Mathproblems

ISSN: 2217-446X, url: http://www.mathproblems-ks.com

Volume 4, Issue 1 (2014), Pages 231-262

Editors: Valmir Krasniqi, Jose Luis Daz-Barrero, Armend Sh. Shabani,Paolo Perfetti, Mohammed Aassila, Mihaly Bencze, Valmir Bucaj, EmanueleCallegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba,Cristinel Mortici, Jozsef Sandor, Ercole Suppa, David R. Stone, Roberto Tauraso,Francisco Javier Garca Capitan.

PROBLEMS AND SOLUTIONS

Proposals and solutions must be legible and should appear on separate sheets, eachindicating the name of the sender. Drawings must be suitable for reproduction.Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourageundergraduate and pre-college students to submit solutions. Teachers can help byassisting their students in submitting solutions. Student solutions should includethe class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerningproposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail.com

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

Problems

88. Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 a, b, c, 1)satisfy the following equality;

cyc

(a b1 ab

a

1 a2)

= 0

Prove that a = b = c.(Note that

cyc

means cyclic sumcyc

f(x, y, z) = f(x, y, z) + f(y, z, x) + f(z, x, y))

89. Proposed by Mohammed Aassila, Strasbourg, France.Let S be the set of positive integers that does not contain the digit 7 in their decimalrepresentation. Prove that

nS

1

n< +.

c2010 Mathproblems, Universiteti i Prishtines, Prishtine, Kosove.231

232

90. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technol-ogy, Damascus, Syria. Let n be a positive integer, prove that

nk=0

(1)bkc n ,

and determine the cases of equality.

91. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca,Romania. Calculate 1

0

10

ln(1 x) ln(1 xy)dxdy.

92. Proposed by D.M. Batinetu-Giurgiu, Matei Basarab National College,Bucharest, Romania, and Neculai Stanciu, George Emil Palade School, Buzau,Romania. Let {an}n0 be a sequence of positive integer numbers such that 5 doesnot divide an for all, n N,and let the sequence {bn}n0 be defined by bn = a2nL2n ,for n N where {Ln}n0 is the sequence of Lucas numbers. Prove that bn is square-free, (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

+n=1

(1)n+1n(

tan1 x x+ x3

3 + (1)n+1 x

2n+1

2n+ 1

).

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

|AP |2b2

+|BP |2c2

+|CP |2a2

is minimal, where a = BC, b = CA and c = AB.

95. Proposed by Li Yin, Department of Mathematics, Binzhou University, BinzhouCity, Shandong Province, 256603, China. An approximation formula of Wallisproduct. For all n N, then

Wn 1pi

(1 lnn

2n

)nwhere Wn :=

(2n1)!!(2n)!! is Wallis product.

233

SolutionsNo problem is ever permanently closed. We will be very pleased considering forpublication new solutions or comments on the past problems.

81. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technol-ogy, Damascus, Syria. Find, in terms of a > 0, the minimum of

a(x2 + y2 + z2) + 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 VergataUniversity, Rome, Italy. Answer: The minimum is 2a if a 1 and a + 1 ifa 1.Proof Let x+ y + z = 3u, xy + yz + zx = 3v2, xyz = w3. Let m be the searchedminimum. We homogenize by writing

a(x2 + y2 + z2)(x+ y + z) + 9xyz

(xy + yz + zx)(x+ y + z)

which in terms of the variables (u, v, w) becomes

a(9u2 6v2)3u+ 9w39uv2

m 0that is

9w3 +R(u, v) 0This is a linear function in w3 and then it holds if and only if it holds for theminimum 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 iszero or when two of the are equals.

In the first case we set z = 0 and get

a(x2 + y2)

xy m

and it is evident that m = 2a.

As for the second case we set x = y and get

a(x2 + y2 + z2)(x+ y + z) + 9xyz

(xy + yz + zx)(x+ y + z) a 1 = (x z)

2(2ax 2x+ az)x(x+ 2z)(2x+ z)

which is positive for any x, z R if and only if a 1. The minimum is thus themaximum 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, Mon-golia. a) Let a 1. Schurs inequality is equivalent to

x2 + y2 + z2 +9xyz

x+ y + z 2(xy + yz + zx).

234

Using the above inequality and the given condition x+ y + z = 1, we have

x2 + y2 + z2 + 9xyz 2(xy + yz + zx) (1)Also clearly, we have x2 + y2 + z2 xy + yz + zx. From this we find that

(a 1)(x2 + y2 + z2) (a 1)(xy + yz + zx). (2)Add (1) and (2) we get

a(x2 + y2 + z2) + 9xyz (a+ 1)(xy + yz + zx)or equivalently

a(x2 + y2 + z2) + 9xyz

xy + yz + zx a+ 1.

This inequality becomes equality when x = y = z = 13 . Hence

min

(a(x2 + y2 + z2) + 9xyz

xy + yz + zx

)= a+ 1.

b) Let 0 < a < 1. From (1) we have,

a(x2 + y2 + z2) + 9axyz 2a(xy + yz + zx). (3)On the other hand we have

a(x2 + y2 + z2) + 9xyz a(x2 + y2 + z2) + 9axyz (4)From (3) and (4) we have

a(x2 + y2 + z2) + 9xyz 2a(xy + yz + zx)or equivalently

a(x2 + y2 + z2) + 9xyz

xy + yz + zx 2a.

with equality when x = y = 12 , and z = 0. Hence

min

{a(x2 + y2 + z2) + 9xyz

xy + yz + zx

}= 2a.

Also solved by Arkady Alt, San Jose, California, USA; Moti Levy, Re-hovot, Israel; and the proposer.

82.Proposed by Anastasios Kotronis, Athens, Greece.Let F (n) :=

k1

1(kn+1)k! . Determine the sequence {cm}m0 such that

limn+n

m(F (n)

m1k=0

cknk

)= cm

where, for m = 0, the sum is considered to be 0.

Solution 1 by Moti Levy, Rehovot, Israel. Actually, the problem deals with

finding asymptotic expansion of F (n) = 10

(ex

n 1) dx.1

kn+ 1=

1

kn

1

1 + 1kn=

j=1

(1)j1kj

1

nj

235

k=1

1

(kn+ 1)k!=

k=1

j=1

(1)j1kjk!

1

nj

=

j=1

1

nj

k=1

(1)j1kjk!

.

Let

{cm}m0 ={

(1)m1k=1

(1)m1kmk!

}m0

thenk=1

1

(kn+ 1)k!=c1n

+c2n2

+c3n3

+ .

The sequence {cm} can be expressed by the generalized hypergeometric function,cm = (1)m1 m+1Fm+1 (1, 1, . . . , 1; 2, 2, . . . , 2; 1) .

For m = 1, in particular,

c1 = 2F2 (1, 1; 2, 2; 1) = + Ei (1) = 1.3179

Solution 2 by Paolo Perfetti, Department of Mathematics, Tor VergataUniversity, Rome, Italy.

F (n).=k1

1

(kn+ 1)k!=k1

1

k!kn

j=0

(1)j(kn)j

The absolute convergence of the series allows us to take the limit n + underthe sum so for m = 0 we have

limn+F (n) = 0, (c0 = 0)

Let m = 1.

nm

(F (n)

m1k=0

cknk

)= nF (n) =

k1

1

k!k

j=0

(1)j(kn)j

whose limit n + is evidently c1 =k=1

1

k!k. By induction we suppose that

cm =

k=1

(1)m1k!km

for any 0 m r. We have

nr+1

(F (n)

rm=0

cmnm

)= nr+1

j=0

(1)jnj+1

k1

1

k!kj+1

rm=1

cmnm

= nr+1

rp=1

(1)p1np

k1

1

k!kp+

p=r+1

(1)p1np

k1

1

k!kp

rm=1

cmnm

and the limit as n yields (1)r

k=1

1

k!kr+1.

236

Solution 3 by Omran Kouba, Higher Institute for Applied sciences andTechnology, Damascus, Syria. Let = {z C : 1}. For z weconsider G(z) defined by the formula

G(z) =

p=1

1

p! (p+ z)

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

G(m)(z) =

p=1

(1)mm!p! (p+ z)m+1

In particular, for |z| < 1 we have

G(z) =

m=0

G(m)(0)

m!zm

Thus for z in the neighborhood of 0, and for every m 0 (with the same conventionas in the statement of the problem,) we have

zG(z) =

m1k=0

ckzk +O(zm)

with c0 = 0 and ck = G(k1)(0)/(k 1)! when k is a positive integer. But F (n) =

1nG(

1n ), so for large n, and for every nonnegative integer m, we have

F (n) =

m1k=0

cknk

+O

(1

nm

).

Moreover, according to (1) we have

ck = (1)k1p=1

1

p! pk.

This yields the desired conclusion. Note that the cks do not seem to have a closedform. but they can, alternatively, be expressed as integrals:

ck =(1)k1(k 1)!

p=1

1

p!

0

tk1eptdt =(1)k1(k 1)!

0

tk1(eet 1

)dt.

Also solved by Moubinool Omarjee, Paris, France, and the proposer.

Editors Comment: The proposer of this problem indicated that it is a general-ization of problem U278 of Mathematical Reflections.

83.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy,Daejeon, South Korea. Let {an} strictly increasing sequence of positive integerssuch that gcd(ai, aj) = 1 for any i, j(i < j). Let bn = an+1 an. Prove that thesequence {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 dividesan. Since ai and aj are relatively prime (for i 6= j) then pi 6= pj . Let A (x) be thenumber 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) pi (x) , (1)where pi (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

aA(X)+1 a1 X X2

=X

2.

Clearly,A(X)n=1

bn = aA(X)+1 a1.

Suppose, on the contrary, that the sequence {bn} is bounded by the positive con-stant M > 0. Then

aA(X)+1 a1 A (X)M,X

2 aA(X)+1 a1 A (X)M. (2)

By equations ((1)) and (2),

pi (X)

X A (X)

X 1

2M

Multiplying both sides by lnX > 0,

pi (X)

X/ lnX 1

2MlnX.

On one hand, the Prime Number Theorem states that limXpi(X)X/ lnX = 1 but on

the other hand limX 12M lnX = ; This is a contradiction, which leads us tothe conclusion that the sequence {bn} is unbounded.

