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Solutions for Mathematical Reflections 6(2006)

Juniors

J31. Find the least perimeter of a right-angled triangle whose sides and

altitude are integers.

Proposed by Ivan Borsenco, University of Texas at Dallas

Solution by Jos Alejandro Samper Casas, Colegio Helvetia de Bogota,Colombia

Solution. The answer for the least possible perimeter is 60. It holdsfor a right-angled triangle (15, 20, 25),whose altitude is 12.

Letx, y,zbe a Pythagorean triple withzthe hypotenuse and leth bethe altitude of the triangle. Let d = gcd(x,y,z) be the greatest commondivisor of x,y,z. We have that x = d a , y = d b and z = d c fora,b,c a primitive Pythagorean triple. Now calculating the area of thetriangle in two ways we obtain that h = xy

z = abd

c . Using the fact that

gcd(ab,c) = 1,we getc|dwhich tells us thatc

d,since both are positive

integers. Because the perimeter is equal to d(a+b+c) c(a+b+c),we can minimize it be taking c = d. Then the altitude of a right-angledtriangle having sides ca,cb,c2 (with c2 the hypotenuse) is ab and is aninteger.

We know that a,b,c is a primitive Pythagorean triple if and only ifthere exist m, n Z+ such that gcd(m, n) = 1,m n (mod 2) m > n >0that satisfy a = m2 n2, b = 2mn, c = m2 +n2. Replacing in (1), wenotice that all we need to find is the minimum value of

p= (m2 +n2)(2m2 + 2mn).

Clearlym > n >0, therefore m 2 andn 1. Thusp (22 + 1)(2 22 + 2 2 1) = 60.

Now the triangle with sides (15, 20, 25) satisfies all the conditions of theoriginal problem and its perimeter is 60. The problem is solved.

Also solved by Ashay Burungale, India

Mathematical Reflections 1 (2007) 1

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J32. Leta and b be real numbers such that

9a2 + 8ab+ 7b2 6.Prove that 7a+ 5b+ 12ab 9.

Proposed by Dr. Titu Andreescu, University of Texas at Dallas

First solution by Ashay Burungale, India

Solution. We start from the initial inequality

9a2 + 8ab+ 7b2 6.By making simple transformations the inequality is equivalent to

2(a b)2 + 7(a 12

)2 + 5(b 12

)2 + 7a+ 5b+ 12ab 9.

Clearly from this we can derive that

7a+ 5b+ 12ab

9.

Second solution by Daniel Campos Salas, Costa Rica

Solution. Leta= x +1

2 and b = y +

1

2. The given condition turns

to

9x + 9x2 + 4x + 4y+ 8xy+ 7y+ 7y2 = 13x + 9x2 + 11y+ 7y2 + 8xy 0.The result follows from the following chain of inequalities:

7a+ 5b+ 12ab = 7x+12 + 5y+12 + 12x+12y+12= 9 + 13x+ 11y+ 12xy

9 + 13x+ 11y+ 8xy+ 4x2 +y2 9 + (13x+ 9x2 + 11y+ 7y2 + 8xy) 9,

and we are done.


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J33. Consider the sequence: 31, 331, 3331,... whose nth term hasn 3s followed by a 1. Prove that this sequence contains infinitely manycomposite numbers.

Proposed by Wing Sit, University of Texas at Dallas

First solution by Jos Alejandro Samper Casas, Colegio Helvetia deBogota, Colombia

Solution. Let an denote the n-th term of the sequence. We willprove that there exist infinitely many j so that 31|aj.We first note that

an= 1 +n

k=1

3 10k =10n+1 7

3 .

Nowa1= 31 = 1027

3 . Fermats Little Theorem tells us that :

31|1030 1 = 31|1030k 1 = 31|1030k+2 102

= 31|1030k+2 102 + (102 7) = 1030k+2 7

Because gcd(31, 3) = 1 we have that 31|1030k+2

7

3 = a30k+1 and we aredone.

Second solution by Aleksandar Ilic, Serbia

Solution. Lets compute n-th term:

an= 33 . . . 33 n

1 = 1 + 3 10 11 . . . 11 n

= 1 + 30 10n 19

=10n+1 7

3 .

