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1 GAZETA MATEMATIC SERIA A REVIST DE CULTUR MATEMATIC ANUL XXVICV) Nr. / 008 O nou demonstra ie a inegalit ii lui Surányi de Marian Tetiva Abstract In this note we present a new proof of Surányi's inequality. Key words: Surányi's inequality, mathematical induction. M.S.C.: 6D5 Dorim s prezent m în aceast not o variant de demonstrare a inegalit ii lui Surányi [], [], [4]). De i mai complicat, ne-am propus s facem aceast demonstra ie pentru c socotim c are o anume elegan i, în plus, metoda folosit atac inegalit ile pe care le demonstreaz adic pe acelea omogene, inegalitatea lui Surányi fiind de acest tip) în chiar esen a lor, ar tând cum ele deriv din identit i c ci i identit i se ob in pe aceast cale, identit i la care altfel cu greu am ajunge). Este vorba de urm toarea Propozi ie. Pentru orice numere pozitive a,a,...,a n are loc inegalitatea lui Surányi) n )a n + a n + + a n n)+na a...a n a + a + + a n )a n + a n + + a n n ). Demonstra ia o facem tot prin induc ie ca în []), dar noi vom presupune inegalitatea demonstrat pentru orice num r p {,,..., n} de variabile i o s ar t m c are loc i pentru n +variabile verificarea pentru n =nu constituie o problem, ba chiar inegalitatea se poate demonstra u or i pentru n =, când devine egalitate, sau pentru n =3, inegalitate ce se nume te a lui Schur, cum se arat i în []). Cu alte cuvinte, avem de demonstrat inegalitatea na n+ + a n+ + + a n+ n )+n +)a a...a n a n+ a + a + + a n + a n+ )a n + a n + + a n n + a n n+), pentru a,a,...,a n,a n+ numere nenegative. Metoda pe care o vom utiliza este numit metoda norm rii în lucrarea [3]; ce înseamn de fapt aceasta? Datorit simetriei, putem presupune, f r a restrânge generalitatea, c a n+ este cel mai mic dintre numerele a,a,...,a n,a n+ ; mai mult, putem presupune c el este strict pozitiv, 93

2 deoarece inegalitatea de mai sus cu a n+ =0deriv f r probleme din inegalitatea lui Cebî ev convinge i-v de asta! Atunci rapoartele a a,,..., a n+ a n+ a n a n+ sunt mai mari sau egale cu i prin urmare pot fi exprimate în forma a a n+ =+x, a a n+ =+x,..., a n a n+ =+x n, unde x,x,...,x n suntnumere nenegative. Inegalitatea de demonstrat se dovede te afiechivalent dup împ r irea cu a n+ n+ i desfacerea parantezelor de tip +x k) n+ i + x k ) n )cu ) n n n n n n ++Cn+ x k + Cn+ x k + + Cn n+ x n k + + k= k= k= k= x n+ k +n +) + x + x x + + ) x x...x n + x x...x n n ++ n x k )n ++Cn k= n k= x k + + C n n n k= x n k + n x n k k= a doua parantez din membrul stâng con ine toate sumele simetrice formate cu x,x,...,x n ), inegalitate despre care trebuie s dovedim c este adev rat pentru orice x,x,...,x n 0 folosind presupunerea de induc ie, conform c reia enun ul propozi iei este valabil pentru orice num r k n de variabile). Aceast inegalitate,,se sparge în inegalit ile corespunz toare pentru componentele omogene de acela i grad) din fiecare membru. Anume, este u or de v zut c termenii liberi de grad 0 din ambii membri deci cei care nu con in pe x,x,...,x n )sunt egali cu n +). La fel, se stabile te f r dificultate c termenii de gradul întâi sunt, în cei doi membri ai inegalit ii, n n +) x k. k= Pentru p n, termenii de grad p sunt, în membrul stâng iar în dreapta nc p n+ k= n +)C p n deoarece un calcul simplu) n x p k +n +) x x...x p, n k= x p k + Cp n n k= x k n k= nc p n+ n +)Cp n =p )Cp n+, x p k ; ) 94

3 inegalitatea de demonstrat pentru componentele de grad p, p n, este p )C p n+ n k= x p k +n +) x x...x p C p n n k= x k n k= x p k. Ei bine, aceast inegalitate este adev rat i se ob ine folosind i) inegalitatea lui Surányi pentru p variabile a a cum presupunerea de induc ie ne permite), cum se va vedea în continuare. Pentru început s observ m c avem C p n+ = n + Cn p, p astfel c inegalitatea de demonstrat se poate scrie i în forma n + n p )Cn p x p k p + p ) x x...x p Cn p De asemenea, avem C p n k= = C p n + Cp n i n C p n k= x p k = x p + xp + + xp p), n k= x k n k= x p k. ) suma din partea dreapt fiind extins asupra tuturor submul imilor cu p elemente ale mul imii celor n variabile x,x,...,x n ; mai precis, aceasta este suma x p i + x p i + + x p i p ), i <i < <i p n pentru care prefer m, pentru comoditate, nota ia mai simpl de mai sus ceea ce vom face mai departe cu fiecare sum de acest tip). Rezult c în membrul stâng al inegalit ii avem de fapt n + p p )C p n n k= x p k + n + p p )x p + xp + + xp p )+px x...x p ), expresie care se minoreaz pe baza inegalit ii lui Surányi aplicat fiec rui grup de p dintre cele n variabile) cu n + p p )C p n n k= x p k + n + p x + x + + x p )x p + x p + + x p p ) sum format, cum spuneam, din cei Cn p termeni asem n tori cu cel scris, câte unul pentru fiecare grupare de p dintre cele n variabile). Acum, un calcul pe care-l l s m cititorului de fapt, mai curând, un ra ionament combinatoric) justific egalitatea x + x + + x p )x p + x p + + x p p )= 95

4 mai folosim inegalitatea = C p n k= n n x p k + Cp n x p k k= n n n x k k= k= n n x k k= k= x p k x p k ; care rezult u or, chiar pentru orice num r real p, cu inegalitatea lui Cebî ev, v. i mai sus, v. i []) i minor m, în cele din urm, membrul stâng al inegalit ii ) cu n + p p )C p n + ) Cp n n + C p n n k= x k n k= x p k. Ce ne-a mai r mas de f cut? Evident, s ar t m c aceast din urm expresie este mai mare sau egal cu membrul drept al inegalit ii de demonstrat ); iar, deoarece fiecare x k este nenegativ, aceasta revine la inegalitatea corespunz toare între coeficien ii produselor n + p n k= x k n k= x p k, deci ne r mâne de ar tat c p )C p n + ) Cp n + C p n n C p n pentru n i p numere naturale, n p. Pentru p = n va ap rea aici num rul Cn n, dar aceasta nu reprezint o problem : conven ional, dar i conform calculelor f cute, acesta se înlocuie te cu 0.) Din identitatea combinatoric de mai sus luând pe x,x,...,x n egale, toate, cu ), sau prin calcul direct, ob inem nc p n + n C p n = p C p n; de aceea avem p )C p n n + ) Cp n + C p n n Atunci n n + p =p ) + pn p +))C p n p )C p n + ) Cp n + C p n = n =p )nc p n + p C p n = =pn p +)C p n. n +)pn p +) Cn p = p deci = pn + n p + Cn p = n C p p n + p n + n p + p C p n p )C p n + ) Cp n + C p n = Cn p + Cp n n n = n C p n + C p n, >Cp n 96