Solution 2 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. Let pn be the smallest prime number that dividesan. Since gcd(ai, aj) = 1 for every distinct i and j we conclude that p1, p2, . . . , pnare distinct primes from the interval [1, an] and consequently, pi(an) n. wherepi(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 absolutepositive constant A such that pi(x) A xln x hence

n pi(an) A anln an

A anlnn

where we used the trivial inequality n an (since, with a0 = 0, we have an =nk=1(ak ak1)

nk=1 1 = n.) Thus, an n lnnA , and in particular

limn

ann

= +. (3)

238

Now, suppose that the sequence {bn} is bounded by some constant M then itfollows immediately that

an a1n

=1

n

n1k=1

bk M

and consequently lim supnann 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 infn

ann lnn

1, and lim supn

bnlnn

> 1.

Solution 3 by Paolo Perfetti, Department of Mathematics, Tor VergataUniversity, Rome, Italy. The sequence {aj} with the smallest entries aj forany j is the sequence of the prime numbers. Any other sequence {aj} is such thataj aj and then asymptotically aj c0j ln j for any j large enough. We argue bycontradiction by assuming that there exists a constant c1 such that bn c1. Thisimplies that

nk=1

aj+1 aj = an+1 a1 nc1

but this contradicts aj c0j 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, BinzhouUniversity, Binzhou City, Shandong Province, China.Let 1 < p 2 and

x (0, pip2 ), prove that

lnx

sinp xx coshp x sinhp x

p sinhp x.

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

239

85.Proposed by D.M. Batinetu-Giurgiu, Matei Basarab National College, Bucharest,Romania, and Neculai Stanciu, George Emil Palade School, Buzau, Romania.Let n be a positive integer. Prove that

( nk=1

F 2kLk

)( nk=1

F 3kL2k

) (Fn+2 1)

5

(Ln+2 3)2

where Fn , respectively Ln represents the nth Fibonacci number respectively the

nth Lucas number.

Solution 1 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia. Using the following well known indentities L1 + L2 + + Ln = Ln+2 3and F1 + F2 + + Fn = Fn+2 1 we have(

nk=1

Lk

)3(

nk=1

F 2kLk

)(nk=1

F 3kL2k

)(

nk=1

Fk

)5. (1)

It is enough to prove the above inequality. Applying the Holders inequality, we get(nk=1

Lk

)3/5(

nk=1

F 2kLk

)1/5(

nk=1

F 3kL2k

)1/5

=

(nk=1

(L3/5k

) 53

)3/5 nk=1

((F 2kLk

)1/5)51/5 nk=1

((F 3kL2k

)1/5)51/5

(

nk=1

L3/5k

F2/5k

L1/5k

F3/5k

L2/5k

)=

nk=1

Fk.

Hence (1) is proved.

Solution 2 by Moti Levy, Rehovot, Israel. A more general version of thisproblem appeared in The Fibonacci Quarterly, Volume 51, Number 4, November2013. (

nk=1

Fm+1kLmk

)(nk=1

F p+1kLpk

) (Fn+2 1)

m+p+2

(Ln+2 3)m+p, m > 0, p > 0. (2)

To prove (2), we use the inequality (3) due to J. Radon,

nk=1

xp+1kypk (nk=1 xk)

p+1

(nk=1 yk)

p , p > 0, and xk 0, yk > 0, for 1 k n. (3)

By Radons inequality,(nk=1

Fm+1kLmk

)(nk=1

F p+1kLpk

) (nk=1 Fk)

p+m+2

(nk=1 Lk)

p+m (4)

240

The following sums are well known,nk=1

Fk = Fn+2 1 (5)nk=1

Lk = Ln+2 3 (6)

The required result is obtained by substituting (5) and (6) in (3).

Solution 3 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. We will use the following lemma:

Lemma. Let p, , be real numbers with p > 1, and 0 < , < 1. Then for everypositive numbers a1, . . . , an and b1, . . . , bn we have:

(a1 + + an)2p(b1 + + bn)2p2

nk=1

a2pk

b2(p1)k

nk=1

a2(1)pk

b2(1)(p1)k

.

Proof. Let x1, . . . , xn and y1, . . . , yn be positive numbers such thatnk=1

xk =

nk=1

yk = 1

Let r > 1 be defined by 1p +1r = 1, then by Holders inequality we have

1 =

nk=1

xk =

nk=1

xk

y1/rk

y1/rk (

nk=1

xpk

yp/rk

) 1p(

nk=1

yk

) 1r

=

(nk=1

xpkyp1k

) 1p

That is, by the Cauchy-Schwarz inequality:

1 nk=1

xpkyp1k

=

nk=1

xpk

y(p1)k

x(1)pk

y(1)(p1)k

nk=1

x2pk

y2(p1)k

nk=1

x2(1)pk

y2(1)(p1)k

for 0 < , < 1. Applying this, with xk = a/a and yk/b with

a = a1 + + an and b = b1 + + bn,we obtain

(a1 + + an)2p(b1 + + bn)2p2

nk=1

a2pk

b2(p1)k

nk=1

a2(1)pk

b2(1)(p1)k

.

Taking, (p, , ) =(52 ,

35 ,

23

), we obtain

(a1 + + an)5(b1 + + bn)3

(nk=1

a3kb2k

)(

nk=1

a2kbk

).

Finally, the desired inequality is obtained, by setting ak = Fk, bk = Lk and notingthat

nk=1

Fk =

nk=1

(Fk+2 Fk+1) = Fn+2 F2 = Fn+2 1,nk=1

Lk =

nk=1

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

241

Remark. Similarly, for m > 2 and 0 < q, p < m, we have(nk=1

FmpkLmqk

)(

nk=1

F pkLq2k

) (Fn+2 1)

m

(Ln+2 3)m2 .

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 1

0

ln(x+

1 x)x

dx.

Solution 1 by Paolo Perfetti, Department of Mathematics, Tor VergataUniversity, Rome, Italy. Let x = t2. The integral reads as 1

0

2 ln(t+

1 t2)dt = 2 10

ln tdt+ 2

10

ln

(1 +

1

t2 1)dt

10

ln t dt = (t ln t t)10= 1

In the second integral we change t = 1/

1 + y2 and it becomes 0

ln(1 + y)y

(1 + y2)3/2dy = ln(1 + y)

1 + y2

0

+

0

1

1 + y

11 + y2

dy

=y=sinh t

0

dt

1 + sinh t=z=et

2

1

1

z2 + 2z 1dz 1

(1

z z1 1

z z2

)1

z1 z2 dz

where z1 = (1 +

2)/2, z2 = (1

2)/2. We get evidently

12

ln1 z11 z2 =

12

ln222 +

2=1

2ln(3 2

2) =

ln(3 + 2

2)2

and the integral finally is 2 ln(3 + 2

2) 2.

Solution 2 by Arkady Alt, San Jose, California, USA.

Let I := 10

ln(x+

1 x)x

dx. The change of variables x = sin2 t shows that

I = 2

pi/20

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

= 2

pi/20

ln (sin t+ cos t) sin t dt. (t pi2 t)

242

Taking the half sum we obtain

I =

pi/20

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

=

2

pi/4pi/4

ln(

2 cos )

cos d (t pi4 + )

=

2

2(ln 2)

pi/4pi/4

cos d I1

+

2

pi/4pi/4

ln(cos ) cos d I2

Clearly, I1 =

2, and

I2 =[

sin ln(cos )]pi/4pi/4

+

pi/4pi/4

sin2

cos d

=

2 ln12

+ 2

1/20

u2

1 u2 du (u = sin )

=

2

2ln 2

2 +

1/20

(1

1 + u+

1

1 u)du

=

2

2ln 2

2 + ln

2 + 12 1

=

2

2ln 2

2 + 2 ln(

2 + 1)

Finally I = 2 + 22 ln(2 + 1).Also solved by Albert Stadler, Switzerland; Omran Kouba, Higher In-stitute for Applied Sciences and Technology, Damascus, Syria; Anas-tasios 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 theproposer.

87.Proposed by Dorlir Ahmeti, University of Prishtina, Republic of Kosova. Leta, b, c be positive real numbers such that a+ b+ c = 3. Prove that

a+b

1 +ab

+

b+c

1 +bc

+

c+a

1 +ca 3.

Solution 1 by AN-anduud Problem Solving Group. The proposed problemis equivalent to the following problem. x, y, z be positive real number such thatx2 + y2 + z2 = 3. Prove that

x+ y

1 + xy+

y + z

1 + yz+

z + x

1 + zx 3. (1)

Applying Holders inequality, we have(x+ y

1 + xy+

y + z

1 + yz+

z + x

1 + zx

)2((x+ y)(1 + xy)2 + (y + z)(1 + yz)2 + (z + x)(1 + zx)2)

((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 + y5 + z5) + 48xyz 25(x3 + y3 + z3). (3)To prove (3) we note that

9cyc

x5 + 48xyz 25cyc

x3 =cyc

(x+ y)(x y)2(x+ y z)2

+cyc

z(x y)2((x+ y 6z)2 + 35xy)

+ 25xyzcyc

(x y)2 0

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

244

MATHCONTEST SECTION

This section of the Journal offers readers an opportunity to solve interesting and el-egant mathematical problems mainly appeared in Math Contest around the worldand most appropriate for training Math Olympiads. Proposals are always wel-comed. The source of the proposals will appear when the solutions be published.

Proposals61. Find all real solutions of the following system of equations:

x2 + y2 + 6x+ 9 +x2 + y2 8y + 16 = 5,

9y2 4x2 = 60.

62. Let P be an interior point to an equilateral triangle ABC. Draw perpendicularsPX,PY and PZ 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 7weighs 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 provethat

Re

(nk=0

akbk

) 1

2n

(n+ 1

n

nk=0

|ak|2 + n2

n 1(

2n 2n

) nk=0

|bk|2)

245

Solutions56. 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 therein 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 cycleof 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 acycle of order 4. The integers n searched are of the form n = 3k + 1 and at thesame time of the form n = 4h + 1. Therefore, n 1 must be multiple of 3 and 4at the same time. That is, a multiple of 12 and n = 12k + 1. According to the

statement 1 12k + 1 201314 and therefore 0 k 20131312

or 0 k 16776.The smallest positive integer, for k = 0 is 1. The biggest, for k = 16776 is 201313and the total number is 16776 + 1 = 16777.

