r. m. m. - 27...tuzson, florin popovici, viorel gh. vodă, ovidiu pop, e. grosswald (usa), leroy f....

ROMANIAN MATHEMATICAL SOCIETYMehedinți Branch

WINTER EDITION 2020

R. M. M. - 27 ROMANIAN MATHEMATICAL

MAGAZINE

ISSN 2501-0099

Romanian Mathematical Society-Mehedinți Branch 2020

1 ROMANIAN MATHEMATICAL MAGAZINE NR. 27

ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

ROMANIAN MATHEMATICAL MAGAZINE

R.M.M.

Nr.27-WINTER EDITION 2020



ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

DANIEL SITARU-ROMANIA EDITOR IN CHIEF

ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT ISSN 1584-4897

GHEORGHE CĂINICEANU-ROMANIA

EDITORIAL BOARD

D.M.BĂTINEȚU-GIURGIU-ROMANIA

CLAUDIA NĂNUȚI-ROMANIA

NECULAI STANCIU-ROMANIA

FLORICĂ ANASTASE-ROMANIA

DAN NĂNUȚI-ROMANIA

IULIANA TRAȘCĂ-ROMANIA

EMILIA RĂDUCAN-ROMANIA

DRAGA TĂTUCU MARIANA-ROMANIA

DANA PAPONIU-ROMANIA

GIMOIU IULIANA-ROMANIA

DAN NEDEIANU-ROMANIA

OVIDIU TICUȘI-ROMANIA

MARIA UNGUREANU-ROMANIA

DANIEL STRETCU-ROMANIA

ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO

DANIEL WISNIEWSKI-USA

EDITORIAL BOARD VALMIR KRASNICI-KOSOVO



CONTENT

HAPPY BIRTHDAY TO EDITOR-IN-CHIEF OF OCTOGON MATHEMATICAL MAGAZINE -

MIHÁLY BENCZE – 65 YEARS! – D.M. Bătinețu- Giurgiu, Daniel Sitaru and Neculai

Stanciu………………………………………………………………………................................................................4

ABOUT JI CHEN’S INEQUALITY – D.M. Bătinețu- Giurgiu...............................................................7

STRUCTURI ALGEBRICE (IV) – Vasile Buruiană………………………………………...……………................8

ABOUT SOME APPLICATIONS OF INTEGRAL CHEBYSHEV’S INEQUALITY – Florică

Anastase.............................................................................................. …………………...…….......12

ABOUT AN INEQUALITY IN TRIANGLE FROM RMM-2019 – Marin Chirciu…………………............18

A NEW DEMONSTRATION OF RAMUS-E.ROUCHE-BLUNDON DOUBLE INEQUALITY AND IT’S CONSEQUENCES – Marian Dincă………………...............................................................……………20 METRIC RELATIONSHIPS IN CHIRIȚĂ’S TRIANGLE – Daniel Sitaru, Claudia Nănuți…………..….25

SPECIAL PROBLEMS IN ROMANIAN MATHEMATICAL CONTESTS - Marian Ursărescu..........29

ABOUT CALCULUS OF SOME AMAZING LIMITS - Florentin Vișescu………………............................30

A SIMPLE PROOF FOR SIERPINSKY’S INEQUALITY - Daniel Sitaru …………..………………………..32

A DEFINITIVE WAY OF SOLVING MATHEMATICS PROBLEMS-Srinivasa Raghava……………….33

ABOUT PROBLEM IX.33-RMM 24,SPRING EDITION 2020-Marin Chirciu…………………………..36

PROPOSED PROBLEMS…………………………………….………………………………………………………………...39

INDEX OF PROPOSERS AND SOLVERS RMM-27 PAPER MAGAZINE.……………………………………104



HAPPY BIRTHDAY TO EDITOR-IN-CHIEF OF OCTOGON MATHEMATICAL

MAGAZINE - MIHÁLY BENCZE – 65 YEARS !

By D.M. Bătineţu-Giurgiu, Daniel Sitaru and Neculai Stanciu

ABSTRACT. This paper is a dedicated to Mihály Bencze who is celebrating his 65 years.

Mihály Bencze

Mihály Bencze, born on 20 November 1954 in Săcele-Négyfalu, Braşov, Romania. He is

teacher of mathematics. He received Ph. D. in mathematics in 2010. He is Editor in Chief of

math journal Gamma (1978 - 1989) which becomes in 1993 Octogon Mathematical

Magazine; also he is editor in chief and member of many other scientific journals all over the

world. The first mathematical journal of the town Braşov (Brassó, Kronstadt) was founded by

Mihály Bencze in the year 1978, named ’’Gamma’’. Mihály Bencze was the editor-in-chief

and the members of the editorial board were the following: Constantin Ionescu-Ţiu, József

Sándor, Alexandru Lupaş, Florentin Smarandache, József Benkő, Alexandru Blaga, Zoltán

Tuzson, Florin Popovici, Viorel Gh. Vodă, Ovidiu Pop, E. Grosswald (USA), Leroy F. Meyers

(USA), Francisco Bellot (Spain). Mihály Bencze contributed to ’’Gamma’’with 9000 Proposed

Problems, 150 Open Question and 250 articles during 12 volumes between 1978-1989. In

April 1989 the ’’Gamma’’ journal was suppressed by the Romanian Communist party for

unclarified reasons, and till 1983 it had a latency period functioning as a prohibited journal.

In 1993, Mihály Bencze, laid down bases of the Octogon Mathematical Magazine written in

english and became the legal succesor of the former ’’Gamma’’ journal. During these years

Octogon Mathematical Magazine is published two times a year in April and October.



Topics covered by Octogon Mathematical Magazine: general, inequalities, history and

biography, mathematical logic and foundations, combinatorics, number theory, general

algebraic systems, real functions, integrations, geometry, and others in all areas of pure and

applied mathematics. Aims and scope: Octogon Mathematical Magazine publishes high

quality original research papers and survey articles, proposed problems and open questions.

The Octogon Mathematical Magazine promote mathematical culture in the world of the

mathematical sciences and to facilitate the exchange of resources.

The publishing process in the Octogon Mathematical Magazine begins with the author's idea

for the work itself, continuing as a collaboration between the author and the editor as the

work is created. As such, editing is a practice that includes creative skills, human relations,

and a precise set of methods. The title of the top editor at many publications may be known

as the editor in chief, executive editor, or simply the editor. A frequent and esteemed

contributor to a magazine may acquire a title of editor at-large or contributing editor.

The top editor sometimes has the title executive editor or editor-in-chief. This person is

generally responsible for the content of the publication. An exception is large newspapers,

who usually have a separate editor for the editorials and opinion pages to separate news

reporting and editorial content. The editor-in-chief sets the publication standards for

performance, and motivates and develops the staff. The editor-in-chief is also responsible

for developing and maintaining the publication budget. In concert with the publisher and the

operating committee, the editor-in-chief is responsible for strategic and operational

planning. The editor-in-chief is effectively the head of this magazine and has considerable

influence on its content. One of the great strengths of the present editorial team of Octogon

is the ability of its members to cooperate on as a whole. The Octogon Mathematical

Magazine has thousand-pages of research papers which often becomes the next hundred-

pages articles and notes in a popular journals, and this keeps the volume of mathematics

manageable, what can be accessible to the students, graduate students or their teachers of a

future academic generation. One might think that such magazine are of interest only the

researchers and problem buffs, but this is not quite so. These articles, notes, proposed

problems and open questions – created by mathematicians, often with considerable

qualifications and experience, are used and adapted to become accessible to a wide range of

students and to foster some interesting investigations to highlight ideas, approaches and



results, even if they cannot be directly imported into general classroom situations, they can

often be reworded. Even if the research cannot take into account all variables and cannot

answer definitively all of the questions, we might ask about mathematics, we can expect

research in mathematics to be helpful in many ways. The research it can: answer questions,

inform us, educate us, prompt new questions, create reflection and discussion, clarify

situations, help make educational decisions, challenge what we currently do as educators,

etc. Above all, despite educational research provides information that the community of

educators can use. Also we can use the research to shift from the ’’Yesterday’’ mind to the

’’Tomorrow’’ mind (see e.g. *3+).. The Octogon Mathematical Magazine is an inclusive

professional community of researchers and teachers who promote research in mathematics.

The Octogon Mathematical Magazine focuses on promoting research-based innovations

related to preparation and continued professional enhancement in mathematics. Target

audiences include teachers, researchers, higher education, faculty members, the secondary,

undergraduate and graduate levels.Mihály Bencze has published thousands of problems,

notes and articles in many scientific journal all over the world and also he has published

many books for Mathematical Contests. Mihály Bencze is: the organizer of Rubik’s contest

(1980) , the president of association Wildt József, the scientific leader of Fulgur publisher of

Braşov (since 1993), the founder of the International Hungarian mathematical Competition,

the founder of the Wildt József International Mathematical Competition. Also he is editor-in-

chief of 20 journals, editor to other 20 magazine and mentor for other 20 journals.

His creative work is amazing: has published over 1000 articles, 20 books and over 20.000

proposed problems and open questions in almost all the journals with columns of Problems

all over the world. The fact that Mihály Bencze, is a example for many pupils, students,

teachers and colleagues was proved in this paper and others to (e.g [6] , [7], [8], [9]). It has

been a great pleasure for us to have developed on ongoing friendship with such a modest

and such a refined and talented mathematician. Another important feature of his character

is the fact that he does not upset anyone, not ever judge people, is always happy and

respect all - and that means for him to be human.We are sure we speak on behalf of all

lovers of mathematics when we say:“Mihály Bencze may you live long, and continue to write

books, to submit articles, proposed problems and open questions to enjoy us”.

Mihály BenczeHappy birthday to our amazing Editor-In-Chief .



ABOUT JI CHEN’S INEQUALITY

By D.M. Bătinețu-Giurgiu – Romania

Ji Chen’s inequality: If ( ), then:

( ) .

( )

( )

( ) /

(J.C)

and this inequality was published in Crux Mathematicorum magazine, vol. 20, Nr. 4, pp. 108, year 1995. This problem was also given to Iran’s Mathematical Olympiad in 1996.

Bellow we give a generalization of this inequality:

Theorem: Let be and , then:

( ( ) ( )

( ) ) . .

( )

( )

( )

/ / (*)

Proof: We have: ∑ ( ) ∑ ( ( )

) ⏞

( ) ∑ √ ⏟

( )

( )∑ (1)

and ∑

( )

.∑

( ) /

:∑

( )

;

( ) ∑ √ ⏟

(

( ) *

( )∑

( ) (2)

From (1) and (2) inequality (*) becomes:

: ∑( )

;: ∑

( ) ( )

;

( )( ):∑

;:∑

( )

;

( )( )

. Q.E.D.

Observation. If in inequality (*) we take we obtain Ji Chen inequality.



STRUCTURI ALGEBRICE (IV)

By Vasile Buruiană – Romania

14. Fie inel integru comutativ. Atunci sunt echivalente:

a) este factorial

b) In nu există șiruri astfel ca și , și orice pereche de elemente din are cmmdc.

c) In nu există șiruri astfel ca și și orice element ireductibil din este prim.

Demonstrație: Pentru a) b) și a) c) este suficient și arătăm că nu există șiruri cu proprietatea anunțată. Dacă prin absurd ar exista astfel de șiruri notând ( ) numărul factorilor ireductibili în care se descompune ( ) ( ) ceea ce este absurd.

b) c) evident din Proprietatea 2.

c) a) Este sufficient să arătăm că orice element nenul și neinversabil din se descompune în produs finit de elemente ireductibile.

Dacă nu ar avea o descompunere finită un element atunci ar exista neasociat cu cu și nu ar avea descompunere finită cu și nu are descompunere finită continuând și luând , se construiește un șir cu și ceea ce contrazice ipoteza c).

