COMPETITIONS
2014
With the cooperation of MARIEAN ANDRONACHE, MIHAIL B ˘ALUN ˘A,
C ˘AT ˘ALIN GHERGHE, RADU GOLOGAN, ANDREI ECKSTEIN,
MARIUS PERIANU, C ˘ALIN POPESCU, DINU S¸ERB ˘ANESCU
Foreword . . . v
1.1. 2014 Romanian Mathematical Olympiad – District Round . . . 1
1.2. 2014 Romanian Mathematical Olympiad – Final Round . . . 19
1.3. Shortlisted problems for the 2014 Romanian NMO . . . 41
1.4. Selection tests for the 2014 BMO and IMO . . . 47
1.5. Selection tests for the 2014 JBMO . . . 65
1.6. 2013 – 2014 Local Mathematical Competitions 1.6.1. The 2013 Danube Mathematical Competition . . . 85
1.6.2. The 2013 Tenth IMAR1 Mathematical Competition . . . 91
1Institute of Mathematics of the Romanian Academy. One-day IMO-type contest.
The 21st volume of the ”Romanian Mathematical Contests” series contains
more than 200 problems, submitted at different stages of the Romanian Mathe-matical Olympiad, other Romanian contests, and some international ones. Most of them are original, but some problems from other sources were used as well during the competition.
Most of the problems are discussed in detail, and alternative solutions or generalizations are given. Some of the solutions belong to students and were given while they sat the contest; we thank them all.
Special thanks are due to C˘alin Popescu, one of the editors, who produced the final form of the majority of the qualification tests for the IMO team. Alexandru Negrescu did a great job in putting all the materials together.
A remarkable effort came from former multiple IMO-participants Tudor P˘adurariu and ¨Omer Cerrahoˇglu, who contributed to the training of this year Romanian IMO Team.
We thank the Ministry of National Education for constant involvement in supporting the Olympiads and the participation of our teams in international events.
Special thanks are due to companies who were involved in sponsoring the Romanian Olympiad and the participation of the Romanian teams to inter-national competitions: BCR Bank, eMAG Foundation and T¸ uca Zbˆarcea & Associates. The printing of this volume has been supported by The Group company.
DISTRICT ROUND
5th
GRADE
Problem 1. Find, with proof, all positive integers abc satisfying b · ac = c · ab + 10.
***
Solution. The given condition rewrites as b(10a + c) = c(10a + b) + 10, from which a(b − c) = 1, and hence a = b − c = 1. The numbers are
110, 121, 132, 143, 154, 165, 176, 187, 198.
Problem 2. Let M be the set of palindromic numbers of the form 5n + 4, where n ≥ 0 is an integer. (A positive integer is a palindromic number if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.)
a) If we write the elements of M in increasing order, which is the 50th
number?
b) Among all numbers in M, written with nonzero digits who sum up to 2014, which is the greatest one and which the smallest one?
Mircea Fianu
Solution. a) The last (and hence, the first) digit of a number from M equals 4 or 9. A direct count shows that M contains 2 one digit numbers, 2 two digit numbers, 20 three digit and 20 four digit numbers, hence the 50th number is the 6th five digit number, that is, 40504.
b) The greatest number in M has the maximum number of digits. Therefore, we put 4 as the first and last digit and complete the decimal representation
with 2006 digits 1, obtaining thus 411 . . . 1
2006 digits 1
4. Similarly, the smallest number in M has the least number of digits. The answer is 9899 . . . 9
220 digits 9
89.
Problem 3. Let A = {1, 3, 32, 33, . . . , 32014}. We obtain a partition of A if
A is written as a disjoint union of nonempty subsets.
a) Prove that there is no partition of A such that the product of elements in each subset is a square.
b) Prove that there exists a partition of A such that the sum of elements in each subset is a square.
Traian Preda
Solution. a) Assume that such a partition exists. Then the product of all elements of A must be a square as well. But this equals 31+2+3+···+2014 = 32015·1007, obviously not a square.
b) Observe that 32n+ 32n+1= (3n· 2)2, hence a possible partition is
A = {1, 3} ∪ {32, 33} ∪ · · · ∪ {32012, 32013} ∪ {32014}.
Problem 4. A 10 digit positive integer is called a cute number if its digits are from the set {1, 2, 3} and every two consecutive digits differ by 1.
a) Prove that exactly 5 digits of a cute number are equal to 2. b) Find the total number of cute numbers.
c) Prove that the sum of all cute numbers is divisible by 1408.
Csap¨o Hajnalka
Solution. a) In the decimal representation of a cute number, the parity of the digits clearly alternates, hence exactly 5 digits are even, that is, equal to 2. b) There are 25= 32 cute numbers of the form 2a2b2c2d2e and another 32
of the form a2b2c2d2e2, therefore the requested number is 64. c) If a1a2. . . a10 is a cute number, then
(4 − a1)(4 − a2) . . . (4 − a10)
is also cute and distinct from the previous one. They add up to 4444444444. Grouping the cute numbers in 32 such pairs, we obtain that their sum equals 32 · 4444444444 = 27· 11 · 101010101 = 1408 · 101010101, hence the conclusion.
6th GRADE
Problem 1. Prove that: a) 1 2 3 + 2 3 3 + 5 6 3 = 1; b) 333+ 433+ 533< 633. Damian Marinescu
Solution. a) A straightforward computation proves the claim. b) It suffices to show that
333 633 + 433 633 + 533 633 < 1, that is, 1 2 33 + 2 3 33 + 5 6 33 < 1. But 1 2 33 + 2 3 33 + 5 6 33 < 1 2 3 + 2 3 3 + 5 6 3 = 1.
Problem 2. We call a nonempty set M good if its elements are positive integers, each having exactly 4 divisors. If the good set M has n elements, we denote by SM the sum of all 4n divisors of its members (the sum may contain
repeating terms). a) Prove that
A = {2 · 37, 19 · 37, 29 · 37} is good and SA= 2014.
b) Prove that if the set B is good and 8 ∈ B, then SB�= 2014.
Relu Ciupea
Solution. a) Any number equal to the product of two distinct primes p, q, has exactly 4 divisors: 1, p, q and pq, hence A is good.
A short computation shows that SA= 2014.
b) A number n having exactly 8 divisors is either the product of two distinct primes, or the cube of a prime. It is not difficult to see that the divisor’s sum of such a number is even, except for n = 8. Therefore, if 8 ∈ B, then SB must
Problem 3. The points M, N, and P are chosen on the sides BC, CA and AB of the triangle ABC such that BM = BP and CM =CN . The perpendi-cular dropped from B onto M P and the perpendiperpendi-cular dropped from C onto M N intersect at I. Prove that the angles IP A and IN C are congruent.
Gabriel Popa A P B M C N I
Solution. Since CM = CN and CI ⊥ MN, the line CI is the perpendicular bisector of the line segment M N , hence IM = IN . Similarly, we have IM = IP . Triangles IM C and IN C are equal, so IM C ≡ IN C, and, in a similar way, we deduce that IM B ≡ IP B. It follows that IP A = IM C, and, finally, that IP A ≡ IN C.
Problem 4. Determine all positive integers a for which there exist exactly 2014 positive integers b such that 2 ≤ a
b ≤ 5.
D. M. B˘atinet¸u-Giurgiu, Neculai Stanciu
Solution. Rewrite 2 ≤ a
b ≤ 5 as 2a ≤ 10b ≤ 5a. It follows that the sequence 2a, 2a + 1, . . . , 5a contains 2014 multiples of 10, hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. We deduce that 2013 · 10 ≤ 5a − 2a < 2015 · 10, which leads to a ∈ {6710, 6711, . . . , 6716}. By inspection, we conclude that the required values of a are: 6710, 6712, and 6713.
7th GRADE
Problem 1. a) Prove that for any real numbers a and b the following inequality holds:
b) Find all positive integers n and p such that:
n2+ 1 p2+ 1 + 45 = 2 (2n + 1) (3p + 1) .
Nicolae Papacu
Solution. a) The given condition rewrites as (ab−6)2+(a−2)2+(b−3)2≥ 0,
obviously true.
b) Similarly, we obtain (np−6)2+(n−2)2+(p−3)2= 5, hence the numbers
(np − 6)2, (n − 2)2, and (p − 3)2 are equal to 0, 1, and 4, in some order. By
inspection, we find (n, p) ∈ {(2, 4), (2, 2)}.
Problem 2. Let a, b, c be real numbers such that: |a − b| ≥ |c|, |b − c| ≥ |a|, |c − a| ≥ |b|.
Prove that one of the numbers a, b, c equals the sum of the other two. George Stoica
Solution. Squaring the first inequality gives (a − b)2 ≥ c2, hence (a − b +
c)(b + c − a) ≥ 0. Multiplying the latter with the other two similar inequalities implies (a + b − c)2(b + c − a)2(c + a − b)2≤ 0, hence one of a, b, c is the sum
of the other two.
Problem 3. Let ABC be a triangle in which m( ˆA) = 135◦. The
perpen-dicular to the line AB erected at A intersects the side [BC] at D, and the angle bisector of ∠B intersects the side [AC] at E. Find the measure of BED.