2

Also solved by Jose Luis Daz-Barrero, BARCELONA TECH, Barcelona,Spain.

57. Let n,m be positive integers. Prove that(1 +

1

n

)n 0, then by QM-AM inequality we have

x4 + y4 + z4 + x2y2z2

4(x2 + y2 + z2 + xyz

4

)2= 1 = x4+y4+z4+x2y2z2

4.Since the lower bound 4 can be attained if x = y = z A = B = C, the desiredminimum is 4.

247

Solution 2 by Jose Luis Daz-Barrero BARCELONA TECH, Barcelona,Spain. Since A+B + C = pi, then we have

x2 + y2 + z2 + xyz = 4 cos2A+ 4 cos2B + 4 cos2(A+B)

8 cosA cosB cos(A+B)= 4(cos2A+ cos2A cos2A cos2A+ sin2A sin2B)= 4[sin2B(cos2A+ sin2A) + cos2B

]= 4

Taking into account AM-QM inequality yields

1 =x2 + y2 + z2 + xyz

4x4 + y4 + z4 + x2y2z2

4

from which follows x4 + y4 + z4 + x2y2z2 4. So, the minimum value of theexpression claimed is 4 and it is attained when 4ABC is equilateral.

Solution 3 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. The answer is 4 and it is attained only whenthe triangle is equilateral.First, note as in the preceding solutions we have

x2 + y2 + z2 + xyz = 4

Thus,

x4 + y4 + z4 + x2y2z2 4 = x4 + y4 + z4 + x2y2z2 2(x2 + y2 + z2 + xyz) + 4= (x2 1)2 + (y2 1)2 + (z2 1)2 + (xyz 1)2 0

with equality if and only if x2 = y2 = z2 = 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 VergataUniversity, Rome, Italy. It is a known standard result that 1 cosA+ cosB+cosC 3/2 and then 2 2 cosA + 2 cosB + 2 cosC 3. The minimum 1 cor-responds to a degenerate isosceles triangle while the maximum to an equilateraltriangle. Clearly we have 2 x + y + z 3. We employ the so called uvwtheory which can be found at The art of problem solving forum. Define three newquantities

x+ y + z = 3u, xy + yz + zx = 3v2, xyz = w3

We have

x4+y4+z4+x2y2z2 = (w3)2+12uw3+81u4108u2v2+18v4 = (w3)2+12uw3+R(u, v)This is a convex increasing parabola if w3 0 whose minimum has negative ab-scissa. It follows that the minimum of the parabola occurs when w = 0 or when wis minimum once fixed the values of u and v. According to the theory, the latteroccurs when x = y (or cyclic). If w = 0 we have for instance z = 0 that is C = pi/2.At x + y fixed, the minimum of x4 + y4, occurs when x = y that is A = B = pi/4

or z = y =

2. This yields

x4 + y4 = 2x4 = 8

248

If z = y and x = a 2y, a 2 3 we get

x4 + y4 + z4 + (xyz)2 3(a

3

)4(a

3

)6=

(3y a)2729

P (y, a)

=(3y a)2

729(324y4 108ay3 + 1458y2 1620ay + 702a2 27a2y2 6a3y a4)

Now we prove that P (y, a) > 0. Indeed

702a2 a4 = a2(702 a2) a2(702 9) = 693a2, 6a3y = 6a2ay 54ay1011y2 + 693a2 > 1674ay = (1620 + 54)ay

so we get

324y4 + 442y2 108ay3 + 27a2y2and this is implied by

324y4 + 442y2 108 3y3 + 27 9y2 324y4 + 199y2 324y3

and this finally follows by the AGM 324y4 + 199y2 507y3. We have showed that

x4 + y4 + z4 + (xyz)2 3(a

3

)4(a

3

)6 0

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 = pi/6 whence x = y = z = 1 and

x4 + y4 + z4 + (xyz)2 4 0The searched minimum is thus min{8, 4} = 4.

Also solved by Jose Gibergans-Baguena, BARCELONA TECH, Barcelona,Spain.

59. Let x, y, z be nonzero complex numbers. Prove that

81

(1

|x|2 +1

|y|2 +1

|z|2)1

3|x+y+z|2+ |2xyz|2+ |2yxz|2+ |2zxy|2

(Training Catalonian Team for OME 2014)

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. More generally, consider n nonzero complexnumbers z1, . . . , zn, and let =

1n (z1 + . . .+ zn). Clearly we have

nk=1

|zk |2 =nk=1

|zk|2 2nk=1

249

Now, the HM -AM inequality shows that

n2

(nk=1

1

|zk|2)1

nk=1

|zk|2 (2)

Combining (1), (2), and rearranging, we obtain

n4

(nk=1

1

|zk|2)1

nnk=1

zk

2

+

nk=1

(n 1)zk j 6=k

zj

2

.

The proposed inequality corresponds to n = 3 and (z1, z2, z3) = (x, y, z).

Solution 2 by Jose Luis Daz-Barrero, BARCELONA TECH, Barcelona,Spain. We havex+ y + z3

2 + 13(2x y z3

2 + 2y x z32 + 2z x y3

2)

=(x+ y + z)(x+ y + z)

9

+1

3

[(2x y z)(2x y z)

9+

(2y x z)(2y x z)9

+(2z x y)(2z x y)

9

]=

1

3

(|x|2 + |y|2 + |z|2)Applying AM-HM inequality, yields

1

3

(|x|2 + |y|2 + |z|2) 31

|x|2 +1

|y|2 +1

|z|2and taking into account that the given inequality is equivalent to

31

|x|2 +1

|y|2 +1

|z|2 1

9(|x+ y + z|2)

+1

27

(|2x y z|2 + |2y x z|2 + |2z x y|2

)from which the statement follows. Notice that equality holds when |x| = |y| = |z|and we are done.

Solution 3 by Arkady Alt, San Jose, California, USA. Since

|x+ y + z|2 = (x+ y + z) (x+ y + z)= |x|2 + |y|2 + |z|2 + (xy + xy) + (yz + yz) + (zx+ zx) ,

|2x y z|2 = (2x y z) (2x y z)= 4 |x|2 + |y|2 + |z|2 2 (xy + xy) + (yz + yz) (zx+ zx) ,

|2y x z|2 = (2y z x) (2y z x)= |x|2 + 4 |y|2 + |z|2 2 (yz + yz) + (zx+ zx) (xy + xy) ,

|2z x y|2 = (2z x y) (2z x y)= |x|2 + |y|2 + 4 |z|2 2 (zx+ zx) + (xy + xy) (zx+ zx) ,

250

then

3 |x+ y + z|2 + |2x y z|2 + |2y z x|2 + |2z x y|2

= 3(|x|2 + |y|2 + |z|2

)+ 6

(|x|2 + |y|2 + |z|2

)= 9

(|x|2 + |y|2 + |z|2

)and the original inequality becomes

81

(1

|x|2 +1

|y|2 +1

|z|2)1

9(|x|2 + |y|2 + |z|2

)or

9

(1

|x|2 +1

|y|2 +1

|z|2)1

(|x|2 + |y|2 + |z|2

),

where latter inequality holds because by Cauchys inequality

(|x|2 + |y|2 + |z|2

)( 1|x|2 +

1

|y|2 +1

|z|2) 9

Also solved by Jose Gibergans-Baguena, BARCELONA TECH, Barcelona,Spain.

60. Compute the following sum:1i1n+1

1

i1+

1i1

251

Then by general Vietas Theorem

ak =

1i1

252

MATHNOTES SECTION

Calculating the limits of some realsequences

D.M.Batinetu-Giurgiu, Neculai Stanciu, Anastasios Kotronis

Abstract. In this note we present new methods to calculate the limits of someparticular sequences and their generalizations that appeared in some problem solv-ing journals.

1. Main Results

For this section we assume that:

(1) {an}n1 is a positive sequence such thatlim

n+ an = a R+ and lim

n+n (an+1 an) = b R,

(2) f : R+ R+ is a monotone nondecreasing function with a continuousderivative f , and that

(3) {xn}n1 is a positive sequence for which there exists a number t R suchthat

limn+

xn+1xnnt+1

= x R+.With the above assumptions, we can deduce the following results:

Proposition 1. limn+n (f(an+1) f(an)) = bf

(a).

Proof. For n N, on each interval [an, an+1], if an+1 6= an the function f satisfiesthe assumptions of Lagranges theorem, so there exists a real number n betweenan and an+1, such that f(an+1) f(an) = (an+1 an) f (n), (and this remainsvalid with n = an if an = an+1,) thus

n (f(an+1) f(an)) = n (an+1 an) f (n).Now, on account of 1 and 2, we have limn n = a, because limn an = a, andf is continuous, therefore,

limn+n (f(an+1) f(an)) = limn+n (an+1 an) limn+ f

(n) = bf (a),

which is the desired result.

Remark. Note that, we do not need the full strength of 2, we only need that fbe a real valued function defined on (0,+) with a continuous derivative. Thisremark will help us in the next proposition.

253

Proposition 2. limn+(f(an+1)f(an)

)n= e

bf(a)f(a) .

Proof. According to the preceding remark, we can Apply Proposition 1, to thefunction ln(f) to conclude that

limnn (ln f(an+1) ln f(an)) = b

f (a)f(a)

.

Now, the Proposition follows using the continuity of the exponential function.

Proposition 3. limn+

nt+1

nxn

= et+1

x .

Proof. Consider vn = nn(t+1)/xn. We have

vn+1vn

=(1 + n)(n+1)(t+1)xn

nn(t+1)xn+1=

xnxn+1

(1 + n)(n+1)(t+1)

nn(t+1)

=xnn

t+1

xn+1

(1 +

1

n

)1+t((1 +

1

n

)n)1+tThus, using 3, we have lim

nvn+1vn

= e1+t

x , and the proposition follows immediately

from the well-known property [17, p. 46]:

limn

vn+1vn

= ` = limn

nvn = `,

for every positive sequence {vn}n1.

For the following propositions we set un =n+1xn+1

(n+ 1)t n

t

nxn

for n 2.

Proposition 4. limn+un = 1, limn+u

nn = e, and lim

n+n(un 1) = 1.