15. Fie inel factorial și un sistem multiplicativ. Atunci este factorial (inelul de fracții pentru )

Demonstrație: Evident este integru. Fie apoi prim în , care nu este inversabil în

; el va fi prim și în . În adevăr fie

astfel ca

(unde

) sau . Cum (dacă este inversabil în )

sau

sau

. Cum orice

, nenul și neinversabil, se

descompune în produs finit de primi din , obținem afirmația.

16. Este posibil ca factorial dar să nu fie factorial (în particular este posibil ca să fie factorial dar să nu fie factorial).

În adevăr fie [ √ ] corpul său de fracții. este factorial ca orice corp, dar ,

am văzut, deja, nu este factorial.

17. Un ideal se numește prim dacă din relație sau . Evident în ideal din (inelul integru cu , neinversabil) este prim este prim.

Fie inel factorial. Atunci orice ideal prim nenul, minimal pentru relația de incluziune, este principal. În adevăr fie un ideal prim minimal, nenul astfel încât . Dacă



∏ , căci este factorial, cu și cum este prim iar minimal

avem , q.e.d.

18. În inelul factorial , - singurele elemente ireductibile sunt , iar în , - singurele elemente ireductibile sunt cu sau cu și

În adevăr orice polinom din , - are rădăcinile complexe, deci se descompune în factori liniari. În cazul lui , - orice alt polinom în afara de cele enumerate este reductibil (căci ori are o rădăcină reală ori are rădăcini complex conjugate și se divide prin ( )( ).

Faptul că este ireductibil în , - dacă și numai dacă rezultă din aceea că dacă s-ar descompune în factori de gr. I ar avea rădăcini reale. Restul este evident.

19. Criteriul de ireductibilitate Eisenstein. Fie factorial cu corpul de fracții și , -,

și prim astfel încât

. Atunci este ireductibil.

Demonstrație. Putem presupune că este primitiv. Dacă ar fi reductibil în , -

și în , - ∑

∑ cu . Din și dar

sau dar nu pe amândouă. Presupunem că și . Fie minim astfel că

(există fiindcă deci nu toți ceficienții lui se divid la ). Dar

sau ceea ce este contradictoriu.

20. Dacă este morfism de inele integre, corpurile de fracții ale

, - , - ce extinde astfel ca ( ) factorial și , - astfel ca ( ) este ireductibil în , - și grad grad ( ) este ireductibil în , -

În adevăr, putem presupune primitiv. Dacă în , - el e reductibil în

, - ( ) ( ) ( ) și cum grad ( ) grad , grad ( ) grad .

( ) este reductibil, ceea ce este contradictoriu.

21. În , - polinomul este ireductibil. În adevăr în , - este ireductibil și ( )( ) ( )( ) ( ) și aplicând criteriul lui Eisenstein pentru și , rezultă afirmația.

La fel în , - este ireductibil. De asemenea în , - , este ireductibil pentru prim. În adevăr fie , - , - cu ( ) , morfism de

inele bijective ( ( )). Este suficient să arătăm că ( ) este

ireductibil. În adevăr



( ) (

*

( )

și cum

și

folosind criteriul Eisenstein,

obținem afirmația.

In , - polinomul ( ) este reductibil.

În adevăr privit ca polinom în , -, - și luînd avem

( ) ( ) ( ) ( ) deci este reductibil.

22. In , - există polinoame ireductibile de orice grad. În adevăr și , folosind criteriul Eisenstein se vede că este ireductibil în , -.

23. Dacă este prim și atunci (Mica teoremă a lui Fermat)

Demonstrație. Prind inducție asupra lui . Pentru , afirmația este evidentă și presupunem că , voind să arătăm că ( ) ( ) . Dar

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) (2)

Pentru * + avem

( ) ( )

și cum este prim,

folosind și ipoteza inductivă suma (2) se divide la

Dacă ( ) căci ( ) și ( ) deci .

Rezultă că în inelul claselor de resturi modulo , avem că polinomul

( )( ) . ( )/ de grad are rădăcini, deci

este identic nul, deci avem ( )( ) ( ( )) ( )

Facând obținem teorema Wilson: Dacă este număr prim atunci

( ) (mod ). Cum (mod) (mod )

(mod ), înmulțind ultimele relații și înlocuind în teorema Wilson

0.

/ 1

( )

(mod ) obținem .

/ ( )

(mod )

24. Elementele ireductibile în , - sunt:

a) elementele prime din astfel încât (mod )



b) factori primi din , - ai elemntelor prime în și (mod )

c) și asociații săi.

Demonstrație: Mai întâi observăm că , -, prim ( ) , decompunerea în în factori ireductibili , prim, astfel ca ⁄ .

In plus nu există doi intregi primi neasociați astfel ca ⁄ ⁄ căci

( ) și , ceea ce arată că este contradictoriu. Această arată că elementele ireductibile din , - trebuie căutate printre divizorii din , - ai elementelor prime din . Fie ⁄ , unde (mod 4) ( ) ( ) deci ( ) și sunt posibile cazurile: ( ) (contradictoriu), ( ) (contradictoriu dacă ) deci ( ) . Luând . Însă * + deci nu putem avea căci (mod ) nu sunt decompozabile în , - a). Dacă ⁄ , unde (mod ) avem mai întâi, conform observației precedente,

că (.

/ *

(mod ) fiindcă

este par.

Deci există astfel ca (mod ), deci ( )( ). Cum

sau . Dacă

sau

, - deci

, ceea ce este absurd

nu poate fi inductibil în , - și factorii săi ireductibili sunt ireductibili în , - deci b)

În fine ( )( ) deci ( ) . Cum ( ) , dacă ar fi reductibil ar rezulta că ( ), deci ( ) ( ) (căci este neinversabil și deci ( ) ( ) și ( ) c).

25. (Reciproca teoremei Wilson ca un criteriu de recunoscut numere prime)

Dacă și ( ) (mod ) atunci este prim.

Demonstrație: Presupunem că este reductibil. Pentru relația (mod ) evident nu are loc, iar dacă avem cazurile:

I. * + ( ) și deci (mod ) ceea ce este absurd.

II. căci și * + ( ) deci iarăși (mod ) ceea ce este absurd.

Rezultă că pentru decompozabil nu este posibilă relația dată, deci este prim.

26. Metoda lui Kronecker de determinare a divizorilor unui polinom , -

și factorial, într-un număr finit de pași:

Mai întâi observăm dacă în , - avem ( ) ( ) ( ) deci

( ) ( ). Deci polinoamele ce divid pe au proprietatea că ( ) ( ) . In plus

unul din grad , grad

. Fie grad

.



Apoi observăm că dacă , distincte reciproc și neasociate atunci

, există un unic polinom , - de grad astfel ca ( )

. În adevăr fie ( ) ( )( ) ( ) ( ) ( )

( ) ( )

unde ( ) este derivata lui adică ( ) ∑ ∏ ( )

Evident ( ) {

* +, grad . Luăm

∑ ( ) ( ) . Existența este demonstrată. Dacă ar fi alt

polinom cu ( ) are grad și rădăcini este polinom nul și deci rezultă unicitatea. Acum procedeul de determinare a divizorilor lui este următorul: Fie ( ) și ( ) în astfel ca . Pentru fiecare construim cu ( ) și verificăm dacă sau . Într-un număr finit de pași găsim toți divizorii sau stabilim că este ireductibil. Deși finite, totuși calculele sunt prohibitive.

ABOUT SOME APPLICATIONS OF INTEGRAL CHEBYSEV’S INEQUALITY

By Florică Anastase-Romania

In memory of my Professor LAURENȚIU MIRCEA PANAITOPOL

Theorem:( Chebysev’s Inequality)

For , - , continuous function with same increasing and , - , ) integrable function. Then:

:∫ ( )

;:∫ ( ) ( ) ( )

; :∫ ( ) ( )

;:∫ ( ) ( )

; ( )

In the case f and g different increasing:

:∫ ( )

;:∫ ( ) ( ) ( )

; :∫ ( ) ( )

;:∫ ( ) ( )

;

Demonstration:

If and are same increasing ( ) , -

( ) ( )( ( ) ( ))( ( ) ( )) , -



( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )∫ ( ) ( ) ( ) ( ) ( )∫ ( ) ( )

( ) ( )∫ ( ) ( ) ( ) ( ) ( )∫ ( )

:∫ ( )

;:∫ ( ) ( ) ( )

; :∫ ( ) ( )

;:∫ ( ) ( )

;

:∫ ( ) ( )

;:∫ ( ) ( )

; :∫ ( ) ( ) ( )

;:∫ ( )

; ( )

For ( ) , -, we get:

) ( )∫ ( ) ( ) .∫ ( )

/ .∫ ( )

/

if and are same increasing;

) ( )∫ ( ) ( ) .∫ ( )

/ .∫ ( )

/

if and are different increasing;

Application 1:

If , - is continuous and increasing. Then:

∫ ( )

∫ ( )

Solution: If or is constant, then we have equalities. Let and variable on , -. Let function ( ) , -. Then and have same increasing on , - and from Chebyshev’s inequality (a):

:∫ ( )

;:∫

; ( )∫ ( )

∫ ( )

( )∫ ( )

∫ ( )

∫ ( )

Application 2: If , - continuous and derivable with ( ) , -. Then:

∫ ( )

∫

( )

Solution:



First: . ( )

/

( ) ( )

( )

is decreasing. Applying Chebyshev’s inequality, we

have:

∫ ( ) ∫ ( )

:∫ ( )

;

:∫

;

( )∫

( )

∫ ( )

Application 3:

If , - , ) are continuous and

are increasing, then:

∫(∫ ( )

∫ ( )

+ ∫ ( )

( )

Solution: From Chebyshev’s inequality:

∫ ( ) ∫ ( )

( ) ∫ ( )

( )

( ) ∫ ( )

⏞

∫ ( )

( ) ∫

( )

( )

∫ ( )

∫ ( )

∫

( )

( ) , -

∫(∫ ( )

∫ ( )

+ ∫:

∫

( )

( )

; ∫ ( )

( )( )

∫ ( )

( )

Application 4: If , - ( ) is decreasing, then:

∫ ( )

∫ ( )

∫ ( )

∫ ( )

Solution:

In Chebyshev’s inequality, let: ( ) ( )

( ) ( ) ( )

:∫ ( )

; :∫ ( )

( )

; :∫ ( )

; :∫ ( )

( )

;

:∫ ( )

; :∫ ( )

; :∫ ( )

; :∫ ( )

;

Application 5: If , - is twice derivable function, ( ) and ⏟

( ) then:



( ) ∫ ( )

Solution: Applying Chebyshev’s inequality function ( ) ( )

∫ ( )

( )

:∫ ( )

;

( ) ∫ ( )

( ) ∫ ( )

:∫ ( )

;

|∫ ( )

|

Application 6: Prove:

∫

√

.

/

Solution:

Let 0

1 ( )

( )

( )

decreasing and ( ) increasing for 0

1.

∫ √

∫

⏞

∫

( )( )

∫

∫

.

/

∫

∫

∫

⏞

√

.

/

Application 7: Prove:

∫

( )( )

Solution:

∫

( )( ) ∫(

* ∫

⏞

∫

⏞

∫

∫

Application 8: If , - , - is twice derivable, ( ) , - are increasing and convexe on , -, then:



) ( ( )) ( ( )) ( ( ) ( )) 4 ( ) ( )

5

) (

*

(

*

Solution:

a) ( ) , - increasing on , - and convexe ⏞

( ( )) ( ( )) ∫( ) ( )

∫ ( ( )) ( )

∫ ( ( )) ∫ ( )

( ) ( )

∫ ( ( ))

( ( ) ( )) :

∫ ( )

; ( ( ) ( )) 4 ( ) ( )

5

b) Let 0

1 0

1 ( ) ( )

( )

( )

⏞

( )

4 (

*5 ( ( )) ( (

* ( )* 4

( ) ( )

5

(

*

(

*

Application 9: If

, then:

∫:

∫

∫ .