Traian Preda A B C I D E
Solution. Let I ∈ (BE) such that IA bisects the angle DAB. We de-duce that ID is the bisector of the angle ADB. A short computation shows that m( DIB) = 135◦, hence triangles ABE and IBD are similar. It follows
thatABIB = BEBD, so that EBAB =BDBI. Thus, triangles ABI and DBE are similar as well and, since m( BED) = m( BAI), we infer that m( BED) = 45◦.
Problem 4. Let ABCD be a square and consider the points K ∈ (AB), L ∈ (BC), and M ∈ (CD) such that KLM is a right isosceles triangle, with the right angle at L. Prove that the lines AL and DK are perpendicular to each other. Bogdan Enescu A D M K B L C
Solution. It is not difficult to observe that △KLB ≡ △LMC, hence KB = LC. Because AB = BC, it follows that AK = BL. But then, △AKD ≡ △BLA, and since AK ⊥ BL and AD ⊥ BA, we deduce that AL ⊥ KD, as well.
8th
GRADE
Problem 1. In the right parallelepiped ABCDA′B′C′D′, with AB = 12√3
cm and AA′ = 18 cm, we consider the points P ∈ [AA′] and N ∈ [A′B′] such
that A′N = 3B′N. Determine the length of the line segment [AP ] such that
for any position of the point M ∈ [BC] , the triangle MNP is right angled at N.
Damian Marinescu
Solution. We have BC ⊥ (ABB′) , therefore BC ⊥ P N. Since P N ⊥ NM,
it follows that P N ⊥ (NBC) , hence P N ⊥ NB, that is, triangle NBP is right angled at N. Let AP = x; we obtain BP2= x2+ 432, P N2= (18 − x)2
+ 243 and BN2= 351. But BP2= P N2+ BN2, and hence x = 13, 5 cm.
Problem 2. For each positive integer n we denote by p (n) the greatest square less than or equal to n.
a) Find all pairs of positive integers (m, n) , with m ≤ n, for which p (2m + 1) · p (2n + 1) = 400.
b) Determine the set n ∈ N∗ n ≤ 100 and p (n + 1) p (n) ∈ N/ . Lucian Dragomir
Solution. a) Since 400 = 1 · 400 = 4 · 100 = 16 · 25, we analyze three cases. If p (2m − 1) = 1, p (2n − 1) = 400, we obtain 1 ≤ 2m − 1 < 4 and 400 ≤ 2n − 1 < 441, hence m ∈ {1, 2} and n ∈ {200, 201, . . . , 219} , giving 40 pairs (m, n) . Similarly, in the second case we obtain 20 pairs, and in the last case, 24 pairs, leading to a total of 84 pairs.
b) Let n ∈ N, and let p (n) = k2; it follows k2 ≤ n ≤ (k + 1)2
− 1, hence k2 < n + 1 ≤ (k + 1)2
. We deduce that p (n + 1) ∈ {k, k + 1} . Then p (n + 1)
p (n) ∈ N if and only if p (n) = k/
2 �= 1 and p (n + 1) = (k + 1)2
, that is, n = (k + 1)2− 1, k ∈ N, k ≥ 2. Since n ≤ 100, it results k ≤ 9. The requested set is {8, 15, 24, 35, 48, 63, 80, 99} .
Problem 3. Let ABCDEF be a regular hexagon with side length a. At point A, the perpendicular AS, with length 2a√3, is erected on the hexagon’s plane. The points M, N, P, Q, and R are the projections of point A onto the lines SB, SC, SD, SE, and SF, respectively.
a) Prove that the points M, N, P, Q, R lie in the same plane.
b) Find the measure of the angle between the planes (M N P ) and (ABC). Petru Braica
Solution. a) Using the Three Perpendiculars Theorem, from SA ⊥ (ABC) and AB ⊥ BD, it results SB ⊥ BD. Since BD ⊥ AB and BD ⊥ SB, it follows that BD ⊥ (SAB) , hence BD ⊥ AM.
Since AM ⊥ SB, it results AM ⊥ (SBD) , hence AM ⊥ SD. We also have SD ⊥ AP, therefore we obtain that SD ⊥ (AMP ). In a similar way one can show that SD ⊥ (ARP ), SD ⊥ (ANP ) and SD ⊥ (AQP ) , therefore the points M, N, P, Q, R lie in the same plane.
C B A M N D E F S P Q R d
b) Because M R � BF, it follows that the intersection between the planes (M N P ) and (ABC) is the line d parallel to BF and passing through A. Since d ⊥ SA and d ⊥ AD, it results that d ⊥ (SAD) , hence d ⊥ AP. Therefore the angle between the planes (M N P ) and (ABC) equals P AD.
Using the Pythagorean Theorem, one obtains SD = 4a, hence m( P DA) = 60◦ and m( P AD) = 30◦.
Problem 4. Let n ≥ 2 be a positive integer. Determine all possible values of the sum
S = ⌊x2− x1⌋ + ⌊x3− x2⌋ + . . . + ⌊xn− xn−1⌋ ,
where x1, x2, . . . , xn are real numbers whose integer parts are 1, 2, . . . , n,
respectively.
Ion Pˆır¸se
Solution. Let a and b be arbitrary reals. We have ⌊b⌋ − ⌊a⌋ − 1 < b − a < ⌊b⌋ − ⌊a⌋ + 1, hence ⌊b⌋ − ⌊a⌋ − 1 ≤ ⌊b − a⌋ ≤ ⌊b⌋ − ⌊a⌋. Applying this to some consecutive terms of the sequence, we obtain 0 ≤ ⌊xk− xk−1⌋ ≤ 1, for
all k = 2, 3, . . . , n. Thus, 0 ≤ S ≤ n − 1. We claim that the set of all possible values of S is {0, 1, 2, . . . , n − 1} . The value S = n − 1 can be obtained, for instance, for xk= k, k = 1, 2, . . . , n . For the value S = p, with 0 ≤ p ≤ n − 2,
one can take, for instance,
xk = k + 1 k + 1, if 1 ≤ k ≤ n − 1 − p k, if n − p ≤ k ≤ n .
9th GRADE
Problem 1. Find the irrational numbers x with the property that x2+ x and x3+ 2x2are integer numbers.
George Stoica
Solution. Denote x2+ x = a and x3+ 2x2= b. Then b − ax = x2= a − x,
hence x (a − 1) = b − a. Since x is an irrational number and a, b are integers, we deduce that a = b = 1, and, finally, x = −1±2√5.
Problem 2. Let ABC be a triangle and let the points D ∈ (BC), E ∈ (AC), F ∈ (AB) be such that
DB DC = EC EA = F A F B.
The halflines (AD, (BE, and (CF intersect the circumcircle of ABC at points M, N and P. Prove that the triangles ABC and M N P share the same centroid if and only if the areas of the triangles BM C, CN A and AP B are equal. Marin Ionescu C B A M N D E F P
Solution. First, observe that −−→AD +−−→BE +−−→CF = 0 (∗) . Denoting the areas of the triangles ABC, BM C, CN A and AP B by s, sa, sb, and sc, respectively,
we have DMAD = sa s, hence AM AD = s+sa s , which implies −−→ AM = s+sa s · −−→ AD, and the analogous equalities. Triangles ABC and M N P share the same centroid if and only if−−→AM +−−→BN +−−→CP = 0, hence if and only if
s + sa s · −−→ AD +s + sc s · −−→ BE +s + sc s · −−→ CF = 0, or, equivalently, sa·−−→AD + sb·−−→BE + sc·−−→CF = 0.
Using (∗) , the latter rewrites as
(sa− sc) ·−−→AD + (sb− sc) ·−−→BE = 0,
and since−−→AD and−−→BE are non collinear vectors, this holds if and only if sa =
sb= sc.
Problem 3. The medians AD, BE and CF of triangle ABC intersect at G. Let P be a point lying in the interior of the triangle, not belonging to any of its medians. The line through P parallel to AD intersects the side BC at A1. Similarly one defines the points B1 and C1. Prove that
−−→ A1D +−−→B1E +−−→C1F = 3 2 −−→ P G. Bogdan Enescu A B C C C B B A D A A P A B A C B 1 C
Solution. Let us draw through point P parallels to the triangle’s sides. If the parallels to AB and AC intersect the side BC at points AB and AC, then
the triangles ABC and P ABAC are similar, and since P A1 is parallel to the
median AD, it follows that A1is the midpoint of the line segment ABAC. With
the notations in the figure, we have
2−−→A1D =−−→A1B +−−→A1C =A−−−→BB +−−−→ACC =−−−→P CB+−−−→P BC.
and the similar equalities. We deduce that the sum 2(−−→A1D +−−→B1E +−−→C1F )
equals
−
→v =−−−→P CB+−−−→P BC+−−−→P AC+−−−→P CA+−−−→P BA+−−−→P AB. But−−−→P CA+−−−→P BA=−→P A and adding up all similar equalities yields
−
→v =−→P A +−−→P B +−−→P C = 3−−→P G, hence the conclusion.
Problem 4. Find all functions f : N∗→ N∗ with the properties:
a) f (m + n) − 1 divides f (m) + f (n), for all m, n ∈ N∗;
b) n2− f (n) is a square, for all n ∈ N∗.