Its easy to see that all numbers are relatively prime with 2, 3 and5, and that array is strictly increasing. Let p be any prime that dividessome member an of the sequence (we can take p = 31).

p | an 10n+1 7(modp).From Fermats Little Theorem we get 10p1 1 modp. Now we can

consider members with indexes of form n +k (p 1), and the problemis solved - because they are composite.

p | an+k(p1) 10n+1+k(p1) 10n+1

10p1k 7(modp).

Also solved by Ashay Burungale, India


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J34. Let ABCbe a triangle and let Ibe its incenter. Prove that atleast one ofIA, IB,IC is greater than or equal to the diameter of theincircle ofABC.

Proposed by Magkos Athanasios, Kozani, Greece

First solution by Andrea Munaro, Italy

Solution. Letr be the inradius of triangleABC. Assume contradic-

tion, thatI A

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Second solution by Courtis G. Chryssostomos, Larissa, Greece

Assume contradiction, that AI

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Fourth solution by Aleksandar Ilic, Serbia

Solution. Let projection incenter I on side AB be point C. Fromsine theorem on triangle AICwe have thatAIsin

2 =r. Analogously,

we getBIsin 2 =r andC Isin

2=r. We will prove inequality

AI BI CI (2r)3,from which at least one ofAI, BI,CI is greater that 2r. Inequality

is equivalent to

1 8sin2

sin

2sin

2.

Function sin x is concave (non-convex) on interval [0, ], so we canuse Jensens inequality and Power Mean:

sin 2 + sin

2 + sin

2

3 sin

2 +

2 +

2

3 = sin

6 =

1

2.

sin 2

+ sin 2

+ sin 2

3 3sin

2sin

2sin

2.

Combining above inequalities, the problem is solved.

Also solved by Daniel Campos Salas, Costa Rica


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J35. Prove that among any four positive integers greater than orequal to 1 there are two, saya and b, such that

(a2 1)(b2 1) + 1ab

3

2


First solution by Daniel Campos Salas, Costa Rica

Solution. Let (a,b,c,d) be these numbers. Since these are greaterthan or equal to 1 there exist,,, , in (0, 90], such that

1

a,1

b,1

c,1

d

= (sin , sin , sin , sin ).

Then,

Na,b =

(a2 1)(b2 1) + 1

ab =

1 1

a2

1 1

b2

+

1

ab=

cos cos + sin sin = cos( ).By the Pigeonhole Principle we have that at least two elements of

the set [,, , ] lie on one of the intervals (0, 30], (30, 60], (60, 90],which implies that there are two of them such that their difference isleast than or equal to 30. Since the function cos x is strictly decreasingon [0, 90], and there exist two elements, say and such that 30 0, we conclude thatNa,b = cos( ) cos30 =

3

2 .


Solution. Let these four numbers be a,b,c and d. From the condition

in the problem we can find numbers x, y, z,t [0,

2 ], such that1

a= cos x,

1

b= cos y,

1

c= cos z, and

1

d= cos t.

(a2 1)(b2 1) + 1ab

=

1 cos2 x

1 cos2 y+cos x cos y= cos(xy).By the Pigeonhole Principle, among four numbers x,y,z, t, there are

two with absolute value less or equal than 6

. Let|x y| 6

then

cos(x y) cos6

=

3

2, and we are done.


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Third solution by David E. Narvaez, Universidad Tecnologica de Panama

Solution. Let the four numbers be n1, n2, n3 and n4. Since sec reaches every real number greater than or equal to 1 for 0 , thereexist four positive numbers 1,2,3 and 4 such that

n1 = sec 1, n2= sec 2, n3= sec 3, n4 = sec 4

and 0

1, 2, 3, 4 0andf(x)>0, forx 4. Thereforef(x) is a strictly increasing functionand should have at most one solution. We claim thatx= 4 is that uniquesolution and clearly it satisfies the property.

It is remained to find ifn = 2 and n= 4 satisfy the condition of theproblem.

Ifn = 2, by Leibnitz Theorem we immediately get that

P A2 +P B2 +P C2 = 2a2.


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Ifn= 4, (we may assume that Pbelongs to the small arc ofBC) thenusing Ptolemy Theorem and the Law of Cosines we get

P A= P B+P C, and a2 =P B2 +P C2 +P B P C.Thus

P A4 +P B4 +P C4 = (P B+P C)4 +P B4 +P C4 =

= 2(P B4 + 3P B3P C+ 2P B2P C2 + 3PBPC3 +P C4) =

= 2(P B2

+P B P C+P C2

)2

= 2a4

.


