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ROMANIAN MATHEMATICAL SOCIETYMehedinți Branch

SPRING EDITION 2020

R. M. M. - 24 ROMANIAN MATHEMATICAL

MAGAZINE

ISSN 2501-0099

Romanian Mathematical Society-Mehedinți Branch 2020

1 ROMANIAN MATHEMATICAL MAGAZINE NR. 24

ROMANIAN MATHEMATICAL SOCIETY

Mehedinți Branch

ROMANIAN MATHEMATICAL MAGAZINE

R.M.M.

Nr.24-2020



ROMANIAN MATHEMATICAL SOCIETY

Mehedinți Branch

DANIEL SITARU-ROMANIA EDITOR IN CHIEF ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT

ISSN 1584-4897 GHEORGHE CĂINICEANU-ROMANIA

EDITORIAL BOARD

DAN NĂNUȚI-ROMANIA EMILIA RĂDUCAN-ROMANIA MARIA UNGUREANU-ROMANIA DANA PAPONIU-ROMANIA GIMOIU IULIANA-ROMANIA DRAGA TĂTUCU MARIANA-ROMANIA CLAUDIA NĂNUȚI-ROMANIA DAN NEDEIANU-ROMANIA GABRIELA BONDOC-ROMANIA OVIDIU TICUȘI-ROMANIA

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

DANIEL WISNIEWSKI-USA EDITORIAL BOARD VALMIR KRASNICI-KOSOVO

ALEXANDER BOGOMOLNY-USA



CONTENT

Identities and inequalities in cyclic quadrilaterals- Claudia Nănuți, Daniel Sitaru............................4

Gakopoulos’ Lemmas- Thanasis Gakopoulos.............................................................................................9

About 1010 Inequality in Triangle – RMM 2018 – Marin Chirciu.......................................................15

About 996 Inequality in Triangle – RMM 2018 – Marin Chirciu.........................................................16

About 1070 Inequality in Triangle – RMM 2018 – Marin Chirciu.......................................................18

Properties of the eigenvalues of some classes of real matrices-Marian Ursărescu……………20 Structuri algebrice (I) - Vasile Buruiană…….……………………………………………………………………….23

Inequalities with cevians (I) – Bogdan Fustei….………………………………………………………………….29

Proposed problems………………………………………….………………………………………………………………...31

Index of proposers and solvers RMM-24 Paper Magazine.………………………………………….………100



IDENTITIES AND INEQUALITIES IN CYCLIC QUADRILATERALS By Claudia Nănuți, Daniel Sitaru – Romania

Abstract: In this paper are proved several metric identities in cyclic quadrilaterals Keywords and phrases: Bretschneider, Brahmagupta; Ptolemy; Girard. Notations: If is a cyclic quadrilateral denote:

= ; = ; = ; = ; = ; = ;Δ – area; – semiperimeter; , , , – exradii; – circumradii. By Bretschneider’s formula:

Δ = ( − )( − )( − )( − ) − cos (1)

+ = + = ; sin = sin ; sin = sin ; cos = − cos ; cos = − cos Replace + = in (1) and we obtain:

Δ = ( − )( − )( − )( − ) (2) which is called Brahmagupta’s formula.

Ptolemy’s theorem states that:

= + (3) Proposition 1: In any cyclic quadrilateral holds:

cos =( )

= − cos (4)

Proof: By the law of cosine in Δ ;Δ :

= + − 2 cos = + − 2 cos − − + = 2( cos − cos( − ))

2 cos ( + ) = − − +

cos =− − +2( + )

Proposition 2: In any cyclic quadrilateral holds:

sin = sin = (5)

sin = sin = (6) Proof:

Δ = [ ] + [ ] =12 sin +

12 sin ⇒

2Δ = sin + sin( − )

2Δ = ( + ) sin ⇒ sin =2+ = sin



Δ = [ ] + [ ] =12 sin +

12 sin

2Δ = sin + sin( − )

2Δ = sin ( + ) ⇒ sin =2Δ+ = sin


cos 2 =( − )( − )

+ = sin 2 ; cos 2 =( − )( − )

+ = sin 2

cos = ( )( ) = sin ; cos = ( )( ) = sin (7) Proof:

cos 2 =1 + cos

2 =1 +

( )

2 =

=2( + ) + − − +

4( + ) =( + ) − ( − )

4( + ) =

=( + − + )( + + − )

4( + ) =(2 − 2 )(2 − 2 )

4( + ) =( − )( − )

+

Proposition 4: In any Δ cyclic quadrilateral holds:

tan 2 =( − )( − )( − )( − ) = cot 2 ; tan 2 =

( − )( − )( − )( − ) = cot 2

tan = ( )( )( )( )

= cot ; tan = ( )( )( )( )

= cot (8)

Proof:

tan 2 =sin

cos=

( )( )

( )( ) =( − )( − )( − )( − ) = tan

−2 = cot 2

Proposition 5: (GIRARD’S IDENTITY) In any cyclic quadrilateral holds:

( + )( + )( + ) = 16 Δ (9) Proof: In Δ : = ; In Δ : = By proposition (2):

⋅= ⇒ = (10)

⋅= ⇒ = (11)

By multiplying (10); (11):

=16Δ

( + )( + )

By Ptolemy’s theorem: = + , hence:

+ =16 Δ

( + )( + )

( + )( + )( + ) = 16 Δ



Proposition 6 (circumradii) In any cyclic quadrilateral holds:

= ( )( )( ) (12) Proof: By Girard’s identity:

( + )( + )( + ) = 16 Δ

=( + )( + )( + )

16Δ

=( + )( + )( + )

4Δ

Proposition 7 (diagonals) In any cyclic quadrilateral holds:

=( + )( + )

+ ; =( + )( + )

+

Proof: By (10); (11):

=4Δ

+=( ) 1

+( + )( + )( + ) =

( + )( + )+

=4Δ

+ =( ) 1

+( + )( + )( + ) =

( + )( + )+

The circles tangent to a side of the cyclic quadrilateral and tangent to the extensions of its two other sides has radii , , , are called exradii. Proposition 8 (exradii) In any cyclic quadrilateral holds:

= =( )( )

(13), = =( )( )

=tan + tan

=Δ

( − )( + ) , =tan + tan

=Δ

( − )( + )

Proof: Denote – centre of excircle tangent to .

tan(∢ ) = = (14)



tan(∢ ) = tan = cot (15)

By (14); (15):

= cot ⇒ = tan (16)

tan(∢ ) = = (17)

tan(∢ ) = tan = cot (18)

By (17); (18):

= cot ⇒ = tan (19)

= = + =( );( )

tan 2 + tan 2 = tan 2 + tan 2 ⇒ =tan + tan

=tan + tan

=Δ

Δ tan + Δ tan=

=( ) Δ

( − )( − )( − )( − ) ⋅ ( )( )( )( )

+ ( − )( − )( − )( − ) ⋅ ( )( )( )( )

=Δ

( − )( − ) + ( − )( − ) =Δ

( − )(2 − − ) =Δ

( − )( + )


+ + + ≥2

Proof: By (13): =

( )( ); =

( )( )

By adding:

+ = + = ⋅( )( )

= ⋅( )( )

(20)

Analogous:

+ = ⋅( )( )

(21)

By adding (20); (21):

+ + + =Δ+ ⋅

+( − )( − ) +

Δ+ ⋅

+( − )( − ) ≥

≥ Δ ⋅ 2 ( )( )( )( )( )( )( )( )

= Δ ⋅ = 2 (21)



+ + + = + + + ≥

≥+ + ++ + + ≥

( ) 22 =

2

Equality holds for = = = ( – square)


+ + + ≥32

Proof: By (13):

= tan 2 + tan 2 ; = tan 2 + tan 2

= tan 2 + tan 2 ; = tan 2 + tan 2

By adding: + + + = 2 tan + tan + tan + tan (22)

Let be : 0, → ℝ; ( ) = tan ; ( ) =

( ) = = = > 0; – convexe

By Jensen’s inequality:

tan 2 + tan 2 + tan 2 + tan 2 ≥ 4 tan+ + +

2 =

= 4 tan = 4 tan = 4 tan = 4 (23) By (22); (23):

+ + + ≥ 2 ⋅ 4 = 8 (24)

+ + + = + + + ≥

≥+ + +

+ + + ≥( ) 8

2 =32

Equality holds for = = = ( – square) BIBLIOGRAPHY:

1. V.Pop,N.Minculete,M.Bencze-“An introduction in quadrilaterals geometry”-EDP-Bucharest-2015

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



GAKOPOULOS’ LEMMAS by Thanasis Gakopoulos-Greece

Lemma 1 Lemma 2

Plan 1

= = ⋅ ⋅ + 1 − ⋅ Lemma 1

= = ⋅ ⋅ + ⋅ Lemma 2

Proof. PLAGIOGONAL system: ≡ , ≡

(0,0), ( , 0), ( + , 0), ( , 0), (0, ), (0, ), ( , ), = , =

( , )

+ + = , + = , + =

: + = 1 (1), : = (2)

(1),(2)→ ( , )

=( + ) + ( + ) − ( + )

( + ) + −

=( + ) −

( + ) + −

=−− = ⋅ ⋅ + 1 − ⋅ →

→ = ⋅ ⋅ + 1 − ⋅

=−− =

1⋅

++ → =

1⋅ ⋅ + ⋅



Lemma 3 Lemma 4

Plan 2

= = ⋅ ⋅ or = = ⋅ ⋅ Lemma 3

= = ⋅ + ⋅ Lemma 4

Note: Lemma 4 is on the internet under the name Lemma Cristea. Proof: From plan 1, if ≡ then = 0 and we have Lemma 3. From Plan 1, if ≡ , then

= 0 and = . So, we have Lemma 4. Applications of Lemma 3 for the proof of theorems. Application 1

= ⋅ ⋅ → =

21

Application 2

= ⋅ ⋅ = ⋅ 1 ⋅ → =+



Application 3

=⋅ cos⋅ cos =

⋅=

− ++ −

= ⋅ cos =⋅

=2

+ −

= ⋅ ⋅ → =− ++ − ⋅

2+ −

Application 4

∥ → =

= ⋅ ⋅ → =

Application 5

∥ → = , =

= ⋅ ⋅ = ⋅ ⋅ → ⋅ ⋅ = 1

(Menelaus theorem: Δ , )



Application 6

, : = ⋅ ⋅

, : = ⋅ ⋅→(÷)

⋅ = ⋅ →

→ ⋅ ⋅ = 1 (Ceva theorem) Application 7

, : = ⋅ ⋅

, : = ⋅ ⋅→( )

→ 2 = ⋅ + ⋅ = ⋅+

+ ⋅+

→

→ 2 = + ⋅ + + ⋅ → 2 = + + + →

→ 2 = 2 + 2 → = + (Van Aubel theorem) Exercises that can be solved with the previous Lemmas.

Exercise 1.

= = . Find: , ,

Proposed by Thanasis Gakopoulos-Greece



Exercise 2. (Arsalan Waves)

: : = 1: 2: 3

: : = 3: 2: 1, = 3 . Find: Exercise 3 (Arsalan Waves)

: : = 3: 4: 3 : : = 4: 3: 4

= 160 Find: , ,

Exercise 4 (Romania 2009)

Δ , = , bisector of , bisector of , centroid of Δ

Prove: ∈ Exercise 5 (Proposed by Thanasis Gakopoulos-Greece)



Δ , ( ) = , ( ) = . Find: = ( , , ). If = ⋅ . Find:

Exercise 6 (Proposed by Thanasis Gakopoulos-Greece)

Δ , incenter of , ∥ , incenter of , , , , touch points

Find: , Exercise 7 (Proposed by Thanasis Gakopoulos-Greece)

Find:

= ( , , ), = ( , , )

Exercise 8 (internet)

= 90°, = , = , = 9, = 5

( ) =?



Exercise 9 (Proposed by Thanasis Gakopoulos-Greece)

= , = , = ∈ ℝ , 1 = 1 = 2 , = , =

Prove: = 1 + Exercise 10 (internet)

=

= , = Find:

ABOUT 1010 INEQUALITY IN TRIANGLE

ROMANIAN MATHEMATICAL MAGAZINE 2018

By Marin Chirciu – Romania 1) In Δ the following relationship holds:

⋅ℎ ≥ 2

486

Proposed by Daniel Sitaru – Romania Solution Using ≥ ℎ we obtain:

⋅ℎ ≥

⋅ ℎℎ = ℎ = 6 = 6

It suffices to prove that:



6 ≥ 2 ⇔ 3 ≥ ⇔ ≥ , which follows from Mitrinovici’s inequality

≥ 27 and Euler’s inequality ≥ 2 . Equality holds if and only if the triangle is equilateral. Remark Let’s find an inequality having on opposite sense: 2) In Δ the following relationship holds:

⋅ℎ ≤ 2 ( + )

Proposed by Marin Chirciu – Romania Solution

Using = cos , ℎ = and cos = ( ), we obtain:

∑ ⋅ = ∑ = ∑ ( )( )

= 8 ∑ ( )( )

(1)

From ( + ) ≥ 4 it follows: ∑ ( )

( )≤ ∑ ( ) = ∑ ( ) = ( ) = (2)

From (1) and (2) it follows ∑ ⋅ ≤ 2 ( + ).Equality holds if and only if the triangle is

equilateral. Remark The double inequality can be written: 3) In Δ the following inequality holds:

6 ≤⋅ℎ ≤ 2 ( + )

Solution See inequalities 1) and 2). Equality holds if and only if the triangle is equilateral.

