Soluţii juniori., unde 1, 2


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1 Soluţii juniori Problema 1 Se consideră suma S x1x x3x4... x015 x016 Este posibil să avem S 016? Răspuns: Da., unde 1,,..., 016 3, 3 Termenii sumei sunt de forma 3 3 1, x x x sau Cristian Lazăr Să spunem că sunt a termeni de primul tip, b termeni de al doilea tip şi c de al treilea tip. Avem ab c 1008 şi a b5 6 c , echivalent cu a 5b 5c 016 6(c b). De aici deducem că b c. Obţinem ab 1008 şi a10b 016. Prin urmare a 756 şi bc 16. Problema Determinați numerele naturale n 1 care au proprietatea că, pentru orice divizor d 1 al lui n, numerele d d Răspuns: ;3;6 n. 1 și d d 1 sunt prime. Lucian Petrescu Arătăm mai întâi că n este liber de pătrate. Dacă prin absurd ar exista d 1, astfel încât d n și d n, atunci conform ipotezei ar rezulta că numărul 4 4 d d 1 d d 1 d d 1, contradicţie. 3 Dar Deci n p1 p ps, cu s * și p1 p ps numere prime. d d 1 d d 1 este prim. Fie p 5 un număr prim, divizor al lui n. Atunci p 1 (mod6) sau p 5 (mod6). Dacă p 1 (mod6) obținem că numărul p p1 3mod6 (și p p1 3) este compus, fals. Dacă p 5 (mod6) rezultă că numărul p p1 3mod6 (și p p1 3) este compus, fals. Așadar, n nu are factori primi 5 cele trei valori ale lui n verificând ipoteza problemei. p de unde rezultă că p ;3, adică n ;3;6 i, toate
2 Problema 3 Se consideră un triunghi ABC, cu AB AC, în care punctul I este intersecţia bisectoarelor şi punctul M este mijlocul laturii [BC]. Dacă IA IM, determinaţi cea mai mică măsură posibilă a unghiului AIM. Răspuns: 150. FieD AI BC. Deoarece AB AC, atunci DBM şi m ACB m ABC. Avem midb mdac m ACB mdab m ABD m ADC, deci unghiul IDB [] este ascuţit. Fie F şi E proiecţiile punctului I pe dreptele AB, respectiv BC. E BD BM. Deducem că Cum IBF IBE (CU) şi IFA IEM (IC), BC rezultă că BA BM şi IBA IBM. Avem mmid midb mimb mdac m ACD miab m ACD Ca urmare, m AIM 180 m ACB.(1) Fie H proiecţia punctului B pe dreapta AC. Rezultă că Deducem că m( ACB) 30. () Din (1) şi () rezultă că m AIM BC BH AB.
3 Problema 4 Un pătrat unitate este îndepărtat din colţul unui pătrat n n, unde n, n. Demonstraţi că suprafaţa rămasă poate fi pavată cu dale formate din 3 sau 5 pătrate unitate de forma celor din figurile de mai jos: [] Vom numi un pătrat descris de enunţ pătrat ciobit. m,3,4,5,6,7 orice pătrat ciobit mmpoate fi pavat ca mai Mai întâi constatăm că, pentru jos. Ȋn plus, dacă * p, orice dreptunghi 6p m, cu m,3,4,5,6,7, precum şi pătratul 6p 6p pot fi pavate cu dreptunghiuri de tipul Pentru oricare n, n 8, există k * astfel încât n 6k m Prin urmare, orice pătrat ciobit n n, unde m,3,4,5,6,7.,,3,4,5,6,7 se va obţine din pătratul ciobit m m care se bordează cu două dreptunghiuri 6k m şi un pătrat 6k 6k astfel: m,
4 The 016 Danube Competition in Mathematics, October 9 th 1. Let ABC be a triangle, D the foot of the altitude from A and M the midpoint of the side BC. Let S be a point on the closed segment DM and let P, Q the projections of S on the lines AB and AC respectively. Prove that the length of the segment P Q does not exceed one quarter the perimeter of the triangle ABC. Adrian Zahariuc Solution. From the cyclic quadrilateral AP SQ, whose circumcircle has diameter AS, P Q = AS sin P AQ. So, the largest value of P Q is obtained when AS is largest, that is when S = M. In the case S = M, denote E, F the midpoints of the segments AB, respectively AC and G, H the midpoints of the segments ME, respectively