Solution. Assume that circumradius is R = 1. Then the side ofequilateral triangle has length a =

3. Consider two special positions

for point P: whenP A and when P A, where A is midpoint forsmaller arc BC. In these cases we have:

P A: P An +P Bn +P Cn = 0n +ABn +ACn = 2 3n2

P A: P An

+P Bn

+P Cn

=AAn

+ABn

+ACn

= 2n

+ 2From condition in the problem we have to find all n, such that 3

n2 =

2n1 + 1. This can be true only for evenn. By checking, we get thatequality holds for n= 2 and n= 4. In case n6, we have 2n1 + 1 >(

3)n by trivial induction.

(

3)n+1 = (

3)n

3< 2n1

3 +

3< 2n + 1.

Now well prove that statement of problem is true in case n = 2 orn= 4. Assume without loss of generality that point P is on smaller arcBC. First apply Ptolemys theorem on quadrilateral ABP C:

AB P C+AC P B=BC P A P B+P C=P AFrom the Law of Cosines on triangleBC Mwe get

P B2 +P C2 2P B P C cos 120o =P B2 +P C2 +P B P C=a2.Lets calculate expression for n = 2:

P A2 +P B2 +P C2 = (P B+P C)2 +P B2 +P C2 =

= 2(P B2 +P C2 +P B P C) = 2a2 = const.


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In the casen= 4 we get:

P A4 +P B4 +P C4 = (P B+P C)4 +P B4 +P C4 =

= 2(P B4 +P C4 + 3P B2 P C2 + 2P B3 P C+ 2P C3 P B) == 2(P B2 +P C2 +P B P C)2 = 2a4 = const.

Also solved by Courtis G. Chryssostomos, Larissa, Greece;

David E. Narvaez, Universidad Tecnologica de Panama


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S35. Let ABC be a triangle with the largest angle at A. On theline AB consider the point D such that A lies between B and D and

AD=AB3

AC2. Prove that C D

3 BC

3

AC2


First solution by Ashay Burungale, India

Solution. Denote AB = c, AC = b, BC = a. If angle A is thegreatest, thena max(b, c). From the Law of Cosines we get

CD2 =AC2 +AD2 2AC AD cosCAD=

=b2 +c6

b4+ 2b c

3

b2

b2 +c2 a22bc

=

b6 +c6 +b2c2(b2 +c2 a2)b4

.

Thus it is enough to prove that

b6 +c6 +b2c2(b2 +c2 a2)b4

3a6

b4 ,

b6 +c6 +b2c2(b2 +c2 a2) 3a6,or

b6 +c6 +b2c2(b2 +c2) 3a6 +a2b2c2.But note that ifa max(b, c), then

b6 +c6 +b2c2(b2 +c2) b6 +c6 + 2a2b2c2 3a6 +a2b2c2,and we are done.

Second solution by Daniel Campos Salas, Costa Rica

Solution. From Stewarts theorem we have

AD BC2 +AB CD2 =BD(AB AD+AC2), or

AB

AB2 BC2

AC2 +CD2

= (AB+AD)(AB AD+AC2) =

AB

1 +

AB2

AC2

AB4 +AC4

AC2

.


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This implies that

CD2 = (AB2 +AC2)(AB4 +AC4) (AB BC AC)2

AC4

= AB6 +AC6 +AB4 AC2 +AB2 AC2(AC2 BC2)

AC4

AB6 +AC6 +AB4 AC2

AC4

3BC6

AC4 ,

that completes the proof.

Also solved by Aleksandar Ilic, Serbia


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S36. LetPbe a point in the plane of a triangleABC, not lying on thelinesAB,BC,or C A. Denote byAb, Ac the intersections of the parallelsthrough A to the lines P B, P C with the line BC. Define analogouslyBa, Bc, Ca, Cb. Prove thathAb, Ac, Ba, Bc, Ca, Cb lie on the same conic.

Proposed by Mihai Miculita, Oradea, Romania

First solution by Ricardo Barosso Campos, Universidad de Sevilla,Spain

Solution. First of all we remind a famous Carnot Theorem:

Let ABC be a triangle and points X1, X2, Y1, Y2, Z1, Z2 such thatX1, X2 BC, Y1, Y2 AC, Z1, Z2 AB. If the relation

AZ1Z1B

AZ2Z2B

BX1X1C

BX2X2C

C Y1Y1A

C Y2Y2A

= 1

holds, then X1, X2, Y1, Y2, Z1, Z2 lie on a conic.