5 i ne-am atins scopul! De fapt probabil cititorul a remarcat deja), diferen a între membrul stâng i cel drept al inegalit ii ), unde am pus x = x = = x n =,este p )nc p n+ +n +)Cp n n C p n = C p n, acesta fiind un calcul mult mai simplu; deoarece inegalit ile folosite devin egalit i pentru x = x = = x n =de fapt, oricând variabilele sunt egale), înseamn c nu am modificat, prin transform rile f cute, aceast diferen, deci rezultatul calculelor de mai sus era previzibil i se justific i pe aceast cale putem spune c am f cut doar o verificare). În sfâr it, pentru termenii de grad n +, avem de demonstrat inegalitatea n n k= x n+ k n n x k k= k= despre care am mai spus c se ob ine imediat cu inegalitatea lui Cebî ev i am încheiat ce ne-am propus. Aceasta a fost demonstra ia; cititorul interesat va putea, f r îndoial, s deduc i singur alte identit i i inegalit i din care d m doar un exemplu oarecum spectaculos, oarecum plicticos). Exerci iu. Ar ta i c, pentru orice numere a, b, c, d are loc identitatea 3a 4 + b 4 + c 4 + d 4 )+4abcd a + b + c + d)a 3 + b 3 + c 3 + d 3 )= x n k, = d 3a d) +3b d) +3c d) a d)b d) a d)c d) b d)c d))+ +d5a d) 3 +5b d) 3 +5c d) 3 +4a d)b d)c d) 3a d)b d)a + b d) 3a d)c d)a + c d) 3b d)c d)b + c d))+ +a b) a d) +a d)b d)+b d) )+ +a c) a d) +a d)c d)+c d) )+ +b c) b d) +b d)c d)+c d) ). Bibliografie [] M. Bencze, About Surányi's Inequality, JIPAM, art. 8, vol. 8, no. 4, 005. [] G. Dospinescu, O teorem uitat inegalitatea lui Surányi, Recrea ii Matematice, nr. /005. [3] M. O. Drimbe, Inegalit i idei i metode, Editura Gil, Zal u, 003. [4] P. Erdös, J. Surányi, Topics in the Theory of Numbers, Springer, 003. Colegiul Na ional Gheorghe Ro ca Codreanu, str. Nicolae B lcescu nr., 7383, Bârlad 97

6 Noi rafin ri ale inegalit ii lui Durrande în tetraedru de Marius Olteanu Abstract In this note we present some refinements of Durrande's inequality. Key words: Durrande's inequality, tetrahedron M.S.C.: 5M04, 5M6, 5M0. Dac R i r reprezint raza sferei circumscris. respectiv înscris tetraedrului ABCD, atunci este cunoscut inegalitatea R 3r, ) care se mai nume te inegalitatea lui Durrande, dup unii autori, sau a lui Euler în tetraedru, dup al i autori, prin analogie cu inegalitatea Euler în triunghi. Consider m c problema rafin rii inegalit ii ) reprezint un domeniu destul de s rac în literatura de specialitate. O abordare a acestui subiect o putem g si în [6] i[7] rela iile 0), 7), 8), 9), 3)). Demonstra ii ale inegalit ii ) pot fi g site în [], [4], [8], [0]. Pornind de la aceste aspecte, în articolul de fa se vor prezenta noi rafin ri ale inegalit ii ), completându-se astfel seria rezultatelor stabilite în [6] i [7]. Vom folosi urm toarele nota ii referitoare la un tetraedru oarecare [ABCD]: V volumul s u, r a raza sferei exînscrise de spe a întâi, care este tangent la fa a BCD analog r b, r c, r d ); h a, m a lungimea în l imii, respectiv a medianei tetraedrului ce con in vârful A analog h b, h c, h d ; m b, m c, m d ); r A, R A raza cercului înscris i respectiv circumscris fe ei BCD opus vârfului A) i analog r B, r C, r D ; R B, R C, R D ); S A aria fe ei BCD analog S B, S C, S D ), S aria total a tetraedrului, a = BC, b = AC, c = AB, l = AD, m = BD, n = CD; d, d, d 3, b, b, b 3 lungimile perpendicularelor comune i respectiv a bimedianelor corespunz toare celor trei perechi de muchii opuse. Pentru început, reamintim pe scurt câteva no iuni teoretice. Se nume te tetraedru circumscriptibil sau Crelle sau,,carcas ) tetraedrul [ABCD] care admite o sfer tangent la fiecare dintre cele ase medii ale tetraedrului sfer hexatangent ). Se nume te tetraedru ortocentric, tetraedrul ale c rui în l imi sunt concurente într-un acela i punct. Asupra propriet ilor celor dou clase de tetraedre nu mai insist m aici, din motive lesne de în eles. Dac se noteaz cu ρ raza sferei hexatangent a unui tetraedru Crelle atunci, conform problemei 88 din [8] sau al rezultatelor din [9] se tie c R 3 ρ, ) iar ρ> 5 r. 3) 98

7 În [] au fost stabilite foarte recent urm toarele rezultate, conform teoremei 3 i a propozi iei 6 men ionate în aceast lucrare: În orice tetraedru Crelle [ABCD] au loc inegalit ile i S A + S B + S C + S D ) 9V ρ 3 r 4) r A + rb + rc + ) rd. 5) Evident, inegalitatea 4) este mai tare decât inegalitatea 3) i, inând seama de inegalitatea ), se ob ine urm toarea rafinare a inegalit ii lui Euler într-un tetraedru Crelle R r ρ 3r, 6) cu egalitate numai dac tetraedrul este regulat. Nu mai red m aici demonstra ia inegalit ii 5) aceasta putând fi consultat pe Internet). Men ion m, îns, c demonstra ia este aceea i i în cazul unui tetraedru oarecare i, inând seama c S = S A + S B + S C + S D, iar 3V = r S, atunci putem afirma c în orice tetraedru [ABCD] între razele cercurilor înscrise fe elor i raza sferei înscris, exist inegalitatea r A + rb + rc + rd r, 7) cu egalitate numai dac [ABCD] este tetraedru regulat. Folosind inegalitatea mediilor i inând seama de rela ia 7), avem urm toarea inegalitate r A + r B + r C + r D 8r. 8) Deoarece R A r a i analoagele), conform inegalit ii lui Euler în triunghi, rezult c R A + R B + R C + R D 4 r A + r B + r C + r D) 3r. 9) Inegalitatea 9) reprezint o rafinare a aplica iei 5., pag. 439, din [3]. Propozi ia. Într-un tetraedru oarecare [ABCD] au loc urm toarele rafin ri ale inegalit ii lui Durrande: a) 4 R r A + r B + r C + r D 4 r; 3 b) 8 9 R ra + r B + r C + r D 8r ; c) 8 ) 3 R r R A + R B + R C + R D 8 r; d) 3 5R 36r ) RA 9 + R B + R C + R D 3r. Demonstra ie. a) Avem 6 r A + r B + r C + r D ) r A r B r C r D 99