/

;

∫ .

/

Solution:

First: ∫

∫

( )

Let be the functions: 0

1 ( )

( ) ∫ ( ) ( ) .

/

and

( ) ∫ ( )

How: ( ) ( )

( ) ( )

( )( )

0

1 is convexe , - and such that:

(( ) ) ( ) ( ) ( ) for

( )

( )

( )

is increasing (analogous

( )

is increasing

( ) decreasing)

Applying Chebyshev’s inequality, we get:



∫ ( )

( ) ∫

( )

( )

∫

( )

∫

( ) ( )

( )

( )

.

/

( )

∫ .

/

Application 10: If and then:

∫ ( )

√

( √ ) ( √ )

Solution:

∫ ( )

√ ∫

√ ( )

From Chebyshev’s inequality, let: ( )

integrable and ( ) √

( ) ( ) are increasing

(

∫

)

(

∫ ( )

√

)

(

∫√

)

(

∫ ( )

)

( )

∫

∫

.

/

( )

∫√

∫

√

( √ )

( )

∫ ( )

∫ .

/

(.

/

*

∫

.

/

∫

.

/

.

/

∫

.

/

.

/

⏞

∫

( )

⏞

∫ ( )

√

( √ )

( )

From (i),(ii),(iii),(iv), we get:

∫

( )

( √ ) ( √ )

∫ ( )

( √ ) ( √ )



Application 11: If , then:

∫

∫

4

√

√ 5

Solution:

Let: 0

1 ( )

( )

derivable with

( ) ( )

( ) ( )

( ) is increasing and decreasing

⏞

∫

∫

∫

( ) ( )

∫

( )

∫

( )

( )

∫

( )

∫

( )

Let

.

( ) /

∫

∫

( )

( )

From (ii), (iii) we get:

|

√

√ . So:

∫

∫

4

√

√ 5

Refferences: Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro

ABOUT AN INEQUALITY IN TRIANGLE FROM RMM-2019

By Marin Chirciu – Romania

1) In – centroid and – circumradii respectively.

Prove that:

Proposed by Marian Ursărescu – Romania

Solution: We prove the following lemma:

Lemma:


Prove that:



Expressing the area of in two ways, we obtain:

, -

and , -

, wherefrom

It follows

and from here

We have:

Let’s get back to the main problem:

Using the Lemma and inequality we obtain:

∑

∑

∑

∑ ∑

( )

, so, ∑

Using the known identity in triangle

.

In order to prove the inequality from the enunciation, it suffices to prove that:

, which follows from Gerretsen’s inequality:

. It suffices to prove that: (Euler’s

inequality). Equality holds if and only if the triangle is equilateral.

Remark: Let’s find an inequality having an opposite sense:


Prove that:

.

/

Proposed by Marin Chirciu – Romania

Solution: Using the Lemma and Tereshin’s inequality

, we obtain:

∑

∑

∑

∑( )( )

( ) ( )

, which follows from

∑( ) ( ) ( ) ( ) , so

∑

( ) ( )

Using the known identity in triangle:

.

In order to prove the inequality from enunciation it suffices to prove that:



( ) ( )

( ) ( ) ( ), which follows from Gerretsen’s

inequality ( )

. It suffices to prove that:

( )

( ( ) ) ( ) ( )

( )( ) , true from Euler’s inequality

. Equality holds if and only if the triangle is equilateral.

Remark: The double inequality can be written:


Prove that:

.

/

Solution

See inequalities 1) and 3). Equality holds if and only if the triangle is equilateral.

Refferences:

Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro

A NEW DEMONSTRATION OF RAMUS-E.ROUCHE-BLUNDON DOUBLE

INEQUALITY AND IT’S CONSEQUENCES

By Marian Dincă – Romania



THEOREM: Let be a triangle that has its circumcircle ( ) and the incircle ( ) Then, it exists two isosceles triangles inscribed to the circle ( ) and circumscribed to the circle ( ).

Proof: Without using Poncelet theorem we use Carnot identity:

We will prove that there exists two angles respectively such that:

.

/ (

*

Such that: . Elementary proof:

.

/ .

/ .

/

or: .

/ .

/

.

/

√ ( )

√ ( )

√ .

/

√

So:

, it follows:

, let be

respectively:

, let be

Let be: , .

/

but: .

/ .

/ .

/

It follows: .

/ .

/

So: .

/ .

/. The function:

( ) .

/ 0

1, ( ) .

/ , so, increasing

( ) ( ), implies: . Also: .

/ .

/

.

/

But: .

/ .

/

So: .

/ .

/.

The function: ( ) .

/ 0

1

( ) .

/ , decreasing, so:

In the end there are two isoscel triangles that satisfy Carnot relationship:

.

/ (

*

And: . Let be an isoscelles triangle with:

.

/ .

/

.

/

we prove that: (i)

( ) (

*

Or:

.

/ (ii)



.

/ , it follows: .

/ , it follows:

.

/ .

/. So:

, it follows:

, then it exists and , - such that:

.

/ ( ) (1)

.

/ ( ) (1’)

, implies: .

/ ( ) .

/ ( )

Or: ( ) .

/ ( ) , or ( ) .

/

It follows:

( ) .

/ ( )

0 .

/ 1

Or: .

/ .

/ .

/

Or: .

/ .

/ , it follows:

Inequality (ii) is equivalent with:

: .

/ ( )

; :

.

/ ( )

;

: .

/ ( )

.

/ ( )

;

4.

/

( )5 4.

/

( )5

(.

/ (

* (

*+

4.

/

( )5 4.

/

( )5

(.

/ (

* (

*+

[ (

*

] [ (

*

] [(

* (

* ]

Let be , - , -

( ) [ (

*

] [ (

*

] [(

* (

* ]

is a convex function defined on a convex set: , - , - 2( )

3

So the pentagon , where: ( ) ( ) .

/ .

/ ( ). Will attain its

maximum in one of these peaks. Any convex function defined on a convex set will attain its maximum in its extreme points.



( ) .

/

(

* .

/

(

*

.

/

(

* .

/ (

(

*

* (

*

.

/ .

/

.

/, or:

(

* .

/

, it follows:

( )

.

/

, -

(

* .

( )

/ .

/ ( )

So: .

/ .

/ ( )

We obtain:

.

/ .

/

.

/ ( )

.

/

( ) .

/ ( ( )) .

/

6

4.

/57

[

] (

*

But: ( ) ( ) and .

/ .

/ being symmetric.

.

/ , it follows: .

/

We prove the other inequality:

, or so:

, it follows:

.

/ we obtain:

Then it exists respectively , -, such that:

( ) (

* ( ) (

*



( ) ( ) (

* ( ) (

*

It follows: , because

( ) (

* ( ) (

*

( ) ( ) (

*

[ (

*]

It follows: .

/ .

/ .

/

Or: .

/ .

/ . It follows:

( ( ) (

*+ ( ( ) (

*+

[ ( ) (

* ( ) (

*]

[ (

*

] [ (

*

]

[( ) (

* ] ( )

, - , -

is concave defined on the convex set , - , - 2( )

3

It will attain its maximum in the points .

/ ( ) ( ) ( ) .

/

Being simetric it remains to verify .

/ ( ) ( )

(

* (

(

*

* (

* (

(

* *

(

* (

*

(

*

Or:

.

/ .

/. Using the same technique:

Or Steffensen in discreete form

(

* (

* (

*

(

*

Ramus-E, Rouche, Blundon for any acute angled triangle: If triangle is acute-angled, that has the circumcircle triangle ( ), and the

inscribed circle ( ) and 0

1. Then it exists two acute-angled triangles that

has the same respectively and it exists the double inequality: where are semiperimeters of the those triangles.

Proof: In the first part of the article we’ve proved that: In the situation that the triangle is acute-angled, we can replace * + with

2

3 the triangle that has the angles having the measure:

, is acute

angled and:

, and



In the same way like in the first part it will follow: .

/ (

* .

/

.

/ .

/ .

/ 4

5. So:

Bibliography: [1] R. Tyrrel Rockafellar: Convex Analysis, Theta Publishing House, 2002. [2] Arthur Engel: Mathematical problems, Solving strategies, GIL Publishing House, 2006. [3] Shan – He Wu, Yu-Ming Chu: Geometric interpretation of Blundon’s inequality and Ciamberlini’s inequality, Journal of inequalities and Applications, December 2014 Bellow we illustrate the proof: O246. Let be a point inside or on the boundry of a convex polygon . Prove that the maximum value of ( ) ∑

is achieved when is a vertex

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by G.R A. 20 Problems Solving Group, Roma, Italy Let be the set of points inside or on the boundry of the convexe polygon. We first note that the map ( ) is convex: if with then, for any , we have that and ( ) ∑ , ( ) ( )-

∑

∑

( ) ( )

Moreover, for any point in is a convex combination of the vertices that is, there exists such that ∑

and ∑

Hence, by the covexity of ,

( ) (∑

+ ∑

( ) ∑

( )

( )

which means that the maximum value of ( ) is attained when is a vertex of .

METRIC RELATIONSHIPS IN CHIRIȚĂ’S TRIANGLE

By Daniel Sitaru, Claudia Nănuți-Romania

Abstract: In the romanian mathematician Marcel Chiriță proved that for any

√ √ √ can be sides in triangle.

In this paper we prove some metric relationships in this kind of triangle.

Proposition 1. Chiriță’s triangle is obtuse.

Proof: Let √ √ √

√( )( )



√( )

( )

√

( )

Proposition 2. Area of Chiriță’s triangle is constant for any real value of .

Proof:

4 ( )

√ 5

( )

( )

( )

√

√

, -

√ √

√

√

√

√

√

√

Proposition 3. A median of Chiriță’s triangle is constant for any real value of .

Proof:

( )

( )

( )

( )

Proposition 4. If then:

√ √ √ √ √

Proof.

( )

( )

( )

( )

( )

( )

( ) √ (1)

( )

( )

( )



( )

( )

( )

( ) √ (2)

It’s known that:

√ (3)

√ (4)

By adding (3); (4): √ √

√ √ √


√ √ √ √( )( )

Proof.

√

√

√

√( )( )

By Mitrinovic’s inequality:

√

√

√

( )( )

√( )( )

√ √ √ √( )( )


.√ √ /

.√ √ /

.√ √ /

Proof 1 (Soumava Chakraborty-Kolkata-India)

.√ √ /

.√ √ /

.√ √ /

⏞( )

(1) ( )

√( )( ) √( )( )



√( )( ) (squaring (1))

( ) *( )( ) ( )( ) ( )( )+

8( )√( )( ) ( )√( )( )

( )√( )( )9

(squaring again) ( ) ( )

{

( )√( )( )

( )√( )( )

( )√( )( )

} (2)

.

/

and .

/

,

of (2) {( ) .

√

/ (√ ) ( ) .

√

/ .

√

/

( )(√ ) .√

/

}

( )

( )

of (2) (2) is true (Proved)

Proof 2.

