harmonic quadrilaterals and other stuff

8/13/2019 Harmonic Quadrilaterals and Other Stuff

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GAZETA MATEMATICASERIA A

ANUL XXIX(CVIII) Nr. 1 2/ 2011

ARTICOLE

On some stability concepts for real functions

Larisa Biris1), Traian Ceausu2) and Mihail Megan3)

Abstract. In this paper we investigate some asymptotic properties forreal functions defined onR+ as exponential stability, uniform exponentialstability, polynomial stability and uniform polynomial stability. Our mainobjectives are to give characterizations for these concepts and to establishconnections between them.

Keywords: Exponential stability, polynomial stability.

MSC : 34D05

1. Exponential stability

The exponential stability property plays a central role in the theory ofasymptotic behaviors for differential equations. In this section we considertwo concepts of exponential stability for the particular case of real functionsdefined on R+.

Let 0 be the set defined by

0= {(y, x) R2+ with y x}.Definition 1. A real functionF : R+ R is called

(i) uniformly exponentially stable (and denote u.e.s), if there areN 1and >0 such that

|F(y)| N e(yx)|F(x)|, for all(y, x) 0;

1)Department of Mathematics, Faculty of Mathematics and Computer Science, WestUniversity of Timisoara, Romania, [email protected]




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2 Articole

(ii) (nonuniformly) exponentially stable (and denote e.s), if there exist

>0 and a nondecreasing functionN : R+ [1, ) such that|F(y)| N(x)e(yx)|F(x)|, for all(y, x) 0.

Remark 1. It is obvious thatu.e.s e.s. The converse implication is nottrue, as shown in

Example 1. LetF : R+ R be the function defined byF(x) =ex cosx3x.

We observe that for all (y, x) 0, we have

|F(y)

|=F(y) =ey cosy3yx cosx+3x

|F(x)

| e2y+4x|F(x)| e2xe2(yx)|F(x)|and hence F is e.s.

If we suppose that ifFis u.e.s, then there exist N 1 and >0 suchthat

ey cosy3y N e(yx)ex cosx3x, for all (y, x) 0.In particular, for y = 2n and x = 2n /2, we obtain

e2n N e(3)/2,which for n

, leads to a contradiction.

A necessary and sufficient condition for uniform exponential stability isgiven by

Proposition 1. A function F : R+ R is uniformly exponentially stableif and only if there exists a decreasing functionf : R+ R+ = (0, ) withlimx

f(x) = 0 and

|F(y)| f(y x)|F(x)|, for all(y, x) 0.Proof. Necessity. It is an immediate verification.

Sufficiency. Let > 1 with f() < 1. Then for all (y, x) 0 thereexistn N and r [0, ) such that y = x + n+ r. Then

|F(y)| f(r)|F(x + n)| f(0)|F(x + n)| f(0)f()|F(x + (n 1))| f(0)fn()|F(x)| ==f(0)en ln f()|F(x)| =f(0)ere(yx)|F(x)| f(0)ee(yx)|F(x)| N e(yx)|F(x)|,

where = ln f()

and N= 1 + f(0)e.


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L. Biris, T. Ceausu, M. Megan, Stability of real functions 3

Definition 2. A function F : R+

R with the property that there are

M 1 and >0 such that|F(y)| M e(yx)|F(x)|, for all(y, x) 0,

is calledwith exponential growth (and we denote e.g).

Remark 2. It is obvious that u.e.s e.g and the converse implication isnot true.

Proposition 2. A function F : R+ R is with exponential growth if andonly if there exists a nondecreasing function: R+ R+ with limx(x) == and

|F(y)| (y x)|F(x)|, for all(y, x) 0.Proof. It is similar with the proof of Proposition 1.

Another characterization of the u.e.s property is given by

Proposition 3. LetF : R+ R be an integrable function on each compactinterval [a, b] R+ (i.e., F is locally integrable onR+) with exponentialgrowth. Then F is uniformly exponentially stable if and only if there existsD 1 with

x

|F(y)|dy D|F(x)|, for allx R+.

Proof. Necessity. It is an immediate verification.Sufficiency. BecauseF is with e.g, it follows that there is a nondecreas-

ing function : R+ [1, ) with|F(y)| (y x)|F(x)|, for all (y, x) 0.

Then for all (y, x) 0 with y x + 1 we have

|F(y)| =y

y1

|F(y)|dzy

y1

(y z)|F(z)|dz

(1)

x

|F(z)|dz D(1)|F(x)|.

If (y, x) 0 with y [x, x + 1), then|F(y)| (y x)|F(x)| D(1)|F(x)|

and hence

|F(y)| D(1)|F(x)|, for all (y, x) 0.


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4 Articole

Thus we obtain

(y x)|F(y)| =y

x

|F(y)|dz D(1)y

x

|F(z)|dz D2(1)|F(x)|,

for all (y, x) 0 and|F(y)| f(y x)|F(x)|, for all (y, x) 0,

where

f(z) =D2(1)

z+ 1 .

Using the preceding proposition we conclude that F is u.e.s. A generalization of the preceding result for the nonuniform case is given

byProposition 4. LetF : R+ R be a locally integrable function onR+ withexponential growth. Then F is exponentially stable if and only if there are >0 andD: R+ [1, ) such that

x

ey |F(y)|dy D(x)ex|F(x)|, for allx 0.

Proof. Necessity. It is a simple verification for (0, ), where is givenby Definition 1 (ii).

Sufficiency. If (y, x) 0 with y x + 1 then

ey |F(y)| =y

y1

ey |F(y)|dzy

y1

(y z)ey |F(z)|

(1)y

y1

e(yz)ez |F(z)|dz (1)ex

ez |F(z)|dz

D(x)e(1)ex |F(x)|.If (y, x) 0 with y [x, x + 1) then

ey |F(y)| e(yx)ex(y x)|F(x)|

e

(1)ex

|F(x)| D(x)e

(1)ex

|F(x)|.Finally, we obtain

|F(y)| N(x)e(yx)|F(x)|, for all (y, x) 0,where

N(x) =(1)eD(x).

In the particular case of u.e.s property we obtain:


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Corollary 1. Let F : R+

R be a locally integrable function onR+ with

exponential growth. Then F is uniformly exponentially stable if and only ifthere existD 1 and >0 such that

x

ey |F(y)|dy Dex|F(x)|, for allx 0.

2. Polynomial stability

In this section we consider two concepts of polynomial stability forreal functions. Our approach is based on the extension of techniques ofexponential stability to the case of polynomial stability. Two illustrating

examples clarify the relations between the stability concepts considered inthis paper.Let 1 be the set defined by

1= {(y, x) R2 with y x 1}.Definition 3. A real functionF : R+ R is called

(i) uniformly polynomially stable (and denote u.p.s) if there existN 1and >0 such that

y|F(y)| N x|F(x)|, for all(y, x) 1;(ii) (nonuniformly) polynomially stable (and denote p.s) if there are a

nondecreasing functionN : R+ [1, ) and >0 such thaty|F(y)| N(x)x|F(x)|, for all(y, x) 1.

Remark 3. If the functionF : R+ R is u.e.s then it is u.p.s. Indeed, ifF is u.e.s then using the monotony of the function

f : [1, ) [e, ), f(t) = et

t

we obtain

y|F(y)| xe(yx)|F(x)| N x|F(x)|,for all(y, x) 1 and henceF is u.p.s. The converse is not true, phenome-non illustrated by

Example 2.The functionF : R+ R, F(x) = 1x3 + 1

, satisfies the inequa-

lity

y3|F(y)| = y3

y3 + 1 2x

3

x3 + 1= 2x3|F(x)|,

for all (y, x) 1. This shows that F is u.p.s. We show that F is not e.sand hence it is not u.e.s. Indeed, if we suppose that F is e.s then there are


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6 Articole

>0 andN : R+

[1,

) such that

ey

y3 + 1 N(x) ex

x3 + 1, for all(y, x) 0.

Forx= 0 andy we obtain a contradiction which proves thatF is note.s.

Remark 4. Similarly, as in Remark3 we can prove that ifF : R+ R ise.s then it is p.s.

The function considered in Example 2 shows that the converse is nottrue.

Remark 5. It is obvious that if the function F : R+ R is u.p.s thenit is p.s. The function F considered in Example 1 shows that the converse

implication is not true. Indeed, F is e.s. and by Remark4 it is p.s.If we suppose thatF is u.p.s then there are >0 andN 1 such that

yey cos y3y N xex cosx3x, for all(y, x) 1.Then forx= 2n /2, y= 2n andn we obtain a contradiction.Remark 6. The preceding considerations prove that between the asymptoticbehaviors defined in this paper we have the following implications:

e.g u.e.s e.s

p.g u.p.s p.sDefinition 4. A function F : R

+ R with the property that there are

M 1 andp >0 such thatxp|F(y)| M yp|F(x)|, for all(y, x) 1

is calledwith polynomial growth (and denote p.g).

A necessary and sufficient condition for polynomial stability is given by

Proposition 5. LetF : R+ R be a locally integrable function onR+ withpolynomial growth. ThenF is polynomially stable with >1 if and only ifthere exists >0 andD: [1, ) [1, ) such that

x y

|F(y)

|dy

D(x)x+1

|F(x)

|, for allx

1.

Proof. Necessity. IfF is p.s with >1 then for (0, 1) we havex

y|F(y)|dy N(x)|F(x)|xx

ydy

N(x) 1 x

+1|F(x)| D(x)x+1|F(x)|, for all x 1,


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where

D(x) = 1 + N(x) 1 .

Sufficiency. If (y, x) 1 and y [x, 2x) then

y+1|F(y)| M y+1y

x

p |F(x)| M x+1 yx

p++1 |F(x)| M2p++1x+1|F(x)|,

where M and p are given by Definition 4.If (y, x) 1 with y 2xand

C=

21

dzzp++2

= 1yp++1

yy2

tp+dt,

then

Cy+1|F(y)| = 1yp

yy2

tp+|F(y)|dt

Mx

t|F(t)|dt M D(x)x+1|F(x)|.

Finally, we obtain

y|F(y)| N(x)x|F(x)|, for all (y, x) 1,where = + 1 > 1 and

N(x) =M(2p++1 + D(x)) 1,which shows that F is p.s. with >1.

From the proof of the preceding result we obtain

Corollary 2. Let F : R+ R be a locally integrable function onR+ withpolynomial growth. Then F is uniformly polynomially stable with > 1 ifand only if there areD 1 and >0 such that

x

y|F(y)|dy Dx+1|F(x)|, for allx 1.

Acknowledgements: This paper was presented at the 14th Annual Conference of

S.S.M.R., 15 17 October 2010, Alba Iulia.


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References

[1] M. Megan, T. Ceausu, A. Minda,On Barreira Valls stabilities of evolution operators inBanach spaces, Seminar of Mathematical Analysis and Applications in Control Theory,West University of Timisoara, nr. 18 (2009), 1 15.

[2] M. Megan, C. Stoica, Concepts of dichotomy for skew-evolution semiflows in Banachspaces, Annals of the Academy of Romanian Scientists, Series on Mathematics and itsApplications, vol. 2, nr. 2 (2010), 125 140.

