p−1)

k=1

yk

for 0 < λ, µ < 1. Applying this, with xk = a/ a ˜ and yk /˜b with a ˜ = a1 + · · · + an

and ˜b = b1 + · · · + bn ,

we obtain

n n 2(1−λ)p X X ak a2λp (a1 + · · · + an )2p k ≤ · . 2µ(p−1) 2(1−µ)(p−1) (b1 + · · · + bn )2p−2 k=1 bk k=1 bk
Taking, (p, λ, µ) = 52 , 35 , 23 , we obtain ! ! n n X X (a1 + · · · + an )5 a3k a2k ≤ · . (b1 + · · · + bn )3 b2k bk k=1

k=1

Finally, the desired inequality is obtained, by setting ak = Fk , bk = Lk and noting that n n X X Fk = (Fk+2 − Fk+1 ) = Fn+2 − F2 = Fn+2 − 1, k=1 n X k=1

Lk =

k=1 n X k=1

(Lk+2 − Lk+1 ) = Ln+2 − L2 = Ln+2 − 3.

241

Remark. Similarly, for m > 2 and 0 < q, p < m, we have ! ! n n X X Fkm−p Fkp (Fn+2 − 1)m ≥ . · m−q q−2 (Ln+2 − 3)m−2 L L k=1 k k=1 k with the same proof. ´ Also solved by G. C. Greubel, Newport News, VA, USA; Angel Plaza, University of De Las Palmas, Grain Canaria, Spain; and the proposer. 86.Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj, Romania. Calculate √ √ Z 1 ln( x + 1 − x) √ dx. x 0 Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Let x = t2 . The integral reads as ! r Z 1 Z 1 Z 1 p 1 2 ln tdt + 2 − 1 dt ln 1 + 2 ln(t + 1 − t )dt = 2 t2 0 0 0 Z

1

1 ln t dt = (t ln t − t) = −1 0 0 p 2 In the second integral we change t = 1/ 1 + y and it becomes Z ∞ Z ∞ y 1 ln(1 + y) ∞ 1 p p ln(1 + y) dy = − dy + 2 )3/2 2 1 + y 0 (1 + y 1+y 1 + y2 0 0 Z ∞ Z ∞ 1 dt = 2 dz = 2 |{z} |{z} 1 + sinh t z + 2z − 1 1 0 y=sinh t z=et  Z ∞ 1 1 1 − dz z − z z − z z − z2 1 2 1 1 √ 2)/2, z2 = (−1 − 2)/2. We get evidently √ √ √ −1 1 − z1 −1 2 − 2 −1 ln(3 + 2 2) √ √ ln √ √ √ = ln = ln(3 − 2 2) = 2 1 − z2 2 2+ 2 2 2 and the integral finally is √ √ 2 ln(3 + 2 2) − 2. where z1 = (−1 +

√

Solution 2 by √ Arkady
San Jose, California, USA. √ Alt, R 1 ln x + 1 − x √ Let I := 0 dx. The change of variables x = sin2 t shows that x Z π/2 I=2 ln (sin t + cos t) cos t dt. 0

Z =2

π/2

ln (sin t + cos t) sin t dt. 0

(t ←

π − t) 2

242

Taking the half sum we obtain Z π/2 ln (sin t + cos t) (cos t + sin t)dt. I= 0

√ Z = 2

ln

√

 2 cos θ cos θ dθ

(t ←

−π/4

√ =

π/4

2 (ln 2) 2

Z

π/4

√

π/4

{z I1

}

|

Z iπ/4 I2 = sin θ ln(cos θ) + √

ln(cos θ) cos θ dθ {z I2

}

2, and π/4



Z

−π/4 √ 1/ 2

−π/4

sin2 θ dθ cos θ

2

1 u 2 ln √ + 2 du 1 − u2 2 0 √ √  Z 1/ 2  √ 1 1 2 =− ln 2 − 2 + + du 2 1+u 1−u 0 √ √ √ 2 2+1 =− ln 2 − 2 + ln √ 2 2−1 √ √ √ 2 =− ln 2 − 2 + 2 ln( 2 + 1) 2 √ √ Finally I = −2 + 2 2 ln( 2 + 1). =

+ θ)

−π/4

−π/4

| Clearly, I1 =

√ Z cos θ dθ + 2

π 4

(u = sin θ)

Also solved by Albert Stadler, Switzerland; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Anastasios Kotronis, Athens, Greece; G. C. Greubel, Newport News, VA, USA; Moti Levy, Rehovot, Israel; Moubinool Omarjee, Paris, France, AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; and the proposer. 87.Proposed by Dorlir Ahmeti, University of Prishtina, Republic of Kosova. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that √ √ √ √ √ √ a+ b b+ c c+ a √ + √ + √ ≥ 3. 1 + ca 1 + ab 1 + bc Solution 1 by AN-anduud Problem Solving Group. The proposed problem is equivalent to the following problem. x, y, z be positive real number such that x2 + y 2 + z 2 = 3. Prove that x+y y+z z+x + + ≥ 3. (1) 1 + xy 1 + yz 1 + zx Applying H¨ older’s inequality, we have  2 x+y y+z z+x + + ((x + y)(1 + xy)2 + (y + z)(1 + yz)2 + (z + x)(1 + zx)2 ) 1 + xy 1 + yz 1 + zx ≥ ((x + y) + (y + z) + (z + x))3 = 8(x + y + z)3 .

(2)

243

Thus (1) would follow from (2) If we prove the next inequality 8(x + y + z)3 ≥ 9((x + y)(1 + xy)2 + (y + z)(1 + yz)2 + (z + x)(1 + zx)2 ) or equivalntly 9(x5 + y 5 + z 5 ) + 48xyz ≥ 25(x3 + y 3 + z 3 ). To prove (3) we note that X X X 9 x5 + 48xyz − 25 x3 = (x + y)(x − y)2 (x + y − z)2 cyc

cyc

(3)

cyc

+

X

z(x − y)2 ((x + y − 6z)2 + 35xy)

cyc

+ 25xyz

X (x − y)2 ≥ 0 cyc

Also solved by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer.

244

MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always welcomed. The source of the proposals will appear when the solutions be published.

Proposals 61. Find all real solutions of the following system of equations: p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5, 9y 2 − 4x2 = 60. 62. Let P be an interior point to an equilateral triangle ABC. Draw perpendiculars P X, P Y and P Z to the sides BC, CA and AB, respectively. Compute the value of BX + CY + AZ PX + PY + PZ 63. How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram, 5 of two grams, and 6 of five grams, respectively? 64. Let A(x) be a polynomial with integer coefficients such that for 1 ≤ k ≤ n + 1, holds: A(k) = 5k Find the value of A(n + 2). 65. Let a0 , a1 , . . . , an and b0 , b1 , · · · , bn be complex numbers. If n ≥ 2, then prove that ! !   n n n X 1 n+1 X n2 2n − 2 X 2 2 Re a k bk ≤ n |ak | + |bk | 2 n n−1 n k=0

k=0

k=0

245

Solutions 56. Find all positive integers n smaller that 201314 such that 3n ≡ 3 (mod 13) and 5n ≡ 5 (mod 13). What are the smallest and the biggest? How many are there in total? (50th Catalonian Mathematical Olympiad) Solution by Eloi Torrent Juste, AULA Escola Europea, Barcelona, Spain. We have that 30 ≡ 1 (mod 13), 31 ≡ 3 (mod 13), 32 ≡ 9 (mod 13), 33 ≡ 1 (mod 13), and so forth. Thus, powers of three when divided by 13 present a cycle of order 3. Likewise, 50 ≡ 1 (mod 13), 51 ≡ 5 (mod 13), 52 ≡ 12 (mod 13), 53 ≡ 8 (mod 13), 54 ≡ 1 (mod 13), etc. So, powers of five when divided by 13 present a cycle of order 4. The integers n searched are of the form n = 3k + 1 and at the same time of the form n = 4h + 1. Therefore, n − 1 must be multiple of 3 and 4 at the same time. That is, a multiple of 12 and n = 12k + 1. According to the 201313 or 0 ≤ k ≤ 16776. statement 1 ≤ 12k + 1 ≤ 201314 and therefore 0 ≤ k ≤ 12 The smallest positive integer, for k = 0 is 1. The biggest, for k = 16776 is 201313 and the total number is 16776 + 1 = 16777. 2 Also solved by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. 57. Let n, m be positive integers. Prove that  n  m+1 1 1 1+ < 1+ n m (Training Sessions of Catalonian Team for OME 2013) Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Applying AM-GM inequality, namely √ n

a1 + a2 + . . . + an , n where a1 , a2 , . . . , an are positive numbers not all the same; we have s n  m+1 1 1 n+m+1 1+ 1− n m+1     1 1 n 1+ + (m + 1) 1 − n m+1 < =1 n+m+1 From the preceding, we have  n  m+1  n  −(m+1) 1 1 1 1 1+ 1− 0, we have a < ln(1 + a) < a. 1+a Applying the upper inequality with a = 1/n and the lower one with a = 1/m, we get     1 1 < 1 < (m + 1) ln 1 + n ln 1 + n m Taking exponentials yields the desired inequality. Also solved by Arkady Alt, San Jose, California, USA, Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy, and Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain. 58. Let x = 2 cos A, y = 2 cos B and z = 2 cos C, where A, B, C are the measures of the angles of an acute triangle ABC. Find the minimum value of x4 + y 4 + z 4 + x2 y 2 z 2

(Training Catalonian Team for OME 2014) Solution 1 by Arkady Alt, San Jose, California, USA. By replacing (cos A, cos B, cos C) in identity cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 with

x y z  , , we obtain 2 2 2 x2 y2 z2 x y z + + +2· · · =1 4 4 4 2 2 2

Thus x2 + y 2 + z 2 + xyz = 4 Since triangle ABC is acute, that is x, y, z > 0, then by QM-AM inequality we have  2 2 x4 + y 4 + z 4 + x2 y 2 z 2 x + y 2 + z 2 + xyz ≥ = 1 =⇒ x4 + y 4 + z 4 + x2 y 2 z 2 ≥ 4 4 4. Since the lower bound 4 can be attained if x = y = z ⇐⇒ A = B = C, the desired minimum is 4.

247

Solution 2 by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH, Barcelona, Spain. Since A + B + C = π, then we have x2 + y 2 + z 2 + xyz = 4 cos2 A + 4 cos2 B + 4 cos2 (A + B) − 8 cos A cos B cos(A + B) = 4(cos2 A + cos2 A − cos2 A cos2 A + sin2 A sin2 B) h i = 4 sin2 B(cos2 A + sin2 A) + cos2 B = 4 Taking into account AM-QM inequality yields r x4 + y 4 + z 4 + x2 y 2 z 2 x2 + y 2 + z 2 + xyz 1= ≤ 4 4 from which follows x4 + y 4 + z 4 + x2 y 2 z 2 ≥ 4. So, the minimum value of the expression claimed is 4 and it is attained when 4ABC is equilateral. Solution 3 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. The answer is 4 and it is attained only when the triangle is equilateral. First, note as in the preceding solutions we have x2 + y 2 + z 2 + xyz = 4 Thus, x4 + y 4 + z 4 + x2 y 2 z 2 − 4 = x4 + y 4 + z 4 + x2 y 2 z 2 − 2(x2 + y 2 + z 2 + xyz) + 4 = (x2 − 1)2 + (y 2 − 1)2 + (z 2 − 1)2 + (xyz − 1)2 ≥ 0 with equality if and only if x2 = y 2 = z 2 = xyz = 1, or equivalently A = B = C = 60◦ . Remark. Note that the condition that ABC is acute is unnecessary. Solution 4 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. It is a known standard result that 1 ≤ cos A + cos B + cos C ≤ 3/2 and then 2 ≤ 2 cos A + 2 cos B + 2 cos C ≤ 3. The minimum 1 corresponds to a degenerate isosceles triangle while the maximum to an equilateral triangle. Clearly we have 2 ≤ x + y + z ≤ 3. We employ the so called ”uvw” theory which can be found at The art of problem solving forum. Define three new quantities x + y + z = 3u, xy + yz + zx = 3v 2 , xyz = w3 We have x4 +y 4 +z 4 +x2 y 2 z 2 = (w3 )2 +12uw3 +81u4 −108u2 v 2 +18v 4 = (w3 )2 +12uw3 +R(u, v) This is a convex increasing parabola if w3 ≥ 0 whose minimum has negative abscissa. It follows that the minimum of the parabola occurs when w = 0 or when w is minimum once fixed the values of u and v. According to the theory, the latter occurs when x = y (or cyclic). If w = 0 we have for instance z = 0 that is C = π/2. 4 4 At x + y fixed, √ the minimum of x + y , occurs when x = y that is A = B = π/4 or z = y = 2. This yields x4 + y 4 = 2x4 = 8

248

If z = y and x = a − 2y, a ≤ 2 ≤ 3 we get  a 4  a 6 (3y − a)2 x4 + y 4 + z 4 + (xyz)2 − 3 − = P (y, a) 3 3 729 (3y − a)2 = (324y 4 − 108ay 3 + 1458y 2 − 1620ay + 702a2 − 27a2 y 2 − 6a3 y − a4 ) 729 Now we prove that P (y, a) > 0. Indeed 702a2 − a4 = a2 (702 − a2 ) ≥ a2 (702 − 9) = 693a2 ,

6a3 y = 6a2 ay ≤ 54ay

1011y 2 + 693a2 > 1674ay = (1620 + 54)ay so we get 324y 4 + 442y 2 ≥ 108ay 3 + 27a2 y 2 and this is implied by 324y 4 + 442y 2 ≥ 108 · 3y 3 + 27 · 9y 2 ⇐⇒ 324y 4 + 199y 2 ≥ 324y 3 and this finally follows by the AGM 324y 4 + 199y 2 ≥ 507y 3 . We have showed that  a 4  a 6 − ≥0 x4 + y 4 + z 4 + (xyz)2 − 3 3 3 and the difference equals zero if x = y = z = a/3. On account of their definition, x, y, z can be equal if and only if A = B = C = π/6 whence x = y = z = 1 and x4 + y 4 + z 4 + (xyz)2 − 4 ≥ 0 The searched minimum is thus min{8, 4} = 4. Also solved by Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain. 59. Let x, y, z be nonzero complex numbers. Prove that  −1 1 1 1 81 + + ≤ 3|x+y +z|2 +|2x−y −z|2 +|2y −x−z|2 +|2z −x−y|2 |x|2 |y|2 |z|2 (Training Catalonian Team for OME 2014) Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. More generally, consider n nonzero complex numbers z1 , . . . , zn , and let µ = n1 (z1 + . . . + zn ). Clearly we have n X

|zk − µ|2 =

k=1

=

n X k=1 n X

|zk |2 − 2

n X

5x . 5x − 3x = 5x + 7 7 7 Which is (a). (b) Let an =

6 7

·

5n n4 .

Clearly, a4 > 1 and, for n ≥ 4, we have  4  4 n an+1 4 256 =5 > 1. ≥5 = an 1+n 1+4 125

Thus an > 1 for n ≥ 4, which is equivalent to (b) Also solved by Arber Igrishta, Mathematical Group Galaktika Shqiptare, Albania, Ioan Viorel Codreanu, Satulung, Maramures, Romania, D.M. Bˇ atinet¸u-Giurghiu, Bucharest, Romania, Neculai Stanciu, Buzˇ au and Titu Zvonaru, Comˇ ane¸sti, Romania, and the proposer. 20. Proposed by Paolo Perfetti, Math. Dept. “ Tor Vergata” University, Rome, Italy. Let a0 = a1 = a2 = 1 and for n ≥ 1 an an+1 an + an−1

an+2 = Find an for any n.

Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let xn = an /an+1 , with this notation we have xn+1 = 1 + xn−1 . Hence x2n − x0 = x2n+1 − x1 =

n X

(x2k − x2k−2 ) = n

k=1 n X

(x2k+1 − x2k−1 ) = n

k=1

262

Thus, x2n = x2n+1 = n + 1 for n ≥ 0. So, for n ≥ 1, we have a2n−2 = x2n−2 x2n−1 = n2 a2n Multiplying, we obtain a2n =

1 , (n!)2

and

a2n+1 =

a2n 1 = . xn n! · (n + 1)!

Finally 1 , bn/2!c · dn/2!e where b·c and d·e are the floor and ceiling functions respectively. an =

Mathproblems ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 4, Issue 2 (2014), Pages 263-302

Editors: Valmir Krasniqi, Jos´ e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Mohammed Aassila, Mih´aly Bencze, Valmir Bucaj, Emanuele Callegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mortici, Jozsef S´ andor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco Javier Garc´ıa Capit´an.

PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]

Solutions to the problems stated in this issue should arrive before October 15, 2014

Problems 96. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Let p ≥ 1 be an integer and let x ∈ R. Prove that   Z x ∞ X x x2 xn np ex − 1 − − − ··· − = ex Qp (t)dt, 1! 2! n! 0 n=1

where Qp is a polynomial of degree p which satisfies the equation Qp+1 (x) = xQ0p (x) + xQp (x) with Q1 (x) = x

97. Proposed by Sava Grozdev and Deko Dekov (Jointly), Bulgaria. Given points U and P in the plane of 4ABC. Let Ua Ub Uc be the cevian triangle of U . Denote by Ra , Rb and Rc the reflections of Ua , Ub and Uc in P , respectively. If the lines ARa , BRb and CRc concur in a point, we say that the Prasolov product of U and P is defined. In this case the intersection point of the lines is the Prasolov product of U and P. Note that if U is the orthocenter of 4ABC and P is the nine-point center of 4ABC, then the Prasolov product is known as the Prasolov point. Prove c

2010 Mathproblems, Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. 263

264

that the Prasolov product is defined, provided U is the Nagel point of 4ABC and P is the Spieker center of 4ABC. The problem could be re-formulated as follow. Let U a be the point at which the A-excircle meets the side BC of 4ABC, and define U b and U c similarly. Let P be incenter of the medial triangle of 4ABC. Denote by Ra , Rb and Rc the reflections of Ua , Ub and Uc in P , respectively. Prove that the lines ARa , BRb and CRc concur in a point. 98. Proposed by Anastasios Kotronis, Athens, Greece. Show that ! n +∞ X 4 n! X nk nk − = + O(n−1 ). nn k! k! 3 k=0

k=n+1

99. Proposed by Li Yin, Department of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603, China. Calculate "  # 2n+1 ∞ Y 4 1 (n − 1)(n + 1) 1+ e2 n (2n − 1)(2n + 1) n=2 100. Proposed by D.M. B˘ atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade”P School, Buzˇ au, n Romania. Let (γn )n≥1 be the sequence defined by γn = − ln n + k=1 k1 and let γ = lim γn . Consider a continuous function f : (0, +∞) → (0, ∞). Find n→∞ Z γn p n lim (2n − 1)!! f (x)dx. n→∞

γ

101. Proposed by Florin Stanescu, Serban Cioculescu school, city Gaesti, jud. Dambovita, Romania. Consider a real function f : [a, b] → R (with a > 0,) having a positive and increasing derivative. Show that for every positive integer n with n ≥ 2 the following inequality holds   Z b (b − a)(bn f (b) − an f (a)) bf (b) − af (a) n − . f (x)dx ≤ n−1 bn − an n a 102. Proposed by Marcel Chirit¸ˇ a, Bucharest, Romania. Let f : (0, +∞) → R be a bounded continuous function. Suppose that the limit lim xα |f (x + 2) − 2f (x + 1) + f (x)| = a ∈ R,

x→∞

exists for some α ∈ [0, 1]. Find the value(s) of a.

265

Solutions No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.

88.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technology, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 ≤ a, b, c, ≤ 1) satisfies the following equality;  X a − b a =0 · 1 − ab 1 − a2 cyc Prove that X a = b = c. X (Note that means ’cyclic sum’ f (x, y, z) = f (x, y, z) + f (y, z, x) + f (z, x, y)). cyc

cyc

Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. It must be a, b, c 6= 1. The equality is X cyc

X 1 1 = 1 − a2 1 − ab cyc

which is, by virtue 0 ≤ a, b, c < 1,

2

∞ XX

cyc k=0 2 2

(a2 )k =

∞ XX

(ak )2 =

cyc k=0

We know that a + b + c ≥ ab + bc + ca, then

∞ XX

(ab)k

cyc k=0

(ak )2 + (bk )2 + (ck )2 ≥ (ab)k + (bc)k + (ca)k

with the equality if and only if a = b = c. The result follows.

Solution 2 by Moshe Goldstein and Moti Levy, Rehovot, Israel. Without loss of generality, we may assume that a ≥ b ≥ c. Suppose that a = b, then X a−b a a c a−c c−a = + 0= 2 2 1 − ab 1 − a 1 − ac 1 − a 1 − ca 1 − c2 cyc (1 + ac) . (1 − ac) (1 − a2 ) (1 − c2 ) It follows that a = c, which implies a = b = c. Now suppose, a 6= b, that is a > b ≥ c. Then a b > ≥ 0, 1 − a2 1 − b2 1 1 > ≥ 0, 1 − ab 1 − bc 1 1 > ≥ 0. 1 − ca 1 − bc = (a − c)

(1) (2) (3)

266

It follows from (1) that  X X a−b c a−b a > . 2 2 1 − ab 1 − a 1 − c 1 − ab cyc cyc Using (2) and (3) we obtain, X a−b

(4)

1 X (a − b) = 0. (5) 1 − ab 1 − bc cyc cyc P a−b a Inequalities (4) and (5) imply that cyc 1−ab 1−a2 > 0, which contradicts the assumption in the problem statement, hence a = b and we are done. >

Solution 3 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. For a, b, c ∈ [0, 1) we consider F (a, b, c) = Noting that

a−b a b−c b c−a c · + · + · 1 − ab 1 − a2 1 − bc 1 − b2 1 − ca 1 − c2

x−y x (1 − xy) − (1 − x2) 1 1 · = = − 2 2 2 1 − xy 1 − x (1 − xy)(1 − x ) 1−x 1 − xy ∞ X = (x2n − xn y n ) n=1

we conclude that

F (a, b, c) = =

∞ X

(a2n + b2n + c2n − an bn − bn cn − cn an )

n=1 ∞ X


1 (an − bn )2 + (bn − cn )2 + (cn − an )2 2 n=1

1 (a − b)2 + (b − c)2 + (c − a)2 2 So, if F (a, b, c) = 0 then a = b = c. ≥

Also solved by Arkady Alt, San Jose, California, USA; D.M. Bˇ atinet¸uGiurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzˇ au, Romania(Jointly); and the proposer. 89. Mohammed Aassila, Strasbourg, France. Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Prove that X1 < +∞. n n∈S

Solution 1 by Henry Ricardo, New York Math Circle, New York, USA. We can write S = S1 + S2 + S3 + · · · . where Si is the sum of all terms of the harmonic series whose denominators contain exactly i digits, all different from 7. Now the number of i-digit numbers that do not contain the digit 7 is 8 · 9i−1 : There are 8 choices for the first digit, excluding 0 and 7, and 9 choices for the remaining i − 1 digits.

267

Furthermore, each number in Si is of the form 1/m, where m is an i-digit number. So m ≥ 10i−1 , which implies that 1/m ≤ 1/10i−1 . Therefore S=

∞ X i=1

Si ≤

∞ X 8 · 9i−1 i=1

10i−1

i−1 ∞  X 9 =8 = 80, 10 i=1

so S converges by comparison. Remark: This method can be used to show convergence of the sum of reciprocals of integers that do not contain the digit k ∈ {0, 1, 2, . . . , 9}. Solution 2 by Moti Levy, Rehovot, Israel. We partition the set S into disjoint subsets {Sk }, S=

∞ [

Sk ,

k=1


where Sk := S ∩ 10k−1 , 10k . Define the sequence {ak }k≥1 by ak =

X 1 . n

n∈Sk

P∞ P Clearly, k=1 ak = n∈S n1 . Now we show by mathematical induction that the number of terms in Sk is less than 9k . The number of terms in S1 is 8 < 91 . Suppose that it is true that the number of terms in Sk is less than 9k , then we have to show that the number of terms in Sk+1 is less than 9k+1 . To show this, we split the set Sk+1 into nine intervals Sk+1 =

9 [

α=1


S ∩ α10k , α10k + 1, . . . , (α + 1)10k .


The seventh intersection S ∩ 7 · 10k , 8 · 10k is empty, of course. The other intervals have the same number of terms, which is equal to the number
of terms of the interval S ∩ 1, 10k . By the induction hypothesis, it is less than 9 + 9 2 + · · · + 9k .
Hence, the number of terms in Sk+1 is less than 8 9 + 92 + · · · + 9k < 9k+1 . P k+1 Each term in Sk+1 is not less than 10k , therefore ak+1 = n∈Sk+1 n1 < 910k . ∞ ∞ X1 X X 9k = ak < = 90. n 10k−1

n∈S

k=1

k=1

Reference: A. J. Kempner, A Curious Convergent Series, Amer. Math. Monthly, 21(2), (Feb. 1914) pp. 48-50. Solution 3 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon1 golia. Denote the number of terms of the given series between 101k and 10k+1 by k nk for k ≥ 0. n0 = 8 and nk+1 = 9nk , k ≥ 0. Hence nk = 8 · 9 .