ABOUT 996 INEQUALITY IN TRIANGLE ROMANIAN MAHTEMATICAL MAGAZINE

2018 By Marin Chirciu – Romania

1) In Δ the following relationship holds:

12 ≤+ℎ ≤

9 √3

Proposed by Mehmet Șahin – Ankara – Turkey Solution

We will prove the following lemma: Lemma 2) In Δ the following relationship holds:

+ℎ =

+ − 2

Proof:



Using ℎ = we obtain: +ℎ =

+=

( + )2 =

2 ( + − 2 )2 =

+ − 2

Using the Lemma, the left-hand inequality can be written:

≥ 12 ⇔ ≥ 14 − , true from Gerretsen’s inequality: ≥ 16 − 5 and Euler’s inequality ≥ 2 .

For right-hand inequality we prove that: ∑ ≤ ⇔ ≤ ⇔ ≤ 6 + 2 − , true from Gerretsen’s

inequality: ≤ 4 + 4 + 3 and Euler’s inequality ≥ 2 .

Then ≤ √ ⇔ ≤ √ (Mitrinovic’s inequality) Equality hold if and only if the triangle is equilateral. Remark If we replace ℎ with we propose: 3) In Δ the following relationship holds:

12 ≤+

≤ 42

− 1

Proposed by Marin Chirciu – Romania Solution We prove the following lemma: Lemma: 4) In Δ the following lemma:

+=

2( − 3 − 6 )

Proof Using ℎ = we obtain:

+=

+=

( + )( − )=

2 ( − 3 − 6 )=

2( − 3 − 6 )

Using Lemma, the left-hand inequality can be written:

≥ 12 ⇔ ≥ 12 + 3 , true from Gerretsen’s inequality:

≥ 16 − 5 and Euler’s inequality ≥ 2 . The right-hand inequality can be written:

2( − 3 − 6 )≤ 4

2− 1 ⇔

+ − 2≤

6⇔ ≤ 4 + 4 + 3

(Gerretsen’s inequality). Equality holds if and only if the triangle is equilateral.

Remark

Between the sums ∑ and ∑ the following relationship holds:

5) In Δ the following relationship holds:



+ℎ ≤

+

Proposed by Marin Chirciu – Romania Solution Using the above Lemmas, we have:

∑ = and ∑ = . We the write the inequality:

≤ ⇔ ≥ 10 + 7 , true from Gerretsen’s inequality: ≥ 16 − 5 and Euler’s inequality ≥ 2 .

Remark We can write the sequence of inequalities: 6) In Δ the following inequality holds:

12 ≤+ℎ ≤

+≤ 4

2− 1

Solution See inequalities 5), 3) and 1). Equality holds if and only if the triangle is equilateral.

ABOUT 1070 INEQUALITY IN TRIANLGE ROMANIAN MATHEMATICAL MAGAZINE

2019 By Marin Chirciu – Romania

1) In Δ , – centroid, – Lemoine’s point. Prove that:

⋅ + ⋅ + ⋅ ≥4√3

Proposed by Marian Ursărescu – Romania Solution We prove the following lemma: Lemma 2) In Δ , – centroid, – Lemoine’s point. Prove that:

⋅ =23 ⋅

( − 6 ) − (4 + )− (4 + )

Proof Using = and = ⋅ we obtain:

⋅ =23 ⋅

2+ + ⋅ =

43( + + ) ⋅ =

=⋅ [ ( )]

⋅ [ ( − 6 ) − (4 + ) ] = ⋅ ( )( )

, which follows

from the known identity in triangle ∑ ⋅ = − 6 − (4 + )

Back to the main problem: Using Lemma, the inequality can be written:



23 ⋅

( − 6 ) − (4 + )− (4 + ) ≥

4 √33

Taking into account Doucet’s inequality 4 + ≥ √3 it suffices to prove that:

⋅ ( )( )

≥ ( ) ⇔ ( − 2 − 14 ) + (4 + ) ≥ 0 (*)

We distinguish the following cases:

Case 1). If ( − 2 − 14 ) ≥ 0, the inequality is obvious.

Case 2). If ( − 2 − 14 ) < 0, the inequality can be rewritten:

(4 + ) ≥ (14 + 2 − ) which follows from Blundon-Gerretsen’s inequality 16 − 5 ≤ ≤ ( )

( )

It remains to prove that:

(4 + ) ≥(4 + )

2(2 − ) (14 + 2 − 16 + 5 ) ⇔ 2 − 3 − 2 ≥ 0 ⇔

( − 2 )(2 + ) ≥ 0, obviously from Euler’s inequality ≥ 2 . Equality holds if and only if the triangle is equilateral. Remark The inequality can be strengthened: 3) In Δ , – centroid, – Lemoine’s point. Prove that:

⋅ + ⋅ + ⋅ ≥43

(4 + )

Solution See (*) from the above proof.Equality holds if and only if the triangle is equilateral. Remark Inequality 3) is stronger than inequality 1). 4) In Δ , – centroid, – Lemoine’s point. Prove that:

⋅ + ⋅ + ⋅ ≥43

(4 + ) ≥4√3

Solution See 3) and Doucet’s inequality 4 + ≥ √3.Equality holds if and only if the triangle is equilateral. Remark Let’s find an inequality having an opposite sense: 5) In Δ , – centroid, – Lemoine’s point. Prove that:

⋅ + ⋅ + ⋅ ≤ 3(4 + )

Solution Using Lemma, the inequality can be written:

23 ⋅

( − 6 ) − (4 + )− (4 + ) ≤ 3

(4 + ) ⇔



⇔ (4 + + 12 − 2 ) ≥ (4 + ) ( − 2 ), which follows from

Gerretsen’s inequality: ( ) ≤ 16 − 5 ≤ ≤ 4 + 4 + 3 .

It remains to prove that:

(4 + )+ 4 + + 12 − 2 (4 + 4 + 3 ) ≥ (4 + ) ( − 2 ) ⇔

⇔ 3 − 8 + 6 − 4 ≥ 0 ⇔ ( − 2 )(3 − 2 + 2 ) ≥ 0, obviously from

Euler’s inequality ≥ 2 .Equality holds if and only if the triangle is equilateral.

Remark The double inequality can be written: 6) In Δ , – centroid, – Lemoine’s point. Prove that:

43

(4 + ) ≤ ⋅ + ⋅ + ⋅ ≤ 3(4 + )

Proposed by Marin Chirciu – Romania Solution See inequalities 3) and 5).Equality holds if and only if the triangle is equilateral. Remark. The following inequalities can be written: 7) In Δ , – centroid, – Lemoine’s point. Prove that:

12 ≤4√3

≤44

(4 + ) ≤ ⋅ + ⋅ + ⋅ ≤ 3(4 + ) ≤

32

Solution See inequalities 4), 6), Euler’s inequality ≥ 2 and Mitrinovic’s inequality ≥ 3 √3. Equality holds if and only if the triangle is equilateral. Reference: 1. Marian Ursărescu, 1070 Inequality in triangle, Romanian Mathematical Magazine, February 2019. 2. Marin Chirciu, Algebraic Inequalities, from beginner to performer, Paralela 45 Publishing House, Pitești, 2014. 3. Marin Chirciu, Inequalities with important lines in triangle from beginner to performer, Paralela 45 Publishing House, Pitești, 2018.

PROPERTIES OF THE EIGENVALUES OF SOME CLASSES OF REAL MATRICES

By Marian Ursărescu – Romania

In this article we will prove some interesting properties about the eigenvalues of some real matrices, followed by applications. For start, we will remember some of the classic results from matrices theory and their determinants. Definition



Let be ∈ (ℝ) and ∈ , (ℝ). If it exists ∈ ℂ such that = , then is called

own vector, and eigenvalue for the matrix .

Observation. The matricial equation ( − ) = is equivalent with the system:

(1)

( − ) + + ⋯+ = 0+ ( − ) + ⋯+ = 0

… … … … … … … … … … … … … … … … … .+ + ⋯+ ( − ) = 0

The system (1) has nonzero solutions ⇔ det( − ) = 0.

Definition The polynom ( ) = det( − ) is called characteristic polynom of the matrix , and the equation ( ) = det( − ) = 0 is called characteristic equation of the matrix . Observation The eigenvalues of matrix are the solutions of the characteristic equation. Theorem. The characteristic polynom has the expression ( ) = − Δ + Δ +⋯+ (−1) Δ , where Δ represents the sum of the principal minors having the order of the matrix − . Observation

1. … = det 2. + + ⋯+ =

3. + + ⋯+ = ∗ Cayley – Hamilton Theorem If ( ) is the characteristic polynom of matrix , then ( ) = . Definitions Let be ∈ (ℝ) 1. Matrix is called symmetric if: = 2. Matrix is called antisymmetric if = − . 3. Matrix is called orthogonal if: ⋅ = (we have denoted = transposed matrix). Theorem 1 The eigenvalues of some real symmetric matrix are real. Proof Let be ∈ (ℝ) with = . Let’s suppose that it exists an eigenvalue such that:

∈ ℂ ⇒ = (1)

We multiply to the left relation (1) with ⇒ = ⋅ (2)

We conjugate relation (1) ⇒ = (3) ∈ (ℝ)

We multiply to the left relationship (3) with ⇒ = ⋅ ⇒

⇒ ( ) = ( ⋅ ) ⇒ = ⋅ (4)

From (2) and (4) ⇒ ⋅ = ⋅ ⇒

− ( ⋅ ) = ⋅ = ⋅ + + ⋯+ ⋅ = ( ) + ( ) + ⋯+ ( ) > 0

⇒



⇒ − = 0 ⇒ = ⇒ ∈ ℝ.

Theorem 2 The eigenvalues of some real antisymmetric matrix are or nonzero or purely imaginary. Proof Let ∈ (ℝ) with = − . For start let’s prove that the only eigenvalues are nonzero. Let ∈ ℝ be an eigenvalue ⇒ = ⇒ by multiplying to the left with ⇒

= (1) ⇒ ( ) = ( ) ⇒ − = (2)

From (1)+(2)⇒ = − ⇒ 2 ⋅ = 0 ⇒ = 0

Let ∈ ℂ be an eigenvalue ⇒ = ⇒ = (3) ⇒ = ⇒ = ⋅ ⇒ ( ) = ( ⋅ ) ⇒

− = ⋅ (4) From (3)+(4)⇒ ⋅ = − ⋅ ⇒

+ ⋅ = 0 ⇒ + = 0 ⇒ = − ⇒ is purely imaginary, namely =

Theorem 3 The eigenvalues of an orthogonal real matrix have an absolute value equal with 1. Proof. Let be ∈ (ℝ) with = . Let be an eigenvalue ⇒ = ⇒ by multiplying to the left with ⇒ = (1)

= ⇒ = ⇒ = ⇒ by multiplying to the left with ⇒ = ⇒ ( ) = ( ⋅ ) ⇒

⇒ = ⋅ (2) (1) + (2) ⇒ ⋅ ⋅ ⋅ = | | ⋅ ( ⋅ )

= | | + | | + ⋯+ | | ⇒

⇒ ⋅ ⋅ = = | | ( ⋅ ) ⇒ ( ⋅ ) = | | ( ⋅ ) ⇒ | | = 1 ⇒ | | = 1

Applications 1. Let ∈ (ℝ) an antisymmetric matrix a) If is odd, then det = 0 b) If is even, then det is a perfect square. Proof a) = 2 + 1 ⇒ ( ) has an odd number of pairs ⇒ at least one is real ⇒ from theorem 2 ⇒ the real root is ⇒ det = … = 0. b) = 2 ⇒ ( ) has an even number of pairs, ( ) ∈ ℝ[ ] ⇒ the roots are complex conjugate, from theorem 2 ⇒ = and = − , = and = − … = and = − ⇒ det = … = ( ⋅ … ) > 0 2. Let ∈ (ℝ) such that = and = 0. Prove that: | | = 3

(Mathematical Gazette) Proof Let , , be the eigenvalues of matrix from theorem 3 ⇒ | | = | | = | | = 1 = 0 ⇔ + + = 0 ⇒ + + = 3 (known identity)⇒

| + + | = 3| || || | ⇒ ( ) = 3



3) Let be ∈ (ℝ). If ⋅ = and = 0 ⇒ ∈ {−2,0,2} (Mathematical Gazette)

Proof Let be , , the eigenvalues of matrix . From theorem 3⇒ | | = | | = | | = 1

= + + , = + + = 0 det( ⋅ ) = det ⇒ (det ) = 1 ⇒ det = ±1

+ + = 0 ⇒ ( + + ) = 2( + + )| | = 1 ⇒ ⋅ = 1, = 1,3

⇒

( + + ) = 21

+1

+1

⇒

( + + ) = 2 + + ⇒ ( ) = 2 det ; ∈ ℝ ⇒ = ⇒ ( − 2 det ) = 0 ⇒ or = ±2

STRUCTURI ALGEBRICE (I)

By Vasile Buruiană – Romania INTRODUCERE

Fie ( ,⋅) un semigrup comutativ unitar cu unitatea . Se spune că ∈ se divide (în ) la

∈ și se notează „ ⋮ ” dacă există ∈ astfel încât = . Se mai spune în acest caz că

e este multiplu al lui sau că este divizor (factor) al lui sau că divide notând în acest

ultim caz „ | ”. Evident că ∀ ∈ avem | (divizibilitatea este reflexivă) și dacă | și |

pentru , , ∈ atunci | (divizibilitatea este tranzitivă), deci relația de divizibilitate în

este o relație de preordine. Se spune că ∈ este inversabil sau unitate ⇔ ∃ ∈ și

= . Pentru , ∈ se spune că sunt asociate și notăm „ ∼ ” ⇔ ( | și | ). Se

spune că ∈ ∖ { } este atom ⇔ ( | , ∈ ⇒ ori = ori = ), iar ∈ − { } se

numește prim dacă din relația | ⇒ | sau | . Un element ∈ ∖ { } se numește

ireductibil sau idecompozabil dacă din | ⇒ ori ∼ ori este unitate, iar ∈ ∖ { } se

numește reductibil sau decompozabil dacă există ∈ , ≁ și neinversabil astfel că | .

