FINAL ROUND
Problem 1. Find the maximum value of
. Asa, c 6 1, we have1+ca 6 a+ca and1+ac 6 c+ac , so
a
1 + c+ c
1 + a 6 a
a + c+ c c + a= 1 and the like. Summing up the three we get the desired result.
9thGRADE
Problem 1. Find the maximum value of
(x3+ 1)(y3+ 1), forx, y∈ R such that x + y = 1.
Solution. Putxy = t; as x + y = 1 we get (x3+ 1)(y3+ 1) = t3− 3t + 2.
Fromx + y = 1 we obtain t = xy 6 x+y2 2
= 14. It is easy to prove that t3− 3t + 2 6 4 for t 6 14, with equality if and only ift =−1.
We infer that(x3+ 1)(y3 + 1) 6 4 for x, y ∈ R with x + y = 1 and (φ3+ 1)(−1/φ3+ 1) = 4, where φ is one of the roots of z2− z − 1 = 0.
Remark. In fact, forx, y∈ R, we have
[x3+ (x + y)3][y3+ (x + y)3] 6 4(x + y)6, with equality if and only ifx2+ 3xy + y2= 0.
Problem 2. Consider the trianglesABC and DBC such that AB = BC, DB = DC and ∠ABD = 90◦. LetM be the midpoint of BC. Points E, F, P are such thatE ∈ (AB), P ∈ (MC), C ∈ (AF ) and ∠BDE = ∠ADP = ∠CDF . Prove thatP is the midpoint of EF and DP ⊥ EF .
Solution. Putu = ∠BDE = ∠M DP = ∠CDF . In the right triangles DBE, DM P, DCP we have
cos ∠BDE = BD
DE, cos ∠M DP = DM
DP , cos ∠CDF =DC DF.
ThusBD = DE cos u, DM = DP cos u, DC = DF cos u.
Moreover, ∠BDC = ∠EDF and ∠BDM = ∠EDP . We get from here that the triangles DBC and DEF are similar and points M, F correspond to each other.
PointM is the midpoint of BC, implying that P is the midpoint of EF . As DM ⊥ BC we conclude that DP ⊥ EF .
Problem 3. Consider quadrilateralsABCD inscribed in a circle of radius r, such that there is a pointP on sidea CD for which CB = BP = P A = AB.
a) Prove that there is a configuration of points A, B, C, D, P for which the above configuration is possible.
b) Prove that for any such configuration we also haveP D = DA = r.
Solution. a) Consider a chordAB such that AB < r√
3 and P in the interior of the circle such that triangleABC is equilateral. Let C be a point on the circle such thatBP = BC and AC∩ BP 6= ∅. Line P C meets again the circle at D.
The configuration thus obtained fulfils the conditions in the statement.
b) Let ∠BP C = ∠BCP = x. As the triangle BP C is isosceles, we get
∠P BC = 180◦ − 2x. The quadrilateral ABCD is cyclic, ∠BCD = x, thus
∠DAB = 180◦− x. Therefore, ∠DAP = 120◦− x, ∠ABC = 240◦− 2x, and consequently, ∠ADC = 2x− 60◦. In the triangleADP we have ∠AP D = 120◦− x, thus DA = DP .
TrianglesABD and P BD are equal, thus ∠ABD = ∠P BD = 30◦. More-over, as ∠ABD = 30◦we getDP = r.
Problem 4. A table tennis competition takes place during 4 days, the number of participants being2n, n > 5. Every participant plays exactly one game daily (it is possible that a pair of participants meet more times). Prove that such a competi-tion can end with exactly one winner and exactly three players on the second place and such that there is no player losing all four matches. How many participants have won a single game and how many exactly two, under the above conditions?
Solution. Denote bynkthe number of participants that won exactlyk games,
52 THE57 ROMANIANMATHEMATICALOLYMPIAD
0 6 k 6 4. Under the given conditions we have
n0= 0, n1+ n2+ n3+ n4= 2n > 10. (1) The total number of games is4n, thus
4n = 1· n1+ 2· n2+ 3· n3+ 4· n4 (counting the winners) (2) 4n = 3· n1+ 2· n2+ 1· n3+ 0· n4 (counting the loosers) (3) thus2n1= 2n3+ 4n4. Substituing in (1) we obtain
n2+ 2n3+ 3n4= 2n. (4)
The other conditions of the problem will imply
n4 n3 n2 n1 n2+ 2n3+ 3n4
− − −− − − −− − − −− − − −− − − −− − − −−
0 0 1 3 1
0 1 0 3 2
1 0 0 3 3
0 1 3 5
1 0 3 6
giving a contradiction.