268

Then X

1 < n

n∈S, n 1 √ √ which evidently holds true. This implies that b n + 1c = b nc. Thus we have √ b nc min{(q+1)2 −1,n}

X

X

q

(−1) =

√ b n+1−1c

q=0 k=q 2 √ b n+1−1c

=

X

X q=0

(−1)q (2q + 1) + (−1)b

q=0 √ b n+1−1c

√

(−1)

q

(q+1)2 −1

X

k=q 2 nc

1+

√ b nc

X

√ (n − (b nc)2 + 1) =

√ √ √ (b n + 1 − 1c + 1) + (−1)b nc (n − (b nc)2 + 1) = √ √ √ √ = (−1)b n+1−1c b n + 1c + (−1)b nc (n − (b nc)2 + 1) = √ √ √ √ = −(−1)b n+1c b n + 1c + (−1)b nc (n − (b nc)2 + 1)

= (−1)

√ √ Since b n + 1c = b nc, we have

(−1)

√ q=b n+1−1c+1

q

n X

k=q 2

1=

270

√ √ √ √ −(−1)b n+1c b n + 1c + (−1)b nc (n − (b nc)2 + 1) ≤ √ √ √ √ ≤ −b n + 1c + (n − (b nc)2 + 1) = n + 1 − (n − (b nc)2 + 1) ≤ √ ≤ n+1−1 and the last step is to show √

√ n + 1 − 1 ≤ d ne

Recall that n = p2 − r thus we need to show p

√ √ p2 − r + 1 ≤ d ne = 1 + b nc = 1 + p − 1

which clearly holds.

Also solved by Arkady Alt, San Jose, California, USA; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria; and the proposer. 91.Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Calculate Z 1Z 1 ln(1 − x) ln(1 − xy)dxdy. 0

0

Solution 1 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. The answer is 3 − 2ζ(3). Let the considered integral be denoted by I. Since Z 1 h (1 − xy) iy=1 (1 − x) ln(1 − xy)dy = − ln(1 − xy) − y ln(1 − x) − 1 =− x x y=0 0

we see that

  ln2 (1 − x) 2 I= ln (1 − x) − ln(1 − x) − dx x 0  Z 1 ln2 x = ln2 x − ln x − dx 1−x 0 Z

1

Now, noting that 3x − 3x ln x + x ln2 x we see that

Z

1

0

Also, since Z

0

1 n

2

x ln xdx =

Z

0

∞


0

= ln2 x − ln x


ln2 x − ln x dx = 3 t2 e−(n+1)t dt =

Γ(3) 2 = (n + 1)3 (n + 1)3

271

we see that Z

1

0

∞ Z 1 X ln2 x dx = xn ln2 xdx 1−x 0 n=0

=

∞ X

2 = 2ζ(3) (n + 1)3 n=0

Finally I = 3 − 2ζ(3). Solution 2 by Anastasios Kotronis, Athens, Greece. It is easy to see that for k a positive integer: X1

n≥1

1 − n n+k



=1+

1 1 1 + + · · · + = Hk 2 3 k

(1)

the k-th Harmonic Number. Now, in what follows, the change of the way of summation and of summationintegration order, whenever it takes place, is justified by the constant sign of the summands-integrands.

I:=

Z

1

0

=− =

Z

Z

1

ln(1 − x) ln(1 − xy) dx dy = −

0

Z

1

X yk Z

0 k≥1

1

k

1

0

X yk X 1 k n

0 k≥1

n≥1

Z

1

xn+k dx dy =

0

Z

Z

1

0

Z

1

Z

1

k

0 k≥1

X yk Z

0 k≥1

1

X (xy)k

XX

0 k≥1 n≥1

k

1

xk

0

ln(1 − x) dx dy

X xn dx dy n

n≥1 k

y dy nk(n + k + 1)

XX 1 1 y k dy = nk(n + k + 1) 0 nk(k + 1)(n + k + 1) k≥1 n≥1 k≥1 n≥1   X  XX X1 1 1 1 1 = − = − k(k + 1)2 n k(k + 1)2 (n + k + 1) k(k + 1)2 n (n + k + 1) n≥1 k≥1 n≥1 k≥1     X X 1 1 1 1 1 − + − =− 2 k + 1 k (k + 1) n (n + k + 1) n≥1 k≥1 X   X 1 X 1 X1 1 1 1 1 =− − − +1− − k+1 k n (n + k + 1) k2 n n+k =

XX

xk ln(1 − x) dx dy =

Z

1

n≥1

k≥1

k≥1

n≥1

:=A+1−B

For A, from (1) and summing by parts we have:  1 1 Hk+1 +∞ X 1 − Hk+1 = − + (Hk+2 − Hk+1 ) k+1 k k 1 k+1 k≥1 k≥1   3 X 1 1 = + − =2 2 k+1 k+2

A=−

X

k≥1

272

For B, from (1), B=

X Hk

k≥1

and furthermore B=

 X 1 X1 1 − k2 n n+k n≥1

k≥1

=

XX

k≥1 n≥1

=2

XX 1 = nk(n + k)

k≥1 n≥1

XX

k≥1 n≥1

=2 =2

X

N ≥1

1 1 + n(n + k)2 k(n + k)2

X 1 n+k−1=N ====== 2 2 n(n + k)

N ≥1

N XX

N ≥1



k2

X

n,k≥1 n+k−1=N



1 n(n + k)2

X X HN +1 − N1+1 HN 1 = 2 = 2 n(N + 1)2 (N + 1)2 (N + 1)2 n=1 N ≥1

N ≥1

X HN +1 1 −2 = 2(B − 1) − 2(ζ(3) − 1) 2 (N + 1) (N + 1)3 N ≥1

where ζ is the Riemann zeta function. So B = 2B − 2ζ(3) and hence B = 2ζ(3). Finally I = 3 − 2ζ(3) ≈ 0.595886. Solver’s Note: A reference to the details presented here is “ M. S. Klamkin,Amer. Math. Monthly, 59 (1952) pp. 471–472”. See also page 6 of the Collected Contributions of M. S. Klamkin to the Amer. Math. Monthly. Solution 3 by Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria. For n ∈ N>0 , we can integrate by parts to get :  n+1 1 Z 1 Z 1 n+1 (t − 1) ln(1 − t) 1 t −1 tn ln(1 − t) dt = − dt n + 1 n + 1 t − 1 0 0 0 n Z −1 X 1 k −Hn+1 = (1) t dt = n+1 n+1 0 k=0

Where Hn is the n-th harmonic number, therefore Z 1Z 1 ∞ Z 1Z 1 k k ∞ X X y x Hk+1 I= ln(1−x) ln(1−xy) dxdy = − ln(1−x) dxdy = . k k(k + 1)2 0 0 0 0 k=1

k=1

Now, we can simplify the latter series to be : A=

∞ X

k=1

Hk+1 = k(k + 1)2

∞ X Hk+1

k=1

k

Hk+1 − k+1

!

−

∞ X Hk

k=2

k2

!

.

The first sum is easy to calculate :  n n  X X Hk+1 Hk+1 1 Hk Hk+1 1 Hn+1 − = + − =2− − . k k+1 k(k + 1) k k+1 n+1 n+1 k=1

k=1

273

Since Hn ∼ ln(n) and ln(n) = o(n) we get : A=

∞ X Hk+1

k=1

k

−

Hk+1 1 Hn+1 = lim 2 − − = 2. k + 1 n→+∞ n+1 n+1

To calculate the other sum we use the following result for m > 0: Z ∞ Z 1 2 1 x=e−t/(m+1) 2 m t2 e−t dt = x ln x dx = 3 (m + 1) (m + 1)3 0 0

(2)

From (1) we get : B=

∞ X Hk

k=1

k2

=−

∞ Z X

k=1

1 k−1

t

k

0

ln(1 − t) dt =

Z

0

1

ln2 (1 − t) dt = t

Z

0

1

ln2 t dt. 1−t

Using the result (2) we get B=

∞ X

k=0

Now, it is clear that :

tk ln2 t dt =

∞ X

k=0

2 = 2ζ(3). (k + 1)3

I = A − (B − 1) = 3 − 2ζ(3). Also solved by Arkady Alt, San Jose, California, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; B.C. Greubel, Newport News, VA, USA; Moti Levy, Rehovot, Israel; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer. 92.Proposed by D.M. B˘ atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇ au, Romania. Let {an }n≥0 be a sequence of positive integer numbers such that 5 does not divide an for all, n ∈ N, and let the sequence {bn }n≥0 be defined by bn = a2n L2n , for n ∈ N where {Ln }n≥0 is the sequence of Lucas numbers. Prove that bn is not a perfect square, for every n ∈ N \ {1}. Solution 1 by Moti Levy, Rehovot, Israel. By its definition, bn is a perfect square if and only if L2n is a perfect square. It was proved in the article, J. H. E. Cohn, ”Square Fibonacci Numbers, Etc.” Fibonacci Quarterly 2 1964, pp. 109-113, that if Lk is perfect square then k = 1 or k = 3. It follows that bn is not a perfect square for every n ∈ N. Solution 2 by Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria. Let mod(m, n) be the remainder on division of m by n. With a quick evaluation we have {mod(n2 , 5)|n ∈ N} = {0, 1, 4, 4, 1, 0, 1, ...} = {0, 1, 4} This is true because of the periodicity of the mod sequence : mod(n + 5k, 5) = mod(n, 5) for any n, k ∈ N.

274

Hence mod(a2n , 5) ∈ {1, 4} for all n ∈ N. Let’s take a look on mod(Ln , 5), Here’s a  2    1 mod(Ln , 5) =  3    4

lemma : if if if if

mod(n, 4) = 0; mod(n, 4) = 1; mod(n, 4) = 2; mod(n, 4) = 3.

(lemma 1)

Proof : for n ∈ {0, 1, 2, 3} it is clearly true. suppose it is true for some integers 4k, 4k + 1, 4k + 2, 4 + 3, then mod(L4k+4 , 5) = mod(L4k+3 + L4k+2 , 5) = mod(4 + 3, 5) = 2 mod(L4k+5 , 5) = mod(L4k+4 + L4k+3 , 5) = mod(2 + 4, 5) = 1 mod(L4k+6 , 5) = mod(L4k+5 + L4k+4 , 5) = mod(1 + 2, 5) = 3 mod(L4k+7 , 5) = mod(L4k+6 + L4k+5 , 5) = mod(3 + 1, 5) = 4 Which means that (lemma 1) is true for any integer n by strong induction. For n = 0, we have mod(b0 , 5) ∈ {2, 3}, which means that b0 ∈ / {0, 1, 4}, then b0 is not a perfect square. For n > 1, we have 2n = 4k where k ∈ N>0 , therefore : mod(L2n , 5) = 2, which implies that mod(bn , 5) = mod(2a2n , 5) ∈ {2, 3} Which means that bn ∈ / {0, 1, 4}, then bn is not a perfect square for n > 1. Conclusion : bn is not a perfect square for any n ∈ N\{1}. Also solved by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria; and the proposer. 93.Proposed by Anastasios Kotronis, Athens, Greece. For x ∈ (−1, 1), evaluate   x2n+1 x3 − · · · + (−1)n+1 . (−1)n+1 n tan−1 x − x + 3 2n + 1 n=1 +∞ X

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let x3 x2n+1 fn (x) = tan−1 x − x + − · · · + (−1)n+1 3 2n + 1 Clearly we have 2n

fn0 (x) =

X 1 1 1 − (−x2 )n+1 (−1)n+1 x2n+2 2 k − (−x ) = − = 1 + x2 1 + x2 1 + x2 1 + x2 k=0

Thus n+1

fn (x) = (−1)

Z

0

x

t2n+2 dt 1 + t2

275

It follows that ∞ X

n+1

(−1)

n=1

1 nfn (x) = 2 1 = 2

Z

x

0

Z

x

0

t3 1 + t2 t3 1 + t2

∞ X

(2n)t

n=1 ∞ X

n=0

t2n

2n−1

!0

!

dt

dt

 0 t3 1 dt 2 1 − t2 0 1+t x t4 dt 2 2 2 0 (1 + t )(1 − t )

1 = 2 Z = Finally, noting that t4 1 = (1 + t2 )(1 − t2 ) 8



Z

x

1 1 2 2 2 + − − + (1 + t)2 (1 − t)2 1 − t 1 + t 1 + t2

we conclude that     ∞ X 1 1−x x n+1 (−1) nfn (x) = + ln + arctan x 4 1 − x2 1+x n=1



which is the desired conclusion.

Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Clearly ∞ X

(−1)

n=1 ∞ X

=

n+1



n tan

(−1)n+1 n

n=1

−1

∞ X

x3 x2n+1 − . . . + (−1)n+1 x−x+ 3 2n + 1 (−1)k+1

k=n+1



=

x2k+1 2k + 1

Since |x| < 1, we can rearrange the series as ∞ X

n=1 k−1 X

n=1

∞ X

(−1)n+1 n

(−1)k+1

k=n+1

∞ k−1 X x2k+1 x2k+1 X = (−1)k (−1)n n 2k + 1 2k + 1 n=1 k=2

(−1)n n is equal to −k/2 if k is even and (k − 1)/2 if k is odd. It follows ∞ X

(−1)k

k=2

=− =−

k−1 x2k+1 X (−1)n n = 2k + 1 n=1

∞ X kx4k+1

k=1 ∞ X k=1

4k + 1

−

∞

∞

k=2

k=2

1X x4k−1 1 X x4k−1 (2k − 1) + = 2 4k − 1 2 4k − 1

∞ ∞ ∞ 1 X x4k+1 1 X 4k−1 3 X x4k−1 x4k+1 + − x + . 4 4k + 1 4 4 4k − 1 k=1

k=2

k=2

276 ∞ X

x4k+1 =

k=1

x5 . 1 − x4

Z 1 x4k+1 4k+1 = y 4k dy = x 4k + 1 0 k=1 k=1 Z x Z 1 y4 1 1+x 1 x4 y 4 dy = dy = −x + ln + arctan x. =x 4 y4 4 1 − x 1 − y 4 1−x 2 0 0 ∞ X

∞ X

Moreover ∞ X

x4k−1 =

k=2 ∞ X

Z

1

Z

1

x7 . 1 − x4

x8 y 6 dy = 1 − (xy)4

Z

x

t6 dt = 4 0 0 0 1−t k=2  Z x 1 t2 1 1 t+1 1 1 1 2 − + + − − t dt = = 2 2 1 + t2 41+t 41−t 4 0 x 1 x2 x 1 1 x2 x x3 = − arctan x + − + ln(1 + x) − ln(1 − x) − − − = 2 2 8 4 4 4 8 4 3 1 1 1 x3 − arctan x + ln(1 + x) − ln(1 − x) − . 2 4 4 3 By summing up the three contributions we obtain x4k−1

y 4k−2 dy =

1 x

1 x5 x 1 1+x 1 1 x7 − + ln + arctan x − + 4 1 − x4 4 16 1 − x 8 4 1 − x4   3 1 1 1 x3 + − arctan x + ln(1 + x) − ln(1 − x) − = 4 2 4 4 3 1 x 1 1 1−x =− − arctan x − ln . 4 1 − x2 4 4 1+x −

Solution 3 by Arkady Alt, USA.  San Jose,California,  3 2n+1 +∞ P x n x n+1 −1 + ... + (−1) . (−1) n tan (x) − x − Let S (x) := 3 2n + 1 n=1   +∞
n
P 1 n+1 = Then S 0 (x) = (−1) n − 1 − x2 + ... + −x2 2 1+x n=1 !
n+1 +∞ +∞
n−1 P P x2(n+1) P 1 − −x2 1 x4 +∞ n+1 (−1) n − = n· = n x2 = 2 2 2 2 1+x 1+x 1+x 1 + x n=1 n=1 n=1 x4 1 1 1 1 1 1 · = − + + 2 + 4 (x2 + 1) 1 + x2 (1 − x2 )2 4 (x − 1) 4 (x + 1) 8 (x − 1)2 8 (x + 1)  x Rx t4 dt 1 1−t t −1 and, therefore, S (x) = 0 = ln + + tan (t) = 2 4 1 + t 1 − t2 (1 + t2 )(1 − t2 ) 0  1 1−x x ln + + tan−1 (x) . 4 1 + x 1 − x2

277

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria; Moti Levy, Rehovot, Israel; and the proposer. 94.Proposed by Sava Grozdev and Deko Dekov (Jointly), Bulgaria. Find a point P in the plane of a given triangle ABC, such that the sum 2

2

2

|BP | |CP | |AP | + + b2 c2 a2 is minimal, where a = BC, b = CA and c = AB. Solution 1 by Moti Levy, Rehovot, Israel.

− − − Lemma 1. Let → r 1, → r 2, . . . , → r nP be n arbitrary points on a plane. Let w1 , w2 , . . . , wn n −c denote the weighted average be n positive weights, such that k=1 wk = 1. Let → vector defined by n X → −c := − wm → r m. m=1

Then the following holds: n X

m=1

2 − − wm |→ z −→ r m| =

n X

m=1

2 2 −c − → − −c − → − wm |→ r m | + |→ z|

Proof.   2 2 2 − − − − − − − − − − wm |→ z −→ r m | = wm (→ z −→ r m ) · (→ z −→ r m ) = wm |→ z | + |→ r m | − 2→ z ·→ rm

  2 2 −c − → − −c − → − −c − → − −c |2 + |→ − −c · → − wm |→ r m | = wm (→ r m ) · (→ r m ) = wm |→ r m | − 2→ rm   2 2 2 − − −c − → − − −c |2 + 2w (→ − → − → − wm |→ z −→ r m | − wm |→ r m | = wm |→ z | − |→ m c − z )· r m

Summing,

n X

m=1

2 − − wm |→ z −→ r m| −



n X

m=1

2 −c − → − wm |→ r m|

n n X X 2 − −c |2 −c − → − − = |→ z | − |→ wm + 2 (→ z )· wm → rm m=1

m=1

2 − −c |2 + 2 (→ −c − → − −c = |→ z | − |→ z )·→ 2 − −c |2 + 2 |→ −c |2 − 2→ − −c = |→ z | − |→ z ·→

2 2 − − −c + |→ −c |2 = |→ −c − → − = |→ z | − 2→ z ·→ z| .

This proves the lemma.



It follows from the lemma that the vector which minimizes the weighted sum Pn → − → − 2 → − m=1 wm | z − r m | is c . Now we apply the lemma to our problem:

278

− − − Let → r A ,→ r B and → r C be vectors from the origin to the triangle vertices, and let the weights be: wA =

1 a2

+

1 b2 1 b2

1 c2 2 2

+

=

c2 a2 a2 b2 + a2 c2 + b2 c2

a b a2 b2 + a2 c2 + b2 c2 b2 c 2 wC = 2 2 , a b + a2 c2 + b2 c2 then the original problem becomes: − Find a vector → r P which minimizes the weighted sum 2 2 2 − − − − − − wA |→ rP −→ r A | + wB |→ rP −→ r B | + wC |→ rP −→ r C| . wB =

The answer is

→ − − − − r P = wA → r A + wB → r B + wC → r C. If the triangle is equilateral then the point P is the centroid. Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We will use the following Lemma: Lemma 2. Consider n points A1 , . . . , An in the plane P, and let λ1 , . . . , λn be n positive numbers. We consider, the real function f : P → R defined by n X f (M ) = λk |Ak M |2 k=1

Let also G be the barycenter of the wighted points ((Ak ; λk ))1≤k≤n , that is G is the unique point G defined by n X −−→ λk GAk = ~0, k=1

then G is the unique point in the plane P where f attains its minimum. Proof. Indeed, f (M ) − f (G) = = = = =

n X

k=1 n X

k=1 n X

−−−→ −−→  λk Ak M 2 − Ak G2

−−−→ −−→ −−−→ −−→ λk Ak M − Ak G · Ak M + Ak G

−−→ −−→ −−→ λk GM · GM + 2Ak G

k=1 n X

k=1 n X k=1

λk λk

! !

n

−−→ X −−→ |GM |2 + 2GM · λ k Ak G k=1

|GM |2 .

This shows that f (M ) ≥ f (G) with equality if and only if M = G, and the lemma follows. 

279

In the proposed problem we have n = 3, A1 = A, A2 = B, A3 = C and λ1 = 1/b2 , λ1 = 1/c2 and λ1 = 1/a2 . Thus the desired minimum is attained at the unique point P which is the barycenter of the weighted points (A; b12 ), (B; c12 ), and (C; a12 ). This point is the first Brocard point Ω, it is the unique point inside ABC such that ∠ΩAB = ∠ΩBC = ∠ΩCA. (when the triangle ABC is labeled in counterclockwise order.) Also solved by the proposer. 95.Proposed by Li Yin, Department of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603, China. An approximation formula of Wallis product. For all n ∈ N, then  n ln n 1 Wn ∼ √ 1 − 2n π where Wn := (2n−1)!! (2n)!! is Wallis product. Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. We use 1 ln(1 − x) = −x − x2 + o(x2 ), 2

ex = 1 + o(1),

√ n! ∼ (n/e)n 2πn(1 + o(1))

We get 1 √ π



ln n 1− 2n

n

      1 1 1 ln n ln n = √ exp n ln 1 − = √ exp n − + o( ) 2n 2n n π π   1 ln n 1 = √ exp − + o(1) = √ √ (1 + o(1)) 2 π π n

√ √ 1 (2n)! (2n)2n e−2n 2π 2n(1 + o(1)) = √ √ (1 + o(1)) Wn = 2n = 2 (n!)2 22n n2n e−2n 2πn π n It follows lim

n→∞ √1 π

Wn 1−


ln n n 2n

=1

Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. On one hand, using Stirling’s formula, we have √ (2n)! 2 πn(2n)2n e−2n 1 ∼ 2n =√ , Wn = 2n (1) 2 (n!)2 2 (2πn)n2n e−2n πn

280

and on the other hand  n    ln n ln n 1− = exp n ln 1 − 2n 2n    2  ln n ln n = exp n − +O 2n n2   2  ln n ln n = exp − +O 2 n   2  1 ln n = √ exp O n n   2  1 ln n =√ 1+O n n Thus  n ln n 1 1 √ 1− ∼√ , 2n π πn and the desired conclusion follows from (1) and (2).

(2)

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria.

281

MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always welcomed. The source of the proposals will appear when the solutions be published.

Proposals 65. Let a, b, c, d be digits such that d > c > b > a ≥ 0. How many numbers of the form 1a1b1c1d1 are multiples of 33? 66. Given trapezoid ABCD with parallel sides AB and CD, let E be a point on line BC outside segment BC, such that segment AE intersects segment CD. Assume that there exists a point F inside segment AD such that ∠EAD = ∠CBF . Denote by I the point of intersection of CD and EF , and by J the point of intersection of AB and EF . Let K be the midpoint of segment EF , and assume that K is different from I and J. Prove that K belongs to the circumcircle of 4ABI if and only if K belongs to the circumcircle of 4CDJ 67. For all positive real numbers a, b, c, d prove the inequality p p p p √ a4 + c4 + a4 + d4 + b4 + c4 + b4 + d4 ≥ 2 2(ad + bc)

68. Consider a sequence of equilateral triangles Tn as represented below:

T1

T2

T3

T4

T5

The length of the side of the smallest triangles is 1. A triangle is called a delta if its vertex is at the top; for example, there are 10 deltas in T3 . A delta is said to be perfect if the length of its side is even. How many perfect deltas are there in T20 ?

282

69. Consider a chess board, with the numbers 1 through 64 placed in the squares as in the diagram below. 1 9 17 25 33 41 49 57

2 10 18 26 34 42 50 58

3 11 19 27 35 43 51 59

4 12 20 28 36 44 52 60

5 13 21 29 37 45 53 61

6 14 22 30 38 46 54 62

7 15 23 31 39 47 55 63

8 16 24 32 40 48 56 64

Assume we have an infinite supply of knights. We place knights in the chess board squares such that no two knights attack one another and compute the sum of the numbers of the cells on which the knights are placed. What is the maximum sum that we can attain? Note. For any 2 × 3 or 3 × 2 rectangle that has the knight in its corner square, the knight can attack the square in the opposite corner.

283

Solutions 61. Find all real solutions of the following system of equations: p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5, 9y 2 − 4x2

=

60.