8 Îns r A r B r C r D ra + rb + rc + rd ) conform inegalit ii dintre media aritmetic i media geometric ). inând seama de inegalit ile 7), 0) i ), avem r A + r B + r C + r D 6 de unde r A + rb + rc + rd 8 r =4 r. ) În continuare, r A = S A, unde p A este semiperimetrul triunghiului BCD. p A Rezult ra = S A p = S A SA A p. 3) A Dar S A = p Ap A a)p A m)p A n) p A [ 3pA a + m + n) 3 Din 3) i 4) avem ] 3 = p4 A 7, S A = p A 3 3 S A p A ) r A S A 3 3. Analog ob inem c rb S B 3 3, r C S C 3 3, r D S D 3 i deci urmeaz c 3 ra + rb + rc + rd 3 S A + SB + SC + S S D) = ) Îns, conform inegalit ii lui Weitzenböck, aplicat triunghiului BCD avem 4 3S A a + m + n ; aceast inegalitate împreun cu celelalte trei similare aplicat fiec reia dintre cele trei fe e ale tetraedrului r mase) implic 4 3 S A + S B + S C + S D) a + b + c + d + l + m + n ), 6) Avem, de asemenea, a + b + c + d + l + m + n 6R 7) conform problemei 79, pag. 3 din [8]). inând seama de rela iile 5), 6) i 7), avem r A + r B + r C + r D 8 9 R. 8) 00

9 De asemenea, avem r A + r B + r C + r D ra + r B + r C + r D. 9) Din 8) i 9) se ob ine r A + r B + r C + r D 4 3 R. 0) Cuplând inegalit ile ), 0) i 8), 8) ob inem punctul a) i respectiv b) al propozi iei. Mai departe, din R A r A i analoagele) i pct. a) avem R A + R B + R C + R D r A + r B + r C + r D ) 8 r. ) Conform problemei 3..54, pag. 36, din [5], putem scrie urm toare inegalitate referitoare la elementele triunghiului BCD 9R A +r A ) 4 3p A, de unde 9R A 4 3p A 8r A. ) Inegalitatea ), aplicat celorlalte trei fe e ale tetraedrului, ne conduce la 9R A + R B + R C + R D ) Îns 4 3p A + p B + p C + p D ) 8 r A + r B + r C + r D ). 3) p A + p B + p C + p D = a + b + c + l + m + n 6 a + b + c + l + m + n 4 6R 4) am inut seama de inegalitatea 7)). Acum, în conformitate cu inegalit ile 3), 4) i pct. a), ob inem 9R A + R B + R C + R D ) R 8 4 r =48 R 7 r =8 6R 9r), ceea ce este echivalent cu R A + R B + R C + R D 8 ) 3 R r. 5) Prin ridicarea inegalit ii ) la p trat i analoagele), avem i analoagele). Rezult 8R A 48p A + 34r A 44 3S A 8 R A + R B + R C + R D) 48 p A + p B + p C + p D) r A + r B + r C + r D) 44 3S. 6) 0

10 Dar p a + m + n) A = 4 p A + p B + p C + p D a + m + n ) i analoagele) i deci a + b + c + l + m + n ) 3 6R =4 R. 7) Totodat, conform aplica iei 5., pag. 499, din [3], se tie c S 4 3r. 8) Atunci, conform rela iilor 6), 7), 8) i a punctului b), ob inem 8 RA + R B + R C + ) R D 48 4R R r, ceea ce este echivalent cu RA + R B + R C + R D 3 5R 36r ). 9) 9 În fine, cuplând inegalit ile ), 5) i 9), 9) ob inem punctul c) i respectiv d) al propozi iei. Egalit ile au loc numai dac [ABCD] este tetraedru regulat. L s m în seama cititorului demonstrarea urm toarelor: Consecin e. În orice tetraedru [ABCD] au loc urm toarele rafin ri ale inegalit ii Durrande: a) R r A r B r C r D r ; b) 8 R ra + rb + rc + rd r ; c) 3 R r d) R A R B R C R D 5R 36r ) RA + RB e) 64r R A R B R C R D 64 + r ; RC + R ) D 3 R r ; r ; f) 4r 4 r A r B r C r D S R4 inegalitate ce reprezint o extindere a problemei 54, pag. 37 din [8]); 9 g) 4R a + b + c + l + m + n 4r inegalitatea constituie o extindere a problemei 83, d), pag. 4 din [8], precum i o extindere la tetraedru a cunoscutei inegalit i din triunghi R a + a b + c r, unde, de aceast dat, R, r reprezint raza cercului circumscris, respectiv înscris triunghiului ABC de laturi a, b, c); h) 36 a b + b c + c l + l m + m n + n a R r inegalitatea constituie o extindere la tetraedru a inegalit ii din triunghi R r a 3 b + b c + c ), autori V. B ndil, M. Lascu, conform problemei 7.39, pag a 33 din [0]); 0

11 5 i) 8r ab + ac + al + am + an + bc + bl + bm + bn + cl + cm + cn + + lm + ln + mn 45 8 R aceast inegalitate constituie o extindere la tetraedru a problemei 4 dat la Nordic Mathematical Contest, edi ia 8, din ). Propozi ia. În orice tetraedru [ABCD] au loc urm toarele rafin ri ale inegalit ii Durrande: ) α 3r a) 4 R pentru orice α [, ); b) 9 r R h a m a ha m a + h b c) 54 3 r3 R 6V m b ) α ) α ) α ) α hb hc hd , m b m c m d + h c m + h d c m d al + bm + cm r ; ) d + d + d 3 b + b + b 3 3R; d) r6 R 4 d + d + d 3 b + b + b 3 4R ; e ) 3 3 R 3 a b c + 3 a m n + 3 b m n + 3 c m n 96r. Demonstra ie. Sunt cunoscute urm toarele rezultate = h a h b h c h d r [4], pag. 39) 30) h a + h b + h c + h d =6r [4], pag. 39) 3) m a + m b + m c + m d = 6 R 3 [8], problema 80 sau [6], pag. 47) 3) m a + m b + m c + m d = 64 9 R [8], problema 80 sau [6], pag. 47) 33) b + b + b 3 = a + b + c + l m + n ) 4 [8], pag. 8), 34) ceea ce implic b + b + b 3 = 4 6R =4R conform inegalit ii 7)) 35) V 8 3r 3 [3], pag. 500, aplica ia 5.3) 36) x + x x n x + x x n ) 37) α α α n α + α α n unde x i R, α i > 0, i =,n, n N []); h a h b h c h d 56r 4 conform inegalit ii mediilor i a rela iei 30)). 38) a) Din h a m a i analoagele) rezult ) α h a 4, m a pentru orice α [, ). 03