[3] M. Megan, T. Ceausu, M.L. Ramneantu,Polynomial stability of evolution operators inBanach spaces, Opuscula Mathematica vol. 31, nr. 2 (2011), 283 292.

[4] A.A. Minda, M. Megan, On (h, k)stability of evolution operators on Banach spaces,Applied Mathematics Letters 72 (2010), 1305 1313.

[5] C. Stoica, M. Megan,On nonuniform exponential stability for skew-evolution operatorsin Banach spaces, to appear in Carpathian J. Math.

Some inequalities about certain arithmetic functions whichuse the e-divisors and the e-unitary divisors

Nicusor Minculete1)

Abstract. The purpose of this paper is to present several inequa-lities about the arithmetic functions (e), (e), (e), (e) and otherwell-known arithmetic functions. Among these, we have the following:

(n)

(n)

(e)(n)

(e)(n),

(n)

(n)

(e)(n)

(e)(n), (n) + 1 (e)(n) +(n) and

(n) + n (e)(n) + (n), for any n 1, where (n) is the number ofnatural divisors ofn, (n) is the number of natural divisors ofn, (n) is

the sum of the divisors ofn,

(n) is the number of unitary divisors of n,(n) is the sum of the unitary divisors ofn and is the ,,core ofn.

Keywords: arithmetic function, exponential divisor, exponential unitarydivisor.

MSC : 11A25

1. Introduction

First we have to mention that the notion of ,,exponential divisor wasintroduced by M. V. Subbarao in [9], in the following way: if n > 1 is an

integer of canonical form n = pa1

1

pa2

2 parr , then the integer d =

r

i=1pbii iscalled an exponential divisor (or e-divisors) of n =

ri=1

paii > 1, if bi|ai for

every i = 1, r. We noted|(e)n. Let(e)(n) denote the sum of the exponentialdivisors ofn and (e)(n) denote the number of exponential divisors ofn.

1)Dimitrie Cantemir University of Brasov, [email protected]


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N. Minculete, Some inequalities 9

For example, if n = 2432, then the exponential divisors of n are the

following:2 3, 2 32, 22 3, 22 32, 24 3 a n d 24 32.

For various properties of the arithmetic functions which use the e-divisors see the monograph ofJ. Sandor and B. Crstici [5].

J. FabrykowskiandM. V. Subbarao in [1] study the maximal order andthe average order of the multiplicative function (e)(n). E.G. Straus and M.V. Subbarao in [8] obtained several results concerning e-perfect numbers (n

is an e-perfect number if(e)(n) = 2n). They conjecture that there is only afinite number of e-perfect numbers not divisible by any given prime p.

In [4], J. Sandorshowed that, ifn is a perfect square, then

2

(n)

(e)

(n) 2(n)

, (1)where (n) and (n) denote the number of distinct prime factors ofn, andthe total number of prime factors ofn, respectively. It is easy to see that,for n= pa11 p

a22 parr >1, we have (n) =r and (n) =a1+ a2+ ... + ar.

In [6], J. Sandor and L. Tothproved the inequality

nk + 1

2

k(n)

(n)

nk, (2)

for all n 1 and k 0, where(n) is the number of the unitary divisors ofn, k(n) is the sum ofkth powers of the unitary divisors ofn.

In [11] L. Toth and N. Minculete presented the notion of ,,exponential

unitary divisors or ,,e-unitary divisors. The integer d =

ri=1

pbii is called

an e-unitary divisor of n =r

i=1

paii > 1 if bi is a unitary divisor of ai, sobi,

aibi

= 1, for every i = 1, r. Let(e)(n) denote the sum of the e-unitary

divisors ofn, and (e)(n) denote the number of the e-unitary divisors ofn.For example, ifn= 2432, then the exponential unitary divisors ofn are thefollowing:

2 3, 2 32, 24 3 and 24 32.

By convention, 1 is an exponential divisor of itself, so that(e)(1) =(e)(1) = 1.

We notice that 1 is not an e-unitary divisor of n > 1, the smalleste-unitary divisor ofn = pa11 p

a22 parr > 1 is p1p2 pr = (n) is called the

,,core ofn.In [3], it is show that

(e)(n) (n) (n), (3)


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where (n) is the function of Dedekind, and

(n) (e)(n)

(e)(n), (4)

for all integers n 1.Other properties of the sum of the exponential divisors ofn and of the

number of the exponential divisors ofn can be found in the papers [2, 6 and10].

2. Inequalities for the functions e, e, e and e

Lemma 1. For everyn 1, there is the following inequalityn(n)

(n), (5)

and for all n = pa11 pa22 parr > 1, with ai 2, () i = 1, r, there is the

following inequalityne(n) (n). (6)

Proof. For a 1 we show that2pa pa +pa1 + ... +p + 1.

This inequality is rewritten as

pa pa1 + ... +p + 1,which by multiplication with p 1 becomes

pa+1 + 1

2pa,

which is true, because pa+1 2pa, for every prime number p and for alla 1.

Therefore,pa(pa) = 2pa pa+pa1+...+p+1 =(pa),sopa(pa) (pa). Because the arithmetic functions and are multiplicative, wededuce the inequality

n(n) (n).Since a 2, it follows that (a) 2, so

pa(a) 2pa pa + ... +p + 1 =(pa),which is equivalent topae(pa) =pa(a) (pa). As the arithmetic functionse and are multiplicative, we get the inequality of the statement.

Remark 1. It may be noted that the lemma readily implies the inequality

n2e(n2) (n2), (7)for alln 1.Theorem 2. For everyn 1, there is the inequality

(n)

(n)

e(n)

e(n). (8)


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Proof. For n = 1, we have

(1)

(1) = 1 =

e(1)

e(1) .Lets consider n = pa11 p

a22 parr , with ai 1, () i = 1, r. According to

lemma 1, we deduce the relation

ai(ai) (ai), () i = 1, r. (9)

From the inequality of S. Sivaramakrishnan and C. S. Venkataraman [7],(n)n(n), () n1, we get (ai)ai(ai), () i = 1, r. This lastinequality combined with inequality (9) implies the inequality

ai (ai)

(ai), () i = 1, r, (10)

which means that ri=1

ai e(n)

e(n), for all n= pa11 p

a22 parr >1, because (11)

e(n) =(a1) ... (ar) and e(n) =(a1) ... (ar).But ai+ 1 2ai, () i = 1, r, and by taking the product, we obtain

the inequalityr

i=1

(ai + 1) 2r r

i=1

ai, which is equivalent to the relation

(n) (n)

r

i=1ai, so

(n)

(n) r

i=1

ai for all n = pa11 p

a22 ...p

arr >1. (12)

Combining relations (11) and (12), we deduce inequality (8). Finally, theproof is completed.

Theorem 3. For everyn 1, there is the equality(n)

(n)

e(n)

e(n). (13)

Proof. We distinguish the following cases:Case I. For n = 1, we have

(1)

(1) = 1 =

e(1)

e(1).

Case II. If n is squarefree, then (n) = (n), and e(n) = n = e(n).Therefore, we obtain the relation of statement.Case III. Lets consider n = pa11 p

a22 parr >1, with ai 2, ()i = 1, r, then

(pa)

(pa) =

1 +p +p2 + ... +pa

1 +pa = 1 +

p +p2 + ... +pa1

1 +pa (14)


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and, because the exponential unitary divisorspd1 ,...,pdq ofpa are among the

exponential divisors pd1 ,...,pds ofpa, we have the inequality

e(pa)

e(pa)=

pd1 +pd2 + ... +pds

pd1 + ... +pdq= 1 +

pd2 + ... +pds1

p + ... +pa

1 + p2 +p3 + ... +pa1

p +pa 1 + p + ... +p

a1

1 +pa .

Therefore, using the inequality (14), we get the relation

e(pa)

e(pa) (p

a)

(pa).

Because the arithmetic functions e, e, and are multiplicative, wededuce the inequality

(n)

(n)

e(n)

e(n), for all n= pa11 p

a22 parr >1, with ai 2, ()i = 1, r.

Case IV. Lets consider n = n1n2, where n1 is squarefree, and n2 =p|n2a2

pa.

Since (n1, n2) = 1, we deduce the following relation

(n)

(n)=

(n1)(n2)

(n1)(n2)

e(n1)e(n2)

e(n1)e(n2)=

e(n)

e(n).

Thus, the demonstration is complete.

Remark 2. In Theorem2 and Theorem3 the equality in relations (8) and(13) holds whenn= 1 orn is squarefree.

Theorem 4. For anyn 1 the following inequality(n) + 1 e(n) + (n) (15)

holds.

Proof. Ifn = 1, then we obtain (1) + 1 = 2 = e(1) + (1).We considern >1. To prove the above inequality, will have to study severalcases, namely:Case I. Ifn = p21p

22...p

2r, then (n) = 3

r and

e(n) =(a1) (a2) ... (ar) =r(2) = 2r =(n),which means that inequality (15) is equivalent to the inequality 3r + 1 2 2r, which is true because, by using Jensens inequality, we have

3r + 1

2

3 + 1

2

r= 2r.

Case II. Ifak= 2, ()k = 1, r, then the numbersn

p1, n

p2, ...,

n

pr,

n

p1p2,...,

n

pipj, ...,

n

pipjpk, ...,

n

p1p2 pr


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are not exponential divisors of n, so they are in a total number of 2r

1,

such that we have the inequality

(n) =d|en

1 +den

1 =e(n) +den

1 e(n) + 2r 1.

Therefore, we have

(n) e(n) + 2r 1, so (n) + 1 e(n) + (n).Case III. If there is at least one aj = 2 and at least one ak = 2, where

j, k {1, 2,...,r}, then without loss of the generality we renumber the primefactors from the factorization ofn and we obtain

n= p21p22 p2spas+1s+1 parr , with as+1, as+2,...,ar= 2.

Therefore, we writen = n1 n2, wheren1 = p21p22 p2s andn2= pas+1s+1 parr ,

which means that (n1, n2) = 1, and

(n) =(n1 n2) =(n1)(n2) (e(n1)+(n1)1)(e(n2)+(n2)1) == (e(n1) + 2

s 1)(e(n2) + 2rs 1) ==e(n1)

e(n2) + e(n1)(2

rs 1) + e(n2)(2s 1) + (2s 1)(2rs 1) e(n)+2rs1+2s1+(2s1)(2rs1) e(n)+2r1 =e(n)+(n)1.Thus, the inequality of the statement is true.

Lemma 5. For any xi > 0 with i {1, 2,...,n}, there is the following ine-quality:

ni=1

(1 + xi+ x2i ) +

ni=1

x2in

i=1(xi+ x2i ) +

ni=1

(1 + x2i ). (16)

Proof. We apply the principle of mathematical induction.

Theorem 6. For anyn 1, the following inequality:(n) + n e(n) + (n) (17)

holds.

Proof. Ifn = 1, then we obtain (1) + 1 = 2 =e(1)+ (1). Lets considern > 1. To prove the above inequality, we will have to study several casesnamely:

Case I. Ifn= p21p

22 p

2r, then we deduce the equalities (n) =

ri=1

(1 +pi+

+p2i ), e(n) =

ri=1(pi + p

2i ) and

(n) =r

i=1

(1 +p2i ), which means that

inequality (17) implies the inequalityr

i=1

(1 +pi+p2i ) +

ri=1

p2ir

i=1

(pi+p2i ) +

ri=1

(1 +p2i ),


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14 Articole

which is true, because we use inequality (16), for n = r and xi = pi.