It remains the casen4= 1, n3= 3, which implies n2= 2n− 9, n1= 5.
For a model, denote bya the winner; by b1, b2, b3those on the second place;
byc one of the 2n−9 winners of exactly two games and by d1, d2, d3, d4, d5those five with only one won game. The remaining2n− 10 players having won two games, will be denoted (forn > 5) by c1, . . . , c2n−10. Finally, byxy we will mean thatx won the game against y.
Day
1 ab1 cd2 b2d3 b3d4 d1d5 cici+1
2 ab2 b1d1 cd3 b3d4 d2d5 cici+1
3 ab3 b1d1 b2d2 d5c d3d4 ci+1ci
4 ac b1d1 b2d2 b3d3 d4d5 ci+1ci, wherei = 1, 3, . . . , 2n− 11.
10thGRADE
Problem 1. Consider a setM with n elements and letP(M) denote all subsets ofM . Find all functions f :P(M) → {0, 1, 2, . . . , n} , satisfying the following two conditions:
a)f (A)6= 0, for any A 6= ∅, and
b) f (A∪ B) = f (A ∩ B) + f (A∆B) , for any A, B ∈ P(M), where A∆B = (A∪ B) (A ∩ B) .
Solution. From condition b) we obtain
f (∅ ∪ ∅) = f (∅ ∩ ∅) + f (∅∆∅) , givingf (∅) = 0.
By b), forA, B ∈ P(M), with A B, we get
f (B) = f (A∪ B) = f(A) + f (B A) . From a) we havef (B A)6= 0, thus f (B) > f(A).
Consequently, for any permutation(α1, α2, . . . , αn) of the set M we have the sequence of inequalities
0 = f (∅) < f ({α1}) < f ({α1, α2}) < · · · < f ({α1, α2, . . . , αn}) . Sincef (A)∈ {0, 1, . . . , n} , for all A ∈ P(M), it follows that
f ({α1, α2, . . . , αj}) = j, for anyj∈ {1, 2, . . . , n}.
Consequently, f (A) = |A|, for any A ∈ P(M). It is easy to see that this function fulfils the given conditions.
Problem 2. Prove that fora, b∈ 0,π4
we have sinna + sinnb
(sin a + sin b)n > sinn2a + sinn2b (sin 2a + sin 2b)n.
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Problem 3. Prove that the sequence given byan= n√ contains an infinity of odd nubers and an infinity of even numbers.
Solution. Let xn =
Suppose, by way of contradiction, that there isk ∈ N such that all elements an, n > k, have the same parity. As 2 6 an+1− an 64, for any n∈ N, we get thatan+1− an∈ {2, 4} , for all n > k.
Ifan+1− an= 2, then xn+1− xn= yn+1− yn = 1, and if an+1− an= 4, thenxn+1− xn= yn+1− yn= 2.
Thusyn− xn= yn+1− xn+1, for all n > k, which gives yn− xn= yk− xk, for alln > k.
Butyn− xn> n√
3− 1 − n√
2, for any n, so n <yk− xk+ 1
√3−√ 2 , for alln > k, a contradiction.
Problem 4. Givenn∈ N, n > 2, find n disjoint sets Ai, 1 6 i 6 n, in the plane, such that:
a) for any diskC and any i ∈ {1, 2, . . . .n} , we have Ai∩ Int (C) 6= ∅; and b) for any lined and any i∈ {1, 2, . . . .n} , the projection of Aiond is not all ofd.
Solution. There are a lot of natural ways to construct such sets. For example, takep1, p2, . . . , pnsquare free positive integers and consider the sets
Ak=
m1
q1√pk
, m2
q2√pk
| m1, m2, q1, q2∈ Z∗
.
It is easy to see that any such set is countable and dense in the plane.
11thGRADE
Problem 1. A is a square matrix with complex entries. Denote by A∗ its adjoint (the matrix formed by the cofactors of the transpose). Prove that if there is an integerm > 1 such that (A∗)m= 0n, then(A∗)2= 0n.