(50th Catalonian Mathematical Olympiad) Solution 1 by Eloi Torrent Juste, AULA Escola Europea, Barcelona, Spain. First we observe that points (x, y) that satisfy the first equation are those that the sum of their distances to A(−3, 0) and B(0, 4) is equal to 5. Moreover, if a point P lies out of the segment AB then AP + P B > AB = 5. This let us to conclude that points (x,  y) solution  of the system must lie on AB. The equation of 4 4 AB is y = x + 4 or x, x + 4 with −3 ≤ x ≤ 0. Substituting these values in 3 3 the second equation, yields 2  4 x + 4 − 4x2 = 60 ⇔ x2 + 8x + 7 = 0 9 3 with roots x = −7 and x = −1. Since only the second lie in [−3, 0], then the unique solution of the given system is (−1, 8/3). Solution 2 by p Arkady Alt, San Jose, p California, USA. Squaring both sides of the equation x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5 we have 2 p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 25 ⇔ 4x − 3y + 12 = 0

Then, from

4x − 3y + 12 = 0 9y 2 − 4x2 = 60 we obtain



⇔

3y = 4x + 12 2 (4x + 12) − 4x2 = 60

⇔

3y = 4x + 12 12 (x + 7) (x + 1) = 0



 8 3  16 (x, y) = −7, 3 (x, y) =





−1,

  8 By substitution immediately follows that only (x, y) = −1, satisfies the given 3 system and it is the desired solution.  Also solved by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. 62. Let P be an interior point to an equilateral triangle ABC. Draw perpendiculars P X, P Y and P Z to the sides BC, CA and AB, respectively. Compute the value of BX + CY + AZ PX + PY + PZ (First BARCELONATECH MATHCONTEST 2014)

284

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. First let us denote the side length of the triangle by a. The area of the triangle can be calculated in two ways and we get √ 3 2 a = a(P X + P Y + P Z), hence 2 √ 3 PX + PY + PZ = a (1) 2 On the other hand. −−→ −−→ −→ −−→ −−→ −→ aBX = BC · BP , aCY = CA · CP , aAZ = AB · AP hence −−→ −−→ −→ −−→ −−→ −→ a(BX + CY + AZ) = BC · BP + CA · CP + AB · AP −−→ −→ −−→ −−→ −→ −−→ −−→ −−→ = (BC + CA + AB) ·BP + CA · CB + AB · AB | {z } ~ 0

−→ −−→ −−→ −−→ = CA · CB + AB · AB 1 = a2 + a2 2

so

3 a 2

(2)

√ BX + CY + AZ = 3. PX + PY + PZ



BX + CY + AZ = Thus, from (1) and (2) we get

Solution 2 by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Joining A, B, C with P we obtain three pairs of right triangles: AZP, AY P ; BZP, BXP and CXP, CY P. If a is the length of the side of 4ABC, then on account of Pithagoras theorem, we have AZ 2 + ZP 2 = (a − CY )2 + P Y 2 BX 2 + P X 2 = (a − AZ)2 + P Z 2

CY 2 + P Y 2 = (a − BX)2 + P X 2

Developing and adding up, yields

3a 2 On the other hand the sum of the areas of triangles AP B, BP C, CAP is the area of 4ABC. That is, √ √ a(P X + P Y + P Z) a2 3 a 3 = ⇔ PX + PY + PZ = 2 4 2 From the preceding immediately follows that √ BX + CY + AZ = 3 PX + PY + PZ and we are done.  BX + CY + AZ =

285

Solution 3 by Arkady Alt, San Jose, California, USA. Let a =√BC = CA = a 3 AB, x = BX, y = CY, z = AZ, u = P X, v = P Y, w = P Z and h = be height 2 of the equilateral triangle ABC, then ah au av az [ABC] = [P BC] + [P CA] + [P AB] ⇔ = + + ⇐⇒ u + v + w = h 2 2 2 2 and x+y+z x+y+z 2 (x + y + z) BX + CY + AZ √ = = = PX + PY + PZ u+v+w h a 3 Applying Pythagorean theorem to chain of right triangles 4P XB, 4P BZ, 4P ZA, 4P AY, 4P XB, 4P Y C, 4P CX, we obtain  2 2  u + x2 = w2 + (a − z) 2 w2 + z 2 = v 2 + (a − y)  2 2 v + y 2 = u2 + (a − x) Adding all equations we get  X
X 2 2 u2 + x2 = w + (a − z) ⇔ 3a2 = 2a (x + y + z) cyc

S0, x + y + z =

cyc

√ 3a BX + CY + AZ and, therefore, = 3. 2 PX + PY + PZ



Also solved by Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain. 63. How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram, 5 of two grams, and 6 of five grams, respectively? (Training Catalonian Team for OME 2014) Solution 1 by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH, Barcelona, Spain. The required number is the number of solutions of x + y + z = 31 with x ∈ {0, 1, 2, 3, 4, 5, 6, 7},

y ∈ {0, 2, 4, 6, 8, 10},

z ∈ {0, 5, 10, 15, 20, 25, 30}

We claim that the number of solutions of this equation equals the coefficient of x31 in the product (1 + x + x2 + . . . + x7 ) (1 + x2 + x4 + . . . + x10 ) (1 + x5 + x10 + . . . + x30 ) Indeed, a term with x31 is obtained by taking some term x from the first parentheses, some term y from the second, and z from the third, in such a way that x+y+z = 31. Each such possible selection of x, y and z contributes 1 to the considered coefficient of x31 in the product. Since, (1 + x + x2 + . . . + x7 ) (1 + x2 + x4 + . . . + x10 ) (1 + x5 + x10 + . . . + x30 ) = 1 + x + . . . + 10x30 + 10x31 + 10x32 + . . . + x46 + x47 , then the number of ways to obtain 31 grams is 10, and we are done.



Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We are looking for the number of triplets (k, l, m) such that k ∈ {0, . . . , 7}, l ∈ {0, . . . , 5}, m ∈ {0, . . . , 6}, k + 2l + 5m = 31.

286

Since k + 2l ≤ 17 we conclude that 5m ≥ 14, so m ∈ {3, 4, 5, 6}. • m = 3, then k + 2l = 16 or k = 2(8 − l) ≥ 2(8 − 5) = 6 this gives the unique solution (k, l, m) = (6, 5, 3). • m = 4, then k + 2l = 11 or k − 1 = 2(5 − l) ≤ 6. So every l ∈ {2, 3, 4, 5} yields a solution, and we get four solutions: (k, l, m) ∈ {(7, 2, 4), (5, 3, 4), (3, 4, 4), (1, 5, 4)}

• m = 5, then k + 2l = 6 or k = 2(3 − l). So every l ∈ {0, 1, 2, 3} yields a solution, and we get four solutions: (k, l, m) ∈ {(6, 0, 5), (4, 1, 5), (2, 2, 5), (0, 3, 5)}

• m = 6, then k + 2l = 1, and this yields the unique solution (k, l, m) = (1, 0, 6). So, the total number of ways to weight 31 grams is 10.  Solution 3 by Arkady Alt, San Jose, California, USA. We have to compute the number of elements of the set S := {(x, y, z) | x, y, z ∈ Z and x + 2y + 5z = 31, 0 ≤ x ≤ 7, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6}

= {(31 − 2y − 5z, y, z) | y, z ∈ Z and 24 ≤ 2y + 5z ≤ 31, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6} .  28 − 2y ≤ Since in integers 24 ≤ 2y+5z ≤ 31 ⇐⇒ 24−2y ≤ 5z ≤ 31−2y ⇐⇒ 5       31 − 2y 31 − 2y 28 − 2y z≤ and for any 0 ≤ y ≤ 5 holds ≤ 6, >0 5 5   5   31 − 2y 21 + 2t 2t + 1 then, denoting t := 5 − y, we obtain = = 4+ , 5 5 5       18 + 2t 2t + 3 28 − 2y = =3+ and 5 5 5      2t + 1 2t + 3 ≤z ≤4+ . S = (31 − 2y − 5z, 5 − t, z) | t, z ∈ Z and 0 ≤ t ≤ 5, 3 + 5 5          5 5 X X 2t + 3 2t + 1 2t + 1 2t + 3 Hence, |S| = 2+ − = 12 + − . 5 5 5 5 t=0 t=0    X  X  5  5  6  X 2t + 1 2t + 3 2t + 1 2t + 1 Noting that − = − = 5 5 5 5 t=0 t=1   t=0     2·0+1 2·6+1 13 − =− = −2 we get |S| = 12 − 2 = 10.  5 5 5 Also solved by Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain. 64. Let A(x) be a polynomial with integer coefficients such that for 1 ≤ k ≤ n + 1, holds: A(k) = 5k Find the value of A(n + 2). (Training UPC Team for IMC 2014) Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.

287

Remarks: • If n ≥ 3 no such polynomial exists. Indeed, if a polynomial P has integer coefficients then for every distinct integers a and b, we have (b−a) | (P (b)− P (a)). So, when n ≥ 3, if there is a polynomial A with integer coefficients such that A(1) = 5 and A(4) = 54 , then 3 must divide A(4) − A(1) = 620 which is absurd. • When n ∈ {0, 1, 2}, (which are the only possible values left for n,) the polynomial A is not uniquely determined by the conditions that it has integer coefficients and that it satisfies A(k) = 5k for 1 ≤ k ≤ n + 1. Indeed, when n = 0 the polynomial A(X) = λ(X − 1) + 5(2 − X), (for an arbitrary λ ∈ Z,) satisfies A(1) = 5 and A(2) = λ. Similarly, when n = 1, the polynomial A(X) = 20X − 15 + λ(X − 1)(X − 2) satisfies A(1) = 5, A(2) = 25 and takes an arbitrary odd value at X = 3. A similar conclusion also holds when n = 2. In view of the above, I propose a modified statement of the problem as follows: Generalization of Proposal 64. Let An (X) be a polynomial of degree n such that A(k) = 5k for 1 ≤ k ≤ n + 1. Find the value of A(n + 2). Solution. First, note that the existence and uniqueness, of An is guaranteed, from general theorems about Lagrange interpolation. Now, given An (X) for some n > 0 we consider 1 Qn (X) = (An (X + 1) − An (X)) 4 Clearly deg Qn = n − 1, and Qn(k) = 5k for 1 ≤ k ≤ n. Hence Qn (X) = An−1 (X). This allows us to conclude that An (X + 1) − An (X) = 4An−1 (X).

Let bn = An (n + 2), from the above formula we see that bn = 5n+1 + 4bn−1 This is equivalent to

 n+1 bn−1 5 = . 4n+1 4n 4 Adding these equalities and noting that b0 = 5 we see that n  k+1 bn 5 X 5 − = 4n+1 4 4 bn

−

k=1

Thus,

bn = 4n+1

n  k+1 X 5

k=0

4

= 5(5n+1 − 4n+1 ).

which is the desired conclusion.  Solution 2 by Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain. We observe that for all k ≥ 1, holds:       k   X k j k k k k 5k = (1 + 4)k = 4 = + 4 + ... + 4 j 0 1 k j=0

288

Now, we consider the polynomial         x−1 x−1 2 x−1 n B(x) = 5 1 + 4+ 4 + ... + 4 , 1 2 n where

  a a(a − 1)(a − 2) . . . (a − n + 1) = n n! for any real a and n ≥ 1. Clearly, deg(B(x)) = n and it is easy to see that A(k) = B(k) for 1 ≤ k ≤ n + 1. So, we conclude that A(x) = B(x) for all x ∈ R. Thus we have         n+1 n+1 2 n+1 n A(n + 2) = B(n + 2) = 5 1 + 4+ 4 + ... + 4 1 2 n             n+1 n+1 2 n+1 n n + 1 n+1 n + 1 n+1 =5 1+ 4+ 4 + ... + 4 + 4 − 4 1 2 n n+1 n+1 = 5(5n+1 − 4n+1 )



Also solved by by Arkady Alt, San Jose, California, USA and Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. 65. Let a0 , a1 , . . . , an and b0 , b1 , · · · , bn be complex numbers. If n ≥ 2, then prove that ! !   n n n X n2 2n − 2 X 1 n+1 X 2 2 |bk | |ak | + Re a k bk ≤ n 2 n n−1 n k=0

k=0

k=0

(Training UPC Team for IMC 2014)

Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. We begin with the following claim: Let α, a0 , a1 , . . . , an and b0 , b1 , · · · , bn be complex numbers, then it holds ! ! n n n X X 1 X Re α |ak |2 + |α|2 |bk |2 ak bk ≤ 2 k=0

k=0

Indeed, n X

k=0

2

|ak | + |α|

2

n X

k=0

2

|bk | − 2Re α =

n X

k=0

n X

k=0

ak bk

k=0

!

=

n X

(ak − αbk )(ak − αbk )

k=0

|ak − αbk |2 ≥ 0

and the inequality claimed follows.

n3 2n−2
n−1 n d holds. Indeed, applying two times the operator x dx
to the binomial identity
Pn Pn (1 + x)n = k=0 nk xk we get nx(1 + x)n−1 = k=0 k nk xk and   n X n−1 2 n−2 2 n nx(1 + x) + n(n − 1)x (1 + x) = k xk k Now, we prove that for all n ≥ 2, the following identity

Pn

k=0

k=0

k2


n 2 k

=

289

Multiplying the first and the third and equating terms of degree n, we obtain      2 n X 2n − 1 2n − 2 2 n k =n + n(n − 1) n−1 n−2 k k=0   (2n − 1)(2n − 2)! (2n − 2)! n3 2n − 2 =n + n(n − 1) = n!(n − 1)! n!(n − 2)! n−1 n as claimed.   n To prove the statement, we put into the claim α = k for 0 ≤ k ≤ n, and we k obtain ! !    2 X n n n X n 1 X 2 2 n 2 k Re ak bk ≤ |ak | + k |bk | k 2 k k=0

k=0

k=0

After adding up the above expressions, we get ! !   n n n X X n3 2n − 2 X 1 2 2 n−1 |bk | |ak | + (n + 1) n2 Re ak bk ≤ n 2 n−1 k=0 k=0 k=0   n X n on account that k = n 2n−1 . The last identity can be easily obtained k k=0   n X n k setting x = 1 in nx(1 + x)n−1 = k x . From the preceding the inequality k k=0 claimed immediately follows and the proof is complete.  Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let un be defined by     n+1 n2 2n − 2 n + 1 2n − 2 un = 4 n · n = n−1 2 n 2 (n − 1) n 4 n−1 It is easy to check that un+1 (n + 2)(2n − 1) n−2 −1= −1= ≥1 un 2n(n + 1) 2n(n + 1) So, the sequence {un }n≥2 is increasing with u2 = 23 ≥ 1. Thus un ≥ 1 for every √ n ≥ 2. Now let us come to our problem. Using the simple inequality X+Y ≥ 4XY we conclude that v v ! u n u n  X n n 2 X u uX √ 1 n+1 X n 2n − 2 2 2 2 t t |a | + |b | ≥ u |a | |bk |2 k k n k 2n n n−1 n k=0 k=0 k=0 k=0 ! n n n X X X ≥ |ak ||bk | ≥ ak bk ≥ Re ak bk . k=0

k=0

where used the Cauchy-Schwarz inequality.

k=0



Also solved by Jos´ e Gibergans-B´ aguena, BARCELONA TECH, Barcelona, Spain.

290

MATHNOTES SECTION On an equation for the sum-of-divisors function ´ zsef Sa ´ ndor Jo Abstract. We study the solutions of the equation σ(22k − 1) = 3 · 22k−1 . Keywords. Sum-of-divisors function, inequalities. AMS Subject Classification. 11A25. 1. Introduction Let σ(n) denote the sum of divisors of the positive integer n. It is well-known that this function is multiplicative, and for prime powers p one has σ(pa ) = pa + . . . + p + 1 =

pa+1 − 1 . p−1

(1)

We define also σ(1) = 1. By studying certain composition function of σ with other arithmetical functions (see [1]) we have encountered the equation σ(22k − 1) = 3 · 22k−1 .

(2)

The aim of this note is to completely determine all solutions of (2) when k is even. For k odd, a conjecture is stated. 2. Main results Theorem. The only even solution of equation (2) is k = 2. Proof. The following auxiliary result will be used: Lemma. Let p be a prime divisor of n. Then σ(n) ≥

p+1 · n. p

(3)

There is equality only for n = p. Proof. Let n = pk · N , where p is prime and (p, N ) = 1. First remark that the inequality pk+1 − 1 p+1 k ≥ ·p p−1 p

(4)

is true. Indeed, a simple computation shows that (4) is equivalent to pk−1 ≥ 1, which is true, as k ≥ 1 and p ≥ 2, with equality only for k = 1. p+1 k Now, since σ(n) = σ(pk ) · σ(N ) ≥ · p · N , inequality (3) follows. p Let now be k an even solution to (2). Let k = 2m (m ≥ 1). Then remark that n = 22k − 1 = 24m − 1 = 16m − 1, which is divisible by 5, for any m ≥ 1.

291

Clearly n is always divisible by 3, as n = 4k − 1. Let n = 3a · 5b · N be the prime factorization of N , where N is not divisible by 15. By applying the Lemma, we can write:    4 a 6 b 24 σ(n) ≥ ·3 · 5 · σ(N ) ≥ n, as σ(N ) ≥ N. 3 5 15 Therefore, the inequality 24 2k 3 · 22k−1 ≥ (2 − 1) (5) 15 must hold true. Written equivalently: 15 2k · 2 ≥ 8(22k − 1), or 22k ≤ 24 so k ≤ 2. 2 As k is even, we get k = 2. Remark. Unfortunately, the above method cannot be applied for k = odd. A such solution of equation (2) is k = 5. Conjecture. The only odd solution k to equation (2) is k = 5. References

[1] J. S´ andor, On the composition of some arithmetic functions, III (in preparation) [For parts I and II, see: Studia Univ. Babe¸s-Bolyai, Math., 34(1989), 7-14; resp. J. Ineq. Appl. Pure Math., 6(2005), no. 2, art. 73 (electronic)].

J´ ozsef S´ andor: Babe¸s-Bolyai University Department of Mathematics, Str. Kog˘ alniceanu nr. 1 400084 Cluj-Napoca, Romania E-mail:[email protected]

292

JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before October 15, 2014

Proposals ´ 26. Proposed by Francisco Javier Garc´ıa Capit´ an, I.E.S. Alvarez Cubero de C´ ordoba, Spain. Let ABC be a rectangle triangle, CD the altitude corresponding to the right angle, and I, Ja , Jb the incenters of the triangles ABC, CAD and CDB, respectively. Find the area of the triangle IJa Jb in terms of the hypothenuse c and the inradius r of the triangle ABC. 27. Proposed by Ercole Suppa, Teramo, Italy. Let 4ABC a triangle with incenter I and circumcenter O. Let D, E, F be the touch points of the incircle (I) with BC, CA, AB respectively. Let V be the center of circle passing through F which touches BC at B. Let W be the center of circle passing through E which touches BC at C. Prove that if AC + AB = 2 · BC then O, I, V , W are collinear. 28. Proposed by Paolo Perfetti, “Tor Vergata” Univ., Rome, Italy. Let a, b, c be real positive numbers. Prove that r r r r 3a2 + 2b2 + 3c2 3b2 + 2c2 + 3a2 3c2 + 2a2 + 3b2 3 (a + b)(b + c)(c + a) + + ≥3 2ac 2ba 2cb abc 29. Proposed by Armend Sh. Shabani, University of Prishtina, Republic of Kosova. Let p be a fixed positive integer, greater than 2. Show that the mapping f : N → N ∪ {0} defined by h n i h 2n i h (p − 1)n i f (n) = + + ... + 2 3 p is neither injective nor surjective. 30. Proposed by D.M. Bˇ atinet¸u-Giurgiu, Bucharest, Romania and Neculai Stanciu, Buzˇ au, Romania. Determine all positive integers numbers a, b, c, d which satisfy the equation a2 + b3 + c4 = d5 .

293

Solutions 21. Proposed by Dorlir Ahmedi, University of Prishtina, Republic of Kosova a a, b, c > 0 and b+c ≥ 32 then prove that

If

a b c + + ≥ 2. b+c c+a a+b Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria The proposed inequality is equivalent to ∆ ≥ 0 where ∆ = a(c + a)(a + b) + b(b + c)(a + b) + c(b + c)(c + a) − 2(b + c)(a + c)(a + b)171 Now, withwe s= b + c prove and p that = bc, we have Therefore should 2 3 ∆ = a3 −c a+2 sa + a(p b +−a s ) + s − 4ps + ≥8 c By assumption a = xs for some x b≥ 3/2, so   or equivalently x ∆ = s3c x3b− xa2 − a(3 + u) + u + + + 4 ≥ 8.  2 b c b c 4p b−c wherecu =b1 − 2 = ∈ [0, 1]. aNowa if P (x) = x3 − x2 − x4 (3 + u) + u then Since + ≥ 2sone hasb + tocprove that + ≥ 6. b c   b c       3 5u 3 12 − u 3 3 0 00 (3) From Cauchy-Schwarz inequality we have P = , P = , P = 7, P =6 2 8 2 4 2 2 a a  b  c + · + ≥4 Thus, b 3c a a k X P (k) (3/2) 3 Therefore P (x) = x− . ak! 3 2 a a k=0 + ≥4· ≥ 4 · ≥ 6. c b + c ∆ ≥ 20 which is the desired inequality. So, P (x) ≥ 0 for every x ≥b 32 . Consequently Note solved that equality holds if and only if ua= 0sand x = 3/2 that (a, b, c) =Stanciu, (2λ, λ, λ) Also by Titu Zvonaru, Comˇ ne¸ ti, Romania andis Neculai for some λ > 0. ”George Emil Palade” School, Buzˇ au, Romania, and the propser

Also solved by Titu Zvonaru, Comˇ ane¸sti, Romania and Neculai Stanciu, 22. Proposed by Emanuele Callegari, Math. Dept., “Tor Vergata” University, “George Emil Palade” School, Buzˇ au, Romania; and the proposer Rome, Italy We have a rectangular chessboard 3 × 20 made of square boxes by Emanuele Callegari, “Tor University, 122. × 1. Proposed We also have 30 dominoes of size 1Math. × 2 or 2Dept., × 1 that we Vergata” want to use to cover Rome, Italy. We have a rectangular chessboard × 20 the chessboard. How many are the different ways3to do made that? of square boxes 1 × 1. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the Solution by the proposer chessboard. How many are the different ways to do that? The answer is 413403. Solution by the proposer. Let Xn be the number of different coverings of a chessboard of the following type coverings of a chessboard of the following type Let Xn be the number of different

3 2n Figure 1 with with little little rectangles rectangles of of dimensions dimensions 22× ×11 or or 11× ×2. 2. Observe that, if the chessboard is of type (2n + 1) × 3, the coverings are trivially zero because the chessboard has an odd number of cells. In order to solve our problem we must compute X10 . A possible strategy could be to look for a recursive formula for Xn but, after a few attempts, we probably realize that it is hard to do it. In a certain sense, it is better to be more ”ambitious” and to search non only Xn , but also the number Yn of different coverings of a chessboard of the following type:

The answer is 413403. Let Xn be the number of different coverings of a chessboard of the following type 3

294

2n Figure 1

with little rectangles of dimensions 2 × 1 or 1 × 2. Observe that, if the chessboard Observe that, if the chessboard is of type (2n+1)×3, the coverings are trivially zero is of type (2n + 1) × 3, the coverings are trivially zero because the chessboard has because the chessboard has an odd number of cells. In order to solve our problem an odd number of cells. we must compute X10 . In order to solve our problem we must compute X10 . A possible strategy could be to look for a recursive formula for X but, after a A possible strategy could be to look for a recursive formula for Xnn but, after a few attempts, we probably realize that it is hard to do it. In a certain sense, it is few attempts, we probably realize that it is hard to do it. In a certain sense, it is better to be more ”ambitious” and to search non only Xn , but also the number Yn better to be more ”ambitious” and to search non only Xn , but also the number Yn of different coverings of a chessboard of the following type: of different coverings of a chessboard of the following type: 3

172 172

2n Figure 2

In fact, In172 fact, in in spite spite of of the the fact fact that that itit isis hard hard to to find find aa recursive recursive formula formula for for X Xn (and (and In fact, in),spite of the fact that it rather is hardsimple to findtoa write recursive formulaformula for Xnn (and also for Y we discover that it is a recursive n also for Yn ), we discover that it is rather simple to write a recursive formula for for alsofact, for in Ynspite ), we of discover simple write a recursive formula for both, simultaneously. In the factthat thatititisisrather hard to find atorecursive formula for Xn (and both, simultaneously. both, simultaneously. To this aim, let us start to place a little rectangle to cover the left upper cell of the also for Y ), we discover that it is rather simple to write a recursive formula for n To this aim, let us start to place a little rectangle to cover the left upper cell of the To thissimultaneously. aim, letchessboard us start to place a little can rectangle to cover the left upper cell of the non-rectangular both, non-rectangular chessboard of of fig.2. fig.2. We We can do do itit in in two two ways: ways: non-rectangular chessboard of fig.2. We can do it in two ways: To this aim, let us start to place a little rectangle to cover the left upper cell of the non-rectangular chessboard of fig.2. We can do it in two ways: 3 3 3 3 3 3 2n 2n 2n 3 2n 4 Figure Figure 2n 3 2n 4 Figure Figure When we start in the way of fig.3 the part of the chessboard that remains uncovered Figure 3 Figure 4remains start the way of fig.3 theXpart of the chessboard thatthe is When of thewe kind of in fig.1, so that it has coverings. On other uncovered hand, if n different When we start in the way of fig.3 the part of the chessboard that remains uncovered is start of the ofinfig.1, so of that it has Xn to different coverings. Onremains the other hand, if weWhen inkind the way of way fig.4 we are forced in the following way we start the fig.3 the part ofcontinue the chessboard that uncovered iswe of the kind of way fig.1,ofso thatwe it are has Xn different coverings. the other if in the fig.4 to continue in theOn following wayhand, is ofstart the kind of fig.1, so that it hasforced Xn different coverings. On the other hand, if we start in the way of fig.4 we are forced to continue in the following way we start in the way of fig.4 we are forced to continue in the following way 3 3 3 2n 2n 5 Figure 2n 5 Figure so that the part of chessboard still uncovered has Yn−1 different coverings. As a Figure 5 has Yn−1 different coverings. As a so that partthat of chessboard still uncovered result we the obtain result obtain so thatwethe part that of chessboard Ystill uncovered has n−1 different coverings. As a Xn + Yn−1 . YYn−1 so that the part of chessboard still has different coverings. As a n =uncovered result we obtain that n =. X n+Y n−1 result we obtain that Yn = Xn + YYn−1 Now, let us. consider the rectangle in fig.1. Now, let us consider the rectangle in fig.1. We can cover the left upper cell in two =X We canlet cover the left the upper cell inYntwo ways: n + Yn−1 . Now, us consider rectangle in fig.1. We can cover the left upper cell in two ways: ways: let us consider the rectangle in fig.1. We can cover the left upper cell in two Now, ways: 3 3 3 3 3 3 2n 2n 2n 6 2n 7 Figure Figure 2n 6 2n 7 Figure Figure If we cover it in the way of fig.6 we are forced to continue in the following way: IfIfwe cover it in the way of fig.6 we are forced to continue in Figure 6 Figure 7following we cover it in the way of fig.6 we are forced to continue inthe the followingway: way: If we cover it in the way of fig.6 we are forced to continue in the following way: 3 3 3 2n 2n Figure 8 2n 8 Figure so that the part of chessboard still uncovered has Yn−1 different coverings. On the 8 so that the part of chessboard stillFigure uncovered has Y different coverings. On the

Now, let consider us consider rectangle in fig.1. cover upper in two Now, let us the the rectangle in fig.1. We We cancan cover the the left left upper cell cell in two ways: ways: 3

3

3

3

2n 2n 2n 2n Figure 6 Figure Figure 6 Figure 7 7 295 If cover we cover in the of fig.6 we are forced to continue in the following If we it initthe wayway of fig.6 we are forced to continue in the following way:way: 3

3

2n 2n Figure Figure 8 8 so that part of chessboard uncovered Y different different coverings. so that that thethe part of chessboard chessboard stillstill uncovered hashas different coverings. OnOn thethe n−1n−1 so the part of still uncovered has YYn−1 coverings. On the other hand, if we start in the way of fig.7, we can continue in two ways: other hand, if we start in the way of fig.7, we can continue in two ways: other hand, if we start in the way of fig.7, we can continue in two ways: 3

3

2n 2n Figure Figure 9 9

3

3

2n 2n Figure Figure 10 10

The region still uncovered in fig.9 has Yn−1 different coverings while the one of fig.10 has Xn−1 . Thus we obtain: Xn = 2Yn−1 + Xn−1 . It is now easy to verify by direct inspection that X1 = 3 and Y1 = 4. Therefore we obtain the recursive formula: ( Xn = 2Yn−1 + Xn−1 (1) Yn = Xn + Yn−1 with initial data:

(

X1 = 3 Y1 = 4.