12 Dac în rela ia 37) alegem: n = 4, x = h a, x = h b, x 3 = h c, x 4 = h d, α = m a, α = m b, α 3 = m c, α 4 = m d, ob inem: ha = ) ha ) ) ha 4 8 h a h b h c h d 3 64r m a m a ma ma 6R = r R 39) am utilizat inegalit ile 3) i 38)). Mai departe, conform inegalit ii Jensen aplicat func iei convexe, f :0, ) 0, ), fx) =x, q [, ], avem ) α ha 4 m a ha m a 4 α 4 r R 4 α =4 ) α 3r, R pentru orice α am utilizat i inegalitatea 39)). b) Avem: ha h a m a h a = = h a r 40) conform identit ii 30)). Apoi, similar demonstra iei punctului a), utiliz m inegalitatea 37), inegalitatea mediilor, inegalitatea 38) i inegalitatea 33) i ob inem succesiv ha m a = h a ) m a ha ) ma 4 4 h a h b h c h d ) ma 64r 9 64 r m a 64 R =9 r R, 4) Cuplând inegalit ile 40) i 4), ob inem concluzia punctului b). c) Conform problemelor 866 i 867, pag. 9 din [6], rezult c d i b i, i =,, 3 i deci d + d + d 3 b + b + b 3. 4) Îns b + b + b 3 3 b + b3 + b 3 3R 43) conform inegalit ii 35)). Se tie c 6V al + bm + ) d + d + d 3 44) cn conform problemei 69, pag. 39, din [8]). În plus, conform inegalit ilor dintre medii, precum i a inegalit ii 4) avem: 04 al + bm + cn abclmn a + b + c + l + m + n) 6

13 3 6 ) = 9 4 6R 8 R. 45) inând seama acum de inegalit ile 36) i 45), ob inem 6V al + bm + ) 54 3 r3 cn R. 46) Cuplând inegalit ile 4), 43), 44) i 46), g sim concluzia de la punctul c). d) Similar punctului c), avem d i b i,pentru orice i {,, 3}; rezult conform inegalit ii 35). De asemenea d + d + d 3 b + b + b 3 4R, d + d + d 3 3 d + d + d 3 ) r3 R ) = r6 R 4 48) am utilizat inegalitatea de la punctul c)). Cuplând inegalit ile 47) i 48), se ob ine concluzia punctului d). e) Partea dreapt a inegalit ii reprezint sub form corectat rezultatul aplica iei 5.4, pag. 500, din [3]. Pentru demonstrarea p r ii stângi, utiliz m de patru ori inegalitatea mediilor precum i inegalitatea 7). A adar, avem 3 a b c + 3 a m n + 3 b n l + 3 c m l [ a + b + c ) + a + m + n ) + b + n + l ) + c + m + l )] = 3 = a + b + c + l + m + n ) 3 3 6R = 3 3 R. Observa ii.. Toate inegalit ile acestei propozi ii devin simultan egalit i numai dac [ABCD] este tetraedru regulat.. Inegalitatea a) generalizeaz pentru orice tetraedru nu numai pentru cele ortocentrice ale c ror ortocentre sunt situate în interiorul acestora) punctul i) al problemei 9, pag. 3, din [8]. Propozi ia 3. În orice tetraedru [ABCD] au loc inegalit ile 3 4 R m a + m b + + m c m b + m c + m + d m c + m d + + m a m d + m a + m b h a + h b + h c + h b + h c + h + d h c + h d + + h a h d + h a + h b ra + rb + rc + ) rd. 05

14 b) Dac [ABCD] este tetraedru ortocentric, atunci are loc urm toarea rafinare a inegalit ii Durrande: 3 4 R m a + m b + m c h a + h b + h c R4 r 6. Demonstra ie. a) Din inegalitatea mediilor i inegalitatea 33), avem m a + m b + m c 6 3m a + m b + m c + m d ) R = 3 4 R. 49) Evident, m a h a i analoagele); rezult m a + m b + m c h a + h b +. 50) h c În continuare: h a + h b + h c 9 h a + h b + h c ) i analoagele); deci h a + h b + h c 3 h a + h b + h c + h d ). 5) Dar, conform solu iei problemei 5086, G.M.-B nr. /004, pag. 488, se cunoa te identitatea ra + rb + rc + rd =4 h + a h + b h + ) c h. 5) d a). Prin cuplarea rela iilor 49), 50), 5) i 5), se ob ine concluzia punctului b) Inegalitatea rezult imediat din punctul a) i din inegalitatea r ra + rb + r c + r d 8 R4 r 6, 53) care constituie fondul problemei O:3, G.M.-B nr.8/006 sau al propozi iei, rela ia 3) din [7], pag. 05. Observa ie. La punctul a) egalit ile se ating pe rând sau simultan, numai dac [ABCD] este tetraedru echifacial, iar la punctul b) numai dac tetraedrul este regulat. Propozi ia 4. În orice tetraedru ortocentric [ABCD] au loc urm toarele rafin ri ale inegalit ii Durrande: a) 4r h + a h + b h + c h d 34 R4 r. 6 06

15 9 b) 4R m + a m + b m + c m d m a h a m b h b m c h c m d h d 34 R4 r 6. Demonstra ie. a) Se aplic identitatea 5) în inegalitatea 53). b) Conform inegalit ii mediilor precum i a inegalit ii 33), rezult m 6 6 a m a 9 64 R = 9 4 R. Totodat, m a m a h a h a 34 R4 r. 6 La ambele puncte, egalit ile se ating pe rând, sau simultan, numai dac tetraedrul este regulat. În încheiere, având în vedere egalitatea din problema 5, pag. 30, din [8], precum i rela ia ra + rb + rc + rd, putem spune c, într-un tetraedru oarecare r [ABCD], în baza celor prezentate mai înainte, au loc urm toarele consecin e: i) 8 ) 3 R r R A + R B + R C + R D 3 a + b + c + l + m + n) 3 r A + r B + r C + r D ) 8 r; ii) ra + rb + rc + ) rd r ra + rb + rc + rd. În acela i timp, propunem cititorului interesat, în baza celor prezentate în acest articol, ca aplica ii directe, s arate c au loc urm toarele rafin ri de tip Durrande, ce completeaz problemele 06, 3, 57, 65, 70, 35, 305-c), 3- d) i 364 din [8]: a) În orice tetraedru [ABCD] au loc urm toarele rafin ri ale inegalit ii Durrande:. 4 6R a + b + c + l + m + n 6 6 abclmn 6 6 al + bm + cn)al + bm cn)al + cn bm) V 6r;. 48r 3m a + m b + m c + m d ) 8 b + b + b 3 6R; r a b c l m n al + bm + cn) a + b + c ) l + m + n ) 64R 4 ; R3 m a + m b + ) m c + m d ha + h b + h c + h d ) 8 V 0 r 3 ; R S A + S B + S C + S D )h a + h b + h c + h d ) 48 V r 3 ; 6. 6 r R l + b + c m a + a + c + m m b + a + b + n m c + + l + m + n m d 4 R; 07