Solution. Fromdet(A∗)m= 0, we have det(A∗) = 0 and det(A) = 0. We claim that the rank ofA∗is at most 1. For if rankA 6 n− 2 then A∗= 0n, and if rankA = n− 1 then AA∗= 0nand by Sylvester’s inequality0 = rank (AA∗) >
rank(A) + rank (A∗)−n = rank (A∗)−1. Suppose now that m > 3 (otherwise, there is nothing to prove). As A has at most rank 1 there is a row matrix X ∈ M1n(C) and a column matrix Y ∈ Mn1(C) such that A = Y X. Denoting XY = a∈ C we obtain 0n = Am = Y (XY )m−1X = am−1Y X = am−1A, implyinga = 0 or A = 0n, soA2= aA = 0n.
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Problem 2. A matrixB∈ Mn(C) will be called a pseudo-inverse of a matrix A∈ Mn(C) if A = ABA and B = BAB.
a) Prove that any square matrix has at least one pseudo-inverse.
b) Characterize the class of matrices with a unique pseudo-inverse.
Solution. a) Denote byr the rank of A. This means that by elementary trans-formations,A goes over to a matrix that has zeroes everywhere except the first r entries on the main diagonal. That is, there exist two invertible matricesP and Q such thatP AQ has 1 on the first r entries on the main diagonal and 0 elsewhere.
We thus have
This means thatB = QDP is a pseudo-inverse of A.
b) IfA = 0nthenB = 0n, and, ifA is invertible then B = A−1. We claim that in any other situationA has an infinite number of pseudo-inverses. Such matrices can be obtained, for instance, by replacingD by any matrix of the form
Dx=
which coincides withD, except that the (1, n)-entry is now any complex x.
Problem 3. Consider two systems of points in the plane:A1, A2, . . . , Anand B1, B2, . . . , Bnhaving different centers of gravity. Prove that there is a pointP in the plane such that
P A1+ P A2+· · · + P An= P B1+ P B2+· · · + P Bn.
Solution. Consider a Cartesian system of coordinates such that the two cen-troids have differentx-coordinates. Suppose coordonates are Ai(ai, a0i) and B(bi, b0i).
We are looking forP on Ox: P (p, 0). Consider f : R→ R, defined by f (p) = P A1+ P A2+· · · + P An− (P B1+ P B2+· · · + P Bn).
We have
p→∞lim f (p) = lim
p→∞
Xn k=1
2p(bk− ak) + a2k− b2k+ a02k − b02k
p(p− ak)2+ a02k +p
(p− bk)2+ b02k = Xn k=1
(bk− ak)
and
p→−∞lim f (p) =− Xn k=1
(bk− ak).
Since f is continuous, by the intermediate value property, f (p) = 0 for some realp.
Remark. The condition on the centroids is necessary only forn > 3. For, if B1, B2, . . . , Bn are the mid-points of[A1A2], [A2A3], . . . , [AnA1], P can exist, unless theAiare collinear.
Problem 4. Consider a functionf : [0,∞) → R, with the property: for any x > 0, the sequence (f (nx))n>0is increasing.
a) Iff is also continuous on [0, 1], does it follow that it is increasing?
b) What iff is continuous on Q+?
Solution. a) The answer is “no”. A counterexample is
f (x) =
( x ifx∈ [0, 1] ∪ Q+; 2x− 1 if x ∈ (1, ∞) \ Q.
b) The answer is “yes”. The following remark is an easy consequence of the hypothesis: if(rn) is an increasing sequence of rational numbers and x is positive, then the sequence(f (rnx)) is increasing.
Suppose, by contradiction, that for somex < y we have f (x) > f (y). Con-sider a rational numbera in (x, y). Find an increasing sequence of rationals (qn) and a decreasing sequence of rationals(rn) such that (qnx)→ a and (rny)→ a.
We get, by continuity f (x) < lim
n→∞f (qnx) = f (a) = lim
n→∞f (rny) < f (y), a contradiction.
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12thGRADE
Problem 1. LetK be a finite field. Prove that the following are equivalent:
a)1 + 1 = 0;
b) for anyf ∈ K[X] with deg f > 1 the polynomial f(X2) is reducible.
Solution. To prove that a) implies b), considerF : K→ K, given by F (x) = x2. IfF (x) = F (y) we get x2= y2, thus(x− y)(x + y) = 0, so (x − y)2= 0, orx = y. This means that F is one-to-one, consequently onto (for K is finite).
If f = Pn reducible, it has a root inK. This means that F is onto and thus one-to-one. As F (1) = F (−1) = 1, we get 1 = −1.