After 10 iterations of (1), we obtain X10 = 413403, which is the answer to the problem. 23.Proposed by Emanuele Callegari, Math. Dept., “Tor Vergata” University, Rome, Italy. Mary has 5 baskets each containing 97 colored balls numbered from 0 to 96. The color of the balls in the first basket is blue, in the second is green and in the third is red. The color of the balls in the fourth and fifth baskets is white. Mary picks up a ball from each basket in such a way that the sum of the five numbers is 96. How many different configurations can occur to Mary? Solution by the proposer The answer is 200425.

We need the Lemma:



m+k−1 Lemma If k and m are strictly positive integers, then there exist exactly k−1 k−uples of non negative integers (n1 , n2 , . . . , nk ) such that n1 + n2 + · · · + nk = m. This is a standard combinatorial result whose proof is omitted. Let (n1 , n2 , n3 , n4 , n5 ) be any choice of balls, i.e for any i = 1, . . . , 5 let ni be the number that is written on the ball that has been picked up from the i-th



296

 100 is the number of different 5−tuples 4 (n1 , n2 , n3 , n4 , n5 ) such that n1 + n2 + n3 + n4 + n5 = 96.

basket. Clearly, thanks to the Lemma,



Two different situations may occur. In the first one (solution of type I) n4 6= n5 while in the second one (solution of type II) n4 = n5 . Let’s call N1 the number of solutions of type I and N2 the number solutions of type II. The sum N1 + N2 is our number. Clearly   100 2N1 + N2 = total number of 5−tuples = (0) 4

Thus, if we find N2 , the relation above will allow us to find also N1 and therefore, also N1 + N2 . Let us find N2 , i.e. the number of 5-tuples of non negative integer of the form (x, y, z, a, a), such that x + y + z + a + a = 96. (1) Note that, thanks to the Lemma, for every fixed  a = 0, 1, . . . , 48, the 3−tuples of 98 − 2a non negative integer (x, y, z) satisfying (1), are . 2 Therefore the number N2 of all (x, y, z, a, a) satisfying (1) is          48  X 98 − 2a 2 4 6 98 N2 = = + + + ··· + . (2) 2 2 2 2 2 a=0 To compute (2), define the quantity         3 5 7 97 S= + + + ··· + . 2 2 2 2

(3)

Thanks to the elementary properties of the binomial coefficients, from (2) and (3) it follows:               2 3 4 5 97 98 99 N2 + S = + + + + ··· + + = (4) 2 2 2 2 2 2 3 and

          2 4 3 6 5 N2 − S = + − + − + ... 2 2 2 2 2     98 97 ... + − = 1 + 3 + 5 + · · · + 97 = 492 . 2 2

As a consequence of (4) and (5) we have: N2 =

1 1 · ((N2 + S) + (N2 − S)) = · 2 2

Thus, from (0) and (6) it follows: 1 1 N1 +N2 = ·((2N1 + N2 ) + N2 ) = · 2 2 which is the result.



100 4

Two incorrect solutions were received.





1 + · 2

99 3





99 3

(5)

 + 492 .



+ 49

2



(6)

= 2000425,

297

24.Proposed by Stanescu Florin, Serban Cioculescu school, Gaesti, jud.Dambovita, Romania. Prove that in a triangle ABC the following inequality holds: 27r ra rb rc p ≤ + + ≤ 2p a b c 2r where ra , rb , rc are the lengths of the rays of the excircles, r is the radius of the circle inscribed in the triangle, and p the semi-perimeter of the triangle. Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria We will use the following lemma: Lemma: for positive numbers x, y, z with x + y + z = 1, we have 1 1 1 1 27 ≤ + + ≤ 2 x(1 − x) y(1 − y) z(1 − z) 2xyz 1 x(1−x)

1 + 1−x is convex on (0, 1), so   x+y+z 27 , f (x) + f (y) + f (z) ≥ 3f = 3 2

Proof. First, the function f (x) =

=

1 x

uv which is the first inequality. On the other hand, since u+v ≤ u+v 4 for every positive u and v, we have zx xy yz + + xyz(f (x) + f (y) + f (z)) = y+z z+x x+y y+z+z+x+x+y 1 ≤ = 4 2 and the upper inequality follows.  Next, we come to the proposed problem. Since

ra (p − a) = rb (p − b) = rc (p − c) = rp = Area(ABC)

we conclude, by setting x = 1 − ap , y = 1 − pb , z = 1 − pc , that

ra rb rc rp rp rp + + = + + a b c a(p − a) b(p − b) c(p − c)   r 1 1 1 = + + p x(1 − x) y(1 − y) z(1 − z)

So, by the lemma we conclude that

ra rb rc r 27r ≤ + + ≤ p a b c 2pxyz But xyz =

r2 by Heron’s formula, and the desired inequality follows. p2

Also solved by Neculai Stanciu, Buz˘ au, Romania, Titu Zvonaru, Comˇ ane¸sti, Romania, and the proposer 25. Proposed by Neculai Stanciu, “George Emil Palade” School, Buzˇ au, Romania and D.M. Bˇ atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania Find all pairs of real numbers (x, y) such that p p p p x2 + 2x + 1 + x2 − 4x + 4 + x2 − 2xy + y 2 + y 2 − 6y + 9 = 4.

298

Solution by the proposer. We note first that p p p p x2 + 2x + 1 + x2 − 4x + 4 + x2 − 2xy + y 2 + y 2 − 6y + 9 = 4 is equivalent to

|x + 1| + |x − 2| + |x − y| + |y − 3| = 4. (1) Since |a| + |b| ≥ |a + b| and | − a| = |a| for every real numbers a and b, we conclude that |x + 1| + |x − y| + |y − 3| ≥ |x + 1 + y − x| + |3 − y| ≥ |1 + y + 3 − y| = 4.

From (1) and (2) we have:

4 = |x + 1| + |x − 2| + |x − y| + |y − 3| ≥ |x − 2|. hence |x − 2| = 0 ⇔ x = 2. For x = 2 equation (1) becomes |2 − y| + |y − 3| = 1. After solving last equation one obtains solutions (x, y) ∈ {(2, a)|2 ≤ a ≤ 3}.

(2)

299

SOLUTION OF A PROBLEM ANOTHER PERSPECTIVE! This section of the Journal presents a solution of a problem, proposed at MathProblems or elsewhere. The presentation of such solution should be done in the style of an article. The articles, should contain ideas and information that readers may find suitable to use in other similar problems.

Dominos on a rectangular board Solution to Junior Problem 22 MathProblems (4)1 (2014) Brendan W. Sullivan Emmanuel College, USA Abstract: We have a rectangular chessboard 3 × 20 made of square boxes 1 × 1. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the chessboard. In this note we address the question of counting the number of different ways to do that? (Problem proposed by Callegari Emanuele, Math. Dept. “Tor Vergata” University, Rome, Italy.) Solution: Let T (2n) be the number of ways to tile a 3 × 2n rectangular chessboard using dominoes. (Note: there are trivially no domino tilings of a 3×k board when k is odd.) This problem seeks a number for T (20). We solve this problem by proving the more general recursive formula T (2n) = 4 · T (2n − 2) − T (2n − 4)

for n ≥ 2, with the additional convention that T (0) = 1. (To motivate this convention, consider: How many ways can we tile an empty board? One way: I just did it!) This will lead us to the answer: T (20) = 413403. First, observe that T (2) = 3:

Now, consider a 3 × 2n board. We will describe all of the cases wherein one can construct a domino tiling of this board. These disjoint cases are based on the dominoes in the (2n − 2)-th and (2n − 1)-th columns of the board; more specifically,

300

these cases depend on how many domino tiles cross the boundary between those two columns.

··· 3 × 2n − 2 portion

boundary

(i) 0 crossings. This means we have a proper tiling of the final two columns, appended to a proper tiling of the first 2n − 2 columns. We know that there are 3 ways to tile the final two columns (since T (1) = 3), and there are T (2n−2) ways to tile the remaining portion. Thus, this case contributes the term 3 · T (2n − 2) to the total number of tilings. (ii) 1 crossing. By exhibiting three cases, we can show that this case actually contributes no proper tilings: each possibility yields an irreconcilable situation in the 2n-th column. Firstly, if the one and only crossing occurs on the top row, we can attempt to tile the surrounding squares in two ways:

···

?

?

··· Secondly, by obvious symmetry, the situations where the one and only crossing occurs on the bottom row yield no tilings, as well. Finally, when the one and only crossing occurs in the middle row, we have one possibility:

···

?

301

Thus, the case of exactly 1 crossing yields no tilings. (iii) 2 crossings. This is the interesting case! First, we observe that if the two crossings occur on the top and bottom row, then this yields no tilings:

?

···

?

Next, we consider the more fruitful cases where the two crossings occur on the top two rows. (By symmetry, this will be analogous to the cases where the two crossings occur on the bottom two rows; thus, the terms we derive presently will then be doubled to account for all the tilings.) What we will observe is that such cases do, indeed, yield proper tilings. Furthermore, each case can be extended arbitrarily far to the left, in blocks of two columns. Thus, we will generate tilings that correspond to this summation of terms: T (2n − 4) + T (2n − 6) + T (2n − 8) + · · · T (2) + T (0). One possibility is as follows:

··· 3 × 2n − 4 portion This amounts to a particular proper tiling of the final 4 columns, which can be appended to any proper tiling of the first 2n − 4 columns. Thus, the number of such tilings is T (2n − 4). We can extend this to the left by converting the leftmost vertical domino into two horizontal ones, like so:

··· 3 × 2n − 6 portion This amounts to a particular proper tiling of the final 6 columns, which can be appended to any proper tiling of the first 2n − 6 columns. Thus, the number of such tilings is T (2n − 6).

302

Observe that this extension process can be continued until we reach the leftmost side of the board. That final case amounts to a particular proper tiling of the entire board, “appended” to the only tiling of the empty board. Therefore, we include the terms of the summation mentioned above to account for all of these tilings. Also, as mentioned above, the cases where the two crossings of the boundary column occur on the bottom two rows are symmetrically analogous. In total, then, we include the terms Pn−2 2 · k=0 T (2k). (iv) 3 crossings. Observe that this case yields no proper tilings:

···

?

Overall, we have now deduced that T (2n) = 3 · T (2n − 2) + 2 ·

n−2 X

T (2k)

k=0

provided n > 1 (otherwise the first part of the case involving 2 crossings above is invalid). The next step is to simplify this recurrence to two terms. Consider the difference T (2n) − T (2n − 2), for n ≥ 1, and observe that we can cancel terms:
T (2n) − T (2n − 2) = 3T (2n − 2) + 2T (2n − 4) + 2T (2n − 6) + · · · + 2T (0)
− 3T (2n − 4) + 2T (2n − 6) + · · · + 2T (0) = 3T (2n − 2) − T (2n − 4)

Adding T (2n − 2) to both sides, we conclude that

T (2n) = 4T (2n − 2) − T (2n − 4)

as claimed, originally. Knowing that T (0) = 1 (by convention), or even just counting to observe that T (2) = 3 and T (4) = 11, we can plug numbers into this recurrence and deduce that n

0

1

2

3

T (2n)

1

3

11 41

4

5

153 571

6

7

2131

7953

8

9

29681 110771

10 413403

Thus, the desired answer is that there are T (20) = 413403 ways to properly tile a 3 × 20 rectangular chessboard using dominoes. Finally, note that the above recurrence is linear, so one can solve for a closed form using standard techniques (i.e. generating functions). This yields & ' √ ! √ √ ! √ n √ n −1 + 3 (2 + 3)n+1 1+ 3 √ √ √ · (2 + 3) + · (2 − 3) = . T (2n) = 2 3 2 3 3+ 3 This can be used somewhat more efficiently for larger n. Plugging in n = 10 (i.e. 2n = 20) yields the same answer as above.

Mathproblems ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 3, Issue 1 (2013), Pages 118–139

Editors: Valmir Krasniqi, Jos´ e Luis D´ıaz-Barrero, Paolo Perfetti, Armend Sh. Shabani, Valmir Bucaj, Mih´aly Bencze, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, J´ ozsef S´andor, David R. Stone, Roberto Tauraso, Francisco Javier Garca Capit´ an, Emanuele Callegari, Ercole Suppa, Cristinel Mortici.

PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]

Solutions to the problems stated in this issue should arrive before June 15, 2013

Problems 58. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. 1) Let a be a real number such that −1 ≤ a ≤ 1. Prove that     ∞ X ∞ X (i − 1)!(j − 1)! i+j a a a = 2Li2 − 2Li2 − − 2Li2 (a), (i + j)! 2−a 2−a i=1 j=1 where Li2 (z) is the Dilogarithm function defined by Li2 (z) = − all z ∈ {z ∈ C : |z| ≤ 1}. 2) Show that ∞ X ∞ X (i − 1)!(j − 1)! i=1 j=1

(i + j)!

= ζ(2),

where ζ(z) represents the Riemann zeta function. c

2010 Mathproblems, Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. 118

R z ln(1 − t) dt for 0 t

119

59. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Romania (Jointly). Let n be a positive integer. Prove that √ p Ln L2n+2 Ln+1 L2n+3 2 + + (Ln + Ln+2 ) ≥ 2 6 · Ln Ln+1 · Ln+2 , Ln+3 Ln + Ln+2 where Ln represents the nth Lucas number defined by L0 = 2, L1 = 1, and for all n ≥ 2, Ln = Ln−1 + Ln−2 . 60. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Determine all functions f : R \ {−2, −1, 0, 1, 2} → R, which satisfy the relation   7+x f (x) + f = ax + b 2−x where a, b ∈ R. 61. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil PaladeP Secondary School, Buzˇ au, Ron mania (Jointly). If a ∈ R+ , b, c, d, xk ∈ R∗+ , Xn = k=1 xk and cXn > d max xk , 1≤k≤n

then prove that n X aXn + bxk k=1

cXn − dxk

≥

(an + b)n . cn − d

62. Proposed by Mih´ aly Bencze, Bra¸sov, Romania Show that if ak , bk , ck ∈ R∗ (k = 2 1, 2), such that a1 + b21 + c21 = a22 + b22 + c22 , then at least one of the equations a1 x2 + 2c2 x + b1 = 0; b1 x2 + 2a2 x + c1 = 0; c1 x2 + 2b2 x + a1 = 0 has real roots. 63. Proposed by Anastasios Kotronis, Athens, Greece. Let k ≥ 3 be an integer number. Show that X n≥0

1 = (2n + 1)(3n + 2) · · · (kn + k − 1)

  k 1 X m−1 k (−1) (m − 1)mk−2 k! m=2 m

 ! m−1 X π π 2`π `π cot − ln m + cos · ln sin . 2 m m m `=1

64. Proposed by Florin Stanescu, Serban School Cioculescu, Gaesti, Romania. Let f : [0, 1] → R be a continuous function. If Z 1 Z 1 2 x f (x)dx = −2 F (x)dx, 0

where F (x) =

Rx 0

1/2

f (t)dt, x ∈ [0, 1], then prove that Z 1 2 Z 1 f 2 (x)dx ≥ 24 f (x)dx . 0

0

65.∗ Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. Let Ωn

120

π n/2 represents the volume of the unit ball in Rn , that is Ωn = , n = 1, 2 . . . , Γ(n/2 + 1) R ∞ x−1 −t where Γ(x) = 0 t e dt, x > 0. For all natural numbers n, show that r r 1 (Ωn+1 ) n+1 n+2 α n + 2 6 β 6 1 n+3 n+3 n (Ωn ) with the best possible constant factors α = 2 and β =

ln( 34 ) √

ln(

π 2 )

= 2.3818 · · ·

121

Solutions No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.

50. Proposed by Anastasios Kotronis, Athens, Greece. Show that +∞ X k=0

(−1)k 1 = + O(n−1 ln−2 n) ln(n + k) 2 ln n

n → +∞.

Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. More generally we have : Proposition. Let f : (a, ∞) → R be a continuously differentiable P∞ convex function such that limx→∞ f (x) = 0. Then, for every x > a, the series k=0 (−1)k f (x + k) does converge, and its sum satisfy the following inequality: ∞

X 1 1 (−1)k f (x + k) ≤ − f 0 (x). 0 ≤ − f (x) + 2 2 k=0

Proof. First, note that the conditions on f imply that f is nonnegative decreasing and that limx→∞ f 0 (x) = 0. Indeed, if for some b ∈ (a, +∞) we have f 0 (b) > 0, then the convexity of f implies that f (x) ≥ f (b) + (x − b)f 0 (b) and this leads to the contradiction limx→∞ f (x) = +∞. So, f 0 ≤ 0 and f must be nonnegative and decreasing. On the other hand, since f 0 is increasing and negative, there exists a real number ` ≤ 0 such that limx→∞ f 0 (x) = `. If ` < 0, then the convexity of f implies that for y > b we have f (y) − f (b) ≤ f 0 (y)(y − b) ≤ `(y − b). Taking the limit as y tend to +∞ we obtain the contradiction −f (b) ≤ −∞. So, we must have ` = 0. Now, for each x ∈ (a, +∞) the sequence {f (k + x)}k≥0 is decreasing to zero, and P the alternating series test proves the convergence of the series (−1)k f (x + k). So, let S(x) be defined, for x ∈ (a, +∞), by S(x) =

∞ X

(−1)k f (x + k) =

k=0

∞ X


f (2k + x) − f (2k + 1 + x)

k=0

On the other hand f (x) = f (x + 2m + 2) +

m X


f (2k + x) − f (2k + 2 + x)

k=0

So, letting m tend to +∞, we obtain f (x) =

∞ X k=0


f (2k + x) − f (2k + 2 + x)

122

Combining the preceding, we get 2S(x) − f (x) =

∞ X


f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x)

k=0

But using the convexity of f we have f (2k+1+x)−f (2k+x) ≥ f 0 (2k+x)

and

f (2k+2+x)−f (2k+1+x) ≤ f 0 (2k+2+x)

Thus f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f 0 (2k + 2 + x) − f 0 (2k + x) On the other hand, again using the convexity of f , we have f (2k + 1 + x) ≤

1 (f (2k + x) + f (2k + 2 + x)) 2

It follows that 0≤

m X


f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f 0 (2m + 2 + x) − f 0 (x)

k=0

So letting m tend to +∞ we obtain using (3) that 0 ≤ 2S(x) − f (x) ≤ −f 0 (x). Applying the Lemma to the function f : (1, +∞) → R, f (x) = 1/ ln(x) we obtain +∞

∀ x > 1,

X (−1)k 1 1 1 ≤ ≤ + 2 ln x ln(x + k) 2 ln x 2x ln2 x k=0

from which the statement follows. Also solved by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Moubinool Omarjee, Paris, France; and the proposer. 51. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Romania (Jointly). Let a ∈ (1, ∞) and b ∈ (0, ∞). Calculate: n X (n + k)a−1 lim n 2 − exp n→∞ (n + k)a + b

!!

k=1

Solution 1 by Moti Levy. The first step is to show that the limit does not dependent of a and b. Indeed, we have n n a−1 X (n + k) 1X = a n (n + k) + b 1+

k=1

k=1

1 k n

+

b n(n+k)a−1

123

If we put en,k =

b , n(n+k)a−1

n

1X 1 n 1+ k=1

n

1X = n ≤

1 n

k=1 n X k=1

1 ≤ k ≤ n, then 0 < en,k ≤

n

k n

−

1X n 1+ k=1

b na

and

1 k n

+

b n(n+k)a−1

! ! n 1 1X en,k

− = k n 1 + nk + en,k 1 + nk 1 + nk + en,k n k=1 ! Z 1 n 1 en,k 1 1 b b 1X b ≤ a

≤ a 2 dx = 2 na . k 2 k 2 n n n 0 (1 + x) 1+ 1+

1 1+

k=1

n

0≤

1 n

n X

1 1+ k=1

k n

−

1 n

n

n X k=1

1 1+

k n

+

b n(n+k)a−1

≤

1 b 2 na

By Taylor’s theorem,  h2 exp (x + h) − exp (x) = exp (x) h + exp (δ) , 0 ≤ δ ≤ h 2 ! ! n n 1X 1 1X 1 − exp exp b n n 1 + nk 1 + nk + n(n+k) a−1 k=1 k=1 ! !   n 2 1 b 1 1 b 1 b 1X 1 ≤ exp + exp (δ) , 0 ≤ δ ≤ . n 2 na 2 2 na 2 na 1 + nk 

k=1

!! n n a−1 X 1X 1 (n + k) − n 2 − exp n 2 − exp a n (n + k) + b 1+ k=1 k=1 ! !! n n a−1 X (n + k) 1X 1 = n exp − exp a k n (n + k) + b 1+ n k=1 k=1 ! !  2 n 1 b 1X 1 1 1 b ≤ n exp + exp (δ) n 2 na 2 2 na 1 + nk

!! k n

k=1

Z1

n

1X 1 n 1+ k=1

k n

≤

1 dx = ln 2 1+x

0

!!

!! n n a−1 X 1X 1 (n + k) n 2 − exp − n 2 − exp a n (n + k) + b 1 + nk k=1 k=1 ! !  2  2 1 b 1 1 b 1 b 1 1 b ≤ n exp (ln 2) exp (δ) = 2n exp (δ) + + 2 na 2 2 na 2 na 2 2 na 1 b2 exp (δ) 4 n2a−1 It follows from the preceding that =

b

na−1

+

n a−1 X (n + k) lim n 2 − exp a n→∞ (n + k) + b k=1

!!

!!

n

= lim n 2 − exp n→∞

1X 1 n 1+ k=1

k n

.

124

  P n The second step is evaluation of limn→∞ n 2 − exp n1 k=1 Let ∆n := ln 2 −

1 n

1 k 1+ n



.