Case II. Ifak= 2, ()k = 1, r, then the numbersn

p1,

n

p2, ...,

n

pr,

n

p1p2, ...,

n

pipj,...,

n

pipjpk,...,

n

p1p2 prare not exponential divisors ofn, so they are in a total number of 2r 1, andtheir sum is (n) n, so that as we have the inequality

(n) =d|en

d +den

d= e(n) +den

d e(n) + (n) n.

Since we have the inequality

(n) =nr

i=11 + 1

pi n

r

i=11 + 1

paii =

(n),

it follows that

(n) e(n) + (n) n, so(n) + n e(n) + (n).Case III. If there is at least one ak = 2, and at last one aj = 2, where

j, k {1, 2,...,r}, then without loss of generality, we renumber the primefactors from the factorization ofn and we obtain

n= p21p22 p2sp2s+1 parr , with as+1, as+2,...,ar= 2.

Hence, we will writen = n1 n2, wheren1 = p21p22 p2s andn2= pas+1s+1 parr ,which means that (n1, n2) = 1, and by simple calculations, it is easy to seethat

(n) =(n1n2) =(n1)(n2) (e(n1)+(n1)n1)(e(n2)+(n2)n2) ==e(n1)

e(n2)+e(n1)(

(n2)n2)+(n1)(e(n2)n2)+(n1)(n2)n1(e(n2) + (n2)) + n1n2 e(n) + n1((n2) n2) + n1(e(n2) n2)+

+(n) n1e(n2) n1(n2) + n1n2 e(n) + (n) n.Thus, the demonstration is complete.

Remark 3. In Theorem4 and Theorem6 the equality in relations (15) and(17) hold, whenn= 1 orn is squarefree.

References

[1] J. Fabrykowski and M. V. Subbarao, The maximal order and the average order ofmultiplicative functione(n), Theorie des Nombres (Quebec, PQ, 1987), 201 206, deGruyter, Berlin-New York, 1989.

[2] N. Minculete, Several inequalities about arithmetic functions which use the e-divisors,Proceedings of The Fifth International Symposium on Mathematical Inequalities-MATINEQ 2008, Sibiu.

[3] N. Minculete, On certain inequalities about arithmetic functions which use the expo-nential divisors(to appear).

[4] J. Sandor,On exponentially harmonic numbers, Scientia Magna, Vol.2 (2006), No. 3,44 47.


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C. Pohoata, Harmonic quadrilaterals revisited 15

[5] J. Sandor and B. Crstici,Handbook of Number Theory II, Kluwer Academic Publishers,

Dordrecht/Boston/London, 2004.[6] J. Sandor and L. Toth ,On certain number-theoretic inequalities, Fibonacci Quart. 28

(1990), 255 258.[7] R. Sivaramakrishnan and C. S. Venkataraman,Problem 5326, Amer. Math. Monthly,

72 (1965), 915.[8] E. G. Straus and M. V. Subbarao,On exponential divisors, Duke Math. J. 41 (1974),

465 471.[9] M. V. Subbarao, On some arithmetic convolutions in The Theory of Arithmetic Func-

tions, Lecture Notes in Mathematics, New York, Springer-Verlag, 1972.[10] L. Toth, On Certain Arithmetic Functions Involving Exponential Divisors, Annales

Univ. Sci. Budapest., Sect. Comp. 24 (2004), 285 294.[11] L. Toth and N. Minculete, Exponential unitary divisors(to appear in Annales Univ.

Sci. Budapest., Sect. Comp.).

Harmonic quadrilaterals revisited

Cosmin Pohoata1)

Abstract. With a plethora of instruments ranging from harmonic divi-sions to various geometric transformations, synthetic projective geometryhas become very popular in olympiad geometry problems nowadays. Inthis note we expand a bit the literature concerned with applications ofthese techniques by discussing a topic of great importance (or better said,spectacularity) which might not be that covered in usual Euclidean geo-metry textbooks: harmonic quadrilaterals.

Keywords: harmonic, harmonic division, harmonic quadrilateral, inver-sion, symmedian.

MSC : 51A05, 51A20, 51A45.

1. Introduction

The name of harmonic quadrilateral dates from the middle of thenineteenth century, belonging to the famous mathematician R. Tucker [1].He gives a slightly different definition from the ones popular these days:

Let A, B, C, D, Mbe five points in plane such that A, B, C, D areconcyclic and M is the midpoint of the segment BD. Denote BD = 2m,i.e. M B = M D = m. Call vectorial inversion and denote (M, m2, BD)the transformation which firstly inverts with respect to the poleMwith con-

stantm2, and secondly takes the image of the inversed object inBD. Then,the quadrilateral ABCD is harmonic if and only if each of A, C could beobtained of the other after a vectorial inversion (M, m2, BD).

Remark. In the above statement it isnt necessary to firstly point outthe concyclity of A, B, C, D. In fact, if we perform a vectorial inversion

1)Tudor Vianu National College, Bucharest, Romania,pohoata [email protected]


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16 Articole

(M, m2, BD) of a pointTin plane, then its image under the transformation,

T, lies on the circumcircle ofT BD.In the following, we will firstly present the most used characterization

these days, a more synthetic view:ConsiderABCDa cyclic quadrilateral inscribed in C(O) andXa point

onC. Let an arbitrary lined intersect the lines XA, XB, XC, XD at M,N, P and Q, respectively. If

N M

N P =

QM

QP , then the quadrilateral ABCD

is called harmonic. In addition, ABCD is a harmonic quadrilateral if andonly ifAB CD= BC DA.

Proof. Denote by T the intersection of AC and BD. Since the pencilX(A,B,C,D) is harmonic, it is clear that the pencil A(A,B,C,D) is har-monic. Hence, if by Swe note the intersection ofBD with the tangent in Aw.r.t.C , then the quadruple (SDTB) is harmonic.

Similarly, the division (ST DB) is harmonic, where by S we denotedthe intersection ofBD with the tangent in C toC. Therefore, S S, i.e.BD, the tangent inA, respectively the tangent in Cw.r.t.C are concurrent.But since

SAD =

SB A, if follows that SA/SB = AD/BA.

Likewise,SC D =SBC, i.e. SD/SC = CD/BC. Hence, ABCD= BC DA.

2. Another classical property

Consider ABCD a cyclic quadrilateral inscribed inC, having O as theintersection of its diagonals. Then, B O is the B -symmedian inABCandDO is the D-symmedian inADCif and only ifABCD is harmonic.


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Proof. As above, one can notice that a cyclic quadrilateral is harmonicif and only the tangents in two opposite points w.r.t. the circumscribed circleconcur on the diagonal determined by the other two points, i.e. BB , DDand AC are concurrent in a point X (where by BB we denote the tangentin B to the circumcircle ofAB C).

On the other hand, it is well-known that in a triangle M N P, havingM1as the foot of the M-symmedian and M2 as the intersection of the tangent

in M to its circumcircle with the side BC, the quadruple (M2, N , M 1, P) isa harmonic division. Hence, BO is the B-symmedian inABC and DO istheD-symmedian inADCif and only ifBB,DD andACare concurrent.

Remark. In the initial statement it isnt really necessary to stress fromthe beginning thatABCD is cyclic. One can give the following enuntiation:

ConsiderABCD a convex quadrilateral inscribed inC, havingO as theintersection of its diagonals. Then,BO is theB-symmedian inABC andDO is theD-symmedian inADC if and only ifABCD is harmonic.

As probably yet noticed, the condition of BO, DO beeing the sym-

medians inABC, respectivelyADC, together with the harmonicity of(X ,A,O,C), imply the coincidence of the polars of X w.r.t. ABC andADC (the common polar beeing the line BD). Hence, the two circlescoincide as well.

Corollary 1. Consider ABC a triangle and T the intersection of thetangents at B and C w.r.t. the circumcircle of ABC. Then AT is the A-symmedian.


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18 Articole

Corollary 2. In the initially described configuration, AO, C O are the

A, respectively C-symmedians inBAD,BC D.

3. An IMO-type alternative definition

Consider ABCa triangle and D a point on its circumcircle. Draw theSimson line ofD w.r.t.ABC. This line cuts its sides BC, CA, AB in P,Q and R, respectively. Then, the quadrilateral ABCD is harmonic if andonly ifP Q= QR.

The above statement is slightly modifying the original one from IMO2003, problem 4:

LetABCD be a cyclic quadrilateral. LetP, Q andR be the feet of theperpendiculars fromD to the linesBC, C AandAB , respectively. Show thatP Q= QR if and only if the bisectors ofABC andADCmeet onAC.

First proof. As we first stated, it is well-known that P,Q,Rare collinearon the Simson line of D. Moreover, since DP C and DQC are rightangles, the points D, P, Q, C are concyclic and so DCA = DP Q =DP R. Similarly, sinceD,Q,R,Aare concyclic, we have DAC= DRP.Therefore,DCA DP R.

Likewise,DAB DQP andDBC DRQ. ThenDA

DC =

DR

DP =

DB QRBC

DB P QBA

=QR

P Q BA

BC.

Thus P Q = QR if and only if DA/DC = BA/BC, whence by theconverse of the bisector theorem, we deduce that it is equivalent with theconcurrence of the bisectors ofABC and ADC on AC.


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Second proof. Because DP

BC, DQ

AC, DR

AB, the circles

with diameters DC and DA contain the pairs of points P, Q and Q, R,respectively. It follows that P DQ is equal to or 180 , where =ACB.

Similarly, QDR is equal to or 180 , where = CAB. Then,by the law of sines, we have P Q= C D sin and QR= AD sin . Hence thecondition P Q= QR is equivalent with CD/AD= sin / sin .

On the other hand, sin / sin = CB/AB by the law of sines again,.ThusP Q= QR if and only ifCD/AD= CB/AB, i.e. AB CD= C B AD.

4. A probably new non-standard characterization

LetC(O) be a given circle and X, Y two points in its plane. It is well-known that there exist exactly two circles, 1 and 2, passing through X, Ytangent to the initial circleC. ConsiderA1 and A2 the mentioned tangencypoints. Draw a third mobile circle, 3, through X, Y touching the referencecircleC at B1 and B2, respectively. Then, the quadrilateral A1B1A2B2 isharmonic.

Proof. For example, perform an inversion with pole X. Hence,C ismapped into an other circle C and1,2into the lines1and2, respectively,not passing through X, representing the tangents from Y toC, i.e. A1, A2are the tangency points ofC with 1, 2.

Since3is mapped into the line3throughY

cutting C atB1 ,B2 , wededuce that the harmonicity ofA1B1A2B2is equivalent with the harmonicityofA1B

1A

2B

2 , which is clear, because A

1A

2 is the polar ofY

w.r.t.C.


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20 Articole

If in the above configuration we consider

Ca circle centered at infinity,

we deduce the following consequence:

Corollary. Considerd a line and X,Ytwo points in its plane. DenotebyA1,A2the points where the two circles passing through X,Yand tangentto d, touch the respective line. Let be an arbitrary circle through X, Y,which touches d at B1 and B2. Then, the quadruple (A1, B1, A2, B2) formsa harmonic division.