Proof. Keeping the notation of the previous proof we have nvn = n

t+1/ nxn, thus

un =

(1 +

1

n

)nvn

n+1vn+1

So, from Proposition 3, we have limn un = 1. Also,

unn =

(1 +

1

n

)n vnvn+1

n+1vn+1and limn unn = e follows also from Proposition 3.Finally, since

n(un 1) = ln(unn) un 1lnun

,

we conclude immediately that limn+n(un 1) = 1.

Proposition 5. If for n 2, Cn =(

n+1xn+1

(n+ 1)t

nxnnt

), then

limn+Cn = xe

(t+1)

254

Proof. We have

Cn =n+1xn+1

(n+ 1)t

nxnnt

=nxn

nt+1 n(un 1),

so on account of Propositions 3 and 4:

limn+Cn = limn+

nxn

nt+1 limn+n(un 1) = xe

(t+1).

This concludes the proof of the Proposition.

Proposition 6. If for n 2, we set Bn =(f(an+1)

n+1xn+1

(n+ 1)t f(an)

nxnnt

),

then

limn+Bn =

x (f(a) + bf (a))et+1

.

Proof. We have

Bn = f(an)nxnnt

(f(an+1)

f(an)un 1

)= f(an)

nxnnt

(tn 1),

where tn :=f(an+1)f(an)

un.

Now, since limn+ tn4

== f(a)f(a) 1 = 1, we obtain limn+ tn1ln tn = 1, and onaccount of Propositions 2, 4:

limn+ t

nn = lim

n+

(f(an+1)

f(an)

)n limn+u

nn = e

bf(a)f(a) e = e

f(a)+bf(a)f(a) .

Writing

Bn = f(an)nxn

nt+1 tn 1

ln tnln tnn

we get

limn+Bn = f(a)

x

et+1 1 ln e f(a)+bf

(a)f(a) =

x (f(a) + bf (a))et+1

.

Remark. Proposition 6 may also be proved as follows:We have

Bn = f(an+1)n+1xn+1

(n+ 1)t f(an)

nxnnt

= f(an+1)n+1xn+1

(n+ 1)t f(an+1)

nxnnt

+ f(an+1)nxnnt f(an)

nxnnt

= f(an+1)

(n+1xn+1

(n+ 1)t

nxnnt

)+

nxn

nt+1 n (f(an+1) f(an))

and Propositions 1, 3 and 5 yield

limn+Bn = f(a)

x

et+1+

x

et+1bf (a) = x (f(a) + bf (a)) e(t+1).

255

2. Applications

On the following we preserve the notation of the previous section.

A1. Evaluate limn+(

n+1

(n+ 1)! nn!)

.

Solution: With xn = n!, n N, an a sequence satisfying the assumptionsin 1., and f : R+R+, f(x) = 1 we have

a = limn+ an, x = limn+

xn+1nxn

= limn+

(n+ 1)!

n!n= 1, t = 0

and from Proposition 6,

Bn =n+1

(n+ 1)! nn!,

so

limn+

(n+1

(n+ 1)! nn!)

= e1.

Bn is known as Traian Lalescus sequence (see [5]). A2. Evaluate limn+

(n+1n+ 1 n+1

(n+ 1)! nn nn!

).

Solution: With xn = n!, n N, an = nn, and f : R+R+, f(x) = x

we have

a = 1, x = limn+

xn+1nxn

= limn+

(n+ 1)!

n!n= 1, t = 0

and from Proposition 1:

b = limn+ (an+1 an)n = limn+

(n+1n+ 1 nn)n

= limn+

(((n+ 1) n+1

n+ 1 n nn) n+1n+ 1)

= 1 1 = 0.Now, from Proposition 6,

Bn =n+1n+ 1 n+1

(n+ 1)! nn n

n!,

so

limn+

(n+1n+ 1 n+1

(n+ 1)! nn n

n!)

= e1.

A3. Let {un}n1, {vn}n1 be real positive sequences, and c R with limn+ un+1nc+1un =

u R+ and limn+ vn+1ncvn = v R+. Calculate limn+(

n+1

un+1vn+1

n

unvn

).

(This application is a generalization of [1].)

Solution: We have

limn+

un+1vnnvn+1un

= limn+

un+1nc+1un

limn+

ncvnvn+1

=u

v R+

and taking f : R+R+, f(x) = 1, xn = unvn , n N and {an}n1 anysequence satisfying the assumptions in 1., we get

a = limn+ an, x = limn+

xn+1nxn

=u

v R+, t = 0

256

and

limn+

(n+1

un+1vn+1

nunvn

)=x

e(f(a) + bf (a)) =

x

e(1 + b 0)

=x

e=

u

ve.

For c = 2, with the notation in [1], it is un = bn, vn = an, n N, an, bn R+ and limn+

an+1n2an

= limn+bn+1n3bn

= a R+, so

limn+

(n+1

bn+1an+1

nbnan

)=

a

ae= e1

and we have solved [1]. A4. Let f, g : R R such that f(x) + g(x) = 1, x R and

Bn(f, g) = nf(x)

((n+1

(n+ 1)!)g(x)

(

nn!)g(x))

.

Calculate limn+Bn(f, g).

Solution: Setting wn(x) =

(n+1(n+1)!nn!

)g(x), we have

limn+wn(x) =

(lim

n+

(n+1

(n+ 1)!

n+ 1

) n

nn! n+ 1

n

)g(x)

=

(1

e e 1

)g(x)= 1

and limn+wn(x)1lnwn(x)

= 1. Furthermore,

limn+ (wn(x))

n=

(lim

n+

(n+1

(n+ 1)!nn!

)n)g(x)

=

(lim

n+(n+ 1)!

n! 1

n+1

(n+ 1)!

)g(x)

=

(lim

n+n+ 1

n+1

(n+ 1)!

)g(x)= eg(x).

Now we note that

Bn(f, g) = nf(x)

(nn!)g(x)

(wn(x) 1) = n(

nn!

n

)g(x)(wn(x) 1)

=

(nn!

n

)g(x) wn(x) 1

lnwn(x)ln (wn(x))

n

so

limn+Bn(f, g) =

(1

e

)g(x) 1 ln

(eg(x)

)= g(x)eg(x).

257

If f(x) = cos2 x, g(x) = sin2 x, x R, then with the notation in [2],

Bn(f, g) = Ln = ncos2x

((n+1

(n+ 1)!)sin2 x

(

nn!)sin2 x)

,

so limn+Bn(f, g) = limn+Ln = sin

2 x e sin2 x and the solution of [2]follows.

Remark: With the methods presented above, one can solve [1] to [16] and manyother problems from various math problem solving journals.

References

[1] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 24, Mathproblems Mathematical journal, Vol.1,

Issue 4, 2011, p.33, http://mathproblems-ks.com/?wpfb_dl=4

[2] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 67, Mathproblems Mathematical journal, Vol.3,Issue 2, 2013, p.140, http://mathproblems-ks.com/?wpfb_dl=10

[3] Maria Batinetu-Giurgiu, On Lalescu Sequences, Octogon Mathematical magazine, Vol.13,

No.1A, April 2005, pp.198-202,[4] D.M.Batinetu-Giurgiu, Siruri Lalescu, Revista Matematica duin Timisoara, Nr.1-2, 1985,

pp.33-38,

[5] Trian Lalescu, Problem 579, Gazeta Matematica, Vol.VI, 1900-1901, pp.33-38,[6] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 692, The Pentagon, Vol.71, No.1, 2011, p.54,

http://www.pentagon.kappamuepsilon.org/pentagon/Vol_71_Num_1_Fall_2011.pdf

[7] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 5208, School Science and Mathematics jour-

nal, April, 2012, p.1, http://www.ssma.org/Websites/ssma/images/Problems%20Section/

April-2012.pdf

[8] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 43, Mathproblems Mathematical journal, Vol.2,

Issue 3, 2012, p.91, http://mathproblems-ks.com/?wpfb_dl=7

[9] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 704, The Pentagon, Vol.71, No.2, 2012, p.42,http://www.pentagon.kappamuepsilon.org/pentagon/Vol_71_Num_2_Spring_2012.pdf

[10] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 11676, The American Mathematical Monthly,

Vol.119, No.9, November 2012, p.801,[11] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 3713, Crux Mathematicorum, Vol.38, No.2, Feb-

ruary 2012, p.63,

[12] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 715, The Pentagon, Vol.72, No.1, 2012, p.44,http://www.pentagon.kappamuepsilon.org/pentagon/Vol_72_Num_1_Fall_2012.pdf

[13] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 234, La Gazeta de la RSME, Vol.16, No.3, 2013,p.502, http://www.rsme.es/gacetadigital/english/vernumero.php?id=92

[14] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 2414, Revista Escolar de la Olimpiada Iberoamer-

icana de Matematica, No.49, 2013, p.502, http://www.oei.es/oim/revistaoim/numero49/Probs241_245.pdf

[15] D.M.Batinetu-Giurgiu, N.Stanciu, Problem 3764, Crux Mathematicorum, Vol.38, No.7, Sep-

tember 2012, p.285.[16] D.M.Batinetu-Giurgiu, N.Stanciu, Problem W3, Jozsef Wildt International Mathematical

Competition, Edition XXIII, 2013, p.285, Octogon Mathematical magazine, Vol.21, No.1,

April 2013, p.229, http://www.uni-miskolc.hu/~matsefi/Octogon/volumes/Wildt_2013_1.pdf

[17] W.J. Kaczor, M.T. Nowak Problems in Mathematical Analysis I, Real Numbers, Sequences

and Series, A.M.S., 2000.

D.M.Batinetu-Giurgiu: Department of Mathematics, Matei Basarab National College,Bucharest, Romania,

Neculai Stanciu: Department of Mathematics, Geogre Emil Palade General School,Buzau, Romania, stanciuneculai@yahoo.com,

Anastasios Kotronis: Athens, Greece, akotronis@gmail.com, www.asymmetry.gr.

258

JUNIOR PROBLEMS

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

Proposals21. Proposed by Dorlir Ahmedi, University of Prishtina, Republic of Kosova Ifa, b, c > 0 and ab+c 3/2 then prove that

a

b+ c+

b

c+ a+

c

a+ b 2

22. Proposed by Callegari Emanuele, Math. Dept. Tor Vergata University,Rome, Italy. 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 coverthe chessboard. How many are the different ways to do that?