We want to prove that Ab, Ac, Ba, Bc, Ca, Cb satisfy this condition.

LetAP

BC=Ap , BP

AC=Bp , CP

AB= Cp.

DenoteAB=c, BC=a, CA= b and

BAp = m, ApC=n, CBp = q, BpA= r, ACp = s, CpB = t.

Using the similarity of trianglesBCpCand BAACwe get

BAcBC

= BA

BCP, BAC= ca

t .

Analogously,

BCa=

ca

m , CAB =

ba

q , CBa =

ab

n, ACb=

cb

r, ABc=

bc

s.

On the other hand we have

BAb= CAb a= baq a= a(b q)

q =

ar

q .

Similarly,

BCb=cq

r, CAc=

as

t , CBc=

bt

s, ABa=

bm

n , ACa=

cn

m.


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Finally we calculate the product

ACaBCa

ACbBCb

BAcCAc

BAbCAb

C BaABa

C BcABc

=

=

cn/m

ca/m cb/r

cq/r

ca/t

as/tar/q

ba/q

ab/n

bm/n bt/s

bc/s

=

=

n

a b

q

c

sr

b

a

mt

c

=

n r tq

s

m

= 1.

Last equality follows from Ceva Theorem, and we are done.


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Second solution by Mihai Miculita, Oradea, Romania

Solution. Denote{Pa} = AP BC,{Pb} = BP AC,{Pc} =P C AB, and with

{Qa} =BaCa BC, {Qb} =AbCb AC, {Qc} =AcBc AB.If (l ,m,n) are barycentric coordinates of point P with respect to

triangleABC, then pointsPa, Pb, Pc have coordinates:

P(0, m , n), Pb(l, 0.n), Pc(l,m, 0).

Observe the following facts

AAb||P B, AbBAbC

=APbAC

; PbA

PbC =

n

l, APb

AC =

n

l+n.

ThereforeAbB

AbC =

n

l+n, Ab(0, l+n, n).

Analogously, we get

Ac(0,

m, l+m), Ba(m+n, 0,

n), Bc(

l, 0, l+m), Ca(m+n,

m, 0), Cb(

l, l+n, 0).

The equation of the line BaCa is

x y z

m+n 0 nm+n m 0

= 0, mn x+n(m+n) y+m(m+n) z= 0.ThusQa is the solution of the system

x= 0 ; mn x+n(m+n) y+m(m+n) z= 0.It follows that y

m z

n, Qa= (0,m,n).Similarly we get the coordinates of the points Qb(l, 0, n) and Qc(l, m, 0).Because the determinant

0 m nl 0 nl m 0

= 0,we deduce that points Qa, Qb, Qc are collinear. Using the reciprocal ofthe Pascal Theorem, we get that points Ab, Ac, Ba, Bc, Ca, Cb lie on thesame conic.


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U32. Let a0, a1, . . . , an and b0, b1, , bn be sequences of complexnumbers. Prove that

Re

nk=0

akbk

1

3n+ 2

nk=0

|ak|2 +9n2 + 6n+ 2

2

nk=0

|bk|2

Proposed by Jose Luis Daz-Barrero, Barcelona, Spain

First solution by Aleksandar Ilic, Serbia

Solution. Letak =xk+ yk and bk =Xk Yk, for 0k n. Wecan assume thatxk, yk, Xk, Yk 0, because we can increase left hand-sideusing absolute values. Now we have to prove inequality:

nk=0

(xkXk+ykYk) 13n+ 2

nk=0

x2k+y

2k

+

9n2 + 6n+ 2

2

nk=0

X2k +Y

2k

.

Because of symmetry we only need to prove:

2(3n+ 2)xkXk

2x2k+ (9n2 + 6n+ 2)X2k

When consider this as quadratic inequality for variable xk, the dis-criminant is negative, and thus parabola is above y = 0.