√ √ √

In with sides :

√

√

( )

√

√

√

√

By Hadwiger – Finsler’s inequality: ∑( ) √

∑( ) √ √

∑( )

.√ √ /

.√ √ /

.√ √ /

Refferences:

1.Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro

2.Mihaly Bencze,Daniel Sitaru,Marian Ursărescu:”Olympic Mathematical Energy”-Studis-

Publishing House-Iași-2018



3.Daniel Sitaru, George Apostolopoulos:”The Olympic Mathematical Marathon”-Publishing

House- Cartea Românească-Pitești-2018

4. Mihaly Bencze,Daniel Sitaru:”Quantum Mathematical Power” -Publishing House Studis,

Iasi-2018

5.Daniel Sitaru,Marian Ursărescu:”Calculus Marathon”-Studis-Publishing House-Iasi-2018

6.Daniel Sitaru, Mihaly Bencze:”699 Olympic Mathematical Challenges”-Publishing House

Studis, Iasi-2017

7.Daniel Sitaru,Marian Ursărescu:”Ice Math-Contests Problems”-Studis-Publishing House-

Iasi-2019

SPECIAL PROBLEMS IN ROMANIAN MATHEMATICAL CONTESTS

By Marian Ursărescu-Romania

ABSTRACT: In this article are presented some problems with matrices and determinants used as subjects in romanian mathematical contests. Lemma. If ( ) are such that: a) , b) Matrix is invertible. c) , , then ( ) Demonstration: Second degree equation with roots is: ( ) ( ) are roots of equation (1)

We have: ( )( ) ⏞( )

( ) ( ) ⏞( )

( ) ( ) ( )( ) ( )( ) ( )( )

,( )( )- ,( )( )-

( ) ( ) [( )( ) ] ( ) ( )

⏞

( ) If ( ) then ( ) and ( ) . Applications: 1) If ( ) such that and is invertible, then . Demonstration: By lemma, we have with and

By lemma ( ) but

2) If ( ) with

( ) and is invertible,

then RMO

Demonstration:In condition of lemma, we have

and such that



.

By lemma: .

/

.

/

.

/

3) If ( ) such that ( ) √ √ and is invertible, then

RMO

Demonstration: In conditions of lemma, we have: √ √

and

√ √

Second degree equation have roots: √ √ (√ )

By lemma ( ) ( ) . But

.

/

Applications:

1) If ( ) such that ( √ )( ) and is invertible,

then . GMB

2) If ( ) such that . √ √ / ( ) (√ )( ) and

is invertible, then RMM

3) If ( ) such that √( √ √ ( ) √ √ √ ( ) and

is invertible, then . RMM

4) If ( ) such that √ √ ( ) √ √ ( ) and is invertible, then

RMM Refferences Romanian Mathematical Magazine-Interactive Mathematical Journal-www.ssmrmh.ro

ABOUT CALCULUS OF SOME AMAZING LIMITS

By Florentin Vișescu-Romania

Lemma: Let be and , - , ) a continuous function, then:

∏4

(

*5

Proof:



∏4

(

*5

∏ (

.

/*

∑ (

.

/*

We prove that

( ) , ).

Consider , ) ( ) ( )

( )

, )

on , ) ( ) ( ) , ) or ( ) , )

consider , ) ( ) ( )

( )

( ) on , ) ( ) ( ) , ) or ( )

, ). So,

( ) , ).We choose

(

*

(

*

4

(

*5

4

(

*5

(

*

∑

(

*

∑4

(

*5

∑ 4

(

*5

∑

(

*

∑

(

* ∫ ( )

∑

(

*

∑

(

*

∫ ( )

Then ∑ (

.

/*

∫ ( )

and so,

∏4

(

*5

∫ ( )

1) ∏ .

.

/

/

2) ∏ 4 √

5

. √( )

/

3) ∏ ( √

*



4) ∑ .

/

(√ )

5) ∏ .

/

A SIMPLE PROOF FOR SIERPINSKY’S INEQUALITY

By Daniel Sitaru – Romania

Abstract: In this paper we give a proof by mathematical induction for Sierpinsky’s

inequality.

SIERPINSKY’S INEQUALITY:

√

For

(√ )

For

:

;

( )

( )

( ) ( )

( ) ( )

( )

(

*



( )

( ) ( )

( ) ( )

( )

( ) suppose true

( ) to prove

( )

( ) ( )

A DEFINITIVE WAY OF SOLVING MATHEMATICS PROBLEMS K. SRINIVASA RAGHAVA

[email protected]

ALL INDIA RAMANUJAN MATHS CLUB

The difficulty in solving mathematics problems was the most common answer in the survey I conducted, when I asked if you have problems with mathematics. For this reason I will try to give some light to this need that we have raised.

STRATEGIES TO SOLVE MATHEMATICS PROBLEMS You get good marks in the exercises and you get stuck in the problems, right? It is normal! It also happened to the most for a while. Clear! If you barely spend time teaching us problem solving in class! Solving problems is the heart of mathematics, and I believe that more emphasis should be placed on teaching methodologies for resolution. What satisfaction when you succeed! Eureka! You may not believe it much, but you can increase your motivation and your ideas by facing appropriate problems.

EXERCISES AND PROBLEMS They are very different things. You are used to doing more exercises, so you get much better. An exercise is usually a mere practical application of a formula or calculation. At a glance you know what you have to do. You already know the way and you just have to apply it. A problem is necessary to read it carefully to understand it well. Here you ignore the road! Walker there is no way; you make your way when you walk. You have to organize and relate your ideas and have a positive and creative attitude. Problems are less defined and more open issues than exercises. Do not focus on memorizing theory and doing repetitive exercises. It is much more productive and fun to spend time solving problems.

mailto:[email protected]

http://soymatematicas.com/tienes-problemas-con-las-matematicas-te-echo-una-mano/



A GOOD FRIEND I think it's about time you start considering problems as friends that offer you the opportunity to progress in an active way in mathematics and to check what you know. Even sometimes, you can enjoy thinking and have "a happy idea" to come up with a simple and elegant solution. Solving a problem requires time and energy. It implies some circumstances like initial frustration, willingness to resolve it, perseverance But solving problems will increase your confidence with mathematics. You will learn more. Worth the effort!

STEPS TO SOLVE MATHEMATICAL PROBLEMS

It is true that solving mathematics problems is a complex activity. It may seem like a messy activity, but it is essential to organize your ideas and be schematic. These tips will help you think better:

You must have a positive attitude. Curiosity and desire to learn. Taste for the challenge

Trust your possibilities. We are what we think. Act calmly, without fear.

Be patient. Do not give up at the slightest difficulty. If you get stuck, think about a new approach to

the problem.

Concentrate. Solving problems is a complex activity and requires attention.

Don't look for short term success. Reaching the solution is a slow process, but when you

notice the progress you will feel great satisfaction.

There are really no definite strategies to solve mathematics problems that will ensure success. But if we can point out some general steps.

1. UNDERSTAND THE PROBLEM Read the sentence calmly. Several times, until you understand it well. Do not miss any interesting facts. What does it consist of? What do you know, what are you asked for? What are the conditions…? This is necessary to face the problem with guarantees of success.

2. PREPARE AN ACTION PLAN When you understand the problem, it is time to choose a strategy to solve it. There are many strategies! I indicate some at the end of the article. It is good that you know them and practice them to improve your ability to solve problems.

3. CARRY ON YOUR PLAN Once you have chosen a strategy, work it decisively and do not abandon it to the first difficulty.



It is possible that things get complicated and you were wrong to choose a strategy. Another test! There are usually several ways to arrive at the solution and we cannot always succeed with the most appropriate one at the first attempt. He left? Are you sure? Review the result and verify that you have reached the solution. Many times we believe we have solved a problem and then it is not so.

4. REFLECT ON THE ENTIRE PROCESS «Every problem I solved became a rule that later helped me solve other problems. » Descartes Have you solved the problem?!! Congratulations!! Have you had a good time entertaining, trying hard, and ended up not solving it? Congratulations too! You learn much more from the problems worked with interest and determination, and not resolved, than from those that are solved almost at first sight. How did you solve it? This stage is very helpful and is often forgotten.

Examine well the path you have followed. How did you get to the solution? Or, why haven't you

reached the solution? What mistakes and successes have you had? What made you intuit that it was

going to go well?

See if you can do it in a simpler way.

Reflect a little on your thinking process and draw consequences for the future. Each person has a

different way of thinking.

How is your thought? Visual or analytical? Everything can be improved. With practice you can go from having a single rigid idea to having several related and original ideas.

STRATEGIES TO SOLVE MATHEMATICS PROBLEMS

Look for similarities with other problems. Does it remind you of any similar situation?

Reduce the complicated by something simpler. Divide and you will win!

Consider particular cases. Keep track! Use very small numbers

Make a drawing or scheme. A picture is worth a thousand words. It incorporates only the important.

Study all possible cases. Can you discard any?

Choose a good notation. You will simplify the problem a lot

Trial and error. If it doesn't work, take another path.

Work backwards. Imagine that the problem is solved and that you are a crab. It is possible that you

can build the solution.

Take advantage of symmetry. You may be able to take advantage of regularities or symmetries.

I hope that from now on solving mathematics problems become a pleasant distraction for you.

Who knows, one day you might say, "I like problems!" Or not?



ABOUT PROBLEM IX.33-RMM-24, SPRING EDITION 2020

By Marin Chirciu – Romania

1) Prove that in any acute-angled triangle the following inequality holds:

∑( )( )

.

/


Solution

We prove the following lemma:

Lemma:

2) In the following inequality holds:

∑( )( )

( ) ( ) ( )

Proof.

Using the formula

, we obtain:

∑( )( )

∑

.

/ .

/

∑( )( )

( ) ( ) ( )

, which follows from

∑( )( )

∑ ( )( )

and

∑ ( )( ) ( ) ( ) ( )

Let’s get to the main problem:

Using the Lemma it suffices to prove that:

( ) ( ) ( )

.

/

(1)

Using Gerretsen’s inequality and Euler’s inequality we have:

( ) ( ) ( )

, ( ) ( ) ( ) -

( ),( )( ) ( )

( ) -



,( )( ) ( ) -

It follows that ∑( )( )

(2)

Using (2) in order to prove (1) it suffices to prove that:

, which follows from Mitrinovic’s inequality

.

It remains to prove that

, obviously from Euler’s inequality


Remark: The inequality can be strengthened:

3) In the following relationship holds:

∑( )( )

.

/


Solution

Using (2) in order to prove the proposed inequality it suffices to prove that:

, which follows from Mitrinovic’s inequality

Equality holds if and only if the triangle is equilateral.

Remark.

Inequality 3) is stronger than inequality 1)

4) In the following inequality holds:

∑( )( )

.

/

.

/

Solution

See inequality 3) and .

/

.

/

, obviously from Euler’s inequality


Remark.

If we replace with we propose:




∑

( )( )

4

5


Solution

We prove the following lemma:

Lemma:


∑( )( )

( ) ( )

Proof.

Using the formula

, we obtain:

∑( )( )

∑

.

/ .

/

∑

( )

( ) ( )

Let’s get back to the main problem.

The left hand inequality:

Using the Lemma it suffices to prove that: ( ) ( )

( ) ( ), which follows from Blundon-Gerretsen’s inequality

( )

( ). It remains to prove that:

( ) ( )

( )( )

( )( ) , obviously from Euler’s inequality .

Equality holds of and only if the triangle is equilateral.

The right hand inequality.

Using the Lemma it suffices to prove that: ( ) ( )

.

/

( ) ( ) , it follows from Gerretsen’s inequality

( )

. It follows to prove that:

( )

( ) ( ) (Euler’s inequality)

Equality holds if and only if the triangle is equilateral.



PROPOSED PROBLEMS

5-CLASS-STANDARD

V.1. Find such that:

Proposed by Petre Stângescu-Romania



V.3. Let be . Find p- prime number such that ( ) ( )( )


V.4. Let be . If ( ) and find ( )( ) then find ( )( )


V.5. Find such that simultaneous are perfect square.

Proposed by Ștefan Marica-Romania



V.7. Let be a number formed from 2019 digits of . Prove that:

a) is not a perfect square. b) is not a perfect cube

c) is not a power having the oreder of a natural number bigger than .