23. Proposed by Callegari Emanuele, Math. Dept. Tor Vergata University,Rome, Italy. Mary has 5 baskets each containing 97 colored balls numbered from0 to 96. The color of the ballsin 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 upa 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?

24. Proposed by Stanescu Florin, Serban Cioculescu school, Gaesti, jud.Dambovita,Romania Prove that in a triangle ABC the following inequality holds:

27r

2p ra

a+rbb

+rcc p

2r

where ra, rb, rc are the lengths of the rays of the excircles, r is the radiusof the circle inscribed in the triangle, and p the semiperimeter of the triangle

25. Proposed by Proposed by Neculai Stanciu, George Emil Palade School, Buzau,Romania and D.M. Batinetu-Giurgiu, Matei Basarab National College, Bucharest,Romania. Find all pairs of real numbers (x, y) such that

x2 + 2x+ 1 +x2 4x+ 4 +

x2 2xy + y2 +

y2 6x+ 9 = 4.

259

Solutions16. Proposed by D.M. BatinetuGiurgiu, Matei Basarab National College,Bucharest, Romania, Neculai Stanciu, George Emil Palade School, Buzau, Ro-mania. Determine all real number x satisfying

1

x 3x+ 2 +1

xx =2

x 2x

Solutions by Codreanu Ioan Viorel , Maramures, Romania, Daniel Vacaru,Pitesti, Romania, Omran Kouba, Damascus, Syria and the proposers(independently). The existence conditions are

x > 0, x 3x+ 2 6= 0, xx 6= 0, x 2x 6= 0

x > 0 x 3x+ 2 = (x 2)(x 1) 6= 0 x 6= 4, x 6= 1

x > 0 xx 6= 0 x 6= 1, x 2x 6= 0 x 6= 4Thus if x (0,+)\{1, 4}, the l.h.s. of the equation becomes

1

(x 2)(x 1) +

1x(x 1) =

x+x 2

x(x 1)(x 2) =

=2(x 1)

x(x 1)(x 2) =

2x(x 2) =

2

x 2x

which equals the r.h.s.

Two incorrect solutions have been submitted.

17. Proposed by D.M. BatinetuGiurgiu, Matei Basarab;; National College,Bucharest, Romania, Neculai Stanciu, George Emil Palade School, Buzau, Ro-mania. Prove that the acute triangle ABC is equilateral if and only if

tan2A

sin2B + cos2 C+

tan2B

sin2 C + cos2A+

tan2 C

sin2A+ cos2B= 9

Solution by the authors. If A = B = C the triangle is equilateral and theequality trivially holds.Suppose now that the equality holds.

9 = U =cyc

tan2A

sin2B + cos2 C

=1

cyc(sin2B + cos2 C)

(cyc

(sin2B + cos2 C))

cyc

tan2A

sin2B + cos2 C

=1

3

(cyc

(sin2B + cos2 C))

cyc

tan2A

sin2B + cos2 C

260

and CauchySchwartzs inequality yields

9 = U 13

(tanA+ tanB + tanC)2

that is 27 = 3U (cyc

tanA)2. Since ABC is acute, the tangent is defined on

(0, pi/2) and then the convexity of tanx, x (0, pi/2), through Jensensinequalitygives

27 = 3U (

cyc

tanA

)2(

3 tanA+B + C

3

)2= 3 tan2

pi

3= 27

Equality holds if and only if A = B = C. The injectivity of the tangent in the given

domain imposes A = B = C =pi

3that is the triangle is equilateral.

Also solved by Titu Zvonaru, Comanesti, Romania and Omran Kouba,Damascus, Syria.

18.Proposed by Ercole Suppa, teramo, Italy Let K be the symmedian point of4ABC, let D = AK BC. Denote by P and Q the intersection points (differentfrom A) of AB and AC with the circumcircles of triangles 4ADC and 4ABDrespectively. Show that PQ is parallel to BC.

Solution by D.M. Batinetu-Giurghiu, Bucharest, Romania, Neculai Stan-ciu, Buzau, Romania and Titu Zvonaru, Comanesti, Romania.Using the power of the point B with respect to the circumcircle of triangle ADCwe obtain BP BA = BD BC, so

BP

BA=BD BCBA2

Similar from the power of the point C with respect to the circumcircle of triangleABD we obtain

CQ

CA=CD BCCA2

Since AD is the symmedian from A we have

BD

DC=AB2

AC2

From last three equation we have BPBA =CQCA , so PQ is parallel to BC.

Remark The equivalence

BP

BA=CQ

CA BD

DC=AB2

AC2

shows that the reverse is true, i.e. if PQ is parallel to BC then AD is the symmedianfrom A.

Also solved by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria, and the proposer.

261

19.Proposed by Armend Shabani, University of Prishtina, Republic of Kosova. Findall integer solutions of the equation 3x + x4 = 5x.

Solution by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria The answer is x {0, 2}.Indeed, if x is a negative integer then clearly 3x + x4 > 3x > 5x and x is not asolution. Moreover, if x is a nonnegative integer solution, then x must be even sinceboth 3x and 5x are odd integers in this case.It is clear that x = 0 and x = 2 are solutions. Now, if x 4 then

5x 3x > 67

5x (a)

6

7 5x > x4 (b)

Combining (a) and (b) we see that for x 4 we have 5x 3x > x4 and x is nota solution to the proposed equation. Thus {0, 2} are the only solutions to thisequation. Let us now prove (a) and (b):

(a) We have(53

)x ( 53)4 > 7, so that5x 3x = 6

75x +

(5x 7 3x

7

)>

6

75x.

Which is (a).

(b) Let an =67 5

n

n4 . Clearly, a4 > 1 and, for n 4, we havean+1an

= 5

(n

1 + n

)4 5

(4

1 + 4

)4=

256

125> 1.

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.Batinetu-Giurghiu, Bucharest, Romania, Neculai Stanciu, Buzau andTitu Zvonaru, Comanesti, 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+2 =anan+1an + an1

Find an for any n.

Solution by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. Let xn = an/an+1, with this notation we have

xn+1 = 1 + xn1. Hence

x2n x0 =nk=1

(x2k x2k2) = n

x2n+1 x1 =nk=1

(x2k+1 x2k1) = n

262

Thus, x2n = x2n+1 = n+ 1 for n 0. So, for n 1, we havea2n2a2n

= x2n2x2n1 = n2

Multiplying, we obtain

a2n =1

(n!)2, and a2n+1 =

a2nxn

=1

n! (n+ 1)! .Finally

an =1

bn/2!c dn/2!e ,where bc and de are the floor and ceiling functions respectively.

Mathproblems

ISSN: 2217-446X, url: http://www.mathproblems-ks.com

Volume 4, Issue 2 (2014), Pages 263-302

Editors: Valmir Krasniqi, Jose Luis Daz-Barrero, Armend Sh. Shabani,Paolo Perfetti, Mohammed Aassila, Mihaly Bencze, Valmir Bucaj, EmanueleCallegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba,Cristinel Mortici, Jozsef Sandor, Ercole Suppa, David R. Stone, Roberto Tauraso,Francisco Javier Garca Capitan.

PROBLEMS AND SOLUTIONS

Proposals and solutions must be legible and should appear on separate sheets, eachindicating the name of the sender. Drawings must be suitable for reproduction.Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourageundergraduate and pre-college students to submit solutions. Teachers can help byassisting their students in submitting solutions. Student solutions should includethe class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerningproposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail.com

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

Problems96. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca,Romania. Let p 1 be an integer and let x R. Prove that

n=1

np(ex 1 x

1! x

2

2! x

n

n!

)= ex

x0

Qp(t)dt,

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

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

c2010 Mathproblems, Universiteti i Prishtines, Prishtine, Kosove.263

264

that the Prasolov product is defined, provided U is the Nagel point of 4ABC andP is the Spieker center of 4ABC. The problem could be re-formulated as follow.Let Ua be the point at which the A-excircle meets the side BC of 4ABC, anddefine Ub and Uc 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. Provethat the lines ARa, BRb and CRc concur in a point.

98. Proposed by Anastasios Kotronis, Athens, Greece. Show that

n!

nn

(nk=0

nk

k!

+k=n+1

nk

k!

)=

4

3+O(n1).

99. Proposed by Li Yin, Department of Mathematics, Binzhou University, BinzhouCity, Shandong Province, 256603, China. Calculate

n=2

[4

e2

(1 +

1

n

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

(2n 1)(2n+ 1)

]

100. Proposed by D.M. Batinetu-Giurgiu, Matei Basarab National College,Bucharest, Romania, and Neculai Stanciu, George Emil Palade School, Buzau,Romania. Let (n)n1 be the sequence defined by n = lnn +

nk=1

1k and let

= limn n. Consider a continuous function f : (0,+) (0,). Find

limn

n

(2n 1)!! n

f(x)dx.

101. Proposed by Florin Stanescu, Serban Cioculescu school, city Gaesti, jud.Dambovita, Romania. Consider a real function f : [a, b] R (with a > 0,) havinga positive and increasing derivative. Show that for every positive integer n withn 2 the following inequality holds b

a

f(x)dx nn 1

((b a)(bnf(b) anf(a))

bn an bf(b) af(a)

n

).

102. Proposed by Marcel Chirita, Bucharest, Romania. Let f : (0,+) R be abounded continuous function. Suppose that the limit

limxx

|f(x+ 2) 2f(x+ 1) + f(x)| = a R,exists for some [0, 1]. Find the value(s) of a.

265

SolutionsNo problem is ever permanently closed. We will be very pleased considering forpublication new solutions or comments on the past problems.

88.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 a, b, c, 1)satisfies the following equality;

cyc

(a b1 ab

a

1 a2)

= 0

Prove that a = b = c.(Note that

cyc

means cyclic sumcyc

f(x, y, z) = f(x, y, z)+f(y, z, x)+f(z, x, y)).