Lucian Dragomir
Solution. It is not difficult to check that f (1) = 1 and f (2) = 3. We prove inductively that f (n) = 2n − 1, for all n. Indeed, assume that for some k > 1, f (k) = 2k − 1. Then f (k + 1) − 1 divides 2k, hence f (k + 1) ≤ 2k + 1. If the inequality is strict, then (k + 1)2− f (k + 1) > (k + 1)2− (2k + 1) = k2, and
since it is a square, we have (k + 1)2− f (k + 1) ≥ (k + 1)2, a contradiction. We deduce that f (k + 1) = 2k + 1, as desired.
10th GRADE
Problem 1. Solve in complex numbers the equation |z − |z + 1|| = |z + |z − 1||.
Dan Negulescu
Solution. Writing the equation as |z − |z + 1||2 = |z + |z − 1||2, and using
|w|2= w · w, for any complex number w, yields the equivalent form (z + z) (|z − 1| + |z + 1| − 2) = 0.
We deduce that either z + z = 2 Re z = 0, hence z = ia, for some real a, or |z − 1| + |z + 1| = 2. In this second case, we deduce that, in the complex plane, the sum of distances from the point z to the points −1 and 1 equals 2, which is possible if and only if z is a real number, lying on the line segment [−1, 1] .
Problem 2. Solve in real numbers the equation x + log2 1 + 5x 3x+ 4x = 4 + log1/2 1 + 25x 7x+ 24x . Cristinel Mortici
Solution. Rewrite the equation as x + log2 1 + 1 (3/5)x+ (4/5)x = 4 + log1/2 1 + 1 (7/25)x+ (24/25)x ,
and observe that the left hand side is an increasing function, while the right hand side is a decreasing one. We conclude that the equation has at most a solution, and it is not difficult to guess it: x = 2.
Problem 3. Let p and n be positive integers, with p ≥ 2, and let a be a real number such that 1 ≤ a < a + n ≤ p. Prove that the set
⌊log2x⌋ + ⌊log3x⌋ + · · · + logpx x ∈ R, a ≤ x ≤ a + n has exactly n + 1 elements.
Ioan B˘aetu
Solution. Let f (x) =pk=2⌊logkx⌋ and let M = {f(x) | x ∈ [a, a+n]}. It is
easy to show that if k ≥ 2 is a positive integer, then ⌊logk⌊x⌋⌋ = ⌊logkx⌋ . This
implies that f (x) = f (⌊x⌋), for all x ∈ [1, ∞), and hence M = {f(x) | x ∈ S}, where S = {⌊a⌋ , ⌊a⌋ + 1, . . . , ⌊a⌋ + n} has n + 1 elements. On the other hand, for s ∈ S, s < ⌊a⌋ + n ≤ p, we have s + 1 ∈ {2, 3, . . . , p}, and
f (s+1)−f(s) = p k=2 (⌊logk(s + 1)⌋−⌊logks⌋) ≥ logs+1(s + 1) − logs+1s = 1, therefore f (s + 1) > f (s), and this proves that M has exactly n + 1 elements.
Problem 4. Find all functions f : Q → Q such that f (x + 3f (y)) = f (x) + f (y) + 2y, for all x, y ∈ Q.
Nicu¸sor Berbecel
Solution. We claim that the solutions are f1(x) = x and f2(x) = −2x/3,
for all real x.
Set x = y − 3f(y) to obtain f(y − 3f(y)) = −2y, y ∈ Q. Replacing y with y − 3f(y) in the initial equation gives f(x − 6y) = f(x) − 2y + 2(y − 3f(y)), hence f (x − 6y) = f(x) − 6f(y), for all x, y ∈ Q. Set now x = y = 0 to get f (0) = 0 and replace x = 6y to obtain f (6y) = 6f (y), for all y ∈ Q.
We derived that f (x−6y) = f(x)−f(6y), which, for u = 6y and v = x−6y, yields f (u + v) = f (u) + f (v), for all u, v ∈ Q. The solution of this classical functional equation is f (x) = xf (1), x ∈ Q. On the other hand, by setting x = y = 1 in the initial equation we obtain 3f2(1) − f(1) − 2 = 0, hence
11th
GRADE
Problem 1. a) Give an example of matrices A and B from M2(R), such
that A2+ B2= 2 3 3 2 .
b) Let A and B be matrices from M2(R), such that A2+ B2=
2 3 3 2
. Prove that AB �= BA.
Vladimir Cerbu, Mihai Piticari
Solution. a) An example is A = 3/2 1 1 1 1 and B = 0 1 −1 0 . b) If the matrices A and B commute, then A2+ B2= (A + iB)(A − iB), hence
|det(A + iB)|2= det(A + iB) det(A − iB) = det(A2+ B2) = det 2 3 3 2 = −5, a contradiction.
Problem 2. a) Let f : R → R be a function such that g : R → R, g(x) = f (x) + f (2x), and h : R → R, h(x) = f(x) + f(4x), are continuous functions. Prove that f is also continuous.
b) Give an example of a discontinuous function f : R → R, with the fol-lowing property: there exists an interval I ⊂ R, such that, for any a in I, the function ga: R → R, ga(x) = f (x) + f (ax), is continuous.
Dorel Mihet¸
Solution. a) Since g and h are continuous, and f (x) = (g(x) − g(2x) + h(x)) /2, x ∈ R, it follows that f is continuous, as well.
b) The function f : R → R, f (x) = −1, x < 0, 0, x = 0, 1, x > 0,
is discontinuous at 0, but ga: R → R, ga(x) = f (x) + f (ax) = 0, is continuous
on R, for all a < 0.
Problem 3. a) Let A be a matrix from M2(C), A �= aI2, for any a ∈ C.
Prove that the matrix X from M2(C) commutes with A, that is, AX = XA, if
and only if there exist two complex numbers α and α′, such that X = αA+α′I 2.
b) Let A, B and C be matrices from M2(C), such that AB �= BA, AC = CA
and BC = CB. Prove that C commutes with all matrices from M2(C).
Adrian Troie
Solution. a) Clearly, if X = αA+α′I2, then X and A commute. Conversely,
let A = � a1 a2 a′ 1 a′2 � and X = � x1 x2 x′ 1 x′2 �
. The equality AX = XA implies
a2x′1= a′1x2, (1)
(a1− a′2)x2+ a2(x′2− x1) = 0, (2)
(a1− a′2)x′1+ a′1(x′2− x1) = 0. (3)
Since A �= aI2, a ∈ C, either one of a2, a′1 is non-zero, or a2= a′1 = 0 and
a1�= a′2.
In the first case, if, for instance, a2�= 0, we obtain
X = x2 a2 A +�x1− a1 a2 x2 � I2,
while in the second case
X = x1− x′2 a1− a′2
A + a1x′2− a′2x1 a1− a′2
I2.
b) We prove that C = γI2, for some complex number γ, hence C commutes
with all matrices in M2(C). Suppose the contrary. Since A and C commute,
there exist α and α′, such that A = αC + α′I
2. Similarly, there exist β and β′,
Problem 4. Let f : N → N∗be a strictly increasing function. Prove that:
a) there exists a decreasing sequence of positive real numbers, (yn)n∈N,
converging to 0, such that yn≤ 2yf (n), for all n ∈ N;
b) if (xn)n∈Nis a decreasing sequence of real numbers, converging to 0, then
there exists a decreasing sequence of real numbers (yn)n∈N, converging to 0,
such that xn≤ yn ≤ 2yf (n), for all n ∈ N.
George Stoica
Solution. a) Since f (0) > 0 and f is strictly increasing, it follows that f (n) > n, for all n ∈ N. Consider the sequence of non-negative integers (nk)k∈N, defined by n0 = 0 and nk = f (nk−1), k ∈ N∗. The properties of
f imply that the sequence is strictly increasing. We define the decreasing se-quence (yn)n∈N, by yn= 2−k, for all n with nk ≤ n < nk+1, k ∈ N, obviously
convergent to 0.
It suffices to prove that yn ≤ 2yf (n), n ∈ N, for nk ≤ n < nk+1, k ∈ N.
Since f strictly increasing, nk+1 = f (nk) ≤ f(n) < f(nk+1) = nk+2, hence
yf (n)= 2−k−1= yn/2.
b) Obviously, xn≥ 0, for all n ∈ N. Using the previously defined sequence
(nk) , we define the decreasing sequence of positive reals (zk)k∈N, as follows:
z0 = x1 and zk = max (xnk, zk−1/2), k ∈ N∗. The monotony of this sequence
follows inductively. Moreover, (zk)k∈Nconverges to 0: if zk = xnk, for infinitely
many k’s, then zk→ 0 because it is decreasing and xn → 0; if zk= xnk, only for
finitely many k’s, then zk= zk−1/2 from some k onwards, and again, zk→ 0.
Finally, we define the sequence (yn)n∈N by yn = zk, nk ≤ n < nk+1,
k ∈ N. Clearly, the sequence decreases to 0. In order to prove the inequalities xn ≤ yn ≤ 2yf (n), n ∈ N, it suffices to check them for nk ≤ n < nk+1, k ∈ N.
Obviously, xn ≤ xnk ≤ zk = yn. On the other hand, nk+1 = f (nk) ≤ f(n) <
f (nk+1) = nk+2, since f is strictly increasing, hence yf (n) = zk+1 ≥ zk/2 =
yn/2.