MC. Then P Q P G + GH + HQ = 1 (EM + EF + F M) = 1 4 P ABC, as desired. Equality occurs if and only if ABC is equilateral.. A bank has a set S of codes for its customers, in the form of sequences of 0 and 1, each sequence being of length n. Two codes are called close if they are different at exactly one position. It is known that each code from S has exactly k close codes in S. a) Show that S has an even number of elements. b) Show that S contains at least k codes. Solution. Start by noticing that we can suppose that S contains the nil code (0, 0,..., 0): otherwise take a code x S and replace S with the set S = x + S = {x + y y S}, where addition is taken mod. Then S and S have the same cardinal and S fulfills the same condition as S. In the sequel we will denote w(x) the number of nonnil components of the code x and S i = {x S w(x) = i}. a) Consider the bipartite graph G = A B, where the vertices are the codes, A = {x S w(x) = even}, B = {x S w(x) = odd} and an edge between two vertices exists if and only if the corresponding codes are close. Count the edges of this graph: from A emerge k A edges and from B emerge k B edges. But k A = k B, hence A and B have the same number of vertices, whence the conclusion. b) Notice first that S 0 = 1. Take now x S i. The set S i 1 has at most i codes close to x and the set S i+1 has at least k i codes close to x. On the other hand, each code from S i+1 has at most i + 1 close codes in the set S i. Consider now the (bipartite) subgraph S i S i+1. We count the number N of its edges twice: first we count the edges emerging from S i to find that N (k i) S i, then we count the edges emerging from S i+1 to find that N (i + 1) S i+1. So, (i + 1) S i+1 (k i) S i. The last inequality yields inductively to S i C i k for every 0 i k, whence S k.
5 3. Let n > 1 be an integer and a 1, a,..., a n be positive integers with sum 1. a) Show that there exists a constant c 1/ so that 1 + (a ) c, where a 0 = 0. b) Show that the best value of c is at least π/4. Solution. For the first part, notice that 1 + (a ) (1 + a 0 + a ). Since (1 + a 0 + a ) (1 + a 0 + a )(1 + a 0 + a ) 1 1 =, 1 + a 0 + a a 0 + a 1 + the relation follows by summing. For the second part we denote x k = a 0 + a and use the inequality x k+1 x k 1 + x k arctan x k+1 arctan x k, valid for 1 x k+1 x k 0. This is equivalent to tan x k+1 x k 1+x k x k+1 x k 1+x k x k+1, and results from tan x x for x (0, π ). Remark. Taking = 1/n, k = 1,,..., n and n, a limit argument shows that the value c = π/4 cannot be improved. 4. Prove that there exist only finitely many positive integers n such that ( n ) n + n is an integer. Adrian Zahariuc Solution. Assume that n is sufficiently large. We claim that there exists a prime number p n such that p does not divide k + n, for any integer k. Lemma. There exist coprime positive integers a, b of different parities such that 3n (a + b ) 4n. Proof. We may take a = 1 and b even maximal such that a + b n, etc. Note that N = 4n (a + b ) 3(mod4), so we can find a prime factor p of N which is of the form 4k + 3, since N 0. We claim that p satisfies the desired property. Recall that for a prime number p 3(mod4), if p divides x + y, then p divides both x and y. Assume that there was some k such that p divides k + n. Then p (k) + (a + b ), hence p a + b, since p 3(mod4). Then, p a and p b, contradicting gcd(a, b) = 1.
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