= (2(3n+ 2)Xk)2 4 2 (9n2 + 6n+ 2)X2k =

= 4X2k

9n2 + 12n+ 4 18n2 12n 4 = 36n2X2k

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This can be separated into

0 13n+ 2

x2 +

9n2 + 6n+ 2

2 w2

xw, (1)

and

0 13n+ 2

y2 +

9n2 + 6n+ 2

2 z2

+yz. (2)

We have that (1) equals

(x w)2 + (x (3n+ 1)w)22(3n+ 2)

,

and (2) equals

(y+z)2 + (y+ (3n+ 1)z)2

2(3n+ 2) .

These inequalities prove our claim and complete the proof.


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U34. Let f : [0, 1] R be a continuous function with f(1) = 0.Prove that there is a c (0, 1) such that

f(c) =

c0

f(x)dx

Proposed by Cezar Lupu, University of Bucharest, Romania

Solution by Aleksandar Ilic, Serbia

Solution. Let F(x) =x0

f(t)dt and F(x) = f(x) and F(0) =f(1) = 0. Consider continuous function g(x) = F(x) f(x). Withoutloss of generality, assume that g(x)> 0 forx (0, 1). Forx = 0 we haveg(0) = f(0) 0, and forx = 1 we have g(1) =F(1) 0.

Leth(x) =exF(x). We can compute first derivative for h(x):

h(x) = exF(x) +exf(x) = ex(F(x) f(x)) = exg(x) 0

So, functionh is decreasing function and h(0) = 0 and h(1) = F(1)e

0. This gives h(x) = 0, and contradiction! Thus there is c(0, 1) suchthatg(c) = 0 or equivalently f(c) =F(c).


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U35. Find all linear maps f :Mn(C) Mn(C) such thatf(XY) =f(X)f(Y) for all nilpotent matrices X andY.

Proposed by Gabriel Dospinescu, Ecole Normale Superieure, Paris

Solution by Jean-Charles Mathieux, Dakar University, Senegal

Solution. We consider Ka zero characteristic field and do not limitthe answer to C.

Denote byfa linear mapf :Mn(K) Mn(K) such thatf(XY) =f(X)f(Y) for all nilpotent matrices X and Y . We show that eitherf = 0 or there exists P GLn(K) such that f(M) = P MP1 for allM Mn(K).

The casef= 0 is obvious so lets assume that f= 0.Denote by Ei,j the matrix ofMn(K) with all coefficients equal to 0

except on row i and column j, where it equals 1 and Fi,j =f(Ei,j).

We proceed in two steps.

Lemma 1. {Fi,j}i,j{1,...,n} is a basis ofMn(K) such that Fi,jFk,l =j,kFi,l.

Lemma 2. If{Mi,j}i,j{1,...,n}is a basis ofMn(K) such thatMi,jMk,l =j,kMi,l, then there exists P GLn(K) such thatMi,j =P Ei,jP1 for alli, j {1, . . . , n}.

Proof of Lemma 1. If{Fi,j}i,j{1,...,n} was not a basis ofMn(K),we could find{i,j}i,j{1,...,n} Kn2 with at least one l,p = 0 and

i,j{1,...,n} i,jFi,j = 0

For all i, j, k,l {1, . . . , n} we have Ei,jEk,l = j,kEi,l so Fi,jFk,l =j,kFi,l. This last relation is obvious if i = j and k = l, becauseEi,j and Ek,l are nilpotent. But also f(Ei,iEk,l) = f(Ei,j

0

Ej0,iEk,l) =

f(Ei,j0i,kEj0,l) =i,kFi,l (where we choose j0=i and j0=l).There is at least one Fk,q= 0 because f= 0,Fk,l

i,j{1,...,n} i,jFi,j

Fp,q =l,pFk,q = 0 which is a contradiction, so{Fi,j}i,j{1,...,n} is a basisofMn(K).

Proof of Lemma 2. Mi,kMk,j =Mi,j , so ImMi,k = ImMi,j , lets denotethis space byVi.

There exist{i,j}i,j{1,...,n} Kn2 such thatIn=

i,j{1,...,n} i,jMi,j .If X Kn, X =ni=1nj=1 i,jMi,jX, andnj=1 i,jMi,jX Vi, soKn = V1+ V2+ + Vn.


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IfXi Vi, andX1 + X2 + + Xn= 0 thenMi,i(X1 + X2 + Xn) =Xi = 0, so K

n = V1 V2 V n.Ifk=j , ImMi,k KerMi,j , so

nk=1,k=jImMi,k = KerMi,j (because

they have same dimension and the previous inclusion).Let Y1 V1, such that Y1= 0. For i= 1, Yi = Mi,1Y1. We have

Mi,jYj =Y1. So{Y1, . . . , Y n} is a basis ofKn.We define P GLn(K) the matrix whose columns are the Yi. We

have P Ei,j =Mi,jP, which concludes the proof of lemma 2.