12th GRADE
Problem 1. For each positive integer n the function fn : [0, n] → R is
and find lim n→∞ 1 n n 0 fn(x)dx. ***
Solution. The function fn is locally constant, hence Riemann integrable.
Next, we have n 0 fn(x)dx = k−1 i=0 i+1 i fn(i)dx = n−1 i=0 arctg i. Applying Stolz-Ces`aro theorem, we obtain
lim
n→∞
arctg 1 + arctg 2 + . . . + arctg n
n = limn→∞arctg n =
π 2.
Problem 2. Let f : [0, 1] → R be a derivable function, with continuous derivative, and let sn= nk=1f
k n
.
Prove that the sequence (sn+1− sn)n∈N∗ converges to
1
0 f (x)dx.
***
Solution. Using the mean value theorem, we obtain sn+1− sn= n+1 k=1 f k n + 1 − n k=1 f k n = f (1) − n k=1 f k n − fn + 1k = f (1) −n(n + 1)1 n k=1 kf′(xk), for some xk, n+1k < xk< kn, k = 1, 2, . . . , n. If f′ ≥ 0, then xkf′(xk) n + 1 ≤ kf′(xk) n(n + 1) ≤ xkf′(xk) n , k = 1, 2, . . . , n, hence 1 n + 1 n k=1 xkf′(xk) ≤ 1 n(n + 1) n k=1 kf′(xk) ≤ 1 n n k=1 xkf′(xk). Because 0 ≤
x1≤ 1/n ≤ x2≤ . . . ≤ xn≤ n/n is a tagged partition of the interval [0, 1], it
follows that lim n→∞ 1 n(n + 1) n k=1 kf′(x k) = 1 0 xf′(x)dx = xf (x)1 0− 1 0 f (x)dx,
hence the conclusion.
If f′ takes negative values, replace f with g : x �→ f(x) + Mx, where
M = sup |f′|. As above, for
tn= n k=1 g k n we have (tn+1− tn)n → 1 0 g(x)dx = 1 0 f (x)dx +M 2 and tn+1− tn= sn+1− sn+ M 2 , therefore the conclusion holds in this case as well.
Problem 3. Let (A, +, ·) be an unit ring with the property: for all x ∈ A, x + x2+ x3= x4+ x5+ x6.
a) Let x ∈ A and let n ≥ 2 be an integer such that xn = 0. Prove that
x = 0.
b) Prove that x4= x, for all x ∈ A.
George Stoica
Solution. a) From the given equality we derive that xn−1= xn(x4+ x3+ x2− x − 1) = 0
and, step by step, that xn−2= xn−3= . . . = x = 0.
b) Rewrite the given equation as
x(x3− 1)(x2+ x + 1) = 0. It follows that
(x4− x)2= x2(x − 1)(x3− 1)(x2+ x + 1) = 0,
hence x4− x = 0.
Problem 4. Let (G, ·) be a group with no elements of order 4, and let f : G → G be a group morphism such that f(x) ∈ {x, x−1}, for all x ∈ G.
Prove that either f (x) = x for all x ∈ G, or f(x) = x−1 for all x ∈ G.
Solution. Assume, by way of contradiction, that there exist a, b ∈ G such that f (a) = a �= a−1 and f (b) = b−1�= b. Then f(ab) = f(a)f(b) = ab−1�= ab,
therefore f (ab) = (ab)−1 = b−1a−1. It follows that ab−1 = b−1a−1, which
implies b−1= ab−1a.
Next, f (ab2) = f (a)f2(b) = ab−2. If f (ab2) = ab2, then ab−2 = ab2,
therefore either b2 = e, which contradicts b �= b−1, or ord(b) = 4, which is
impossible. Thus, f (ab2) = (ab2)−1= b−2a−1, hence ab−2= b−2a−1. It follows
FINAL ROUND
5th
GRADE
Problem 1. Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers.
Marian Ciuperceanu
Solution. Let the three odd consecutive numbers be 2p + 1, 2p + 3 and 2p + 5, where p is a positive integer. Then one of these numbers is divisible by 3: if p = 3k, with integer k, then 2p + 3 = 2(3k) + 3 = 6k + 3 = M3; if p = 3k + 1, with integer k, then 2p + 1 = 2(3k + 1) + 1 = 6k + 3 = M3; if p = 3k + 2, with integer k, then 2p + 5 = 2(3k + 2) + 5 = 6k + 9 = M3.
In all the cases the product P is a multiple of 3, therefore P = 3a = (a − 1) + a + (a + 1), with integer a.
Problem 2. We will say that a positive integer n is subject to an interesting change if it is multiplied by 2 and the result is increased by 4, a special change if it is multiplied by 3 and the result is increased by 9 and an awesome change if it is multiplied by 4 and the result is increased by 16.
a) Show that there exists a positive integer which after three changes, the first – interesting, the second – special and the third – awesome, becomes 2020. b) Find all positive integers with the property that after two changes of different types, selected among the three above, becomes 2014.
Lucian Dragomir
Solution. a) Before the last change the number must be (2020 − 16) : 4 = 501; the previous number must be (501 − 9) : 3 = 164 and the required number is (164 − 4) : 2 = 80.
b) An interesting change produces a multiple of 2, a special change gives a multiple of 3 and an awesome change yields a multiple of 4. Since 2014 is neither a multiple of 3, nor a multiple of 3, the last change must be an interesting one. So, the second number must be (2014 − 4) : 2 = 1005. Since this number is odd, the first change must be special, and the initial number is (1005 − 9) : 3 = 332.
Problem 3. Show that there exists a multiple of 2013 which ends in 2014. Mihaela Berindeanu
Solution. Consider the numbers a1, a2, . . . , a2014, where ak is the 4k- digit
number whose digits are k groups of the form 2014, 1 ≤ k ≤ 2014. Since there are only 2013 possible remainders after the division by 2013, two different a-s mua-st give the a-same remainder in thia-s divia-sion, hence there exia-sta-s i, j, with 1 ≤ j < i ≤ 2014, so that ai− aj is divisible by 2013. Since
ai− aj= 20142014 . . . 2014 i−j times 2014 00 . . . 0 4j times = 20142014 . . . 2014 i−j times 2014 ·100 . . . 0 4j times = ai−j· 104j
and 104j and 2103 are relatively prime, 2013 must divide a i−j.
Problem 4. One hundred boxes are labeled from 1 to 100. Each box has at most 10 stones.The difference of the numbers of stones for every two boxes labeled with consecutive numbers is 1. The boxes labeled 1, 4, 7, 10, . . . , 100 contain a total of 301 stones. Find the maximum possible number of stones contained by the 100 boxes.
Gabriel Popa
Solution. Since the difference of the number of stones in every two con-secutive boxes is 1, two concon-secutive boxes contain at most 19 stones. We know the number of stones in the 34 boxes #1, #4, #7, . . . , #100 and if we group the remaining 66 boxes in pairs of consecutive boxes, we obtain at most 301 + 33 · 19 = 928 stones.
This value is indeed obtained if we have 10 groups of the form (9, 10, 9, 8, 9, 10), then 6 groups of the form (9, 10, 9, 10, 9, 10) and, at the end, 9, 10, 9, 8.
6th
GRADE
Problem 1. Denote A = {1000, 1001, 1002, . . . , 2014}. Find the maximum number of elements of a subset of A which contains only perfect squares pairwise relatively prime.
Marcel Neferu
Solution. If n ∈ A and n = p2, then 1000 ≤ p2≤ 2014, that is 32 ≤ p ≤ 44.
The largest subset of A whose elements are perfect squares is
B = 322, 332, 342, 352, 362, 372, 382, 392, 402, 412, 422, 432, 442 .
We must choose among them the maximum number of pairwise prime num-bers. Consider the partition of B into the sets C1= {322, 342, 362, 382, 402, 422,
442}, C
2 = {332, 392}, C3 = {352}, C4 = {372}, C5 = {412}, C6 = {432}. If
we choose 7 or more elements of B, then two of them are in the same Ci, so
they are not co-prime. So we cannot take more than 6 elements; an example is {322, 332, 352, 372, 412, 432}.
Problem 2. An integer n > 1 will be called p-periodic if n1 is a repeating decimal fraction, whose shortest period has length p and begins immediately after the decimal point. For instance, 19 = 0.111 . . . is 1-periodic and 111 = 0.090909 . . . is 2-periodic.
a) Find all p-periodic numbers n so that the first digit of the period of 1 n is
not nil.
b) Find the largest 4-periodic prime.
Marius Perianu
Solution. a) The first digit of the period is at least 1 if and only if 1 n ≥
1 10,
that is n ≤ 10. Since the prime factors of n must be different from 2 and 5, it follows that n ∈ {3, 7, 9}; indeed, in this case 1
3 = 0,(3), 1
7 = 0,(142857) and 1
9 = 0,(1).
b) Let p be the largest 4-periodic prime; then 1 p =
m
9999, with 1 ≤ m ≤ 9998.
This yields pm = 9999 = 32· 11 · 101. So, the candidate for the largest such
prime is p = 101, which is acceptable, because 1
101 = 0,(0099).
Problem 3. If n is a positive integer, we will call a triple (x, y, z) of positive integers of type n if x + y + z = n. Denote s(n) the number of triples of type n.
b) Find the smallest positive n so that s(n) > 2014.