Pn

1 k=1 1+ k n

then ! n 1X 1 = exp n 1 + nk k=1

exp (ln 2) − exp

!

n

1X 1 n 1+ k=1

k n

1 ∆n + ∆2n exp δ 2



and !!

n

lim n 2 − exp

n→∞

1X 1 n 1+ k=1

k n

!

n

= lim exp n→∞

1X 1 n 1+ k=1

k n

 1 2 n∆n + n∆n exp δ . 2

1 So now we estimate ∆n . Since the function 1+x is convex then the area of the triangle formed by the secant is greater than ∆n . That is, ! ! n−1 n−1 1 1 X 1 1 1 X
n
= − ∆n ≤ k k k+1 2n 2n 1 + 1 + k+1 1 + 1 + n n n n k=0 k=0 ! ! ! n−1 n X 1 X 1 3 1 1 ≤ 2
2 = 2
2 + 2n 2n 4 1 + nk 1 + nk k=0 k=1 ! Z 1 n 3 1 1X 1 1 dx 1 3 3 + + 2 = = ≤ + 2.
2 2 2 2n n 8n 2n 0 (1 + x) 8n 4n 8n 1+ k k=1

n

1 1+x

Since the function is convex then the area of the triangle formed by the tangent is less than ∆n . Namely, ! ! n n n−1 1 1 3 1 X 1 1X 1 1X 1 ∆n ≥ 2 =
=

− k 2 k 2 k 2 2n 2n n 2n n 4n k=1 1 + n k=1 1 + n k=0 1 + n ! Z 1 1 dx 3 1 3 ≥ = − 2. 2 − 4n 2n 4n 8n 0 (1 + x) ∆n − 1 ≤ 3 . 4n 8n2 !! n 1X 1 lim n 2 − exp n→∞ n 1 + nk k=1 !  n 1X 1 1 2 = lim exp n∆n + n∆n exp δ n→∞ n 2 1 + nk k=1 !  n 1X 1 1 2 = lim exp lim n∆ + lim n∆ exp δ n n n→∞ n→∞ n→∞ 2 n 1 + nk k=1   1 1 = exp (ln 2) +0 = . 4 2 Solution 2 by Anastasios Kotronis, Athens, Greece. We use the fact that Hn = ln n + γ +

1 + O(n−2 ) 2n

n → +∞.

(1)

125

Note that for m > 1, it is X k≥n+1

1 = O(n1−m ), km

(2)

Applying Ces` aro–Stolz criterium, we have lim nm−1

n→+∞

−(n + 1)−m 1 = lim m n→+∞ (n + 1)1−m − n1−m k k≥n+1
−1 = lim m − 1 + O(n−1 ) X

n→+∞

→

1 . m−1

Now for a > 1 and b ∈ R, Sa,b,n

  n n X X 1 b (n + k)a−1 −2a = 1− + O(n ) := (n + k)a + b n+k (n + k)a k=1

k=1

= H2n − Hn − b

n X k=1

1 + O(n−2a ) (n + k)a+1

n X

1 + O(n−2 ) (n + k)a+1 k=1 ! 2n n X X 1 1 1 = ln 2 − −b − + O(n−2 ) 4n k a+1 k a+1 k=1 k=1   X X 1 1  1 − b − + O(n−2 ) = ln 2 − 4n k a+1 k a+1 k≥n+1 k≥2n+1   1 = ln 2 − + O n− min{a,2} . 4n = ln 2 −

1 −b 4n

On account of the above 

  1 1 a>1 1 − min{a,2} + O(n ) = +O(n− min{a−1,1} ) −→ . n (2 − exp Sa,b,n ) = n 2 − 2 1 − 4n 2 2 Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Moubinool Omarjee, Paris, France; Adrian Naco, Polytechnic University, Tirana, Albania; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer. 52. Proposed by Yuanzhe Zhou, The School of Physics and Technology (SPT) at Wuhan University. Let a, b, c > 0, prove that,       b c 3 a 2 2 2 cos + cos + cos > b+c a+c a+b 2

126

Solution by Moti Levy. Setting x =

a b c , y= , z= , a+b+c a+b+c a+b+c

then the inequality becomes       x y z 3 cos2 + cos2 + cos2 > , 1−x 1−y 1−z 2 Using the well-known trigonometric identity cos2 α =   2x , the inequality becomes f (x) = cos 1−x f (x) + f (y) + f (z) > 0,

1 2

x + y + z = 1. +

1 2

cos 2α , and setting

x + y + z = 1.

1 2,

The function f (x) is concave for 0 < x < so we can find a piecewise linear lower bound for f (x). Let g (x) be defined by  −2 (1 − cos 2) x + 1, 0 < x < 12 g (x) = 1 −1, 2 ≤x≤1 then f (x) ≥ g (x) for 0 < x ≤ 1. WLOG, we may assume that 0 < x ≤ y ≤ z ≤ 1. Then it will be suffice to consider the following two cases: 1) z < 21 and 2) z ≥ 21 . Case 1) z <

1 2

:

f (x) + f (y) + f (z) ≥ (−2 (1 − cos 2) x + 1) + (−2 (1 − cos 2) y + 1) + (−2 (1 − cos 2) z + 1) = −2 (1 − cos 2) (x + y + z) + 3 = −2 (1 − cos 2) + 3 > 0 Case 2) z ≥

1 2

: x+y <

1 2

f (x) + f (y) + f (z) ≥ −2 (1 − cos 2) x + 1 − 2 (1 − cos 2) y + 1 − 1 ≥ −2 (1 − cos 2) (x + y) + 2 − 1 = 1 − (1 − cos 2) (x + y) ≥ 1 −

1 (1 − cos 2) > 0. 2

Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; and the proposer. 53. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. The Stirling numbers of the first kind denoted by s(n, k), are the special by the generating function z(z − 1)(z − 2) · · · (z − n + 1) = Pn numbers defined k k=0 s(n, k)z . Let n and m be nonnegative integers with n > m − 1. Prove that ( Z 1 (−1)n · n! · ζ(n + 1), m = 1, lnn x Pm−1 dx = n! n+m−1 i m (−1) · (m−1)! · i=1 (−1) · s(m − 1, i) · ζ(n + 1 − i), m ≥ 2, 0 (1 − x) where ζ denotes the Riemann zeta function. Solution 1 by Anastasios Kotronis, Athens, Greece. By repeated integration by parts one can easily show that for any non negative integer k and any positive integer n, we have Z

k

n

x ln x dx = x

k+1



 lnn x n lnn−1 x n(n − 1) lnn−2 x (−1)n n! − + − ... + +c k+1 (k + 1)2 (k + 1)3 (k + 1)n+1

127

Furthermore, we will use that (1 − x)−m =

X k + m − 1 k≥0

k

xk

x ∈ (−1, 1)

Now, we have  Z 1 Z 1X X k + m − 1  Z 1 lnn x k+m−1 k n dx= x xk lnn x dx ln x dx = m k k 0 (1 − x) 0 k≥0 0 k≥0 X k + m − 1 (−1)n n! = k (k + 1)n+1 k≥0

−k−1 X k+m−1 X m−1 m−1 n n+m−1 = (−1) n! = (−1) n! (k + 1)n+1 (k + 1)n+1 k≥0

k≥0

But for m ≥ 2, from the definition of Stirling numbers, we have 

 −k − 1 (−k − 1)(−k − 1 − 1)(−k − 1 − 2) · · · (−k − 1 − (m − 1 − 1)) = m−1 (m − 1)! m−1

X 1 s(m − 1, i)(−k − 1)i , (m − 1)! i=0
and for m = 1, −k−1 m−1 = 1, so, Z 1 n ln x dx = (−1)n n!ζ(n + 1) 1 −x 0 =

and for m ≥ 2: Z 0

1

lnn x dx (1 − x)m

=

(−1)n+m−1

X m−1 X (−1)i s(m − 1, i) n! (m − 1)! (k + 1)n−i+1 i=0 k≥0

=

(−1)n+m−1

m−1 X X n! 1 (−1)i s(m − 1, i) (m − 1)! i=0 (k + 1)n−i+1 k≥0

=

(−1)n+m−1

n! (m − 1)!

m−1 X

(−1)i s(m − 1, i)ζ(n − i + 1)

i=1

and we get what we wanted. Solution 2 by Omran Kouba, Higher Institute for AppliedR Sciences and 1 Technology, Damascus, Syria. First let us calculate the integral 0 xα lnβ (1/x) dx for α > −1 and β > 0. Using the change of variables x → e−t we have Z 1 Z ∞ Z ∞ 1 Γ(β + 1) β α β −(α+1)t x ln (1/x) dx = t e dt = uβ e−u dt = . β+1 (α + 1) (α + 1)β+1 0 0 0 Now, for x ∈ [0, 1), we have ∞

X (− ln x)n = xk (− ln x)n . 1−x k=0

128

Since all the terms are positive we can integrate on [0, 1] and exchange the order of summation Z 1 n ∞ Z 1 ∞ X X ln x 1 = (−1)n n!ζ(n + 1) dx = xk lnn x dx = (−1)n n! (k + 1)n+1 0 1−x 0 k=0

k=0

This gives the case m = 1. Now suppose that m ≥ 2. For x ∈ [0, 1) we have   ∞  ∞  X X −m 1 m−1+j j j = (−x) = x (1 − x)m j m−1 j=0 j=0

(1)

Pm−1 On the other hand, by definition we have i=0 s(m−1, i)z i = z(z−1)(z−2) · · · (z− m + 2). Replacing z by −(j + 1), we obtain m−1 X

s(m − 1, i)(−1)i (j + 1)i = (−1)m−1 (j + 1)(z + 2) · · · (j + m − 1)

i=0

or   m−1 m−1+j (−1)m−1 X s(m − 1, i)(−1)i (j + 1)i = (m − 1)! i=0 m−1 Replacing in (1) we obtain m−1 ∞ X (−1)m−1 X 1 i = (−1) s(m − 1, i) (j + 1)i xj (1 − x)m (m − 1)! i=0 j=0

It follows that Z 0

1

  Z 1 m−1 ∞ X lnn x (−1)m−1 X n dx = (−1)i s(m − 1, i)  (j + 1)i xj ln xdx (1 − x)m (m − 1)! i=0 0 j=0   m−1 ∞ X n! (−1)m−1 X i i n  (−1) s(m − 1, i)  (j + 1) (−1) = (m − 1)! i=0 (j + 1)n+1 j=0   m−1 ∞ n+m−1 X X (−1) n! 1  = (−1)i s(m − 1, i)  n+1−i (m − 1)! (j + 1) i=0 j=0 =

m−1 (−1)n+m−1 n! X (−1)i s(m − 1, i)ζ(n + 1 − i) (m − 1)! i=0

which is the desired result, since s(m − 1, 0) = 0 for m ≥ 2. (In fact, the above formula includes the case m = 1 since s(0, 0) = 1.) Remark. The same proof shows that, more generally, for a real β and an integer m such that 1 ≤ m ≤ β we have Z 1 β m−1 ln (1/x) (−1)m−1 Γ(β + 1) X dx = (−1)i s(m − 1, i)ζ(β + 1 − i) m (m − 1)! 0 (1 − x) i=0 Also solved by Moti Levi; and the proposer.

129

54. Proposed by Moubinool OMARJEE, Paris, France. Let f : [0, +∞[→ R be a measurable function such that g(t) = et f (t) ∈ L1 (R+ ) ; the space of Lebesgue integrable functions. Find      n1 Z nt + 1 nt − 1 lim f (t) 4 sinh sinh dt n→+∞ R 2 2 + where sinh(x) =

ex −e−x 2

Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. We prove that nt − 1 nt + 1 sinh ≤ etn 2 2 = x, we have   √ √  √ √ 1 x e √ −√ e x− √ √ ≤x e x e x sinh

Indeed, putting etn

from which follows e2 − x(e2 + 1) ≤ 0 This holds true since x ≥ 1. Now define 

nt − 1 nt + 1 sinh fn (t) = f (t) 4 sinh 2 2

 n1

|fn | ≤ et |f (t)| ∈ L1 (R+ ) by the Lebesgue integrability of et f (t). The dominated convergence theorem allows us to take the limit under integral getting 1 nt + 1 nt − 1 n lim f (t) 4 sinh sinh dt = n→+∞ R 2 2 + 1  Z Z nt − 1 n nt + 1 sinh dt = f (t)et dt f (t) lim 4 sinh n→+∞ 2 2 R+ R+ Z



Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Anastasios Kotronis, Athens, Greece; Moti Levi; and the proposer. 55. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Let x, y, z, α be real positive numbers. Show that if X nx3 + (n + 1)x =α x2 + 1 cycl

then

X1 2α 27n3 > + 2 x 3 9n α + α3 cycl

where n is a natural number.

130

Solution by Moti Levi. Let x = y = z = 1 and α = 301.5; then n = 100 satisfies the X condition. 3 1 2α 27n3 2×301.5 + 9×100227×100 x = 3 and 3 + 9n2 α+α3 = 3 ×301.5+301.53 = 201. 50. cycl

Clearly 3 > 201. 50 is false, which implies that the claim in Problem 55 is not true. Author’s Comment: I assume that either I did not understand the problem statement or there are typo errors. 56. Proposed byJos´e Luis D´ıaz–Barrero, BARCELONA TECH, Barcelona, Spain. Let n ≥ 2 be a positive integer. Find all possible values of the number k such that 2 2 2 2 2 Fn+2 ) Fn+2 (1 + Fn2 Fn+1 ) Fn2 ) Fn2 (1 + Fn+1 F 2 (1 + Fn+2 + = k + n+1 Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn

Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let 2 2 Fn+2 Fn+1 Fn2 + − Bn Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn   1 1 1 2 2 = Fn2 Fn+1 Fn+2 + − Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn so, that k = An + Bn . Clearly,

An =

Bn =

2 2 Fn+2 Fn2 Fn+1 (Fn + Fn−1 − Fn+1 ) = 0 Fn−1 Fn Fn+1

On the other hand An =

3 2 + Fn+2 Fn−1 Fn3 − Fn+1 . Fn−1 Fn Fn+1

Now 3 2 2 2 Fn3 − Fn+1 + Fn+2 Fn−1 = −Fn−1 (Fn2 + Fn+1 + Fn Fn+1 ) + Fn+2 Fn−1
2 2 2 = Fn−1 Fn+2 − Fn − Fn+1 − Fn Fn+1 2 = Fn−1 (Fn+1 + Fn )2 − Fn2 − Fn+1 − Fn Fn+1

= Fn−1 (2Fn Fn+1 − Fn Fn+1 ) = Fn−1 Fn Fn+1 Thus An = 1 and consequently k = 1. Also solved by Moti Levi; and the proposer.




131

MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always welcomed. The source of the proposals will appear when the solutions be published.

Proposals 41. Find a cubic polynomial which zeros are the square of the zeros of p(x) = x3 + 2x2 + 3x + 4. 42. Let n be an odd positive integer. A knight is placed at random in one of the n2 squares of a chessboard. It is possible that knight back to its initial position after passing once through each square? 43. Compute

sin 1◦ + sin 2◦ + . . . + sin 133◦ + sin 134◦ cos 1◦ + cos 2◦ + . . . + cos 133◦ + cos 134◦

44. Let n, m be positive integers. Find the value of the following expression    n+m 1 X k X n m 7 23n i j k=0

i+j=k

45. Let AGB, BHC and CKA be equilateral triangles constructed on the sides of any triangle ABC. Prove that the centers of the circumcircles of those equilateral triangles D, E, F themselves form an equilateral triangle.

132

Solutions 36. Find all positive integers n such that 17n−1 + 19n−1 divides 17n + 19n . (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´ anchez (student), CFIS, BARCELONA TECH, Barcelona, Spain. We have that 17(17n−1 + 19n−1 ) < 17n + 19n < 19(17n−1 + 19n−1 ) as can be easily checked. Since 17n−1 + 19n−1 divides 17n + 19n , then 17n +19n = 18(17n−1 +19n−1 ) = (17+1)17n−1 +(19−1)19n−1 = 17n +17n−1 + 19n − 19n−1 from which follows 17n−1 = 19n−1 .The preceding is only possible for n = 1 for which 17n−1 + 19n−1 = 2 that divides 17n + 19n = 36, and we are done. 2 Also solved by Bruno Salguerico, Viveiro, Spain ; Ioan Viorel Codeanu, Satulung, Rumania ; Jos´ e Gibergnas-B´ aguena, BARCELONA TECH, Barcelona, Spain ; Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. 37. Let α, β and γ be three distinct complex numbers. Show that they are collinear if, and only if, Im(αβ + βγ + γα) = 0. (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. The equation of the line through γ = (c, d), β = (a, b), z = (x, y) is   x−c y−d z−γ = ⇐⇒ Im =0 a−c b−d β−γ   α−γ = 0, or Therefore,α, β, γ are collinear if, and only if, Im β−γ
  Im αβ − γβ − αγ + |γ|2 (α − γ)(β − γ) Im = =0 |β − γ|2 |β − γ|2 which is equivalent to
Im αβ − γβ − αγ = 0 Since for any complex number z = x + iy is Im(−z) = −y = Im z, then
Im αβ − γβ − αγ = Im(αβ) + Im(−γβ + Im(−αγ) = Im(αβ + βγ + γα) = 0 and we are done. 2 Also solved by Jos´ e Gibergnas-B´ aguena, BARCELONA TECH, Barcelona, Spain, Daniel V´ acaru, Pitesti, Rumania, Bruno Salguerico, Viveiro, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.

133

38. Through the midpoint M of a chord P Q of a circle, any other chords AB and CD are drawn; chords AD and BC meet P Q at points X and Y respectively. Prove that M is the midpoint of XY. (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´ e Gibergnas-B´ aguena, BARCELONA TECH, Barcelona, Spain. We begin by dropping perpendiculars x1 = XX 00 and y1 = Y Y 00 from X and Y to AB, x2 = XX 0 and y2 = Y Y 0 from X and Y to CD.

C

A x1 P

X

Y'

x'' M

y2

Q Y y1

x2

x'

Y'' B

D

Figure 1. Butterfly’s Theorem Letting a = P M = M Q, x = XM, y = M Y, and observing the pairs of similar triangles: 4M XX 00 ∼ 4M Y Y 00 , 4M XX 0 ∼ 4M Y Y 0 , 4AXX 00 ∼ 4CY Y 0 ,4DXX 0 ∼ 4BY Y 00 , yields x x1 = , y y1

x2 x = , y y2

x1 AX , = y2 CY

x2 XD = y1 YB

from which follows x2 x1 x2 x1 x2 AX · XD P X · XQ = = · = = y2 y1 y2 y2 y1 CY · Y B PY · Y Q (a − x)(a + x) a 2 − x2 (a2 − x2 ) + x2 a2 = 2 = 2 =1 = 2 2 2 2 (a + y)(a − y) a −y (a − y ) + y a and x = y, as we wanted to prove. =

2 Also solved by Bruno Salguerico, Viveiro, Spain and Jos´ e Luis D´ıazBarrero, BARCELONA TECH, Barcelona, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. 39. The students of a University Course in Mathematics take their exams in Calculus, Algebra, Physics and Geometry. It is known that 73% passed Calculus exam, 82% passed Algebra, 77% passed Physics and 89% passed Geometry. At least, how many students have passed the exam of all four subjects?

134

(Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´ anchez (student), CFIS, BARCELONA TECH, Barcelona, Spain. First, we calculate the number of students that passed Calculus and Algebra. That is, N (C ∪ A) = N (C) + N (A) − N (C ∩ A) and N (C ∩ A) = N (C) + N (A) − N (C ∪ A) Since N (C) = 73, N (A) = 82, then N (C∩A) will be minimum when N (C∪A) = 100 that it is the maximum value that it may attain. So, N (C ∩ A) = 73 + 82 − 100 = 55 Now, we calculate the minimum number of students that pass the exam of Calculus, Algebra and Physics. Let us denote by N (B) = N (C ∩ A) the number of student that passed the exam of Calculus and Algebra under the hypothesis that N (C∪A) = 100. Then N (B ∪ P ) = N (B) + N (P ) − N (B ∩ P ) and N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P ) The minimum value of N (B ∩ P ) will be got when N (B ∪ P ) = 100. So, N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P ) = 55 + 77 − 100 = 32 Finally, we compute the number of students that passed the four exams. Let us denote by B ∩ P = C ∩ A ∩ P = D. Then, N (D ∪ G) = N (D) + N (G) − N (D ∩ G) and N (D ∩ G) = N (D) + N (G) − N (B ∪ P ) The minimum value of N (D ∩G) corresponds to the maximum value of N (B ∪P ) = 100. So, N (D ∩ G) = N (D) + N (G) − N (B ∪ P ) = 32 + 89 − 100 = 21 2

and we are done.

Also solved by Bruno Salguerico, Viveiro, Spain and Jos´ e Luis D´ıazBarrero BARCELONA TECH, Barcelona, Spain. 40. Let α, β, γ be the angles of an acute triangle ABC. Prove that  2/3 tan2 α 1 1 X +3 ≥2 3 tan β tan γ tan α + tan β + tan γ cyclic

(Mediterranean Mathematical Olympiad 2012)

135

Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Using the well-known identity tan α tan β tan γ = tan α + tan β + tan γ and rearranging terms, we can write the inequality claimed as √ tan3 α + tan3 β + tan3 γ 3 3 tan α tan β tan γ + ≥2 3 tan α tan β tan γ tan α + tan β + tan γ Now, applying two times AM-GM inequality, we have √ 3 3 tan α tan β tan γ tan3 α + tan3 β + tan3 γ + 3 tan α tan β tan γ tan α + tan β + tan γ  √   √  3 3 3 3 tan α + tan β + tan γ 3 tan α tan β tan γ 3 3 tan α tan β tan γ = +3 −2 3 tan α tan β tan γ tan α + tan β + tan γ tan α + tan β + tan γ s 9(tan3 α + tan3 β + tan3 γ) ≥4 4 −2 (tan α + tan β + tan γ)3 So, it will be suffice to prove that s 3 3 3 4 9(tan α + tan β + tan γ) ≥1 (tan α + tan β + tan γ)3 The preceding is equivalent to see that 9(tan3 α + tan3 β + tan3 γ) ≥ (tan α + tan β + tan γ)3 It trivially holds from r 3 3 3 tan α + tan β + tan γ 3 tan α + tan β + tan γ ≤ 3 3 Equality holds when tan α = tan β = tan γ. That is when 4ABC is equilateral and we are done. 2 Also solved by Bruno Salguerico, Viveiro, Spain and Jos´ e GibergnasB´ aguena, BARCELONA TECH, Barcelona, Spain.

136

MATHNOTES SECTION The Inequality Ionescu - Weitzenb¨ ock ˇ tinetu-Giurgiu and Neculai Stanciu D. M. Ba In Memory of Ion Ionescu

Abstract. The aim of this note is to prove that the Weitzenb¨ oc’s inequality must be named the Ionescu-Weitzenb¨ oc’s inequality.

1. Introduction In Romanian Mathematical Gazette, Vol. III (15 September 1897 – 15 August 1898), No. 2 , 15 October 1897, on page 52, Ion Ionescu, the founder of Mathematical Gazette, published the following problem: *273. Prove that there is no triangle for which the inequality √ 4S 3 > a2 + b2 + c2 is satisfied. The solution of the problem 273, appeared in Gazeta Matematicˇa, Vol. III (15 September 1897–15 August 1898), No. 12, 15 August 1898, on pages 281-283, as follows: Problema 273. Prove that there is no triangle for which the inequality √ 4S 3 > a2 + b2 + c2 is satisfied. Solution by N.G. Muzicescu (Student, Ia¸si). Let ABC be a triangle, and we construct around the side BC the equilateral triangles BCA0 and BCA00 ; so M = AA0 ∩ BC will be the middle of these two lines. Let AA0 = d and AA00 = d0 . In triangle AA0 A00 , and applying well–known theorems we deduce that: d2 + d02 = 2 · AM 2 + 2 · A00 M 2

(1)

d2 − d02 = 4 · A00 M 2 · M H;

(2)

and from the triangle ABC, we have b2 + c2 = 2 · AM 2 +

a2 . 2

(3)

137

On the other hand, because A00 M is the height of the equilateral triangle with length of the side a, we have: 1 √ (4) A00 M = a 3. 2 By the relations (1), (2), (3), and (4) we deduce: d2 + d02 = a2 + b2 + c2 , (5) √ 2 02 d − d = 2a · M H 3, (6) but 2a · M H represents 4 times the area of the triangle ABC, because M H is evidently equal to the height of this triangle; making the substitutions given by (5) and (6) we get √ 2d02 = a2 + b2 + c2 − 4S 3, which in turn gives 2

2

2

√

A

A H B M

C

A

a + b + c ≥ 4S 3, so the given inequality is impossible.

 

Solution by I. Moscuna (Bucharest) and I. Penescu (Bucharest). We have √ 1 cos 30◦ S = bc sin A, a2 = b2 + c2 − 2bc cos A, 3 = cot 30◦ = 2 sin 30◦ Replacing, simplifying with 2 and letting the second member by only b2 + c2 we have   cos 30◦ + cos A > b2 + c2 , bc sin A · sin 30◦ which yields 2bc sin(A + 30◦ ) > b2 + c2 . On the other hand we have (b − c)2 ≥ 0, so 2bc ≤ b2 + c2 .