Their converses are also true. Precisely, if we considerA1, A2 similarlyas above and this time B1, B2 are two points onC such that A1B1A2B2 is aharmonic quadrilateral (degenerated or not), then the points X, Y, B1, B2lie on a same circle.

5. The Brocard points of a harmonic quadrilateral

In this section we present the existence of the Brocard points of a har-monic quadrilateral, proved by F. G. W. Brown [2], more as an other cha-racterization of this particular cyclic quadrilateral.


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Consider ABCD a cyclic quadrilateral. If there exists a pointX such

that the angles XAD, XBA, XCB, XDCare equal, then quadrila-teral ABCD is harmonic.

Proof. Denote the sides BC, C D, DA, AB bya, b, c, d; the diagonalsDB, AC bye, f; the area of the quadrilateral by Q; the circumradius by Rand each of the angles X AD, X BA, X CB, X DC by . Then

AXB = (A ) = A.Similarly, BX C= B, CXD= C, DX A= D.Now, since

AX

sin

= d

sin AXB

= d

sin( A) =

d

sin Aand

AX

sin(D ) = c

sin AXD =

c

sin( D) = c

sin D,

we deduce thatsin(D )

sin =

d sin D

c sin A,

or

cot = d

c sin A+ cot D.

Similarly,

cot w=

c

b sin D + cot C, ()and the cyclic analogues.

Hence,

cot w= d

c sin A+ cot D=

c

b sin D+ cot C=

b

a sin C+cot B =

a

d sin B+cot A

or

d

c sin A a

d sin B+

b

a sin C c

b sin D= cot A cot B+ cot C cot D.

Since ABCD is cyclic, 1/ sin A= 1/ sin C, cot A= cot C, etc; there-fore d

c+ b

a 1

sin A= c

b+ a

d 1

sin B.

But

(ab + cd)sin A= (ad + bc)sin B = 2Q;

hence ac = bd, i.e. ABCD is harmonic.

Its converse is also valid and can be proved on the same idea. More-over, there is a second Brocard point X, such that XAB = XBC =


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22 Articole

= XCD = XDA = . By simple manipulations on () and using

Ptolemys theorem, we obtain

cot =8R2Q

e2f2 .

Similarly, we deduce that

cot = 8R2Q

e2f2 .

Hence, = .Further, we will compute the distance X X. For this, letPbe the foot

of the perpendicular from X to B C and C P =x, P X=y. Then

x= C Xcos = b sin cos sin C = b(ab + cd)cot 2Q(1 + cot2 ) = bRc sin2and

y= x tan = b(ab + cd)

2Q(1 + cot2 ) =

2bR

c sin2 .

Similarly, ifx = PB, y = PX, where P is the foot of the perpen-dicular fromX to B C, then

x =d(ad + bc)cot

2Q(1 + cot2 )

and

y = d(ad + bc)

2Q(1 + cot2 ) .

Since by the law of cosinesX X2 =X B2+XB22XB XB cos XBXand by

sin B =ab + cd

2bd tan =

a2 + d2

2ad tan ,

using thatac = bd, i.e. ef = 2ac= 2bd, it follows that

XX = efbd cos cos2

2RQ =

b2d2 cos cos2RQ

.

6. Remarks on an Iranian concurrence problem

This problem firstly became popular on the Mathlinks forum for itsdifficulty. It is a nice result involving a concurrence of three lines in triangle.

LetAB Cbe a triangle. The incircle ofABC touches the sideBC atA, and the lineAA meets the incircle again at a pointP. Let the linesCPandBPmeet the incircle of triangleABC again atN andM, respectively.Prove that the linesAA, BN andCMare concurrent.

Iran National Olympiad 2002


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Proof. Consider Q the intersection of M N with BC. By Ceva andMenelaus theorem, it is clear that P A, BN and CMconcur if and only ifthe quadruple (Q ,B,A, C) is harmonic. So, the problem reduces to provingthat Q Q, where Q is the intersection of the tangent in P to the incirclewith B C.

Since (Q, B , D , C ) is a harmonic division, we deduce that the pencilP(Q, B , D , C ) is harmonic and by intersecting it with the incircle, it followsthat the quadrilateral P M AN is harmonic. Hence, the lines M N, the tan-gent in P, respectively, the tangent in A to the incircle are concurrent, i.e.Q Q.

We leave the readers a more difficult extended result (Figure 8.):

I. Suppose that in the configuration described above the linesAA, BNandCMare concurrent in a pointX. Similarly, one can prove the excirclerelated problem, meaning, the lines AA, BN1, CM1 are concurrent in apointX1, where byA

we denoted the tangency point of theA-excircle withthe sideBCand byN1,M1 the second intersections ofC P

, respectivelyBP

with theA-excircle, whereP is the second intersection of the lineAA withtheA-excircle. Prove that the linesP P, BC andXY are concurrent.

Moreover, we can go further and observe the next concurrence, whichremains as an other proposed problem for those who are interested (Figure9.):

II. Consider that in the above configuration the linesP P,BCandX Yconcur in a point Xa. Similarly, we deduce the existence of points Yb, Zc,with same right asXa, on the linesC A, respectivelyAB . Then, the trianglesABC and XaYbZc are perspective (i.e. the lines AXa, BYb, CZc concur);their perspector is X(75), i.e. the isotomic conjugate of the incenter.

The last remark, regarding the determination of the perspector, wascommunicated by Darij Grinberg, using computer dynamic geometry.


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24 Articole

Figure 8.

Figure 9.

In the following we leave again the readers, this time, a slightly moregeneral statement of the original problem:

III. Let ABC be a triangle. The incircle ofABC touches the sideBC at A, and let P be an arbitrary point on (AA). Let the linesCP andBPmeet the incircle of triangleABC again atN, N andM, M,respectively. Prove that the linesAA, B N andCM, respectivelyAA, B N,CM, are concurrent. (i.e. the linesBC, M N, MN, BC are concurrent,where B, C are the tangency points of the incircle with CA, respectivelyAB.)


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Figure 10

As a remark, it is clear that an extension similar to II of the aboveresult is usually not true.

7. On a perpendicularity deduced projectively

This problem is a result involving a perpendicularity as a consequenceof a projective fact. Moreover, we will give a pure by synthetic proof, basedon a simple angle chasing.

Figure 11

Let (O) be a circle and A a point outside it. Denote by B, C thepoints where the tangents from A w.r.t to (O) meet the circle, D the point


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26 Articole

on(O) for whichO

(AD),Xthe foot of the perpendicular from B to CD,

Ythe midpoint of the line segmentB Xand by Zthe second intersection ofDY with (O). Prove that Z A ZC(Figure 11).

BMO Shortlist 2007, proposed by the author of this paper

First proof. Let H the second intersection of CO with (O). ThusDC DH, soDHBX.

Because Y is the midpoint of (BX), we deduce that the division(B,Y,X, ) is harmonic, so also is the pencil D(B , Y , X , H ), and by in-tersecting it with(O), it follows that the quadrilateral HBZC is harmonic.

Hence the pencil C(HBZC) is harmonic, so by intersecting it with theline HZ, it follows that the division (AZT H) is harmonic, where A, T arethe intersections ofH Zwith the tangent in C, respectively withB C.

So, the line CH is the polar ofA w.r.t. (O), but CH BC is thepolar ofAas well, soA A, hence the pointsH,Z,Aare collinear, thereforeZA ZC.

Second proof. Let E be the midpoint of BC. So EY CD andEY BX. Since Y EB = DCB = DZB, we deduce that Y, E,Z, B lie on a same circle. Thus ZB C = ZE A (OA is tangent to thecircle Y EZ B).

But because ZB C = ZC A, we have that ZC A = ZE A, i.e.the points C, A, Z, E are concyclic. Hence CZA = CEA = 90, i.e.ZA ZC.

On the idea of the second solution, we give a more general statement:

IV. The tangents from a point A to the circle (O) touch it at thepoints B, C. For a point E (BC) denote: the point D on for whichE (AD); the second intersection X of the line CD with the circumcircleof the triangleBE D; the pointY on the lineBX for whichEY C D; thesecond intersectionZ ofDY with the circle; the intersectionTof the lineAC with BX; the second intersection H of the line AZ with the circle .Prove that the quadrilaterals AZEC, AZ YT are cyclic and BH AED.(Figure 12)

Proof. Since ZB E = ZB C = ZDC = ZY E, we have thatZBE = ZY E, i.e. the quadrilateral Z BY E is cyclic. Hence EB Y == EB X = EDX = EDC = Y ED. So EB Y = Y ED, meaningthat the lineAEDis tangent to the circumcircle of the quadrilateralZ BY E .

Thus, ZE A = ZB E = ZB C = ZCA, i.e. the quadrilateralAZEC is cyclic. So, ZY B = ZE B = ZAC = ZAT, whence thequadrilateral AZ Y T is also cyclic.

Finally, we have AEC= AZC= CDH. Hence, BHAED .


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Figure 12

Observe that when Eis the midpoint of the segment BCwe obtain theoriginal proposed problem.

8. On a classic locus problem as a recent IMO Team PreparationTest exercise

Consider the isosceles triangle ABC with AB = AC, and M the mid-point ofBC. Find the locus of the pointsPinterior to the triangle, for whichBP M+ CP A= .

IMAR Contest 2006

Figure 13


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28 Articole

First proof(by Dan Schwarz). We will start with the following claim:

Lemma. Prove that if, for a pointPinterior to the triangle, ABP == BC P, thenBP M+ CP A= .

Proof. Notice that the configuration is fully symmetrical:

ABP = BC P iff ACP = CBP

BP M+ CP A= iff CP M+ BP A=

and also that P AMguarantees the result, therefore we may assume w.l.o.g.(for constructions sake) that P is interior to the triangle ABM. The givenangle equality is readily seen to be equivalent with P , where is the arcinterior to ABCof the circumcircleK of triangle BC I, with I the incenterofAB C. It is also immediately seen that AB and ACare tangent toK.

Consider the Apollonius circleA for points A, M and ratio BA/BM;denote by U and V the points where AM meetsA. Clearly, B, C A.Then BA/BM = UA/UM = V A/V M , from which follows BA2/BM2 == U A V A/UM V M = U A V A/BM2, by symmetry and the power ofa point relation, so BA2 = U A V A, hence AB is tangent toA, again bythe power of a point relation. Therefore the two circlesK andA coincide.Now, prolong AP until it meetsK again at Q, and prolong P M until itmeetsK again at R. Then BA/BM = PA/PM = QA/QM, thereforeBA2/BM2 = P A QA/PM QM = BA2/P M QM, by the power of apoint relation, so BM2 = P M QM. But BM2 = P M RM, again by thepower of a point relation, so QM = RM, hence BQ = C R. It follows thatBP Q= CP R, and this is enough to yield BP M+ CP A= .

Alternatively, one can calculate ratios using the symmetrical points P

and Q with respect to AM, and denoting by N the meeting point ofP Q

andPQ. It can be obtained thatAM=AN, thereforeM Nand the resteasily follows as above.