Pentru , ∈ se numește cmmdc al lor (dacă există) un element ∈ , notat = ( , )

având proprietățile: | , | și ∀ ∈ cu | , | ⇒ | , la un cmmmc al lor (dacă

există) se numește un element ∈ având proprietățile ⋮ , ⋮ și ∀ ∈ cu

⋮ , ⋮ ⇒ ⋮ . Asemănător, acestea se definesc și pentru mai mult de două

elemente. Un semigrup ( ,⋅) comutativ unitar cu unitatea se numește atomic

⇔[(∀ ∈ ⇒ există o descompunere finită = … , ≥ 0 și atomi pentru



= 1, … , ) și ( = 0 ⇔ = ) și descompunerea este unică in afara ordinii și asocierii

factorilor]. Un semigrup ( ,⋅) se numește cu simplificarea ⇔[ ∀ , , ∈ și = ⇒

= ]. Dacă este inel integru comutativ și unitar, este evident că ( ∗ = ∖ {0},∘) este

semigrup cu simplificare și că | , | ⇒ |( ± ). O mulțime ⊂ se numește ideal

⇔[∀ , ∈ ⇒ − ∈ și ∀ ∈ , ∈ ⇒ ∈ ]. Idealul = = { | ∈ } se

numește principal (generat de ). Idealul generat de o mulțime constă din intersecția

tuturor idealelor lui care îl conțin pe și constă din toate sumele finite ∑ cu

∈ , ∈ . Notăm ( , … ) = ∑ idealul generat de = { , … , }. Un inel

integru comutativ unitar se numește euclidian ⇔ ∃ : ∗ → ℕ astfel că ∀ , ∈ ∗ cu

≠ 0,∃ , ∈ astfel că = + și ori = 0 ori ( ) < ( ). Un inel integru comutativ

și unitar se numește principal ⇔ orice ideal al său este principal.

Exemple și observații.

1) (ℕ∗,⋅) este semigrup cu simplificarea, singurul element inversabil fiind 1. In acest

semigrup relația de divizibilitate este și antisimetrică (deci de ordine). Intr-adevăr, dacă

, ∈ ℕ∗ și | , | ⇒ ∃ , ∈ ℕ∗ astfel ca = , = ⇒ = deci = 1 ⇒

= 1 = deci = .

2) Dacă , ∈ ℕ∗, și | ⇒ 1 ≤ ≤ . Într-adevăr dacă = , ∈ ℕ∗ ⇒ ≥ 1. Dacă

= 1 ⇒ = sau ≥ 2 fie ∈ ℕ∗ astfel ca succesorul său = + 1 = ⇒

= = = ( + 1) = + ≥ , deci 1 ≤ ≤

3) Dacă ∈ ℕ∗; > 1 atunci:

este reductibil ⇔[∃ , ∈ ℕ∗, 1 < < , 1 < < și = .

Demonstrație: „⇒” Dacă este reductibil ⇒ ∃ ≠ 1, ≠ și | . Fie = ⇒ 1 ≤ ≤

și 1 ≤ ≤ și 1 < < . Dar = 1 ⇒ = , iar dacă = ⇒ = 1, ceea ce este

imposibil.

„⇐” Fie = , 1 < < , 1 < < . Evident nu avem niciuna din relațiile ∼ , ∼ ,

inversabil, inversabil ⇒ este reductibil.

4) Dacă ∈ ℕ∗ , > 1 ⇒ ∃ un divizor ireductibil al lui . Intr-adevăr, luând

= { ∈ ℕ∗| | , > 1}, avem că ≠ ∅ căci | ⇒ ∃ = cel mai mic elemnt al lui ,

> 1. Dacă ar fi reductibil și ar exista astfel că | , 1 < < . Deci | | , < ,



1 < ⇒ ∈ ceea ce contrazice minimalitatea lui ⇒ este ireductibil și de aici

afirmația.

5) În (ℕ∗,⋅) sunt echivalente afirmațiile:

a) este prim; b) este ireductibil

Demonstrație: a) ⇒b). Dacă prin absurd ar fi prim și reductibil ar rezulta că ∃ , ∈ ℕ∗

astfel încât = , 1 < < , 1 < < . Cum = ⋅ 1 = ⇒ | ⇒ | sau

| ⇒ ≤ sau ≤ , ceea ce este contradictoriu

b)⇒ a) Presupunem prin absurd că există ∈ ℕ∗ ireductibil și cu proprietatea că | și

∤ , ∤ ; alegem , astfel ca sa fie minimal cu proprietatea de mai sus. Evident ≠ 1

căci atunci | . Rezultă că 1 < . Dacă am avea ≥ ⇒ ℕ

.î ∃ , ∈ ℕ astfel ca

= + , 0 < < , deci ∤ și deci și , deoarece = + , au proprietatea că

| , ∤ , ∤ ; însă < cee ace contrazice minimalitatea lui . Vom avea deci

1 < < (și analog 1 < < ). Dacă ∈ ℕ∗ și = nu putem avea = 1. Deci

> 1 ⇒ ∃ ireductibil, | . Cum = < ⋅ = ⇒ < deci ≤ < și cum

| iar era ireductibil minimal ce nu era prim ⇒ | sau | . Fie = , =

(admițând că | ). Din = ⇒ = , deci | și ∤ , ∤ , < ⇒ <

și iarăși se contrazice minimalitatea lui . Se infirmă astfel existența numerelor naturale

ireductibile care nu sunt prime.

6) Teorema fundamentală a aritmeticii

∀ ∈ ℕ∗ , > 1 ⇒ există , , … , prime (ireductibile) unice în afara unei permutări,

astfel că = , … ,

Demonstrație: Existența Fie = { | > 1, ≠ } ≠ ∅

⇒ ∃ = cel mai mic element al lui . Evident elementele lui nu sunt prime (ireductibile)

⇒ nu este ireductibil ⇒ este decomposibil ⇒ ∃ , ∈ cu = și 1 < < , 1 < < .

Cum , ∉ căci este minimal ⇒ și pot fi reprezentate ca produse finite elemente

prime, deci și = , ceea ce este absurd. Rămâne deci adevărată afirmația existenței.

Prima demonstrație a unicității (Euclid): Fie = … = … un element cu două

descompuneri în factori primi. Cum | … ⇒ există și îl putem presupune pe = ,



astfel că | . Datorită ireductibilității factorilor ⇒ = ⇒ … = … (1).

Dacă = 1; ≥ 2 atunci , … , sunt inversabile, ceea ce este absurd ⇒ = 1.

Deci pentru un număr > 1 care are o descompunere cu = 1 factori primi, afirmația este

adevărată. Folosim inducția după , și presupunem că afirmația este adevărată pentru

numerele > 1 având o descompunere cu − 1 factori. Din (1), având o astfel de

descompunere ⇒ − 1 = − 1 ⇒=ș

= , = 1, … , deci afirmația este adevărată și

pentru .

A doua demonstrație a unicității (Zermelor, Hasse, Lord Cherwell)

Evident 2 și 3 admit o descompunere unică. Dacă ∃ = … = … cel mai mic

număr cu două descompuneri, putem presupune ≤ ⋯ ≤ , ≤ ⋯ ≤ . Nu putem

avea existența unor , cu = , căci atunci = < ar avea două descompuneri

distincte și s-ar contrazice minimalitatea lui . Fie < și fie = … <

… = . Cum | , | ⇒ | − = ( − ) … = . Cum evident

∤ − și ∤ … ⇒ mai are o descompunere în care nu apare . Dar <

contrazice iarăși minimalitatea lui cu două descompuneri, fiind astfel infirmată existența

numerelor naturale cu două descompuneri distincte în factori ireductibile (primi). Deci orice

număr natural are o descompunere unică.

7) Rezultă din cele de mai sus că (ℕ∗,⋅) este un semigrup atomic în care atomii sunt chiar

elementele prime (ireductibile). El are o infinitate de atomi (dacă , … , are fi toate

numerele prime două câte două distincte, atunci … + 1 nu se divide prin minimul

dintre , … , și este distinct de fiecare dintre … deci = … + 1 este de

asemenea prim, ceea ce este contradictoriu)

8) ({ }, ≥ 0,⋅) unde este prim în ℕ∗ este un semigrup atomic cu un singur atom, iar

({1},⋅) este și el semigrup atomic fără atomi. (ℕ, +) este de asemenea semigrup atomic cu

singurul atom 1, singurul element inversabil fiind 0. Ultimele două semigrupuri atomice sunt

izomorfe prin ( ) = . Cu (ℕ, +) relația | echivalează cu ≤ .

9) Fie mulțimea atomilor din (ℕ∗,⋅). În locul scrierii unice = … , dacă factorii identici

se grupează sub același exponent vom mai scrie = … sau = ∏ ( )∈ unde



( ) == 1, … ,

dacă = și ( ) = 0 pentru ≠ , = 1, … , . Adică = ∏ ( )∈ unde

( ) sunt unici și aproape toți, cu excepția unui număr finit, sunt nuli. Dacă

= ∏ ( )∈ ; = ∏ ( )

∈ , = ∏ ( ) ( )∈ și avem și

| ⇔ ( ) ≤ ( )∀ ∈ . Ultima echivalență rezultă din aceea că dacă = cu

= ∏ ( )∈ avem ∏ ( ) = ∏ ( ) ( ) ⇔ ( ) = ( ) + ( ) ⇔

( ) ≤ ( );∀ ∈ .

10) Dacă = … este descompunerea unică, atunci numărul divizorilor lui este

( ) = ( + 1)( + 1) … ( + 1), după cum rezultă din observația precedentelor.

Evident este prim (atom, ireductibil)⇔ ( ) = 2 ⇔[ se divide doar prin și prim 1]

11) Fie ∈ ℕ∗, > 1. Atunci este prim ⇔ ∑ − = 2, unde [ ] este partea

întreagă a lui , adică cel mai mare întreg din stânga lui sau care coincide cu .

Într-adevăr, dacă | ⇒ ∃ ∈ ℕ cu = ⇔ = deci = iar − 1 = − 1 =

( − 1) + ( − 1) și deci = − 1 ⇒ − = − ( − 1) = 1. Dacă însă

∤ , există , ∈ ℕ astfel că = + , 0 < < , deci = și

− 1 = + ( − 1) ⇒ = deci − = − = 0. Deci

− = 1 ă |0 ă ∤ ⇒ ∑ − = numărul divizorilor lui = ( ). Deja

am văzut că este prim ⇔ ( ) = 2, de unde afirmația făcută.

12) ∀ , ∈ (ℕ∗,⋅) ⇒ ∃ cmmdc și ∃ cmmmc al lor. Într-adevăr doar = … ,

= … luănd = … și = … unde = min( , ) , = 1, … , ,

= max( , ) , = 1, … și o simplă verificare o a definițiilor arată că = ( , ),

= [ , ]. Observăm doar | ⇒ ( , ) = și [ , ] = , iar dacă ∤ putem căuta

masura lor comună maximă, adică cmmdc, luănd = + , 0 < < și constatînd că

( , ) = ( , ). Repetând procesul pentru și , rezultă algoritmul lui Euclid cu care putem

găsi cmmdc: scriem șirul de împărțiri cu rest

= + , 0 < <

= + , 0 ≤ <

= + , 0 ≤ <



⋮

ș.a.m.d. până ce obținem obligatoriu un rest nul (aceasta se va întâmpla într-un număr finit

de pași fiindcă > > ⋯ nu poate fi infinit). Atunci ( , ) va fi ultimul rest nenul din acest

algoritm și în plus [ , ] =( , )

.

13) ℤ este inel euclidian. Intr-adevăr, mai întâi arătăm că ∀ , ∈ ℤ, ≠ 0 ⇒ ∃ , ∈ ℤ, unice

astfel că = + , 0 ≤ < | | (teorema împărțirii cu rest în ℤ)

Existența: Pentru = 0 luăm = = 0. Pentru ≠ 0 și > 0, luăm mulțimea

= { − | ∈ ℤ} și în ea pe = − = cel mai mic element pozitiv. Aceste și =

verifică afirmația. Într-adevăr nu putem avea = − ≥ , căci ar rezulta

− ( + 1) > 0 și cum − ( + 1) < − = se contrazice minimalitatea lui

⇒ 0 ≤ < . Procedăm asemănător și pentru < 0. În privința unicității:

dacă = + = +

0 ≤ < | |0 ≤ < | | și ≠ , ≠ ⇒ | || − | = | − | și cum

| − | ≥ 1 ⇒ | | ≤ | − | ceea ce este absurd.

Luând în fine : ℤ∗ → ℕ pun ( ) = | |, cu cele de mai sus rezultă imediat că ℤ este

euclidian.

14) În inelul ℤ sunt adevărate afirmațiile:

a) ∀ , ∈ ℤ∗, idealul = { + | , ∈ ℤ} este principal și generatorul acestuia, , este

cmmdc al numerelor , .

b) (Lema Euclid) | și ( , ) = 1 ⇒ | . Elementele ireductibile de ℤ sunt prime.

c) ∀ ∈ ℤ se scrie unic sub forma = (−1) ( ) ∏ ( )∈ unde este mulțimea

elementelor prime pozitive din ℤ, ( ) = 0, > 01, < 0 și ( ) = 0 cu excepția

eventual a unui număr finit de elemente din ( ∈ , ∈ ℤ).