Nicu¸sor Berbecel
Solution. Let us count the triples (x, y, z) of positive integers such that x + y + z = n, where n ≥ 3 is a given integer. One can assign to x any value from 1 to n − 2. If x = 1, then, taking into account that z ≥ 1, y can take n − 2 values: from 1 to n − 2. If x = 2, then y can take n − 3 values, . . . , if x = n − 2, then y can take only the value 1. This shows that the number of triples of type n is
s(n) = 1 + 2 + 3 + . . . + (n − 3) + (n − 2) = (n − 2)(n − 1)2 .
a) If there exists n such that s(n) = 14, then (n − 2)(n − 1) = 28, which is impossible: if n ≤ 6, then (n−2)(n−1) ≤ 20 and if n ≥ 7, then (n−2)(n−1) ≥ 30.
b) We have to find the smallest n such that (n − 2)(n − 1) > 4028. Since (n − 1)2 > (n − 2)(n − 1), n − 1 must be at least the smallest perfect square
larger than 4028.
Since 4028 = 22· 1007 and 210= 322= 1024 > 1007 > 31.52, it follows that
n − 1 ≥ 64, that is n ≥ 65. From (65 − 2)(65 − 1) = 4032 > 4028 follows that n = 65.
Problem 4. In triangle ABC points M, N ∈ (AB), P, Q ∈ (BC) and S, R ∈ (AC) are taken such that AM = CR, AN = CS, M QB ≡ RQC and
N P B ≡ SP C.
Show that if M Q + QR = N P + P S, then triangle ABC is isosceles. Mircea Fianu, Traian Preda
Solution. Denote R′ and S′ the reflection of R, respectively S, across the
line BC. Then RQ = R′Q and RQC ≡ R′QC. Since M QB ≡ RQC, it follows
that M QB ≡ R′QC, and because B, Q, C are collinear, so are M , Q, R′.
This yields M Q + QR = M Q + QR′ = M R′. In the same way N P + P S = N P + P S′= N S′. Now M Q + QR = N P + P S implies M R′= N S′.
The symmetry also implies RCQ ≡ R′CQ and SCP ≡ S′CP , so RCQ =
CR = CR′ and CS = CS′ and, since CR = AM and CS = AN , M N = R′S′. A M S R C N B P Q S' R'
Therefore △NS′M ≡ △R′M S′ (case S-S-S), whence �N M S′ ≡ �M S′R′.
This shows that AB � CS′, therefore �ABC ≡ �BCS′. Now �BCS′ ≡ �BCA
implies �ABC ≡ �ACB, q.e.d.
7th GRADE
Problem 1. Find all primes p and q, with p ≤ q, so that p (2q + 1) + q (2p + 1) = 2 p2+ q2 .
Lucian Dragomir
Solution. The equality can be written p + q = 2 (p − q)2, which shows that p is odd.
If p ≥ 5, then p and q leave remainder 1 or 2 when divided by 3.
We will show that in this case the equality is impossible. Indeed, if p and q leave the same remainder mod 3, then 3 | 2 (p − q)2 and 3 ∤ p + q; if p and q leave different remainders, then 3 ∤ 2 (p − q)2and 3 | p + q.
Finally, if p = 3, then q = 5.
Problem 2. Construct outside the square ABCD the rhombus BCM N , with ∠BCM an obtuse angle. The straight lines BM and AN meet at P . Prove that DM ⊥ CP and triangle DP M is right and isosceles.
Andrei Bud
Solution. Isosceles triangles CDM and CBM yield �CM D ≡ �CDM and �
m( BDM ) + m( DBM ) = m( DM B) + 45◦+ m( CDM ) + 45◦+ m( CBM ) =
90◦+ 2m( DM B), whence m( DM B) = 45◦.
Since quadrilateral ADM N is a parallelogram, m( M P N ) = m( DM B) = 45◦. The point P is on the perpendicular bisector of the segment [CN ], so
M P N ≡ M P C, whence m( N P C) = 90◦, that is CP ⊥ AN, implying CP ⊥
DM . C N M P B A D
In the isosceles triangle CDM the straight line CP is the perpendicular from C onto DM , so it is the perpendicular bisector of the segment [DM ]. This shows that triangle DP M is isosceles with vertex P . So, from m( DM P ) = 45◦
follows m( DP M ) = 90◦.
Problem 3. Find all positive integers n so that 17n+ 9n2
= 23n+ 3n2
. Marius Perianu
Solution. Clearly n = 0 and n = 1 are solutions. If n ≥ 2, then n2≥ 2n, hence 9n2
− 3n2
= 3n2
(3n2
− 1) ≥ 32n 32n− 1 =
81n− 9n. Since 81n− 9n > 23n− 17n, the equation does not have solutions
n ≥ 2.
Problem 4. Construct outside the square ABCD the right isosceles tri-angle ABE, with hypotenuse [AB]. Denote N the midpoint of the segment [AD] and {M} = CE ∩ AB, {P } = CN ∩ AB, {F } = P E ∩ MN. Take on the straight line F P the point Q so that [CE is the bisector of the angle ∠QCB. Prove that M Q ⊥ CF .
Sorin Furtun˘a
Solution. Clearly P A = P B/2, so P A = AB = BC. This leads to △AP E ≡ △BCE (SAS), which implies CE = P E and BEC ≡ AEP . Now
m( BEC) + m( AEC) = 90◦ yields m( AEC) + m( AEP ) = 90◦, so m( CEP ) =
90◦. This shows that ECP is a right isosceles triangle, so m( CP E) = 45◦.
P D Q A N B C F E M
Also △CEQ ≡ △P EM (AAS), whence EQ = EM, so EMQ is a right isosceles triangle. This gives m( EQM ) = 45◦ = m( EP C), therefore M Q �
CP . Menelaus’ Theorem for the triangle CP E and the transversal N F yields
CN N P ·
F P F E ·
M E
M C = 1. If we denote R the orthogonal projection of E onto AB,
then △CBM ∼ △ERM, whence M EM C = ER BC =
1 2.
From N P = CN follows F PF E = 2, hence P E = EF , therefore CE is altitude and median in △CF P , that is this triangle is isosceles. This gives m( F CP ) = 90◦, that is CP ⊥ CF , whence MQ ⊥ CF .
8th GRADE
Problem 1. Let a, b, c ∈ (0, ∞). Prove the inequality a −√bc a + 2 (b + c)+ b −√ca b + 2 (c + a)+ c −√ab c + 2 (a + b) ≥ 0. Nicolae Papacu
Solution. The AM-GM inequality leads to a −√bc a + 2 (b + c)≥ a −b+c2 a + 2 (b + c) = 2a − b − c 2 (a + 2b + 2c) .
So, the left-hand part of the given inequality is at least 2a − b − c 2 (a + 2b + 2c)+ 2b − c − a 2 (2a + b + 2c)+ 2c − a − b 2 (2a + 2b + c) not = S.
Denote a + 2b + 2c = 5x, 2a + b + 2c = 5y and 2a + 2b + c = 5z. Then a = −3x + 2y + 2z, b = 2x − 3y + 2z, c = 2x + 2y − 3z and 2a − b − c 2 (a + 2b + 2c) = −10x + 5y + 5z 10x = 1 2 y x+ z x− 2 . This and the two similar relations leads to
S = 1 2 y x+ z x− 2 +1 2 x y + z y − 2 +1 2 x z + y z− 2 =1 2 x y + y x + x z + z x + z y + y z − 6 ≥ 0, whence the conclusion.
Problem 2. The cube ABCDA′B′C′D′ has edges of length a. Take the
points E ∈ (AB) and F ∈ (BC) so that AE + CF = EF .
a) Find the measure of the angle of the planes (D′DE) and (D′DF ).
b) Compute the distance from D′ to the straight line EF .
Sorin Peligrad, Adrian T¸ urcanu
D A B C H A' B' C' D' E F P
Solution. a) Extend the segment BA with AH ≡ F C. Then ∆DAH ≡ ∆DCF (SAS), whence HDA ≡ F DC, so m( HDF ) = 90◦. Then [DH] ≡
[DF ], leads to ∆DHE ≡ ∆DF E (SSS) and from here m( F DE) = m( HDE) = 45◦. Since F D ⊥ DD′and ED ⊥ DD′, it follows that m((D′DE) , (D ′DF )) =
b) Denote P the orthogonal projection of the point D onto EF . The Three Perpendiculars Theorem yields D′P ⊥ EF , hence d (D′, EF ) = D′P . The
congruence ∆DHE ≡ ∆DF E gives DP = AD = a. Finally, from Pythagoras’ Theorem, D′P = a√2.
Problem 3. Find the smallest integer n for which the set A = {n, n + 1, n + 2, . . . , 2n} contains five elements a < b < c < d < e so that
a c = b d = c e. Mircea Fianu
Solution. Let p, q ∈ N∗, (p, q) = 1 so that a c = b d = c e = p q. Obviously,
p < q. Since a, b and c are divisible by p and c, d, e are divisible by q, there exists m ∈ N∗ so that c = mpq.