Since f is linear we have proven that f(M) = P MP1

for all MMn(K).


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U36. Let n be an even number greater than 2. Prove that if thesymmetric group Sn contains an element of order m, then GLn2(Z)contains an element of order m.

Proposed by Jean-Charles Mathieux, Dakar University, Senegal

No solutions proposed


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Olympiad

O31. Letn is a positive integer. Prove that

nk=0

n

k

n+k

k

=

nk=0

2k

n

k

2

Proposed by Jean-Charles Mathieux, Dakar University, Senegal

First solution by Zhao Bin, HUST, China

Solution. We start from the right side of the equality. First of all,because

2k =ki=0

k

i

We have

n

k=02k

n

k2

=n

k=0k

i=0 k

in

k2

=n

i=0n

k=i k

in

k2

.

Then using well-known identitiesk

i

n

k

=

n

i

n in k

and

nk=i

n in k

n

k

=

2n i

n

.

The first inequality goes directly from the binomial theorem, while thesecond can be obtained from the combinational way or comparing the

the coefficient ofxn in the identity:

(1 +x)ni(1 +x)n = (1 +x)2ni.

Thus we haven

k=0

2k

n

k

2=

ni=0

nk=i

k

i

n

k

2=

ni=0

n

i

nk=i

n in k

n

k

=

=ni=0

n

i

2n i

n

.


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Then by substituting k = n i, we easily getni=0

n

i

2n i

n

=

nk=0

n

k

n+k

k

.

Thus we have proved the desired identity.

Second solution by Jose Hernandez Santiago, UTM, Oaxaca, Mexico.

Solution. Let m N with n m. The desired identity is a specialcase of the more general result (see Wilf, Herbert S. Generatingfunc-tionology. p. 127),

mk=0

m

k

n+k

m

=

mk=0

2k

m

k

n

k

. (3)

Settingm = n in (1) we get,

n

k=0 2k

n

k2

=n

k=0 2k

n

kn

k=

nk=0

n

k

n+k

n

=n

k=0

n

k

n+k

n+k n

=n

k=0

n

k

n+k

k

.

Next two solutions were kindly presented by Dr. Scott H. Brown,Auburn University Montgomery.

Third solution (Benjamin and Bataille)

Solution. Let [n] denote the set {1, 2, 3, . . . , n} andD denote the setof ordered pairs (A, B) whereA is a subset ofn andB is an n-subset of[2n] that is disjoint from A. We can select elements for D in two ways:


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O32. 18. Let a, b, c >0. Prove that a2

4a2 +ab+ 4b2+

b2

4b2 +bc+ 4c2+

c2

4c2 +ca+ 4a2 1

Proposed by Bin Zhao, HUST, China

First solution by Ho Phu Thai, Da Nang, Vietnam

Solution. By Cauchy-Schwarz inequality a2

4a2 +ab+ 4b2+

b2

4b2 +bc+ 4c2+

c2

4c2 +ca+ 4a2

(8a2 + 8b2 + 8c2 +ab+bc+ca)

cyc

a2

(4a2 +ab+ 4b2)(4a2 +ac+ 4c2).

We will prove

cyc

a2

(4a2 +ab+ 4b2)(4a2 +ac+ 4c2) 1

8a2 + 8b2 + 8c2 +ab+bc+ca .

This is equivalent to

64sym

a4b2 + 4sym

a4bc+ 4sym

a3b3 + 3sym

a3b2c 66a2b2c2,

which is clearly true from AM-GM inequality.

Second solution by Vasile Cartoaje, University of Ploiesti, Romania

Solution. By the Cauchy-Schwarz inequality we have a2

4a2 +ab+ 4b2+

b2

4b2 +bc+ 4c2+

c2

4c2 +ca+ 4a2

a2(4a2 +ab+ 4b2)(4c2 +ca+ 4a2)

(4c2 +ca+a2) =

=

a2(4b2 +bc+ 4c2) (4c2 +ca+ 4a2)

(4a2 +ab+ 4b2)(4b2 +bc+ 4c2)(4c2 +ca+ 4a2)=

A

B,


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whereA= (8

b2c2 +abc

a)(8

a2 +

bc),

B= (4a2 +ab+ 4b2)(4b2 +bc+ 4c2)(4c2 +ca+ 4c2).