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

cyc

1

1 a2 =cyc

1

1 abwhich is, by virtue 0 a, b, c < 1,

cyc

k=0

(a2)k =cyc

k=0

(ak)2 =cyc

k=0

(ab)k

We know that a2 + b2 + c2 ab+ bc+ ca, then(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. Withoutloss of generality, we may assume that a b c.Suppose that a = b, then

0 =cyc

a b1 ab

a

1 a2 =a c1 ac

a

1 a2 +c a1 ca

c

1 c2

= (a c) (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

1 a2 >b

1 b2 0, (1)1

1 ab >1

1 bc 0, (2)1

1 ca >1

1 bc 0. (3)

266

It follows from (1) thatcyc

a b1 ab

a

1 a2 >(

c

1 c2)cyc

a b1 ab . (4)

Using (2) and (3) we obtain,cyc

a b1 ab >

1

1 bccyc

(a b) = 0. (5)

Inequalities (4) and (5) imply thatcyc

ab1ab

a1a2 > 0, which contradicts the as-

sumption in the problem statement, hence a = b and we are done.

Solution 3 by Omran Kouba, Higher Institute for Applied sciences andTechnology, Damascus, Syria. For a, b, c [0, 1) we consider

F (a, b, c) =a b1 ab

a

1 a2 +b c1 bc

b

1 b2 +c a1 ca

c

1 c2Noting that

x y1 xy

x

1 x2 =(1 xy) (1 x2)

(1 xy)(1 x2) =1

1 x2 1

1 xy

=

n=1

(x2n xnyn)

we conclude that

F (a, b, c) =

n=1

(a2n + b2n + c2n anbn bncn cnan)

=1

2

n=1

((an bn)2 + (bn cn)2 + (cn an)2)

12

(a b)2 + (b c)2 + (c a)2

So, if F (a, b, c) = 0 then a = b = c.

Also solved by Arkady Alt, San Jose, California, USA; D.M. Batinetu-Giurgiu, Matei Basarab National College, Bucharest, Romania andNeculai Stanciu, George Emil Palade School, Buzau, 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 decimalrepresentation. Prove that

nS

1

n< +.

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 theharmonic 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 9i1 : Thereare 8 choices for the first digit, excluding 0 and 7, and 9 choices for the remainingi 1 digits.

267

Furthermore, each number in Si is of the form 1/m, where m is an i-digit number.So m 10i1, which implies that 1/m 1/10i1. Therefore

S =

i=1

Si i=1

8 9i110i1

= 8

i=1

(9

10

)i1= 80,

so S converges by comparison.

Remark: This method can be used to show convergence of the sum of reciprocalsof 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 disjointsubsets {Sk},

S =

k=1

Sk,

where Sk := S [10k1, 10k

). Define the sequence {ak}k1 by

ak =nSk

1

n.

Clearly,k=1 ak =

nS

1n .

Now we show by mathematical induction that the number of terms in Sk is lessthan 9k.The number of terms in S1 is 8 < 9

1.Suppose that it is true that the number of terms in Sk is less than 9

k, then we haveto show that the number of terms in Sk+1 is less than 9

k+1. To show this, we splitthe set Sk+1 into nine intervals

Sk+1 =

9=1

S [10k, 10k + 1, . . . , (+ 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 numberof terms of the interval S [1, 10k) . By the induction hypothesis, it is less than9 + 92 + + 9k.Hence, the number of terms in Sk+1 is less than 8

(9 + 92 + + 9k) < 9k+1.

Each term in Sk+1 is not less than 10k, therefore ak+1 =

nSk+1

1n 1which evidently holds true. This implies that bn+ 1c = bnc. Thus we have

bncq=0

min{(q+1)21,n}k=q2

(1)q =bn+11c

q=0

(1)q(q+1)21k=q2

1 +

bncq=bn+11c+1

(1)qn

k=q2

1 =

=

bn+11cq=0

(1)q(2q + 1) + (1)bnc(n (bnc)2 + 1) =

= (1)bn+11c(bn+ 1 1c+ 1) + (1)b

nc(n (bnc)2 + 1) =

= (1)bn+11cbn+ 1c+ (1)b

nc(n (bnc)2 + 1) =

= (1)bn+1cbn+ 1c+ (1)b

nc(n (bnc)2 + 1)

Since bn+ 1c = bnc, we have

270

(1)bn+1cbn+ 1c+ (1)bnc(n (bnc)2 + 1) bn+ 1c+ (n (bnc)2 + 1) = n+ 1 (n (bnc)2 + 1) n+ 1 1

and the last step is to show

n+ 1 1 dne

Recall that n = p2 r thus we need to showp2 r + 1 dne = 1 + bnc = 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; andthe proposer.

91.Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca,Romania.Calculate 1

0

10

ln(1 x) ln(1 xy)dxdy.

Solution 1 by Omran Kouba, Higher Institute for Applied sciences andTechnology, Damascus, Syria. The answer is 3 2(3).Let the considered integral be denoted by I. Since 1

0

ln(1 xy)dy =[ (1 xy)

xln(1 xy) y

]y=1y=0

= (1 x)x

ln(1 x) 1

we see that

I =

10

(ln2(1 x) ln(1 x) ln

2(1 x)x

)dx

=

10

(ln2 x lnx ln

2 x

1 x)dx

Now, noting that (3x 3x lnx+ x ln2 x) = ln2 x lnx

we see that 10

(ln2 x lnx) dx = 3

Also, since 10

xn ln2 xdx =

0

t2e(n+1)tdt =(3)

(n+ 1)3=

2

(n+ 1)3

271

we see that 10

ln2 x

1 x dx =n=0

10

xn ln2 xdx

=

n=0

2

(n+ 1)3= 2(3)

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

n1

(1

n 1n+ k

)= 1 +

1

2+

1

3+ + 1

k= Hk (1)

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

I : =

10

10

ln(1 x) ln(1 xy) dx dy = 10

10

k1

(xy)k

kln(1 x) dx dy

= 10

k1

yk

k

10

xk ln(1 x) dx dy = 10

k1

yk

k

10

xkn1

xn

ndx dy

=

10

k1

yk

k

n1

1

n

10

xn+k dx dy =

10

k1

n1

yk

nk(n+ k + 1)dy

=k1

n1

1

nk(n+ k + 1)

10

yk dy =k1

n1

1

nk(k + 1)(n+ k + 1)

=k1

n1

(1

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

)=k1

1

k(k + 1)2

n1

(1

n 1

(n+ k + 1)

)

= k1

(1

k + 1 1k

+1

(k + 1)2

)n1

(1

n 1

(n+ k + 1)

)

= k1

(1

k + 1 1k

)n1

(1

n 1

(n+ k + 1)

)+ 1

k1

1

k2

n1

(1

n 1n+ k

): = A+ 1B

For A, from (1) and summing by parts we have:

A = k1

(1

k + 1 1k

)Hk+1 = Hk+1

k

+1

+k1

1

k + 1(Hk+2 Hk+1)

=3

2+k1

(1

k + 1 1k + 2

)= 2

272

For B, from (1),

B =k1

Hkk2

and furthermore

B =k1

1

k2

n1

(1

n 1n+ k

)

=k1

n1

1

nk(n+ k)=k1

n1

(1

n(n+ k)2+

1

k(n+ k)2

)= 2

k1

n1

1

n(n+ k)2n+k1=N====== 2

N1

n,k1

n+k1=N

1

n(n+ k)2

= 2N1

Nn=1

1

n(N + 1)2= 2

N1

HN(N + 1)2

= 2N1

HN+1 1N+1(N + 1)2

= 2N1

HN+1(N + 1)2

2N1

1

(N + 1)3= 2(B 1) 2((3) 1)

where is the Riemann zeta function.So B = 2B 2(3) and hence B = 2(3). Finally I = 3 2(3) 0.595886.

Solvers Note: A reference to the details presented here is M. S. Klamkin,Amer.Math. Monthly, 59 (1952) pp. 471472. See also page 6 of the Collected Contri-butions 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 : 1

0

tn ln(1 t) dt =[

(tn+1 1) ln(1 t)n+ 1

]10

1n+ 1

10

tn+1 1t 1 dt

=1n+ 1

nk=0

10

tk dt =Hn+1n+ 1

(1)

Where Hn is the n-th harmonic number, therefore

I =

10

10

ln(1x) ln(1xy) dxdy = k=1

10

10

ykxk

kln(1x) dxdy =

k=1

Hk+1k(k + 1)2

.

Now, we can simplify the latter series to be :

A =

k=1

Hk+1k(k + 1)2

=

( k=1

Hk+1k Hk+1k + 1

)( k=2

Hkk2

).

The first sum is easy to calculate :

nk=1

Hk+1k Hk+1k + 1

=

nk=1

(1

k(k + 1)+Hkk Hk+1k + 1

)= 2 1

n+ 1 Hn+1n+ 1

.

273

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

A =

k=1

Hk+1k Hk+1k + 1

= limn+ 2

1

n+ 1 Hn+1n+ 1

= 2.

To calculate the other sum we use the following result for m > 0: 10

xm ln2 x dxx=et/(m+1)

=1

(m+ 1)3

0

t2et dt =2

(m+ 1)3(2)

From (1) we get :

B =

k=1

Hkk2

= k=1

10

tk1

kln(1 t) dt =

10

ln2(1 t)t

dt =

10

ln2 t

1 t dt.

Using the result (2) we get

B =

k=0

tk ln2 t dt =

k=0

2

(k + 1)3= 2(3).

Now, it is clear that :

I = A (B 1) = 3 2(3).

Also solved by Arkady Alt, San Jose, California, USA; AN-anduud Prob-lem Solving Group, Ulaanbaatar, Mongolia; B.C. Greubel, NewportNews, VA, USA; Moti Levy, Rehovot, Israel; Paolo Perfetti, Depart-ment of Mathematics, Tor Vergata University, Rome, Italy; and theproposer.

92.Proposed by D.M. Batinetu-Giurgiu, Matei Basarab National College, Bucharest,Romania, and Neculai Stanciu, George Emil Palade School, Buzau, Romania.Let {an}n0 be a sequence of positive integer numbers such that 5 does not dividean for all, n N, and let the sequence {bn}n0 be defined by bn = a2nL2n , for n Nwhere {Ln}n0 is the sequence of Lucas numbers. Prove that bn is not a perfectsquare, for every n N \ {1}.