Proposed by Ene Cristina, Nicu Elena – Romania

V.8. Prove that the number

Proposed by Iuliana Trașcă, Chiriță Aurel –Romania

V.9. Find the numbers having the form for which the number

is a perfect square. Proposed by Constantin Ionică – Romania

V.10. Find the natural nonzero numbers the divided to give quotient and the reminder

, and divided to give the quotient and the reminder .

Proposed by Constantin Ionică – Romania



V.11. Dan has written on the board consecutive natural numbers and after that he deleted a

number. Find the deleted number, if the sum of the remained numbers is equal to .


V.12. Prove that is divisible with . Proposed by Daniel Stretcu – Romania

V.13. Solve in natural numbers the following equation: .

Proposed by Daniel Stretcu – Romania

V.14. Find the natural numbers having the following form , knowing that:

. Proposed by Daniel Stretcu – Romania

V.15. Prove that can be written as a sum of three perfect squares different and

nonzero, for any natural nonzero number. Proposed by Marin Chirciu – Romania

All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical

Magazine-Interactive Journal.

6-CLASS-STANDARD

VI.1. Solve for natural numbers: .


VI.2. If p-prime number then can not be a perfect cube.


VI.3. Find and -prime number such that: be a perfect square.


VI.4. Find - prime numbers; such that be a perfect square

. Proposed by Petre Stângescu-Romania

VI.5. Find such that: .

Proposed by Gheorghe Calafeteanu-Romania



VI.6. Find such that

. Proposed by Gheorghe Calafeteanu-Romania

VI.7. Find a such that √ . Proposed by Ștefan Marica-Romania

VI.8. Find such that √ . Proposed by Ștefan Marica-Romania

VI.9. Find { }. Proposed by Ștefan Marica-Romania

VI.10. Let be the natural numbers, and ( ) ( )

. Prove that if

, then . Proposed by Constantin Ionică – Romania

VI.11. Find the pairs of nonzero natural numbers with for which

, - ( ) , where( ) is biggest common divisor and , - is the biggest

common divisor of the numbers and . Proposed by Constantin Ionică – Romania

VI.12. Find the numbers knowing that and are directly proportional with and ,

and are directly proportional with and , and .


VI.13. Prove that the number is divisible with , for any natural nonzero

number . Proposed by Daniel Stretcu – Romania

VI.14. Let be . Prove that . Proposed by Daniel Stretcu – Romania

VI.15. Prove that the number is divisible with 11.


VI.16. Prove that it doesn’t exist the natural numbers such that

.


VI.17. Find the natural numbers with three digits which divided to and give the

quota , respectively . Proposed by Daniel Stretcu – Romania

VI.18. Prove that: if and only if .


VI.19. Find the quota of to the number .




VI.20. Prove that for any natural nonzero number , among the elements of the set

* + there is at least a power of .




7-CLASS-STANDARD

VII.1. In obtuse ; ; . Find perimeter. Proposed by Petre Stângescu-Romania

VII.2. In ( )

. Find . Proposed by Petre Stângescu-Romania

VII.3. In ( ) . If ( √ )( )( ) then find ( ) ( ) . Proposed by Petre Stângescu-Romania VII.4 If in ( ) and ( ) then .


VII.5. In ( ) the following relationship holds:

(√ )


VII.6. In a bicentric issoscelles trapezian perimeter is . If is circumcentre

and ( ) find the sides of trapezian. Proposed by Gheorghe Calafeteanu-Romania

VII.7. In ( ) ( ) * + Prove that:

. Proposed by Gheorghe Calafeteanu-Romania



VII.8. In . Find area and

semiperimeter. Proposed by Ștefan Marica-Romania VII.9. Solve for natural numbers:

{

√ √

√ √

√ √ √


VII.10. Solve for natural numbers: . Proposed by Ștefan Marica-Romania

VII.11 Find depending on (all natural numbers) such that:

( ) Proposed by Ștefan Marica-Romania

VII.12. Find the natural numbers with and prime numbers, knowing that:

√ √ √ Proposed by Constantin Ionică – Romania

VII.13. Find the natural nonzero numbers , knowing that:

and

.


VII.14. Find such that the equation:

has an unique solution. Proposed by Daniel Stretcu – Romania

VII.15. The real strictly positive numbers verify the relationship . Prove

that the number

is a perfect square natural number.


VII.16. In a medical laboratory, a biologist finds that he has unitary cells of a

square, that are infected and then, after a time, the cells that have at least two neibours cells infected (having a common side) they also infect. Can the infection spread in all the square?

Proposed by Daniel Stretcu – Romania VII.17. A natural number formed from digits has the first digits . Prove that the

number cannot be a perfect square. Proposed by Daniel Stretcu – Romania

VII.18. Solve for in such that:

Proposed by Naren Bhandari-Nepal



VII.19. Find the last digit of:

⏟

Proposed by Seyran Ibrahimov-Azerbaijan



8-CLASS-STANDARD

VIII.1. Solve for integers: ( )


VIII.2. Solve for integers:


VIII.3. Find: ⏟

( ). Proposed by Petre Stângescu-Romania

VIII.4. If then: . Find ⏟

( ).


VIII.5. Solve for natural numbers: ( ) ( ) .


VIII.6. Let be the set * +

a) Write the set in the form of an interval.

b) Find the natural number for which contains exactly integer numbers.


VIII.7. Let be the number , where .

a) Prove that can be written as a difference of two perfect squares.

b) If , prove that . Proposed by Constantin Ionică – Romania



VIII.8. Let be positive real numbers such that * + . Prove that:

√ ( )

√

( )

√

( )

Proposed by Nguyen Viet Hung – Vietnam

VIII.9. Solve in real numbers the following equation:

√ √ √


VIII.10. Let be non-negative real numbers such that . Prove that:

a) ( ) ( )

b) ( ) ( )


VIII.11. Prove that if * + then:

( )

( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu,Alecu Orlando – Romania


( )

( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu,Cătălin Spiridon – Romania


( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu,Elena Alexie – Romania

VIII.14. If then:

Proposed by D.M. Bătinețu – Giurgiu,Gabriela Vasile – Romania

VIII.15. If then:

( )

( )( )

Proposed by D.M. Bătinețu – Giurgiu,Gheorghe Boroica – Romania



VIII.16. If then:

( )( )( )( )

( ) ( ) √

Proposed by Daniel Sitaru – Romania

VIII.17. If are positive integers such that:

.

Find the unique solution for keeping as small as possible if it exists.


VIII.18. If then:

( ) (

* (

* (

*

Proposed by Daniel Sitaru,Elena Iacob Meda – Romania

VIII.19. If then in the following relationship holds:

∑4

(

*

5

4 ( )

√ 5

Proposed by Daniel Sitaru,Alina Georgiana Ghiță – Romania VIII.20. If then:

4 ( )

5 ( )4

( )

5 ( )

Proposed by Daniel Sitaru – Romania VIII.21. Solve for natural numbers:


VIII.22. Find such that:




VIII.23. If then:


VIII.24. Solve:

4 , -

5

4 , -

5




9-CLASS-STANDARD

IX.1. Let such that and . Prove that:

√

√

√

√


IX.2. Let be and such that . Prove that:

( )

( )

( )

( )


IX.3. Let , such that and . Prove that:

(

√ *

(

√ *

(

√ *

√( )




IX.4. In the following relationship holds:

√ ( )



, where


IX.6. Let be such that

, where . Prove:

.

/

.

/

.

/

, where .


IX.7. If then:

( )


IX.8. If then:


IX.9. If such that and , find the minimum of the expression:


IX.10. Let . Find the maximum of the expression:

√

√

√

√

√

√ , where is fixed.



√

√

√ 4

√

√

√

5





√

√

√

4√

√

√

5


IX.13. Let be such that

. Prove that:


IX.14. Let such that and . Prove that:

√

√

√ √



(

* (

* (

* ( )


IX.16. Let such that and

. Prove that:


IX.17. Let be such that and . Prove that:

√

√

√

√


IX.18. Let . Prove that:

∑4

5

( )


IX.19. Prove that in any triangle with the standard notations:

( ) ( )


IX.20. Prove that in any triangle ,



( )


IX.21. In – orthocenter the following relationship holds:

( )


IX.22. If in ( ) ( ) ( ) * + then:


IX.23. Let be the circumcevian triangle of Nagel’s point of . Prove that:

, -

, - (

*



(

* (

* (

* √ .

/



∑

( )

(

*



( √ ) ( √ )

( √ )

√

Proposed by D.M. Bătinețu – Giurgiu,Dan Grigorie – Romania

IX.27. If then:

√( )( )

√( )( )

√( )( )


IX.28. If then:



∑

∑( )

Proposed by D.M. Bătinețu – Giurgiu,Nicolae Radu – Romania


√

√

√

√

Proposed by D.M. Bătinețu – Giurgiu,Mihaela Dăianu – Romania


√

Proposed by D.M. Bătinețu – Giurgiu,Luiza Cremeneanu – Romania

IX.31. If then:

∑.

/

∑.

/

∑

Proposed by D.M. Bătinețu – Giurgiu,Neculai Stanciu – Romania

IX.32. If then:

√

√

√

√ ( √ √ )


IX.33. If then:

4

5 4

54

5 ∏( )

∏

Proposed by D.M. Bătinețu – Giurgiu,Alina Tigae – Romania

IX.34. If √ √ √ then:

(

√ *

(

√ *

(

√ *

Proposed by D.M. Bătinețu – Giurgiu,Amelia Curcă Năstăselu– Romania

IX.35. If then:

Proposed by D.M. Bătinețu – Giurgiu,Camelia Dană – Romania

IX.36. If * + then:



∑.

/

∑(

*

Proposed by D.M. Bătinețu – Giurgiu,Patricia Anicuța Bețiu – Romania

IX.37. If then in the following relationship holds:

:∑

;:∑

( )

;

( )

Proposed by D.M. Bătinețu – Giurgiu,Gheorghe Stoica – Romania


∑( )

Proposed by D.M. Bătinețu – Giurgiu,Nicolae Mușuroia – Romania


( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu,Cătălin Pană – Romania


∑ ( ) ( )

√


IX.41. If then in the following relationship holds: ( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu,Claudiu Ciulcu – Romania

IX.42. If then in the following relationship holds: ( )

( )

( )

√

Proposed by D.M. Bătinețu – Giurgiu,Roxana Vasile – Romania

IX.43. In Nagel’s cevian, the following relationship holds:

:√

√ ;∑√

Proposed by Bogdan Fuștei – Romania



IX.44. In – Nagel’s cevian the following relationship holds:

√ ∑

∑√


IX.45. In – Nagel’s cevian, – Gergonne’s cevian the following relationship holds:

∑ ( √ )

– area , -


IX.46. In – incenter, the following relationship holds:

:∑( )

;:∑( )

; √ :∑( )

;


IX.47. In – Nagel’s cevian, – Gergonne’s cevian the following relationship holds:

(

* ∑4

5


IX.48. In – Gergonne’s cevian the following relationship holds:

∑( )

Proposed by Bogdan Fuștei – Romania IX.49. In – Nagel’s cevian the following relationship holds:

∑ √

√ (

*

Proposed by Bogdan Fuștei – Romania IX.50. In the following relationship holds:

√

√

√

√

Proposed by Bogdan Fuștei – Romania IX.51. In – Nagel’s cevian, – Gergonne’s cevian the following relationship holds:

(

* ∑

√





∑

√

( √ )

√



∑

:∑

;:∑

√

;


IX.54. In – Nagel’s cevian, the following relationship holds:

(

*



∏( )

∑


IX.56. If then:

√

( )

√

( )

Proposed by Daniel Sitaru,Delia Popescu – Romania

IX.57.