It suffices to prove thatA B. We have

A= 64(

a2)(

b2c2) + 8(

bc)(

b2c2)+

+8abc( a2)( a) +abc( bc)( a) == 64

b2c2(b2+c2)+195a2b2c2+17abc

bc(b+c)+8abc

a3+8

b3c3.

and

B= 64

(b2+c2)+16

bc(a2+b2)(c2+a2)+4abc

a(b2+c2)+a2b2c2 =

= 64

b2c2(b2+c2)+129a2b2c2+20abc

bc(b+c)+16abc

a3+16

b3c3.

Therefore

A B= 3abc bc(b+c) + 8abc a3 + 8 b3c3 66a2b2c2 == 3abc[

bc(b+c)6abc]+8abc(

a33abc)+8(

b3c33a2b2c2) 0,

because bc(b+c) 6abc=

a(b c)2 0,

and by the AM-GM inequality,a3 3abc,

b3c3 3a2b2c2.

The equality occurs if and only ifa = b = c.


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O33. 23. Let ABCbe a triangle with cicrumcenter O and incenterI. Consider a point Mlying on the small arc BC. Prove that

MA+ 2OI MB+M C MA 2OIProposed by Hung Quang Tran, Ha Noi University, Vietnam

Solution by Hung Quang Tran, Ha Noi University, Vietnam

Solution. Using the Law of Cosines in the triangle AOM we get2AO OM cos AMO =M A2 +AO2 MO2 =M A2, 2MO MA=MA2, Analogously

2MO MA= M B2, 2MO MC=M C2

From here we have

MB+MC MA= 2MO

MA

MA+

M B

MB+

M C

MC

From the Cauchy- Scwartz Inequality we get

MO MAMA

+ M BMB

+M CMC

2MO MAMA

+ M BMB

+M CMC

MO MA

MA+

M B

MB+

M C

MC

We have M O= R, we will calculate MA

MA+

M B

MB+

M C

MC .

MA

MA+

MB

MB+

MC

MC2 = 3+2

MB MCMB MC +

M B MAMB MA+

MA MAMC MA

=

= 3 + 2(cos(MB,MC) cos(MB,MA) cos(M C , M A)) =3 2(cos A+ cos B+ cos C) = 3 2 R +r

R =

OI2

R2 .

Thus we have

MO MAMA

+M B

MB+

M C

MC =OI.

It follows that

MA+ 2OI MB+M C MA 2OI ,and the problem is solved.


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h Z[X]. If we take x = m such that f(m2) is a perfect square, weget N

f(m2) = h(m). This relations holds for infinitely manym thus

f(x2) = g(x)2. As g has no non-zero roots, its roots must group intopairs of opposite roots (because the roots off(x2) do). Thusg(x) =r(x2)so f(x) =r2(x), and clearly r(x) Z[X] (we can use lemma 1 to prover(x) Q[X] then we deducer(x) Z[x]). Now let us solve the problem.

Letf(x) =P(x) =

n

i=1(x wn), then P(x2 +ax) = n

i=1(y2 ui),

where ui =

a2

4 +wi and y = x+

a

2 . Thus Q(x) =ni=1(x2 ui) isirreducible. The Q(x) can be written as f(x) g(x) wheref(x) =u(x2)h(x), g(x) = v(x2)h(x).

From all such representations f(x) g(x), pick up one for which deg(f)is the least possible. It follows that gcd(f(x), g(x)) is either 1 orf(x), which means that either h(x) = 1 or u(x)|v(x). Ifu(x)|v(x) andu= 1, then gcd(f(x), g(x))= 1. Thus gcd(f(x), g(x)) = f(x), hencef(x)2|Q(x). This readily implies thatPhas a double root (ifa is suffi-ciently big), so (P, P) = 1 and therefore Pis irreducible. It follows thatwe have h(x) = 1 or u(x) = 1.

a) If h(x) = 1, then f(x) = u(x2), g(x) = v(x2), u(x)v(x) =f(x+ a2) so f is reducible.

b) Ifu(x) = 1,then f(x) =h(x), g(x) =h(x)v(x2), h(x)h(x)v(x2) = Q(x). If h(x)h(x) = w(x2) we get w(x)v(x) =f(x + a2), thusf is reducible unlessv = 1. In this case we have Q(x) =h(x)h(x). ThereforeQ(0) =h(0)2 and thusP(a2

4) is a perfect square.