Solution 1 by Moti Levy, Rehovot, Israel. By its definition, bn is a perfectsquare 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 ork = 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 ofm 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.Lets take a look on mod(Ln, 5), Heres a lemma :

mod(Ln, 5) =

2 if mod(n, 4) = 0;

1 if mod(n, 4) = 1;

3 if mod(n, 4) = 2;

4 if mod(n, 4) = 3.

(lemma 1)

Proof : for n {0, 1, 2, 3} it is clearly true. suppose it is true for some integers4k, 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 isnot a perfect square.For n > 1, we have 2n = 4k where k N>0, therefore : mod(L2n , 5) = 2, whichimplies 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 andTechnology, Damascus, Syria; and the proposer.

93.Proposed by Anastasios Kotronis, Athens, Greece. For x (1, 1), evaluate

+n=1

(1)n+1n(

tan1 x x+ x3

3 + (1)n+1 x

2n+1

2n+ 1

).

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria.Let

fn(x) = tan1 x x+ x

3

3 + (1)n+1 x

2n+1

2n+ 1

Clearly we have

f n(x) =1

1 + x2

2nk=0

(x2)k = 11 + x2

1 (x2)n+1

1 + x2=

(1)n+1x2n+21 + x2

Thus

fn(x) = (1)n+1 x0

t2n+2

1 + t2dt

275

It follows thatn=1

(1)n+1nfn(x) = 12

x0

t3

1 + t2

( n=1

(2n)t2n1)dt

=1

2

x0

t3

1 + t2

( n=0

t2n

)dt

=1

2

x0

t3

1 + t2

(1

1 t2)dt

=

x0

t4

(1 + t2)(1 t2)2 dt

Finally, noting that

t4

(1 + t2)(1 t2) =1

8

(1

(1 + t)2+

1

(1 t)2 2

1 t 2

1 + t+

2

1 + t2

)we conclude that

n=1

(1)n+1nfn(x) = 14

(x

1 x2 + ln(

1 x1 + x

)+ arctanx

)which is the desired conclusion.

Solution 2 by Paolo Perfetti, Department of Mathematics, Tor VergataUniversity, Rome, Italy. Clearly

n=1

(1)n+1n(

tan1 x x+ x3

3 . . .+ (1)n+1 x

2n+1

2n+ 1

)=

=

n=1

(1)n+1n

k=n+1

(1)k+1 x2k+1

2k + 1

Since |x| < 1, we can rearrange the series asn=1

(1)n+1n

k=n+1

(1)k+1 x2k+1

2k + 1=

k=2

(1)k x2k+1

2k + 1

k1n=1

(1)nn

k1n=1

(1)nn is equal to k/2 if k is even and (k 1)/2 if k is odd. It follows

k=2

(1)k x2k+1

2k + 1

k1n=1

(1)nn =

= k=1

kx4k+1

4k + 1 1

2

k=2

(2k 1) x4k1

4k 1 +1

2

k=2

x4k1

4k 1 =

= k=1

x4k+1 +1

4

k=1

x4k+1

4k + 1 1

4

k=2

x4k1 +3

4

k=2

x4k1

4k 1 .

276

k=1

x4k+1 =x5

1 x4 .

k=1

x4k+1

4k + 1=

k=1

x4k+1 10

y4kdy =

= x

10

x4y4

1 x4y4 dy = x0

y4

1 y4 dy = x+1

4ln

1 + x

1 x +1

2arctanx.

Moreover

k=2

x4k1 =x7

1 x4 .

k=2

x4k1 10

y4k2dy =1

x

10

x8y6

1 (xy)4 dy = x0

t6

1 t4 dt =

=

x0

(1

2 1

2

1

1 + t2+

1

4

t2

1 + t+

1

4

1

1 t t+ 1

4 t2

)dt =

=x

2 1

2arctanx+

x2

8 x

4+

1

4ln(1 + x) 1

4ln(1 x) x

2

8 x

4 x

3

3=

12

arctanx+1

4ln(1 + x) 1

4ln(1 x) x

3

3.

By summing up the three contributions we obtain

14

x5

1 x4 x

4+

1

16ln

1 + x

1 x +1

8arctanx 1

4

x7

1 x4 +

+3

4

(1

2arctanx+

1

4ln(1 + x) 1

4ln(1 x) x

3

3

)=

= 14

x

1 x2 1

4arctanx 1

4ln

1 x1 + x

.

Solution 3 by Arkady Alt, San Jose, California, USA.

Let S (x) :=+n=1

(1)n+1 n(

tan1 (x)(x x

3

3+ ...+ (1)n x

2n+1

2n+ 1

)).

Then S (x) =+n=1

(1)n+1 n(

1

1 + x2 (1 x2 + ...+ (x2)n)) =

+n=1

(1)n+1 n(

1

1 + x2 1

(x2)n+11 + x2

)=

+n=1

nx2(n+1)

1 + x2=

x4

1 + x2

+n=1

n(x2)n1

=

x4

1 + x2 1

(1 x2)2 =1

4 (x 1) 1

4 (x+ 1)+

1

8 (x 1)2 +1

8 (x+ 1)2 +

1

4 (x2 + 1)

and, therefore, S (x) = x0

t4dt

(1 + t2) (1 t2)2 =1

4

(ln

1 t1 + t

+t

1 t2 + tan1 (t)

)x0

=

1

4

(ln

1 x1 + x

+x

1 x2 + tan1 (x)

).

277

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Haroun Meghaichi, University of Science and Techonology, HouariBoumediene, Algeria; Moti Levy, Rehovot, Israel; and the proposer.

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

|AP |2b2

+|BP |2c2

+|CP |2a2

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 n be n arbitrary points on a plane. Let w1, w2, . . . , wnbe n positive weights, such that

nk=1 wk = 1. Let

c denote the weighted averagevector defined by

c :=n

m=1

wmr m.

Then the following holds:

nm=1

wm |z r m|2 =n

m=1

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

Proof.

wm |z r m|2 = wm (z r m) (z r m) = wm(|z |2 + |r m|2 2z r m

)wm |c r m|2 = wm (c r m) (c r m) = wm

(|c |2 + |r m|2 2c r m

)wm |z r m|2 wm |c r m|2 = wm

(|z |2 |c |2

)+ 2wm (

c z ) r mSumming,

nm=1

wm |z r m|2 n

m=1

wm |c r m|2

=(|z |2 |c |2

) nm=1

wm + 2 (c z )

nm=1

wmr m

= |z |2 |c |2 + 2 (c z ) c= |z |2 |c |2 + 2 |c |2 2z c= |z |2 2z c + |c |2 = |c z |2 .

This proves the lemma.

It follows from the lemma that the vector which minimizes the weighted sumnm=1 wm |z r m|2 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 theweights be:

wA =1b2

1a2 +

1b2 +

1c2

=c2a2

a2b2 + a2c2 + b2c2

wB =a2b2

a2b2 + a2c2 + b2c2

wC =b2c2

a2b2 + a2c2 + b2c2,

then the original problem becomes:Find a vector r P which minimizes the weighted sum

wA |r P r A|2 + wB |r P r B |2 + wC |r P r C |2 .The answer is r P = wAr A + wBr B + wCr C .If the triangle is equilateral then the point P is the centroid.

Solution 2 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. We will use the following Lemma:

Lemma 2. Consider n points A1, . . . , An in the plane P, and let 1, . . . , n be npositive numbers. We consider, the real function f : P R defined by

f(M) =

nk=1

k|AkM |2

Let also G be the barycenter of the wighted points ((Ak;k))1kn, that is G is theunique point G defined by

nk=1

kGAk = ~0,

then G is the unique point in the plane P where f attains its minimum.Proof. Indeed,

f(M) f(G) =nk=1

k

(AkM

2 AkG2)

=

nk=1

k

(AkM AkG

)(AkM +

AkG

)=

nk=1

kGM

(GM + 2

AkG

)=

(nk=1

k

)|GM |2 + 2GM

nk=1

kAkG

=

(nk=1

k

)|GM |2.

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

279

In the proposed problem we have n = 3, A1 = A, A2 = B, A3 = C and 1 = 1/b2,

1 = 1/c2 and 1 = 1/a

2. Thus the desired minimum is attained at the uniquepoint P which is the barycenter of the weighted points (A; 1b2 ), (B;

1c2 ), and (C;

1a2 ).

This point is the first Brocard point , it is the unique point inside ABC such that

AB = BC = CA.

(when the triangle ABC is labeled in counterclockwise order.)

Also solved by the proposer.

95.Proposed by Li Yin, Department of Mathematics, Binzhou University, BinzhouCity, Shandong Province, 256603, China. An approximation formula of Wallisproduct. For all n N, then

Wn 1pi

(1 lnn

2n

)nwhere Wn :=

(2n1)!!(2n)!! is Wallis product.

Solution 1 by Paolo Perfetti, Department of Mathematics, Tor VergataUniversity, Rome, Italy. We use

ln(1 x) = x 12x2 + o(x2), ex = 1 + o(1), n! (n/e)n

2pin(1 + o(1))

We get

1pi

(1 lnn

2n

)n=

1pi

exp

{n ln

(1 lnn

2n

)}=

1pi

exp

{n

( lnn

2n+ o(

1

n)

)}=

1pi

exp

{ lnn

2+ o(1)

}=

1pin

(1 + o(1))

Wn =(2n)!

22n(n!)2=

(2n)2ne2n

2pi

2n(1 + o(1))

22nn2ne2n2pin=

1pin

(1 + o(1))

It follows

limn

Wn1pi

(1 lnn2n

)n = 1Solution 2 by Omran Kouba, Higher Institute for Applied Sciences andTechnology, Damascus, Syria. On one hand, using Stirlings formula, we have

Wn =(2n)!

22n(n!)2 2pin(2n)2ne2n

22n(2pin)n2ne2n=

1pin

, (1)

280

and on the other hand(1 lnn

2n

)n= exp

(n ln

(1 lnn

2n

))= exp

(n

( lnn

2n+O

(ln2 n

n2

)))= exp

( lnn

2+O

(ln2 n

n

))=

1n

exp

(O(

ln2 n

n

))=

1n

(1 +O

(ln2 n

n

))Thus

1pi

(1 lnn

2n

)n 1

pin, (2)

and the desired conclusion follows from (1) and (2).