( )

( )

( )

( )

( )

( )

( )

( )

( ) ( )

Prove that:

Proposed by Daniel Sitaru,Delia Schneider – Romania

IX.58. Let be a triangle and the lengths of the medians. Then, the following

inequality holds:

√

, where is the semiperimeter of . When do we have

equality?

Proposed by Gheorghe Alexe and George – Florin Șerban– Romania



IX.59. Find the positive numbers and which satisfy the relationship:

( )( )( )


IX.60. If , with . Prove the following inequality:


IX.61. Prove that in any triangle with the sides with the following

inequality holds: ( )

( )

( )

( )


IX.62. Prove that in any triangle with the sides the following inequality holds: ( )

( )

( )

( ) ( )

( )


IX.64. If in then:

Proposed by Daniel Sitaru,Dorina Goiceanu – Romania

IX.65. Applications of Tarek’s Lemmas. In – Nagel’s cevians. Prove that:

Proposed by Daniel Sitaru,Constantina Prunaru – Romania

IX.66. In – Gergonne’s cevians, – Nagel’s cevians. Prove that:


IX.67. In the following relationship holds: √ √ √ √

√


IX.68. If in √ then: √

√

√


IX.69. If then:



∑

∑

∑

( )


IX.70. Prove that in any the following inequality holds:

( )

√

( )

√

( )

√ ( )√


IX.71. If with and , prove that:


IX.72. If and prove that:


IX.73. Solve in the following equation:

( )√ √ ( ) √ , where


IX.74. Let be , such that and . Prove that:



( )

( )

( )


IX.76. Let be , such that and

. Prove that:



( )

( )

( ) (

*




IX.78. In the following relationship hold: √∑ ∑ ∑


IX.79. Let be and

. Prove that:

∑


IX.80. Let be . Solve in real numbers the system of equations: {

( )

( )

( )



∑(

*

∑(

*


IX.82. Find such that:


IX.83. If then:




10-CLASS-STANDARD



X.1. Solve in the following equations:

a) √

b) √

c) ( ) √

.

Proposed by Lucian Tuțescu, Mirea Mihaela Mioara – Romania

X.2. Let be the sequence of real numbers ( ) having the property that

( ) ( ) . Prove that .

Proposed by Cristina Ene, Elena Nicu– Romania

X.3. Let be . Solve the following equation:

( )


X.4. In the following relationship holds:

( ) , where


X.5. Let be . Prove that:



∑

( )


X.7. Let be positive real numbers such that and . Prove that:



∑ ∑

, where .





∑



, where


X.11. Let such that . Prove that:

( )

( )

( )



( )

.

/ ∑

( )( )

, where


X.13 Let be the inradii of circumcevian triangle of incenter in acute . Prove that:

√



∑

√ 4 (

*

5


X.15. Let be the excentral triangle of . Prove that:


X.16. – centroid, – Lemoine’s point. Prove that:

, -

, - .

/




X.17. In – Nagel’s point ( ) ( )

( ). Prove that:


X.18. Let be the circumcevian triangle of Lemonine’s point of .

Prove that:

(

*

, -

, - (

*


X.19. In – orthocenter the following relationship holds:

( )



Proposed by D.M. Bătinețu – Giurgiu,Gheorge Stoica – Romania

X.21. If then ( ) : √ √

Proposed by D.M. Bătinețu – Giurgiu,Gheorghe Stoica – Romania

X.22. If then:

Proposed by D.M. Bătinețu – Giurgiu,Virginia Grigorescu – Romania


( )

( )

( )

√



√

Proposed by D.M. Bătinețu – Giurgiu,Dan Nănuți – Romania

X.25. If then in the following relationship holds:

( )

( )

( )

( ) ( )

Proposed by D.M. Bătinețu – Giurgiu,Claudia Nănuți – Romania




4

5

4

5

4

5

.

/

Proposed by D.M. Bătinețu – Giurgiu,Ștefan Marica – Romania


∑

( )

( )( )

Proposed by D.M. Bătinețu – Giurgiu,Cristina Spiridon – Romania

X.28. In – centroid;

the following relationship

holds:

∑

√

Proposed by D.M. Bătinețu – Giurgiu – Romania


√

Proposed by D.M. Bătinețu – Giurgiu,Lavinia Trincu– Romania


( )

√

Proposed by D.M. Bătinețu – Giurgiu,Mihaela Nascu – Romania


∑( ) ( ) ( )

√

Proposed by D.M. Bătinețu – Giurgiu,Mihaela Stăncele – Romania


√


X.33. If denote √

;

√

√

then the following

relationship holds:

√

Proposed by D.M. Bătinețu – Giurgiu,Ileana Duma – Romania



X.34. If then the following relationship holds in :

∑

√

√ ( )

Proposed by D.M. Bătinețu – Giurgiu,Ileana Stanciu – Romania


∑



∑

√

( ) √

Proposed by D.M. Bătinețu – Giurgiu,Daniela Beldea – Romania

X.37. If , different in pairs, ,

( ) ( ) ( ) ∑

Prove that: .


X.38 If –internal bisectors then:


X.39 In acute – orthocenter, – altitudes. Prove that:

( ) ( )


X.40. Prove that it doesn’t exists ( ) such that: ( )


X.41. In Brocard angle, the following relationship holds:

( ) ( ) ( )


X.42. , are different in pairs,



( ) ( ) ( ) ∑

Prove that: .


X.43. In acute the following relatioinship holds:

( ) ( )


X.44. In acute the following relationship holds:

∑

4 (

*

5


X.45. Let be the median triangle of . If ( ) ( ) ( ) collinears,

– incenter. Prove that:

( )

( )

( )


X.46. In – internal bisectors, ( ) ( ) ( ).

Prove that: , - .

/ , - , -


X.47. Let be the inradii of circumcevian triangle of incenter in acute . Prove that:

( )



( )


X.49. In – Brocard’s angle, the following relationship holds:

√ (

√ (

*+




X.50. In – incenter, – Nagel’s point, the following relationship holds:

(

*∑


X.51. In – Nagel’s cevian, – Gergonne’s cevian, the following relationship holds:

∑

√ √

∑


X.52. In any triangle we have:

√

√

√

√ ∑ √

√


X.53. In – incenter, – Nagel’s point, – Nagel’s cevian the following

relationship holds:

∑

√


X.54. If in – nine – point center then: √ √ √ √

Proposed by Daniel Sitaru,Nineta Oprescu – Romania

X.55. If then:

( ) ( ) ( )

Proposed by Daniel Sitaru,Claudia Nănuți – Romania

X.56. If

then:


X.57. If then:



∑(

.

/ ∑.

/

,


X.58. If then:

√


X.59. If then:

4( )√ .

/√

5

√

:√( ) √.

/

;


X.60. If then:

( √ ) √

( √ ) √

( √ ) √

Proposed by Daniel Sitaru,Mădălina Giurgescu – Romania

X.61. Solve for real numbers: √

√ √

Proposed by Daniel Sitaru,Mădălina Giurgescu– Romania

X.62. If then: ( √ √ ) .

√

√ /

When does equality holds?

Proposed by Daniel Sitaru,Laura Zaharia – Romania

X.63. If then:

∏

√ √

∏

√ √

∏

√ √


X.64. If sides in a cyclic quadrilateral with – area, – semiperimeter then:

.

/(

*(

* (

* (√ )




X.65. If in ( ) then: ( )( )( )


X.66. , - – pyramid, ( ) ( ) ( )

( ) ( ) ( ) – radii of circumsphere, – radii of insphere. Prove that:

∑(

*


X.67.

Prove that:

Proposed by Daniel Sitaru,Tatiana Cristea – Romania

X.68. Prove that: .

/ is divisible by * +

Proposed by Prem Kumar-India

X.69. In – Bevan’s point, – excenters, ( )

( ) ( ) the following relationship holds:

( )

Proposed by Mehmet Sahin-Turkey

X.70. In – Bevan’s point, – excenters, ( ),

( ) ( ) the following relationship holds:

(, - , -)


X.71. In – Bevan’s point, – excenters, ( ),

( ) ( ) – area, the following relationship holds:

(

*

( )




X.72. In – excenters the following relationship holds:

(

*

(

*

(

*

( )


X.73. In – Bevan’s point, – excenters, ( )

( ) ( ) – area, the following relationship holds:

( )

( )

( )

( )


X.74. Find that in any triangle the following inequality holds: ∑

∑

When does the equality holds?

Proposed by Gheorghe Alexe and George – Florin Șerban-Romania

X.75. Prove that in any triangle the following inequality holds:

∑

∑

Proposed by Gheorghe Alexe and George – Florin Șerban-Romania

X.76. * + different in pairs ( ) ( ) ( ). If

∑

(

)

then .


X.77. Let be the intouch triangle of . Let be the orthic triangle of

. Prove that: [ ]

, -


X.78. In acute denote – radii of circle tangent simultaneous to incircle and to sides

. Define analogous . Prove that:


X.79. * +, different in pairs,

( ) ( ) ( ). Prove that:



∑ ( )

( )

( ) ( ) ( )


X.80. In acute – inradii of orthic triangle. Prove that:


X.81. In acute – Lemoine’s point the following relationship holds:

( ) (

*




X.83. If , different in pairs, ,

( ) ( ) ( ) ∑

. Prove that: .


X.84. In – Brocard’s point the following relationship holds:

( ) ( ) ( )


X.85. In acute – orthocenter, – altitudes. Prove that:

( ) ( ) ( )


X.86. * + different in pairs ( ) ( ) ( )

( )

( )

( )

. Prove that: .


X.87. In acute – Lemoine’s point, – bisectors of

( ) ( ) ( ). Prove that: , -

, - .

/


X.88. * +, different in pairs,



. Find:


X.89. , different in pairs, ; ( ) ( ) ( )

∑


X.90. In – internal bisectors, ( ) ( ) ( ).

Prove that: , - .

/ , - , -



∑

∑ (

*

√


X.92. In – centroid, – symedians in

( ) ( ) ( ). Prove that: , -

, - .

/


X.93. * +, different in pairs, ( ) ( ) ( )

∑

( ) √ ( ) ( ) ( )


X.94. , different in pairs, ( ) ( ) ( )

∑ ( )

√


X.95. If in – Nagel’s cevians then:

√


X.96. If in acute – incentre, – centroid, – orthocenter, – Nagel’s point then:

Proposed by Daniel Sitaru,Iulia Sanda – Romania



X.97. In – Nagel’s cevians, – Gergonne’s cevians. Find .


X.98. If then: ( ) ( ) ( ) ( )


X.99. Prove that in any triangle the following inequality holds:

∑

(

*


X.100. Prove that in any triangle the following relationship holds:

∑


X.101. Let be , such that . Prove that:

, where is a natural number.


X.102. Let be , such that . Prove that:

, where is a natural number. Proposed by Marin Chirciu – Romania


.

/

.

/

.

/




11-CLASS-STANDARD



XI.1. ( ) ( ( ) ) . Find:

( ) (( ) )


XI.2. Find all functions such that:

( ) ( ) ( ) ( )


XI.3. ( ) . Find: ( )


XI.4 , - ( ) ( ) – continuous. Prove that:

, - ( ) ( ) √ ( )


XI.5. Find ( ) such that: (

+


XI.6. ( ) (

+. Find: ( ) ( )


XI.7. If , then:

∑ ( ) ( )

( ) ( ) ( )

Proposed by Florică Anastase – Romania

XI.8. If , then:

√

√

√

√

√

√


XI.9. If , then: ( ) .

/ ( )


XI.10. If ∑

(

) , find:

.

/




XI.11. If .