(7) (8)

◦

The inequality (7) is satisfied for A < 150 , because otherwise the first member would be negative and could not be greater like the second member which is positive. Assuming this condition is satisfied we can divide (7) by (8). Therefore sin(A + 30◦ ) > 1,

(9)

which is impossible, since the sine function cannot be greater than the unity. So the inequality (7) and therefore the given inequality are absurd for any triangle. If (8) and (9) become equalities then so does (7). So we must have A + 30◦ = 90◦ , i.e. A = 60◦ and b = c, i.e. the triangle must be equilateral. Hence the given inequality becomes equality for any equilateral triangle. Solution by Miss. Maria Rugesu (Student Ia¸si) and by Mrs. Th. M. Vladimirescu (Rˆımnicul Vˆılcei); G. G. Urechilˇ a and I. Sichitiu (Sc. Sp. De Art. ¸si Geniu) and Corneliu P. Ionescu (Student Cl, VI Lic. Galat¸i).

138

The given inequality becomes successively 4

p √ p(p − a)(p − b)(p − c) · 3 > a2 + b2 + c2 , 16p(p − a)(p − b)(p − c)3 > (a2 + b2 + c2 )2 ,

6p(2p − 2a)(2p − 2b)(2p − 2c) > (a2 + b2 + c2 )2 , 3(a + b + c)(b + c − a)(a + c − b)(a + b − c) > (a2 + b2 + c2 )2 ,    3 (b + c)2 − a2 a2 − (b − c)2 > (a2 + b2 + c2 )2 ,    3 2bc + c2 + b2 − a2 2bca2 − b2 − c2 > (a2 + b2 + c2 )2 ,   3 4b2 c2 − (a2 − b2 − c2 )2 > (a2 + b2 + c2 )2 ,   3 2a2 b2 + 2a2 c2 + 2b2 c2 − a4 − b4 − c4 > a4 + b4 + c4 + 2a2 b2 + 2a2 c2 + 2b2 c2 , 4a2 b2 + 4a2 c2 + 4b2 c2 − 4a4 − 4b4 − 4c4 > 0,   2 2a4 + 2b4 + 2c4 − 2a2 b2 − 2a2 c2 − 2b2 c2 < 0,   2 (a2 − b2 )2 + (a2 − c2 )2 + (b2 − c2 )2 < 0. But the last inequality is impossible, because all the terms on the left-hand side are positive. Hence, the given inequality is impossible for all triangles. The last inequality becomes equality if and only if a = b = c, i.e. when the triangle is equilateral. Also solved, in different ways, by Mrs. A. Iliovici, I. Nicolaescu, Emil G. Nit¸escu, V. V. Cambureanu and C. Vintilˇ a. In the year 1919, Roland Weitzenb¨ ock published in Mathematische Zeitschrift, Vol. 5, No. 1-2, pp. 137-146 the article Uber eine Ungleichung in der Dreiecksgeometrie, where he proved that: In any triangle ABC, with usual notations, holds the inequality √ a2 + b2 + c2 ≥ 4 3S We observe that the inequality of Ion Ionescu is the same as the inequality of Weitzenb¨ ock, and therefore from this moment the inequality of Weitzenb¨ ock must be named the inequality Ionescu–Weitzenb¨ ock. The inequality Ionescu-Weitzenb¨ ock was given as a problem at the third IMO, Vespr´em, Hungary, 8–15, July, 1961. Eleven proofs of the Ionescu–Weitzenb¨ ock ’s inequality were presented by Arthur Engel in his book Problem solving strategies, Springer Verlag, 1998. We discovered the above while we were working on an article about Weitzenb¨ ock ’s inequality. Moreover, we have 23 demonstrations and 10 generalizations other than those published of the Ionescu–Weitzenb¨ ock ’s inequality. Department of Mathematics, Matei Basarab National College, Bucharest, Romania ˇ u, RoDepartment of Mathematics, George Emil Palade School, Buza mania

139

JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before June 15, 2013

Proposals 1. Proposed by D.M. B˘ atinet¸u–Giurgiu, “Matei Basarab” National College, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School, Buz˘ au, Romania. Let ABC be a triangle with side lengths a, b, c, area S, circumradius R and inradius r, respectively. Prove that √ a3 b3 c3 4 3 + + ≥ S b·R+c·r c·R+a·r a·R+b·r R+r 2. Proposed by D.M B˘ atinet¸u–Giurgiu, “Matei Basarab” National College, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School, Buz˘ au, Romania. Prove that if a, b, c are real positive numbers, then   a 1 1 1 ≤ + a2 + bc 4 b c ´ 3. Proposed by Francisco Javier Garc´ıa Capit´ an, I.E.S. Alvarez Cubero de Priego 0 0 0 de C´ ordoba, Spain. Let AA , BB , CC be any three diameters of the same circle. Let us consider the intersection points P = A0 C ∩ BC 0 , Q = AC 0 ∩ B 0 C and T = P Q ∩ AB. Show that T C is tangent to the circle. (Dedicated to Juan Bosco Romero M´arquez) 4. Proposed by Paolo Perfetti, Math. Dept. “Tor Vergata” Univ., Rome, Italy. Without using calculus, find with proofs the coordinates of all the extremum points of x2 x4 + x3 + 4x + 16 5. Proposed by Ercole Suppa, Teramo, Italy. In ∆ABC let D be the midpoint of BC, let P be an arbitrary point on AD, let E = BP ∩ AC, F = CP ∩ AB and let E1 , F1 be the second intersection of BP, CP with the circumcircle, respectively. Show that E, E1 , F1 , F are concyclic.

Mathproblems ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 3, Issue 2 (2013), Pages 140–170

Editors: Valmir Krasniqi, Jos´ e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Valmir Bucaj, Mih´aly Bencze, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Jozsef S´andor, David R. Stone, Roberto Tauraso, Francisco Javier Garca Capit´ an, Emanuele Callegari, Ercole Suppa, Cristinel Mortici, Mohammed Aassila.

PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]

Solutions to the problems stated in this issue should arrive before September 15, 2013

Problems 66. Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. If xk > 0, 1 ≤ k ≤ n, then show that v u n n n n X X X u X x2k x4k 6 x3k tn x6k . k=1

k=1

k=1

k=1

67. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Romania (Jointly). Let x ∈ R and (Ln (x))n≥2 be the sequence defined by  sin2 x  √ sin2 x  p 2 n n+1 Ln (x) = ncos x (n + 1)! − n! Calculate lim Ln (x). n→∞

c

2010 Mathproblems, Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. 140

141

68. Proposed by Vasile Cirtoaje, Ploiesti, Romania. Let a and b be positive real 1 numbers such that a + b = 2. If k ≥ , then prove that 2 kb

aa bb

ka

≥ 1.

69. Proposed by Mih´ aly Bencze, Bra¸sov, Romania. Solve the following equation: m m Y X X m(m + 1) 2 2 log a = (m + 1) log a log log xi log xj + xk , log xk + 2 k=1

1≤i 0. 70. Proposed by Anastasios Kotronis, Athens, Greece. Let an = Show that r r 1 π −3/2 π −1/2 2 −1 n + n + n + O(n−2 ). an = 2 3 12 2

Pn

k=0

n k




k!n−k .

71. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Solve the following system of equations 1 1 1   x + y+z = a 1 1 1 y + z+x = b  1 1 1 z + x+y = c where a, b, c ∈ < − {0}. 72∗ . Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. Let 1 < p < ∞, we can generalize the inverse of arcsin as follows: Z x 1 0≤x≤1 arcsinp (x) = dt (1 − tp )1/p 0 and Z 1 πp 1 = arcsinp (1) = dt. p 1/p 2 0 (1 − t )  πp  The inverse of arcsinp (x) on 0, 2 is called the generalized sine function and denoted by sinp . The generalized hyperbolic cosine function is defined as cosp (x) = d dx sinp (x). Calculate Z π2p ln cosp (x) ln sinp (x) Ip = dx. cosp (x) sinp (x) 0

142

Solutions No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.

58. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. 1) Let a be a real number such that −1 ≤ a ≤ 1. Prove that ∞ X ∞ X (i − 1)!(j − 1)!



 a a = 2Li2 − 2Li2 − − 2Li2 (a) (i + j)! 2−a i=1 j=1 Rz dt for all where Li2 (z) is the Dilogarithm function defined by Li2 (z) = − 0 ln(1−t) t z ∈ {z ∈ C : |z| ≤ 1}. i+j

a 2−a





2) Show that ∞ X ∞ X (i − 1)!(j − 1)!

= ζ(2),

(i + j)!

i=1 j=1

where ζ(z) represents the Riemann zeta function. Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. It is well-known that (i − 1)!(j − 1)! Γ(i)Γ(j) Γ(i + j) Γ(i)Γ(j) 1 = = (i + j)! Γ(i + j) Γ(i + j + 1) Γ(i + j) i + j Moreover Γ(i)Γ(j) = Γ(i + j)

Z

1

xi−1 (1 − x)j−1 dx

0

and then ∞ X ∞ X (i − 1)!(j − 1)! i=1 j=1

(i + j)!

a

i+j

=

∞ X ∞ Z X

1

xi−1 (1 − x)j−1 dx

i=1 j=1

0

∞

1

ai+j i+j

and ∞ X ∞ Z X

1

xi−1 (1 − x)j−1 dx

0

i=1 j=1

∞

XX ai+j = i+j i=1 j=1

Z

xi−1 (1 − x)j−1 dx

0

Z

a

y i+j−1 dy

0

The geometric series yields Z

1

Z dx

0

or

a

dy 0

xy 1 y(1 − x) 1 = x(1 − xy) 1 − x 1 − y(1 − x) y

Z

1

Z dx

0

a

dy 0

y 1 1 − xy 1 − y(1 − x)

143

 Z 1 Z 1 Z a y 1 1 y 1 dy dx = dy + dx 1 − xy 1 − y(1 − x) 2−y 0 1 − xy 1 − y + xy 0 0 0 Z a ln(1 − y) dy dy = −2 2−y 0

Z

a

Thus we have ∞ X ∞ X (i − 1)!(j − 1)!

(i + j)!

i=1 j=1

ai+j = −2

Z

a

dy 0

ln(1 − y) dy = G(a) 2−y

Differentiating 

   a a − 2Li2 − − 2Li2 (a) 2−a 2−a and keeping in mind the minus sign, we get F (a) = 2Li2

F 0 (a) = −2

 ln 1 −

a 2−a

a 2−a

 2 +2 (2 − a)2

 ln 1 +

a 2−a

−a 2−a

 −2 ln(1 − a) ln(1 − a) +2 = −2 2 (2 − a) a 2−a

which is G0 (a). Now we show G(1) = ζ(2), and F (1) = ζ(2) and this proves both 1) and 2). We assume that Li2 (1) = ζ(2) and Li2 (−1) = −ζ(2)/2. Integrating by parts Z 1 1 Z 1 ln(2 − y) ln(1 − y) G(1) = −2 dy = 2 ln(2 − y) ln(1 − y) +2 dy |{z} = 2−y 1−y 0 0 0 1−y=−x Z −1 ln(1 − x) 2 dx = −Li2 (−1) = ζ(2) x 0 On the other hand F (1) = 2Li2 (1) − 2Li2 (−1) − 2Li2 (1) = ζ(2) Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Moti Levy, Israel ; Hun Min Park,Advanced Institute of Science and Technology, Dajeon, South Korea; and the proposer 59. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Romania (Jointly). Let n be a positive integer. Prove that √ p Ln L2n+2 Ln+1 L2n+3 2 + + (Ln + Ln+2 ) ≥ 2 6 · Ln Ln+1 · Ln+2 , Ln+3 Ln + Ln+2 where Ln represents the nth Lucas number defined by L0 = 2, L1 = 1, and for all n ≥ 2, Ln = Ln−1 + Ln−2 .

144

Solution by Angel Plaza, University of Las Palmas de Gran Canaria, Spain. By the AM-GM inequality, s   Ln L2n+2 Ln+1 L2n+3 Ln+1 L2n+3 1 Ln L2n+2 + ≥ · 2 Ln+3 Ln + Ln+2 Ln+3 Ln + Ln+2 s p Ln+3 = · Ln Ln+1 · Ln+2 Ln + Ln+2 p > Ln Ln+1 · Ln+2 .  p √ (Ln + Ln+2 )2 6 − 1 · Ln Ln+1 · Ln+2 . So it is enough to prove that ≥ 2 p Again, by the AM-GM inequality, Ln Ln+1 < Ln L2n+1 = Ln+2 2 . So, it is sufficient to prove that  √ 6 − 1 L2n+2 (Ln + Ln+2 )2 ≥  2 √ 2Ln + Ln+1 ≥ 6−1 Ln + Ln+1 2  √ Ln 6 − 1. ≥ 1+ Ln + Ln+1 Since


2 Ln 1 ≥ , and 1 + 14 = Ln + Ln+1 4

25 16

>

5 2

−1 >

√

6 − 1, the conclusion follows.

Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Moti Levy, Israel ; and the proposer. 60. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Determine all functions f : R \ {−2, −1, 0, 1, 2} → R, which satisfy the relation   7+x f (x) + f = ax + b 2−x where a, b ∈ R. Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongo7+x . Then lia. Consider the function g(x) = 2−x (g ◦ g)(x) = g(g(x)) =

7+ 7 + g(x) = 2 − g(x) 2−

7+x 2−x 7+x 2−x

=

2x − 7 , x+1

7+x 2 · 2−x −7 2g(x) − 7 = 7+x =x g(x) + 1 2−x + 1   7+x f (x) + f = ax + b 2−x In (1) if we do the change of variables 2x − 7 7+x ,x → , x→ 2−x x+1

(g ◦ g ◦ g)(x) = g(g(g(x))) =

(1)

145

then



     7+x 2x − 7 7+x f +f =a· +b 2−x x+1 2−x     2x − 7 2x − 7 f + f (x) = a · +b x+1 x+1 Now adding up (1) and (3) and subtracting (2), yields   2x − 7 7 + x − +b 2f (x) = a · x + x+1 2−x from which follows ax3 + (2a + b)x2 − (5a + b)x + 21a − 2b f (x) = 2(x + 1)(x − 2)

(2) (3)

Also solved by Moti Levy, Israel ; Adrian Naco, Polytechnic University, Tirana, Albania; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Fozi M. Dannan, Damascus, Syria; Moubinool OMARJEE, Paris, France; Dorlir Ahmeti, University of Prishtina, Republic of Kosova; and the proposer. 61. Proposed by D.M. B˘ atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Ron o Pn ∗ mania (Jointly). If a ∈ R+ , b, c, d, xk ∈ R+ , Xn = k=1 xk and cXn > d max xk , 1≤k≤n

then

n X aXn + bxk k=1

cXn − dxk

≥

(an + b)n cn − d

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let f be the function defined on the interval [0, cn/d) by an + bx f (x) = cn − dx Clearly, for x ∈ [0, cn/d) we have f 00 (x) = 2dn(bc+ad) (cn−dx)3 > 0. Thus f is convex. Now, let tk = nxk /Xn for k = 1, 2, . . . , n. By assumption tk ∈ [0, cn/d) for every k, therefore, using the convexity of f , we find that   n n X X aXn + bxk t1 + · · · + tn (an + b)n = f (tk ) ≥ nf = nf (1) = cXn − dxk n cn − d k=1

k=1

which is the desired inequality. Solution 2 by Moti Levy, Israel . Let ρk = 1. We re-write the inequality using ρk , n

a b c k=1 d

bX d

+ ρk b ≥n − ρk d

Let f (x) =

b d

a b c d

a b c d

xk Xn

+ −

+x −x

then 0 < ρk < 1 and

Pn

k=1

Pn n

k=1

n

ρk ρk

Pn

k=1

ρk =

146

then the inequality in terms of the function f is ! n 1X ρk . n k=1 k=1
0 The function f (x) is convex in the interval 0, dc since its derivative f (x) = ad+bc is monotonically non-decreasing in the interval. The inequality follows (c−dx)2 from Jensen’s inequality for convex functions. n

1X f (ρk ) ≥ f n

Also solved by Ioan Viorel Codreanu, Satulung, Maramure¸s, Romania; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia and the proposer. 62. Proposed by Mih´ aly Bencze, Bra¸sov, Romania. Show that if ak , bk , ck ∈ R∗ (k = 2 1, 2), such that a1 + b21 + c21 = a22 + b22 + c22 , then at least one of the equations a1 x2 + 2c2 x + b1 = 0; b1 x2 + 2a2 x + c1 = 0; c1 x2 + 2b2 x + a1 = 0 has real roots. Solution by Juan Gabriel Alonso (E.S.O. First Course, student, Garo´ e ´ Atlantic School), and Angel Plaza, University of Las Palmas de Gran Canaria, Spain (Jointly). We argue by contradiction. Let us suppose that none of the given equations has any real root. Then, all the discriminants are negative. So, 4c22 − 4a1 b1 < 0; 4a22 − 4b1 c1 < 0; 4b22 − 4c1 a1 < 0 from where
2 a22 + b22 + c22 − 2 (a1 b1 + b1 c1 + c1 a1 ) < 0,
2 a21 + b21 + c21 − 2 (a1 b1 + b1 c1 + c1 a1 ) < 0. (a1 − b1 )2 + (b1 − c1 )2 + (c1 − a1 )2 < 0, which indeed it is a contradiction. Also solved by Moti Levy, Israel ; Dorlir Ahmeti, University of Prishtina, Republic of Kosova; Ioan Viorel Codreanu, Satulung, Maramure¸s, Romania; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Muhammad Thoriq, Department of Mathematics, Gadjah Mada University, Yogyakarta, Indonesia; ANanduud Problem Solving Group, Ulaanbaatar, Mongolia; Roberto De la Cruz Moreno, Mathematics Research Center, Bellaterra Campus, Barcelona, Spain; and the proposer. 63. Proposed by Anastasios Kotronis, Athens, Greece. Let k ≥ 3 be an integer. Show that X n≥0

1 (2n + 1)(3n + 2) · · · (kn + k − 1)

  k 1 X m−1 k = (−1) (m−1)mk−2 k! m=2 m

 ! m−1 X π π 2`π `π cot − ln m + cos · ln sin 2 m m m `=1

147

Solution by Moti Levy, Israel . k Y k! m−1 1 Let un = (2n+1)(3n+2)···(kn+k−1) = n+bm , where bm = m . m=2 Y Pk Let Q (n) , (n + bm ) . The partial fractions of un are m=2

am n+bm ,

where the

m

coefficients {am } are given by am = k X d (Q (n)) = dn m=2

k Y

1 d dn (Q(n))|n=−bm

.

(n + bj )

j=2, j6=m

d (Q (n))|n=−bm = dn

k Y j=2, j6=m k Y

=

k Y

(bj − bm ) =



j=2, j6=m



j=2, j6=m

j−m mj



 =

k Y

j=2, j6=m

 j−1 m−1 − j m   k 1   Y 1 m j j=2, j6=m

k Y

m m (−1) (m − 2)! (k − m)! k! 1 1 m! (k − m)! 1 m m = (−1) = (−1) k−2 k−2 m k! m−1 m (m − 1)

=

(j − m)

j=2, j6=m

1

mk−2

k m




Thus m

am = (−1) mk−2 (m − 1)

  k m

(4)

Pk

m=2 am = 0 follows from the convergence of our series that limn→∞ nun = 0, or otherwise the series will be greater than harmonic series and thus diverges. Pk Pk Pk nam Hence 0 = limn→∞ nun = limn→∞ m=2 n+b = limn→∞ m=2 1+ambm = m=2 am . m n X The Digamma function ψ (z) can be used to evaluate the infinite sum un . n≥0

It is known that X

un = −

k X

am ψ (bm )

(5)

m=2

n≥0

The Digamma function for rational arguments is given by Gauss’s Digamma Theorem,        q π πp X 2πlp πl p = −γ − ln 2q − cot + cos ln sin , 0 < p < q. ψ q 2 q q q l=1

 ψ (bm ) = ψ

m−1 m

 m

π π (m − 1) X = − (γ + ln 2) − ln m − cot + cos 2 m l=1

Since cot

(m − 1) π π = − cot m m



2πl (m − 1) m



 ln sin



πl m



148

and

 cos

2l (m − 1) π m



 = cos

2lπ m

 ,

then m

ψ (bm ) = − (γ + ln 2) − ln m +

X π π cot + cos 2 m l=1

Pk

Using the fact that X

un = −

n≥0



2lπ m



   πl ln sin . m

(γ + ln 2) am = 0, we obtain    !  m−1 X π π lπ 2lπ cot − ln m + ln sin cos 2 m m m

m=2

k X

am

m=2

(6)

l=1

and by substituting (4) in (6), the required result follows       ! k m−1 X X X k π 2lπ π lπ m−1 k−2 un = (−1) m (m − 1) cos cot − ln m + . ln sin m 2 m m m m=2 n≥0

l=1

Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria and the proposer. 64. Proposed by Florin Stanescu, Serban School Cioculescu, Gaesti, Romania. Let f : [0, 1] → R be a continuous function. If Z 1 Z 1 2 x f (x)dx = −2 F (x)dx, 0

where F (x) =

Rx 0

1/2

f (t)dt, x ∈ [0, 1], then prove that Z 1 2 Z 1 2 f (x)dx ≥ 24 f (x)dx 0

0

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Note that  1 Z 1 Z 1 −2 F (x) dx = (1 − 2x)F (x) − (1 − 2x)f (x) dx 1/2

x=1/2

so Z

1

Z

2

1

x f (x) dx = −F (1) − 0

1/2

(1 − 2x)f (x) dx 1/2

or equivalently Z 1/2 Z 2 (1 + x )f (x) dx + 0

1 2

Z

(1 + (1 − x) )f (x) dx =

1/2

g(x)f (x) dx = 0 0

where ( 1 + x2 g(x) = 1 + (1 − x)2

if x ∈ [0, 12 ) if x ∈ [ 12 , 1]

Now, for an arbitrary λ ∈ R, we have Z 1 Z 1 f (x)dx = (1 + λg(x))f (x)dx 0

0

1

149

So, using the Cauch-Schwarz inequality, we obtain 2 Z 1 Z 1 Z 2 (1 + λg(x)) dx · f (x)dx ≤

f 2 (x) dx

0

0

0

1

But 1

Z

13 283 2 283 4 λ+ λ = + 6 240 849 240

(1 + λg(x))2 (x) dx = 1 +

0

 2 260 . λ+ 283

4 Thus, the minimum value that this integral may take is 849 and it corresponds to 260 λ = − 283 . Choosing this value for λ we conclude that Z 1 2 Z 1 4 f (x)dx ≤ f 2 (x) dx 849 0 0

which is much stronger than the proposed inequality. Moreover, this is the best 260 possible inequality since we have equality if f (x) = 1 − 283 g(x). Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. First we note that, Z 1Z x Z 1 Z 21 Z 1 Z 1 Z 1 F (x)dx = f (y)dy = dx dyf (y) f (y)dy + dx = 1 2

1 2

1 2

1 2

Z

Z

1 2

0

1 2

0

y

1

f (x)(1 − x)dx

f (x)dx + 1 2

0

This yields 1

Z

1 2

Z

2

x f (x)dx = − 0

1

Z f (x)dx − 2

f (x)(1 − x)dx, 1 2

0

and then Z

1

Z

1

f (x)(x − 1)2 dx −

f (x)dx = − 1 2

0

1 2

Z

f (x)x2 .

0

We get by Cauchy–Schwarz and the fact that the integrands in the second line are positive 1

Z

2 f (x)dx =

 Z ≤

f (x)(x − 1) dx +

1

f 2 (x)dx

Z



1 2

f (x)dx  0

Z

!2

1 2

f (x)x

! 21

1

(x − 1)4 dx

Z

1 2

+

x4 dx

0

! 21

1 4

(x − 1) dx 1 2

which implies our inequality.

2

≤

0

1 2

0

=

Z

2

1 2

0

Z

1

Z

Z +

! 21 2

1

f 2 (x)dx

 =

0 1 2

! 21 2 4

x dx 0

Z

Z

 = 0

1

f 2 (x)dx

1 40

150

Also solved by Moti Levy, Israel ; Moubinool Omarjee, Paris, France and the proposer. 65∗ . Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. Let Ωn π n/2 the volume of the unit ball in Rn , that is Ωn = Γ(n/2+1) , n = 1, 2 . . . . where R ∞ x−1 −t Γ(x) = 0 t e dt , x > 0. For all positive integers n, Prove that r r 1 (Ωn+1 ) n+1 n+2 α n + 2 6 β 6 1 n+3 n+3 (Ωn ) n with the best possible constant factors α = 2 and β =

ln( 34 ) √

ln(

π 2 )

= 2.3818 · · · .

Solution by Moti Levy, Israel . We will use the following results: Lemma 1: For x > 0, 1 (x + 1)

2

+

1 x+

1

0

> ψ (x + 1) >

3 2

(x + 1)

2

+

1 x+1+

6 π2

(7)

Proof. See Theorem 1 in ”Feng Qi and Bai-Ni Guo, Sharp Inequalities for polygamma functions. arXiv:0903.1984v1, May 1, 2009”. Lemma 2. The function 1

(Γ (x + 1)) x hv (x) = v (x + 1) is strictly increasing for x ≥ 1 and v ≤ vc = decreasing for x ≥ 0 and v ≥ 1.