16 b) Dac [ABCD] ese tetraedru ortocentric având ortocentrul situat în interiorul tetraedrului, atunci are loc urm toarea rafinare a inegalit ii Durrande 34 R4 r m a h a m b h b m c h c m d h d 4Rr c) 4 R GA + GB + GC GD 9 a + b + c + l + m + ) n 4 3r, unde G reprezint centrul de greutate al tetraedrului [ABCD], iar A, B, C, D, al doilea punct de intersec ie al dreptelor AG, BG, CG i respectiv DG cu sfera circumscris tetraedrului inegalitatea completeaz problema 049* din revista Crux Mathematicorum, autor Jan Ciak, Polonia sau 379 din [8]). Bibliografie [] T. Andreescu, M. Lascu, Asupra unei inegalit i, G. M. - B nr. 9-0/00. [] D. Brânzei, C. Coca, S. Ani a, Planul i spa iul euclidian, Editura Academiei R.S.R., Bucure ti, 986. [3] M. Dinc, M. Bencze, About Inequalities, Octogon Mathematical Magazine, vol., no., A, october 004. [4] M. Miculi a, D. Brânzei, Analogii triunghi-tetraedru, Editura Paralela 45, Pite ti, 000. [5] N. Minculete, Egalit i i inegalit i geometrice în triunghi, Editura Eurocarpatica, Sfântu Gheorghe, 003 [6] M. Olteanu, Rafin ri ale inegalit ii lui Durande în tetraedru, G. M. - B, nr. 8, 006 i G. M. - B, nr., 006 partea a II-a). [7] M. Olteanu, Asupra unor inegalit i în tetraedru, G. M. - B, nr. 3, 006. [8] M. Olteanu, Inegalit i în tetraedru culegere de probleme, Editura Universitar Conspress, Bucure ti, 003. [9] M. Olteanu, Asupra unei inegalit i în tetraedre Crelle, G. M. - A, nr. 3, 003. [0] L. Panaitopol, O demonstra ie a inegalit ii lui Durrande, G. M. - B nr. 5-6, 00. [] Gh. i eica, Probleme de geometrie, edi ia a VI-a, Editura Tehnic, Bucure ti, 98. [] Y.-D. Wu and Z,-H. Zang, The Edge-Tangent Sphere of a Circumscriptible Tetrahedron, Forum Geometricum, volume 7, 007, pp. 9-4, edi ie electronic. Inginer, S. C. Hidroconstruc ia S.A. Bucure ti, sucursala,,olt-superior Râmnicu-Vâlcea 08

17 EXAMENE I CONCURSURI IMC 007, Blagoevgrad, Bulgaria de Dan Schwarz I was fortunate to accompany the Romanian team at the 4 th edition of the International Mathematical Competition IMC), held in Blagoevgrad, Bulgaria, between August 4 and August 0, 007. This is a very arduous mathematical competition, consisting of problems of different degrees of difficulty, asked over days of 5 hours working time each. It is sat by undergraduates from as many as 40 universities and colleges over the 4 editions), ranging from powerful in mathematics!) eastern European countries like Romania, Bulgaria, Hungary, Poland, Russia, Ukraine, Belarus, to western European countries like France, Germany, Spain, The Netherlands; famous places like Princeton and Oxford; but also exotic to us) places like Iran, Indonesia, Mongolia, Brazil. The venue was the American University in Bulgaria and its campus, with outstanding bed and board conditions, nested in the heart of a friendly, touristoriented little city. The low living cost and high quality of goods and services in Bulgaria were affordable and enjoyable even to limited-revenue students. The weather was unfortunately rainy for most of the time, butawelcome change from the heat-wave we all experienced, and conducive toworking on math problems! After problem selection, examination, and final contestation solving and corrections, two Grand First Prizes were awarded, to Russian Alexander Efimov, and our own Andrei Negu hailing from Princeton). They both solved all but the last problem from the set! Moldovan Simion Filip Princeton) had an outstanding performance too, being very close to one of top places. Another Romanian, Claudiu Raicu University of Bucharest), grabbed one of the First Prizes, while Octavian Ganea Poli Bucharest) missed one by a hair. He and Drago Fr il University of Bucharest), as well as Marius Pachi ariu Princeton), had to settle for Second Prizes, while others were awarded lesser ones. The best placed girl was Iris Smit University of Amsterdam). In all, there were 50 contestants. We all want to thank the rectors of Universities, as well as the private sponsors that made the trip to IMC 007 available to quite a large number of Romanian students. As they say: train tickets xxx RON; competition fees xxx RON; the unique experience, opportunity tomeet fellow mathematicians and bathe in a melting pot of cultures priceless! All that is left is to enjoy the diversity of problems, and elegance and technique of the solutions presented in the sequel! Day, August 6, problems, 5 hours working time). Problem. Let f be a polynomial of degree n with integer coefficients, and let p>nbe prime. Suppose that fk) is divisible by p for every integer k. Prove that all coefficients of f are divisible by p. ) ) In the problem submitted, n was and p was 5. A.N.) 09

18 0 i<j n Solution. XII th grade) Consider fx) as a polynomial in Z p [x]. Then fx) has p roots, while its degree is less than p. Therefore f 0, so all of its coefficients are divisible by p. Alternative Solutions. ) XI th grade) Have fx) =a 0 + a x + + a n x n. Then fk) =pv k for all k =0,,...,n can be seen as a linear system of n + equations in n +unknowns the a i coefficients). Its determinant isvandermonde, of value = j i) coprime with p. It follows that the unknowns are uniquely determined, so their values coïncide with a i, and so will be divisible by p. X th grade) We have x i) n i k fx) = fk) k i) = pf 0x), k=0 i k a particular form of Lagrange's interpolation polynomial, therefore the coefficients will be divisible by p. IX th /VIII th grade) We will prove that, if a polynomial fx) with integer coefficients, of degree deg f = n<p,issuchthatp divides fx k ) for n + integers x k, 0 k n, from {0,,...,p }, then all its coefficients are divisible by p. The proof goes by induction, the case n =0being trivial. We have fx) =x x n )gx) +fx n ), where clearly gx) Z[x], of degree deg g = n. Then p divides fx i ) fx n )=x i x n )gx i ), hence gx) inherits the property, and by induction hypothesis, gx) =phx), andso fx) =p x x n )hx)+ fx ) n). p Alternatively, to avoid induction altogether, we can take f n f, and f k x) =x x k )f k x) +f k x k ), for k = n, n,...,. Since f 0 will be constant, and p will divide f 0 x 0 ), the result follows from bottom up. Problem. Let n be an integer. What are the minimal and maximal possible ranks of an n n matrix whose n entries are precisely the integers from to n? Solution. The rank cannot be less than. The row and column crossing at the entry will contain n entries, hence the highest is M n, wlog on the row. The highest entry on the column being m n, the minor completed by x n ) [ ] M m x will have a determinant of absolute value mm x nn ) n = nn ) 0. The rank can be. Takeasi th row,,...,n)+ni ),,...,), therefore all rows lie in the two-dimensional subspace spanned by vectors,,..., n) and,,...,). Since we have justshown it cannot be less than, the rank will be. ) Given, so that only less and less knowledge is required, by Dan Schwarz. A.N.) 0

19 The rank can be as large as n. Use odd numbers on the diagonal, and even numbers above the diagonal; modulo this is a lower triangular matrix of determinant, thus odd. Therefore the matrix is nonsingular, whence its rank is n. Alternative Solution. ) Take as first row thevector,,...,n), andasi th row i>) thevector,,...,n)+nσ 0 i ),σ i ),...,σ n i )), where σ is the cycle n,...,). After subtracting the first row from all others, then the first column from n the last one, the determinant is seen to be ) n n )n n P ω k ), where P x) =+x + +n )x n, and ω k = e kπi/n ) the known formula for a circulant determinant obtained by cyclically permuting row,,..., n )). But P ω 0 )=P) = nn )/, while, for x, P x) = d dx + x + + xn )= d dx hence, for k>0, Alternatively, So x n x ) k=0 = n )xn nx n + x ) P ω k )= n )ωn k nωn k + ω k ) = n )ω k ) ω k ) = n ω k. n ω k )P ω k )=n ) ωk i = n. n k=0 i=0 P ω k )= ) n n ) n n/, and the value of the determinant is n ) n n n /, different from zero, hence the rank is n. Problem 3. Call a polynomial P x,...,x k ) good if there exist real matrices A,...,A k such that k ) P x,...,x k )=det x i A i. Find all values of k for which all homogeneous polynomials with k variables of degree are good. i= Solution. The possible values of k are and. If k =, then P x) =αx and we can choose ) 0 A =. 0 α ) Given during the competition by Romanian contestant Claudiu Raicu. Itsinterest also resides in being related to a solution to Problem 4 of Day. A.N.)