/, then:

. ( )

/

. ( )

/

( ( )) ( ( ))

Proposed by Florică Anastase – Romania XI.12. If , then:

∏4

5

(∏

+

∑

Proposed by Florică Anastase – Romania XI.13. If ( ) or ( ) , then:

√:∏

; :∑

;

(

√(∑

) (∑ ( ) )

)

Proposed by Florică Anastase – Romania XI.14. If ( - then prove:

∑( )( )

( )( ( )( ))


XI.15. If ( ) ( ) ( ) are sequences of real numbers such that:

∑.

/

∑.

/ (

*

∑

(

( )

( )

( )*

Find: ( )


XI.16. If ( ) then:

∑ 4( )( )

√ 5

√∏ ( ) ∑ ( )


XI.17. If ,

( )

:

;

then: ( ) ( ) ( ) √


XI.18. If then:



|

| |

| ( )

|

|

Proposed by Daniel Sitaru,Doina Cristina Călina – Romania

XI.19. If .

/ then:

( ) ( ) ( ) √


XI.20. If

then: .

( )

/ .

( )

/ .

( )

/ .

( )

/


XI.21 If then: .

/ .

/ .

/

Proposed by Daniel Sitaru,Corina Ionescu – Romania

XI.22. Find all continuous function satisfying: ( ) ( ( )) and

( ) for all real and real .


XI.23. Prove that:

√

√


XI.24. Let ( ) ( )

* +. Prove that if is prime then:

.

/ ( ) for


XI.25. Find:

(

∑

+




XI.26. Consider the following function defined for all real numbers . ( )

How many integers are there in the range of ?


XI.27. Prove that .

/ is odd iff is a non-negative integral power of .


XI.28. If ( ) ( ) ( ) then find:


XI.29 For (.∏

/ .∑ .

/

/*

, find:

:

:

.

/;

;


XI.30. ( ) . Find: ( ) (( ) )


XI.31. If ( ) then find: ( )


XI.32. ( ) . Find: ( )


XI.33. ( ) . Find: ( )


XI.34. If ( ) then: ( ) ( )


XI.35 ( ) ( ) . Find: ( )


XI.36. Find all continuous functions such that:

( ) .

√ / – fixed


XI.37. If ( ) ( ) then:




XI.38. If ( ) then:

( ( ) (

* *


XI.39. ( ) ( ) ( ) . Find: ( )


XI.40. Prove that in any triangle, the following inequality holds:

(

* ( )

with the usual notations in triangle.

Proposed by Radu Diaconu – Romania


∑(

*

* +


Proposed by Radu Diaconu, Emil Popa – Romania


(

* ( )


Proposed by Radu Diaconu – Romania

XI.43. In the following relationship holds:

( ) .

/ .

/

Proposed by Daniel Sitaru,Cristian Moanță – Romania

XI.44. If then:



:∑:√

√ ;

; ( )( )( )

Proposed by Daniel Sitaru,Dana Cotfasă – Romania

XI.45. If then:

( )

( )

( )

( )

Proposed by Daniel Sitaru,Eugenia Turcu – Romania

XI.46. If then:

√ √

√ √

Proposed by Daniel Sitaru – Romania XI.47. Find:

(. /

:

∑:

. /

.

/;

;,


XI.48. If in

( ) ( ) ( )

then:

Proposed by Daniel Sitaru – Romania XI.49.

(

:

∑

;

(

∑

+

)

((

∑

+:

∑

;,

Find:

(∏

+


XI.50. {

⁄ * + * +

( ) }



Find: .

/


XI.51. Find:

:

.( )

( ) ( )

/

( ) ( ) ( );


XI.52. If then:

. ( )

( )/. ( )

( )/ . ( )

( )/


XI.53. Solve for real numbers:

( )

Proposed by Rovsen Pirguliyev-Azerbaijan

XI.54. Find such that: √( )( ) (√ √ )


XI.55. If

* + then:

∑

√∏


XI.56. Prove without softs:

∫(

( )

( ) *

√


XI.57. If

then:






12-CLASS-STANDARD

XII.1. If and , find:

∫

( √ √ √ √

,

Proposed by Florică Anastase – Romania XII.2. Prove:

∫( )

√

√( √ ) ( √ )


XII.3. For , prove that ( ) such that:

∫

√( )

√

( )

√

√

Proposed by Florică Anastase – Romania XII.4. Prove that:

∫

√

Proposed by Florică Anastase – Romania XII.5. Prove that:



∫

√

.

/


XII.6. ( ( ))( ( )) ( ) . Find:

( ) ∫4( ) ( ) ( ) ( )

( ( ))( ( ))5


XII.7. Find:

(

:∑√

∑

√

∑(√ √ )

√

;,

Proposed by Daniel Sitaru – Romania XII.8. Find:

(

( )∑4 4

55

+


XII.9. If

then:

∫4 ( ) ( )

5

( )

Proposed by Daniel Sitaru,Ionuț Ivănescu – Romania

XII.10. If √

then:

∫∫:

.

/;

( )( )

.

/


XII.11. If are all distinct such that and

( ) ∑ ∑

∑ ∑ ( )

where then prove that:

( ) (∏( )

+

Proposed by Daniel Sitaru – Romania XII.12. If then:



∫

∫

Proposed by Daniel Sitaru – Romania XII.13. If then:

(

∫

)

(

∫

)

( ∫

,

Proposed by Daniel Sitaru – Romania XII.14. Let * + be the sequence of all positive integers which do not contain the digits

zero. Prove that ⌈

⌉ 0

1, where ( ).

Note: ∑ . 0

1 /

, -


XII.15. Integrate: ∫ ( √ )


XII.16. If , - ( ) , continuous then:

∫:∑(( ( ) ( ))

( ) ( ) ( ) ( )+

;

( )


XII.17. Find:

∑ (∑( (

* (

**

+


XII.18. Find:

:

∑

( )( )( )

;


XII.19. , - , continuous, ∫ ( )

. Find:

∑ :

:

∑ (

* (

* (

*

;;




XII.20. Solve the following system of congruences.

{

( )

( )

( )

( )

Proposed by Naren Bhandari-Nepal XII.21. Find the number of ordered quadruples of positive integers ( ) such that the following holds:

, and are primes.

Proposed by Naren Bhandari-Nepal XII.22. Find:

∫ √

√

Proposed by Radu Diaconu – Romania XII.23. If then:

:∏:∫. √ /

;

;

∏(

*


(:∫(

*

;:∫(

*

;:∫(

*

;,


(

∫ 4

5

√ ∑ .

/

∫ 4

5

√ ∑ .

/

)


XII.26. Find:



:∑∑∑∑

( )( )( )( )

;


XII.27. Find:

:∫(

*

;


XII.28. If

then:

( ) ( ):∫

;

∫ .

/

:∫

;:∫

;


XII.29.

( ) ∫4

5

( ) ∫4

5

Prove that:

( ) ( )

0

1


XII.30. If

then: ∫

( )

Find: ∫ (( ) )


XII.31. Find (( )) .

/ such that:

( ) ( )


XII.32. Find:



∫ .0

( ) ( )

1/

, - great integer function.

Proposed by Jalil Hajimir-Canada

XII.33. Prove:

∫√ .

/




UNDERGRADUATE PROBLEMS

U.1. Find:

:∫ ( )

√

;

Proposed by Ajao Yinka-Nigeria U.2. Find:

∫ ( )


∫


:∫ ( )

√

;

Proposed by Ajao Yinka-Nigeria



U.5. Find:

∫

( )

Proposed by Ajao Yinka-Nigeria U.6. ( ) . Find:

∑( )

Proposed by Daniel Sitaru – Romania U.7. , - ( ) – continuous. Find ⏟

such that:

∫∫∫(( ( ) ( )) ( )

( ( ) ( )) ( )

( ( ) ( )) ( )

∫ ( )


U.8. If then:

∫∫ √ ( )

( ) √ ( )

Proposed by Daniel Sitaru,Marian Voinea – Romania

U.9. If then:

∫∫∫4

√ 5

(

*

( )

Proposed by Daniel Sitaru,Marian Voinea – Romania

U.10. If then:

( )( )∫∫4

5


U.11. Evaluate the following sum:

∑

( )


U.12. Prove that . ∑

( ) /

In fact, its value is Niven’s constant that is equal to Proposed by Prem Kumar-India

U.13. Prove that:



( )∏

√ ( )

The product on the left-hand side extends over all prime numbers . Proposed by Prem Kumar-India

U.14. Prove that

∏

Proposed by Prem Kumar-India U.15. Prove that:

( ) ( ∑

( )

+


U.16. Find all values of for which .

/ is not divisible by .

Proposed by Prem Kumar-India U.17. Evaluate the following sum:

∑

( )


U.18. For any real number and a positive integer prove that:

| ( )

( )

( )

( )

|

Proposed by K. Srinivasa Raghava-AIRMC-India

U.19. Prove the following:

∫ 4

( ) 5

( )

∫ 4

( ) 5

( )

( ( ( )) )

– Catalan constant Proposed by K. Srinivasa Raghava-AIRMC-India

U.20. Evaluate the integral in a closed – form:

∫ ( )

( )

Proposed by K. Srinivasa Raghava-AIRMC-India U.21. Prove the relation:

∫ .√ √ /

√

∫

( ) ( )



( ) ( ) ( )

.

/ ( ) ( )

where ( ) ( ) ∫ ( )


U.22. Compute the integral in a closed – form:

∫ 4√

5

Proposed by K. Srinivasa Raghava-AIRMC-India U.23. If

( ) ∫ ( )

( )

then, ∑ , ( )-( )

, ( )-( )

, ( )-( ) – Laplace transform of ( ) with variable , ( )-( ) – Mellin transform of ( ) with variable

Proposed by K. Srinivasa Raghava-AIRMC-India U.24. Let – Harmonic Number

( ) ∑ ∑( )

( )( )( )

then find the shortest solutions to the following values ( ) ( )

( ) ( )

( ) ( )

( ) ( )


U.25. For any complex number ( )

∫

( ) .

/

∫

( )

( )

( )

( )


U.26. Prove the following relation

∫ (

√

*

( ( )

* ( )

4 4 √ √ 5 5


U.27. Prove the integral relation and evaluate in a closed form:



∫∫ .

/

4 ( )

( ) 5 ∫

( ( ))

√


U.28. Find the Principal value of the integral:

∫ ( )

.

/

( )

.

/ ( )


U.29. Let be the angles of a triangle then show that:

.

/ .

/ .

/

.

/ .

/ .

/ .

/


U.30. Let be real numbers and if: ( ) ( ) ( )( )

and ( ) ( ) ( )( ) then find the value of the expression

4

5

(

*

Proposed by K. Srinivasa Raghava-AIRMC-India U.31. Prove the following relation:

∫ ( )√ ( )

(√ ( ))

√√ √

4 4√ √

5 (

√ *5


U.32. Evaluate the integral in a closed form:

:∫ ( )

( )

( ) ;


U.33. Find the shortest way to prove that:

∫ (√ )

( ) ( )

( )

where is the Catalan Constant Proposed by K. Srinivasa Raghava-AIRMC-India



U.34. If we define the function:

( ) ∫ ( )

( )

( )

then evaluate the integral in a closed – form:

∫ :∫ ( )

√

;


U.35. Find:

:∫ 4 ( )

5

;

Proposed by Mohamed Arahman Jama-Somalia

U.36. Find:

( )

. .

//


U.37. Let be the solution of the following equation: ( )

.

/ ( ) .

Find:

:

( )∫

( )

;


U.38. Find:

:∫( )( )

;


U.39. Find:

∫(∑(

( ) *

+




U.40. Find:

∑(

( ( )

* *


U.41. Find:

∑4( ) ( )

( ) 5


U.42. Find:

∫ :∑ :

(∑

(

*+;

;


U.43. For ( ) * +. Prove that:

∫. ( √ )

( ) /

Proposed by Mohammed Bouras – Maroc

U.44. Prove that:

∫ .( )

( )

/


U.45.