Demonstrație: a) Fie = + cel mai mic elemnt pozitiv din . Evident ∈ . Luând

apoi ∀ ∈ ⇒ ∃ , unice cu = + cu 0 ≤ < . Dar 0 < < cum = − ∈

se contrazice minimalitatea lui ⇒ = 0 deci ∈ ⇒ ⊂ ⊂ ⇒ = ℤ. Apoi cum

| ⋅ 1 + , | ⋅ 0 + 1 ⇒ | , iar dacă | , ⇒ | + = ⇒ =cmmdc. În

particular | , | = 1 ⇔ ∃ , ∈ ℤ cu + = 1

b) Cum ( , ) = 1 ⇒ ∃ , cu + = 1 ⇒ | + = și afirmația este dovedită

două luăm ireductibil, | și ∤ căci atunci ( , ) = 1 ⇒ | deci este prim.



c) Observăm mai întâi că ∀ > 0 și ∀ prim pozitiv ⇒ ∃ , unic astfel că | și ∤ .

Aceasta este evident căci puterile lui sunt crescătoare. Numărul ∈ ℕ, unic, se numește

ordinul lui în și se notează ( ) și verifică proprietățile ( ) = 0 ⇔ ∤ și

( ) = ( ) + ( )

Existența scrierii lui rezultă din teorema fundamentală a aritmeticii aplicată lui sau − .

Cum în ℤ∗ singurele unității sunt ±1, elementele asociate sunt egale sau diferă pun semn și

unicitatea rezultă aplicând funcția lui = (−1) ( ) ∏ ( ) și găsind că

( ) = ( ). Remarcăm că (ℤ∗,⋅) este semigrup cu simplificarea, dar nu este atomic

fiindcă orice element are divizorii ±1, ± .

INEQUALITIES WITH CEVIANS (I)

By Bogdan Fustei-Romania In ∆ABC holds:

2 ( − )( − ) ≤ ( − ) + ( − ), −semiperimeter

p-b+ p-c=a; ( − )( − ) = hence 4 ≤ (and analogous).

By =( -r)( + ) (and analogous) follows: ≤ . By summing: ∑ ≤ −

(Adil Abdullayev ,Marian Ursărescu RMM TRIANGLE MARATHON problem 214).

Hence: ≤ (and analogous);

+ + = 4 + , ∑ ≤ −

= (and analogous); = = ≥ (and analogous).

∑ ≥ + + , ≥ , ≥ ∑

By ≤ (and analogous) → ≤ (and analogous).

By summing:

≤3

4 ,1≤

14 ,

1≤

14

1+

1+

1=

14

1

+1

+1

=1;

By ≥ (andanalogous) → ≥ (and analogous) ;



1ℎ +

1ℎ +

1ℎ =

1;

By summing:

≥ ∑ , ≥

2 ≥ℎℎ ℎ ; ≤ 4 → ≤ 4 ; ≤ 2 ;

4 ( + + ) ≤ + + ↔ 4 (4 + ) ≤ + + ;

cos +cos + cos ≤ √ (O.Bottema-Geometric Inequalitys 1969);

3√3 ≤ ( ) →3√32 ≤ 2 ;

∑ cos ≤ ; = cos → + = cos

By summing:

cos 2 + cos 2 + cos 2 =12

+ ,

12

+≤ 2

2 ≥ + + , ≥12

2 + +

Panaitopol’s inequality :

2 ≥ ℎ → ℎ ≥ 2 ; = 2 ℎ ;

bc≥4r , bc= + → + ≥4r => ≥ (4 − )

= = ; = ;

( ) ≥ (4 − )(4 − )(4 − ), ≥ (4 − )(4 − )(4 − )

≥(4 − )(4 − )(4 − )

, 2 ≥ + +

By summing:

≥ + + + ( )( )( ) , ≥ ∑

Hence:

≥ (∑ + ( )( )( ) ; ≥ 4 → ≥

4 ≥4

; = 4 = 4 ; = ; 4 ≥

By summing:



34 ≥ => ≥

43 ;

= → = ∙ =( )

; ≥ ∑( )

PROPOSED PROBLEMS

5-CLASS-STANDARD

V.1. Alcătuiți o frază care să dea un sens literar expresiei: ∪ , , , , Ț ,

Proposed by Carina – Maria Viespescu– Romania

V.2. Aflați cel mai mic număr natural cu cel puțin două cifre distincte și ultima cifră nenulă, cu proprietatea că mutând ultima cifră a numărului în fața primei cifre, obținem un divizor al său. Proposed by Petre Stângescu – Romania

V.3. Find all pairs ( , ) of positive integers for which 8 = 2 + 8 + 12 + 8 + 5

Proposed by Pedro Pantoja-Brazil

V.4. Să se afle trei numere naturale consecutive < < cu proprietățile:

1) = , 2) – număr prim. Proposed by Ștefan Marica – Romania

V.5. Să se afle numărul cu proprietățile: 1) = ⋅ +

2) ( + ) = . Proposed by Ștefan Marica – Romania

V.6. a) Să se demonstreze că pentru (∀) ∈ ℕ∗, (∃) , ∈ ℕ∗ astfel încât

+ = ( + 1) + (*). b) Să se demonstreze că dacă , , ∈ ℕ∗ verifică relația (*), atunci + + 1 este pătrat perfect.

Proposed by Petre Stângescu – Romania

V.7. Find last digit of the number: Ω = . Proposed by Sorin Pîrlea – Romania



V.8. Determinați cel mai mic număr de 4 cifre care împărțit la 43 dă câtul egal cu restul.

Proposed by Dana Cotfasă – Romania

V.9. Let be ∈ ℕ, ≥ 3, odd. a) Find the reminder of the division of the number 4 to 5

b) Write the number 4 that the sum of four natural consecutive odd numbers.

c) Prove that the number 4 can be written as a difference between two perfect squares.

Proposed by Marin Chirciu – Romania

V.10. Let be a nonzero natural number. Solve the equation.

( + 1) + ( + 2) + ⋯+ ( + 2 + 1) = (2 + 1)


V.11. Prove that if 2009 can be written as a sum of squares of two nonzero natural numbers, for any ∈ ℕ∗.

Proposed by Marin Chirciu, Octavian Stroe – Romania

V.12. Compare the numbers: = 50 , = 353 , = 354 . Generalization.


V.13. Let be < nonzero natural numbers. Find ∈ ℕ such that the fractions:

( ) and are equivalents. Proposed by Marin Chirciu – Romania

V.14. Because 15 = 3 + 5 + 7, then number 15 can be written as a sum of at least two odd consecutive numbers. a) 2018 can be written as a sum of at least two odd consecutive numbers? If yes, then give an example of how it can be written. If no, the prove why not.

b) 2019 can be written as a sum of at least two odd consecutive numbers? If yes, the give an example of it can be written. If not, then prove why not.

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.15. In an apartment block, there are apartments with a room and with two rooms. If 75% of the residents live in apartments with two rooms, find what is the percentage of apartments with two rooms.




V.16. Find all the numbers which verify the properties:

i) { , , , , , } = {1,2,3,4,5,6}, ii) < 1, iii) < 1, iv) > 1

Proposed by Neculai Stanciu, Titu Zvonaru -Romania

V.17. a) Prove that it exists an infinity of natural nonzero consecutive numbers , , , , such that + + + + is a perfect square and + + is a perfect cube;

b) Prove that it exists an infinity of nonzero, consecutive natural numbers , , , , such that + + + + is a perfect cube and + + is a perfect square.


V.18. a) Find all lowest nonzero, consecutive, natural numbers , , , , , , such that

+ + + + + + is a perfect square and + + + + is a perfect cube;

b) Find all the lowest nonzero, consecutive natural numbers , , , , , , such that

+ + + + + + is a perfect cube and + + + + is a perfect square.


V.19. Prove that there is an infinity of pairs ( , ) of integers which verify the relationship

3 + 4 + 6 = 0. Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.20. Prove that there is an infinity of pairs ( , ) of integers which verify the relationship

15 + 16 + 20 = 0. Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.21. Prove that the number 100. .01 with 2012 zeros is composed.


V.22. Find the rest of the division of the number 1 + 2 + ⋯+ 2011 to the number 2013.


V.23. Find the rest of the division of the number 1 + 2 + ⋯+ 2013 to the number 2015.

Proposed by Titu Zvonaru-Romania



V.24. Find the digit such that: ( + + + ⋯+ … ): = 123456789.


V.25. Prove that there aren’t natural numbers formed from three digits of 1 and the rest of the digits 0 which can be written as a sum of two perfect squares.

Proposed by Neculai Stanciu, Titu Zvonaru-Romania

V.26. Prove that the number 11 + 11 + 11 + 11 + 1 has an even number of divisors. Proposed by Titu Zvonaru, Neculai Stanciu -Romania

V.27. Find the numbers of three digits which divided to their reversed give the quotient 5 and the reminder a prime number. Proposed by Titu Zvonaru -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. Fie prim, ∈ ℕ∗ cu ( , ) = 1, ∈ ℕ, ≥ 2. Să se demonstreze că ecuația:

( ) − = ⋅ , nu are soluții în ℕ.


VI.2. Fie ∈ ℕ, ≥ 3, ∈ ℕ∗, ≤ − 1, , … , cifre cu și ≠ 0. Să se demonstreze că: … ≠ … ⋅ …


VI.3. Fie ∈ ℕ∗ și ecuația + = (*). Să se demonstreze că ecuația (*) are exact trei

soluții în ℕ∗ dacă și numai dacă este număr prim.




VI.4. Fie , ∈ ℕ cu 7 = 9 + 5. Demonstrați că + și nu pot fi pătrate perfecte.


VI.5. Să se afle suma a 20 de numere naturale distincte două câte două, știind că suma oricăror trei dintre ele este cel puțin 24, iar suma oricăror cinci dintre ele este cel mult 120.


VI.6. Să se arate că 2019( + ) nu este pătrat perfect, (∀) , ∈ ℕ∗


VI.7. Aflați număr prim, ∈ ℕ, ≥ 2, pentru care există numerele naturale , , … , distincte două câte două cu: + + ⋯+ = 2017 (*)


VI.8. Să se afle numărul cu proprietatea: = ⋅ + ⋅

Proposed by Ștefan Marica – Romania

VI.9. Să se afle numerele prime ; ; și știind că au loc relațiile:

1) + = + , 2) + = + = 10


VI.10. Find , , , ∈ ℕ; – prime number such that = + + and + + is divisible with . Proposed by Roxana Vasile, Sanda Iulia - Romania

VI.11. Aflați numărul de triunghiuri având laturile exprimate prin numere naturale iar cea mai mare latură este 11. Proposed by Doina Cristina Călina– Romania

VI.12. Prove that if and are natural numbers, prime between them, such that:

1 + + + ⋯+ = , then 2011 divides .

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.13. Prove that if and are natural numbers, prime between them, such that:

⋅ 1 − + − + ⋯− + + + = , then 2011 divides .

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania



VI.14. Find all the triangles that have the side’s lengths prime numbers and the square of the area natural number. Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.15. Prove that the equation + = has an infinity of natural solutions.


VI.16. Prove that:

(9 + 2 ⋅ 5 + 5) ∈ ℕ, where ∈ ℕ


VI.17. Solve for integers:

5 − ⋅ 5 − 6 = 601


VI.18. Prove that there is an infinity of triplets of natural numbers ( , , ) for which the number 4 + 4 + 4 is a perfect square. Proposed by Marin Chirciu – Romania

VI.19. Let be , ∈ ℝ∗ such that + = = . Calculate ( + ) .


VI.20. Solve for integers: 4 − ⋅ 5 − 3 = 55.


VI.21. If − = − , where ≠ 0 and ≠ , calculate the product .


VI.22. Prove that: (5 + 2 ⋅ 3 + 1) ∈ ℕ, where ∈ ℕ.


VI.23. If 0 < ≤ 1, = 1,2011, prove that:

2010 + 1+

2010 + 1+ ⋯+

2010 + 1≥ 2011




VI.24. Find the numbers 0 < ≤ 1, = 1,2015, which satisfy the relationship:

2014 + 1+

2014 + 1+ ⋯+

2014 + 1= 2015

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

VI.25. Prove the inequality:

12017 1 +

12 +

13 + ⋯+

12017 >

12018 1 +

12 +

13 + ⋯+

12018

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

VI.26. Find all the pairs of integers ( , ) which verify the equality: 2 + 2 = .

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

VI.27. Prove that there is an infinity of pairs ( , , , ) of integers which verify the relationship: (3 + 4 )(15 + 16 ) = 120


VI.28. If , ∈ ℝ verify the relationship + + = 2, then prove that:

min(2 + 3 + 2 ) = max(7 − 8 )

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



7-CLASS-STANDARD

VII.1. If , , ∈ ℝ, + + = 3 then: | + ( + ) + | ≤ 4

Proposed by Daniel Sitaru – Romania



VII.2. Find , ∈ ℕ, – prime such that: ! = + + + ⋯+

Proposed by Seyran Ibrahimov-Azerbaijan

VII.3. Aflați ∈ ℤ astfel încât expresia = + 2 + 9 − 8 să fie cub perfect.


VII.4. Aflați ∈ ℕ și număr prim știind că:

= ( + ) − (2 + 1) + 1 ∈ ℚ


VII.5. Să se demonstreze că pentru (∀) ∈ ℕ, ≥ 2, = ≠ ℚ


VII.6. Dacă =√

+√

+ ⋯+√

, ∈ ℕ∗, calculați [ ]


VII.7. Să se demonstreze că √66 + 106 + 34 ≠ ℚ, (∀) ∈ ℕ


VII.8. Arătați că 7( + ) ∉ ℚ, (∀) , ∈ ℕ∗


VII.9. Dacă , ∈ ℕ∗ cu √2019 > , arătați că

√2019 > +73

100


VII.10. If , , > 0 then:

2( + ) + 3( + ) + 6( + ) ≥6 + +3 + 4 + 5




VII.11 Să se afle numărul determinat de relația: + + = ⋅


VII.12. If = 3 − √3 then find Ω = .