20 and If k =, then P x) =αx + βy + γxy and we canchoose ) 0 A = 0 α 0 β A = γ If k 3 we will show that P x,...,x k )= ). k x i is not good. Suppose that it is; since the first columns of the matrices are linearly dependent there are k of them, more than their dimension, which is ), there must exist a non-trivial relation. Then the determinant has zero first column, hence its value is 0, while the value of P is non-zero, a contradiction. Problem 4. Let G be a finite group. For arbitrary sets U, V, W G, denote by N UV W the number of triples x, y, z) U V W for which xyz is the unity. Suppose that G is partitioned into three nonempty sets A, B and C. Prove that N ABC = N CBA. Solution. Condensed form.) One gets that N UV G = U V = U V et al. Then one gets the equality N UV W = N VWU = N WUV. Finally, for a partition of W into three nonempty sets W, W and W 3, one obtains i= N UV W = N UV W + N UV W + N UV W3. Applying these three formulae, the statement follows. Problem 5. Let n be apositive integer, and a,...,a n be arbitrary integers. n Suppose that a function f : Z R satisfies fk + a i l)=0whenever k and l are integers and l 0. Prove that f 0. Solution. ) Denote Ek, l) := i= n fk + a i l). i= Since fk + Cl)+a i l)=fk +a i + C)l), we can translate the integers a i with any integer C, therefore we may assume wlog 0=a a a n = a. We obviously have a Ek, l) = λ j fk + jl)=0, where λ j = {i ; a i = j}. Denote j=0 RX) := a λ j X j, ) Given during the competition by Princeton contestant Andrei Negu. A.N.) j=0

21 so a R) = λ j = n 0. j=0 Define the subset of the polynomial ring R[X] m I := P X) = m b j X j ; b j fk + jl)=0, k, l Z,l 0. j=0 j=0 This is clearly a subspace of the real vector space R[X]. P X) I implies XPX) I, since we can write Furthermore, m m b j fk +j +)l) = b j fk + l)+jl)=0. j=0 j=0 Hence I is a non-zero ideal, since RX) I. Thus I is generated as an ideal) by some non-zero polynomial Q since R[X] is principal). If Q is monomial, then the definition of I implies f 0, so we can assume Q has a complex) zero c 0. Again, by definition of I, QX M ) Ifor every positive integer M, since m b j fk +Mj)l) = j=0 m b j fk + jml)) = 0, j=0 hence QX) divides QX M ). This shows that all numbers c, c,...,c M,... are zeroes of Q. Since Q has only finitely many roots, we must have c N =for some N ; in particular Q) = 0, which implies P ) = 0 for all P I. This contradicts the fact that R) 0. Alternative Solution. ) As before, assume 0=a a a n = a. Now, if the values of f are known for any contiguous sequence of a integers, all the others can be calculated from the given relation, so the Q-linear subspace spanned by the values of f in R is of finite dimension. A canonical projection map π onto Q is linear in any element of the basis, therefore the function πof : Z Q also satisfies the functional equation. This finite dimension business is not really necessary, as we could use a Hamel basis of R over Q). The same observation now entitles us to derive allvalues of πof from a finite set of rational numbers, therefore computing a common denominator N for them lets the image of n!nπof) to be part of Z, so n!nπof) :Z Z, satisfying the functional equation. Take now an arbitrary prime p>n!nn. Modulo p, the given relation induces a period of some length t. Taking l = t, the relation reduces to n!nnπof)k) 0 mod p), therefore p divides πof)k). Since p was taken arbitrary large, this in turn forces πof)k) =0,andasany projection map is zero, so will be f. ) Composite from several contributions, and put together by Dan Schwarz. A.N.) 3

22 Alternative Solution. ) For a logical proposition P we will use the notation [[ P ]] for a variable that has value if P is true, and 0 if P is false. For integer C> max n a j we have a j + C 0for any j, so j= n n ) 0= fk a a j + C)) + a i a j + C)) j= i= n n n = fk + Ca i a )) + a j a i a )) = nfk) [[a i a ]] i= j= therefore fk) =0for all integers k. Remark. The ideal I from the first solution is nothing but the family of all characteristic polynomials associated with some linear recurrence relation, with variable coefficients, derived from the given functional equation. The key feature of substituting X M for X amounts to just changing the value of l in the given equation. Some contestants actually went this way, studying with various degrees of success) the common possible roots of such characteristic polynomials using the resultant determinant, or some other tools). Using the ideal demonstrates the depth packed in the fact that R[X] is a principal ring. Problem 6. How many non-zero coefficients can a polynomial P z) C[z] have, if its coefficients have integer moduli, and P z) for any complex number z of unit modulus? ) Solution. We will show that the answer is 0, or. These values are possible, for example polynomials P 0 z) =0, P z) =,andp z) =+z. Consider an arbitrary polynomial P z) =a 0 + a z + + a n z n + a n z n, with a 0 a n 0otherwise, if a 0 =0,divideitbysomepower of z). Take P 0 z) = a 0 a 0 P a0 a 0 ) /n a n z), a n of free term a 0 and leading coefficient a n, while all other coefficients still have integer moduli. Obviously P 0 fulfills the conditions, therefore we maywlogassume both a 0 and a n to be positive integers. Take Qz) =a z + + a n z n ; our goal is to show Q 0. Consider ω = e πi/n, a primitive root of order n of the unit. Notice that n n n n Qω k )= a i ω k ) i = k=0 k=0 i= i= a i n i= ω i ) k =0. Taking the average of polynomial P z) at the points ω k we obtain n P ω k )= n n k=0 k=0 n a 0 + Qω k )+a n ω k ) n )=a 0 + a n k=0 ) Given during the competition by Bulgarian contestant Emil Baronov. A.N.) ) In the problem submitted, P z) was given of integer coefficients. A.N.) 4

23 and n P ω k ) n P ω k ) n n = a 0 + a n. k=0 k=0 This clearly implies a 0 = a n =, and P ω k ) = +Qω k ) =for all k, therefore all values of Qω k ) must lie on the circle +z =, while their sum is 0. This is only possible if Qω k )=0for all k, but then Q has n roots, while deg Q n, so Q 0 and P z) =+z n, with only two non-zero coefficients. Remark. From Parseval's formula integrating P z) = P z)p z) on the unit circle) it is obtained that n i=0 a i = π π 0 P e it ) dt π 0 π 4dt =4. Hence there cannot be more than four non-zero coefficients. Even four are too many, as equality above is impossible. For only three monomials, the argument from the solution may be applied in a simplified form. Let us also notice that when P z) has exactly two non-zero coefficients, both coefficients are of unit modulus, and the bound in P z) is reached, that is sup z = P z) =. Day, August 7, problems, 5 hours working time). Problem. Let f : R R be acontinuous function. Suppose that for any c>0, the graph of f can be moved to the graph of cf using only translations) and/or rotations). Does this imply that fx) =ax + b for some real numbers a and b? Solution. No. The function fx) =ae bx, a, b R, also has this property, since cae bx )=ae bx+ln c)/b), i.e. cfx) =fx + C), with C =lnc)/b, and thus we can superimpose the two graphs by a translation. Remark. The problem proved to be quite taxing, as many competitors did not manage to find an example. Trying to find it by exhibiting some constraints, and solving them, leads to complications. Good guesswork and some street-smartness in looking for a known basic function, as this was a Problem ) was rewarded! Problem. Let x, y and z be integers such that S = x 4 + y 4 + z 4 is divisible by 9. Show that S is divisible by 9 4. Solution. We claim that 9 then divides all three x, y and z. Assume the contrary, wlog that 9 does not divide x. Since Z 9 is a field 9 is a prime), there is some integer w such that wx mod 9). Then wx) 4 +wy) 4 +wz) 4 is also divisible by 9,sowe can assume x mod 9). There are only eight bi-quadratic residues modulo 9, listed below {0,, 7, 6, 0, 3, 4, 5}. 5