Returning to the problem, we claim the locus is the arc (defined inthe above), together with the (open) segment (AM). Clearly P (AM)fills the bill, so from now on we will assume P / (AM). Also, as above,we will assume Pinterior to the triangle AB M(otherwise we work with thesymmetrical relations).

Assume P / , equivalent to BP C= ABC; then AB and ACare not tangent toK. TakeB and C to be the tangency points onK fromA, and M to be the midpoint of BC. We are now under the conditionsfrom the lemma (for triangle ABC), therefore BP M + CP A = .But BP M = BP M+(BP B+ M P M) and CP A= CP A CP C, where = 1 if BP C < ABC, respectively =1 ifB P C > ABC. We have BP B = CP C from the symmetry ofthe configuration, and BP M+ CP A= given; these relations therefore


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imply M P M = 0, which can only happen whenM,M andPare collinear,

i.e. P AM, which was ruled out from the start in this part of the proof.The contradiction thus reached confirms our claim.

Second proof. Denote the point D as the intersection of the line APwith the circumcircle ofB P C and S= DP BC.

Since SP C= 180 CP A, it follows that BP S= CP M.From the Steiner theorem applied to triangle BP C for the isogonals

P S and P M,SB

SC =

P B2

P C2.

On other hand, using the law of sines, we obtain

SBSC

= DBDC

sin SDBsin SDC

= DBDC

sin P CBsin P BC

= DBDC

P BP C

.

Thus by the above relations, it follows that DB/DC = P B/PC, i.e.the quadrilateralPBDC is harmonic. Therefore the pointA = BB CClies on the line P D (where by XX we denoted the tangent in X to thecircumcircle ofB P C).

If A = A, then lines AB and AC are always tangent to the circleBP C, and so the locus of P is the circle BI C, where I is the incenter ofABC. Otherwise, ifA =A, thenA =AMP SBB CC, due to the factthat A lies on P D and A = P S AM, and by maintaining the conditionthat A =A, we obtain thatP S AM, therefore P lies on (AM).

Next, after two synthetic approaches, we continue with a direct trigono-metric solution, actually the one which all three contestants who solved theproblem used.

Third solution. Consider m(BP M) = and m(CP M) = som(AP C) = 180 and m(AP B) = 180 . Also let m(CBP) =u,m(BC P) =v and denote AB = AC=l.

By the law of sines, applied to trianglesABPandACP, l

sin(180 ) =

= AP

sin(B

u)

and l

sin(180

)

= AP

sin(C

v)

. Hence

sin

sin =

sin(C v)sin(B u) . ()

Again, by the law of sines, this time applied in to the triangles BP M

and C P M, P M

sin u=

BM

sin and

P M

sin v =

CM

sin . Hence

sin

sin =

sin u

sin v. ()


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30 Articole

From () and () we deduce that

sin u

sin v =

sin(C

v)

sin(B u) , i.e. cos(B2u) cos B = cos(C 2v) cos C. Since B =C, it follows that cos(B 2u) == cos(B 2v).

IfB 2u= B 2v, then we have u= v i.e. P lies on (AM). Alterna-tively, ifB 2u = 2v B, then we have B = u+v, i.e. P lies on the arcBI C.

9. The orthotransversalline of a triangle refreshed as a Mathema-tical Reflections problem

LetABCbe a triangle and Pbe an arbitrary point inside the triangle.LetA, B , C be respectively the intersections ofAP and BC,BP and C A,CP andAB . ThroughPwe draw a line perpendicular to P Athat intersectsBCat A1. We define B1and C1analogously. Let P

be the isogonal conjugateof the point Pwith respect to triangle ABC. Then A1, B1, C1 all lie on asame line l that is perpendicular to P P.

Math. Reflections, 4/2006, problem O13, proposed by Khoa Lu Nguyen

In [7], Bernard Gibertnames the lineA1B1C1 the orthotransversallineofP.

Figure 14

First proof. Given four collinear points X, Y, Z, T, let (X ,Y,Z ,T)denote, we mean the cross-ratio of four points X, Y, Z, T. Given fourconcurrent lines x, y, z, t, let (x,y ,z ,t) denote, we mean the cross-ratio offour lines x, y , z , t. We first introduce the following claim:


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Lemma. Let ABC be a triangle and P be the isogonal conjugate of

an arbitrary pointP with respect to ABC. Then the six projections fromPandP to the sides of triangleABC lie on a circle with center the midpointofP P.

Proof. LetPa, Pb, Pc be the projections from Pto the sides B C, C A,AB. Similarly, let Pa , P

b, P

c be the projections from P

to the sides BC,CA, AB . CallO the midpoint ofP P. We need to show that Pa,Pb, Pc, P

a,

Pb, Pc lie on a circle with center O.Consider the trapezoid P PPaPa that has m(P PaP

a) = m(P

PaPa) == 90 andO the midpoint of (P P). Hence, O must lie on the perpendicularbisector ofPaP

a. By a similar argument, we obtain thatO also lies on the

perpendicular bisector ofPbPb and PcP

c.

Because P is the isogonal conjugate of P with respect to ABC, wehave BAP = PAC, or PcAP = P

APb. Hence, we obtain AP Pc == PbP

A. On the other hand, since quadrilaterals APcP Pb and APcP

Pbare cyclic, it follows that AP Pc = APbPc and P

bP

A = PbPcA =

= PbPcPc. Thus, APbPc = P

bP

cPc. This means that PbPcP

cP

b is

inscribed in a circle. We notice that the center of this circle is the intersectionof the perpendicular bisectors of PbP

b and PcP

c, which is O. In the same

manner, we obtain thatPcPaPcP

a is inscribed in a circle with center O. Thus

these two circles are congruent as they have the same center and pass througha common point Pc. Therefore, these six projections all lie on a circle withcenter O.

Back to the problem, let Pa, Pb, Pc be the projections from P to thesides BC, CA, AB and O be the midpoint of P P. Then (O) is thecircumcircle of triangle PaPbPc. We will show a stronger result. In fact, l isthe polar ofPw.r.t. the circle (O). We will prove thatA1 lies on the polarofPw.r.t. the circle (O), and by a similar argument so does B1, respectivelyC1.

LetB2 and C2be the intersections of the lineP A1 withAC andAB,

respectively. Denote by M and Nthe intersections of the line P A1 with thecircle (O). It is suffice to prove that (A1, P , M , N ) = 1.

ConsiderXthe projection fromPtoBC. Then five pointsP,Pb,Pc,X,

A lie on the circle (a) with diameter P A. Since (AA1, AA, AC, AB) == 1, we obtain thatP PcXPb is a harmonic quadrilateral. This yields thatPbPc and the tangents at P and X of the circle (a) are concurrent at apoint U. Since the tangent at P of the circle (a) is P A1, it follows thatU is the concurrence point ofPbPc, P A1 and the perpendicular bisector of(P X). Consider the right triangleP XA1 atX. SinceUlies onP A1 and theperpendicular bisector of (P X), we deduce that Uis the midpoint of (P A1).Therefore U lies on the line PbPc.


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32 Articole

Since APb

AB2 = A

Pc

AC2 = A

P2, it follows that PbPcC2B2

is cyclic. Hence, U B2 U C2 = U Pb U Pc. On the other hand, we haveU Pb U Pc= U M U N, because PbPcM Nis cyclic.

Thus, we obtain U B2 U C2= U M U N.Again, since (AA1, A

, AC, AB) =1, we have (A1, P , B2, C2) == 1. But, because U is the midpoint ofA1P, it follows that U B2 U C2 ==U A21. Hence U A

21= U M U N, i.e. (A1, P , M , N ) = 1.

Second proof for the collinearity ofA1, B1, C1. (by Darij Grinberg,after Jacques Hadamard [6]) LetKbe an arbitrary circle centered at P;the polars of the points A, B, C, A1, B1, C1 w.r.t.K are called a, b, c, a1,b1, c1. After the construction of polars, we have a

P A, b

P B, c

P C,

a1 P A1,b1 P B1andc1 P C1. Since the pointA1 lies on BC, the polara1 passes through the intersectionb c. Froma1 P A1 andP A1 P A, wehave a1P A; froma P A, thus we geta1 a. Hence, a1 is the line passingthrough the intersection b c and orthogonal to a. Analogously, b1 is theline passing through the intersectionc a and orthogonal to b, and c1 is theline passing through the intersection a b and orthogonal to c. Therefore,a1, b1 and c1 are the altitudes of the triangle formed by the lines a, b and c;consequently, the lines a1, b1 and c1 concur. From this, we derive that thepoints A1, B1 and C1 are collinear.

Following the ideas presented above, we leave the readers a similarcollinearity:

V. Consider ABC a triangle and P a point in its plane. Let R, S,T, X, Y, Z be the midpoints of the segments BC, CA, AB, P A, P B andP C, respectively. Draw the lines through the X, Y, Z, orthogonal to P A,P B, respectively P C; these lines touch ST, T R, RS at A1, B1 and C1,respectively. Prove that the pointsA1, B1, C1 lie on the same line.

Figure 15


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10. Bellavitis theorem on balanced quadrilaterals

We call ABCD a balanced[8] quadrilateral if and only if AB CD == B C DA. Particulary, if ABCD is cyclic we come back to the harmonicquadrilateral. However, this class contains other quadrilaterals, for exampleall kites.

Let the lengths of the sides AB,BC,C Dand DA of a (convex) quadri-lateralABCDbe denoted bya,b,c and drespectively. Similarly, the lengthsof the quadrilaterals diagonalsACand BD will be denoted byeandf. LetEbe the point of intersection of the two diagonals. The magnitude ofDABwill be referred to as , with similar notation for the other angles of thequadrilateral. The magnitudes ofDAC, ADB etc. will be denoted byB, C a.s.o. (see Figure 16.). Finally, the magnitude of CED will bereferred to as .

Figure 16.

Theorem. (Bellavitis 1854) If a (convex) quadrilateral ABCD isbalanced, then

B+ C+ D+ A= A+ B+ C+ D = 180.

Note that the convexity condition is a necessary one. The second equa-lity sign does not hold for non-convex quadrilaterals.

Proof. (by Eisso J. Atzema [8]) Although in literature it is knownthat Giusto Bellavitis himself gave a proof using complex numbers, a trigono-metric proof of his theorem follows from the observation that by the law ofsines for any balanced quadrilateral we have

sin B sin D = sin B = sin D,or

cos(B+ D) cos(B D) = cos (B+ D) cos(B D).


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34 Articole

That is,

cos(B+ D) cos(B + B) = cos (B+ D) cos(B + B),or

cos(B+ D) + cos (+ ) = cos (B+ D) + cos (+ ).

By cycling through, we also have

cos(C+ A) + cos ( + ) = cos (C+ A) + cos ( + ).

Since cos ( + ) = cos (+ ), adding these two equations gives

cos(B+ D) + cos (C+ A) = cos (B+ D) + cos (C+ A),

or

cos1

2(

C+

B+

A+

D)

cos1

2(

B+

D C

A

)

= cos1

2(B+ C+ D+ A) cos1

2(B+ D C A).

Now, note that

B+ D C A= 360 2 and likewise

B+ D C A= 2 .Finally,

1

2(C+ B+ A+ D) +

1

2(B+ C+ D+ A) = 180

.