By setting k(x) = 4mf(x4

) for a suitable m, the conditions of lemma 2are satisfied. Hence k is a perfect square, so must be P.

In any case we get that f=P is reducible.

Remark. The key assertion that iff Z[X] is irreducible and f(0)is not a perfect square thenf(x2) is also irreducible was a problem fromRomanian TST 2003.


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O35. Let 0 < a < 1. Find, with proof, the greatest real numberb0such that ifb < b0 and (An [0;1])nN are finite unions of disjoint seg-ments with total lengtha, then there are two different i, j Nsuch thatAi

Aj is a union of segments with total length at least b. Generalizethis result to numbers greater than 2: if k N find the least b0 suchthat whenever b < b0 and (An [0;1])nN are finite unions of disjointsegments with total lengtha, then there are k differenti1, i2, . . . , ik Nsuch that Ai1

Ai2 . . .

Aik is a union of segments with total length

at leastb.

Proposed by Iurie Boreico, Moldova

Solution by Iurie Boreico, Moldova

Solution. We will use the following famous result:n

k

>

nk k2nk1k!

Indeed,n

k

= n(n1)...(nk+1)

k! . Using the following inequalityxy (x+ 1)(y 1), forx > y,we conclude that

n(n 1) . . . (n k+ 1) > nk1(n 1 2 . . . (k 1)),from which we deduce the result.

Let us solve the general problem. Pick up some n. Suppose [0, 1] is di-vided into intervals of lengths x1, x2, x3, . . . , xmby the sets A1, A2, . . . , An.Letxi belong toki ofA1, A2, . . . , An, andk1x2+ k2x2+ . . . + knxn= na.

Suppose that each k or the Ai intersect in a set of measure at mostb. Consider the sum

1i1

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which translates into

mi=1

kki xi (mi=1

kixi)k =nkak.

Therefore the sum in contest is at least nk

k!(ak k2

n). But weve prove it is

at mostbn

k

< b n

k

k!. Thus ifb < ak, we get a contradiction for sufficiently

bign.

Let us prove that any number greater than ak fails. Indeed, considerb > ak and pick up a rational number p

qsuch thata < p

q < k

b. Consider

Ai be the set of all real numbers from [0; 1] which written in base qhavethe i-th digit after zero from the set{0, 1, . . . , p 1}. Its obvious thatAi are a union of segments, that|Ai| = pq > a and|Ai1

Ai2. . . Aik | =

(pq

)k < b so this is the required counter-example.

The answer is thus ak.


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O36. Leta1, a2,..,an and b1, b2,...,bn be real numbers and let xij bethe number of indicesk such thatbk max(ai, aj). Suppose thatxij >0for any i and j. Prove that we can find an even permutationf and an

odd permutationg such thatni=1

xif(i)xig(i)

n.

Proposed by Gabriel Dospinescu, Ecole Normale Superieure, Paris

Solution by Iurie Boreico, Moldova

Solution. Ifn = 2 then we must prove that x1,1+ x2,2 2x1,2 whichis true because clearly x1,1 x1,2 andx2,2 x1,2.

Now assume n 3. By AM-GM it suffices to find an even permuta-tionfand an odd permutation g such that

ni=1

xi,f(i)ni=1

xj,g(j).

Setxi() =xi,(i) for a permutation .We must find an even permutationfand an odd permutationg such

thatni=1 xi(f) ni=1 xi(g). If X() =ni=1 xi() we must haveX(f) X(g). This will follow directly from the relationfX(f) =

gX(g) where f runs over all even permutations and g over all oddpermutations (recall that the number od even permutation equals thenumber of odd permutations).

To prove this relation, it suffices to prove that

fxi(f) =

gxi(g).This fact follows from the following result:

For any 1 i, j n, the number of even permutations f such thatf(i) =j equals the number of odd permutations g such thatg(i) =j.

Indeed if we pick up k, l such that j, k, l are pairwise different, thenthe permutations with(i) =j group into pairs , () where is the

transposition exchanging k with l. As , () have different signs, wededuce there is one even permutation and one odd permutation in eachpair, hence the result.

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