π 2 +1 π 2 +3

∼ = 0.844 60 and is monotonic

v Proof. Let gv (x) = ln hv (x) and fv (x) = x2 dg dx .

gv (x) =

1 ln Γ (x + 1) − v ln (x + 1) , x 0

fv (x) = − ln (Γ (x + 1)) + xψ (x + 1) − 0 dfv (x) v (x + 2) = x ψ (x + 1) − 2 dx (x + 1)

vx2 . x+1 ! (8)

We will show that dfv (x) > 0 for x ≥ 1 and v ≤ vc dx Substitute the right hand of Lemma 1 in (8) , dfv (x) >x dx

1

(9)

1 v (x + 2) 2 + 6 − 2 x + 1 + π2 (x + 1) (x + 1) x =
Pv (x) , 2 (x + 1) x + 1 + π62

!

where 

6 Pv (x) = (x + 1) + (1 − v (x + 2)) x + 1 + 2 π 2



151

It is clear that f or v < vc and x ≥ 1

Pv (x) > Pvc (x)

dPvc (x) 4 6 = 2 x− 2 2 dx π +3 π (π + 3) Pvc (1) = 0 dPvc (1) dx

> 0 and Pvc (1) = 0, then Pvc (x) ≥ 0 for x ≥ 1 and consequently > 0 for x ≥ 1 and v ≤ vc . 2 −4 ∼ The fact fvc (1) = 1 − 12 ππ2 +1 > 0 and (8) imply that gv (x) is +3 − γ = 4.868 × 10 strictly increasing for x ≥ 1 and v ≤ vc . It follows that hv (x) is strictly increasing for x ≥ 1 and v ≤ vc . We substitute the left hand of Lemma 1 in (8) to show that

Since

dfv (x) dx

dfv (x) −3/4 be a real number. Show that s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − 2 6 3 2 6 3 is an integer and determine its value. 22. Let a1 , a2 , a3 , a4 be nonzero real numbers defined by ak = (1 ≤ k ≤ 4), α, β ∈ R. Calculate 1 + a21 + a22 a1 + a2 + 1/a4 a1 + a2 + 1/a4 2 + 1/a24 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4

sin(kβ + α) , sin kβ

1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23



23. Let A be a set of positive integers. If the prime divisors of elements in A are among the prime numbers p1 , p2 , . . . , pn and |A| > 3 · 2n + 1, then show that it contains one subset of four distinct elements whose product is the fourth power of an integer. 24. Find all triples (x, y, z) of real numbers such that  12x − 4z 2 = 25,  24y − 36x2 = 1,  20z − 16y 2 = 9. 25. Let a, b, c be the lengths of the sides of a triangle ABC with inradius r and cicumradius R. Prove that s    a b c 1 r 3 ≤ ≤ 2r + R b + c + 2a c + a + 2b a + b + 2c 4

60

Solutions 16. A number of three digits is written as xyz in base 7 and as zxy in base 9. Find the number in base 10. (III Spanish Math Olympiad (1965-1966)) Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. To write a number in the system of base 7 the only integers used are 0, 1, 2, 3, 4, 5, 6, and to write it in base 9 only the integers form 0 to 8 are used. Consequently, x, y, z ∈ {0, 1, 2, 3, 4, 5, 6}. The number xyz in base 7 represents the number x · 72 + y · 7 + z in base 10. On the other hand, the number zxy in base 9 represents the number z · 92 + y · 9 + x in base 10. Therefore, x · 72 + y · 7 + z = z · 92 + y · 9 + x from which follows 8(3x − 5z) = y. Since x, y, z are integers ranging form 0 to 6, then y = 0 and 3x = 5y. From the preceding, we have x = z = 0 or x = 5 and z = 3. In the first case, we obtain the number 000 which does not have three digits; and from the second, we get the number 503 in base 7 and 305 in base 9. Both numbers in base 10 are the number 248 and this is the answer. 2 Also solved by Bruno Salgueiro Fanego, Viveiro, Spain. 17. A regular convex polygon of L + M + N sides must be colored using three colors: red, yellow and blue, in such a way that L sides must be red, M yellow and N blue. Give the necessary and sufficient conditions, using inequalities, to obtain a colored polygon with no two consecutive sides of the same color. (XI Spanish Math Olympiad (1973-1974)) Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Let K = L + M + N. We distinguish two cases: (a) K is even, then it must be K K K M≤ , N≤ L≤ , 2 2 2 That is, L + M ≥ N, L + N ≥ M and M + N ≥ L. (b) If K is odd, then it must be K −1 K −1 K −1 , 0 M > 0 and M + N > L > 0. We claim that these conditions that are necessary are also sufficient. Indeed, WLOG we can assume that L ≥ M ≥ N, independently of the parity of K. We begin coloring in red sides first, third, fifth,... from the starting point in a circular sense until complete L red sides. Then, remains to be colored L − 1 non consecutive sides and K − (2L − 1) = M + N − L + 1 ≥ 1 consecutive sides. Since L ≥ M, then M + N − L + 1 ≤ N + 1, therefore this set of consecutive sides can not be colored alternatively yellow-blue-yellow, etc., without painting two consecutive sides of the same color. Finally, the non consecutive L − 1 sides are colored yellow or blue until to complete the M yellow sides and the N blue sides, and our claim is proven.

61

2 Also solved by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain. 18. Let ABDC be a cyclic quadrilateral inscribed in a circle C. Let M and N be the midpoints of the arcs AB and CD which do not contain C and A respectively. If M N meets side AB at P, then show that AC + AD AP = BP BC + BD (IMAC 2011) Solution by Ivan Geffner Fuenmayor, Technical University of Catalonia (BARCELONA TECH), Barcelona, Spain. Applying Ptolemy’s theorem to the inscribed quadrilateral ACN D, we have AD · CN + AC · N D = AN · CD Since N is the midpoint of the arc CD, then we have CN = N D = x, and

C N D

A

P

B

M Figure 2. Problem 18 AN · CD = (AC + AD)x. Likewise, considering the inscribed quadrilateral CN DB we have BN · CD = (BD + BC)x. Dividing the preceding expressions, yields AC + AD AN = BN BD + BC Applying the bisector angle theorem, we have AN AP = BN BP This completes the proof. 2

62

Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; and Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain. 19. Place n points on a circle and draw in all possible chord joining these points. If no three chord are concurrent, find (with proof ) the number of disjoint regions created. (IMAC-2011) Solution 1 by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain. First, we prove that if a convex region crossed by L lines with P interior points of intersection, then the number of disjoint regions created is RL = L+P +1. To prove the preceding claim, we argue by Mathematical Induction on L. Let R be an arbitrary convex plane. For each L ≥ 0, let A(L) be the statement n region in the
o L that for each P ∈ 1, 2, . . . , 2 , if L lines that cross R, with P intersection points inside R, then the number of disjoint regions created inside R is RL = L + P + 1. When no lines intersect R, then P = 0, and so, R0 = 0 + 0 + 1 = 1 and A(0) holds. Fix some K ≥ 0 and suppose that A(K) holds for K lines and some P ≥ 0 with RK = K + P + 1 regions. Consider a collection C of K + 1 lines each crossing R (not just touching), choose some line ` ∈ C, and apply A(K) to C\{`} with some P intersection points inside R and RK = K + P + 1 regions. Let S be the number of lines intersecting ` inside R. Since one draws a (K + 1)−st line `, starting outside R, a new region is created when ` first crosses the border of R, and whenever ` crosses a line inside of R. Hence the number of new regions is S + 1. Hence, the number of regions determined by the K + 1 lines is, on account of A(K), RK+1 = RK + S + 1 = (K + P + 1) + S + 1 = (K + 1) + (P + S) + 1, where P +S is the total number of intersection points inside R. Therefore, A(K +1) holds and by the PMI the claim is proven. Finally, since the circle is convex and any intersection point is determined  by a   n n unique 4−tuple of points, then there are P = intersection points and L = 4 2     n n chords and the number of regions is R = + + 1. 4 2 2 Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let us denote by an the number of disjoint regions created. Clearly a1 = 1, a2 = 2 and a3 = 4. Suppose that we have an regions obtained on placing the n points A1 , . . . , An in this order, and let us add the n + 1st point A0 on the arc An A1 that does not contain any other point. The chords A0 A1 and A0 An add two regions. And for 1 < k < n, there are (k −1)(n−k) points of itersection of the chord A0 Ak with the other chords. Hence, the chord A0 Ak passes through (k − 1)(n − k) + 1 regions. Consequently, drawing the chord

63

A0 Ak adds (k − 1)(n − k) + 1 new regions. Thus an+1 = an +

n X

((k − 1)(n − k) + 1)

k=1

But n X

((k − 1)(n − k) + 1)

=

−

k=1

n X

k 2 + (n + 1)

k=1

= =

n X

k + (1 − n)n

k=1

n(n + 1)(2n + 1) n(n + 1)2 + − n(n − 1) − 2   6   n+2 n −2 3 2

So, an

 n−1 X k  k+2 −2 3 2 k=1 k=1     n−1 n−1 X X k + 1 k  k+3 k+2 = 1+ − −2 − 4 4 3 3 k=1 k=1         n+2 n n n = 1+ −2 = + +1 4 3 4 2

=

1+

n−1 X

2

which is the required number of regions.

Also solved by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. 20. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that s  3

1+a b+c

 1−a  bc

1+b c+a

 1−b  ca

1+c a+b

 1−c ab ≥ 64

(J´ozsef Wildt Competition 2009) Solution by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona,   1+x 1 . Spain. Consider the function f : (0, 1) → R defined by f (x) = ln x 1−x ∞ ∞ X X x2k kx2k−1 Since f (x) = 2 for |x| < 1, then f 0 (x) = 4 (|x| < 1) and 2k + 1 2k + 1 k=0 k=0 ∞ X k(2k − 1)x2k−2 00 f (x) = 4 (|x| < 1). Therefore, f 0 (x) > 0 and f 00 (x) > 0 for all 2k + 1 k=0 x ∈ (0, 1), and f is increasing and convex.

64

 Applying Jensen’s inequality, we have f

a+b+c 3

 ≤

f (a) + f (b) + f (c) or equiva3

lently, 3 ln a+b+c



3 + (a + b + c) 3 − (a + b + c)



"  1/a  1/b  1/c # 1 1+a 1+b 1+c ≤ ln + ln + ln 3 1−a 1−b 1−c

Taking into account that a + b + c = 1 and the properties of logarithms, we get s  1/a  1/b  1/c 1+a 1+b 1+c 3 ≥8 (5) 1−a 1−b 1−c 1 1 1 WLOG we can assume that a ≥ b ≥ c. We have, ≤ ≤ and g(a) ≥ g(b) ≥ a b c  1+x . Applying g(c), where g is the increasing function defined by g(x) = ln 1−x rearrangement’s inequality, we get 1 1 1 1 1 1 g(a) + g(b) + g(c) ≥ g(a) + g(b) + g(c) b c a a b c or  1/b  1/c  1/a  1/a  1/b  1/c 1+a 1+b 1+c 1+a 1+b 1+c ≥ 1−a 1−b 1−c 1−a 1−b 1−c From the preceding and (5) we obtain s s  1/b  1/c  1/a  1/b  1/c  1/a 1+a 1+b 1+c 1+a 1+b 1+c 3 3 = b+c c+a a+b 1−a 1−b 1−c s  1/a  1/b  1/c 1+a 1+b 1+c 3 ≥ ≥8 1−a 1−b 1−c Likewise, applying rearrangement inequality again, we get 1 1 1 1 1 1 g(a) + g(b) + g(c) ≥ g(a) + g(b) + g(c) c a b a b c and s s  1/c  1/a  1/b  1/c  1/a  1/b 1+a 1+b 1+c 1+a 1+b 1+c 3 3 = b+c c+a a+b 1−a 1−b 1−c s  1/a  1/b  1/c 1+a 1+b 1+c 3 ≥ ≥8 1−a 1−b 1−c Multiplying up the preceding inequalities yields, s  1+1  1+1  1+1 1+a b c 1+b c a 1+c a b 3 ≥ 64 b+c c+a a+b from which the statement follows. Equality holds when a = b = c = 1/3, and we are done. 2 Also solved by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain.

65

MATHNOTES SECTION On a Discrete Constrained Inequality ´ ly Bencze and Jose ´ Luis D´ıaz-Barrero Miha Abstract. In this note a constrained inequality is generalized and some refinements and applications of it are also given. 1. Introduction In [1] the following problem was posed: Let a, b, c be positive real numbers such that a + b + c = 1. Prove that   b c 3 a + + ≥ (6) (ab + bc + ca) 2 b + b c2 + c a2 + a 4 A solution to the preceding proposal and some related results appeared in [2]. Our aim in this short paper is to generalize it and to give some of its applications. 2. Main Results In the sequel some generalizations and refinements of (6) are given. We begin with Theroem 1. Let x and ak , bk , (1 ≤ k ≤ n) be positive real numbers. Then ! n ! !2 n n X X X ak ak ak ≥ (x + bk )2 x + bk k=1 k=1 k=1 !4 , n !2 n X X ≥ ak ak (x + bk ) k=1

Proof. Setting ~u = inequality, we have n X k=1

ak x + bk

√ √ a1 a2 an x+b1 , x+b2 , . . . , x+bn

√

!2 =



k=1

and ~v =

!2 √ n X ak √ ak ≤ x + bk

k=1

√

n X k=1

√ √
a1 , a2 , . . . , an into CBS !

ak

n X k=1

ak (x + bk )2

!

and  the isq proven. To proveRHS inequality we set  q LHS inequality q p p p a1 a2 an v= a1 (x + b1 ), a2 (x + b2 ), . . . , an (x + bn ) , , . . . , ~u = x+b1 x+b2 x+bn and ~ into CBS inequality again and we get !2 ! n ! n n X X X ak ak ≤ ak (x + bk ) x + bk k=1

k=1

from which the statement immediately follows.

k=1



A constrained inequality that can be derived immediately from the preceding result is given in the following

66

Pn Corllary 1. Let ak , bk , (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. Then holds: !2  ! n n n X X X a a k k ≥2 + ak (1 + bk )2  (1 + bk )2 1 + bk k=1

k=1

k=1

Proof. Setting x = 1 in Theorem 1, we get ! n n X X 2 ak (1 + bk ) k=1

k=1

ak (1 + bk )2

and n X

!

n X

ak (1 + bk )2

k=1

k=1

ak 1 + bk

! ≥1

!2 ≥1

Adding up the preceding inequalities the statement follows.  Theroem 2. Let 0 ≤ y < z and ak , bk , (1 ≤ k ≤ n) be positive real numbers. Then ! ! n n X X ak ak (y + bk )(z + bk ) k=1 k=1   i aj   b2a−b n 2 X Y j i ak (y + bj )(z + bi )  ≥ + log (y + bk )(z + bk ) (y + bi )(z + bj ) 1≤i 0. So, for all (x1 , . . . , xn ) ∈ (0, +∞)n is f (xn1 ) + f (xn2 ) + . . . + f (xnn ) = f (x1 x2 · · · xn ) n as desired. (b): f (1) 6= 0. In this case we obtain the conclusion by applying the preceding case to the function f : (0, ∞) → R defined by f (x) = f (x) − f (1).

73

Also solved by Islam Foniqi, University of Prishtina, Department of Mathematics, Prishtin¨ e, Republic of Kosova; Adrian Naco, Department of Mathematics, Polytechnic University of Tirana, Albania; Florin Stanescu, Serban Cioculescu School, Gaesti, Dambovita, Romania; and the proposer 30. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzˇ au, Romania. Evaluate Z 2012 x sinn t dt, lim x→0 2011 x tm where n, m ∈ N. Solution by Anastasios Kotronis, Athens, Greece. More generally, let 0 < a < b and n, m ∈ N. We consider the following cases: (1) For n − m ≥ −1 we have Z bx Z bx
n sinn t dt = tn−m 1 + O(t2 ) m t ax ax Z bx
= tn−m 1 + O(t2 ) ax bx

Z =

=

=


tn−m + O tn−m+2 dt ax   bx bx    n−m+1 n−m+3  t t ,  n−m+1 + O  n−m+3   ax

n−m≥0

ax

     bx  bx   ln |t| + O t2 , n − m = −1 ax ax  n−m+1 n−m+1 b −a  xn−m+1 + O(xn−m+3 ), n − m ≥ 0  n−m+1

  b ln a + O(x2 ), ( 0, n−m≥0 x→0 −−−→ . b ln a , n − m = −1

n − m = −1

Carrying out the change of variable t = −y, we get Z bx Z b(−x) sinn t sinn y n−m+1 dt = (−1) dy, tm ym ax a(−x)

(1)

(2) For n − m ≤ −2 we distinguish two cases: • If n − m is odd, then for some 0 < ε < 1 and while x → 0+ we have sin t (1 − ε)n sinn t 1 ≤ 1 ⇒ m−n ≤ m ≤ m−n and t t t t Z bx n n−m+1 n−m+1 b − a sin t bn−m+1 − an−m+1 (1 − ε)n ≤ ≤ (n − m + 1)xm−n−1 tm (n − m + 1)xm−n−1 ax (1 − ε) ≤

74

R bx n t Thus limx→0+ ax sin tm dt = +∞ and from (1) Z bx Z bx sinn t sinn t lim dt = lim dt = +∞ m t tm x→0− ax x→0+ ax • If n − m is even, then similarly while x → 0+ we have Z bx sinn t bn−m+1 − an−m+1 bn−m+1 − an−m+1 ≤ ≤ m−n−1 m (n − m + 1)x t (n − m + 1)xm−n−1 ax R bx n t Thus limx→0+ ax sin tm dt = +∞ and from (1) Z bx Z bx n sin t sinn t lim− dt = lim − dt = −∞ tm tm x→0 x→0+ ax ax and the limit does not exist. Finally, collecting yields  = 0, n−m≥0    Z bx  n b = ln a , n − m = −1 sin t lim dt x→0 ax  tm = +∞, n − m ≤ −2 and n − m = odd    does not exist, n − m ≤ −2 and n − m = even (1 − ε)n

Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer 31. (Correction) Proposed by Valmir Bucaj, Texas Lutheran University, Seguin, TX. If the vertices of a polygon, in clockwise order, are:         1 1 2 1 3 1 4 1 , √ ,√ , √ ,√ , √ ,√ ,..., √ ,√ a1 an+1 a2 a2n a3 a2n−1 a4 a2n−2     n−1 1 n 1 , , , , √ √ √ √ an−1 an+3 an an+2 where (an )n≥1 is a decreasing geometric progression, show that the area of this polygon is   6 1 (n + 3)(n − 2) n(n + 1) A= √ − √ +√ √ 4 a1 a2n+2 a2n an+2 Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and   Technology, Damascus, Syria. Let P1 = √1a1 , √a1n+1 and for 2 ≤ k ≤ n let   1 Pk = √kak , √a2n+2−k . Then the area A of the polygon (P1 , P2 , . . . , Pn ) is given by n−1 −−−−−→ −−−→ 1X det(P1 Pk+1 , P1 Pk ) 2 k=2 n−1 k+1 1 1 X √ak+1 − √a1 = √ 1 a2n+1−k − √a1n+1 2

A=

k=2

− √1a1 1 √ √ − a2n+2−k an+1 √k ak 1

75

Noting that ak a2n+1−k = a1 a2n and ak+1 a2n+2−k = a1 a2n+2 we conclude that n−1 X

  k+1 k 1 k+1 k A= −√ −√ −√ √ √ a1 a2n+2 a1 a2n an+1 ak+1 ak k=2   1 1 1 −√ −√ √ a1 a2n+2−k a2n+1−k   1 n 2 n(n + 1) − 6 n(n − 1) − 2 − −√ = √ √ √ −√ 2 a1 a2n+2 2 a1 a2n an+1 an a2   1 1 1 −√ −√ √ a1 a2n an+2 But an+1 an = a1 a2n and a2 an+1 = a1 an+2 . So, the above formula simplifies to   1 (n + 3)(n − 2) n(n + 1) 6 A= √ − + √ √ √ 4 a1 a2n+2 a2n an+2 as claimed. Also solved by Adrian Naco, Department of Mathematics, Polytechnic University of Tirana, Albania; and the proposer. 32. (Correction) Proposed by Mih´ aly Bencze, Bra¸sov, Romania. Let f : [a, b] → R, with 0 ≤ a < b, be a two times differentiable function such that f 00 and f 0 are continuous. If m = min f 00 (x) and M = max f 00 (x), then prove that x∈[a,b] 2

x∈[a,b]

2

m (b − a ) M (b2 − a2 ) ≤ bf 0 (b) − af 0 (a) − f (b) + f (a) ≤ 2 2 Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. We have Z f (b) − f (a) = a

b

b Z f (x)dx = xf (x) − 0

b

0

a

xf 00 (x) dx

a

from which follows 0

Z

0

b

bf (b) − af (a) − f (b) + f (a) =

xf 00 (x) dx

a

Moreover, 00

Z

b

Z

00

xdx ≤

min f (x) x∈[a,b]

b

Z

xf (x)dx ≤ max f (x)

a

a

and then b2 − a2 ≤ m 2

00

Z a

x∈[a,b]

b

xf 00 (x)dx ≤ M

b

xdx a

b2 − a2 2

concluding the proof. Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Angel Plaza, Department of Mathematics, Universidad de Las Palmas de Gran Canaria, Spain; Florin Stanescu, Serban Cioculescu School, Gaesti, Dambovita, Romania;Adrian Naco,

76

Department of Mathematics, Polytechnic University of Tirana, Albania; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; and the proposer. 33. Proposed by Ovidiu Furdui, Cluj, Romania. Find the value of Z π/2 √ n sinn x + cosn x dx lim n→∞

0

Solution
by Moubinool Omarjee, Paris France. Let fn (x) = fn π2 − x . Then, Z π2 Z π4 fn (x)dx = 2 fn (x) dx 0

√ n

sinn x + cosn x =

0

Let us denote by π 4

Z yn =

Z fn (x)dx =

0

π 4

n

1

cos x (1 + (tan x) ) n dx 0

  n 1 The functions gn (x) = cos x (1 + (tan x) ) n are continuous on 0; π4 and the se π quence (gn ) converges to the function x 7→ cos x on 0; 4 as can be easily checked. Furthermore, 1   1 π n  n |gn (x)| ≤ cos x 1 + tan ≤ cos x · 2 n ≤ cos x · 2 ≤ 2 4

So, by the Dominated Convergence theorem, we have √ Z π4 2 lim yn = cos x dx = n→∞ 2 0 Finally, Z lim

n→∞

π 2

√ n

√ sinn x + cosn x dx = 2 ·

0

2 √ = 2 2

Also solved by Albert Stadler, Switzerland; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Anastasios Kotronis, Athens, Greece; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Adrian Naco, Department of Mathematics, Polytechnic University of Tirana, Albania; and the proposer 34. Proposed by Mih´ aly Bencze, Bra¸sov, Romania. Solve the following equation r r q q √ √ √ 2 + 2 + . . . + 2 + x + 2 − 2 + . . . + 2 + x = x 2, | {z } | {z } n-times

n-times

where n ≥ 3. Solution by Islam Foniqi, University of Prishtina, Department of Mathematics, Prishtin¨ e, Republic of Kosova. We see that x has to be in (0, 2), so we

77

p can take x = 2 cos y where 0 < y < π2 . Using the formula 2(1 + cos 2a) = 2 cos a, we have r q √ y 2 + 2 + · · · + 2 + x = 2 cos n−1 2 {z } | (n−1)−times

and the given equation becomes r r √ y y 2(1 + cos n−1 ) + 2(1 − cos n−1 ) = 2 2 cos y 2 2 or √ y y cos n + sin n = 2 cos y 2 2 which can be written as y π cos( n − ) = cos y 2 4 Now 2yn − π4 = y or 2yn − π4 = −y which is equivalent to y(1 − 21n ) = − π4 or y(1+ 21n ) = π4 , with 0 < y < π2 . The equation y(1− 21n ) = − π4 does not have solution because n ≥ 3 and 0 < y < π2 , but the equation y(1 + 21n ) = π4 has the solution n n π π y = 4(22n +1) which is clearly between 0 and π2 . Therefore, x = 2 cos y = 2 cos 4(22n +1) when n ≥ 3. Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; and the proposer 35. Proposed by Florin Stˇ anescu, School Cioculescu Serban, Gˇ ae¸sti, jud. Dambovita, Romania. Let P (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 be a polynomial with positive real coefficients of degree n ≥ 3, such that P 0 has only real zeros. If 0 ≤ a < b show that Rb 1  0  dx 1 P (b) P 0 (b) − P 0 (a) a P 0 (x) ≥ ln ≥ Rb 1 0 b−a P (a) P (b) − P (a) dx 00 a P (x)

Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. The roots of P 0 (x) = 0 are evidently all negative and P 0 (x) = nan

n−1 Y

(x − ξk ),

ξk < 0, 1 ≤ k ≤ n − 1

k=1

Thus we have n−1 P 00 (x) X 1 = P 0 (x) x − ξk k=1

that it is a decreasing function as well as P 01(x) and P 001(x) . Now  0  Z b 00 P (b) P (x) ln = dx 0 0 P (a) a P (x) Z b 00 Z b Z b P (x) 1 1 dx · dx ≤ (b − a) dx 0 00 0 a P (x) a P (x) a P (x) on account of Chebyshev’s inequality for integrals applied to decreasing (increasing) functions. This proves the LHS inequality.