Proposed by Tatiana Cristea – Romania



8-CLASS-STANDARD

VIII.1. I have a ≥ 4 digits positive integers which shows the following property:

9 × … = …

8 × … = …

that is when it’s multiplied by 9,8 the digits are reversed and , ∈ ℕ. Let , be general formula for above largest number respectively then for = prove that =

Proposed by Naren Bhandari-Nepal

VIII.2. Let , , > 0 and √ + √ + √ = 3√ . Prove:

2√21

√ +( , , )

−1

√ + −( , , )

≤ 3

Proposed by Nguyen Van Nho-Vietnam

VIII.3. Let > 0,∀ = 1; ( ≥ 2) and ∑ = . Prove: ∑

∑≤ √2 .




VIII.4. Let , , > 0. Prove:

++ ( + 2 )( + 2 )

++

+ ( + 2 )( + 2 )+

++ ( + 2 )( + 2 )

≥32


VIII.5. If , , > 0 then:

( + )( + ) ≥ 2 + 2( + )( + )( + )

Proposed by Nguyen Van Canh-Vietnam

VIII.6. If , , > 0 and = 1 then:

1− 2 + + 3 +

1− 2 + + 3 +

1− 2 + + 3 ≤ 1


VIII.7. If 0 < , ≤ 1 then:

+ ≥ 2 − − + 1


VIII.8. If , , ≥ 0 and + + = 1 then:

( + + )( + + ) ≥ + + +


VIII.9. Solve in ℝ:

+ +

3 + 4 + 4 ++

3 + − 8 +

4 + 4 +=

12


VIII.10. If , , ∈ (0,1) then: + +( )

+( )

+( )

≥ √

Proposed by D.M. Bătinețu – Giurgiu; Constantina Prunaru – Romania



VIII.11. If , ≥ 0; + > 0; , , > 0 then: + + ≥ ( )

Proposed by D.M. Bătinețu – Giurgiu; Oana Preda – Romania


( + ) + ( + ) + ( + ) ≥ 4 3 ( + + ) + ( − ) + ( − ) + ( − )

Proposed by D.M. Bătinețu – Giurgiu; Ramona Nălbaru – Romania


( + ) ( + )( + ) + ( + ) +

( + ) ( + )( + ) + ( + ) +

( + ) ( + )( + ) + ( + ) ≥ 2 3 ( + + )

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

VIII.14. If , , > 0, different in pairs then:

( + + )1

( − ) +1

( − ) +1

( − ) ++ +

>814

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

VIII.15. If , , > 0; + = ; ∈ ℕ then:

√ + √√

+√ + √√

+√ − √√

+√ − √√

≥ 4

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

VIII.16. If , , > 0 then: + + + + + ≥ 2 3 ( + + )

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

VIII.17. If , , , , , > 0 then: + + ≥

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

VIII.18. If , , , , , , > 0 then: ( )

+( )

+( )

≥ ( ) ( )( )

Proposed by D.M. Bătinețu – Giurgiu; Dana Cotfasă – Romania



VIII.19. Să se demonstre că expresia:

= 2[(2 + 1) + 2 ][(2 + 1) + (2 + 1)] + 4 + 2

nu poate fi pătrat perfect oricare ar fi , , , , , ∈ ℕ.


VIII.20. Aflați toate valorile lui ∈ ℕ∗ pentru care: !+

!+

!+ ⋯+

!=


VIII.21. If , , , , , , > 0 then:

+( + ) + ( + ) +

+( + ) + ( + ) +

+( + ) + ( + ) ≥

3+


VIII.22. If , , , , > 0 then:

+ √+

+ √+

+ √≥

+ ++


VIII.23. If , , > 0, + + = then:

1 + + 1 + + 1 + ≥ 1

Proposed by Vasile Mircea Popa – Romania

VIII.24. If , , > 0, + + = 12 then:

14( + + ) ≤ 9 + 96

Proposed by Iuliana Trașcă – Romania

VIII.25. If , , > 0, + + = 6 then:

31 − 6 − 6+ + − 5 +

31 − 6 − 6+ + − 5 +

31 − 6 − 6+ + − 5 ≤ 7

Proposed by Iuliana Trașcă – Romania



VIII.26. If , , , are sides of convexe quadrilater then:

( + + − )( + + − )( + + − )( + + − )( + )( + )( + )( + ) < 1


VIII.27. If , , > 0, = 1 then: ∑ + +( )

− 6 ≥ 0


VIII.28. Să se afle cel puțin o pereche de numere prime ; cu proprietățile:

− = , ( + + ) − ( + + ) = , − = 1, − = 2.


VIII.29. Piramida patrulateră regulată are: = √7; = √ . Să se afle:

1) ( ; ) unde ∈ și = . 2) unde ⊥ ; ∈ .


VIII.30. Find the last 674 digists of the number: Ω = ; [∗] - great integer function.

Proposed by Nicolae Tomescu, Lucian Tuțescu – Romania

VIII.31. If , , > 0 then: max ; ; ≥ 1

Proposed by Doina Cristina Călina – Romania



9-CLASS-STANDARD



IX.1. In Δ the following relationship holds: sin = (1 + cos ) + 2 sin

Proposed by Mustafa Tarek-Egypt

IX.2. In Δ the following relationship holds:

√ + √ + √ ≥ 24√3( − 2 )


IX.3. Let ∈ 0; . Prove:

(1 + sin ) + (1 + cos ) + (1 + sin ) + (1 + cos ) < 6


IX.4. Let , , ∈ (−1; +∞) ∧ + + = 3. Prove:

1( + + 1) +

1( + + 1) +

1( + + 1) ≥

13



cos−4 + cos

−4 + cos

−4 ≤

√ + +2

Proposed by Mustafa Tarek-Cairo-Egypt

IX.6. In Δ , – incenter the following relationship holds:

⋅ ℎ( + ℎ ) +

⋅ ℎ( + ℎ ) +

⋅ ℎ( + ℎ ) = 1



( + ℎ )⋅ ℎ +

( + ℎ )⋅ ℎ +

( + ℎ )⋅ ℎ ≥ 9


IX.8. Solve for natural numbers:



= (1 − + )

= √

Proposed by Urfan Aliyev-Baku-Azerbaijan

IX.9. Let , , ≥ 0 ∧ + + = 1. Prove:

2 + 1+ + 1

( , , )

≤ 3

Proposed by Nguyen Van Nho- Vietnam

IX.10. If , > 0, ≥ then:

sin+

sin>

1

cos + sin



ℎ + ℎℎ =

− ℎ ℎ ℎ

Proposed by Bogdan Fustei-Romania


23( + + )

≤ 2



sin cot 2 + sin cot 2 + sin cot 2 ≤ 3 −12

Proposed by Marian Ursărescu-Romania

IX.14. In Δ , – incentre the following relationship holds:



ℎ + ℎ + ℎ ≤3√3

2


IX.15. Solve in ℝ:

√ − + 1 + √ − 3 + 1 + √ + + 1 + √ + 3 + 1+ 2 = 2


IX.16. In Δ , – incentre the following relationship holds:

+ℎ +

+ℎ +

+ℎ ≤

+ +


IX.17.

= 6 tan327 + 4 sin

227 −

1496646325× 10

This form is accurate up to 10th decimal place. More, interestingly with further simplification we can obtain

tan327 + 4 sin

227 −

49888215 × 10 = 6


IX.18. Find all functions :ℝ → ℝ satisfying:

( + ) ≥ ( + 1) ( ),∀ , ∈ ℝ, 1 ≤ ∈ ℕ


IX.19. In any triangle holds:

+ + 42 ≤ + + ≤ 4 +




IX.20. In all triangle holds: 1. max ∑ ;∑ ≤ 4 + , 2. min ∑ ;∑ ≥


IX.21. If , , > 0 then in Δ the following relationship holds:

2 sin + 2 sin + 2 sin+ + ≤

+ +


IX.22. Find all functions :ℝ → ℝ:

= ( ) ( ) ,∀ , ∈ ℝ, , ∈ ℕ − {0}− fixed


IX.23. Let be Δ and , , the middles of , , arcs (made with the circumcenter). Prove that:

⋅ ⋅⋅ ⋅ ≤ cos

−2 cos

−2 cos

−2

Proposed by Marian Ursărescu – Romania

IX.24. Prove that in any acute-angled triangle the following inequality holds:

cos sin(sin ) + cos (sin(sin ) + cos ⋅ sin(sin )) ≤32 sin

√34


IX.25. Let be Δ , , and the internal bisectors. Prove that:

⋅ + ⋅ + ⋅ ≥ 18


IX.26. Find the functions :ℕ∗ → ℕ∗ having the property:

(1)− (2) + (3) … + (−1) ( ) =



= ( + − 1) (1) − (2) + ⋯+ (−1) ( ) ;∀ ≥ 1


IX.27. Find all :ℝ → ℝ satisfying: ( ) = 4 ( )− 3 ,∀ ∈ ℝ


IX.28. In Δ , ≥ ≥ , + ≥ 3 . Prove that: 4 − 9 ≥ 0.


IX.29. Prove that in any Δ holds the following inequality: + + 6∑ ≥ 16


IX.30. Let be Δ and , , the interior bisectors (concurrent in ). Prove that:

⋅ + ⋅ + ⋅ ≥643 ⋅


IX.31. Prove that in any acute-angled Δ the following inequality holds:

+ + ≥ 4 ( + )


IX.32. Prove that in any Δ the following inequality holds:

8 cos 2 − cos 2 ≤ 15


IX.33. Prove that in any Δ the following inequality holds:

(ℎ + ℎ )(ℎ + ℎ )ℎ ℎ ≥ 768 ⋅


IX.34. Prove that in any acute-angled triangle the following inequality holds:



ℎ+ℎ

+ℎ

≥12



( + ) sinℎ + ℎ +

( + ) sinℎ + ℎ +

( + ) sinℎ + ℎ = 3


IX.36. In Δ the following relationship holds: + + = 3



6 + ℎ + ℎ + ℎ ≥ 2( + + )



+ℎ + ℎ

≥ 6 sin 2



ℎ + ℎ ≥+ +

4 ≥ ℎ + ℎ


IX.40. If in Δ , – incentre then:

1( + + ) ≥

1( + ) +

1( + ) +

1( + )





+ + ≥ 6√3 ; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Gabriel Tică – Romania


+cos 2 +

+cos 2 +

+cos 2 ≥

4 − 94

Proposed by D.M. Bătinețu-Giurgiu, Lavinia Trinu – Romania


+ + ≥ 6√3 ; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Alexandrina Năstase – Romania

IX.44. If , > 0 then in Δ the following relationship holds:

( + + )( )

+( )

+( )

≥( )

; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Cătălina Pană – Romania

IX.45. In Δ the following relationship holds: + + ≥ √


IX.46. If , > 0 then in Δ the following relationship holds:

( + ) + ( + ) + ( + ) ≥48

+


IX.47. If ≥ 0 then in Δ the following relationship holds:

( ) + ( ) + ( ) ≥ 3 + + 4

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



IX.48. If , , , > 0 then in Δ the following relationship holds:

+ ( − − 4 )+

++ ( − − 4 )

++

+ ( − − 4 )+

≥(2 + 3 )

+

Proposed by D.M. Bătinețu-Giurgiu – Romania, Martin Lukarevski – Macedonia

IX.49. If , , > 0 then:

3( + + )+ + ≤ 2( + + )

14 + + +

1+ 4 + +

1+ + 4 ≤ 9

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

IX.50. If : (0,∞) → (0,∞) is such that: ( ) + ( ) ≤ 2 , (∀) , > 0 then

( ) + ( ) + ( ) ≤ 33

+ + ; (∀) , , > 0


IX.51. In Δ the following relationship holds: + + ≥

Proposed by D.M. Bătinețu-Giurgiu – Romania

IX.52. Prove that:

cos317 + cos

517 − cos

617 + cos

717 =

1 + √174


IX.53. Let Δ be the circumcevian triangle of incenter in Δ .

If { } = ∩ , { } = ∩ , { } = ∩ then:

+ + ≤(4 + )

3


IX.54. Solve for ∈ 0, : sin + cos + tan + cot + (sec + csc ) = 2 1 + √2




IX.55. In Δ ; cos = ; cos = . Find cos 3 .

Proposed by Ileana Duma – Romania

IX.56. Exists :ℝ → ℝ; ( ) = + + ; , , ∈ ℝ; ≠ 0 such that:

(2 ) = ( + + 1), (∀) ∈ ℝ ?