∫√ √

√ √ √ √ √ √ √ √

⏟

– Fibonacci number;

Prove that

is faster to give golden number then




U.46. If

then:

( ) √ ∫∫∫

4

( )5


U.47. Evaluate:

∫ ( )

Proposed by Abdul Mukhtar-Nigeria U.48.

∫√ ( ) ( )

√ ( ) ( ) ( )

Proposed by Abdul Mukhtar-Nigeria U.49. Prove:

∫ (

√ *

4 ( )

5 (

*


∫∑

( )

( ) ( )

Proposed by Abdul Mukhtar-Nigeria

U.51. Find:

:∫ ( ( ) )

;

Proposed by Abdul Mukhtar-Nigeria U.52. Evaluate:



∫ ( (

**


∫ ( ) ( ( ) ( ))


∫ ( )

∫ ( )

∫ ( )

( )

( )

where ( ) is the Riemann zeta function.

Proposed by Abdul Hafeez Ayinde-Nigeria

U.55.

∫( ( ) ( ))


U.56. Find:

∫4

√ 5


U.57. ( ) ∫ .

/

( ). Find:

∫4 ( )

√ 5


U.58. Find:



( ) ∫ (

( )( )*


U.59. Prove without the use of Harmonic series that:

∫ ( )

( )


U.60. Prove that:

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

( )

where is catalan’s constant.


U.61. A mixture of integrals and summations:

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫

∑

( )

∫

( )

∫ ( )

∑

∫

( )

∑

( )

∫

( ) ( )

( )

Where ( ) is the Riemann zeta function, is Catalan’s constant, is the Harmonic Number.


U.62.



∫ . √ /

Find .

/ and prove that:

(( ) ) (

* ( ) ( )


U.63. Inspired by Prof. Daniel Sitaru

Prove that:

∑(∑ (

( )( ) *

+

( )√

√


U.64. Define for all ( ) √ √ √ √ ( )

⏟

then prove that:

∫ ( )

( )

√ √ √ √ √

√ √


U.65. For all let:

( )

∫

Prove that:

∑ ( ) ( ) ( )

( ) ( )

( )

( )




U.66.

( ) ∑ . .

//

.

/

Prove that:

4

5

( ) ( ) ( )

Proposed by Daniel Sitaru – Romania U.67. Find:

∑(4

( )

5

( ) +


U.68. If then:

∫∫ (

*

( )∫

∫∫ (

*

( )∫

Proposed by Daniel Sitaru – Romania U.69. If , different in pairs then:

( ) ( ) ( ) :∏:√

√

;

;

( ) ∑(

4

5

+


U.70.

∫ ( )

( )




U.71. Let ( ) ( ) and ( ) ∫ ( )

prove that ∑ ( )

.

√ /


U.72.

∫ ( )

( ( ) ( ))


U.73. Find:

:∫ 4

5

;


U.74.

∫ ( ) ( ( ) ( ))


U.75. Find:

∫ (

√ *


U.76. Prove that:

∫ ( )

(

*


U.77. Prove:

∫∫ ∫

6, -

, - 7




U.78. Prove:

∫∫∫.

/


U.79. Find:

( ) ∫ ∫ ( , -)

, - - great integer function.


U.80. Find the points on the surface that are closest to the origin.


U.81. Prove without softs:

∫∫∫4 ( ) ( ) ( )

5

√


U.82.

∫∫ , -

, -

, -

, - is the greatest integer part of


U.83. Find:

∫4 ( ) ( )

5

Proposed by Cornel Ioan Vălean – Romania

U.84. Find:

∫4 ( )

5

Proposed by Cornel Ioan Vălean – Romania





ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-Winter 2020

PROBLEMS FOR JUNIORS

JP.271. If then:

( ) ( ) ( )


JP.272. If then:

√ √ √ √

Proposed by Hung Nguyen Viet – Vietnam

JP.273. If then:

( )


JP.274. If then:

( )( ) ( )


JP.275. If in then:

√


JP.276. In the following relationship holds:





(

*

(

*

(

*


JP.278. Solve for real numbers ( ; fixex):

√

√ ( )



( )

( )

( )



√ √

√

( )


JP.281. If then:

( )

√

( )

√

( )

√ √


JP.282. If then:

( ) ( )


JP.283. If then:

∑

∑ ( )


JP.284. In acute the following relationship holds:

√ √ √

√ √ √ √ .

/





√


PROBLEMS FOR SENIORS

SP.271. If

then:


SP.272. In the following relationship holds:

4

5

Proposed by George Apostolopoulos – Greece

SP.273. If then:

When does the equality holds?


SP.274. If in

then the following relationship holds:



(

*

(

*

(

*


SP.276. If then:

∑( )( )

( )

∑





(

*

∑:√

√

;


SP.278. Let be 0

1 ( )

. Find Imf.


SP.279. If in – Brocard angle then the following relationship holds:

√

√

Proposed by Vasile Jiglău – Romania

SP.280. If * + * + * +

then:

, - * + , - * + , - * + (, - , - , - )

* + , - , - great integer function


SP.281. If .

/ then:

(

4√

5

:√

;

)

(

4√

5

:√

;

)

:√

√

;


SP.282. If in – orthocenter; bisectors of angles respectively

( ) ( ) ( ) then the following relationship holds:

, -

, - .

/


SP.283. Find such that:

√

√

√

√





( )

( )

( )




UNDERGRADUATE PROBLEMS

UP.271. If

then:

∫∫∫. .

/ .

/ .

//

( ) ( )

Proposed by Daniel Sitaru – Romania UP.272. Prove without softs:

∫∫∫( (√ ))

Proposed by Florentin Vișescu – Romania

UP.273. In acute the following relationship hold:

(√ ) (√ ) (√ )

Proposed by Florentin Vișescu – Romania

UP.274. If

. /

. /

( )

. / then find:

4 √

5

Proposed by Florică Anastase – Romania UP.275. Find:

:

∑∑∑∑( )

;

Proposed by Daniel Sitaru – Romania UP.276. Find:

:∑∑∑∑(

*

;




UP.277. Find:

(∑ (

√( ) *

,


UP.278. If then:

∫∫( ( ))

( )


UP.279. If ( ) ( ) . ( )

( )/ and exists

(( ( ))

* then:

:(( ( ))

( ( ))

* ( ( ))

;

Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania

UP.280. Let be sides in ( ) ( ) ( ) sequence of positive numbers such that:

( )

(

*

(

*

Prove that:

4 √

√ √

5

√


UP.281. If ( ) ( ) .

/ √

√ √ √ √

√ then find:

4( )

√

√

5

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania

UP.282. If ( ) ( ) ( ) ( )

(

*

(

*

.

/ then:



(

√

( )

√

)

( )


UP.283. In the following relationship holds:

( )

( )

( )



( ) ∑

( )( )

√



4

5






INDEX OF AUTHORS RMM-27

Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele

1 DANIEL SITARU-ROMANIA 44 AJAO YINKA-NIGERIA

2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 45 PREM KUMAR-INDIA

3 CLAUDIA NĂNUȚI-ROMANIA 46 MEHMET ȘAHIN-TURKEY

4 NECULAI STANCIU-ROMANIA 47 SEYRAN IBRAHIMOV-AZERBAIJAN

5 MARIAN URSĂRESCU-ROMANIA 48 NGUYEN VIET HUNG-VIETNAM

6 BOGDAN FUȘTEI-ROMANIA 49 SRINIVASA RAGHAVA-INDIA

7 FLORICĂ ANASTASE-ROMANIA 50 NAREN BHANDARI-NEPAL

8 MARIN CHIRCIU-ROMANIA 51 MOHAMMED BOURAS-MAROC

9 VASILE BURUIANĂ-ROMANIA 52 MOHAMED ARAHMAN JAMA-SOMALIA

10 MARIAN DINCĂ-ROMANIA 53 VASILE JIGLĂU-ROMANIA

11 FLORENTIN VIȘESCU-ROMANIA 54 JALIL HAJIMIR-CANADA

12 ȘTEFAN MARICA-ROMANIA 55 ROVSEN PIRGULIYEV-AZERBAIJAN

13 ENE CRISTINA-ROMANIA 56 GEORGE APOSTOLOPOULOS-GREECE

14 GHEORGHE CALAFETEANU-ROMANIA 57 ABDUL MUKHTAR-NIGERIA

15 ELENA NICU-ROMANIA 58 ABDUL HAFEEZ AYINDE-NIGERIA

16 PETRE STÂNGESCU-ROMANIA 29 MĂDĂLINA GIURGESCU-ROMANIA

17 IULIANA TRAȘCĂ-ROMANIA 60 ALINA GEORGIANA GHIȚĂ-ROMANIA

18 AUREL CHIRIȚĂ -ROMANIA 61 ALINA TIGAE-ROMANIA

19 CONSTANTIN IONICĂ-ROMANIA 62 AMELIA CURCĂ NĂSTĂSELU-ROMANIA

20 DANIEL STRETCU-ROMANIA 63 LAVINIA TRINCU-ROMANIA

21 MIREA MIHAELA MIOARA-ROMANIA 64 CAMELIA DANĂ-ROMANIA

22 GHEORGHE ALEXE-ROMANIA 65 MARIAN VOINEA-ROMANIA

23 GEORGE-FLORIN ȘERBAN-ROMANIA 66 CĂTĂLIN PANĂ-ROMANIA

24 LUCIAN TUȚESCU-ROMANIA 67 CLAUDIU CIULCU-ROMANIA

25 RADU DIACONU-ROMANIA 68 CRISTINA SPIRIDON-ROMANIA

26 EMIL POPA-ROMANIA 69 CONSTANTINA PRUNARU-ROMANIA

27 ALECU ORLANDO-ROMANIA 70 CORINA IONESCU-ROMANIA

28 CRISTIAN MOANȚĂ-ROMANIA 71 IULIA SANDA-ROMANIA

29 DAN GRIGORIE-ROMANIA 72 LAURA ZAHARIA-ROMANIA

30 DANA COTFASĂ-ROMANIA 73 LUIZA CREMENEANU-ROMANIA

31 DANIELA BELDEA-ROMANIA 74 MIHAELA DĂIANU-ROMANIA

32 CATALIN SPIRIDON-ROMANIA 75 MIHAELA NASCU-ROMANIA

33 DELIA POPESCU-ROMANIA 76 MIHAELA STĂNCELE-ROMANIA

34 DELIA SHNEIDER-ROMANIA 77 NICOLAE RADU-ROMANIA

35 DOINA CRISTINA CĂLINA-ROMANIA 78 NINETA OPRESCU-ROMANIA

36 DORINA GOICEANU-ROMANIA 79 TATIANA CRISTEA-ROMANIA

37 ELENA ALEXIE-ROMANIA 80 IOAN CORNEL VĂLEAN-ROMANIA

38 ELENA IACOB MEDA-ROMANIA 81 PATRICIA ANICUȚA BEȚIU-ROMANIA

39 EUGENIA TURCU-ROMANIA 82 ROXANA VASILE-ROMANIA

40 GABRIELA VASILE-ROMANIA 83 VIRGINIA GRIGORESCU-ROMANIA

41 ILEANA DUMA-ROMANIA 84 GHEORGHE BOROICA-ROMANIA

42 ILEANA STANCIU-ROMANIA 85 NICOLAE MUȘUROIA-ROMANIA

43 IONUȚ IVĂNESCU-ROMANIA 86 GHEORGHE STOICA-ROMANIA NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected]

mailto:[email protected]

r. m. m. - 27...tuzson, florin popovici, viorel gh. vodă, ovidiu pop, e. grosswald (usa), leroy f....

Documents