Let us find the minimal value of the difference e − a, for given p, q. This is obtained when a, b, c are consecutive multiples of p and c, d, e are consecutive multiples of q, that is a = mpq − 2p and e = mpq + 2q. From ce =
mpq mpq+2q =
p q
follows m (q − p) = 2, hence m ∈ {1, 2}.
Condition n ≤ a < e ≤ 2n ≤ 2a implies 2a ≥ e, that is 2mpq − 4p ≥ mpq + 2q, or mpq ≥ 4p + 2q. (∗)
If m = 1, then q − p = 2, hence q = p + 2. Relation (∗) yields (p − 2)2≥ 8, whence p ≥ 5. For p = 5 and q = 7 we get a = 25, b = 30, c = 35, d = 42, e = 49 and, since n ≤ a < e ≤ 2n, n = 25.
If m = 2, then q − p = 1, so q = p + 1. Relation (∗) yields (p − 1)2 ≥ 2, whence p ≥ 3. For p = 3 and q = 4 we get a = 18, b = 21, c = 24, d = 28, e = 32 and, since n ≤ a < e ≤ 2n, n ∈ { 16, 17, 18 }. Therefore, nmin= 16.
Problem 4. Prove that three discs of radius 1 cannot cover entirely a square surface of side 2, but they can cover more than 99.75% of it.
Traian Preda
Solution. Denote ABCD the square and S1, S2, S3 the discs.
Suppose that S1, S2 and S3 cover the whole square. Since there are three
discs and the square has four vertices, one of the discs must cover two vertices of the square, say S1 covers A, B. Then [AB] is a diameter for S1, so S1
one of the other discs, say S2. Then S2 cannot cover any point from (AD),
therefore (AD) ⊂ S3. In this case [AD] is a diameter of S3, so S3cannot cover
any point from (BC), therefore S2 must cover (BC). This shows that [BC]
must be a diameter of S2. But, in this case, no point from (CD) is covered –
contradiction.
b) Take M ∈ (AC) so that AM = 2. Denote P and R the orthogonal projections of M onto AB and onto AD. Let T ∈ BC and U ∈ DC be such that P T = RU = 2. D A B C P M R
U
T
X
We take as S1, S2, S3 the discs of diameters [AM ], [P T ], [RS]. Denote
X the point on [AC] for which XT ⊥ BC (and XU ⊥ CD). Then S1 covers
the square surface AP M R , S2 covers the pentagonal surface BP M XT and
S3 covers the pentagonal surface DRM XU , so the points not covered by the
three discs are inside the square CU XT . It is enough to prove that
area[CU XT ] < 0.25% · area[ABCD],
which is equivalent to CT < BC/20 = 0.1, or BT > 1.9. Simple considerations yield AP =√2, BP = 2 −√2, BT2= 4 − 2 −√2 2 = 4√2 − 2. We are left
to prove BT > 1.9, that is 4√2 − 2 > 1.92, or √2 > 1.4025, which is true.
9th
GRADE
Problem 1. Let n be a natural number. Find the integers x, y, z such that x2+ y2+ z2= 2n(x + y + z) .
Petre Simion, Mircea T¸ eca
Solution. If n = 0, using the inequalities x2 ≥ x and its analogues, we
If n ≥ 1, then 2 divides x2+ y2+ z2, and hence, either the three numbers
are even, or one is even and the others are odd. In the former case, if we take x = 2x1+ 1, y = 2y1+ 1, z = 2z1, we get 4 x2 1+ x1+ y12+ y1+ z12 + 2 = 4 (x1+ y1+ z1+ 1) , which is a contradiction.
Let us consider the case when the numbers x, y, z are even. For x = 2x1,
y = 2y1, z = 2z1, we get x21+ y12+ z12= 2n−1(x1+ y1+ z1) , and thus, if n = 1,
then x, y, z ∈ {0, 2}.
For n > 1, using the same argument, we deduce that if x = 2nxn, y = 2nyn,
z = 2nz
n, then xn, yn, zn ∈ Z and x2n+ yn2+ zn2 = xn+ yn+ zn, whence xn,
yn, zn∈ {0, 1} , and thus x, y, z ∈ {0, 2n}.
Problem 2. Let a be a natural odd number which is not a perfect square. If m and n are strictly positive integers, prove that
a) {m (a +√a)} �= {n (a −√a)} , b) [m (a +√a)] �= [n (a −√a)].
Vasile Pop
Solution. a) As ma, na are natural numbers, the equality implies {m√a} = {−n√a}.
Two numbers have the same fractional part if and only if their difference is an integer, whence (m + n)√a ∈ Z, which is absurd.
b) Again, let us as suppose that there is a natural number N , for which there exist two not equal numbers m, n which are different from zero, such that N = [m(a +√a] = [n(a −√a]. Then N ≤ m(a +√a) < N + 1 and N ≤ n(a −√a) < N + 1; moreover, the inequalities are strict, because the terms in the middle are irrational numbers.
We rewrite the inequalities as N a +√a < m < N + 1 a +√a, N a −√a < n < N + 1 a −√a, and thus, by addition, we get N 2
a − 1 < m + n < (N + 1) 2 a − 1. From here, N < a − 1
2 (m + n) < N + 1, which is a contradiction, because the term in the middle is natural (a is odd).
Remark. We could get the problem by using Beatty Theorem, which states that if α and β are positive irrational numbers and 1
α+ 1
β = 1 then the sets [nα] and [nβ] determine a partition of N.
Problem 3. Let P and Q be the midpoints of the diagonals BD and AC of the quadrilateral ABCD. Consider the points M ∈ (BC), N ∈ (CD), R ∈ (P Q) and S ∈ (AC) such that BMM C =DN
N C = P R RQ =
AS
SC = k. Prove that the centroid of the triangle AM N lies on the segment [RS].
Andrei Bud
Solution. Let G be the centroid of the triangle AM N . Then we have −→ GR = −−→ GP + k−−→GQ 1 + k = −−→ GB +−−→GD + k(−→GA +−−→GC) 2 (1 + k) and −→ GS = −→ GA + k−−→GC 1 + k . On the other hand,
0 =−→GA +−−→GM +−−→GN =−→GA + −−→ GB + k−−→GC 1 + k + −−→ GD + k−−→GC 1 + k ,
from where, we deduce that (1 + k)−→GA +−−→GB + 2k−−→GC +−−→GD = 0, whence we get−→GS + 2−→GR = 0, and thus G, R and S are collinear points.
Problem 4. Let ABCD a cyclic quadrilateral, inscribed in the circle of diameter AC. We suppose that there exist the points E ∈ (CD) and F ∈ (BC) such that the lines AE and DF are perpendicular on the lines AF and BE respectively. Prove that AB = AD.
Bogdan Enescu
Solution. From the perpendicularity conditions, we have −→ AE ·−−→DF = 0 ⇔−→AE · (−→AF −−−→AD) = 0, −→ AF ·−−→BE = 0 ⇔−→AF · (−→AE −−−→AB) = 0, whence −→ AE ·−→AF =−→AE ·−−→AD, −→AE ·−→AF =−→AF ·−−→AB,
and thus−→AE ·−−→AD =−→AF ·−−→AB. Equivalently, we have −−→AD +−−→DE
·−−→AD = −−→AB +−−→BF ·−−→AB ⇔ AD2+−−→AD ·−−→DE = AB2+−−→AB ·−−→BF .
But ∠ADE = ∠ABF = 90◦, and hence −−→AD ·−−→DE =−−→AB ·−−→BF = 0, that is
AB = AD.
10thGRADE
Problem 1. Let n be a strictly positive integer. For any natural number k, we denote by a(k) the number of natural divisors d of k such that k ≤ d2≤ n2.
Compute the sum
n2
k=1
a(k).
***
Solution. We will get the result by doubly-counting the elements of the set S(n) = {(k, d) | d divides k, k ≤ d2≤ n2, 1 ≤ k ≤ n2}.
For 1 ≤ k ≤ n2, let A(k) be the set of the natural divisors d of k such that
k ≤ d2≤ n2.
A natural number d ∈ {1, 2, . . . , n} belongs to the sets A(d), A(2d), . . . , A(d2) and only to them.
It follows that the contribution of each d in the sum
n2 k=1 a(k) = n2 k=1 |A(k)| is 1 + 1 + · · · + 1 d terms = d. Thus, n2 k=1 a(k) = n d=1 d = n(n + 1)/2.
Problem 2. Consider the function f : N∗ → N∗ which satisfies the
prop-erties:
a) f (1) = 1,
b) f (p) = 1 + f (p − 1) for any prime number p,
c) f (p1p2· · · pn) = f (p1) + f (p2) + · · · + f(pn) for any prime numbers, not
Prove that 2f (n)≤ n3≤ 3f (n), for any natural number n, n ≥ 2.
George Stoica
Solution. As 2 is a prime number, we have f (2) = 1 + f (1) = 2, whence 22≤ 23≤ 32. Then f (3) = 1 + f (2) = 3, and thus 23< 33= 33. We will prove
the inequalities
3 log3n ≤ f(n) ≤ 3 log2n
by mathematical induction on n. Let n be a natural number, n ≥ 3, and let n = p1p2· · · pk be its prime factorization (not necessarily distinct factors). If
n is not a prime number, i.e. k ≥ 2, using the induction hypothesis, f(n) = f (p1) + f (p2) + · · · + f(pk) ≥ 3
k
i=1
log3pi = 3 log3n and f (n) ≤ 3 k
i=1
log2pi=
3 log2n.