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Haroun Meghaichi, University of Science and Techonology, HouariBoumediene, Algeria.

281

MATHCONTEST SECTION

This section of the Journal offers readers an opportunity to solve interesting and el-egant mathematical problems mainly appeared in Math Contest around the worldand most appropriate for training Math Olympiads. Proposals are always wel-comed. 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 theform 1a1b1c1d1 are multiples of 33?

66. Given trapezoid ABCD with parallel sides AB and CD, let E be a point on lineBC outside segment BC, such that segment AE intersects segment CD. Assumethat there exists a point F inside segment AD such that EAD = CBF . Denoteby I the point of intersection of CD and EF , and by J the point of intersectionof AB and EF . Let K be the midpoint of segment EF , and assume that K isdifferent from I and J . Prove that K belongs to the circumcircle of 4ABI if andonly if K belongs to the circumcircle of 4CDJ67. For all positive real numbers a, b, c, d prove the inequality

a4 + c4 +a4 + d4 +

b4 + c4 +

b4 + d4 2

2(ad+ bc)

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

1T

2T

3T

4T

5T

The length of the side of the smallest triangles is 1. A triangle is called a delta ifits vertex is at the top; for example, there are 10 deltas in T3. A delta is said to beperfect 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 squaresas in the diagram below.

1 2 3 4 5 6 7 89 10 11 12 13 14 15 1617 18 19 20 21 22 23 2425 26 27 28 29 30 31 3233 34 35 36 37 38 39 4041 42 43 44 45 46 47 4849 50 51 52 53 54 55 5657 58 59 60 61 62 63 64

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

283

Solutions61. Find all real solutions of the following system of equations:

x2 + y2 + 6x+ 9 +x2 + y2 8y + 16 = 5,

9y2 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 thosethat the sum of their distances to A(3, 0) and B(0, 4) is equal to 5. Moreover, ifa point P lies out of the segment AB then AP + PB > AB = 5. This let us toconclude that points (x, y) solution of the system must lie on AB. The equation of

AB is y =4

3x + 4 or

(x,

4

3x+ 4

)with 3 x 0. Substituting these values in

the second equation, yields

9

(4

3x+ 4

)2 4x2 = 60 x2 + 8x+ 7 = 0

with roots x = 7 and x = 1. Since only the second lie in [3, 0], then the uniquesolution of the given system is (1, 8/3).

Solution 2 by Arkady Alt, San Jose, California, USA. Squaring both sides

of the equationx2 + y2 + 6x+ 9 +

x2 + y2 8y + 16 = 5 we have(

x2 + y2 + 6x+ 9 +x2 + y2 8y + 16

)2= 25 4x 3y + 12 = 0

Then, from

4x 3y + 12 = 09y2 4x2 = 60

} 3y = 4x+ 12

(4x+ 12)2 4x2 = 60

} 3y = 4x+ 12

12 (x+ 7) (x+ 1) = 0

}we obtain

(x, y) =

(1, 8

3

)(x, y) =

(7, 16

3

)By substitution immediately follows that only (x, y) =

(1, 8

3

)satisfies the given

system and it is the desired solution.

Also solved by Jose Luis Daz-Barrero, BARCELONA TECH, Barcelona,Spain.

62. Let P be an interior point to an equilateral triangle ABC. Draw perpendicularsPX,PY and PZ 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 andTechnology, Damascus, Syria. First let us denote the side length of thetriangle by a. The area of the triangle can be calculated in two ways and we get32 a

2 = a(PX + PY + PZ), hence

PX + PY + PZ =

3

2a (1)

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)

~0

BP +CA CB +AB AB

=CA CB +AB AB

= a21

2+ a2

so

BX + CY +AZ =3

2a (2)

Thus, from (1) and (2) we get

BX + CY +AZ

PX + PY + PZ=

3.

Solution 2 by Jose Luis Daz-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 onaccount of Pithagoras theorem, we have

AZ2 + ZP 2 = (a CY )2 + PY 2

BX2 + PX2 = (aAZ)2 + PZ2

CY 2 + PY 2 = (aBX)2 + PX2Developing and adding up, yields

BX + CY +AZ =3a

2

On the other hand the sum of the areas of triangles APB,BPC,CAP is the areaof 4ABC. That is,

a(PX + PY + PZ)

2=a2

3

4 PX + PY + PZ = a

3

2

From the preceding immediately follows that

BX + CY +AZ

PX + PY + PZ=

3

and we are done.

285

Solution 3 by Arkady Alt, San Jose, California, USA. Let a = BC = CA =

AB, x = BX, y = CY, z = AZ, u = PX, v = PY,w = PZ and h =a

3

2be height

of the equilateral triangle ABC, then

[ABC] = [PBC] + [PCA] + [PAB] ah2

=au

2+av

2+az

2 u+ v + w = h

andBX + CY +AZ

PX + PY + PZ=x+ y + z

u+ v + w=x+ y + z

h=

2 (x+ y + z)

a

3Applying Pythagorean theorem to chain of right triangles4PXB,4PBZ,4PZA,4PAY,4PXB,4PY C,4PCX, we obtain

u2 + x2 = w2 + (a z)2w2 + z2 = v2 + (a y)2v2 + y2 = u2 + (a x)2

Adding all equations we getcyc

(u2 + x2

)=cyc

(w2 + (a z)2

) 3a2 = 2a (x+ y + z)

S0, x+ y + z =3a

2and, therefore,

BX + CY +AZ

PX + PY + PZ=

3.

Also solved by Jose Gibergans-Baguena, BARCELONA TECH, Barcelona,Spain.

63. How many ways are there to weigh of 31 grams with a balance if we have 7weighs of one gram, 5 of two grams, and 6 of five grams, respectively?

(Training Catalonian Team for OME 2014)

Solution 1 by Jose Luis Daz-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 coefficientof 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 andTechnology, 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(85) = 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 computethe 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} .Since in integers 24 2y+5z 31 242y 5z 312y

[28 2y

5

]

z [

31 2y5

]and for any 0 y 5 holds

[31 2y

5

] 6,

[28 2y

5

]> 0

then, denoting t := 5 y, we obtain[

31 2y5

]=

[21 + 2t

5

]= 4 +

[2t+ 1

5

],[

28 2y5

]=

[18 + 2t

5

]= 3 +

[2t+ 3

5

]and

S =

{(31 2y 5z, 5 t, z) | t, z Z and 0 t 5, 3 +

[2t+ 3

5

] z 4 +

[2t+ 1

5

]}.

Hence, |S| =5t=0

(2 +

[2t+ 3

5

][

2t+ 1

5

])= 12 +

5t=0

([2t+ 1

5

][

2t+ 3

5

]).

Noting that

5t=0

([2t+ 1

5

][

2t+ 3

5

])=

5t=0

[2t+ 1

5

]

6t=1

[2t+ 1

5

]=[

2 0 + 15

][

2 6 + 15

]=

[13

5

]= 2 we get |S| = 12 2 = 10.

Also solved by Jose Gibergans-Baguena, 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 andTechnology, Damascus, Syria.

287

Remarks:

If n 3 no such polynomial exists. Indeed, if a polynomial P has integercoefficients then for every distinct integers a and b, we have (ba) | (P (b)P (a)). So, when n 3, if there is a polynomial A with integer coefficientssuch that A(1) = 5 and A(4) = 54, then 3 must divide A(4) A(1) = 620which 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 hasinteger coefficients and that it satisfies A(k) = 5k for 1 k n + 1.Indeed, when n = 0 the polynomial A(X) = (X 1) + 5(2X), (for anarbitrary Z,) satisfies A(1) = 5 and A(2) = . Similarly, when n = 1,the polynomial A(X) = 20X 15 + (X 1)(X 2) satisfies A(1) = 5,A(2) = 25 and takes an arbitrary odd value at X = 3. A similar conclusionalso 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 suchthat 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, fromgeneral theorems about Lagrange interpolation. Now, given An(X) for some n > 0we consider

Qn(X) =1

4(An(X + 1)An(X))

Clearly degQn = n 1, and Qn(k) = 5k for 1 k n. Hence Qn(X) = An1(X).This allows us to conclude that

An(X + 1)An(X) = 4An1(X).Let bn = An(n+ 2), from the above formula we see that

bn = 5n+1 + 4bn1

This is equivalent to

bn4n+1

bn14n

=

(5

4

)n+1.

Adding these equalities and noting that b0 = 5 we see that

bn4n+1

54

=

nk=1

(5

4

)k+1Thus,

bn = 4n+1

nk=0

(5

4

)k+1= 5(5n+1 4n+1).

which is the desired conclusion. Solution 2 by Jose Gibergans-Baguena, BARCELONA TECH, Barcelona,Spain. We observe that for all k 1, holds:

5k = (1 + 4)k =

kj=0

(k

j

)4j =

(k

0

)+

(k

1

)4 + . . .+

(k

k

)4k

288

Now, we consider the polynomial

B(x) = 5

(1 +

(x 1

1

)4 +

(x 1

2

)42 + . . .+

(x 1n

)4n),

where (a

n

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

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 wehave

A(n+ 2) = B(n+ 2) = 5

(1 +

(n+ 1

1

)4 +

(n+ 1

2

)42 + . . .+

(n+ 1

n

)4n)

= 5

(1 +

(n+ 1

1

)4 +

(n+ 1

2

)42 + . . .+

(n+ 1

n

)4n +

(n+ 1

n+ 1

)4n+1

(n+ 1

n+ 1

)4n+1

)= 5(5n+1 4n+1)

Also solved by by Arkady Alt, San Jose, California, USA and Jose LuisDaz-Barrero, BARCELONA TECH, Barcelona, Spain.

65. Let a0, a1, . . . , an and b0, b1, , bn be complex numbers. If n 2, then provethat

Re

(nk=0

akbk

) 1

2n

(n+ 1

n

nk=0

|ak|2 + n2

n 1(

2n 2n

) nk=0

|bk|2)

(Training UPC Team for IMC 2014)

Solution by Jose Luis Daz-Barrero, BARCELONA TECH, Barcelona,Spain. We begin with the following claim:

Let , a0, a1, . . . , an and b0, b1, , bn be complex numbers, then it holds

Re

(

nk=0

akbk

) 1

2

(nk=0

|ak|2 + ||2nk=0

|