24 We need y 4 z 4 mod 9), but the reduced residues of z 4 are {8, 7,,, 8, 5, 4, 3}, and the two sets have no common elements. Remark. The problem is entirely computational. The trick of reducing to x mod 9) is only needed to simplify the computations, as after obtaining the list of bi-quadratic residues, one can establish a list of all sums of two, and compare it with the original. Problem 3. Let C be a nonempty closed bounded subset of the real line, and f : C C be a nondecreasing continuous function. Show that there exists a point p C such that fp) =p. ) Solution. C being bounded, inf and sup taken on nonempty subsets of it are finite and, since C is closed, belong to it. Let a = in f x and also let A = {x C ; x fx)}. Clearly a A, hence x C A is nonempty. Take p =sup x A x, so p C. For every x A we have x p, hence x fx) fp), since f is nondecreasing. This implies p =sup x A p A. x fp), hence But then fp) ffp)), so fp) A, and so fp) sup x = p. Putting x A both inequalities together, one gets fp) =p. Alternative Solution. Also taking b =supx C, one can extend f to x C f : [a, b] [a, b] through fx) = λ x fy), where sup x y C fy) + λ x ) in f x y C λ x [0, ] is such thatx = λ x sup y + λ x ) in f y. The function f is easily x y C x y C seen to be nondecreasing and, if f continuous, also continuous). The problem is thus reduced to the case when C is an interval, where having continuity available solves the problem, but lacking it presents the same difficulty as the original. Problem 4. Let n> be apositive integer, and A n =a ij ) i,j=,...,n be the n n matrix where a ij =if i = j, a ij =if i j ± mod n), while a ij =0 otherwise. Find det A n. ) Solution. For n 4 the values can be computed by hand: det A =4, det A 3 =4and det A 4 =9. We will then concern ourselves with n 5. Notice that A n = B n, where B n =b ij ) i,j=,...,n is the n n matrix where b ij =if i j ± mod n), while b ij =0otherwise. To finddet B n, expand the determinant with respect with the first row, then expand both terms with respect to the first column, to get the formal expression + ) n ) ) n 3 ) n 4 ). ) The condition f continuous is superfluous; without it, the statement isknown as Knaster's theorem. A.N.) ) In the problem submitted, n was postulated odd. A.N.) 6

25 When n is odd, the second and third matrices obtained are lower/upper triangular, while in the first and the fourth matrices we haverow row 3 + ±row n = = 0, sodet B n = 0 ) + 0) = and thus det A n =4. When n is even, the values of these terms are harder to find, but then the problem was submitted for n odd! We will find the correct value in the alternative solution to follow. Alternative Solution. ) For n 5, the matrix A n is circulant, therefore its n determinant isdet A n = P ω k ), where P x) =+x + x 4 =+x ), and k=0 ω k = e kπi/n the known formula for a circulant determinant obtained by cyclically permuting row, 0,, 0,, 0,...,0)). As a matter of fact, this formula itself suggests that A n = Bn, where B n is also circulant, obtained by cyclically permuting row, 0,, 0,...,0)). Since ω k are the roots of the polynomial Q n x) =x n, it follows that n n + ωk )= n i ω k ) i ω k )=Q n i)q n i) =i n ) i) n ) in k=0 k=0 k=0 these computations i = ). For n odd, it amounts to ) n i n )i n +)= ) n i n ) =, hence det A n =4. For n even, it amounts to ) n i n ), which is 0 when n 0 mod 4), and 4 when n mod 4), therefore det A n =0for n 0 mod 4), while det A n =6 for n mod 4). Problem 5. For positive integers k and m, find the least number nk, m) for which there exist real nk, m) nk, m) matrices A,...,A k such that all three of the following conditions hold: A m i = 0 for all i k, A i A j = A j A i for all i, j k, and A A i A k ) m 0. ) Solution. Since A A i A k ) m 0, there must exist a vector v 0 such that A A i A k ) m v 0. The more then v p,...,p i,...,p k ) := A p Api i A p k k )v 0, for all 0 p i m. For consistency, we also use the notation v 0,...,0,...,0) := v. Let us show that these m k vectors are linearly independent, therefore the dimension nk, m) is at least m k. Assume the contrary, and consider a non-trivial relation N λ j v Sj = 0 j= of least number N of terms, where S j =p j,...,p ij,...,p kj ), with 0 p ij m. ) Given with omissions) during the competition by Princeton contestant Simion Filip see related issues at Problem of Day ). A.N.) ) In the problem submitted, m was. A. N.) 7

26 N There must exist an index i such that max p ij > min N p ij, otherwise N is, j= j= absurd. Take p = m N max j= p ij and apply A p i to the relation; at least one term will vanish, and at least one term will not. We obtain therefore a relation with less than N number of terms, contradiction. This also offers a way to build a model for nk, m) =m k. Consider m k vectors forming a basis, and index them with k-tuples p,...,p i,...,p k ), with 0 p i m. Define A i v p,...,p i,...,p k ) = 0 if p i = m, and A i v p,...,p i,...,p k ) = v p,...,p i+,...,p k ) if 0 p i >m. It is readily checked that all three conditions hold. Alternative Solution. ) Since A k is nilpotent of order m, the dimension of the image IA m k ) is at most nk, m)/m. Then nk, m) m k follows by induction on k. Consider the linear operators λ i associated with matrices A i. The matrices B i i>k) associated with the linear operators π k oλ i oι k verify all three conditions ι k is the canonical injection from Iλ m k ) into the full space, while π k is the canonical projection from the full space onto Iλ m k )). This reduces the situation to the identical one for k. Alternative Solution. ) The model is that of the real vector space made of all polynomials P in k variables x i, with degree in each variable at most m. Partial differentiation with respect to x i is a linear operator, which will be denoted A i, thus A i P := P Again, it is readily checked that all three condi-. x i tions hold: A m i makes all polynomials vanish, A i and A j obviously commute, and A...A k ) m x m...x m k =m )!) k 0. Problem 6. Let f 0 be a polynomial with real coefficients. Define the sequence f 0,f,...,f n,... of polynomials by f 0 = f and f n+ = f n + f n for every n 0. Prove that there exists a number N such that, for every n N, allroots of f n are real. Solution. 3) The proof is extremely technical. One can definitely prove the case deg f =as the cases for degree 0 and are trivial). Also, noticing that e x f n x)) = e x f n x)+e x f nx) =e x f n+ x) sheds some light on the phenomenon. We will refrain in this presentation from reproducing the very) difficult official solution. Liceul Interna ional de Informatic, Bucure ti ) Given during the competition by Princeton contestant Simion Filip. A.N.) ) Given during the competition by Brazilian contestant Fábio Moreira. A.N.) 3) Nobody managed to tackle this during the competition only very few scores of, or 3 points out of 0 were recorded). A.N.) 8