It follows that

cos1

2(C+ B+ A+ D) cos( +1

2(+ )) =

= cos12

(C+ B+ A+ D) cos( 12

(+ )),

or

cos1

2(C+ B+ A+ D) cos()cos1

2(+ ) = 0.

Clearly, if neither of the last two factors is equal to zero, the first factorhas to be zero and we are done. The last factor, however, will be zero if andonly ifABCD is cyclic. It is easy to see that any such quadrilateral has theangle property ofBellavitis theorem. Therefore, in the case thatABCD is

cyclic,Bellavitis theorem is true. Consequently, we may assume thatABCDis not cyclic and that the third term does not vanish. Likewise, the secondfactor only vanishes in case ABCD is orthodiagonal. For such quadrilaterals,we know that a2 +c2 = b2 +d2. In combination with the initial conditionac= bd, this implies that each side has to be congruent to an adjacent side.In other words, ABCD has to be a kite. Again, it is easy to see that in thatcase Bellavitis theorem is true. We can safely assume thatABCD is not akite and that the second term does not vanish either.


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Moreover, for a synthetic solution one can see Nikolaos Dergiades proof

in [9]. Also, a solution using inversion exists. See [11].

References

[1] E. J. Atzema, A Theorem by Giusto Bellavitis on a Class of Quadrilaterals, ForumGeometricorum, (2006), 181 185.

[2] F.G.W. Brown,Brocard points for a quadrilateral, The Mathematical Gazette, Vol. 9,No. 129. (May, 1917), 83 85.

[3] N. Dergiades, A synthetic proof and generalization of Bellavitis theorem, Forum Ge-ometricorum, (2006) 225 227.

[4] D. Djukic, V. Jankovic, I. Matic, N. Petrovic, The IMO Compendium, Springer, (May12, 2005).

[5] B. Gibert, Orthocorrespondence and Orthopivotal Cubics, Forum Geometricorum,(2003), 1 27.

[6] D. Grinberg,Problem: The ortho-intercepts line.[7] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium,

(1998) vol. 129, 1 285.[8] E. M. Langley, A Note on Tuckers Harmonic Quadrilateral, The Mathematical

Gazette, Vol. 11, No. 164. (May, 1923), 306 309.[9] V. Nicula, C. Pohoata, Diviziune armonica, GIL Publishing House, Zalau (2007).

[10] C. Pohoata, Harmonic Division and Its Applications, Mathematical Reflections, No.4 (2007).

[11] See also http://www.faculty.evansville.edu/ck6/encyclopedia/ETC.html;http://www.mathlinks.ro/viewtopic.php?t=88077,Mathlinks Forum.


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Problems and Solutions from SEEMOUS 2011 Competition 37

Bucuresti), 14 medalii de argint: Titiu Radu (Universitatea din Bucuresti),

Barzu Mihai(Departamentul de Informatica, Universitatea ,,Alexandru IoanCuza, Iasi), Burtea Cosmin (Universitatea ,,Alexandru Ioan Cuza, Iasi),Filip Laurian(Universitatea din Bucuresti), Fodor Dan(Universitatea ,,Ale-xandru Ioan Cuza, Iasi), Hlihor Petru(Universitatea din Bucuresti), SarbuPaul (Universitatea Tehnica ,,Gheorghe Asachi, Iasi), Popescu Roxana-Irina (Universitatea din Bucuresti), Mesaros Ionut(Universitatea Tehnica,Cluj-Napoca), Tuc a Laurentiu (Universitatea Politehnica Bucuresti), Glc aDragos (Departamentul de Informatica, Universitatea ,,Alexandru IoanCuza, Iasi), Beltic Marius Jimy Emanuel (Universitatea ,,Alexandru IoanCuza, Iasi), Sasu Robert (Universitatea Politehnica Bucuresti), Plesca Iu-lia (Universitatea ,,Alexandru Ioan Cuza, Iasi) si 19 medalii de bronz:

Cervicescu Virgil (Academia Tehnica Militara, Bucuresti), Raicea Marina(Academia Tehnica Militara, Bucuresti), Mihaila Stefan (Departamentul deInformatica, Universitatea ,,Alexandru Ioan Cuza, Iasi),Munteanu Alexan-dra-Irina (Universitatea din Bucuresti), Vasile Mihaela Andreea (Univer-sitatea Politehnica Bucuresti), Mocanu Maria-Cristina (Departamentul deInformatica, Universitatea ,,Alexandru Ioan Cuza, Iasi), Petre Luca (Uni-versitatea Politehnica Bucuresti), Rublea Alina (Universitatea PolitehnicaBucuresti), Bobes Maria Alexandra (Universitatea ,,Babes Bolyai, Cluj-Napoca), Genes Cristian (Universitatea Tehnica ,,Gheorghe Asachi, Iasi),Kolumban Jozsef (Universitatea ,,Babes-Bolyai, Cluj-Napoca), Craus Sabi-na(Universitatea ,,Alexandru Ioan Cuza, Iasi), Damanian Lavinia-Mariana(Departamentul de Informatica, Universitatea ,,Alexandru Ioan Cuza, Iasi),

Moldovan Dorin Vasile (Universitatea Tehnica, Cluj-Napoca), Mincu Di-ana (Universitatea Politehnica Bucuresti), Birghila Corina (Universitatea,,Ovidius, Constanta), Vlad Ilinca (Universitatea Tehnica ,,Gh. Asachi,Iasi), Sav Adrian-Gabriel (Universitatea ,,Alexandru Ioan Cuza, Iasi), Al-ban Andrei(Universitatea din Bucuresti).

Echipa Universitatii Politehnica din Bucuresti a fost formata din 8studenti, dintre care 6 au reusit sa obtina medalii.

Proba de concurs a constat n rezolvarea a 4 probleme pe durata a 5ore. Vom prezenta mai jos solutiile problemelor, precum si comentariile derigoare.

2. Probleme, Solutii si ComentariiProblema 1. Pentru un ntreg datn1, fief : [0, 1]R o functie

crescatoare. Demonstrati ca

10

f(x)dx (n + 1)1

0

xnf(x)dx.

Gasiti toate functiile continue crescatoare pentru care egalitatea are loc.


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38 Articole

Solutia 1. Pentru n dat si x, y [0, 1] integram pe intervalul [0, 1] nraport cu x si n raport cu y inegalitatea evidenta

(xn yn)(f(x) f(y)) 0, (1)si obtinem

10

10

(xn yn)(f(x) f(y))dxdy 0,

sau1

0 xnf(x)dx

1

0 yndy

1

0 f(x)dx 1

0 xndx

1

0 f(y)dy+1

0 ynf(y)dy

0,

care ne da inegalitatea ceruta.Daca f este continua, cand egalitatea din enunt are loc, n (1) are loc

de asemenea egalitate, deci functiaftrebuie sa fie constanta.

Solutia a 2-a. Prin schimbare de variabila avem

(n + 1)

10

xnf(x)dx=

10

f( n+1

t)dt.

Cumf este crescatoare, rezulta ca f(x) f( n+1x), x [0, 1], de undeprin integrare obtinem inegalitatea ceruta.

Solutia a3-a. Problema este cazul particular al inegalitatii luiCebasev:Fief1, f2, . . . , f n: [a, b] R functii integrabile, pozitive, monotone.1)Dacaf1, f2, . . . , f n sunt fie toate monoton crecatoare, fie toate mono-

ton descrescatoare, atunci

ba

f1(x)dx b

a

f2(x)dx b

a

fn(x)dx (b a)n1b

a

f1(x)f2(x) . . . f n(x)dx.

2)Dacaf1, f2, . . . , f nsunt de monotonii diferite, atunci inegalitatea estede semn contrar.

In cazul nostru, pentru n = 2 si f1(x) = f(x), f2(x) = xn obtinem

concluzia din enunt.

Solutia a 4-a. Notam C=

10

f(x)dx si fie g : [0, 1] R,

g(x) =f(x) C. Deci1

0

g(x)dx= 0.


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Astfel, inegalitatea noastra este echivalenta cu 0 1

0

xng(x)dx. Cumg

este crescatoare, rezulta ca existad = sup{x [0, 1] :g(x) 0}. Vom obtine1

0

xng(x)dx= d

0

xn|g(x)|dx +1

d

xn|g(x)|dx.

Cum

10

g(x)dx= 0, avem

d0

|g(x)|dx =1

d

|g(x)|dx. Inmultind cu dn,

rezulta urmatorul sir de inegalitati evidente

d0

xn|g(x)|dx d

0

dn|g(x)|dx=1

d

dn|g(x)|dx 1

d

xn|g(x)|dx,

de unde concluzia problemei noastre. Comentariu. Problema a fost considerata de multi concurenti drept

una foarte usoara. Cu toate acestea, au existat destui care n-au reusit saobtina punctajul maxim. De asemenea trebuie remarcat si faptul ca au exis-tat concurenti care nu si-au mai amintit sau pur si simplu nu stiau inegalitatealui Cebasev.

Problema 2. Fie A = (aij) o matrice reala cu n linii si n coloaneastfel nc at An

= 0 si a

ijaji

0 pentru orice i, j. Demonstrati ca existadoua numere nereale printre valorile proprii ale lui A.

Ivan Feshchenko, Kiev, Ucraina

Solutie. Fie 1, . . . , n toate valorile proprii ale acestei matrice. Poli-nomul caracteristic asociat matriceiA este

P() = det(In A) =n a1n1 + . . . + (1)nan,

unde a1 =n

i=1

aii, an= det A si ai suma minorilor principali de ordin i ai lui

A; de exemplu, a2=i,j

0 aijaji 0

.

Din ipoteza rezulta caaii = 0, decin

k=1

k = 0. Suma i


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40 Articole

conjugata complexa a acestei valori este valoare proprie pentru A, deci ceea

ce trebuia demonstrat. Comentariu. Problema a fost considerata de dificultate medie. Mai

mult de jumatate dintre concurenti n-au reusit sa o rezolve complet. Dinpacate multi dintre studentii romani au ntampinat dificultati mari n abor-darea acestei probleme. Acest lucru arata ca este nevoie de mai multa AlgebraLiniara n programele noastre din universitati.

Problema 3. Fie vectoriia, b, c Rn. Aratati ca(||a|| b, c)2 + (||b|| a, c)2 ||a||||b||(||a||||b|| + | a, b |)||c||2,

undex, y este produsul scalar al vectorilor x, y six2 = x, x.Dan Schwarz, Bucuresti, Romania

Solutia 1. (Solutia trigonometrica.) Vom demonstra mai ntai ca, dativectoriiu, v Rn, cu||u|| = ||v|| = 1 (deciu, v Sn1), atunci

sup||x||=1

u, x2 + v, x2

= 1 + | u, v |.

Consideram reprezentarea vectoriala unica x= x+x cux span(u, v)si x span(u, v). Atunci

1 = ||x||2 = x + x, x + x== ||x||2 + 2 x, x+ ||x||2 = ||x||2 + ||x||2 (relatia lui Pitagora),

deci

||x

|| 1.Avem

| u, x

|=

| u, x

| | u, y

|, undey = 0 dacax = 0 si

y = x

||x|| daca x = 0 (deci||y|| = 1). Analog| v, x | = | v, x | | v, y |.