78

The RHS inequality is actually R b 00 Z b 00 P (x) dx P (x) dx ≥ (b − a) Rab 0 a P (x) P 0 (x) dx a

That is, b

b

Z b P 00 (x) dx ≥ (b − a) P 00 (x) dx 0 a a P (x) a and this follows also by applying Chebyshev’s result again with P 0 (x) increasing 00 (x) is decreasing. while PP 0 (x) Z

P 0 (x)dx ·

Z

Comment by the Editor. This problem has appeared as part of the following paper by the same author: Aplicat¸i ale inegalitˇa¸tii lui Cebi¸sev in formˇa integralˇa, Gazeta Matematicˇ a, Seria B, Anul CXVII, nr. 3 (2012) 113–121. Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; and the proposer

79

MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always welcomed. The source of the proposals will appear when the solutions be published.

Proposals 26. Let f : [0, 1] → [0, 1] be a differentiable function such that |f 0 (x)| = 6 1 for all x ∈ [0, 1]. Show that there exist two unique points α, β ∈ [0, 1] such that f (α) = α and f (β) = 1 − β. 27. Prove that the equation

√ √ √ (x + y 3)4 + (z + t 3)4 = 7 + 6 3

does not have rational solutions. 28. Find all polynomials p(x) with real coefficients such that p(a + b − 2c) + p(b + c − 2a) + p(c + a − 2b) = 3p(a − b) + 3p(b − c) + 3p(c − a) for all a, b, c ∈ R. 29. Equation x3 − 2x2 − x + 1 = 0 has three real roots a > b > c. Find the value of ab2 + bc2 + ca2 . 30. Let {an }n≥1 be a strictly increasing sequence of positive integers such that ∞ X 1 an+1 = +∞. Prove that is an irrational number. lim n→+∞ a1 a2 . . . an a n=1 n

80

Solutions 21. Let a > −3/4 be a real number. Show that s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − 2 6 3 2 6 3 is an integer and determine its value. (XXVI Spanish Math Olympiad 1989-1990) Solution by Bruno Salgueiro Fanego, Viveiro, Spain; Islam Foniqi, University of Prishtina, Department of Mathematics, Prishtin¨ e, Kosov¨ e; and Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Putting s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 x= + and y = − 2 6 3 2 6 3 and adding up and multiplying up the preceding expressions, we get x + y = a + 1 √ √ and xy = −a3 /27, respectively. Now, we call z = 3 x + 3 y and cubing, yields √ √ √
z 3 = x + y + 3 3 xy 3 x + 3 y = a + 1 − az or equivalently, z 3 + az − (a + 1) = 0 ⇔ (z − 1)(z 2 + z + a + 1) = 0 Since the discriminant of z 2 + z + a + 1 = 0 is δ = −(3 + 4q) < 0, then it does not have real roots. So, s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − =1 2 6 3 2 6 3 2

and we are done.

Also solved by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Adrian Naco, Department of Mathematics Faculty of Mathematical Engineering and Physical Engineering Polytechnic University of Tirana, Albania; and Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Alexandros Sygkelakis, Iraklion Crete, Greece. 22. Let a1 , a2 , a3 , a4 be nonzero real numbers defined by ak = k ≤ 4), α, β ∈ R. Calculate 1 + a21 + a22 a1 + a2 + 1/a4 a1 + a2 + 1/a4 2 + 1/a24 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4

sin(kβ + α) , (1 ≤ sin kβ

1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23



(J´ozsef Wildt Mathematics Competition 2005)

81

Solution by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain. First, we evaluate a1 a1 1 a2 1 a2 1 a4 1 a4 ∆ = 1 1/a4 1 = a4 a2 1 a3 a2 1 a3 1 = [a1 (a3 − a4 ) + a2 (a4 − a2 ) + a4 (a2 − a3 )] a4 Taking into account that ak − ah

sin(kβ + α) sin(hβ + α) − sin kβ sin hβ sin(kβ + α) sinh β − sin(hβ + α) sin kβ = sin kβ sin hβ 1 cos[(k − h)β + α] − cos[(k − h)β − α] = 2 sin kβ sin hβ sin(k − h)β sin α = − sin kβ sin hβ =

we have   − sin α sin 3β sin(2β + α) − sin 2β sin(β + α) sin β sin(4β + α) − sin(4β + α) sin 2β sin 3β sin 2β sin 3β   − sin α sin(4β + α) sin β sin(4β + α) sin β = − = 0. sin(4β + α) sin 2β sin 3β sin 2β sin 3β Now, it is easy to see by direct calculations that 1 + a21 + a22 a1 + a2 + 1/a4 1 + a2 (a1 + a3 ) a1 + a2 + 1/a4 2 + 1/a24 a2 + a3 + 1/a4 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23 2 a1 1 a2 = 1 1/a4 1 = 0 a2 1 a3 and we are done. 2 ∆=

Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; and Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain 23. Let A be a set of positive integers. If the prime divisors of elements in A are among the prime numbers p1 , p2 , . . . , pn and |A| > 3 · 2n + 1, then show that it contains one subset of four distinct elements whose product is the fourth power of an integer. (Training Sessions of Catalonia Team for OME 2012) Solution 1 by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. To each element a in A we associate an n-tuple (x1 , x2 , . . . , xn ), where xi

82

is 0 if the exponent of pi in the prime factorization of a is even, and 1 otherwise. These n-tuples are the “pigeons”. The “holes” are the 2n possible choices of 00 s and 10 s. Hence, by the PHP, every subset of 2n + 1 elements of A contains two distinct elements say a11 , a12 with the same associated n-tuple, and the product √ of these two elements is then a square, as all exponents are even. Thus, a11 a12 is an integer. Likewise, from the remaining numbers in A we can choose a21 and √ a22 such that a21 a22 is an integer. Similarly, from the set A, which has at least 3 · 2n + 1 elements, we can select 2n + 1 such pairs or more. Consider the 2n + 1 integer numbers that are the square roots of products of the two elements of each pair. That is, √ √ √ √ a11 a12 , a21 a22 , a31 a32 , . . . , a2n +1,1 a2n +1,2 Since all the previous numbers have the same divisors, the previous arguments give √ √ us two of them ai1 ai2 , aj1 aj2 with the same 0 − 1 n-tuple. The product of them is a perfect square √ √ ai1 ai2 · aj1 aj2 = x2 , where x is an integer. So, ai1 ai2 aj1 aj2 = x4 and we are done. Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. As usual, for a positive integer a and a prime p we will write νp (a) to denote the largest integer α such that pα divides a. We will also use the notation p (a) to denote νp (a) mod 2. Let P (2) (A) denote the collection of subsets of A each consisting of two elements P (2) (A) = {B ⊂ A : |B| = 2} We will say that a subset G of P (2) (A) is good if it satisfies the following two properties (i): If B and C are two distinct elements of G then B ∩ C = ∅. (ii): If B = {α, β} belongs to G then π(B) = αβ is a perfect square. Consider a good subset G0 of P (2) (A) of maximal cardinality among good subsets of P (2) (A). We will show that |G0 | > 2n . Indeed, suppose that, on the contrary, |G0 | ≤ 2n . Then the subset A0 = ∪B∈G0 B has 2|G0 | ≤ 2n+1 elements, and consequently |A \ A0 | ≥ 3 · 2n + 1 − 2n+1 = 2n + 1. In particular, the map Φ : A \ A0 → {0, 1}n , k 7→ (p1 (k), p2 (k), . . . , pn (k)) can not be injective, and there must be two distinct elements a and b in A \ A0 such that Φ(a) = Φ(b). This implies that ab is a perfect square and consequently G1 = G0 ∪ {a, b} is a good subset of P (2) (A) with |G1 | = 1 + |G0 |. This contradicts the maximality of |G0 |. This contradiction shows that |G0 | > 2n . Now, since |G0 | > 2n = |{0, 1}n |, the map p p p
Ψ : G0 → {0, 1}n , B 7→ p1 ( π(B)), p2 ( π(B)), . . . , pn ( π(B)) 0 can not be injective, and there must be two distinct elements B = {x, p y} and pB = √ 0 {z, t} in G0 , such that Ψ(B) = Ψ(B ). This means that xyzt = π(B) π(B 0 ) is the square of an integer, or equivalently that xyzt is a forth power. This is the desired conclusion.

Remark. Note that we only assumed that |A| ≥ 3 · 2n + 1.

2

83

Also solved by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain 24. Find all triples (x, y, z) of real numbers such that  12x − 4z 2 = 25,  24y − 36x2 = 1,  20z − 16y 2 = 9. (Training Sessions for COM-2011) Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Setting 6x = t, 4y = u, 2z = v we get 2t − v 2 = 25,

6u − t2 = 1,

10v − u2 = 9

Since 1 + t2 ≥ 2t then we have 6u ≥ 25 + v 2 . Moreover 10v = 9 + u2 ≥ 6u ≥ 25 + v 2

=⇒ (v − 5)2 ≤ 0

which implies v = 5. This in turn implies t = 25 from the first equation and u = 313/3 from the second but the triple (t, u, v) = (25, 313/3, 5) does not satisfy the third equation 10v − u2 = 9. So, the given system has no solutions. Solution 2 by Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain; and Ioan Viorel Codreanu, Satulung, Maramure¸s, Romania. First, we write the given system in the most convenient form  12x = 25 + 4z 2 ,  24y = 1 + 36x2 ,  20z = 9 + 16y 2 . Subtracting 20z to both members of the first equation, 12x to the second, and 24y to the third, we obtain   12x − 20z = (5 − 2z)2 ,  12x − 20z = 25 + 4z 2 − 20z,  24y − 12x = 1 + 36x2 − 12x, ⇔ 24y − 12x = (1 − 6x)2 ,   2 20z − 24y = 9 + 16y − 24y. 20z − 24y = (3 − 4y)2 . Adding up the last three equations, yields (1 − 6x)2 + (3 − 4y)2 + (5 − 2z)2 = 0 which is possible only when  1 − 6x = 0,  3 − 4y = 0,  5 − 2z = 0. But, triple (1/6, 3/4, 5/2) does not satisfy the system and therefore it does not have solution. 2 Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Albert Stadler, Herrliberg, Switzerland; Iv´ an Geffner Fuenmayor, BARCELONA TECH, Barcelona, Spain; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Adrian Naco, Department of Mathematics Faculty of Mathematical Engineering and Physical Engineering

84

Polytechnic University of Tirana, Albania; and Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. 25. Let a, b, c be the lengths of the sides of a triangle ABC with inradius r and cicumradius R. Prove that s    a r b c 1 3 ≤ ≤ 2r + R b + c + 2a c + a + 2b a + b + 2c 4 (Training Sessions of Spanish Math Team for IMO 2011) Solution 1 by Islam Foniqi, University of Prishtina, Department of Mathematics, Prishtin¨ e, Kosov¨ e. Let L be the left hand side, M the middle and P the right side of the inequality. a) First we prove that L ≤ M . Indeed, r 1 2r + R 1 3 (b + c + 2a) (c + a + 2b) (a + b + 2c) ⇔ ≥ L≤M ⇔ ≥ L M r a b c From AM − GM we have r 1 b + c + 2a c + a + 2b a + b + 2c 3 (b + c + 2a) (c + a + 2b) (a + b + 2c) ≤ ( + + ) a b c 3 a b c So its enough to prove that 2r + R b + c + 2a c + a + 2b a + b + 2c 3R a+b b+c c+a 3· ≥ + + ⇔ ≥ + + r a b c r c a b Since a+b+c abc S4ABC = r = , 2 4R then we obtain abc Rr = , 2(a + b + c) and using the well-known inequality 9R2 ≥ (a2 + b2 + c2 ) we get 9R2 a2 + b2 + c2 (a2 + b2 + c2 )(a + b + c) 3R = ≥ =2 abc r 3Rr 3abc 3 2(a+b+c) Now it is enough to prove that (a2 + b2 + c2 )(a + b + c) a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 3abc abc which can be written as 2

2(a2 + b2 + c2 )(a + b + c) ≥ 3(a2 (b + c) + b2 (c + a) + c2 (a + b)) This inequality is equivalent to 2(a3 + b3 + c3 ) ≥ a2 (b + c) + b2 (c + a) + c2 (a + b) Using Schur inequality a2 (b + c) + b2 (c + a) + c2 (a + b) ≤ 3abc + a3 + b3 + c3 it is enough to prove that 2(a3 + b3 + c3 ) ≥ 3abc + a3 + b3 + c3 ⇔ a3 + b3 + c3 ≥ 3abc

85

This clearly holds on account of the identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) and the fact that a2 + b2 + c2 ≥ ab + bc + ca as it is well-known. b) Now we prove that M ≤ P. Let us denote x = a + b + c. On account of the AM − GM inequality we have 1 a b c 1 a(b + x)(c + x) + b(c + x)(a + x) + c(a + x)(b + x) M≤ ( + + )= 3 a+x b+x c+x 3 (a + x)(b + x)(c + x) To prove that M ≤ P it is enough to see that 3(a + x)(b + x)(c + x) ≥ 4(a(b + x)(c + x) + b(c + x)(a + x) + c(a + x)(b + x)) Using the fact that x = a + b + c

the last inequality becomes

3

2(a + b + c) ≥ 9abc + 5(a + b + c)(ab + bc + ca) But 3(ab + bc + ca) ≤ (a + b + c)2

so we get

3

2(a + b + c) − 9abc − 5(a + b + c)(ab + bc + ca) 1 5 ≥ 2(a + b + c)3 − 9abc − (a + b + c)3 = (a + b + c)3 − 9abc ≥ 0 3 3 Finally, from a) and b) we have proved that L ≤ M ≤ P as required. The equality in both sides holds if and only if the triangle is equilateral. Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. First, if a + b + c is denoted by u then (u + a)(u + b)(u + c) = 2u3 + u(ab + bc + ca) + abc. √ √ Using the AM-GM inequality, we have u ≥ 3 3 abc and ab + bc + ca ≥ 3( 3 abc)2 , so (u + a)(u + b)(u + c) ≥ 54abc + 9abc + abc = 64abc. This is equivalent to the upper inequality : s    a b c 1 3 ≤ b + c + 2a c + a + 2b a + b + 2c 4 with equality, if and only if a = b = c. Note that we have proved this inequality for any positive real numbers a, b and c. On the other hand, if s denotes the semi-perimeter of ABC then we know that a + b + c = 2s,

ab + bc + ca = s2 + r2 + 4rR,

abc = 4rRs.

So, if  L=

2a + b + c a



a + 2b + c a



a + b + 2c a



then (2s + a)(2s + b)(2s + c) 16s3 + 2s(ab + bc + ca) + abc = abc abc 16s2 + 2s(s2 + r2 + 4rR) 9s2 + r2 = 1+ =3+ 4rRs 2rR Now, using the fundamental inequality, we have p s2 ≤ 2R2 + 10rR − r2 + 2(R − 2r) R2 − 2rR ≤ 4R2 + 6rR − r2 . L =

86

(See for instance [1,Theorem A, page 2].) We conclude that L≤3+

36R2 + 54rR − 8r2 60rR + 36R2 − 8r2 = 2rR 2rR

It follows that  3 2r + R −L ≥ r

R4 + 6rR3 − 6r2 R2 − 22r3 R + 4r4 r3 R (R − 2r)(R3 + 8R2 r + 9Rr2 ) + r2 (R − 2r)2 ≥ 0, = r3 R where we used the well-known inequality R ≥ 2r. Therefore we have proved that L−1/3 ≥ r/(2r + R) which is the desired inequality. [1] D.S. Mitrovni´c, J. E. Peˇcari´c, and V. Volenec. Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989. Solution 3 by Jos´ e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. To prove the LHS inequality we can assume that a + b + c = 1 on account t of the homogeneity. Consider the function f : (0, +∞) → R define by f (t) = 1+t : 0 −2 00 −3 Then, we have f (t) = (1 + t) and f (t) = −2(1 + t) < 0 for all t > 0. So, f is concave and applying Jensen’s inequality, we have   a+b+c 1 f ≥ (f (a) + f (b) + f (c)) 3 3

That is, (a + b + c)/3 1 = 4 1 + (a + b + c)/3

≥ ≥

=

  a 1 b c + + 3 1+a 1+b 1+c s    a b c 3 1+a 1+b 1+c s    b c a 3 b + c + 2a c + a + 2b a + b + 2c

on account of AM-GM inequality. On the other hand, we have that b c R + ≤ (cyclic) c b r In fact, using the duality principle, there exist three positive numbers x, y, z such that a = y + z, b = z + x and c = x + y for which r (x + y)(y + z)(z + x) xyz p R= and r = x+y+z 4 (x + y + z)xyz Then, R (x + y)(y + z)(z + x) = r 4xyz Now, we will see that

and

b c z+x x+y + = + c b x+y z+x

(x + y)(y + z)(z + x) z+x x+y ≥ + 4xyz x+y z+x

87

or equivalently, y+z 1 1 + ≥ 2 4xyz (x + y) (z + x)2 which follows immediately from the fact that (x + y)2 ≥ 4xy and (z + x)2 ≥ 4zx. From the preceding and taking into account GM-HM inequalities, we have s    a b c 3 b + c + 2a c + a + 2b a + b + 2c  −1 b + c + 2a c + a + 2b a + b + 2c ≥3 + + a b c −1       r b b c c a a ≥ + + + + + ≥3 6+ b a c b a c 2r + R Equality holds when a = b = c because in this case R = 2r and we are done. 2 Also solved by Ioan Viorel Codreanu, Satulung, Maramure¸s, Romania; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Iv´ an Geffner Fuenmayor, BARCELONA TECH, Barcelona, Spain and Jos´ e Gibergans B´ aguena, BARCELONA TECH, Barcelona, Spain; Bruno Salgueiro Fanego, Viveiro, Spain.

88

MATHNOTES SECTION Note on an Algebraic Inequality Vandanjav Adiyasuren and Bold Sanchir Abstract. In this note a constrained inequality is generalized. 1. Introduction In [1] the following problem was posted: Let n be a positive integer. Find the largest constant cn > 0 such that, for all positive real numbers a1 , ..., an , 2  1 1 1 1 1 1 + · · · + + ≥ c + · · · + + n a21 a2n (a1 + · · · + an )2 a1 an a1 + · · · + an A solution to the preceding proposal and some related results appeared in [2]. Our aim in this short note is to generalize it. 2. Main Results Theorem 1. For all positive numbers a1 , . . . , an and for all positive integer p > 1, the following inequality holds:  p 1 1 1 1 1 1 +...+ p + ≥ cn (p) + ... + + , (2) ap1 an (a1 + . . . + an )p a1 an a1 + . . . + an where cn (p) = 

(n3 +1)p 2p−1 n p−1

p−1 +1

. (n2 +1)p

Proof. Denote A=

1 1 1 1 1 + ... + + , B= + ... + a1 an a1 + . . . + an a1 an

By applying H¨ older’s Inequality, we get 1 1 1 p + ···+ p + a1 an (a1 + . . . + an )p  p 1 1 1 1 ≥ + ··· + + a1 an n(a1 + . . . + an ) (1 + . . . + 1 +  p np 1 n(n − 1) n−1 = A + B + B 2p−1 n2 + 1 n(n2 + 1) (n p−1 + 1)p−1 n

(3)

1

p

)p−1

n p−1

Applying Cauchy-Schwarz Inequality, we have 1 1 n2 + ··· + ≥ a1 an a1 + . . . + an From (3), (4) we get (2).

(4) 

89

Corollary 1. Let a1 , . . . , an be positive numbers. Then  2 1 1 1 n3 + 1 1 1 1 + ... + 2 + ≥ 2 + ... + + a21 an (a1 + . . . + an )2 (n + 1)2 a1 an a1 + . . . + an (5) Proof. Choosing p = 2 in (2), we get (5).



Theorem 2. Let n, m, k be positive integers. For all positive real numbers a1 , . . . , an and α, β > 0 with kβ − mα > 0, k > m, p > 1, we hve X X 1 1 + p (ai1 + . . . + aim ) (ai1 + . . . + aik )p 1≤i1 1 2 2

Theorem 5. Let α ≥ 0, δ and χ be integers such that χ = ..., −3, −2, −1, 0, 1, 2, 3, ..., z(α) z(α) (χ − 21 ) × 2 < δ ≤ (χ + 12 ) × 2 ; then it holds     α+δ α +χ (11) = 2z(α) 2z(α)   Proof. For convenience,we use the symbol v(α, δ) to denote 2α+δ z(α) , use the symbol z(α)

z(α)

Iχ to denote the interval (χ − 12 ) × 2 < δ ≤ (χ + 21 ) × 2 and keep using I = z(α). Note that the condition of the theorem 5 is the same as that of the theorem 4; hence the conclusions drawn in the theorem 4 can be directly adopted here. According to thetheorem 2, it is merely necessary to prove that v(α, δ) takes an identical value 2αI + χ, which merely depends upon α and χ, on the whole interval Iχ . Also by the theorem 2, this is a job to compute the values of v(α, δ) at five points δ = (χ − 12 ) × 2I , δ = (χ − 12 ) × 2I + 1, δ = χ × 2I , δ = (χ + 21 ) × 2I and δ = (χ + 12 ) × 2I + 1, as follows. (i). When δ = (χ − 12 ) × 2I , it yields   j α+(χ− 12 )×2I k v(α, δ) = α+δ = I 2 2I (12)       = 2αI + χ − 21 = χ + 2αI − 21 = χ − 1 + 2αI (ii). When δ = (χ − 12 ) × 2I + 1, it yields v(α, δ) = =

 α+1 2I I

 α+δ  2I

+χ−

1 2

= 

j

α+(χ− 12 )×2I +1 2I

=χ+

 α+1 2I

−

1 2

k



=χ+

α

(13)

2I

(iii). When δ = χ × 2 , it yields     j k jαk α+δ α + χ × 2I α v(α, δ) = = = + χ = χ + 2I 2I 2I 2I

(14)

140

(iv). When δ = (χ + 12 ) × 2I , it yields v(α, δ) = =

α

2I

+

 α+δ 

1 2

2I

=

j

α+(χ+ 12 )×2I 2I

k

     + χ = χ + 2αI + 12 = χ + 2αI

(v). When δ = (χ + 12 ) × 2I + 1, it yields v(α, δ) =

 α+δ 

=χ+1+

2I

 α+1 2I

= −

j

1 2

α+(χ+ 12 )×2I +1 2I



=χ+1+

k

α

(15)

(16)

2I

Obviously, j the k results in (13),(14) and (15) show that v(α, δ) does take an identical value of 2αI + χ on the whole interval Iχ , and the value merely depends upon α

and χ since I = z(α) depends on α . Note that the results j k(12) and (16) also imply

that, for a fixed χ , v(α, δ) takes an identical value of 2αI + χ − 1 on the interval j k left to Iχ , and takes an identical value 2αI + χ + 1 on the interval right to Iχ . j k Hence, as χ changes, it is true that v(α, δ) does take the value of 2αI + χ on every interval Iχ . This ends the proof of the theorem 5. By the theorem 5, the relationship between the function v(α, δ) and the variable δ can be illustrated by figure 1. Through the figure, it gets to know that the length of the interval Iχ is 2I . By expressing the interval Iχ in its equivalent form Iχ = (−2I−1 + χ × 2I , 2I−1 + χ × 2I ] it shows that the function v(α, δ) changes with the interval I0 = (−2I−1 , 2I−1 ] to be a basic unit. We call the interval I0 a principal interval since the property of the function v(α, δ) on Iχ can be translated from the property on it.

Figure 1. Relationship between the variable δ and the function v(α, δ). Finally, we state and prove the following Theorem 6. For arbitrary integers α ≥ 0 and δ , it holds

141



α+δ 2z(α)



j α k δ − 1 1 = z(α) + z(α) + 2 2 2

(17)

Proof. Keep using the symbol I = z(α) and suppose δ ∈ Iχ without loss of generality; then by the theorem 5, it yields I I 1 1 (χ − ) × 2 < δ ≤ (χ + ) × 2 (18) 2 2 Performing a simple transformation on (18) leads to 1 1 δ δ − ≤χ< I + I 2 2 2 2 That is 1 1 δ δ − ≤χ< I − +1 2I 2 2 2 Referring to definition of the ceiling function yields   δ 1 χ= I − 2 2 By the properties (P6) and (P3) of the lemma 1, it follows           δ 1 δ − 2I−1 δ − 2I−1 − 1 δ−1 1 δ−1 1 − = = + 1 = − + 1 = + 2I 2 2I 2I 2I 2 2I 2 l m j k δ  I−1 I−1 I − 12 = δ−22I = δ−2 2I+2 −1 2Ij k   I−1 = δ−2 2I −1 + 1 = δ−1 + 12 2I

Hence the theorem 6 holds.

4. Computer Test Computations related to this paper can be tested on personal computers. The C-language program is as follows. \#include"math.h" int GetI(int a)

/* Find the smallest I that fits $0\leq a mod 2^i < 2^{(i-1)}$ */

\{ int i,I; int \_2i\_1, \_2i; for(i=1;;i++) \{ \_2i\_1 = 1$\ll$(i-1);\quad /* Compute $2^{(i-1)}$*/ \_2i=\_2i\_1$\ll${1}; \quad if(a\%\_2i$