27 PUNCTE DE VEDERE Starea actual a înv mântului superior românesc de Lauren iu Modan Abstract In our work,,actual Estate of the Romanian University Education, we propose realising a study based on a careful analysis of the teaching and learning present system, from our country society. For this, our starting point is the precarious Mathematics education in the contemporary Romania, and equally, the searching since 005, about Bologna Program, of the French professor in Amiens University, respectively, Bernard Kalaora. Key words: teaching Romanian system, learning Romanian system, Mathematics education. M.S.C.: 0A85, 0A99.. Tradi ia universitar româneasc Începuturile normalit ii culturale ale unei na ii coincid, în general, cu momentul apari iei primei sale universit i. Cu cât întemeierile colilor academice au fost mai timpurii în istoria unei ri, cu atât mai prolifice vor fi câ tigurile na iei respective, care, prin r bdare, a beneficiat de o îndelung perioad de c utare, de cizelare i de a ezare a unor fapte, c rora numai trecerea vremii le asigur lefuirea spre giuvaer! Este cazul unor culturi solide ca ale Fran ei, Italiei, i Germaniei, care, în fond, i-au pus amprenta pe dezvoltarea ulterioar a umanit ii. La noi, în ciuda existen ei unor institu ii superioare de educare, ceva mai vechi, ca Academia Mih ilean din Ia i, coala de la Sfântul Sava, din Bucure ti, i Colegiul Piari tilor din Cluj, nu putem emite preten ia de a le numi universit i! Primele l ca e academice române ti, în adev ratul sens al cuvântului, apar târziu, în 860, la Ia i i respectiv în 864, la Bucure ti, prin întemeierea celor dou prime universit i, în urma hrisoavelor semnate de Al. I. Cuza, domnul Unirii v. [3]). Acestora, din 87, li se adaug i universitatea din Cluj. Primii doctori români, mai ales cei coli i la Paris i Roma, indiferent de domeniu, s-au întors în majoritate, cu mult entuziasm, în ar, deci i s contribuie la punerea temeliilor tiin ei i ale culturii na ionale. Cu to ii erau c l uzi i de formularea magic, potrivit c reia,,cum arat ast zi coala, a a va ar ta mâine ara, din spusele lui Spiru Haret, primul doctor român în Matematici, care, renun ând la o str lucit carier universitar la Paris, a revenit în ar, pentru adeveni în timp,,omul colii sau,,marele ministru, dup cum îl va evoca, deseori, Nicolae Iorga v.[3]). Munca temeinic i serioas, a pionierilor înv mântului universitar românesc, a început s dea roade în primii ani ai secolului al XX-lea... Iar între cele dou r zboaie mondiale, truda lor a izbutit chiar s impun universit ile române ti! Cultivarea sentimentului pentru lucrul bine f cut, transmis de mae tri în sufletele discipolilor, a reu it s reziste i peste timpurile tulburi ce au venit dup impunerea for at a ideologiei sovietice, când, aproape o genera ie întreag, de profesori universitari, a fost distrus în închisorile comuniste... i ne-am bucurat constatând c nu 9

28 se pierduse totul i c semin ele bine s dite de c tre ace ti întemeietori de drumuri, în tiin i cultur, au reu it s reziste, i germinând, s contribuie la... continuitatea firescului! i, de i mai rar ca în perioada de glorie na ional, respectiv aceea a României Mari, s-au pus iar i c r mizi solide pentru spiritualitatea na iei, chiar i în vremurile bântuite de s r cia ideologiei proletare a egalitarismului... De altfel, trebuie s subliniem c întreaga perioad comunist, dintre 947 i 989, a fost constituit dintr-o alternan de etape mai luminoase, cu altele foarte întunecoase, datorate luptei dintre vechii universitari, cu o forma ie solid de specialitate i cu o vast cultur, realizate dup modelul Europei vestice, dar i între cei noi, reîntor i de la studii, mai mult sau mai pu in coerente, f cute în Rusia sovietic, i impu i apoi de puterea proletar, sus inut de Moscova. Într-o bun parte a sa, deceniul al 7-lea al veacului trecut s-a bucurat de prestigiul unor personalit i ce au profesat în înv mântul nostru superior, i care în ciuda unor compromisuri f cute cu regimul, i-au impus totu i un standard i o calitate de nivel interna ional. Pentru Matematic, trebuie s amintim c în aceast perioad istoric, România avea realiz ri ce erau imediat citate dup cele ale ru ilor i polonezilor, la vremea aceea, for e de necontestat ale domeniului. Calitatea moral, cel mai adesea îndoielnic a persoanelor formate în fosta Uniune Sovietic i promovate apoi în universit i, al turi de intrarea în via a activ i impunerea în lumea academic, de c tre partidul unic, a unor,,produse ale discriminatoarei reforme a înv mântului din 948, care sus inea doar non-valoarea, totul cumulat i cu dispari ia natural a ultimilor mari fo ti profesori, forma i între cele dou r zboaie mondiale, au dus treptat, la o adev rat s r cire spiritual a înv mântului superior românesc! Astfel, cu un num r dominant de pseudo-speciali ti formatori, pentru care politicul f cea regula, în majoritatea universit ilor predecembriste disp ruse respectul fa de morala cre tin, fa de valoare în general, i fa de viitorul na iei, în special! Iar acestea func ionau dup un compromis tacit, sus inut de programe deseori neadecvate, înso ite de o exigen în continu sc dere, conducând în general, spre lipsa unei perspective reale a înv mântului universitar din ultimii ani, ai deceniului al 8-lea, din secolul deja apus.... Perioada universitar postdecembrist Prim vara adus în sufletele românilor, de c tre însoritul decembrie 989, nu a atras dup sine i normalitatea atât de necesar, în via a academic i tiin ific a rii! Dup rezolvarea intereselor carieriste ale marii majorit i a unor mode ti înv cei universitari, care au devenit apoi, cale de câ iva ani, adev ra i globe-trotteri, erija i în... somit i academice, f r oper de baz, i care, f r nici o pricepere i cu o capacitate limitat de a discerne alegerea utilului pentru institu ia ce o reprezentau, au încercat copierea a tot felul de modele: american, anglo-saxon, austriac, fran uzesc, italian, scandinav... Relativ repede, a ap rut o bulversare total a sistemului universitar românesc, în care s-au n scut înainte de vreme, specialit i noi, dar f r a avea mai întâi, dasc lii necesari, în care s-au înfiin at universit i noi, în ora e unde nici licee cu tradi ie nu existau, sau chiar în localit i unde cei cu studii superioare abia se num rau în câteva duzini... i culmea ridicolului, în ara în care pân -n 989, num rul inginerilor ne tiutori de meserie, era de 4 ori mai mare ca al mai trilor, în ara cu o suprafa medie, dar cu pu ini locuitori, s-a decretat... 0

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