Maximul este asadar obtinut cand x span(u, v).Deci problema noastra vectoriala a fost transferata n spat iul de dimen-

siune 2, cu vectorii unitariu,v six. Capetele vectorilor usi v mpart cer-culS1 n patru arce, fiecare masurand cel mult (si posibil 0, candu = v);capatul lui x va fi ntr-unul din ele. Fie masura acelui arc si , , cu + = , masurile arcelor dintre capatul luix si capetele acelui arc. Atunciu, x2 + v, x2 = cos2 +cos2 , din binecunoscuta interpretare geometricaa produsului scalar (independent de dimensiunea spatiului). Atunci

u, x2

+ v, x2

= cos

2

+ cos

2

= 1 +

1

2 (cos 2 + cos 2) == 1 + cos( + )cos( ) 1 + | cos | = 1 + | u, v |,

cu egalitate n cazurile evidente:

cand = = /2, deci cand| u, x | = | v, x |, asadarx= (u v)/|| u v||,

unde semnele sunt astfel ncat 0


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cand = /2, deci cand

u, v

= 0, i.e. u

v, pentru orice

x span(u, v).

Atunci, pentru orice a, b, c nenuli, luam u = a

||a|| , v = b

||b|| , x= c

||c|| ,asadar

u, x2 = 1||a| | | |c||2a, c2,v, x2 = 1||b| | | |c||2

b, c2

,u, v =

a, b

||a| | | |b|| ,deci relatia demonstrata devine

1

||a

||a, c2 + 1

||b||

b, c

2

1 +

|

a, b

|

||a| | | |

b||

||c||2,

echivalent cu inegalitatea ceruta, de asemenea adevarata pentru a, b sau cegale cu zero.

Solutia a2-a.(Solutie cu forme patratice) Pentru ||x|| = ||u|| = ||v|| = 1avem

0 ||x + u + v||2 = x + u + v,x + u + v ==2 + 2 + 2 + 2 x, u + 2x, v + 2u, v ,

o forma patratica care ia valori pozitive pentru orice parametri reali , , ,deci corespunzand unei matrice pozitiv semidefinita

1 x, u x, vx, u 1 u, vx, v u, v 1

.Minorii principali de ordinul 1 sunt asadar pozitivi, ceea ce ne da semi-

pozitivitatea normei; minorii principali de ordinul 2 sunt pozitivi, ceea cene da inegalitatea Cauchy-Buniakowski-Schwarz, 1 u, v2; de asemenea,determinatul matricii este pozitiv

= 1 (u, v2 + x, u2 + x, v2) + 2 u, v x, u x, v 0,care poate fi scris

x, u2 + x, v2 1 + | u, v | | u, v |(1 + | u, v | 2| x, u x, v |).Darx, u2 + x, v2 2| x, u x, v |, care nlocuit n relatie ne da

(1 | u, v |)(1 + | u, v | 2| x, u x, v |) 0.Apoi, fie 1 =| u, v |, cand 1 +| u, v | = 2 2| x, u x, v | (din

inegalitatea Cauchy-Buniakowski-Schwarz), fie 1 >| u, v |, ceea ce implica1 + | u, v | 2| x, u x, v |. Asadar, ntotdeauna

x, u2 + x, v2 1 + | u, v |.


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42 Articole

Solutia a 3-a (Solutia cu multiplicatorii lui Lagrange). Definim

L(x, ) = u, x2 + v, x2 (||x||2 1)si consideram sistemul

L

xi= 2ui u, x + 2vi v, x 2xi = 0, pentru 1 i n,

siL

= ||x||2 1 = 0.

Avem

0 = 1

2

n

i=1 xiL

xi

=

u, x

n

i=1 uixi+ v, xn

i=1 vixi n

i=1 x2i =

= u, x2 + v, x2 ||x||2 = u, x2 + v, x2 .Pe de alta parte,

0 = 1

2

ni=1

uiL

xi= u, x

ni=1

u2i + v, xni=1

viui ni=1

xiui =

= u, x ||u||2 + v, x u, v u, x = u, x + u, v v, x u, x ,si analog pentru v, deci obtinem sistemul de doua ecuatii (n variabilele u, xsiv, x)

(1 ) u, x + u, v v, x = 0,u, v u, x + (1 ) v, x = 0.Determinantul al matricei acestui sistem este = (1 )2 u, v2 ,

si daca este nenul, unica solutie este cea trivialau, x =v, x = 0, candexpresia noastra atinge un minim evident egal cu zero (atunci = 0, aceastanseamna u, v = 1, care s-ar traduce nu = vsix span(u, v)). Asadar,suntem interesati de = 0 (pentru toate celelalte puncte critice), duc and la= 1u, v, deci = 1+ | u, v | ntr-un punct de maxim si = 1| u, v |ntr-un punct critic. Exista cateva cazuri particulare.

Cand u= v, deciu, v = 1, situatia este simpla. Avem maximul 2cand x= u si minimul 0 cand x u.

Candu v, deci u, v = 0, atunci(deci si expresia) este 1 n punctelede maxim, pentru oricex span(u, v), si(deci si expresia) este 0 n punctelede minim, pentru orice x span(u, v).

Solutia a 4-a. Vom reduce problema la cazul R3. Consideram o bazaortonormala pentru care primii trei vectori genereaza subspatiul vectorialgenerat de

a,b,c

. O transformare ortogonala pastreaza norma si produsul

scalar, deci este suficient sa demonstram inegalitatea n noua baza n care a,b, c au primele trei componente nenule.


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In R3 vom folosi un argument de geometrie. Mai ntai, observam ca

orice schimbare de semn a vectorilor a, b, c nu schimba semnul inegalitatiiinitiale.

Fie = (b, c), = (c, a) si = (a, b). Aranjam semnele astfel

ncat ,

0,

2

si consideram triedrul OABC avand laturile OA, OB,

OCdeterminate de vectorii a,b,c astfel ncat proiectand vectorulc pe planulOAB, proiectia sa sa ramana n interiorul unghiului OAB.

Este cunoscut faptul ca, daca x, y, z satisfac x + y = z, atuncicos2 x+ cos2 y + cos2 z 2cos x cos y cos z = 1. Astfel, avem de aratat cacos2 + cos2 1 + | cos |. Se observa ca, daca unul dintre unghiurile sau este egal cu

2, atunci inegalitatea este evidenta .

Proiectam punctul Cpe planul OAB n punctul P si consideram1=P OB, 1 = P OA. Proiectam P pe OA, respectiv OB n punctele A1,respectiv B1. In triunghiurile dreptunghice OB1C si OB1P avem

cos2 = OB21OC2

OB21

OP2= cos2 1.

Analog avem cos2 cos2 1. Din 1+ 1= obtinemcos2 1+ cos

2 1= 1 + 2 cos 1cos 1cos cos2 .Aratam ca 1 + 2 cos 1cos 1cos cos2 1 + cos .Daca cos = 0, atunci inegalitatea este evidenta. Daca cos = 0,

inegalitatea este echivalenta cu

2cos 1cos 1 1 + cos = cos(1+ 1) = 1 + cos 1cos 1 sin 1sin 1,care n final va da cos(1 1) 1.

Solutia a 5-a (Nicolae Beli). Asemanator cu solutia 4, presupunem

,

0,

2

.

Trebuie aratat ca

| cos | cos2 + cos2 1 =12

(cos2 + cos 2) = cos( + )cos( ).

Cum | | 2

, avem cos( ) 0. Prin urmare, putem presupune

ca cos( ), cos(+ )> 0. Insa , , sunt unghiurile triedrului OABC(care poate fi si degenerat), deci avem 0 + . Rezulta ca0 < cos( + ) cos , care mpreuna cu 0 < cos() 1 implicacos( + )cos( ) cos .

Comentariu. Aceasta problema s-a dovedit a fi cea mai grea problemadin concurs, doar cativa reusind sa o rezolve complet. Dintre romani, doarVictor Padureanu a rezolvat problema. In rezolvarea problemei de fata s-aufolosit aceleasi ingrediente ca ntr-o alta problema, un pic mai veche, data la


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44 Articole

ultimul test de selectie a lotului olimpic al Romaniei din anul 2007. Ea suna

cam asa:Problema nrudita. Pentru n 2 ntreg pozitiv, se considera nu-

merele realeai, bi, i= 1, n, astfel ncatni=1

a2i =ni=1

b2i = 1,ni=1

aibi = 0.

Sa se arate ca ni=1

ai

2+

ni=1

bi

2 n.

Pentru solutii si comentarii recomandam cititorilor Romanian Mathe-matical Competitions 2007, paginile 97100.

Problema 4. Fie f : [0, 1] R o functie crescatoare de clasaC2.

Definim sirurile date de Ln = 1

n

n1k=0

f

k

n

si Un =

1

n

nk=1

f

k

n

, n 1.

Intervalul[Ln, Un] se mparte n trei segmente egale. Demonstrati ca, pentru

n suficient de mare, numarulI=

10

f(x)dx apartine segmentului din mijloc

dintre aceste trei segmente egale.Alexander Kukush, Kiev, Ucraina

Solutia 1. Enuntam si demonstram mai ntai urmatoarea lema :

Lema. Pentruf C2[0, 1]: Ln= I f(1) f(0)2n

+O 1

n2

, n .

Demonstratie. Notam C=f(1) f(0). Consideram

I Ln=n1k=0

k+1n

kn

f(x) f

kn

dx=

=n1

k=0k+1n

knf

k

nx k

n+1

2f(xkn)x

k

n2

dx==

1

2n2

n1k=0

fk

n

+ rn, (1)

unde

|rn| 12

max|f| n1k=0

1

3

x k

n

3k+1nkn

constn2

.


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Aici

x

kn kn ,k+ 1n sunt punctele intermediare din teorema luiTaylorn vecinatatea punctului

k

n.

Analog

10

fdx 1n

n1k=0

f

k

n

=

n1k=0

k+1n

kn

f(xkn)x kn

dx= O

1

n

.

Deci n partea dreapta a relatiei (1) avem

1

n

n1

k=0 f

kn 1

0

f

dx= C, n ,

si eroarea n partea dreapta a relatiei (1) este de ordin O

1

n2

.

Deci din (1) avem

I Ln= C2n

+ O 1

n2

, n .

Acum ne ntoarcem la solutia problemei noastre si avemUn= Ln+C

n.

Fie xn= Ln+kC

3n, k = 1, 2. Atunci

xn= I+C

n

k3 1

2

+ O

1n2

.

Pentru k = 1 avem k

3 1

2 0, deci xn > I pentru n suficient de mare.

Asadar, pentru n suficient de mare

Ln+ C

3n< I < Ln+

2C

3N.

Solutia a 2-a. Fie F(t) =

t

0

f(x)dt. Atunci avem

I=

10

f(x)dx= F(1) F(0) =

=F(1) F

n 1n

+ F

n 1

n

F

n 2

n

+ . . . + F

1

n

F(0).