Proposed by Camelia Dană, Cristian Moanță – Romania



10-CLASS-STANDARD

X.1. In Δ , , , – circumradii of Δ ,Δ ,Δ

, , – excenters, – Bevan’s point. Prove that:

+ + ≥92

Proposed by Mehmet Sahin-Turkey

X.2. In Δ , – incentre the following relationship holds:

ℎ − 2 ≥ 1 + 8+ + + + +

+ + + ℎ + ℎ + ℎ − 3

Proposed by Bogdan Fustei – Romania

X.3. In Δ , , , – excenters, = (40) – Bevan’s point

, , – circumradii in Δ ,Δ ,Δ . Prove that:



ℎ+ℎ

+ℎ

= 2( + + )


X.4. In Δ the following relationship holds:

≥32 ⋅min ( − ) , ( − ) , ( − )



+ + ≥2

− 1

Proposed by Adil Abdullayev-Baku-Azerbaijan


+ + ≤+ +

3

Proposed by Adil Abdullayev-Baku-Azerbaijan

X.7. In Δ , , , – circumradii of Δ ,Δ ,Δ , , , – excenters, – Bevan’ s point. Prove that:

1=

2 −2 , =

4, + + =

[ ]− 2[ ]2


X.8. Solve for natural numbers:

⎩⎨

⎧ = −

1( ) =


X.9. Solve for real numbers:



+ =( ) − ( + + )

( + + )( + ) =


X.10. In Δ , – Lemoine’s point. Prove that:

+ + ≤√33 sin + sin + sin


X.11. In Δ , – Mittenpunkt, , , – excenters. Prove that:

[ ]

cos=

[ ]

cos=

[ ]

cos


X.12. Solve for natural numbers:

( + 3) = (5 − − )

( + ) = (8 + 3 )


X.13. Prove that if – is the circumcenter of external triangle of Δ (Bevan’s point) then:

+ + = 12 − − − 4

+ + = 4 ([ ] − 3{ })


X.14. In Δ , , , – circumradii of Δ ,Δ ,Δ , , , – excenters, – Bevan’s point. Prove that:

1=

2 −2 , =

4, + + =

[ ] − 2[ ]2


X.15. Prove that exact value in radical form of



sin 3∘ =14 8 − 10 − 2√5 − √3 − √15 =

14 8 − √10 √10 − √2 − √3 − √15

In radian form

sin 3∘ = sin 60 =12 1 − sin

815 = − cos

3160 ⋅ sin 60


X.16. In any triangle , show that:

(4 + )(2 − ) ≥ max{ℎ ;ℎ ;ℎ }


X.17. Find all functions :ℝ → ℝ such that:

( + ) ≥ ( ) ( ) ≥ 2018



+ + ≥94


X.19. Let be Δ and , and the internal bisectors of Δ . Find:

≤ + +32


X.20. Let be Δ , the intersection point of the symmedians and , , the intersection points of the cevians , , and with the circumcenter of Δ . Prove that:

+ + ≥( + + )

+ +




X.21. Let be Δ and the intersection point of the symmedians. Prove that:

+ + ≥( + + )+ +


X.22. If , , > 0 then in Δ the following relationship holds:

sin + sin + sin ≤( + + )

4


X.23. Prove that in any acute-angled triangle the following inequality holds:

min( cot , cot , cot ) ≤ 2 ≤ max( cot , cot , cot )


X.24. Prove that in any Δ the following inequality holds:

cos+

cos+

cos≤

332 ⋅


X.25. In Δ , – incenter the following relationship holds:

+ + ≥ + 3 2 − √3


X.26. In Δ , – incentre, , , – circumradii in Δ ,Δ ,Δ the following relationship holds:

ℎ + ℎ + ℎ ≤3√3

2


X.27. In Δ the following relationship holds: + + ≤ + +





ℎ + ℎ + ℎ +2

+2

+2

≤+ +



+ + ≥ 2 ⋅+ℎ



cot+ +

cot+ +

cot+ = 3



ℎ − 2 + ℎ − 2 + ℎ − 2 ≥4 +



2+

2+

2≤

+− +

+− +

+−



ℎ + ℎℎ +

ℎ + ℎℎ +

ℎ + ℎℎ =

2⋅

ℎ⋅

ℎ





−( − )( − ) +

−( − )( − ) +

−( − )( − ) =

2 −√



+ +≥

++

++

+



13 2 ℎ ℎ + ℎ ℎ + ℎ ℎ ≥

+ ++ +


X.37. Let Δ be the circumcevian triangle of – incenter in Δ . Prove that:

+ + ≥ +52


X.38. If ∈ ℕ; , , > 0 then in Δ the following relationship holds:

+ 3 ⋅ ∑ ( ) ≥ 4( + 1) + + ; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Constantin Basarab – Romania

X.39. If ∈ ℕ then in Δ the following relationship holds:

+13 cot 2 ≥ 6 ( + 1)

Proposed by D.M. Bătinețu-Giurgiu, Marian Voinea – Romania

X.40. In Δ the following relationshp holds: ∑( )( )

≥ 36

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




+2

(4 + ) − 2 ≤ (4 + )

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


cot 2 + cot 2 + cot 2 ≥ 8(4 + ) ≥ 8√3

Proposed by D.M. Bătinețu-Giurgiu, Alina Țigae – Romania


( )+

( )+

( )≥ √ ; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Anicuța Betiu – Romania


( + ) + + ( + ) + ( + ) + ≥ 48


X.45. If ∈ ℕ then in Δ the following relationship holds:

1( + ) +

1( + ) +

1( + ) ≥

9( + 1)4(( ) + ( ) + ( ) ) + 12


X.46. If , , , , , > 0 then:

( + ( + ) )( + ) +

( + ( + ) )( + ) +

( + ( + ) )( + ) ≥

≥32

( + + ) − 2( + + )

Proposed by D.M. Bătinețu-Giurgiu – Romania, Martin Lukarevski – Macedonia




tan + tan tan + tan

1 + tan 1 + tan⋅

1

tan + tan≥

94


X.48. Prove that:

cos419 + cos

619 − cos

919 =

√193 cos

13 cos

72√19

−16


X.49. In acute Δ the following relationship holds:

√tan + √tan + √tan − √cot − √cot − √cot ≥ 3√3 −3√3


X.50. Prove that for any acute or right triangle the following inequalities holds:

3 √3 − 1√2

≤ √sin + √sin + √sin − √cos − √cos − √cos ≤ 1



cos + cos + cos +32 ≤ 2 sin 2 + sin 2 + sin 2



6 sin + 3 sin 2 + 2 sin 3 > 0


X.53. Prove that for all , , ∈ (0,∞) and any natural number ≥ 3 , we have:



1√ + + ≥

3√( + + ) + − 1



+ + ≥ ℎ + √ ℎ + ℎ , , , ≥ 0


X.55. If , , ∈ ℂ − {0} – different in pairs, | | = | | = | | = 1, ( ), ( ), ( )

( )| |

+ ( )| |

+ ( )| |

+ 3√3 = 0 then: = =


X.56. In Δ let Ω be area of intouch triangle, Ω area of circumcevian triangle of incenter. Prove that: Ω = Ω



maxsin 2

cos( − ) ,sin 2

cos( − ) ,sin 2

cos( − ) ≥ 2√3


X.58. Let Δ be the circumcevian triangle of othocenter in acute Δ . Prove that:

⋅ + ⋅ + ⋅ ≥ 21

+1


X.59. ∈ ℝ, ≠ 1, ( + + ), ( + + ), ( + + )

( ), ( ), ( ), , , ∈ ℂ. Prove that: = = ⇒ = =


X.60. If , , ∈ ℂ − {0} – different in pairs, | | = | | = | | = 1, ( ), ( ), ( )



( )| |

+ ( )| |

+ ( )| |

= 3 then: = = .


X.61 … – regular polygon, ∈ ℕ, ≥ 3, ( ), ( ), … , ( ), ∈ ℂ, ∈ 1,

If (0) – centre of polygon and exists , ∈ 1, , ≠ such that ⋅ + ⋅ = 0 then:

is divisible with 4.


X.62. Prove that = cos , = cos , = cos are roots of equation:

4 − 4 ( + ) + ( + − 4 ) + (2 + ) − = 0


X.63. Let , , be centers of circles tangent each one at two sides of triangle and extangent to circumcircle. If – incenter then: ⋅ + ⋅ + ⋅ ≥ 36


X.64. BĂTINEȚU’S INEQUALITY – 1

If , , > 0 then in Δ the following relationship holds:

+⋅ +

+⋅ +

+⋅ ≥ 8√3 ⋅


X.65. BĂTINEȚU’S INEQUALITY – 2

If , , > 0 then in Δ the following relationship holds:

+⋅ +

+⋅ +

+⋅ ≥ 32 ⋅


X.66. Let be a ring with identity. For each ∈ we define: ≔ { ∈ : = 1}

Show that if ∈ and | | ≥ 2 then the function : → defined by

( ) = + − 1 is injective but not surjective. Proposed by Oleg Țurcan – Portugal



X.67. If , , > 0 then:

√+ + ≤ + +


X.68. If , , , > 0, + + + = 1 then:

1−

1−

1−

1− ≥

25516


X.69. If , , , are sides in a cyclic quadrilateral, , , , – exradii, – semiperimeter then:

+ + + ≥32


X.70. If , , , are sides in a cyclic quadrilateral, , , , – exradii, – semiperimeter then:

+ + + ≥2



+ + ≤ + +


X.72. If , , > 0, + + √ = 3 then:

− √ + − + − √ + 2 − 1 ≥ 0




X.73. Let be :ℝ → ℝ; ( ) = sin + √ + 4. If (lg(log 10)) = 5 then prove that (lg(lg 3)) = 3

Proposed by Simona Radu, Simona Miu– Romania


+ + ( + + ) ≥ 96 ; ( – area)

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

X.75. If , , ≥ 0 then in Δ the following relationship holds:

+ + ≥ ; ( – area)

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



11-CLASS-STANDARD

XI.1. Solve for real numbers:

2 + 3 + 4 + 192 + 648 + 1536 = 48 + 72 + 96 + 3 ⋅ 24

Proposed by Rovsen Pirguliyev-Azerbaijan

XI.2. Find:

Ω = lim→

1− 90 +1

,Ω = lim→

4 − 3 +1

( + )



Ω = lim→

5− 4 log 2 − 6 +1

(2 + 1) ,Ω = lim→

1 − 8 +1

(2 − 1)


XI.3. Find:

Ω = lim→

1− log(1 + ) , =

25

2 −


XI.4 Prove without softs: 2018√ < √

Proposed by Rovsen Pirguliyev-Azerbaijan

XI.5. Find all ROLLE functions : [0,1] → ℝ such that:

(0) = (1) =20192018

2017 ( ) + 2018 ( ) ≤ 2019,∀ ∈ (0,1)


XI.6. Let , , denote the angle of triangle . Find the maximum value of:

= cos cos + cos


XI.7. Find:

lim→

ln(1 + sinh ) − ln (1 + sinh ), ∈ ℕ∗, ≥ 2


XI.8. Find:

lim → + + ⋯+ − , where > 1, ∈ ℕ∗




XI.9. Find:

lim→

1+ 1 + + 1


XI.10. Let be , ∈ (ℝ) such that = , det = , > 0 and det( + ) = 0. Find: det( − + )


XI.11. Let be = ∫ ( ) ( ) , ∈ ℕ∗ , ≥ 2. Find such that ∈ ℚ.


XI.12. Find the continuous functions :ℝ → (0, +∞) having the property:

( ) ⋅ (2 ) ⋅ (4 ) = 2 ,∀ ∈ ℝ


XI.13. Let be = − + … and = − + … Find:

lim→

+ +3


XI.14. Let be , ∈ (ℝ) such that 5 − √5( + ) = 3 + √5( − ) and

− is invertible. Prove that ⋮ 5.


XI.15. In Δ the following relationship holds:

cos( − ) + cos( − ) + cos( − ) ≤12

++

++

+


XI.16. Let , , , > 0. Find the maximum value of:



= ( + )( + ) + ( + )( + ) + ( + )( + ) + ( + )( + )


XI.17. Find all functions :ℝ → ℝ continuous in = 0 such that:

(2018 ) = (2019 ) +


XI.18. Let be , , ∈ ℝ, ≠ 0. If det( + + ) ≥ 0,∀ ∈ (ℝ), then:

− 4 ≤ 0. Proposed by Marian Ursărescu – Romania

XI.19. Does it exist , ∈ (ℚ) such that: ( − ) + = ?


XI.20. Find the continuous functions : (0, +∞) → ℝ having the property:

=( ) + ( )

2 ,∀ , > 0


XI.21. Find: lim → − ( )

( ), ∈ ℕ∗


XI.22. Let be the sequences > 0, = 1 + − 1,∀ ∈ ℕ and > 0,

= + , ∈ ℕ, ≥ 2. Find: lim → ⋅


XI.23. Solve the equation: + (2 − 1) = 21 1 00 1 10 0 1

, where ∈ (ℝ) and

∈ ℕ, ≥ 2




XI.24. Prove that in any Δ the following inequality holds:

+ + ≥ , , , the measures in radians.


XI.25. ∈ (ℝ), det( + 3 + 3 ) = det( + 4 ) = 0. Find: Ω = det


XI.26. ∈ (ℝ), det = 1, = ( ), det( + ) = 0. Find:

Ω = ( ∗), ∗ - adjoint of , =


XI.27. ∈ (ℤ), det( + + ) = det( + ). Find: Ω = , ∈ ℕ, ≥ 2


XI.28. ∈ (ℝ), det( + 2 + 2 ) = det( + ) = 0. Find: Ω = det


XI.29 If ∈ ℕ; ≥ 2; , ∈ (0,∞); ∈ 1, then:

1+ 2 − + 3 + + 4 >

1( + 3) ⋅

Proposed by D.M. Bătinețu-Giurgiu, Delia Popescu – Romania

XI.30. If > 0; ∈ 1, ; ∈ ℕ; ≥ 2; ∈ ℝ then:

≥ (2 − ) ; =

Proposed by D.M. Bătinețu-Giurgiu, Delia Goiceanu – Romania

XI.31. If = ∑ ( ) then find: Ω = lim → 1 + −( ) ⋅

Proposed by D.M. Bătinețu-Giurgiu, Doina Cristina Călina – Romania



XI.32. If ( ) ⊂ ℝ; lim → = ∈ ℝ; (∃) > 0 such that lim → ( − ) = ∈ℝ then find: Ω = lim → (1 − + )

Proposed by D.M. Bătinețu-Giurgiu, Aurel Chiriță – Romania

XI.33. If , , ∈ 0, then in Δ the following relationship holds:

( ) + ( ) + ( ) > 8√3 ; ( – area)

Proposed by D.M. Bătinețu-Giurgiu, Aurel Chiriță – Romania

XI.34. If , , > 0; , , ∈ 0, then:

( + ) tansin + sin +

( + ) tansin + sin +

( + ) tansin + sin ≥ 2 3 ( + + )

Proposed by D.M. Bătinețu-Giurgiu, Nicolae Oprea – Romania

XI.35. If in Δ , ≤ ≤ then the following relationship holds:

( + + ) + + ≤( + )