If n is a prime number, i.e. k = 1, then n − 1 is an even number and f (n) = 1+f (n−1) = 1+f 2n − 1 2 = 1+f (2)+f n − 1 2 = 3+f n − 1 2 . Then f (n) = 3 + f n − 1 2
≤ 3 + 3 log2n − 12 = 3 log2(n − 1) ≤ 3 log2n
and
f (n) = 3 + f n − 1 2
≥ 3 + 3 log3n − 12 = 3 log33(n − 1)2 ≥ 3 log3n.
Problem 3. Let n be a strictly positive integer and let A = {1, 2, . . . , n}. Find the number of the increasing functions f : A → A which satisfy the property
|f(x) − f(y)| ≤ |x − y|, for any x, y ∈ A.
Vladimir Cerbu, Mihai Piticari
Solution. We remark that f (k + 1) −f(k)not= ak∈ {0, 1}, k = 1, 2, . . . , n−1.
Moreover, as ak ∈ {0, 1}, we have |f(x) − f(y)| ≤ |x − y| for any x, y ∈ A.
completely determined by the n-tuple (f (0), a1, a2, . . . , an−1) ∈ A × {0, 1} ×
· · · × {0, 1} which satisfies the condition
f (0) + a1+ · · · + an−1≤ n.
If we fix f (0) = a and f (n) = b, we have b − a = a1+ · · · + an−1, and thus
the number of the functions f , with the required property, is equal with the number of choices of b − a terms equal to 1 in the sum a1+ · · · + an−1, that is
n(a, b) = Cn−1b−a.
Thus, the number of functions is 0≤a≤b≤n n(a, b) = 0≤a≤b≤n Cn−1b−a = n−1 k=0 (n − k)Cn−1k .
Summing up we have a number of functions equal to
n−1 k=0 nCn−1k − n−1 k=1 kCn−1k = n2n−1− n−1 k=1 (n − 1)Cn−2k−1= = n2n−1− (n − 1)2n−2= (n + 1)2n−2
Problem 4. Let n be an integer with n ≥ 2, and let a0, a1, a2, . . . , an
be complex numbers with an �= 0. Prove that the following statements are
equivalent:
(P) |anzn+ an−1zn−1+ · · · + a1z + a0| ≤ |an+ a0|, for every complex number
z of modulus 1,
(Q) a1= a2= · · · = an−1= 0 and
a0
an ∈ [0, ∞).
Dan S¸tefan Marinescu
Solution. (Q)⇒(P). If a1= a2= · · · = an−1= 0 and
a0
an ∈ [0, ∞), then
|anzn+ an−1zn−1+ · · · + a1z + a0| = |anzn+ a0| ≤
≤ |anzn| + |a0| = |an| + |a0| = |an+ a0|,
(P)⇒(Q). Let g(z) = an−1zn−1+ · · · + a1z, z ∈ C and w = a0+ an. For
any ε ∈ Un = {z ∈ C | zn = 1} we have |an+ g(ε) + a0| ≤ |an+ a0|, that is
|w + g(ε)| ≤ |w| or |g(ε)|2+ wg(ε) + ¯wg(ε) ≤ 0. As ε∈Un εk= 0, k = 1, 2, . . . , n−1, we have ε∈Un g(ε) = 0, whence, summing up, ε∈Un |g(ε)|2+wg(ε)+ ¯wg(ε) = ε∈Un
|g(ε)|2≤ 0, which implies that g(ε) = 0, for every ε ∈ Un.
From the above, we get ak = 1
n
ε∈Un
g(ε)
εk = 0, k = 1, 2, . . . , n − 1.
Thus, we get |anzn+a0| ≤ |an+a0| for every complex number z of modulus
1. If we denote c = a0
an and t = z
n, we have |t + c| ≤ |1 + c| for every complex
number z of modulus 1. Let P , M , A be the points in the complex plane with complex coordinates t, −c, respectively 1. Then P M ≤ MA, for every point P on the unit circle, therefore the unit circle is interior to the circle with center M and radius M A. As the point A belongs to both circles, we get that they are tangent in the point A, and thus the points M , O and A are collinear and M A ≥ OA = 1. We get that −c ≤ 0, and thus c ∈ [0, ∞).
11th GRADE
Problem 1. For a non-negative integer n the n-th iterate of a function f : R → R is fn = f ◦ · · · ◦ f
n times
, and f0 is the identity function. Determine the
continuous functions f : R → R, that satisfy simultaneously the conditions: a) The function f0+ f1 is increasing.
b) There is a positive integer m such that the function f0+ · · · + fm is
decreasing.
Dorel Mihet¸
Solution. We shall prove that all such functions are of the form f (x) = −x + c, where c is a real constant. It is clear that such functions verify the given conditions.
We shall first prove that f is one to one. Let x and y be real numbers such that f (x) = f (y) and define gn = f0+ · · · + fn, n ∈ N. Because g1 is
(x − y)2= (g
1(x) − g1(y))(gm(x) − gm(y)) ≤ 0, so x = y.
The fact that f is one to one and continuous implies that it is strictly monotonic, so all its iterates of the form f2k are increasing. As g
1is increasing,
from the equality
gn= n/2−1 � k=0 g1◦ f2k+ fn, if n is even, (n−1)/2 � k=0 g1◦ f2k, if n odd,
gnis strictly increasing for even n and increasing for odd n. As gmis decreasing
we deduce that m is odd and gm is constant. Finally, as g1 is increasing and
all even iterates of f are increasing, we conclude that g1is constant. This gives
the result.
Problem 2. Determine all differentiable functions f : R → R, that satisfy the equality f ◦ f = f.
Mihai Piticari, Marcel T¸ ena
Solution. We shall show that only the identity function and the constant functions satisfy the conditions of the problem. It is clear that these functions verify indeed the conditions.
Because f is continuous, its range {f(x) | x ∈ R} is an interval I ⊆ R. If I is degenerate at a point then f is constant.
If I is non-degenerate, let a = inf I < sup I = b, where a, b ∈ R. By the given condition we deduce that the restriction of f to the interval (a, b) is the identity:
f (x) = x, a < x < b. (1) We shall show that a = −∞ and b = +∞, i.e. I = R and f is the identity function. Suppose a is a finite number. By the continuity of f in a and by (1) we get f (a) = a, so
f′(a) = fd′(a) = limx→a x>a f (x) − f(a) x − a = limx→a x>a x − a x − a = 1. (2) On the other side f has a minimum at a, because
so, by the Theorem of Fermat, f′(a) = 0, in contradiction with (2). We
con-clude a = −∞. Analogously b = +∞.
Problem 3. Consider a positive integer n and A, B two matrices in Mn(C)
such that A2+ B2= 2AB. Prove that:
a) The matrix AB − BA is not invertible.
b) If the rank of A − B is 1, then matrices A and B commute.
Leonard Giugiuc, C˘at˘alin Gherghe
Solution. a) The given relation can be written in each of the two forms: (A − B)2= AB − BA, (1) A(A − B) = (A − B)B. (2) Suppose AB − BA is non singular. By (1), A − B is also non-singular, that is B = (A − B)−1A(A − B), by (2). Then A − B = A − (A − B)−1A(A − B),
from where we get In= A(A − B)−1− (A − B)−1A.
Considering traces we obtain n = tr In= tr
A(A − B)−1− (A − B)−1A = 0, a contradiction.
So AB − BA is singular; moreover by (1) the matrix A − B is also singular. b) It is known that for a matrix X of rank 1 in Mn(C), then X2= (tr X) X
(each row of X is proportional with a nonzero fixed row X). Using (1) we get AB − BA = (A − B)2= (tr (A − B))(A − B),
so 0 = tr (AB − BA) = (tr (A − B))2, i.e., tr (A − B) = 0. We conclude,
AB − BA = On, i.e., AB = BA.
Problem 4. Let A be an invertible matrix in M4(R), such that tr A =
tr A∗ �= 0, where A∗ is the adjugate of A. Prove that the matrix A2+ I 4
is singular if and only if there exists a nonzero matrix B in M4(R), so that
AB = −BA.
Solution. We show first that if A is a matrix from Mn(R) such that A2+ In
is singular, then there exists a matrix B in Mn(R), such that AB = −BA.
As A2+ I
n is singular, i is a proper value of A, and −i is a proper value of
the transpose Aτ. There exist two proper vectors x and y in M
n,1(C), so that
Ax = ix and Aτy = −iy. Because x and y are nonzero, B = xyτ is a nonzero
matrix in Mn(C).
The following relations show that A and B anticommute:
AB = Axyτ= ixyτ= −x(−iy)τ= −x (Aτy)τ= −xyτA = −BA. Taking conjugates and using that A is in Mn(R), we get that the conjugate
¯
B of B anticommutes with A. This proves that any linear combination with complex coefficients of the matrices B and ¯B anticommutes with A. Conse-quently, B or i(B − ¯B) is a nonzero matrix in Mn(R) that anticommutes with
A. This proves the first part of the problem.
We shall prove the converse. Let B be a nonzero matrix in M4(R) that
anticommutes with A. Then
AkB = (−1)kBAk, k ∈ N. (∗) Consider the characteristic polynomial f of A,
f = λ4−(tr A)λ3+aλ2−(tr A∗)λ+det A = λ4−(tr A)λ3+aλ2−(tr A)λ+det A, where a is a real number (the latter form of f is a consequence of tr A = tr A∗).