46/68

46 Articole

Din formula lui Taylor de ordinul 2, vom obtine (pentru punctele

0, 1n

, 2n

, . . . , n 1n

):

F(x) =F

k

n

+

x k

n

F

k

n

+

1

2

x k

n

2F(),

cu

k

n, x

, k = 0, n 1, sau

F(x) =F

k

n

+

x k

n

f

k

n

+

1

2

x k

n

2f().

Pentru x = 1

n, . . . , 1 avem

Fk+ 1

n

F

kn

=

1

nfk

n

+

1

2n2f(k),

cu k

k

n,k+ 1

n

, k = 0, n 1.

Insumand aceste relatii pentru k = 0, 1, n 1 obtinemI=Ln+

1

2nn,

unde n este suma Riemann pentru

10

f(x)dx= f(1) f(0).

Intervalul din mijloc este2

3Ln+

1

3Un,

1

3Ln+

2

3Un

= [un, vn].

Daca f(0) =f(1), atunci f este constanta, ceea ce nu se poate.Presupunem ca f(1)> f(0).

Avemn(Iun) = n3

(LnUn) +12

n=1

3(f(0)f(1))+1

2n si trecand

la limita cand n , obtinem n(I un) 16

(f(1) f(0)) > 0, de undepentru n suficient de mare, I > un.

In acelasi mod, avem

n(vn

I) =2n

3

(Un

Ln)

1

2

n

1

6

(f(1)

f(0))> 0.

Remarca. RelatiaI=Ln+ 1

2nn mai poate fi dedusa, din urmatorul

fapt:Fief : [0, 1] R derivabila cu derivata integrabila pe [0, 1]. Atunci

limn

n

10

f(x)dx 1n

nk=1

f

k

n

= f(0) f(1)2

.


47/68


Demonstratie. Fie = 0, 1n, 2n , . . . ,n 1n , 1 o diviziune a interva-lului [0, 1]. Notam

Mk = supx[k1n ,

kn ]

f(x), mk = inf x[k1n ,

kn ]

f(x), k= 1, n.

Atunci avem

En= n

I 1

n

nk=1

f

k

n

= n

nk=1

kn

k1n

f(x)dx 1n

f

k

n

=

=n

nk=1

kn

k1n

f(x) fkndx .

Aplicand teorema lui Lagrange pe intervalul

x,

k

n

, rezulta ca exista

ck(x)

k 1n

,k

n

astfel ncatf(x) f

k

n

=

x k

n

f(ck(x)).

AtunciEn= nn

k=1

kn

k1n

x k

n

f(ck(x))dx.

Dar mk f(ck(x)) Mk, deci avem 12n

nk=1

Mk En 12n

nk=1

mk.

Trecand la limita obtinem limn

En= f(0) f(1)

2 .

Comentariu. Problema a fost considerata una foarte dificila, iar putiniconcurenti au reusit sa o rezolve n ntregime, cu toate ca problema putea fiabordata si de un elev foarte bine pregatit de clasa a XII-a. Dintre concurentiiromani, doar cativa au reusit sa obtina mai mult de jumatate din punctaj.

3. Concluzii

Olimpiada pentru studenti SEEMOUS din acest an a fost o reusita pen-

tru studentii romani, care s-au descurcat onorabil. Multi dintre concurentiau aratat destul de bine pregatiti, dovada fiind medaliile obtinute. Pe de altaparte trebuie remarcat si faptul ca parcurgerea unor notiuni din anul I defacultate n liceu poate constitui un real avantaj pentru cei care erau foartebuni la matematica n liceu.

In ceea ce priveste studentii Universitatii Politehnica din Bucuresti,putem spune ca au avut o comportare decenta. Din pacate niciunul dintre ein-a reusit sa obtina medalie de aur, cu toate caLaurentiu Tuc aa fost foarte


48/68


49/68

Proposed problems 49

PROBLEMS

Authors should submit proposed problems to [email protected] or [email protected]. Files shold be in PDF or DVI format. Once a problemis accepted and considered for publication, the author will be asked to submit theTeX file also. The referee process will usually take between several weeks and twomonths. Solutions may also be submitted to the same e-mail adress. For this issue,solutions should arrive beforeNovember 15, 2011.

Editors: Mihai Cipu, Radu Gologan, Calin Popescu, Dan Radu

Assistant Editor: Cezar Lupu

PROPOSED PROBLEMS

323. Let C be the set of the circles in the plane and L be the set of thelines in the plane. Show that there exist bijective maps f, g :C L suchthat for any circle C C, the line f(C) is tangent at C and the line g(C)contains the center ofC.

Proposed by Marius Cavachi, Ovidius University of Constanta,

Romania.

324. Consider the set

K :=

f

4

20 , 6

500

| f(X, Y) Q[X, Y]

.

(a) Show that Kis a field with respect to the usual addition and multi-plication of real numbers.

(b) Find all the subfields of K.(c) If one considers K as a vector space QK over the field Q in the

usual way, find the dimension of QK.(d) Exhibit a vector space basis of QK.

Proposed by Toma Albu, Simion Stoilow Institute of Mathematics

of the Romanian Academy, Bucharest, Romania.

325. We call toroidal chess board a regular chess board (of arbitrarydimension) in which the opposite sides are identified in the same direction.Show that the maximum number of kings on a toroidal chess board of dimen-sions m n (m, nN) such that each king attacks no more than six otherkings is less than or equal to

4mn

5 and the inequality is sharp.

Proposed by Eugen Ionascu, Columbus State University, Columbus,

GA, USA.

326. For t > 0 define H(t) =n=0

tn

n!(n + 1)!. Show that

limt

t3/4H(t)

exp(2

t)=

1

2

.


50/68

50 Problems

Proposed by Moubinool Omarjee, Jean Lurcat High School, Paris,

France.

327. Let f : [a, b] R be a convex and continuous function. Provethat:

a)M (a; b) + (b a)f

a + b

2

M

3a + b

4 ;

3b + a

4

;

b) 3M

2a + b

3 ;

2b + a

3

+ M (a; b) 4M

3a + b

4 ;

3b + a

4

.

HereM(x, y) = 1y x

yx

f(t)dt.

Proposed by Cezar Lupu, Polytechnic University of Bucharest,

Bucharest, Romania, and Tudorel Lupu, Decebal High School, Constanta,

Romania.

328. Given any positive integers m, n, prove that the set1, 2, 3, . . . , mn+1

can be partitioned into m subsets A1, A2, . . . , Am, each

of size mn, such thata1A1

ak1 =

a2A2

ak2 =. . .=

amAm

akm, for all k= 1, 2, . . . , n.

Proposed by Cosmin Pohoata, student Princeton University,

Princeton, NJ, USA.

329. Letp

11 be prime number. Show that, if

(p1)/2j=1

1

j6 =

a

b,

with a, b relatively prime, then p divides a.Proposed by Marian Tetiva, Gheorghe Rosca Codreanu National

College, Barlad, Romania.

330. Determine all nonconstant monic polynomialsf Z[X] such that(f(p)) =f(p 1) for all natural prime numbers p. Here (n) is the Eulertotient function.

Proposed by Vlad Matei, student Cambridge University, Cambridge,

UK.331. LetBn ={(x1, . . . , xn) Rn | xi xi+2 for 1 i n 2} and

letB =n1 Bn. OnBwe define the relation as follows. If x, y B,x = (x1, . . . , xm) and y = (y1, . . . , yn), we say that x y ifm n and forany 1 i nwe have either xi yi or 1< i < mand xi+ xi+1 yi1+ yi.Prove that (B, ) is a partially ordered set.

Proposed by Nicolae Constantin Beli, Simion Stoilow Institute of

Mathematics of the Romanian Academy, Bucharest, Romania.


51/68

Proposed problems 51

332. For a positive integer n =

s

i=1

pii denote by (n) :=

s

i=1

i the

total number of prime factors of n (counting multiplicities). Of course, by

default (1) = 0. Define now(n) := (1)(n), and consider the sequenceS := ((n))n1. You are asked to prove the following claims on S:

a) It contains infinitely many terms (n) = (n + 1);b) It is not ultimately periodic;c) It is not ultimately constant over an arithmetic progression;d) It contains infinitely many pairs (n) =(n + 1);d) It contains infinitely many terms (n) =(n + 1) = 1;e) It contains infinitely many terms (n) =(n + 1) = 1.Proposed by Dan Schwarz, Bucharest, Romania.

333. Show that there do not exist polynomials P, Q R[X] such thatlog logn0

P(x)

Q(x)dx=

1

p1+

1

p2+ . . . +

1

pn, n 1,

where pn is the nth prime number.Proposed by Cezar Lupu, Polytechnic University of Bucharest,

Bucharest and Cristinel Mortici, Valahia University of Targoviste,

Targoviste, Romania.

334. Let a, b be two positive integers with a even and b 3 (mod 4).Show thatam + bm does not divide an

bn for any even m, n

3.

Proposed by Octavian Ganea, student Ecole Polytechnique Federale

de Lausanne, Lausanne, Switzerland.

335. Letmandnbe positive integers with m nand letA Mm,n(R)and B Mn,m(R) be matrices such that rank A= rank B =m. Show thatthere exists C Mn(R) such that A C B =Im, where Im denotes the mbym unit matrix.

Proposed by Vasile Pop, Technical University Cluj-Napoca, Cluj-

Napoca, Romania.

336. Show that the sequence (an)n1 defined by an= [2n

2]+[3n

3],n 1, contains infinitely many odd numbers and infinitely many even num-

bers. Here [x] is the integer part ofx.Proposed by Marius Cavachi, Ovidius University of Constanta,Constanta, Romania.


52/68

52 Problems

SOLUTIONS

295. Determine all nonconstant polynomialsP Z[X] such that P(p)is square-free for all prime numbers p.

Proposed by Vlad Matei, student University of Bucharest,

Bucharest, Romania.

Solution by the author. We will need the following observation.Lemma. For all nonconstant polynomialsf Z[X] the set

A= {qprime| pprime, q|f(p)}is infinite.

Proof of the lemma. Assume the contrary. Let A ={q1, q2, . . . , q n}and letM =q1q2 qn. Letk be an arbitrary natural number. According toDirichlets theorem, the arithmetic progression Mkr +1 withr N containsan infinity of prime numbers.

Let us note that f(Mkr+ 1) f(1)(modqki), i = 1, . . . , n. Belowwe will assume that the leading coefficient of f is positive, the other caseis similar consideringf. We know that lim

x(f(x) x) =, thus for r

sufficiently large f(Mkr+ 1)Mkr. Since A is finite, it follows that thereis an index j with qkj |f(Mkr+ 1). We would have that qkj |f(1). Now forksufficiently large qki > |f(1)|,i= 1, . . . , n, and we are done unless f(1) = 0.In this case the lemma is obvious, since f(p) = (p

1)g(p), with g

Z[X].

First of all, we claim that f(0) = 0. Assume the contrary f(0)= 0.We can choose infinitely many primesqsuch that f(q) =d p1p2 pm with

pi= pj for 1 i= j m and (f(0), f(q)) = d. We know from Taylorexpansion of polynomials that

f(q+ tpi) =f(q) + f(q) tpi + f(q) (tpi)

2

2! + f(q) + f(q) tpi (mod p2i )

for 1 i m.If we would have pnf

(q) for a certain index n, then we could cho

harmonic quadrilaterals and other stuff

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