Ian2012_Articol Zvonaru-Stanciu, Pt. Ian.2012 (Mateinfo.ro)

7/29/2019 Ian2012_Articol Zvonaru-Stanciu, Pt. Ian.2012 (Mateinfo.ro)

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Revista Electronic Mateinfo.ro ISSN 2065 6432 Ianuarie 2012 www.mateinfo.ro

Soluii pentru patru probleme (J 208, J 209, S207, S209)din Mathematical Reflections nr. 5 / 2011

altele dect cele din Mathematical Reflections nr. 6 / 2011

Titu Zvonaru1i Neculai Stanciu2

Solution by:Titu Zvonaru, Comneti, Romania and Neculai Stanciu,

George Emil PaladeSecondary School, Buzu, Romania.

We well-known that

222

2

cba

bcmAK a

++= (by Stewart theorem we obtain that aa m

cb

bcs

+=

22

2and the by Van

Aubell theorem or Menelaus theorem we deduce that222

2

cba

bcmAK a

++= ) , so we have to

prove that

( ) ( )222

22

4

932 a

RbcmaRbcm aa , and because

=22

43 ama , and applying C-B-S we obtain that

( )( ) ( )( ) ( )( ) = 2222222

4

3acbmcbmbc aa , so is enough(sufficcient) to prove that

( )( ) ( ) 222222

2222 93

4

9

4

3aRcba

Racb .

But we known that 22 9Ra , and we have done if we prove that 42222223 acbaacb , which is true.

We have equality if and only if ABC is equilateral.

1Comneti2c. gen. George Emil Palade , Buzu
http://www.mateinfo.ro/http://www.mateinfo.ro/http://www.mateinfo.ro/


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Solution by Titu Zvonaru, Comneti, Romania and Neculai Stanciu, George EmilPalade Secondary school, Buzu, Romania.

By Cauchy-Buniacovski Schwarz or Bergstrm s inequality (or Lema2T , or Titu

Andreescus inequality) we have

[ ])(2

)()()(

)(

)(

)(

)(

)(

)()()()(2333666555

cabcab

baaccb

bac

ba

acb

ac

cba

cb

c

ba

b

ac

a

cb

++

+++++

+

++

+

++

+

+=

++

++

+

so it is sufficient to prove that[ ]

( )cabcabcabcab

baaccb++

++

+++++

9

32

)(2

)()()(2333

( )cabcabbaaccb +++++++3

8)()()( 333

))((8)()()(3 333 cbacabcabbaaccb +++++++++

We use =cyclic

and we have successively that

( ) ( ) ++++ abcabbaabbaa 383323 22223 abcabbaa 246 223 ++ ,

which is true by AM-GM inequality ( abca 33 , ,32 abcba abcab 32 ).

We have equality if and only if3

1=== cba .


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Solution by: Titu Zvonaru, Comneti, Romania and Neculai Stanciu,George Emil Palade Secondary School, Buzu , Romania.

We denote by zabycaxbc === 1,1,1 and then we have 0=++ zyx .

We use the identities

( )( )zxyzxyzyxzyxxyzzyx ++++=++ 222333 3 and

( )( ) ++=++++223 xyyxxyzzxyzxyzyx

and we deduce that

xyzzyx 3333 =++ and =+ xyzxyyx 322 .We have

xyz

yxxy

z

xy

y

zx

x

yz

bc

cba

=

+

+

=

22

1

)(

and because ( )( )( ) ( )( ) =+= yxxyxzyzxzyxyxzzyyx 222 we obtain that

=

xyz

xzzyyx

bc

cba ))()((

1

)(.

Then we calculate

xy

z

zx

y

yz

x

cba

bc

+

+

=

)(

1, where we taking account that

+++=++32232))(())(())(( yxyzzyzxxyzxyxzxyzzxyyzyxyzxx

xyzxyzxyyxxyzxyzzxzxy 932232322 =++=+++ , and we obtain that

))()((

9

)(

1

xyzxyz

xyz

xy

z

zx

y

yz

x

cba

bc

=

+

+

=

, and in finnaly we have

239

))()((

9))()((

)(

1

1

)(==

=

xyzxyz

xyz

xyz

xzzyyx

cba

bc

bc

cba.


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Solution by Titu Zvonaru, Comneti, Romania and Neculai Stanciu,George Emil Palade Secondary School, Buzu, Romania.

We have )()())(( 22 sabscababbassababbsas +=++= and by

Rrrscabcab 422 ++=++

we obtain that

=+=

)(1))(( 222 absbaabcsabcc

bsas

( )[ ])4()(2431 222222 RrrsscbaabcRrrsabcsabc

+++++++=

( )222432222244 4882161 RrssrsRrRrsrsrRrsabcsabc

++++++=

( )23224222 481641 RrsRrrRrrsRrsabc

+++++=

( ) ( )[ ]22

2222

2

4816 rRsabc

rrRrRsabc

r

++=+++=

and the inequality to prove becomes

( )[ ] ( )222222 44

2444

44

21 rRs

rR

rRssrRs

Rs

r

rR

rR

R

sr++

+

+++

+

+

( )3222 4)516()4(4

243 rRrRsrR

rR

rRs ++

+

+ .

Because222 344 rRrRs ++ (see 5.8. from Geometric Inequalities, O. Bottema, Groningen, 1969) is

sufficient to prove that

( )( ) ( )322 4515344 rRrRrRrR +++ 3223322223 124864152020486464 rRrrRRrRrrRRrrRR +++++

( ) 024016164 2322 + rRrrRrrR , which is true.We have equality if and only if the triangle is equilateral.

Ian2012_Articol Zvonaru-Stanciu, Pt. Ian.2012 (Mateinfo.ro)

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