By Hamilton-Cayley, f (A) = O4. Taking into account (∗), we successively
obtain:
O4= f (A)B = B
A4+ (tr A)A3+ aA2+ (tr A)A + (det A)I 4 = B f (A) + 2(tr A)(A2+ I 4)A = 2(tr A)B(A2+ I 4)A.
Because tr A �= 0 and A is invertible, we obtain that B(A2+ I
4) = O4. As B
is nonzero we conclude that A2+ I
4is singular.
12th GRADE
Problem 1. Let (A, +, ·) be a ring with unity. For a ∈ A define functions sa : A → A and da : A → A by sa(x) = ax, da(x) = xa, for all x ∈ A.
a) Suppose A is finite. Prove that for any a ∈ A, sa is injective if and only
if da is injective.
b) Give an example of an ring that contains an element a such that exactly one of the functions sa and da is injective.
Dorel Mihet¸
Solution. a) Suppose sa is one to one. As A is finite sa is bijective, thus
there exists b ∈ A such that ab = 1. As a consequence, if da(x) = da(y), we
get in succession (x − y)a = 0, (x − y)ab = 0, that is x − y = 0, which proves the injectivity of da. The proof of the converse goes on the same lines.
b) To construct an example, consider S = {(xn)n∈N | xn ∈ R} and the
ring of additive functions f : S → S, endowed with the operations of addition and composition. For a one can consider the function defined by a ((xn)n) =
(xn+1)n. As a is surjective, da is injective. It is clear that sais not one to one.
Problem 2. Let I, J be intervals and consider ϕ : J → R a continuous function which is nonzero on J. Let f, g : I → J be two differentiable functions such that f′= ϕ ◦ f and g′= ϕ ◦ g. Prove that if there exists x0∈ I such that
f (x0) = g(x0), than f and g coincide.
***
Solution. As ϕ is non-zero and is continuous, the function 1/ϕ is correctly defined and continuous. Consider an anti-derivative F : J → R. The given relation for f can be then written (F ◦ f)′(x) = 1, for all x ∈ I. Thus there
exists a ∈ R such that F (f(x)) = x + a, for all x ∈ I. In the same manner, there exists b ∈ R such that F (g(x)) = x + b, for all x ∈ I. We deduce x0+ a = F (f (x0)) = F (g(x0)) = x0+ b, that is a = b. So F (f (x)) = F (g(x)),
for all x ∈ J. (*)
On the other side, from F′(x) �= 0, for all x ∈ J, F′ has a constant sign on
J, so F is injective. So, (*) implies f = g.
Problem 3. Let f : [1, +∞) → (0, +∞) be a continuous function having the following properties:
(i) The function g : [1, +∞) → (0, +∞) given by g(x) = f (x)x has limit at +∞,
(ii) The function h : [1, +∞) → (0, +∞) given by h(x) = x1 x 1 f (t) dt has finite limit at +∞.
a) Show that lim
x→+∞g(x) = 0.
b) Show that lim
x→+∞ 1 x2 x 1 f2(t) dt = 0. Mihai Piticari
Solution. a) Let ℓ = lim
x→+∞g(x). If ℓ ∈ (0, +∞), let a > 0 be such that
g(x) > ℓ/2 for x ≥ a. Then h(x) = 1 x a 1 f (t) dt + x a f (t) dt ≥ 1x a 1 f (t) dt + ℓ 2x x a t dt = = 1 x a 1 f (t) dt + ℓ(x 2− a2) 4x x→+∞−→ +∞,
in contradiction with (ii). In the same way we can prove that ℓ = +∞ is in contradiction with (ii), so ℓ = 0.
b) Remark that 1xf (t)dt > 0, for x > 1, so
lim x→+∞ 1 x2 x 1 f2(t)dt = lim x→+∞ x 1 f 2(t)dt x 1xf (t)dt· x 1 f (t)dt x = = λ lim x→+∞ x 1 f 2(t)dt x 1xf (t)dt = λ limx→+∞ u(x) v(x), where λ = lim x→+∞h(x), u(x) = x 1 f2(t)dt, v(x) = x x 1 f (t)dt.
To show that lim
x→+∞
u(x)
v(x) = 0 we shall use l’Hospital Rule: • u and v are differentiable,
• lim
x→+∞v(x) = +∞ (this follows from v(x) ≥ m(x − 1) for x ≥ 2, where
m = infx∈[1,2]f (x)), • v′(x) =
x
• relations u ′(x) v′(x) = f2(x) xf (x) + x 1 f (t) dt = g(x) ·f (x) + h(x)f (x) ∈ (0, g(x)) and lim x→+∞g(x) = 0 imply limx→+∞ u′(x) v′(x) = 0.
Problem 4. Suppose (G, ·) is a finite group with unity e, a is an element in G\{e} and p is a prime number such that xp+1= a−1xa, for all x ∈ G.
a) Show that there is k ∈ N∗such that ord(G) = pk.
b) Prove that H = {x ∈ G | xp= e} is a subgroup of G and
(ord(H))2> ord(G).
Ioan B˘aetu
Solution. a) If x, y ∈ G, than (xy)p+1 = a−1xya = a−1xaa−1ya =
= xp+1yp+1. We can write x(yx)py = xp+1yp+1, then (yx)p = xpyp. For
x = a we get ap= e so by the preceding equality (ya)p= yp.
Multiplying at left by ya we obtain yayp= (ya)p+1= yp+1a, that is ayp=
ypa, for all y ∈ G. From the hypothesis we have yp(p+1) = a−1ypa = yp, so
yp2
= e, for all y ∈ G. Because p is a prime, every element of the group has order 1, p or p2 and by the Cauchy theorem we deduce ord(G) = pk, for a
k ∈ N∗.
b) For x, y ∈ H, we have (xy)p = ypxp = e, that is xy ∈ H, proving that
H is a stable part of G, and, as it is finite, H is a subgroup. Consider f : G → G, given by f(x) = xp. Because e = xp2
= (xp)p, the
image of f is contained in H. Moreover, x, y ∈ G and f(x) = f(y) imply xp(y−1)p = e, so (y−1x)p = e, that is y−1x ∈ H. This gives x ∈ Hy. We
conclude that for every element in Imf , the number of its pre-images in G is exactly ord(H), so |Imf| = ord(H)ord(G).
Because a �= e we get a /∈ Imf: for if not a = bp for some b ∈ G. This
would imply bp+1 = b−pbbp, that is e = bp= a a contradiction. As a ∈ H, we
SMALL JUNIORS
1. a) Prove that, for every positive integers a, b, k such that a < b, a b ≤ a + k b + k. b) Prove that 1 100+ 4 101+ 7 102 + . . . + 148 149 ≥ 25.
Gabriel Vrˆınceanu, Ion Cicu
2. A 4 × 4 magic square has in its first line the non-nil digits a, b, c, d, written in this order (from left to right), and each of the other lines contains the same numbers, written in different orders. It is known that the sum of the eight four-digit numbers obtained by reading the columns downwards and by reading the lines from left to right equals 59994. Find the largest and the smallest possible value for abcd.
Gabriel Vrˆınceanu
3. Find how many pairs (a, b), with a and b non-nil digits, have the property that ab is irreducible and the decimal fractions ab and b(b+1)a+b are finite.
Gabriel Vrˆınceanu
4. Triangle ABC has a right angle in A and AD ⊥ BC, with D ∈ (BC). Points E ∈ (AB) and F ∈ (AC) are taken so that (DE is the bisector of the angle ADB and (DF is the bisector of the angle ADC. Prove that triangle AEF is isosceles.
Petru Braica
5. The isosceles triangle ABC has AB = AC and points M and N are taken on (BC) such that M is between B and N . Prove that the following properties are equivalent:
i) m(∠MAN) = 1
2m(∠BAC);
ii) the segments [BM ], [M N ] and [M C] are the sides of a triangle in which the opposite angle of [M N ] has measure 180◦− m(∠A).
Mihai Dicu
6. Denote O the center of the square ABCD. The bisector of the angle ∠OAB meets OB in N and BC in P . Prove that P C = 2ON .
George Stoica
7. Find all positive integers n with at least two digits, n’s digits are pairwise distinct and n equals the product of the sum of its digits with one of its digits.
Cosmin Manea, Drago¸s Petric˘a
8. Define an= 18 77 . . . 7
n digits
889 for every non-negative integer n. a) Prove that an is divisible by 13, for every n ∈ N.
b) Denote by cn the quotient of the division of an by 13. Find all n ∈ N so
that s(an) = 2s(cn), where s(m) represents the sum of the digits of the number
m.
Marius Burtea
9. If n is a composite number, denote Dn its largest proper divisor. A
number n will be called squarish if the number Dn+ Dn+1is a perfect square.
a) Show that 35, 76 and 755 are squarish.
b) Show that there are infinitely many squarish numbers.
Marius Burtea
JUNIORS
1. Solve in positive integers the equation 2n= 3m+ 23.