4 ⋅

Proposed by D.M. Bătinețu-Giurgiu, Cristina Miu – Romania

XI.36. If , , , , , > 0; ( ) ; ( ) ⊂ ℝ sequences such that:

= + ; = + ; ∈ ℕ then:

+ ≤

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

XI.37. If , , > 0; ∈ 0, then:

sin+

sin+

sin+ 3√

tan> 6√


XI.38. If , , ∈ 0, then: + + > + +




XI.39. If , , ∈ (0, ) then in Δ the following relationship:

1sin +

1sin +

1sin >

√3; ( ≥ 0)


XI.40. If , ∈ ℕ; ≥ 2; , ∈ 0, ; (∀) ∈ 1, then:

+tansin >

( + 1)(∑ )∑



( + − ) ≥(2 )

⋅ ⋅


XI.42. In acute or right Δ the following relationship holds:

sin( ) +

sin( ) +

sin( ) >

9−

4( ) + ( ) + ( )


XI.43. If 0 < < then:

sin+

tan sin+

tan sin+

tan> 8



+ + + + + ≤ 4




XI.45. Let be the th prime number. Find:

Ω = lim→

+ − 23 + 5 − 8


XI.46. In Δ the following holds:

6 sin + 3 sin 2 + 2 sin 3 > 0


XI.47. If 0 ≤ , , ≤ then:

(sin + sin + sin ) ≥ 6 8


XI.48 Find: , , ⊂ ℝ, lim → = , lim → = , lim → = , , , ∈ℝ∗, , , ∈ ℕ∗


XI.49. Find:

Ω = lim→

∑ ∑ ∑∑ ( )


XI.50. Find:

Ω = lim→

tan⋅ ⋅√


XI.51. If ∈ (ℝ), det ≠ 0, ( ) = 3 ( ) then: ( ) = 3 ⋅ det ⋅ ( )




XI.52. If ∈ (ℝ), ( ∗) = 0, ∗ - adjoint of then:

(det ) + ( ) + + 1 ≥ 2 det + ⋅ det


XI.53. , > 0, = , = + , ∈ ℕ, ∈ ℕ∗, ≥ 2. Find:

Ω = lim→

( )


XI.54. If ∈ (ℝ), det = 1, det( + ) = 0 then: ( ∗) = (( )∗),

∗ - adjoint of , – trace of .


XI.55. Let , , be positive real numbers such that: = ( = 1,2, … )

Prove that: − + − + − ≥ 3( )

( = 1,2, … )

Proposed by Kunihiko Chikaya-Japan

XI.56. ( ) = ( ) . Find:

Ω = lim→

⋅ ( )− 672

Proposed by Lazaros Zachariadis-Greece

XI.57. Evaluate the limit:

lim→

tan( − 1)− log(sin( ) + 1)

√ + 1 − 1

Proposed by John Horton Conway-France

XI.58 Solve for real numbers:



⎩⎪⎪⎨

⎪⎪⎧

= tan37 − 4 sin 7

10 + 95−

= 2019

Proposed by Urfan Aliyev-Azerbaijan

XI.59 Resolve:

2018 √ + + 2019 √ + + 1 2018 √ + + √ + + 2019√ + + 2019 √ + + 2018 2018 √ + + √ + + 2019

= ( + )( + )

≥ 1; , > 0. Proposed by Jhoaw Carlos-Bolivia

XI.60 In Δ the following relationship holds:

cot 2 cot 2 + tan 2 tan 2 + cot 2 cot 2 − tan 2 tan 2 ≥1064

27


XI.61 If , , > 0 then:

+ + ≥+2 +

+2 +

+2


XI.62 In acute Δ the following relationship holds:

sin ⋅ sin ⋅ sin > ⋅ ⋅ ⋅ cos 2 ⋅ cos 2 ⋅ cos 2


XI.63 If 0 < < < < < then:

csc 2 ⋅ csc 2 ⋅ csc 2 <sinsin




XI.64 Find = − , , ∈ ℝ, + ≠ 0 such that: − 3 + 4 − 3 + =




12-CLASS-STANDARD

XII.1. If 0 < ≤ then:

1+

1+

1≥ √3 ⋅ log


XII.2. If 0 < ≤ , : [ , ] → (0,1], – continuous then:

( ) + ( )( ) ( ) + ( ) ( ) + ( ) ≤ 2( − ) ( )


XII.3. If ≥ 0 then:

( + 1) +( (1 − ) + 1)

( + 1) ≥ ( + 1)


XII.4. Find:



1cos √tan + 2

Proposed by Abdul Mukhtar-Nigeria

XII.5. Find:

√cot (tan + 2)


XII.6. Find: tan

√tan (tan + 1)


XII.7. Find:

2 + 44 sin(2 )− 2 cos(2 ) + 3 cos( )− 3 cos(3 )8 + 9 sin( )− sin(3 )

Proposed by Nader Al Homsi-Amman-Jordan

XII.8. Find:

√1 + ln (1 + ln ) + ln√1 + ln (ln ) + ln + 1


XII.9. Find: sec (1 + sin )

+ sin − cos − 1


XII.10. Find: ∫ (csc sec ⋅ cos tan sin cot )


XII.11. Find :ℝ → ℝ, which admits primitives with the property

( − ) ⋅ ( − ) = − ,∀ ∈ ℝ, where > 0 and is a primitive of .




XII.12. If , > 0; < ; + = ∈ 0, then find:

( − + ) log(1 + tan ⋅ tan )


XII.13. Find:

Ω = lim→

1log 1 +


XII.14. Find:

Ω = lim→

⎝

⎛ arctan1

⎠

⎞


XII.15. Find:

Ω = lim→

(3 + 1)


XII.16. Find:

Ω = lim→

( + 1)


XII.17. Find:



Ω = lim→

sin( )


XII.18. Find:

Ω = lim→

⎝

⎛ ⋅ tan1

⎠

⎞


XII.19. Find:

Ω = lim→

( + 1)


XII.20. Find:

Ω = lim→

1 ⋅ ⋅( + )( + )( + )


XII.21. ∗ = + + 0.5− [ + + 0.5], ∘ = + + 0.2 − [ + + 0.2]

Prove that: ([0,1),∗) ≅ ([0,1),∘) as abelian groups. [∗] - great integer function.


XII.22. Find:

Ω( , ) = tan( ) tan( ) tan ( + ) , , > 0,0 < < 2( + )


XII.23. Λ = { ⁄ ∈ ℝ, :ℝ → ℝ, ( ) = + }, ∗ =



Θ = { ⁄ ∈ ℝ∗, :ℝ → ℝ, ( ) = }, ∘ =

Prove that: (Λ,∗) ≇ (Θ,∘) as abelian groups.


XII.24. ∗ = 2 , ∘ = 3

Find an isomorphism between abelian groups (0,∞),∗ and (0,∞),∘


XII.25 Prove that if 0 < < then:

√1 +> ( − ) + ln


XII.26. Find:

Ω =

⎝

⎛ cos 7 + 7 cos 5 + 21 cos 3 + 35 cos(4 + sin )(cos 6 + 6 cos 4 + 15 cos 2 + 10)

⎠

⎞


XII.27. Find:

Ω = lim→

sin


XII.28. If the equation:

+ + + 2 + + + 1 = 0, , , ∈ ℝ

has all roots real numbers then: | − | ≥ 8.




XII.29. The equation:

+ + + + + + + − 1 = 0, , , , , ∈ ℝ

has all roots real numbers. Prove that: | + − | ≥ 16


XII.30. Solve the following system:

⎩⎪⎪⎨

⎪⎪⎧ + + + =

5612

( ) + ( ) + ( ) + ( ) + ( ) + ( ) =89

12

( ) + ( ) + ( ) + ( ) =89

12

( ) =1

Proposed by Jhoaw Carlos-Bolivia

XII.30. Prove without softs:

⋅ 1 + <

Proposed by Kunihiko Chikaya-Japan

XII.31. If 0 < ≤ < then:

tancos ≥ log

1 + tan1 + tan


XII.32. ℤ = { ⁄ ∈ ℤ}, ∈ ℤ. Prove that: (2ℤ ∩ 5ℤ ∩ 11ℤ, +) ≅ (3ℤ∩ 7ℤ ∩ 13ℤ, +) as abelian groups.


XII.33. Find:



Ω =√1 −

89 + √89 + +

Proposed by D.M. Bătinețu – Giurgiu; Dan Grigore – Romania

XII.34. Find:

Ω =2 tan − tan

( + tan )

Proposed by D.M. Bătinețu – Giurgiu; Claudia Preda – Romania

XII.35. If ∈ 0, ; > 1; :ℝ → ℝ continuous and even function then find:

Ω = ( ) + 1

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

XII.36. If > 0; , :ℝ → (0,∞),ℎ:ℝ → ℝ are continuous functions; , even and ℎ even function then prove that:

( )( ) + ( ) + ℎ ( ) + ℎ( )

=( )( )

Proposed by D.M. Bătinețu – Giurgiu; Clarisa Cavachi – Romania



UNDERGRADUATE PROBLEMS



U.1. Prove this sharp inequality:

2 +23 1 −

− 1 + 2 cos√

< 0

Proposed by K. Srinivasa Raghava – AIRMC – India

U.2. If

1+ ( ) + ( + ) + =

cosh( )cos( )

(−1)+ ( ) + ( + ) +

then show that = √2 + 3


U.3. Evaluate in closed – form:

sin ( )√


U.4. Prove that: 4

+1!− 1

3!+ 2

5!− 3

7!+ 4

9!− 5

11!+ ⋯ =

64(63− 7 )


U.5. Prove the following relationship:

⎝

⎛ log(tan( ))sin ( ) + cos ( )

⎠

⎞ =√3

− −√3

+23−

log(3)6 − log(2)

2√3

+log(3)√3

where ( ) is Poly – Logarithm


U.6. If, for any complex number , ( ) > 0



( ) =ln( )

+

then prove that: 3 ( ) + ( ) = 0 at the point = 2


U.7. Prove the relationship:

ln( ) sech( ) tanh( ) =13

1−4

+ ln( ) −11 ln(2)

60+

7 (3)+ 10 (−3)− 4 ln( )

( ) – is derivative of Zeta function, – Glaisher constant, – Euler’s Constant


U.8. Find:

Ω = lim→

⎝

⎛ (1 + 2 csc )(1 + 2 cot )

⎠

⎞

Proposed by Mohamed Arahman Jama-Somalia

U.9. Find:

logΓ( )

Γ(1 − ) logΓ( )

Γ(1 + )

Proposed by Obidah Al Sharafy-Yemen

U.10. Find:

Ω = lim→

⋅ log( + 1) ⋅ tan ( )

√1−


U.11. Find:



Ω = lim→

1+ 1 + log(1 + )


U.12. Find:

Ω = lim→

log ⋅ log(1 − ) ⋅ (1 + log(1 − ))


U.13.

⋅ log Γ( ) ⋅ log Γ(1 − ) = +log(2 )

+ +log(2 )

+(2)( + log(2 ))

⋅ +(2)⋅

Find: Ω = + + + + +


U.14. Find:

Ω =cos ⋅ cos(2 )

sin , ∈ 0, 4


U.15. Find:

Ω = lim→

⎝

⎛ loglog −log +

⎠

⎞


U.16. Prove the following closed form:

4‼‼!!

+(−1)

(4 + 3)! =1

2 √√√

1√2

− 4 + sin1√2

+ 4




U.17. Prove that:

− 1( )

+ −1

( )

= (2)

where (. ) is the Euler totient function of ( ) is the number ∈ ℕ such that 1 ≤ < and g.c.d. ( , ) = 1.


U.18. Prove that:

12 +

(−1)+ 1 −

(−1) (−1)+ 1 = ln 2


U.19. Show that:

+ 1−√

= 3 , where ⌊⋅⌋ is the floor function.


U.20. Prove the following:

1 + + + +( − 1)! ! =

75203


U.21. Let ∑ ( ≥ 1) be a convergent series of real positive numbers. Prove that:

⋅ ⋅sin( )

⋅ 10 ⋅ ( − 1)! < +∞

Proposed by Artan Ajredini-Serbie

U.22. Prove that:



2 − 2 − (−1) ( + )( + )! =

11

Proposed by Srinivasa Raghava-AIRMC-India

U.23. Find:

Ω = lim→

log− √3 + 1


U.24. Let for any > 0

( ) = sin (1 + )

Prove the relationship:

1 − ( ) = cot√2

( )

Proposed by Srinivasa Raghava-AIRMC-India

U.25. ( ) = + log + ∑⋅( )!

. Prove that:

cos+ ( ) ≥ ( ) − ( ), 0 < ≤


U.26. Prove:

lim→

1 + ( ) − 3 1 + ( ) ( ) + ( )

Γ( ) = 3 − 6 + 3 − 2


U.27.



=(2 − 1)‼(2 + 2)‼

Find:

Ω = lim→

( + ) − ( )


U.28. Evaluate:

2log(tan )

where, ( ) are Euler Polynomials

Proposed by Arafat Rahman Akib-Bangladesh

U.29. Find the limit:

lim→

1 +1

10 ⋅ !


U.30. Find:

lim→

arccos( )1

tan ln1

cos


U.31.Find:

lim→

arcsec( ) ln(1 − )2 − 2 + 1




U.32. Find:

−1

1 + +1 − coth( )


U.33. Find:

ln√1 + cos

cos


U.34. Prove that:

ln ( )( )

cos( )

(1 + sin( )) 1− sin( )= −8

Proposed by Nader Al Homsi- Jordan

U.35.

…2

2 − 1 … = 1

Find the value of such that above equality holds true, where ⌊⋅⌋ denotes the floor function.


U.36. Prove:

Ω =( ) ⋅ sin( )

= − { + log(2)}

Proposed by Obidah Al Sharafy- Yemen

U.37. Prove:

Ω = { + (1 + )} =32



where ( ) DiGamma function


U.38. Prove:

Ω =( ) Γ( )

Γ( ) −( ) ( )

( )

−1 + Γ( ) −1 + ( )( )

=

=1

√ ( − 1)+

12√ ( − 1)

Where ( )DiGamma function, ( ) Polylogarithmic, =

√

, =

√


U.39. Prove:

Ω =cos( ( ) + )

{ + ( ( ) + )} =(1 + )2

where ( ) polyGamma function


U.40. Find:

Ω =log( + 1)+ + 1


U.41. Find:

Ω =⋅ log( + 1)

+ + 1




U.42. Find:

Ω( ) = ( + + 1)(1 + ) , > 0


U.43. Find:

Ω( ) = ( − + 1)(1 + ) , > 0


U.44. Find:

Ω = lim→

1− 1

arcsin

√


U.45. Find: Ω( ) = ∫ ( )( ), > 0


U.46. If , ∈ ℝ then:

8 (cos cos cos( + )) + ( − ) ≥ 0


U.47. Calculate:

arctan( )

√1 +−

12

arctan( )

√1 +−

1√1 −

arctanh1 −1 +

Proposed by Cornel Ioan Vălean – Romania

U.48. Ω( , , ) = ∫⋅( ) ⋅

( ⋯ ), , , ∈ ℕ∗. Find:



Ω(17,02,2008)

Proposed by Feti Sinani-Kosovo

U.49. Find:

Ω = lim→

log(1 + 2 )

√

Proposed by Max Wong-Hong Kong

U.50. Find:

sin(ln ) ( )

Proposed by Nawar Alasadi-Iraq

U.51. If 0 < ≤ then:

( + )( + )( + )

≥( + )( − )

2 3 cos 7 − cos37


U.52. If 0 < ≤ < then:

8sin ⋅ sin ⋅ sin( + )

( − − ) ≤ (3 − 3 )√3


U.53. If 0 < ≤ < 2 then:

16( − )

sin sin sin

sin sin sin sin+ cot

+ +2

≤ log2 sin






ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-autumn 2019

PROBLEMS FOR JUNIORS

JP.196. Let , , be the sides in a triangle such that = 1. Find the minimum value of:

=√ + − 1

+√ + − 1

+√ + − 1

+3( + + )

+ +

When does it occurs?

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam

JP.197. Solve for real numbers:

6 2 − 2 + 1 + 4 3 − 2 = 2 − 5 + 13


JP.198. Prove that in any Δ the following inequality holds:

min( , , ) ≤ 4 ( + ) ≤ max( , , )


JP.199. Let be a pyramid with the base parallelogram and any point which bolongs to the side such that: = . Through the vertex and the point we consider a variable plane which intersects the segment in and the segment in . Prove that:

≥2+ 1


JP.200. Let be :ℝ → ℝ such that: ( ) + ( ) ≥ 2 ; (∀) , ∈ ℝ. Prove that:

( ) + ( ) + ⋯+ ( ) ≥+ + ⋯+

; (∀) ≥ 2;



(∀) , , … , ∈ ℝ.


JP.201. If , , > 0 then:

( + )+ + 2 +

( + )+ + 2 +

( + )+ + 2 ≥ 2 3 ( + + )


JP. 202. Let , , be positive real numbers such that + + + 2 = 1.

Prove that:

√2 + 16 + 7+√2 + 16 + 7

+√2 + 16 + 7

≥3

20


JP.203. If , , > 0; ⋅ ⋅ = 1 then:

− √ + − √ + − √ ≥ 0


JP.204. In Δ the following relationship holds:

cos cos

tan+

cos cos

tan+

cos cos

tan>


JP.205. Let , , be positive real numbers. Prove that:

++

23

++

23

++

23 ≥

83

Proposed by George Apostolopoulos – Messolonghi – Greece

JP.206. Let be a triangle with inradius and circumradius . Let ℎ ,ℎ , ℎ the altitudes to sides , , respectively and let , , the exradii to , , respectively. Prove that:



4≤

ℎ⋅ +

ℎ⋅ +

ℎ⋅ ≤ 2


JP.207. Let , , be the lengths of the sides of a triangle with inradius and circumradius , and let , , the exradii to , , respectively. Prove that:

6 ≤ + + + + + ≤2 −


JP.208. Prove that in any triangle the following inequality holds:

tan + tan≤


JP.209. If , , , ∈ ℝ then:

+ + | − | ≤ 2( + )( + )


JP.210. Let , , be positive real numbers such that + + = 3. Prove that:

+ +9 ≤

1+ ( + ) +

1+ ( + ) +

1+ ( + ) ≤

13


PROBLEMS FOR SENIORS

SP.196. Find:

lim→

1, ∈ ℕ∗, ≥ 2




SP.197. If , , ≥ 0 then:

7√

5√ + 3√+

7√

5√ + 3√+

7√

5√ + 3√≤

√7√5 + √3

+√7

√5 + √3+

√7√5 + √3


SP.198. If , , , ∈ ℝ; + = + = 10 then:

(10 − − 3 )(10 − − )(10 − − 3 ) < 10125


SP.199. If > 1, ∈ ℕ then:

1log 2

2 − 1<

1 ⋅ 3 ⋅ 7 ⋅… ⋅ (2 − 1)(2 )!


SP.200. If , , , ∈ ℝ then:

2| − |( + ) + ( + ) ≤ ( − ) + ( + )( + )√2


SP.201. Find:

Ω = lim→

tan1

2( + 1) tan2 + 4 + 1

2( + 1)


SP.202. Prove that in any triangle , the following relationship holds:

+ + ≥ 3 +−+ +

−+ +

−+

Proposed by Nguyen Viet Hung – Hanoi – Vietnam

SP.203. Let , , be positive real numbers such that ( + )( + )( + ) = 8. Prove that:



1+ + +

1+ + ≥

23


SP.204. Let , , be positive real numbers such that + + = 3. Prove that:

+ 2 + + 2 + + 2 ≥ 1


SP.205. In Δ , , , are length’s of Nagel’s cevians. Prove that:

≥ , , , – exradii of triangle.


SP.206. Prove that in any triangle the following inequality holds:

−2 + 17 ≤ tan 2 ≤6

( − 5 )



9(8 − 23 ) ≤ cot 2 ≤8132

(13 − 88 )



36 ≤ sec 2 ≤ 9

Proposed by Marin Chiricu – Romania


27 ≤ csc 2 ≤4

(4 − 37 )

Proposed by Marin Chiricu – Romania

SP.210. Let be an acute-angled triangle. If + + = and



cos + cos + cos = ; ( , , – the measures in radians), then Δ is equilateral.


UNDERGRADUATE PROBLEMS

UP.196. Let be , > 0, ≠ such that lim → = lim → = , ∈ ℕ∗. Find:

lim→

−−


UP.197. Let be :ℝ → (0,∞) continuous such that for , , > 0, fixed values

( ) + ( ) + ( ) = ( ) ( ) ( ), (∀) , , ∈ ℝ

Prove that:

( ) ≥( − )( + + )√ + +

3 ; (∀)0 < ≤


UP.198. Let be a positive integer. Evaluate:

lim→

1 − (cos ) cos( )


UP.199. Given the triangle . The internal angle bisectors from , , meet sides , , at , , respectively. Prove that:

tan2

+ tan2

+ tan2

+cos ,

cos+

cos ,

cos+

cos ,

cos= 0


UP.200. If 0 < ≤ then:



( + + + )+ + √ + √

≤( + ) ( − )

4


UP.201. Calculate the integral:

arctan− + 1

It is required to express the integral value with the usual mathematical constants and Ψ , where Ψ ( ) is the trigamma function.


UP.202. Prove that:

Ψ5

12 =32− 6√3

3 + 40 − 10Ψ13

Ψ1112 =

32 + 6√33 − 40 − 10Ψ

13

where Ψ ( ) is the trigamma function and is the Catalan’s constant.


UP.203. Given a triangle with incenter . The lines , , meet the sides , , at , , and meet the circumcircle at the second points , , respectively. Prove that:

(a) + + = 2,

(b) + + = − 1


UP.204. Let ( ) be a positive real sequence such that lim → = ∈ ℝ∗ , where

is a positive integer. Compute:

lim→

1[ ⋅ ]



where ∈ ℝ; we denote by [ ] the integer part of .


UP.205. Compute:

lim→

lim→

Γ( + 2) ( ) − Γ( + 1)

where ( ) , = 0, = 1, = + ,∀ ∈ ℕ is the Fibonacci sequence.


UP.206. Compute:

lim→

lim→

Γ( + 2) ( ) − Γ( + 1) ,

where ( ) , = 0, = 1, = + ,∀ ∈ ℕ is the Fibonnaci sequence.


UP.207. Let be ∈ (ℝ) such that det = −1. Prove that:

( + + 1) ≥ 3( ⋅ − 1)


UP.208. Let be an acute – angled triangle and , , the points in which the heights of the triangle intersect the circumcircle of Δ . Prove that:

≤2


UP.209. Demonstrate the followig inequality:

+ + + + ⋯+ + ≤ + 1

where , , … , are strictly positive real numbers which satisfy the relationship

+ + ⋯+ =




UP.210. Prove that for any acute triangle the following inequality holds:

cot + cot + cot + √3 ≥ 2 tan 2 + tan 2 + tan 2






INDEX OF AUTHORS RMM-24

Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele 1 DANIEL SITARU-ROMANIA 42 SRINIVASA RAGHAVA-INDIA 2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 43 NGUYEN VAN NHO-VIETNAM 3 CLAUDIA NĂNUȚI-ROMANIA 44 HUNG NGUYEN VIET-VIETNAM 4 NECULAI STANCIU-ROMANIA 45 MOHAMED ARAHMAN JAMA-SOMALIA 5 MARIAN URSĂRESCU-ROMANIA 46 ARAFAT RAHMAN AKIB-BANGLADESH 6 BOGDAN FUSTEI-ROMANIA 47 OBIDAH AL SHARAFY-YEMEN 7 DAN NĂNUȚI-ROMANIA 48 GEORGE APOSTOLOPOULOS-GREECE 8 MARIN CHIRCIU-ROMANIA 49 HOANG LE NHAT TUNG-VIETNAM 9 TITU ZVONARU-ROMANIA 50 MEHMET SAHIN-TURKEY

10 TUTESCU LUCIAN-ROMANIA 51 ROVSEN PIRGULIEV-AZERBAIJAN 11 PETRE STĂNGESCU-ROMANIA 52 NADER AL HOMSI-JORDAN 12 VASILE MIRCEA POPA-ROMANIA 53 ADIL ABDULLAYEV-AZERBAIJAN 13 LUIZA CREMENEANU-ROMANIA 54 IOAN CORNEL VALEAN-ROMANIA 14 IULIANA TRAȘCĂ-ROMANIA 55 NAWAR ALASADI-IRAQ 15 CAMELIA DANĂ-ROMANIA 56 MARTIN LUKAREVSKI-MACEDONIA 16 LAVINIA TRINU-ROMANIA 57 MAX WONG-HONG KONG 17 ALINA ȚIGAE-ROMANIA 58 KUNIHIKO CHIKAYA-JAPAN 18 SIMONA RADU-ROMANIA 59 SEYRAN IBRAHIMOV-AZERBAIJAN 19 SIMONA MIU-ROMANIA 60 JHOAW CARLOS-BOLIVIA 20 BETIU ANICUTA-ROMANIA 61 NAREN BHANDARI-NEPAL 21 CĂTĂLINA PANĂ-ROMANIA 62 URFAN ALIYEV-AZERBAIJAN 22 GABRIEL TICĂ-ROMANIA 63 FETI SINANI-KOSOVO 23 VIRGINIA GRIGORESCU-ROMANIA 64 ABDUL MUKHTAR-NIGERIA 24 CLAUDIU CIULCU-ROMANIA 65 ARTAN AJREDINI-SERBIE 25 MIHAELA STĂNCELE-ROMANIA 66 CLARISA CAVACHI-ROMANIA 26 RAMONA NĂLBARU-ROMANIA 67 CLAUDIA PREDA-ROMANIA 27 OANA PREDA-ROMANIA 68 DAN GRIGORE-ROMANIA 28 CONSTANTINA PRUNARU-ROMANIA 69 OLEG ȚURCAN-PORTUGAL 29 TATIANA CRISTEA-ROMANIA 70 CRISTIAN MOANȚĂ-ROMANIA 30 DOINA CRISTINA CĂLINA-ROMANIA 71 ALEXANDRINA NĂSTASE-ROMANIA 31 SANDA IULIA-ROMANIA 72 ILEANA DUMA-ROMANIA 32 ROXANA VASILE-ROMANIA 73 CONSTANTIN BASARAB-ROMANIA 33 VASILE BURUIANĂ-ROMANIA 74 MARIAN VOINEA-ROMANIA 34 OCTAVIAN STROE-ROMANIA 75 ALECU ORLANDO-ROMANIA 35 DANA COTFASĂ-ROMANIA 76 NGUYEN VAN CANH-VIETNAM 36 SORIN PĂRLEA-ROMANIA 77 MUSTAFA TAREK-EGYPT 37 ȘTEFAN MARICA-ROMANIA 78 THANASIS GAKOPOULOS-GREECE 38 CARINA MARIA VIESPESCU-ROMANIA 79 PEDRO PANTOJA-BRAZIL 39 DELIA POPESCU-ROMANIA 80 AUREL CHIRIȚĂ-ROMANIA 40 DELIA GOICEANU-ROMANIA 81 NICOLAE OPREA-ROMANIA 41 CRISTINA MIU-ROMANIA 82 AMELIA CURCĂ NĂSTĂSELU-ROMANIA

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r. m. m. - 24 · 2019. 11. 29. · draga tĂtucu mariana-romania claudia nĂnuȚi-romania dan...

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